125:T7ba,**Set-up.** Tension positive; elongation positive; SI units throughout ($E$ in Pa, lengths in m, forces in N). The bar is statically indeterminate (two reactions, one equation $R_A + R_B = 900$ kN). Release B, treat $R_B$ (upwards) as an unknown load, and require the total change in length to be zero. **(a) Loads only (left problem).** Number the portions from the bottom: 1 = KB, 2 = CK, 3 = DC, 4 = AD. With B free, the internal force in each portion is the sum of the loads below it: $$P_1 = 0, \quad P_2 = P_3 = 600\ \mathrm{kN}, \quad P_4 = 600 + 300 = 900\ \mathrm{kN}$$ $$\delta_L = \frac{1}{E}\left(\frac{0 \times 0.15}{400\times10^{-6}} + \frac{600\times10^3 \times 0.15}{400\times10^{-6}} + \frac{600\times10^3 \times 0.15}{250\times10^{-6}} + \frac{900\times10^3 \times 0.15}{250\times10^{-6}}\right) = \frac{1.125\times10^9}{E}$$ $$\delta_L = \frac{1.125\times10^9}{200\times10^9} = 5.625\times10^{-3}\ \mathrm{m} = 5.625\ \mathrm{mm}$$ **(b) Redundant only (right problem).** An upward $R_B$ at B compresses the whole bar: portion CB (area 400 mm$^2$, length 0.30 m) and portion AC (area 250 mm$^2$, length 0.30 m) both carry $-R_B$: $$\delta_R = -\frac{R_B}{E}\left(\frac{0.30}{400\times10^{-6}} + \frac{0.30}{250\times10^{-6}}\right) = -\frac{1.95\times10^3 R_B}{E}$$ Compatibility (close fit at both supports): $$\delta_L + \delta_R = 0 \;\Rightarrow\; R_B = \frac{1.125\times10^9}{1.95\times10^3} = 5.77\times10^5\ \mathrm{N} = 577\ \mathrm{kN}$$ Equilibrium: $R_A = 300 + 600 - 577 = 323$ kN (upwards). **Check (back-substitution).** With the reactions known the internal forces are $N_{AD} = R_A = 323$ kN, $N_{DC} = N_{CK} = 323 - 300 = 23$ kN, $N_{KB} = 23 - 600 = -577$ kN. The elongations are $+0.969$, $+0.069$, $+0.043$ and $-1.082$ mm, which sum to zero. Each portion deforms even though the bar as a whole does not. Units: $\mathrm{N\,m/(m^2\,Pa)} = \mathrm{m}$. Marks (likely): method and FBD 2, $\delta_L$ 3, $\delta_R$ 2, reactions 3. 126:T4b9,**Set-up.** Tension positive. Cut through each portion and draw the FBD of the part to the RIGHT of the cut, which avoids the unknown wall reaction. **Internal forces.** - CD: only $P_D$ acts on the right part, so $P_3 = P_{CD} = +130$ kN (tension). - BC: $P_D$ to the right and $P_C$ to the left, $P_2 = P_{BC} = 130 - 200 = -70$ kN (compression). - AB: add $P_B$ to the right, $P_1 = P_{AB} = -70 + 350 = +280$ kN (tension). Check: the wall reaction is $R_A = 280$ kN to the left, and $350 - 200 + 130 - 280 = 0$. **Deformation.** $$\delta = \frac{1}{E}\left(\frac{P_{AB} l_{AB}}{A_{AB}} + \frac{P_{BC} l_{BC}}{A_{BC}} + \frac{P_{CD} l_{CD}}{A_{CD}}\right)$$ $$\delta = \frac{1}{200\times10^9}\left(\frac{280\times10^3 \times 0.3}{5.8\times10^{-4}} + \frac{-70\times10^3 \times 0.3}{5.8\times10^{-4}} + \frac{130\times10^3 \times 0.4}{1.9\times10^{-4}}\right)$$ Portion by portion: $+0.724$ mm, $-0.181$ mm, $+1.368$ mm, so $$\delta = 1.91\ \mathrm{mm}\ \text{(elongation)}$$ Check: units N m/(m$^2$ Pa) = m; the slender portion CD contributes most, as its area is about a third of the others. Remember $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$. Marks (likely): internal forces 3, deformation 3. 127:T644,**Set-up.** The links are two-force members, so they apply vertical forces $F_{AB}$ at B and $F_{CD}$ at D. Assume both pull upward on the bar (links in tension); a negative result means compression. **Forces in the links.** $$\sum M_B = 0: \quad F_{CD}(0.2) - 30(0.6) = 0 \;\Rightarrow\; F_{CD} = +90\ \mathrm{kN}\ \text{(tension)}$$ $$\sum M_D = 0: \quad -F_{AB}(0.2) - 30(0.4) = 0 \;\Rightarrow\; F_{AB} = -60\ \mathrm{kN}\ \text{(compression)}$$ Check $\sum F_y$: $-60 + 90 - 30 = 0$. **(a)** Link AB shortens: $$\delta_{AB} = \frac{F_{AB} L_{AB}}{E A} = \frac{-60\times10^3 \times 0.3}{70\times10^9 \times 500\times10^{-6}} = -0.514\ \mathrm{mm}$$ A is fixed above B, so shortening moves B up: $\delta_B = 0.514$ mm upwards. **(b)** Link CD lengthens: $$\delta_{CD} = \frac{90\times10^3 \times 0.4}{200\times10^9 \times 600\times10^{-6}} = +0.300\ \mathrm{mm}$$ so D moves 0.300 mm downwards. **(c)** The bar stays straight. B' is 0.514 mm above the original line and D' is 0.300 mm below it, so the displaced bar crosses the original line at a point H between B and D. With $x = HD$, similar triangles give $$\frac{0.514}{0.300} = \frac{200 - x}{x} \;\Rightarrow\; x = 73.7\ \mathrm{mm}$$ $$\frac{\delta_E}{0.300} = \frac{400 + 73.7}{73.7} \;\Rightarrow\; \delta_E = 1.93\ \mathrm{mm}\ \text{downwards}$$ Check (linear interpolation): downward displacement varies linearly along the bar, $-0.514$ at B and $+0.300$ at D, so at E, 0.4 m beyond D: $0.300 + (0.300 + 0.514)(0.4/0.2) = 1.929$ mm. Marks (likely): link forces 3, $\delta_B$ and $\delta_D$ with directions 3, $\delta_E$ 3. 128:T8ca,**Set-up.** A pushed down makes the bar rotate about B, so C and D rise and stretch both rods: the rods are in tension and pull DOWN on the bar with $F_{CE}$ and $F_{DF}$. The pin B gives $B_x$ and $B_y$. Unknowns $B_y$, $F_{CE}$, $F_{DF}$ (and $B_x = 0$) against two useful equilibrium equations: statically indeterminate (degree 1). **(a)** Moments about B (lengths in m, forces in kN): $$100(0.338) - F_{CE}(0.225) - F_{DF}(0.375) = 0 \;\Rightarrow\; 0.225F_{CE} + 0.375F_{DF} = 33.80 \qquad (1)$$ Compatibility: the rigid bar rotates about B, so $\delta_C = \dfrac{225}{375}\delta_D = 0.600\,\delta_D$ and $\delta_A = \dfrac{338}{375}\delta_D = 0.901\,\delta_D$. Each rod end moves by the rod elongation, $\delta_C = \dfrac{F_{CE}L_{CE}}{A_{CE}E}$, $\delta_D = \dfrac{F_{DF}L_{DF}}{A_{DF}E}$, so $$F_{CE} = 0.6\,\frac{L_{DF}}{L_{CE}}\frac{A_{CE}}{A_{DF}}F_{DF} = 0.6 \times \frac{562}{450}\left(\frac{10}{15}\right)^2 F_{DF} = 0.3330\,F_{DF} \qquad (2)$$ Solving (1) and (2): $$F_{DF} = \frac{33.80}{0.225(0.3330) + 0.375} = 75.12\ \mathrm{kN}, \qquad F_{CE} = 25.02\ \mathrm{kN}$$ (both tension). Then $B_y = 100 + 25.02 + 75.12 = 200.14$ kN upwards. **(b)** $A_{DF} = \pi(0.015)^2/4 = 1.767\times10^{-4}\ \mathrm{m^2}$: $$\delta_D = \frac{75.12\times10^3 \times 0.562}{1.767\times10^{-4} \times 200\times10^9} = 1.19\times10^{-3}\ \mathrm{m} = 1.19\ \mathrm{mm}\ \text{(up)}$$ $$\delta_A = 0.901\,\delta_D = 1.08\ \mathrm{mm}\ \text{(down)}$$ Check: $\delta_C$ from rod CE, $\dfrac{25.02\times10^3 \times 0.45}{\pi(0.01)^2/4 \times 200\times10^9} = 0.717$ mm, equals $0.600\,\delta_D = 0.717$ mm. **(c)** Bar as a beam, $x$ from A (m), moment in kN m with the slide convention: - AB: $M = -100x$, reaching $-33.8$ at B ($x = 0.338$). - BC: $M = -100x + 200.14(x - 0.338) = 100.14x - 67.65$; at C ($x = 0.563$) $M = -11.27$. - CD: $M = 75.12x - 53.56$; zero at D ($x = 0.713$), as it must be at a free end. $M$ is linear between the point forces, so the extreme is at a load point: $|M|_{max} = 100 \times 0.338 = 33.8$ kN m at B. Check from the right of B: $F_{CE}(0.225) + F_{DF}(0.375) = 33.8$ kN m, by equation (1). Marks (likely): FBD and indeterminacy 3, equilibrium 2, compatibility 4, forces 2, displacements 3, moment 4. 129:T4fa,**Set-up.** Hydrostatic compression: $\sigma_x = \sigma_y = \sigma_z = -p$. Tensile stress and extension positive. **(a)** $\epsilon_x = \delta_x / l_x = -0.04/100 = -4.0\times10^{-4}$. Generalised Hooke's law with equal stresses gives the same strain in every direction: $$\epsilon_x = \epsilon_y = \epsilon_z = \epsilon = -\frac{p}{E}(1 - 2\nu)$$ $$\delta_y = \epsilon l_y = -4.0\times10^{-4} \times 50 = -0.020\ \mathrm{mm}, \qquad \delta_z = \epsilon l_z = -4.0\times10^{-4} \times 75 = -0.030\ \mathrm{mm}$$ (both shortenings). **(b)** $$p = -\frac{E\epsilon}{1 - 2\nu} = -\frac{200\times10^9 \times (-4.0\times10^{-4})}{1 - 0.58} = 1.905\times10^8\ \mathrm{Pa} = 190.5\ \mathrm{MPa}\ (1904.8\ \mathrm{bar})$$ **(c)** $$k = \frac{E}{3(1-2\nu)} = \frac{200}{3 \times 0.42} = 158.7\ \mathrm{GPa}, \qquad e = -\frac{p}{k} = -\frac{190.5}{158.7\times10^3} = -1.200\times10^{-3}$$ $$\Delta V = eV = -1.200\times10^{-3} \times (100 \times 50 \times 75) = -450\ \mathrm{mm^3}$$ Check: $e = 3\epsilon = 3(-4.0\times10^{-4}) = -1.2\times10^{-3}$, the same; the exact volume change $V[(1+\epsilon)^3 - 1] = -449.8\ \mathrm{mm^3}$ confirms the small-strain result. Marks (likely): strains and edge changes 2, pressure 2, bulk modulus and dilatation 2, volume change 1. 12a:T5f9,**Set-up.** Cut through a straight portion. The internal loads at the centroid C are equivalent to a centric force and a couple: $$P = 712\ \mathrm{N}, \qquad M = Pd = 712 \times 0.016 = 11.39\ \mathrm{N\,m}$$ Superpose the uniform stress from $P$ and the linear stress from $M$. **(a)** With $c = 6.5$ mm: $$A = \pi c^2 = 1.327\times10^{-4}\ \mathrm{m^2}, \qquad \sigma_0 = \frac{P}{A} = 5.36\ \mathrm{MPa}$$ $$I = \frac{\pi c^4}{4} = 1.402\times10^{-9}\ \mathrm{m^4}, \qquad \sigma_m = \frac{Mc}{I} = \frac{11.39 \times 0.0065}{1.402\times10^{-9}} = 52.8\ \mathrm{MPa}$$ As in the back of a C-clamp, the couple puts the inner fibre (nearer the load line) in tension and the outer fibre in compression: $$(\sigma_t)_{max} = \sigma_0 + \sigma_m = 58.2\ \mathrm{MPa}, \qquad (\sigma_c)_{max} = \sigma_0 - \sigma_m = -47.5\ \mathrm{MPa}$$ **(b)** On the neutral axis $\sigma_x = 0$: $$\frac{P}{A} - \frac{My_0}{I} = 0 \;\Rightarrow\; y_0 = \frac{P}{A}\frac{I}{M} = \frac{5.36\times10^6 \times 1.402\times10^{-9}}{11.39} = 6.60\times10^{-4}\ \mathrm{m} = 0.660\ \mathrm{mm}$$ from the centroid, on the outer (compressed) side, away from the load line. Check: symbolically $y_0 = \dfrac{I}{Ad} = \dfrac{c^2}{4d} = \dfrac{6.5^2}{64} = 0.660$ mm. The neutral axis no longer passes through the centroid because the tensile axial stress must be cancelled by bending compression, which only happens on the compressed side. Marks (likely): equivalent force-couple 2, axial and bending stresses 3, combination 2, neutral axis 2. 12b:T6e1,**Set-up.** Positive shear and bending as on the formula sheet (positive moment sags). Pin A and roller B give vertical reactions $R_A$, $R_B$ (no horizontal load, so no horizontal reaction). **Reactions.** Resultant of the load: $$P = \int_0^L w_0\sin\frac{\pi x}{L}\,dx = -\frac{w_0 L}{\pi}\left[\cos\frac{\pi x}{L}\right]_0^L = \frac{2w_0L}{\pi}$$ acting at midspan by symmetry, so $R_A = R_B = \dfrac{P}{2} = \dfrac{w_0L}{\pi}$. **(a)** Shear, starting from $V_A = R_A$: $$V(x) = R_A - \int_0^x w_0\sin\frac{\pi x'}{L}\,dx' = \frac{w_0L}{\pi} + \frac{w_0L}{\pi}\left[\cos\frac{\pi x'}{L}\right]_0^x = \frac{w_0L}{\pi}\cos\frac{\pi x}{L}$$ Moment, starting from $M_A = 0$ (pin): $$M(x) = \int_0^x V\,dx' = \frac{w_0L^2}{\pi^2}\sin\frac{\pi x}{L}$$ Checks (sympy): $dV/dx = -w$, $dM/dx = V$; $V(L) = -w_0L/\pi = -R_B$ and $M(L) = 0$ at the roller. **(b)** $V$ is a half cosine: $+w_0L/\pi$ at A, zero at midspan, $-w_0L/\pi$ at B (antisymmetric). $M$ is a half sine: zero at both supports, maximum $w_0L^2/\pi^2$ at midspan (symmetric), and positive (sagging) everywhere. The maximum of $M$ is where $V = 0$. **(c)** The largest normal stress is at the section of largest $|M|$, i.e. at $x = L/2$, on the top fibre (compression) and bottom fibre (tension): $$\sigma_{max} = \frac{M_{max}c}{I} = \frac{(w_0L^2/\pi^2)(a/2)}{a^4/12} = \frac{6w_0L^2}{\pi^2a^3}$$ Check: units $\mathrm{(N/m)\,m^2/m^3} = \mathrm{Pa}$. For comparison a uniform load with the same resultant ($w = 2w_0/\pi$) gives $M_{max} = wL^2/8 = w_0L^2/(4\pi) = 0.0796\,w_0L^2$, slightly below $w_0L^2/\pi^2 = 0.1013\,w_0L^2$, as expected since the sine load is concentrated towards midspan. Marks (likely): resultant and reactions 3, $V(x)$ 3, $M(x)$ 3, plots 3, maximum stress 3. 12c:T621,**Set-up.** Positive shear and positive (sagging) moment as on the formula sheet. Cut at $x$ and use the FBD of the part between A and the cut. **Sub-problem 1: uniform load $W$ only.** By symmetry $R_A' = R_B' = WL/2$. $$V'(x) = \frac{WL}{2} - Wx, \qquad M'(x) = \frac{WL}{2}x - \frac{Wx^2}{2}$$ $V'$ falls linearly from $WL/2$ to $-WL/2$; $M'$ is a parabola with maximum $WL^2/8$ at $L/2$. **Sub-problem 2: point load $P$ at midspan only.** $R_A'' = R_B'' = P/2$. $$V''(x) = \frac{P}{2}\ (0 < x < L/2), \qquad V''(x) = -\frac{P}{2}\ (L/2 < x < L)$$ $$M''(x) = \frac{P}{2}x\ (0 < x < L/2), \qquad M''(x) = \frac{P}{2}(L - x)\ (L/2 < x < L)$$ $M''$ is a triangle with peak $PL/4$ at midspan. **Sum (the given problem).** Reactions $R_A = R_B = (WL + P)/2$. $$V(x) = \frac{WL + P}{2} - Wx\ (0 < x < L/2), \qquad V(x) = \frac{WL - P}{2} - Wx\ (L/2 < x < L)$$ $$M(x) = -\frac{Wx^2}{2} + \frac{WL + P}{2}x\ (0 < x < L/2), \qquad M(x) = -\frac{Wx^2}{2} + \frac{WL - P}{2}x + \frac{PL}{2}\ (L/2 < x < L)$$ **Diagrams.** Shear: straight line of slope $-W$ from $P/2 + WL/2$ at A down to $P/2$ just left of midspan, a jump of $-P$ at midspan, then from $-P/2$ down to $-P/2 - WL/2$ at B. Moment: zero at both ends, rising to a peak $M_{max} = \dfrac{WL^2}{8} + \dfrac{PL}{4}$ at midspan (a kink there, where $V$ changes sign), symmetric. Checks (sympy): both $M$ expressions give $L(WL + 2P)/8$ at $x = L/2$ (continuous), and the second gives 0 at $x = L$; $dM/dx = V$ in each interval. Marks (likely): reactions 2, each sub-problem 3, combination and diagrams 4. 12d:T4fe,**Set-up.** $x$ from A; positive shear and sagging moment as on the formula sheet. **(a) Reactions.** The load resultant is $20 \times 6 = 120$ kN at $x = 3$ m. $$\sum M_A = 0: \quad R_C(9) - 120(3) = 0 \;\Rightarrow\; R_C = 40\ \mathrm{kN}, \qquad R_A = 120 - 40 = 80\ \mathrm{kN}$$ **Shear.** $V_A = 80$ kN. On AB ($0 < x < 6$): $V = 80 - 20x$, so $V_B = -40$ kN, crossing zero at $x_D = 4$ m. On BC there is no load, so $V = -40$ kN constant, matching $-R_C$ at C. **Moment.** $M_A = 0$. On AB: $M = 80x - 10x^2$ (parabola), so $M_D = 80(4) - 10(16) = 160$ kN m and $M_B = 80(6) - 10(36) = 120$ kN m. On BC: $M = M_B + V(x - 6) = 120 - 40(x - 6) = 360 - 40x$, falling linearly to zero at C. Check: areas under the shear diagram: $+\tfrac12(4)(80) = 160$ from A to D, $-\tfrac12(2)(40) = -40$ from D to B, $-40 \times 3 = -120$ from B to C; running total $160 \to 120 \to 0$. **(b)** The maximum moment is where $V = 0$: $dM/dx = 80 - 20x = 0$ at $x = 4$ m, $M_{max} = 160$ kN m. $$\sigma_{max} = \frac{M_{max}}{S} = \frac{160\times10^3}{1270\times10^{-6}} = 1.26\times10^8\ \mathrm{Pa} = 126\ \mathrm{MPa}$$ Check: $1\ \mathrm{cm^3} = 10^{-6}\ \mathrm{m^3}$; units N m/m$^3$ = Pa. Marks (likely): reactions 2, shear diagram 3, moment diagram 3, location 1, stress 2. 12e:T44e,**Set-up.** By symmetry the neutral axis is at mid-depth. The nails must carry the horizontal shear between the upper flange and the web, so the area "beyond the cut" is the upper flange. **First moment of the upper flange.** Its centroid is $(a + b)/2 = 60$ mm from the neutral axis: $$Q = \bar yA = \frac{a + b}{2}\,ab = 0.060 \times (0.100 \times 0.020) = 1.20\times10^{-4}\ \mathrm{m^3}$$ **Second moment of the whole section** (web about its own centroid, flanges by the parallel axis theorem): $$I = \frac{1}{12}a^3b + 2\left[\frac{1}{12}ab^3 + \left(\frac{a+b}{2}\right)^2 ab\right] = 1.667\times10^{-6} + 2(6.67\times10^{-8} + 7.20\times10^{-6}) = 16.20\times10^{-6}\ \mathrm{m^4}$$ **Shear flow and nail force.** $$q = \frac{VQ}{I} = \frac{500 \times 1.20\times10^{-4}}{16.20\times10^{-6}} = 3704\ \mathrm{N/m}$$ $$F = qz = 3704 \times 0.025 = 92.6\ \mathrm{N}$$ Check: units N m$^3$/m$^4$ = N/m. The same nail force applies at the lower flange by symmetry. The flanges contribute almost 90 % of $I$, which is why the I shape is efficient. Marks (likely): $Q$ 2, $I$ 3, $q$ 2, $F$ 1. 12f:T564,**Set-up.** $t = 0.020$ m is the joint width. $\bar y$ distances are measured from the neutral axis. **Shear at n-n.** The loading is symmetric, so $R_A = R_B = (P_1 + P_2)/2 = 3.0$ kN. Between A and $P_1$ the FBD of the left part gives $V = R_A = 3.0$ kN. **Joint u.** Area beyond the cut: the top flange, $A_1 = 100 \times 20 = 2000\ \mathrm{mm^2}$, centroid at $\bar y_1 = 120 - 68.3 - 10 = 41.7$ mm above the neutral axis: $$Q_1 = A_1\bar y_1 = 2000 \times 41.7 = 83.4\times10^3\ \mathrm{mm^3} = 83.4\times10^{-6}\ \mathrm{m^3}$$ $$\tau_u = \frac{VQ_1}{It} = \frac{3000 \times 83.4\times10^{-6}}{8.63\times10^{-6} \times 0.020} = 1.45\times10^6\ \mathrm{Pa} = 1450\ \mathrm{kPa}$$ **Joint l.** Area beyond the cut: the bottom flange, $A_2 = 60 \times 20 = 1200\ \mathrm{mm^2}$, centroid at $\bar y_2 = 68.3 - 10 = 58.3$ mm below the neutral axis: $$Q_2 = 1200 \times 58.3 = 70.0\times10^{-6}\ \mathrm{m^3}, \qquad \tau_l = \frac{3000 \times 70.0\times10^{-6}}{8.63\times10^{-6} \times 0.020} = 1.22\times10^6\ \mathrm{Pa} = 1216\ \mathrm{kPa}$$ **Checks.** Given data: $\bar y = (2000 \times 110 + 1600 \times 60 + 1200 \times 10)/4800 = 68.33$ mm and $I = 8.627\times10^{-6}\ \mathrm{m^4}$ (parallel axis theorem), both as stated. Joint u carries more because the wider top flange has the larger first moment. Marks (likely): shear force 2, $Q$ values 3, stresses 3. 130:T60b,**Set-up.** Reactions $P/2$ at each support. The shear diagram is $+P/2$ on AC and $-P/2$ on CB, so $V_{max} = P/2$; the moment diagram is a triangle with $M_{max} = PL/4$ at C. **(a)** The shear stress is parabolic over the depth with its maximum on the neutral axis; the bending stress is linear with its maximum at $y = h/2$: $$\tau_{max} = \frac{3}{2}\frac{V_{max}}{A} = \frac{3}{4}\frac{P}{bh}, \qquad \sigma_{max} = \frac{M_{max}}{I}\frac{h}{2} = \frac{(PL/4)(h/2)}{bh^3/12} = \frac{3}{2}\frac{PL}{bh^2}$$ $$\frac{\tau_{max}}{\sigma_{max}} = \frac{3P/(4bh)}{3PL/(2bh^2)} = \frac{h}{2L}$$ For a slender beam ($L \gg h$) the shear stress is much smaller than the bending stress. **(b)** $$h = 2L\,\frac{\tau_{max}}{\sigma_{max}} = 2(3.5)\frac{1.10}{31.0} = 0.248\ \mathrm{m}$$ $$b = \frac{3}{4}\frac{P}{h\,\tau_{max}} = \frac{0.75 \times 60\times10^3}{0.248 \times 1.10\times10^6} = 0.165\ \mathrm{m}$$ Check: $\sigma_{max} = 1.5 \times 60\times10^3 \times 3.5/(0.1647 \times 0.2484^2) = 31.0$ MPa, as specified. **(c)** Shear governs this part: $\tau_{max} = \dfrac{3P}{4bh} \le \tau_{all}$, so $$P_{max} = \frac{4}{3}\tau_{all}bh = \frac{4}{3} \times 1.6\times10^6 \times 0.1647 \times 0.2484 = 8.73\times10^4\ \mathrm{N} = 87.3\ \mathrm{kN}$$ Check: $P_{max}/P = 1600/1100 = 1.455$ and $60 \times 1.455 = 87.3$ kN (stresses scale with load). At this load the bending stress rises to $31.0 \times 1.455 = 45.1$ MPa; if an allowable bending stress were also given it would have to be checked. Marks (likely): (a) 4, (b) 3, (c) 2. 131:T51c,**Set-up.** Cut at $x$ and keep the part to the right (free end, no reactions needed): $P$ acts at distance $L - x$ from the cut, hogging the beam, so $$M(x) = -P(L - x), \qquad 0 \le x \le L$$ **(a)** $M < 0$ everywhere, so the curvature $y'' = M/EI$ is negative: the elastic curve is concave downwards over the whole length. With zero slope and deflection at A it leaves the wall horizontally and curves down, with the largest deflection at B. **(b)** Integrate once, with $\theta_A = \theta(0) = 0$ giving $C_1 = 0$: $$\theta(x) = \frac{1}{EI}\int_0^x M\,dx' + C_1 = \frac{1}{EI}\left[-PLx + \frac{Px^2}{2}\right]$$ Integrate again, with $y_A = y(0) = 0$ giving $C_2 = 0$: $$y(x) = \frac{1}{EI}\left[-\frac{PLx^2}{2} + \frac{Px^3}{6}\right]$$ **(c)** $y$ decreases monotonically ($\theta < 0$ for $0 < x \le L$), so the maximum deflection is at B: $$y(L) = -\frac{PL^3}{3EI}, \qquad \delta = |y(L)| = \frac{PL^3}{3EI}\ \text{(downwards)}$$ with end slope $\theta(L) = -\dfrac{PL^2}{2EI}$. Check: $EI\,y'' = -PL + Px = M(x)$ (back-substitution); units $\mathrm{N\,m^3/(N\,m^2)} = \mathrm{m}$; doubling $L$ multiplies $\delta$ by 8, the familiar cubic dependence of cantilever stiffness on length. Marks (likely): $M(x)$ 2, sketch 2, integration with boundary conditions 4, maximum deflection 2. 132:T49b,**Set-up.** Between A and C, the moment and boundary conditions are identical to a cantilever of length $L$ with an end load $P$, so the left half deforms exactly like that beam. Between C and B there is no load: $M = 0$, zero curvature, so CB stays straight and simply continues the slope at C. **(a)** AC: concave downwards, starting horizontal at A. CB: a straight line tangent to the curve at C, sloping down to B. Maximum deflection at B. **(b)** At C ($x = L$ in the length-$L$ result): $$y_C = -\frac{PL^3}{3EI}, \qquad \theta_C = -\frac{PL^2}{2EI}$$ Then for the straight segment CB: $$y_B \approx y_C + L\theta_C = -\frac{PL^3}{3EI} - \frac{PL^2}{2EI}L = -\frac{5PL^3}{6EI}$$ so $\delta_{max} = \dfrac{5PL^3}{6EI}$ downwards, at B. Check (independent): integrating $EI y'' = -P(L - x)$ on $0 \le x \le L$ and $EI y'' = 0$ on $L \le x \le 2L$ with continuity of $y$ and $\theta$ at C (sympy) gives the same $-5PL^3/(6EI)$; this is also the standard cantilever result $y = -\dfrac{Pa^2}{6EI}(3x - a)$ at $x = 2L$, $a = L$. The free end deflects 2.5 times more than C. Marks (likely): reasoning about the unloaded half 2, sketch 2, $y_C$ and $\theta_C$ 2, $y_B$ 2. 133:T531,**Set-up.** Reactions $R_A = R_B = wL/2$ (symmetry). Cut at $x$, keep the left part: $$M(x) = \frac{wL}{2}x - \frac{wx^2}{2}$$ **(a)** $M \ge 0$ on the whole span, so the elastic curve is concave upwards everywhere (sagging), zero at both supports and symmetric about midspan, where the deflection is largest. **(b)** Integrate twice: $$y(x) = \frac{1}{EI}\left[\frac{wLx^3}{12} - \frac{wx^4}{24}\right] + C_1x + C_2$$ Both conditions are on the deflection: $y(0) = 0 \Rightarrow C_2 = 0$; $y(L) = 0 \Rightarrow C_1 = -\dfrac{wL^3}{24EI}$. $$y(x) = \frac{1}{EI}\left[\frac{wLx^3}{12} - \frac{wx^4}{24} - \frac{wL^3x}{24}\right], \qquad \theta(x) = \frac{1}{EI}\left[\frac{wLx^2}{4} - \frac{wx^3}{6} - \frac{wL^3}{24}\right]$$ **(c)** By symmetry $\theta = 0$ at $x = L/2$, where the deflection is largest: $$\delta = |y(L/2)| = \frac{5wL^4}{384EI}, \qquad \theta_A = \theta(0) = -\frac{wL^3}{24EI}\ \mathrm{rad}$$ ($\theta_A$ negative: the beam slopes down from A.) Check (sympy): $\theta(L/2) = 0$ and $\theta(L) = +wL^3/(24EI) = -\theta_A$ (antisymmetric slope, as symmetry requires); $EIy'' = M$ by back-substitution. Unlike the cantilever, here neither constant comes from a slope condition: both come from the two deflection conditions. Marks (likely): $M(x)$ 2, sketch 2, constants 3, $\delta$ and $\theta_A$ 3. 134:T89d,**Reactions.** Take $R_A$ upwards and $M_A$ counterclockwise at A. $$\sum F_y = 0: \quad R_A + 2wa - \tfrac23 wa = 0 \;\Rightarrow\; R_A = -\tfrac43 wa\ \ (\text{i.e. } \tfrac43 wa \text{ downwards})$$ $$\sum M_A = 0: \quad M_A + (2wa)(a) - \tfrac23 wa(3a) = 0 \;\Rightarrow\; M_A = 0$$ **Bending moment** (left part for AB, right part for BC): $$M(x) = \frac{wx^2}{2} - \frac{4}{3}wax \quad (0 < x < 2a), \qquad M(x) = -\frac{2}{3}wa(3a - x) \quad (2a < x < 3a)$$ Both give $M(2a) = -\tfrac23 wa^2$ (continuous, no point couple at B). **Portion AB.** With $\theta(0) = 0$ and $y(0) = 0$ (fixed end) both constants vanish: $$\theta(x) = \frac{w}{EI}\left[\frac{x^3}{6} - \frac{2}{3}ax^2\right], \qquad y(x) = \frac{w}{EI}\left[\frac{x^4}{24} - \frac{2}{9}ax^3\right]$$ $$\theta_B = -\frac{4}{3}\frac{wa^3}{EI}, \qquad y_B = -\frac{10}{9}\frac{wa^4}{EI}$$ **Portion BC.** Integrate the second $M(x)$; fix the constants by continuity of slope and deflection at B: $$\theta(x) = \frac{wa}{EI}\left[-2ax + \frac{x^2}{3} + \frac{4}{3}a^2\right], \qquad y(x) = \frac{wa}{EI}\left[-ax^2 + \frac{x^3}{9} + \frac{4}{3}a^2x - \frac{2a^3}{3}\right]$$ (check: at $x = 2a$ these give $-\tfrac43 wa^3/EI$ and $-\tfrac{10}{9}wa^4/EI$). **(a)** $$\theta_C = \theta(3a) = -\frac{5}{3}\frac{wa^3}{EI}, \qquad y_C = y(3a) = -\frac{8}{3}\frac{wa^4}{EI}\ \text{(downwards)}$$ **(b)** $M$ on AB is an upward-concave parabola, zero at A, minimum $-\tfrac89 wa^2$ at its vertex $x = 4a/3$ (where $V = 0$), $-\tfrac23 wa^2$ at B; on BC it rises linearly to zero at C. $M \le 0$ everywhere, so $y$ is concave downwards along the whole beam: it leaves A horizontally and bends down to C. **(c)** The maximum deflection is at the free end or where $\theta = 0$. On AB, $\theta = 0$ only at $x = 0$ and $x = 4a$ (outside AB); on BC the roots are $x = (3 \pm \sqrt5)a = 0.76a$ and $5.24a$ (both outside BC). So $\theta \ne 0$ inside the beam and the maximum deflection is at C: $\delta_{max} = \dfrac{8}{3}\dfrac{wa^4}{EI}$. Check (sympy): $EIy'' = M$ on each segment and both continuity conditions hold. Marks (likely): reactions 2, $M(x)$ 3, AB integration 3, BC with continuity 4, answers 2, sketches 3, maximum 3. 135:T938,**Set-up.** Reactions: $R_{Ax}$, $R_{Ay}$ (up), $M_A$ (counterclockwise at A) and $R_{By}$ (up): four unknowns, three equilibrium equations, so the beam is statically indeterminate (degree 1). Write the elastic curve in terms of the unknown reactions, then use the extra support condition $y_B = 0$. **(a) Equilibrium.** $$\sum F_x = 0: R_{Ax} = 0; \qquad \sum F_y = 0: R_{Ay} + R_{By} - wL = 0 \quad (1)$$ $$\sum M_A = 0: \quad M_A + R_{By}L - \frac{wL^2}{2} = 0 \quad (2)$$ **Moment** (left part, $M_A$ counterclockwise acting on the left end produces a hogging moment): $$M(x) = -M_A + R_{Ay}x - \frac{wx^2}{2}$$ **Elastic curve** with $\theta(0) = 0$, $y(0) = 0$ (fixed end, so both constants are zero): $$\theta(x) = \frac{1}{EI}\left[-M_Ax + \frac{R_{Ay}x^2}{2} - \frac{wx^3}{6}\right], \qquad y(x) = \frac{1}{EI}\left[-\frac{M_Ax^2}{2} + \frac{R_{Ay}x^3}{6} - \frac{wx^4}{24}\right]$$ **Compatibility** $y(L) = 0$: $$-\frac{M_A}{2} + \frac{R_{Ay}L}{6} - \frac{wL^2}{24} = 0 \quad (3)$$ Solving (1), (2), (3): $$R_{Ay} = \frac{5}{8}wL, \qquad R_{By} = \frac{3}{8}wL, \qquad M_A = \frac{1}{8}wL^2$$ Check: $R_{Ay} + R_{By} = wL$; the fixed end, being stiffer, takes the larger share. **(b)** With these reactions $$M(x) = \frac{w}{8}\left(-4x^2 + 5Lx - L^2\right) = -\frac{w}{8}(L - x)(L - 4x)$$ a downward-concave parabola: $M(0) = -wL^2/8$ (hogging), zero at $x = L/4$ (point of inflection of the elastic curve), maximum $+\tfrac{9}{128}wL^2$ at $x = 5L/8$ (where $V = 0$), and zero at B. The elastic curve leaves A horizontally, is concave down for $x < L/4$, concave up beyond, and returns to zero at B with a nonzero slope. **(c)** Maximum deflection where $\theta = 0$ inside the span: $$\theta(x) = \frac{w}{EI}\left[-\frac{L^2x}{8} + \frac{5Lx^2}{16} - \frac{x^3}{6}\right] = -\frac{wx}{48EI}\left(8x^2 - 15Lx + 6L^2\right) = 0$$ $$x_\delta = \frac{15 - \sqrt{33}}{16}L = 0.5785L$$ (the other root, $1.297L$, lies outside the beam). There $|y| = 0.00542\,wL^4/EI \approx wL^4/(185EI)$. Check (sympy): $y(L) = 0$, $EIy'' = M$, and the quadratic roots are $(15 \pm \sqrt{33})L/16$. The maximum deflection lies beyond midspan, towards the roller, because the clamped end holds its half of the beam up. Marks (likely): indeterminacy argument 2, equilibrium 2, elastic curve 4, compatibility and reactions 4, sketch 3, position 3. 136:T699,**Set-up.** $\sigma_x = +100$, $\sigma_y = -20$ (compression is negative), $\tau_{xy} = +80$ MPa; $\theta$ counterclockwise from $x$. **(a)** $$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{160}{120} = 1.333 \;\Rightarrow\; 2\theta_p = 53.1^\circ$$ $$\theta_{p1} = 26.6^\circ, \qquad \theta_{p2} = \theta_{p1} + 90^\circ = 116.6^\circ$$ **(b)** $$\sigma_{max,min} = \frac{100 - 20}{2} \pm \sqrt{\left(\frac{100 + 20}{2}\right)^2 + 80^2} = 40 \pm 100$$ $$\sigma_{max} = 140\ \mathrm{MPa}, \qquad \sigma_{min} = -60\ \mathrm{MPa}$$ Which is which: substitute $\theta = 26.6^\circ$ into $\sigma_{x'} = 40 + 60\cos2\theta + 80\sin2\theta = 40 + 60(0.6) + 80(0.8) = 140$ MPa, so the face at $26.6^\circ$ carries $\sigma_{max}$ and the face at $116.6^\circ$ carries $\sigma_{min} = -60$ MPa. Shear on both is zero (check: $-60(0.8) + 80(0.6) = 0$). **(c)** $$\tan2\theta_s = -\frac{\sigma_x - \sigma_y}{2\tau_{xy}} = -\frac{120}{160} = -0.75 \;\Rightarrow\; \theta_{s1} = -18.4^\circ, \quad \theta_{s2} = 71.6^\circ$$ $$\tau_{max} = \sqrt{60^2 + 80^2} = 100\ \mathrm{MPa}, \qquad \sigma' = \frac{\sigma_x + \sigma_y}{2} = 40\ \mathrm{MPa}$$ Sign: $\tau_{x'y'}(-18.4^\circ) = -60\sin(-36.9^\circ) + 80\cos(-36.9^\circ) = +100$ MPa (positive on the face rotated by $\theta_{s1}$), and $-100$ MPa at $\theta_{s2} = 71.6^\circ$. Checks: $\theta_{p1} - \theta_{s1} = 45^\circ$ (maximum-shear planes lie at $45^\circ$ to the principal planes); $\sigma_{max} + \sigma_{min} = 80 = \sigma_x + \sigma_y$; Mohr's circle centre 40 MPa, radius 100 MPa. Marks (likely): principal angles 2, principal stresses 3 (with assignment 1), shear planes 2, $\tau_{max}$ and $\sigma'$ 2. 137:T7d3,**Internal loads at H.** Take the part of the axle from the wheel to the section through H. The two 3 kN forces form a couple: $M = 3000 \times 0.15 = 450$ N m, the same at every section between the body loads (zero shear there). Check: $3 \times 0.35 - 3 \times 0.20 = 0.45$ kN m. The upward force at the outer end sags the axle, so the top fibre (H) is in compression. The torque is $T = 350$ N m. **Stresses at H** (surface element, plane stress: no load on the free surface): $$|\sigma_x| = \frac{Mr}{I} = \frac{450 \times 0.016}{\pi(0.016)^4/4} = 139.9\ \mathrm{MPa}, \quad \sigma_x = -139.9\ \mathrm{MPa}$$ $$|\tau_{xy}| = \frac{Tr}{J} = \frac{350 \times 0.016}{\pi(0.016)^4/2} = 54.4\ \mathrm{MPa}, \quad \tau_{xy} = -54.4\ \mathrm{MPa}, \qquad \sigma_y = 0$$ **(a)** $$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2(-54.4)}{-139.9} = 0.778 \;\Rightarrow\; \theta_{p} = 18.9^\circ \text{ and } 108.9^\circ$$ $$\sigma_{max,min} = \frac{-139.9}{2} \pm \sqrt{\left(\frac{-139.9}{2}\right)^2 + 54.4^2} = -69.9 \pm 88.6$$ $$\sigma_{max} = 18.7\ \mathrm{MPa}, \qquad \sigma_{min} = -158.5\ \mathrm{MPa}$$ Which face carries which: $\sigma_{x'}(18.9^\circ) = -69.9 - 69.9\cos37.9^\circ - 54.4\sin37.9^\circ = -158.5$ MPa, so the principal axis at $18.9^\circ$ carries $\sigma_{min}$ (compression, close to the axial direction, as the bending stress dominates) and the axis at $108.9^\circ$ carries $\sigma_{max} = +18.7$ MPa. **(b)** $$\tau_{max} = \sqrt{\left(\frac{-139.9}{2}\right)^2 + 54.4^2} = 88.6\ \mathrm{MPa} = \frac{\sigma_{max} - \sigma_{min}}{2}$$ on planes at $45^\circ$ to the principal axes ($\theta_s = -26.1^\circ$ and $63.9^\circ$), carrying normal stress $-69.9$ MPa. Checks: $\sigma_{max} + \sigma_{min} = -139.9 = \sigma_x + \sigma_y$; $J = 2I$ for a circle, so $\tau/\sigma = T/(2M) = 350/900$ (0.389, and $54.4/139.9 = 0.389$). Marks (likely): internal loads 3, stresses at H 4, principal axes 3, principal stresses 3, assignment 2, $\tau_{max}$ 2. 138:T5de,**Method.** Two requirements: no buckling with the safety factor, $P_{cr} = \dfrac{\pi^2EI}{L^2} \ge kP$, and no overstress, $P/(\pi r^2) \le \sigma_{all}$. Size for buckling first, then check the stress; if the stress check fails, size for stress instead. **(a) $P_1 = 380$ kN.** $$P_{cr} = kP_1 = 2.5 \times 380 = 950\ \mathrm{kN}, \qquad I = \frac{P_{cr}L^2}{\pi^2E} = \frac{950\times10^3 \times 2^2}{\pi^2 \times 200\times10^9} = 1.925\times10^{-6}\ \mathrm{m^4}$$ $$\frac{\pi r^4}{4} = 1.925\times10^{-6} \;\Rightarrow\; r = 39.6\ \mathrm{mm}$$ Stress check: $\sigma = \dfrac{380\times10^3}{\pi(0.0396)^2} = 77.3\ \mathrm{MPa} < 100$ MPa. Acceptable: buckling governs, $r_{min} = 39.6$ mm (choose, for example, 40 mm). **(b) $P_2 = 800$ kN.** $$P_{cr} = 2000\ \mathrm{kN}, \qquad I = 4.053\times10^{-6}\ \mathrm{m^4} \;\Rightarrow\; r = 47.7\ \mathrm{mm}$$ Stress check: $\sigma = \dfrac{800\times10^3}{\pi(0.0477)^2} = 112.1\ \mathrm{MPa} > 100$ MPa: NOT acceptable. Size for the stress instead: $$\frac{P_2}{\pi r^2} = \sigma_{all} \;\Rightarrow\; r = \sqrt{\frac{800\times10^3}{\pi \times 100\times10^6}} = 50.5\ \mathrm{mm}$$ With $r = 50.5$ mm the buckling safety factor is $\pi^2EI/(L^2P_2) = 3.14 > 2.5$, so both requirements hold. Checks: $I \propto P$ so $r \propto P^{1/4}$: $39.6 \times (800/380)^{1/4} = 47.7$ mm; units $\mathrm{N\,m^2/Pa} = \mathrm{m^4}$. Marks (likely): (a) $P_{cr}$ 1, $I$ and $r$ 3, stress check 2; (b) buckling size 2, failed check 2, stress sizing 2. 139:T431,**Set-up.** $A = \pi r^2 = 5.027\times10^{-3}\ \mathrm{m^2}$, $\rho = \sqrt{I/A} = r/2 = 20$ mm. The allowable working stress is the smaller of the buckling stress divided by $k$ and $\sigma_{all}$. **(a) Pinned-pinned,** $L_e = L = 2$ m, slenderness $L_e/\rho = 2/0.02 = 100$: $$\frac{\sigma_{cr}}{k} = \frac{1}{2.5}\frac{\pi^2 \times 200\times10^9}{100^2} = \frac{197.4}{2.5} = 78.96\ \mathrm{MPa} < \sigma_{all}$$ Buckling governs: $P_{max} = 78.96\times10^6 \times 5.027\times10^{-3} = 396.9$ kN. **(b) Fixed-fixed,** $L_e = 0.5L = 1$ m, slenderness 50: $$\frac{\sigma_{cr}}{k} = \frac{789.6}{2.5} = 315.8\ \mathrm{MPa} > \sigma_{all}$$ Now the material strength governs: $P_{max} = \sigma_{all}A = 100\times10^6 \times 5.027\times10^{-3} = 502.7$ kN. Checks: halving $L_e$ multiplies the Euler load by 4, so the factored buckling load in (b) is $4 \times 396.9 = 1588$ kN, far above the strength limit of 502.7 kN; $\rho = r/2$ follows from $\sqrt{(\pi r^4/4)/(\pi r^2)}$. Marks (likely): $\rho$ and slenderness 2, each case 3, choosing the governing limit 2. 13a:T59b,**Set-up.** The column can buckle in either plane of symmetry; each plane has its own end conditions and its own second moment of area. Bending in the $xy$ plane (lateral deflection along $y$) is resisted by $I_z$, which involves the side $a$ cubed; bending in the $xz$ plane by $I_y$, which involves $b$ cubed. $A = ab$. **(a) $xy$ plane.** Base fixed, top held in $y$ but free to rotate: fixed-pinned, $L_e = 0.7L$. $$I_z = \frac{ba^3}{12}, \qquad \rho_z = \sqrt{\frac{I_z}{A}} = \frac{a}{\sqrt{12}}, \qquad \frac{L_e}{\rho_z} = \frac{0.7L\sqrt{12}}{a} = 2.425\frac{L}{a}$$ **(b) $xz$ plane.** Base fixed, top free to move in $z$: fixed-free, $L_e = 2L$. $$I_y = \frac{ab^3}{12}, \qquad \rho_y = \frac{b}{\sqrt{12}}, \qquad \frac{L_e}{\rho_y} = \frac{2L\sqrt{12}}{b} = 6.928\frac{L}{b}$$ **(c)** The critical stress $\sigma_{cr} = \pi^2E/(L_e/\rho)^2$ is set by the LARGER slenderness ratio. For a given area, the most efficient section makes the column equally likely to buckle in both planes, i.e. equal slenderness ratios (any other ratio wastes material in the stronger plane): $$\frac{0.7L}{a/\sqrt{12}} = \frac{2L}{b/\sqrt{12}} \;\Rightarrow\; \frac{a}{b} = \frac{0.7}{2} = 0.35$$ Check: with $a/b = 0.35$, (a) gives $2.425L/(0.35b) = 6.928L/b$, equal to (b). The section is deeper ($b$) in the direction where the top is unrestrained, as intuition suggests. Marks (likely): (a) 3, (b) 3, (c) ratio 2, justification 2. 13f:X 0:{"P":null,"c":["","practice"],"q":"","i":false,"f":[[["",{"children":["practice",{"children":["__PAGE__",{},"$undefined","$undefined",4608]},"$undefined","$undefined",4608]},"$undefined","$undefined",4624],[["$","$1","c",{"children":[[["$","link","0",{"rel":"stylesheet","href":"/_next/static/immutable/chunks/2oqtt1w9attoc.css","precedence":"next","crossOrigin":"$undefined","nonce":"$undefined"}],["$","script","script-0",{"src":"/_next/static/immutable/chunks/3vnkbo3uor9pe.js","async":true,"nonce":"$undefined"}],["$","script","script-1",{"src":"/_next/static/immutable/chunks/3u-moss-j0vod.js","async":true,"nonce":"$undefined"}]],["$","html",null,{"lang":"en","children":["$","body",null,{"children":[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"WebSite\",\"name\":\"Revisica\",\"url\":\"https://revisica.net\",\"description\":\"Free revision for Lancaster University second year engineering: past paper and tutorial questions with worked solutions, practice sets, mock exams, quizzes, steam tables and lecture study packs.\",\"potentialAction\":{\"@type\":\"SearchAction\",\"target\":\"https://revisica.net/search?q={search_term_string}\",\"query-input\":\"required name=search_term_string\"}}"}}],["$","$L2",null,{"children":["$","$L3",null,{"children":[["$","$L4",null,{}],["$","$L5",null,{}],["$","$L6",null,{}],["$","main",null,{"className":"page","children":["$","div",null,{"className":"wrap","children":["$","$L7",null,{"parallelRouterKey":"children","error":"$undefined","errorStyles":"$undefined","errorScripts":"$undefined","template":["$","$L8",null,{}],"templateStyles":"$undefined","templateScripts":"$undefined","notFound":[[["$","title",null,{"children":"404: This page could not be found."}],["$","div",null,{"style":{"fontFamily":"system-ui,\"Segoe UI\",Roboto,Helvetica,Arial,sans-serif,\"Apple Color Emoji\",\"Segoe UI Emoji\"","height":"100vh","textAlign":"center","display":"flex","flexDirection":"column","alignItems":"center","justifyContent":"center"},"children":["$","div",null,{"children":[["$","style",null,{"dangerouslySetInnerHTML":{"__html":"body{color:#000;background:#fff;margin:0}.next-error-h1{border-right:1px solid rgba(0,0,0,.3)}@media (prefers-color-scheme:dark){body{color:#fff;background:#000}.next-error-h1{border-right:1px solid rgba(255,255,255,.3)}}"}}],["$","h1",null,{"className":"next-error-h1","style":{"display":"inline-block","margin":"0 20px 0 0","padding":"0 23px 0 0","fontSize":24,"fontWeight":500,"verticalAlign":"top","lineHeight":"49px"},"children":404}],["$","div",null,{"style":{"display":"inline-block"},"children":["$","h2",null,{"style":{"fontSize":14,"fontWeight":400,"lineHeight":"49px","margin":0},"children":"This page could not be found."}]}]]}]}]],[]],"forbidden":"$undefined","unauthorized":"$undefined"}]}]}],["$","footer",null,{"className":"site-footer","children":["$","div",null,{"className":"wrap footer-inner","children":[["$","div",null,{"className":"footer-text","children":[["$","p",null,{"children":"Revisica is an independent, student-run revision site. 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The displacement $x(t)$ of the mass\nis measured downwards from its static equilibrium position, and an external force $F(t)$ acts on the mass in the\ndirection of $x$. (The spring stiffness is written $K'$ so that it is not confused with the $K$ of the\ngeneralised form.)\n","parts":[{"label":"a","text":"List the assumptions needed for a linear model.","grade":"$undefined"},{"label":"b","text":"Use Newton's second law to develop the model relating $F(t)$ to $x(t)$.","grade":"$undefined"},{"label":"c","text":"Put the model into the generalised second order form and identify the output, the input (forcing function),\n$K$, the natural frequency $\\omega_n$, the damping ratio $\\zeta$ and the steady state gain.\n","grade":"$undefined"},{"label":"d","text":"What is the implication of the damping ratio you found for the free response?","grade":"$undefined"}],"solution":"$d","confidence":"HIGH","solutionNote":"Matches Lecture 2 slides 4-7 and 15 (also the old-spec Control 1.3 \"very quick exercise\", slides 13-14, which is the same example)."},{"id":"engr5001-ex-model-02","module":"ENGR5001","week":1,"label":"Lecture 2 example, LC circuit","kind":"tutorial","level":2,"minutes":10,"tags":["modelling","generalised-forms","lc-circuit","mechanistic-models","damping-ratio"],"source":{"name":"ENGR5001 Lecture 2 worked example 2, LC circuit (slides 9-14)","path":"ENGR5001 - Control and Robotics/Week 1/ENGR5001_P1_lecture2.pdf"},"grade":"2:2","formulaSheet":"- Inductor $V_L = L\\dfrac{di}{dt}$; capacitor $V_C = \\dfrac{q}{C} = \\dfrac{1}{C}\\displaystyle\\int i\\,dt$; Kirchhoff's voltage law (learn this)\n- Generalised second order form $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$ (learn this)\n","text":"For the LC circuit in the figure (an inductor $L$ in series with a capacitor $C$, input voltage $V_i$ applied\nacross the pair, output voltage $V_o$ taken across the capacitor, loop current $i$):\n","figures":[{"src":"/figures/engr5001-ex-model-02.png","caption":"LC circuit: inductor L in series with capacitor C, input V_i, output V_o across C, loop current i."}],"parts":[{"label":"a","text":"State the assumptions.","grade":"$undefined"},{"label":"b","text":"Develop the linear model relating $V_i$ to $V_o$.","grade":"$undefined"},{"label":"c","text":"Convert the model into the generalised form and identify the input, output, $K$, natural frequency, damping\nratio and steady state gain. Which mechanical model has the same form?\n","grade":"$undefined"}],"solution":"$e","confidence":"HIGH","solutionNote":"Matches Lecture 2 slides 9-14 (the same example is the old-spec Control 1.3 exercise on slide 20, solution on slide 26)."},{"id":"engr5001-ex-model-03","module":"ENGR5001","week":1,"label":"Lecture 2 example, mass-spring-damper","kind":"tutorial","level":2,"minutes":8,"tags":["modelling","generalised-forms","mass-spring-damper","mechanistic-models","damping-ratio"],"source":{"name":"ENGR5001 Lecture 2 worked example 3, mass-spring-damper (slides 18-20)","path":"ENGR5001 - Control and Robotics/Week 1/ENGR5001_P1_lecture2.pdf"},"grade":"2:2","formulaSheet":"- Newton's second law; spring force $K'x$; viscous damper force $C\\dfrac{dx}{dt}$ (learn this)\n- Generalised second order form $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$, steady state gain $K/\\omega_n^2$ (learn this)\n","text":"A mass $M$ hangs from a fixed support by a spring of stiffness $K'$ and a viscous damper of coefficient $C$ in\nparallel. An external force $F(t)$ acts on the mass, and $x(t)$ is the displacement of the mass from its static\nequilibrium position, measured in the direction of $F$.\n","parts":[{"label":"a","text":"State the additional assumption needed compared with the undamped harmonic oscillator, and derive the model.\n","grade":"$undefined"},{"label":"b","text":"Put the model into generalised second order form, giving $K$, $\\omega_n$, $\\zeta$ and the steady state gain in\nterms of $M$, $C$ and $K'$.\n","grade":"$undefined"},{"label":"c","text":"How does the time response differ from that of the harmonic oscillator?","grade":"$undefined"}],"solution":"$f","confidence":"HIGH","solutionNote":"Matches Lecture 2 slides 18-20 (also old-spec Control 1.2 slide 7 and 1.3 slides 16-17)."},{"id":"engr5001-ex-model-04","module":"ENGR5001","week":1,"label":"Lecture 2 example, thermal system (liquid cooling)","kind":"tutorial","level":2,"minutes":8,"tags":["modelling","generalised-forms","mechanistic-models","time-constant","steady-state-gain"],"source":{"name":"ENGR5001 Lecture 2 worked example 4, thermal system (slides 23-25)","path":"ENGR5001 - Control and Robotics/Week 1/ENGR5001_P1_lecture2.pdf"},"grade":"2:2","formulaSheet":"- Newton's law of cooling: rate of change of temperature proportional to the excess temperature (learn this)\n- Generalised first order form $\\tau\\dfrac{dx}{dt} + x = Ku(t)$ (learn this)\n","text":"A container of liquid at temperature $\\theta$ stands in surroundings at temperature $\\theta_S$ (for example a cup of\ncoffee cooling in a room).\n","parts":[{"label":"a","text":"State the modelling assumption (Newton's law of cooling) and write it as a differential equation with a model\ncoefficient $K'$. Identify the output and the input.\n","grade":"$undefined"},{"label":"b","text":"Put the model into the generalised first order form and give the time constant and steady state gain. What\nsign must $K'$ have, and why?\n","grade":"$undefined"}],"solution":"$10","confidence":"HIGH","solutionNote":"Matches Lecture 2 slides 23-25 (model constant renamed K' as on slide 25)."},{"id":"engr5001-ex-model-05","module":"ENGR5001","week":1,"label":"Lecture 2 example, single tank hydraulic system","kind":"tutorial","level":2,"minutes":10,"tags":["modelling","generalised-forms","single-tank","mechanistic-models","time-constant"],"source":{"name":"ENGR5001 Lecture 2 worked example 5, single tank hydraulic system (slides 26-28)","path":"ENGR5001 - Control and Robotics/Week 1/ENGR5001_P1_lecture2.pdf"},"grade":"2:2","formulaSheet":"- Conservation of volume (mass) for an incompressible fluid: $A\\dfrac{dh}{dt} = Q_i - Q_o$ (learn this)\n- Linear outflow assumption $Q_o = K'h$ (learn this)\n- Generalised first order form $\\tau\\dfrac{dx}{dt} + x = Ku(t)$ (learn this)\n","text":"A tank of constant cross-sectional area $A$ is filled at volumetric flow rate $Q_i$ and drains through an orifice\nnear the base at flow rate $Q_o$. The liquid depth above the orifice is $h$ (see figure).\n","figures":[{"src":"/figures/engr5001-ex-model-05.png","caption":"Single tank of area A, inflow Q_i at the top, head h, outflow Q_o through an orifice."}],"parts":[{"label":"a","text":"State the assumptions used in the lecture model.","grade":"$undefined"},{"label":"b","text":"Derive a linear model relating the outflow $Q_o$ to the inflow $Q_i$.","grade":"$undefined"},{"label":"c","text":"Put it into generalised first order form and give the time constant and steady state gain.","grade":"$undefined"}],"solution":"**(a) Assumptions** (slide 26): flow rate out is proportional to head, $Q_o = K'h$; no energy dissipated in the\norifice; incompressible fluid. Also: constant tank area $A$, tank never empties or overflows.\n\n**(b)** From the first assumption $h = Q_o/K'$, so $\\dfrac{dh}{dt} = \\dfrac{1}{K'}\\dfrac{dQ_o}{dt}$.\nVolume balance (rate of change of stored volume = flow in - flow out):\n$$\\frac{dV}{dt} = A\\frac{dh}{dt} = Q_i - Q_o \\;\\Rightarrow\\; \\frac{A}{K'}\\frac{dQ_o}{dt} = Q_i - Q_o \\;\\Rightarrow\\; \\frac{A}{K'}\\frac{dQ_o}{dt} + Q_o = Q_i$$\n\n**(c)** Compare with $\\tau\\dot x + x = Ku$: $x = Q_o$, $u = Q_i$, $\\tau = A/K'$, $K = 1$.\nGain 1: at steady state $dh/dt = 0$, so everything that flows in flows out. A wider tank (bigger $A$) or a smaller\norifice (smaller $K'$) gives a slower response.\n\n**Checks.** Units: $K'$ in m$^3$s$^{-1}$/m = m$^2$/s, so $A/K'$ is m$^2$/(m$^2$/s) = s. In terms of the level,\n$\\tau\\dot h + h = (1/K')Q_i$: same time constant, steady state gain $1/K'$ (m per m$^3$/s).\n","confidence":"HIGH","solutionNote":"Matches Lecture 2 slides 26-28. Two wording issues on the slides: the volume balance $A\\,dh/dt = Q_i - Q_o$ is\nlabelled \"Torricelli's law\", but it is conservation of mass (continuity); Torricelli's law is the ideal orifice\nresult $Q_o \\propto \\sqrt{h}$ (no losses), which is nonlinear. A linear law $Q_o = K'h$ is really a linearised\n(or laminar, dissipative) orifice, so the assumption \"no energy dissipated in orifice\" sits uneasily with it.\nNeither affects the model or the answers.\n"},{"id":"engr5001-ex-model-06","module":"ENGR5001","week":1,"label":"Old spec 1.2 example, two tank hydraulic system","kind":"tutorial","level":2,"minutes":15,"tags":["modelling","generalised-forms","mechanistic-models","damping-ratio","single-tank"],"source":{"name":"ENGR202 (old spec) Control 1.2 Mechanistic Models, two tank hydraulic system (slides 14-17; generalised form in 1.3 slides 24-25)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 1 - Control 1.2 Mechanistic Models (old spec ENGR202).pdf"},"grade":"2:1","formulaSheet":"- Volume balance $A\\dfrac{dh}{dt} = Q_{in} - Q_{out}$; linear orifice $Q = Kh$ (learn this)\n- Generalised second order form $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$ (learn this)\n","text":"Two tanks are connected in cascade (see figure). Tank 1 (area $A_1$, head $h_1$) is filled at flow rate $Q_1$ and\ndrains through an orifice at flow rate $Q_2$ into tank 2 (area $A_2$, head $h_2$), which drains through its own\norifice at flow rate $Q_3$. Assume, in addition to the single tank assumptions, that each outflow is proportional to\nits own head: $Q_2 = K_1h_1$ and $Q_3 = K_2h_2$.\n","figures":[{"src":"/figures/engr5001-ex-model-06.png","caption":"Two tanks in cascade: Q_1 into tank 1 (area A_1, head h_1), Q_2 from tank 1 into tank 2 (area A_2, head h_2), Q_3 out of tank 2."}],"parts":[{"label":"a","text":"Derive a first order model for each tank.","grade":"$undefined"},{"label":"b","text":"Eliminate $Q_2$ to obtain a single model relating $Q_3$ to $Q_1$.","grade":"$undefined"},{"label":"c","text":"Put the model into generalised second order form, giving $\\omega_n$, $\\zeta$ and the steady state gain. Show\nthat the system can never oscillate.\n","grade":"$undefined"}],"solution":"$11","confidence":"HIGH","solutionNote":"Matches old-spec Control 1.2 slides 14-17 and 1.3 slides 24-25. The no-oscillation result in (c) is added here; it is not on the slides."},{"id":"engr5001-ex-model-07","module":"ENGR5001","week":1,"label":"Old spec 1.4 example, linearised free pendulum","kind":"tutorial","level":2,"minutes":8,"tags":["modelling","mechanistic-models","generalised-forms"],"source":{"name":"ENGR202 (old spec) Control 1.4 Nonlinear Systems, example linearised model for a free pendulum (slides 11-15)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 1 - Control 1.4 Nonlinear Systems (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Linearisation about an operating point $x_0$: $g(x) \\approx g(x_0) + \\left.\\dfrac{dg}{dx}\\right|_{x = x_0}(x - x_0)$ (first order Taylor series) (learn this)\n","text":"A free pendulum consists of a mass $M$ on a light rod of length $L$, pivoted at the top; $\\theta$ is the angle of\nthe rod from the vertical (see figure). The restoring torque about the pivot due to gravity is\n$$T = MLg\\sin\\theta$$\nwhich is a continuous nonlinear function of $\\theta$.\n","figures":[{"src":"/figures/engr5001-ex-model-07.png","caption":"Free pendulum: mass M on a rod of length L, angle theta from the vertical."}],"parts":[{"label":"a","text":"Linearise the torque about the operating condition $\\theta_0 = 0$ using a Taylor series.","grade":"$undefined"},{"label":"b","text":"When is the linear model valid? Find the percentage error in the linearised torque at $\\theta = 10$ degrees.\n","answer":[{"label":"error at 10 deg","value":0.51,"units":"%","tol":0.05}],"grade":"$undefined"}],"solution":"$12","confidence":"HIGH","solutionNote":"Part (a) matches old-spec Control 1.4 slide 15. The numerical error in (b) and the equation of motion check are added here (the slide only states that the model is valid for small angles)."},{"id":"engr5001-ex-time-01","module":"ENGR5001","week":2,"label":"Old spec 2.4 numerical example, first order impulse response","kind":"tutorial","level":1,"minutes":5,"tags":["time-response","generalised-forms","time-constant"],"source":{"name":"ENGR202 (old spec) Control 2.4 Time Response, numerical example (slides 5-8)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 2 - Control 2.4 Time Response (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Unit impulse response of $\\tau\\dot x + x = Ku$ (zero initial conditions): $x(t) = \\dfrac{K}{\\tau}e^{-t/\\tau}$ (learn this)\n","text":"A first order system $\\tau\\dfrac{dx}{dt} + x = Ku(t)$ has a time constant of 2 s and unity steady state gain. It\nstarts at rest and a unit impulse is applied at $t = 0$.\n","parts":[{"label":"a","text":"Write down the model and its unit impulse response $x(t)$.","grade":"$undefined"},{"label":"b","text":"Find $x(0^+)$ and $x$ at $t = 2$ s, and sketch the response.","answer":[{"label":"x(0+)","value":0.5},{"label":"x(2 s)","value":0.1839}],"grade":"$undefined"}],"solution":"**(a)** Model: $2\\dfrac{dx}{dt} + x = u$. The impulse response of the first order model (from the general solution\nwith the impulse integrated from $0^-$ to $0^+$) is the transient response with initial condition $x(0) = K/\\tau$:\n$$x(t) = \\frac{K}{\\tau}e^{-t/\\tau} = 0.5e^{-t/2}$$\n\n**(b)** $x(0^+) = K/\\tau = 0.5$; $x(2) = 0.5e^{-1} = 0.184$ (36.8% of the initial value after one time constant);\n$x(4) = 0.068$, $x(8) = 0.009$. Sketch: zero for $t < 0$, a jump to 0.5 at $t = 0$, then an exponential decay to\nzero, essentially finished after $4\\tau = 8$ s. The tangent at $t = 0^+$ (slope $-0.25$ per s) hits zero at\n$t = \\tau = 2$ s.\n\n**Checks.** Substitution: $2(-0.25e^{-t/2}) + 0.5e^{-t/2} = 0$ for $t > 0$, as required with no input. The impulse\nresponse is the derivative of the step response $K(1 - e^{-t/\\tau})$: $\\frac{d}{dt} = \\frac{K}{\\tau}e^{-t/\\tau}$.\nThe area under it is $K = 1$, the steady state gain. A positive time constant gives decay (stable); a negative one\nwould give exponential growth.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 2.4 slides 7-8 ($x(t) = 0.5e^{-t/2}$). The value at $t = 2$ s is added as a check."},{"id":"engr5001-ex-time-02","module":"ENGR5001","week":2,"label":"Old spec 2.4 exercise, key points of the first order step response","kind":"tutorial","level":1,"minutes":4,"tags":["time-response","step-response","time-constant","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 2.4 Time Response, exercise (slide 11; answer slide 12)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 2 - Control 2.4 Time Response (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- First order unit step response $x(t) = K\\left(1 - e^{-t/\\tau}\\right)$ (learn this)\n","text":"For a unit step input, the output of the first order system $\\tau\\dfrac{dx}{dt} + x = Ku$ (starting from rest) is\n$$x(t) = K\\left(1 - e^{-t/\\tau}\\right)$$\nComplete the table: the output $x(t)$ at $t = 0$, at $t = \\tau$ and as $t \\to \\infty$, as a multiple of $K$. Then\nexplain how these points are used to estimate $K$ and $\\tau$ from an experimental step response.\n","answer":[{"label":"x(0) / K","value":0},{"label":"x(tau) / K","value":0.632,"tol":0.01},{"label":"x(infinity) / K","value":1}],"solution":"| $t$ | $x(t)$ |\n|---|---|\n| 0 | $K(1 - e^0) = 0$ |\n| $\\tau$ | $K(1 - e^{-1}) = 0.632K$ |\n| $\\to\\infty$ | $K(1 - e^{-\\infty}) = K$ |\n\n**Use.** The output settles at the steady state gain, so for a step of size $u_0$,\n$K = x(t \\to \\infty)/u_0$. The time constant is the time (measured from the step) for the output to complete\n63.2% of its change. Further points: $x(2\\tau) = 0.865K$, $x(3\\tau) = 0.950K$, $x(4\\tau) = 0.982K$, so the response\nis about 98% complete after $4\\tau$. The initial slope is $K/\\tau$: the tangent at $t = 0$ reaches $K$ at\n$t = \\tau$.\n\n**Checks.** $e^{-1} = 0.3679$ (Python). Substituting $x(t)$ into the ODE: $\\tau\\frac{K}{\\tau}e^{-t/\\tau} + K - Ke^{-t/\\tau} = K$ for $u = 1$.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 2.4 slide 12 (which labels all three rows \"x(0) =\", a slip; the values 0, 0.6321K and K are correct)."},{"id":"engr5001-ex-time-03","module":"ENGR5001","week":2,"label":"Old spec 2.4 worked example, axial fan model from a step test","kind":"tutorial","level":2,"minutes":8,"tags":["time-response","graphical-estimation","step-response","time-constant","steady-state-gain","data-based-models"],"source":{"name":"ENGR202 (old spec) Control 2.4 Time Response, axial fan worked example (slides 18-20)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 2 - Control 2.4 Time Response (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- $K = \\dfrac{x(t \\to \\infty)}{u(t \\to \\infty)}$ (for a step from zero) (learn this)\n- At $t = \\tau$ after the step: $x = 0.632\\,x(t \\to \\infty)$ (learn this)\n","text":"The figure shows a laboratory step test on an axial fan. The input voltage is stepped from 0 to 40 V at about\n$t = 5$ s, and the airflow (m$^3$/hour) rises from 0 towards a steady value. The fan is to be modelled by the\nfirst order system $\\tau\\dfrac{dx}{dt} + x = Ku(t)$, with $x$ the airflow and $u$ the input voltage.\n","figures":[{"src":"/figures/engr5001-ex-time-03.png","caption":"Axial fan step test: input voltage steps from 0 to 40 V at about t = 5 s; airflow (m^3/hour) rises to about 60."}],"parts":[{"label":"a","text":"Estimate the steady state gain $K$.","answer":[{"label":"K","value":1.5,"units":"(m^3/hour)/V","tol":0.05}],"grade":"$undefined"},{"label":"b","text":"Estimate the time constant $\\tau$, and write down the model.","answer":[{"label":"tau","value":10,"units":"s","tol":0.1}],"grade":"$undefined"},{"label":"c","text":"Using the model, what airflow is expected at steady state for a constant input of 20 V? Sketch the response to\na step from 0 to 20 V.\n","answer":[{"label":"x(infinity) for 20 V","value":30,"units":"m^3/hour","tol":0.05}],"grade":"$undefined"}],"solution":"$13","confidence":"HIGH","solutionNote":"Matches old-spec Control 2.4 slides 19-20 ($K = 1.5$, $\\tau = 10$ s, $10\\dot x + x = 1.5u$, 30 for an input of 20). The slide's own markers for the 63% point have been removed from the figure."},{"id":"engr5001-ex-laplace-01","module":"ENGR5001","week":3,"label":"Old spec 3.2 worked example, mass-spring-damper step response by Laplace","kind":"tutorial","level":2,"minutes":15,"tags":["laplace-tf","partial-fractions","step-response","mass-spring-damper","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 3.2 Worked Example, mass-spring-damper (slides 4-18)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.2 Worked Example (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- $\\mathcal{L}[\\dot x] = sX(s) - x(0)$, $\\mathcal{L}[\\ddot x] = s^2X(s) - sx(0) - \\dot x(0)$; unit step $\\to 1/s$; $e^{-\\alpha t} \\leftrightarrow \\dfrac{1}{s + \\alpha}$ (learn this)\n- Final Value Theorem $\\lim_{t\\to\\infty} f(t) = \\lim_{s\\to 0} sF(s)$ (learn this)\n","text":"The mass-spring-damper system is modelled by\n$$M\\frac{d^2x}{dt^2} + C\\frac{dx}{dt} + Kx = u(t)$$\n","parts":[{"label":"a","text":"Take Laplace transforms (keeping the initial conditions), then set the initial conditions to zero and find the\nTransfer Function $X(s)/U(s)$.\n","grade":"$undefined"},{"label":"b","text":"For $M = 1$, $C = 3$ and $K = 2$, find the response $x(t)$ to a unit step input using partial fractions and\ninverse Laplace transforms. Sketch the response.\n","answer":[{"label":"A (coefficient of 1/s)","value":0.5},{"label":"B (coefficient of 1/(s+1))","value":-1},{"label":"C (coefficient of 1/(s+2))","value":0.5},{"label":"x(1 s)","value":0.1998}],"grade":"$undefined"},{"label":"c","text":"Find the steady state value of $x$ using the Final Value Theorem, and confirm it using the steady state gain of\nthe Transfer Function.\n","answer":[{"label":"x(infinity)","value":0.5}],"grade":"$undefined"}],"solution":"$14","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.2 slides 4-18 ($x(t) = 1/2 - e^{-t} + e^{-2t}/2$, final value 0.5)."},{"id":"engr5001-ex-tf-01","module":"ENGR5001","week":3,"label":"Old spec 3.3 example, steady state gain of a Transfer Function","kind":"tutorial","level":1,"minutes":3,"tags":["laplace-tf","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 3.3 Transfer Functions, another steady state gain example (slide 10)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.3 Transfer Functions (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Steady state gain: set $s = 0$ in the Transfer Function; $x(t \\to \\infty) = \\text{gain} \\times u_0$ for a constant input $u_0$ (stable system) (learn this)\n","text":"A system has the Transfer Function model\n$$X(s) = \\frac{s + 12}{s^2 + 6s + 3}U(s)$$\nFind the steady state gain, and the steady state value of the output when the input is held constant at $u = 10$.\n","answer":[{"label":"steady state gain","value":4},{"label":"x(infinity) for u = 10","value":40}],"solution":"Set $s = 0$: steady state gain $= \\dfrac{0 + 12}{0 + 0 + 3} = 4$. Then\n$$x(t \\to \\infty) = 4 \\times u(t \\to \\infty) = 4 \\times 10 = 40$$\n\n**Checks.** The result is only meaningful if the system is stable: the poles are the roots of $s^2 + 6s + 3 = 0$,\n$s = -3 \\pm \\sqrt 6 = -0.55$ and $-5.45$, both negative, so it is. Final Value Theorem with a step of size 10:\n$\\lim_{s\\to 0}s\\cdot\\dfrac{s + 12}{s^2 + 6s + 3}\\cdot\\dfrac{10}{s} = 40$. In ODE terms\n$\\ddot x + 6\\dot x + 3x = \\dot u + 12u$; at steady state all derivatives vanish, so $3x = 12u$.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.3 slide 10 (gain 4, output 40)."},{"id":"engr5001-ex-tf-02","module":"ENGR5001","week":3,"label":"Old spec 3.3 \"exam question (1 mark)\", steady state displacement","kind":"tutorial","level":1,"minutes":2,"tags":["laplace-tf","steady-state-gain","mass-spring-damper"],"source":{"name":"ENGR202 (old spec) Control 3.3 Transfer Functions, exam question (1 mark) (slide 11)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.3 Transfer Functions (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- $x(t \\to \\infty) = \\text{steady state gain} \\times u_0$ (learn this)\n","text":"The displacement $x$ represents the distance in metres from an arbitrary reference point. If the steady state gain\nof the mass-spring-damper system is 0.01 and a constant external force of 8 N is applied, what is the displacement\n$x$ at steady state? (1 mark)\n","answer":[{"label":"x at steady state","value":0.08,"units":"m"}],"solution":"$$$x(t \\to \\infty) = 0.01 \\times 8 = 0.08\\ \\text{m}$$\nCheck: the steady state gain of a mass-spring-damper $1/(Ms^2 + Cs + K')$ is $1/K'$, so a gain of 0.01 m/N means a\nspring stiffness of 100 N/m, and $8/100 = 0.08$ m. Units: (m/N) x N = m.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.3 slide 11 (0.08 m)."},{"id":"engr5001-ex-tf-03","module":"ENGR5001","week":3,"label":"Old spec 3.3 exercise, Transfer Function of a 4th order equation","kind":"tutorial","level":1,"minutes":4,"tags":["laplace-tf","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 3.3 Transfer Functions, exercise (slide 12)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.3 Transfer Functions (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Zero initial conditions: $\\dfrac{d^n x}{dt^n} \\to s^nX$ (learn this)\n- Steady state gain: set $s = 0$ (learn this)\n","text":"Find the Transfer Function of the following differential equation, in which $y$ is the output, $r$ is the input,\n$\\alpha_1$ to $\\alpha_4$ are four estimated coefficients, and the other coefficients are exactly known. Then find the\nsteady state gain. Work with the notation given; there is no need to force this into a generalised form.\n$$\\frac{d^4y}{dt^4} + \\alpha_1\\frac{d^3y}{dt^3} + \\frac{d^2y}{dt^2} + \\alpha_2\\frac{dy}{dt} + 5y = \\alpha_3\\frac{dr}{dt} + \\alpha_4 r$$\n","solution":"Laplace transform with zero initial conditions ($s$ as the differentiation operator):\n$$s^4Y + \\alpha_1s^3Y + s^2Y + \\alpha_2sY + 5Y = \\alpha_3sR + \\alpha_4R$$\n$$\\frac{Y(s)}{R(s)} = \\frac{\\alpha_3 s + \\alpha_4}{s^4 + \\alpha_1 s^3 + s^2 + \\alpha_2 s + 5}$$\nSteady state gain (set $s = 0$): $\\dfrac{\\alpha_4}{5}$.\n\n**Checks.** At steady state all derivatives are zero, so the ODE gives $5y = \\alpha_4 r$ directly. It is fourth\norder (highest derivative of the output) with one zero at $s = -\\alpha_4/\\alpha_3$. The gain is only meaningful if\nthe system is stable; note the $s^2$ coefficient is fixed at 1, so stability depends on the estimated $\\alpha_1$ and\n$\\alpha_2$ (check with Hurwitz).\n","confidence":"HIGH","solutionNote":"No answer is printed in the deck; the solution follows the method of slides 4-9."},{"id":"engr5001-ex-poles-01","module":"ENGR5001","week":3,"label":"Old spec 3.4 example, pole of a first order system","kind":"tutorial","level":1,"minutes":3,"tags":["poles-stability","laplace-tf","time-constant"],"source":{"name":"ENGR202 (old spec) Control 3.4 Poles, Zeros and Stability, first order example (slide 14)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.4 Poles, Zeros and Stability (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Characteristic equation: denominator of the Transfer Function $= 0$; its roots are the poles (learn this)\n- Stable if all poles have negative real parts (learn this)\n","text":"Find the Transfer Function, characteristic equation and pole of the system below, and determine its stability.\n$$10\\frac{dx}{dt} + x = u$$\n","answer":[{"label":"pole","value":-0.1}],"solution":"Laplace with zero initial conditions: $10sX + X = U \\Rightarrow (10s + 1)X = U$, so\n$$X = \\frac{1}{10s + 1}U$$\nCharacteristic equation $10s + 1 = 0$, pole $s = -1/10 = -0.1$. The pole is real and negative, so the system is\n**stable**; its transient is $e^{-0.1t}$, a decaying exponential with time constant $\\tau = -1/p = 10$ s.\n\n**Check.** In generalised form $\\tau\\dot x + x = Ku$ with $\\tau = 10$ s and $K = 1$; a positive time constant is\nequivalent to a negative pole $-1/\\tau$.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.4 slide 14 (pole -0.1, stable)."},{"id":"engr5001-ex-poles-02","module":"ENGR5001","week":3,"label":"Old spec 3.4 example, poles and zeros of a second order system","kind":"tutorial","level":2,"minutes":6,"tags":["poles-stability","laplace-tf","zeros"],"source":{"name":"ENGR202 (old spec) Control 3.4 Poles, Zeros and Stability, second order example (slides 15-17)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.4 Poles, Zeros and Stability (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Zeros: roots of the numerator; poles: roots of the denominator (learn this)\n- Quadratic formula $s = \\dfrac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$ (learn this)\n- Complex poles $\\sigma \\pm j\\omega$ give a mode $e^{\\sigma t}\\sin(\\omega t + \\phi)$ (learn this)\n","text":"Find the poles and zeros, plot the poles on the complex $s$-plane, determine the stability and sketch the response\nmode.\n$$\\frac{d^2x}{dt^2} - \\frac{dx}{dt} + 0.5x = \\frac{du}{dt} + 2u$$\n","answer":[{"label":"zero","value":-2},{"label":"real part of poles","value":0.5},{"label":"imaginary part of poles (+)","value":0.5}],"solution":"**Transfer Function.** $s^2X - sX + 0.5X = sU + 2U$:\n$$X = \\frac{s + 2}{s^2 - s + 0.5}U$$\n**Zero:** $s + 2 = 0 \\Rightarrow s = -2$.\n**Poles:** $s^2 - s + 0.5 = 0$:\n$$s = \\frac{1 \\pm \\sqrt{1 - 4(0.5)}}{2} = \\frac{1}{2} \\pm \\frac{\\sqrt{-1}}{2} = 0.5 \\pm 0.5j$$\n**Plot:** a complex conjugate pair of crosses at $0.5 \\pm 0.5j$ in the right half plane; a circle at $-2$.\n**Stability:** the poles have positive real parts, so the system is **unstable** and oscillatory.\n**Response mode:** $e^{0.5t}\\sin(0.5t + \\phi)$, a growing oscillation of angular frequency 0.5 rad/s (period\n$2\\pi/0.5 = 12.6$ s) whose envelope doubles every $\\ln 2/0.5 = 1.4$ s.\n\n**Checks.** $(s - 0.5)^2 + 0.25 = s^2 - s + 0.5$; numpy roots agree. The negative coefficient of $dx/dt$ already\nsignals instability (negative damping). The zero is in the left half plane and does not affect stability.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.4 slide 16 (zero -2, poles 0.5 +/- 0.5j, unstable oscillatory)."},{"id":"engr5001-ex-poles-03","module":"ENGR5001","week":3,"label":"Old spec 3.4 example, dominant poles of a third order system","kind":"tutorial","level":2,"minutes":10,"tags":["poles-stability","dominant-poles","laplace-tf","step-response"],"source":{"name":"ENGR202 (old spec) Control 3.4 Poles, Zeros and Stability, dominant poles example (slides 18-19)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.4 Poles, Zeros and Stability (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Dominant pole(s): the pole(s) closest to the imaginary axis, which decay most slowly (learn this)\n- Steady state gain: set $s = 0$ (learn this)\n","text":"A stable system has the Transfer Function\n$$G(s) = \\frac{4a}{(s^2 + 2s + 4)(s + a)}, \\qquad a > 0$$\n","parts":[{"label":"a","text":"Find the three poles and the steady state gain.","answer":[{"label":"real part of complex poles","value":-1},{"label":"imaginary part of complex poles (+)","value":1.732},{"label":"steady state gain","value":1}],"grade":"$undefined"},{"label":"b","text":"Which pole(s) dominate the transient response when $a = 4$, and when $a = 0.25$? Describe the step response in\neach case.\n","grade":"$undefined"}],"solution":"$15","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.4 slides 18-19 (the complex pair prints as \"-1 +/- j)3\" because the square root sign is lost; it is $-1 \\pm j\\sqrt 3$)."},{"id":"engr5001-ex-poles-04","module":"ENGR5001","week":3,"label":"Old spec 3.4 example, effect of a zero","kind":"tutorial","level":2,"minutes":8,"tags":["poles-stability","zeros","non-minimum-phase","laplace-tf","step-response"],"source":{"name":"ENGR202 (old spec) Control 3.4 Poles, Zeros and Stability, zeros example (slides 20-21)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.4 Poles, Zeros and Stability (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Zeros do not affect stability; a zero with positive real part makes the system non-minimum phase (learn this)\n- Unit impulse $\\delta(t)$ is the derivative of the unit step (learn this)\n","text":"Consider\n$$X = \\frac{8(1 + as)}{s^2 + 2s + 8}U$$\n","parts":[{"label":"a","text":"Find the poles, the zero (in terms of $a$) and the steady state gain.","answer":[{"label":"imaginary part of poles (+)","value":2.646},{"label":"steady state gain","value":1}],"grade":"$undefined"},{"label":"b","text":"Show that the numerator acts as a modified input, and describe this modified input when $u$ is a unit step.\nHence explain how $a$ changes the step response. When is the system non-minimum phase?\n","grade":"$undefined"}],"solution":"$16","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.4 slides 20-21. The numerical peak and dip values are from a simulation added here; the slide shows the curves only."},{"id":"engr5001-ex-poles-05","module":"ENGR5001","week":3,"label":"Old spec 3.4 example, third order system with poles on the imaginary axis","kind":"tutorial","level":2,"minutes":6,"tags":["poles-stability","marginal-stability","dominant-poles","third-order"],"source":{"name":"ENGR202 (old spec) Control 3.4 Poles, Zeros and Stability, third order example (slide 22)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.4 Poles, Zeros and Stability (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Poles on the imaginary axis (non-repeated), the rest in the left half plane: marginally stable (learn this)\n","text":"Find the poles and zeros of the system below and determine its stability. Which poles dominate?\n$$X = \\frac{5}{s^3 + s^2 + 5s + 5}U$$\n","answer":[{"label":"real pole","value":-1},{"label":"imaginary part of complex poles (+)","value":2.236}],"solution":"**Zeros:** none (the numerator is the constant 5; \"5 = 0\" has no solution).\n**Poles:** factorise by grouping: $s^3 + s^2 + 5s + 5 = s^2(s + 1) + 5(s + 1) = (s + 1)(s^2 + 5)$.\n$s + 1 = 0 \\Rightarrow s = -1$; $s^2 + 5 = 0 \\Rightarrow s = \\pm j\\sqrt 5 = \\pm 2.236j$.\n**Stability:** a pair of poles lies on the imaginary axis (not repeated) and the other pole is in the left half\nplane, so the system is **marginally stable**.\n**Dominant poles:** the complex pair $\\pm j\\sqrt 5$ (zero real part, so it never decays): after the $e^{-t}$ mode\nhas died away the output oscillates indefinitely at $\\sqrt 5 = 2.24$ rad/s.\n\n**Checks.** numpy roots give $-1$ and $\\pm 2.236j$. Hurwitz shortcut: all coefficients present and positive, but\n$\\Delta_2 = a_1a_2 - a_0a_3 = 1 \\times 5 - 1 \\times 5 = 0$, the boundary case, consistent with marginal stability.\nNote the \"steady state gain\" $5/5 = 1$ is not reached: a step response keeps oscillating about 1.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.4 slide 22 (poles -1 and +/- j sqrt 5, marginally stable, complex pair dominant)."},{"id":"engr5001-ex-hurwitz-01","module":"ENGR5001","week":3,"label":"Old spec 3.5 example, fourth order Hurwitz determinant","kind":"tutorial","level":2,"minutes":10,"tags":["hurwitz","poles-stability","stability-shortcuts"],"source":{"name":"ENGR202 (old spec) Control 3.5 Hurwitz Stability Criterion, fourth order example (slides 4-6)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.5 Hurwitz Stability Criterion (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Hurwitz determinant for $a_0s^n + a_1s^{n-1} + \\dots + a_n = 0$: rows $(a_1, a_3, a_5, \\dots)$, $(a_0, a_2, a_4, \\dots)$, then the same pair shifted one column right, and so on; stable if all principal minors $\\Delta_1, \\dots, \\Delta_n > 0$ (learn this)\n- Necessary condition: all coefficients present and of the same sign (learn this)\n","text":"A system has the characteristic equation\n$$2.5s^4 + 1.3s^3 + 0.02s^2 - s + 5 = 0$$\nWrite down the 4th order Hurwitz determinant and use its principal minors to determine the stability.\n","answer":[{"label":"Delta_1","value":1.3},{"label":"Delta_2","value":2.526},{"label":"Delta_3","value":-10.98}],"solution":"$17","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.5 slides 5-6 (the slide rounds $\\Delta_2$ to 2.53; $\\Delta_3 = -10.98$, unstable)."},{"id":"engr5001-ex-hurwitz-02","module":"ENGR5001","week":3,"label":"Old spec 3.5 example, Hurwitz shortcut for a third order system","kind":"tutorial","level":1,"minutes":6,"tags":["hurwitz","poles-stability","stability-shortcuts"],"source":{"name":"ENGR202 (old spec) Control 3.5 Hurwitz Stability Criterion, Hurwitz shortcut example (slides 7-8)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.5 Hurwitz Stability Criterion (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Hurwitz criterion; shortcut: it is enough that either $\\Delta_1, \\Delta_3, \\Delta_5, \\dots > 0$ or $\\Delta_2, \\Delta_4, \\dots > 0$ (with all coefficients positive) (learn this)\n- Third order: stable if all coefficients are positive and $a_1a_2 > a_0a_3$ (learn this)\n","text":"Use the Hurwitz criterion, and its shortcut, to determine the stability of a system with characteristic equation\n$$3s^3 + 3s^2 + 9s + 6 = 0$$\n","answer":[{"label":"Delta_1","value":3},{"label":"Delta_2","value":9},{"label":"Delta_3","value":54}],"solution":"$$a_0 = 3$, $a_1 = 3$, $a_2 = 9$, $a_3 = 6$:\n$$H_3 = \\begin{vmatrix} a_1 & a_3 & 0 \\\\ a_0 & a_2 & 0 \\\\ 0 & a_1 & a_3 \\end{vmatrix} = \\begin{vmatrix} 3 & 6 & 0 \\\\ 3 & 9 & 0 \\\\ 0 & 3 & 6 \\end{vmatrix}$$\nEither $\\Delta_1 = 3 > 0$ and $\\Delta_3 = 6\\Delta_2 = 54 > 0$, or (simpler) just\n$$\\Delta_2 = \\begin{vmatrix} 3 & 6 \\\\ 3 & 9 \\end{vmatrix} = 3 \\times 9 - 6 \\times 3 = 9 > 0$$\nEither way the system is **stable**.\n\n**Checks.** numpy roots: $-0.715$ and $-0.142 \\pm 1.666j$, all in the left half plane (the complex pair is lightly\ndamped, close to the imaginary axis). All coefficients positive, so the necessary condition also holds.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.5 slides 7-8 ($\\Delta_1 = 3$, $\\Delta_2 = 9$, $\\Delta_3 = 54$, stable)."},{"id":"engr5001-ex-hurwitz-03","module":"ENGR5001","week":3,"label":"Old spec 3.5 examples, stability short cuts","kind":"tutorial","level":2,"minutes":10,"tags":["hurwitz","stability-shortcuts","poles-stability","marginal-stability"],"source":{"name":"ENGR202 (old spec) Control 3.5 Hurwitz Stability Criterion, stability short cut examples (slides 9-10)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.5 Hurwitz Stability Criterion (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Necessary conditions for stability: all coefficients exist (non-zero) and have the same sign (learn this)\n- If these hold, the Hurwitz criterion (or the poles) is still needed (learn this)\n","text":"Using the stability short cuts where possible (and the Hurwitz criterion or the poles where they are not enough),\ndecide whether each characteristic equation is stable, marginally stable or unstable.\n","parts":[{"label":"a","text":"$$s^4 + 3s^3 + 5s^2 + 10 = 0$","grade":"$undefined"},{"label":"b","text":"$$s^5 + 4s^4 + 2s^3 + 9s^2 + 3s = 0$","grade":"$undefined"},{"label":"c","text":"$$5s^2 - 2s + 2 = 0$","grade":"$undefined"},{"label":"d","text":"$$-3s^3 - 2s^2 - 4s - 10 = 0$ (careful!)","grade":"$undefined"},{"label":"e","text":"$$(s + 2)(s + 3) = 0$","grade":"$undefined"}],"solution":"$18","confidence":"HIGH","solutionNote":"The slide lists the five equations without answers. Its rule \"if $a_n = 0$ the system is marginally stable\" is\nonly true when the other roots are all in the left half plane; for example (b), which has $a_n = 0$, the remaining\nquartic has a right half plane pair, so (b) is unstable, not marginally stable.\n"},{"id":"engr5001-ex-hurwitz-04","module":"ENGR5001","week":3,"label":"Old spec 3.5 example, Hurwitz conditions for the mass-spring-damper","kind":"tutorial","level":1,"minutes":4,"tags":["hurwitz","poles-stability","mass-spring-damper"],"source":{"name":"ENGR202 (old spec) Control 3.5 Hurwitz Stability Criterion, mass-spring-damper example (slide 12)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.5 Hurwitz Stability Criterion (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Hurwitz determinant and principal minors (learn this)\n","text":"Use the Hurwitz criterion to find the conditions on the parameters for the mass-spring-damper, with characteristic\nequation $Ms^2 + Cs + K = 0$ ($M > 0$), to be stable. Interpret the result.\n","solution":"$$a_0 = M$, $a_1 = C$, $a_2 = K$:\n$$H_2 = \\begin{vmatrix} C & 0 \\\\ M & K \\end{vmatrix}$$\n$\\Delta_1 = C > 0$ and $\\Delta_2 = CK - 0 \\cdot M = CK > 0$. So stability needs $C > 0$ and (therefore) $K > 0$.\n\nInterpretation: positive damping makes the oscillations decay rather than continue indefinitely ($C = 0$ is the\nharmonic oscillator, marginally stable), and the spring must restore rather than push the mass away ($K < 0$ gives a\nreal positive pole). We already knew this physically.\n\n**Check.** Poles $s = \\dfrac{-C \\pm \\sqrt{C^2 - 4MK}}{2M}$: both have negative real parts exactly when $C > 0$ and\n$K > 0$ (if $K < 0$ the square root exceeds $C$ and one root is positive).\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.5 slide 12 ($\\Delta_1 = C > 0$, $\\Delta_2 = CK > 0$)."},{"id":"engr5001-ex-block-01","module":"ENGR5001","week":3,"label":"Old spec 3.6 examples, blocks in series","kind":"tutorial","level":1,"minutes":4,"tags":["block-diagrams","laplace-tf","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 3.6 Block Diagram Analysis, rule 1 and rule 2 examples (slides 10-13)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.6 Block Diagram Analysis (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Series rule: $X = G_1(s)G_2(s)U$ (learn this)\n","text":"Reduce each pair of blocks in series to a single Transfer Function from $U$ to $X$, and give its steady state gain.\n","parts":[{"label":"a","text":"A gain of 5 followed by $\\dfrac{0.8}{5s + 1}$.","answer":[{"label":"steady state gain","value":4}],"grade":"$undefined"},{"label":"b","text":"$$\\dfrac{1}{s + 2}$ followed by $\\dfrac{0.8}{5s + 1}$.","answer":[{"label":"steady state gain","value":0.4}],"grade":"$undefined"}],"solution":"**(a)** Series rule: $X = 5 \\times \\dfrac{0.8}{5s + 1}U = \\dfrac{4}{5s + 1}U$. Steady state gain 4 (first order,\ntime constant still 5 s).\n\n**(b)** $X = \\dfrac{1}{s + 2}\\cdot\\dfrac{0.8}{5s + 1}U = \\dfrac{0.8}{(s + 2)(5s + 1)}U = \\dfrac{0.8}{5s^2 + 11s + 2}U$.\nSteady state gain $0.8/2 = 0.4$ (the product of the individual gains $0.5 \\times 0.8$).\n\n**Checks.** $(s + 2)(5s + 1) = 5s^2 + s + 10s + 2 = 5s^2 + 11s + 2$ (numpy polymul). The order of blocks in series\ndoes not matter ($G_1G_2 = G_2G_1$), and the overall gain is always the product of the gains.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.6 slides 12-13 ($4/(5s+1)$ and $0.8/(5s^2 + 11s + 2)$)."},{"id":"engr5001-ex-block-02","module":"ENGR5001","week":3,"label":"Old spec 3.6 example, wind turbine blocks in series","kind":"tutorial","level":2,"minutes":6,"tags":["block-diagrams","laplace-tf","wind-turbine","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 3.6 Block Diagram Analysis, wind turbine simulation example (slides 14-15)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.6 Block Diagram Analysis (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Series rule: $X = G_1(s)G_2(s)G_3(s)U$ (learn this)\n","text":"A wind turbine simulation includes an actuator, an aerodynamic gain and an induction lag in series:\n$$\\frac{25.9}{s + 25.9}, \\qquad 8090, \\qquad \\frac{11.25s + 1}{7.5s + 1}$$\nDetermine the Transfer Function from the rotor pitch angle to the input to the powertrain, as a ratio of\npolynomials with a monic (leading coefficient 1) denominator, and give its steady state gain.\n","answer":[{"label":"numerator s coefficient","value":314296.5},{"label":"numerator constant","value":27937.5},{"label":"denominator s coefficient","value":26.03},{"label":"denominator constant","value":3.453},{"label":"steady state gain","value":8090}],"solution":"Series rule:\n$$G(s) = \\frac{25.9 \\times 8090 \\times (11.25s + 1)}{(s + 25.9)(7.5s + 1)} = \\frac{2\\,357\\,223.75s + 209\\,531}{7.5s^2 + 195.25s + 25.9}$$\nusing $25.9 \\times 8090 = 209\\,531$, $209\\,531 \\times 11.25 = 2\\,357\\,223.75$ and\n$(s + 25.9)(7.5s + 1) = 7.5s^2 + (1 + 194.25)s + 25.9$. Divide top and bottom by 7.5:\n$$G(s) = \\frac{314\\,296.5s + 27\\,937.5}{s^2 + 26.03s + 3.453}$$\nSteady state gain: $G(0) = 209\\,531/25.9 = 8090$ (the actuator and induction lag both have unity gain, so only the\naerodynamic gain remains).\n\n**Checks.** numpy polymul reproduces the coefficients ($27\\,937.47$ and $3.4533$ before rounding). Poles $-25.9$\nand $-1/7.5 = -0.133$, zero $-1/11.25 = -0.089$: the zero partly cancels the slow induction lag pole, so the lag\nblock mostly adds a lead-like jump of $11.25/7.5 = 1.5$ at high frequency.\n","confidence":"HIGH","solutionNote":"Matches old-spec Control 3.6 slides 14-15 (to the slide's rounding, 27937.5 and 3.45)."},{"id":"engr5001-ex-block-03","module":"ENGR5001","week":3,"label":"Old spec 3.6 example, closed-loop control of vehicle speed","kind":"tutorial","level":1,"minutes":6,"tags":["block-diagrams","closed-loop-tf","proportional-control","vehicle-speed","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 3.6 Block Diagram Analysis, closed-loop control of vehicle speed (slides 5-7 and 16-19)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.6 Block Diagram Analysis (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Unity negative feedback: $X = \\dfrac{C(s)G(s)}{1 + C(s)G(s)}D$ (learn this)\n","text":"The speed $X$ of a vehicle is controlled by a proportional controller $U = K_p(D - X)$, where $D$ is the desired\nspeed and $U$ the engine force. The vehicle model is $X = \\dfrac{0.8}{5s + 1}U$ (unity negative feedback of the\nspeed).\n","parts":[{"label":"a","text":"Use the rules of block diagram analysis to find the Closed-Loop Transfer Function $X/D$.","grade":"$undefined"},{"label":"b","text":"For $K_p = 5$, find the closed-loop steady state gain and time constant.","answer":[{"label":"steady state gain","value":0.8},{"label":"closed-loop time constant","value":1,"units":"s"}],"grade":"$undefined"}],"solution":"**(a)** Series rule: forward path $C(s)G(s) = \\dfrac{0.8K_p}{5s + 1}$. Unity negative feedback rule:\n$$\\frac{X}{D} = \\frac{\\dfrac{0.8K_p}{5s + 1}}{1 + \\dfrac{0.8K_p}{5s + 1}} = \\frac{0.8K_p}{5s + 1 + 0.8K_p}$$\n\n**(b)** Steady state gain ($s = 0$): $\\dfrac{0.8K_p}{1 + 0.8K_p} = \\dfrac{4}{5} = 0.8$ for $K_p = 5$. Rewrite as\n$\\dfrac{0.8K_p/(1 + 0.8K_p)}{\\frac{5}{1 + 0.8K_p}s + 1}$, so the closed-loop time constant is\n$5/(1 + 0.8K_p) = 5/5 = 1$ s, five times faster than the open-loop 5 s.\n\n**Checks.** Algebra route: $X = \\frac{0.8}{5s + 1}K_p(D - X) \\Rightarrow (5s + 1 + 0.8K_p)X = 0.8K_pD$, the same.\nLimits: $K_p \\to 0$ gives gain 0 (no control); large $K_p$ gives gain $\\to 1$ and a very fast response (high gain\ncontrol), but always a steady state error with proportional action alone (20% here).\n","confidence":"HIGH","solutionNote":"The deck sets up this block diagram (slides 5-7) and gives the general result $CG/(1 + CG)$ (slides 18-19) without substituting the vehicle numbers; part (b) is added here."},{"id":"engr5001-ex-laplace-02","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck examples, transforms from the definition","kind":"tutorial","level":2,"minutes":12,"tags":["laplace-tf","linearity","time-delay"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, \"some examples\" (pages 6-9), cosh example and \"try L[sinh at]\" (page 9)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"2:2","formulaSheet":"- Definition $F(s) = \\mathcal{L}[f(t)] = \\displaystyle\\int_0^\\infty e^{-st}f(t)\\,dt$ (learn this)\n- $\\displaystyle\\int e^{\\beta t}dt = \\dfrac{1}{\\beta}e^{\\beta t}$; $\\cosh at = \\dfrac{e^{at} + e^{-at}}{2}$, $\\sinh at = \\dfrac{e^{at} - e^{-at}}{2}$ (learn this)\n- Linearity $\\mathcal{L}[\\alpha f + \\beta g] = \\alpha\\mathcal{L}[f] + \\beta\\mathcal{L}[g]$ (learn this)\n","text":"Using the definition of the Laplace transform (and its linearity), find:\n","parts":[{"label":"a","text":"$$\\mathcal{L}[1]$ (for $t > 0$)","grade":"$undefined"},{"label":"b","text":"$$\\mathcal{L}[e^{at}]$, $a$ a real constant, stating the condition on $s$; hence $\\mathcal{L}[e^{-at}]$","grade":"$undefined"},{"label":"c","text":"$$\\mathcal{L}[u(t - t_0)]$, where $u(t - t_0)$ is the unit step starting at $t = t_0 > 0$","grade":"$undefined"},{"label":"d","text":"$$\\mathcal{L}[\\cosh at]$","grade":"$undefined"},{"label":"e","text":"$$\\mathcal{L}[\\sinh at]$ (the \"try this\" homework)","grade":"$undefined"}],"solution":"$19","confidence":"HIGH","solutionNote":"(a)-(d) match the ENGR201 deck; (e) is the deck's homework and has no printed answer."},{"id":"engr5001-ex-laplace-03","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck example, transform by linearity","kind":"tutorial","level":1,"minutes":3,"tags":["laplace-tf","linearity"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, example f(t) = 5e^(-2t) - 3 sin(4t) (page 11)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"third","formulaSheet":"- $\\mathcal{L}[e^{at}] = \\dfrac{1}{s - a}$, $\\mathcal{L}[\\sin\\omega t] = \\dfrac{\\omega}{s^2 + \\omega^2}$; linearity (learn this)\n","text":"Find the Laplace transform of $f(t) = 5e^{-2t} - 3\\sin(4t)$ for $t \\ge 0$, and evaluate $F(s)$ at $s = 1$ as a\nnumerical check.\n","answer":[{"label":"F(1)","value":0.9608}],"solution":"By linearity:\n$$F(s) = 5\\mathcal{L}[e^{-2t}] - 3\\mathcal{L}[\\sin 4t] = \\frac{5}{s + 2} - 3\\cdot\\frac{4}{s^2 + 16} = \\frac{5}{s + 2} - \\frac{12}{s^2 + 16}$$\nAt $s = 1$: $5/3 - 12/17 = 1.6667 - 0.7059 = 0.9608$.\n\n**Checks.** sympy gives the same transform. Direct numerical integration of $\\int_0^\\infty e^{-t}f(t)\\,dt$ also\ngives 0.9608.\n","confidence":"HIGH","solutionNote":"Matches the ENGR201 deck ($5/(s + 2) - 12/(s^2 + 16)$); the value at $s = 1$ is added as a check."},{"id":"engr5001-ex-laplace-04","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck example 1, piecewise signal built from steps","kind":"tutorial","level":2,"minutes":5,"tags":["laplace-tf","time-delay","linearity"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, Example 1 (page 14)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"2:2","formulaSheet":"- $\\mathcal{L}[u(t - c)] = \\dfrac{e^{-cs}}{s}$ (learn this)\n","text":"The signal $f(t)$ in the figure is 0 for $t < 1$, equal to 1 for $1 \\le t < 2$, equal to 2 for $2 \\le t < 3$, and 0\nagain for $t \\ge 3$. Construct $f(t)$ using unit step (Heaviside) functions and find its Laplace transform. As a\ncheck, evaluate $F(s)$ at $s = 1$.\n","figures":[{"src":"/figures/engr5001-ex-laplace-04.png","caption":"f(t): 0 until t = 1, then 1 until t = 2, then 2 until t = 3, then 0."}],"answer":[{"label":"F(1)","value":0.4036}],"solution":"Build it from the jumps: up 1 at $t = 1$, up another 1 at $t = 2$, down 2 at $t = 3$:\n$$f(t) = u(t - 1) + u(t - 2) - 2u(t - 3)$$\nCheck the pieces: $1 \\le t < 2$: 1; $2 \\le t < 3$: $1 + 1 = 2$; $t \\ge 3$: $1 + 1 - 2 = 0$. Then by linearity\n$$F(s) = \\frac{e^{-s}}{s} + \\frac{e^{-2s}}{s} - \\frac{2e^{-3s}}{s}$$\nAt $s = 1$: $e^{-1} + e^{-2} - 2e^{-3} = 0.3679 + 0.1353 - 0.0996 = 0.4036$.\n\n**Checks.** Direct integration $\\int_1^2 e^{-t}dt + 2\\int_2^3 e^{-t}dt = 0.4036$ (scipy quad agrees). Final value\ntheorem: $sF(s) \\to 1 + 1 - 2 = 0$ as $s \\to 0$, matching $f(\\infty) = 0$.\n","confidence":"HIGH","solutionNote":"Matches the ENGR201 deck Example 1; the value at $s = 1$ is added as a check."},{"id":"engr5001-ex-laplace-05","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck examples, Laplace transform of derivatives","kind":"tutorial","level":1,"minutes":6,"tags":["laplace-tf"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, \"example to verify\" and Example 3 (pages 18-19)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"third","formulaSheet":"- $\\mathcal{L}[f'(t)] = sF(s) - f(0)$; $\\mathcal{L}[f''(t)] = s^2F(s) - sf(0) - f'(0)$ (learn this)\n- $\\mathcal{L}[\\sin t] = \\dfrac{1}{s^2 + 1}$, $\\mathcal{L}[\\cos t] = \\dfrac{s}{s^2 + 1}$, $\\mathcal{L}[t^n] = \\dfrac{n!}{s^{n+1}}$ (learn this)\n","text":"Use the derivative rules to find the transforms below, and verify each by differentiating first and using the\ntable directly.\n","parts":[{"label":"a","text":"$$\\mathcal{L}[f'(t)]$ for $f(t) = \\cos t$","grade":"$undefined"},{"label":"b","text":"$$\\mathcal{L}[f'(t)]$ and $\\mathcal{L}[f''(t)]$ for $f(t) = t^2$","grade":"$undefined"}],"solution":"**(a)** Rule: $\\mathcal{L}[f'] = s\\mathcal{L}[\\cos t] - \\cos 0 = s\\cdot\\dfrac{s}{s^2 + 1} - 1 = \\dfrac{s^2 - (s^2 + 1)}{s^2 + 1} = -\\dfrac{1}{s^2 + 1}$.\nDirect: $f'(t) = -\\sin t$, so $\\mathcal{L}[f'] = -\\dfrac{1}{s^2 + 1}$. Verified.\n\n**(b)** $F(s) = \\mathcal{L}[t^2] = 2/s^3$, $f(0) = 0$, $f'(0) = 0$.\n$$\\mathcal{L}[f'] = s\\cdot\\frac{2}{s^3} - 0 = \\frac{2}{s^2}, \\qquad \\mathcal{L}[f''] = s^2\\cdot\\frac{2}{s^3} - 0 - 0 = \\frac{2}{s}$$\nDirect: $f' = 2t \\to 2/s^2$; $f'' = 2 \\to 2/s$. Verified.\n\n**Checks.** sympy transforms of $-\\sin t$, $2t$ and $2$ agree. The rule shows why the Transfer Function method needs\nzero initial conditions: then differentiation is simply multiplication by $s$.\n","confidence":"HIGH","solutionNote":"Matches the ENGR201 deck (pages 18-19)."},{"id":"engr5001-ex-laplace-06","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck example, solving a second order ODE","kind":"tutorial","level":2,"minutes":12,"tags":["laplace-tf","partial-fractions","step-response"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, Laplace transforms of differential equations example (page 24)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"2:2","formulaSheet":"- $\\mathcal{L}[y'] = sY - y(0)$, $\\mathcal{L}[y''] = s^2Y - sy(0) - y'(0)$; constant $c \\to c/s$; $e^{-at} \\leftrightarrow \\dfrac{1}{s + a}$ (learn this)\n","text":"Solve, using Laplace transforms,\n$$\\frac{d^2y}{dt^2} + 6\\frac{dy}{dt} + 8y = 2, \\qquad y(0) = y'(0) = 0$$\n","parts":[{"label":"a","text":"Find $Y(s)$.","grade":"$undefined"},{"label":"b","text":"Find $y(t)$ by partial fractions, and give $y(1)$ and the final value.","answer":[{"label":"y(1)","value":0.1869},{"label":"y(infinity)","value":0.25}],"grade":"$undefined"}],"solution":"**(a)** Transform each term: $\\mathcal{L}[y''] = s^2Y$, $\\mathcal{L}[y'] = sY$ (zero initial conditions),\n$\\mathcal{L}[2] = 2/s$:\n$$(s^2 + 6s + 8)Y = \\frac{2}{s} \\;\\Rightarrow\\; Y(s) = \\frac{2}{s(s^2 + 6s + 8)} = \\frac{2}{s(s + 2)(s + 4)}$$\n\n**(b)** $\\dfrac{2}{s(s + 2)(s + 4)} = \\dfrac{A}{s} + \\dfrac{B}{s + 2} + \\dfrac{C}{s + 4}$, cover-up:\n$A = \\dfrac{2}{2 \\times 4} = \\dfrac{1}{4}$, $B = \\dfrac{2}{(-2)(2)} = -\\dfrac{1}{2}$, $C = \\dfrac{2}{(-4)(-2)} = \\dfrac{1}{4}$.\n$$y(t) = \\frac{1}{4} - \\frac{1}{2}e^{-2t} + \\frac{1}{4}e^{-4t}$$\n$y(1) = 0.25 - 0.5(0.1353) + 0.25(0.0183) = 0.187$; final value $0.25$.\n\n**Checks.** $y(0) = 0.25 - 0.5 + 0.25 = 0$, $y'(0) = 1 - 1 = 0$. Final value theorem:\n$\\lim_{s\\to 0}sY = 2/8 = 0.25$ (also $8y = 2$ at steady state). sympy inverse transform and a scipy step response of\n$2/(s^2 + 6s + 8)$ agree.\n","confidence":"HIGH","solutionNote":"The deck stops at $Y(s) = 2/(s(s^2 + 6s + 8))$ and asks how to get back to the time domain; part (b) completes it using the deck's partial fraction method."},{"id":"engr5001-ex-laplace-07","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck tutorial Q7, water tank struck by a gust (impulse)","kind":"tutorial","level":2,"minutes":12,"tags":["laplace-tf","modelling","mass-spring-damper","time-delay"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, Q7 of tutorial sheet, water tank (pages 28-29)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"2:2","formulaSheet":"- $\\mathcal{L}[\\delta(t - c)] = e^{-cs}$; $\\mathcal{L}[x''] = s^2X - sx(0) - x'(0)$ (learn this)\n- Shift: $\\mathcal{L}^{-1}[e^{-cs}F(s)] = f(t - c)u(t - c)$; $\\mathcal{L}^{-1}\\left[\\dfrac{b}{(s - a)^2 + b^2}\\right] = e^{at}\\sin bt$ (learn this)\n","text":"A water tank of mass $m = 1$ is supported by a column with stiffness constant $k = 5$ and damping constant $c = 2$\n(a mass-spring-damper in horizontal motion, displacement $x(t)$). At the instant $t = \\pi$ the tank is struck by a\ngust of wind giving an impulse $p = 10$. Also $x(0) = 0$ and $x'(0) = 0$.\n","parts":[{"label":"a","text":"Write the differential equation governing the motion of the tank.","grade":"$undefined"},{"label":"b","text":"Find the Laplace transform $X(s)$ of the solution.","grade":"$undefined"},{"label":"c","text":"Find $x(t)$, and evaluate it at $t = \\pi + 0.5$.","answer":[{"label":"x(pi + 0.5)","value":2.552}],"grade":"$undefined"}],"solution":"$1a","confidence":"HIGH","solutionNote":"Parts (a)-(b) match the deck. On page 29 the deck first writes $x'' + 2x' + x = 10\\delta(t - \\pi)$ (the $5x$ term\nis printed as $x$) before using $5x$ correctly in the next line; the answer $X(s) = 10e^{-\\pi s}/(s^2 + 2s + 5)$ is\nright. Part (c) is added here.\n"},{"id":"engr5001-ex-laplace-08","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck examples, inverse transforms with distinct real poles","kind":"tutorial","level":1,"minutes":8,"tags":["laplace-tf","partial-fractions"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, inverse transform example and partial fractions case 1 (pages 35-36)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"third","formulaSheet":"- $\\mathcal{L}^{-1}\\left[\\dfrac{1}{s + a}\\right] = e^{-at}$; linearity of $\\mathcal{L}^{-1}$ (learn this)\n- Partial fractions: one term $\\dfrac{A}{s + p}$ per distinct linear factor (learn this)\n","text":"Find the inverse Laplace transforms:\n","parts":[{"label":"a","text":"$$\\mathcal{L}^{-1}\\left[\\dfrac{1}{(s + a)(s + b)}\\right]$, where $a \\ne b$ are constants","grade":"$undefined"},{"label":"b","text":"$$\\mathcal{L}^{-1}\\left[\\dfrac{s + 1}{(s + 2)(s + 3)}\\right]$, and its value at $t = 1$","answer":[{"label":"f(1)","value":-0.03576}],"grade":"$undefined"}],"solution":"**(a)** $\\dfrac{1}{(s + a)(s + b)} = \\dfrac{A}{s + a} + \\dfrac{B}{s + b}$, so $A(s + b) + B(s + a) = 1$: equating\ncoefficients, $A + B = 0$ and $Ab + Ba = 1$, giving $B = -A = \\dfrac{1}{a - b}$. Hence\n$$\\frac{1}{(s + a)(s + b)} = \\frac{1}{b - a}\\left(\\frac{1}{s + a} - \\frac{1}{s + b}\\right) \\;\\Rightarrow\\; \\mathcal{L}^{-1} = \\frac{e^{-at} - e^{-bt}}{b - a}$$\n\n**(b)** $\\dfrac{s + 1}{(s + 2)(s + 3)} = \\dfrac{A}{s + 2} + \\dfrac{B}{s + 3}$: $s + 1 = A(s + 3) + B(s + 2)$, so\n$A + B = 1$, $3A + 2B = 1$, giving $A = -1$, $B = 2$ (cover-up: $A = \\frac{-2 + 1}{-2 + 3} = -1$,\n$B = \\frac{-3 + 1}{-3 + 2} = 2$).\n$$f(t) = -e^{-2t} + 2e^{-3t}, \\qquad f(1) = -0.1353 + 0.0996 = -0.0358$$\n\n**Checks.** sympy `apart` and inverse transforms agree. Initial value: $f(0) = 1 = \\lim_{s\\to\\infty}sF(s)$. In (a),\nswapping $a$ and $b$ leaves the result unchanged, as it must.\n","confidence":"HIGH","solutionNote":"Matches the ENGR201 deck (pages 35-36); the value at $t = 1$ is added as a check."},{"id":"engr5001-ex-laplace-09","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck example, partial fractions with a repeated pole","kind":"tutorial","level":2,"minutes":8,"tags":["laplace-tf","partial-fractions"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, partial fractions case 2 (page 37)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"2:2","formulaSheet":"- Repeated factor $(s + 1)^3$: terms $\\dfrac{A}{s + 1} + \\dfrac{B}{(s + 1)^2} + \\dfrac{C}{(s + 1)^3}$ (learn this)\n- $\\mathcal{L}^{-1}\\left[\\dfrac{n!}{(s - a)^{n+1}}\\right] = t^ne^{at}$ (learn this)\n","text":"Find $\\mathcal{L}^{-1}[X(s)]$ for\n$$X(s) = \\frac{s^2 + 3s + 4}{(s + 1)^3}$$\nand evaluate the result at $t = 1$.\n","answer":[{"label":"A","value":1},{"label":"B","value":1},{"label":"C","value":2},{"label":"x(1)","value":1.1036}],"solution":"$$$\\frac{s^2 + 3s + 4}{(s + 1)^3} = \\frac{A}{s + 1} + \\frac{B}{(s + 1)^2} + \\frac{C}{(s + 1)^3}$$\nRecombine: $s^2 + 3s + 4 = A(s + 1)^2 + B(s + 1) + C = As^2 + (2A + B)s + (A + B + C)$. Equate coefficients:\n$A = 1$; $2A + B = 3 \\Rightarrow B = 1$; $A + B + C = 4 \\Rightarrow C = 2$.\n$$X(s) = \\frac{1}{s + 1} + \\frac{1}{(s + 1)^2} + \\frac{2}{(s + 1)^3}$$\nUsing $t^ne^{at} \\leftrightarrow n!/(s - a)^{n+1}$ with $a = -1$: $\\frac{1}{(s + 1)^2} \\to te^{-t}$ and\n$\\frac{2}{(s + 1)^3} = \\frac{2!}{(s + 1)^3} \\to t^2e^{-t}$:\n$$x(t) = e^{-t} + te^{-t} + t^2e^{-t} = (1 + t + t^2)e^{-t}, \\qquad x(1) = 3e^{-1} = 1.104$$\n\n**Checks.** sympy gives the same. Initial value theorem: $x(0) = 1 = \\lim_{s\\to\\infty}sX(s)$. Shortcut for\nrepeated poles: substitute $p = s + 1$, so the numerator is $(p - 1)^2 + 3(p - 1) + 4 = p^2 + p + 2$, giving\n$1/p + 1/p^2 + 2/p^3$ at once.\n","confidence":"HIGH","solutionNote":"Matches the ENGR201 deck (page 37). Note that ENGR5001 Part 3 slides only tabulate distinct real poles; the $t^ne^{at}$ pair is from the ENGR201 table."},{"id":"engr5001-ex-laplace-10","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck example, partial fractions with a quadratic factor","kind":"tutorial","level":2,"minutes":12,"tags":["laplace-tf","partial-fractions","step-response"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, partial fractions case 3 (pages 38-39)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"2:1","formulaSheet":"- Quadratic factor with no real roots: term $\\dfrac{Bs + C}{s^2 + ps + q}$; complete the square (learn this)\n- $\\mathcal{L}^{-1}\\left[\\dfrac{s - a}{(s - a)^2 + b^2}\\right] = e^{at}\\cos bt$, $\\mathcal{L}^{-1}\\left[\\dfrac{b}{(s - a)^2 + b^2}\\right] = e^{at}\\sin bt$ (learn this)\n","text":"Find $\\mathcal{L}^{-1}[X(s)]$ for\n$$X(s) = \\frac{1}{s(s^2 + 4s + 13)}$$\nand evaluate the result at $t = 0.5$. (This is the unit step response of $1/(s^2 + 4s + 13)$.)\n","answer":[{"label":"x(0.5)","value":0.0561}],"solution":"$$s^2 + 4s + 13$ has no real factors ($16 - 52 < 0$), so\n$$\\frac{1}{s(s^2 + 4s + 13)} = \\frac{A}{s} + \\frac{Bs + C}{s^2 + 4s + 13}$$\n$A(s^2 + 4s + 13) + s(Bs + C) = 1 \\Rightarrow (A + B)s^2 + (4A + C)s + 13A = 1$, so $A = 1/13$, $B = -1/13$,\n$C = -4/13$:\n$$X(s) = \\frac{1}{13}\\cdot\\frac{1}{s} - \\frac{1}{13}\\cdot\\frac{s + 4}{s^2 + 4s + 13}$$\nComplete the square: $s^2 + 4s + 13 = (s + 2)^2 + 3^2$, and $s + 4 = (s + 2) + 2$:\n$$X(s) = \\frac{1}{13}\\cdot\\frac{1}{s} - \\frac{1}{13}\\cdot\\frac{s + 2}{(s + 2)^2 + 3^2} - \\frac{2}{39}\\cdot\\frac{3}{(s + 2)^2 + 3^2}$$\n$$x(t) = \\frac{1}{13} - \\frac{1}{13}e^{-2t}\\cos 3t - \\frac{2}{39}e^{-2t}\\sin 3t$$\nAt $t = 0.5$: $0.07692 - 0.07692(0.3679)(0.0707) - 0.05128(0.3679)(0.9975) = 0.0561$.\n\n**Checks.** $x(0) = 1/13 - 1/13 = 0$; final value $1/13$ = steady state gain of $1/(s^2 + 4s + 13)$; the scipy\nstep response matches $x(t)$ to $10^{-15}$; sympy gives the same expression.\n","confidence":"HIGH","solutionNote":"Matches the ENGR201 deck (pages 38-39), $A = 1/13$, $B = -1/13$, $C = -4/13$ and the same $x(t)$."},{"id":"engr5001-ex-laplace-11","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck example, railway buffer hit by a 5 s pulse","kind":"tutorial","level":3,"minutes":20,"tags":["laplace-tf","partial-fractions","time-delay","mass-spring-damper","step-response"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, railway buffer example (pages 45-53)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"2:1","formulaSheet":"- $\\mathcal{L}[u(t - c)] = \\dfrac{e^{-cs}}{s}$; shift theorem $\\mathcal{L}^{-1}[e^{-cs}F(s)] = f(t - c)u(t - c)$ (learn this)\n- $\\mathcal{L}^{-1}\\left[\\dfrac{s - a}{(s - a)^2 + b^2}\\right] = e^{at}\\cos bt$, $\\mathcal{L}^{-1}\\left[\\dfrac{b}{(s - a)^2 + b^2}\\right] = e^{at}\\sin bt$ (learn this)\n","text":"A damped spring constrained to move in one direction, such as a railway buffer, is subjected to a force of 1 unit\napplied for 5 seconds (a rectangular pulse starting at $t = 0$). The equation of motion is\n$$\\frac{d^2x}{dt^2} + 4\\frac{dx}{dt} + 20x = F(t)$$\nand the buffer starts with no displacement and no velocity.\n","parts":[{"label":"a","text":"Write the pulse force $F(t)$ using unit step (Heaviside) functions.","grade":"$undefined"},{"label":"b","text":"Find $X(s)$.","grade":"$undefined"},{"label":"c","text":"Solve for the displacement $x(t)$.","grade":"$undefined"},{"label":"d","text":"Find the static deflection under the 1 unit force, and compare it with the maximum deflection (and when it\noccurs). Evaluate $x(6)$.\n","answer":[{"label":"static deflection","value":0.05},{"label":"maximum deflection","value":0.0604,"tol":0.02},{"label":"time of maximum","value":0.785,"units":"s","tol":0.03},{"label":"x(6 s)","value":-0.00698,"tol":0.03}],"grade":"$undefined"}],"solution":"$1b","confidence":"HIGH","solutionNote":"The final $x(t)$ and graph match the deck. Slide errors: the statement gives a damping coefficient of 2 but the\nequation (used throughout) has $4\\,dx/dt$, i.e. damping 4 for unit mass; the first line of page 48 writes the\ndenominator as $s(s^2 + 2s + 10)$, a typo for $s(s^2 + 4s + 20)$; the force is called a \"unit impulse of duration\n5 seconds\" but it is a rectangular pulse of height 1 (impulse 5 units); and the quadratic-factor method is cited as\n\"Case 2\" when it is Case 3. None of these change the answer.\n"},{"id":"engr5001-ex-laplace-12","module":"ENGR5001","week":3,"label":"ENGR201 Laplace deck example, free vibration of a two-mass system","kind":"tutorial","level":3,"minutes":20,"tags":["laplace-tf","partial-fractions","modelling"],"source":{"name":"ENGR201 (old spec) Laplace transforms deck, more examples, free vibration of a spring-mass system (pages 60-63)","paper":"ENGR201","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Engineering Analysis Laplace transforms, inverse transforms and ODEs (old spec ENGR201).pdf"},"grade":"first","formulaSheet":"- $\\mathcal{L}[y''] = s^2Y - sy(0) - y'(0)$ (learn this)\n- $\\mathcal{L}^{-1}\\left[\\dfrac{\\alpha}{s^2 + \\alpha^2}\\right] = \\sin\\alpha t$, $\\mathcal{L}^{-1}\\left[\\dfrac{s}{s^2 + \\alpha^2}\\right] = \\cos\\alpha t$ (learn this)\n","text":"Two equal masses $m = 1$ are connected in a line by three springs of stiffness $k = 2$ (wall-spring-mass-spring-\nmass-spring-wall). Their displacements $y_1$ and $y_2$ satisfy\n$$m\\frac{d^2y_1}{dt^2} + 2ky_1 - ky_2 = 0, \\qquad m\\frac{d^2y_2}{dt^2} + 2ky_2 - ky_1 = 0$$\nwith initial conditions $y_1(0) = y_2(0) = 1$ and $y_1'(0) = -y_2'(0) = \\sqrt 6$. Use Laplace transforms to find\nthe motion of the two masses.\n","parts":[{"label":"a","text":"Transform the equations and write them as two simultaneous equations in $Y_1(s)$ and $Y_2(s)$.","grade":"$undefined"},{"label":"b","text":"Solve for $Y_1$ and $Y_2$, expand in partial fractions and invert.","answer":[{"label":"y1(1)","value":0.7941},{"label":"y2(1)","value":-0.4822}],"grade":"$undefined"}],"solution":"$1c","confidence":"HIGH","solutionNote":"Matches the deck (pages 60-63), where $\\sqrt 6$ appears as \"6\" in places because the root sign is lost in the PDF. Graded first because coupled ODEs go beyond the single-ODE examples in the ENGR5001 Part 3 slides."},{"id":"engr5001-ex-pid-01","module":"ENGR5001","week":4,"label":"Old spec 4.2 worked example, open and closed loop control of a DC motor","kind":"tutorial","level":2,"minutes":12,"tags":["feedback-objectives","dc-motor","open-loop","closed-loop","proportional-control","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 4.2 Example DC Motor, worked example and exercise (slides 5-18)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 4 - Control 4.2 Example DC Motor (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- $x(t \\to \\infty) = K \\times u_0$; open-loop proportional control $U = K_pD$ (learn this)\n- Closed loop with $U = K_p(D - X)$: $X = \\dfrac{HK_p}{1 + HK_p}D$ (learn this)\n","text":"A DC motor has input voltage $U$ and shaft speed $X$ (measured as a voltage by a tachometer), modelled as\n$\\tau\\dot x + x = Ku$, i.e. $X = H(s)U$ with $H(s) = K/(\\tau s + 1)$. In an open-loop test with no load disturbance,\na constant input $U = 0.5$ V gives a steady motor speed $X = 20$ V, which is the desired speed $D = 20$.\n","parts":[{"label":"a","text":"What is the steady state gain $K$ of the motor?","answer":[{"label":"K","value":40}],"grade":"$undefined"},{"label":"b","text":"An open-loop proportional controller $U = K_pD$ is used. What control gain $K_p$ gives the desired speed at\nsteady state? Why is this sensitive to load disturbances?\n","answer":[{"label":"Kp (open loop)","value":0.025}],"grade":"$undefined"},{"label":"c","text":"The loop is now closed: $U = K_p(D - X)$. Derive the closed-loop relationship between $D$ and $X$ by algebra,\nand find the closed-loop steady state gain with $K_p = 0.025$ and with $K_p = 2.5$.\n","answer":[{"label":"CL gain, Kp = 0.025","value":0.5},{"label":"CL gain, Kp = 2.5","value":0.99}],"grade":"$undefined"}],"solution":"$1d","confidence":"HIGH","solutionNote":"Parts (a)-(b) match slides 9-11 ($K_p = 0.025$, $K = 40$). The closed-loop algebra matches slides 15-17; the numerical closed-loop gains in (c) are added here."},{"id":"engr5001-ex-pid-02","module":"ENGR5001","week":4,"label":"Old spec 4.4 and 4.5 examples, proportional and integral action on a second order plant","kind":"tutorial","level":2,"minutes":15,"tags":["pid-actions","proportional-control","integral-action","closed-loop-tf","steady-state-error","block-diagrams"],"source":{"name":"ENGR202 (old spec) Control 4.4 Proportional Action (slides 4-14) and 4.5 Integral Action (slides 4-8)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 4 - Control 4.4 Proportional Action (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Series rule and negative feedback rule $\\dfrac{G_1}{1 + G_1G_2}$ (learn this)\n- Final Value Theorem $\\lim_{t\\to\\infty}x(t) = \\lim_{s\\to 0}sX(s)$ (learn this)\n","text":"The plant is the generalised second order system with unity steady state gain,\n$$X = \\frac{\\omega_n^2}{s^2 + 2\\zeta\\omega_n s + \\omega_n^2}U$$\nin a unity negative feedback loop with set point $D$, error $E = D - X$ and controller $U = C(s)E$.\n","parts":[{"label":"a","text":"Proportional action $C(s) = k_p$. Find the closed-loop Transfer Function, the closed-loop characteristic\nequation, and the steady state error for a unit step in $D$. Evaluate the error for $k_p = 0.5$ and $k_p = 2$.\n","answer":[{"label":"error, kp = 0.5","value":0.6667},{"label":"error, kp = 2","value":0.3333}],"grade":"$undefined"},{"label":"b","text":"Integral action $C(s) = k_I/s$. Find the closed-loop Transfer Function, characteristic equation and steady\nstate value for a unit step. What is the price paid?\n","answer":[{"label":"steady state output (I action)","value":1}],"grade":"$undefined"}],"solution":"$1e","confidence":"HIGH","solutionNote":"Matches 4.4 slides 7-12 and 4.5 slides 5-8. The 4.5 steady state gain slide (7) repeats the proportional Transfer Function in its header by mistake, but its calculation (unity) uses the integral one correctly. The stability limit $k_I < 2\\zeta\\omega_n$ is added here (it is the $k_p = 0$ case of the 5.1 PI result)."},{"id":"engr5001-ex-pid-03","module":"ENGR5001","week":4,"label":"Old spec 4.6 example, derivative action on a second order plant","kind":"tutorial","level":2,"minutes":10,"tags":["pid-actions","closed-loop-tf","damping-ratio","velocity-feedback","block-diagrams"],"source":{"name":"ENGR202 (old spec) Control 4.6 Derivative and Velocity, derivative action example (slides 5-13)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 4 - Control 4.6 Derivative and Velocity (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Derivative action $u(t) = k_d\\dfrac{de}{dt}$, i.e. $U = k_ds(D - X)$ (learn this)\n- Negative feedback rule; Final Value Theorem (learn this)\n","text":"Derivative action $u(t) = k_d\\dfrac{d}{dt}\\big(d(t) - x(t)\\big)$ is applied, with unity negative feedback, to the\nplant\n$$X = \\frac{\\omega_n^2}{s^2 + 2\\zeta\\omega_n s + \\omega_n^2}U$$\n","parts":[{"label":"a","text":"Find the closed-loop Transfer Function and characteristic equation.","grade":"$undefined"},{"label":"b","text":"Show how the control gain changes the damping ratio. For a plant with $\\zeta = 0.1$ and $\\omega_n = 2$ rad/s,\nwhat $k_d$ gives a closed-loop damping ratio of 0.7?\n","answer":[{"label":"kd","value":0.6,"units":"s"}],"grade":"$undefined"},{"label":"c","text":"Find the steady state output for a unit step in the set point, and explain the practical problem with a step\nin the set point. What is used instead?\n","answer":[{"label":"steady state output","value":0}],"grade":"$undefined"}],"solution":"$1f","confidence":"HIGH","solutionNote":"Parts (a) and (c) match 4.6 slides 8-13 ($\\zeta_{cl} = \\zeta + 0.5\\omega_nk_d$, steady state 0). The numerical design in (b) is added here. The velocity feedback exercise from the same deck is engr5001-ox-4-6."},{"id":"engr5001-ex-pid-04","module":"ENGR5001","week":5,"label":"Old spec 5.1 example, PI control stability and steady state","kind":"tutorial","level":2,"minutes":12,"tags":["pid-actions","pi-control","hurwitz","closed-loop-tf","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 5.1 PI Control, closed loop, Hurwitz and steady state (slides 7-12)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.1 PI Control (old spec ENGR202).pdf"},"grade":"2:1","formulaSheet":"- PI control law $u(t) = k_pe(t) + k_I\\displaystyle\\int e\\,dt$, i.e. $C(s) = k_p + \\dfrac{k_I}{s} = \\dfrac{k_I + k_ps}{s}$ (learn this)\n- Third order Hurwitz: $a_0s^3 + a_1s^2 + a_2s + a_3 = 0$ stable if all $a_i > 0$ and $a_1a_2 > a_0a_3$ (learn this)\n","text":"PI control $C(s) = k_p + k_I/s$ is applied, with unity negative feedback, to the plant\n$X = \\dfrac{\\omega_n^2}{s^2 + 2\\zeta\\omega_n s + \\omega_n^2}U$.\n","parts":[{"label":"a","text":"Find the closed-loop Transfer Function and characteristic equation.","grade":"$undefined"},{"label":"b","text":"Use the Hurwitz criterion to find the stability condition on the gains. For $\\zeta = 0.5$, $\\omega_n = 1$ rad/s\nand $k_p = 1$, what is the largest $k_I$ for stability?\n","answer":[{"label":"maximum kI","value":2}],"grade":"$undefined"},{"label":"c","text":"Find the steady state output for a unit step in the set point.","answer":[{"label":"steady state output","value":1}],"grade":"$undefined"}],"solution":"$20","confidence":"HIGH","solutionNote":"Parts (a)-(c) match 5.1 slides 9-11 (condition $2\\zeta\\omega_n(1 + k_p) > k_I$, unity gain). The numerical case in (b) is added here."},{"id":"engr5001-ex-pid-05","module":"ENGR5001","week":5,"label":"Old spec 5.2 examples, PD and PV control","kind":"tutorial","level":2,"minutes":12,"tags":["pid-actions","pd-control","pv-control","controller-design","pole-placement","closed-loop-tf"],"source":{"name":"ENGR202 (old spec) Control 5.2 PD, PV and PID Control, PD and PV examples (slides 4-9)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.2 PD, PV and PID Control (old spec ENGR202).pdf"},"grade":"2:1","formulaSheet":"- PD: $u = k_pe + k_d\\dfrac{de}{dt}$; PV: $u = k_pe - k_v\\dfrac{dx}{dt}$ (learn this)\n- Compare the closed-loop characteristic equation with $s^2 + 2\\zeta_{cl}\\omega_{cl}s + \\omega_{cl}^2$ (learn this)\n","text":"The plant is $X = \\dfrac{\\omega_n^2}{s^2 + 2\\zeta\\omega_n s + \\omega_n^2}U$ with unity negative feedback and error\n$e = d - x$.\n","parts":[{"label":"a","text":"PD control $u = k_pe + k_d\\,de/dt$. Find the closed-loop Transfer Function, characteristic equation and steady\nstate output for a unit step.\n","grade":"$undefined"},{"label":"b","text":"PV control $u = k_pe - k_v\\,dx/dt$ (velocity feedback of the output in an inner loop). Find the closed-loop\nTransfer Function and steady state output, and say how it differs from PD.\n","grade":"$undefined"},{"label":"c","text":"For a plant with $\\zeta = 0.5$ and $\\omega_n = 1$ rad/s, choose PV gains so that the closed loop has natural\nfrequency 2 rad/s and damping ratio 0.7. What is the steady state output for a unit step?\n","answer":[{"label":"kp","value":3},{"label":"kv","value":1.8,"units":"s"},{"label":"steady state output","value":0.75}],"grade":"$undefined"}],"solution":"$21","confidence":"HIGH","solutionNote":"Parts (a)-(b) match 5.2 slides 5-9. The pole-placement design in (c) is added here in the style of the ENGR202 past-paper PV design questions."},{"id":"engr5001-ex-pid-06","module":"ENGR5001","week":5,"label":"Old spec 5.2 example, PID control stability","kind":"tutorial","level":2,"minutes":12,"tags":["pid-actions","hurwitz","closed-loop-tf","steady-state-gain"],"source":{"name":"ENGR202 (old spec) Control 5.2 PD, PV and PID Control, PID example (slides 10-12)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.2 PD, PV and PID Control (old spec ENGR202).pdf"},"grade":"2:1","formulaSheet":"- PID: $C(s) = k_p + \\dfrac{k_I}{s} + k_ds$ (learn this)\n- Third order Hurwitz: stable if all coefficients positive and $a_1a_2 > a_0a_3$ (learn this)\n","text":"PID control $u = k_pe + k_I\\int e\\,dt + k_d\\,de/dt$ is applied with unity negative feedback to the plant\n$X = \\dfrac{\\omega_n^2}{s^2 + 2\\zeta\\omega_n s + \\omega_n^2}U$.\n","parts":[{"label":"a","text":"Find the closed-loop Transfer Function and characteristic equation, and check the steady state gain.","answer":[{"label":"steady state gain","value":1}],"grade":"$undefined"},{"label":"b","text":"Use the Hurwitz determinant to find the stability condition relating the gains. For $\\zeta = 0.5$,\n$\\omega_n = 1$ rad/s, $k_p = 1$ and $k_d = 1$, find the largest $k_I$ for stability.\n","answer":[{"label":"maximum kI","value":4}],"grade":"$undefined"}],"solution":"$22","confidence":"HIGH","solutionNote":"Part (a) and the Hurwitz determinant match 5.2 slides 11-12 (the deck stops at the determinant); the explicit condition and the numerical case are added here."},{"id":"engr5001-ex-freq-01","module":"ENGR5001","week":5,"label":"Old spec 5.3 worked example, first order response to a harmonic input by Laplace","kind":"tutorial","level":3,"minutes":15,"tags":["frequency-response","laplace-tf","partial-fractions","magnitude-phase"],"source":{"name":"ENGR202 (old spec) Control 5.3 Frequency Response Revisited, first order response to a harmonic input (slides 5-7)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.3 Frequency Response Revisited (old spec ENGR202).pdf"},"grade":"2:1","formulaSheet":"- $\\mathcal{L}[\\sin\\omega t] = \\dfrac{\\omega}{s^2 + \\omega^2}$ (learn this)\n- First order frequency response $M = \\dfrac{K}{\\sqrt{1 + \\tau^2\\omega^2}}$, $\\phi = -\\tan^{-1}(\\omega\\tau)$ (learn this)\n","text":"The first order system $\\tau\\dfrac{dx}{dt} + x = Ku(t)$, initially at rest ($x(0) = 0$), is driven by the harmonic\ninput $u(t) = a\\sin(\\omega t)$.\n","parts":[{"label":"a","text":"Find $X(s)$.","grade":"$undefined"},{"label":"b","text":"Show that $x(t)$ is the sum of a decaying transient and a steady state sinusoid, and find the coefficient of\nthe transient term (use the partial fraction for the pole at $s = -1/\\tau$ and the condition $x(0) = 0$ as a\ncheck).\n","grade":"$undefined"},{"label":"c","text":"For $K = 2$, $\\tau = 4$ s, $a = 1$ and $\\omega = 0.5$ rad/s, find the steady state amplitude and phase of the\noutput and the transient coefficient.\n","answer":[{"label":"steady state amplitude aM","value":0.8944},{"label":"phase","value":-63.43,"units":"deg"},{"label":"transient coefficient","value":0.8}],"grade":"$undefined"}],"solution":"$23","confidence":"HIGH","solutionNote":"Slide 6 gives the transient term as $\\dfrac{Ka\\omega}{(1/\\tau^2 + \\omega^2)}e^{-t/\\tau}$, which is too large by a\nfactor $\\tau$: the correct coefficient is $\\dfrac{Ka\\omega}{\\tau(1/\\tau^2 + \\omega^2)} = \\dfrac{Ka\\omega\\tau}{1 + \\omega^2\\tau^2}$.\nConfirmed by the residue, by the condition $x(0) = 0$ and by simulation (the slide's version misses by 2.4 in the\ntest case of part (c)). The steady state part on the slide is correct.\n"},{"id":"engr5001-ex-freq-02","module":"ENGR5001","week":5,"label":"Old spec 5.3 numerical example, Bode diagram of a first order system","kind":"tutorial","level":2,"minutes":10,"tags":["frequency-response","bode-diagram","magnitude-phase"],"source":{"name":"ENGR202 (old spec) Control 5.3 Frequency Response Revisited, Bode diagram numerical example (slides 13-14)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.3 Frequency Response Revisited (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- $M = |G(j\\omega)|$, $\\phi = \\mathrm{Arg}\\,G(j\\omega)$; $|z_1/z_2| = |z_1|/|z_2|$, $\\mathrm{Arg}(z_1/z_2) = \\mathrm{Arg}\\,z_1 - \\mathrm{Arg}\\,z_2$ (learn this)\n- Corner frequency $\\omega = 1/\\tau$, where $\\log_{10}M = \\log_{10}K - 0.15$ and $\\phi = -45^\\circ$ (learn this)\n","text":"For the first order system $G(s) = \\dfrac{1}{1 + 3s}$, find general expressions for the magnitude $M$ and phase\n$\\phi$ of the frequency response, then complete a table of $M$, $\\log_{10}M$ and $\\phi$ (radians and degrees) at\n$\\omega = 0$, $1/3$, 1, 10 rad/s and $\\omega \\to \\infty$, and describe the Bode diagram.\n","answer":[{"label":"M at omega = 1/3","value":0.7071},{"label":"M at omega = 1","value":0.3162},{"label":"phase at omega = 1","value":-71.57,"units":"deg"},{"label":"M at omega = 10","value":0.03331},{"label":"phase at omega = 10","value":-88.09,"units":"deg"}],"solution":"$$G(j\\omega) = \\dfrac{1 + j0}{1 + 3j\\omega}$, so\n$$M = \\frac{1}{\\sqrt{1 + 9\\omega^2}}, \\qquad \\phi = \\tan^{-1}\\frac{0}{1} - \\tan^{-1}\\frac{3\\omega}{1} = -\\tan^{-1}(3\\omega)$$\n\n| $\\omega$ (rad/s) | $M$ | $\\log_{10}M$ | $\\phi$ (rad) | $\\phi$ (deg) |\n|---|---|---|---|---|\n| 0 | 1 | 0 | 0 | 0 |\n| 1/3 | 0.7071 | $-0.15$ | $-0.785$ | $-45$ |\n| 1 | 0.3162 | $-0.50$ | $-1.249$ | $-71.6$ |\n| 10 | 0.0333 | $-1.48$ | $-1.537$ | $-88.1$ |\n| $\\infty$ | 0 | $-\\infty$ | $-1.571$ | $-90$ |\n\nBode diagram: $\\log_{10}M$ is flat at $\\log_{10}K = 0$ at low frequency, is 0.15 down (3 dB) at the corner\nfrequency $\\omega = 1/\\tau = 1/3$ rad/s, and then falls with slope $-1$ on log-log axes (20 dB per decade). The phase\ngoes from 0 to $-90^\\circ$, passing $-45^\\circ$ at the corner.\n\n**Checks.** At high frequency $M \\approx 1/(3\\omega)$: at $\\omega = 10$, $1/30 = 0.0333$. Values computed in Python\n(also in dB: $-3.0$, $-10.0$, $-29.5$ dB).\n","confidence":"HIGH","solutionNote":"Matches the table on 5.3 slide 14 (the slide rounds the phases to $-72^\\circ$ and $-88^\\circ$)."},{"id":"engr5001-ex-fdesign-01","module":"ENGR5001","week":5,"label":"Old spec 5.4 example, closed-loop stability from the Bode diagram","kind":"tutorial","level":3,"minutes":15,"tags":["frequency-design","bode-diagram","gain-margin","phase-margin","poles-stability"],"source":{"name":"ENGR202 (old spec) Control 5.4 Using the Frequency Response, application to Bode diagrams example (slides 12-16 and 22)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.4 Using the Frequency Response (old spec ENGR202).pdf"},"grade":"2:1","formulaSheet":"- Unity feedback around $KG(s)$: closed-loop characteristic equation $1 + KG(s) = 0$; marginal stability when $|KG(j\\omega)| = 1$ at the frequency where $\\mathrm{Arg}\\,KG(j\\omega) = -180^\\circ$ (learn this)\n- Gain margin $= 1/|KG(j\\omega_{pc})|$ at the phase crossover; phase margin $= 180^\\circ + \\mathrm{Arg}\\,KG(j\\omega_{gc})$ at the gain crossover (learn this)\n","text":"A unity feedback control system has the open-loop Transfer Function\n$$KG(s) = \\frac{K}{s(s + 1)^2}$$\n","parts":[{"label":"a","text":"Find the closed-loop Transfer Function and characteristic equation. Show that for $K = 2$ the closed loop is\nmarginally stable, giving the closed-loop poles.\n","grade":"$undefined"},{"label":"b","text":"Using the open-loop frequency response, find the phase crossover frequency and $|KG|$ there, and hence the\nvalue of $K$ at the stability limit. Classify $K = 0.1$, 2 and 10.\n","answer":[{"label":"phase crossover frequency","value":1,"units":"rad/s"},{"label":"critical K","value":2}],"grade":"$undefined"},{"label":"c","text":"For $K = 0.1$, find the gain margin (in dB) and the phase margin.","answer":[{"label":"gain margin","value":26.02,"units":"dB"},{"label":"phase margin","value":78.7,"units":"deg","tol":0.01}],"grade":"$undefined"}],"solution":"$24","confidence":"HIGH","solutionNote":"Parts (a)-(b) match 5.4 slides 12-16 (marginal at K = 2, unstable at 10, stable at 0.1). The slide marks GM and PM on the K = 0.1 Bode plot (slide 22) without numbers; the values in (c) are computed here."},{"id":"engr5001-ex-fdesign-02","module":"ENGR5001","week":5,"label":"Old spec 5.4 example, Nyquist stability and Hurwitz limit for a third order loop","kind":"tutorial","level":3,"minutes":15,"tags":["frequency-design","nyquist","hurwitz","gain-margin","phase-margin"],"source":{"name":"ENGR202 (old spec) Control 5.4 Using the Frequency Response, application to Nyquist diagrams example (slides 17-20)","paper":"ENGR202","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.4 Using the Frequency Response (old spec ENGR202).pdf"},"grade":"2:1","formulaSheet":"- Critical point for neutral stability on the Nyquist diagram: $-1 + 0j$ (learn this)\n- Third order Hurwitz: stable if all coefficients positive and $a_1a_2 > a_0a_3$ (learn this)\n","text":"A unity feedback system has open-loop Transfer Function\n$$KG(s) = \\frac{K}{(s + 4)(s^2 + 4s + 8)}$$\n","parts":[{"label":"a","text":"Find the closed-loop characteristic equation and use the Hurwitz criterion to find the range of $K > 0$ for\nstability.\n","answer":[{"label":"critical K","value":160}],"grade":"$undefined"},{"label":"b","text":"Find where the Nyquist plot of $KG(j\\omega)$ crosses the negative real axis (frequency and value in terms of\n$K$), and confirm the critical gain. Describe the Nyquist plots for $K = 80$, 160 and 200.\n","answer":[{"label":"phase crossover frequency","value":4.899,"units":"rad/s"}],"grade":"$undefined"},{"label":"c","text":"For $K = 80$, find the gain margin (dB) and the phase margin.","answer":[{"label":"gain margin","value":6.02,"units":"dB"},{"label":"phase margin","value":30.3,"units":"deg","tol":0.02}],"grade":"$undefined"}],"solution":"$25","confidence":"HIGH","solutionNote":"Parts (a)-(b) match 5.4 slides 17-20 (unstable for K > 160; K = 80 stable, 200 unstable). The characteristic\nequation on slide 17 is printed as $s^3 + 8s + 24s + (32 + K) = 0$; the second term should be $8s^2$. The axis\ncrossing and margins in (b)-(c) are computed here.\n"},{"id":"engr201-2017-b3","module":"ENGR5001","week":3,"label":"ENGR201 2017 B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["laplace-tf","partial-fractions","step-response","modelling","generalised-forms","time-constant","steady-state-gain"],"source":{"name":"ENGR201 Engineering Analysis 2017 B3","paper":"ENGR201","year":2017,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR201_Exam_2017.pdf","tag":"VALID"},"text":"It is the year 1971. A steel container containing nuclear waste, of combined weight $W$ (N), is dumped into the\nocean at time $t = 0$. Let $v(t)$ (m/s) be the velocity with which the container moves towards the bottom of the\nocean at time $t$ (s). Let $B$ be the buoyancy force of the water and $D_f$ the drag force, both in newtons and\nboth acting against the motion of the container. If $v$ is not too large, the drag force is proportional to the\nspeed:\n\n$$D_f = k\\,v(t) \\qquad (4)$$\n\nwhere $k$ is a constant.\n","formulaSheet":"- Newton's second law $m\\dfrac{dv}{dt} = \\sum F$, weight $W = mg$ (learn this)\n- Laplace of a derivative $\\mathcal{L}\\left[\\dfrac{dv}{dt}\\right] = sV(s) - v(0)$ (learn this; printed in the ENGR201 Laplace table)\n- Pairs: $1 \\leftrightarrow \\dfrac{1}{s}$, $e^{-\\alpha t} \\leftrightarrow \\dfrac{1}{s + \\alpha}$ (learn this; printed in the ENGR201 Laplace table)\n- Partial fractions $\\dfrac{1}{s(s + a)} = \\dfrac{1}{a}\\left(\\dfrac{1}{s} - \\dfrac{1}{s + a}\\right)$ (learn this)\n- Final Value Theorem $\\lim_{t \\to \\infty} f(t) = \\lim_{s \\to 0} sF(s)$ (learn this)\n","parts":[{"label":"a","text":"Show that $v(t)$ is the solution of the differential equation\n\n$$m\\frac{dv(t)}{dt} + k\\,v(t) = W - B, \\qquad v(t = 0) = 0 \\qquad (5)$$\n\nwhere $m$ is the combined mass of the container (kg). Explain the origin of each term in your answer.\n","marks":5,"grade":"third"},{"label":"b","text":"Use the method of Laplace transforms to solve equation (5) for $v(t)$. From this solution, derive an\nequation for the steady state velocity of the container.\n","marks":8,"grade":"2:2"},{"label":"c","text":"Let $y(t)$ be the depth below the ocean surface attained by the container at time $t$. Find $y(t)$ by\nintegrating $v(t)$ and using the condition $y(0) = 0$.\n","marks":5,"grade":"2:2"},{"label":"d","text":"The container will break if its velocity exceeds 12 m/s when it hits the sea bed. Given that\n$W = 3254$ N, $B = 3090$ N and $k = 0.637$ kg/s, show that the container will break if the depth of the\nocean where it is dumped exceeds 153 m.\n","marks":7,"grade":"2:1","answer":[{"label":"terminal velocity v_T","value":257.46,"units":"m/s"},{"label":"time to reach 12 m/s","value":24.85,"units":"s"},{"label":"depth at which v = 12 m/s","value":150.3,"units":"m"}]}],"solution":"$26","confidence":"HIGH","solutionNote":"No official solution available. The critical depth works out at 150.3 m (g = 9.81 m/s^2; 150.5 m with g = 9.8), not exactly the 153 m quoted in the paper; the paper's statement is still true (any depth above 150.3 m, so certainly above 153 m, breaks the container) and at 153 m the impact speed is 12.1 m/s."},{"id":"engr201-2018-b3","module":"ENGR5001","week":3,"label":"ENGR201 2018 B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["laplace-tf","rlc-circuit","modelling","poles-stability","damping-ratio","step-response"],"source":{"name":"ENGR201 Engineering Analysis 2018 B3","paper":"ENGR201","year":2018,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR201_2018.pdf","tag":"VALID"},"text":"The quartz crystal is a small electronic component used in timing applications and sensing. It comprises a\nthin piece of quartz, sandwiched between two vapour-deposited electrodes, which oscillates in response to an\nr.f. excitation. The electrical characteristics of the crystal can be represented by the equivalent circuit\nin Figure B3: a resistor $R$, inductor $L$ and capacitor $C$ in series, driven by a voltage $v(t)$.\n","figures":[{"src":"/figures/engr201-2018-b3.png","caption":"Figure B3: series R, L, C equivalent circuit of the quartz crystal, driven by v(t)."}],"formulaSheet":"- Component laws $v_R = Ri$, $v_L = L\\dfrac{di}{dt}$, $v_C = \\dfrac{1}{C}\\displaystyle\\int i\\,dt$; Kirchhoff's voltage law (learn this)\n- Laplace with zero initial conditions: $\\dfrac{di}{dt} \\to sI$, $\\displaystyle\\int_0^t i\\,dt \\to \\dfrac{I}{s}$; unit step $\\to \\dfrac{1}{s}$ (learn this)\n- $e^{-\\alpha t}\\sin\\omega t \\leftrightarrow \\dfrac{\\omega}{(s + \\alpha)^2 + \\omega^2}$ (learn this; printed in the ENGR201 Laplace table)\n- Second order form $s^2 + 2\\zeta\\omega_n s + \\omega_n^2$, damped frequency $\\omega_d = \\omega_n\\sqrt{1 - \\zeta^2}$ (learn this)\n","parts":[{"label":"a","text":"Show that the differential equation describing the rate of voltage change with time in terms of current\nfor the quartz crystal is\n\n$$\\frac{dv}{dt} = L\\frac{d^2i}{dt^2} + R\\frac{di}{dt} + \\frac{i}{C}$$\n\nexplaining the origin of each term in your answer.\n","marks":4,"grade":"third"},{"label":"b","text":"Using Laplace transformation, derive an expression describing the current response $I(s)$ in terms of\nthe voltage stimulus $V(s)$, given that the crystal is initially quiescent.\n","marks":8,"grade":"2:2"},{"label":"c","text":"If the quartz crystal is subject to an impulse voltage stimulus in the form of a unit step function at\n$t = 0$, determine an expression for the current response $i(t)$.\n","marks":8,"grade":"2:1"},{"label":"d","text":"Given that $R = 10\\ \\Omega$, $L = 1$ mH and $C = 4$ pF, determine the time at which the current reaches\n$1/e$ of its initial value following the impulse. Approximately how many cycles will the quartz\noscillator have undergone at this time?\n","marks":5,"grade":"2:2","answer":[{"label":"time to 1/e","value":0.2,"units":"ms"},{"label":"number of cycles","value":503,"units":"cycles","tol":0.03}]}],"solution":"$27","confidence":"MEDIUM","solutionNote":"No official solution available. Part (c) is ambiguously worded (\"impulse voltage stimulus in the form of a unit step\"); the solution takes v(t) as a unit step (so dv/dt is an impulse) and gives the Dirac-impulse alternative. Both give the same decay time (0.2 ms) and about 503 cycles in (d)."},{"id":"engr201-2019-b3","module":"ENGR5001","week":3,"label":"ENGR201 2019 B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["laplace-tf","partial-fractions","modelling","frequency-response","time-response","time-constant"],"source":{"name":"ENGR201 Engineering Analysis 2019 B3","paper":"ENGR201","year":2019,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR201_Exam_ 2019.pdf","tag":"VALID"},"text":"The circuit shown in Figure B3-1 consists of a resistor $R$ and an inductor $L$ connected in series with a\nvoltage source $V(t)$.\n","figures":[{"src":"/figures/engr201-2019-b3.png","caption":"Figure B3-1: inductor L and resistor R in series with a voltage source V(t)."}],"formulaSheet":"- $v_R = Ri$, $v_L = L\\dfrac{di}{dt}$, Kirchhoff's voltage law (learn this)\n- $\\mathcal{L}\\left[\\dfrac{di}{dt}\\right] = sI(s) - i(0)$ (learn this; printed in the ENGR201 Laplace table)\n- $\\sin\\omega t \\leftrightarrow \\dfrac{\\omega}{s^2 + \\omega^2}$, $\\cos\\omega t \\leftrightarrow \\dfrac{s}{s^2 + \\omega^2}$, $e^{-\\alpha t} \\leftrightarrow \\dfrac{1}{s + \\alpha}$ (learn this; printed in the ENGR201 Laplace table)\n- Partial fractions with an irreducible quadratic: $\\dfrac{A}{Ls + R} + \\dfrac{Bs + C}{s^2 + 1}$ (learn this)\n","parts":[{"label":"a","text":"Write down a differential equation describing the rate of change of current $i(t)$ with time, explaining\nthe origin of each term in your answer.\n","marks":7,"grade":"third"},{"label":"b","text":"Using Laplace transformation methods, derive an expression describing the current response $I(s)$ in\nterms of the voltage stimulus $V(s)$, given that $i(0) = 0$.\n","marks":8,"grade":"third"},{"label":"c","text":"If the voltage source is given by $V\\sin(t)$, derive an expression for the current response $i(t)$.\n","marks":10,"grade":"2:2"}],"solution":"$28","confidence":"HIGH","solutionNote":"No official solution available; i(t) verified symbolically (sympy inverse Laplace and back-substitution)."},{"id":"engr201-2021-q2","module":"ENGR5001","week":3,"label":"ENGR201 2021 Q2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["laplace-tf","partial-fractions","poles-stability","time-delay","step-response"],"source":{"name":"ENGR201 Engineering Analysis 2021 (Michaelmas) Q2","paper":"ENGR201","year":2021,"question":"Q2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR201 Exam (Michaelmas) 2021.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2021/ENGR201 Part A answers.pdf","tag":"VALID"},"text":"Consider the initial value problem of a dynamic system subjected to the exponential input $e^{2t}$:\n\n$$\\frac{d^2y}{dt^2} - 2\\frac{dy}{dt} + y = e^{2t}$$\n\nwith initial conditions $y(0) = \\dfrac{dy(0)}{dt} = 0$.\n","formulaSheet":"- $\\mathcal{L}[y'] = sY - y(0)$, $\\mathcal{L}[y''] = s^2Y - sy(0) - y'(0)$ (learn this; printed in the ENGR201 Laplace table)\n- $e^{at} \\leftrightarrow \\dfrac{1}{s - a}$, $t\\,e^{at} \\leftrightarrow \\dfrac{1}{(s - a)^2}$ (learn this; printed in the ENGR201 Laplace table)\n- Impulse $\\delta(t - a) \\leftrightarrow e^{-as}$; delay $H_a(t)g(t - a) \\leftrightarrow e^{-as}G(s)$ (learn this; printed in the ENGR201 Laplace table)\n- Repeated-root partial fractions $\\dfrac{A}{s - 2} + \\dfrac{B}{s - 1} + \\dfrac{C}{(s - 1)^2}$ (learn this)\n","parts":[{"label":"a","text":"Use the Laplace transform to solve the initial value problem (show your workings).","marks":18,"grade":"2:2","answer":[{"label":"y(1)","value":1.9525,"units":"-"}]},{"label":"b","text":"If the system is subjected to an additional impulse input $\\delta(t - 2)$, use the Laplace transform again\nto solve the problem (show your workings).\n","marks":5,"grade":"2:1","answer":[{"label":"y(3)","value":325.8,"units":"-"}]},{"label":"c","text":"Will the solution from (b) be discontinuous at $t = 2$? Explain your answer.","marks":2,"grade":"2:1"}],"solution":"$29","confidence":"HIGH","solutionNote":"Matches the official answers (y = -e^t - t e^t + e^{2t}; extra term (t-2)e^{t-2}u(t-2); continuous at t = 2)."},{"id":"engr201-2022-a2","module":"ENGR5001","week":3,"label":"ENGR201 2022 A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["laplace-tf","partial-fractions","step-response","mass-spring-damper","damping-ratio","generalised-forms"],"source":{"name":"ENGR201 Engineering Analysis 2021-22 Part A, A2","paper":"ENGR201","year":2022,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR201 Exam Part A 2021-22.pdf","tag":"VALID"},"text":"An impact protection device is subjected to a constant unit impact force $P(t)$ from $t = 0$. The response of\nthe device is described by the equation of motion\n\n$$\\frac{d^2x(t)}{dt^2} + 2\\frac{dx(t)}{dt} + 50x(t) = P(t) \\qquad \\text{(A2)}$$\n\nwhere $x(t)$ is the displacement of the device from its initial position and $t$ is time in seconds. The device\nhas no initial displacement or velocity.\n","formulaSheet":"- $\\mathcal{L}[x'] = sX - x(0)$, $\\mathcal{L}[x''] = s^2X - sx(0) - x'(0)$; unit step $\\to 1/s$ (learn this; printed in the ENGR201 Laplace table)\n- $e^{at}\\cos\\omega t \\leftrightarrow \\dfrac{s - a}{(s - a)^2 + \\omega^2}$, $e^{at}\\sin\\omega t \\leftrightarrow \\dfrac{\\omega}{(s - a)^2 + \\omega^2}$ (learn this; printed in the ENGR201 Laplace table)\n- Step function $H_a(t) \\leftrightarrow e^{-as}/s$ (learn this; printed in the ENGR201 Laplace table)\n- Generalised second order form $\\ddot x + 2\\zeta\\omega_n\\dot x + \\omega_n^2 x = Ku$ (learn this)\n","parts":[{"label":"a","text":"Find the Laplace transform of the displacement of the device.","marks":15,"grade":"2:2"},{"label":"b","text":"Find the displacement as a function of time.","marks":8,"grade":"2:2","answer":[{"label":"final displacement","value":0.02,"units":"-"},{"label":"peak displacement","value":0.03277,"units":"-"}]},{"label":"c","text":"If the impact force lasts only for 3 seconds, write down and plot the force pulse using the step\n(Heaviside) function.\n","marks":2,"grade":"third"}],"solution":"$2a","confidence":"HIGH","solutionNote":"No official solution available; x(t) verified symbolically and numerically (peak 0.03277 at 0.449 s)."},{"id":"engr201-2023-a2","module":"ENGR5001","week":3,"label":"ENGR201 2023 A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"first","tags":["laplace-tf","partial-fractions","step-response","modelling","poles-stability","steady-state-gain"],"source":{"name":"ENGR201 Engineering Analysis 2023 Section A, A2","paper":"ENGR201","year":2023,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR201 Exam 2023 Section A.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2023/ENGR201 ( Part A) 2023 Exam Numerical answers.pdf","tag":"VALID"},"text":"Consider the RL circuit shown in Figure A2. $i_1$ and $i_2$ are the currents in each loop, which satisfy the\ndifferential equations\n\n$$L_1\\frac{di_1}{dt} + R_2(i_1 - i_2) + R_1 i_1 = v(t) \\qquad \\text{(A3a)}$$\n$$L_2\\frac{di_2}{dt} + R_2(i_2 - i_1) = 0 \\qquad \\text{(A3b)}$$\n\nAssume $L_1 = 0.8$ H, $L_2 = 1$ H, $R_1 = 1.4\\ \\Omega$, $R_2 = 1\\ \\Omega$ and $v(t) = 100$ V for $t > 0$. The\ninitial conditions are $i_1(0) = 0$ and $i_2(0) = 0$.\n","figures":[{"src":"/figures/engr201-2023-a2.png","caption":"Figure A2: two-loop circuit; loop 1 (current i1) contains v(t), R1, L1 and the shared R2; loop 2 (current i2) contains R2 and L2."}],"formulaSheet":"- $\\mathcal{L}\\left[\\dfrac{di}{dt}\\right] = sI(s) - i(0)$; constant $100 \\to \\dfrac{100}{s}$ (learn this; printed in the ENGR201 Laplace table)\n- $e^{at} \\leftrightarrow \\dfrac{1}{s - a}$ (learn this; printed in the ENGR201 Laplace table)\n- Cover-up rule for distinct poles: residue at $s = p$ is $\\left[(s - p)F(s)\\right]_{s = p}$ (learn this)\n- Final value theorem $\\lim_{t \\to \\infty} i(t) = \\lim_{s \\to 0} sI(s)$ (learn this)\n","parts":[{"label":"a","text":"Find the Laplace transforms of the currents.","marks":10,"grade":"2:1"},{"label":"b","text":"Find the currents by the inverse Laplace transform.","marks":12,"grade":"first","answer":[{"label":"i1 at t = 1 s","value":45.26,"units":"A"},{"label":"i2 at t = 1 s","value":21.24,"units":"A"}]},{"label":"c","text":"Explain what happens to the currents when $t$ increases.","marks":3,"grade":"2:2","answer":[{"label":"i1 = i2 as t -> infinity","value":71.43,"units":"A"}]}],"solution":"$2b","confidence":"HIGH","solutionNote":"Matches the official numerical answers (I1 = 125(s+1)/(s(s+0.5)(s+3.5)), I2 = 125/(s(s+0.5)(s+3.5)), i1 = 500/7 - (125/3)e^{-t/2} - (625/21)e^{-7t/2}, i2 = 500/7 - (250/3)e^{-t/2} + (250/21)e^{-7t/2}). Graded first because coupled (simultaneous) ODEs go beyond the single-ODE examples in the ENGR5001 Part 3 slides."},{"id":"engr201-2024-a2","module":"ENGR5001","week":3,"label":"ENGR201 2024 A2(a)-(b)","kind":"past-paper","level":2,"marks":13,"minutes":16,"grade":"2:2","tags":["laplace-tf","partial-fractions","step-response","rlc-circuit","damping-ratio","generalised-forms"],"source":{"name":"ENGR201 Engineering Analysis 2024 A2 (a)-(b)","paper":"ENGR201","year":2024,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR201 Main -Exam Paper - 2024 Combined.pdf","tag":"PARTIAL"},"text":"Figure A2-1 shows a typical RLC circuit. After applying Kirchhoff's law, the charge $q(t)$ can be solved from\nthe differential equation\n\n$$2\\frac{d^2q(t)}{dt^2} + 16\\frac{dq(t)}{dt} + 50q(t) = 300 \\qquad \\text{(A1)}$$\n\nAt $t = 0$ the charge $q(0)$ on the capacitor is zero and the current $I(0) = \\dfrac{dq(0)}{dt}$ of the\ncircuit is also zero.\n\n(Parts (c) and (d) of the original question, Newton iteration for the time of maximum current, are not on the\nENGR5001 syllabus and are left out.)\n","figures":[{"src":"/figures/engr201-2024-a2.png","caption":"Figure A2-1: series RLC circuit driven by a voltage source V(t), current I(t)."}],"formulaSheet":"- $\\mathcal{L}[q'] = sQ - q(0)$, $\\mathcal{L}[q''] = s^2Q - sq(0) - q'(0)$; constant $c \\to c/s$ (learn this; printed in the ENGR201 Laplace table)\n- $e^{at}\\cos\\omega t \\leftrightarrow \\dfrac{s - a}{(s - a)^2 + \\omega^2}$, $e^{at}\\sin\\omega t \\leftrightarrow \\dfrac{\\omega}{(s - a)^2 + \\omega^2}$ (learn this; printed in the ENGR201 Laplace table)\n- Current $I = \\dfrac{dq}{dt}$, so $I(s) = sQ(s) - q(0)$ (learn this)\n","parts":[{"label":"a","text":"Use the Laplace transform to find the charge $Q(s)$ in the s-domain.","marks":8,"grade":"third"},{"label":"b","text":"Use the inverse Laplace transform to find the charge $q(t)$ and the current $I(t)$ in the time domain.","marks":5,"grade":"2:2","answer":[{"label":"final charge q(infinity)","value":6,"units":"C"},{"label":"I at t = 0.1 s","value":9.905,"units":"A"}]}],"solution":"$2c","confidence":"HIGH","solutionNote":"No official solution available; q(t) and I(t) verified symbolically (sympy) and by back-substitution."},{"id":"engr201-2025-q2c","module":"ENGR5001","week":3,"label":"ENGR201 2025 Q2(c)","kind":"past-paper","level":2,"marks":28,"minutes":25,"grade":"2:2","tags":["laplace-tf","partial-fractions","step-response","steady-state-gain","heating-control"],"source":{"name":"ENGR201 Engineering Analysis 2024-25 Q2(c)","paper":"ENGR201","year":2025,"question":"Q2(c)","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR201 Main Exam Paper 2024-25.pdf","tag":"PARTIAL"},"text":"The differential equation below represents a temperature regulation system in a heat exchanger. $u(t)$ is a\nstep (Heaviside) function denoting the power supplied to the heating element, and $y(t)$ is the temperature\noutput, with initial conditions $y(0) = 0$ and $\\dfrac{dy(0)}{dt} = 0$.\n\n$$\\frac{d^2y(t)}{dt^2} + 5\\frac{dy(t)}{dt} + 6y(t) = u(t)$$\n\n(Parts (a) and (b) of the original question, on parametric curves and gradients, are not on the ENGR5001\nsyllabus.)\n","formulaSheet":"- $\\mathcal{L}[y'] = sY - y(0)$, $\\mathcal{L}[y''] = s^2Y - sy(0) - y'(0)$; unit step $\\to 1/s$ (learn this)\n- $e^{-\\alpha t} \\leftrightarrow \\dfrac{1}{s + \\alpha}$ (learn this)\n- Final value theorem $\\lim_{t \\to \\infty} y(t) = \\lim_{s \\to 0} sY(s)$, or steady state gain = TF at $s = 0$ (learn this)\n","parts":[{"label":"i","text":"Use the Laplace transform to find the temperature $Y(s)$ in the s-domain.","marks":13,"grade":"third"},{"label":"ii","text":"Use the inverse Laplace transform to find the temperature $y(t)$.","marks":12,"grade":"2:2","answer":[{"label":"y at t = 1","value":0.1156,"units":"-"}]},{"label":"iii","text":"What is the stabilised temperature of the heat exchanger?","marks":3,"grade":"third","answer":[{"label":"y(infinity)","value":0.16667,"units":"-"}]}],"solution":"$2d","confidence":"HIGH","solutionNote":"No official solution available; y(t) verified symbolically and by the final value theorem."},{"id":"engr202-2017-a1","module":"ENGR5001","week":5,"label":"ENGR202 2017 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["laplace-tf","generalised-forms","poles-stability","frequency-response","controller-design","time-constant","pole-placement","oil-fractionator"],"source":{"name":"ENGR202 Summer 2017 Q A1","paper":"ENGR202","year":2017,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR202_Exam_2017rev2.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR202 Numerical Answers 2017.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Generalised forms: $\\tau\\dfrac{dx}{dt} + x = Ku(t)$ and $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$ (given in the question)\n- Frequency response: $M = |G(s)|_{s = j\\omega}$, $\\phi = \\mathrm{Arg}(G(s))|_{s = j\\omega}$ (given in the question)\n- Laplace with zero initial conditions: $\\dfrac{dx}{dt} \\to sX(s)$, $\\dfrac{d^2x}{dt^2} \\to s^2X(s)$ (learn this)\n- Steady state gain: set $s = 0$ (learn this)\n- Pole placement: match the closed-loop characteristic equation to $s^2 + 2\\zeta\\omega_n s + \\omega_n^2$ (learn this)\n","text":"The relationship between the pressure $X(s)$ and the flow rate control input $U(s)$ in an oil fractionator\nis described by the following Transfer Function model:\n\n$$X(s) = \\frac{b_1}{s + a_1}\\,U(s) \\qquad \\text{(1a)}$$\n\nUsing standard notation, generalised first and second order differential equations are:\n\n$$\\tau\\frac{dx}{dt} + x = Ku(t) \\quad \\text{and} \\quad \\frac{d^2x}{dt^2} + 2\\zeta\\omega_n\\frac{dx}{dt} + \\omega_n^2 x = Ku(t) \\qquad \\text{(1b)}$$\n","parts":[{"label":"a","text":"Showing your working and indicating any assumptions made, develop the Transfer Functions forms of the\ngeneralized first and second order differential equations (1b), expressed in terms of $\\tau$, $K$,\n$\\omega_n$ and $\\zeta$.\n","marks":5,"grade":"$undefined"},{"label":"b","text":"Assume $a_1 = 5$ and $b_1 = 2$ for this part of the question. Determine the pole and stability condition of\nthe oil fractionator model (1a). Determine the time constant and steady state gain. For a flow rate\n$u(t > 0) = 5$, what is the steady state pressure.\n","marks":6,"answer":[{"label":"pole","value":-5},{"label":"time constant","value":0.2},{"label":"steady state gain","value":0.4},{"label":"steady state pressure","value":2}],"grade":"$undefined"},{"label":"c","text":"The steady state response of the oil fractionator model (1a) for $u(t) = \\sin(\\omega t)$ when $t > 0$, is\n$M\\sin(\\omega t + \\phi)$. Determine relationships for $M$ and $\\phi$ as a function of frequency $\\omega$ and\nthe model parameters $a_1$ and $b_1$.\n\nNote that for a Transfer Function $G(s)$, then: $M = |G(s)|_{s = j\\omega}$ and $\\phi = \\mathrm{Arg}(G(s))|_{s = j\\omega}$\n","marks":7,"grade":"$undefined"},{"label":"d","text":"An automatic control system is used to regulate the pressure in the oil fractionator model. The\nclosed-loop Transfer Function is shown below, where $D(s)$ is the set point and $(k_1, k_2)$ are gains,\n\n$$X(s) = \\frac{k_1 b_1 s + k_2 b_1}{s^2 + (a_1 + k_1 b_1)s + k_2 b_1}\\,D(s)$$\n\nWhat is the stready state gain of the closed-loop Transfer Function? Comment on the significance of your\nanswer. Assuming that $a_1 = 5$ and $b_1 = 2$, design a control system (i.e. determine values for the\ncontrol gains) such that the closed-loop system has a damping ratio $\\zeta = 1.2$ and natural frequency\n$\\omega_n = 0.1$.\n","marks":7,"answer":[{"label":"closed-loop steady state gain","value":1},{"label":"k_1","value":-2.38},{"label":"k_2","value":0.005}],"grade":"$undefined"}],"solution":"$2e","confidence":"HIGH","solutionNote":"Agrees with the official answers for (b) (pole -5, $\\tau = 0.2$, $K = 0.4$, pressure 2), the closed-loop gain\n1.0 and $k_2 = 0.005$. The official $k_1 = -2.44$ is wrong: it comes from $5 + 2k_1 = \\zeta\\omega_n = 0.12$\ninstead of $2\\zeta\\omega_n = 0.24$. With $k_1 = -2.38$ the closed-loop poles give exactly $\\zeta = 1.2$,\n$\\omega_n = 0.1$ (checked numerically); with $-2.44$ they give $\\zeta = 0.6$.\n"},{"id":"engr202-2017-a2","module":"ENGR5001","week":5,"label":"ENGR202 2017 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["feedback-objectives","poles-stability","block-diagrams","pid-actions","hurwitz","dominant-poles","proportional-control","pi-control"],"source":{"name":"ENGR202 Summer 2017 Q A2","paper":"ENGR202","year":2017,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR202_Exam_2017rev2.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR202 Numerical Answers 2017.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Poles: roots of the characteristic equation; stable if all have negative real parts; dominant = closest to the imaginary axis (learn this)\n- Closed loop with feedback element: $\\dfrac{X}{V} = \\dfrac{C H}{1 + C H F}$ (learn this)\n- Hurwitz for $a_0 s^2 + a_1 s + a_2 = 0$: $H_2 = \\begin{vmatrix} a_1 & 0 \\\\ a_0 & a_2 \\end{vmatrix}$, need $\\Delta_1 = a_1 > 0$ and $\\Delta_2 = a_1 a_2 > 0$ (given in the question)\n- PI controller $C(s) = K_1 + \\dfrac{K_2}{s}$ (learn this)\n","text":"Consider the control system in Figure A2 in which $H(s)$ is the plant, $C(s)$ is the control algorithm and\n$F(s)$ is a feedback element.\n","figures":[{"src":"/figures/engr202-2017-a2.png","caption":"Figure A2: Feedback control system with controller C(s), plant H(s) and feedback element F(s)."}],"parts":[{"label":"a","text":"What are the name of the signals $V(s)$, $U(s)$ and $X(s)$? Using Figure A2 as an example, explain the\ndifference between open and closed-loop control. Give a practical example of both open and closed-loop\ncontrol.\n","marks":5,"grade":"$undefined"},{"label":"b","text":"Assume the following Transfer Function for this part of the question:\n\n$$H(s) = \\frac{2.5}{(s^2 + 3s + 2.26)(s + 2)}$$\n\nDetermine the poles of $H(s)$ and sketch their position on the $s$-plane. What is the stability of $H(s)$.\nIdentify the dominant poles and sketch the likely form of the time response mode. Note: this part of the\nquestion concerns $H(s)$ only.\n","marks":6,"answer":[{"label":"real pole","value":-2},{"label":"complex pair real part","value":-1.5},{"label":"complex pair imaginary part (+/-)","value":0.1}],"grade":"$undefined"},{"label":"c","text":"Using either standard algebra or the rules of block diagram manipulation, determine the closed-loop\nTransfer Function for Figure A2 i.e. the relationship between $V(s)$ and $X(s)$. Next, an engineer\nimplements a simple proportional control system based on $C(s) = K$ and $F(s) = 1$ where $K$ is a scalar\ncontrol gain. In this case, show how for large values of $K$ the error $E(s) = V(s) - X(s)$ approaches zero.\nWhat are the disadvantages of this control system?\n","marks":6,"grade":"$undefined"},{"label":"d","text":"Still using Figure A2, now assume the following Transfer Functions:\n\n$$H(s) = \\frac{1}{s + 10} \\qquad C(s) = K_1 + \\frac{K_2}{s} \\qquad F(s) = 1$$\n\nWhat is the name of this controller? Using these definitions of $H(s)$, $C(s)$ and $F(s)$, determine the\nclosed-loop characteristic equation and the associated Hurwitz determinant. Use your answer to find the\nminimum values of $K_1$ and $K_2$ that ensure stability of the closed-loop system.\n\nNote: given $a_0 s^2 + a_1 s + a_2 = 0$ then $H_2 = \\begin{vmatrix} a_1 & 0 \\\\ a_0 & a_2 \\end{vmatrix}$\n","marks":8,"answer":[{"label":"K_1 must exceed","value":-10},{"label":"K_2 must exceed","value":0}],"grade":"$undefined"}],"solution":"$2f","confidence":"HIGH","solutionNote":"Matches the official numerical answers (poles $-2$ and $-1.5 \\pm 0.1j$; $K_1 > -10$, $K_2 > 0$)."},{"id":"engr202-2017-a3","module":"ENGR5001","week":5,"label":"ENGR202 2017 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["poles-stability","block-diagrams","pid-actions","feedback-objectives","frequency-design","pv-control","stability-shortcuts","heating-control"],"source":{"name":"ENGR202 Summer 2017 Q A3","paper":"ENGR202","year":2017,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR202_Exam_2017rev2.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR202 Numerical Answers 2017.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Stability shortcuts: all coefficients must exist and have the same sign; $a_n = 0$ gives a pole at the origin; for a second order equation, all coefficients positive is necessary AND sufficient (learn this)\n- Generalised second order characteristic equation $s^2 + 2\\zeta\\omega_n s + \\omega_n^2 = 0$ (given in the question)\n- Negative feedback rule $\\dfrac{G_1}{1 + G_1 G_2}$ (learn this)\n- Gain margin and phase margin read from the Bode diagram of the open-loop $C(s)H(s)$ (learn this)\n","text":"Consider the following model for the heating system in a room,\n\n$$X(s) = \\frac{b_0}{s^2 + a_1 s + a_2}\\,U(s) \\qquad \\text{(3a)}$$\n\nwhere $X(s)$ represents the temperature and $U(s)$ is the heater input. A Proportional-Velocity (PV) control\nalgorithm takes the following form,\n\n$$U(s) = k_p\\left(R(s) - X(s)\\right) - k_v sX(s) \\qquad \\text{(3b)}$$\n\nwhere $R(s)$ is the command input and $(k_p, k_v)$ are control gains.\n","parts":[{"label":"a","text":"Write down the open-loop characteristic equation for the heating system model (3a) i.e. before the\napplication of PV control. Without calculating the poles, and without using the Hurwitz determinant, state\nthe stability condition for each of the following sets of coefficients. Give the reason for your answer in\neach case.\n\nCase 1: $a_1 = -1.2$, $a_2 = 3$\n\nCase 2: $a_1 = -1.2$, $a_2 = 0$\n\nCase 3: $a_1 = 0$, $a_2 = 2.5$\n\nCase 4: $a_1 = 1.2$, $a_2 = -3$\n","marks":5,"grade":"$undefined"},{"label":"b","text":"Draw a block diagram of the heating control system i.e. including the model (3a) and PV control algorithm\n(3b). Using either standard algebra or the rules of block diagram manipulation, derive the closed-loop\nTransfer Function. What is the steady state gain of the closed-loop Transfer Function?\n","marks":8,"grade":"$undefined"},{"label":"c","text":"Using the room temperature control problem as an example, describe four typical objectives of control. Use\nthe PV control system and associated closed-loop Transfer Function to support your answer where possible\ne.g. explain how this control system addresses or fails to address the temperature control objectives. For\ninformation, the characteristic equation of a generalised 2nd order system is defined as follows:\n$s^2 + 2\\zeta\\omega_n s + \\omega_n^2 = 0$.\n","marks":8,"grade":"$undefined"},{"label":"d","text":"Briefly explain how Bode diagrams can be used to analyse the stability of closed-loop control systems. No\nnumerical or algebraic analysis is required to answer this question.\n","marks":4,"grade":"$undefined"}],"solution":"$30","confidence":"HIGH","solutionNote":"No official numerical answers (the answer sheet says N/A for A3). Cases in (a) checked with numpy roots; the closed-loop Transfer Function checked with sympy."},{"id":"engr202-2018-a1","module":"ENGR5001","week":5,"label":"ENGR202 2018 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["modelling","generalised-forms","time-response","controller-design","pid-actions","mass-spring-damper","critical-damping","pv-control"],"source":{"name":"ENGR202 Summer 2018 Q A1","paper":"ENGR202","year":2018,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR202_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR202 206a.pdf","tag":"VALID"},"grade":"2:2","formulaSheet":"- Newton's second law $\\sum F = M\\ddot x$; spring force $Kx$; viscous damper force $C\\dot x$ (learn this)\n- Generalised second order form $\\ddot x + 2\\zeta\\omega_n\\dot x + \\omega_n^2 x = Ku(t)$ (given in the question)\n- Mass-spring-damper: $\\omega_n = \\sqrt{K/M}$, $\\zeta = \\dfrac{C}{2\\sqrt{MK}}$, steady state gain $1/K$ (learn this)\n- Critical damping $\\zeta = 1$ (learn this)\n","text":"Answer ALL parts (a) - (e).\n","figures":[{"src":"/figures/engr202-2018-a1.png","caption":"Figure A1: Mass-spring-damper system (mass M, spring K, damper C, force F(t), displacement x downwards)."}],"parts":[{"label":"a","text":"For the mass-spring-damper system shown in Figure A1, derive a linear time invariant mathematical model\nexpressing the displacement x as a function of the mass M, spring stiffness K and damping parameter C, when\nan external force F(t) is applied to the mass. List the assumptions you have made.\n","marks":3,"grade":"$undefined"},{"label":"b","text":"Put the mass-spring-damper model into the standard second order form shown below. Find the steady state\ngain, damping ratio, natural frequency and forcing function.\n\n$$\\frac{d^2x}{dt^2} + 2\\zeta\\omega_n\\frac{dx}{dt} + \\omega_n^2 x = Ku(t)$$\n","marks":5,"grade":"$undefined"},{"label":"c","text":"Sketch the response of the mass-spring-damper system for the case that $C = 0$ and explain the\nsignificance of this result. On the same plot, sketch the response of a critically damped system. For the\ncase that $M = K = 1$, determine the value of C that yields a critically damped system.\n","marks":5,"answer":[{"label":"C for critical damping","value":2}],"grade":"$undefined"},{"label":"d","text":"The following differential equation represents the closed-loop dynamics of a proportional-velocity (PV)\ncontrol system:\n\n$$\\frac{d^2x}{dt^2} + \\left(0.3 + k_v\\right)\\frac{dx}{dt} + 0.5 + k_p = 0.4r(t)$$\n\nwhere $x$ is the output and $r(t)$ is the set point, while $k_v$ and $k_p$ are control gains. Design a\ncontroller (i.e. determine the values of the control gains) that yields closed-loop damping $\\zeta = 1$ and\n$\\omega_n = 1.2$ rads/s. Find the steady state gain of the closed-loop system, and state whether or not this\ncontrol system achieves steady state tracking of the set point.\n","marks":6,"answer":[{"label":"k_v","value":2.1},{"label":"k_p","value":0.94},{"label":"closed-loop steady state gain","value":0.2778}],"grade":"$undefined"},{"label":"e","text":"With the aid of sketches and/or any of the equations and analysis considered above, discuss the strengths\nand weaknesses of PV control in comparison to proportional-integral (PI) control.\n","marks":6,"grade":"$undefined"}],"solution":"$31","confidence":"HIGH","solutionNote":"Matches the official numerical answers (forcing function $F(t)$, $\\omega_n = \\sqrt{K/M}$,\n$\\zeta = C/(2\\sqrt{MK})$, steady state gain $1/K$, $C = 2$, $k_v = 2.1$, $k_p = 0.94$). The printed equation in\n(d) omits the $x$ after $0.5 + k_p$; it is read as $(0.5 + k_p)x$, which is also what the official gains imply.\n"},{"id":"engr202-2018-a2","module":"ENGR5001","week":5,"label":"ENGR202 2018 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["feedback-objectives","poles-stability","modelling","laplace-tf","block-diagrams","pid-actions","robot-manipulator","dead-zone","stability-shortcuts","pi-control"],"source":{"name":"ENGR202 Summer 2018 Q A2","paper":"ENGR202","year":2018,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR202_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR202 206a.pdf","tag":"VALID"},"grade":"2:2","formulaSheet":"- Laplace with zero initial conditions: $dx/dt \\to sX$ (learn this)\n- Stable if all poles have negative real parts; a single pole at the origin is marginally stable; repeated poles on the imaginary axis are unstable (learn this)\n- Necessary conditions: all coefficients exist and have the same sign (learn this)\n- Steady state gain $\\lim_{s \\to 0} G(s)$ (learn this)\n- Unity feedback: $\\dfrac{CG}{1 + CG}$ (learn this)\n","text":"Answer ALL parts (a) - (d).\n","figures":[{"src":"/figures/engr202-2018-a2.png","caption":"Figure A2: PI control system, controller (k1 s + k2)/s and plant 2/(s^2 + 5s + 3) with unity negative feedback."}],"parts":[{"label":"a(i)","text":"Consider one of the joints of a robotic manipulator, for which the joint angle and an applied voltage\nrepresent the output and input respectively.\n\nDefine the terms open-loop and closed-loop control, using the robotic manipulator as an example.\n","marks":3,"grade":"$undefined"},{"label":"a(ii)","text":"Briefly explain the term stability. What are the likely practical consequences of using an unstable robotic\nmanipulator control system?\n","marks":3,"grade":"$undefined"},{"label":"a(iii)","text":"Using the robotic manipulator as an example, explain the term dead-zone nonlinearity.\n","marks":2,"grade":"$undefined"},{"label":"b","text":"A simplified model for the robotic manipulator (without control) is as follows:\n\n$$\\frac{dx}{dt} = Ku(t)$$\n\nwhere $x$ is the joint angle and $u(t)$ is the input. Determine the Transfer Function form of this model.\nFind the pole, plot its position on the complex plane and state the stability condition. Write down an\nexpression for how to determine the steady state gain of this model, and comment on the physical\ninterpretation of your answer.\n","marks":6,"answer":[{"label":"pole","value":0}],"grade":"$undefined"},{"label":"c","text":"State the stability condition of the following characteristic equations, giving the reason for your answer\nin each case.\n\n$$s^3 + s + 1 = 0$$\n$$s^4 + 2s^3 - s^2 + 2s + 0.01 = 0$$\n$$s^3 + 2s^2 + 3s = 0$$\n$$(s + 2)(s + 3) = 0$$\n$$s^2 = 0$$\n$$(s - 0.5)(s + 0.7)(s + 1) = 0$$\n","marks":3,"grade":"$undefined"},{"label":"d(i)","text":"Figure A2 shows a Proportional-Integral (PI) control system.\n\nIs the proportional gain $k_1$ or $k_2$?\n","marks":1,"grade":"$undefined"},{"label":"d(ii)","text":"Using algebra or the rules of block diagram manipulation, show that the Closed Loop Transfer Function is\n\n$$X(s) = \\frac{2(k_1 s + k_2)}{s^3 + 5s^2 + (2k_1 + 3)s + 2k_2}\\,V(s)$$\n","marks":4,"grade":"$undefined"},{"label":"d(iii)","text":"Find the steady state gain of the Closed Loop Transfer Function and comment on the significance of your\nanswer.\n","marks":3,"answer":[{"label":"closed-loop steady state gain","value":1}],"grade":"$undefined"}],"solution":"$32","confidence":"HIGH","solutionNote":"Matches the official numerical answers (pole $s = 0$; gain $K/s$ as $s \\to 0$, i.e. unbounded; closed-loop\ngain 1). Part (c) classifications checked with numpy roots.\n"},{"id":"engr202-2018-a3","module":"ENGR5001","week":4,"label":"ENGR202 2018 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["block-diagrams","feedback-objectives","laplace-tf","time-response","frequency-response","partial-fractions","step-response"],"source":{"name":"ENGR202 Summer 2018 Q A3","paper":"ENGR202","year":2018,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR202_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR202 206a.pdf","tag":"VALID"},"grade":"2:2","formulaSheet":"- Laplace pairs: unit step $u(t) \\leftrightarrow \\dfrac{1}{s}$; $e^{-\\alpha t} \\leftrightarrow \\dfrac{1}{s + \\alpha}$ (given in the question)\n- Frequency response: $M = |G(s)|_{s = j\\omega}$, $\\phi = \\mathrm{Arg}(G(s))|_{s = j\\omega}$ (given in the question)\n- Closed loop with feedback element: $\\dfrac{X}{V} = \\dfrac{CH}{1 + CHF}$; characteristic equation $1 + CHF = 0$ (learn this)\n- Partial fractions: $\\dfrac{A}{s(s + a)} = \\dfrac{A/a}{s} - \\dfrac{A/a}{s + a}$ (learn this)\n","text":"The following information is provided for this question:\n\n| Time Function | Laplace Transform |\n|---|---|\n| Unit step $u(t)$ | $\\dfrac{1}{s}$ |\n| $\\exp(-\\alpha t)$ | $\\dfrac{1}{s + \\alpha}$ |\n\nThe frequency response of a Transfer Function $G(s)$ is given by:\n\n$$M = |G(s)|_{s = j\\omega} \\quad \\text{and} \\quad \\phi = \\mathrm{Arg}(G(s))|_{s = j\\omega}$$\n","parts":[{"label":"a","text":"Draw a block diagram of a generalized control system, in which $H(s)$ is the model, $C(s)$ is the control\nalgorithm and $F(s)$ is a feedback element. On your diagram, label the set point $V(s)$, control input\n$U(s)$ and output $X(s)$.\n","marks":2,"grade":"$undefined"},{"label":"b","text":"Using either standard algebra or the rules of block diagram manipulation, determine the Closed Loop Transfer\nFunction and write down the associated characteristic equation.\n","marks":4,"grade":"$undefined"},{"label":"c","text":"List and explain, with examples of your own choice, the four common objectives of control system design as\nconsidered in the course. Where relevant, support your answer with the block diagram from part (a) and the\nTransfer Function from part (b).\n","marks":9,"grade":"$undefined"},{"label":"d(i)","text":"Consider the model: $H(s) = \\dfrac{5}{2s + 1}$\n\nUsing the Laplace transforms stated above, determine the unit step response of the model.\n","marks":5,"answer":[{"label":"final value x(t -> infinity)","value":5},{"label":"exponent coefficient (x = 5(1 - e^(-a t)), a =)","value":0.5}],"grade":"$undefined"},{"label":"d(ii)","text":"Determine the frequency response of the model i.e. the magnitude $M$ and phase $\\phi$.\n","marks":5,"grade":"$undefined"}],"solution":"$33","confidence":"HIGH","solutionNote":"Matches the official numerical answer ($x(t) = 5(1 - e^{-0.5t})$)."},{"id":"engr202-2019-b1","module":"ENGR5001","week":5,"label":"ENGR202 2019 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["laplace-tf","poles-stability","block-diagrams","pid-actions","controller-design","pd-control","pole-placement","unstable-plant"],"source":{"name":"ENGR202 Summer 2019 Q B1","paper":"ENGR202","year":2019,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR202_Exam_2019 (1).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/ENGR202_exam2019_Part_B_numerical.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Laplace with zero initial conditions: $\\dot x \\to sX$, $\\ddot x \\to s^2X$ (learn this)\n- PD controller $U(s) = (K_P + K_D s)(V(s) - X(s))$ (given in the question)\n- Second order characteristic equation $s^2 + 2\\zeta\\omega_n s + \\omega_n^2 = 0$ (given in the question)\n- Unity feedback: $\\dfrac{CG}{1 + CG}$ (learn this)\n","text":"In this question, the differential equation of an uncontrolled system is given by the following mathematical\nmodel:\n\n$$\\frac{d^2x}{dt^2} - 0.2\\frac{dx}{dt} + 0.05x = 2.3u(t)$$\n\nwhere $x(t)$ and $u(t)$ are the output and input, respectively.\n","parts":[{"label":"a","text":"Determine the Transfer Function form of the uncontrolled system and hence show that the characteristic\nequation is given by $s^2 - 0.2s + 0.05 = 0$.\n","marks":3,"grade":"$undefined"},{"label":"b","text":"Using the characteristic equation stated in part (a), determine the poles of the uncontrolled system and the\nstability condition, and sketch the general form of the time response.\n","marks":3,"answer":[{"label":"pole real part","value":0.1},{"label":"pole imaginary part (+/-)","value":0.2}],"grade":"$undefined"},{"label":"c","text":"Draw the block diagram of a proportional-derivative (PD) control system, as applied to the Transfer Function\ndetermined in part (a). Hint: expressed as a Transfer Function, the PD controller takes the following\nstandard form:\n\n$$U(s) = \\left(K_P + K_D s\\right)\\left(V(s) - X(s)\\right)$$\n\nwhere $V(s)$ represents the set point.\n","marks":3,"grade":"$undefined"},{"label":"d","text":"For the control system in part (c), design a PD controller (i.e. determine values of $K_D$ and $K_P$) such\nthat the closed-loop damping ratio $\\zeta = 1.1$ and the natural frequency $\\omega_n = 0.3$ rad/s. Hint: the\ngeneral form of the characteristic equation for a second order system is: $s^2 + 2\\zeta\\omega_n s + \\omega_n^2 = 0$.\nWhat is the significance of designing this controller such that $\\zeta \\ge 1$?\n","marks":12,"answer":[{"label":"K_D","value":0.3739},{"label":"K_P","value":0.01739}],"grade":"$undefined"},{"label":"e","text":"What are the potential disadvantages of the PD control system selected above? Suggest alternatives,\nexplaining the reasons for your answer.\n","marks":4,"grade":"$undefined"}],"solution":"$34","confidence":"HIGH","solutionNote":"Matches the official numerical answers (poles $0.1 \\pm 0.2j$; $K_D = 0.3739$, $K_P = 0.0174$). The overshoot\ncaused by the PD zero is an extra observation checked by simulation (scipy), not part of the official answer.\n"},{"id":"engr202-2019-b2","module":"ENGR5001","week":5,"label":"ENGR202 2019 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["poles-stability","laplace-tf","frequency-response","frequency-design","nyquist","gain-margin","phase-margin","air-conditioning"],"source":{"name":"ENGR202 Summer 2019 Q B2","paper":"ENGR202","year":2019,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR202_Exam_2019 (1).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/ENGR202_exam2019_Part_B_numerical.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Poles, stability, steady state gain $G(0)$ (learn this)\n- Frequency response: $M = |G(s)|_{s = j\\omega}$, $\\varphi = \\mathrm{Arg}(G(s))|_{s = j\\omega}$ (given in the question)\n- $|j\\omega + a| = \\sqrt{\\omega^2 + a^2}$, $\\mathrm{Arg}(j\\omega + a) = \\tan^{-1}(\\omega/a)$ (learn this)\n- Nyquist diagram: $\\mathrm{Im}\\,G(j\\omega)$ against $\\mathrm{Re}\\,G(j\\omega)$ as $\\omega$ goes from 0 to $\\infty$; critical point $-1 + 0j$ (learn this)\n- Gain margin and phase margin (learn this)\n","text":"Consider the following Transfer Function model derived for an air conditioning unit:\n\n$$X(s) = \\frac{6.7}{(s + 2)(s + 3)}\\,U(s)$$\n\nwhere $X(s)$ and $U(s)$ represent the outlet and inlet temperature, respectively.\n","parts":[{"label":"a","text":"Write down the characteristic equation of the model. Determine the poles, stability condition and steady\nstate gain of the model. Plot the position of the poles on the complex s-plane. For a unit step input of\n$20^\\circ$ C, what is the steady state outlet temperature? Comment on the physical interpretation of all these\nresults.\n","marks":7,"answer":[{"label":"dominant pole","value":-2},{"label":"other pole","value":-3},{"label":"steady state gain","value":1.1167},{"label":"steady state outlet temperature","value":22.33,"units":"degC"}],"grade":"$undefined"},{"label":"b","text":"Briefly explain what is meant by the frequency response of a system.\n","marks":3,"grade":"$undefined"},{"label":"c","text":"Using standard notation from the lectures, and noting that $M = |G(s)|_{s = j\\omega}$ and\n$\\varphi = \\mathrm{Arg}(G(s))|_{s = j\\omega}$, determine the frequency response of the air conditioning unit\nTransfer Function model i.e. derive expressions that show how $M$ and $\\varphi$ vary with $\\omega$.\n","marks":6,"grade":"$undefined"},{"label":"d","text":"Using a carefully annotated sketch, including axis labels, explain how a Nyquist Diagram can be used to\ndetermine the stability of a system. Your sketch should show appropriate traces for (i) a typical stable and\n(ii) unstable system. Your written answer should briefly explain what is plotted in the Nyquist Diagram. Hint:\nthe sketch is an approximate graph drawn in your answer book and does not require graph paper nor use of the\nequations derived in part (c).\n","marks":5,"grade":"$undefined"},{"label":"e","text":"Explain how the frequency response can be used to analyse and set design criteria for the stability of a\nclosed-loop control system.\n","marks":4,"grade":"$undefined"}],"solution":"$35","confidence":"HIGH","solutionNote":"Matches the official numerical answers (poles $-2$, $-3$; gain 1.1167; steady state temperature $22.3^\\circ$C)."},{"id":"engr202-2019-b3","module":"ENGR5001","week":1,"label":"ENGR202 2019 Q B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["generalised-forms","modelling","mechanistic-models","single-tank","lc-circuit","harmonic-oscillator","linearity"],"source":{"name":"ENGR202 Summer 2019 Q B3","paper":"ENGR202","year":2019,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR202_Exam_2019 (1).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/ENGR202_exam2019_Part_B_numerical.pdf","tag":"VALID"},"grade":"2:2","formulaSheet":"- Generalised first order model $\\tau\\dfrac{dx}{dt} + x = Ku(t)$; second order $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2x = Ku(t)$ (given in the question)\n- Tank: $A\\dfrac{dh}{dt} = Q_i - Q_o$ (conservation of volume) (learn this)\n- Inductor $V_L = L\\dfrac{di}{dt}$; capacitor $i = C\\dfrac{dV_C}{dt}$; Kirchhoff's voltage law (learn this)\n- Steady state: $x(t \\to \\infty) = K\\,u$ (first order); $K/\\omega_n^2$ times the input (second order) (learn this)\n","text":"Generalised first order model: $\\tau\\dfrac{dx}{dt} + x = Ku(t)$\n\nGeneralised second order model: $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$\n","figures":[{"src":"/figures/engr202-2019-b3.png","caption":"Figure B3-1: Single tank system (area A, head h, inflow Q_i, outflow Q_o through an orifice)."},{"src":"/figures/engr202-2019-b3-2.png","caption":"Figure B3-2: LC (inductor capacitor) circuit."}],"parts":[{"label":"a","text":"In the generalised first order model, what are the names of the variables $\\tau$ and $K$? For the first order\nmodel with $K = 5$ and a steady input of 2 units, determine the value of $x(t \\to \\infty)$. For the second order\nmodel, sketch the time response mode if $\\zeta = 0$ and explain the physical significance of this result.\n","marks":5,"answer":[{"label":"x(t -> infinity)","value":10}],"grade":"$undefined"},{"label":"b","text":"For the tank system shown in Figure B3-1, derive an ordinary differential equation expressing the outflow\nrate $Q_o$, as a function of the inflow rate $Q_i$ and area $A$. You may assume that $Q_o$ is proportional to the\nhead of water $h$. List any other assumptions you have made to develop this model. Convert the model into the\ngeneralised first order form. Determine the steady state gain and comment on the physical significance of\nthis result.\n","marks":5,"answer":[{"label":"steady state gain","value":1}],"grade":"$undefined"},{"label":"c","text":"For the LC circuit shown in Figure B3-2, derive a model expressing the voltage $V_o(t)$ in terms of $L$, $C$ and\nthe applied voltage $V_i(t)$. List the assumptions you have made. Convert the model into a generalised second\norder form. Determine the forcing function, output, damping ratio and natural frequency of the model.\n","marks":7,"answer":[{"label":"damping ratio","value":0}],"grade":"$undefined"},{"label":"d","text":"Using the LC circuit as an example, explain what is meant by the following terms: (i) linear system;\n(ii) mechanistic model; (iii) open-loop system. Briefly explain how generalised model forms can be useful when\nanalysing dynamic systems.\n","marks":8,"grade":"$undefined"}],"solution":"$36","confidence":"HIGH","solutionNote":"Matches the official numerical answer ($x(t \\to \\infty) = K u = 10$); the other parts are derivations following the Part 1 modelling slides."},{"id":"engr202-2021-q1","module":"ENGR5001","week":5,"label":"ENGR202 2021 Q1","kind":"past-paper","level":3,"marks":19,"minutes":23,"tags":["poles-stability","laplace-tf","generalised-forms","block-diagrams","pid-actions","damping-ratio","pi-control","steady-state-gain"],"source":{"name":"ENGR202 Summer 2021 Q1","paper":"ENGR202","year":2021,"question":"Q1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR202 Exam Control 2021.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2021/ENGR202_exam2021_Part_B_numerical.pdf","tag":"VALID"},"grade":"2:2","formulaSheet":"- Poles: roots of the characteristic equation (denominator $= 0$); zeros: roots of the numerator (learn this)\n- Stable if ALL poles have negative real parts; dominant pole = the one closest to the imaginary axis (learn this)\n- Steady state gain: $G(0)$, i.e. set $s = 0$ in the Transfer Function (learn this)\n- Second order characteristic equation: $s^2 + 2\\zeta\\omega_n s + \\omega_n^2 = 0$ (learn this)\n- PI control: $U(s) = \\left(k_P + \\dfrac{k_I}{s}\\right)\\left(V(s) - X(s)\\right)$ (learn this)\n- Negative feedback rule: $\\dfrac{G_1}{1 + G_1 G_2}$ (learn this)\n","text":"Consider the following Transfer Function:\n\n$$X(s) = \\frac{3.5s + 1}{(s + 1.2)(s + 2)}\\,U(s)$$\n\nwhere $X(s)$ and $U(s)$ are the output and input, respectively.\n","parts":[{"label":"a","text":"What are the pole(s) and zero(s) of this system? Plot the poles on the complex plane and label the\ndominant pole. Giving the reason for your answer, what is the stability? Determine the steady state\ngain. For the case that the input is a time-invariant value of 5, what is the value of the output at\nsteady state.\n","marks":7,"answer":[{"label":"zero","value":-0.2857},{"label":"dominant pole","value":-1.2},{"label":"other pole","value":-2},{"label":"steady state gain","value":0.4167},{"label":"x at steady state","value":2.083}],"grade":"$undefined"},{"label":"b","text":"Determine the damping ratio of this system. Comment on the physical meaning of the damping ratio and\nrelate your answer to the poles determined in part (a).\n","marks":4,"answer":[{"label":"omega_n","value":1.549,"units":"rad/s"},{"label":"zeta","value":1.033}],"grade":"$undefined"},{"label":"c","text":"Draw the block diagram of a standard proportional-integral (PI) control system, as applied to the\nTransfer Function above. Label all the blocks and signals.\n","marks":2,"grade":"$undefined"},{"label":"d","text":"Showing your working, derive the closed-loop Transfer Function for the controller in Part (c). What is\nthe order of the control system? What is the significance of this result? Determine the steady state\ngain and comment on the significance of your answer.\n","marks":6,"answer":[{"label":"closed-loop order","value":3},{"label":"closed-loop steady state gain","value":1}],"grade":"$undefined"}],"solution":"$37","confidence":"HIGH","solutionNote":"Matches the official numerical answers (zero -0.286, poles -1.2 and -2, dominant -1.2, gain 0.417, output 2.1,\nnatural frequency 1.55, damping 1.03, third order, closed-loop gain 1). The answer sheet's row labels are\noffset by one question.\n"},{"id":"engr202-2021-q2","module":"ENGR5001","week":4,"label":"ENGR202 2021 Q2","kind":"past-paper","level":3,"marks":16,"minutes":19,"tags":["hurwitz","frequency-response","controller-design","pid-actions","proportional-control","steady-state-error","time-constant"],"source":{"name":"ENGR202 Summer 2021 Q2","paper":"ENGR202","year":2021,"question":"Q2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR202 Exam Control 2021.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2021/ENGR202_exam2021_Part_B_numerical.pdf","tag":"VALID"},"grade":"2:2","formulaSheet":"- Hurwitz: for $a_0 s^3 + a_1 s^2 + a_2 s + a_3 = 0$, stable if $\\Delta_1 = a_1 > 0$, $\\Delta_2 = a_1 a_2 - a_0 a_3 > 0$, $\\Delta_3 = a_3\\Delta_2 > 0$ (learn this)\n- Frequency response: $M = |G(s)|_{s = j\\omega}$, $\\phi = \\mathrm{Arg}(G(s))|_{s = j\\omega}$ (learn this)\n- $|j\\omega + a| = \\sqrt{\\omega^2 + a^2}$, $\\mathrm{Arg}(j\\omega + a) = \\tan^{-1}(\\omega/a)$ (learn this)\n- First order: $\\tau\\dfrac{dx}{dt} + x = Ku$, $X(s) = \\dfrac{K}{\\tau s + 1}U(s)$ (learn this)\n- Proportional control: $U = k_P(V - X)$; closed loop $\\dfrac{k_P G}{1 + k_P G}$ (learn this)\n","text":"Answer ALL parts (a) - (c).\n","parts":[{"label":"a","text":"Use the Hurwitz method to determine the stability condition of the following characteristic equation:\n$s^3 + 4s^2 + s + 2 = 0$.\n","marks":3,"answer":[{"label":"Delta_2","value":2},{"label":"Delta_3","value":4}],"grade":"$undefined"},{"label":"b","text":"Determine the frequency response for the following model i.e. showing your working, derive equations\nfor the magnitude and phase.\n\n$$G(s) = \\frac{1}{(s + 1)(s + 2)(s + 5)}$$\n","marks":4,"grade":"$undefined"},{"label":"c","text":"A first order system has a time constant of 8 s and a steady state gain of 7. Design a proportional\ncontrol system such that the closed loop time constant is just 4 s. For a step in the set point of\nmagnitude 10, what is the steady state error of this control system? Show all your working.\n","marks":9,"answer":[{"label":"k_P","value":0.1429},{"label":"steady state error","value":5}],"grade":"$undefined"}],"solution":"$38","confidence":"HIGH","solutionNote":"Matches the official numerical answers (stable; $k_P = 0.1429$; steady state error 5)."},{"id":"engr202-2021-q3","module":"ENGR5001","week":2,"label":"ENGR202 2021 Q3","kind":"past-paper","level":3,"marks":15,"minutes":18,"tags":["modelling","time-response","mechanistic-models","data-based-models","graphical-estimation"],"source":{"name":"ENGR202 Summer 2021 Q3","paper":"ENGR202","year":2021,"question":"Q3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR202 Exam Control 2021.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- First order model: $\\tau\\dfrac{dx}{dt} + x = Ku(t)$; unit step response $x(t) = K\\left(1 - e^{-t/\\tau}\\right)$ (learn this)\n- Graphical estimates: $K = \\dfrac{x(t\\to\\infty)}{u(t\\to\\infty)}$; $\\tau$ = time to reach 63% of the final change (learn this)\n","text":"Answer BOTH parts (a) and (b).\n","parts":[{"label":"a","text":"Mechanistic and graphical methods were both used to develop mathematical models in this module. With\nthe support of examples, briefly describe the key features of these two approaches.\n","marks":6,"grade":"$undefined"},{"label":"b","text":"Evaluate the advantages and disadvantages of the two methods from part (a). Suggest how they could\ncomplement each other.\n","marks":9,"grade":"$undefined"}],"solution":"$39","confidence":"HIGH","solutionNote":"No official solution available (descriptive question); the answer follows the module's modelling slides (Parts 1 and 2)."},{"id":"engr202-2022-b1","module":"ENGR5001","week":5,"label":"ENGR202 2022 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["generalised-forms","frequency-response","time-response","laplace-tf","feedback-objectives","pid-actions","bode-diagram","wind-turbine"],"source":{"name":"ENGR202 Summer 2022 Q B1","paper":"ENGR202","year":2022,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR202 Exam Part B 2021-22.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2022/ENGR202_exam2022_Part_B_numerical.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Generalised first order form: $\\tau\\dfrac{dx}{dt} + x = Ku(t)$ (learn this)\n- Laplace transform with zero initial conditions: $\\dfrac{dx}{dt} \\to sX(s)$ (learn this)\n- First order frequency response: $M = \\dfrac{K}{\\sqrt{1 + \\omega^2\\tau^2}}$, $\\phi = -\\tan^{-1}(\\omega\\tau)$; corner (break) frequency $\\omega = 1/\\tau$ (learn this)\n- Decibels: $20\\log_{10}M$ (learn this)\n- Four control objectives: stability, tracking, transient behaviour, disturbance rejection (learn this)\n","text":"An engineering system is represented by the following model:\n\n$$\\frac{dx}{dt} + 2.5x = 0.7u \\qquad (1)$$\n","parts":[{"label":"a","text":"What is the order of this model? Convert the model into an appropriate generalised form and state the\ntime constant and steady state gain of the model. Sketch and annotate a Bode diagram for this system.\nGraph paper is not required for this sketch.\n","marks":6,"answer":[{"label":"time constant","value":0.4,"units":"s"},{"label":"steady state gain","value":0.28},{"label":"low-frequency magnitude","value":-11.06,"units":"dB","tol":0.05},{"label":"corner frequency","value":2.5,"units":"rad/s"}],"grade":"$undefined"},{"label":"b","text":"Briefly describe a graphical method of estimating both the time constant and steady state gain of a\nfirst order model from step response experimental data. Comment on potential limitations of this\napproach and suggest an alternative.\n","marks":5,"grade":"$undefined"},{"label":"c","text":"Develop the Transfer Function form of the model (1). State any assumptions made.\n","marks":2,"grade":"$undefined"},{"label":"d","text":"An engineer is designing a control system for a constant speed wind turbine. The output is the\ngenerator reaction torque. When the wind speed reaches a predetermined value, the wind turbine\ngenerates 'rated' power, which is associated with a known value of the generator reaction torque. For\nactive pitch control systems, the pitch angle of the blades is adjusted in real time. The system\nshould minimise load transients and smooth the power generated.\n\nState the four standard objectives of control and discuss what these could mean in the context of a\ncontrol system for generator reaction torque.\n","marks":6,"grade":"$undefined"},{"label":"e","text":"A wind turbine system shows highly oscillatory dynamics when the pitch angle of the blades is adjusted\nin open loop. Briefly explain why PID control might be a suitable choice to meet the control objectives\ndeveloped in part (d). Speculate on any potential limitations of PID control in this context.\n","marks":6,"grade":"$undefined"}],"solution":"$3a","confidence":"HIGH","solutionNote":"Matches the official numerical answers ($0.4\\dot x + x = 0.28u$, time constant 0.4, gain 0.28, $X(s) = 0.28U(s)/(0.4s + 1)$)."},{"id":"engr202-2022-b2","module":"ENGR5001","week":5,"label":"ENGR202 2022 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["block-diagrams","laplace-tf","poles-stability","controller-design","pid-actions","hurwitz","pv-control"],"source":{"name":"ENGR202 Summer 2022 Q B2","paper":"ENGR202","year":2022,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR202 Exam Part B 2021-22.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2022/ENGR202_exam2022_Part_B_numerical.pdf","tag":"VALID"},"grade":"first","formulaSheet":"- Steady state gain $G(0)$; poles from the characteristic equation; quadratic formula $s = \\dfrac{-a_1 \\pm \\sqrt{a_1^2 - 4a_2}}{2}$ (learn this)\n- Negative feedback rule: $\\dfrac{G_1}{1 + G_1 G_2}$; series rule: $G_1 G_2$ (learn this)\n- Pole placement: divide the characteristic equation by the $s^2$ coefficient and match to $s^2 + 2\\zeta\\omega_n s + \\omega_n^2$ (learn this)\n- Hurwitz for $s^3 + a_1 s^2 + a_2 s + a_3$: $\\Delta_1 = a_1$, $\\Delta_2 = a_1 a_2 - a_3$, $\\Delta_3 = a_3\\Delta_2$, all $> 0$ (learn this)\n","text":"This question concerns the model and control system shown in Figure B2-1.\n","figures":[{"src":"/figures/engr202-2022-b2.png","caption":"Figure B2-1: Control system with two negative feedback loops."}],"parts":[{"label":"a","text":"Briefly define and explain what $V(s)$, $X(s)$ and $U(s)$ refer to. Describe one practical example to\nillustrate your answer.\n","marks":3,"grade":"$undefined"},{"label":"b","text":"Write down the open loop Transfer Function of the control model shown in Figure B2-1. Develop algebraic\nexpressions to determine the steady state gain and the poles of the control model. Use these\nexpressions to calculate numerical values of both the gain and the poles for the case that\n$a_1 = 5.5$, $a_2 = 7.5$ and $b_1 = 5$. Determine the stability condition and dominant pole. Finally,\nfor the same numerical example, what is the steady state value of $X(s)$ for an input magnitude of 5?\n","marks":8,"answer":[{"label":"steady state gain","value":0.6667},{"label":"dominant pole","value":-2.5},{"label":"other pole","value":-3},{"label":"x at steady state","value":3.333}],"grade":"$undefined"},{"label":"c","text":"Show that the Closed-Loop Transfer Function associated with Figure B2-1 takes the following form:\n\n$$X(s) = \\frac{s + b_1}{(1 + k_V)s^2 + (a_1 + k_P + k_V b_1)s + a_2 + k_P b_1}\\,V(s)$$\n","marks":5,"grade":"$undefined"},{"label":"d","text":"For the case that $a_1 = 5.5$, $a_2 = 7.5$ and $b_1 = 5$, determine the characteristic equation for the\nClosed-Loop Transfer Function. Design a control system based on Figure B2-1 that achieves a closed-loop\ndamping of 1.0 and a natural frequency of 0.5, i.e. determine suitable values of $k_P$ and $k_I$.\n","marks":6,"answer":[{"label":"k_P","value":-1.4877},{"label":"k_V","value":-0.7531}],"grade":"$undefined"},{"label":"e","text":"Use the Hurwitz method to determine the stability condition of the following, different,\ncharacteristic equation: $s^3 + \\alpha s^2 + \\beta s + \\gamma = 0$.\n","marks":3,"grade":"$undefined"}],"solution":"$3b","confidence":"HIGH","solutionNote":"Matches the official numerical answers (gain 0.666, poles -2.5 and -3, steady state output 3.33,\n$k_V = -0.7531$, $k_P = -1.4877$). As printed, the numerator in (c) omits a factor $k_P$ and (d) says $k_I$\nwhere $k_V$ is meant; both are explained in the solution.\n"},{"id":"engr202-2023-b1","module":"ENGR5001","week":3,"label":"ENGR202 2023 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["modelling","time-response","generalised-forms","laplace-tf","poles-stability","step-response","wave-energy"],"source":{"name":"ENGR202 Summer 2023 Q B1","paper":"ENGR202","year":2023,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR202 Exam 2023 v2.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2023/ENGR202_exam2223_Part_B_numerical.doc","tag":"VALID"},"grade":"2:2","formulaSheet":"- Generalised forms: $\\tau\\dfrac{dx}{dt} + x = Ku$ and $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$ (given in the question)\n- First order unit step response: $x(t) = K\\left(1 - e^{-t/\\tau}\\right)$ (given in the question)\n- Second order steady state gain: $K/\\omega_n^2$ (learn this)\n- Complex poles: $s = -\\zeta\\omega_n \\pm j\\omega_n\\sqrt{1 - \\zeta^2}$ (learn this)\n","text":"This question concerns first and second order models. Using standard notation, the general forms of these\nmodels are stated below.\n\n$$\\tau\\frac{dx}{dt} + x = Ku \\quad \\text{and} \\quad \\frac{d^2x}{dt^2} + 2\\zeta\\omega_n\\frac{dx}{dt} + \\omega_n^2 x = Ku(t)$$\n\nThe unit step response of a first order system is as follows: $x(t) = K\\left(1 - e^{-t/\\tau}\\right)$\n","parts":[{"label":"a","text":"Briefly explain the difference between the data-based and mechanistic approaches to modelling.\n","marks":3,"grade":"$undefined"},{"label":"b","text":"For an engineering system of your choice (e.g. the simple thermal or hydraulic systems from the\nlectures), develop a first order, linear, mechanistic model, stating all the notation and assumptions\nmade.\n","marks":4,"grade":"$undefined"},{"label":"c","text":"Sketch the unit step response of a first order system and use it to fully define the following terms:\n(i) initial conditions, (ii) steady state gain and (iii) time constant. Link these definitions to your\nanswer from part (b). Comment on how this information could be used to estimate a model from\nexperimental data, briefly explaining the likely experimental protocol.\n","marks":8,"answer":[{"label":"x(tau)/K","value":0.632}],"grade":"$undefined"},{"label":"d","text":"The following model represents the power take off component of a wave energy convertor, in which $p$\nis the generator reaction torque and $f(t)$ is the force of the damper (arbitrary scaled units for the\npurpose of this question).\n\n$$2\\frac{d^2p}{dt^2} + 0.8\\frac{dp}{dt} + 0.46p = f(t) \\qquad \\text{(B1-1)}$$\n\nConvert this model into the appropriate generalised form and hence determine the damping and natural\nfrequency of the model.\n","marks":3,"answer":[{"label":"omega_n","value":0.4796},{"label":"zeta","value":0.417}],"grade":"$undefined"},{"label":"e","text":"Develop the Transfer Function form of the power take off model (B1-1), stating the assumption made.\nWrite down the characteristic equation. Determine the poles and steady state gain of the system. Plot\nthe position of the poles on the complex $s$-plane. What is the stability of the model? Link your answer\nto the damping and natural frequency as appropriate.\n","marks":7,"answer":[{"label":"pole real part","value":-0.2},{"label":"pole imaginary part (+/-)","value":0.4359},{"label":"steady state gain","value":2.174}],"grade":"$undefined"}],"solution":"$3c","confidence":"HIGH","solutionNote":"Agrees with the official poles ($-0.2 \\pm 0.4359j$), gain 2.17 and $\\omega_n = 0.4796$. The official damping\nratio 0.8696 is wrong: it uses $0.4 = 2\\zeta\\omega_n^2$ instead of $0.4 = 2\\zeta\\omega_n$. The correct value\n$\\zeta = 0.417$ is confirmed by the official poles themselves ($\\zeta = 0.2/0.4796$) and by the simulated overshoot.\n"},{"id":"engr202-2023-b2","module":"ENGR5001","week":5,"label":"ENGR202 2023 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["feedback-objectives","block-diagrams","hurwitz","pid-actions","pi-control","control-objectives","medical-control"],"source":{"name":"ENGR202 Summer 2023 Q B2","paper":"ENGR202","year":2023,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR202 Exam 2023 v2.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2023/ENGR202_exam2223_Part_B_numerical.doc","tag":"VALID"},"grade":"2:1","formulaSheet":"- Negative feedback rule: $\\dfrac{G_1}{1 + G_1 G_2}$; series rule $G_1 G_2$ (learn this)\n- Hurwitz determinant for $s^3 + a_1 s^2 + a_2 s + a_3 = 0$: $H_3 = \\begin{vmatrix} a_1 & a_3 & 0 \\\\ 1 & a_2 & 0 \\\\ 0 & a_1 & a_3 \\end{vmatrix}$ (given in the question)\n- Stability needs $\\Delta_1 = a_1 > 0$, $\\Delta_2 = a_1 a_2 - a_3 > 0$, $\\Delta_3 = a_3\\Delta_2 > 0$ (learn this)\n- PI controller: $\\dfrac{k_1 s + k_2}{s} = k_1 + \\dfrac{k_2}{s}$ (learn this)\n","text":"In medicine, during an operation, the patient is continually monitored and a combination of anaesthetic\ndrugs are adjusted to ensure normal vital signs are maintained during loss of consciousness. There is some\nresearch into automatic anaesthetic drug delivery. However, this is a challenging control problem, due to\ninter-patient variability, poor understanding of drug mechanisms and difficulties in sensing.\n\nFigure B2-1 shows a Proportional-Integral (PI) control system applied to one aspect of the anaesthetic\ndrug delivery control problem. Here, $X(s)$ represents the measured concentration of a drug in the\npatient's blood and $U(s)$ is the applied drug dosage. The value of $X(s)$ subsequently impacts on the\npatient's level of consciousness.\n","figures":[{"src":"/figures/engr202-2023-b2.png","caption":"Figure B2-1: PI Control."}],"parts":[{"label":"a","text":"In regard to Figure B2-1:\n\n(i) What does $V(s)$ represent?\n\n(ii) Is Figure B2-1 an example of open-loop or closed-loop control? Explain your answer.\n\n(iii) What is $k_2$ usually called?\n","marks":3,"grade":"$undefined"},{"label":"b","text":"Using the rules of block diagram manipulation, derive the Transfer Function for the relationship between\n$V(s)$ and $X(s)$, and hence show that the characteristic equation takes the following form:\n\n$$s^3 + 2.6s^2 + (1.73 + 4.3k_1)s + 4.3k_2 = 0$$\n","marks":4,"grade":"$undefined"},{"label":"c","text":"Using the characteristic equation stated in part (b), derive and hence sketch the stability\nrelationship between $k_1$ and $k_2$. Hint: for a third order system, i.e.\n$s^3 + a_1 s^2 + a_2 s + a_3 = 0$, use the following Hurwitz determinant:\n\n$$H_3 = \\begin{vmatrix} a_1 & a_3 & 0 \\\\ 1 & a_2 & 0 \\\\ 0 & a_1 & a_3 \\end{vmatrix}$$\n","marks":4,"answer":[{"label":"k2 limit at k1 = 0","value":1.046},{"label":"slope dk2/dk1 of boundary","value":2.6},{"label":"k1 intercept (k2 = 0)","value":-0.4023}],"grade":"$undefined"},{"label":"d","text":"Four standard objectives of control relate to stability, tracking, transient behaviour and disturbance\nrejection. Carefully define and explain each of these terms, referring to relevant mathematical concepts\nwhere possible. In addition, speculate on how these control objectives, and possibly other control\nobjectives that you can think of, could apply in the case of the anaesthetic drug delivery control\nproblem. Finally, comment on the limitations of the proposed PI control system in this context.\n","marks":14,"grade":"$undefined"}],"solution":"$3d","confidence":"HIGH","solutionNote":"Agrees with the official answer $1.046 + 2.6k_1 > k_2$. The official \"equivalently $k_1 > 0.3846 - 4.023k_2$\"\nis a misprint; rearranging correctly gives $k_1 > 0.3846k_2 - 0.4023$. The official sheet also omits the\nsecond condition $k_2 > 0$ from $\\Delta_3$.\n"},{"id":"engr202-2024-b1","module":"ENGR5001","week":5,"label":"ENGR202 2024 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["feedback-objectives","block-diagrams","poles-stability","controller-design","pid-actions","dc-motor","pole-placement"],"source":{"name":"ENGR202 Summer 2024 Q B1","paper":"ENGR202","year":2024,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR202 Main Exam 2024.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Negative feedback rule $\\dfrac{G_1}{1 + G_1 G_2}$; series rule $G_1 G_2$ (learn this)\n- Steady state gain: set $s = 0$ (learn this)\n- Pole placement: desired poles $p_{1,2}$ give the characteristic polynomial $(s - p_1)(s - p_2)$; match coefficients (learn this)\n- $s = -\\zeta\\omega_n \\pm j\\omega_n\\sqrt{1 - \\zeta^2}$; $\\omega_n = |s|$ (learn this)\n","text":"Figure B1-1 consists of some control actions and a model. Using standard terminology, $X(s)$ is the\noutput, $U(s)$ is the control input and $V(s)$ is the set point. Also, $K_I$ and $K_p$ are control gains,\nwhilst $A(s)$ and $B(s)$ are polynomials representing the model.\n","figures":[{"src":"/figures/engr202-2024-b1.png","caption":"Figure B1-1: Block diagram of a control system."}],"parts":[{"label":"a","text":"Giving your reason, state if Figure B1-1 is an example of open-loop or closed-loop control. What are the\nadvantages and disadvantages open-loop and closed-loop control?\n","marks":8,"grade":"$undefined"},{"label":"b","text":"Develop the Transfer Function for the relationship between $X(s)$ and $V(s)$ in Figure B1-1. Determine\nthe steady state gain of this Transfer Function and comment on your answer.\n","marks":8,"answer":[{"label":"closed-loop steady state gain","value":1}],"grade":"$undefined"},{"label":"c","text":"In the case that $B(s) = 3s + 1$ and $A(s) = (s + \\alpha)(s + \\beta)$, where $\\alpha$ and $\\beta$ are\ncoefficients, determine the pole(s) and zero(s) of the model (without control).\n","marks":2,"answer":[{"label":"zero","value":-0.3333}],"grade":"$undefined"},{"label":"d","text":"Using a different model, one that represents a DC motor in a laboratory experiment, an engineer has\ndetermined the following Transfer Function for the control system in Figure B1-1. Here, $X(s)$ is the\nmotor speed.\n\n$$X(s) = \\frac{2.3K_I}{s^2 + (4.6 + 2.3K_p)s + 2.3K_I}\\,V(s)$$\n\nDesign a controller such that that the poles of the characteristic equation are a complex conjugate\npair $-3 \\pm 2j$.\n","marks":3,"answer":[{"label":"K_p","value":0.6087},{"label":"K_I","value":5.652}],"grade":"$undefined"},{"label":"e","text":"In terms of the response mode and stability, what is the significance of the pole positions stated in\npart (d) above? When the controller from part (d) is implemented in the lab, the control system starts\nto yield an unstable oscillatory response to a change in the set point. Speculate as to what might have\nhappened.\n\nHint: even if you were unsure how to answer part (d), you can still attempt this question part.\n","marks":4,"answer":[{"label":"zeta","value":0.832},{"label":"omega_n","value":3.606,"units":"rad/s"}],"grade":"$undefined"}],"solution":"$3e","confidence":"HIGH","solutionNote":"No official solution available. Gains checked by back-substitution and the step response simulated (scipy)."},{"id":"engr202-2024-b2","module":"ENGR5001","week":5,"label":"ENGR202 2024 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["modelling","generalised-forms","laplace-tf","frequency-response","hurwitz","poles-stability","pid-actions","bode-diagram","pi-control"],"source":{"name":"ENGR202 Summer 2024 Q B2","paper":"ENGR202","year":2024,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR202 Main Exam 2024.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Generalised second order form: $\\ddot x + 2\\zeta\\omega_n\\dot x + \\omega_n^2 x = Ku(t)$ (given in the question)\n- Laplace with zero initial conditions: $\\dot x \\to sX$, $\\ddot x \\to s^2 X$ (learn this)\n- Second order steady state gain $K/\\omega_n^2$ (learn this)\n- Hurwitz for $a_0 s^4 + a_1 s^3 + a_2 s^2 + a_3 s + a_4$: $\\Delta_1 = a_1$, $\\Delta_2 = a_1 a_2 - a_0 a_3$, $\\Delta_3 = a_3\\Delta_2 - a_1^2 a_4$, $\\Delta_4 = a_4\\Delta_3$ (learn this)\n- Necessary conditions for stability: all coefficients exist and have the same sign (learn this)\n","text":"The generalized form of a second order model is stated below:\n\n$$\\frac{d^2x}{dt^2} + 2\\zeta\\omega_n\\frac{dx}{dt} + \\omega_n^2 x = Ku(t) \\qquad \\text{(B2-1)}$$\n","parts":[{"label":"a","text":"For a second order engineering example of your choice (such as one of the examples from the lectures),\ndevelop a linear dynamic model from first principles, defining all the terms and stating all the\nassumptions made. Convert your answer into generalized second order form, and hence identify the forcing\nfunction, steady state gain, damping and natural frequency.\n","marks":7,"grade":"$undefined"},{"label":"b","text":"Convert the generalised second order model (B2-1) into Transfer Function form, stating any assumptions\nmade. Briefly discuss how Transfer Functions can be useful in the context of control system design.\n","marks":4,"grade":"$undefined"},{"label":"c","text":"Sketch the general form of a Bode diagram for a second order system, labelling all the axis (including\nfrequency, phase and magnitude) and identifying key features.\n","marks":6,"grade":"$undefined"},{"label":"d","text":"For the following characteristic equation, what is (i) the inequality defining the necessary condition\nfor stability and (ii) the value of the coefficient $\\alpha$ such that the system is marginally stable:\n$s^4 + 8s^3 + 2s^2 + 3s + \\alpha = 0$.\n","marks":2,"answer":[{"label":"alpha for marginal stability","value":0.6094}],"grade":"$undefined"},{"label":"e","text":"In the context of a practical example of your choice, briefly summarise the relative advantages of PI\ncontrol and its potential limitations.\n","marks":6,"grade":"$undefined"}],"solution":"$3f","confidence":"HIGH","solutionNote":"No official solution available. Hurwitz result checked by computing the roots either side of $\\alpha = 39/64$."},{"id":"engr202-2025-q2","module":"ENGR5001","week":5,"label":"ENGR202 2025 Q2","kind":"past-paper","level":3,"marks":50,"minutes":60,"tags":["modelling","laplace-tf","poles-stability","generalised-forms","feedback-objectives","block-diagrams","controller-design","pid-actions","pv-control","heating-control"],"source":{"name":"ENGR202 Summer 2025 Q2","paper":"ENGR202","year":2025,"question":"Q2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR202 Main Exam 2024-25.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Steady state gain $G(0)$ (set $s = 0$) (learn this)\n- Poles: roots of the denominator; zeros: roots of the numerator; stable if all poles have negative real parts (learn this)\n- $s^2 + 2\\zeta\\omega_n s + \\omega_n^2$: $\\omega_n = \\sqrt{a_2}$, $\\zeta = a_1/(2\\omega_n)$ (learn this)\n- Negative feedback rule $\\dfrac{G_1}{1 + G_1 G_2}$; series rule $G_1 G_2$ (learn this)\n","text":"This question concerns the heating system in a human occupied office building. The relationship between\nthe indoor temperature $X(s)$ and the applied voltage to a heater $U(s)$ is approximated by Equation (2-1):\n\n$$X(s) = \\frac{s + b_1}{s^2 + a_1 s + a_2}\\,U(s) \\qquad \\text{Equation (2-1)}$$\n\nFigure Q2 shows a control system with two negative feedback loops.\n","figures":[{"src":"/figures/engr202-2025-q2.png","caption":"Figure Q2: Control system with two negative feedback loops."}],"parts":[{"label":"a(i)","text":"The following questions are about the model given by Equation (2-1).\n\nEquation (2-1) was obtained by considering heat transfer and energy balance equations (details not\nimportant). Use this example to briefly explain the difference between data-based and mechanistic\nmodels. Suggest potential limitations of Equation (2-1), for example in its ability to represent real\nworld temperature data and/or if it is subsequently used to design a control system.\n","marks":6,"grade":"$undefined"},{"label":"a(ii)","text":"State an algebraic equation for determining the steady state gain of Equation (2-1). Use your answer to\ndetermine the steady state indoor temperature for $u(t > 0) = 5$ V, assuming initial conditions\n$u(0) = 0$, $x(0) = 20^\\circ\\mathrm{C}$, and coefficients $a_1 = 6$, $a_2 = 8$ and $b_1 = 10$.\n","marks":6,"answer":[{"label":"steady state gain","value":1.25,"units":"degC/V"},{"label":"steady state temperature","value":26.25,"units":"degC"}],"grade":"$undefined"},{"label":"a(iii)","text":"Again with $a_1 = 6$, $a_2 = 8$ and $b_1 = 10$, what are the pole(s) and zero(s) of Equation (2-1).\nPlot and label the pole(s) on the complex $s$-plane. State the stability condition, explaining the\nreason for your answer.\n","marks":6,"answer":[{"label":"zero","value":-10},{"label":"dominant pole","value":-2},{"label":"other pole","value":-4}],"grade":"$undefined"},{"label":"a(iv)","text":"Still using $a_1 = 6$, $a_2 = 8$ and $b_1 = 10$, determine the natural frequency and damping ratio of\nthe model given by Equation (2-1). Comment on what the damping ratio tells us about the dynamics of this\nmodel.\n","marks":6,"answer":[{"label":"omega_n","value":2.828,"units":"rad/s"},{"label":"zeta","value":1.061}],"grade":"$undefined"},{"label":"b(i)","text":"The following questions are about the control system shown in Figure Q2.\n\nWhat is $V(s)$ usually called and what does it represent? Briefly suggest how it might be chosen for the\ntemperature control application.\n","marks":4,"grade":"$undefined"},{"label":"b(ii)","text":"Develop the closed-loop Transfer Function for Figure Q2 and hence show that the closed-loop\ncharacteristic equation is as follows:\n\n$$(1 + K_2)s^2 + (a_1 + K_1 + K_2 b_1)s + a_2 + K_1 b_1 = 0 \\qquad \\text{Equation (2-2)}$$\n\nIn addition, calculate the steady state gain of the closed-loop Transfer Function and comment on your\nanswer.\n","marks":9,"grade":"$undefined"},{"label":"b(iii)","text":"Suggest and briefly describe two methods for designing the above controller, i.e., methods to determine\nsuitable values of $K_1$ and $K_2$ in Figure Q2.\n","marks":6,"grade":"$undefined"},{"label":"b(iv)","text":"Figure Q2 represents an initial attempt at designing a controller for this problem. A more experienced\ncontrol engineer states that Figure Q2 should be revised to include a more suitable control algorithm.\nSpeculate on the reasons for their statement and suggest a more appropriate algorithm. Hints: your\nanswer might refer to the model dynamics from part (a); the closed loop Transfer Function from part (b);\nand practical aspects of the temperature control application.\n","marks":7,"grade":"$undefined"}],"solution":"$40","confidence":"MEDIUM","solutionNote":"No official solution available. MEDIUM only because of a(ii): the question gives $x(0) = 20^\\circ$C with\n$u(0) = 0$, which only makes sense if the Transfer Function describes changes from the initial state, giving\n$26.25^\\circ$C; if the setter intended absolute values the answer would be $6.25^\\circ$C. Confirm with the\nlecturer. All other numbers were checked by two methods.\n"},{"id":"engr270-2026-q1","module":"ENGR5001","week":3,"label":"ENGR270 2026 Q1","kind":"past-paper","level":3,"marks":45,"minutes":54,"tags":["generalised-forms","modelling","laplace-tf","poles-stability","frequency-response","bode-diagram","marginal-stability","wave-energy"],"source":{"name":"ENGR270 Summer 2026 Q1","paper":"ENGR270","year":2026,"question":"Q1","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR270 In-person Main Exam Paper 2025-26.pdf","tag":"VALID"},"grade":"2:1","formulaSheet":"- Generalised second order form: $\\ddot x + 2\\zeta\\omega_n\\dot x + \\omega_n^2 x = Ku(t)$ (learn this)\n- For $m\\ddot x + b\\dot x + kx = F$: $\\omega_n = \\sqrt{k/m}$, $\\zeta = \\dfrac{b}{2\\sqrt{km}}$, steady state gain $1/k$ (learn this)\n- Steady state gain $G(0)$; poles from the characteristic equation (learn this)\n- A linear system driven by a sine at frequency $\\omega$ responds at steady state with a sine of the SAME frequency, scaled by $M = |G(j\\omega)|$ and shifted by $\\phi = \\mathrm{Arg}\\,G(j\\omega)$ (learn this)\n- $\\omega = 2\\pi f$ (learn this)\n- Stability shortcuts: all coefficients must exist and have the same sign; a single pole at the origin ($a_n = 0$) gives marginal stability (learn this)\n","text":"A second-order model for a point absorber Wave Energy Converter (WEC) captures the oscillatory motion of\nthe device in response to wave excitation and hydrodynamic forces. By making some linearising\nassumptions, the model is as follows:\n\n$$m\\frac{d^2x}{dt^2} + b\\frac{dx}{dt} + kx = F(t) \\qquad \\text{(Q1-1)}$$\n\nwhere $x$ is the buoy displacement and $F(t)$ is the wave excitation force. Units are not stated in this\nquestion, and coefficients represent scaled values. However, in physical terms, the coefficients $m$, $b$\nand $k$ are linked to the equivalent mass, damping coefficient and hydrostatic restoring stiffness,\nrespectively.\n","parts":[{"label":"a","text":"Convert the model (Q1-1) into the generalised second order differential equation form. Identify the\ninput and output variables. Determine the damping ratio and natural frequency in terms of the model\ncoefficients. Briefly comment on potential limitations of this model e.g. if used to design control\nsystems for the WEC.\n","marks":10,"grade":"$undefined"},{"label":"b","text":"For $m = 12048$, $b = 4200$ and $k = 876$, show that the Transfer Function for equation (Q1-1) takes the\nform given by equation (Q1-2). State any assumptions made in developing the Transfer Function.\n\n$$X(s) = \\frac{0.000083}{s^2 + 0.349s + 0.073}\\,F(s) \\qquad \\text{(Q1-2)}$$\n","marks":4,"grade":"$undefined"},{"label":"c","text":"Determine the steady state gain of the Transfer Function (Q1-2). If the wave force takes a constant\nvalue $F_0$ (unlikely in practice), write down an expression for the steady state displacement of the\nbuoy.\n","marks":2,"answer":[{"label":"steady state gain","value":0.00114}],"grade":"$undefined"},{"label":"d","text":"Write down the characteristic equation associated with equation (Q1-2) and determine the poles. Sketch\nthese poles on the complex $s$-plane. What is the stability of the model and the response mode, i.e. the\nnature of the transient response? What is the steady state gain? Comment on the physical interpretation\nof all these results, linking to your answer to part (a) if possible.\n","marks":10,"answer":[{"label":"pole real part","value":-0.1745},{"label":"pole imaginary part (+/-)","value":0.2063},{"label":"zeta","value":0.646},{"label":"omega_n","value":0.2696,"units":"rad/s"}],"grade":"$undefined"},{"label":"e","text":"An oncoming series of waves has a frequency of 0.09 Hz. In simulation, this is represented by a sine\ninput. What is the frequency of oscillations of the model at steady state? Sketch the Bode diagram for a\ngeneralised second order system with a resonant peak. Hint: label the axis and any important points on\nyour sketch, but there is no need to include any numerical values in your sketch. Briefly speculate on\nthe significance of the Bode diagram in the context of wave energy conversion.\n","marks":10,"answer":[{"label":"steady state frequency","value":0.09,"units":"Hz"}],"grade":"$undefined"},{"label":"f","text":"Unrelated to the above WEC model, define stability. With the help of a time response sketch, explain\nmarginal stability. Give the stability condition (stable, unstable or marginally stable) for systems\nwith the following characteristic equations: (i) $s^5 + s^3 + s^2 + \\alpha s + \\beta = 0$,\n(ii) $s^3 + s^2 + \\alpha s = 0$, (iii) $s = 0$, (iv) $s^3 + s^2 - s + 5 = 0$ and\n(v) $(s + 3.72)(s + 0.08)(s + 0.12) = 0$, where $\\alpha$ and $\\beta$ are positive coefficients; in each\ncase, give the reason for your answer.\n","marks":9,"grade":"$undefined"}],"solution":"$41","confidence":"HIGH","solutionNote":"No official solution available. Poles, gains and stability cases checked with numpy roots, and the step and frequency responses simulated (scipy)."},{"id":"engr270-2026-q2","module":"ENGR5001","week":5,"label":"ENGR270 2026 Q2","kind":"past-paper","level":3,"marks":55,"minutes":66,"tags":["feedback-objectives","time-response","controller-design","pid-actions","block-diagrams","ics-security","disturbance-rejection","time-delay","integral-action"],"source":{"name":"ENGR270 Summer 2026 Q2","paper":"ENGR270","year":2026,"question":"Q2","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR270 In-person Main Exam Paper 2025-26.pdf","tag":"VALID"},"grade":"first","formulaSheet":"- First order $G(s) = \\dfrac{K}{\\tau s + 1}$: unit step response $K(1 - e^{-t/\\tau})$, 63.2% at $t = \\tau$ (learn this)\n- Time delay $T$: the same response shifted right by $T$ (Transfer Function factor $e^{-sT}$) (learn this)\n- Unity feedback: $\\dfrac{Y}{R} = \\dfrac{CG}{1 + CG}$ (learn this)\n- Final value theorem: $y(\\infty) = \\lim_{s \\to 0} sY(s)$; a constant $d$ has transform $d/s$ (learn this)\n- Integral action $C(s) = K/s$; PID $C(s) = K_P + \\dfrac{K_I}{s} + K_D s$ (learn this)\n","text":"$42","figures":[{"src":"/figures/engr270-2026-q2.png","caption":"Figure Q2-1: Control system."}],"parts":[{"label":"a","text":"Explain why Figure Q2-1 is called unity feedback closed-loop control. Speculate on how an open-loop\ncontroller could be designed for this plant. What are the advantages of closed-loop compared to\nopen-loop control in this context?\n","marks":8,"grade":"$undefined"},{"label":"b(i)","text":"For this question part only, i.e. (b) (i), (b) (ii) and (b) (iii), assume $D(s) = 0$ and the plant model\n$G(s) = 2/(40s + 1)$.\n\nFor zero initial conditions, sketch the response of the plant model to a unit step, i.e. $u(t) = 1$ for\n$t > 0$. Use your sketch to define and explain the terms steady state gain and time constant. Add to\nyour sketch the unit step response of a model with the same time constant and steady state gain, but\nwhich has a time delay of 10 s.\n","marks":7,"answer":[{"label":"steady state gain","value":2},{"label":"time constant","value":40,"units":"s"},{"label":"output at t = tau (no delay)","value":1.264}],"grade":"$undefined"},{"label":"b(ii)","text":"For $C(s) = K$ where $K$ is the control gain, determine the closed-loop Transfer Function, i.e. the\nrelationship between $Y(s)$ and $R(s)$. Design a controller to ensure the closed-loop time constant is\n10 s.\n","marks":5,"answer":[{"label":"K","value":1.5},{"label":"closed-loop steady state gain","value":0.75}],"grade":"$undefined"},{"label":"b(iii)","text":"Briefly explain the advantages of using PID control, in comparison to the simple controller from\n(b) (ii).\n","marks":3,"grade":"$undefined"},{"label":"c","text":"Develop the closed loop Transfer Functions showing how $Y(s)$ is a function of both external signals,\ni.e. $R(s)$ and $D(s)$. Hint: use straightforward algebra, starting with $Y(s) = G(s)Q(s)$ or, for\nnotational brevity when writing your answer, you can abbreviate as follows: $Y = GQ$ and similar.\n","marks":5,"grade":"$undefined"},{"label":"d","text":"Using your answer from part (c) and assuming the disturbance is time invariant, i.e. a constant value,\nshow that simple integral control action successfully mitigates for the aerator failure. Hint: integral\naction implies $C(s) = K/s$ where $K$ is the control gain.\n","marks":3,"grade":"$undefined"},{"label":"e","text":"The fault in the airflow system may have arisen deliberately, i.e. by one or more individuals with\nmalicious intent. From a security perspective, identify other potential vulnerabilities in Figure Q2-1\nthat could impact on the control of DO and hence the quality of the released effluent.\n","marks":12,"grade":"$undefined"},{"label":"f","text":"As manager of the wastewater treatment plant, briefly explain how you could secure the plant from both\ncyber and physical intrusion.\n","marks":12,"grade":"$undefined"}],"solution":"$43","confidence":"HIGH","solutionNote":"No official solution available. Numerical parts checked by simulation (closed-loop time constant and gain) and\nsympy (final value theorem for the disturbance). The security parts follow the module's ICS security\nlectures; mark allocation within (e) and (f) is estimated.\n"},{"id":"engr5001-saq1-1","module":"ENGR5001","week":1,"label":"Part 1 SAQ 1","kind":"exercise","level":2,"minutes":12,"tags":["modelling","generalised-forms","rlc-circuit","mechanistic-models","damping-ratio"],"source":{"name":"ENGR5001 Part 1 Self-Assessment Questions, Q1","path":"ENGR5001 - Control and Robotics/Week 1/ENGR5001_Part1_SAQ.pdf"},"grade":"2:2","formulaSheet":"- Inductor $V_L = L\\dfrac{di}{dt}$; resistor $V_R = Ri$; capacitor $i = C\\dfrac{dV_C}{dt}$; Kirchhoff's voltage law (learn this)\n- Generalised second order form $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$, steady state gain $K/\\omega_n^2$ (learn this)\n- Generalised first order form $\\tau\\dfrac{dx}{dt} + x = Ku(t)$ (learn this)\n","text":"Develop a linear time invariant model showing the relationship between the input voltage $V_i$ and output\nvoltage $V_o$ for the RLC network below (inductor $L$ and resistor $R$ in series, output taken across the\ncapacitor $C$).\n\nWhat assumptions have you made? Put the equation obtained into the standard form, indicating the steady state\ngain, damping ratio, natural frequency, forcing function and/or time constant.\n","figures":[{"src":"/figures/engr5001-saq1-1.png","caption":"RLC network: L and R in series from the input V_i, output V_o across C."}],"solution":"$44","confidence":"HIGH","solutionNote":"No official solution in the folder; the derivation follows the LC circuit and mass-spring-damper examples in the Part 1 slides."},{"id":"engr5001-saq1-2","module":"ENGR5001","week":1,"label":"Part 1 SAQ 2","kind":"exercise","level":2,"minutes":15,"tags":["modelling","generalised-forms","mass-spring-damper","base-excitation","suspension-seat","mechanistic-models"],"source":{"name":"ENGR5001 Part 1 Self-Assessment Questions, Q2","path":"ENGR5001 - Control and Robotics/Week 1/ENGR5001_Part1_SAQ.pdf"},"grade":"2:1","formulaSheet":"- Newton's second law $\\sum F = m\\ddot x$; spring force $k \\times$ (extension); viscous damper force $c \\times$ (relative velocity) (learn this)\n- Generalised second order form $\\dfrac{d^2x}{dt^2} + 2\\zeta\\omega_n\\dfrac{dx}{dt} + \\omega_n^2 x = Ku(t)$ (learn this)\n- Mass-spring-damper: $\\omega_n = \\sqrt{k/m}$, $\\zeta = \\dfrac{c}{2\\sqrt{km}}$ (learn this)\n","text":"For a person sitting on the suspension seat below, develop a linear time invariant mathematical model showing the\nrelationship between the floor movement $y$ and the seat position $x$. (The seat and person are modelled as a\nmass $m$ supported on the floor by a spring $k$ and damper $c$ in parallel; $x$ and $y$ are vertical\ndisplacements.)\n\nWhat assumptions have you made? Put the equation obtained into the standard form, indicating the steady state\ngain, damping ratio, natural frequency, forcing function and/or time constant.\n","figures":[{"src":"/figures/engr5001-saq1-2.png","caption":"Person on a suspension seat, modelled as mass m on spring k and damper c, floor displacement y, seat displacement x."}],"solution":"$45","confidence":"HIGH","solutionNote":"No official solution in the folder; this is the standard base-excitation model and follows the mass-spring-damper slides."},{"id":"engr5001-ox-2-4","module":"ENGR5001","week":2,"label":"Old spec 2.4 exercise","kind":"exercise","level":2,"minutes":10,"tags":["time-response","modelling","generalised-forms","graphical-estimation","vehicle-speed","step-response"],"source":{"name":"ENGR202 (old spec) Control 2.4 Time Response, exercise","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 2 - Control 2.4 - exercise answers (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Newton's second law; friction force $b\\dot y$ proportional to velocity (learn this)\n- Generalised first order form $\\tau\\dot x + x = Ku(t)$ (learn this)\n- $K = \\dfrac{x(t \\to \\infty)}{u(t \\to \\infty)}$; at $t = \\tau$ the output has completed 63% of its change (learn this)\n","text":"A vehicle of mass $m$ has displacement $y$ from a reference point, so its speed is $x = \\dot y$. The engine\nimparts a force $u(t)$, friction is proportional to velocity (force $b\\dot y$) and the rotational inertia of the\nwheels is negligible.\n\nIn an experiment, the engine force is stepped from 0% (zero force) to 100% (full power) at $t = 10$ s. The\nresulting vehicle speed, in miles per hour, is shown in the figure.\n","figures":[{"src":"/figures/engr5001-ox-2-4.png","caption":"Vehicle speed (mph) against time (s) after a step from 0 to 100% engine power at t = 10 s."}],"parts":[{"label":"a","text":"Develop a linear first order mechanistic model for the vehicle speed and express it in generalised first\norder form.\n","grade":"$undefined"},{"label":"b","text":"Estimate the steady state gain and time constant of the model from the graph, and hence write down the\nestimated model.\n","answer":[{"label":"steady state gain","value":0.8,"units":"mph per %","tol":0.05},{"label":"time constant","value":5,"units":"s","tol":0.1}],"grade":"$undefined"}],"solution":"$46","confidence":"HIGH","solutionNote":"Matches the official exercise answer ($K = 0.8$, $\\tau = 5$ s, $5\\dot x + x = 0.8u$)."},{"id":"engr5001-ox-3-4a","module":"ENGR5001","week":3,"label":"Old spec 3.4 exercise 1","kind":"exercise","level":2,"minutes":8,"tags":["poles-stability","laplace-tf","zeros","non-minimum-phase","dominant-poles"],"source":{"name":"ENGR202 (old spec) Control 3.4 Poles, Zeros and Stability, Exercise 1","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.4 - exercise answers (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Laplace with zero initial conditions: $\\dot x \\to sX$, $\\ddot x \\to s^2X$, $\\dot u \\to sU$ (learn this)\n- Poles: roots of the denominator; zeros: roots of the numerator; stable if all poles have negative real parts (learn this)\n- Dominant pole: the one closest to the imaginary axis (learn this)\n","text":"For the system below, find the poles and zeros, plot these on the complex $s$-plane, determine the stability and\nsketch the response mode.\n\n$$\\frac{d^2x}{dt^2} + 8\\frac{dx}{dt} + 14x = 2\\frac{du}{dt} - u$$\n","answer":[{"label":"zero","value":0.5},{"label":"dominant pole","value":-2.586},{"label":"other pole","value":-5.414}],"solution":"$47","confidence":"HIGH","solutionNote":"Matches the official exercise answer (zero 0.5; poles -2.6 and -5.4; stable but non-minimum phase; dominant pole -2.6)."},{"id":"engr5001-ox-3-4b","module":"ENGR5001","week":3,"label":"Old spec 3.4 exercise 2","kind":"exercise","level":1,"minutes":4,"tags":["poles-stability","laplace-tf","marginal-stability","integrator","piling-rig"],"source":{"name":"ENGR202 (old spec) Control 3.4 Poles, Zeros and Stability, Exercise 2 (hydraulic piling rig)","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 3 - Control 3.4 - exercise answers (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Laplace with zero initial conditions: $\\dot x \\to sX$ (learn this)\n- Pole at the origin (single): marginally stable (learn this)\n","text":"Hydraulic piling rig. The input is the applied voltage $u$ and the output $x$ is the angle of the mast from the\nvertical. The model is\n\n$$\\frac{dx}{dt} = Ku(t)$$\n\nDetermine the stability and interpret the results.\n","answer":[{"label":"pole","value":0}],"solution":"$48","confidence":"HIGH","solutionNote":"Matches the official exercise answer (pole at the s-plane origin, marginally stable)."},{"id":"engr5001-ox-4-1a","module":"ENGR5001","week":4,"label":"Old spec 4.1 toaster exercise","kind":"exercise","level":1,"minutes":4,"tags":["feedback-objectives","open-loop","closed-loop","disturbances"],"source":{"name":"ENGR202 (old spec) Control 4.1 Feedback, toaster exercise","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 4 - Control 4.1 - exercise answers (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Open loop: input set without measuring the output; closed loop: output measured and fed back to correct the input (learn this)\n","text":"Toaster exercise: does a standard kitchen toaster use open-loop or closed-loop control? Explain your answer.\n","solution":"A standard kitchen toaster uses **open-loop** control.\n- It works on a timer: the dial setting effectively sets how long the bread is heated (even if it is not\n labelled in seconds). When the time is up, the toast pops up.\n- Nothing measures how well the bread is actually toasted (the output), so nothing adjusts the heating time.\n- It is therefore subject to disturbances it cannot correct: frozen bread, different bread types and thickness,\n and the toaster warming up over repeated use, so the toast may be burnt or under-cooked.\n\nA closed-loop toaster would need a sensor measuring browning (e.g. an optical colour sensor or surface\ntemperature) and would switch off when the desired level is reached; it would handle the disturbances above but\ncosts more, which is probably why most toasters are open loop.\n\nCheck against the definitions: there is no feedback path from output to input, so it is open loop.\n","confidence":"HIGH","solutionNote":"Matches the official exercise answer."},{"id":"engr5001-ox-4-1b","module":"ENGR5001","week":4,"label":"Old spec 4.1 robot exercise","kind":"exercise","level":2,"minutes":8,"tags":["feedback-objectives","block-diagrams","robot-manipulator","disturbances","set-point"],"source":{"name":"ENGR202 (old spec) Control 4.1 Feedback, robot exercise","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 4 - Control 4.1 - exercise answers (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Closed-loop block diagram: set point, summing junction, controller, control input, plant, output, feedback sensor; disturbances enter at the plant (learn this)\n","text":"$49","solution":"$4a","confidence":"HIGH","solutionNote":"Matches the official exercise answer (other answers with similar information are accepted there too)."},{"id":"engr5001-ox-4-3","module":"ENGR5001","week":4,"label":"Old spec 4.3 exercise","kind":"exercise","level":2,"minutes":6,"tags":["block-diagrams","feedback-objectives","closed-loop-tf","dc-motor"],"source":{"name":"ENGR202 (old spec) Control 4.3 Control Objectives, exercise","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 4 - Control 4.3 - exercise answers (old spec ENGR202).pdf"},"grade":"third","formulaSheet":"- Series rule: $G_1 G_2$; negative feedback rule: $\\dfrac{G_1}{1 + G_1 G_2}$ (learn this)\n- Steady state gain: set $s = 0$; characteristic equation: denominator $= 0$ (learn this)\n","text":"Develop the Closed Loop Transfer Function for the DC motor speed control system in the figure, where $D$ is the\ndesired motor speed (volts), $C(s)$ the controller, $U$ the input signal (volts), $H(s)$ the motor, $X$ the actual\nmotor speed (volts) and $F(s)$ the feedback elements. Either solve using algebra or (saves time) use the rules of\nblock diagram analysis. What condition gives unity steady state gain, and what is the characteristic equation?\n","figures":[{"src":"/figures/engr5001-ox-4-3.png","caption":"DC motor speed control loop with controller C(s), motor H(s) and feedback elements F(s)."}],"solution":"$4b","confidence":"HIGH","solutionNote":"Matches the official exercise answer."},{"id":"engr5001-ox-4-6","module":"ENGR5001","week":4,"label":"Old spec 4.6 exercise","kind":"exercise","level":2,"minutes":8,"tags":["pid-actions","block-diagrams","velocity-feedback","damping-ratio","closed-loop-tf"],"source":{"name":"ENGR202 (old spec) Control 4.6 Derivative and Velocity, exercise","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 4 - Control 4.6 - exercise answers (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Negative feedback rule: $\\dfrac{G_1}{1 + G_1 G_2}$ (learn this)\n- Velocity feedback: $u(t) = d(t) - k_v\\dfrac{dx}{dt}$, i.e. $U = D - k_v sX$ (given in the question)\n- Generalised second order form $s^2 + 2\\zeta\\omega_n s + \\omega_n^2$ (learn this)\n","text":"Draw the block diagram for velocity feedback (instead of derivative action) applied to the second order plant,\nand use the block diagram to derive the Closed Loop Transfer Function.\n\n$$X(s) = \\frac{\\omega_n^2}{s^2 + 2\\zeta\\omega_n s + \\omega_n^2}U(s) \\qquad u(t) = d(t) - k_v\\frac{dx(t)}{dt}$$\n","solution":"$4c","confidence":"HIGH","solutionNote":"Matches the official exercise answer."},{"id":"engr5001-ox-5-3","module":"ENGR5001","week":5,"label":"Old spec 5.3 exercise","kind":"exercise","level":2,"minutes":12,"tags":["frequency-response","frequency-design","nyquist","magnitude-phase","third-order"],"source":{"name":"ENGR202 (old spec) Control 5.3 Frequency Response Revisited, exercise","path":"ENGR5001 - Control and Robotics/$ Old Spec/Week 5 - Control 5.3 - exercise answers (old spec ENGR202).pdf"},"grade":"2:2","formulaSheet":"- Frequency response: $M = |G(j\\omega)|$, $\\phi = \\mathrm{Arg}\\,G(j\\omega)$ (learn this)\n- $|j\\omega + a| = \\sqrt{\\omega^2 + a^2}$, $\\mathrm{Arg}(j\\omega + a) = \\tan^{-1}(\\omega/a)$ (learn this)\n- Arguments of a product add; of a quotient subtract (learn this)\n","text":"Determine the frequency response of the following 3rd order system i.e. magnitude and phase\n\n$$G(s) = \\frac{1}{(s + 4)(s + 5)(s + 7)}$$\n\nWork out the values of magnitude and phase for the following frequencies: 0, 1, 3, 10, infinity.\n","answer":[{"label":"M at omega = 0","value":0.007143},{"label":"M at omega = 1","value":0.006727},{"label":"phase at omega = 1","value":-33.48,"units":"deg"},{"label":"M at omega = 3","value":0.004504},{"label":"phase at omega = 3","value":-91.03,"units":"deg"},{"label":"M at omega = 10","value":0.0006803},{"label":"phase at omega = 10","value":-186.6,"units":"deg"}],"solution":"$4d","confidence":"HIGH","solutionNote":"The official table agrees at $\\omega = 0$, 10 and infinity, but its rows for $\\omega = 1$ and $\\omega = 3$ are wrong:\nit lists $M = 0.0071$, $-11^\\circ$ at $\\omega = 1$ and $M = 0.0067$, $-34^\\circ$ at $\\omega = 3$ (the values that\nactually belong to about $\\omega = 0.3$ and $\\omega = 1$). The official Nyquist sketch itself crosses the imaginary\naxis at about $-0.0045$, which is the $\\omega = 3$ value computed here ($M = 0.0045$, $\\phi = -91^\\circ$), so these\nanswers are trusted.\n"},{"id":"engr263-2017-a1","module":"ENGR5002","week":10,"label":"ENGR263 2017 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["convective-mass-transfer","diffusion","mass-transfer","stagnant-gas","film-theory","mass-transfer-coefficient","absorption"],"source":{"name":"ENGR263 Process of Mass and Heat Transfers 2017 Section A Q A1","paper":"ENGR263","year":2017,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR263_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR263 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (b).\n","formulaSheet":"- Flux with bulk flow: $N_{A,z} = -cD_{AB}\\dfrac{dy_A}{dz} + y_A(N_{A,z} + N_{B,z})$ (on the formula sheet)\n- Diffusion through a stagnant gas: $N_A = \\dfrac{PD_{AB}}{RT\\delta}\\,\\dfrac{p_{A1} - p_{A2}}{p_{B,lm}} = \\dfrac{PD_{AB}}{RT\\delta}\\ln\\dfrac{P - p_{A2}}{P - p_{A1}}$, $p_{B,lm} = \\dfrac{p_{B2} - p_{B1}}{\\ln(p_{B2}/p_{B1})}$ (on the formula sheet)\n- Mass transfer coefficient: $N_A = k_c(c_{A1} - c_{A2})$, ideal gas $c_A = p_A/(RT)$ (on the formula sheet)\n- Film theory: stagnant film $k_c = \\dfrac{D_{AB}}{\\delta\\,x_{B,lm}}$; equimolar or dilute $k_c = \\dfrac{D_{AB}}{\\delta}$ (learn this)\n","parts":[{"label":"a(i)","text":"What arguments are used for eliminating the flux $N_B$ of component B when a component A diffuses in a\nstagnant medium B?\n","marks":3,"grade":"third"},{"label":"a(ii)","text":"Suggest an experimental method in which the diffusivity $D_{AB}$ of component A in gaseous B could be measured.","marks":3,"grade":"third"},{"label":"a(iii)","text":"Give conditions, in which the concentration of a diffusing component follows: 1. An exponential profile in\nthe film of interphase (1 mark). 2. A linear profile in the film of interphase (1 mark).\n","marks":2,"grade":"third"},{"label":"b(i)","text":"In a packed column, operating at approximately $1.5 \\times 10^5\\ \\mathrm{Pa}$ and 303 K, air-H$_2$S mixture is\nscrubbed with water. It is assumed that the whole of the resistance to mass transfer may be regarded as\nlying within a thin laminar film on the gas side of the gas-liquid interphase. At some intermediate point,\nthe partial pressure of H$_2$S in the gas phase has been reduced to 5000 Pa while the partial pressure of\nH$_2$S in equilibrium with the aqueous solution is 1233 Pa.\n\nDerive an expression at any position in the column for the rate of absorption of H$_2$S by using the first\nprinciples of molecular diffusion.\n","marks":5,"grade":"2:1"},{"label":"b(ii)","text":"Derive expression of mass transfer coefficient ($k_c$) based on concentration. Calculate the transfer flux\nof absorption of H$_2$S ($\\mathrm{mol\\,m^{-2}\\,s^{-1}}$), assuming that value of mass transfer coefficient\nis $0.96\\ \\mathrm{m\\,s^{-1}}$.\n","marks":5,"answer":[{"label":"N_A","value":1.44,"units":"mol m^-2 s^-1","tol":0.02}],"grade":"$undefined"},{"label":"b(iii)","text":"Calculate the thickness of the hypothetical gas film if the diffusivity of H$_2$S in air is\n$0.24\\ \\mathrm{cm^2\\,s^{-1}}$. (Data: $R = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$.)\n","marks":4,"answer":[{"label":"p_B,lm","value":146900,"units":"Pa","tol":0.01},{"label":"delta","value":0.0255,"units":"mm","tol":0.02}],"grade":"$undefined"},{"label":"b(iv)","text":"What would be the thickness of the hypothetical gas film if concentration of H$_2$S was assumed negligible\ncompared to concentration of air?\n","marks":3,"answer":[{"label":"delta","value":0.025,"units":"mm","tol":0.02}],"grade":"$undefined"}],"solution":"$4e","confidence":"HIGH","solutionNote":"Agrees with the official numerical answers for (b)(iii) 0.0256 mm (ours 0.02553 mm) and (b)(iv) 0.025 mm.\nFor (b)(ii) the official sheet gives 1.45 mol m^-2 s^-1; recomputing 0.96 x 3767 / (8.314 x 303) gives\n1.436 (1.44), so the official value appears slightly mis-rounded (about 1 % high); the answer box accepts\n1.44 +/- 2 %, which includes 1.45 at the edge.\n"},{"id":"engr263-2017-a2","module":"ENGR5002","week":9,"label":"ENGR263 2017 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["diffusion","convective-mass-transfer","mass-transfer","stagnant-gas","diffusivity","sherwood","multicomponent"],"source":{"name":"ENGR263 Process of Mass and Heat Transfers 2017 Section A Q A2","paper":"ENGR263","year":2017,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR263_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR263 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (b).\n","formulaSheet":"- Diffusion through a stagnant gas: $N_A = \\dfrac{cD_{AB}}{\\delta}\\ln\\dfrac{1 - y_{A2}}{1 - y_{A1}} = \\dfrac{PD_{AB}}{RT\\delta}\\,\\dfrac{p_{A1} - p_{A2}}{p_{B,lm}}$, $c = P/(RT)$ (on the formula sheet)\n- Multicomponent stagnant mixture: $\\dfrac{1}{D_{A,mix}} = \\sum_{j \\ne A}\\dfrac{y'_j}{D_{Aj}}$, $y'_j = \\dfrac{y_j}{\\sum_{j \\ne A} y_j}$ (on the formula sheet; also given in the question)\n- Sherwood number $\\mathrm{Sh} = \\dfrac{k_cd}{D_{AB}}$ (on the formula sheet; also given in the question)\n- Radial diffusion from a sphere: $W_A = 4\\pi r^2N_A = -4\\pi r^2 D_{AB}\\dfrac{dc_A}{dr} = \\text{constant}$ (on the formula sheet)\n","parts":[{"label":"a(i)","text":"What is the difference between the diffusivity or diffusion coefficient ($\\mathrm{m^2\\,s^{-1}}$) and the\nmass transfer coefficient?\n","marks":3,"grade":"third"},{"label":"a(ii)","text":"Prove that for equimolecular counter diffusion from a sphere to a surrounding stationary infinite medium,\nthe Sherwood number based on the diameter $d$ of the sphere is equal to 2. (Data: the Sherwood number is\ndefined as $\\mathrm{Sh} = k_cd/D_{AB}$, where $k_c$ is the mass transfer coefficient based on\nconcentration ($\\mathrm{m\\,s^{-1}}$) and $D_{AB}$ is the diffusivity of component A in the medium B\n($\\mathrm{m^2\\,s^{-1}}$).)\n","marks":5,"grade":"2:1"},{"label":"b(i)","text":"Ammonia gas is diffusing at a constant rate through a layer of 1 mm thickness of stagnant air. Conditions\nare fixed so that the gas contains 50 % by volume of ammonia at one boundary of the stagnant layer. The\nammonia diffusing to the other boundary is quickly absorbed and the concentration is then negligible. The\ntemperature is 295 K and the pressure is atmospheric, and under these conditions the diffusivity of\nammonia in air is $0.18\\ \\mathrm{cm^2\\,s^{-1}}$.\n\nCalculate the flux of mass transfer of ammonia ($\\mathrm{mol\\,m^{-2}\\,s^{-1}}$).\n","marks":9,"grade":"2:2","answer":[{"label":"c","value":41.3,"units":"mol/m^3","tol":0.01},{"label":"N_A","value":0.515,"units":"mol m^-2 s^-1","tol":0.02}]},{"label":"b(ii)","text":"Assuming that the non-diffusing gas (air) is replaced by a mixture of methane (B) and hydrogen (C) in\nvolume ratio 2:1, re-calculate the flux of mass transfer of ammonia, if the diffusivity of ammonia in\nmethane and the diffusivity of ammonia in hydrogen are $0.184\\ \\mathrm{cm^2\\,s^{-1}}$ and\n$0.690\\ \\mathrm{cm^2\\,s^{-1}}$, respectively. (Data: the diffusivity of a component $i$ in a\nmulticomponent mixture follows the Stefan-Maxwell model\n$D_{i,mix} = 1/\\sum_{j \\ne i} (y'_j/D_{ij})$, $y'_j = y_j/\\sum_{j \\ne i} y_j$.)\n","marks":8,"grade":"2:2","answer":[{"label":"D_mix","value":0.2435,"units":"cm^2/s","tol":0.01},{"label":"N_A","value":0.697,"units":"mol m^-2 s^-1","tol":0.02}]}],"solution":"$4f","confidence":"MEDIUM","solutionNote":"Disagrees with the official numerical answers: official (b)(i) 0.39 and (b)(ii) 0.53 mol m^-2 s^-1; ours\n0.515 and 0.697. The official ratio 0.53/0.39 = 1.36 matches our diffusivity ratio 0.2435/0.18 = 1.35, so the\ndisagreement is only in the base flux. 0.39 is close to the pure-Fick (no bulk flow) value 0.372 (or 0.40 if\nT = 273 K is used), which ignores the stagnant-air drift term; the question states the air is stagnant, so the\nlog (Stefan) form is the correct model and we trust 0.515 (verified by the log and log-mean forms in Python).\n"},{"id":"engr263-2018-a1","module":"ENGR5002","week":9,"label":"ENGR263 2018 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["diffusion","mass-transfer","stagnant-gas","diffusivity","evaporation","pressure-temperature-correction"],"source":{"name":"ENGR263 Processes of Heat and Mass Transfers 2018 Section A Q A1","paper":"ENGR263","year":2018,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR263_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR263.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (g).\n\nAn open cylindrical tank of 6 m diameter contains benzene at $22^\\circ\\mathrm{C}$. The tank is exposed to the\natmosphere in such a manner that the liquid is covered with a stagnant film of vapour estimated to be 5 mm\nthick. The concentration of benzene in the bulk of the vapour phase is assumed to be negligible. The saturated\nvapour pressure of benzene at $22^\\circ\\mathrm{C}$ is 100 mmHg.\n\nData: $1\\ \\mathrm{atm} = 1.013 \\times 10^5\\ \\mathrm{Pa} = 760\\ \\mathrm{mmHg}$, $R = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$.\nThe effect of pressure and temperature changes from ($P_1$, $T_1$) to ($P_2$, $T_2$) on gaseous diffusivity is\n$D_{AB,T_2,P_2} = D_{AB,T_1,P_1}\\left(\\dfrac{P_1}{P_2}\\right)\\left(\\dfrac{T_2}{T_1}\\right)^{3/2}$ (T in K).\n","formulaSheet":"- Diffusion through a stagnant gas: $N_A = \\dfrac{PD_{AB}}{RT\\delta}\\ln\\dfrac{P - p_{A2}}{P - p_{A1}} = \\dfrac{PD_{AB}}{RT\\delta}\\,\\dfrac{p_{A1} - p_{A2}}{p_{B,lm}}$ (on the formula sheet)\n- Pressure and temperature correction $D_{AB,T_2,P_2} = D_{AB,T_1,P_1}\\dfrac{P_1}{P_2}\\left(\\dfrac{T_2}{T_1}\\right)^{3/2}$ (on the formula sheet; also given in the question)\n- Rate = flux x area; mass rate = molar rate x $M$ (learn this)\n","parts":[{"label":"a","text":"Calculate the flux of mass transfer of benzene $N_{benzene}$ in $\\mathrm{mol/(m^2\\,s)}$ if the diffusivity of\nbenzene in air at $22^\\circ\\mathrm{C}$ is $0.094\\ \\mathrm{cm^2/s}$ (no need to derive the model of the flux).\n","marks":10,"grade":"2:2","answer":[{"label":"N_A","value":0.01095,"units":"mol m^-2 s^-1","tol":0.02}]},{"label":"b","text":"Calculate the rate of diffusion of benzene in mol/day.","marks":3,"answer":[{"label":"rate","value":26750,"units":"mol/day","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Calculate the rate of diffusion of benzene in kg/day, if the molecular weight of benzene is 0.078 kg/mol.","marks":2,"answer":[{"label":"rate","value":2086,"units":"kg/day","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"What is the value of the loss of benzene from this tank in pounds sterling per day, if benzene is worth 0.25 GBP/kg?","marks":3,"answer":[{"label":"loss","value":522,"units":"GBP/day","tol":0.02}],"grade":"$undefined"},{"label":"e","text":"If a floating roof is used to decrease the losses, what is the diameter of the roof needed to reduce the loss\nin rate of benzene by 95 %?\n","marks":2,"answer":[{"label":"d_roof","value":5.85,"units":"m","tol":0.01}],"grade":"$undefined"},{"label":"f","text":"An increase of the total pressure would have a positive effect on the loss. What is the total pressure\nrequired to save 50% of the loss in benzene rate?\n","marks":3,"grade":"2:1","answer":[{"label":"P_2","value":1468,"units":"mmHg","tol":0.02}]},{"label":"g","text":"A decrease of temperature would have a positive effect on the losses. What is the saving in losses in\npercent (%) of benzene rate, if temperature drops by 5 %? Assuming the saturated vapour pressure of benzene\nremains constant.\n","marks":2,"grade":"2:1","answer":[{"label":"saving","value":2.5,"units":"%","tol":0.05}]}],"solution":"$50","confidence":"MEDIUM","solutionNote":"Compared with the official numerical answers (2018, handwritten): (a) 0.011 mol/m^2 s agrees (ours 0.01095);\n(b) 26 858.3 mol/day, (c) 2094.9 kg/day, (d) 523.7 GBP/day agree within 0.5 % (the official values carry a\nslightly larger unrounded flux, 0.01100). Disagreements: (e) the official answer is 1.34 m, which is the\ndiameter of a circle with the 5 % area left uncovered (6 x sqrt(0.05)); the roof itself must cover 95 % of the\nsurface, so we give 5.85 m and mention 1.34 m. (g) the official answer is a 9.1 % saving; with p_sat fixed,\nN is proportional to D/T, i.e. to T^0.5, giving 2.5 % (we could not reproduce 9.1 % with any reading of T,\nincluding T^1.5 alone, 7.4 %). We trust 2.5 %; confidence MEDIUM until the intended reading is confirmed.\nNo official answer for (f).\n"},{"id":"engr263-2018-a3","module":"ENGR5002","week":9,"label":"ENGR263 2018 Q A3(a)","kind":"past-paper","level":3,"marks":15,"minutes":18,"grade":"2:2","tags":["diffusion","convective-mass-transfer","mass-transfer","diffusivity","wilke-chang","knudsen","film-theory","penetration-theory"],"source":{"name":"ENGR263 Processes of Heat and Mass Transfers 2018 Section A Q A3","paper":"ENGR263","year":2018,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR263_2018.pdf","tag":"PARTIAL","leftOut":"Part (b) (10 marks), fitting the Lennard-Jones parameters sigma and epsilon to D(T) data for water vapour in\noxygen using a collision-integral chart, is marked \"unsure\" in the audit (chart-based parameter fitting is\nnot in the current slides), so it is left out.\n"},"text":"Answer ALL parts (a) - (b). [Part (b), 10 marks, fitting Lennard-Jones parameters from diffusivity data with a\ncollision-integral chart, is omitted.]\n","formulaSheet":"- Wilke-Chang (liquids): $\\dfrac{D_{AB}\\mu_B}{T} = \\dfrac{7.4 \\times 10^{-8}(\\Phi_BM_B)^{1/2}}{V_A^{0.6}}$ (on the formula sheet)\n- Effective diffusivity in a porous solid: $D_{eff} = \\dfrac{\\varepsilon}{\\tau}D_{AB}$; Knudsen diffusivity $D_K = \\dfrac{d_{pore}}{3}\\sqrt{\\dfrac{8RT}{\\pi M_A}}$ (learn this)\n- Film theory $k_c = D_{AB}/\\delta$; penetration theory $k_c = 2\\sqrt{D_{AB}/(\\pi t_e)}$ (learn this)\n- Unsteady diffusion into a semi-infinite medium $\\dfrac{c_A - c_{A0}}{c_{As} - c_{A0}} = \\mathrm{erfc}\\left(\\dfrac{z}{2\\sqrt{D_{AB}t}}\\right)$ (on the formula sheet)\n","parts":[{"label":"a(i)","text":"How does the diffusion coefficient of liquid species vary with the molecular weight, temperature and\nviscosity of the liquid medium?\n","marks":3,"grade":"$undefined"},{"label":"a(ii)","text":"How does the diffusivity vary with the porosity of a solid medium (1 mark), with the tortuosity of a solid\nmedium (1 mark), with the diameter of the pores (1 mark)?\n","marks":3,"grade":"$undefined"},{"label":"a(iii)","text":"On which side of the gas/liquid interphase boundary would the mass transfer resistance take place in the\nfollowing processes? An evaporation of a liquid through a stagnant gas phase (1 mark); an absorption of a\ngas by a liquid of low velocity (1 mark); a continuous distillation column (1 mark).\n","marks":3,"grade":"$undefined"},{"label":"a(iv)","text":"Discuss the relevance of the mass transfer coefficient for unit operations.","marks":3,"grade":"$undefined"},{"label":"a(v)","text":"A thin liquid film is falling down one side of a vertical surface wall of an absorption column at\nsteady-state conditions. Discuss qualitatively (no need for relevant mathematical models) the two cases of\nlong-residence time and short-residence time and their effects on profiles of concentration through the\nliquid film.\n","marks":3,"grade":"2:1"}],"solution":"$51","confidence":"HIGH","solutionNote":"Descriptive part only; no official answers for (a) (the 2018 answer sheet gives only the (b) chart result,\nsigma = 2.6 A). Content follows the standard treatment (Wilke-Chang, effective/Knudsen diffusivity, film and\npenetration theory).\n"},{"id":"engr263-2019-a1","module":"ENGR5002","week":10,"label":"ENGR263 2019 Q A1(b)-(d)","kind":"past-paper","level":3,"marks":15,"minutes":18,"grade":"2:1","tags":["convective-mass-transfer","diffusion","mass-transfer","equimolar-counter-diffusion","stagnant-gas","film-theory","mass-transfer-coefficient"],"source":{"name":"ENGR263 Process of Mass Transfer 2019 Q A1","paper":"ENGR263","year":2019,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR263_Exam_2019 (1).pdf","tag":"PARTIAL","leftOut":"(a) (5 marks), the Raoult's-law vapour composition, is ENGR5003 week 6 material and is in that bank as\nengr263-2019-a1a; its result is quoted in the stem below. (e) (5 marks), reading the ethanol-water azeotrope\nfrom the VLE diagram and alternative separation methods, is distillation and off-syllabus.\n"},"text":"A binary mixture of ethanol and water in the reboiler of a batch distillation column (a liquid in contact with\nits vapour across a thin vapour-liquid film; Figure A1-1 of the paper shows $y_w = 0.7$ in the bulk vapour and\n$x_w = 0.6$ in the liquid) presents the following data:\n\n- Total pressure $P = 101\\,300\\ \\mathrm{Pa}$\n- Temperature $T = 85^\\circ\\mathrm{C}$\n- Mole fraction of water in the liquid phase $x_w = 0.6$\n- Mole fraction of water in the vapour phase $y_w = 0.70$\n- Saturated vapour pressure of ethanol $P_e^{sat} = 135\\,087\\ \\mathrm{Pa}$\n- Thickness of vapour-liquid interphase film $\\delta = 0.002\\ \\mathrm{m}$\n- Binary diffusivity $D_{e-w} = 1.6 \\times 10^{-5}\\ \\mathrm{m^2/s}$\n- Gas constant $R = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$\n\n[Part (a) asked for the ethanol vapour mole fraction in equilibrium with the liquid, assuming ideal phases\n(Raoult's law): $y_e^* = x_eP_e^{sat}/P = 0.4 \\times 135\\,087/101\\,300 = 0.533$. Use it below. Part (e), on the\nazeotrope, is omitted.]\n","formulaSheet":"- Flux with bulk flow $N_{A,z} = -cD_{AB}\\dfrac{dy_A}{dz} + y_A(N_{A,z} + N_{B,z})$ (on the formula sheet)\n- Equimolar counter-diffusion through a film (plane surface): $N_A = \\dfrac{cD_{AB}}{\\delta}(y_{A1} - y_{A2}) = D_{AB}\\dfrac{c_{A1} - c_{A2}}{\\delta}$ (on the formula sheet)\n- Stagnant B: $N_A = \\dfrac{cD_{AB}}{\\delta}\\ln\\dfrac{1 - y_{A2}}{1 - y_{A1}}$ (on the formula sheet)\n- $N_A = k_c\\Delta c_A = k_y\\Delta y_A$; film theory $k_c = D_{AB}/\\delta$, $k_y = ck_c$; $c = P/(RT)$ (learn this)\n","parts":[{"label":"b","text":"Calculate the flux of mass transfer ($\\mathrm{mol\\,m^{-2}\\,s^{-1}}$) of ethanol from the liquid phase to the\nvapour phase in the reboiler.\n","marks":5,"grade":"2:2","answer":[{"label":"c","value":34,"units":"mol/m^3","tol":0.01},{"label":"N_e","value":0.0635,"units":"mol m^-2 s^-1","tol":0.02}]},{"label":"c","text":"Calculate the mass transfer coefficient of ethanol from the liquid phase to the vapour phase of the reboiler.","marks":5,"grade":"2:2","answer":[{"label":"k_c","value":0.008,"units":"m/s","tol":0.02},{"label":"k_y","value":0.272,"units":"mol m^-2 s^-1","tol":0.02}]},{"label":"d","text":"The batch distillation process is replaced by a simple evaporation process in a stagnant vapour phase,\nwhere the condensation of the less volatile component (water) is not taking place. Calculate the flux of\nmass transfer ($\\mathrm{mol\\,m^{-2}\\,s^{-1}}$) of ethanol in the reboiler.\n","marks":5,"grade":"2:1","answer":[{"label":"N_e","value":0.11,"units":"mol m^-2 s^-1","tol":0.02}]}],"solution":"$52","confidence":"HIGH","solutionNote":"No official numerical answers for ENGR263 2019 exist in the folder. All values computed in Python and checked\nby the log-mean form and back-substitution. Part (a) is the separate ENGR5003 item engr263-2019-a1a.\n"},{"id":"engr263-2019-a2","module":"ENGR5002","week":9,"label":"ENGR263 2019 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["diffusion","convective-mass-transfer","mass-transfer","stagnant-gas","diffusivity","sherwood","schmidt","penetration-theory","multicomponent"],"source":{"name":"ENGR263 Process of Mass Transfer 2019 Q A2","paper":"ENGR263","year":2019,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR263_Exam_2019 (1).pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (b).\n","formulaSheet":"- Schmidt number $\\mathrm{Sc} = \\dfrac{\\mu}{\\rho D_{AB}} = \\dfrac{\\nu}{D_{AB}}$, Sherwood number $\\mathrm{Sh} = \\dfrac{k_cL}{D_{AB}}$ (on the formula sheet)\n- Diffusion through a stagnant gas: $N_A = \\dfrac{PD_{AB}}{RT\\delta}\\ln\\dfrac{P - p_{A2}}{P - p_{A1}}$ (on the formula sheet)\n- $D_{AB} \\propto T^{3/2}/P$ (on the formula sheet)\n- Multicomponent stagnant mixture $\\dfrac{1}{D_{A,mix}} = \\sum_{j \\ne A}\\dfrac{y'_j}{D_{Aj}}$ (on the formula sheet; also given in the question)\n- Film theory $k_c = D_{AB}/\\delta$; penetration theory $k_c = 2\\sqrt{D_{AB}/(\\pi t_e)}$ (learn this)\n","parts":[{"label":"a(i)","text":"The dimensionless Sherwood number (Sh) is a mass transfer coefficient that is function of Schmidt number\n(Sc). Explain the definition of Schmidt number by using concepts of the boundary layers of momentum and\nconcentration.\n","marks":3,"grade":"third"},{"label":"a(ii)","text":"Explain the difference between the penetration theory and the two-phase (two-film) theory of mass transfer.","marks":2,"grade":"third"},{"label":"a(iii)","text":"What is the difference between the diffusivity or diffusion coefficient and the mass transfer coefficient?","marks":2,"grade":"third"},{"label":"a(iv)","text":"Prove that for equimolecular counter diffusion from a sphere to a surrounding stationary, infinite medium,\nthe Sherwood number based on the diameter of the sphere is equal to 2.\n","marks":3,"grade":"2:1"},{"label":"b(i)","text":"Oxygen (A) is diffusing through carbon monoxide (B) under steady-state conditions while carbon monoxide is\nnot diffusing. The total pressure is one atmosphere and the temperature is $0^\\circ\\mathrm{C}$. The partial\npressures of oxygen at the interface of the two gases and the bulk of carbon monoxide are 100 mmHg and\n50 mmHg, respectively. The thickness of the interface between the two gases is 0.2 cm and the diffusivity\n$D_{AB}$ is $0.185\\ \\mathrm{cm^2\\,s^{-1}}$. (Data: 1 atmosphere = 760 mmHg = $1.013 \\times 10^5$ Pa.)\n\nCalculate the flux of diffusion ($\\mathrm{mol\\,m^{-2}\\,s^{-1}}$) of oxygen in carbon monoxide.\n","marks":5,"grade":"2:2","answer":[{"label":"N_A","value":0.0301,"units":"mol m^-2 s^-1","tol":0.02}]},{"label":"b(ii)","text":"If the total pressure was doubled, would the flux of mass transfer increase, decrease or remain the same?","marks":5,"answer":[{"label":"N_A (2P, same p_A)","value":0.0143,"units":"mol m^-2 s^-1","tol":0.03}],"grade":"$undefined"},{"label":"b(iii)","text":"Assuming that the non-diffusing gas (carbon monoxide) is replaced by a mixture of methane (B) and hydrogen (C)\nin volume ratio 2:1, re-calculate the flux of diffusion ($\\mathrm{mol\\,m^{-2}\\,s^{-1}}$) if the diffusivity of\noxygen in methane is $D_{O_2-CH_4} = 0.184\\ \\mathrm{cm^2\\,s^{-1}}$ and the diffusivity of oxygen in hydrogen is\n$D_{O_2-H_2} = 0.690\\ \\mathrm{cm^2\\,s^{-1}}$. (Data:\n$D_{i,mix} = 1/\\sum_{j \\ne i}(y'_j/D_{ij})$, $y'_j = y_j/\\sum_{j \\ne i}y_j$.)\n","marks":5,"grade":"2:2","answer":[{"label":"D_mix","value":0.2435,"units":"cm^2/s","tol":0.01},{"label":"N_A","value":0.0397,"units":"mol m^-2 s^-1","tol":0.02}]}],"solution":"$53","confidence":"HIGH","solutionNote":"No official numerical answers for ENGR263 2019 exist in the folder. All numbers computed in Python and checked\nby two forms of the stagnant-film equation. (b)(ii) is ambiguous (fixed partial pressures vs fixed mole\nfractions); both readings are given, and the answer box uses the fixed-partial-pressure reading.\n"},{"id":"engr263-2021-a1","module":"ENGR5002","week":9,"label":"ENGR263 2021 Q A1","kind":"past-paper","level":3,"marks":16,"minutes":19,"grade":"2:2","tags":["diffusion","mass-transfer","diffusivity","wilke-chang","equimolar-counter-diffusion"],"source":{"name":"ENGR263 Process of Mass Transfer Summer 2021 (open book) Q A1","paper":"ENGR263","year":2021,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR263 Exam 2021.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2021/ENGR263_Numerical_answers_20_21.pdf","tag":"VALID"},"text":"Answer BOTH parts (a) and (b).\n","formulaSheet":"- Wilke-Chang: $D_{AB} = \\dfrac{7.4 \\times 10^{-8}(\\Phi_BM_B)^{1/2}T}{\\mu_BV_A^{0.6}}$ with $D$ in $\\mathrm{cm^2/s}$, $\\mu_B$ in cP, $V_A$ in $\\mathrm{cm^3/mol}$, $T$ in K, $M_B$ and $\\Phi_B$ of the solvent (on the formula sheet)\n- Equimolar counter-diffusion through a film: $N_A = -N_B = \\dfrac{cD_{AB}}{\\delta}(y_{A1} - y_{A2}) = \\dfrac{D_{AB}}{RT\\delta}(p_{A1} - p_{A2})$, $c = P/(RT)$ (on the formula sheet)\n","parts":[{"label":"a","text":"Compare estimates of the diffusion coefficient of an ethanol-water mixture at $10^\\circ\\mathrm{C}$ (283 K),\nunder conditions that (1) ethanol is the solute and water is the solvent and (2) water is the solute and\nethanol is the solvent. Ethanol has the molecular formula C$_2$H$_5$OH and a molecular weight of 46 g/gmole,\nand water (H$_2$O) has a molecular weight of 18 g/gmole. At $10^\\circ\\mathrm{C}$, the liquid viscosity of\nwater is $1.306 \\times 10^{-3}$ Pa s (1.306 cP), and the liquid viscosity of ethanol is\n$1.394 \\times 10^{-3}$ Pa s (1.394 cP). The molecular volume of water is $18.9\\ \\mathrm{cm^3/gmole}$ and the\nmolar volume of ethanol (EtOH) $59.2\\ \\mathrm{cm^3/gmole}$. The association parameter of the solvent\n$\\Phi_{water} = 2.6$ and $\\Phi_{EtOH} = 1.5$.\n","marks":6,"answer":[{"label":"D (EtOH in water)","value":0.00000948,"units":"cm^2/s","tol":0.02},{"label":"D (water in EtOH)","value":0.0000214,"units":"cm^2/s","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"In a rectification column, toluene is diffusing from gas to liquid and benzene from liquid to gas at\ntemperature of $30^\\circ\\mathrm{C}$ and pressure $101.3\\ \\mathrm{kN/m^2}$ under conditions of equal molar\ncounter diffusion. At one point in the column, the molal concentrations of toluene on the two sides of a\ngas film 0.5 mm thick are 70% and 20%, respectively. Assuming the diffusivity of benzene-toluene vapour\nunder the operating conditions to be $0.38 \\times 10^{-5}\\ \\mathrm{m^2/s}$, estimate the rate of diffusion\nof toluene and benzene in kg/hr across an area of $0.01\\ \\mathrm{m^2}$. The molecular weights of toluene\nand benzene are 92 and 78 g/mol, respectively. The ideal gas constant is equal to\n$8.314\\ \\mathrm{m^3\\,Pa\\,K^{-1}\\,mol^{-1}}$.\n","marks":10,"answer":[{"label":"N","value":0.153,"units":"mol m^-2 s^-1","tol":0.02},{"label":"toluene","value":0.506,"units":"kg/h","tol":0.02},{"label":"benzene","value":0.429,"units":"kg/h","tol":0.02}],"grade":"$undefined"}],"solution":"$54","confidence":"HIGH","solutionNote":"Matches the official 2020-21 numerical answers: (a) 9.47e-6 and 2.13e-5 cm^2/s (ours 9.48e-6 and 2.14e-5,\nrounding), (b) 0.506 kg/h toluene and 0.429 kg/h benzene (ours identical).\n"},{"id":"engr263-2021-a2","module":"ENGR5002","week":9,"label":"ENGR263 2021 Q A2(c)","kind":"past-paper","level":3,"marks":12,"minutes":14,"grade":"2:2","tags":["diffusion","mass-transfer","stagnant-gas","diffusivity","pressure-temperature-correction"],"source":{"name":"ENGR263 Process of Mass Transfer Summer 2021 (open book) Q A2","paper":"ENGR263","year":2021,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR263 Exam 2021.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2021/ENGR263_Numerical_answers_20_21.pdf","tag":"PARTIAL","leftOut":"(a) distillation vs evaporation (2 marks) and (b) precipitating dextran (2 marks) are separation-process\nquestions that are not on the current syllabus.\n"},"text":"Answer All parts (a) - (c). [Parts (a) and (b), 4 marks on distillation/evaporation and precipitation, are\nomitted.]\n\nOxygen is diffusing through a stagnant layer of methane 10 mm thick. The temperature is $20^\\circ\\mathrm{C}$\nand the pressure $100\\ \\mathrm{kN/m^2}$. The concentrations of oxygen on the two sides of the film are 15% and\n5% by volume. The diffusivity of oxygen in methane at $20^\\circ\\mathrm{C}$ and $100\\ \\mathrm{kN/m^2}$ is\n$2.046 \\times 10^{-5}\\ \\mathrm{m^2/s}$.\n","formulaSheet":"- Diffusion through a stagnant gas: $N_A = \\dfrac{PD_{AB}}{RT\\delta}\\ln\\dfrac{1 - y_{A2}}{1 - y_{A1}}$ (on the formula sheet)\n- Pressure correction: $D_{AB} \\propto 1/P$ at fixed $T$ (on the formula sheet)\n- $R = 8314\\ \\mathrm{J\\,kmol^{-1}\\,K^{-1}}$ (learn this)\n","parts":[{"label":"c(i)","text":"Calculate the rate of diffusion of oxygen in $\\mathrm{kmol/(m^2\\,s)}$.","marks":5,"answer":[{"label":"N_A","value":0.00000934,"units":"kmol m^-2 s^-1","tol":0.02}],"grade":"$undefined"},{"label":"c(ii)","text":"What will be the rate of diffusion if the total pressure is raised to $200\\ \\mathrm{kN/m^2}$, other conditions remaining the same?","marks":7,"answer":[{"label":"N_A","value":0.00000934,"units":"kmol m^-2 s^-1","tol":0.02}],"grade":"$undefined"}],"solution":"$55","confidence":"HIGH","solutionNote":"The official 2020-21 numerical answers give 9.33e-9 kmol/m^2 s for both (i) and (ii). Our value is\n9.34e-6 kmol/m^2 s (= 9.34e-3 mol/m^2 s): the official figure is a factor 1000 out, almost certainly a\nkmol/mol slip, since 8314 J/kmol K with SI data gives kmol directly (checked by two forms in Python). The\nofficial answer does confirm that (ii) equals (i), i.e. the fixed-mole-fraction reading.\n"},{"id":"engr263-2022-a2","module":"ENGR5002","week":10,"label":"ENGR263 2022 Q A2(a)-(b)","kind":"past-paper","level":3,"marks":11,"minutes":13,"grade":"2:1","tags":["convective-mass-transfer","diffusion","mass-transfer","mass-transfer-coefficient","stagnant-gas","evaporation","film-theory"],"source":{"name":"ENGR263 Process of Mass Transfer 2021-22 Q A2","paper":"ENGR263","year":2022,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR263 Exam 2021-22.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2022/ENGR263_Numerical_answers_21_22.pdf","tag":"PARTIAL","leftOut":"(c) (14 marks), oxygen through a stagnant CH4/H2/CO2 mixture, is word for word the ENGR271 2026 Q3 already\nin the bank (engr271-2026-q3), so it is not converted twice.\n"},"text":"Answer ALL parts (a) - (c). [Part (c), oxygen diffusing through a stagnant CH$_4$/H$_2$/CO$_2$ mixture, is the\nsame question as ENGR271 2026 Q3 (already in the bank) and is omitted here.]\n","formulaSheet":"- Mass transfer coefficient $N_A = k_c(c_{A,s} - c_{A,\\infty})$ (on the formula sheet)\n- Diffusion through a stagnant gas: $N_A = \\dfrac{PD_{AB}}{RT\\delta}\\ln\\dfrac{P - p_{A2}}{P - p_{A1}}$ (on the formula sheet)\n- $1\\ \\mathrm{atm} = 760\\ \\mathrm{mm\\,Hg}$; $1\\ \\mathrm{dm^3} = 1000\\ \\mathrm{cm^3}$ (learn this)\n","parts":[{"label":"a","text":"Starch in the human diet is digested in the stomach and small intestine into monosaccharides such as\nglucose, which are the only forms that can be absorbed. For one set of experiments, the rate of intestine\nuptake of glucose is $1.6 \\times 10^{-10}\\ \\mathrm{mol/(cm^2\\,s)}$ from a solution containing\n$2.65 \\times 10^{-4}\\ \\mathrm{mol/dm^3}$ glucose. Calculate the mass transfer coefficient if the amount of\nglucose in blood is very small.\n","marks":3,"grade":"third","answer":[{"label":"k_c","value":0.000604,"units":"cm/s","tol":0.02}]},{"label":"b","text":"The rate of evaporation of water from a flat surface maintained at a temperature of $60^\\circ\\mathrm{C}$ is\n$2.91 \\times 10^{-4}\\ \\mathrm{kg/(s\\,m^2)}$. What will be the rate of evaporation of benzene from a similar\nflat surface, but maintained at $26^\\circ\\mathrm{C}$, if the effective film thicknesses are the same in both\ncases? It is given: molecular weight of water 18 g/mol; vapour pressure of water at 60 C = 149 mm Hg; vapour\npressure of benzene at 26 C = 99.5 mm Hg; diffusivity of air-water vapour at 60 C =\n$2.6 \\times 10^{-5}\\ \\mathrm{m^2/s}$; diffusivity of air-benzene vapour at 26 C =\n$0.98 \\times 10^{-5}\\ \\mathrm{m^2/s}$; atmospheric pressure in both cases = $1.013 \\times 10^5\\ \\mathrm{N/m^2}$;\nthe ideal gas constant is equal to $8.314\\ \\mathrm{m^3\\,Pa\\,K^{-1}\\,mol^{-1}}$.\n","marks":8,"answer":[{"label":"delta","value":12.8,"units":"mm","tol":0.02},{"label":"N_benzene","value":0.00000436,"units":"kmol m^-2 s^-1","tol":0.02}],"grade":"$undefined"}],"solution":"$56","confidence":"HIGH","solutionNote":"Matches the official 2021-22 numerical answers: (a) 6e-4 cm/s (ours 6.04e-4); (b) 4.38e-6 kmol/m^2 s (ours\n4.36e-6, 0.4 % lower, rounding of intermediate values). Note for the omitted part (c): the official 2021-22\nanswer is 6.803e-6 kmol/m^2 s, whereas the bank entry engr271-2026-q3 (same question) gives 3.07e-6 with\nD corrected by (T2/T1)^1.5 (P1/P2); correcting for temperature only gives 6.05e-6 and no correction 5.44e-6,\nso the official value appears to omit the pressure correction (we could not reproduce 6.803e-6 exactly).\n"},{"id":"engr263-2023-a1","module":"ENGR5002","week":9,"label":"ENGR263 2023 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["diffusion","mass-transfer","diffusivity","wilke-chang","droplet-evaporation","diffusion-in-solids","solubility"],"source":{"name":"ENGR263 Process Transfer of Mass Summer 2023 (open book) Q A1","paper":"ENGR263","year":2023,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR263 Exam Paper 2023_Final_version_2023_01_24.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR263 Exam Paper 2023_Final_version_2023_01_24 answers.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n","formulaSheet":"- Wilke-Chang: $D_{AB} = \\dfrac{7.4 \\times 10^{-8}(\\Phi_BM_B)^{1/2}T}{\\mu_BV_A^{0.6}}$ ($D$ in $\\mathrm{cm^2/s}$, $\\mu_B$ in cP, $V_A$ in $\\mathrm{cm^3/mol}$) (on the formula sheet)\n- Le Bas molar volume: $V_A = \\sum$ atomic volumes (learn this)\n- Droplet evaporation time $\\theta = \\dfrac{\\rho_Ar_s^2RT}{2M_AD_{AB}P\\ln[(P - p'_{A,b})/(P - p'_{A,s})]}$ (given in the question)\n- Steady diffusion through a plane solid wall $N_A = D_{AB}\\dfrac{c_{A1} - c_{A2}}{\\delta}$ (on the formula sheet)\n- Gas solubility in a solid: $c_A = \\dfrac{S\\,p_A}{22.4\\ \\mathrm{L/mol}}$ with $S$ in volume (STP) of gas per volume of solid per atm (learn this)\n","parts":[{"label":"a","text":"Estimate the diffusivity of ethyl alcohol (C$_2$H$_5$OH) in dilute aqueous solution at $20^\\circ\\mathrm{C}$.\nFor water as solvent, $\\Phi_B = 2.26$; molecular weight of water, $M_B = 18.02$; viscosity of water,\n$\\mu_B = 0.001005\\ \\mathrm{kg/(m\\,s)}$; atomic volumes of C, H and O are 14.8, 3.7 and 7.4\n$\\mathrm{cm^3/mol}$, respectively.\n","marks":8,"grade":"2:2","answer":[{"label":"V_A","value":59.2,"units":"cm^3/mol","tol":0.01},{"label":"D_AB","value":0.0000119,"units":"cm^2/s","tol":0.02}]},{"label":"b","text":"$57","marks":8,"answer":[{"label":"theta","value":89700,"units":"s","tol":0.03}],"grade":"$undefined"},{"label":"c","text":"Hydrogen gas is diffusing through an unglazed neoprene rubber with a wall thickness of 50 mm at\n$25^\\circ\\mathrm{C}$ and pressure of 1 atm. The solubility of hydrogen in the rubber has been estimated to\nbe $0.053\\ \\mathrm{cm^3}$ (at STP, $V_{mol} = 22.4$ litres at STP) of H$_2$ per $\\mathrm{cm^3}$ of neoprene.\nThe diffusivity of hydrogen through the rubber wall is $1.8 \\times 10^{-6}\\ \\mathrm{cm^2/s}$. Calculate the\nmass diffusion rate ($\\mathrm{g/(m^2\\,s)}$) of hydrogen per square meter of the wall.\n","marks":9,"answer":[{"label":"c_A1","value":0.00000237,"units":"mol/cm^3","tol":0.02},{"label":"N_A","value":8.52e-9,"units":"mol m^-2 s^-1","tol":0.02},{"label":"mass rate","value":1.72e-8,"units":"g m^-2 s^-1","tol":0.02}],"grade":"$undefined"}],"solution":"$58","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers (2023): V_A = 59.2, D = 1.19e-5 cm^2/s; theta = 8.97e4 s (24.9 h);\nc_A1 = 2.366e-6 mol/cm^3, N_A = 8.52e-9 mol/m^2 s, 1.72e-8 g/m^2 s. All numbers recomputed in Python and agree.\nTwo remarks on the worked answers' commentary (not the numbers): (b) says lower total pressure helps because\n\"D ~ 1/P dominates the P in the numerator\", but D P is independent of P; the benefit comes from the\nlog driving-force term, as stated above. (c) its \"independent check\" against Geankoplis Example 6.5-1 compares\nfluxes for walls of different thickness and pressures and does not actually confirm the value; the direct\ncalculation stands. No official solutions exist.\n"},{"id":"engr263-2023-a2","module":"ENGR5002","week":9,"label":"ENGR263 2023 Q A2(a)-(b)","kind":"past-paper","level":3,"marks":16,"minutes":19,"grade":"2:1","tags":["diffusion","mass-transfer","stagnant-gas","stefan-tube","multicomponent","diffusivity"],"source":{"name":"ENGR263 Process Transfer of Mass Summer 2023 (open book) Q A2","paper":"ENGR263","year":2023,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR263 Exam Paper 2023_Final_version_2023_01_24.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR263 Exam Paper 2023_Final_version_2023_01_24 answers.pdf","tag":"PARTIAL","leftOut":"(c) (9 marks), McCabe-Thiele design of a continuous distillation column, is distillation and off-syllabus.\n"},"text":"Answer ALL parts (a) - (c). [Part (c), a McCabe-Thiele distillation design (9 marks), is omitted.]\n","formulaSheet":"- Diffusion through a stagnant gas: $N_A = \\dfrac{D_{AB}P}{RT(z_2 - z_1)}\\ln\\dfrac{P - p_{A2}}{P - p_{A1}}$ (on the formula sheet)\n- Multicomponent stagnant mixture $\\dfrac{1}{D_{A,mix}} = \\sum_{j \\ne A}\\dfrac{y'_j}{D_{Aj}}$, $y'_j$ on an A-free basis (on the formula sheet)\n- $R = 82.06\\ \\mathrm{cm^3\\,atm\\,mol^{-1}\\,K^{-1}} = 8314\\ \\mathrm{J\\,kmol^{-1}\\,K^{-1}}$ (learn this)\n","parts":[{"label":"a","text":"Water is held at the bottom of a narrow metal tube at a constant temperature of 318 K. The dry ambient air\noutside the tube is at 1 atm and 318 K. Water evaporates and diffuses through the air in the tube, and the\ndiffusion path $z_2 - z_1$ is 30 cm long (from the water surface at $z_1$ to the open top at $z_2$).\nCalculate the rate of evaporation at steady state in $\\mathrm{mol/(s\\,cm^2)}$. The diffusivity of water\nvapour (A) in air (B) at 1 atm and 318 K is $0.282\\ \\mathrm{cm^2/s}$. Vapour pressure of water at 318 K is\n9.5 kPa. Assume that air is insoluble in water.\n","marks":6,"grade":"2:2","answer":[{"label":"N_A","value":3.55e-8,"units":"mol s^-1 cm^-2","tol":0.02}]},{"label":"b","text":"At 298 K and total pressure of 1 atm, methane (A) is diffusing at a steady state through non-diffusing argon\n(B) and helium (C). At position $z_1 = 0$, the partial pressures of the gases (in atm) are $p_{A1} = 0.4$,\n$p_{B1} = 0.4$ and $p_{C1} = 0.2$, and at position $z_2 = 0.005$ m, $p_{A2} = 0.1$, $p_{B2} = 0.6$ and\n$p_{C2} = 0.3$. The binary diffusivities are $D_{AB} = 2.02 \\times 10^{-5}\\ \\mathrm{m^2/s}$,\n$D_{AC} = 6.75 \\times 10^{-5}\\ \\mathrm{m^2/s}$ and $D_{BC} = 7.29 \\times 10^{-5}\\ \\mathrm{m^2/s}$. Calculate the\nflux of methane $N_A$ ($\\mathrm{kmol/(m^2\\,s)}$).\n","marks":10,"answer":[{"label":"D_A,mix","value":0.00002636,"units":"m^2/s","tol":0.02},{"label":"N_A","value":0.0000874,"units":"kmol m^-2 s^-1","tol":0.02}],"grade":"$undefined"}],"solution":"$59","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers (2023): (a) 3.55e-8 mol/s cm^2, (b) D_A,mix = 2.636e-5 m^2/s and\nN_A = 8.74e-5 kmol/m^2 s. All numbers recomputed in Python and agree. No official solutions exist.\n"},{"id":"engr263-2024-a2","module":"ENGR5002","week":9,"label":"ENGR263 2024 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["diffusion","mass-transfer","diffusivity","multicomponent","pressure-temperature-correction","liquid-diffusion","stagnant-gas"],"source":{"name":"ENGR263 Process Transfer of Mass 2024 Q A2","paper":"ENGR263","year":2024,"question":"A2","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2024/ENGR263 In-person Main Exam Paper Data 2024.pdf","solutionsPath":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2024/ENGR263 In-person Main Exam Paper Data 2024 answers.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (b).\n","formulaSheet":"- Multicomponent mixture $\\dfrac{1}{D_{1,mix}} = \\sum_{j \\ne 1}\\dfrac{y'_j}{D_{1j}}$, $y'_j = \\dfrac{y_j}{\\sum_{j \\ne 1}y_j}$ (on the formula sheet)\n- $D_{AB,T_2,P_2} = D_{AB,T_1,P_1}\\left(\\dfrac{P_1}{P_2}\\right)\\left(\\dfrac{T_2}{T_1}\\right)^{3/2}\\dfrac{\\Omega_{D,T_1}}{\\Omega_{D,T_2}}$ (on the formula sheet)\n- Flux with bulk flow $N_A = -cD_{AB}\\dfrac{dx_A}{dz} + x_A(N_A + N_B)$; total concentration $c = \\rho/M_{ave}$ (on the formula sheet)\n- Non-diffusing B in a liquid film: $N_A = \\dfrac{D_{AB}c_{av}}{z}\\ln\\dfrac{1 - x_{A2}}{1 - x_{A1}} = \\dfrac{D_{AB}c_{av}}{z\\,x_{BM}}(x_{A1} - x_{A2})$ (learn this)\n","parts":[{"label":"a","text":"Calculate the diffusion coefficient of ammonia (NH$_3$) in a mixture with the following composition:\n$y_{N_2} = 0.7$, $y_{O_2} = 0.1$ and $y_{NH_3} = 0.2$. The mixture is at $150^\\circ\\mathrm{C}$ and 2 bar. The\nfollowing data are provided for $T = 293$ K and $P = 1$ bar: $D_{N_2-NH_3} = 2.441 \\times 10^{-5}\\ \\mathrm{m^2/s}$\nand $D_{O_2-NH_3} = 2.563 \\times 10^{-5}\\ \\mathrm{m^2/s}$.\n","marks":12,"grade":"2:2","answer":[{"label":"D_mix (293 K, 1 bar)","value":0.00002456,"units":"m^2/s","tol":0.02},{"label":"D_mix (150 C, 2 bar)","value":0.0000213,"units":"m^2/s","tol":0.02}]},{"label":"b","text":"A very thin film of ethanol (A)-water (B) solution, with thickness of 0.4 cm, is at temperature of\n$20^\\circ\\mathrm{C}$ and one surface of the film is in contact with an organic liquid where the water is\ninsoluble. The ethanol concentration in the interface with the organic liquid is 6.8% w/w and in the other\nside of the film is 16.8% w/w, with densities of $0.9881\\ \\mathrm{g/cm^3}$ and $0.9712\\ \\mathrm{g/cm^3}$\nrespectively. If the diffusion coefficient of the ethanol in the organic liquid is\n$0.74 \\times 10^{-5}\\ \\mathrm{cm^2/s}$, then calculate in the steady state conditions the flux $N_A$. The\nmolecular weights of ethanol and water are $M_A = 46.05$ and $M_B = 18.02$ g/mol respectively.\n","marks":13,"grade":"2:1","answer":[{"label":"x_A1 (16.8%)","value":0.0732,"units":"-","tol":0.01},{"label":"x_A2 (6.8%)","value":0.0278,"units":"-","tol":0.01},{"label":"c_av","value":0.0505,"units":"mol/cm^3","tol":0.01},{"label":"N_A","value":4.47e-7,"units":"kmol m^-2 s^-1","tol":0.02}]}],"solution":"$5a","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers (2024) on every number: D_mix = 2.456e-5 m^2/s at 293 K, 1 bar and\n2.13e-5 m^2/s at 150 C, 2 bar; x_A1 = 0.0732, x_A2 = 0.0278, c_av = 0.05047 mol/cm^3, x_BM = 0.949,\nN_A = 4.47e-7 kmol/m^2 s (all recomputed in Python). One small slip in the worked answers' commentary: it says\nusing the unnormalised fractions 0.7 and 0.1 \"gives D_m = 2.95e-5\"; recomputing gives 3.07e-5 m^2/s at the\nreference state (2.66e-5 after correction), not 2.95e-5. The question's phrase \"diffusion coefficient of the\nethanol in the organic liquid\" is read, as in the worked answers, as the diffusivity of ethanol in the aqueous\nfilm. No official solutions exist.\n"},{"id":"engr263-2025-q2","module":"ENGR5002","week":9,"label":"ENGR263 2025 Q2","kind":"past-paper","level":3,"marks":52,"minutes":62,"grade":"2:1","tags":["diffusion","mass-transfer","stagnant-gas","diffusivity","multicomponent"],"source":{"name":"ENGR263 Mass Transfer Summer 2025 Q2","paper":"ENGR263","year":2025,"question":"Q2","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2025_/ENGR263 Main Exam Paper 2024-25.pdf","solutionsPath":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2025_/ENGR263 Main Exam Paper 2024-25 answers.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (b).\n","formulaSheet":"- Flux with bulk flow $N_{A,z} = -cD_{AB}\\dfrac{dy_A}{dz} + y_A(N_{A,z} + N_{B,z})$ (on the formula sheet)\n- Diffusion through a stagnant gas: $N_A = \\dfrac{D_{AB}P}{RTz}\\ln\\dfrac{P - p_{A2}}{P - p_{A1}} = \\dfrac{D_{AB}P}{RTz}\\,\\dfrac{p_{A1} - p_{A2}}{p_{BM}}$ (on the formula sheet in the current ENGR271 sheet; on the 2025 ENGR263 sheet it had to be derived from the flux equation)\n- Multicomponent stagnant mixture $\\dfrac{1}{D_{A,m}} = \\sum_{j \\ne A}\\dfrac{y'_j}{D_{Aj}}$, $y'_j$ on an A-free basis (on the formula sheet)\n","parts":[{"label":"a","text":"Oxygen (A) is diffusing through carbon monoxide (B) via a rectangular plane surface between positions 1 and\n2 under steady-state conditions, with the carbon monoxide non-diffusing. The total pressure is\n$1 \\times 10^5\\ \\mathrm{N/m^2}$ and the temperature $0^\\circ\\mathrm{C}$. The partial pressure of oxygen at\nplanes 1 and 2 is 13000 and 6500 $\\mathrm{N/m^2}$, respectively. The two planes are 2 mm apart. The\ndiffusivity for this mixture is $1.87 \\times 10^{-5}\\ \\mathrm{m^2/s}$. Calculate the rate of diffusion of\noxygen in kmol/s through each square meter of the two planes. The following data are provided:\n$R = 8314\\ \\mathrm{N\\,m/(kmol\\,K)}$.\n","marks":22,"grade":"2:2","answer":[{"label":"N_A","value":0.0000297,"units":"kmol m^-2 s^-1","tol":0.02}]},{"label":"b","text":"Calculate the rate of diffusion of oxygen (A) under steady-state conditions and the same experimental\nconditions as the above part (a), but this time assuming that the non-diffusing gas is a mixture of methane\n(B) and hydrogen (C) under a volume ratio 2:1. Comment on this result with respect to those in part (a).\nThe diffusivities of oxygen to hydrogen and methane are estimated to be $D_{O_2-H_2} = 6.99 \\times 10^{-5}$\nand $D_{O_2-CH_4} = 1.86 \\times 10^{-5}\\ \\mathrm{m^2/s}$, respectively.\n","marks":30,"grade":"2:1","answer":[{"label":"D_A,m","value":0.0000246,"units":"m^2/s","tol":0.02},{"label":"N_A","value":0.0000391,"units":"kmol m^-2 s^-1","tol":0.02}]}],"solution":"$5b","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers (2025) on every number: (a) prefactor 4.117e-4, log term 0.0720,\nN_A = 2.97e-5 kmol/m^2 s, p_BM = 90 210 Pa; (b) D_A,m = 2.46e-5 m^2/s, N_A = 3.91e-5 kmol/m^2 s, ratio 1.32\n(all recomputed in Python). No official solutions exist.\n"},{"id":"engr262-2018-b1","module":"ENGR5002","week":10,"label":"ENGR262 2018 Q B1(a)-(b)","kind":"past-paper","level":3,"marks":12,"minutes":14,"grade":"2:2","tags":["convective-mass-transfer","mass-transfer","two-film-theory","interfacial-composition","mass-transfer-coefficient","absorption"],"source":{"name":"ENGR262 Particle Technology and Separation 2018 Section B Q B1","paper":"ENGR262","year":2018,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR262_2018.pdf","tag":"PARTIAL","leftOut":"(c) minimum liquid flow rate (7 marks) and (d) number of theoretical stages (6 marks) are absorption-column\ndesign (operating lines, stage stepping), not on the current ENGR5002 syllabus.\n"},"text":"Answer ALL parts (a) - (d). [Parts (c)-(d), absorption-column design, are omitted.]\n\nThe equilibrium distribution of a solute A between gas and water at a particular temperature is given as\n$y_A = 1.2x_A$, where $x_A$ and $y_A$ stand for mole fractions of solute A in the liquid and in the gas phases,\nrespectively. At a certain point in a mass transfer device, the concentration of solute A in the bulk air is\n0.04 mole fraction and that in the bulk aqueous phase is 0.025. At the same point, the local individual mass\ntransfer coefficients in the gas and liquid phases are $k_y = 7.2\\ \\mathrm{kmol\\,h^{-1}\\,m^{-2}}(\\Delta y)^{-1}$\nand $k_x = 4.6\\ \\mathrm{kmol\\,h^{-1}\\,m^{-2}}(\\Delta x)^{-1}$.\n","formulaSheet":"- Two-film theory, flux continuity at the interface: $N_A = k_y(y - y_i) = k_x(x_i - x)$, with $y_i = mx_i$ (interface at equilibrium) (learn this)\n- Overall coefficients: $\\dfrac{1}{K_y} = \\dfrac{1}{k_y} + \\dfrac{m}{k_x}$, $N_A = K_y(y - y^*)$, $y^* = mx$ (learn this)\n","parts":[{"label":"a","text":"In which direction does the transport of solute A occur (from gas to liquid or from liquid to gas)? Explain\nyour answer.\n","marks":7,"grade":"third","answer":[{"label":"y* (equilibrium with bulk liquid)","value":0.03,"units":"-","tol":0.01}]},{"label":"b","text":"Determine interfacial concentrations in mole fractions in both the gas-phase and the liquid-phase. (Present\nyour results on a y-x graph.)\n","marks":5,"answer":[{"label":"x_i","value":0.0304,"units":"-","tol":0.02},{"label":"y_i","value":0.0365,"units":"-","tol":0.02},{"label":"N_A","value":0.025,"units":"kmol h^-1 m^-2","tol":0.02}],"grade":"$undefined"}],"solution":"$5c","confidence":"HIGH","solutionNote":"No official numerical answers for ENGR262 2018 Section B in the folder (the 2018 ENGR262 answer sheet covers\nSection A only). Values computed in Python and cross-checked with the overall-coefficient route.\n"},{"id":"engr262-2022-a2","module":"ENGR5002","week":10,"label":"ENGR262 2022 Part B Q A2(b)","kind":"past-paper","level":3,"marks":7,"minutes":8,"grade":"2:1","tags":["convective-mass-transfer","mass-transfer","two-film-theory","overall-mass-transfer-coefficient","absorption"],"source":{"name":"ENGR262 Separation Processes 2021-22 Part B Q A2","paper":"ENGR262","year":2022,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR262 Exam Part B 2021-22.pdf","tag":"PARTIAL","leftOut":"(a)(i)-(iii) (18 marks), selection and design of solid-liquid extractors, are separation-process design and\nnot on the current ENGR5002 syllabus.\n"},"text":"Answer ALL parts (a) - (b). [Part (a), 18 marks on solid-liquid extraction, is omitted.]\n","formulaSheet":"- Individual coefficients $N_A = k_G(p_{A,G} - p_{A,i}) = k_L(c_{A,i} - c_{A,L})$ (learn this)\n- Linear equilibrium (Henry) $p_A = Hc_A$; $p_A^* = Hc_{A,L}$ (learn this)\n- Result: $\\dfrac{1}{K_G} = \\dfrac{1}{k_G} + \\dfrac{H}{k_L}$ (learn this)\n","parts":[{"label":"b","text":"For an absorption process where the equilibrium relationship is a straight line over the range of\ncompositions considered, derive an expression relating the overall mass transfer coefficient, $K_G$, to the\nindividual mass transfer coefficients, $k_G$ and $k_L$, using a suitably labelled driving force diagram.\n($K_G$: overall mass transfer coefficient based on the gaseous phase; $k_L$: individual (interfacial) mass\ntransfer coefficient based on the liquid phase; $k_G$: individual (interfacial) mass transfer coefficient\nbased on the gaseous phase.)\n","marks":7,"grade":"$undefined"}],"solution":"$5d","confidence":"HIGH","solutionNote":"Derivation only; no official answers exist for this paper. Standard two-resistance (Whitman) result.\n"},{"id":"engr262-2025-q3","module":"ENGR5002","week":10,"label":"ENGR262 2025 Q3(b)","kind":"past-paper","level":3,"marks":9,"minutes":11,"grade":"2:2","tags":["convective-mass-transfer","mass-transfer","two-film-theory","interfacial-composition","absorption"],"source":{"name":"ENGR262 Particle Technology and Separation Summer 2025 Q3","paper":"ENGR262","year":2025,"question":"Q3","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2025_/ENGR262 Main Exam 2024-25.pdf","tag":"PARTIAL","leftOut":"(a) minimum water-to-air ratio (8 marks) and (c) outlet liquid composition at a given water rate (8 marks) are\nabsorption-column design and not on the current ENGR5002 syllabus.\n"},"text":"[Parts (a) and (c), absorption-column design, are omitted.]\n\nAn exhaust gas stream (100 kmol/h) from a chemical plant contains 15 mol% of a pollutant while the rest is\nair. An absorption column using pure water as the absorbent operates isothermally at $30^\\circ\\mathrm{C}$. The\nmeasured vapour-liquid equilibrium data at 30 C are:\n\n| Mole fraction of pollutant in liquid, $x$ | 0.020 | 0.060 | 0.100 | 0.140 | 0.180 | 0.220 | 0.260 |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n| Mole fraction of pollutant in vapour, $y$ | 0.010 | 0.031 | 0.053 | 0.075 | 0.099 | 0.124 | 0.149 |\n","formulaSheet":"- Two-film theory: $N_A = k_y(y - y_i) = k_x(x_i - x)$, with $(x_i, y_i)$ on the equilibrium curve (learn this)\n- Graphical construction: line of slope $-k_x/k_y$ from the bulk point $(x, y)$ to the equilibrium curve (learn this)\n","parts":[{"label":"b","text":"After taking samples at a certain point in the column, the technician measured the mole fraction of\npollutant in bulk gas and bulk liquid: $y = 0.11$ and $x = 0.15$, respectively. Find the interface mole\nfraction of pollutant in liquid and gas phases at that point in the column if the mass transfer coefficient\nfor pollutant in the gas phase is $3 \\times 10^{-3}\\ \\mathrm{kmol_A\\,s^{-1}\\,m^{-2}(mole\\ fraction)^{-1}}$ and in\nthe liquid phase is $4 \\times 10^{-1}\\ \\mathrm{kmol_A\\,s^{-1}\\,m^{-2}(mole\\ fraction)^{-1}}$. (Present your\nresults on a y-x graph.)\n","marks":9,"answer":[{"label":"x_i","value":0.1502,"units":"-","tol":0.01},{"label":"y_i","value":0.081,"units":"-","tol":0.03}],"grade":"$undefined"}],"solution":"$5e","confidence":"HIGH","solutionNote":"No official solutions or numerical answers exist for ENGR262 2025. Values computed in Python with linear and\nspline interpolation of the equilibrium table (they agree to 0.0002); the answer box tolerance allows for\ngraph reading.\n"},{"id":"engr217-2017-a1","module":"ENGR5002","week":2,"label":"ENGR217 2017 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["buoyancy-stability","metacentric-height","free-surface-effect","relative-density","stability"],"source":{"name":"ENGR217 Summer 2017 Q A1","paper":"ENGR217","year":2017,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR217_Exam_2017.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR217 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n","parts":[{"label":"a","text":"A car ferry of displacement 12500 tonne carries oil fuel in 8 tanks each 8 m long and 4 m broad,\nthe tanks arranged with their long dimension in the fore and aft directions.\nThe specific gravity of the oil is 0.87, and the density of the sea water in which the ship is\nfloating is $1090\\ \\mathrm{kg/m^3}$.\nBy how much is the metacentric height of the ship reduced by the oil contained in the tanks and is\nthis serious?\n","marks":10,"answer":[{"label":"i per tank","value":42.67,"units":"m^4","tol":0.02},{"label":"GM reduction (8 tanks)","value":23.76,"units":"mm","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"The car deck on the above car ferry is approximately rectangular, 20 m wide by 150 m long, and the\nmetacentric height in roll is 0.94 m.\nIf sea water of density $1090\\ \\mathrm{kg/m^3}$ is accidentally taken into the car deck, in quantity\nsufficient to cover the deck even when the boat heels, will the vessel become stable or unstable?\n","marks":10,"answer":[{"label":"V displaced","value":11468,"units":"m^3","tol":0.02},{"label":"GM reduction","value":8.72,"units":"m","tol":0.02},{"label":"GM effective","value":-7.78,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"If on the above car ferry, the car deck is divided down the middle, into two areas each of\n150 m x 10 m, would the vessel still be unstable if one of these areas took in water?\n","marks":5,"answer":[{"label":"GM reduction (one half)","value":1.09,"units":"m","tol":0.02},{"label":"GM effective","value":-0.15,"units":"m","tol":0.05}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Buoyancy / displacement: $F_B = \\rho_{sw} g V$, so $V = M/\\rho_{sw}$ (on the formula sheet)\n- Relative density: $RD = \\rho/\\rho_w$ (on the formula sheet)\n- Second moment of a rectangular free surface about its own fore-and-aft centreline: $i = \\dfrac{l b^3}{12}$ (on the formula sheet, shape table)\n- Free-surface effect (virtual rise of G): $\\Delta GM = \\dfrac{\\rho_{liq} \\sum i}{\\rho_{sw} V}$ (learn this)\n- Stability criterion: stable if $GM > 0$ (M above G) (learn this)\n","solution":"$5f","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers. (a) agrees: 42.67 m^4, 2.97 x 10^-3 m per\ntank, 23.76 mm total, not serious. (b) agrees: 11467.89 m^3, 100000 m^4, 8.72 m, unstable.\n(c) DISAGREES: the official sheet gives 12500 m^4 (agrees) but then about 0.125 m and \"stable\".\n0.125 is 12500/100000, the ratio of the two second moments (dimensionless), not a change in GM; the\ncorrect reduction is i/V = 12500/11468 = 1.09 m, which is also 8.72 m / 8 because i scales with the\nbreadth cubed. 1.09 m exceeds 0.94 m, so the vessel is still (marginally) unstable. Trust this\nsolution; confirm with a lecturer if this part matters to you.\n"},{"id":"engr217-2017-a2","module":"ENGR5002","week":4,"label":"ENGR217 2017 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["bernoulli","pipe-flow","continuity","mass-flow-rate","energy-equation","head-loss","friction-factor"],"source":{"name":"ENGR217 Summer 2017 Q A2","paper":"ENGR217","year":2017,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR217_Exam_2017.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR217 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\n[Old data sheet: the friction factor is the Fanning factor $f = \\tau_w / (\\tfrac{1}{2}\\rho u^2)$ with\n$H_L = \\dfrac{4 f L}{D}\\dfrac{u^2}{2g}$; the current module uses the Darcy factor $f_D = 4f$.]\n","figures":[{"src":"/figures/engr217-2017-a2.png","caption":"Figure A2a: tapered circular pipe, 200 mm diameter at inlet (left) reducing to 75 mm at outlet (right)."},{"src":"/figures/engr217-2017-a2-2.png","caption":"Figure A2c: Y-pipe with inlet 1 dividing into branches 2 and 3."},{"src":"/figures/engr217-2017-a2-3.png","caption":"Figure A2d: inclined tube from point 1 up to point 2, 75 m long along the tube, rising 25 m."}],"parts":[{"label":"a","text":"Water is flowing from left to right through the tapered circular pipe shown below (see Figure A2a).\nThe mean velocity of the water at the outlet is 4.5 m/s. What is the mean velocity of the water at\nthe inlet?\n","marks":3,"answer":[{"label":"v inlet","value":0.633,"units":"m/s","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"Liquid of density $920\\ \\mathrm{kg/m^3}$ is flowing along a cylindrical pipe of internal diameter\n250 mm. A container of mass 2.5 kg is used to catch the liquid emerging from the pipe. After a period\nof 20.0 s the total mass of the container and the liquid is 50.0 kg. Find:\n\n(i) The mass flow rate of the liquid (3 marks)\n\n(ii) The volume flow rate of the liquid (3 marks)\n\n(iii) The mean velocity of flow of the liquid along the pipe (3 marks)\n","marks":9,"answer":[{"label":"mdot","value":2.375,"units":"kg/s","tol":0.02},{"label":"Q","value":0.00258,"units":"m^3/s","tol":0.02},{"label":"v","value":0.0526,"units":"m/s","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Water enters the Y-pipe (see Figure A2c) at point 1 at a rate 5.4 L/s, and leaves through branches 2\nand 3. The flow rate through branch 3 is 1.8 L/s. Find the mean velocity of the flow through\nbranch 2, whose internal diameter is 50 mm.\n","marks":5,"answer":[{"label":"v2","value":1.83,"units":"m/s","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"A fluid of density $1180\\ \\mathrm{kg/m^3}$ flows from point 1 to point 2 along the 50 mm internal\ndiameter tube at a velocity of 3.8 m/s (see Figure A2d). The friction factor is $f = 0.0068$. The\ngauge pressure at point 2 is to be $2.0 \\times 10^5\\ \\mathrm{N\\,m^{-2}}$. Find the gauge pressure\nneeded at point 1.\n","marks":8,"answer":[{"label":"friction pressure drop","value":347600,"units":"Pa","tol":0.02},{"label":"p1 gauge","value":837000,"units":"Pa","tol":0.02}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Continuity: $Q = vA$, $\\dot m = \\rho Q = \\rho A v$ (on the formula sheet)\n- Mass balance at a junction: $\\sum \\dot m_{in} = \\sum \\dot m_{out}$ (on the formula sheet)\n- Energy equation with head loss: $\\dfrac{v_1^2}{2g} + z_1 + \\dfrac{p_1}{\\rho g} = \\dfrac{v_2^2}{2g} + z_2 + \\dfrac{p_2}{\\rho g} + h_L$ (on the formula sheet)\n- Darcy-Weisbach: $h_L = f_D \\dfrac{L}{D}\\dfrac{v^2}{2g}$ (on the formula sheet)\n- Fanning to Darcy: $f_D = 4 f_{Fanning}$ (learn this)\n","solution":"$60","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: all agree. (a) 0.63 m/s; (b) 2.375 kg/s,\n2.58 x 10^-3 m^3/s, 0.053 m/s; (c) 1.83 m/s; (d) friction term 3.48 x 10^5 N/m^2 and\np1 = 8.37 x 10^5 N/m^2. The official answer confirms the old f = 0.0068 is a Fanning factor\n(Darcy 0.0272).\n"},{"id":"engr217-2017-a3","module":"ENGR5002","week":2,"label":"ENGR217 2017 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["surface-forces","curved-surface","hydrostatics","centre-of-pressure","triangle-of-forces","equilibrium"],"source":{"name":"ENGR217 Summer 2017 Q A3","paper":"ENGR217","year":2017,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR217_Exam_2017.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR217 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (e).\n\nThe ends of a water tank (see Figure A3 below) are of radius 20 m and 36 m long, in the form of\ncircular quadrants. We wish to determine the force exerted by the water on one end of the tank. To\ncalculate the force by direct summation or integration is laborious. It is far simpler to choose a\nsuitable volume of liquid (one pressing against the end of the tank and otherwise bounded by planes)\nand write down the conditions of equilibrium for it.\n\nTake the quadrant-shaped volume of water. The area of the quadrant is $\\tfrac{1}{4}\\pi r^2$. The\nforces on this water consist of its weight W, the force F acting on the vertical plane which forms the\nleft hand boundary, and the force R exerted by the inner surface of the tank wall.\n\n[Data Book: water $\\rho = 1000\\ \\mathrm{kg\\,m^{-3}}$.]\n","figures":[{"src":"/figures/engr217-2017-a3.png","caption":"Figure A3: water tank whose end is a circular quadrant of radius 20 m, 36 m long."}],"parts":[{"label":"a","text":"What is the volume of water in the quadrant shaped end of the tank?","marks":3,"answer":[{"label":"V","value":11310,"units":"m^3","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"What is the weight W?","marks":2,"answer":[{"label":"W","value":110.9,"units":"MN","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Find the force acting on the left of this volume of water.","marks":8,"answer":[{"label":"F","value":70.63,"units":"MN","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"Draw the triangle of forces and indicate magnitudes and direction.","marks":2,"answer":[{"label":"R","value":131.5,"units":"MN","tol":0.02},{"label":"angle of R to horizontal","value":57.5,"units":"deg","tol":0.02}],"grade":"$undefined"},{"label":"e","text":"Find where the force R acts.","marks":10,"answer":[{"label":"x of W from plane","value":8.49,"units":"m","tol":0.02},{"label":"depth of F","value":13.33,"units":"m","tol":0.02},{"label":"R meets wall: x from plane","value":10.74,"units":"m","tol":0.02},{"label":"R meets wall: depth","value":16.87,"units":"m","tol":0.02}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Weight: $W = \\rho g V$ (on the formula sheet, $F_g = mg$)\n- Force on a plane surface: $F = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\, l_{CG}}$, rectangle $I_G = \\dfrac{b h^3}{12}$ (on the formula sheet)\n- Quadrant centroid from a straight edge: $\\dfrac{4r}{3\\pi}$ (on the formula sheet, semicircle row of the shape table)\n- Curved surface by equilibrium of a fluid volume: $R = \\sqrt{W^2 + F^2}$, $\\tan\\theta = W/F$ (learn this)\n- Three non-parallel forces in equilibrium are concurrent (learn this)\n","solution":"$61","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: agree. Official: V = 11304 m^3 and\nW = 110892.24 kN (they used pi = 3.14; with full pi, 11310 m^3 and 110.95 MN), mean pressure\n98.1 kN/m^2 and F = 70632 kN, and for (e) the lever arm of W 8.468 m (true value 4r/(3 pi) = 8.488 m;\nprobably a slip), Ix = 24000 m^4 and q = 3.33 m. The official sheet does not list R; the answer above\nadds R = 131.5 MN at 57.5 deg through the centre of curvature. The same question is repeated as\nENGR217 2019 Q B2.\n"},{"id":"engr217-2018-b1","module":"ENGR5002","week":4,"label":"ENGR217 2018 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["pipe-flow","reynolds","moody","friction-factor","poiseuille","laminar-turbulent"],"source":{"name":"ENGR217 Summer 2018 Q B1","paper":"ENGR217","year":2018,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR217_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/217 - 2018 answers.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n\n[A Moody diagram is needed for (b). The old data sheet defines the Fanning friction factor,\n$H_L = \\dfrac{4 f L}{D}\\dfrac{u^2}{2g}$; the current module uses the Darcy factor $f_D = 4f$.]\n","parts":[{"label":"a","text":"A pipeline 1200 mm in diameter carries a flow of methane gas at $15^\\circ\\mathrm{C}$ and gauge\npressure 3.5 bar. The density of the gas is $2.70\\ \\mathrm{kg/m^3}$ and its viscosity is\n$1.15 \\times 10^{-5}\\ \\mathrm{N\\,s/m^2}$. The mean velocity of flow of the gas is 0.90 m/s.\nIs the flow laminar or turbulent?\n","marks":5,"answer":[{"label":"Re","value":254000,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"Find the friction factor for the condition given above, and estimate the pressure drop in a 250 m\nlength of the pipeline. The roughness value of the inner surface of the pipeline is $k = 0.35$ mm.\n","marks":10,"answer":[{"label":"f Darcy","value":0.0172,"units":"-","tol":0.05},{"label":"dp","value":3.92,"units":"Pa","tol":0.05}],"grade":"$undefined"},{"label":"c","text":"Diesel fuel flows along a pipe 6 m long, whose bore (internal diameter) is 5 mm. The density and\nviscosity of the fuel are $850\\ \\mathrm{kg/m^3}$ and $1.8 \\times 10^{-3}\\ \\mathrm{N\\,s/m^2}$\nrespectively. If the pressure drop over the length of the pipe is $1.3\\ \\mathrm{kN/m^2}$, estimate the\nvolume flow rate of fuel along the pipe. (Hint: Poiseuille's equation:\n$Q = \\Delta P\\, \\pi R^4 / (8 \\mu L)$.)\n","marks":10,"answer":[{"label":"Q","value":0.00000185,"units":"m^3/s","tol":0.02},{"label":"v","value":0.094,"units":"m/s","tol":0.02},{"label":"Re check","value":222,"units":"-","tol":0.02}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Reynolds number: $Re = \\dfrac{\\rho v D}{\\mu}$; laminar $Re < 2400$, turbulent $Re > 4000$ (on the formula sheet)\n- Darcy-Weisbach: $\\Delta p = f_D \\dfrac{L}{D}\\dfrac{\\rho v^2}{2}$ (on the formula sheet)\n- Moody chart / Colebrook: $\\dfrac{1}{\\sqrt{f_D}} = -2\\log_{10}\\left(\\dfrac{k/D}{3.7} + \\dfrac{2.51}{Re\\sqrt{f_D}}\\right)$ (learn this; a Moody chart would be supplied)\n- Hagen-Poiseuille: $Q = \\dfrac{\\Delta p\\, \\pi R^4}{8 \\mu L}$, equivalently $\\Delta p = \\dfrac{32 \\mu L v}{D^2}$ (on the formula sheet)\n- Fanning to Darcy: $f_D = 4 f_{Fanning}$ (learn this)\n","solution":"$62","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: (a) Re = 2.54 x 10^5, turbulent, agrees.\n(b) official dp = 4.01 N/m^2 against 3.92 here: the 2 % difference is the Moody-chart reading\n(official reading about f_D = 0.0176; Colebrook gives 0.0172), within chart accuracy. (c) official\nQ = 1.85 cm^3/s and Re = 222, agrees. The ENGR266 2018 paper carries the same question\n(word-for-word identical fluids section).\n"},{"id":"engr217-2018-b2","module":"ENGR5002","week":2,"label":"ENGR217 2018 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["buoyancy-stability","metacentric-height","heel-test","centre-of-buoyancy","free-surface-effect","relative-density"],"source":{"name":"ENGR217 Summer 2018 Q B2","paper":"ENGR217","year":2018,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR217_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/217 - 2018 answers.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n","figures":[{"src":"/figures/engr217-2018-b2.png","caption":"Figure B2 (b): semi-submerged block, a 12.0 m cube floating with 6.0 m submerged; G is a distance y below the geometric centre."}],"parts":[{"label":"a","text":"To find the metacentric height in roll of his 9.9 tonne vessel, a yachtsman uses a heavy container\nof mass 150 kg. When placed at one side of the deck, the container causes the yacht to heel by\n$4.2^\\circ$. The container is then moved across the deck to the opposite side, and produces a heel of\n$4.2^\\circ$ the other way. The overall width of the deck is 9.5 m.\nWhat is the metacentric height in roll?\n","marks":8,"answer":[{"label":"GM","value":0.98,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"A block in the form of a cube of side 12.0 m floats with half its height submerged as shown in\nFigure B2 (b). The mass of the block is not uniformly distributed: the centre of gravity G is below\nthe geometrical centre by a distance y.\n\n(i) What is the height of the centre of buoyancy B above the bottom of the block? (2 marks)\n\n(ii) Find the value of the metacentric radius, BM, and so find the height of the metacentre M above\nthe bottom of the block. (5 marks)\n\n(iii) If the block is to float stably as shown, what is the minimum distance y by which the centre\nof gravity G is below the geometric centre of the block? (2 marks)\n","marks":9,"answer":[{"label":"height of B","value":3,"units":"m","tol":0.02},{"label":"BM","value":2,"units":"m","tol":0.02},{"label":"height of M","value":5,"units":"m","tol":0.02},{"label":"y min","value":1,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"A ship of displacement 7000 tonne carries oil fuel in four tanks 10.0 m long and 5.0 m broad, the\ntanks arranged with their length in the fore-and-aft direction (lengthwise of the ship). The specific\ngravity of the oil is 0.84, and the density of sea-water in which the ship is floating is\n$1020\\ \\mathrm{kg/m^3}$. By how much is the metacentric height of the ship reduced by the oil\ncontained in the tanks?\n","marks":8,"answer":[{"label":"GM reduction","value":50,"units":"mm","tol":0.02}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Buoyancy: $F_B = \\rho g V_{displaced}$ (on the formula sheet)\n- Relative density: $RD = \\rho/\\rho_w$ (on the formula sheet)\n- Rectangle second moment: $I = \\dfrac{b d^3}{12}$ (on the formula sheet, shape table)\n- Metacentric radius: $BM = \\dfrac{I}{V}$; $GM = BM - BG$ (learn this)\n- Heel (inclining) test: $GM = \\dfrac{w x}{W \\tan\\theta}$ (learn this)\n- Free-surface effect: $\\Delta GM = \\dfrac{\\rho_{liq} \\sum i}{\\rho_{sw} V}$ (learn this)\n","solution":"$63","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: all agree. (a) GM = 0.98 m; (b) B 3.0 m\nabove the bottom, BM = 2 m, M 5 m above the bottom, y at least 1 m; (c) reduction 50 mm, not serious.\nThe ENGR266 2018 paper carries the same question (word-for-word identical fluids section).\n"},{"id":"engr217-2018-b3","module":"ENGR5002","week":4,"label":"ENGR217 2018 Q B3","kind":"past-paper","level":3,"marks":16,"minutes":19,"tags":["external-flow","drag","drag-coefficient","power","wind-tunnel"],"source":{"name":"ENGR217 Summer 2018 Q B3","paper":"ENGR217","year":2018,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR217_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/217 - 2018 answers.pdf","tag":"PARTIAL","leftOut":"Parts (a) (7 marks) and (b) (2 marks) are omitted: (a) asks for the wind-tunnel air speed of a\none-fifth scale model by Reynolds-number similarity and (b) asks whether a 75 m/s tunnel can still\ngive a fair drag coefficient. Dynamic similarity / dimensional analysis is not on the current ENGR5002\nsyllabus (tagged CHECK in the audit). Parts (c)-(e) do not need their answers; the stem data they use\nis kept below.\n"},"text":"Answer ALL parts (c) - (e). [Parts (a) and (b), on dynamic similarity, are omitted.]\n\nA road haulage truck has an approximate rectangular frontal area 2.5 m wide by 3.8 m high. To\ninvestigate the drag on the truck, a one-fifth scale model is to be tested in a wind tunnel. We can\nassume that the density and viscosity of the air are the same for the model as for the full size truck.\nThe truck travels at 120 km/h. [From part (b): the wind tunnel available is only capable of generating\nan air speed of 75 m/s.]\n\n[Data Book: air at $15^\\circ\\mathrm{C}$, $\\rho = 1.22\\ \\mathrm{kg\\,m^{-3}}$.]\n","parts":[{"label":"c","text":"For the one fifth scale model tested at an air speed of 75 m/s the drag force is 700 N. In these\nconditions, what is the drag coefficient?\n","marks":5,"answer":[{"label":"C_D","value":0.537,"units":"-","tol":0.03}],"grade":"$undefined"},{"label":"d","text":"So if we assume that the drag coefficient for the full size truck has approximately the same value\nas (c) above, what drag force will it experience when traveling at 120 km/h?\n","marks":5,"answer":[{"label":"F_D at 120 km/h","value":3.46,"units":"kN","tol":0.02}],"grade":"$undefined"},{"label":"e","text":"If the vehicle speed is reduced to 100 km/h, calculate the associated reduction in power required to\novercome the drag force, commenting on the result.\n","marks":6,"answer":[{"label":"P at 120 km/h","value":115.2,"units":"kW","tol":0.02},{"label":"P at 100 km/h","value":66.7,"units":"kW","tol":0.02},{"label":"power reduction","value":48.5,"units":"kW","tol":0.02}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Drag force: $F_D = C_D \\tfrac{1}{2}\\rho v^2 A$, A = frontal area for road vehicles (learn this; it was on the old data sheet)\n- Power to overcome drag: $P = F_D v$ (learn this)\n","solution":"$64","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: (c) official C_D = 0.55 against 0.537\nhere, because the official answer uses air density 1.2 kg/m^3 and rounds (0.546 to 0.55); the Data\nBook value is 1.22. (d) official 3.48 kN against 3.46 kN (from carrying the rounded 0.55; with an\nunrounded C_D the force is 3457 N whatever density is used). (e) the official sheet lists only\nP = 115.88 kW (the drag power at 120 km/h; 115.2 kW here), not the reduction. Agreement within\nrounding. Parts (a)-(b) official: model speed 166.5 m/s; Re matters little for bluff bodies so 75 m/s\nis fine. The ENGR266 2018 paper carries the same question (word-for-word identical fluids section).\n"},{"id":"engr217-2019-b1","module":"ENGR5002","week":1,"label":"ENGR217 2019 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["surface-forces","hydrostatics","centroid","second-moment-of-area","parallel-axis","centre-of-pressure"],"source":{"name":"ENGR217 Summer 2019 Q B1","paper":"ENGR217","year":2019,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR217_Exam_2019 (2).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/217 - 2019 answer.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\n[Data Book: water $\\rho = 1000\\ \\mathrm{kg\\,m^{-3}}$.]\n","figures":[{"src":"/figures/engr217-2019-b1.png","caption":"Figure B1-1: vertical tank end made of an 11.6 m wide x 6.6 m deep rectangle (water surface at its top edge) above a triangle 3.6 m deep with its apex downward; C1, C2 and C are the centroids, O-O the horizontal axis through C."}],"parts":[{"label":"a","text":"Find the depths $y_1$ and $y_2$, to the centroids $C_1$ and $C_2$ of the rectangle and triangle shown\nin Figure B1-1 below, and then find the depth $y$ to the centroid C of the whole figure.\n","marks":5,"answer":[{"label":"y1","value":3.3,"units":"m","tol":0.02},{"label":"y2","value":7.8,"units":"m","tol":0.02},{"label":"y","value":4.264,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"The plane figure shown forms the vertical end of the water tank. What is the magnitude of the force\nexerted on the plane by the water?\n","marks":5,"answer":[{"label":"F","value":4.076,"units":"MN","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Find the second moment of area of the figure about the horizontal axis O-O through the centroid C.\n(This will involve the use of the parallel axis theorem in both the triangle and the rectangle).\n","marks":10,"answer":[{"label":"I_OO","value":625.2,"units":"m^4","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"Finally find the position of the centre of pressure P of the complete figure.","marks":5,"answer":[{"label":"P below C","value":1.505,"units":"m","tol":0.02},{"label":"depth of P","value":5.769,"units":"m","tol":0.02}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Force on a plane surface: $F = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\, l_{CG}}$ (on the formula sheet)\n- Shape table: rectangle $A = bh$, $I_G = bh^3/12$; triangle $A = bh/2$, centroid $h/3$ from the base, $I_G = bh^3/36$ (on the formula sheet)\n- Composite centroid: $\\bar y = \\dfrac{\\sum A_i y_i}{\\sum A_i}$ (learn this)\n- Parallel axis theorem: $I = I_G + A d^2$ (learn this)\n","solution":"$65","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: all agree. Official y = 4.264 m,\nF = 4075.90 kN (4076 kN here; rounding), I_C = 625.164 m^4, q = 1.505 m. The ENGR266 2019 paper carries\nthe same question (word-for-word identical fluids section).\n"},{"id":"engr217-2019-b2","module":"ENGR5002","week":2,"label":"ENGR217 2019 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["surface-forces","curved-surface","hydrostatics","centre-of-pressure","triangle-of-forces","equilibrium"],"source":{"name":"ENGR217 Summer 2019 Q B2","paper":"ENGR217","year":2019,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR217_Exam_2019 (2).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/217 - 2019 answer.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (e).\n\nThe ends of a water tank shown in Figure B2-1 below, are of radius 20 m and 36 m long, in the form of\ncircular quadrants. We wish to determine the force exerted by the water on one end of the tank. To\ncalculate the force by direct summation or integration is laborious. It is far simpler to choose a\nsuitable volume of liquid (one pressing against the end of the tank and otherwise bounded by planes)\nand write down the conditions of equilibrium for it.\n\nTake the quadrant-shaped volume of water. The area of the quadrant is $\\tfrac{1}{4}\\pi r^2$. The\nforces on this water consist of its weight W, the force F acting on the vertical plane which forms the\nleft hand boundary, and the force R exerted by the inner surface of the tank wall.\n\n[Data Book: water $\\rho = 1000\\ \\mathrm{kg\\,m^{-3}}$.]\n","figures":[{"src":"/figures/engr217-2019-b2.png","caption":"Figure B2-1: water tank whose end is a circular quadrant of radius 20 m, 36 m long."}],"parts":[{"label":"a","text":"What is the volume of water in the quadrant shaped end of the tank?","marks":3,"answer":[{"label":"V","value":11310,"units":"m^3","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"What is the weight W?","marks":2,"answer":[{"label":"W","value":110.9,"units":"MN","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Find the force acting on the left of this volume of water.","marks":8,"answer":[{"label":"F","value":70.63,"units":"MN","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"Draw the triangle of forces and indicate magnitudes and direction.","marks":2,"answer":[{"label":"R","value":131.5,"units":"MN","tol":0.02},{"label":"angle of R to horizontal","value":57.5,"units":"deg","tol":0.02}],"grade":"$undefined"},{"label":"e","text":"Find where the force R acts.","marks":10,"answer":[{"label":"x of W from plane","value":8.49,"units":"m","tol":0.02},{"label":"depth of F","value":13.33,"units":"m","tol":0.02},{"label":"R meets wall: x from plane","value":10.74,"units":"m","tol":0.02},{"label":"R meets wall: depth","value":16.87,"units":"m","tol":0.02}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Weight: $W = \\rho g V$ (on the formula sheet, $F_g = mg$)\n- Force on a plane surface: $F = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\, l_{CG}}$, rectangle $I_G = \\dfrac{b h^3}{12}$ (on the formula sheet)\n- Quadrant centroid from a straight edge: $\\dfrac{4r}{3\\pi}$ (on the formula sheet, semicircle row of the shape table)\n- Curved surface by equilibrium of a fluid volume: $R = \\sqrt{W^2 + F^2}$, $\\tan\\theta = W/F$ (learn this)\n- Three non-parallel forces in equilibrium are concurrent (learn this)\n","solution":"$66","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: agree. Official V = 11304 m^3 and\nW = 110892.24 kN (pi taken as 3.14; 11310 m^3 and 110.95 MN with full pi), F = 70632 kN (horizontal),\nW vertical, and for (e) q = 3.33 m (centre of pressure of F below the centroid of the vertical plane).\nThe official sheet does not list R; the answer above adds R = 131.5 MN at 57.5 deg through the centre\nof curvature. This is a repeat of ENGR217 2017 Q A3 (same numbers), and the ENGR266 2019 paper carries\nthe same question (word-for-word identical fluids section).\n"},{"id":"engr217-2019-b3","module":"ENGR5002","week":4,"label":"ENGR217 2019 Q B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["pipe-flow","reynolds","laminar-turbulent","moody","friction-factor","iteration","flow-rate"],"source":{"name":"ENGR217 Summer 2019 Q B3","paper":"ENGR217","year":2019,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR217_Exam_2019 (2).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/217 - 2019 answer.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n\n[A Moody diagram is needed for (c). Data Book: water $\\rho = 1000\\ \\mathrm{kg\\,m^{-3}}$. The old data\nsheet uses the Fanning friction factor, $H_L = \\dfrac{4 f L}{D}\\dfrac{u^2}{2g}$; the current module uses\nthe Darcy factor $f_D = 4f$.]\n","parts":[{"label":"a","text":"When a fluid flows steadily along a pipe, there are two possible patterns of flow. Who originally\ninvestigated these flow patterns? Describe these two flow patterns, and explain how you could\npredict which pattern of flow is likely to occur in a given situation.\n","marks":6,"grade":"$undefined"},{"label":"b","text":"Hot water at $50^\\circ\\mathrm{C}$ is flowing along a pipe 800 mm diameter at a mean speed of 2.3 m/s.\nAt this temperature the kinematic viscosity of the water is $\\nu = 5.55 \\times 10^{-7}\\ \\mathrm{m^2/s}$.\nWill the flow be laminar or turbulent?\n","marks":4,"answer":[{"label":"Re","value":3320000,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Water at $12^\\circ\\mathrm{C}$ is supplied to a 40 mm diameter steel pipe at 9 bar gauge pressure. The\npipe is 125 m long, the roughness is $k = 0.02$ mm, and at the end of the pipe the water is discharged\nto atmosphere. Estimate the volume flow rate of the water.\n(Kinematic viscosity for the water is $\\nu = 1.22 \\times 10^{-6}\\ \\mathrm{m^2/s}$).\n","marks":15,"answer":[{"label":"f Darcy","value":0.019,"units":"-","tol":0.05},{"label":"v","value":5.51,"units":"m/s","tol":0.05},{"label":"Q","value":0.00692,"units":"m^3/s","tol":0.05}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Reynolds number: $Re = \\dfrac{\\rho v D}{\\mu} = \\dfrac{v D}{\\nu}$; laminar $Re < 2400$, turbulent $Re > 4000$ (on the formula sheet)\n- Darcy-Weisbach: $\\Delta p = f_D \\dfrac{L}{D}\\dfrac{\\rho v^2}{2}$ (on the formula sheet)\n- Moody chart / Colebrook: $\\dfrac{1}{\\sqrt{f_D}} = -2\\log_{10}\\left(\\dfrac{k/D}{3.7} + \\dfrac{2.51}{Re\\sqrt{f_D}}\\right)$ (learn this; a Moody chart would be supplied)\n- Fanning to Darcy: $f_D = 4 f_{Fanning}$ (learn this)\n","solution":"$67","confidence":"HIGH","solutionNote":"Compared after solving with the official numerical answers: (a) laminar Re < 2000, turbulent above,\nagrees; (b) Re = 3315315, turbulent, agrees; (c) official Q = 6.81 L/s against 6.92 L/s here. The\n1.6 % difference is within Moody-chart reading accuracy (6.81 L/s corresponds to reading\nf_D of about 0.0196, Fanning 0.0049, against Colebrook's 0.0190); tolerance set to 5 %. The ENGR266 2019\npaper carries the same question (word-for-word identical fluids section).\n"},{"id":"engr217-266-2021-q1","module":"ENGR5002","week":4,"label":"ENGR217/266 2021 Q1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["pipe-flow","reynolds","moody","friction-factor","poiseuille","laminar-turbulent"],"grade":"2:2","formulaSheet":"- Reynolds number: $Re = \\dfrac{\\rho v D}{\\mu} = \\dfrac{vD}{\\nu}$ (on the formula sheet)\n- Pipe-flow regimes: laminar $Re < 2400$, turbulent $Re > 4000$ (on the formula sheet)\n- Darcy-Weisbach: $\\Delta p = f \\dfrac{L}{D} \\dfrac{\\rho v^2}{2}$, $h_L = f \\dfrac{L}{D}\\dfrac{v^2}{2g}$ (on the formula sheet)\n- Moody chart / Colebrook: $\\dfrac{1}{\\sqrt f} = -2\\log_{10}\\left(\\dfrac{k/D}{3.7} + \\dfrac{2.51}{Re\\sqrt f}\\right)$ (learn this; the chart itself would be supplied)\n- Fanning to Darcy: $f_{Darcy} = 4 f_{Fanning}$, old sheet $H_L = \\dfrac{4 f_F L u^2}{2 g D}$ (learn this)\n- Hagen-Poiseuille: $Q = \\dfrac{\\pi R^4 \\Delta p}{8 \\mu L}$, $\\Delta p = \\dfrac{32 \\mu L v}{D^2}$ (on the formula sheet)\n- Flow rate: $Q = vA$ (on the formula sheet)\n","source":{"name":"ENGR217/266 Summer 2021 Q1","paper":"ENGR217/266","year":2021,"question":"Q1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR217-266 Exam 2021.pdf","tag":"VALID"},"text":"Answer all parts (a) - (c).\n\n[Data Book values, not printed in the question: ethyl alcohol at $15^\\circ\\mathrm{C}$,\n$\\rho = 789\\ \\mathrm{kg\\,m^{-3}}$, $\\mu = 1.323 \\times 10^{-3}\\ \\mathrm{N\\,s\\,m^{-2}}$. A Moody diagram is needed for (b).]\n","parts":[{"label":"a","text":"A pipeline 600 mm in diameter carries a flow of ethyl alcohol at $15^\\circ\\mathrm{C}$ and gauge pressure\n3.5 bar. The mean velocity of flow of the liquid is $4.50\\ \\mathrm{m\\,s^{-1}}$. Referring as appropriate to\nthe Data Book please estimate if the flow is laminar or turbulent. Please comment further on the value\nobtained.\n","marks":5,"answer":[{"label":"Re","value":1610000,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"Find the friction factor for the condition given above, and estimate the pressure drop in a 125 m length\nof the pipeline. The roughness value of the inner surface of the pipeline is $k = 0.35$ mm.\n","marks":10,"answer":[{"label":"f (Darcy)","value":0.0176,"units":"-","tol":0.05},{"label":"dp over 125 m","value":29.3,"units":"kPa","tol":0.05}],"grade":"$undefined"},{"label":"c","text":"An n-Octane liquid, flows along a 1500 m long pipe, whose bore (internal diameter) is 0.2 m. The values\nfor the density and dynamic viscosity of the fuel are $702\\ \\mathrm{kg\\,m^{-3}}$ and\n$0.582 \\times 10^{-3}\\ \\mathrm{N\\,s\\,m^{-2}}$ respectively. If the pressure drop over the length of the pipe\n$2.0\\ \\mathrm{N\\,m^{-2}}$, estimate the volume flow rate of fuel along the pipe. (Hint: since the bore is\nsmall, assume the flow is laminar and use Poiseuille's equation ($Q = \\Delta p\\,\\pi R^4 / (8 \\mu l)$) to\nfind the flow rate. Then find the flow velocity, and hence the Reynolds number, to check that the flow is\nindeed laminar.)\n","marks":10,"answer":[{"label":"Q","value":0.00009,"units":"m^3/s","tol":0.02},{"label":"v","value":0.00286,"units":"m/s","tol":0.02},{"label":"Re","value":691,"units":"-","tol":0.02}],"grade":"$undefined"}],"solution":"$68","confidence":"HIGH","solutionNote":"No official solution for 2021. Solved independently; f from Colebrook (the Moody-chart value). Method\ncross-checked on the near-identical ENGR217 2018 Q B1: the same method gives Re = 2.54e5 (official 2.54e5),\nQ = 1.85 cm^3/s (official 1.85) and dp = 3.92 Pa (official 4.01, a Moody-chart reading of f, within chart\naccuracy). Data Book values (rho = 789, mu = 1.323e-3 for ethyl alcohol at 15 deg C) confirmed from the\nFluid Mechanics Data Book appendix printed in the ENGR217 2018/2019 papers. The old data sheet uses the\nFanning factor ($f_F = f/4 = 0.0044$); the solution uses the current Darcy notation.\n"},{"id":"engr217-266-2021-q2","module":"ENGR5002","week":2,"label":"ENGR217/266 2021 Q2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["buoyancy-stability","metacentric-height","centre-of-buoyancy","free-surface-effect","relative-density"],"grade":"2:1","formulaSheet":"- Heel (inclining) test: $GM = \\dfrac{w\\,x}{W \\tan\\theta}$ (learn this)\n- Metacentric radius: $BM = \\dfrac{I}{V}$, $I$ = second moment of the waterplane area about the roll axis (learn this)\n- Metacentric height: $GM = OB + BM - OG$; stable if $GM > 0$ (learn this)\n- Rectangle: $I_G = \\dfrac{b h^3}{12}$, centroid at $h/2$ (on the formula sheet)\n- Buoyancy: $F_B = \\rho_W g V_{displaced}$ (on the formula sheet)\n- Free-surface effect: reduction in $GM = \\dfrac{\\rho_{liquid} \\sum i}{\\rho_{sea} V}$, $i = \\dfrac{l b^3}{12}$ per tank (learn this)\n- Relative density: $RD = \\rho/\\rho_w$ (on the formula sheet)\n","source":{"name":"ENGR217/266 Summer 2021 Q2","paper":"ENGR217/266","year":2021,"question":"Q2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR217-266 Exam 2021.pdf","tag":"VALID"},"text":"Answer all parts (a) - (c).\n","figures":[{"src":"/figures/engr217-266-2021-q2.png","caption":"Figure 2: floating block 11 m x 11 m x 14 m, half submerged, with G a distance y below the geometric centre."}],"parts":[{"label":"a","text":"A yachtsman uses a heavy container of a total mass of 60 kg in order to find the metacentric height in\nroll of a 6.5 tonne vessel. When placed at one side of the deck, the container causes the yacht to heel\nby $3.2^\\circ$. Next, the container is moved across the deck to the opposite side and produces a heel of\n$3.2^\\circ$ the other way. The overall width of the deck is 18 m. What is the metacentric height in roll?\n","marks":8,"answer":[{"label":"GM","value":1.49,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"The dimensions of a solid body are 11 m by 11 m by 14 m long and is floating with half its height being\nsubmerged as shown on Figure 2. There is not a uniform distribution of the mass of this solid body and\nits centre of gravity G is below the geometrical centre by a distance y.\n\n(i) What is the height of the centre of buoyancy B above the bottom of the block? Please discuss the\ninfluence of the relative values of the centre of buoyancy. (2 marks)\n\n(ii) Find the value of the metacentric radius, BM, and so find the height of the metacentre M above the\nbottom of the block. Please comment on the significance of the metacenter M. (5 marks)\n\n(iii) If the block is to float stably as shown, what is the minimum distance y by which the centre of\ngravity G is below the geometric centre of the block? Please comment further on the significance of the\nvalue obtained and its relative effect on stability. (2 marks)\n","marks":9,"answer":[{"label":"B above bottom","value":3.5,"units":"m","tol":0.02},{"label":"BM","value":1.44,"units":"m","tol":0.02},{"label":"M above bottom","value":4.94,"units":"m","tol":0.02},{"label":"y min","value":2.06,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"A ship of displacement 36000 tonne carries oil fuel in four tanks 16.0 m long and 6.0 m broad, the tanks\narranged with their length in the fore and aft direction. The specific gravity of the oil is 0.84, and\nthe density of seawater in which the ship is floating is $1020\\ \\mathrm{kg\\,m^{-3}}$. For a ship of this\nsize, the metacentric height is at least 500 mm. What is the reduction of the metacentric height of the\nship because of the oil that is contained in the tanks and how serious is this reduction? Please discuss\nfurther your conclusions.\n","marks":8,"answer":[{"label":"GM reduction","value":26.9,"units":"mm","tol":0.02}],"grade":"$undefined"}],"solution":"$69","confidence":"HIGH","solutionNote":"No official solution for 2021. Solved independently, then the identical method was run on the\nnear-identical ENGR217 2018 Q B2 data and reproduces all its official numerical answers: GM = 0.98 m,\nB = 3.0 m, BM = 2 m, M = 5 m, y > 1 m, free-surface reduction 50 mm. Figure 2 shows the block as\n11 m x 11 m in plan and 14 m high with a 7 m draft, which is how the \"14 m long\" in the text is read.\nW in (a) is taken as the 6.5 t displacement (container included); using 6.56 t changes GM by 1%.\n"},{"id":"engr217-266-2022-a1","module":"ENGR5002","week":1,"label":"ENGR217/266 2022 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["hydrostatics","surface-forces","gate","centre-of-pressure","free-body-diagram","moments"],"grade":"2:1","formulaSheet":"- Hydrostatic pressure: $p(z) = p_0 + \\rho g z$ ($z$ = depth) (on the formula sheet)\n- Force on a plane surface: $F = p_{CG} A = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\, l_{CG}}$ (on the formula sheet)\n- Rectangle: $A = bh$, $I_G = \\dfrac{b h^3}{12}$ (on the formula sheet)\n- Moment equilibrium about the hinge: $\\sum M_O = 0$ (learn this)\n","source":{"name":"ENGR217/266 Summer 2022 Q A1","paper":"ENGR217/266","year":2022,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR217-266 Exam Part A 2021-22.pdf","tag":"VALID"},"text":"A tank is divided in two compartments as shown in Figure A1-1. The left compartment contains a fluid of\nunknown density $\\rho_f$ (darker area in the Figure A1-1), confined on the left by an L-shaped gate. A beam\nholding a mass of weight W is hinged to the gate as shown in the Figure A1-1. The rest of the gate is filled\nwith water (light blue region, seawater density $\\rho_{sw}$ is $1035\\ \\mathrm{kg\\,m^{-3}}$).\n","figures":[{"src":"/figures/engr217-266-2022-a1.png","caption":"Figure A1-1: two-compartment tank with an L-shaped gate, hinged beam (50 cm arm) and weight W; levels 30, 90, 50 and 100 cm."}],"parts":[{"label":"a","text":"Using the fluids levels indicated in the figure, determine the density of the unknown fluid $\\rho_f$ occupying the leftmost portion of the tank.","marks":6,"answer":[{"label":"rho_f","value":575,"units":"kg/m^3","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"Using the value of the density $\\rho_f$ computed in A1-(a), determine the force exerted by the fluid on the gate (assume that the width of the gate, i.e. its size in the direction normal to the paper is 2 m).","marks":6,"answer":[{"label":"F","value":4569,"units":"N","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Determine the location of the centre of pressure on the gate.","marks":6,"answer":[{"label":"cp depth below surface","value":0.6,"units":"m","tol":0.02},{"label":"cp distance below hinge","value":0.9,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"Draw a free body diagram incorporating the weight W and the hydrostatic force exerted by the fluid in the tank into the gate.","marks":4,"grade":"$undefined"},{"label":"e","text":"Determine the minimum value of the weight W that keeps the gate closed (gate does not rotate about the hinge).","marks":3,"answer":[{"label":"W min","value":8224,"units":"N","tol":0.02}],"grade":"$undefined"}],"solution":"$6a","confidence":"HIGH","solutionNote":"No official solution available. Geometry read from Figure A1-1: the gate is the vertical leg from the\nhinge down to the stop at the interface level (30 cm dry + 90 cm wetted by the unknown fluid); below that\nthe left wall is fixed tank wall, so only the unknown fluid loads the gate. Gate self-weight neglected\n(not given). Results checked by pressure-prism and numerical integration of the pressure moment.\n"},{"id":"engr217-266-2022-a2","module":"ENGR5002","week":3,"label":"ENGR217/266 2022 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["bernoulli","continuity","mass-balance","tee-junction"],"grade":"2:2","formulaSheet":"- Flow rate: $Q = vA$, $A = \\pi D^2/4$ (on the formula sheet)\n- Mass balance: $\\sum \\dot m_{in} - \\sum \\dot m_{out} = \\dfrac{dm_{CV}}{dt}$, $\\dot m = \\rho Q$ (on the formula sheet)\n- Bernoulli (no losses, $h_L = 0$): $\\dfrac{p_1}{\\rho g} + \\dfrac{v_1^2}{2g} + z_1 = \\dfrac{p_2}{\\rho g} + \\dfrac{v_2^2}{2g} + z_2$ (on the formula sheet)\n","source":{"name":"ENGR217/266 Summer 2022 Q A2","paper":"ENGR217/266","year":2022,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR217-266 Exam Part A 2021-22.pdf","tag":"VALID"},"text":"Water flows through a horizontal tee pipe junction as shown in Figure A2-1. The pipe containing section (1)\nis orthogonal to the pipe joining sections (2) and (3). Each of the three pipes joining the junction has an\ninternal diameter of 2.5 m. The water density can be assumed to be $\\rho_w = 1000\\ \\mathrm{kg\\,m^{-3}}$. The\nflow rate through section (3) is $Q_3 = 55\\ \\mathrm{m^3/s}$, the average velocity at section (1) is\n$V_1 = 5.5\\ \\mathrm{m/s}$ where the pressure is $p_1 = 450\\ \\mathrm{kPa}$.\n","figures":[{"src":"/figures/engr217-266-2022-a2.png","caption":"Figure A2-1: horizontal tee junction; inflow at section (1), outflow at (3), section (2) on the y axis."}],"parts":[{"label":"a","text":"Determine the water volume flow rate $Q_1$ at section (1) and the average velocity $V_3$ at section (3).","marks":6,"answer":[{"label":"Q1","value":27,"units":"m^3/s","tol":0.02},{"label":"V3","value":11.2,"units":"m/s","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"Use a mass balance to find the water volume flow rate $Q_2$ at section (2) and the corresponding average velocity $V_2$.","marks":5,"answer":[{"label":"Q2","value":28,"units":"m^3/s","tol":0.02},{"label":"V2","value":5.7,"units":"m/s","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Is the flow directed towards the positive or the negative y direction at (2)?","marks":4,"grade":"$undefined"},{"label":"d","text":"State the hypotheses that should be verified by the fluid flow for the Bernoulli's equation to be applicable.","marks":4,"grade":"$undefined"},{"label":"e","text":"Assuming that the hypotheses behind Bernoulli's equation are verified by the flow in the junction (see question A2-(d)), consider two streamlines joining respectively section (1) with sections (2) and (3) to determine the pressure at (2) and at (3).","marks":6,"answer":[{"label":"p2","value":448.9,"units":"kPa","tol":0.02},{"label":"p3","value":402.4,"units":"kPa","tol":0.02}],"grade":"$undefined"}],"solution":"$6b","confidence":"HIGH","solutionNote":"No official solution available. Solved independently; continuity checked by summing flows and the\nBernoulli pressures checked by applying the equation between (2) and (3) directly.\n"},{"id":"engr217-266-2023-a1","module":"ENGR5002","week":1,"label":"ENGR217/266 2023 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["hydrostatics","surface-forces","manometer","gate","centre-of-pressure","free-body-diagram"],"grade":"2:1","formulaSheet":"- Hydrostatic pressure: $p(z) = p_0 + \\rho g z$; equal pressure at equal levels in a continuous static fluid (on the formula sheet)\n- Gauge pressure: $p_g = p - p_{atm}$ (on the formula sheet)\n- Force on a plane surface: $F = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure along the plane: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\, l_{CG}}$ (on the formula sheet)\n- Rectangle: $I_{xc} = \\dfrac{b a^3}{12}$ (also given in the question) (on the formula sheet)\n- Depth of a point on an inclined plane: $z = l \\sin\\theta$ (learn this)\n- Moment equilibrium about the hinge: $\\sum M_A = 0$ (learn this)\n","source":{"name":"ENGR217/266 Summer 2023 Q A1","paper":"ENGR217/266","year":2023,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR217-266 - Fluid Mechanics Exam Paper FINAL 2023.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2023/ENGR217-266 - Fluid Mechanics - Numerical Answers 2023.pdf","tag":"VALID"},"text":"The U-tube manometer shown in Figure A1-1 has the two ends exposed to gauge pressures $p_{left}$ and\n$p_{right}$. The blue area is occupied by water (take water density to be $\\rho_w = 1000\\ \\mathrm{kg\\,m^{-3}}$),\nwhile the grey one is filled with a fluid of unknown properties.\n","figures":[{"src":"/figures/engr217-266-2023-a1.png","caption":"Figure A1-1: U-tube manometer (14 cm, 3.5 cm, 8.5 cm, 12.5 cm levels). Figure A1-2: gate hinged at A, inclined at 60 deg, held by a horizontal cable (1.8 m wetted, 2.4 m long)."}],"parts":[{"label":"a","text":"Supposing $p_{left}$ and $p_{right}$ to be equal to the atmospheric pressure, determine the density of the unknown fluid.","marks":3,"answer":[{"label":"rho unknown","value":1300,"units":"kg/m^3","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"Assuming the left end to be open ($p_{left}$ is the atmospheric pressure, i.e., $p_{left,gauge} = 0$) and $p_{right} = 15$ Pa (gauge), show that the unknown fluid has the same density as the water.","marks":2,"answer":[{"label":"p_right for rho = 1000","value":147,"units":"Pa","tol":0.02},{"label":"rho if p_right = 15 Pa","value":1269,"units":"kg/m^3","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"The rectangular gate shown in Figure A1-2 weighs 350 N. The gate, which has $b = 1.2$ m wide and\n$a = 2.4$ m long, is hinged in A and held in place by a horizontal cable. Taking water density to be\n$\\rho_w = 1000\\ \\mathrm{kg\\,m^{-3}}$, determine the hydrostatic force exerted by the water on the gate.\n","marks":4,"answer":[{"label":"F","value":16516,"units":"N","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"The location of the centre of pressure on the gate (i.e. the moment of inertia of a rectangle is $I_{xc} = b a^3/12$).","marks":4,"answer":[{"label":"l_cp from surface (along gate)","value":1.2,"units":"m","tol":0.02},{"label":"cp distance from hinge A","value":0.6,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"e","text":"Sketch the free body diagram.","marks":4,"grade":"$undefined"},{"label":"f","text":"Write down the momentum equilibrium about the hinge.","marks":4,"grade":"$undefined"},{"label":"g","text":"Determine the force exerted by the cable on the gate.","marks":4,"answer":[{"label":"cable force","value":4869,"units":"N","tol":0.02}],"grade":"$undefined"}],"solution":"$6c","confidence":"HIGH","solutionNote":"Compared with the official numerical answers after solving. Agreement on (a) 1300 kg/m^3, (d) y_cp = 1.2 m,\n(f) the same moment equation, and (c), (g) to within 0.1%: official F = 16527.89 N and F_c = 4872.23 N\ncome from rounding the centroid depth to 0.78 m; the exact 0.9 sin 60 deg = 0.7794 m gives 16516 N and\n4869 N. Disagreement on (b): the official answer says rho = 1000 kg/m^3 gives p_right = 15 Pa, but\nrho_w g (0.14 - 0.125) = 1000 x 9.81 x 0.015 = 147 Pa; 15 is rho h without g (i.e. 15 mm of water).\nWith the printed 15 Pa the unknown density would be 1269 kg/m^3, not 1000. I trust 147 Pa (a 1.5 cm water\nhead cannot be 15 Pa); the question statement contains this slip and is transcribed as printed.\n"},{"id":"engr217-266-2023-a2","module":"ENGR5002","week":3,"label":"ENGR217/266 2023 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["bernoulli","momentum","continuity","nozzle","bend","control-volume"],"grade":"2:1","formulaSheet":"- Mass flow rate: $\\dot m = \\rho A v$, $Q = vA$ (on the formula sheet)\n- Bernoulli (no losses, here $z_1 = z_2$): $\\dfrac{p_1}{\\rho g} + \\dfrac{v_1^2}{2g} + z_1 = \\dfrac{p_2}{\\rho g} + \\dfrac{v_2^2}{2g} + z_2$ (on the formula sheet)\n- Steady-flow momentum balance (vector, per direction): $\\sum F_{ext} = \\sum(\\dot m v)_{out} - \\sum(\\dot m v)_{in}$ (on the formula sheet)\n- Gauge pressure: $p_g = p - p_{atm}$ (on the formula sheet)\n","source":{"name":"ENGR217/266 Summer 2023 Q A2","paper":"ENGR217/266","year":2023,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR217-266 - Fluid Mechanics Exam Paper FINAL 2023.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2023/ENGR217-266 - Fluid Mechanics - Numerical Answers 2023.pdf","tag":"VALID"},"text":"Water circulates through the horizontal elbow shown in Figure A2 with a constant mass flow rate of\n$140\\ \\mathrm{kg\\,s^{-1}}$. The water reaching section 2 (the nozzle exit) is discharged at atmospheric\npressure. Taking the water density to be $\\rho_w = 1000\\ \\mathrm{kg\\,m^{-3}}$:\n","figures":[{"src":"/figures/engr217-266-2023-a2.png","caption":"Figure A2: the elbow and nozzle combination (section 1 diameter 30 cm, nozzle exit section 2 diameter 16 cm, flow reversed by 180 deg)."}],"parts":[{"label":"a","text":"Use the mass conservation principle to obtain the mean normal velocities at section 1 and 2 (the nozzle outlet).","marks":4,"grade":"third","answer":[{"label":"v_1","value":1.981,"units":"m/s","tol":0.02},{"label":"v_2","value":6.963,"units":"m/s","tol":0.02}]},{"label":"b","text":"By neglecting viscous losses, write down Bernoulli's equation between section 1 and section 2.","marks":4,"grade":"third"},{"label":"c","text":"Use the above derived Bernoulli's equation to determine the pressure at section 1.","marks":4,"grade":"2:2","answer":[{"label":"p_1 (gauge)","value":22.28,"units":"kPa","tol":0.02}]},{"label":"d","text":"Explain why the y component of the force $R_y$ required to hold the elbow in place is zero.","marks":6,"grade":"2:2"},{"label":"e","text":"Select an appropriate control volume and apply the momentum equation in the x direction to determine the x component of the force required to hold the elbow.","marks":7,"grade":"2:1","answer":[{"label":"R_x","value":-2827,"units":"N","tol":0.02}]}],"solution":"$6d","confidence":"HIGH","solutionNote":"Official numerical answers (linked PDF): $v_1 = 1.981$ m/s, $v_2 = 6.930$ m/s, $p_1 = 22\\,050.27$ Pa,\n$F = \\dot m(V_2 - V_1) - p_1A_1 = -865.65$ N. Part (a) $v_1$ agrees. The official $v_2 = 6.930$ m/s is an\narithmetic slip ($140/(1000 \\times \\pi \\times 0.08^2) = 6.963$ m/s, and $v_2/v_1$ must equal $(30/16)^2 = 3.516$);\nthis carries into $p_1$ (22.05 kPa official vs 22.28 kPa here, 1 percent). Part (e) disagrees: the official\nexpression uses $\\dot m(V_2 - V_1)$, which ignores that the nozzle jet leaves in $-x$ (the arrow on $V_2$\npoints left), so the outlet momentum flux must enter with a negative sign, giving $-\\dot m(V_1 + V_2)$. With\nthe official own $v_2$ and $p_1$ the correct expression gives $-2806$ N, still nowhere near $-865.65$ N.\nThis solution ($R_x = -2.83$ kN, acting right to left) is trusted: it follows the standard reversing-elbow\nresult and is checked by the physical argument in (e). Both agree on the direction (right to left).\nENGR217 2023 Section A is the same question.\n"},{"id":"engr217-2024-a1","module":"ENGR5002","week":5,"label":"ENGR217 2024 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["pipe-flow","hydrostatics","pumps-turbines","siphon","moody","manometer","pump-head","relative-density"],"grade":"2:1","formulaSheet":"- Continuity: $Q = vA$ (on the formula sheet)\n- Bernoulli with head loss: $\\dfrac{v_1^2}{2g} + z_1 + \\dfrac{p_1}{\\rho g} = \\dfrac{v_2^2}{2g} + z_2 + \\dfrac{p_2}{\\rho g} + h_L$ (on the formula sheet)\n- Darcy-Weisbach: $h_L = f\\,\\dfrac{L}{D}\\,\\dfrac{v^2}{2g}$, $f$ = Darcy factor (on the formula sheet)\n- Reynolds number: $Re = \\dfrac{\\rho v D}{\\mu}$ (on the formula sheet)\n- Moody chart / Colebrook: $\\dfrac{1}{\\sqrt f} = -2\\log_{10}\\left(\\dfrac{k/D}{3.7} + \\dfrac{2.51}{Re\\sqrt f}\\right)$ (learn this; the Moody chart is supplied with the paper)\n- Hydrostatics in a manometer: $p = p_0 + \\rho g z$, $RD = \\rho/\\rho_w$ (on the formula sheet)\n- Pump head: $H = \\left(\\dfrac{p}{\\rho g} + \\dfrac{v^2}{2g} + z\\right)_{out} - \\left(\\dfrac{p}{\\rho g} + \\dfrac{v^2}{2g} + z\\right)_{in}$ (learn this; printed on the 2024 data sheet, not on the current sheets)\n","source":{"name":"ENGR217 Summer 2024 Q A1","paper":"ENGR217","year":2024,"question":"A1","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2024/ENGR217 In-person Main Exam 2024 combined FINAL.pdf","tag":"VALID"},"text":"Answer all parts (a) - (b).\n\n[Data sheet values, not printed in the question: water at $15^\\circ\\mathrm{C}$, $\\rho = 1000\\ \\mathrm{kg\\,m^{-3}}$,\n$\\mu = 1.14 \\times 10^{-3}\\ \\mathrm{N\\,s\\,m^{-2}}$. The roughness of drawn tubing is not given; a typical\ntextbook value is $k \\approx 0.0015$ mm.]\n","figures":[{"src":"/figures/engr217-2024-a1.png","caption":"Figure A1-1: centrifugal pump with a differential mercury manometer between suction (1) and discharge (2)."}],"parts":[{"label":"a","text":"A farmer wishes to siphon $15^\\circ\\mathrm{C}$ water from a lake 1300 m from a field positioned a distance\nof 2.5 m below the surface of the lake. What size of drawn tubing should be selected if\n$0.5\\ \\mathrm{m^3}$ of water is required each minute? Use the Moody diagram and neglect all losses except\nthat due to wall friction.\n","marks":12,"grade":"2:1","answer":[{"label":"D (minimum)","value":0.143,"units":"m","tol":0.05},{"label":"f (Darcy)","value":0.0197,"units":"-","tol":0.05}]},{"label":"b","text":"The suction and discharge of a pump are connected to a differential mercury manometer. The diameter of\nthe suction pipe is 250 mm and the diameter of the discharge pipe is 200 mm. The specific gravity of the\npumped liquid is $S_1 = 0.82$ and the specific gravity of mercury is $S_2 = 13.6$. The manometer reading\nis $H = 650$ mm. Determine the pressure head rise across the pump if the discharge is\n$Q = 0.12\\ \\mathrm{m^3/s}$. As shown in Figure A1-1, the centreline of the suction pipe to the suction\nside of the mercury level on the manometer is distance $z_1$ and the centreline of the discharge pipe to\nthe discharge side of the mercury level of the manometer is distance $z_2$.\n","marks":13,"grade":"2:1","answer":[{"label":"piezometric head rise","value":10.13,"units":"m of liquid","tol":0.02},{"label":"total head rise H","value":10.57,"units":"m of liquid","tol":0.02}]}],"solution":"$6e","confidence":"MEDIUM","solutionNote":"No official solution available. ENGR266 2024 Section A is the same question. \"Specific gravity\" is\nrelative density. Note: the roughness of drawn tubing ($k = 0.0015$ mm) is in fact printed in the roughness\ntable on the 2024 data-sheet Moody chart, so the bracketed note in the stem matches the paper's own value.\nMEDIUM because there is no official answer and the wording of (b) is ambiguous: \"pressure head rise\" could mean\nthe piezometric head rise $H(S_2/S_1 - 1) = 10.13$ m or the total pump head including the velocity heads,\n10.57 m; the elevation difference of the centrelines is not given, so the pure $\\Delta p/\\gamma$ cannot be found\nunless the centrelines are level. Confirm with a lecturer which quantity is wanted. Part (a) also assumes the\ntube length equals the 1300 m distance; the answer (about 143 mm, select 150 mm) is insensitive to small changes\n($D \\propto L^{1/5}$).\n"},{"id":"engr217-2024-a2","module":"ENGR5002","week":6,"label":"ENGR217 2024 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["momentum","pumps-turbines","external-flow","bernoulli","nozzle","affinity-laws","lift","drag"],"grade":"2:1","formulaSheet":"- Continuity: $Q = v_1A_1 = v_2A_2$ (on the formula sheet)\n- Bernoulli: $\\dfrac{p_1}{\\rho g} + \\dfrac{v_1^2}{2g} + z_1 = \\dfrac{p_2}{\\rho g} + \\dfrac{v_2^2}{2g} + z_2$ (on the formula sheet)\n- Steady-flow momentum balance: $\\sum F_{ext} = (\\dot m v)_{out} - (\\dot m v)_{in}$ (on the formula sheet)\n- Pump similarity (affinity) laws: $\\dfrac{Q_2}{Q_1} = \\dfrac{N_2}{N_1}\\left(\\dfrac{D_2}{D_1}\\right)^3$, $\\dfrac{H_2}{H_1} = \\left(\\dfrac{N_2}{N_1}\\right)^2\\left(\\dfrac{D_2}{D_1}\\right)^2$ (learn this; printed on the 2024 data sheet, not on the current sheets)\n- Lift and drag: $F_L = C_L\\,\\tfrac{1}{2}\\rho u^2 A$, $F_D = C_D\\,\\tfrac{1}{2}\\rho u^2 A$ (learn this)\n- Power to overcome drag: $P = F_D\\,u$ (learn this)\n- Weight: $F_g = mg$ (on the formula sheet)\n","source":{"name":"ENGR217 Summer 2024 Q A2","paper":"ENGR217","year":2024,"question":"A2","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2024/ENGR217 In-person Main Exam 2024 combined FINAL.pdf","tag":"VALID"},"text":"Answer all parts (a) - (c).\n","parts":[{"label":"a","text":"A nozzle and hose are attached to the ladder of a fire truck. What force is needed to hold a nozzle\nsupplied by a 100 mm diameter hose with a pressure of 2800 kPa? The nozzle outlet diameter is 50 mm.\n","marks":10,"grade":"2:1","answer":[{"label":"v_1 (hose)","value":19.32,"units":"m/s","tol":0.02},{"label":"v_2 (jet)","value":77.29,"units":"m/s","tol":0.02},{"label":"F (holding force)","value":13.19,"units":"kN","tol":0.02}]},{"label":"b","text":"A manufacturer desires to triple both the discharge and head on a geometrically similar pump. Determine\nthe change in rotational speed and diameter under the new conditions.\n","marks":10,"grade":"2:2","answer":[{"label":"N2/N1","value":1.316,"units":"-","tol":0.02},{"label":"D2/D1","value":1.316,"units":"-","tol":0.02}]},{"label":"c","text":"An aircraft with a mass, including payload, of 1500 kg and density $0.45\\ \\mathrm{kg/m^3}$ is designed to\ncruise at a speed of 100 m/s. The effective wing area is approximately $20\\ \\mathrm{m^2}$. Determine the\nlift coefficient and the power required by the aerofoil during cruising. Assume a conventional aerofoil\nhaving a drag coefficient $C_D$ of 0.0065.\n","marks":5,"grade":"third","answer":[{"label":"C_L","value":0.327,"units":"-","tol":0.02},{"label":"P","value":29.25,"units":"kW","tol":0.02}]}],"solution":"$6f","confidence":"HIGH","solutionNote":"No official solution available. ENGR266 2024 Section A is the same question. The 2024 data sheet gave the\npump affinity laws ($Q \\propto N D^3$, $H \\propto N^2 D^2$ in dimensionless form) for part (b). Part (a) assumes\nthe 2800 kPa is gauge pressure at the nozzle inlet and water density 1000 kg/m^3; part (b) gives\n$N_2/N_1 = D_2/D_1 = 3^{1/4} = 1.316$. All three parts checked by an independent route (closed-form nozzle force,\nback-substitution and power ratio, $L/D$ sanity check).\n"},{"id":"engr217-2025-a1","module":"ENGR5002","week":4,"label":"ENGR217 2025 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["pipe-flow","moody","reynolds","poiseuille","laminar-turbulent"],"grade":"2:1","formulaSheet":"- Bernoulli with head loss: $\\dfrac{v_1^2}{2g} + z_1 + \\dfrac{p_1}{\\rho g} = \\dfrac{v_2^2}{2g} + z_2 + \\dfrac{p_2}{\\rho g} + h_L$ (on the formula sheet)\n- Darcy-Weisbach: $\\Delta p = f\\,\\dfrac{L}{D}\\,\\dfrac{\\rho v^2}{2}$, $h_L = f\\,\\dfrac{L}{D}\\,\\dfrac{v^2}{2g}$ ($f$ = Darcy factor) (on the formula sheet)\n- Reynolds number: $Re = \\dfrac{\\rho v D}{\\mu} = \\dfrac{vD}{\\nu}$; laminar below about 2300, turbulent above 4000 (on the formula sheet)\n- Moody chart / Colebrook: $\\dfrac{1}{\\sqrt f} = -2\\log_{10}\\left(\\dfrac{k/D}{3.7} + \\dfrac{2.51}{Re\\sqrt f}\\right)$ (learn this; the Moody chart is supplied with the paper)\n- Hagen-Poiseuille: $\\Delta p = \\dfrac{8\\mu L Q}{\\pi R^4}$, i.e. $Q = \\dfrac{\\pi\\,\\Delta p\\,R^4}{8\\mu L}$ (on the formula sheet)\n- Continuity: $Q = vA$ (on the formula sheet)\n","source":{"name":"ENGR217 Summer 2025 Q A1","paper":"ENGR217","year":2025,"question":"A1","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2025_/ENGR217 Main Exam 2024-25.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/217 - 2019 answer.pdf","tag":"VALID"},"text":"Answer all parts (a) - (c).\n","parts":[{"label":"a","text":"Water at $12^\\circ\\mathrm{C}$ is supplied to a 40 mm diameter steel pipe at 9 bar gauge pressure. The pipe\nis 125 m long, the roughness is $k = 0.02$ mm, and at the end of the pipe the water is discharged to\natmosphere. Estimate the volume flow rate of the water. (Kinematic viscosity for the water is\n$\\nu = 1.22 \\times 10^{-6}\\ \\mathrm{m^2/s}$).\n","marks":15,"grade":"2:1","answer":[{"label":"f (Darcy)","value":0.019,"units":"-","tol":0.05},{"label":"Q","value":6.92,"units":"L/s","tol":0.05}]},{"label":"b","text":"Diesel fuel flows along a pipe 12 m long, whose bore (internal diameter) is 8 mm. The density and\nviscosity of the fuel are $850\\ \\mathrm{kg/m^3}$ and $1.8 \\times 10^{-3}\\ \\mathrm{N\\,s/m^2}$ respectively.\nIf the pressure drop over the length of the pipe $1.5\\ \\mathrm{kN/m^2}$, estimate the volume flow rate of\nfuel along the pipe. (Hint: use Poiseuille's equation.)\n","marks":6,"grade":"third","answer":[{"label":"Q","value":0.00000698,"units":"m^3/s","tol":0.02}]},{"label":"c","text":"Hot water at $50^\\circ\\mathrm{C}$ is flowing along a pipe 800 mm diameter at a mean speed of 2.3 m/s. At\nthis temperature the kinematic viscosity of the water is $\\nu = 5.55 \\times 10^{-7}\\ \\mathrm{m^2/s}$. Will\nthe flow be laminar or turbulent?\n","marks":4,"grade":"third","answer":[{"label":"Re","value":3315000,"units":"-","tol":0.02}]}],"solution":"$70","confidence":"HIGH","solutionNote":"No official 2025 solution. Parts (a) and (c) repeat ENGR217 2019 Q B3(c) and B3(b); the linked 2019 numerical\nanswers give $Q = 6.81$ L/s for (a) and $Re = 3\\,315\\,315$ (turbulent) for (c). (The earlier note in this entry\nread the handwritten 6.81 as 6.31; the answer sheet says 6.81.) Agreement: (c) matches exactly. For (a) this\nsolution gives 6.92 L/s with Colebrook $f = 0.0190$; the official 6.81 L/s corresponds to a Moody-chart\nreading of $f \\approx 0.0196$ (Fanning 0.0049). The 1.6 percent difference is chart-reading accuracy, inside the\n5 percent tolerance; either value earns the marks. Part (b) has no official answer; it is checked by\nback-substitution and the laminar check. ENGR266 2025 Section A Q1 is the same question.\n"},{"id":"engr217-2025-a2","module":"ENGR5002","week":1,"label":"ENGR217 2025 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"tags":["surface-forces","centroid","second-moment-of-area","parallel-axis","centre-of-pressure"],"grade":"2:2","formulaSheet":"- Rectangle: $A = bh$, centroid $h/2$ from the top, $I_G = \\dfrac{bh^3}{12}$ (on the formula sheet)\n- Triangle: $A = \\tfrac{1}{2}bh$, centroid $h/3$ from the base, $I_G = \\dfrac{bh^3}{36}$ (on the formula sheet)\n- Composite centroid: $\\bar y = \\dfrac{\\sum A_i y_i}{\\sum A_i}$ (learn this)\n- Parallel axis theorem: $I = I_G + A d^2$ (learn this)\n- Hydrostatic force on a plane: $F = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\,l_{CG}}$ (on the formula sheet)\n","source":{"name":"ENGR217 Summer 2025 Q A2","paper":"ENGR217","year":2025,"question":"A2","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2025_/ENGR217 Main Exam 2024-25.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/217 - 2019 answer.pdf","tag":"VALID"},"text":"Answer all parts (a) - (d).\n\n[Figure summary, not printed: a rectangle 11.6 m wide and 6.6 m deep, its top edge at the free water\nsurface, with a triangle of the same base and 3.6 m depth below it.]\n","figures":[{"src":"/figures/engr217-2025-a2.png","caption":"Figure A2-1: the rectangle (11.6 m x 6.6 m) and triangle (3.6 m deep) forming the end of a water tank."}],"parts":[{"label":"a","text":"Find the depths $y_1$ and $y_2$, to the centroids $C_1$ and $C_2$ of the rectangle and triangle shown in Figure A2-1 below, and then find the depth $y$ to the centroid C of the whole figure.","marks":5,"grade":"third","answer":[{"label":"y1","value":3.3,"units":"m","tol":0.02},{"label":"y2","value":7.8,"units":"m","tol":0.02},{"label":"y","value":4.264,"units":"m","tol":0.02}]},{"label":"b","text":"The plane figure shown in Figure A2-1 forms the vertical end of the water tank. What is the magnitude of the force exerted on the plane by the water?","marks":5,"grade":"third","answer":[{"label":"F","value":4076,"units":"kN","tol":0.02}]},{"label":"c","text":"Find the second moment of area of the Figure A2-1 about the horizontal axis O-O through the centroid C.\n(Hint: use parallel axis theorem in both the triangle and the rectangle).\n","marks":10,"grade":"2:2","answer":[{"label":"I_C","value":625.2,"units":"m^4","tol":0.02}]},{"label":"d","text":"Finally find the position of the centre of pressure P of the complete Figure A2-1.","marks":5,"grade":"2:2","answer":[{"label":"q (P below C)","value":1.505,"units":"m","tol":0.02},{"label":"depth of P","value":5.769,"units":"m","tol":0.02}]}],"solution":"$71","confidence":"HIGH","solutionNote":"No official 2025 solution, but the question repeats ENGR217 2019 Q B1 exactly; the linked 2019 numerical\nanswers give $y = 4.264$ m, $F = 4075.90$ kN, $I_C = 625.164\\ \\mathrm{m^4}$ and $q = 1.505$ m (centre of\npressure below C). ENGR266 2025 Section A Q2 is the same question. Agreement on all four: this solution gives\n$y = 4.264$ m, $F = 4076.2$ kN (the official 4075.90 uses $y$ rounded to 4.264 m), $I_C = 625.16\\ \\mathrm{m^4}$,\n$q = 1.505$ m (P at 5.769 m depth).\n"},{"id":"engr5002-w1-ex1","module":"ENGR5002","week":1,"label":"Week 1 Exercise 1","kind":"exercise","level":1,"minutes":4,"tags":["hydrostatics","absolute-pressure","relative-density"],"source":{"name":"Week 1 Exercises, Exercise 1 (Pressure in a Tank)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"A large tank is filled with crude oil of relative density ($RD = 0.90$). The tank is vented at the top, so\nthe free surface of the oil is exposed to the atmosphere at pressure $p_a$. The height level of the oil in\nthe tank is $H = 12\\ \\mathrm{m}$.\n\nCalculate the absolute pressure at the bottom of the tank.\n\nData: relative density of crude oil $RD = 0.90$; density of water $\\rho_w = 1000\\ \\mathrm{kg/m^3}$;\ngravitational acceleration $g = 9.81\\ \\mathrm{m/s^2}$; atmospheric pressure $p_a = 101.3\\ \\mathrm{kPa}$.\n","figures":[{"src":"/figures/engr5002-w1-ex1.png","caption":"Vented tank of crude oil, depth H = 12 m, free surface at p_a."}],"grade":"third","formulaSheet":"- Relative density: $RD = \\dfrac{\\rho}{\\rho_w}$ (on the formula sheet)\n- Pressure at depth: $p(z) = p(z=0) + \\rho g z$ (on the formula sheet)\n- Gauge pressure: $p_g = p - p_{atm}$ (on the formula sheet)\n","answer":[{"label":"p_abs bottom","value":207.2,"units":"kPa"},{"label":"p_gauge bottom","value":105.9,"units":"kPa"}],"solution":"**Assumptions.** Oil at rest and incompressible; the vent keeps the free surface at $p_a$; $z$ measured\ndownwards from the free surface.\n\n**Density.** $\\rho_{oil} = RD\\,\\rho_w = 0.90 \\times 1000 = 900\\ \\mathrm{kg\\,m^{-3}}$.\n\n**Hydrostatics.** Integrating $dp/dz = \\rho g$ (z downwards) from the surface to the bottom:\n\n$$p_{bottom} = p_a + \\rho_{oil}\\, g\\, H$$\n\nGauge part: $\\rho_{oil} g H = 900 \\times 9.81 \\times 12 = 105\\,948\\ \\mathrm{Pa} = 105.9\\ \\mathrm{kPa}$.\n\n$$p_{bottom,abs} = 101.3 + 105.9 = 207.2\\ \\mathrm{kPa}\\ \\ (207\\,248\\ \\mathrm{Pa})$$\n\n**Checks.** Units: $\\mathrm{kg\\,m^{-3}} \\times \\mathrm{m\\,s^{-2}} \\times \\mathrm{m} = \\mathrm{N\\,m^{-2}}$.\nSpecific-weight route: $\\gamma = 0.90 \\times 9810 = 8829\\ \\mathrm{N\\,m^{-3}}$, $\\gamma H = 105\\,948$ Pa, same.\nMagnitude: 12 m of water is about 1.2 bar, so 12 m of a slightly lighter oil gives just under that, and the\nabsolute pressure is roughly two atmospheres. Common slip: quoting the gauge value (105.9 kPa) when the\nquestion asks for absolute.\n","confidence":"HIGH","solutionNote":"No official solution available; single-step result checked by two routes."},{"id":"engr5002-w1-ex2","module":"ENGR5002","week":1,"label":"Week 1 Exercise 2","kind":"exercise","level":2,"minutes":8,"tags":["hydrostatics","manometer","relative-density","multiple-fluids"],"source":{"name":"Week 1 Exercises, Exercise 2 (Multiple Fluid Hydrostatics)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"The U-tube shown in the figure contains oil in the left leg, water in both legs, and a trapped air bubble.\n\nThe following heights are given:\n\n- Oil column: $h_1 = 0.75$ m\n- Air pocket: $h_2 = 0.15$ m\n- Water column in the left leg: $h_3 = 0.30$ m\n- Water column in the right leg: $h_4 = 0.90$ m\n- The horizontal distance between the two legs is 0.60 m.\n\nUsing the hydrostatic balance, determine the relative density of the oil.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$; $p_a = 101.3\\ \\mathrm{kPa}$.\n","figures":[{"src":"/figures/engr5002-w1-ex2.png","caption":"U-tube, both ends open: left leg oil (h1) over an air pocket (h2) over water (h3); right leg water (h4). h3 and h4 are measured from the bottom of the tube."}],"grade":"2:2","formulaSheet":"- Pressure at depth: $p(z) = p(z=0) + \\rho g z$ (on the formula sheet)\n- Relative density: $RD = \\dfrac{\\rho}{\\rho_w}$ (on the formula sheet)\n- Equal pressure at equal level in one continuous static fluid (learn this; it is the basis of every manometer)\n","answer":[{"label":"RD_oil","value":0.8,"units":"-"}],"solution":"$72","confidence":"HIGH","solutionNote":"No official solution available; back-substitution balances exactly."},{"id":"engr5002-w1-ex3","module":"ENGR5002","week":1,"label":"Week 1 Exercise 3","kind":"exercise","level":2,"minutes":10,"tags":["surface-forces","centre-of-pressure","inclined-surface","circular-plate"],"source":{"name":"Week 1 Exercises, Exercise 3 (Aquarium Porthole)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"An aquarium has a circular porthole of diameter $D = 1.0$ m mounted on a sloping wall inclined at an angle\n$\\theta = 60^\\circ$ to the horizontal. The centroid of the porthole is located at a depth of\n$z_{CG} = 1.5$ m below the free surface of water. The density of water is $\\rho = 1000\\ \\mathrm{kg/m^3}$.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$; $p_a = 101.3\\ \\mathrm{kPa}$; $D = 1.0$ m;\n$z_{CG} = 1.5$ m; $\\theta = 60^\\circ$.\n","figures":[{"src":"/figures/engr5002-w1-ex3.png","caption":"Aquarium wall sloping at 60 deg to the horizontal with a 1.0 m diameter porthole, centroid 1.5 m below the free surface."}],"grade":"2:2","formulaSheet":"- Hydrostatic force on a plane surface: $F = p_{CG} A = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure along the plane: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\, l_{CG}}$ (on the formula sheet)\n- Circle: $A = \\dfrac{\\pi d^2}{4}$, $I_G = \\dfrac{\\pi d^4}{64}$ (on the formula sheet, shape table)\n- Slant distance and depth: $z = l \\sin\\theta$ (learn this)\n","parts":[{"label":"a","text":"Compute the resultant hydrostatic force acting on the porthole.","answer":[{"label":"F","value":11.56,"units":"kN"}],"grade":"$undefined"},{"label":"b","text":"Determine the location of the centre of pressure below the free surface.","answer":[{"label":"z_CP","value":1.531,"units":"m"}],"grade":"$undefined"}],"solution":"$73","confidence":"HIGH","solutionNote":"No official solution available; both parts confirmed by direct numerical integration."},{"id":"engr5002-w1-ex4","module":"ENGR5002","week":1,"label":"Week 1 Exercise 4","kind":"exercise","level":3,"minutes":18,"tags":["surface-forces","centre-of-pressure","inclined-surface","gate","moments","free-body-diagram"],"source":{"name":"Week 1 Exercises, Exercise 4 (Slanted Gate)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"A rectangular gate of width $W = 1.5$ m (out of plane) is installed in a vertical water reservoir wall. The\ngate has a slanted length $L = 1.5$ m and is hinged at the bottom point $B$. The gate is inclined at an angle\n$\\theta = 30^\\circ$ with respect to the horizontal. The total water depth is $H = 3.0$ m.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$; $p_a = 101.3\\ \\mathrm{kPa}$.\n","figures":[{"src":"/figures/engr5002-w1-ex4.png","caption":"Gate AB, 1.5 m long at 30 deg to the horizontal, hinged at B on the reservoir bed (depth H = 3.0 m) and resting at A against the bottom of the wall; water above the gate, air at p_a below it."}],"grade":"2:1","formulaSheet":"- Hydrostatic force on a plane surface: $F = \\rho g z_{CG} A$ (on the formula sheet)\n- Centre of pressure along the plane: $l_{CP} = l_{CG} + \\dfrac{I_G}{A\\, l_{CG}}$ (on the formula sheet)\n- Rectangle: $A = bh$, $I_G = \\dfrac{b h^3}{12}$ (on the formula sheet, shape table)\n- Components on an inclined plate: $F_H = F\\sin\\theta$, $F_V = F\\cos\\theta$ (equivalently $F_H = \\rho g z_{CG} A_{proj}$, $F_V$ = weight of fluid above) (learn this)\n- Moment equilibrium about the hinge: $\\sum M_B = 0$ (learn this)\n","parts":[{"label":"a","text":"Calculate the resultant hydrostatic force acting on the gate (magnitude and direction).","answer":[{"label":"F","value":57.94,"units":"kN"},{"label":"angle below horizontal","value":60,"units":"deg"}],"grade":"$undefined"},{"label":"b","text":"Calculate the horizontal and vertical components of the resultant force.","answer":[{"label":"F_H","value":28.97,"units":"kN"},{"label":"F_V","value":50.18,"units":"kN"}],"grade":"$undefined"},{"label":"c","text":"Find the centre of pressure depth $z_{CP}$.","answer":[{"label":"z_CP","value":2.643,"units":"m"}],"grade":"$undefined"},{"label":"d","text":"Determine the horizontal force $R$ exerted by the wall at point $A$.","answer":[{"label":"R","value":55.18,"units":"kN"}],"grade":"$undefined"}],"solution":"$74","confidence":"HIGH","solutionNote":"No official solution available. Gauge pressure used throughout (atmosphere on both sides); gate weight\nneglected because it is not given. All values checked by direct numerical integration.\n"},{"id":"engr5002-w1-ex5","module":"ENGR5002","week":1,"label":"Week 1 Exercise 5","kind":"exercise","level":2,"minutes":8,"tags":["buoyancy-stability","archimedes","flotation","relative-density"],"source":{"name":"Week 1 Exercises, Exercise 5 (Floating Aluminium Cube)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"A cube of aluminium of side $a = 0.4$ m floats in oil as shown in the figure. The relative density of\naluminium is $RD_{Al} = 2.70$, and the relative density of oil is $RD_{oil} = 0.80$.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$; $p_a = 101.3\\ \\mathrm{kPa}$.\n","figures":[{"src":"/figures/engr5002-w1-ex5.png","caption":"Cube of side a = 0.4 m shown floating in oil with submerged depth D."}],"grade":"2:2","formulaSheet":"- Buoyancy force: $F_B = \\rho_W g V_{displaced}$ (here $\\rho$ of the oil) (on the formula sheet)\n- Gravity force: $F_g = m g$ (on the formula sheet)\n- Relative density: $RD = \\rho/\\rho_w$ (on the formula sheet)\n- Floating equilibrium: $F_B = F_g$, so $D/a = RD_{body}/RD_{liquid}$ for a uniform prism (learn this)\n","parts":[{"label":"a","text":"Determine whether the cube will float or sink.","answer":[{"label":"weight W","value":1695,"units":"N"},{"label":"max buoyancy","value":502.3,"units":"N"}],"grade":"$undefined"},{"label":"b","text":"If it floats, calculate the submerged depth $D$.","answer":[{"label":"D required","value":1.35,"units":"m"}],"grade":"$undefined"},{"label":"c","text":"What happens in the case of a cube of wood of the same dimension ($RD_{wood} = 0.6$)?","answer":[{"label":"D wood","value":0.3,"units":"m"}],"grade":"$undefined"}],"solution":"$75","confidence":"HIGH","solutionNote":"No official solution available. The figure shows the cube floating, but with RD 2.70 in oil of RD 0.80 it sinks; part (b) is answered by showing the required depth exceeds the cube size."},{"id":"engr5002-w1-ex6","module":"ENGR5002","week":1,"label":"Week 1 Exercise 6","kind":"exercise","level":3,"minutes":20,"tags":["hydrostatics","ideal-gas","compressible-hydrostatics","atmosphere","mixtures"],"source":{"name":"Week 1 Exercises, Exercise 6 (Pressure Variation in a Gas), all parts","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"At sea level, the pressure of air is $p_0 = 101325$ Pa. Assume the atmosphere is at a uniform temperature\nof $T = 300$ K. The universal gas constant is $R = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$, and gravitational\nacceleration is $g = 9.81\\ \\mathrm{m\\,s^{-2}}$.\n\nAssume that air has the following molar composition:\n\n- 78% $\\mathrm{N_2}$ ($M_{N_2} = 28.02\\ \\mathrm{g\\,mol^{-1}}$)\n- 21% $\\mathrm{O_2}$ ($M_{O_2} = 32.00\\ \\mathrm{g\\,mol^{-1}}$)\n- 1% Ar ($M_{Ar} = 39.95\\ \\mathrm{g\\,mol^{-1}}$)\n","parts":[{"label":"a","text":"Calculate the average molar mass of the mixture using the given composition.","answer":[{"label":"M_avg","value":28.98,"units":"g/mol"}],"grade":"$undefined"},{"label":"b","text":"Derive an expression for $p(z)$ assuming density $\\rho$ is constant and equal to the one at sea level.","answer":[{"label":"rho_0","value":1.177,"units":"kg/m^3"}],"grade":"$undefined"},{"label":"c","text":"Derive an expression for $p(z)$ assuming $\\rho = \\dfrac{p M_w}{R T}$ (ideal gas law, with $T$ = const).","answer":[{"label":"scale height RT/(Mg)","value":8775,"units":"m"}],"grade":"$undefined"},{"label":"d","text":"Calculate the pressure in both cases at altitudes of 100 m, 500 m, 1 km, and 5 km.","answer":[{"label":"p const-rho 100 m","value":100170,"units":"Pa"},{"label":"p const-rho 500 m","value":95551,"units":"Pa"},{"label":"p const-rho 1 km","value":89778,"units":"Pa"},{"label":"p const-rho 5 km","value":43589,"units":"Pa"},{"label":"p isothermal 100 m","value":100177,"units":"Pa"},{"label":"p isothermal 500 m","value":95713,"units":"Pa"},{"label":"p isothermal 1 km","value":90411,"units":"Pa"},{"label":"p isothermal 5 km","value":57313,"units":"Pa"}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Differential hydrostatic equation: $\\dfrac{dp}{dz} = -\\rho g$ ($z$ upwards) (on the formula sheet)\n- Ideal-gas density: $\\rho = \\dfrac{p M}{R T}$ (given in the question; learn this)\n- Mixture molar mass: $M_{ave} = \\sum x_i M_i$ (on the ENGR271 2026 exam formula sheet, mass transfer section)\n- Isothermal atmosphere: $p = p_0 \\exp\\!\\left(-\\dfrac{M g z}{R T}\\right)$ (learn the derivation; not on the sheet)\n","solution":"$76","confidence":"HIGH","solutionNote":"No official solution available. Supersedes engr5002-l1-e6-1 (part 1 only), which is kept as a drill.\nDerivation checked by the small-z limit; all values computed in Python.\n"},{"id":"engr5002-w1-ex7","module":"ENGR5002","week":1,"label":"Week 1 Exercise 7","kind":"exercise","level":2,"minutes":10,"tags":["hydrostatics","multiple-fluids","absolute-pressure","connected-vessels"],"source":{"name":"Week 1 Exercises, Exercise 7 (Hydrostatics with Multiple Fluids)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"A connected vessel contains water and air at $T = 20^\\circ\\mathrm{C}$. The compartment on the left-hand side\nis opened to the atmosphere $p_0 = 101{,}325$ Pa.\n\nGeometric data (see figure):\n\n- Left bay: air space height above the left free surface $h_1 = 0.40$ m; water depth below that free\n surface $h_2 = 0.60$ m.\n- Middle bay: the free surface (point $B$) lies $h_3 = 0.30$ m below the top;\n- Right bay: distance from the bottom to the intermediate level is $h_6 = 0.20$ m.\n- The water level difference between the central and the right bay is $h_4 = 0.50$ m\n\nDetermine the pressure at points B, C and D.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$.\n","figures":[{"src":"/figures/engr5002-w1-ex7.png","caption":"Three bays joined at the bottom: left bay open (surface A), middle bay sealed with surface B, right bay sealed with C at the top and D at the bottom."}],"grade":"2:2","formulaSheet":"- Pressure at depth: $p(z) = p(z=0) + \\rho g z$ (on the formula sheet)\n- Same pressure at the same level in one continuous static fluid (learn this)\n- Pressure in a small trapped gas volume is uniform ($\\rho_{air} g h$ negligible) (learn this)\n","answer":[{"label":"p_B","value":100344,"units":"Pa"},{"label":"p_C","value":105249,"units":"Pa"},{"label":"p_D","value":107211,"units":"Pa"}],"solution":"$77","confidence":"HIGH","solutionNote":"No official solution available. Absolute pressures; D checked by three independent routes."},{"id":"engr5002-w1-ex8","module":"ENGR5002","week":1,"label":"Week 1 Exercise 8","kind":"exercise","level":2,"minutes":8,"tags":["hydrostatics","manometer","differential-manometer","relative-density"],"source":{"name":"Week 1 Exercises, Exercise 8 (A differential manometer)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"A differential U-tube manometer filled with mercury ($RD_{Hg} = 13.6$) is connected to a pipe carrying oil\n($RD_{oil} = 0.91$).\n\n- Point A is located just upstream of the orifice.\n- Point B is located just downstream of the orifice.\n- The vertical distance between the mercury levels is 10 cm.\n\nCalculate the pressure difference between points A and B.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$.\n","figures":[{"src":"/figures/engr5002-w1-ex8.png","caption":"Pipe with an orifice; tappings at A (upstream) and B (downstream) lead through oil-filled loops to a mercury U-tube; mercury levels differ by 10 cm, lower on the A side; y unknown."}],"grade":"2:2","formulaSheet":"- Pressure at depth: $p(z) = p(z=0) + \\rho g z$ (on the formula sheet)\n- Relative density: $RD = \\rho/\\rho_w$ (on the formula sheet)\n- Differential manometer: $p_A - p_B = (\\rho_m - \\rho) g\\,\\Delta h$ for A and B at the same level (learn this; derive it)\n","answer":[{"label":"p_A - p_B","value":12.45,"units":"kPa"}],"solution":"$78","confidence":"HIGH","solutionNote":"No official solution available. Result independent of the unknown y."},{"id":"engr5002-w1-ex9","module":"ENGR5002","week":1,"label":"Week 1 Exercise 9","kind":"exercise","level":3,"minutes":15,"tags":["surface-forces","centre-of-pressure","multiple-fluids","specific-weight"],"source":{"name":"Week 1 Exercises, Exercise 9 (Force on a Panel Between Two Fluids)","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"A vertical rectangular plate AB (total area $A = 2\\ \\mathrm{m^2}$) separates water and glycerine. Specific\nweights: $\\gamma_w = 9.81\\ \\mathrm{kN\\,m^{-3}}$, $\\gamma_g = 12.36\\ \\mathrm{kN\\,m^{-3}}$.\n\nGeometry (refer to the figure):\n\n- The water-free surface is 2.0 m higher than the glycerine free surface.\n- Point A on the plate is 1.0 m below the water free surface\n- Point B is 3.0 m below the water free surface\n- Hence the plate height between A and B is 2.0 m\n- The bottom is 6.0 m below the water free surface, and the glycerine depth is 4.0 m.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$.\n","figures":[{"src":"/figures/engr5002-w1-ex9.png","caption":"Vertical plate AB between water (left) and glycerine (right): h1 = 2 m (water surface to glycerine surface), h2 = 1 m (glycerine surface to A), h3 = 2 m (A to B), h4 = 1 m (B to bottom)."}],"grade":"2:1","formulaSheet":"- Hydrostatic force on a plane surface: $F = \\rho g z_{CG} A = \\gamma z_{CG} A$ (on the formula sheet)\n- Centre of pressure (vertical plate, $l = z$): $z_{CP} = z_{CG} + \\dfrac{I_G}{A z_{CG}}$ (on the formula sheet)\n- Rectangle: $I_G = \\dfrac{b h^3}{12}$ (on the formula sheet, shape table)\n- Resultant of two parallel forces: $F_{net} z_{net} = F_1 z_1 - F_2 z_2$ (learn this)\n","parts":[{"label":"a","text":"Calculate the net hydrostatic force on the panel.","answer":[{"label":"F_water","value":78.48,"units":"kN"},{"label":"F_glycerine","value":49.44,"units":"kN"},{"label":"F_net","value":29.04,"units":"kN"}],"grade":"$undefined"},{"label":"b","text":"Determine the depth of the centre of pressure from the free surface of water.","answer":[{"label":"z_CP net","value":3.941,"units":"m"}],"grade":"$undefined"},{"label":"c","text":"Determine the distance of centre of pressure from the centroid.","answer":[{"label":"distance above centroid","value":0.0585,"units":"m","tol":0.05}],"grade":"$undefined"}],"solution":"$79","confidence":"HIGH","solutionNote":"No official solution available. The bullet text places A and B 1.0 m and 3.0 m below the \"water\" free\nsurface, which contradicts the figure and the stated 6.0 m / 4.0 m depths; the solution follows the figure\n(A and B are 1.0 m and 3.0 m below the glycerine surface). Results verified by integrating the net pressure.\n"},{"id":"engr5002-w1-ex10","module":"ENGR5002","week":1,"label":"Week 1 Exercise 10","kind":"exercise","level":2,"minutes":8,"tags":["buoyancy-stability","archimedes","hydrometer","density"],"source":{"name":"Week 1 Exercises, Exercise 10 (hydrometer; printed title \"Force on a Panel Between Two Fluids\")","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"A hydrometer weighs $W = 0.0216$ N and has a cylindrical stem of diameter $d = 2.80$ mm. When floating in\nwater, its reference immersion depth (measured along the stem) is $h_w = 12.0$ mm. When placed in an unknown\nliquid, the immersion depth increases by $\\Delta h = 6.5$ mm.\n\nCalculate the density of the unknown liquid.\n\nData: $\\rho_w = 1000\\ \\mathrm{kg/m^3}$; $g = 9.81\\ \\mathrm{m/s^2}$.\n","figures":[{"src":"/figures/engr5002-w1-ex10.png","caption":"Hydrometer (bulb plus thin stem) floating in water (left) and, deeper by Delta h, in the unknown liquid (right)."}],"grade":"2:2","formulaSheet":"- Buoyancy force: $F_B = \\rho g V_{displaced}$ (on the formula sheet)\n- Gravity force: $F_g = m g$ (here $W$ is given) (on the formula sheet)\n- Floating equilibrium $F_B = W$ in each liquid (learn this)\n","answer":[{"label":"rho_liquid","value":982,"units":"kg/m^3"}],"solution":"$7a","confidence":"MEDIUM","solutionNote":"No official solution available. The data are over-specified: W and d give 982 kg/m^3 (used here, physically\nconsistent), whereas a pure-cylinder reading of h_w and Delta h would give 649 kg/m^3 but contradicts W.\nConfirm with the lecturer which interpretation the sheet intends. The printed title of this exercise repeats\nExercise 9's title by mistake.\n"},{"id":"engr5002-l1-k1","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"K1","kind":"drill","level":1,"grade":"third","answer":[{"label":"Kn","value":0.00000136,"units":"-"}],"minutes":3,"tags":["knudsen","continuum"],"formulas":["knudsen"],"text":"Air at atmospheric conditions ($\\lambda = 68\\ \\mathrm{nm}$) flows in a pipe of diameter 50 mm.\nCalculate $Kn$ and state the regime.\n","solution":"$$Kn = \\lambda / L = 68 \\times 10^{-9}\\ \\mathrm{m} / 0.050\\ \\mathrm{m} = 1.36 \\times 10^{-6}$.\n\nSince $1.36 \\times 10^{-6} \\ll 0.001$: **continuum regime**.\n\nCheck: units m/m, dimensionless; magnitude sensible (a 5 cm pipe is about 700 000 mean free\npaths across).\n","confidence":"HIGH"},{"id":"engr5002-l1-k2","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"K2","kind":"drill","level":2,"grade":"third","answer":[{"label":"L_min","value":68,"units":"um (micrometres)"}],"minutes":4,"tags":["knudsen","continuum"],"formulas":["knudsen"],"text":"For the same air ($\\lambda = 68\\ \\mathrm{nm}$), what is the smallest channel dimension for which\nthe continuum hypothesis is valid?\n","solution":"Continuum requires $Kn < 0.001$, so\n$L > \\lambda / 0.001 = 68 \\times 10^{-9} / 0.001 = 6.8 \\times 10^{-5}\\ \\mathrm{m} = 68\\ \\mu\\mathrm{m}$.\n\nCheck: substituting back, $68 \\times 10^{-9} / 68 \\times 10^{-6} = 0.001$, on the boundary as\nexpected. So standard fluid mechanics holds for air in anything larger than roughly a tenth of a\nmillimetre; micro-channels below that need care.\n","confidence":"HIGH"},{"id":"engr5002-l1-k3","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"K3","kind":"drill","level":2,"grade":"third","answer":[{"label":"Kn","value":1.5,"units":"-"}],"minutes":4,"tags":["knudsen","continuum"],"formulas":["knudsen"],"text":"At about 120 km altitude the mean free path of air is roughly 3 m. A satellite of characteristic\nlength 2 m passes through. Find $Kn$, state the regime, and say whether the methods of this\nmodule could predict its drag.\n","solution":"$$Kn = 3 / 2 = 1.5$, which lies between 0.1 and 10: **transition regime**.\n\nThe continuum hypothesis fails (molecules travel about one body length between collisions), so\nthe methods of this module (continuum drag coefficients, boundary layers) cannot be used.\n\nConfidence HIGH for the calculation; the 3 m value of $\\lambda$ is an approximate [EXTERNAL]\nfigure supplied in the question.\n","confidence":"HIGH"},{"id":"engr5002-l1-d1","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"D1","kind":"drill","level":1,"grade":"third","answer":[{"label":"rho","value":860,"units":"kg/m^3"}],"minutes":2,"tags":["density"],"formulas":["density"],"text":"A sample of $0.25\\ \\mathrm{m^3}$ of oil has a mass of 215 kg. Find its density.","solution":"$$\\rho = m / V = 215 / 0.25 = 860\\ \\mathrm{kg\\,m^{-3}}$.\n\nCheck: typical oil range 800-920, plausible.\n","confidence":"HIGH"},{"id":"engr5002-l1-d2","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"D2","kind":"drill","level":2,"grade":"third","answer":[{"label":"m","value":680,"units":"kg"}],"minutes":3,"tags":["density","relative-density"],"formulas":["density","relative-density"],"text":"Find the mass of 50 L of mercury (RD = 13.6).","solution":"$$\\rho = 13.6 \\times 1000 = 13\\,600\\ \\mathrm{kg\\,m^{-3}}$; $V = 50\\ \\mathrm{L} = 0.050\\ \\mathrm{m^3}$;\n$m = \\rho V = 13\\,600 \\times 0.050 = 680\\ \\mathrm{kg}$.\n\nCheck: 50 L of water would be 50 kg, and mercury is 13.6 times denser: $50 \\times 13.6 = 680$ kg.\n","confidence":"HIGH"},{"id":"engr5002-l1-d3","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"D3","kind":"drill","level":2,"grade":"third","answer":[{"label":"m","value":2550,"units":"kg"},{"label":"W","value":25,"units":"kN"}],"minutes":4,"tags":["density","specific-weight"],"formulas":["density","specific-weight"],"text":"A rectangular tank $2.0\\ \\mathrm{m} \\times 1.5\\ \\mathrm{m} \\times 1.0\\ \\mathrm{m}$ is full of\ndiesel ($\\rho = 850\\ \\mathrm{kg\\,m^{-3}}$). Find the mass and the weight of the diesel.\n","solution":"$$V = 2.0 \\times 1.5 \\times 1.0 = 3.0\\ \\mathrm{m^3}$; $m = 850 \\times 3.0 = 2550\\ \\mathrm{kg}$;\n$W = m g = 2550 \\times 9.81 = 25\\,016\\ \\mathrm{N} \\approx 25.0\\ \\mathrm{kN}$.\n\nCheck via specific weight: $\\gamma V = (850 \\times 9.81)(3.0) = 8338.5 \\times 3.0 = 25\\,016$ N.\nSame.\n","confidence":"HIGH"},{"id":"engr5002-l1-s1","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"S1","kind":"drill","level":1,"grade":"third","answer":[{"label":"gamma","value":10.06,"units":"kN/m^3"}],"minutes":2,"tags":["specific-weight"],"formulas":["specific-weight"],"text":"Find the specific weight of seawater, $\\rho = 1025\\ \\mathrm{kg\\,m^{-3}}$.","solution":"$$\\gamma = \\rho g = 1025 \\times 9.81 = 10\\,055\\ \\mathrm{N\\,m^{-3}} \\approx 10.06\\ \\mathrm{kN\\,m^{-3}}$.\n\nCheck: slightly above water's $9.81\\ \\mathrm{kN\\,m^{-3}}$, as it should be.\n","confidence":"HIGH"},{"id":"engr5002-l1-s2","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"S2","kind":"drill","level":2,"grade":"third","answer":[{"label":"rho","value":1260,"units":"kg/m^3"},{"label":"RD","value":1.26,"units":"-"}],"minutes":3,"tags":["specific-weight","relative-density"],"formulas":["specific-weight","relative-density"],"text":"Glycerine has $\\gamma = 12.36\\ \\mathrm{kN\\,m^{-3}}$ (Week 1 Exercise 9 data). Find its density and RD.","solution":"$$\\rho = \\gamma / g = 12\\,360 / 9.81 = 1260\\ \\mathrm{kg\\,m^{-3}}$ (1259.9);\n$RD = 1260 / 1000 = 1.26$.\n\nCheck: glycerine is denser than water, RD > 1.\n","confidence":"HIGH"},{"id":"engr5002-l1-s3","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"S3","kind":"drill","level":2,"grade":"third","answer":[{"label":"weight per m^2","value":24,"units":"kPa"}],"minutes":4,"tags":["specific-weight","hydrostatic-pressure"],"formulas":["specific-weight"],"text":"A vertical column of liquid with $\\gamma = 8.0\\ \\mathrm{kN\\,m^{-3}}$ is 3.0 m tall. What is the\nweight of liquid resting on each square metre of the column's base? (This is your first\nhydrostatic pressure calculation; Lecture 2 builds on it.)\n","solution":"Weight per unit base area $= \\dfrac{\\gamma V}{A} = \\dfrac{\\gamma A h}{A} = \\gamma h\n= 8.0 \\times 10^3 \\times 3.0 = 24\\,000\\ \\mathrm{N\\,m^{-2}} = 24\\ \\mathrm{kPa}$.\n\nUnits: $\\mathrm{N\\,m^{-3}} \\times \\mathrm{m} = \\mathrm{N\\,m^{-2}} = \\mathrm{Pa}$, a pressure.\nThis is exactly the hydrostatic gauge pressure at the base, $p = \\rho g h$, from Lecture 2.\n","confidence":"HIGH"},{"id":"engr5002-l1-r1","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"R1","kind":"drill","level":1,"grade":"third","answer":[{"label":"rho","value":900,"units":"kg/m^3"}],"minutes":1,"tags":["relative-density"],"formulas":["relative-density"],"text":"Crude oil has RD = 0.90 (Week 1 Exercise 1). Find its density.","solution":"$$\\rho = RD \\times \\rho_w = 0.90 \\times 1000 = 900\\ \\mathrm{kg\\,m^{-3}}$.","confidence":"HIGH"},{"id":"engr5002-l1-r2","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"R2","kind":"drill","level":2,"grade":"third","answer":[{"label":"RD","value":1.36,"units":"-"}],"minutes":2,"tags":["relative-density","density"],"formulas":["relative-density","density"],"text":"A 1.00 L bottle holds 1.36 kg of a liquid. Find its RD.","solution":"$$\\rho = 1.36\\ \\mathrm{kg} / 1.00 \\times 10^{-3}\\ \\mathrm{m^3} = 1360\\ \\mathrm{kg\\,m^{-3}}$;\n$RD = 1360 / 1000 = 1.36$ (no units).\n\nCheck: 1 L of water is 1 kg, so 1.36 kg in 1 L is 1.36 times water.\n","confidence":"HIGH"},{"id":"engr5002-l1-r3","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"R3","kind":"drill","level":2,"grade":"third","minutes":3,"tags":["relative-density"],"formulas":["relative-density"],"text":"Oil (RD = 0.8) and water are poured into the same open tank and do not mix. Which liquid ends up\non top, and why? Is this consistent with the arrangement in ENGR271 2026 Q1?\n","solution":"The oil floats on top. With immiscible liquids at rest, the denser one settles to the bottom\nbecause that arrangement has the lowest potential energy and is the only stable one (a heavier\nlayer on top would be pushed down by its own weight and the lighter one would rise).\n$RD_{oil} = 0.8 < 1$, so oil is less dense. This matches ENGR271 2026 Q1, where oil is the upper\nlayer above the water.\n","confidence":"HIGH"},{"id":"engr5002-l1-e6-1","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"E6.1","kind":"exercise","level":2,"grade":"third","minutes":8,"tags":["ideal-gas","density","mixtures"],"formulas":["ideal-gas-density"],"source":{"name":"Week 1 Exercises, Exercise 6 part 1","path":"ENGR5002 - Fluid Mechanics and Mass Transfer/Week 1/Exercises Week 1.pdf"},"text":"Air at sea level has $p_0 = 101\\,325\\ \\mathrm{Pa}$ and uniform temperature $T = 300\\ \\mathrm{K}$;\n$R = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$. Its molar composition is 78% $\\mathrm{N_2}$\n($M = 28.02\\ \\mathrm{g\\,mol^{-1}}$), 21% $\\mathrm{O_2}$ ($32.00\\ \\mathrm{g\\,mol^{-1}}$) and 1% Ar\n($39.95\\ \\mathrm{g\\,mol^{-1}}$).\n","parts":[{"label":"a","text":"Calculate the average molar mass of the mixture.","answer":[{"label":"M_avg","value":28.98,"units":"g/mol"}],"grade":"$undefined"},{"label":"b","text":"[Added, needed for Exercise 6 parts 2-4] Using $\\rho = p M_w / (R T)$, calculate the density of air at sea level.","answer":[{"label":"rho_0","value":1.177,"units":"kg/m^3"}],"grade":"$undefined"}],"solution":"$7b","confidence":"HIGH"},{"id":"engr5002-l1-x1","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"X1","kind":"adapted","level":2,"grade":"third","marks":5,"minutes":6,"tags":["relative-density","specific-weight","density","hydrostatic-pressure"],"formulas":["relative-density","specific-weight"],"source":{"name":"ENGR271 2025-26 Q1 (data and RD step only)","paper":"ENGR271","year":2026,"question":"Q1","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR271 In-person Main Exam 2025-26 - FINAL FOR SPA v2.pdf","tag":"PARTIAL","leftOut":"Parts (a)-(d) (absolute and gauge pressure, interface condition, piezometer heights) test Lecture 2 hydrostatics and are in that pack."},"text":"An open vertical tank contains a 1.0 m layer of water ($\\rho_w = 1000\\ \\mathrm{kg\\,m^{-3}}$)\nwith a 1.0 m layer of oil of relative density $RD_{oil} = 0.8$ floating on top. Take\n$g = 9.81\\ \\mathrm{m\\,s^{-2}}$.\n","parts":[{"label":"i","text":"Calculate the density and the specific weight of the oil and of the water.","marks":3,"answer":[{"label":"rho_oil","value":800,"units":"kg/m^3"},{"label":"gamma_oil","value":7848,"units":"N/m^3"},{"label":"gamma_water","value":9810,"units":"N/m^3"}],"grade":"$undefined"},{"label":"ii","text":"Hence calculate the weight of each layer resting on $1\\ \\mathrm{m^2}$ of horizontal area.","marks":2,"answer":[{"label":"oil layer per m^2","value":7848,"units":"N"},{"label":"water layer per m^2","value":9810,"units":"N"}],"grade":"$undefined"}],"solution":"**(i)** $\\rho_{oil} = RD \\times \\rho_w = 0.8 \\times 1000 = 800\\ \\mathrm{kg\\,m^{-3}}$;\n$\\gamma_{oil} = 800 \\times 9.81 = 7848\\ \\mathrm{N\\,m^{-3}}$ ($7.85\\ \\mathrm{kN\\,m^{-3}}$).\n$\\rho_w = 1000\\ \\mathrm{kg\\,m^{-3}}$; $\\gamma_w = 1000 \\times 9.81 = 9810\\ \\mathrm{N\\,m^{-3}}$.\n\n**(ii)** Weight per $\\mathrm{m^2}$ of a layer of thickness $h$ is $\\gamma h$:\noil $7848 \\times 1.0 = 7848\\ \\mathrm{N}$ (7.85 kPa); water $9810 \\times 1.0 = 9810\\ \\mathrm{N}$ (9.81 kPa).\n\nCheck: the bottom of the tank would therefore carry $7848 + 9810 = 17\\,658$ Pa gauge, which is\nwhat you will find in Lecture 2 from $p = p_0 + \\rho g z$ applied layer by layer. Unit check:\n$\\mathrm{N\\,m^{-3}} \\times \\mathrm{m} = \\mathrm{N\\,m^{-2}}$.\n\n**Likely marks:** 1 for converting RD to density, 1 for each specific weight with units, 1 each\nfor the layer weights. A full-marks answer states the formula used, carries units, and gives 3\nsignificant figures.\n","confidence":"HIGH"},{"id":"engr5002-l1-x2","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"X2","kind":"adapted","level":2,"grade":"third","answer":[{"label":"rho_1","value":820,"units":"kg/m^3"},{"label":"gamma_1","value":8044,"units":"N/m^3"},{"label":"rho_2","value":13600,"units":"kg/m^3"},{"label":"gamma_2","value":133400,"units":"N/m^3"}],"marks":3,"minutes":4,"tags":["relative-density","specific-weight"],"formulas":["relative-density","specific-weight"],"source":{"name":"ENGR217 2024 Q A1(b) (data only)","paper":"ENGR217","year":2024,"question":"A1(b)","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2024/ENGR217 In-person Main Exam 2024 combined FINAL.pdf","tag":"PARTIAL","leftOut":"The pump head-rise calculation (manometer plus Bernoulli, Lectures 2 and W3-4)."},"text":"A pumped liquid has specific gravity $S_1 = 0.82$ and the manometer fluid is mercury with\n$S_2 = 13.6$. State what \"specific gravity\" means in the current module's notation, and find the\ndensity and specific weight of both fluids.\n","solution":"\"Specific gravity\" is the old name for relative density: $RD = SG = \\rho / \\rho_w$, dimensionless.\n\nPumped liquid: $\\rho_1 = 0.82 \\times 1000 = 820\\ \\mathrm{kg\\,m^{-3}}$,\n$\\gamma_1 = 820 \\times 9.81 = 8044\\ \\mathrm{N\\,m^{-3}}$ ($8.04\\ \\mathrm{kN\\,m^{-3}}$).\n\nMercury: $\\rho_2 = 13.6 \\times 1000 = 13\\,600\\ \\mathrm{kg\\,m^{-3}}$,\n$\\gamma_2 = 13\\,600 \\times 9.81 = 133\\,400\\ \\mathrm{N\\,m^{-3}}$ ($133\\ \\mathrm{kN\\,m^{-3}}$).\n\n**Likely marks:** these conversions are worth about 1-2 marks inside the original 15+ mark\nquestion; losing them through an RD slip would carry an error through the whole answer.\n","confidence":"HIGH"},{"id":"engr5002-l1-g1","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"G1","kind":"generated","level":3,"grade":"third","marks":5,"minutes":6,"tags":["continuum","knudsen","density"],"formulas":["knudsen","density"],"source":{"name":"Generated, based on ENGR271 2026 Q1(b) \"explain\" style","paper":"ENGR271","year":2026,"question":"Q1(b) style"},"text":"State the continuum hypothesis and explain, with reference to the Knudsen number, when it is\nvalid. Explain why the density at a point is defined as the limit of $\\delta m / \\delta V$ as\n$\\delta V$ becomes small, and why that volume cannot literally shrink to zero.\n","solution":"$7c","confidence":"HIGH"},{"id":"engr5002-l1-g2","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"G2","kind":"generated","level":3,"grade":"2:2","marks":10,"minutes":12,"tags":["knudsen","continuum","no-slip"],"formulas":["knudsen"],"source":{"name":"Generated, based on ENGR271 2026 Q2(c) \"calculate and justify\" style","paper":"ENGR271","year":2026,"question":"Q2(c) style"},"text":"Air at 101.3 kPa and 20 deg C has a mean free path $\\lambda = 68\\ \\mathrm{nm}$. At constant\ntemperature, $\\lambda$ is inversely proportional to pressure. Air flows through a tube of\ninternal diameter 10 mm connected to a vacuum system.\n","parts":[{"label":"a","text":"Calculate the Knudsen number at atmospheric pressure and state the flow regime.","marks":3,"answer":[{"label":"Kn","value":0.0000068,"units":"-"}],"grade":"$undefined"},{"label":"b","text":"Determine the pressure below which the continuum hypothesis is no longer valid for this tube.","marks":4,"answer":[{"label":"p_crit","value":689,"units":"Pa"}],"grade":"$undefined"},{"label":"c","text":"The system is pumped down to 10 Pa. Calculate $Kn$, state the regime, and explain whether the no-slip condition and the standard pipe-flow results of this module can be used.","marks":3,"answer":[{"label":"Kn","value":0.069,"units":"-"}],"grade":"$undefined"}],"solution":"$7d","confidence":"HIGH"},{"id":"engr5002-l1-g3","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"G3","kind":"generated","level":3,"grade":"third","marks":5,"minutes":6,"tags":["fluid-definition","shear-stress"],"formulas":[],"source":{"name":"Generated, based on ENGR271 2026 Q1(b) style","paper":"ENGR271","year":2026,"question":"Q1(b) style"},"text":"A block of steel and a layer of water, each of height $h$, are fixed at their base and subjected\nto the same constant tangential force $F$ on their top surface. With the aid of sketches,\ndescribe how each responds, and hence give a definition of a fluid. State what this implies\nabout the stresses acting in a fluid at rest.\n","solution":"$7e","confidence":"HIGH"},{"id":"engr5002-l1-g4","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"G4","kind":"generated","level":3,"grade":"2:2","marks":10,"minutes":12,"tags":["density","specific-weight","relative-density"],"formulas":["density","specific-weight","relative-density"],"source":{"name":"Generated, based on ENGR271 2026 Q1(a) calculation style","paper":"ENGR271","year":2026,"question":"Q1(a) style"},"text":"A sealed cylindrical drum has internal diameter 0.58 m and internal height 0.85 m. It weighs\n0.25 kN when empty and 2.20 kN when completely full of an oil.\n","parts":[{"label":"a","text":"Calculate the density of the oil.","marks":4,"answer":[{"label":"rho","value":885,"units":"kg/m^3"}],"grade":"$undefined"},{"label":"b","text":"Calculate its specific weight.","marks":2,"answer":[{"label":"gamma","value":8683,"units":"N/m^3"}],"grade":"$undefined"},{"label":"c","text":"Calculate its relative density, and state whether it would sit above or below water in a tank if the two did not mix.","marks":2,"answer":[{"label":"RD","value":0.885,"units":"-"}],"grade":"$undefined"},{"label":"d","text":"The oil is instead blended with an equal volume of a miscible liquid of RD = 1.00. Assuming the volumes are additive, find the RD of the blend.","marks":2,"answer":[{"label":"RD blend","value":0.943,"units":"-"}],"grade":"$undefined"}],"solution":"$7f","confidence":"HIGH"},{"id":"engr5002-l1-g5","module":"ENGR5002","week":1,"lecture":"engr5002-l1","label":"G5","kind":"generated","level":3,"grade":"2:2","marks":10,"minutes":12,"tags":["ideal-gas","density","relative-density","specific-weight","mixtures"],"formulas":["ideal-gas-density","relative-density","specific-weight"],"source":{"name":"Generated, based on Week 1 Exercise 6 and ENGR271 2026 Q3 mixture style","paper":"ENGR271","year":2026,"question":"Q3 style"},"text":"Natural gas in a storage vessel is modelled as an ideal gas mixture of 90% methane\n($M = 16.04\\ \\mathrm{g\\,mol^{-1}}$) and 10% ethane ($M = 30.07\\ \\mathrm{g\\,mol^{-1}}$) by mole, at\n5.0 bar absolute and 15 deg C. Take $R = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$.\n","parts":[{"label":"a","text":"Calculate the average molar mass of the gas.","marks":3,"answer":[{"label":"M","value":17.44,"units":"g/mol"}],"grade":"$undefined"},{"label":"b","text":"Calculate its density using $\\rho = pM/(RT)$.","marks":3,"answer":[{"label":"rho","value":3.64,"units":"kg/m^3"}],"grade":"$undefined"},{"label":"c","text":"Calculate its relative density (with respect to water) and specific weight.","marks":2,"answer":[{"label":"RD","value":0.00364,"units":"-"},{"label":"gamma","value":35.7,"units":"N/m^3"}],"grade":"$undefined"},{"label":"d","text":"Explain why the density of a gas cannot always be treated as constant in hydrostatic calculations, whereas the density of a liquid usually can.","marks":2,"grade":"$undefined"}],"solution":"$80","confidence":"HIGH"},{"id":"engr271-2026-q1","module":"ENGR5002","week":1,"label":"ENGR271 2026 Q1","kind":"past-paper","level":3,"grade":"2:2","formulaSheet":"- Relative density: $RD = \\dfrac{\\rho_A}{\\rho_w}$ (on the formula sheet)\n- Pressure at depth: $p(z) = p_0 + \\rho g z$ ($z$ = depth below the free surface) (on the formula sheet)\n- Gauge pressure: $p_g = p - p_{atm}$ (on the formula sheet)\n- Pressure continuous across a static fluid-fluid interface; layer by layer $p = p_{top} + \\rho g h$ (learn this)\n- Piezometer: open column of height $H$ above the tapping, $p = p_{atm} + \\rho g H$ (learn this; it is the pressure-at-depth formula read backwards)\n","marks":30,"minutes":36,"tags":["hydrostatic-pressure","relative-density","gauge-pressure","piezometer","interface"],"formulas":["relative-density","specific-weight"],"source":{"name":"ENGR271 Summer 2026 Q1","paper":"ENGR271","year":2026,"question":"Q1","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR271 In-person Main Exam 2025-26 - FINAL FOR SPA v2.pdf","tag":"VALID"},"text":"An open vertical tank contains two immiscible liquids: an oil floating on a water layer.\nThe tank bottom is taken as the elevation $z = 0$ m. The water layer is 1.0 m deep (density of\nwater $\\rho_w = 1000\\ \\mathrm{kg\\,m^{-3}}$). Above the water there is a layer of oil of 1.0 m\ndepth. The relative density of the oil with respect to water is $RD_{oil} = 0.8$. The tank is\nopen to the atmosphere, so the free surface is at atmospheric pressure. Take the atmospheric\npressure as $p_{atm} = 101\\,300$ Pa and $g = 9.81\\ \\mathrm{m\\,s^{-2}}$.\n\nTwo vertical, transparent tubes (piezometers) are connected to the tank, as shown in Fig. Q1-1\n(see the paper). One piezometer is connected to the tank wall at an elevation of $z = 1.5$ m\nabove the bottom, and the other at an elevation of $z = 0.5$ m.\n","parts":[{"label":"a","text":"Calculate the absolute pressure and the gauge pressure at point A.","marks":10,"answer":[{"label":"p_A abs","value":105224,"units":"Pa"},{"label":"p_A gauge","value":3924,"units":"Pa"}],"grade":"$undefined"},{"label":"b","text":"Under hydrostatic conditions, what condition must hold for the pressure at the interface between the two immiscible fluids (oil and water)?","marks":5,"grade":"$undefined"},{"label":"c","text":"Determine the elevation $z_1$ reached by the liquid surface in point C. Consider that there is only oil in the tube. The figure is not to scale and piezometer levels are not indicative. Determine $z_1$ and justify your calculation using hydrostatic principles.","marks":5,"answer":[{"label":"z_1","value":2,"units":"m"}],"grade":"$undefined"},{"label":"d","text":"Determine the elevation $z_2$ reached by the liquid surface in point D. Consider that there is only water in the tube. Determine $z_2$ and justify your calculation using hydrostatic principles.","marks":10,"answer":[{"label":"z_2","value":1.8,"units":"m"}],"grade":"$undefined"}],"figures":[{"src":"/figures/engr271-2026-q1.png","caption":"Figure Q1-1: water-oil tank with piezometers connected at elevations 0.5 m and 1.5 m."}],"solution":"$81","confidence":"HIGH"},{"id":"engr271-2026-q2","module":"ENGR5002","week":4,"label":"ENGR271 2026 Q2","kind":"past-paper","level":3,"grade":"2:1","formulaSheet":"- Volume flow rate: $Q = vA$; mass flow rate $\\dot m = \\rho Q$ (on the formula sheet)\n- Reynolds number: $Re = \\dfrac{\\rho v L}{\\mu}$; pipe laminar $Re < 2400$, turbulent $Re > 4000$ (on the formula sheet)\n- Entrance length: laminar $L_E/D = 0.065\\,Re$; turbulent $L_E/D = 1.359\\,Re^{0.25}$ (on the formula sheet)\n- Darcy head loss: $h_L = f \\dfrac{L}{D} \\dfrac{v^2}{2g}$ (on the formula sheet)\n- Minor loss: $h_L = K_L \\dfrac{v^2}{2g}$ (on the formula sheet; also printed in part (d))\n- Hagen-Poiseuille: $\\Delta p = \\dfrac{8 \\mu L Q}{\\pi R^4} = \\dfrac{32 \\mu L v_{avg}}{D^2}$ (laminar only) (on the formula sheet)\n- Bernoulli with head loss: $\\dfrac{v_1^2}{2g} + z_1 + \\dfrac{p_1}{\\rho g} = \\dfrac{v_2^2}{2g} + z_2 + \\dfrac{p_2}{\\rho g} + h_L$ (on the formula sheet)\n- Momentum balance: $(\\dot m \\vec v)_{in} - (\\dot m \\vec v)_{out} + \\sum \\vec F_{ext} = \\dfrac{d}{dt}(m \\vec v)_{CV}$, with $\\sum \\vec F_{ext} = \\vec F_g + \\vec F_p + \\vec F_\\mu + \\vec F_{others}$ (on the formula sheet)\n","marks":40,"minutes":48,"tags":["continuity","bernoulli","pipe-losses","reynolds","momentum","nozzle"],"formulas":[],"source":{"name":"ENGR271 Summer 2026 Q2","paper":"ENGR271","year":2026,"question":"Q2","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR271 In-person Main Exam 2025-26 - FINAL FOR SPA v2.pdf","tag":"VALID"},"text":"Water flows from a pressurised tank through a pipe and a converging nozzle before discharging\nto the atmosphere. The tank free surface (section 1) is at elevation $z_1 = 10$ m. The pipe\nrises to the nozzle, and both section 2 (pipe-nozzle junction) and section 3 (nozzle exit) are\nat the same elevation $z_2 = z_3 = 20$ m (Figure Q2-1 in the paper).\n\nPipe: length $L = 100$ m, diameter $D = 0.10$ m, Darcy friction factor $f = 0.020$, area\n$A_2 = \\pi D^2/4$. Nozzle: exit diameter $d = 0.050$ m, head-loss coefficient $K = 0.80$, exit\narea $A_3 = \\pi d^2/4$. Volumetric flow rate $Q = 0.010\\ \\mathrm{m^3/s}$. Water:\n$\\rho = 1000\\ \\mathrm{kg/m^3}$, $\\mu = 1.0 \\times 10^{-3}$ Pa s. $g = 9.81\\ \\mathrm{m/s^2}$.\n","parts":[{"label":"a","text":"Calculate the average velocity in section 3 (nozzle exit).","marks":5,"answer":[{"label":"v_3","value":5.09,"units":"m/s"}],"grade":"$undefined"},{"label":"b","text":"Calculate the flow rate and the fluid velocity in section 2.","marks":5,"answer":[{"label":"Q_2","value":0.01,"units":"m^3/s"},{"label":"v_2","value":1.27,"units":"m/s"}],"grade":"$undefined"},{"label":"c","text":"Calculate the Reynolds number in the pipe (excluding the nozzle). Is the flow laminar or turbulent? Using your engineering judgment, explain if considering the flow \"fully developed\" is a good approximation in this case.","marks":5,"answer":[{"label":"Re","value":127000,"units":"-"}],"grade":"$undefined"},{"label":"d","text":"Calculate the head loss in the nozzle using $h_{L,nozzle} = K\\,\\dfrac{v_3^2}{2g}$.","marks":5,"answer":[{"label":"h_L nozzle","value":1.06,"units":"m"}],"grade":"$undefined"},{"label":"e","text":"Calculate the head loss in the pipe. Then discuss and demonstrate whether the Hagen-Poiseuille law can be used to calculate the head loss in this specific case.","marks":5,"answer":[{"label":"h_f pipe","value":1.65,"units":"m"}],"grade":"$undefined"},{"label":"f","text":"Determine the gauge pressure required at the tank's free surface to sustain the given flow rate. Then comment on which component of the system (the pipe or the nozzle) contributes most to the total energy loss.","marks":5,"answer":[{"label":"p_1 gauge","value":138,"units":"kPa"}],"grade":"$undefined"},{"label":"g","text":"Draw the free-body diagram of the forces around the nozzle (include only the pressure forces at sections 2 and 3 and the force exerted by the nozzle walls on the fluid; neglect gravity and other body forces). Apply the momentum balance equation and determine the force required to restrain the nozzle.","marks":10,"answer":[{"label":"restraining force","value":139,"units":"N"}],"grade":"$undefined"}],"figures":[{"src":"/figures/engr271-2026-q2.png","caption":"Figure Q2-1: pressurised tank connected to a pipe and a converging nozzle."}],"solution":"$82","confidence":"HIGH"},{"id":"engr271-2026-q3","module":"ENGR5002","week":9,"label":"ENGR271 2026 Q3","kind":"past-paper","level":3,"grade":"2:1","formulaSheet":"- Pressure and temperature correction: $D_{AB,T_2,P_2} = D_{AB,T_1,P_1} \\left(\\dfrac{P_1}{P_2}\\right)\\left(\\dfrac{T_2}{T_1}\\right)^{3/2} \\dfrac{\\Omega_{D,T_1}}{\\Omega_{D,T_2}}$ (take the $\\Omega$ ratio as 1 when no collision data are given) (on the formula sheet)\n- Multicomponent mixture: $\\dfrac{1}{D_{1-mix}} = \\sum_{j \\ne 1} \\dfrac{y'_j}{D_{1-j}}$, $y'_j = \\dfrac{y_j}{\\sum_{j \\ne 1} y_j}$ (on the formula sheet)\n- Diffusion through a stagnant gas: $N_A = \\dfrac{P D_{AB}}{R T \\delta} \\dfrac{p_{A1} - p_{A2}}{p_{B,lm}}$, $p_{B,lm} = \\dfrac{p_{B2} - p_{B1}}{\\ln(p_{B2}/p_{B1})}$ (equivalent to the $\\ln$ form used in the solution) (on the formula sheet)\n- Unit conversions: $1\\ \\mathrm{cm^2/s} = 10^{-4}\\ \\mathrm{m^2/s}$; $R = 8314\\ \\mathrm{J\\,kmol^{-1}\\,K^{-1}}$ (learn this)\n","answer":[{"label":"N_A","value":0.00000307,"units":"kmol m^-2 s^-1","tol":0.05}],"marks":30,"minutes":36,"tags":["mass-transfer","diffusion","stagnant-gas","diffusivity"],"formulas":[],"source":{"name":"ENGR271 Summer 2026 Q3","paper":"ENGR271","year":2026,"question":"Q3","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR271 In-person Main Exam 2025-26 - FINAL FOR SPA v2.pdf","tag":"VALID"},"text":"Oxygen is diffusing through a stagnant gas mixture containing 50% methane, 30% hydrogen and 20%\ncarbon dioxide by volume. The total pressure is $2 \\times 10^5\\ \\mathrm{N/m^2}$ and the\ntemperature is 20 deg C. The partial pressures of oxygen at two planes 25 mm apart are\n$23 \\times 10^3$ and $9 \\times 10^3\\ \\mathrm{N/m^2}$ respectively. Estimate the rate of diffusion\nof oxygen.\n\nGiven: at 0 deg C and 1 atm, the diffusivities of oxygen with respect to methane, hydrogen and\ncarbon dioxide are $D_{O_2-CH_4} = 0.184\\ \\mathrm{cm^2/s}$, $D_{O_2-H_2} = 0.69\\ \\mathrm{cm^2/s}$\nand $D_{O_2-CO_2} = 0.139\\ \\mathrm{cm^2/s}$.\n","solution":"$83","confidence":"HIGH"},{"id":"engr5003-ex-work-01","module":"ENGR5003","week":1,"label":"Lecture 2 exercise - piston static pressure","kind":"tutorial","level":1,"minutes":5,"tags":["ideal-gas-first-law","absolute-pressure","free-piston"],"source":{"name":"ENGR5003 Lecture 2 student exercise 1 (slide 12)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/ENGR5003-Week1-Lecture2.pdf"},"text":"A gas is contained in a vertical cylinder by a moveable, frictionless piston. The piston has a mass of 4 kg and a\ncross-sectional area of $35\\ \\mathrm{cm^2}$. A compressed spring above the piston exerts a force of 60 N on the\npiston. If atmospheric pressure is 95 kPa, determine the pressure inside the cylinder.\n","answer":[{"label":"p","value":123.4,"units":"kPa"}],"grade":"third","formulaSheet":"- Pressure as force per unit area: $p = F/A$ (learn this)\n- Static force balance on the piston: $pA = p_{atm}A + mg + F_{spring}$ (learn this: draw the free-body diagram)\n","solution":"$84","confidence":"HIGH","solutionNote":"No worked solution on the slide; standard force balance (the same exercise is in the old-spec L3 deck)."},{"id":"engr5003-ex-work-02","module":"ENGR5003","week":1,"label":"Lecture 2 exercise - drive-shaft power","kind":"tutorial","level":1,"minutes":3,"tags":["ideal-gas-first-law","shaft-work","power"],"source":{"name":"ENGR5003 Lecture 2 student exercise 2 (slide 12)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/ENGR5003-Week1-Lecture2.pdf"},"text":"Determine the power transmitted through the drive shaft of a car when the torque applied at the clutch is\n$250\\ \\mathrm{N\\,m}$ and the shaft rotates at 6000 rpm.\n","answer":[{"label":"P","value":157.1,"units":"kW"}],"grade":"third","formulaSheet":"- Shaft work: $dW = T\\,d\\theta$ (on the formula sheet, forms of mechanical work)\n- Shaft power: $\\dot W = T\\omega = 2\\pi n T$ with $n$ in rev/s (learn this)\n","solution":"Shaft work is torque times angle turned, $dW = T\\,d\\theta$ (Lecture 2, slide 9). Dividing by time gives the power\n$\\dot W = T\\,d\\theta/dt = T\\omega$.\n\nAngular speed: $\\omega = 2\\pi \\times \\dfrac{6000}{60} = 2\\pi \\times 100 = 628.3\\ \\mathrm{rad/s}$.\n\n$$\\dot W = T\\omega = 250 \\times 628.3 = 157\\,080\\ \\mathrm{W} \\approx 157.1\\ \\mathrm{kW}$$\n\nChecks: units $\\mathrm{N\\,m} \\times \\mathrm{rad/s} = \\mathrm{J/s} = \\mathrm{W}$. Work per revolution is\n$2\\pi T = 1571\\ \\mathrm{J}$; at 100 rev/s that is 157.1 kW, the same. About 210 hp, plausible for a performance car at\nhigh revs. Trap: using rpm directly ($250 \\times 6000$) instead of converting to rad/s.\n","confidence":"HIGH","solutionNote":"No worked solution on the slide; checked by work-per-revolution."},{"id":"engr5003-ex-work-03","module":"ENGR5003","week":1,"label":"Lecture 2 exercise - surface-tension work on a bubble","kind":"tutorial","level":1,"minutes":5,"tags":["ideal-gas-first-law","surface-tension","boundary-work"],"source":{"name":"ENGR5003 Lecture 2 student exercise 3 (slide 12)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/ENGR5003-Week1-Lecture2.pdf"},"text":"As a spherical bubble of ammonia vapour rises in liquid ammonia, its diameter changes from 5 mm to 25 mm. If the\nsurface tension of ammonia is $0.02\\ \\mathrm{N/m}$, calculate the work done by the system to enlarge this bubble.\n","answer":[{"label":"W","value":0.0377,"units":"mJ"}],"grade":"third","formulaSheet":"- Surface-tension work: $dW = \\sigma_s\\,dA$ (on the formula sheet, forms of mechanical work)\n- Surface area of a sphere: $A = \\pi D^2$ (learn this)\n","solution":"Creating new liquid-vapour interface needs work against surface tension, $dW = \\sigma_s\\,dA$ (Lecture 2, slide 10). A\nvapour bubble inside a liquid has ONE interface (unlike a soap film, which has two), so the area is that of one sphere,\n$A = 4\\pi r^2 = \\pi D^2$.\n\n$$A_1 = \\pi (0.005)^2 = 7.854 \\times 10^{-5}\\ \\mathrm{m^2}, \\qquad A_2 = \\pi (0.025)^2 = 1.9635 \\times 10^{-3}\\ \\mathrm{m^2}$$\n$$\\Delta A = 1.885 \\times 10^{-3}\\ \\mathrm{m^2}$$\n$$W = \\sigma_s\\,\\Delta A = 0.02 \\times 1.885 \\times 10^{-3} = 3.77 \\times 10^{-5}\\ \\mathrm{J} = 0.0377\\ \\mathrm{mJ}$$\n\nThis is the energy that the system (the expanding bubble) must supply to stretch the interface.\n\nChecks: units $\\mathrm{N/m} \\times \\mathrm{m^2} = \\mathrm{N\\,m} = \\mathrm{J}$. The final area dominates (25 times the\ninitial), so $W \\approx \\sigma_s \\pi D_2^2 = 3.93 \\times 10^{-5}\\ \\mathrm{J}$ is a quick upper bound. Trap: counting two\nsurfaces (that is for a liquid film, e.g. the wire-ring problem in old Tutorial 1) would double the answer.\n","confidence":"HIGH","solutionNote":"No worked solution on the slide; single interface assumed for a vapour bubble in its own liquid."},{"id":"engr5003-ex-exergy-01","module":"ENGR5003","week":1,"label":"Lecture 3 Example 1 - energy vs exergy losses in a steam plant","kind":"tutorial","level":1,"minutes":10,"tags":["second-law","exergy","rankine-cycle","entropy-production"],"source":{"name":"ENGR5003 Lecture 3 Example 1 (slide 21)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/ENGR5003-Week1-Lecture3.pdf"},"text":"Two different analyses of the same steam power plant give the following loss breakdowns (energy = exergy + anergy).\n\n| Equipment | Energy loss / energy input (%) | Exergy loss / exergy input (%) |\n|---|---|---|\n| Boiler and super-heater | 9 | 49 (combustion 29.7, heat transfer 14.9, flue 0.68, others 3.72) |\n| Turbine | approx. 0 | 4 |\n| Condenser | 47 | 1.5 |\n| Pump | approx. 0 | 1.0 |\n| Others | 3 | 5.5 |\n| In total | 59 | 61 |\n","parts":[{"label":"a","text":"Explain why the condenser accounts for most of the ENERGY loss but almost none of the EXERGY loss, while the\nboiler is the other way round.\n","grade":"$undefined"},{"label":"b","text":"Use the totals to estimate the plant's energy (first-law) efficiency and its exergy (second-law) efficiency.","answer":[{"label":"eta_energy","value":0.41},{"label":"eta_exergy","value":0.39}],"grade":"$undefined"},{"label":"c","text":"What are strategies for reducing the exergy and energy losses?","grade":"$undefined"}],"grade":"third","formulaSheet":"- Exergy of heat $q$ supplied at mean temperature $T_{mH}$: $e_{x,Q} = q\\left(1 - \\dfrac{T_0}{T_{mH}}\\right) = q - T_0\\Delta s$ (learn this, Lecture 3 slide 4)\n- Maximum (Carnot) work from heat: $w_{max} = q\\left(1 - \\dfrac{T_0}{T}\\right)$ (learn this, Lecture 3 slide 20)\n- Entropy increase principle: $\\Delta S_{iso} = S_g \\ge 0$ (on the formula sheet)\n","solution":"$85","confidence":"HIGH","solutionNote":"The slide gives only the table and the question; explanation follows the Lecture 3 exergy definitions."},{"id":"engr5003-ex-exergy-02","module":"ENGR5003","week":1,"label":"Lecture 3 Example 2 - physical and chemical exergy of hydrogen","kind":"tutorial","level":1,"minutes":8,"tags":["second-law","exergy","entropy-production","combustion"],"source":{"name":"ENGR5003 Lecture 3 Example 2 (slide 22)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/ENGR5003-Week1-Lecture3.pdf"},"text":"Figure 1 follows hydrogen from storage to exhaust: $\\mathrm{H_2}$ at 1 MPa, 298 K is throttled to 0.1 MPa, mixed\nwith air, burnt (products at 2 MPa, 1500 K), cooled to 0.1 MPa, 298 K and finally exhausted to the atmosphere\n(ambient $T_0$, $p_0$; mole fractions $\\mathrm{N_2}$ 75.6%, $\\mathrm{O_2}$ 20.3%, $\\mathrm{H_2O}$ 3%,\n$\\mathrm{CO_2}$ 0.03%, Ar 0.91%). Each bar shows the exergy of the stream split into reaction (chemical), diffusion\n(concentration), mechanical (pressure) and thermal parts; the white part of a bar is exergy that has been lost.\n","figures":[{"src":"/figures/engr5003-ex-exergy-02.png","caption":"Figure 1: exergy of the hydrogen stream at each stage, split into reaction, diffusion, mechanical and thermal exergy (Lecture 3, slide 22)."}],"parts":[{"label":"a","text":"For each step (pressure drop, mixing, combustion, cooling, exhaust), state which kind of exergy is lost and why.","grade":"$undefined"},{"label":"b","text":"Which step destroys the most exergy, and why does the stream still have exergy after cooling to 0.1 MPa and 298 K?","grade":"$undefined"},{"label":"c","text":"Suggest how the losses in the first three steps could be avoided or reduced.","grade":"$undefined"}],"grade":"third","formulaSheet":"- Exergy: the maximum useful work obtainable as a system comes into equilibrium with its surroundings (learn this, Lecture 3)\n- Exergy destroyed $= T_0 S_g$, with $S_g \\ge 0$ for any real process (learn this; $\\Delta S_{iso} = S_g \\ge 0$ is on the formula sheet)\n","solution":"$86","confidence":"HIGH","solutionNote":"The slide gives only the figure; interpretation follows the Lecture 3 exergy definitions (qualitative, no numbers to check)."},{"id":"engr5003-ex-brayton-01","module":"ENGR5003","week":2,"label":"Lecture 4 Exercise 1 - ideal Brayton cycle","kind":"tutorial","level":1,"minutes":10,"tags":["gas-turbines","brayton-cycle","thermal-efficiency","net-work","sfee"],"source":{"name":"ENGR5003 Lecture 4 Exercise 1 (slides 20-21)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 2/ENGR5003-Week2-Lecture4.pdf"},"text":"Consider an ideal Brayton cycle (isentropic compression and isentropic expansion). Air may be treated as an ideal gas\nwith $\\gamma = 1.4$ and $c_p = 1005\\ \\mathrm{J/(kg\\,K)}$. The inlet state is $T_1 = 300\\ \\mathrm{K}$. The compressor\npressure ratio is $r_p = 8$. The maximum (turbine inlet) temperature after combustion is $T_3 = 1200\\ \\mathrm{K}$.\nStates are numbered as in Figure 1.\n","figures":[{"src":"/figures/engr5003-ex-brayton-01.png","caption":"Figure 1: ideal Brayton cycle on a T-s diagram: 1-2 isentropic compression, 2-3 heat addition at constant p, 3-4 isentropic expansion, 4-1 heat rejection at constant p."}],"parts":[{"label":"a","text":"Calculate $T_2$ and $T_4$ (isentropic).","answer":[{"label":"T2","value":543.43,"units":"K","tol":0.005},{"label":"T4","value":662.45,"units":"K","tol":0.005}],"grade":"$undefined"},{"label":"b","text":"Calculate the net work per kg of air (J/kg).","answer":[{"label":"w_net","value":295580,"units":"J/kg"}],"grade":"$undefined"},{"label":"c","text":"Calculate the thermal efficiency $\\eta_{th}$.","answer":[{"label":"eta_th","value":0.448,"tol":0.01}],"grade":"$undefined"}],"grade":"third","formulaSheet":"- Adiabatic (isentropic) ideal gas: $\\dfrac{T_2}{T_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- SFEE with negligible KE and PE: $q - w = \\Delta h = c_p\\Delta T$ (on the formula sheet)\n- Brayton efficiency: $\\eta_{th} = 1 - \\dfrac{T_1}{T_2} = 1 - r_p^{-(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Thermal efficiency: $\\eta_{th} = \\dfrac{w_{net}}{q_{in}}$ (on the formula sheet)\n","solution":"$87","confidence":"HIGH","solutionNote":"Matches the slide's step-by-step solution (T2 = 543.43 K, T4 = 662.45 K, w_net = 2.9558e5 J/kg, 44.8%)."},{"id":"engr5003-ex-charging-01","module":"ENGR5003","week":2,"label":"Lecture 6 worked example 1 - charging an insulated vessel","kind":"tutorial","level":1,"minutes":6,"tags":["charging-discharging","unsteady-flow","flow-work","rigid-vessel"],"source":{"name":"ENGR5003 Lecture 6 worked example 1 (slides 10-12)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 2/ENGR5003-Week2-Lecture6.pdf"},"text":"An insulated vessel of negligible heat capacity contains 1 kg of air at 360 K. A compressor pumps in 2 kg of air at a\ntemperature of 320 K. Find the final temperature of the air in the vessel. Assume $\\gamma = 1.4$ for air.\n","answer":[{"label":"T2","value":418.7,"units":"K"}],"grade":"third","formulaSheet":"- Unsteady flow energy equation: $\\Delta U = \\left(h + \\tfrac{1}{2}C^2 + gz\\right)_{in}\\Delta m_{in} - \\left(h + \\tfrac{1}{2}C^2 + gz\\right)_{out}\\Delta m_{out} + \\Delta Q - \\Delta W$ (on the formula sheet)\n- Ideal gas: $u = c_vT$, $h = c_pT$, $\\gamma = c_p/c_v$ (on the formula sheet)\n","solution":"$88","confidence":"HIGH","solutionNote":"Matches the slide (419 K)."},{"id":"engr5003-ex-charging-02","module":"ENGR5003","week":2,"label":"Lecture 6 worked example 2 - isothermal discharge of a heated tank","kind":"tutorial","level":2,"minutes":8,"tags":["charging-discharging","unsteady-flow","rigid-vessel","ideal-gas"],"source":{"name":"ENGR5003 Lecture 6 worked example 2 (slides 13-15)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 2/ENGR5003-Week2-Lecture6.pdf"},"text":"Air is discharged from a tank of volume $2\\ \\mathrm{m^3}$ until the pressure drops from 5 bar to 2 bar. The air in the\ntank is maintained at a constant temperature of $52\\ ^\\circ\\mathrm{C}$ by an electrical heating element in the tank.\nDetermine the electrical heat input during this process. Take $R_g = 0.287\\ \\mathrm{kJ/(kg\\,K)}$.\n","parts":[{"label":"a","text":"Find the initial and final mass of air in the tank and the mass discharged.","answer":[{"label":"m1","value":10.72,"units":"kg"},{"label":"m2","value":4.29,"units":"kg"},{"label":"delta m","value":6.43,"units":"kg"}],"grade":"$undefined"},{"label":"b","text":"Determine the electrical heat input.","answer":[{"label":"Q_e","value":600,"units":"kJ"}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Ideal gas: $pV = mR_gT$ (on the formula sheet)\n- Unsteady flow energy equation: $\\Delta U = h_{in}\\Delta m_{in} - h_{out}\\Delta m_{out} + \\Delta Q - \\Delta W$ (KE, PE neglected) (on the formula sheet)\n- Ideal gas: $u = c_vT$, $h = c_pT$, $c_p - c_v = R_g$ (on the formula sheet)\n","solution":"$89","confidence":"HIGH","solutionNote":"Matches the slide (600 kJ). Slide typo on slide 14 writes m2 = 10.79 x 2/5; it should be 10.72 x 2/5 = 4.29 kg (the result 4.29 is right)."},{"id":"engr5003-ex-throttling-01","module":"ENGR5003","week":2,"label":"Lecture 6 worked example 3 - R134a capillary-tube throttling","kind":"tutorial","level":1,"minutes":6,"tags":["steady-flow","throttling","r134a","dryness-fraction","isenthalpic"],"source":{"name":"ENGR5003 Lecture 6 worked example 3 (slides 16-18)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 2/ENGR5003-Week2-Lecture6.pdf"},"text":"A refrigerant (R134a) enters a capillary tube of a refrigerator as a saturated liquid at 0.8 MPa and is throttled to a\npressure of 0.12 MPa. Determine the temperature of the initial state and the temperature and dryness fraction of the\nfinal state.\n\nR134a saturation data (pressure table):\n\n| p (kPa) | T (C) | $v_f$ (m3/kg) | $v_g$ (m3/kg) | $h_f$ (kJ/kg) | $h_{fg}$ (kJ/kg) | $h_g$ (kJ/kg) |\n|---|---|---|---|---|---|---|\n| 120 | -22.3 | 0.0007324 | 0.1621 | 22.5 | 214.5 | 237.0 |\n| 800 | 31.3 | 0.0008459 | 0.0256 | 95.5 | 171.8 | 267.3 |\n","answer":[{"label":"T1","value":31.3,"units":"C","tol":0.01},{"label":"T2","value":-22.3,"units":"C","tol":0.01},{"label":"x2","value":0.34,"tol":0.02}],"grade":"third","formulaSheet":"- SFEE with $q = w = 0$ and negligible KE, PE: $h_1 = h_2$ (throttling is isenthalpic) (on the formula sheet as the SFEE)\n- Dryness fraction: $h = (1 - x)h_f + xh_g$, so $x = \\dfrac{h - h_f}{h_g - h_f}$ (on the formula sheet)\n","solution":"$8a","confidence":"HIGH","solutionNote":"Matches the slide (31.3 C, -22.3 C, x = 0.34)."},{"id":"engr5003-ex-realgas-01","module":"ENGR5003","week":3,"label":"Lecture 7 example - nitrogen vessel by ideal gas, van der Waals and Redlich-Kwong","kind":"tutorial","level":2,"minutes":15,"tags":["real-gas-steam","van-der-waals","redlich-kwong","compressibility-factor","ideal-gas-validity"],"source":{"name":"ENGR5003 Lecture 7 example (slides 16-18)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 3/ENGR5003-Week3-Lecture7.pdf"},"text":"Nitrogen is filled in a vessel with a volume of $0.425\\ \\mathrm{m^3}$. The pressure and temperature of the nitrogen\nare 16.21 MPa and 189 K respectively. Determine the mass of the nitrogen by (a) the ideal gas model, (b) the van der\nWaals equation and (c) the Redlich-Kwong (R-K) equation.\n\nData: $M = 28.01\\ \\mathrm{kg/kmol}$, $R = 8314.5\\ \\mathrm{J/(kmol\\,K)}$. Van der Waals constants for nitrogen:\n$a = 1.361 \\times 10^5\\ \\mathrm{Pa\\,m^6/kmol^2}$ ($= 0.1361\\ \\mathrm{Pa\\,m^6/mol^2}$),\n$b = 0.0385\\ \\mathrm{m^3/kmol}$. For R-K use the critical point $T_c = 126.2\\ \\mathrm{K}$, $p_c = 3.39\\ \\mathrm{MPa}$.\n","parts":[{"label":"a","text":"Ideal gas model.","answer":[{"label":"m_ideal","value":122.8,"units":"kg"}],"grade":"$undefined"},{"label":"b","text":"Van der Waals equation $\\left(p + \\dfrac{a}{V_m^2}\\right)(V_m - b) = RT$.","answer":[{"label":"V_m","value":0.0811,"units":"m^3/kmol"},{"label":"m_vdW","value":146.9,"units":"kg"}],"grade":"$undefined"},{"label":"c","text":"Redlich-Kwong equation $p = \\dfrac{RT}{V_m - b} - \\dfrac{a}{T^{0.5}V_m(V_m + b)}$ with\n$a = \\dfrac{0.42748R^2T_c^{2.5}}{p_c}$ and $b = \\dfrac{0.08664RT_c}{p_c}$.\n","answer":[{"label":"V_m","value":0.08023,"units":"m^3/kmol","tol":0.01},{"label":"m_RK","value":148.4,"units":"kg"}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Ideal gas: $pV = nRT$, $m = nM$ (on the formula sheet)\n- Van der Waals: $p = \\dfrac{RT}{V_m - b} - \\dfrac{a}{V_m^2}$ (learn this; Lecture 7, marked \"for students who are interested\")\n- Redlich-Kwong: $p = \\dfrac{RT}{V_m - b} - \\dfrac{a}{T^{0.5}V_m(V_m + b)}$, $a = \\dfrac{0.42748R^2T_c^{2.5}}{p_c}$, $b = \\dfrac{0.08664RT_c}{p_c}$ (learn this; Lecture 7)\n- Compressibility factor: $Z = \\dfrac{pv}{R_gT}$ (learn this)\n","solution":"$8b","confidence":"HIGH","solutionNote":"Slide errors: (1) the constants are printed as a = 0.1361e6 (MPa m6)/mol2 and b = 0.0385e3 m3/mol; the values actually used (with R = 8314.5) are per kmol, a = 1.361e5 Pa m6/kmol2, b = 0.0385 m3/kmol. (2) The R-K a is printed as 0.13864 Pa m6 K0.5 mol-2; it is 1.560 Pa m6 K0.5 mol-2 (1.560e6 per kmol2), although the slide's V_m = 0.080238 is consistent with the correct value. (3) The R-K mass is printed as 148.84 kg; 0.425/0.080238 x 28.01 = 148.36 kg. (4) The slide concludes van der Waals is more accurate than R-K; against reference data (149.5 kg) R-K is closer."},{"id":"engr5003-ex-steam-01","module":"ENGR5003","week":3,"label":"Lecture 8 examples - linear interpolation in the saturation table","kind":"tutorial","level":1,"minutes":4,"tags":["real-gas-steam","steam-tables","interpolation"],"source":{"name":"ENGR5003 Lecture 8 examples (slide 16)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 3/ENGR5003-Week3-Lecture8.pdf"},"text":"Saturated water, pressure table extract:\n\n| p (bar) | $T_s$ (C) | $v_g$ (m3/kg) | $u_f$ (kJ/kg) | $u_g$ (kJ/kg) | $h_f$ (kJ/kg) | $h_{fg}$ (kJ/kg) | $h_g$ (kJ/kg) |\n|---|---|---|---|---|---|---|---|\n| 9 | 175.4 | 0.2149 | 742 | 2581 | 743 | 2031 | 2774 |\n| 10 | 179.9 | 0.1944 | 762 | 2584 | 763 | 2015 | 2778 |\n","parts":[{"label":"a","text":"Find the saturation temperature $t_s$ at a pressure of 9.8 bar.","answer":[{"label":"t_s","value":179,"units":"C","tol":0.005}],"grade":"$undefined"},{"label":"b","text":"Find $h_g$ at 9.8 bar.","answer":[{"label":"h_g","value":2777.2,"units":"kJ/kg","tol":0.001}],"grade":"$undefined"}],"grade":"third","formulaSheet":"- Linear interpolation: $\\dfrac{y - y_1}{y_2 - y_1} = \\dfrac{x - x_1}{x_2 - x_1}$ (learn this, Lecture 8-9)\n","solution":"9.8 bar lies between the tabulated 9 and 10 bar rows. Linear interpolation assumes the property varies linearly between\nthem; the fraction of the interval is\n$$f = \\frac{9.8 - 9}{10 - 9} = 0.8$$\n\n**(a)** $t_s = 175.4 + 0.8(179.9 - 175.4) = 175.4 + 3.6 = 179.0\\ ^\\circ\\mathrm{C}$\n\n**(b)** $h_g = 2774 + 0.8(2778 - 2774) = 2774 + 3.2 = 2777.2\\ \\mathrm{kJ/kg}$\n\nChecks: both answers lie between the table rows and closer to the 10 bar values, as they should. IAPWS-IF97 gives\n$t_s = 179.01\\ ^\\circ\\mathrm{C}$ and $h_g = 2776.3\\ \\mathrm{kJ/kg}$ at 9.8 bar; the 0.9 kJ/kg difference comes from this\ntable's rounding to whole kJ/kg (IAPWS gives 2773.0 and 2777.1 at 9 and 10 bar), not from the interpolation.\n","confidence":"HIGH","solutionNote":"Matches the slide (t_s = 179 C, h_g = 2777.2 kJ/kg)."},{"id":"engr5003-ex-steam-02","module":"ENGR5003","week":3,"label":"Lecture 8 example - double interpolation in the superheated table","kind":"tutorial","level":1,"minutes":6,"tags":["real-gas-steam","steam-tables","interpolation","superheated-steam"],"source":{"name":"ENGR5003 Lecture 8 double interpolation example (slides 17-18)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 3/ENGR5003-Week3-Lecture8.pdf"},"text":"Find the specific enthalpy of superheated steam at 18.5 bar and $432\\ ^\\circ\\mathrm{C}$, using the superheated table\nvalues below (h in kJ/kg).\n\n| p (bar) | h at 400 C | h at 450 C |\n|---|---|---|\n| 15 | 3256 | 3364 |\n| 20 | 3248 | 3357 |\n","answer":[{"label":"h","value":3320,"units":"kJ/kg","tol":0.001}],"grade":"third","formulaSheet":"- Linear interpolation: $\\dfrac{y - y_1}{y_2 - y_1} = \\dfrac{x - x_1}{x_2 - x_1}$ (learn this, Lecture 8-9)\n","solution":"Neither 18.5 bar nor 432 C is in the table, so interpolate twice: first in pressure at each tabulated temperature, then\nin temperature at 18.5 bar.\n\n**1st interpolation (pressure)**, fraction $f_p = (18.5 - 15)/(20 - 15) = 0.7$:\n$$h(18.5, 400) = 3256 + 0.7(3248 - 3256) = 3250.4\\ \\mathrm{kJ/kg}$$\n$$h(18.5, 450) = 3364 + 0.7(3357 - 3364) = 3359.1\\ \\mathrm{kJ/kg}$$\n\n**2nd interpolation (temperature)**, fraction $f_T = (432 - 400)/(450 - 400) = 0.64$:\n$$h(18.5, 432) = 3250.4 + 0.64(3359.1 - 3250.4) = 3320.0\\ \\mathrm{kJ/kg}$$\n\nChecks: (i) doing temperature first gives $h(15, 432) = 3325.1$, $h(20, 432) = 3317.8$ and then\n$3325.1 + 0.7(3317.8 - 3325.1) = 3320.0$, the same (bilinear interpolation does not depend on the order).\n(ii) IAPWS-IF97 at 1.85 MPa, 432 C gives 3320.7 kJ/kg, within 0.02%. (iii) Enthalpy falls slightly with pressure\nand rises about 2.2 kJ/kg per K at constant pressure, both as expected for superheated steam.\n","confidence":"HIGH","solutionNote":"Matches the slide (3250.4, 3320 kJ/kg)."},{"id":"engr5003-ex-steam-03","module":"ENGR5003","week":3,"label":"Old spec L5 worked example 1 - boiling water at constant pressure","kind":"tutorial","level":1,"minutes":8,"tags":["real-gas-steam","steam-tables","boundary-work","first-law","closed-vs-open-system"],"source":{"name":"ENGR217 Lecture 5 steam tables worked example 1 (slides 6-11)","paper":"ENGR217","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L5 Steam tables - worked examples (old spec ENGR217).pdf"},"text":"1 kg of water at $T_1 = 30\\ ^\\circ\\mathrm{C}$ and $p = 1$ bar (100 kPa) is heated at constant pressure until it becomes\ndry saturated steam. Determine the heat added, the work that the expanding steam does on the atmosphere and therefore\nthe change in internal energy. Use steam tables.\n","parts":[{"label":"a","text":"Heat added.","answer":[{"label":"Q","value":2549.3,"units":"kJ","tol":0.005}],"grade":"$undefined"},{"label":"b","text":"Work done by the expanding steam on the atmosphere.","answer":[{"label":"W","value":169.3,"units":"kJ","tol":0.005}],"grade":"$undefined"},{"label":"c","text":"Change in internal energy.","answer":[{"label":"delta U","value":2380,"units":"kJ","tol":0.005}],"grade":"$undefined"}],"grade":"third","formulaSheet":"- First law: $\\Delta U = Q - W$ (heat in and work out positive) (on the formula sheet)\n- Enthalpy: $h = u + pv$ (on the formula sheet)\n- Constant-pressure boundary work: $W = p(V_2 - V_1)$ (on the formula sheet as $dW = p\\,dV$)\n- Constant-pressure heat for a closed system: $Q = m\\Delta h$ (learn this: $Q = \\Delta U + p\\Delta V = \\Delta H$)\n","solution":"$8c","confidence":"HIGH","solutionNote":"Matches the slide (2549.3 kJ, 169.3 kJ, 2380 kJ)."},{"id":"engr5003-ex-steam-04","module":"ENGR5003","week":3,"label":"Old spec L5 worked example 2 - heating wet steam in a rigid tank","kind":"tutorial","level":2,"minutes":12,"tags":["real-gas-steam","steam-tables","rigid-tank","dryness-fraction","interpolation"],"source":{"name":"ENGR217 Lecture 5 steam tables worked example 2 (slides 12-16)","paper":"ENGR217","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L5 Steam tables - worked examples (old spec ENGR217).pdf"},"text":"A rigid tank of volume $2.5\\ \\mathrm{m^3}$ contains 5 kg of wet steam at $75\\ ^\\circ\\mathrm{C}$. How much heat is\nrequired to completely convert this into dry saturated steam? What would be the final pressure and temperature?\n\nSaturated water data: at 75 C, $v_f = 0.001026$, $v_g = 4.1291\\ \\mathrm{m^3/kg}$, $u_f = 313.99$,\n$u_g = 2475.3\\ \\mathrm{kJ/kg}$. Pressure table: at 350 kPa, $T_s = 138.86\\ ^\\circ\\mathrm{C}$, $v_g = 0.52422\\ \\mathrm{m^3/kg}$,\n$u_g = 2548.5\\ \\mathrm{kJ/kg}$; at 375 kPa, $T_s = 141.30\\ ^\\circ\\mathrm{C}$, $v_g = 0.49133\\ \\mathrm{m^3/kg}$,\n$u_g = 2550.9\\ \\mathrm{kJ/kg}$.\n","parts":[{"label":"a","text":"Find the initial dryness fraction and specific internal energy.","answer":[{"label":"x1","value":0.121,"tol":0.01},{"label":"u1","value":575.7,"units":"kJ/kg","tol":0.005}],"grade":"$undefined"},{"label":"b","text":"Find the final pressure and temperature when the steam is just dry saturated.","answer":[{"label":"p2","value":368.4,"units":"kPa","tol":0.005},{"label":"T2","value":140.66,"units":"C","tol":0.002}],"grade":"$undefined"},{"label":"c","text":"Find the heat required.","answer":[{"label":"Q","value":9873,"units":"kJ","tol":0.005}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- First law: $\\Delta U = Q - W$, with $W = 0$ for a rigid tank (on the formula sheet)\n- Wet steam: $v = (1 - x)v_f + xv_g$, $u = (1 - x)u_f + xu_g$ (on the formula sheet)\n- Linear interpolation: $\\dfrac{y - y_1}{y_2 - y_1} = \\dfrac{x - x_1}{x_2 - x_1}$ (learn this)\n","solution":"$8d","confidence":"HIGH","solutionNote":"Matches the slide (x = 0.1211, u1 = 575.7, p2 = 368.4 kPa, T2 = 140.66 C, Q = 9872.8 kJ)."},{"id":"engr5003-ex-steam-05","module":"ENGR5003","week":3,"label":"Old spec L5 worked example 3 - isentropic expansion of dry steam, closed vs open","kind":"tutorial","level":1,"minutes":8,"tags":["real-gas-steam","steam-tables","isentropic-expansion","closed-vs-open-system","dryness-fraction"],"source":{"name":"ENGR217 Lecture 5 steam tables worked example 3 (slides 17-19)","paper":"ENGR217","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L5 Steam tables - worked examples (old spec ENGR217).pdf"},"text":"Dry saturated steam at 5 bar is expanded isentropically (constant entropy) to a final pressure of 1 bar. Determine the\ndryness fraction at 1 bar. What is the specific work done if:\n","parts":[{"label":"a","text":"the expansion is in a cylinder (i.e. a closed system)?","answer":[{"label":"x2","value":0.911,"tol":0.005},{"label":"w","value":240.7,"units":"kJ/kg","tol":0.01}],"grade":"$undefined"},{"label":"b","text":"the expansion is through a turbine (i.e. an open system)?","answer":[{"label":"w","value":273.8,"units":"kJ/kg","tol":0.01}],"grade":"$undefined"}],"grade":"third","formulaSheet":"- Wet steam: $s = (1 - x)s_f + xs_g$ (and the same for $u$, $h$) (on the formula sheet)\n- Closed adiabatic process: $w = -\\Delta u = u_1 - u_2$ (on the formula sheet)\n- SFEE, adiabatic, negligible KE and PE: $w = h_1 - h_2$ (on the formula sheet)\n","solution":"$8e","confidence":"HIGH","solutionNote":"Matches the slide (x = 0.911, 240.7 and 273.8 kJ/kg)."},{"id":"engr5003-ex-engine-01","module":"ENGR5003","week":4,"label":"Lecture 11 worked example - realistic petrol engine efficiency","kind":"tutorial","level":2,"minutes":18,"tags":["ic-engines","otto-cycle","engine-efficiencies","engine-losses","variable-specific-heat"],"source":{"name":"ENGR5003 Lecture 11 worked example (slides 12-17)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 4/ENGR5003-Week4-Lecture11.pdf"},"text":"The p-V diagram of a petrol engine at maximum power output (full throttle) with a compression ratio of 10 and a\ngravimetric air/fuel ratio of 15 is analysed on the basis of an Otto cycle with the following modifications to\napproximate real conditions:\n- for compression from state 1 to 2: $\\gamma = 1.4$;\n- for heat addition from state 2 to 3: $c_v = 0.880\\ \\mathrm{kJ/(kg\\,K)}$;\n- for expansion from state 3 to 4: $\\gamma = 1.3$;\n- $p_3$ is reduced by 20% to allow for incomplete combustion and heat-transfer losses;\n- 15% of the work output is lost to friction and to driving essential auxiliaries (e.g. the water pump).\n\nEstimate the brake thermal efficiency of the engine given $T_1 = 288\\ \\mathrm{K}$, calorific value of the fuel\n43 710 kJ/kg and air gas constant $R = 0.287\\ \\mathrm{kJ/(kg\\,K)}$.\n","parts":[{"label":"a","text":"Find $T_2$ after compression.","answer":[{"label":"T2","value":723.4,"units":"K","tol":0.005}],"grade":"$undefined"},{"label":"b","text":"Find the specific heat added per kg of mixture and the temperature $T_3$ after heat addition, with the 20% pressure reduction.","answer":[{"label":"q_in","value":2732,"units":"kJ/kg","tol":0.005},{"label":"T3","value":3062,"units":"K","tol":0.005}],"grade":"$undefined"},{"label":"c","text":"Find $T_4$ after expansion.","answer":[{"label":"T4","value":1535,"units":"K","tol":0.005}],"grade":"$undefined"},{"label":"d","text":"Find the compression and expansion work, the brake work and the brake thermal efficiency.","answer":[{"label":"w_c","value":312.4,"units":"kJ/kg"},{"label":"w_e","value":1461.3,"units":"kJ/kg"},{"label":"w_brake","value":929.7,"units":"kJ/kg"},{"label":"eta_brake","value":0.34,"tol":0.01}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Adiabatic process: $TV^{\\gamma-1} = \\text{const}$, so $T_2/T_1 = r^{\\gamma-1}$ (on the formula sheet)\n- Constant-volume heat addition: $q = c_v\\Delta T$ (on the formula sheet)\n- $c_v = \\dfrac{R_g}{\\gamma - 1}$; adiabatic work $w = -\\Delta u = c_v\\Delta T$ (on the formula sheet)\n- Ideal gas at constant volume: $p_3/T_3 = $ const (Gay-Lussac) (on the formula sheet)\n- Heat released per kg of mixture: $q_{in} = \\dfrac{CV}{AFR + 1}$ (learn this, Lecture 11)\n- Brake work = indicated work minus friction; $\\eta_b = w_b/q_{in}$ (learn this)\n","solution":"$8f","confidence":"HIGH","solutionNote":"Slide arithmetic error on slide 17: 0.85 x 1461 - 312.3 is printed as 919 kJ/kg (hence 33.6%); it is 929.5 kJ/kg, giving a brake thermal efficiency of 34.0%. All other slide values (723 K, 2732 kJ/kg, 3827 K, 3062 K, 1535 K, 312.3 and 1461 kJ/kg) agree."},{"id":"engr5003-ex-engine-02","module":"ENGR5003","week":4,"label":"Old spec L15 Question 2 - petrol engine power band","kind":"tutorial","level":2,"minutes":25,"tags":["ic-engines","otto-cycle","volumetric-efficiency","mechanical-efficiency","power-curve","engine-efficiencies"],"source":{"name":"ENGR217 Lecture 15 review of reciprocating engines, Question 2 (slides 9-17)","paper":"ENGR217","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 4 - Thermo L15 - Review of reciprocating engines (old spec ENGR217).pdf"},"text":"Consider a petrol engine with a volumetric airflow of $1.1N$ litres/second for $N \\le 50$ and 55 litres/second for\n$N > 50$ ($N$ is the engine speed in rev/s), a swept volume of 1.6 litres and a compression ratio of 10. The fuel takes\n0.01 seconds to completely combust, and during this time the heat addition is $Q_{in} = 200t$ kJ, where $t$ is the time\nin seconds since ignition. The gases are discharged through the exhaust valve halfway through the cycle, at which point\nany unburnt air/fuel mixture is no longer able to supply heat to the engine. As in the lecture, take one cycle per\nrevolution (cycle time $1/N$) and $\\gamma = 1.4$.\n","parts":[{"label":"a","text":"Determine the net power output as the engine speed varies from 0 to 200 rev/s (assume the thermal efficiency does not vary with speed).","answer":[{"label":"P_net/N (N <= 50)","value":1.204,"units":"kW per rev/s","tol":0.01},{"label":"P_net (N > 50)","value":60.19,"units":"kW","tol":0.01}],"grade":"$undefined"},{"label":"b","text":"Determine how the volumetric efficiency varies with engine speed from 0 to 200 rev/s.","answer":[{"label":"eta_v (N <= 50)","value":0.6875,"tol":0.005},{"label":"eta_v at N = 100","value":0.1719,"tol":0.01}],"grade":"$undefined"},{"label":"c","text":"Determine the output power of the engine, assuming the mechanical efficiency is 0.75, as a function of engine speed, and determine the power band for the engine (where output power is more than half the peak power).","answer":[{"label":"P_peak","value":31,"units":"kW"},{"label":"N_low","value":25,"units":"rev/s"},{"label":"N_high","value":100,"units":"rev/s"}],"grade":"$undefined"},{"label":"d","text":"Is it reasonable to assume an ideal Otto cycle in this case?","grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Otto efficiency: $\\eta_{th} = 1 - \\dfrac{1}{r^{\\gamma-1}}$ (on the formula sheet)\n- Volumetric efficiency: $\\eta_v = \\dfrac{\\dot V}{V_sN}$ (learn this, Lecture 10-11)\n- Power output: $P = \\eta_{th}\\eta_v\\eta_mQ_{in}N$ with $Q_{in}$ the heat for a full swept-volume charge (learn this)\n","solution":"$90","confidence":"MEDIUM","solutionNote":"Disagrees with the slide in (c) for N > 50. The slide writes P_out = 0.602 x (1718.75/N^2) x 0.75 x 100 = 77.6e3/N^2 kW and a power band of 25 to 70.71 rev/s (1500-4243 rpm). That counts the incomplete-combustion factor twice: it is already in its eta_v from (b) (55 x Q_in/Q_max) and again in Q_in = 100/N. Using it once (either route above) gives 1552/N kW and a band of 25 to 100 rev/s (1500-6000 rpm). (a), (b), the peak 31.0 kW and N_low = 25 rev/s agree with the slide. MEDIUM because the question's model is loosely specified (one cycle per revolution, meaning of Q_max); confirm the intended upper band limit with the lecturer."},{"id":"engr5003-ex-rankine-01","module":"ENGR5003","week":5,"label":"Lecture 13 worked example 1 - ideal Rankine cycle","kind":"tutorial","level":2,"minutes":18,"tags":["gas-turbines","rankine-cycle","steam-tables","t-s-diagram","back-work-ratio","thermal-efficiency"],"source":{"name":"ENGR5003 Lecture 13 worked example 1 (slides 9-10)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 5/ENGR5003-Week5-Lecture13.pdf"},"text":"An ideal Rankine cycle operates with water as the working fluid. Steam enters the turbine as superheated vapour at\n3 MPa and $350\\ ^\\circ\\mathrm{C}$. It expands isentropically to a condenser pressure of 10 kPa, where it condenses into a\nsaturated liquid. The liquid is then pumped back to the boiler. Determine:\n","parts":[{"label":"a","text":"Sketch the T-s diagram and label all four states.","grade":"$undefined"},{"label":"b","text":"The turbine work, pump work and net work output.","answer":[{"label":"w_turbine","value":980,"units":"kJ/kg","tol":0.005},{"label":"w_pump","value":3.02,"units":"kJ/kg"},{"label":"w_net","value":977,"units":"kJ/kg","tol":0.005}],"grade":"$undefined"},{"label":"c","text":"The heat input and thermal efficiency (also give the back work ratio).","answer":[{"label":"q_in","value":2921.2,"units":"kJ/kg","tol":0.005},{"label":"eta_th","value":0.334,"tol":0.01},{"label":"r_bw","value":0.0031,"tol":0.05}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- SFEE, adiabatic, negligible KE and PE: $w = h_{in} - h_{out}$; $q = h_{out} - h_{in}$ (on the formula sheet)\n- Wet steam: $s = (1 - x)s_f + xs_g$, $h = (1 - x)h_f + xh_g$ (on the formula sheet)\n- Pump work for an incompressible liquid: $w_p = v_f(p_2 - p_1)$ (learn this)\n- Back work ratio $r_{bw} = w_{in}/w_{out}$; $\\eta_{th} = w_{net}/q_{in}$ (on the formula sheet)\n","solution":"$91","confidence":"HIGH","solutionNote":"Slide error: the pump work is printed as 197.4 - 191.8 = 5.6 kJ/kg (so w_net = 973.9, q_in = 2918.7 and back work ratio 0.57%). With w = v_f dp = 0.00101 x 2990 = 3.02 kJ/kg (IAPWS h2 = 194.83 kJ/kg) the values are w_net = 977.0, q_in = 2921.2 and r_bw = 0.31%; the efficiency is 33.4% either way. The slide's h4 = 2136.6 differs from 2136.1 by table rounding."},{"id":"engr5003-ex-brayton-02","module":"ENGR5003","week":5,"label":"Lecture 13 worked example 2 - ideal Brayton cycle with air tables","kind":"tutorial","level":2,"minutes":15,"tags":["gas-turbines","brayton-cycle","back-work-ratio","variable-specific-heat","thermal-efficiency"],"source":{"name":"ENGR5003 Lecture 13 worked example 2 (slides 11-14)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 5/ENGR5003-Week5-Lecture13.pdf"},"text":"A gas-turbine power plant operating on an ideal Brayton cycle (Figure 1) has a pressure ratio of 8. The gas temperature\nis 300 K at the compressor inlet and 1300 K at the turbine inlet. Calculate the gas temperatures at the exits of the\ncompressor and the turbine. Also calculate the back work ratio, the cycle work and the thermal efficiency.\n\nUse the ideal-gas air table (Cengel Table A-17), where $P_r$ is the relative pressure, with\n$P_{r2}/P_{r1} = p_2/p_1$ for an isentropic process:\n\n| T (K) | h (kJ/kg) | $P_r$ |\n|---|---|---|\n| 300 | 300.19 | 1.386 |\n| 540 | 544.35 | 11.10 |\n| 760 | 778.18 | 39.27 |\n| 780 | 800.03 | 43.35 |\n| 1300 | 1395.95 | 330.9 |\n\nThen repeat with the cold-air-standard assumptions ($\\gamma = 1.4$, $c_p = 1.005\\ \\mathrm{kJ/(kg\\,K)}$) and compare.\n","figures":[{"src":"/figures/engr5003-ex-brayton-02.png","caption":"Figure 1: ideal Brayton cycle on a T-s diagram (1-2 compressor, 2-3 heat addition at p2, 3-4 turbine, 4-1 heat rejection at p1)."}],"parts":[{"label":"a","text":"Temperatures at the compressor and turbine exits (air tables).","answer":[{"label":"T2","value":540,"units":"K","tol":0.005},{"label":"T4","value":770.3,"units":"K","tol":0.005}],"grade":"$undefined"},{"label":"b","text":"Back work ratio, net (cycle) work and thermal efficiency (air tables).","answer":[{"label":"r_bw","value":0.403,"tol":0.01},{"label":"w_net","value":362.4,"units":"kJ/kg"},{"label":"eta_th","value":0.426,"tol":0.01}],"grade":"$undefined"},{"label":"c","text":"The same quantities with constant specific heats (cold-air standard).","answer":[{"label":"T2","value":543.4,"units":"K","tol":0.005},{"label":"T4","value":717.7,"units":"K","tol":0.005},{"label":"r_bw","value":0.418,"tol":0.01},{"label":"w_net","value":340.6,"units":"kJ/kg"},{"label":"eta_th","value":0.448,"tol":0.01}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- SFEE, adiabatic, negligible KE and PE: $w = \\Delta h$ (on the formula sheet)\n- Isentropic relation with relative pressure: $\\dfrac{P_{r2}}{P_{r1}} = \\dfrac{p_2}{p_1}$ (learn this, Lecture 13; tables provided)\n- Back work ratio $r_{bw} = w_{in}/w_{out}$, $w_{net} = w_{out}(1 - r_{bw})$ (on the formula sheet)\n- Brayton efficiency (constant $c_p$): $\\eta = 1 - r_p^{-(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Linear interpolation (learn this)\n","solution":"$92","confidence":"HIGH","solutionNote":"Matches the slide (T2 = 540 K, T4 = 770.26 K, h4 = 789.39, r_bw = 0.403, w_net = 362.4 kJ/kg, 42.6%). Slide 13 has a typo (1395.95 - 739.39 written for h3 - h4; 789.39 is meant). Cengel's table gives h = 1395.97 at 1300 K; the slide's 1395.95 is used here (0.02 kJ/kg difference)."},{"id":"engr5003-ex-hx-01","module":"ENGR5003","week":5,"label":"Lecture 14 example - jet engine with a regenerative heat exchanger","kind":"tutorial","level":2,"minutes":18,"tags":["gas-turbines","brayton-cycle","regeneration","heat-exchanger","jet-engine","thermal-efficiency"],"source":{"name":"ENGR5003 Lecture 14 example heat exchanger question (slides 5-11)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 5/ENGR5003-Week5-Lecture14.pdf"},"text":"A jet engine is treated as a Brayton cycle with a pressure ratio of 8. The temperature and pressure of the inlet gases\nare 250 K and 50 kPa. The temperature after combustion is 1300 K, and you can assume $\\gamma = 1.4$ and\n$c_p = 1.0045\\ \\mathrm{kJ/(kg\\,K)}$. For an ideal heat exchanger between the exhaust (state 4) and the compressed air\n(state 2), the Lecture 14 models are (Figure 1): cocurrent flow $T_2' = T_4' = \\tfrac{1}{2}(T_2 + T_4)$; countercurrent\nflow $T_2' = T_4$ and $T_4' = T_2$.\n","figures":[{"src":"/figures/engr5003-ex-hx-01.png","caption":"Figure 1: cocurrent flow (both streams leave near the mean temperature) and countercurrent flow (each stream can leave near the other's inlet temperature) (Lecture 14, slide 5)."}],"parts":[{"label":"a","text":"Determine the temperature at each stage of the cycle as well as the thermal efficiency. How much specific heat addition is required?","answer":[{"label":"T2","value":452.86,"units":"K","tol":0.005},{"label":"T4","value":717.66,"units":"K","tol":0.005},{"label":"eta_th","value":0.448,"tol":0.01},{"label":"q_in","value":851,"units":"kJ/kg"}],"grade":"$undefined"},{"label":"b","text":"A heat exchanger is now used to recycle some of the heat from the exhaust to the compressed air. Determine the\nthermal efficiency and the specific heat exchanged when the heat exchanger uses (i) cocurrent flow and\n(ii) countercurrent flow.\n","answer":[{"label":"q_HX cocurrent","value":133,"units":"kJ/kg"},{"label":"eta cocurrent","value":0.531,"tol":0.01},{"label":"q_HX countercurrent","value":266,"units":"kJ/kg"},{"label":"eta countercurrent","value":0.652,"tol":0.01}],"grade":"$undefined"},{"label":"c","text":"Find the maximum pressure ratio for which a heat exchanger would still be useful.","answer":[{"label":"r_p,max","value":17.9}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Adiabatic (isentropic) ideal gas: $\\dfrac{T_2}{T_1} = r_p^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Brayton efficiency: $\\eta = 1 - \\dfrac{T_1}{T_2}$; $\\eta_{th} = 1 - \\dfrac{q_{out}}{q_{in}}$ (on the formula sheet)\n- Isobaric heat: $q = c_p\\Delta T$ (on the formula sheet)\n- Heat exchanger models: cocurrent $T_2' = T_4' = \\tfrac{1}{2}(T_2 + T_4)$; countercurrent $T_2' = T_4$, $T_4' = T_2$ (learn this, Lecture 14)\n","solution":"$93","confidence":"HIGH","solutionNote":"Matches the slides (452.86 K, 717.66 K, 850.95 kJ/kg, 44.8%; 132.99 kJ/kg and 53.1%; 265.99 kJ/kg and 65.2%; 17.9)."},{"id":"engr5003-ex-turbo-01","module":"ENGR5003","week":5,"label":"Lecture 14 example - turbocharged Otto engine","kind":"tutorial","level":2,"minutes":18,"tags":["gas-turbines","ic-engines","otto-cycle","turbocharger","thermal-efficiency"],"source":{"name":"ENGR5003 Lecture 14 example turbocharger question (slides 13-18)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 5/ENGR5003-Week5-Lecture14.pdf"},"text":"A petrol engine is modelled as an Otto cycle. The air temperature and pressure at intake are 290 K and 101 kPa, and the\ncycle has a compression ratio of 4. The temperature after combustion is 1150 K. You may assume $\\gamma = 1.4$ and\n$c_v = 0.7175\\ \\mathrm{kJ/(kg\\,K)}$.\n","parts":[{"label":"a","text":"Determine the temperature at each stage of the cycle, the heat addition and rejection, and the thermal efficiency.","answer":[{"label":"T2","value":504.9,"units":"K","tol":0.005},{"label":"T4","value":660.5,"units":"K","tol":0.005},{"label":"q_in","value":462.85,"units":"kJ/kg"},{"label":"q_out","value":265.83,"units":"kJ/kg"},{"label":"eta_th","value":0.426,"tol":0.01}],"grade":"$undefined"},{"label":"b","text":"A turbocharger is used to increase the thermal efficiency; it provides a pressure ratio of 2 on the air before the\ncompression stroke. Following the Lecture 14 method: (i) determine the temperature of the exhaust gases after the\nturbocharger and the heat exchanged; (ii) determine the new temperature $T_3'$, given that the specific heat addition\nis the same as before; (iii) determine the new thermal efficiency.\n","answer":[{"label":"T1'","value":353.5,"units":"K","tol":0.005},{"label":"w_turbo","value":45.57,"units":"kJ/kg"},{"label":"T4'","value":597,"units":"K","tol":0.005},{"label":"T3'","value":1260.6,"units":"K","tol":0.005},{"label":"eta' (lecture method)","value":0.524,"tol":0.01}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Adiabatic process: $T_2/T_1 = r^{\\gamma-1}$; $T_2/T_1 = (p_2/p_1)^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Constant-volume heat: $q = c_v\\Delta T$ (on the formula sheet)\n- Adiabatic work: $w = \\dfrac{p_2v_2 - p_1v_1}{\\gamma - 1} = c_v\\Delta T$ (on the formula sheet)\n- Otto efficiency: $\\eta = 1 - r^{-(\\gamma-1)}$ (on the formula sheet)\n- Ideal turbocharger: work on the intake air = heat taken from the exhaust (learn this, Lecture 14)\n","solution":"$94","confidence":"MEDIUM","solutionNote":"Numbers match the slides (504.92 K, 660.5 K, 462.85 and 265.83 kJ/kg, 42.6%; 353.51 K, 45.57 kJ/kg, 596.99 K, 615.5 K, 1260.58 K, 220.26 kJ/kg, 52.4%). MEDIUM because the slide's 52.4% is not self-consistent: with the new T3' the exhaust leaves the cylinder at 724.0 K, and after the turbo turbine it is at 660.5 K, so the cycle efficiency is still 42.6%. Confirm with the lecturer which answer is expected."},{"id":"engr5003-ex-cycle-01","module":"ENGR5003","week":1,"label":"Old spec L7 worked example - isochoric / isentropic / isothermal cycle","kind":"tutorial","level":2,"minutes":15,"tags":["ideal-gas-first-law","second-law","p-v-diagram","thermal-efficiency","entropy-change","isothermal-compression"],"source":{"name":"ENGR217 Lecture 7 worked example (slides 17-23)","paper":"ENGR217","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L7 Isentropic and isothermal processes (old spec ENGR217).pdf"},"text":"Air (an ideal gas with constant $c_v$) is heated in a closed volume to increase its temperature from $T_1$ to $T_2$. The\nair then expands reversibly and adiabatically (i.e. isentropically) until its temperature returns to $T_1$. It finally\nreturns to its initial state (pressure and volume) by an isothermal compression.\n","parts":[{"label":"a","text":"Draw the p-V diagram for this sequence of events.","grade":"$undefined"},{"label":"b","text":"Show that the thermal efficiency of this cycle is\n$\\eta_{th} = \\dfrac{\\text{net work}}{\\text{heat supplied}} = 1 - \\dfrac{T_1}{T_2 - T_1}\\ln\\dfrac{T_2}{T_1}$.\n","grade":"$undefined"},{"label":"c","text":"Find the change in entropy (per unit mass) during the isothermal compression in terms of $c_v$, $T_1$ and $T_2$.","grade":"$undefined"},{"label":"d","text":"Evaluate the efficiency and the entropy change in (c) for $T_1 = 300\\ \\mathrm{K}$, $T_2 = 600\\ \\mathrm{K}$, $c_v = 0.718\\ \\mathrm{kJ/(kg\\,K)}$, and compare with a Carnot engine between the same temperatures.","answer":[{"label":"eta_th","value":0.307,"tol":0.01},{"label":"delta s","value":-0.498,"units":"kJ/(kg K)"},{"label":"eta_Carnot","value":0.5}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- First law: $\\Delta U = Q - W$; ideal gas $dU = mc_v\\,dT$ (on the formula sheet)\n- Adiabatic process: $TV^{\\gamma-1} = \\text{const}$; adiabatic work $W = -\\Delta U$ (on the formula sheet)\n- Isothermal ideal gas: $W = mR_gT\\ln(V_2/V_1)$, $\\Delta U = 0$ so $Q = W$ (on the formula sheet)\n- $R_g = c_v(\\gamma - 1)$ (on the formula sheet)\n- Reversible heat at constant $T$: $\\Delta S = Q/T$ (on the formula sheet)\n- Carnot efficiency: $\\eta = 1 - T_C/T_H$ (learn this)\n","solution":"$95","confidence":"HIGH","solutionNote":"Efficiency derivation matches the slides. Slide error in the entropy result: slide 23 writes Delta S = -c_v ln(T1/T2), which is positive for T2 > T1, while its own note says the entropy change is negative. The correct result is Delta S = -c_v ln(T2/T1) = c_v ln(T1/T2) (per unit mass), i.e. the slide's ratio is inverted. Part (d) is added for numerical practice (not on the slide)."},{"id":"engr5003-ex-proprel-01","module":"ENGR5003","week":6,"label":"Old spec L3 exercises 1-2 - expansivity and compressibility of a van der Waals gas","kind":"tutorial","level":2,"minutes":15,"tags":["property-relations","van-der-waals","equation-of-state","partial-derivatives"],"source":{"name":"ENGR217 Lecture 3 student exercises 1-2 (slide 36)","paper":"ENGR217","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 1 - Thermo L3 Heat capacity, enthalpy and cycles (old spec ENGR217).pdf"},"text":"The volumetric expansivity and isothermal compressibility are defined as\n$$\\beta = \\frac{1}{v}\\left(\\frac{\\partial v}{\\partial T}\\right)_p, \\qquad \\alpha = -\\frac{1}{v}\\left(\\frac{\\partial v}{\\partial p}\\right)_T$$\nA substance has the equation of state $p = \\dfrac{RT}{v - b} - \\dfrac{a}{v^2}$, where $a$ and $b$ are constants.\n","parts":[{"label":"a","text":"Derive expressions for the volumetric expansivity and the isothermal compressibility of this substance.","grade":"$undefined"},{"label":"b","text":"Show that for this equation of state $\\beta = \\alpha\\left(\\dfrac{\\partial p}{\\partial T}\\right)_v$.","grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Cyclic relation: $\\left(\\dfrac{\\partial x}{\\partial y}\\right)_z\\left(\\dfrac{\\partial y}{\\partial z}\\right)_x\\left(\\dfrac{\\partial z}{\\partial x}\\right)_y = -1$ (on the chemical thermodynamics formula sheet)\n- Reciprocal relation: $\\left(\\dfrac{\\partial z}{\\partial x}\\right)_y\\left(\\dfrac{\\partial x}{\\partial z}\\right)_y = 1$ (on the chemical thermodynamics formula sheet)\n- Definitions of $\\beta$ and $\\alpha$ (given in the question; learn them)\n","solution":"$96","confidence":"HIGH","solutionNote":"No worked solution in the deck; derivation verified with sympy and the ideal-gas limit."},{"id":"engr203-2018-b1","module":"ENGR5003","week":8,"label":"ENGR203 2018 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["fins","convection","radiation-heat-exchangers","natural-convection","grashof","radiation"],"source":{"name":"ENGR203 Power and Heat 2018 Section B Q B1","paper":"ENGR203","year":2018,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR203_2018.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (f).\n","formulaSheet":"- Grashof number $\\mathrm{Gr} = \\dfrac{\\beta g \\rho^2 \\Delta T L^3}{\\mu^2}$, Prandtl number $\\mathrm{Pr} = \\dfrac{\\mu c_p}{k}$, Nusselt number $\\mathrm{Nu} = \\dfrac{hL}{k}$; natural convection $\\mathrm{Nu} = F(\\mathrm{Gr}, \\mathrm{Pr})$ (on the formula sheet)\n- Newton's law of cooling $q = h\\,\\Delta T$ (on the formula sheet)\n- Rod / fin with negligible tip loss: $\\dot Q = (T_1 - T_0)\\dfrac{hP}{b}\\tanh bL$, $T(x) = T_0 + (T_1 - T_0)\\dfrac{\\cosh b(L - x)}{\\cosh bL}$, $b = \\sqrt{\\dfrac{hP}{kA_c}}$ (on the formula sheet)\n- Radiation to large surroundings $\\dot Q_{net} = \\varepsilon A \\sigma (T_{obj}^4 - T_{sur}^4)$, $\\sigma = 5.6704 \\times 10^{-8}\\ \\mathrm{W\\,m^{-2}\\,K^{-4}}$ (on the formula sheet)\n- For a plate fin of width $w$ and thickness $t$: $P = 2(w + t) \\approx 2w$, $A_c = wt$, so $b \\approx \\sqrt{2h/(kt)}$ (learn this)\n","parts":[{"label":"a","text":"Explain using equations from the data sheet how the Grashof number is used to estimate heat transfer and to\nwhich mode of heat transfer it is relevant.\n","marks":6,"grade":"third"},{"label":"b","text":"A closed rectangular metal tank resting on a perfectly insulating floor and completely filled with cooling\noil is to have its heat dissipation rate by convection increased by a factor of 2 by the addition of fins to\nthe side walls and the top surface. The surface temperature of the tank without fins is\n$105^\\circ\\mathrm{C}$ and the ambient air temperature is $25^\\circ\\mathrm{C}$. The heat transfer coefficient\n$h = 13\\ \\mathrm{W\\,m^{-2}\\,K^{-1}}$ for all surfaces and at all temperatures. (This value applies to the\ntank and the fin surfaces both on the sides and the top.) Calculate the heat loss per square metre from the\ntank before the fins are added and also the heat loss per square metre of tank after the fins are added\n(i.e. multiply by 2).\n","marks":3,"grade":"third","answer":[{"label":"q before","value":1040,"units":"W/m^2","tol":0.02},{"label":"q after","value":2080,"units":"W/m^2","tol":0.02}]},{"label":"c","text":"For the tank described in part (b) the fins are to be 6 mm thick with a pitch of 100 mm (the pitch is the\ncentre to centre spacing). The thermal conductivity $k$ for the metal of the fins will be\n$56\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$. Assuming that when the fins are added, the surface temperature of the tank\n(i.e. the fin base temperature) drops to $90^\\circ\\mathrm{C}$ and neglecting radiation losses, estimate the\nlength of the fins.\n","marks":8,"grade":"2:1","answer":[{"label":"b","value":8.8,"units":"1/m","tol":0.02},{"label":"L","value":0.092,"units":"m","tol":0.06}]},{"label":"d","text":"With respect to part (c) estimate the temperature of the fin tips.","marks":2,"answer":[{"label":"T_tip","value":73.3,"units":"deg C","tol":0.03}],"grade":"$undefined"},{"label":"e","text":"The tank is located in a building such that the walls which surround it are at $25^\\circ\\mathrm{C}$. If,\nbefore the fins are attached, the tank was painted with a paint whose emittance was 0.7 what would have been\nthe net radiant heat loss per square metre from the tank's walls (or top)?\n","marks":3,"answer":[{"label":"q_rad","value":498,"units":"W/m^2","tol":0.02}],"grade":"$undefined"},{"label":"f","text":"Now suppose that the tank's length, width and height, before the fins are added, are all one metre exactly.\nPlace limits on the amount by which radiative loss from the tank changes (expressed as a fraction or a\npercentage) when the fins are added assuming they also are painted with the same paint as used in part (e).\n","marks":3,"grade":"first","answer":[{"label":"Q_rad before (5 faces)","value":2490,"units":"W","tol":0.02},{"label":"lower limit ratio","value":0.52,"units":"-","tol":0.1},{"label":"upper limit ratio","value":2.15,"units":"-","tol":0.1}]}],"solution":"$97","confidence":"MEDIUM","solutionNote":"No official numerical answers for Section B of ENGR203 2018 exist in the folder (the 2018 ENGR203 answer sheet\ncovers Section A only). Parts (a)-(e) are standard and were checked in Python (back-substitution and\nsensitivity runs). Part (f) is open-ended: the limits depend on the model chosen, so its answer boxes use a\nwide tolerance. The temperature-difference unit K and the per-pitch fin balance follow the data-sheet rod\nformula; the old data sheet prints sigma with units W m^-1 K^-4, which should read W m^-2 K^-4.\n"},{"id":"engr203-2018-b2","module":"ENGR5003","week":9,"label":"ENGR203 2018 Q B2(b)-(c)","kind":"past-paper","level":3,"marks":19,"minutes":23,"grade":"2:1","tags":["convection","radiation-heat-exchangers","natural-convection","heat-exchanger","lmtd","cylinder-cross-flow"],"source":{"name":"ENGR203 Power and Heat 2018 Section B Q B2","paper":"ENGR203","year":2018,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR203_2018.pdf","tag":"PARTIAL","leftOut":"Part (a) (6 marks), describe an experiment to measure the specific heat capacity of a liquid and its\naccuracy limits, is laboratory measurement technique that is not in the current ENGR5003 syllabus, so it is\nleft out.\n"},"text":"Answer ALL parts (a) - (c). [Part (a), 6 marks, on measuring the specific heat capacity of a liquid, is not on\nthe current syllabus and is omitted here.]\n","formulaSheet":"- $\\mathrm{Gr} = \\dfrac{\\beta g \\rho^2 \\Delta T L^3}{\\mu^2}$, $\\mathrm{Pr} = \\dfrac{\\mu c_p}{k}$, $\\mathrm{Nu} = \\dfrac{hL}{k}$ (on the formula sheet)\n- Horizontal cylinder in a liquid: $\\mathrm{Nu} = 0.53(\\mathrm{Gr}\\,\\mathrm{Pr})^{0.25}$ with $L = D$ (given in the question)\n- $\\dot Q = hA\\,\\Delta T$ (on the formula sheet)\n- Tube-wall resistance between two fluids: $R_{th} = \\dfrac{1}{L}\\left[\\dfrac{1}{h_1\\pi D_1} + \\dfrac{\\ln(D_2/D_1)}{2\\pi k} + \\dfrac{1}{h_2\\pi D_2}\\right]$, $UA = 1/R_{th}$ (on the formula sheet)\n- Stream energy balance $\\dot Q = \\dot m c_p \\Delta T$ (on the formula sheet)\n- LMTD: $\\dot Q = UA\\,\\Delta T_{mean}$, $\\Delta T_{mean} = \\dfrac{\\Delta T_{min} - \\Delta T_{max}}{\\ln(\\Delta T_{min}/\\Delta T_{max})}$; usable for cross flow when one stream's temperature change is small (on the formula sheet)\n","parts":[{"label":"b","text":"A 1.8 kW immersion heater for heating water in a large tank is to be made from a straight horizontal\nsheathed element of diameter 11 mm. (A sheathed element is an electrical resistance wire embedded in\nmagnesium oxide powder and enclosed in a stainless steel tube, there will be one in your kettle bent to a\ncompact shape.) The tank has a thermostat that controls the water temperature to $55^\\circ\\mathrm{C}$.\nIf the surface temperature of the sheathed element is not to exceed $99^\\circ\\mathrm{C}$, calculate the\nlength of tube required assuming that convective losses from the element can be determined by the\nnon-dimensional equation $\\mathrm{Nu} = 0.53(\\mathrm{Gr} \\times \\mathrm{Pr})^{0.25}$.\n\n(For water at $77^\\circ\\mathrm{C}$ take $\\mu = 0.370 \\times 10^{-3}\\ \\mathrm{kg\\,m^{-1}\\,s^{-1}}$,\n$c_p = 4.196\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$, $\\rho = 974\\ \\mathrm{kg\\,m^{-3}}$,\n$k = 0.666\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$, $\\beta = 0.615 \\times 10^{-3}\\ \\mathrm{K^{-1}}$,\n$g = 9.81\\ \\mathrm{m\\,s^{-2}}$.)\n","marks":9,"grade":"2:2","answer":[{"label":"Gr","value":2450000,"units":"-","tol":0.02},{"label":"Pr","value":2.33,"units":"-","tol":0.02},{"label":"Nu","value":25.9,"units":"-","tol":0.02},{"label":"h","value":1568,"units":"W/(m^2 K)","tol":0.02},{"label":"L","value":0.755,"units":"m","tol":0.02}]},{"label":"c","text":"Water is passed through an aluminium tube of length 12 m, bore 22 mm and wall thickness 1 mm. The tube is\ncooled by air at $15^\\circ\\mathrm{C}$ blown perpendicularly to its outside surface. (Assume that any folds\nin the tube do not affect air side heat transfer to adjacent segments.) Using the LMTD method and data\nbelow estimate the water outlet temperature.\n\n| Quantity | Value |\n| --- | --- |\n| Water inlet temperature | $T_{w1} = 90^\\circ\\mathrm{C}$ |\n| Mass flow rate of water | $\\dot m_w = 0.05\\ \\mathrm{kg\\,s^{-1}}$ |\n| Mass flow rate of air | $\\dot m_a = 0.7\\ \\mathrm{kg\\,s^{-1}}$ |\n| Heat capacity of water | $c_{pw} = 4190\\ \\mathrm{J\\,kg^{-1}\\,K^{-1}}$ |\n| Heat capacity of air | $c_{pa} = 1006\\ \\mathrm{J\\,kg^{-1}\\,K^{-1}}$ |\n| Water side heat transfer coefficient | $h_w = 2200\\ \\mathrm{W\\,m^{-2}\\,K^{-1}}$ |\n| Air side heat transfer coefficient | $h_a = 140\\ \\mathrm{W\\,m^{-2}\\,K^{-1}}$ |\n","marks":10,"grade":"2:1","answer":[{"label":"UA","value":118.4,"units":"W/K","tol":0.02},{"label":"T_w2","value":59.3,"units":"deg C","tol":0.03},{"label":"Q","value":6.43,"units":"kW","tol":0.06}]}],"solution":"$98","confidence":"HIGH","solutionNote":"No official numerical answers for ENGR203 2018 Section B exist in the folder (the 2018 answer sheet covers\nSection A only). The method for (c) (counter-flow LMTD with the small air temperature rise included) is the one\nthat reproduces the official answer to the matching 2017 question B3(d) (80.33 C), so the same method is used\nhere. Aluminium k = 205 W/m K is assumed (not given; it has no effect on the answer).\n"},{"id":"engr203-2018-b3","module":"ENGR5003","week":7,"label":"ENGR203 2018 Q B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["conduction","convection","thermal-resistance","composite-wall","boundary-conditions"],"source":{"name":"ENGR203 Power and Heat 2018 Section B Q B3","paper":"ENGR203","year":2018,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR203_2018.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n","figures":[{"src":"/figures/engr203-2018-b3.png","caption":"Fig. B3-1: plan view (from the top) of one stud and the insulation either side, between two 12 mm plasterboard sheets; heat flows through the 90 mm depth."}],"formulaSheet":"- Conduction resistance of a flat layer $R_{th} = \\dfrac{x}{kA}$; convection resistance $R_{th} = \\dfrac{1}{hA}$; $\\Delta T = \\dot Q R_{th}$ (on the formula sheet)\n- Series resistances add: $R = R_1 + R_2 + \\dots$; parallel resistances: $\\dfrac{1}{R} = \\dfrac{1}{R_1} + \\dfrac{1}{R_2} + \\dots$ (learn this)\n- Steady conduction $\\nabla \\cdot (k\\nabla T) = 0$; general conduction equation $\\rho c_p \\dfrac{\\partial T}{\\partial t} = k\\nabla^2 T + q_v$ (on the formula sheet)\n- Energy: $1\\ \\mathrm{kWh} = 3.6\\ \\mathrm{MJ}$ (learn this)\n","parts":[{"label":"a","text":"The temperature distribution $T(x,y,z)$ in a solid with thermal conductivity $k(x,y,z)$ can be determined by\nsolving the equation $\\nabla \\cdot (k\\nabla T) = 0$. Describe possible types of boundary condition and\ncombinations thereof that ensure that this equation has a unique solution.\n","marks":6,"grade":"2:1"},{"label":"b","text":"An interior wall is constructed from vertical wooden studding. Studs are spaced horizontally. The studding\nis 50 mm wide and 90 mm deep and runs from floor to ceiling. Spaces between the studding are insulated with\nglass fibre wool. The studding is covered on both sides with 12 mm thick plasterboard. A segment of the wall\nconstruction as described above is illustrated by the cross-section drawn in Fig. B3-1.\n\nThermal conductivities are as follows, for the studding $1.4\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$, for the\nplasterboard $1.6\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$ and for the glass fibre insulation\n$0.07\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$.\n\n(i) Calculate the thermal resistance of the wall given that it is 2.4 m high, 5.4 m wide and has 9 full\nheight studs. (8 marks)\n\n(ii) If the interior wall separates two thermostatically controlled rooms at temperatures of\n$18^\\circ\\mathrm{C}$ and $23^\\circ\\mathrm{C}$, expressing your answer in kWh, how much heat flows through the\nwall in 24 hours. (2 marks)\n","marks":10,"grade":"2:2","answer":[{"label":"R_wall","value":0.0396,"units":"K/W","tol":0.11},{"label":"Q","value":126,"units":"W","tol":0.11},{"label":"E (24 h)","value":3.03,"units":"kWh","tol":0.11}]},{"label":"c","text":"A car rear window is de-misted with a thin film heater on its interior surface. For glass 6 mm thick,\ndetermine the power requirement per square metre to maintain an inner surface temperature of\n$12^\\circ\\mathrm{C}$ when the interior air temperature is $22^\\circ\\mathrm{C}$, the interior heat transfer\ncoefficient is $7\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$ [sic: read W m$^{-2}$ K$^{-1}$], the exterior air temperature is $-6^\\circ\\mathrm{C}$ and the\nexterior heat transfer coefficient is $30\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$ [sic]. Take the thermal conductivity of\nthe glass as $k_{glass} = 1.44\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$.\n","marks":9,"grade":"2:2","answer":[{"label":"q_out","value":480,"units":"W/m^2","tol":0.02},{"label":"q_heater","value":410,"units":"W/m^2","tol":0.02}]}],"solution":"$99","confidence":"HIGH","solutionNote":"No official numerical answers for ENGR203 2018 Section B exist in the folder. All numbers verified in Python.\n(b): the answer depends slightly on the composite-wall model (isothermal planes 0.0396 K/W, 3.03 kWh;\nadiabatic paths 0.0439 K/W, 2.73 kWh), so the answer boxes accept both (tolerance 11 %). (c): the question\nprints h in W m^-1 K^-1, a misprint for W m^-2 K^-1.\n"},{"id":"engr203-2017-b1","module":"ENGR5003","week":8,"label":"ENGR203 2017 Q B1(a)(ii), (b)","kind":"past-paper","level":3,"marks":14,"minutes":17,"grade":"2:1","tags":["convection","conduction","cylinder-cross-flow","forced-convection","fourier-law"],"source":{"name":"ENGR203 Power and Heat 2017 Section B Q B1 (same question as ENGR263 2017 B1)","paper":"ENGR203","year":2017,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR203 Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR203 Numerical Answers 2017.pdf","tag":"PARTIAL","leftOut":"(a)(i) (4 marks, draw Searle's bar apparatus) and (a)(iii) (7 marks, uncertainty budget for the measured\nconductivity) are laboratory measurement and uncertainty analysis, not on the current ENGR5003 syllabus.\n"},"text":"Answer ALL parts (a) - (b). [Parts (a)(i) and (a)(iii), on drawing Searle's bar apparatus and its measurement\nuncertainty, are not on the current syllabus and are omitted here.]\n","formulaSheet":"- Fourier's law $\\dot Q = -kA\\dfrac{dT}{dx}$ (on the formula sheet)\n- $\\dot Q = hA\\,\\Delta T$; $\\mathrm{Re} = \\dfrac{\\rho VL}{\\mu}$, $\\mathrm{Pr} = \\dfrac{\\mu c_p}{k}$, $\\mathrm{Nu} = \\dfrac{hL}{k}$ (on the formula sheet)\n- Gas past a cylindrical tube, turbulent: $\\mathrm{Nu} = 0.35\\,\\mathrm{Re}^{0.565}\\mathrm{Pr}^{0.333}$ for $2000 < \\mathrm{Re} < 40\\,000$, $\\mathrm{Re\\,Pr} > 0.2$; streamline: $\\mathrm{Nu} = 0.35 + 0.47\\,\\mathrm{Re}^{0.52}\\mathrm{Pr}^{0.3}$ for $0.1 < \\mathrm{Re} < 2000$ (characteristic length = outer diameter) (on the formula sheet)\n","parts":[{"label":"a(ii)","text":"A perfectly insulated Searle's bar of diameter 30 mm is used to determine the thermal conductivity of\nNickel. If the separation between the thermocouples in the bar is approximately 200 mm, the approximate\ntemperature rise between the thermocouples is $50^\\circ\\mathrm{C}$ and the approximate thermal conductivity\nis $90\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$ what is the approximate heat flow along the bar?\n","marks":2,"grade":"third","answer":[{"label":"Q","value":15.9,"units":"W","tol":0.02}]},{"label":"b","text":"An electrically heated coil is suspended inside an evacuated fused silica tube of length 800 mm in a manner\nsuch that there is no convective or conductive heat transfer from the coil to the tube. The tube has an\nouter diameter of 25 mm. The coil emits 4 kW of infra-red radiation and 16% of this infra-red is absorbed by\nthe tube. A fan is used to keep the exterior of the tube cool. Using formulae in the data sheet, determine\nthe free stream velocity for air at $20^\\circ\\mathrm{C}$ required to keep the surface temperature of the\ntube below $210^\\circ\\mathrm{C}$.\n\nFor air at $110^\\circ\\mathrm{C}$ take the viscosity of air $\\mu = 22.2 \\times 10^{-6}\\ \\mathrm{kg\\,m^{-1}\\,s^{-1}}$,\nthe heat capacity of air $c_p = 1.0123\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$, the density of air\n$\\rho = 0.9213\\ \\mathrm{kg\\,m^{-3}}$, the thermal conductivity of air $k = 0.03194\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$.\n","marks":12,"grade":"2:1","answer":[{"label":"h","value":53.6,"units":"W/(m^2 K)","tol":0.02},{"label":"Nu","value":42,"units":"-","tol":0.02},{"label":"Re","value":5880,"units":"-","tol":0.03},{"label":"V","value":5.67,"units":"m/s","tol":0.03}]}],"solution":"$9a","confidence":"HIGH","solutionNote":"Matches the official numerical answers: (a)(ii) 15.9 W, (b) 5.67 m/s (ours 5.667 m/s). The official sheet\nalso gives (a)(iii) +/-1.9 %, which belongs to the omitted uncertainty part.\n"},{"id":"engr203-2017-b2","module":"ENGR5003","week":7,"label":"ENGR203 2017 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"first","tags":["conduction","radiation-heat-exchangers","heat-generation","radiation","lumped-transient","fuse-wire"],"source":{"name":"ENGR203 Power and Heat 2017 Section B Q B2 (same question as ENGR263 2017 B2)","paper":"ENGR203","year":2017,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR203 Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR203 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (f).\n\nA Constantan fuse wire has a diameter of 0.6 mm. Assume Constantan has the following properties, resistivity\n$= 49 \\times 10^{-8}\\ \\Omega\\,\\mathrm{m}$, emissivity $= 0.12$, thermal conductivity\n$= 20\\ \\mathrm{W\\,m^{-1}\\,K^{-1}}$, specific heat $= 390\\ \\mathrm{J\\,kg^{-1}\\,K^{-1}}$, density\n$8900\\ \\mathrm{kg\\,m^{-3}}$.\n\nFor parts (a) and (b) assume that the wire cools by radiation losses alone to an effective black body\nsurrounding of temperature $30^\\circ\\mathrm{C}$.\n","formulaSheet":"- Radiation to large surroundings $\\dot Q_{net} = \\varepsilon A\\sigma(T_{obj}^4 - T_{sur}^4)$, $\\sigma = 5.6704 \\times 10^{-8}\\ \\mathrm{W\\,m^{-2}\\,K^{-4}}$ (on the formula sheet)\n- General conduction equation $\\rho c_p\\dfrac{\\partial T}{\\partial t} = k\\nabla^2 T + q_v$ (on the formula sheet)\n- Heat stored $\\Delta Q = mc_p\\Delta T$ (on the formula sheet)\n- Wire resistance per unit length $R' = \\rho_e/A_c$; Joule heating $\\dot Q = I^2R$; volumetric generation $q_v = I^2\\rho_e/A_c^2$ (learn this)\n- 1-D conduction with uniform generation, both ends at $T_0$: $T(x) = T_0 + \\dfrac{q_v}{2k}x(L - x)$, $T_{max} - T_0 = \\dfrac{q_vL^2}{8k}$ (learn this)\n","parts":[{"label":"a","text":"Determine the wire temperature when it carries an r.m.s. current of 3 amps.","marks":5,"grade":"2:2","answer":[{"label":"T","value":778.8,"units":"deg C","tol":0.01}]},{"label":"b","text":"Determine the current required to heat the wire to its melting temperature $1210^\\circ\\mathrm{C}$.","marks":4,"grade":"2:2","answer":[{"label":"I","value":5.98,"units":"A","tol":0.01}]},{"label":"c","text":"Neglecting all heat losses and assuming heat capacity, electrical resistance and emissivity do not change\nwith temperature then starting at $30^\\circ\\mathrm{C}$ and applying the current computed in (b) determine\nthe time until the fuse melts.\n","marks":4,"grade":"2:2","answer":[{"label":"t","value":18.7,"units":"s","tol":0.02}]},{"label":"d","text":"If the wire could not lose heat by radiation or convection but only by conduction to its ends derive a\ndifferential equation that gives temperature along the wire.\n","marks":5,"grade":"2:1"},{"label":"e","text":"Using this differential equation determine the shortest length of the wire which would just melt at its\ncentre when it carries a current of 3 amps and the ends are held at $30^\\circ\\mathrm{C}$.\n","marks":4,"grade":"2:1","answer":[{"label":"q_v","value":55200000,"units":"W/m^3","tol":0.02},{"label":"L","value":0.0585,"units":"m","tol":0.02}]},{"label":"f","text":"Write down a partial differential equation for the temperature of the wire as a function of time when it\ncarries current $I$ and cools by both thermal radiation and conduction to its ends.\n","marks":3,"grade":"first"}],"solution":"$9b","confidence":"HIGH","solutionNote":"Matches the official numerical answers for (a) 778.8 C (official also quotes 1051.4 K; ours 1051.9 K, the\ndifference being 273 vs 273.15 in the conversion), (b) 5.98 A and (e) 0.0585 m. For (c) the official answer is\n18.62 s; our value is 18.69 s with the unrounded current 5.979 A (18.68 s with 5.98 A). The 0.4 % gap is\nwithin rounding of intermediate values; the answer box accepts both.\n"},{"id":"engr203-2017-b3","module":"ENGR5003","week":10,"label":"ENGR203 2017 Q B3(b)-(d)","kind":"past-paper","level":3,"marks":20,"minutes":24,"grade":"2:1","tags":["radiation-heat-exchangers","convection","heat-exchanger","lmtd","thermal-resistance"],"source":{"name":"ENGR203 Power and Heat 2017 Section B Q B3 (same question as ENGR263 2017 B3)","paper":"ENGR203","year":2017,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR203 Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR203 Numerical Answers 2017.pdf","tag":"PARTIAL","leftOut":"(a) (5 marks), describe how a thermocouple is constructed and used, is instrumentation and not on the\ncurrent ENGR5003 syllabus.\n"},"text":"Answer ALL parts (a) - (d). [Part (a), 5 marks, on thermocouple construction, is not on the current syllabus\nand is omitted here.]\n","formulaSheet":"- Tube-wall resistance between two fluids: $R_{th} = \\dfrac{1}{L}\\left[\\dfrac{1}{h_1\\pi D_1} + \\dfrac{\\ln(D_2/D_1)}{2\\pi k} + \\dfrac{1}{h_2\\pi D_2}\\right]$, $UA = 1/R_{th}$ (on the formula sheet)\n- Stream energy balances $\\dot Q = \\dot m_h c_{ph}\\Delta T_h = \\dot m_c c_{pc}\\Delta T_c$ (on the formula sheet)\n- LMTD: $\\dot Q = UA\\,\\Delta T_{mean}$, $\\Delta T_{mean} = \\dfrac{\\Delta T_{min} - \\Delta T_{max}}{\\ln(\\Delta T_{min}/\\Delta T_{max})}$; valid for parallel and counter flow, and for cross flow if one stream's temperature change is small (on the formula sheet)\n- Effectiveness $\\varepsilon = \\dfrac{\\dot Q}{C_{min}(T_{hi} - T_{ci})}$, $\\mathrm{NTU} = UA/C_{min}$ (on the formula sheet)\n","parts":[{"label":"b","text":"For a counter flow, liquid to liquid heat exchanger describe in words the key factors which limit the rate\nof energy transfer from one liquid stream to the other.\n","marks":4,"grade":"third"},{"label":"c","text":"When would you expect the Logarithmic Mean Temperature Difference (LMTD) method to give a reasonably\naccurate estimate for heat transfer in a cross flow heat exchanger.\n","marks":2,"grade":"third"},{"label":"d","text":"Water is passed through an aluminium tube of length 8 m, bore 20 mm and wall thickness 1 mm. The tube is\ncooled by air at $15^\\circ\\mathrm{C}$ blown perpendicularly to its outside surface. Any folds in the tube do\nnot affect air side heat transfer to adjacent segments. Using the LMTD method and data below estimate the\nwater outlet temperature.\n\n| Quantity | Value |\n| --- | --- |\n| Water inlet temperature | $T_{w1} = 85^\\circ\\mathrm{C}$ |\n| Mass flow rate of water | $\\dot m_w = 0.3\\ \\mathrm{kg\\,s^{-1}}$ |\n| Mass flow rate of air | $\\dot m_a = 0.8\\ \\mathrm{kg\\,s^{-1}}$ |\n| Heat capacity of water | $c_{pw} = 4190\\ \\mathrm{J\\,kg^{-1}\\,K^{-1}}$ |\n| Heat capacity of air | $c_{pa} = 1006\\ \\mathrm{J\\,kg^{-1}\\,K^{-1}}$ |\n| Water side heat transfer coefficient | $h_w = 2400\\ \\mathrm{W\\,m^{-2}\\,K^{-1}}$ |\n| Air side heat transfer coefficient | $h_a = 180\\ \\mathrm{W\\,m^{-2}\\,K^{-1}}$ |\n","marks":14,"grade":"2:1","answer":[{"label":"UA","value":91.9,"units":"W/K","tol":0.02},{"label":"T_w2","value":80.3,"units":"deg C","tol":0.005},{"label":"Q","value":5.88,"units":"kW","tol":0.06}]}],"solution":"$9c","confidence":"HIGH","solutionNote":"Matches the official numerical answer for (d): 80.33 C (ours 80.32 C with the counter-flow LMTD including the\nair temperature rise; 80.07 C if the air were held at 15 C). Aluminium k = 205 W/m K assumed (not given; any\nvalue 160-237 changes T_w2 by under 0.01 K).\n"},{"id":"engr263-2025-q1c","module":"ENGR5003","week":6,"label":"ENGR263 2025 Q1(c)","kind":"past-paper","level":3,"marks":23,"minutes":14,"grade":"2:2","tags":["phase-reaction-equilibrium","vle","raoult","bubble-point"],"source":{"name":"ENGR263 Mass Transfer Summer 2025 Q1(c) (same question as ENGR263 2021 A3(a))","paper":"ENGR263","year":2025,"question":"Q1","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2025_/ENGR263 Main Exam Paper 2024-25.pdf","solutionsPath":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2025_/ENGR263 Main Exam Paper 2024-25 answers.pdf","tag":"PARTIAL","leftOut":"Q1(a) relative volatility (15 marks) and Q1(b) azeotropic vs extractive distillation (10 marks) are\ndistillation, which is not on the current ENGR5003 or ENGR5002 syllabus. Only the Raoult's-law part (c) is\nconverted.\n"},"text":"[Q1(a)-(b), on relative volatility and azeotropic / extractive distillation, are not on the current syllabus\nand are omitted. The same part (c) was set as ENGR263 2021 A3(a), 8 marks.]\n","formulaSheet":"- Raoult's law (ideal liquid, ideal vapour): $y_iP = x_iP_i^{sat}$ (learn this)\n- Dalton: $P = \\sum_i p_i$, $y_i = p_i/P$ (learn this)\n- Bubble point of a binary at fixed $P$: $x_AP_A^{sat} + (1 - x_A)P_B^{sat} = P$ (learn this)\n- $1\\ \\mathrm{atm} = 760\\ \\mathrm{mm\\,Hg}$ (on the formula sheet)\n","parts":[{"label":"c","text":"Calculate the composition of the vapour phase in equilibrium at $60^\\circ\\mathrm{C}$ with a liquid mixture\nconsisting of 40% mol benzene and 60% mol toluene. Also, determine the composition of a mixture of benzene\nand toluene that will boil at a temperature of $90^\\circ\\mathrm{C}$ under a pressure of 760 mm Hg. Liquid\nmixtures of benzene and toluene obey Raoult's Law. The saturated vapour pressures of benzene and toluene at\n$60^\\circ\\mathrm{C}$ are 385 mm Hg and 140 mm Hg, respectively. At $90^\\circ\\mathrm{C}$, these values are\n1013 and 408 mm Hg, respectively.\n","marks":23,"answer":[{"label":"P (60 C)","value":238,"units":"mm Hg","tol":0.01},{"label":"y_benzene (60 C)","value":0.647,"units":"-","tol":0.01},{"label":"y_toluene (60 C)","value":0.353,"units":"-","tol":0.01},{"label":"x_benzene (90 C)","value":0.582,"units":"-","tol":0.01},{"label":"x_toluene (90 C)","value":0.418,"units":"-","tol":0.01}],"grade":"$undefined"}],"solution":"$9d","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers (2025): y_B = 0.647, y_T = 0.353, P = 238 mm Hg, x_B = 0.582,\ny_B = 0.776; every number re-computed in Python and all agree. The official 2020-21 numerical answers for the\nidentical 2021 A3(a) give xA = 0.583, xB = 0.417 (ours 0.5818, i.e. 0.582; the official last digit is a\nrounding difference).\n"},{"id":"engr263-2019-a1a","module":"ENGR5003","week":6,"label":"ENGR263 2019 Q A1(a)","kind":"past-paper","level":3,"marks":5,"minutes":6,"grade":"third","tags":["phase-reaction-equilibrium","vle","raoult"],"source":{"name":"ENGR263 Process of Mass Transfer 2019 Q A1(a)","paper":"ENGR263","year":2019,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR263_Exam_2019 (1).pdf","tag":"PARTIAL","leftOut":"Parts (b)-(d) (diffusion flux, mass transfer coefficient, stagnant-film flux) are ENGR5002 mass transfer and\nare in that bank as engr263-2019-a1; part (e) (azeotrope reading of the ethanol-water VLE diagram,\nalternative separation methods) is distillation and is off-syllabus.\n"},"text":"A binary mixture of ethanol and water in the reboiler of a batch distillation column presents the following\ndata:\n\n- Total pressure $P = 101\\,300\\ \\mathrm{Pa}$\n- Temperature $T = 85^\\circ\\mathrm{C}$\n- Mole fraction of water in the liquid phase $x_w = 0.6$\n- Mole fraction of water in the vapour phase $y_w = 0.70$\n- Saturated vapour pressure of ethanol $P_e^{sat} = 135\\,087\\ \\mathrm{Pa}$\n","formulaSheet":"- Raoult's law (ideal liquid and ideal vapour): $y_iP = x_iP_i^{sat}$ (learn this)\n","parts":[{"label":"a","text":"Calculate the composition of ethanol in the vapour phase in equilibrium with the liquid phase mixture,\nassuming both the vapour and liquid phases are ideal mixtures.\n","marks":5,"answer":[{"label":"y_e*","value":0.533,"units":"-","tol":0.01}],"grade":"$undefined"}],"solution":"**(a)** Ideal liquid (Raoult) and ideal vapour (Dalton): $y_eP = x_eP_e^{sat}$, with\n$x_e = 1 - x_w = 0.4$:\n$$y_e^* = \\frac{x_eP_e^{sat}}{P} = \\frac{0.4 \\times 135\\,087}{101\\,300} = 0.533$$\nThe vapour in equilibrium with the liquid (at the interface) would hold 53.3 mol% ethanol, compared with\n$1 - 0.70 = 0.30$ in the bulk vapour, so ethanol transfers from the liquid to the vapour (this is the\ndriving force used in parts (b)-(d), ENGR5002).\n\n**Check:** ethanol is the more volatile component ($P_e^{sat} > P$ at 85 C), so $y_e^* > x_e$ is expected.\nConsistency note: with water's vapour pressure at 85 C (57.87 kPa, IAPWS-IF97, not given), ideal Raoult\nwould give a bubble pressure of only $0.4 \\times 135.1 + 0.6 \\times 57.9 = 88.8\\ \\mathrm{kPa}$, below 101.3 kPa: real\nethanol-water deviates strongly (positively) from Raoult's law, which is why it forms an azeotrope. The\nquestion asks only for the ideal estimate.\n\n*Likely marks: 2 Raoult's law, 1 $x_e = 0.4$, 2 result.*\n","confidence":"HIGH","solutionNote":"No official numerical answers for ENGR263 2019 in the folder; single-step Raoult calculation verified in\nPython. The given y_w = 0.70 is the bulk vapour, not the equilibrium value.\n"},{"id":"engr217-2021-q1","module":"ENGR5003","week":4,"label":"ENGR217 2021 Q1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["ic-engines","gas-turbines","otto-cycle","engine-efficiencies","isentropic-efficiency","back-work-ratio","turbocharger"],"source":{"name":"ENGR217 Summer 2021 Q1","paper":"ENGR217","year":2021,"question":"Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2021 Exam paper.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2021 Exam solution.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nA Formula 1 car uses a 2.4 litre, 4-stroke, 8-cylinder petrol engine with a total mass of 95 kg.\nAt 19,000 rpm, the engine produces 550 kW of power. The overall engine efficiency is 33% and the\nbreathing and mechanical efficiencies can be assumed to be 85% and 90% respectively. We will model\nthis engine as an Otto cycle operating on an ideal gas and you may assume that\n$R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.4$, where $\\gamma$ is the adiabatic index.\n","formulaSheet":"- Otto cycle efficiency: $\\eta_{th} = 1 - \\dfrac{T_1}{T_2} = 1 - \\dfrac{1}{r^{\\gamma-1}}$ (on the formula sheet)\n- Overall efficiency: $\\eta_o = \\eta_{th}\\,\\eta_b\\,\\eta_m$ (learn this)\n- Power: $P = W_{net}\\,N\\,\\eta_b\\,\\eta_m$ with $N = \\dfrac{\\mathrm{rpm}}{120}$ power strokes per second per cylinder for a 4-stroke (learn this)\n- Brake work: $W_b = \\eta_m W_{net}$ (learn this)\n- Ideal gas: $pV = mR_gT$; $c_v = \\dfrac{R_g}{\\gamma-1}$ (on the formula sheet)\n- Adiabatic process: $TV^{\\gamma-1} = \\text{constant}$ (on the formula sheet)\n- Isochoric heat addition: $\\Delta Q = m c_v (T_3 - T_2)$; $\\eta_{th} = W_{net}/Q_{in}$ (on the formula sheet)\n- Adiabatic work: $\\Delta W = -\\Delta U = m c_v \\Delta T$ (on the formula sheet)\n- Isentropic efficiencies: $\\eta_c = \\dfrac{W_{c,s}}{W_{c,a}}$, $\\eta_e = \\dfrac{W_{e,a}}{W_{e,s}}$ (learn this)\n- Back work ratio: $r_{bw} = W_{in}/W_{out}$ (given in the question)\n","parts":[{"label":"a","text":"Determine the compression ratio, brake work and net work output per cylinder, per cycle.","marks":10,"grade":"2:2","answer":[{"label":"r","value":4.101,"units":"-"},{"label":"W_b per cylinder","value":0.511,"units":"kJ"},{"label":"W_net per cylinder","value":0.568,"units":"kJ"}]},{"label":"b","text":"Given that at the intake, the air pressure and temperature are $p_1 = 101\\ \\mathrm{kPa}$ and\n$T_1 = 303\\ \\mathrm{K}$. Determine the temperature at each stage in the process, the work done during\ncompression and expansion and the required heat addition per cylinder per cycle.\n","marks":8,"grade":"2:2","answer":[{"label":"T_2","value":532.9,"units":"K"},{"label":"T_3","value":5796,"units":"K"},{"label":"T_4","value":3296,"units":"K"},{"label":"W_comp","value":0.0575,"units":"kJ"},{"label":"W_exp","value":0.625,"units":"kJ"},{"label":"Q_in","value":1.316,"units":"kJ"}]},{"label":"c","text":"Let's now assume that the isentropic efficiency of the compression and expansion strokes are 70% and\n85% respectively. Determine the back work ratio of the engine for the adiabatic assumption and with the\nisentropic efficiencies stated. You may assume the heat addition and compression ratio remains the same.\nRecall that the back work ratio is given as: $r_{bw} = \\dfrac{W_{in}}{W_{out}}$\n","marks":4,"grade":"2:1","answer":[{"label":"r_bw ideal","value":0.0919,"units":"-"},{"label":"r_bw real","value":0.1545,"units":"-"}]},{"label":"d","text":"Discuss the advantages and disadvantages of using a turbocharger on this engine.","marks":3,"grade":"third"}],"solution":"$9e","confidence":"HIGH","solutionNote":"Matches the official solution: r = 4.101, W_net = 0.568 kJ, W_b = 0.511 kJ, T_2 = 532.9 K, Q_in = 1.316 kJ,\nW_comp = 0.057 kJ, W_exp = 0.625 kJ, r_bw = 0.0912 and 0.154. The official T_3 = 5802 K and T_4 = 3299 K differ\nby 0.1% (rounded mass); the official also misprints c_v as 0.7125 in places but uses 0.7175.\n"},{"id":"engr217-2021-q2","module":"ENGR5003","week":2,"label":"ENGR217 2021 Q2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["charging-discharging","second-law","ideal-gas-first-law","unsteady-flow","entropy-change","kinetic-energy"],"source":{"name":"ENGR217 Summer 2021 Q2","paper":"ENGR217","year":2021,"question":"Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2021 Exam paper.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2021 Exam solution.pdf","tag":"VALID"},"text":"Answer All parts (a) - (d).\n\nA 12 litre cryogenic tank initially contains air at 1 bar and a temperature of 293 K. The tank is then\ncharged by a high pressure gas main at an initial temperature of 285 K to a final pressure of 250 bar.\nYou may assume that the air is an ideal gas and the tank is rigid. You may also assume that\n$R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.4$, where $\\gamma$ is the adiabatic index.\n","formulaSheet":"- Unsteady flow energy equation: $\\Delta U = \\left(h_{in} + \\tfrac{C_{in}^2}{2} + gZ_{in}\\right)\\Delta m_{in} - \\left(h_{out} + \\tfrac{C_{out}^2}{2} + gZ_{out}\\right)\\Delta m_{out} + \\Delta Q - \\Delta W$ (on the formula sheet)\n- Ideal gas: $pV = mR_gT$; $dU = mc_v\\,dT$; $h = c_pT$; $c_v = \\dfrac{R_g}{\\gamma-1}$, $c_p = \\dfrac{\\gamma R_g}{\\gamma-1}$ (on the formula sheet)\n- Isochoric process: $\\Delta W = 0$, $\\Delta Q = dU = mc_v\\Delta T$; Gay-Lussac $p_1/T_1 = p_2/T_2$ (on the formula sheet)\n- Entropy change: $\\Delta s = c_v \\ln\\dfrac{T_2}{T_1} + R_g \\ln\\dfrac{v_2}{v_1}$ (on the formula sheet)\n- SFEE across the valve: $h_1 + \\tfrac{C_1^2}{2} = h_2 + \\tfrac{C_2^2}{2}$ (on the formula sheet, as the SFEE)\n","parts":[{"label":"a","text":"Determine the initial and final mass of gas in the tank as well as the final temperature, assuming an\nadiabatic charging process. You should treat this as an unsteady flow problem.\n","marks":8,"grade":"2:2","answer":[{"label":"m_1","value":0.01427,"units":"kg"},{"label":"m_2","value":2.624,"units":"kg"},{"label":"T_2","value":398.4,"units":"K"}]},{"label":"b","text":"The tank is allowed to cool back to 293 K. Determine the heat rejected and the pressure change of the\nprocess. Given that no work is done during the cooling process, determine the entropy change, given that\n$\\Delta Q = m c_p\\, dT = T\\, dS$.\n","marks":8,"grade":"2:2","answer":[{"label":"Q_rejected","value":198.5,"units":"kJ"},{"label":"Delta p","value":-6615,"units":"kPa"},{"label":"Delta S","value":-0.5785,"units":"kJ/K"}]},{"label":"c","text":"The tank is then discharged through a valve. We can neglect the flow velocity of the gas in the tank, but\nafter the valve, the gas flow velocity is 150 m/s. Determine the specific enthalpy of the gas leaving the\ntank and the temperature change. You may assume there is no heat exchange or work done during the\ndischarging process.\n","marks":5,"grade":"2:1","answer":[{"label":"h after valve","value":283.1,"units":"kJ/kg"},{"label":"Delta T","value":-11.2,"units":"K"}]},{"label":"d","text":"From the calculations in part (c), discuss whether it is reasonable or not to consider the kinetic energy\nterm and justify your answer. What is the maximum flow velocity that can be achieved?\n","marks":4,"grade":"2:1","answer":[{"label":"C_max","value":767,"units":"m/s","tol":0.05}]}],"solution":"$9f","confidence":"MEDIUM","solutionNote":"(a) and (b) agree with the official solution to within 1% (official T_2 = 398.42 K, m_2 = 2.65 kg, Q = 199.0 kJ,\nDelta p = -66.15 bar, Delta S = -0.58 kJ/K; it used 101 kPa for 1 bar). Parts (c) and (d) DISAGREE: the official\nsolution writes h_out = u_1 - C^2/2 = c_v T_1 - C^2/2 = 197.5 kJ/kg and then T = 196.6 K (Delta T = -96.4 K), and\nC_max = sqrt(2 c_v T_1) = 646 m/s. That mixes internal energy and enthalpy: for flow through a valve the SFEE\non the formula sheet conserves h + C^2/2, giving h = 283.1 kJ/kg, Delta T = -11.2 K and C_max = 767 m/s. Trust the\nSFEE answer; but the marker may follow the official scheme, so confirm with the lecturer which form is expected.\n"},{"id":"engr217-2022-a1","module":"ENGR5003","week":4,"label":"ENGR217 2022 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["ic-engines","otto-cycle","engine-efficiencies","fuel-economy"],"source":{"name":"ENGR217 Summer 2022 Q A1","paper":"ENGR217","year":2022,"question":"A1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2022 Exam paper.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2022 Exam solution.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nA 700 kg sports car is powered with a 1.6 litre, 4-stroke, 6-cylinder engine. At 5,000 rpm, the engine\nproduces 100 kW of power. The engine has an overall efficiency of 46% and the breathing and mechanical\nefficiencies are 90% and 82% respectively. We will model this as an Otto cycle operating on an ideal gas\nand you may assume that $R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.4$, where $\\gamma$ is the\nadiabatic index.\n","formulaSheet":"- Otto cycle efficiency: $\\eta_{th} = 1 - \\dfrac{T_1}{T_2} = 1 - \\dfrac{1}{r^{\\gamma-1}}$ (on the formula sheet)\n- Overall efficiency: $\\eta_o = \\eta_{th}\\,\\eta_b\\,\\eta_m$ (learn this)\n- Power: $P = W_{net}\\,N\\,\\eta_b\\,\\eta_m$ with $N = \\dfrac{\\mathrm{rpm}}{120}$ for a 4-stroke (learn this)\n- Brake work: $W_b = \\eta_m W_{net}$ (learn this)\n- Ideal gas $pV = mR_gT$, $c_v = \\dfrac{R_g}{\\gamma-1}$, $TV^{\\gamma-1} = \\text{constant}$ (on the formula sheet)\n- $\\eta_{th} = \\dfrac{W_{net}}{Q_{in}} = 1 - \\dfrac{Q_{out}}{Q_{in}}$; isochoric $\\Delta Q = mc_v\\Delta T$ (on the formula sheet)\n- Adiabatic work $\\Delta W = -\\Delta U = mc_v\\Delta T$ (on the formula sheet)\n- Fuel economy: distance per litre $=$ speed $\\times$ time to burn 1 litre (learn this)\n","parts":[{"label":"a","text":"Determine the compression ratio, brake work and net work output per cylinder, per cycle.","marks":10,"grade":"2:2","answer":[{"label":"r","value":11.48,"units":"-"},{"label":"W_b per cylinder","value":0.444,"units":"kJ"},{"label":"W_net per cylinder","value":0.542,"units":"kJ"}]},{"label":"b","text":"At the intake, the air pressure and temperature are $p_1 = 101\\ \\mathrm{kPa}$ and $T_1 = 288\\ \\mathrm{K}$.\nDetermine the temperature at each stage in the process, the work done during compression and expansion\nand the heat rejection per cylinder per cycle.\n","marks":8,"grade":"2:2","answer":[{"label":"T_2","value":764.5,"units":"K"},{"label":"T_3","value":4484,"units":"K"},{"label":"T_4","value":1689,"units":"K"},{"label":"W_comp","value":0.1114,"units":"kJ"},{"label":"W_exp","value":0.6534,"units":"kJ"},{"label":"Q_out","value":0.3276,"units":"kJ"}]},{"label":"c","text":"Suggest two ways in which the thermal efficiency of the engine could be improved and briefly explain how\neach method works.\n","marks":2,"grade":"third"},{"label":"d","text":"The air-fuel mix consists of 1.6% petrol by mass and petrol has a density of 0.8 kg/litre. Assuming the\ncar travels at 70 mph (112 km/hour) in 6th gear at 5000 rpm, determine the fuel economy of the engine in\nkm/litre.\n","marks":5,"grade":"2:1","answer":[{"label":"fuel economy","value":19.1,"units":"km/L"}]}],"solution":"$a0","confidence":"HIGH","solutionNote":"Matches the official solution and the Week 5 L15 worked example: r = 11.482, W_net = 0.542 kJ, W_b = 0.444 kJ,\nQ_out = 0.328 kJ, T_2 = 764.55 K, W_comp = 111.4 J, W_exp = 653.9 J, fuel economy 19.07 km/L (official T_3 = 4481 K\nand T_4 = 1687.8 K differ by under 0.1% from rounding). The official method ignores the breathing efficiency in (d);\nincluding it gives 21.2 km/L.\n"},{"id":"engr217-2022-a2","module":"ENGR5003","week":5,"label":"ENGR217 2022 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"first","tags":["gas-turbines","brayton-cycle","work-ratio","regeneration","mass-flow"],"source":{"name":"ENGR217 Summer 2022 Q A2","paper":"ENGR217","year":2022,"question":"A2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2022 Exam paper.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2022 Exam solution.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nA gas power plant operates a gas turbine, which is modelled as an ideal Brayton cycle. At the compressor\ninlet, a mixture of natural gas and air is drawn in at a pressure of 5 bar at $32^\\circ\\mathrm{C}$. After\ncombustion, the temperature of the gas reaches $2800^\\circ\\mathrm{C}$. You may assume that the working\nfluid is an ideal gas, with $R_g = 0.301\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.42$.\n","formulaSheet":"- $c_p = \\dfrac{\\gamma R_g}{\\gamma - 1}$; isobaric heat addition $q = c_p(T_3 - T_2)$ (on the formula sheet)\n- Adiabatic process: $\\dfrac{p_2}{p_1} = \\left(\\dfrac{T_2}{T_1}\\right)^{\\gamma/(\\gamma-1)}$ (on the formula sheet)\n- Brayton efficiency: $\\eta_{th} = 1 - \\dfrac{T_1}{T_2} = 1 - r_p^{-(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Work ratio: $r_w = \\dfrac{w_{out} - w_{in}}{w_{out}} = 1 - \\dfrac{T_1}{T_3}r_p^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- $W_{net} = Q_{in} - Q_{out}$, $\\eta_{th} = W_{net}/Q_{in}$ (on the formula sheet)\n- Mass flow (continuity): $\\dot m = \\rho A C = \\dfrac{AC}{v}$, with $pv = R_gT$ (learn this; $pv = R_gT$ is on the sheet)\n- SFEE for the heat exchanger: $\\dot Q = \\dot m(h_{out} - h_{in}) = \\dot m c_p \\Delta T$ (on the formula sheet)\n","parts":[{"label":"a","text":"Determine the required pressure ratio if the specific heat addition is 1800 kJ/kg. What is the resulting\nthermal efficiency and work ratio?\n","marks":7,"grade":"2:2","answer":[{"label":"r_p","value":135.8,"units":"-"},{"label":"eta_th","value":0.766,"units":"-"},{"label":"r_w","value":0.5756,"units":"-"}]},{"label":"b","text":"The gas inlet duct has a cross-sectional area of $3.5\\ \\mathrm{m^2}$, determine the mass flow and flow\nvelocity of the gas mixture in the inlet duct, given that the turbine provides a net power output of 10 MW.\n","marks":4,"grade":"2:2","answer":[{"label":"m_dot","value":7.252,"units":"kg/s"},{"label":"C","value":0.381,"units":"m/s"}]},{"label":"c","text":"A regenerative heat exchanger is added to increase the thermal efficiency of the turbine. 10% of the\nexcess heat from the exhaust is lost to the surroundings before the heat exchanger and after the heat\nexchanger, the compressed gas and exhaust are at thermal equilibrium ($T_2' = T_4'$). Determine this\nequilibrium temperature, $T_2'$, the specific heat exchanged as well as the new thermal efficiency and\nwork ratio.\n","marks":10,"grade":"first","answer":[{"label":"T_2'","value":991,"units":"K"},{"label":"q exchanged","value":-319,"units":"kJ/kg"},{"label":"eta_th new","value":0.651,"units":"-"},{"label":"r_w","value":0.5756,"units":"-"}]},{"label":"d","text":"Explain under what conditions a regenerative heat exchanger wouldn't be able to increase the thermal\nefficiency of the turbine. At what is the pressure ratio and thermal efficiency when a regenerative heat\nexchanger stops improving efficiency? You may assume that the specific heat addition does not change and\nneglect any heat loss for the exhaust gases prior to the heat exchanger.\n","marks":4,"grade":"first","answer":[{"label":"r_p limit","value":39.16,"units":"-"},{"label":"eta_th limit","value":0.662,"units":"-"}]}],"solution":"$a1","confidence":"MEDIUM","solutionNote":"The official exam solution is internally inconsistent: it finds T_2 from q = 1800 kJ/kg but then uses\nq = 2500 kJ/kg in (b)-(d) (official: m_dot = 7.92 kg/s, C = 0.412 m/s, T_2' = 1006.9 K, q_x = 397 kJ/kg,\neta = 0.697, r_p = 61.5, eta = 0.704). The lecturer's Week 6 L19 review (old-spec slides) works the printed\n1800 kJ/kg throughout and agrees with this solution for (a) r_p = 135.8, eta = 0.766, r_w = 0.5756; (b) 7.25 kg/s,\n0.381 m/s; (c) T_2' = 991.0 K, q_x = -318.96 kJ/kg, r_w unchanged; (d) r_p = 39.16, eta = 0.662. The one remaining\ndisagreement is the new efficiency in (c): L19 uses 1 - T_1/T_2' = 0.692, which is only valid without\nregeneration; w_net/q_in' = 0.651 is correct. Confirm with the lecturer which value the marking scheme expects.\n"},{"id":"engr217-2023-a1","module":"ENGR5003","week":4,"label":"ENGR217 2023 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["ic-engines","otto-cycle","work-ratio","engine-efficiencies","brake-power"],"source":{"name":"ENGR217 Summer 2023 Q A1","paper":"ENGR217","year":2023,"question":"A1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2023 Exam paper.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2023 Exam solution.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nA small hatchback car has a 4-cylinder, 1 litre petrol engine, which outputs 52 kW of power at 4000 rpm\nand can be modelled as an Otto cycle. You may assume $R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.4$,\nwhere $\\gamma$ is the adiabatic index.\n","formulaSheet":"- Adiabatic process: $pV^\\gamma = \\text{constant}$, $TV^{\\gamma-1} = \\text{constant}$ (on the formula sheet)\n- Gay-Lussac (isochoric): $\\dfrac{p_2}{T_2} = \\dfrac{p_3}{T_3}$ (on the formula sheet)\n- Isochoric heat: $\\Delta Q = mc_v\\Delta T$, $c_v = \\dfrac{R_g}{\\gamma-1}$ (on the formula sheet)\n- Otto efficiency $\\eta_{th} = 1 - \\dfrac{1}{r^{\\gamma-1}}$; $\\eta_{th} = 1 - \\dfrac{Q_{out}}{Q_{in}}$; $W_{net} = Q_{in} - Q_{out}$ (on the formula sheet)\n- Work ratio: $r_w = \\dfrac{W_{net}}{W_{out}}$ (given in the question; Brayton form on the sheet)\n- Brake power: $P_b = P\\,\\eta_v\\,\\eta_m$ (learn this)\n- $P = Fv = mv\\dfrac{dv}{dt}$ (given in the question)\n","parts":[{"label":"a","text":"At the intake, the temperature and pressure are $p_1 = 101\\ \\mathrm{kPa}$ and $T_1 = 22^\\circ\\mathrm{C}$.\nThe compression ratio is 3.2 and the maximum temperature of the cycle is $900^\\circ\\mathrm{C}$. Determine\nthe temperature and pressure at all stages of the cycle.\n","marks":5,"grade":"2:2","answer":[{"label":"p_2","value":514.7,"units":"kPa"},{"label":"T_2","value":470,"units":"K"},{"label":"p_3","value":1284.6,"units":"kPa"},{"label":"p_4","value":252.1,"units":"kPa"},{"label":"T_4","value":736.7,"units":"K"}]},{"label":"b","text":"Determine the specific heat addition and heat rejection of the cycle, as well as the thermal efficiency\nand work ratio, which is given as: $r_w = \\dfrac{W_{net}}{W_{out}}$\n","marks":7,"grade":"2:2","answer":[{"label":"q_in","value":504.5,"units":"kJ/kg"},{"label":"q_out","value":316.8,"units":"kJ/kg"},{"label":"eta_th","value":0.372,"units":"-"},{"label":"r_w","value":0.599,"units":"-"}]},{"label":"c","text":"Given that the volumetric and mechanical efficiencies for the engine are 70% and 78% respectively,\ndetermine the brake power output at the road wheels of the car. Given that the car weighs 1000 kg,\ndetermine the time it would take the car to accelerate from stationary to 100 kph, assuming a constant\npower output from the engine, given that power can be expressed as: $P = Fv = mv\\dfrac{dv}{dt}$\n","marks":7,"grade":"2:2","answer":[{"label":"P_b","value":28.39,"units":"kW"},{"label":"t","value":13.59,"units":"s"}]},{"label":"d","text":"On a particularly cold winter morning, the ambient air temperature is $-5^\\circ\\mathrm{C}$. Discuss how\nthis might affect the isentropic, volumetric and mechanical efficiencies of the engine, when the engine\nstarts up cold and once the car and engine are up to full operating temperature.\n","marks":6,"grade":"2:1"}],"solution":"$a2","confidence":"HIGH","solutionNote":"Matches the official solution: p_2 = 514.67 kPa, T_2 = 470 K, p_3 = 1284.64 kPa, p_4 = 252.1 kPa, T_4 = 736.7 K,\nq_in = 504.51 kJ/kg, q_out = 316.82 kJ/kg, eta = 0.372, r_w = 0.599, P_b = 28.39 kW, t = 13.59 s. Part (d) points\nfollow the official marking scheme.\n"},{"id":"engr217-2023-a2","module":"ENGR5003","week":2,"label":"ENGR217 2023 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["charging-discharging","steady-flow","real-gas-steam","unsteady-flow","diffuser","ideal-gas-validity"],"source":{"name":"ENGR217 Summer 2023 Q A2","paper":"ENGR217","year":2023,"question":"A2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2023 Exam paper.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2023 Exam solution.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nA carbon dioxide fire extinguisher consists of a 5 litre rigid tank, pressurised to 55 bar at a\ntemperature of $20^\\circ\\mathrm{C}$, which is the same as the ambient temperature. The fire extinguisher is\ndischarged for 5 seconds and afterwards the pressure of the vessel is 10 bar. The external pressure may be\nassumed to be 101 kPa, and you may assume $R_g = 0.189\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.29$, where\n$\\gamma$ is the adiabatic index.\n","formulaSheet":"- Unsteady flow energy equation (with $\\Delta m_{in} = 0$, $\\Delta Q = \\Delta W = 0$) (on the formula sheet)\n- Adiabatic discharge: $\\dfrac{T_2}{T_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Mass left in the tank: $\\dfrac{m_2}{m_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{1/\\gamma}$ (learn this; derived in Lecture 6)\n- Ideal gas $pV = mR_gT$; $c_v = \\dfrac{R_g}{\\gamma - 1}$; isochoric $\\Delta Q = mc_v\\Delta T$ (on the formula sheet)\n- Continuity: $\\dot m = \\rho AC = \\dfrac{pAC}{R_gT}$ (learn this)\n- SFEE: $h_1 + \\tfrac{C_1^2}{2} = h_2 + \\tfrac{C_2^2}{2}$ (on the formula sheet)\n","parts":[{"label":"a","text":"Treating this as an unsteady flow problem, determine the change in mass of the tank as well as the final\ntemperature of the gas discharged from the tank, as well as the final temperature of the gas inside the\ntank. Please state all assumptions made.\n","marks":10,"grade":"2:1","answer":[{"label":"Delta m","value":-0.364,"units":"kg"},{"label":"T discharged (final)","value":199.8,"units":"K","tol":0.05},{"label":"T_2 tank","value":199.8,"units":"K"}]},{"label":"b","text":"The tank is allowed to warm back up to ambient temperature. Determine the pressure of the tank and the\namount of heat added to the tank from the surroundings to increase the temperature.\n","marks":6,"grade":"2:2","answer":[{"label":"p_3","value":1467,"units":"kPa"},{"label":"Q","value":8.05,"units":"kJ"}]},{"label":"c","text":"The output of the tank has a cross-sectional area of $1\\ \\mathrm{cm^2}$, which is connected to a nozzle,\nwhich acts as a diffuser, the output of which has a cross-sectional area of $10\\ \\mathrm{cm^2}$. Assuming\nthe mass flow rate during the discharge is approximately constant, determine the flow velocity at the\nentrance and exit of the nozzle, as well as the temperature and pressure of the carbon dioxide at the exit\nof the nozzle. State all assumptions made. You may assume the flow velocity is much lower than the\nacoustic velocity.\n","marks":7,"grade":"2:1","answer":[{"label":"C_in","value":7.33,"units":"m/s","tol":0.05},{"label":"C_out","value":0.733,"units":"m/s","tol":0.05},{"label":"T_exit","value":293.18,"units":"K"},{"label":"p_exit","value":5503,"units":"kPa"}]},{"label":"d","text":"For the temperature of discharged gas calculated from part (a), discuss whether it is appropriate to\nconsider carbon dioxide as an ideal gas in this case and state whether we would expect the real\ntemperature to be higher, lower or the same.\n","marks":2,"grade":"2:2"}],"solution":"$a3","confidence":"MEDIUM","solutionNote":"Agrees with the official solution on the method and on T_2 = 199.83 K, C_in = 7.33 m/s, C_out = 0.73 m/s and the\nanswers to (d). Small differences: the official solution converts 55 bar and 10 bar using 101 kPa per bar, giving\nm_1 = 0.501 kg, Delta m = -0.368 kg, p_3 = 1481.7 kPa and Q = 8.13 kJ (this solution uses 100 kPa per bar: 0.496 kg,\n-0.364 kg, 1467 kPa, 8.05 kJ). DISAGREEMENT: the official gives the final discharged-gas temperature as 59.4 K\n(-213.7 C) from a formula with no derivation; that is colder than the gas left in the tank and than isentropic\nexpansion to 101 kPa (119.3 K), so it cannot be right for an ideal gas. This solution gives 199.8 K at the valve\n(and quotes the 253.6 K mass average and the 119.3 K isentropic jet limit). The official question version omits the\nexit temperature and pressure in (c). Confirm with the lecturer what \"temperature of the gas discharged\" means.\n"},{"id":"engr217-2025-q1","module":"ENGR5003","week":4,"label":"ENGR217 2025 Q1","kind":"past-paper","level":3,"marks":20,"minutes":24,"grade":"2:2","tags":["ic-engines","second-law","diesel-cycle","carnot-efficiency","mean-effective-pressure"],"source":{"name":"ENGR217 Summer 2025 Q1","paper":"ENGR217","year":2025,"question":"Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam paper-main.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam solution-main.pdf","tag":"VALID"},"text":"The compression ratio of a certain Diesel cycle (Figure Q1-1) is 19. The heat absorption\n$q_1 = 800\\ \\mathrm{kJ/kg}$ for 1 kg of air inputting. Given that the state at the beginning of compression\nis $t_1 = 25^\\circ\\mathrm{C}$, $p_1 = 100\\ \\mathrm{kPa}$, calculate:\n","figures":[{"src":"/figures/engr217-2025-q1.png","caption":"Figure Q1-1: Schematic of a Diesel cycle (T-s diagram)."}],"formulaSheet":"- Ideal gas: $pv = R_g T$, $R_g = 0.287\\ \\mathrm{kJ/(kg\\,K)}$ for air (on the formula sheet)\n- $c_v = \\dfrac{R_g}{\\gamma - 1}$, $c_p = \\dfrac{\\gamma R_g}{\\gamma - 1}$, $\\gamma = 1.4$ (on the formula sheet)\n- Isentropic (adiabatic reversible) process: $pv^\\gamma = \\mathrm{const}$, $Tv^{\\gamma-1} = \\mathrm{const}$ (on the formula sheet)\n- Constant-pressure heat addition: $q = c_p\\,\\Delta T$ (on the formula sheet)\n- Constant-volume heat rejection: $q = c_v\\,\\Delta T$ (on the formula sheet)\n- $\\eta_{th} = \\dfrac{w_{net}}{q_{in}} = 1 - \\dfrac{q_{out}}{q_{in}}$ (on the formula sheet)\n- Diesel: $\\eta = 1 - \\dfrac{1}{r^{\\gamma-1}}\\left[\\dfrac{r_c^\\gamma - 1}{\\gamma (r_c - 1)}\\right]$ (on the formula sheet)\n- Cut-off ratio $r_c = v_3/v_2$, compression ratio $r = v_1/v_2$ (learn this)\n- Carnot efficiency $\\eta_C = 1 - \\dfrac{T_L}{T_H}$ (temperatures in K) (learn this)\n- Mean effective pressure $\\mathrm{MEP} = \\dfrac{w_{net}}{v_{max} - v_{min}} = \\dfrac{w_{net}}{v_1 - v_2}$ (learn this)\n","parts":[{"label":"a","text":"The pressure, temperature, and specific volume at each point in the cycle","marks":9,"answer":[{"label":"v_1","value":0.8557,"units":"m^3/kg","tol":0.02},{"label":"p_2","value":6170,"units":"kPa","tol":0.02},{"label":"T_2","value":968.1,"units":"K","tol":0.02},{"label":"v_2","value":0.04504,"units":"m^3/kg","tol":0.02},{"label":"p_3","value":6170,"units":"kPa","tol":0.02},{"label":"T_3","value":1765,"units":"K","tol":0.02},{"label":"v_3","value":0.08208,"units":"m^3/kg","tol":0.02},{"label":"p_4","value":231.7,"units":"kPa","tol":0.02},{"label":"T_4","value":690.9,"units":"K","tol":0.02},{"label":"v_4","value":0.8557,"units":"m^3/kg","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"The thermal efficiency of the cycle, and compare it with the Carnot cycle efficiency with the same\ntemperature limits\n","marks":6,"answer":[{"label":"eta_th","value":0.648,"units":"-","tol":0.02},{"label":"eta_Carnot","value":0.831,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"The mean effective pressure","marks":5,"answer":[{"label":"MEP","value":639.2,"units":"kPa","tol":0.02}],"grade":"$undefined"}],"solution":"$a4","confidence":"HIGH","solutionNote":"Compared with the official solution: v1 = 0.8557, v2 = 0.0450, p2 = 6169.6 kPa, v3 = 0.0820, p4 = 231.5 kPa,\neta_th = 0.648 and MEP = 639.4 kPa all agree. The official T2 = 967.35 K, T3 = 1763.37 K and T4 = 690.25 K\ndiffer from ours (968.1, 1764.6, 690.9 K) by under 0.1 % only because it rounds v2 to 0.0450 and uses\ncp = 1.005 instead of gamma Rg/(gamma-1) = 1.0045; both within tolerance. Disagreement: the official Carnot\nefficiency is printed as 0.848, but its own numbers give 1 - 298.15/1763.37 = 0.831; 0.848 is an arithmetic\nslip, so trust 0.831.\n"},{"id":"engr217-2025-q2","module":"ENGR5003","week":2,"label":"ENGR217 2025 Q2","kind":"past-paper","level":3,"marks":30,"minutes":36,"grade":"2:1","tags":["charging-discharging","second-law","unsteady-flow","entropy-change","kinetic-energy"],"source":{"name":"ENGR217 Summer 2025 Q2","paper":"ENGR217","year":2025,"question":"Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam paper-main.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam solution-main.pdf","tag":"VALID"},"text":"A 12 litre cryogenic tank initially contains air at 1 bar at a temperature of 293 K. The tank is then\ncharged by a high pressure gas main at a temperature of 300 K to a final pressure of 250 bar. You may assume\nthat the air is an ideal gas and the tank is rigid. You may also assume that\n$R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.4$.\n","formulaSheet":"- Ideal gas: $pV = mR_gT$ (on the formula sheet)\n- $c_v = \\dfrac{R_g}{\\gamma-1}$, $c_p = \\dfrac{\\gamma R_g}{\\gamma-1}$, $dU = mc_v\\,dT$, $dH = mc_p\\,dT$ ($u = c_vT$, $h = c_pT$) (on the formula sheet)\n- Unsteady flow: $dU = \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{in} dm_{in} - \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{out} dm_{out} + dQ - dW$ (on the formula sheet)\n- SFEE: $\\dot Q - \\dot W = \\dot m\\left[\\left(h + \\tfrac{C^2}{2} + gZ\\right)_{out} - \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{in}\\right]$ (on the formula sheet)\n- Isochoric process: $dW = 0$, $dQ = dU = mc_v\\,dT$ (on the formula sheet)\n- Entropy change: $\\Delta s = c_v\\ln\\dfrac{T_2}{T_1} + R_g\\ln\\dfrac{v_2}{v_1}$ (on the formula sheet)\n- Reversible heat transfer $dQ = T\\,dS$ (given in the question)\n- Rigid tank charged from a main: $m_2u_2 - m_1u_1 = h_{in}(m_2 - m_1)$ (learn this derivation)\n","parts":[{"label":"a","text":"Determine the initial and final mass of gas in the tank as well as the final temperature, assuming an\nadiabatic charging process. You should treat this as an unsteady flow problem\n","marks":8,"answer":[{"label":"m_1","value":0.01427,"units":"kg","tol":0.02},{"label":"m_2","value":2.493,"units":"kg","tol":0.02},{"label":"T_2","value":419.3,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"The tank is allowed to cool back to 293 K. Determine the heat rejected and the pressure change of the\nprocess. Given that no work is done during the cooling process, determine the entropy change, given that\n$\\Delta Q = m c_p\\, dT = T\\, dS$.\n","marks":8,"answer":[{"label":"Q_rejected","value":225.9,"units":"kJ","tol":0.02},{"label":"dp","value":-75.29,"units":"bar","tol":0.02},{"label":"dS","value":-0.641,"units":"kJ/K","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"The tank is then discharged through a valve. We can neglect the flow velocity of the gas in the tank, but\nafter the valve, the gas flow velocity is 150 m/s. Determine the specific enthalpy of the gas leaving the\ntank and the temperature change. You may assume there is no heat exchange or work done during the\ndischarging process\n","marks":8,"answer":[{"label":"h_out","value":283.1,"units":"kJ/kg","tol":0.02},{"label":"dT","value":-11.2,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"From the calculations in part (c), discuss whether it is reasonable or not to consider the kinetic energy\nterm and justify your answer. What is the maximum flow velocity that can be achieved\n","marks":6,"answer":[{"label":"C_max","value":767,"units":"m/s","tol":0.02}],"grade":"$undefined"}],"solution":"$a5","confidence":"MEDIUM","solutionNote":"$a6"},{"id":"engr217-2025-q1-resit","module":"ENGR5003","week":1,"label":"ENGR217 2025 resit Q1","kind":"past-paper","level":3,"marks":20,"minutes":24,"grade":"2:1","tags":["ideal-gas-first-law","adiabatic-process","free-piston","molar-mass"],"source":{"name":"ENGR217 Resit 2025 Q1","paper":"ENGR217","year":2025,"question":"Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam paper-resit.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam solution-resit.pdf","tag":"VALID"},"text":"The cylinder-piston system's walls and piston are made of rigid adiabatic material. Side A and Side B\ncontain N2 and O2, respectively, which have the same temperature, pressure, and volume initially. Given that\n$T_{A1} = T_{B1} = 300\\ \\mathrm{K}$, $p_{A1} = p_{B1} = 0.1\\ \\mathrm{MPa}$, $V_{A1} = V_{B1} = 0.5\\ \\mathrm{m^3}$.\nThe piston can move freely without friction within the cylinder. After the electric heater on side A is\npowered on, it slowly heats side A until $p_{A2} = 0.22\\ \\mathrm{MPa}$. Assume that both gases are ideal\ngases and calculate using constant specific heats: (see Figure Q1-1).\n($M_{N_2} = 28.0 \\times 10^{-3}\\ \\mathrm{kg/mol}$, $M_{O_2} = 32.0 \\times 10^{-3}\\ \\mathrm{kg/mol}$,\n$R = 8.3145\\ \\mathrm{J/(mol\\cdot K)}$, $c_{V,N_2} = 742.1\\ \\mathrm{J/(kg\\cdot K)}$,\n$c_{V,O_2} = 649.6\\ \\mathrm{J/(kg\\cdot K)}$, adiabatic index $\\gamma = 1.4$)\n","figures":[{"src":"/figures/engr217-2025-q1-resit.png","caption":"Figure Q1-1: Schematic of a piston based cylinder."}],"formulaSheet":"- Ideal gas: $pV = mR_gT$ with $R_g = R/M$ (and $pV = nRT$) (on the formula sheet)\n- Adiabatic (reversible) process: $pV^\\gamma = \\mathrm{const}$, $\\dfrac{T_2}{T_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- First law: $dU = dQ - dW$ (heat in +, work out +), $dU = mc_v\\,dT$ (on the formula sheet)\n- Adiabatic work: $W = -\\Delta U$; polytropic work magnitude $\\dfrac{p_2V_2 - p_1V_1}{n - 1}$ with $n = \\gamma$ (on the formula sheet)\n- Free frictionless piston: equal pressures on both sides at every instant (learn this)\n","parts":[{"label":"a","text":"$$T_{B2}$ and $V_{B2}$","marks":7,"answer":[{"label":"T_B2","value":375.8,"units":"K","tol":0.02},{"label":"V_B2","value":0.2847,"units":"m^3","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"$$V_{A2}$ and $T_{A2}$","marks":7,"answer":[{"label":"V_A2","value":0.7153,"units":"m^3","tol":0.02},{"label":"T_A2","value":944.2,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"$$Q$ and $W_A$ (work done by side A on side B)","marks":6,"answer":[{"label":"Q","value":299.9,"units":"kJ","tol":0.02},{"label":"W_A","value":31.58,"units":"kJ","tol":0.02}],"grade":"$undefined"}],"solution":"$a7","confidence":"HIGH","solutionNote":"Matches the official solution: TB2 = 375.8 K, VB2 = 0.2847 m^3, VA2 = 0.7153 m^3, TA2 = 944.15 K (ours 944.2 K),\nQ = 299.99 kJ (ours 299.9 kJ) and WA = 31.58 kJ. The tiny differences come from the official using\nM_N2 = 28.01 g/mol (mA = 0.5615 kg) instead of the printed 28.0. The official solution also has a typo\n\"pA2 = 0.1 MPa\" for pB1 and writes 8.1345 for R when deriving cv; these do not affect its numbers.\n"},{"id":"engr217-2025-q2-resit","module":"ENGR5003","week":2,"label":"ENGR217 2025 resit Q2","kind":"past-paper","level":3,"marks":30,"minutes":36,"grade":"2:2","tags":["ideal-gas-first-law","gas-turbines","second-law","brayton-cycle","entropy-production"],"source":{"name":"ENGR217 Resit 2025 Q2","paper":"ENGR217","year":2025,"question":"Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam paper-resit.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/2025 Exam solution-resit.pdf","tag":"VALID"},"text":"Answer all parts (a) - (c).\n","formulaSheet":"- Boyle (isothermal) $p_1V_1 = p_2V_2$; Charles (isobaric) $\\dfrac{V_1}{T_1} = \\dfrac{V_2}{T_2}$; Gay-Lussac (isochoric) $\\dfrac{p_1}{T_1} = \\dfrac{p_2}{T_2}$ (on the formula sheet)\n- $c_p = \\dfrac{\\gamma R_g}{\\gamma - 1}$, $q = c_p\\,\\Delta T$ at constant pressure (on the formula sheet)\n- Isentropic: $\\dfrac{p_2}{p_1} = \\left(\\dfrac{T_2}{T_1}\\right)^{\\gamma/(\\gamma-1)}$ (on the formula sheet)\n- Brayton: $\\eta = 1 - \\dfrac{T_1}{T_2} = 1 - r_p^{-(\\gamma-1)/\\gamma}$, $r_w = \\dfrac{w_{out} - w_{in}}{w_{out}} = 1 - \\dfrac{T_1}{T_3}r_p^{(\\gamma-1)/\\gamma}$ (on the formula sheet; also printed in the question)\n- Entropy of the isolated system: $\\Delta S_{iso} = \\Delta S_{sys} + \\Delta S_{env} \\ge 0$, entropy production $S_g = \\Delta S_{iso}$ (on the formula sheet)\n- Entropy change of a solid (incompressible): $\\Delta S = mc\\ln\\dfrac{T_2}{T_1}$ (learn this)\n- Entropy change of surroundings at constant $T_0$: $\\Delta S_{env} = \\dfrac{Q_{env}}{T_0}$ (learn this)\n","parts":[{"label":"a","text":"A fixed mass of a perfect gas undergoes the following three processes: An expansion at constant\ntemperature from a pressure and volume of 11 bar and $0.005\\ \\mathrm{m^3}$ respectively to a volume of\n$0.012\\ \\mathrm{m^3}$. Cooling at constant pressure until the volume is $0.007\\ \\mathrm{m^3}$. Constant\nvolume heating until the gas reaches its original pressure of 11 bar. Determine the unknown pressure and\ntemperature at the end of each process given that the initial temperature is $250^\\circ\\mathrm{C}$.\n","marks":10,"grade":"third","answer":[{"label":"p_2","value":4.583,"units":"bar","tol":0.02},{"label":"T_2","value":523.15,"units":"K","tol":0.02},{"label":"p_3","value":4.583,"units":"bar","tol":0.02},{"label":"T_3","value":305.2,"units":"K","tol":0.02},{"label":"T_4","value":732.4,"units":"K","tol":0.02}]},{"label":"b","text":"A gas power plant operates a gas turbine, which is modelled as an ideal Brayton cycle. At the compressor\ninlet, a mixture of natural gas and air is drawn in at a pressure of 5 bar at $32^\\circ\\mathrm{C}$. After\ncombustion, the temperature of the gas reaches $3500^\\circ\\mathrm{C}$. You may assume that the working\nfluid is an ideal gas, with $R_g = 0.301\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.42$. Determine the required\npressure ratio if the specific heat addition is 1800 kJ/kg. What is the resulting thermal efficiency and\nwork ratio\n($c_p = \\dfrac{\\gamma R_g}{\\gamma - 1}$, $\\eta_T = 1 - \\dfrac{1}{r_p^{(\\gamma-1)/\\gamma}} = 1 - \\dfrac{T_1}{T_2}$,\n$r_p$ is the pressure ratio, work ratio $r_w = \\dfrac{w_{net}}{w_{out}} = 1 - \\dfrac{T_1}{T_3} r_p^{(\\gamma-1)/\\gamma}$)\n","marks":10,"grade":"2:2","answer":[{"label":"r_p","value":580.5,"units":"-","tol":0.02},{"label":"eta_th","value":0.848,"units":"-","tol":0.02},{"label":"r_w","value":0.469,"units":"-","tol":0.02}]},{"label":"c","text":"A house use sensible heat of concrete plate to realize self-heating. Given the size of the plate as\n$5\\ \\mathrm{m} \\times 8\\ \\mathrm{m} \\times 0.3\\ \\mathrm{m}$, the density as $2300\\ \\mathrm{kg/m^3}$, heat\ncapacity as $0.65\\ \\mathrm{kJ/(kg\\cdot K)}$. Please calculate the entropy production if the concrete plate\ndecrease its temperature from $28^\\circ\\mathrm{C}$ to $18^\\circ\\mathrm{C}$ in the evening (ambient\ntemperature $t_0 = 18^\\circ\\mathrm{C}$)\n","marks":10,"grade":"2:2","answer":[{"label":"S_gen","value":10.35,"units":"kJ/K","tol":0.02}]}],"solution":"$a8","confidence":"HIGH","solutionNote":"(a) Agrees with the official p2 = p3 = 4.583 bar and T2 = 523 K; the official T3 = 305.92 K and T4 = 734.05 K\ncontain small arithmetic slips (523.15 x 0.007/0.012 = 305.17 K, and T4 = T1 x 0.007/0.005 = 732.41 K exactly);\nours are within 0.25 %. (b) Agrees: T2 = 2004.46 K, rp = 580.6, eta = 84.8 % and rw = 0.4688 (the official's\n\"0.5051 = 84.8 %\" is a typo; 1 - 305.15/2004.46 = 0.848). (c) Disagreement: the official handwritten solution uses\na 5 K drop (23 C to 18 C), giving Q = 89 700 kJ and Sg = 2.62 kJ/K, but the printed question says 28 C to 18 C.\nWith the printed 28 C we get Q = 179 400 kJ and Sg = 10.35 kJ/K; the method is identical, so trust 10.35 kJ/K\nfor the question as printed (2.62 kJ/K is the answer to the 23 C version).\n"},{"id":"engr272-2026-q1-mock","module":"ENGR5003","week":6,"label":"ENGR272 week 6 mock Q1","kind":"past-paper","level":3,"marks":50,"minutes":60,"grade":"2:1","tags":["property-relations","maxwell-relations","van-der-waals","gibbs-relation","entropy-pressure"],"source":{"name":"ENGR272 Week 6 mock exam Q1 (Dr Andalibi)","paper":"ENGR272","year":2026,"question":"Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/ENGR272_mock_exam_questions_RA.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/ENGR272_mock_exam_questions_RA.pdf","tag":"VALID"},"text":"A pure substance is modelled by the van der Waals equation of state:\n\n$P = \\dfrac{R_u T}{v - b} - \\dfrac{a}{v^2}$\n\nwhere $a$ and $b$ are constants, $v$ is the molar volume, and $R_u$ is the universal gas constant. You may\nuse the Maxwell relations provided in the formula sheet.\n","formulaSheet":"- Exact differential: $dz = M\\,dx + N\\,dy$ with $\\left(\\dfrac{\\partial M}{\\partial y}\\right)_x = \\left(\\dfrac{\\partial N}{\\partial x}\\right)_y$ (on the formula sheet)\n- Reciprocity: $\\left(\\dfrac{\\partial x}{\\partial z}\\right)_y = \\dfrac{1}{(\\partial z/\\partial x)_y}$; cyclic relation $\\left(\\dfrac{\\partial x}{\\partial y}\\right)_z\\left(\\dfrac{\\partial y}{\\partial z}\\right)_x\\left(\\dfrac{\\partial z}{\\partial x}\\right)_y = -1$ (on the formula sheet)\n- Gibbs relation $dg = v\\,dP - s\\,dT$ (on the formula sheet; also given in the question)\n- Maxwell relation $\\left(\\dfrac{\\partial s}{\\partial P}\\right)_T = -\\left(\\dfrac{\\partial v}{\\partial T}\\right)_P$ (on the formula sheet)\n- van der Waals equation $P = \\dfrac{R_uT}{v - b} - \\dfrac{a}{v^2}$ (given in the question; otherwise learn this)\n","parts":[{"label":"a","text":"Starting from the Gibbs relation $dg = v\\,dP - s\\,dT$, derive an expression for\n$\\left(\\dfrac{\\partial s}{\\partial P}\\right)_T$ in the form of a Maxwell relation. Explain briefly why this\nderivative is useful in thermodynamic property calculations.\n","marks":10,"grade":"$undefined"},{"label":"b","text":"Using the van der Waals equation of state, determine an explicit expression for\n$\\left(\\dfrac{\\partial v}{\\partial T}\\right)_P$ in terms of specific volume and temperature as the state\nvariables. Show all steps clearly.\n","marks":12,"grade":"$undefined"},{"label":"c","text":"Using your results from parts (a) and (b), obtain an expression for\n$\\left(\\dfrac{\\partial s}{\\partial P}\\right)_T$ for a van der Waals gas.\n","marks":10,"grade":"$undefined"},{"label":"d","text":"Consider a van der Waals gas with $a = 0.90\\ \\mathrm{Pa\\,m^6\\,mol^{-2}}$,\n$b = 4.0 \\times 10^{-5}\\ \\mathrm{m^3\\,mol^{-1}}$, $R_u = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$. At\n$T = 400\\ \\mathrm{K}$ and $P = 5.0\\ \\mathrm{MPa}$ the molar volume is\n$v = 4.86 \\times 10^{-5}\\ \\mathrm{m^3\\,mol^{-1}}$. Calculate the numerical value of\n$\\left(\\dfrac{\\partial s}{\\partial P}\\right)_T$ at this state. Give your answer in units\n$\\mathrm{J\\,mol^{-1}\\,K^{-1}\\,Pa^{-1}}$.\n","marks":10,"answer":[{"label":"(ds/dP)_T","value":-3.3e-8,"units":"J/(mol K Pa)","tol":0.02}],"grade":"$undefined"},{"label":"e","text":"Comment briefly on the physical meaning of the sign of $\\left(\\dfrac{\\partial s}{\\partial P}\\right)_T$ for\nreal gases.\n","marks":8,"grade":"$undefined"}],"solution":"$a9","confidence":"HIGH","solutionNote":"Parts (a)-(c) agree with the official answers (same Maxwell relation and the same expression\n-Ru / [RuT/(v-b) - 2a(v-b)/v^3]). Disagreement in (d): the official answer -2.23e-8 J/(mol K Pa) comes from a\npower-of-ten slip, writing 2a(v-b)/v^3 = 1.3485e7 Pa; the correct value is 1.3485e8 Pa (1.548e-5/1.148e-13), which\ngives -8.314/(3.867e8 - 1.349e8) = -3.30e-8 J/(mol K Pa). Verified symbolically and numerically (sympy), and by\nre-solving the vdW equation for v at 5 MPa (v = 4.862e-5, giving -3.32e-8). Trust -3.30e-8. (e) agrees.\n"},{"id":"engr272-2026-q2-mock","module":"ENGR5003","week":8,"label":"ENGR272 week 6 mock Q2","kind":"past-paper","level":3,"marks":50,"minutes":60,"grade":"2:1","tags":["fins","convection","infinite-fin","churchill-bernstein","cylinder-cross-flow"],"source":{"name":"ENGR272 Week 6 mock exam Q2 (Dr Andalibi)","paper":"ENGR272","year":2026,"question":"Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/ENGR272_mock_exam_questions_RA.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/ENGR272_mock_exam_questions_RA.pdf","tag":"VALID"},"text":"A very long aluminium rod ($k = 180\\ \\mathrm{W/(m\\cdot K)}$) with a diameter of 5 mm has one end maintained\nat $100^\\circ\\mathrm{C}$ (the base). The surface of the rod is exposed to ambient air at $20^\\circ\\mathrm{C}$,\nflowing at a speed of 5 m/s (see the figure).\n","figures":[{"src":"/figures/engr272-2026-q2-mock.png","caption":"Long aluminium rod (base at 100 C) in cross-flow of air at 20 C."}],"formulaSheet":"- Fourier's law $\\dot Q_{cond} = -kA_c\\dfrac{dT}{dx}$ (on the formula sheet)\n- Newton's law of cooling $\\dot Q_{conv} = hA_s(T_s - T_\\infty)$ (on the formula sheet)\n- Infinitely long fin: $\\dot Q_{long\\ fin} = -kA_c\\left.\\dfrac{dT}{dx}\\right|_{x=0} = \\sqrt{hpkA_c}\\,(T_b - T_\\infty)$, $\\dfrac{T(x) - T_\\infty}{T_b - T_\\infty} = e^{-mx}$ (on the formula sheet)\n- Fin parameter $m = \\sqrt{\\dfrac{hp}{kA_c}}$; for a pin fin $p = \\pi D$, $A_c = \\pi D^2/4$, so $m = \\sqrt{4h/(kD)}$ (learn this)\n- $\\mathrm{Re} = \\dfrac{VD}{\\nu}$, $\\mathrm{Nu} = \\dfrac{hD}{k}$ (on the formula sheet)\n- Churchill-Bernstein: $\\mathrm{Nu}_{cyl} = 0.3 + \\dfrac{0.62\\,\\mathrm{Re}^{1/2}\\mathrm{Pr}^{1/3}}{\\left[1 + (0.4/\\mathrm{Pr})^{2/3}\\right]^{1/4}}\\left[1 + \\left(\\dfrac{\\mathrm{Re}}{282\\,000}\\right)^{5/8}\\right]^{4/5}$ for $\\mathrm{Re\\,Pr} > 0.2$, properties at $T_f = (T_s + T_\\infty)/2$ (on the formula sheet)\n- Air properties at 1 atm (Cengel Table A-15) (not given: table needed)\n","parts":[{"label":"a","text":"Assuming steady-state operations, write an energy balance on a differential control volume of the rod and\nshow that the profile of temperature excess ($\\theta(x) = T(x) - T_\\infty$) in the form of\n$\\theta(x) = C_1 e^{mx} + C_2 e^{-mx}$ with $m^2 = \\dfrac{hp}{kA_c}$ satisfies the energy balance.\nHint: energy balance for this system can be written as: (Rate of heat conduction into the element at $x$)\n= (Rate of heat conduction from the element at $x + \\Delta x$) + (Rate of heat convection from the element).\n","marks":10,"grade":"$undefined"},{"label":"b","text":"Assuming the rod is infinitely long, write the necessary boundary conditions and determine integration\nconstants $C_1$ and $C_2$, and write the final form of temperature profile equation.\n","marks":10,"grade":"$undefined"},{"label":"c","text":"Knowing the temperature profile equation from part (b), derive an equation for steady state heat transfer\nrate from the entire rod.\n","marks":5,"grade":"$undefined"},{"label":"d","text":"Assuming the convective heat transfer between the rod and its ambient can be approximated as flow over\ncylinders, estimate the heat transfer coefficient. To calculate the film temperature, use the average rod\ntemperature (i.e., average of the base and the tip temperatures).\n","marks":15,"answer":[{"label":"Re","value":1469,"units":"-","tol":0.05},{"label":"Nu","value":19.64,"units":"-","tol":0.05},{"label":"h","value":104.5,"units":"W/(m^2 K)","tol":0.05}],"grade":"$undefined"},{"label":"e","text":"At what length does the excess temperature at the rod's tip drop to 1% of excess at the base?","marks":10,"answer":[{"label":"m","value":21.55,"units":"1/m","tol":0.05},{"label":"L","value":0.214,"units":"m","tol":0.05}],"grade":"$undefined"}],"solution":"$aa","confidence":"HIGH","solutionNote":"Agrees with the official worked answers: same derivations in (a)-(c), same film temperature (40 C) and Table A-15\nair properties (rho = 1.127 kg/m^3, k = 0.02662 W/m K, mu = 1.918e-5 kg/m s, Pr = 0.7255), Re = 1469,\ncorrelation terms 18.8 and 1.03, h = 105 W/m^2 K (ours 104.5 before rounding), m = 21.6 1/m (ours 21.55) and\nL = 0.2 m (ours 0.214 m; the official rounds to 1 s.f.). Tolerance 0.05 because air properties come from a table.\n"},{"id":"engr272-2026-q3-mock","module":"ENGR5003","week":9,"label":"ENGR272 week 6 mock Q3","kind":"past-paper","level":3,"marks":50,"minutes":60,"grade":"2:1","tags":["convection","conduction","internal-flow","constant-heat-flux","sieder-tate","heat-generation"],"source":{"name":"ENGR272 Week 6 mock exam Q3 (Dr Andalibi)","paper":"ENGR272","year":2026,"question":"Q3","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/ENGR272_mock_exam_questions_RA.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Exam Papers and Solutions/ENGR272_mock_exam_questions_RA.pdf","tag":"VALID"},"text":"A system for heating water from an inlet temperature of $T_{m,i} = 20^\\circ\\mathrm{C}$ to outlet temperature\nof $T_{m,o} = 60^\\circ\\mathrm{C}$ involves passing the water through a thick-walled tube having inner and outer\ndiameters of 20 and 40 mm (see schematic below). The outer surface of the tube is well insulated, and\nelectrical heating within the wall provides for a uniform generation rate of $10^6\\ \\mathrm{W/m^3}$.\n","figures":[{"src":"/figures/engr272-2026-q3-mock.png","caption":"Schematic of the internally heated, externally insulated tube."},{"src":"/figures/engr272-2026-q3-mock-2.png","caption":"Table A-9: Properties of saturated water (provided with the question)."}],"formulaSheet":"- Energy balance on the flowing water: $\\dot Q = \\dot m c_p (T_{m,o} - T_{m,i})$ (on the formula sheet)\n- Constant surface heat flux: $\\dot q_s A_s = \\dot m c_p (T_{m,o} - T_{m,i})$, $A_s = \\pi D_i L$ (given in the question)\n- Newton's law of cooling: $\\dot q_s = h(T_s - T_m)$ (on the formula sheet)\n- Heat flux from wall generation (insulated outside): $\\dot q_s\\,\\pi D_i L = \\dot e_{gen}\\,\\dfrac{\\pi}{4}(D_o^2 - D_i^2)L$ (learn this)\n- $\\mathrm{Re} = \\dfrac{V_mD}{\\nu}$, so for a circular tube $\\mathrm{Re} = \\dfrac{4\\dot m}{\\pi D\\mu}$; laminar below 2300, turbulent above 10 000 (on the formula sheet; the $4\\dot m/(\\pi D\\mu)$ form: learn this)\n- $\\mathrm{Nu} = \\dfrac{hD}{k}$ (on the formula sheet)\n- Sieder-Tate: $\\mathrm{Nu} = 0.027\\,\\mathrm{Re}^{0.8}\\mathrm{Pr}^{1/3}\\left(\\dfrac{\\mu}{\\mu_s}\\right)^{0.14}$ for $0.7 \\le \\mathrm{Pr} \\le 16\\,700$, $\\mathrm{Re} \\ge 10\\,000$, $L/D > 60$; properties at $T_b$, $\\mu_s$ at $T_s$ (on the formula sheet)\n","parts":[{"label":"a","text":"Write an overall energy balance for the steady state heat exchange between flowing water and the tube\nwalls by considering the entire tube content as the control volume. This equation should relate the\nchange in the heat content of water as a result of convective thermal exchange with the tube wall.\n","marks":5,"grade":"$undefined"},{"label":"b","text":"For the case of constant surface heat flux ($\\dot{q}_s = \\mathrm{constant}$), the overall energy balance\nreads as $\\dot{q}_s A_s = \\dot{m} c_p (T_{m,o} - T_{m,i})$ where $A_s$ is the surface area of inner tube\nwall. For a water mass flow rate of 0.12 kg/s, how long must the tube be to achieve the desired outlet\ntemperature? Use properties of saturated water at the average of inlet and outlet temperatures (see table\nbelow).\n","marks":10,"answer":[{"label":"q_s","value":15000,"units":"W/m^2","tol":0.02},{"label":"L","value":21.3,"units":"m","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"By estimating the local convection heat transfer coefficient, find the inner surface temperature of the\ntube at the outlet of the pipe. Assume the flow is fully developed at the outlet and that the viscosity of\nwater is almost constant at this cross-section.\n","marks":25,"answer":[{"label":"Re","value":11699,"units":"-","tol":0.02},{"label":"Nu","value":79,"units":"-","tol":0.02},{"label":"h","value":2493,"units":"W/(m^2 K)","tol":0.02},{"label":"T_s,o","value":66,"units":"deg C","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"Verify if the assumption in part (c) has a significant impact on the estimated value for the inner wall\ntemperature at the outlet.\n","marks":10,"answer":[{"label":"(mu/mu_s)^0.14","value":1.061,"units":"-","tol":0.02},{"label":"h","value":2644,"units":"W/(m^2 K)","tol":0.02},{"label":"T_s,o","value":65.7,"units":"deg C","tol":0.02}],"grade":"$undefined"}],"solution":"$ab","confidence":"HIGH","solutionNote":"Agrees with the official worked answers throughout: q_s = 15 000 W/m^2, L = 21.3 m, Re = 11 699, Nu = 79,\nh = 2492 W/m^2 K, Ts,o = 66 C; in (d) (mu/mu_s)^0.14 = 1.06, Nu = 83.8, h = 2644 W/m^2 K, Ts,o = 65.7 C (the\nofficial takes mu_s = 0.43e-3 at about 65 C; we interpolate 0.427e-3 at 66 C, same result). All properties are from\nthe printed Table A-9. The schematic label Ts,o = 70 C appears to be a leftover from the textbook version of the problem (where Ts,o was given) and is not data.\nUsing local outlet properties (60 C) instead of the 40 C mean would give Ts,o = 65.0 C.\n"},{"id":"engr217-2024-b1","module":"ENGR5003","week":4,"label":"ENGR217 2024 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["ic-engines","diesel-cycle","engine-efficiencies","isentropic-efficiency","power-curve"],"source":{"name":"ENGR217 Summer 2024 Q B1","paper":"ENGR217","year":2024,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR217 In-person Main Exam 2024 combined FINAL.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nA heavy goods lorry uses a 13.4 litre, 4-stroke, 6 cylinder inline diesel engine that generates 412 kW of\npower at 1700 rpm, with a cut off ratio of 6.3. The overall engine efficiency is 32.5% and the breathing and\nmechanical efficiencies can be assumed to be 75% and 85% respectively. We will model this engine as the\nDiesel cycle, operating on an ideal gas and you may assume that $R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and\n$\\gamma = 1.4$, where $\\gamma$ is the adiabatic index.\n","figures":[{"src":"/figures/engr217-2024-b1.png","caption":"Figure A1-1 (as printed): power curve for a petrol or diesel engine."}],"formulaSheet":"- Diesel efficiency: $\\eta_{th} = 1 - \\dfrac{1}{r^{\\gamma-1}}\\left[\\dfrac{r_c^\\gamma - 1}{\\gamma(r_c - 1)}\\right]$ (on the formula sheet)\n- Overall efficiency $\\eta_o = \\eta_{th}\\eta_b\\eta_m$; power $P = W_{net}N\\eta_b\\eta_m$, $N = \\mathrm{rpm}/120$; $W_b = \\eta_mW_{net}$ (learn this)\n- $\\eta_{th} = W_{net}/Q_{in}$; isobaric heat addition $\\Delta Q = mc_p\\Delta T$; $c_p = \\dfrac{\\gamma R_g}{\\gamma-1}$ (on the formula sheet)\n- Adiabatic process: $TV^{\\gamma-1} = \\text{constant}$; ideal gas $pV = mR_gT$ (on the formula sheet)\n- Isentropic efficiencies: $\\eta_{s,c} = \\Delta T_s/\\Delta T_{real}$, $\\eta_{s,e} = \\Delta T_{real}/\\Delta T_s$ (given in the question)\n","parts":[{"label":"a","text":"Determine the thermal efficiency, compression ratio, brake work and net work per cylinder, per cycle.","marks":10,"grade":"2:2","answer":[{"label":"eta_th","value":0.51,"units":"-"},{"label":"r","value":20.41,"units":"-"},{"label":"W_b per cylinder","value":6.463,"units":"kJ"},{"label":"W_net per cylinder","value":7.603,"units":"kJ"}]},{"label":"b","text":"Given that the compression and expansion strokes have isentropic efficiencies of 90% and 85%\nrespectively, determine the actual temperature of each stage of the process, given that at the intake,\nthe air pressure and temperature are $p_1 = 101\\ \\mathrm{kPa}$ and $T_1 = 288\\ \\mathrm{K}$. You may assume\nthat the heat addition per cycle is the same as for the isentropic process. The equations for the\nisentropic efficiencies are given as:\n$\\eta_{s,\\mathrm{compression}} = \\dfrac{\\Delta h_s}{\\Delta h_{real}} = \\dfrac{\\Delta T_s}{\\Delta T_{real}}$ and\n$\\eta_{s,\\mathrm{expansion}} = \\dfrac{\\Delta h_{real}}{\\Delta h_s} = \\dfrac{\\Delta T_{real}}{\\Delta T_s}$\n","marks":8,"grade":"first","answer":[{"label":"T_2 actual","value":1037,"units":"K"},{"label":"T_3 actual","value":6138,"units":"K","tol":0.06},{"label":"T_4 actual","value":4181,"units":"K","tol":0.05}]},{"label":"c","text":"Without detailed calculations, briefly describe why petrol and diesel engines have a power curve (such as\nthat shown in Figure A1-1). Discuss how the thermal, breathing and mechanical efficiencies vary with RPM,\nas well as the fuel burn rate of the air/fuel mixture.\n","marks":5,"grade":"2:1"},{"label":"d","text":"Discuss why a diesel engine is preferable for a heavy goods vehicle over petrol engines.","marks":2,"grade":"third"}],"solution":"$ac","confidence":"MEDIUM","solutionNote":"No official solution available. Part (a) is unambiguous (checked by back-substitution). Part (b) is\nover-specified: the heat input can be taken from the ideal cycle defined by r and r_c (13.98 kJ, used here,\nfollowing \"heat addition is the same as for the isentropic process\") or from W_net/eta_th = 14.91 kJ, and the\nexpansion volume ratio from the given r_c or from the actual T_3/T_2; T_4 ranges 4100-4400 K depending on these\nchoices. Confirm the intended route with the lecturer.\n"},{"id":"engr217-2024-b2","module":"ENGR5003","week":2,"label":"ENGR217 2024 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"first","tags":["charging-discharging","unsteady-flow","isentropic-efficiency","flow-work"],"source":{"name":"ENGR217 Summer 2024 Q B2","paper":"ENGR217","year":2024,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR217 In-person Main Exam 2024 combined FINAL.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nA 20 litre scuba tank initially contains air at 1 bar and a temperature of 300 K. The tank is charged from a\nhigh pressure air supply, which is at an initial temperature of 315 K, until the scuba tank has a final\npressure of 200 bar. You may assume that air is an ideal gas and that $R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and\n$\\gamma = 1.4$, where $\\gamma$ is the adiabatic index.\n","formulaSheet":"- Unsteady flow energy equation: $\\Delta U = \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{in}\\Delta m_{in} - \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{out}\\Delta m_{out} + \\Delta Q - \\Delta W$ (on the formula sheet)\n- Ideal gas $pV = mR_gT$, $u = c_vT$, $h = c_pT$, $c_v = \\dfrac{R_g}{\\gamma-1}$, $c_p = \\dfrac{\\gamma R_g}{\\gamma-1}$ (on the formula sheet)\n- Isochoric cooling: $\\Delta Q = mc_v\\Delta T$, $p/T = \\text{constant}$ (on the formula sheet)\n- Adiabatic discharge: $\\dfrac{T_2}{T_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{(\\gamma-1)/\\gamma}$, $\\dfrac{m_2}{m_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{1/\\gamma}$ (first on the formula sheet; mass ratio learn this, Lecture 6)\n- Flow work: $W_{flow} = pV$ (per unit mass $pv = R_gT$) (on the formula sheet)\n- Isentropic efficiency $\\eta_s = \\Delta h_s/\\Delta h_{actual}$ (given in the question)\n","parts":[{"label":"a","text":"Treating this as an unsteady flow problem, determine the change in mass of the tank as well as the final\ntemperature, given that this can be assumed to be an adiabatic process. State all assumptions made.\n","marks":8,"grade":"2:2","answer":[{"label":"Delta m","value":3.145,"units":"kg"},{"label":"T_2","value":440,"units":"K"}]},{"label":"b","text":"The charging process has an isentropic efficiency ($\\eta_s = \\Delta h_s / \\Delta h_{actual}$) of 70%, what\nwould be the final temperature of the tank after charging? Use this with the unsteady flow energy equation\nto estimate the heat lost during this process?\n","marks":6,"grade":"first","answer":[{"label":"T_2 actual","value":500,"units":"K","tol":0.05},{"label":"Q (UFEE)","value":120.3,"units":"kJ","tol":0.05}]},{"label":"c","text":"The tank is allowed to cool back to 300 K from your answer to part (b) and then discharged adiabatically\nthrough a valve to provide air to a diver. We can neglect kinetic and potential energy in this case and\nassume that the pressure is 2 bar during a dive. Determine the heat lost during cooling and the flow work\ndone during the discharging process, assuming that the final pressure in the tank is 2 bar. What is the\nfinal temperature of the discharged gas?\n","marks":8,"grade":"first","answer":[{"label":"Q lost in cooling","value":400,"units":"kJ","tol":0.05},{"label":"W_flow","value":168.6,"units":"kJ","tol":0.05},{"label":"T final discharged","value":93.1,"units":"K","tol":0.05}]},{"label":"d","text":"From the final temperature of the tank from part (a), discuss whether this would increase or decrease if\nthe gas constant, $R_g$, or the adiabatic index, $\\gamma$, increased. What happens if $p_2$ is much larger\n$p_2$? What about when the temperature of the high pressure air line is much higher than the ambient\ntemperature?\n","marks":3,"grade":"2:1"}],"solution":"$ad","confidence":"MEDIUM","solutionNote":"No official solution available. (a) and (d) are standard (Lecture 6 charging result). Part (b) is ambiguous:\napplying eta_s = dh_s/dh_actual as for a compressor gives T = 500 K and a UFEE heat term of +120 kJ (energy gained\nby the gas, not lost), while applying it the other way gives 398 K and -106 kJ (heat lost, matching the wording).\nPart (c) depends on (b). Confirm the intended reading with the lecturer. Part (d) is printed as \"if p2 is much\nlarger p2\"; read as the final pressure much larger than the initial pressure p1.\n"},{"id":"engr266-2021-q1","module":"ENGR5003","week":2,"label":"ENGR266 2021 Q1","kind":"past-paper","level":3,"marks":19,"minutes":23,"grade":"2:1","tags":["steady-flow","real-gas-steam","sfee","compressibility-factor","virial-correlation"],"source":{"name":"ENGR266 Summer 2021 Q1","paper":"ENGR266","year":2021,"question":"Q1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR266 Exam 2021.pdf","tag":"PARTIAL","leftOut":"Parts (b)-(c) need the Pitzer second-virial correlation and the generalised compressibility chart, which\nare not in the current slides or formula sheets (only Z, van der Waals and Redlich-Kwong are taught).\n"},"text":"Answer ALL parts (a) - (c).\n","formulaSheet":"- SFEE: $\\dot Q - \\dot W = \\dot m\\left[(h + \\tfrac{C^2}{2} + gZ)_{out} - (h + \\tfrac{C^2}{2} + gZ)_{in}\\right]$ (on the formula sheet)\n- Liquid enthalpy change: $\\Delta h = c_p \\Delta T$ (from $dQ = m c_p dT$, $c_p = (\\partial h/\\partial T)_p$) (on the formula sheet)\n- Compressibility factor: $Z = \\dfrac{PV}{RT}$, reduced properties $T_r = T/T_c$, $P_r = P/P_c$ (learn this)\n- [EXTERNAL] Pitzer second-virial correlation: $Z = 1 + \\hat B \\dfrac{P_r}{T_r}$, $\\hat B = B^0 + \\omega B^1$, $B^0 = 0.083 - \\dfrac{0.422}{T_r^{1.6}}$, $B^1 = 0.139 - \\dfrac{0.172}{T_r^{4.2}}$ (learn this; not in the current materials)\n- [EXTERNAL] Lee-Kessler corresponding states: $Z = Z^0 + \\omega Z^1$ with $Z^0, Z^1$ read from the Lee-Kessler tables (learn this; not in the current materials)\n","parts":[{"label":"a","text":"Water in a tank at $87^\\circ\\mathrm{C}$ is pumped into a second tank that is located 20 m above the initial\ntank at a rate of $6.5\\ \\mathrm{kg\\,s^{-1}}$. The pump does $500\\ \\mathrm{kJ\\,s^{-1}}$ of work and the water\nloses $300\\ \\mathrm{kJ\\,s^{-1}}$ of heat during the transfer. Assuming that the specific heat of water is\n$C_P = 4.186\\ \\mathrm{kJ\\,kg^{-1}}$, calculate the temperature of the water in the second tank, stating any\nassumptions made.\n","marks":7,"grade":"2:2","answer":[{"label":"T_2","value":94.3,"units":"C","tol":0.02}]},{"label":"b(i)","text":"Calculate the compressibility factor for methanol at a temperature of $393^\\circ\\mathrm{C}$ and a pressure\nof 404.85 bar using the Pitzer Correlation for the Second Virial Coefficient. For methanol the critical\ntemperature and pressure are $239.5^\\circ\\mathrm{C}$ and 80.97 bar respectively and the acentric factor\nis 0.354.\n","marks":6,"grade":"2:2","answer":[{"label":"Z_virial","value":0.3628,"units":"-","tol":0.02}]},{"label":"b(ii)","text":"Calculate the same compressibility factor using the theorem of corresponding states.","marks":2,"grade":"2:2","answer":[{"label":"Z_LK","value":0.766,"units":"-","tol":0.05}]},{"label":"c","text":"Comment on differences between the two numbers and discuss their reliability.","marks":4,"grade":"2:1"}],"solution":"$ae","confidence":"MEDIUM","solutionNote":"No official solution available. Part (a) is on-syllabus and checked by back-substitution (HIGH). Parts (b)-(c)\nuse the Pitzer and Lee-Kessler generalised correlations, which are not in the current materials ([EXTERNAL]);\nthe Lee-Kessler $Z^0$, $Z^1$ were computed from the Lee-Kessler equation rather than read from printed tables, so\na table reading may differ in the third decimal place. Confirm with the lecturer whether \"theorem of\ncorresponding states\" means $Z^0$ only (0.736) or $Z^0 + \\omega Z^1$ (0.766).\n"},{"id":"engr266-2021-q2","module":"ENGR5003","week":3,"label":"ENGR266 2021 Q2","kind":"past-paper","level":3,"marks":15,"minutes":18,"grade":"2:2","tags":["real-gas-steam","second-law","refrigeration","r134a","coefficient-of-performance","isentropic-efficiency"],"source":{"name":"ENGR266 Summer 2021 Q2","paper":"ENGR266","year":2021,"question":"Q2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR266 Exam 2021.pdf","tag":"PARTIAL","leftOut":"The vapour-compression refrigeration cycle and its COP are not in the current slides; the component\nanalysis (R134a tables, isentropic compressor, throttling, evaporator energy balance) is on-syllabus.\nPart (d) on refrigerant choice is background knowledge.\n"},"text":"Answer All parts (a) - (d). You will need HFC-134a (R134a) property tables.\n","formulaSheet":"- SFEE per unit mass (each component, $\\Delta KE$, $\\Delta PE$ neglected): $q - w = h_{out} - h_{in}$ (on the formula sheet)\n- Throttle valve (adiabatic, no work): $h_4 = h_3$ (from the SFEE, on the formula sheet)\n- Refrigerator COP: $COP_R = \\dfrac{q_L}{w_{in}} = \\dfrac{h_1 - h_4}{h_2 - h_1}$ (learn this)\n- Compressor isentropic efficiency: $\\eta_c = \\dfrac{h_{2s} - h_1}{h_2 - h_1}$ (learn this)\n- Linear interpolation in the R134a tables (learn this; tables provided)\n","parts":[{"label":"a","text":"A refrigerator employing the vapour-compression cycle operates with a condenser at 750 kPa and an\nevaporator at 200 kPa. Assuming that HFC-134a is the working fluid and that the compressor operates\nisentropically calculate the coefficient of performance.\n","marks":6,"grade":"2:2","answer":[{"label":"COP","value":5.55,"units":"-","tol":0.05}]},{"label":"b","text":"For the same refrigeration cycle as in (a) calculate the work done by the compressor and the coefficient\nof performance if the compressor does not operate isentropically and is only 80% efficient.\n","marks":3,"grade":"2:2","answer":[{"label":"w_c","value":34.31,"units":"kJ/kg","tol":0.05},{"label":"COP","value":4.44,"units":"-","tol":0.05}]},{"label":"c","text":"Calculate the rate of heat extraction from the evaporator if the mass flow rate of HFC-134a is\n$0.4\\ \\mathrm{kg\\,s^{-1}}$.\n","marks":2,"grade":"third","answer":[{"label":"Q_L","value":60.92,"units":"kW","tol":0.05}]},{"label":"d","text":"Discuss how the choice of coolant may affect operation of the refrigerator and what other factors might\nlimit choice of coolant?\n","marks":4,"grade":"2:2"}],"solution":"$af","confidence":"HIGH","solutionNote":"No official solution available. Table values from the HFC-134a tables in the ENGR266 data sheets (same tables as\nin the 2019 and 2024 papers); a reference equation of state (CoolProp) agrees to 0.1% for the COP. Cengel's tables\ngive the same enthalpy differences (different reference state only).\n"},{"id":"engr266-2021-q3","module":"ENGR5003","week":6,"label":"ENGR266 2021 Q3","kind":"past-paper","level":3,"marks":16,"minutes":19,"grade":"2:1","tags":["phase-reaction-equilibrium","vle","margules","antoine","reaction-equilibrium","equilibrium-constant"],"source":{"name":"ENGR266 Summer 2021 Q3","paper":"ENGR266","year":2021,"question":"Q3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR266 Exam 2021.pdf","tag":"PARTIAL","leftOut":"Part (a) needs Margules activity coefficients (modified Raoult's law), which are not on the current formula\nsheet (only ideal Raoult's and Henry's laws). Part (b), reaction equilibrium from K, is on-syllabus.\n"},"text":"Answer All parts (a) - (b).\n","formulaSheet":"- [EXTERNAL] Modified Raoult's law: $y_i P = x_i \\gamma_i P_i^{sat}$, so $P = x_1\\gamma_1 P_1^{sat} + x_2\\gamma_2 P_2^{sat}$ (learn this; only ideal Raoult $y_i P = x_i P_i^{sat}$ is on the formula sheet)\n- [EXTERNAL] Margules equations: $\\ln\\gamma_1 = x_2^2\\left[A_{12} + 2(A_{21} - A_{12})x_1\\right]$, $\\ln\\gamma_2 = x_1^2\\left[A_{21} + 2(A_{12} - A_{21})x_2\\right]$ (learn this; not in the current materials)\n- Equilibrium constant for ideal gases: $K = \\prod_i \\left(\\dfrac{y_i P}{P^\\circ}\\right)^{\\nu_i}$ (on the formula sheet as $K_P = \\prod P_i^{\\nu_i}$)\n- Mole balance with extent of reaction: $n_i = n_{i0} + \\nu_i\\varepsilon$, $y_i = n_i/\\sum n_i$ (learn this)\n","parts":[{"label":"a","text":"The Methanol(1)/Water(2) system in vapour liquid equilibrium (VLE) can be well described using the\ncombination of Margules equations with parameters $A_{12} = 0.6828$ and $A_{21} = 0.4753$ and the Antoine\nequations below:\n\n$\\ln P_1^{sat} = 16.5785 + \\dfrac{3{,}638.27}{T/^\\circ\\mathrm{C} + 239.50}$\n\n$\\ln P_2^{sat} = 16.3872 + \\dfrac{2{,}885.70}{T/^\\circ\\mathrm{C} + 230.17}$\n\nCalculate the pressure of the system if Methanol(1)/Water(2) is held in VLE at $60^\\circ\\mathrm{C}$ and the\nmole fractions of methyl-ethyl ketone in the liquid and vapour phase are $x_1 = 0.2167$ and $y_1 = 0.6268$.\n","marks":6,"grade":"2:1","answer":[{"label":"P","value":42.46,"units":"kPa","tol":0.05}]},{"label":"b(i)","text":"The gas phase reaction between C and H2O can be described as:\n$C_{(g)} + H_2O_{(g)} \\rightarrow CO_{(g)} + H_{2(g)}$.\nThe reaction takes place at 950 K where the equilibrium constant $K = 1$ and all species can be considered\nas ideal gases. Assuming a pressure of 1 bar and that the reactants consist of 1 mole of C and 2 moles of\nH2O, calculate the fraction of steam that reacts.\n","marks":8,"grade":"2:2","answer":[{"label":"frac_steam","value":0.3333,"units":"-","tol":0.02}]},{"label":"b(ii)","text":"How does the fraction of steam that reacts change if the pressure is increased to 20 bar and all other\nconditions remain the same and why?\n","marks":2,"grade":"third","answer":[{"label":"frac_steam_20bar","value":0.3333,"units":"-","tol":0.02}]}],"solution":"$b0","confidence":"MEDIUM","solutionNote":"No official solution available. Part (b) is on-syllabus and checked by back-substitution (HIGH). Part (a) uses\nMargules/modified Raoult ([EXTERNAL]) and the paper's Antoine equations contain typos (\"+\" signs and 2,885.70\nfor water); the corrected equations reproduce the given $y_1$ to within 2%, which supports the correction. Confirm\nwith the lecturer that the intended water constant is 3885.70.\n"},{"id":"engr266-2022-q1","module":"ENGR5003","week":3,"label":"ENGR266 2022 Q1","kind":"past-paper","level":3,"marks":20,"minutes":24,"grade":"2:1","tags":["real-gas-steam","second-law","equation-of-state","entropy-generation","heat-engine"],"source":{"name":"ENGR266 Summer 2022 Q1","paper":"ENGR266","year":2022,"question":"Q1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR266 Exam Part B 2021-22.pdf","tag":"VALID"},"text":"Answer BOTH parts (a) and (b).\n","formulaSheet":"- Ideal gas: $PV = nRT$ (per mole $P = RT/V$) (on the formula sheet)\n- Redlich-Kwong: $P = \\dfrac{RT}{V - b} - \\dfrac{a}{T^{0.5}V(V + b)}$, $a = \\dfrac{0.42748R^2T_c^{2.5}}{P_c}$, $b = \\dfrac{0.08664RT_c}{P_c}$ (learn this; Week 3 slides)\n- van der Waals: $P = \\dfrac{RT}{V - b} - \\dfrac{a}{V^2}$, $a = \\dfrac{27R^2T_c^2}{64P_c}$, $b = \\dfrac{RT_c}{8P_c}$ (learn this; Week 3 slides give $p_{cr} = a/27b^2$, $T_{cr} = 8a/27Rb$)\n- Compressibility factor: $Z = PV/RT$ (learn this)\n- Entropy generation for a heat engine between two reservoirs: $S_{gen} = \\dfrac{Q_C}{T_C} - \\dfrac{Q_H}{T_H} \\ge 0$ (from $dS_{iso} = dS_{sys} + dS_{env} \\ge 0$) (on the formula sheet)\n- Energy balance for a cycle: $W = Q_H - Q_C$ ($W_{net} = Q_{in} - Q_{out}$) (on the formula sheet)\n- Carnot efficiency: $\\eta = 1 - T_C/T_H$ (learn this)\n","parts":[{"label":"a","text":"You are responsible for designing a heat engine. To achieve the required efficiency, the engine's hot\nreservoir acts at $475^\\circ\\mathrm{C}$ where the molar volume is $0.0001\\ \\mathrm{m^3}$. Using your\nknowledge of thermodynamics select an appropriate material to construct your engine based on the Maximum\nAllowable Operating Pressures presented in Table 1. Present all your working and justify any assumptions\nmade.\n\nTable 1: Maximum Allowable operating Pressures for a selection of generic materials (MPa):\nA 20; B 30; C 35; D 42; E 65; F 74.\n\nThe critical temperature and pressure for the working fluid are $374^\\circ\\mathrm{C}$ and 22.12 MPa\nrespectively and the acentric factor is 0.187.\n","marks":16,"grade":"2:1","answer":[{"label":"P_RK","value":35.83,"units":"MPa","tol":0.05}]},{"label":"b","text":"Calculate the work done by the engine if the cold reservoir is at room temperature\n($25^\\circ\\mathrm{C}$), the total entropy generated during the cycle is $30{,}000\\ \\mathrm{kJ\\,K^{-1}}$ and\nthe heat deposited from the hot reservoir is 160,000 kJ.\n","marks":4,"grade":"2:2","answer":[{"label":"W","value":87293,"units":"kJ","tol":0.02}]}],"solution":"$b1","confidence":"MEDIUM","solutionNote":"No official solution available. Part (a) is HIGH: R-K and vdW both give 34-36 MPa and the IAPWS-95 reference\nvalue is 36.8 MPa, so material D. Part (b) as printed ($30{,}000\\ \\mathrm{kJ\\,K^{-1}}$) gives a negative work, so\nthe answer assumes $30\\ \\mathrm{kJ\\,K^{-1}}$ (i.e. 30,000 J/K); confirm the intended entropy value with the lecturer.\n"},{"id":"engr266-2022-q2","module":"ENGR5003","week":3,"label":"ENGR266 2022 Q2","kind":"past-paper","level":3,"marks":18,"minutes":22,"grade":"2:1","tags":["real-gas-steam","second-law","refrigeration","r134a","throttling","dryness-fraction","p-h-diagram"],"source":{"name":"ENGR266 Summer 2022 Q2","paper":"ENGR266","year":2022,"question":"Q2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR266 Exam Part B 2021-22.pdf","tag":"PARTIAL","leftOut":"Parts (a) and (c) rely on the vapour-compression refrigeration cycle and its COP, not in the current slides.\nPart (b), quality after throttling from R134a tables, is on-syllabus (Week 2 throttling, Week 3 tables).\n"},"text":"Answer All parts (a) - (c).\n\nA refrigerator employs the vapour-compression cycle with HFC-134a as its working fluid. The refrigerator\nmaintains the inside at a temperature of $4^\\circ\\mathrm{C}$ and is situated in a kitchen at\n$20^\\circ\\mathrm{C}$.\n","formulaSheet":"- SFEE per unit mass for each component: $q - w = h_{out} - h_{in}$ (on the formula sheet)\n- Throttle: $h_4 = h_3$ (from the SFEE, on the formula sheet)\n- Refrigerator COP: $COP_R = \\dfrac{h_1 - h_4}{w_{in}}$ (learn this)\n- Wet-mixture properties: $v = (1 - x)v_f + x v_g$, so $x = \\dfrac{v - v_f}{v_g - v_f}$ (on the formula sheet)\n- Refrigerant mass flow: $\\dot m = \\dfrac{\\dot Q_L}{h_1 - h_4}$ (learn this)\n- Carnot refrigerator COP: $COP = \\dfrac{T_L}{T_H - T_L}$ (learn this)\n","parts":[{"label":"a","text":"Calculate the Coefficient of Performance for this refrigerator if the compressor does\n$40\\ \\mathrm{kJ\\,kg^{-1}}$ of work.\n","marks":4,"grade":"2:2","answer":[{"label":"COP","value":4.34,"units":"-","tol":0.05}]},{"label":"b","text":"Calculate the vapour fraction of HFC-134a if the volume after the throttling process is\n$0.01\\ \\mathrm{m^3\\,kg^{-1}}$.\n","marks":4,"grade":"2:2","answer":[{"label":"x","value":0.1544,"units":"-","tol":0.05}]},{"label":"c","text":"Discuss with the help of a P-H diagram what will happen to the coefficient of performance and the\nrequired flow rate of the working fluid if the temperature of the kitchen increases.\n","marks":10,"grade":"2:1"}],"solution":"$b2","confidence":"MEDIUM","solutionNote":"No official solution available. Numbers assume the evaporator is at the inside temperature ($4^\\circ$C) and the\ncondenser at the kitchen temperature ($20^\\circ$C) with saturated states at exit; if the lecturer intended\ntemperature approaches, (a) changes. The volume given in (b) does not match the quality implied by (a) (0.113);\nthe answer uses the given volume as asked.\n"},{"id":"engr266-2023-a1","module":"ENGR5003","week":3,"label":"ENGR266 2023 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["real-gas-steam","second-law","refrigeration","r134a","redlich-kwong","t-s-diagram"],"source":{"name":"ENGR266 Summer 2023 Q A1","paper":"ENGR266","year":2023,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR266 Thermodynamics Exam Paper 2023.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2023/ENGR266 Numerical Answers.pdf","tag":"PARTIAL","leftOut":"Part (a) asks for the purpose and operation of a refrigeration cycle, which is not in the current slides.\nParts (b)-(d) (R134a tables, Redlich-Kwong) are on-syllabus.\n"},"text":"Answer ALL parts (a) - (d).\n","figures":[{"src":"/figures/engr266-2023-a1.png","caption":"Figure A1-1: A thermodynamic cycle on a temperature - entropy phase diagram."}],"formulaSheet":"- Wet-mixture property: $s = (1 - x)s_f + x s_g$, $x = m_g/m_{total}$ (on the formula sheet)\n- Redlich-Kwong: $P = \\dfrac{RT}{V - b} - \\dfrac{a}{T^{0.5}V(V + b)}$, $a = \\dfrac{0.42748R^2T_c^{2.5}}{P_c}$, $b = \\dfrac{0.08664RT_c}{P_c}$ (learn this; Week 3 slides)\n- Energy balance for a refrigeration cycle: $Q_H = Q_L + W_{in}$, $COP_R = Q_L/W_{in}$ (learn this)\n","parts":[{"label":"a","text":"In your own words describe the purpose and operation of the thermodynamic cycle presented in Figure A1-1.\n","marks":8,"grade":"2:1"},{"label":"b","text":"Assuming the cycle is using HFC-134a, use the data tables to calculate the liquid fraction at point 4 in\nthe cycle assuming it has an entropy of $1.4\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ and a temperature of\n$-35^\\circ\\mathrm{C}$?\n","marks":6,"grade":"2:2","answer":[{"label":"liquid fraction","value":0.383,"units":"-","tol":0.02}]},{"label":"c","text":"Calculate the pressure at point 1 using the Redlich-Kwong equation of state if the temperature is\n$150^\\circ\\mathrm{C}$ and the volume is $0.006\\ \\mathrm{m^3\\,mol^{-1}}$. The critical temperature for\nHFC-134a is $101.08^\\circ\\mathrm{C}$ and the critical pressure is 4060.3 kPa.\n","marks":7,"grade":"2:2","answer":[{"label":"P_1","value":566.6,"units":"kPa","tol":0.02}]},{"label":"d","text":"What are the deficiencies of the Redlich-Kwong equation of state and what can be done to address these?\n","marks":4,"grade":"2:2"}],"solution":"$b3","confidence":"HIGH","solutionNote":"Matches the official numerical answers: 1b liquid fraction 0.383 (ours 0.383) and 1c 566-577 kPa (ours 566.6 kPa).\nPart (b) needs the IIR-reference R134a tables supplied with the ENGR266 paper; Cengel-reference tables would make\npoint 4 superheated.\n"},{"id":"engr266-2023-a2","module":"ENGR5003","week":6,"label":"ENGR266 2023 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["phase-reaction-equilibrium","vle","txy-diagram","excess-volume","activity-coefficient","reaction-equilibrium"],"source":{"name":"ENGR266 Summer 2023 Q A2","paper":"ENGR266","year":2023,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR266 Thermodynamics Exam Paper 2023.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2023/ENGR266 Numerical Answers.pdf","tag":"PARTIAL","leftOut":"Part (b) (excess volume) and part (c) (modified Raoult's law with activity coefficients) go beyond the\ncurrent formula sheet, which only has ideal Raoult's and Henry's laws. Parts (a) and (d) are on-syllabus.\n"},"text":"Answer ALL parts (a) - (d).\n","formulaSheet":"- [EXTERNAL] Excess property: $V^E = V - \\sum_i x_i V_i$ (learn this; not on the current formula sheet)\n- [EXTERNAL] Modified Raoult's law: $y_i P = x_i\\gamma_i P_i^{sat}$, so $P = \\sum_i x_i\\gamma_i P_i^{sat}$ (learn this; the sheet only has ideal Raoult $y_i P = x_i P_i^{sat}$)\n- Antoine equation as given: $\\ln P^{sat}/\\mathrm{kPa} = A - \\dfrac{B}{T/^\\circ\\mathrm{C} + C}$ (given in the question)\n- Equilibrium constant for ideal gases: $K = \\prod_i\\left(\\dfrac{y_i P}{P^\\circ}\\right)^{\\nu_i}$ (on the formula sheet as $K_P = \\prod P_i^{\\nu_i}$)\n- Extent of reaction: $n_i = n_{i0} + \\nu_i\\varepsilon$, $y_i = n_i/\\sum n_i$ (learn this)\n","parts":[{"label":"a","text":"Sketch a T x y diagram and label the liquid, vapour and liquid-vapour regions as well as the bubble and\ndew curves\n","marks":5,"grade":"third"},{"label":"b","text":"Calculate the excess volume for the propanol(1)/water(2) mixture with a mole fraction of 0.16656 propanol\nat $25^\\circ\\mathrm{C}$ if the mixture's volume is $27.016\\ \\mathrm{cm^3\\,mol^{-1}}$. The volume for\npropanol and water at $25^\\circ\\mathrm{C}$ are $75.168\\ \\mathrm{cm^3\\,mol^{-1}}$ and\n$18.068\\ \\mathrm{cm^3\\,mol^{-1}}$ respectively.\n","marks":4,"grade":"third","answer":[{"label":"V^E","value":-0.563,"units":"cm3/mol","tol":0.02}]},{"label":"c","text":"Calculate the pressure required to hold the chloroform(1)/methanol(2) system in vapour-liquid equilibrium\nat $36^\\circ\\mathrm{C}$ if the mole fractions of chloroform in the liquid and vapour phase are 0.44 and\n0.657 respectively. The activity coefficients are $\\gamma_1 = 1.6423$ and $\\gamma_2 = 0.9534$ and the\nsaturation pressures can be found using the Antoine equations below:\n\n$\\ln P^{sat}_{Chloroform}/\\mathrm{kPa} = 13.7324 - \\dfrac{2548.74}{T/^\\circ\\mathrm{C} + 218.552}$\n\n$\\ln P^{sat}_{Methanol}/\\mathrm{kPa} = 16.5785 - \\dfrac{3638.27}{T/^\\circ\\mathrm{C} + 239.500}$\n","marks":4,"grade":"2:2","answer":[{"label":"P","value":45.37,"units":"kPa","tol":0.02}]},{"label":"d","text":"The cracking of pure n-butane via the reaction below is performed at 750 K, where the equilibrium\nconstant, $K$, for the reaction is 6.0.\n\n$C_4H_{10} \\rightarrow C_2H_4 + C_2H_6$\n\nCalculate the mole fractions for all species if the reaction starts from 1 mole of C4H10 and is carried\nout in the gas phase where the pressure is 2 bar and standard pressure is 1 bar. You can assume all\nspecies are ideal gasses.\n","marks":12,"grade":"2:1","answer":[{"label":"y_C4H10","value":0.0718,"units":"-","tol":0.02},{"label":"y_C2H4","value":0.4641,"units":"-","tol":0.02},{"label":"y_C2H6","value":0.4641,"units":"-","tol":0.02}]}],"solution":"$b4","confidence":"HIGH","solutionNote":"Matches the official numerical answers: 2b $-0.56\\ \\mathrm{cm^3\\,mol^{-1}}$ (ours $-0.563$), 2c 45.374 kPa (ours\n45.37 kPa), 2d $y_{C_4H_{10}} = 0.072$, $y_{C_2H_4} = y_{C_2H_6} = 0.464$ (ours 0.0718, 0.4641, 0.4641).\nParts (b)-(c) use excess properties and modified Raoult's law, which are beyond the current formula sheet ([EXTERNAL]).\n"},{"id":"engr266-2024-b1","module":"ENGR5003","week":6,"label":"ENGR266 2024 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["real-gas-steam","property-relations","second-law","refrigeration","r134a","residual-volume"],"source":{"name":"ENGR266 Summer 2024 Q B1","paper":"ENGR266","year":2024,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR266 In-person Main Exam 2024 combined Final with formula sheets 13 March.pdf","tag":"PARTIAL","leftOut":"Parts (a)-(b) rely on the vapour-compression refrigeration cycle and its COP, which are not in the current\nslides (the R134a table work is on-syllabus). Part (c), residual volume, is on the current formula sheet.\n"},"text":"Answer ALL parts (a) - (c).\n\nThe flow diagram below (Figure B1-1) outlines the vapour compression refrigeration cycle.\n","figures":[{"src":"/figures/engr266-2024-b1.png","caption":"Figure B1-1: Flowchart showing the vapour compression cycle."}],"formulaSheet":"- SFEE per unit mass for each component: $q - w = h_{out} - h_{in}$; throttle $h_1 = h_4$ (on the formula sheet)\n- Vapour-compression COP (flowchart numbering): $COP = \\dfrac{h_2 - h_1}{h_3 - h_2}$ (learn this)\n- Isentropic compressor: $s_3 = s_2$ (learn this)\n- Residual volume: $V^R = V - V^{ig} = V - \\dfrac{RT}{P} = \\dfrac{RT}{P}(Z - 1)$ (on the formula sheet as $v^R = \\frac{RT}{P}(Z - 1)$)\n- Molar from specific volume: $V = v\\,M$ (learn this)\n","parts":[{"label":"a","text":"Draw the vapour compression cycle on a Pressure-Volume diagram, making sure to indicate the locations of\nthe numbered states in the flowchart above. Indicate the direction of the cycle, where heat enters and\nleaves the system and where any work is done.\n","marks":10,"grade":"2:2"},{"label":"b","text":"Assuming the refrigerator employs HFC-134a as its working fluid, calculate the Coefficient of Performance\nfor a refrigerator with a target temperature of $-9^\\circ\\mathrm{C}$ and the condenser operates at 750 kPa.\nYou may assume that the compressor operates isentropically.\n","marks":7,"grade":"2:2","answer":[{"label":"COP","value":5.77,"units":"-","tol":0.05}]},{"label":"c","text":"For the same refrigerator, determine the temperature and the residual volume of HFC-134a in the\nsuperheated vapour phase in $\\mathrm{m^3\\,mol^{-1}}$. The molecular mass of HFC-134a is\n$102.03\\ \\mathrm{g\\,mol^{-1}}$.\n","marks":8,"grade":"2:1","answer":[{"label":"T_3","value":34.2,"units":"C","tol":0.02},{"label":"V^R","value":-0.000522,"units":"m3/mol","tol":0.05}]}],"solution":"$b5","confidence":"HIGH","solutionNote":"No official solution available. Table values from the HFC-134a tables printed with this paper; a reference\nequation of state agrees to 0.3% (COP 5.76, $V^R = -5.23\\times10^{-4}\\ \\mathrm{m^3\\,mol^{-1}}$). The evaporator is\nassumed to run at the target temperature ($-9^\\circ$C).\n"},{"id":"engr266-2025-b-q1","module":"ENGR5003","week":6,"label":"ENGR266 2025 Section B Q1","kind":"past-paper","level":3,"marks":18,"minutes":22,"grade":"2:1","tags":["property-relations","residual-volume","lee-kessler","r134a"],"source":{"name":"ENGR266 Summer 2025 Section B Q1","paper":"ENGR266","year":2025,"question":"B-Q1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR266 Main Exam 2024-25.pdf","tag":"PARTIAL","leftOut":"The Lee-Kessler generalised correlation (and its tables) is not in the current materials; the residual\nvolume from the R134a data tables and its interpretation are on-syllabus (residual properties are on the\ncurrent formula sheet).\n"},"text":"Calculate the residual molar volume of 1,1,1,2-tetrafluotoethant (HFC-134a) at a temperature of\n$104.85^\\circ\\mathrm{C}$ and a pressure of 2400 kPa using the generalized correlation of Lee and Kessler. The\ncritical temperature and pressure are $101.08^\\circ\\mathrm{C}$ and 4060 kPa respectively, the acentric factor\nis 0.32684 and its molecular mass is 102.03 amu. Compare this to the residual molar volume calculated from\nthe datatables. What does this tell us about the interactions between the molecules in HFC-134a.\n","formulaSheet":"- Residual volume: $V^R = \\dfrac{RT}{P}(Z - 1) = V - \\dfrac{RT}{P}$ (on the formula sheet)\n- Reduced properties: $T_r = T/T_c$, $P_r = P/P_c$ (learn this)\n- [EXTERNAL] Lee-Kessler generalised correlation: $Z = Z^0 + \\omega Z^1$, $Z^0$ and $Z^1$ from the Lee-Kessler tables at $(T_r, P_r)$ (learn this; correlation and tables not in the current materials)\n- Molar volume from tables: $V = vM$ (learn this)\n","answer":[{"label":"V^R (Lee-Kessler)","value":-0.000314,"units":"m3/mol","tol":0.05},{"label":"V^R (tables)","value":-0.000313,"units":"m3/mol","tol":0.05}],"solution":"$b6","confidence":"MEDIUM","solutionNote":"No official solution available. The data-table residual volume is HIGH (tables and a reference equation of state\nagree). The Lee-Kessler part is [EXTERNAL]: $Z^0$, $Z^1$ were computed from the Lee-Kessler equation (validated\nagainst argon, $Z^0$ within 0.2%), so a student interpolating printed tables may get $Z$ about 0.758-0.765.\nConfirm with the lecturer whether Lee-Kessler is still examinable.\n"},{"id":"engr266-2025-b-q2","module":"ENGR5003","week":1,"label":"ENGR266 2025 Section B Q2","kind":"past-paper","level":3,"marks":12,"minutes":14,"grade":"2:2","tags":["second-law","stirling-cycle","carnot-efficiency","isothermal-process"],"source":{"name":"ENGR266 Summer 2025 Section B Q2","paper":"ENGR266","year":2025,"question":"B-Q2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR266 Main Exam 2024-25.pdf","tag":"VALID"},"text":"The Stirling engine cycle consists of the following steps: 1. Isochoric heating; 2. Isothermal expansion;\n3. Isochoric cooling; 4. Isothermal contraction.\n","formulaSheet":"- Carnot efficiency: $\\eta = 1 - \\dfrac{T_C}{T_H}$ (learn this; Week 1 slides: Stirling has the Carnot efficiency)\n- Isothermal ideal gas: $\\Delta U = 0$ so $Q = W$, $W = nRT\\ln\\dfrac{V_2}{V_1}$ (the sheet prints $dW = -mR_gT\\ln(V_2/V_1)$, sic) (on the formula sheet)\n- Boyle's law (isothermal): $P_1V_1 = P_2V_2$, so $\\dfrac{V_2}{V_1} = \\dfrac{P_1}{P_2}$ (on the formula sheet)\n","parts":[{"label":"i","text":"Draw the Stirling cycle on pressure-volume diagram and indicate the direction of the cycle","marks":6,"grade":"third"},{"label":"ii","text":"If the efficiency of Stirling cycle is 0.43 and heat is absorbed during isothermal expansion at\n$300^\\circ\\mathrm{C}$, calculate the heat that is released during the isothermal contraction stage where\nthe pressure increases to 6.3 bar from an initial pressure of 2 bar. You can assume that the thermal\nefficiency for the Stirling cycle is the same as for the Carnot cycle.\n","marks":6,"grade":"2:2","answer":[{"label":"T_C","value":326.7,"units":"K","tol":0.02},{"label":"Q_out","value":3117,"units":"J/mol","tol":0.02}]}],"solution":"$b7","confidence":"HIGH","solutionNote":"No official solution available. The question gives no amount of gas, so the heat is given per mole\n($3.12\\ \\mathrm{kJ\\,mol^{-1}}$ released); checked against $\\eta = 1 - Q_C/Q_H$.\n"},{"id":"engr266-2025-b-q4","module":"ENGR5003","week":6,"label":"ENGR266 2025 Section B Q4","kind":"past-paper","level":3,"marks":14,"minutes":17,"grade":"2:2","tags":["phase-reaction-equilibrium","reaction-equilibrium","equilibrium-constant","extent-of-reaction"],"source":{"name":"ENGR266 Summer 2025 Section B Q4","paper":"ENGR266","year":2025,"question":"B-Q4","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR266 Main Exam 2024-25.pdf","tag":"VALID"},"text":"For the cracking reaction shown below, calculate the fraction of propone, $C_3H_8$, that reacts if the\nequilibrium constant, $K = 2$. You can assume that the reaction takes place in the gas phase at standard\npressure, but you should state any further assumptions that you make.\n\n$C_3H_{8(g)} \\leftrightarrow C_2H_{4(g)} + CH_{4(g)}$\n","formulaSheet":"- Equilibrium constant for ideal gases: $K = \\prod_i\\left(\\dfrac{y_iP}{P^\\circ}\\right)^{\\nu_i}$ (on the formula sheet as $K_P = \\prod P_i^{\\nu_i}$)\n- Extent of reaction: $n_i = n_{i0} + \\nu_i\\varepsilon$, $y_i = n_i/\\sum n_i$ (learn this)\n","answer":[{"label":"fraction reacted","value":0.8165,"units":"-","tol":0.02}],"solution":"$b8","confidence":"HIGH","solutionNote":"No official solution available. Checked by back-substitution into K.\n"},{"id":"engr217-2017-b1","module":"ENGR5003","week":4,"label":"ENGR217 2017 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["ic-engines","ideal-gas-first-law","otto-cycle","thermal-efficiency","pv-diagram"],"source":{"name":"ENGR217 Summer 2017 Q B1","paper":"ENGR217","year":2017,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR217_Exam_2017.pdf","tag":"VALID"},"text":"Answer ALL parts.\n\nIn a certain ideal heat engine, a fixed mass of air is taken repeatedly through the cycle of processes shown\nin Figure B1 (Pressure Volume (PV) Diagram). Processes 1-2 and 3-4 are adiabatic compression and expansion:\nwork is done, but there is no heat flow in or out. In process 2-3, heat $Q_{23}$ is added, raising the\npressure and temperature of the air to $p_3$ and $T_3$ respectively. In process 4-1, heat $Q_{41}$ is\nremoved, to return the air to state 1, the original state.\n","figures":[{"src":"/figures/engr217-2017-b1.png","caption":"Figure B1 - Pressure Volume (PV) Diagram."}],"formulaSheet":"- First law for a cycle: $W_{net} = Q_{in} - Q_{out}$ (on the formula sheet)\n- Thermal efficiency: $\\eta_{th} = \\dfrac{W_{net}}{Q_{in}} = 1 - \\dfrac{Q_{out}}{Q_{in}}$ (on the formula sheet)\n- Adiabatic (isentropic) ideal gas: $TV^{\\gamma-1} = \\text{const}$, so $T_2/T_1 = r^{\\gamma-1}$ (on the formula sheet)\n- Constant-volume heat transfer: $dQ = m c_v\\, dT$ (on the formula sheet)\n- $c_v = \\dfrac{R_g}{\\gamma-1}$ with $R_g = 0.287\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ for air (on the formula sheet)\n- Otto efficiency: $\\eta = 1 - \\dfrac{1}{r^{\\gamma-1}}$ (on the formula sheet)\n","parts":[{"label":"a","text":"In this heat engine, the mass of air contained in the cylinder is $500 \\times 10^{-6}\\ \\mathrm{kg}$. The\nheat added $Q_{23}$ per unit mass of air is 1450 kJ/kg, and heat rejected in process 4-1 $Q_{41}$ is\n1160 kJ/kg. Find the work delivered per cycle.\n","marks":8,"grade":"third","answer":[{"label":"W_cycle","value":145,"units":"J","tol":0.02}]},{"label":"b(i)","text":"The sequence of processes that take place in a spark-ignition engine is represented approximately by the\nOtto cycle, again illustrated by Figure B1, in which the addition of heat in process 2-3 takes place at\nconstant volume, and similarly the heat rejection in process 4-1.\n\nA certain spark-ignition engine has compression ratio $r = 8.6$. In the engine cycle, air is compressed\nadiabatically from state 1 ($p_1 = 0.95\\ \\mathrm{bar}$, $T_1 = 20^\\circ\\mathrm{C}$) to state 2. Then, in\nprocess 2-3, 2500 kJ/kg of heat is added at constant volume, raising the temperature to $T_3$ and the\npressure to $p_3$. The air is then expanded isentropically to state 4, doing work on the pistons as it\ndoes so. Finally, heat is removed, to return the air to state 1, the original state.\n\ni. Find the temperatures $T_2$, $T_3$, and $T_4$, in $^\\circ\\mathrm{C}$. Take $\\gamma$, the ratio of\nspecific heats for air, as 1.4.\n","marks":9,"answer":[{"label":"T_2","value":420.1,"units":"C","tol":0.02},{"label":"T_3","value":3904,"units":"C","tol":0.02},{"label":"T_4","value":1493,"units":"C","tol":0.02}],"grade":"$undefined"},{"label":"b(ii)","text":"ii. Find the amount of heat rejected in process 4-1, per kg of working fluid.\n","marks":4,"answer":[{"label":"q_41","value":1057,"units":"kJ/kg","tol":0.02}],"grade":"$undefined"},{"label":"b(iii)","text":"iii. Hence, or otherwise, find the work output of the engine per cycle.\n","marks":2,"answer":[{"label":"w_net","value":1443,"units":"kJ/kg","tol":0.02},{"label":"W_cycle (m = 500 mg)","value":721.4,"units":"J","tol":0.02}],"grade":"$undefined"},{"label":"b(iv)","text":"iv. Estimate the thermal efficiency of the engine. Indicate whether this efficiency is likely to be\nachieved in practice, explaining your answer.\n","marks":2,"answer":[{"label":"eta_th","value":0.577,"units":"-","tol":0.02}],"grade":"$undefined"}],"solution":"$b9","confidence":"HIGH","solutionNote":"No official answers for this question: the ENGR217 2017 numerical answers sheet covers only the fluids section\n(A1-A3). Values checked by two routes each (Otto identities and cycle first law). Note (b)(iii) gives no\nmass for the spark-ignition engine; the answer is given per kg and per cycle assuming the 500 mg charge of (a).\n"},{"id":"engr217-2017-b2","module":"ENGR5003","week":3,"label":"ENGR217 2017 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["real-gas-steam","steady-flow","steam-tables","dryness-fraction","boiler","steam-turbine","sfee"],"source":{"name":"ENGR217 Summer 2017 Q B2","paper":"ENGR217","year":2017,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR217_Exam_2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d)\n","formulaSheet":"- SFEE: $\\dot Q - \\dot W = \\dot m\\left[\\left(h + \\tfrac{C^2}{2} + gZ\\right)_{out} - \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{in}\\right]$ (on the formula sheet)\n- Wet steam: $x = m_g/m_{total}$, $h = (1-x)h_f + x h_g = h_f + x h_{fg}$ (on the formula sheet)\n- Thermal efficiency: $\\eta = \\dfrac{W_{net}}{Q_{in}}$ (on the formula sheet)\n- Steam property values $h_f$, $h_g$, $h_{fg}$ at the given pressures and temperatures (learn this: read from steam tables, provided in the exam)\n","parts":[{"label":"a","text":"A boiler fired by natural gas, and containing water and steam, operates continuously to supply\n0.25 kg/s of dry saturated steam at 40 bar to a turbine. Feed water at $30^\\circ\\mathrm{C}$ is pumped into\nthe boiler at the same mass flow rate, so as to maintain steady conditions.\n\nEstimate the rate of heat supply needed to operate the boiler, assuming no losses. (Extracts from steam\ntables are provided in the Appendix.)\n","marks":9,"answer":[{"label":"Q_boiler","value":668.8,"units":"kW","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"In practice, the steam leaves the boiler slightly wet, and when it enters the turbine it is at pressure\n40 bar and dryness fraction 0.96. The steam exhausts from the turbine at pressure 0.5 bar and dryness\nfraction 0.82. Estimate the power delivered by the turbine, assuming negligible heat losses. You may also\nneglect the kinetic and potential energies of the steam.\n","marks":9,"answer":[{"label":"h_in","value":2732.4,"units":"kJ/kg","tol":0.02},{"label":"h_out","value":2230.4,"units":"kJ/kg","tol":0.02},{"label":"W_turbine","value":125.5,"units":"kW","tol":0.05}],"grade":"$undefined"},{"label":"c","text":"More energy could be extracted from the steam by taking the exhaust steam to a lower value of dryness\nfraction. Explain what places a practical limit on the wetness of the steam leaving the turbine.\n","marks":4,"grade":"$undefined"},{"label":"d","text":"Find the efficiency of the boiler and turbine as a system for producing mechanical power, and suggest\nhow the efficiency might be improved.\n","marks":3,"answer":[{"label":"eta","value":0.188,"units":"-","tol":0.05}],"grade":"$undefined"}],"solution":"$ba","confidence":"HIGH","solutionNote":"No official answers for this question: the ENGR217 2017 numerical answers sheet covers only the fluids section.\nThe steam-table appendix is missing from the archived paper; IAPWS-IF97 values are used (identical to standard\ntables to the precision quoted). The basis of the heat input in (d) is ambiguous (668.8 kW from (a) gives 18.8%;\nthe wet-steam heat input gives 19.3%), hence the wider tolerance.\n"},{"id":"engr217-2017-b3","module":"ENGR5003","week":2,"label":"ENGR217 2017 Q B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["steady-flow","ideal-gas-first-law","compressor","sfee","kinetic-energy","cooling-jacket"],"source":{"name":"ENGR217 Summer 2017 Q B3","paper":"ENGR217","year":2017,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR217_Exam_2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c)\n","formulaSheet":"- Ideal gas: $pV = mR_gT$, so $\\dot m = \\dfrac{p\\dot V}{R_g T}$, with $R_g = 0.287\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ (on the formula sheet)\n- SFEE: $\\dot Q - \\dot W = \\dot m\\left[\\left(h + \\tfrac{C^2}{2} + gZ\\right)_{out} - \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{in}\\right]$ (on the formula sheet)\n- Ideal gas enthalpy change: $dh = c_p\\,dT$, $c_p = \\dfrac{\\gamma R_g}{\\gamma-1}$ (on the formula sheet)\n- Heat picked up by the cooling water: $\\dot Q = \\dot m c\\,\\Delta T$ (on the formula sheet as $dQ = m c_p dT$)\n","parts":[{"label":"a","text":"An air compressor delivers a volume $0.90\\ \\mathrm{m^3}$ per minute of air at 6.5 bar pressure (absolute)\nand $20^\\circ\\mathrm{C}$. Calculate the mass flow rate of air through the compressor, in kg/s.\n","marks":7,"grade":"third","answer":[{"label":"mdot","value":0.1159,"units":"kg/s","tol":0.02}]},{"label":"b","text":"The compressor operates steadily, with air entering the compressor at 1.0 bar and temperature\n$18^\\circ\\mathrm{C}$, and exiting at 6.5 bar and $140^\\circ\\mathrm{C}$. The velocity of the air at inlet\nto the compressor is 90 m/s, and the velocity at the outlet is 35 m/s. If no heat is lost, estimate the\npower required to drive the compressor.\n","marks":11,"answer":[{"label":"W_in","value":13.8,"units":"kW","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"To reduce the driving power required and to lower the temperature of the delivered air, the compressor\nis cooled by a water jacket through which water is pumped at a rate 0.04 kg/s. The water enters at\n$25^\\circ\\mathrm{C}$ and it leaves at $48^\\circ\\mathrm{C}$. The compressor continues to pass air at the\nsame mass flow rate and the entry temperature is unchanged, but the temperature of the air at exit is now\n$66^\\circ\\mathrm{C}$. Estimate the power now consumed by the compressor, taking the specific heat of\nwater as $4.19\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$, and assuming the same air velocities as before.\n","marks":7,"answer":[{"label":"Q_water","value":3.855,"units":"kW","tol":0.02},{"label":"W_in","value":9.04,"units":"kW","tol":0.02}],"grade":"$undefined"}],"solution":"$bb","confidence":"HIGH","solutionNote":"No official answers for this question: the ENGR217 2017 numerical answers sheet covers only the fluids section.\nEach part verified by unit analysis and back-substitution; (c) also by an energy balance on the difference\nbetween the cooled and uncooled cases. Result insensitive to $c_p = 1.0045$ vs 1.005.\n"},{"id":"engr217-2018-a1","module":"ENGR5003","week":3,"label":"ENGR217 2018 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["real-gas-steam","steady-flow","steam-tables","t-v-diagram","isentropic-expansion","dryness-fraction","closed-vs-open-system"],"source":{"name":"ENGR217 Summer 2018 Q A1","paper":"ENGR217","year":2018,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR217_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR217 (SM).pdf","tag":"VALID"},"text":"Answer ALL parts (a) and (b)\n\nProperty tables including the steam tables are provided in the Appendix\n","formulaSheet":"- Wet steam: $x = m_g/m_{total}$; $u, h, s, v = (1-x)f + x\\,g$, i.e. $s = s_f + x s_{fg}$ (on the formula sheet)\n- First law, closed system: $dU = dQ - dW$ (heat in +ve, work out +ve) (on the formula sheet)\n- SFEE: $\\dot Q - \\dot W = \\dot m\\left[\\left(h + \\tfrac{C^2}{2} + gZ\\right)_{out} - \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{in}\\right]$ (on the formula sheet)\n- Isentropic process: $s_2 = s_1$ (learn this)\n- Steam property values at 20 bar and 1 bar (learn this: read from the steam tables provided)\n","parts":[{"label":"a","text":"Draw a property diagram for the phase-change process of water at a constant pressure of 1 atm. Show\nchange in the specific volume with temperature. (4 marks)\n\nGive a phase description, and label the processes or states such as ice, melting, subcooled water,\nsaturated water, evaporation, saturated steam and superheated steam. (3 marks)\n\nHow will increase in pressure to 1 MPa affect the boiling temperature of water, the specific volume of\nthe saturated water and the specific volume of the saturated steam? (2 marks)\n","marks":9,"grade":"third"},{"label":"b","text":"Dry saturated steam at a pressure of 20 bar is expanded *isentropically* to a final pressure of 1 bar.\nDetermine the dryness fraction at 1 bar.\n","marks":10,"answer":[{"label":"x_2","value":0.8316,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"b(i)","text":"Find the specific work done by the steam if the expansion is\n\n(i) in a cylinder (consider as *closed system*),\n","marks":3,"answer":[{"label":"w_closed","value":445.2,"units":"kJ/kg","tol":0.02}],"grade":"$undefined"},{"label":"b(ii)","text":"(ii) through a turbine (consider as *open system*).\n","marks":3,"answer":[{"label":"w_open","value":503.5,"units":"kJ/kg","tol":0.02}],"grade":"$undefined"}],"solution":"$bc","confidence":"HIGH","solutionNote":"Official numerical answers: $x = 0.8314$, $u_2 = 2153.6$, $h_2 = 2294.4\\ \\mathrm{kJ/kg}$, $w_{closed} = 445.5$\nand $w_{open} = 503.9\\ \\mathrm{kJ/kg}$, using the same 1 bar table values ($u_f = 417.4$, $u_g = 2505.6$,\n$h_f = 417.51$, $h_g = 2675.0$, $s_f = 1.3028$, $s_g = 7.3589$). These agree with this solution to within 0.1%;\nthe official $x$ is slightly lower than $(6.3390 - 1.3028)/6.0562 = 0.8316$ (probably rounding), which\nshifts the works by about 0.3 kJ/kg. Both sets of numbers earn the marks.\n"},{"id":"engr217-2018-a2","module":"ENGR5003","week":2,"label":"ENGR217 2018 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["steady-flow","gas-turbines","nozzle","sfee","continuity","brayton-cycle","t-s-diagram","pv-diagram","regeneration"],"source":{"name":"ENGR217 Summer 2018 Q A2","paper":"ENGR217","year":2018,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR217_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR217 (SM).pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n","formulaSheet":"- Ideal gas: $pv = R_gT$, so $\\rho = \\dfrac{p}{R_gT}$ (on the formula sheet)\n- Continuity: $\\dot m = \\rho A C$ (learn this)\n- SFEE: $\\dot Q - \\dot W = \\dot m\\left[\\left(h + \\tfrac{C^2}{2} + gZ\\right)_{out} - \\left(h + \\tfrac{C^2}{2} + gZ\\right)_{in}\\right]$ (on the formula sheet)\n- Ideal gas enthalpy change: $dh = c_p\\,dT$ (on the formula sheet)\n- Brayton cycle: $\\eta = 1 - r_p^{-(\\gamma-1)/\\gamma}$, work ratio $r_w = 1 - \\dfrac{T_1}{T_3} r_p^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n","parts":[{"label":"a(i)","text":"Air enters a nozzle at a pressure of 5 bar, $200^\\circ\\mathrm{C}$ and a velocity of $30\\ \\mathrm{m\\,s^{-1}}$,\nit exits at 1 bar and $180\\ \\mathrm{m\\,s^{-1}}$. The nozzle inlet area is $100\\ \\mathrm{cm^2}$. The gas\nconstant is $R = 0.287\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$, and the specific heat is\n$c_p = 1.005\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$.\n\nDetermine for *steady adiabatic flow*:\n\n(i) the mass flow rate of air;\n","marks":4,"answer":[{"label":"mdot","value":1.105,"units":"kg/s","tol":0.02}],"grade":"$undefined"},{"label":"a(ii)","text":"(ii) the air exit temperature;\n","marks":4,"answer":[{"label":"T_2","value":457.5,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"a(iii)","text":"(iii) the nozzle exit area.\n","marks":4,"answer":[{"label":"A_2","value":80.6,"units":"cm^2","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"A gas turbines operating on the ideal Brayton cycle, draw and label:\n\n(i) pressure against volume (P-*V*) diagram, (2 marks)\n\n(ii) temperature against entropy (T-*s*) diagram. (2 marks)\n\nShow on the T-*s* diagram and explain in terms of the work input and output, how the actual gas-turbine\ncycle differs from the ideal Brayton cycle. (4 marks)\n","marks":8,"grade":"third"},{"label":"c","text":"List three main components of a gas turbine system operating in an ideal cycle. (3 marks)\n\nHow can the cycle efficiency and the work output be improved with addition of new components to the\nsystem? (2 marks)\n","marks":5,"grade":"third"}],"solution":"$bd","confidence":"HIGH","solutionNote":"Official numerical answers: $\\dot m = 1.107\\ \\mathrm{kg/s}$, $T = 457.3\\ \\mathrm{K}$, $A = 80.6\\ \\mathrm{cm^2}$.\nThis solution gives 1.105 kg/s, 457.5 K and 80.6 cm^2. The 0.2% differences in (i) and (ii) are rounding\n(457.3 K corresponds to taking $T_1 = 473\\ \\mathrm{K}$); the method is identical and all values agree within\nthe answer tolerance.\n"},{"id":"engr217-2018-a3","module":"ENGR5003","week":4,"label":"ENGR217 2018 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["ic-engines","ideal-gas-first-law","otto-cycle","isothermal-compression","thermal-efficiency","indicator-diagram","engine-losses"],"source":{"name":"ENGR217 Summer 2018 Q A3","paper":"ENGR217","year":2018,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR217_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR217 (SM).pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (e)\n\nA petrol engine operates on an air-standard Otto cycle, with a compression ratio of 11. At the start of\ncompression, the pressure and the temperature of the air are 1 bar and 303 K respectively.\n\nThe compression and expansion of this engine can be assumed as adiabatic. The adiabatic law is given as\n\n$PV^{\\gamma} = \\text{constant}$,\n\n$TV^{\\gamma-1} = \\text{constant}$,\n\n$P^{1-\\gamma}T^{\\gamma} = \\text{constant}$, where the variables carry their usual meanings.\n","formulaSheet":"- Adiabatic ideal gas: $pV^\\gamma = \\text{const}$, $TV^{\\gamma-1} = \\text{const}$ (on the formula sheet, and printed in the question)\n- Constant-volume heat transfer: $dQ = m c_v\\, dT$; isochoric $p/T = \\text{const}$ (Gay-Lussac) (on the formula sheet)\n- Isothermal ideal gas: $dU = 0$ so $dQ = dW$, $|W| = m R_g T \\ln(V_1/V_2)$ (on the formula sheet)\n- Thermal efficiency: $\\eta_{th} = 1 - \\dfrac{Q_{out}}{Q_{in}}$ (on the formula sheet)\n- $R_g = 0.287\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ for air (on the formula sheet)\n","parts":[{"label":"a","text":"Find the pressure and the temperature of the air at the end of the compression stroke for the isentropic\ncompression with $\\gamma = 1.4$.\n","marks":3,"answer":[{"label":"p_2","value":28.7,"units":"bar","tol":0.02},{"label":"T_2","value":790.7,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"b","text":"Determine the maximum pressure and the temperature of the cycle, when the heat supplied at constant volume\nis $1350\\ \\mathrm{kJ\\,kg^{-1}}$, and the heat capacity (or the specific heat) is\n$c_v = 0.88\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ while heat is added.\n","marks":4,"answer":[{"label":"p_3","value":84.4,"units":"bar","tol":0.02},{"label":"T_3","value":2325,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Consider the isentropic expansion with $\\gamma = 1.33$, and the heat capacity $c_v$ equal to\n$0.74\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ while heat is rejected. Find the heat rejected and the efficiency.\n","marks":4,"answer":[{"label":"q_out","value":555.5,"units":"kJ/kg","tol":0.02},{"label":"eta_th","value":0.589,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"Assume the compression process to be *isothermal* rather than isentropic, for the same heat addition as\nabove, $\\gamma = 1.33$ for expansion, $c_v = 0.88\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ for heat addition and\n$c_v = 0.74\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ for heat rejection, determine the new thermal efficiency.\n","marks":5,"grade":"first","answer":[{"label":"eta_th","value":0.555,"units":"-","tol":0.05}]},{"label":"e","text":"Draw the indicator pressure versus volume (P-V) diagrams of:\n\n(i) a real cycle,\n\n(ii) the Otto cycle.\n\nList the main losses neglected in an air-standard Otto cycle, analysing the P-V diagrams.\n","marks":9,"grade":"$undefined"}],"solution":"$be","confidence":"MEDIUM","solutionNote":"Official numerical answers: (a) $p_2 = 28.7\\ \\mathrm{bar}$, $T_2 = 790.7\\ \\mathrm{K}$; (b) 84.40 bar; (c) 58.9%;\n(d) 71%. (a)-(c) agree exactly. (d) disagrees: the official 71% equals $1 - q_{41}/q_{23} = 0.710$, i.e. it\nomits the heat rejected during the isothermal compression ($q = w = R_g T_1 \\ln r = 208.5\\ \\mathrm{kJ/kg}$).\nSince $dU = 0$ for an isothermal ideal gas, that heat must be rejected, so this solution trusts 55.5%\n(first law; also the efficiency should fall, not rise, compared with (c)). Confirm with the lecturer which\nanswer they expect, and which $R_g$ (0.287 assumed; not printed) for the isothermal work.\n"},{"id":"engr217-2019-a1","module":"ENGR5003","week":3,"label":"ENGR217 2019 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["real-gas-steam","ideal-gas-first-law","steam-tables","t-v-diagram","rigid-tank","dryness-fraction","interpolation","boundary-work"],"source":{"name":"ENGR217 Summer 2019 Q A1","paper":"ENGR217","year":2019,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR217_Exam_2019 (2).pdf","tag":"VALID"},"text":"Answer ALL parts (a) and (c)\n","formulaSheet":"- Wet steam: $x = m_g/m_{total}$; $u, v = (1-x)f + x\\,g$ (on the formula sheet)\n- First law, closed system: $dU = dQ - dW$ (heat in +ve, work out +ve); rigid tank $dW = 0$ so $Q = \\Delta U$ (on the formula sheet)\n- Boundary work at constant pressure: $W = \\int p\\,dV = p\\,(V_2 - V_1)$ (on the formula sheet as $dW = p\\,dV$, sign as printed)\n- Linear interpolation in the steam tables (learn this)\n- Steam property values (saturated temperature and pressure tables, superheated table) (learn this: read from the tables provided)\n","parts":[{"label":"a","text":"On a graph of temperature against volume sketch lines of constant pressure for water. Draw and label the\nsaturated liquid line and the saturated vapour line. Draw and label constant temperature lines through\nthe critical point and the triple point. Label a region where water and vapour co-exist and a region ice\nand vapour can co-exist.\n","marks":7,"grade":"third"},{"label":"b(i)","text":"A rigid tank of volume $3\\ \\mathrm{m^3}$ contains 4 kg of wet steam at $85^\\circ\\mathrm{C}$.\n\n(i) Determine the dryness fraction\n","marks":4,"answer":[{"label":"x_1","value":0.2651,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"b(ii)","text":"(ii) If the water is heated slowly until it is completely vaporised (i.e. becomes dry steam), using steam\ntables determine the temperature and pressure at this instant.\n","marks":4,"answer":[{"label":"T_2","value":126,"units":"C","tol":0.02},{"label":"p_2","value":239.7,"units":"kPa","tol":0.02}],"grade":"$undefined"},{"label":"b(iii)","text":"(iii) Using steam tables determine how much heat was required for the heating process?\n","marks":4,"answer":[{"label":"Q","value":6457,"units":"kJ","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"A mass of 2 kg of saturated water vapour at 400 kPa is heated at constant pressure until the temperature\nreaches $350^\\circ\\mathrm{C}$. Calculate the work done by the steam during this process.\n","marks":6,"answer":[{"label":"W","value":201.1,"units":"kJ","tol":0.02}],"grade":"$undefined"}],"solution":"$bf","confidence":"HIGH","solutionNote":"No official answers for this question: the 2019 \"217 - 2019 answer\" sheet covers only the fluids section\n(B1-B3). All values from the printed Cengel tables, checked against IAPWS-IF97 (agreement within 0.3%, the\ndifference being linear interpolation) and by back-substitution.\n"},{"id":"engr217-2019-a2","module":"ENGR5003","week":4,"label":"ENGR217 2019 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["ic-engines","ideal-gas-first-law","otto-cycle","isothermal-compression","thermal-efficiency","pv-diagram"],"source":{"name":"ENGR217 Summer 2019 Q A2","paper":"ENGR217","year":2019,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR217_Exam_2019 (2).pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (f).\n\nIn an air standard Otto cycle the compression ratio is 8 and the pressure and temperature of the air at the\nstart of compression are 1 bar and 300 K.\n","formulaSheet":"- Adiabatic ideal gas: $pV^\\gamma = \\text{const}$, $TV^{\\gamma-1} = \\text{const}$ (on the formula sheet)\n- Constant-volume heat transfer: $dQ = m c_v\\, dT$; isochoric $p/T = \\text{const}$ (Gay-Lussac) (on the formula sheet)\n- Isothermal ideal gas: $dU = 0$ so $dQ = dW$, $|W| = m R_g T \\ln(V_1/V_2)$ (on the formula sheet)\n- Thermal efficiency: $\\eta_{th} = 1 - \\dfrac{Q_{out}}{Q_{in}}$; Otto $\\eta = 1 - \\dfrac{1}{r^{\\gamma-1}}$ (on the formula sheet)\n- $R_g = 0.287\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ for air (on the formula sheet)\n","parts":[{"label":"a","text":"Sketch and label the processes of an Otto cycle on a PV diagram.\n","marks":4,"grade":"third"},{"label":"b","text":"Taking $\\gamma = 1.4$ during the isentropic compression compute the pressure and temperature at the end of\nthe compression stroke.\n","marks":3,"answer":[{"label":"p_2","value":18.38,"units":"bar","tol":0.02},{"label":"T_2","value":689.2,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"The heat added at constant volume in the cycle is $1080\\ \\mathrm{kJ\\,kg^{-1}}$. Taking the heat capacity as\n$0.880\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ while this heat is added, determine the maximum pressure and\ntemperature of the cycle.\n","marks":4,"answer":[{"label":"p_3","value":51.11,"units":"bar","tol":0.02},{"label":"T_3","value":1916.5,"units":"K","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"If $\\gamma$ and the heat capacity are unchanged during the isentropic expansion and heat rejection\nrespectively, determine the efficiency.\n","marks":3,"answer":[{"label":"eta_th","value":0.565,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"e","text":"Taking $\\gamma = 1.33$ during the isentropic expansion and taking the heat capacity as\n$0.74\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ while heat is rejected determine the heat rejected and hence the\nefficiency.\n","marks":6,"answer":[{"label":"q_out","value":492,"units":"kJ/kg","tol":0.02},{"label":"eta_th","value":0.544,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"f","text":"If it were possible to make the compression process isothermal rather than isentropic, and assuming the\nsame heat addition as before, $\\gamma = 1.33$ for expansion, $c_v = 0.880\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$\nfor heat addition and $c_v = 0.74\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ for heat rejection, compute the new\nthermal efficiency.\n","marks":5,"grade":"first","answer":[{"label":"eta_th","value":0.513,"units":"-","tol":0.05}]}],"solution":"$c0","confidence":"MEDIUM","solutionNote":"No official answers for this question: the 2019 \"217 - 2019 answer\" sheet covers only the fluids section.\n(a)-(e) are standard and double-checked (Otto identities, pressure-temperature consistency at state 4).\n(f) is MEDIUM: the very similar 2018 A3(d) official answer (71%) ignored the heat rejected during isothermal\ncompression; by the same method this question would give 67.9%. This solution counts that heat (51.3%),\nwhich the first law requires. Confirm with the lecturer which answer is expected and the $R_g$ to use.\n"},{"id":"engr217-2019-a3","module":"ENGR5003","week":5,"label":"ENGR217 2019 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["gas-turbines","steady-flow","brayton-cycle","isentropic-efficiency","work-ratio","variable-specific-heat","regeneration"],"source":{"name":"ENGR217 Summer 2019 Q A3","paper":"ENGR217","year":2019,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR217_Exam_2019 (2).pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c)\n","formulaSheet":"- Isentropic ideal gas: $\\dfrac{T_2}{T_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- SFEE per unit mass for each component: $q - w = h_{out} - h_{in}$, $dh = c_p\\,dT$ (on the formula sheet)\n- Brayton: $\\eta = 1 - r_p^{-(\\gamma-1)/\\gamma}$, work ratio $r_w = \\dfrac{w_{out} - w_{in}}{w_{out}} = 1 - \\dfrac{T_1}{T_3} r_p^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- $\\eta_{th} = W_{net}/Q_{in}$ (on the formula sheet)\n- Isentropic efficiencies: $\\eta_C = \\dfrac{T_{2s} - T_1}{T_{2a} - T_1}$, $\\eta_T = \\dfrac{T_3 - T_{4a}}{T_3 - T_{4s}}$ (learn this)\n","parts":[{"label":"a","text":"A gas turbine operating on the air standard Brayton cycle has a pressure ratio of 9, a compressor inlet\ntemperature of 300 K and a turbine inlet temperature of 1400 K. Calculate the compressor and turbine\noutlet temperatures, the heat added, the work ratio (i.e. $(W_T - W_C)/W_T$) and the thermal efficiency.\n(For the air standard cycle take $c_p = 1.005\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$ and $\\gamma = 1.4$)\n","marks":14,"grade":"2:2","answer":[{"label":"T_2","value":562,"units":"K","tol":0.02},{"label":"T_4","value":747.3,"units":"K","tol":0.02},{"label":"q_in","value":842.2,"units":"kJ/kg","tol":0.02},{"label":"r_w","value":0.599,"units":"-","tol":0.02},{"label":"eta_th","value":0.466,"units":"-","tol":0.02}]},{"label":"b","text":"In practice the high temperatures during heating and in the turbine give a significant change to the\nspecific heats and also the turbine and compressor are not perfectly efficient. Making the following\nassumptions:-\n\n- during heating and in the turbine $c_p = 1.157\\ \\mathrm{kJ\\,kg^{-1}\\,K^{-1}}$\n- during expansion the ratio of specific heats becomes $\\gamma = 1.33$\n- isentropic turbine efficiency is 0.85\n- isentropic compressor efficiency is 0.80\n- heat input increases to maintain the turbine inlet temperature of 1400 K\n- air standard cycle parameters still apply to compression\n\ncalculate the heat input, the cycle efficiency and the work ratio.\n","marks":8,"answer":[{"label":"q_in","value":893.7,"units":"kJ/kg","tol":0.02},{"label":"eta_th","value":0.279,"units":"-","tol":0.02},{"label":"r_w","value":0.431,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"How might re-generation be used to enhance efficiency.\n","marks":3,"grade":"third"}],"solution":"$c1","confidence":"HIGH","solutionNote":"No official answers for this question: the 2019 \"217 - 2019 answer\" sheet covers only the fluids section.\n(a) checked against the formula-sheet Brayton efficiency and work-ratio expressions and by $q_{in} - q_{out} =\nw_{net}$; (b) checked by back-substituting the isentropic efficiencies. Interpretation in (b): the heat input\nuses $c_p = 1.157$ from the actual compressor exit temperature to 1400 K, as the bullet list states.\n"},{"id":"engr266-2018-a1","module":"ENGR5003","week":3,"label":"ENGR266 2018 Q A1","kind":"past-paper","level":3,"marks":21,"minutes":25,"grade":"2:2","tags":["real-gas-steam","r134a","p-v-diagram","p-t-diagram","redlich-kwong","ideal-gas"],"source":{"name":"ENGR266 Summer 2018 Q A1","paper":"ENGR266","year":2018,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR266_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/NUMERICAL ANSWERS 266.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d)\n\n**For this question you may need to use the attached data table for 1,1,1,2-tetrafluoroethane (HFC-134a).**\n","figures":[{"src":"/figures/engr266-2018-a1.png","caption":"Figure A1: A general P-T diagram."}],"formulaSheet":"- Ideal gas: $PV = nRT$, $R = 8.314\\ \\mathrm{J\\,mol^{-1}\\,K^{-1}}$ (on the formula sheet)\n- Redlich-Kwong: $P = \\dfrac{RT}{V - b} - \\dfrac{a}{T^{1/2} V (V + b)}$, $a = 0.42748\\dfrac{R^2 T_c^{2.5}}{P_c}$, $b = 0.08664\\dfrac{R T_c}{P_c}$ (learn this)\n- Equivalent generic-cubic form: $a(T) = 0.42748\\,\\alpha(T_r)\\dfrac{R^2T_c^2}{P_c}$ with $\\alpha = T_r^{-1/2}$ (learn this)\n- R134a saturated $v_f$, $v_g$ at the given temperature (learn this: read from the table provided)\n","parts":[{"label":"a","text":"Represent the three isothermal processes indicated by the red lines ($T_1$, $T_2$ and $T_3$) in the P-T\ndiagram below (Figure A1) on a P-V diagram. Label the liquid and vapour regions, the critical point and the\nsaturated liquid and saturated vapour curves.\n","marks":6,"grade":"third"},{"label":"b","text":"Calculate the change in the specific volume of HFC-134a during vapourisation at $43^\\circ\\mathrm{C}$.\n","marks":2,"answer":[{"label":"dv_fg","value":0.01746,"units":"m^3/kg","tol":0.05}],"grade":"$undefined"},{"label":"c(i)","text":"Calculate the pressure of 1 mole of HFC-134a in a rigid container of volume $0.001\\ \\mathrm{m^3}$ at\n$175^\\circ\\mathrm{C}$ according to:\n\n(i) The ideal gas equation.\n","marks":2,"answer":[{"label":"P_ideal","value":3725.9,"units":"kPa","tol":0.02}],"grade":"$undefined"},{"label":"c(ii)","text":"(ii) The Redlich-Kwong equation of state.\n\nThe critical temperature ($T_c$) for HFC-134a is $101.08^\\circ\\mathrm{C}$ and the critical pressure\n($P_c$) is 4060.3 kPa.\n","marks":9,"answer":[{"label":"P_RK","value":3117.5,"units":"kPa","tol":0.02}],"grade":"$undefined"},{"label":"d","text":"What is the origin of the discrepancy between the pressures calculated using the ideal gas equation and the\nRedlich-Kwong equation of state?\n","marks":2,"grade":"$undefined"}],"solution":"$c2","confidence":"HIGH","solutionNote":"Official numerical answers: (b) $\\Delta v = 0.0184 - 0.0009 = 0.0175\\ \\mathrm{m^3/kg}$; (c)(i) 3725.92 kPa;\n(c)(ii) 3117.51 kPa (same $T_r$, $\\alpha = 0.914$, $a = 0.931$, $b = 0.000066$). All agree. The attached\nR134a table is missing from the archived paper, so (b) uses standard property data (0.01746 vs the official\n0.0175 from 3 s.f. table values). Note the printed marks for this question sum to 21, not 25.\n"},{"id":"engr266-2018-a2","module":"ENGR5003","week":1,"label":"ENGR266 2018 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["second-law","ideal-gas-first-law","carnot-cycle","refrigeration","cop","entropy-change","t-s-diagram","pv-diagram"],"source":{"name":"ENGR266 Summer 2018 Q A2","paper":"ENGR266","year":2018,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR266_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/NUMERICAL ANSWERS 266.pdf","tag":"PARTIAL","leftOut":"Nothing is left out: all parts (a)-(e) are transcribed. Tagged PARTIAL because the coefficient of\nperformance of a refrigerator (Carnot refrigeration cycle) is not explicitly in the current slides; the\ncurrent slides cover the Carnot engine, entropy and ideal-gas processes, from which the reversed cycle\nfollows directly.\n"},"text":"Answer ALL parts (a) - (e).\n","formulaSheet":"- Adiabatic ideal gas: $TV^{\\gamma-1} = \\text{const}$, $pV^\\gamma = \\text{const}$ (on the formula sheet)\n- $C_V = \\dfrac{R}{\\gamma - 1}$; adiabatic work $W = C_V\\,\\Delta T$ (from $dU = dQ - dW$ with $dQ = 0$) (on the formula sheet)\n- Isothermal ideal gas: $dU = 0$, $Q = W = RT\\ln(V_2/V_1) = RT\\ln(P_1/P_2)$ (on the formula sheet)\n- Entropy change of an ideal gas: $\\Delta S = C_p\\ln(T_2/T_1) - R\\ln(P_2/P_1)$; reversible $dS = dQ/T$ (on the formula sheet)\n- Carnot refrigerator: $COP_R = \\dfrac{Q_L}{W_{in}} = \\dfrac{T_L}{T_H - T_L}$ (learn this; not explicitly in the current slides)\n","parts":[{"label":"a","text":"Draw a Carnot refrigeration cycle on a *P-V* diagram indicating the direction of the cycle, where heat\nenters and is rejected from the system as well as the values of the polytropic indices for each of the\nprocesses.\n","marks":6,"grade":"2:2"},{"label":"b(i)","text":"In a Carnot refrigerator the working fluid enters the compressor with an initial molar volume of\n$0.002\\ \\mathrm{m^3\\,mol^{-1}}$ at a pressure of 100 kPa and a temperature of $-21^\\circ\\mathrm{C}$. In the\ncompressor the molar volume of the working fluid is reduced to $0.0013\\ \\mathrm{m^3\\,mol^{-1}}$. Assume\nthe working fluid is an ideal gas with a heat capacity ratio $C_P/C_V = 1.5$, calculate:\n\n(i) The coefficient of performance (COP) for this Carnot cycle.\n","marks":6,"answer":[{"label":"T_H","value":312.75,"units":"K","tol":0.02},{"label":"COP","value":4.161,"units":"-","tol":0.02}],"grade":"$undefined"},{"label":"b(ii)","text":"(ii) The work done by the compressor.\n","marks":6,"answer":[{"label":"W_comp","value":1007.7,"units":"J/mol","tol":0.02}],"grade":"$undefined"},{"label":"c","text":"Draw the Carnot refrigeration cycle on a *T-S* diagram.\n","marks":2,"grade":"third"},{"label":"d","text":"What is the total change in entropy during the Carnot refrigeration cycle?\n","marks":1,"grade":"third"},{"label":"e","text":"Calculate the entropy change during the reversible isothermal expansion process if the pressure, when the\nworking fluid starts the process, is 110 kPa. Assume the working fluid is an ideal gas.\n","marks":4,"answer":[{"label":"dS","value":0.792,"units":"J/(mol K)","tol":0.02}],"grade":"$undefined"}],"solution":"$c3","confidence":"HIGH","solutionNote":"Official numerical answers: $T_2 = 312.75\\ \\mathrm{K}$, COP = 4.161, $W = 1007.72\\ \\mathrm{J/mol}$,\n$Q = 199.80\\ \\mathrm{J/mol}$ and $\\Delta S = 0.79\\ \\mathrm{J\\,K^{-1}\\,mol^{-1}}$. All agree. The official sheet\nprints $-23^\\circ\\mathrm{C}$ in its COP and $\\Delta S$ lines but the numbers were computed with $-21^\\circ\\mathrm{C}$\n(a typo there). The printed state 1 data are inconsistent with the ideal gas law (see the data warning in\n(b)(ii)); the temperature-based route used here and officially is the intended one. Refrigerator COP is not\nexplicitly in the current slides (reason for the PARTIAL tag).\n"},{"id":"engr266-2019-a1","module":"ENGR5003","week":3,"label":"ENGR266 2019 Q A1","kind":"past-paper","level":3,"marks":16,"minutes":19,"grade":"2:2","tags":["real-gas-steam","t-s-diagram","phase-diagram","triple-point","r134a","entropy","ideal-gas-assumptions"],"source":{"name":"ENGR266 Summer 2019 Q A1","paper":"ENGR266","year":2019,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR266_Exam_2019 (1).pdf","tag":"PARTIAL","leftOut":"Part (c) (9 marks: molar volume of hexane from the ideal gas law and the Pitzer generalised correlation for the\nsecond virial coefficient) is left out: the generalised (Pitzer) correlation is not in the current slides or\nformula sheets (CHECK in the audit).\n"},"text":"Answer ALL parts (a) - (d)\n","figures":[{"src":"/figures/engr266-2019-a1.png","caption":"Figure A1-1: A general temperature entropy diagram."}],"formulaSheet":"- Wet-mixture properties: $v = (1 - x)v_f + x v_g$, $s = (1 - x)s_f + x s_g$, $x = m_g/m_{total}$ (on the formula sheet)\n- Ideal gas law: $pV = nRT$ (on the formula sheet)\n- Phase rule (triple point: $P = 3$, $C = 1$ gives $F = 0$): $P + F = C + 2$ (on the formula sheet)\n","parts":[{"label":"a","text":"Sketch the temperature-entropy diagram below and label the solid, solid-liquid, solid-vapour, liquid,\nliquid-vapour and vapour regions. Also indicate the location of the critical point and the triple-point\nline.\n","marks":8,"grade":"2:2"},{"label":"b","text":"Use the data-table provided in Appendix 1 - Thermodynamics for Chemical Engineering: Formula Sheet to\ncalculate the entropy of 1,1,1,2-tetrafluroethane (HFC-134a) at $36^\\circ\\mathrm{C}$ with a volume of\n$0.0184\\ \\mathrm{m^3\\,kg^{-1}}$.\n","marks":6,"grade":"2:2","answer":[{"label":"s","value":1.613,"units":"kJ/(kg K)","tol":0.02}]},{"label":"d","text":"What are the two key approximations made in the ideal gas law?","marks":2,"grade":"third"}],"solution":"$c4","confidence":"HIGH","solutionNote":"No official solution available (the 2019 Numerical Answers folder has no ENGR266 answers). Part (b) uses the\nHFC-134a table printed in the paper's appendix; a reference equation of state agrees to 0.03%.\n"},{"id":"engr266-2018-a3","module":"ENGR5003","week":6,"label":"ENGR266 2018 Q A3","kind":"past-paper","level":3,"marks":10,"minutes":12,"grade":"2:2","tags":["phase-reaction-equilibrium","vle","pxy-diagram","bubble-point","dew-point"],"source":{"name":"ENGR266 Summer 2018 Q A3","paper":"ENGR266","year":2018,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR266_2018.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/NUMERICAL ANSWERS 266.pdf","tag":"PARTIAL","leftOut":"Parts (c) (5 marks, fugacity of a pure gas from $Z(P)$ via $\\ln\\phi = \\int_0^P (Z-1)\\,dP/P$), (d) (2 marks,\ndefinition of the excess Gibbs energy) and (e) (8 marks, excess Gibbs energy of benzene/water from modified\nRaoult's law activity coefficients and Antoine equations) are left out: fugacity, excess Gibbs energy and\nactivity coefficients are not in the current slides or formula sheets (audit tag CHECK).\n"},"text":"Answer ALL parts (a) - (e)\n","formulaSheet":"- Phase rule: $P + F = C + 2$ (two-phase binary at fixed $T$: one degree of freedom, so each $P$ fixes $x_1$ and $y_1$) (on the formula sheet)\n- Reading a P-x-y diagram: bubble point from the bubble (liquid, $x_1$) curve, the coexisting vapour from the dew ($y_1$) curve at the same $P$ (tie line) (learn this)\n","parts":[{"label":"a","text":"Plot the following vapour liquid equilibrium data for the acetone(1) / methanol(2) mixture on the graph\npaper provided and add labels to show the liquid and vapour regions and indicate which lines correspond to\nthe bubble and dew curves.\n\n| Pressure /kPa | $x_1$ | $y_1$ |\n|---|---|---|\n| 17 | 0.0 | 0.0 |\n| 20 | | 0.22 |\n| 23 | 0.21 | 0.38 |\n| 27 | 0.46 | 0.59 |\n| 30 | 0.77 | 0.81 |\n| 31 | 1 | 1 |\n","marks":6,"grade":"third"},{"label":"b(i)","text":"The acetone(1) / methanol(2) binary mixture contains a mole fraction of acetone of 0.3.\n\n(i) What is the mole fraction of acetone in the first bubble that is formed when the pressure crosses the\nbubble curve.\n","marks":2,"answer":[{"label":"y_1 (first bubble)","value":0.46,"units":"-","tol":0.1}],"grade":"$undefined"},{"label":"b(ii)","text":"(ii) What is the mole fraction of acetone in the first droplet that is formed when the pressure crosses the\ndew curve.\n","marks":2,"answer":[{"label":"x_1 (first droplet)","value":0.15,"units":"-","tol":0.1}],"grade":"$undefined"}],"solution":"$c5","confidence":"HIGH","solutionNote":"The official numerical answers sheet gives only A3(c) and A3(e) (left out here), so (b) could not be compared;\nreadings were checked by linear and smooth (PCHIP) interpolation of the data, which agree to about 0.02, and\nby the tie-line ordering $x_1 < z_1 < y_1$. Answers depend on hand-drawn curves, hence the 10% tolerance.\n"},{"id":"engr266-2019-a2","module":"ENGR5003","week":2,"label":"ENGR266 2019 Q A2","kind":"past-paper","level":3,"marks":15,"minutes":18,"grade":"2:2","tags":["radiation-heat-exchangers","gas-turbines","heat-exchanger-energy-balance","brayton-cycle","jet-engine","p-v-diagram","adiabatic-compression"],"source":{"name":"ENGR266 Summer 2019 Q A2","paper":"ENGR266","year":2019,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR266_Exam_2019 (1).pdf","tag":"PARTIAL","leftOut":"Part (b) (10 marks: area-velocity relation $\\frac{du}{dx} = -\\frac{u}{A}\\left(\\frac{1}{1 - M^2}\\right)\\frac{dA}{dx}$ and\nsubsonic/supersonic flow in converging and diverging nozzles) is left out: compressible nozzle flow and the\narea-velocity relation have been dropped from the current syllabus and formula sheet (OBSOLETE in the audit).\n"},"text":"Answer ALL parts (a) - (c).\n","formulaSheet":"- Heat exchanger energy balance (adiabatic exchanger, each stream): $\\dot Q = \\dot m c_p \\Delta T$, heat lost by the hot stream = heat gained by the cold stream ($dQ = mc_p dT$ on the formula sheet; the exchanger balance: learn this)\n- Adiabatic process of an ideal gas: $pV^\\gamma = $ const, $\\gamma = c_p/c_v$ (on the formula sheet)\n- Adiabatic / polytropic work: $W = \\dfrac{p_2V_2 - p_1V_1}{n - 1}$ (as printed: positive for compression, i.e. work done on the gas) (on the formula sheet)\n- First law for an adiabatic process: $\\Delta U = -W_{by} = n c_v\\Delta T$ (on the formula sheet)\n- Ideal gas: $pV = nRT$ (on the formula sheet)\n","parts":[{"label":"a","text":"Oil is cooled from an initial temperature of $96^\\circ\\mathrm{C}$ to $30^\\circ\\mathrm{C}$ by water in an adiabatic\nconcentric heat exchanger. The flow rates for the oil and water are $0.7\\ \\mathrm{kg\\,s^{-1}}$ and\n$0.4\\ \\mathrm{kg\\,s^{-1}}$ respectively. If water enters the heat exchanger with a temperature of\n$20^\\circ\\mathrm{C}$, what is the temperature of the water when it leaves the heat exchanger?\n\nAssume the following: heat capacities are constant as a function of temperature,\n$C_p(\\mathrm{Oil}) = 2131\\ \\mathrm{J\\,kg^{-1}\\,K^{-1}}$ and $C_p(\\mathrm{Water}) = 4178\\ \\mathrm{J\\,kg^{-1}\\,K^{-1}}$.\n","marks":6,"grade":"2:2","answer":[{"label":"T_w,out","value":78.9,"units":"C","tol":0.02}]},{"label":"c(i)","text":"An air breathing jet engine cycle consists of the following steps:\n1. Adiabatic compression\n2. Isobaric expansion\n3. Adiabatic expansion\n4. Isobaric compression\n\nDraw this engine cycle on a pressure-volume diagram and indicate the direction of the cycle and where heat\nenters and leaves the system.\n","marks":5,"grade":"2:2"},{"label":"c(ii)","text":"Calculate the work done during the adiabatic compression stage of the cycle if an ideal gas with volume of\n$0.0102\\ \\mathrm{m^3\\,mol^{-1}}$ experiences an increase in pressure from 0.35 MPa to 2 MPa. For the ideal gas\nassume that $C_p = 5/2R$ and $C_v = 3/2R$.\n","marks":4,"grade":"2:2","answer":[{"label":"W_in","value":5.398,"units":"kJ/mol","tol":0.02}]}],"solution":"$c6","confidence":"HIGH","solutionNote":"No official solution available (the 2019 Numerical Answers folder has no ENGR266 answers). Both calculations are\nchecked by an independent energy balance.\n"},{"id":"engr266-2019-a3","module":"ENGR5003","week":6,"label":"ENGR266 2019 Q A3","kind":"past-paper","level":3,"marks":12,"minutes":14,"grade":"2:2","tags":["phase-reaction-equilibrium","vle","txy-diagram","bubble-dew-curves","lever-rule"],"source":{"name":"ENGR266 Summer 2019 Q A3","paper":"ENGR266","year":2019,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR266_Exam_2019 (1).pdf","tag":"PARTIAL","leftOut":"Parts (c)-(d) (13 marks: excess Gibbs energy and vapour composition from the Margules equations, and the meaning\nof activity coefficients) are left out: Margules equations, excess Gibbs energy and activity coefficients are not\nin the current slides or formula sheets (CHECK in the audit).\n"},"text":"Answer ALL parts (a) - (d)\n","formulaSheet":"- Lever rule on a T-x-y diagram: liquid fraction $\\dfrac{L}{F} = \\dfrac{y_1 - z_1}{y_1 - x_1}$, vapour fraction $\\dfrac{V}{F} = \\dfrac{z_1 - x_1}{y_1 - x_1}$ (from the component balance $z_1 = (L/F)x_1 + (V/F)y_1$) (learn this)\n- Phase rule (binary, two phases at fixed P: one degree of freedom, so T fixes $x_1$ and $y_1$): $P + F = C + 2$ (on the formula sheet)\n","parts":[{"label":"a","text":"Plot the following vapour liquid equilibrium data for acetonitrile(1)/1-butanol(2) on the graph paper\nprovided and label the liquid, vapour and liquid-vapour region. Also label the bubble curve and the dew\ncurve.\n\nTable A3-1 vapour liquid equilibrium data for acetonitrile(1)/1-butanol(2).\n\n| Temperature /K | $x_1$ | $y_1$ |\n|---|---|---|\n| 390.85 | 0 | 0 |\n| 379.25 | 0.08 | 0.394 |\n| 370.15 | 0.2 | 0.619 |\n| 361.65 | 0.564 | 0.791 |\n| 357.25 | 0.78 | 0.889 |\n| 354.75 | 1 | 1 |\n","marks":8,"grade":"2:2"},{"label":"b","text":"What is the liquid fraction present in the acetonitrile(1)/1-butanol(2) system at a 1-butanol fraction of 0.5\nat 370 K?\n","marks":4,"grade":"2:2","answer":[{"label":"L/F","value":0.29,"units":"-","tol":0.05}]}],"solution":"$c7","confidence":"HIGH","solutionNote":"No official solution available (the 2019 Numerical Answers folder has no ENGR266 answers). Part (b) depends on\ngraph reading; linear and smooth-curve interpolation agree to 0.01 (tol 0.05).\n"},{"id":"engr5003-w1-q1","module":"ENGR5003","week":1,"label":"Week 1 Q1","kind":"exercise","level":1,"minutes":12,"tags":["ideal-gas-first-law","second-law","p-v-diagram","t-s-diagram","processes"],"source":{"name":"ENGR5003 Week 1 workshop sheet Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/Week1 questions.pdf"},"text":"Describe the following processes: how to draw it in P-V and T-s diagrams.\n","parts":[{"label":"a","text":"Adiabatic","grade":"$undefined"},{"label":"b","text":"Isobaric","grade":"$undefined"},{"label":"c","text":"Isentropic","grade":"$undefined"},{"label":"d","text":"Isochoric","grade":"$undefined"},{"label":"e","text":"Isenthalpic","grade":"$undefined"},{"label":"f","text":"Isothermal","grade":"$undefined"}],"grade":"third","formulaSheet":"- Adiabatic (reversible) process: $pV^\\gamma = \\text{const}$, $TV^{\\gamma-1} = \\text{const}$ (on the formula sheet)\n- Isothermal ideal gas: $pV = \\text{const}$ (Boyle) (on the formula sheet)\n- Ideal gas entropy change: $s_2 - s_1 = c_v \\ln\\dfrac{T_2}{T_1} + R_g \\ln\\dfrac{v_2}{v_1} = c_p \\ln\\dfrac{T_2}{T_1} - R_g \\ln\\dfrac{p_2}{p_1}$ (on the formula sheet)\n- Enthalpy of an ideal gas: $\\Delta h = c_p \\Delta T$ (on the formula sheet)\n- Slope of a constant-property line on T-s: $(\\partial T/\\partial s)_p = T/c_p$, $(\\partial T/\\partial s)_v = T/c_v$ (learn this; follows from $T\\,ds = c_p\\,dT$ at constant $p$)\n","solution":"$c8","confidence":"HIGH","solutionNote":"No official solution; standard ideal-gas results from the Week 1 lectures and formula sheet."},{"id":"engr5003-w1-q2","module":"ENGR5003","week":1,"label":"Week 1 Q2","kind":"exercise","level":2,"minutes":6,"tags":["ideal-gas-first-law","second-law","isothermal","entropy-change","boundary-work"],"source":{"name":"ENGR5003 Week 1 workshop sheet Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/Week1 questions.pdf"},"text":"Consider an ideal gas undergoing an isothermal process from $V_1 = 1\\ \\mathrm{m^3}$ to $V_2 = 5\\ \\mathrm{m^3}$ and\n$T = 400\\ \\mathrm{K}$. Assume that $R_g = 0.287\\ \\mathrm{kJ/kg/K}$.\n","parts":[{"label":"a","text":"Determine the specific work output during this process.","answer":[{"label":"w","value":184.76,"units":"kJ/kg"}],"grade":"$undefined"},{"label":"b","text":"What is the change of specific entropy of this process?","answer":[{"label":"delta s","value":0.4619,"units":"kJ/(kg K)"}],"grade":"$undefined"}],"grade":"third","formulaSheet":"- Isothermal ideal gas work: $w = R_gT\\ln\\dfrac{V_2}{V_1}$ (on the formula sheet, written as $W = -mR_gT\\ln(V_2/V_1)$ for work done on the gas)\n- Ideal gas entropy change: $s_2 - s_1 = c_v\\ln\\dfrac{T_2}{T_1} + R_g\\ln\\dfrac{v_2}{v_1}$ (on the formula sheet)\n- Isothermal ideal gas: $\\Delta U = 0$ so $Q = W$ (on the formula sheet)\n","solution":"Assumptions: closed system, reversible isothermal expansion of an ideal gas; work out positive.\n\n**(a)** Boundary work per kg with $p = R_gT/v$:\n$$w = \\int_1^2 p\\,dv = R_gT\\ln\\frac{v_2}{v_1} = R_gT\\ln\\frac{V_2}{V_1}$$\n(the volume ratio equals the specific volume ratio for a fixed mass).\n$$w = 0.287 \\times 400 \\times \\ln 5 = 114.8 \\times 1.6094 = 184.8\\ \\mathrm{kJ/kg}$$\nPositive: the gas expands and does work on the surroundings. Units: $\\mathrm{kJ\\,kg^{-1}K^{-1}} \\times \\mathrm{K} = \\mathrm{kJ/kg}$.\n\n**(b)** With $T_2 = T_1$ the temperature term vanishes:\n$$\\Delta s = c_v\\ln 1 + R_g\\ln\\frac{v_2}{v_1} = 0.287\\ln 5 = 0.4619\\ \\mathrm{kJ/(kg\\,K)}$$\n\nChecks: (i) numerical integration of $\\int_1^5 R_gT/v\\,dv$ gives 184.76 kJ/kg; (ii) an isothermal ideal gas has $\\Delta u = 0$,\nso $q = w$, and for a reversible process $q = T\\Delta s = 400 \\times 0.4619 = 184.8\\ \\mathrm{kJ/kg}$, matching (a). The entropy\nrises because heat flows in. Marks: about 2 for (a), 2 for (b).\n","confidence":"HIGH","solutionNote":"No official solution; verified by numerical integration and by q = T delta s."},{"id":"engr5003-w1-q3","module":"ENGR5003","week":1,"label":"Week 1 Q3","kind":"exercise","level":2,"minutes":12,"tags":["second-law","rankine-cycle","t-s-diagram","thermal-efficiency","heat-from-ts-area"],"source":{"name":"ENGR5003 Week 1 workshop sheet Q3","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 1/Week1 questions.pdf"},"text":"For the following diagram of a Rankine cycle (Figure Q3), estimate the thermal efficiency, given that\n$\\Delta Q = m\\int T\\,ds$.\n\nRemember that $\\eta_{th} = \\dfrac{W_{net}}{Q_{in}} = 1 - \\dfrac{Q_{out}}{Q_{in}}$. Also for the isobaric heat addition\nprocess ($2 \\to 3$), assume that the function is:\n$$T = \\begin{cases} 107.5\\,s + 6.25, & s \\le 2.5 \\\\ 275, & 2.5 < s \\le 6.0 \\end{cases}$$\nand for the isobaric heat rejection ($4 \\to 1$), $T = 30$, $s = [0.5, 6.25]$. Here temperatures are in Celsius.\n","figures":[{"src":"/figures/engr5003-w1-q3.png","caption":"Figure Q3: Rankine cycle on a T-s diagram for water (50 bar boiler, 0.06 bar condenser), states 1-4 marked."}],"answer":[{"label":"q_in","value":2799.8,"units":"kJ/kg"},{"label":"q_out","value":1743.1,"units":"kJ/kg"},{"label":"eta_th","value":0.3774}],"grade":"2:2","formulaSheet":"- Heat from a reversible path: $q = \\int T\\,ds$ with $T$ in kelvin (given in the question)\n- Thermal efficiency: $\\eta_{th} = \\dfrac{W_{net}}{Q_{in}} = 1 - \\dfrac{Q_{out}}{Q_{in}}$ (on the formula sheet)\n","solution":"$c9","confidence":"HIGH","solutionNote":"No official solution; integrals checked analytically and numerically (scipy quad)."},{"id":"engr5003-w2-q1","module":"ENGR5003","week":2,"label":"Week 2 Q1","kind":"exercise","level":2,"minutes":10,"tags":["steady-flow","sfee","nozzles","first-law","flow-work"],"source":{"name":"ENGR5003 Week 2 workshop sheet Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 2/Week2 questions.pdf"},"text":"Consider the Steady Flow Energy Equation.\n","parts":[{"label":"a","text":"Given that $dq = dw$ and neglecting gravitational potential energy, how does flow velocity vary with specific\nenthalpy and therefore temperature.\n","grade":"$undefined"},{"label":"b","text":"Neglecting kinetic and potential energy, show that the SFEE is equivalent to the 1st Law of thermodynamics.\n","grade":"$undefined"}],"grade":"2:2","formulaSheet":"- SFEE (per unit mass): $q - w = (h_2 - h_1) + \\tfrac12(c_2^2 - c_1^2) + g(z_2 - z_1)$ (on the formula sheet)\n- Enthalpy: $h = u + pv$; ideal gas $\\Delta h = c_p\\Delta T$ (on the formula sheet)\n- First law (closed system): $\\Delta u = q - w$ (on the formula sheet)\n- Flow work: $w_{flow} = pv$ (on the formula sheet)\n","solution":"$ca","confidence":"HIGH","solutionNote":"No official solution; derivation follows the Lecture 4 SFEE and the formula sheet."},{"id":"engr5003-w2-q2","module":"ENGR5003","week":2,"label":"Week 2 Q2","kind":"exercise","level":3,"minutes":25,"tags":["gas-turbines","brayton-cycle","thermal-efficiency","work-ratio","heat-exchanger","regeneration"],"source":{"name":"ENGR5003 Week 2 workshop sheet Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 2/Week2 questions.pdf"},"text":"Consider the Brayton cycle. Given that the pressure ratio, $r_p = \\dfrac{p_2}{p_1} = \\dfrac{p_3}{p_4} = 16$,\n$p_1 = 101\\ \\mathrm{kPa}$, $V_1 = 4.5\\ \\mathrm{m^3}$ and $V_3 = 4.2\\ \\mathrm{m^3}$. Assume we have 10 kg of an ideal gas with\n$R_g = 0.287\\ \\mathrm{kJ/kg/K}$ and $\\gamma = 1.4$.\n","parts":[{"label":"a","text":"Determine the pressure, temperature and volume at all 4 states.","answer":[{"label":"T1","value":158.36,"units":"K"},{"label":"p2","value":1616,"units":"kPa"},{"label":"V2","value":0.6211,"units":"m^3"},{"label":"T2","value":349.69,"units":"K"},{"label":"T3","value":2364.9,"units":"K"},{"label":"V4","value":30.43,"units":"m^3"},{"label":"T4","value":1071,"units":"K"}],"grade":"$undefined"},{"label":"b","text":"Derive and calculate the thermal efficiency of the cycle.","answer":[{"label":"eta_th","value":0.5471}],"grade":"$undefined"},{"label":"c","text":"What is the work ratio of the process and explain why the work ratio affects the practical efficiency of the cycle.\n","answer":[{"label":"r_w","value":0.8521}],"grade":"$undefined"},{"label":"d","text":"A heat exchanger is used to increase $T_2$ up to 1000 K and $T_4 = 1000\\ \\mathrm{K}$. How does this affect the\nthermal efficiency?\n","grade":"first","answer":[{"label":"eta_th with HX","value":0.808}]}],"grade":"2:1","formulaSheet":"- Ideal gas: $pV = mR_gT$ (on the formula sheet)\n- Isentropic: $pV^\\gamma = \\text{const}$, $\\dfrac{T_2}{T_1} = \\left(\\dfrac{p_2}{p_1}\\right)^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- $c_p = \\dfrac{\\gamma R_g}{\\gamma - 1}$ (on the formula sheet)\n- Brayton efficiency: $\\eta_{th} = 1 - \\dfrac{1}{r_p^{(\\gamma-1)/\\gamma}}$ (on the formula sheet; the derivation is in Lecture 4)\n- Work ratio: $r_w = \\dfrac{W_{net}}{W_{out}} = 1 - \\dfrac{T_1}{T_3}r_p^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Heat exchanger: heat gained by the compressed air = heat lost by the exhaust (learn this, Lecture 14)\n","solution":"$cb","confidence":"MEDIUM","solutionNote":"No official solution. Parts (a)-(c) are fully checked (back-substitution and two efficiency routes). Part (d) is worded\nambiguously (\"increase T2 up to 1000 K and T4 = 1000 K\"); the energy-consistent reading (exhaust preheats the compressed\nair to 1000 K, W_net unchanged) gives 80.8%. Confirm the intended reading with the lecturer.\n"},{"id":"engr5003-w2-q3","module":"ENGR5003","week":2,"label":"Week 2 Q3","kind":"exercise","level":2,"minutes":15,"tags":["charging-discharging","unsteady-flow","ideal-gas-first-law","rigid-vessel"],"source":{"name":"ENGR5003 Week 2 workshop sheet Q3","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 2/Week2 questions.pdf"},"text":"Consider a rigid vessel of $0.15\\ \\mathrm{m^3}$ with an initial pressure of 0.6 bar at 320 K. High pressure air at 271 K is\ncharged into the vessel until the pressure is 25 bar. Assume $\\gamma = 1.4$ and that the charging process is adiabatic.\n","parts":[{"label":"a","text":"Determine the final temperature of the system.","answer":[{"label":"T2","value":377.7,"units":"K"}],"grade":"$undefined"},{"label":"b","text":"What mass of gas was added?","answer":[{"label":"m added","value":3.361,"units":"kg"}],"grade":"$undefined"},{"label":"c","text":"The gas in the vessel is allowed to cool down to 320 K again. How much heat is rejected during this cooling process?\nYou may assume $R_g = 0.287\\ \\mathrm{kJ/kg/K}$\n","answer":[{"label":"Q rejected","value":143.3,"units":"kJ"}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Unsteady flow energy equation: $Q - W = (m_2u_2 - m_1u_1) - \\sum m_{in}\\left(h + \\tfrac{c^2}{2} + gz\\right)_{in} + \\sum m_{out}\\left(h + \\tfrac{c^2}{2} + gz\\right)_{out}$ (on the formula sheet)\n- Ideal gas: $u = c_vT$, $h = c_pT$, $\\gamma = c_p/c_v$, $pV = mR_gT$ (on the formula sheet)\n- $c_v = \\dfrac{R_g}{\\gamma - 1}$ (on the formula sheet)\n- Isochoric closed system: $Q = \\Delta U = mc_v\\Delta T$ (on the formula sheet)\n","solution":"$cc","confidence":"HIGH","solutionNote":"No official solution; energy balance back-substituted and (c) checked by two routes."},{"id":"engr5003-w3-q1","module":"ENGR5003","week":3,"label":"Week 3 Q1","kind":"exercise","level":3,"minutes":20,"tags":["real-gas-steam","steam-tables","dryness-fraction","rigid-tank","interpolation","heat-capacity"],"source":{"name":"ENGR5003 Week 3 workshop sheet Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 3/Week3 questions.pdf"},"text":"1 kg of saturated water at a pressure of 75 kPa, with a dryness fraction, $x = 0.25$, is placed into a rigid tank. Heat is\nadded to the system until the water becomes a saturated vapour.\n","parts":[{"label":"a","text":"Determine the volume of the tank.","answer":[{"label":"V","value":0.5551,"units":"m^3"}],"grade":"$undefined"},{"label":"b","text":"Estimate the temperature and pressure of the saturated vapour.","answer":[{"label":"T","value":136.7,"units":"C"},{"label":"p","value":329.3,"units":"kPa"}],"grade":"$undefined"},{"label":"c","text":"Determine the heat required during this process.","answer":[{"label":"Q","value":1634,"units":"kJ"}],"grade":"$undefined"},{"label":"d","text":"Estimate $c_v$ and $c_p$ as well as the adiabatic index.","grade":"first","answer":[{"label":"c_v","value":36.3,"units":"kJ/(kg K)","tol":0.05},{"label":"c_p","value":39.5,"units":"kJ/(kg K)","tol":0.05},{"label":"gamma","value":1.086,"tol":0.02}]}],"grade":"2:1","formulaSheet":"- Wet steam: $v = (1 - x)v_f + xv_g$, $u = (1 - x)u_f + xu_g$, $h = (1 - x)h_f + xh_g$ (on the formula sheet)\n- First law, closed rigid system: $Q = \\Delta U = m(u_2 - u_1)$ (on the formula sheet)\n- Linear interpolation: $\\dfrac{y - y_1}{y_2 - y_1} = \\dfrac{x - x_1}{x_2 - x_1}$ (learn this, Lecture 9)\n- Heat capacities: $du = c_v\\,dT$, $dh = c_p\\,dT$, $\\gamma = c_p/c_v$ (on the formula sheet)\n","solution":"$cd","confidence":"MEDIUM","solutionNote":"No official solution. Parts (a)-(c) checked against IAPWS-IF97 and by table interpolation (agree within 0.1%). Part (d)\nasks for an \"estimate\" without a method in the slides; the finite-difference values above (c_v = Delta u/Delta T,\nc_p = Delta h/Delta T over the process) are one reasonable reading. Confirm the intended method with the lecturer.\n"},{"id":"engr5003-w3-q2","module":"ENGR5003","week":3,"label":"Week 3 Q2","kind":"exercise","level":3,"minutes":20,"tags":["real-gas-steam","steam-tables","dryness-fraction","isenthalpic","compression-ratio"],"source":{"name":"ENGR5003 Week 3 workshop sheet Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 3/Week3 questions.pdf"},"text":"5 kg of saturated steam at $150\\ ^\\circ\\mathrm{C}$ is compressed until the pressure doubles.\n","parts":[{"label":"a","text":"What is the initial volume of steam?","answer":[{"label":"V1","value":1.9625,"units":"m^3"}],"grade":"$undefined"},{"label":"b","text":"What is the dryness fraction after compression, given that this was an isenthalpic compression?","answer":[{"label":"x2","value":0.9855,"tol":0.005}],"grade":"$undefined"},{"label":"c","text":"What is the compression ratio $\\left(\\dfrac{V_1}{V_2}\\right)$?","answer":[{"label":"V1/V2","value":1.956}],"grade":"$undefined"},{"label":"d","text":"The steam cools back to $150\\ ^\\circ\\mathrm{C}$, how much heat is dissipated?","grade":"first","answer":[{"label":"Q","value":10568,"units":"kJ"}]}],"grade":"2:1","formulaSheet":"- Wet steam: $v = (1 - x)v_f + xv_g$, $h = (1 - x)h_f + xh_g$ (on the formula sheet)\n- Dryness fraction: $x = \\dfrac{h - h_f}{h_g - h_f}$ (on the formula sheet)\n- Constant-pressure heat for a closed system: $Q = m\\Delta h$ (learn this: $Q = \\Delta U + p\\Delta V = \\Delta H$)\n","solution":"$ce","confidence":"MEDIUM","solutionNote":"No official solution. Parts (a)-(c) checked against IAPWS-IF97 state lookups. Part (d) does not say how the steam is\ncooled; the constant-pressure answer (10.6 MJ) is given, with the constant-volume alternative (4.70 MJ) in the solution.\nConfirm the intended path with the lecturer.\n"},{"id":"engr5003-w3-q3","module":"ENGR5003","week":3,"label":"Week 3 Q3","kind":"exercise","level":2,"minutes":15,"tags":["real-gas-steam","steam-tables","interpolation","dryness-fraction","superheated-steam"],"source":{"name":"ENGR5003 Week 3 workshop sheet Q3 (Lecture 9 steam table examples)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 3/Week3 questions.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/Week 3/ENGR5003-Week3-Lecture9.pdf"},"text":"In-class examples show: how to check steam tables? Please download the steam tables ('Properties and Steam Tables') from\nthe Moodle and complete the data checking by yourself in workshop.\n\nThe in-class examples (Lecture 9 worked example) are:\n","parts":[{"label":"a","text":"Given $p = 100\\ \\mathrm{kPa}$ and $u = 2000\\ \\mathrm{kJ/kg}$, determine $T$, $v$, $h$, $s$, $x$.","answer":[{"label":"x","value":0.758,"tol":0.005},{"label":"T","value":99.61,"units":"C","tol":0.005},{"label":"v","value":1.284,"units":"m^3/kg"},{"label":"h","value":2128.4,"units":"kJ/kg"},{"label":"s","value":5.893,"units":"kJ/(kg K)"}],"grade":"$undefined"},{"label":"b","text":"Given $p = 2250\\ \\mathrm{kPa}$ and $s = 6\\ \\mathrm{kJ/kg/K}$, find $h$.","answer":[{"label":"h","value":2655.3,"units":"kJ/kg"}],"grade":"$undefined"},{"label":"c","text":"Given $p = 10\\ \\mathrm{bar}$ and $h = 3000\\ \\mathrm{kJ/kg}$, find $T$.","answer":[{"label":"T","value":276.2,"units":"C","tol":0.01}],"grade":"$undefined"},{"label":"d","text":"Given $T = 40.3\\ ^\\circ\\mathrm{C}$ and $v = 10\\ \\mathrm{m^3/kg}$, find $p$.","answer":[{"label":"p","value":7.52,"units":"kPa"}],"grade":"$undefined"}],"grade":"third","formulaSheet":"- Wet steam: $u = (1 - x)u_f + xu_g$ (and the same for $v$, $h$, $s$) (on the formula sheet)\n- Linear interpolation: $\\dfrac{y - y_1}{y_2 - y_1} = \\dfrac{x - x_1}{x_2 - x_1}$ (learn this, Lecture 9)\n","solution":"$cf","confidence":"HIGH","solutionNote":"Matches the Lecture 9 worked answers (x = 0.758, v = 1.284, h = 2128.4, s = 5.893; h = 2655.29; T = 276.2 C; p = 7.52 kPa) and IAPWS-IF97."},{"id":"engr5003-w4-q1","module":"ENGR5003","week":4,"label":"Week 4 Q1","kind":"exercise","level":2,"minutes":12,"tags":["ic-engines","otto-cycle","thermal-efficiency","net-work"],"source":{"name":"ENGR5003 Week 4 workshop sheet Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 4/Week4-questions.pdf"},"text":"Consider a petrol engine operating on the Otto cycle. The compression ratio is 5 and 1 litre of air is drawn into the\nengine at 300 K and 0.85 bar. The specific heat addition per cycle is 1000 kJ/kg.\n","parts":[{"label":"a","text":"Determine the thermal efficiency of the cycle.","answer":[{"label":"eta_th","value":0.4747}],"grade":"$undefined"},{"label":"b","text":"Determine the maximum temperature and pressure of the cycle.","answer":[{"label":"T3","value":1964.8,"units":"K"},{"label":"p3","value":2783.5,"units":"kPa"}],"grade":"$undefined"},{"label":"c","text":"What is the net work output of the cycle?","answer":[{"label":"w_net","value":474.7,"units":"kJ/kg"},{"label":"W_net","value":0.4686,"units":"kJ"}],"grade":"$undefined"}],"grade":"2:2","formulaSheet":"- Otto efficiency: $\\eta_{th} = 1 - \\dfrac{1}{r_v^{\\gamma-1}}$ (on the formula sheet)\n- Isentropic: $\\dfrac{T_2}{T_1} = r_v^{\\gamma-1}$, $\\dfrac{p_2}{p_1} = r_v^{\\gamma}$ (on the formula sheet)\n- Isochoric heat addition: $q_{in} = c_v(T_3 - T_2)$, $c_v = \\dfrac{R_g}{\\gamma - 1}$ (on the formula sheet)\n- Ideal gas: $pV = mR_gT$; Gay-Lussac $p/T$ = const at constant volume (on the formula sheet)\n","solution":"$d0","confidence":"HIGH","solutionNote":"No official solution; three independent routes agree (formula, temperatures, heat rejection)."},{"id":"engr5003-w4-q2","module":"ENGR5003","week":4,"label":"Week 4 Q2","kind":"exercise","level":3,"minutes":20,"tags":["ic-engines","otto-cycle","diesel-cycle","cut-off-ratio","mean-effective-pressure","pumping-work"],"source":{"name":"ENGR5003 Week 4 workshop sheet Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 4/Week4-questions.pdf"},"text":"Consider a petrol engine and a Diesel engine. The petrol engine has a compression ratio of 8 and the diesel engine has a\ncut-off ratio of 4. Given that for both cycles, 0.8 litres of air is drawn into the engine at 0.7 bar and 280 K:\n","parts":[{"label":"a","text":"What is the compression ratio of the Diesel engine, given that both engines have the same thermal efficiency?\n","answer":[{"label":"r_v diesel","value":19.23}],"grade":"$undefined"},{"label":"b","text":"Given that the specific heat rejected per cycle is 350 kJ/kg, how much heat is added each cycle and what is the net\nwork output?\n","answer":[{"label":"q_in","value":804.1,"units":"kJ/kg"},{"label":"w_net","value":454.1,"units":"kJ/kg"},{"label":"Q_in per cycle","value":0.5603,"units":"kJ"},{"label":"W_net per cycle","value":0.3164,"units":"kJ"}],"grade":"$undefined"},{"label":"c","text":"If the specific pumping work is 100 kJ/kg and the swept volume is 0.7 litres, what is the indicated work and the mean\neffective pressure?\n","answer":[{"label":"W_i","value":0.2468,"units":"kJ"},{"label":"MEP","value":352.5,"units":"kPa"}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Otto efficiency: $\\eta_{th} = 1 - \\dfrac{1}{r_v^{\\gamma-1}}$ (on the formula sheet)\n- Diesel efficiency: $\\eta_{th} = 1 - \\dfrac{1}{r_v^{\\gamma-1}}\\left[\\dfrac{r_c^{\\gamma} - 1}{\\gamma(r_c - 1)}\\right]$ (on the formula sheet)\n- $\\eta_{th} = 1 - \\dfrac{q_{out}}{q_{in}}$, $w_{net} = q_{in} - q_{out}$ (on the formula sheet)\n- Indicated work = net work - pumping work; $MEP = \\dfrac{W_{indicated}}{V_s}$ (learn this, Lectures 10-11)\n","solution":"$d1","confidence":"HIGH","solutionNote":"No official solution; r_d checked by root-finding and closed form; MEP units and swept volume checked."},{"id":"engr5003-w4-q3","module":"ENGR5003","week":4,"label":"Week 4 Q3","kind":"exercise","level":3,"minutes":25,"tags":["ic-engines","otto-cycle","engine-efficiencies","volumetric-efficiency","mechanical-efficiency","fuel-economy"],"source":{"name":"ENGR5003 Week 4 workshop sheet Q3 (worked in Lecture 12)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 4/Week4-questions.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/Week 4/ENGR5003-Week4-Lecture12.pdf"},"text":"A petrol engine has a volumetric airflow of 70 litres/s, a swept volume of 1.3 litres at an engine speed of 4000 rpm. This\nengine has a brake work of 1.3 kJ per cycle and a net work output of 1.5 kJ per cycle. The compression ratio of the cycle\nis 8 and we will assume $\\gamma = 1.4$.\n","parts":[{"label":"a","text":"Determine the thermal, volumetric and mechanical efficiencies of the engine.","answer":[{"label":"eta_th","value":0.5647},{"label":"eta_v","value":0.8077},{"label":"eta_m","value":0.8667}],"grade":"$undefined"},{"label":"b","text":"Determine the power output of the engine at 4000 rpm.","answer":[{"label":"P","value":70,"units":"kW"}],"grade":"$undefined"},{"label":"c","text":"Given that the fuel has a calorific value of 43,710 kJ/kg, and a density, $\\rho = 800\\ \\mathrm{kg/m^3}$, determine the\nmass and volume of fuel consumed per cycle and per second.\n","answer":[{"label":"m_f per cycle","value":0.00006077,"units":"kg"},{"label":"V_f per cycle","value":7.596e-8,"units":"m^3"},{"label":"m_f per second","value":0.004051,"units":"kg/s"},{"label":"V_f per second","value":0.000005064,"units":"m^3/s"}],"grade":"$undefined"},{"label":"d","text":"If one revolution of the engine corresponds to 0.48 m distance travelled by a car, determine the speed of the car and\nthe fuel economy in km/litre. How far can the car travel on a 40 litre tank of fuel?\n","answer":[{"label":"speed","value":32,"units":"m/s"},{"label":"fuel economy","value":6.32,"units":"km/L"},{"label":"range","value":252.8,"units":"km"}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Otto efficiency: $\\eta_{th} = 1 - \\dfrac{1}{r_v^{\\gamma-1}}$ (on the formula sheet)\n- Volumetric (breathing) efficiency: $\\eta_v = \\dfrac{\\dot V_a}{V_sN}$ (learn this, Lecture 11)\n- Mechanical efficiency: $\\eta_m = \\dfrac{W_b}{W_{net}}$ (learn this, Lecture 11)\n- Overall efficiency: $\\eta_o = \\eta_{th}\\eta_v\\eta_m$ (learn this, Lecture 11)\n- Fuel energy: $Q_{in} = m_f \\times CV$ (learn this)\n","solution":"$d2","confidence":"HIGH","solutionNote":"Matches the Lecture 12 worked solution (eta_th 0.565, eta_v 0.808, eta_m 0.867, P = 70 kW, 6.08e-5 kg and 7.60e-8 m^3 per\ncycle, 4.05e-3 kg/s, 32 m/s, 252.77 km, 6.32 km/L). The slide's per-second volume line has a typo (7.596e-5 for 7.596e-8);\nthe result 5.06e-6 m^3/s is correct. The slides take N = rpm/60 (one cycle per revolution).\n"},{"id":"engr5003-w5-q1","module":"ENGR5003","week":5,"label":"Week 5 Q1","kind":"exercise","level":3,"minutes":25,"tags":["gas-turbines","brayton-cycle","regeneration","heat-exchanger","isentropic-efficiency","work-ratio"],"source":{"name":"ENGR5003 Week 5 workshop sheet Q1 (worked in Lecture 13, worked example 3)","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 5/Week5-questions.pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/Week 5/ENGR5003-Week5-Lecture13.pdf"},"text":"In a gas turbine plant, air is compressed from 1.01 bar and $15\\ ^\\circ\\mathrm{C}$ through a pressure ratio of 4:1. It is then\nheated to $650\\ ^\\circ\\mathrm{C}$ in a combustion chamber and expanded through a turbine back to atmospheric pressure of 1.01 bar.\n","parts":[{"label":"a","text":"Calculate the cycle efficiency and compare the efficiency when the system is with and without heat regeneration\n","answer":[{"label":"eta with regen","value":0.239},{"label":"eta without regen","value":0.147}],"grade":"$undefined"},{"label":"b","text":"and the ratio of the net work to positive work (work ratio) if an ideal regenerative heat exchanger is used. The\nisentropic efficiencies of the turbine and compressor are 0.85 and 0.8 respectively. For compression, $\\gamma = 1.4$ and\nfor expansion $\\gamma = 1.33$. Assume that $R_g = 0.287\\ \\mathrm{kJ/kg/K}$\n","answer":[{"label":"w_net","value":88.4,"units":"kJ/kg"},{"label":"r_w","value":0.335}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Isentropic: $\\dfrac{T_{2s}}{T_1} = r_p^{(\\gamma-1)/\\gamma}$ (on the formula sheet)\n- Isentropic efficiencies: $\\eta_C = \\dfrac{T_{2s} - T_1}{T_2 - T_1}$, $\\eta_T = \\dfrac{T_3 - T_4}{T_3 - T_{4s}}$ (learn this, Lecture 13)\n- $c_p = \\dfrac{\\gamma R_g}{\\gamma - 1}$ (on the formula sheet)\n- SFEE for compressor/turbine: $w = \\Delta h = c_p\\Delta T$ (on the formula sheet)\n- Ideal (countercurrent) regenerator: $T_2' = T_4$ (learn this, Lecture 14)\n- Work ratio: $r_w = W_{net}/W_{out}$ (on the formula sheet)\n","solution":"$d3","confidence":"MEDIUM","solutionNote":"Lecture 13 worked example 3 gives 24.1% with regeneration, 14.8% without and r_w = 0.337; the small differences come from\nits rounding (T2 = 463 K, T4 = 694 K, w_T = 265.0, w_net = 89.2). The slide prints \"89.2/230\" but its 24.1% is 89.2/370.5.\nNumbers are verified, but the heat input depends on the lecture's convention of mixing two c_p values (h = c_p T per stage);\na single c_p gives 38.5% and 19.1%. Confirm with the lecturer which convention the exam expects.\n"},{"id":"engr5003-w5-q2","module":"ENGR5003","week":5,"label":"Week 5 Q2","kind":"exercise","level":3,"minutes":35,"tags":["real-gas-steam","rankine-cycle","steam-tables","isentropic-efficiency","back-work-ratio","thermal-efficiency"],"source":{"name":"ENGR5003 Week 5 workshop sheet Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/Week 5/Week5-questions.pdf"},"text":"Consider a Rankine cycle, where 5 kg of saturated water ($x = 0$) at 0.05 bar and $32.87\\ ^\\circ\\mathrm{C}$ is compressed to\n50 bar. The water is heated under constant pressure until it is converted into dry saturated steam. The pressurised steam\nis forced through a turbine, allowing the pressure to drop back to 0.05 bar.\n","parts":[{"label":"a","text":"Determine the entropy, pressure, temperature and volume of the water at each state of the cycle, assuming that the\ncompression and expansion processes are isentropic.\n","answer":[{"label":"s1 (= s2)","value":0.4763,"units":"kJ/(kg K)"},{"label":"T3","value":263.94,"units":"C"},{"label":"s3 (= s4)","value":5.9737,"units":"kJ/(kg K)"},{"label":"v3","value":0.03945,"units":"m^3/kg"},{"label":"x4","value":0.6943,"tol":0.005},{"label":"v4","value":19.57,"units":"m^3/kg"},{"label":"V4 (5 kg)","value":97.85,"units":"m^3"}],"grade":"$undefined"},{"label":"b","text":"Determine the entropy, pressure, temperature and volume of the water at each state of the cycle, assuming that the\ncompression and expansion processes have isentropic efficiencies 0.75 and 0.8 respectively.\n","answer":[{"label":"h2","value":144.45,"units":"kJ/kg"},{"label":"h4","value":2014.9,"units":"kJ/kg"},{"label":"x4","value":0.7747,"tol":0.005},{"label":"s4","value":6.61,"units":"kJ/(kg K)"},{"label":"v4","value":21.84,"units":"m^3/kg"}],"grade":"$undefined"},{"label":"c","text":"What is the back work ratio and thermal efficiencies for the isentropic assumption and assuming the isentropic\nefficiencies stated in part b)?\n","answer":[{"label":"r_bw isentropic","value":0.00515},{"label":"eta isentropic","value":0.3655},{"label":"r_bw real","value":0.00858},{"label":"eta real","value":0.2916}],"grade":"$undefined"}],"grade":"2:1","formulaSheet":"- Wet steam: $s = (1 - x)s_f + xs_g$, and the same for $v$, $h$ (on the formula sheet)\n- SFEE per component: $w_T = h_3 - h_4$, $w_P = h_2 - h_1$, $q_{in} = h_3 - h_2$ (on the formula sheet)\n- Pump work for an incompressible liquid: $w_P = v_f(p_2 - p_1)$ (learn this)\n- Isentropic efficiencies: $\\eta_P = \\dfrac{h_{2s} - h_1}{h_2 - h_1}$, $\\eta_T = \\dfrac{h_3 - h_4}{h_3 - h_{4s}}$ (learn this, Lecture 13)\n- Back work ratio $r_{bw} = \\dfrac{w_{in}}{w_{out}}$; $\\eta_{th} = \\dfrac{w_{net}}{q_{in}}$ (on the formula sheet)\n","solution":"$d4","confidence":"HIGH","solutionNote":"No official solution; all states from IAPWS-IF97 lookups, cross-checked with v dp pump work and the efficiency computed two ways."},{"id":"engr217-t1-p1","module":"ENGR5003","week":1,"label":"Old spec Tutorial 1 Problem 1","kind":"tutorial","level":1,"minutes":5,"grade":"third","tags":["ideal-gas-first-law","surface-tension","boundary-work"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 1","paper":"ENGR217","question":"Problem 1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"A clean wire ring of internal diameter 30 mm is repeatedly dipped into a liquid with surface tension\n0.075 N/m and withdrawn. Sometimes after withdrawing the ring from the liquid, a film of liquid has formed\nacross the ring.\n","formulaSheet":"- Surface-tension work: $\\delta W = -\\sigma_s\\,dA$ (work done by the system), i.e. work $\\sigma_s\\,\\Delta A$ must be done on a film to create area $\\Delta A$ (learn this, Week 1 Lecture 2)\n- Circle area $A = \\pi r^2$ (learn this)\n","parts":[{"label":"a","text":"How much work must be done to form the film?","answer":[{"label":"work to form the film","value":0.000106,"units":"J"}],"grade":"$undefined"},{"label":"b","text":"In order to increase the chance of a film forming across the ring, should the surface tension be increased or decreased?","grade":"$undefined"}],"solution":"**(a)** The film is a thin disc spanning the ring. It has TWO surfaces in contact with air (front and back),\nso the new liquid-air area created is\n$$\\Delta A = 2\\pi r^2 = 2\\pi(0.015)^2 = 1.414 \\times 10^{-3}\\ \\mathrm{m^2}$$\nCreating surface area requires work against surface tension (surface energy per unit area $\\sigma_s$):\n$$W = \\sigma_s\\,\\Delta A = 0.075 \\times 1.414 \\times 10^{-3} = 1.06 \\times 10^{-4}\\ \\mathrm{J}$$\nIn the sign convention of the lectures (work done BY the system positive) this is $W_{by} = -1.06 \\times 10^{-4}$ J:\nwork is done ON the liquid. Units: (N/m)(m$^2$) = N m = J. Common slip: using only one face gives half the answer.\n\n**(b)** Decrease it. The work needed to stretch the liquid into a film is proportional to $\\sigma_s$; a larger\nsurface tension makes the film more energetically costly and more likely to pull back into drops, so a lower\nsurface tension (e.g. adding soap) makes it easier for a film to form and survive.\n","confidence":"HIGH","solutionNote":"Matches the deck (1.06e-4 J of work on the film; reduce the surface tension)."},{"id":"engr217-t1-p2","module":"ENGR5003","week":1,"label":"Old spec Tutorial 1 Problem 2","kind":"tutorial","level":1,"minutes":8,"grade":"third","tags":["ideal-gas-first-law","ideal-gas","rigid-vessel"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 2","paper":"ENGR217","question":"Problem 2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"The pressure in a car tyre depends on the temperature of the air in the tyre. When the air temperature is\n10 C, the gauge pressure is 2.4 bar. The tyre creates an internal volume of 0.025 $\\mathrm{m^3}$ against the wheel\nand this volume is independent of pressure. Assume atmospheric pressure is 1 bar and that the air is an\nideal gas ($R_g = 0.287$ kJ/(kg K)).\n","formulaSheet":"- Ideal gas: $pV = mR_gT$, so for a fixed mass $\\dfrac{p_1V_1}{T_1} = \\dfrac{p_2V_2}{T_2}$ (on the formula sheet)\n- Absolute pressure = gauge pressure + atmospheric pressure; temperatures in kelvin (learn this)\n","parts":[{"label":"a","text":"Determine the increase in pressure when the air temperature increases to 40 C.","answer":[{"label":"p2 (gauge)","value":2.76,"units":"bar"},{"label":"pressure increase","value":0.36,"units":"bar","tol":0.03}],"grade":"$undefined"},{"label":"b","text":"Determine the mass of air that must be removed from the tyre to restore the gauge pressure to 2.4 bar at 40 C.","answer":[{"label":"initial mass","value":0.1046,"units":"kg"},{"label":"mass removed","value":0.01002,"units":"kg"}],"grade":"$undefined"}],"solution":"$d5","confidence":"HIGH","solutionNote":"Matches the deck (3.76 bar absolute, +0.36 bar; deck gives 10.06e-3 kg because it rounds m1 to 0.105 kg; unrounded 10.02e-3 kg)."},{"id":"engr217-t1-p3","module":"ENGR5003","week":1,"label":"Old spec Tutorial 1 Problem 3","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["ideal-gas-first-law","isothermal-compression","boundary-work","first-law"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 3","paper":"ENGR217","question":"Problem 3","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"1 kg of air at an absolute pressure $p_1 = 2$ bar and temperature $T_1 = 300$ K is isothermally compressed in a\ncylinder to half its original volume ($R_g = 0.287$ kJ/(kg K)).\n","formulaSheet":"- Isothermal ideal gas work: $W = mR_gT\\ln\\dfrac{V_2}{V_1}$ (work done by the gas) (on the formula sheet)\n- First law (closed system) $\\Delta U = Q - W$; isothermal ideal gas $\\Delta U = 0$ (on the formula sheet)\n- Ideal gas $pV = mR_gT$ (on the formula sheet)\n- Work done by the atmosphere on the outside of the piston $W_{atm} = p_0\\,|\\Delta V|$ (learn this)\n","parts":[{"label":"a","text":"Calculate the work done on the air and the heat transfer from the air.","answer":[{"label":"work done on the air","value":59.7,"units":"kJ"},{"label":"heat rejected by the air","value":59.7,"units":"kJ"}],"grade":"$undefined"},{"label":"b","text":"What is the net work needed to compress the air if the atmospheric pressure is 1 bar?","answer":[{"label":"work done by the atmosphere","value":21.5,"units":"kJ"},{"label":"net work needed","value":38.2,"units":"kJ"}],"grade":"$undefined"}],"solution":"$d6","confidence":"HIGH","solutionNote":"Matches the deck (59.7 kJ on the air, 59.7 kJ rejected, 21.5 kJ from the atmosphere, 38.2 kJ net)."},{"id":"engr217-t1-p4","module":"ENGR5003","week":3,"label":"Old spec Tutorial 1 Problem 4","kind":"tutorial","level":1,"minutes":8,"grade":"third","tags":["steam-tables","real-gas-steam","first-law","boundary-work"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 4","paper":"ENGR217","question":"Problem 4","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"A mass of 0.3 kg of saturated water is completely vaporised at a constant pressure of 75 kPa. Using steam\ntables, determine the volume change and the energy transferred to the water.\n","formulaSheet":"- Saturation table at a given pressure: $v_f$, $v_g$, $u_{fg}$, $h_{fg}$ (steam tables provided)\n- First law, closed system at constant pressure: $Q = \\Delta U + p\\Delta V = \\Delta H$ (on the formula sheet as $\\Delta u = q - w$, $h = u + pv$)\n","answer":[{"label":"volume change","value":0.6648,"units":"m^3"},{"label":"internal energy increase","value":633.5,"units":"kJ"},{"label":"heat transferred","value":683.4,"units":"kJ"}],"solution":"Saturated liquid ($x = 0$) becomes saturated vapour ($x = 1$) at constant pressure, so use the saturated water\nPRESSURE table at 75 kPa: $T_{sat} = 91.76$ C, $v_f = 0.001037$, $v_g = 2.2171\\ \\mathrm{m^3/kg}$,\n$u_{fg} = 2111.8$ kJ/kg, $h_{fg} = 2278.0$ kJ/kg.\n\n**Volume change.**\n$$\\Delta V = m(v_g - v_f) = 0.3 \\times (2.2171 - 0.001037) = 0.6648\\ \\mathrm{m^3}$$\n**Energy.** The internal energy rises by\n$$\\Delta U = m\\,u_{fg} = 0.3 \\times 2111.8 = 633.5\\ \\mathrm{kJ}$$\nbut the expanding vapour also pushes back the surroundings, doing boundary work\n$W = p\\Delta V = 75 \\times 0.6648 = 49.9$ kJ. The heat that must be transferred to the water is therefore\n$$Q = \\Delta U + W = 633.5 + 49.9 = 683.4\\ \\mathrm{kJ} = m\\,h_{fg} = 0.3 \\times 2278.0$$\nCheck: $h_{fg} - u_{fg} = p\\,v_{fg} = 75 \\times 2.2161 = 166.2$ kJ/kg, and $2278.0 - 2111.8 = 166.2$. Units: kPa x m$^3$ = kJ.\n","confidence":"HIGH","solutionNote":"The deck answers 0.66485 m^3 (matches) and gives the \"energy transferred\" as Delta U = m u_fg = 633.5 kJ. That is the internal-energy increase; the heat that must be supplied at constant pressure is m h_fg = 683.4 kJ, because 49.9 kJ leaves again as boundary work. Both are given here."},{"id":"engr217-t1-p5","module":"ENGR5003","week":3,"label":"Old spec Tutorial 1 Problem 5","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["steam-tables","superheated-steam","boundary-work"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 5","paper":"ENGR217","question":"Problem 5","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"4 kg of saturated water vapour at 300 kPa is heated at constant pressure until the temperature reaches 250 C.\nUsing the steam tables, calculate the work done by the steam during the process.\n","formulaSheet":"- Boundary work at constant pressure: $W = \\int p\\,dV = p\\,(V_2 - V_1)$ (on the formula sheet as $dW = p\\,dV$)\n- Saturated and superheated steam tables (provided)\n","answer":[{"label":"volume change","value":0.7625,"units":"m^3"},{"label":"work done by the steam","value":228.8,"units":"kJ"}],"solution":"State 1: saturated vapour at 300 kPa (pressure table): $T_{sat} = 133.5$ C, $v_1 = v_g = 0.60582\\ \\mathrm{m^3/kg}$.\nState 2: 300 kPa and 250 C is above $T_{sat}$, so superheated (superheated table, 0.3 MPa):\n$v_2 = 0.79645\\ \\mathrm{m^3/kg}$.\n\nConstant pressure, so\n$$\\Delta V = m(v_2 - v_1) = 4 \\times (0.79645 - 0.60582) = 0.7625\\ \\mathrm{m^3}$$\n$$W = p\\,\\Delta V = 300\\ \\mathrm{kPa} \\times 0.7625\\ \\mathrm{m^3} = 228.8\\ \\mathrm{kJ}$$\nPositive: the steam expands and does work on the piston. Check: an ideal-gas estimate\n$W = mR_g\\Delta T = 4 \\times 0.4615 \\times 116.5 = 215$ kJ is about 6% lower, reasonable since steam near\nsaturation is not ideal. (For interest, the heat input is $m(h_2 - h_g) = 972$ kJ.)\n","confidence":"HIGH","solutionNote":"Matches the deck (0.76252 m^3, 228.8 kJ)."},{"id":"engr217-t1-p6","module":"ENGR5003","week":3,"label":"Old spec Tutorial 1 Problem 6","kind":"tutorial","level":2,"minutes":15,"grade":"2:2","tags":["steam-tables","real-gas-steam","dryness-fraction","interpolation","phase-diagram"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 6","paper":"ENGR217","question":"Problem 6","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"Using the steam tables, complete the following table for water (blanks marked \"?\").\n\n| | T (C) | p (kPa) | u (kJ/kg) | x | Phase description |\n|---|---|---|---|---|---|\n| (a) | ? | 200 | ? | 0.4 | ? |\n| (b) | 150 | ? | 1800 | ? | ? |\n| (c) | ? | 1200 | 2955.5 | ? | ? |\n| (d) | 75 | 500 | ? | ? | ? |\n| (e) | ? | 850 | ? | 0 | ? |\n","formulaSheet":"- Wet steam: $u = (1 - x)u_f + xu_g$, so $x = \\dfrac{u - u_f}{u_g - u_f}$ (on the formula sheet)\n- Phase test: compare the given $u$ (or $T$) with $u_f$, $u_g$ (or $T_{sat}$) at the given pressure or temperature (learn this)\n- Compressed liquid: $u \\approx u_f(T)$ (learn this)\n","parts":[{"label":"a","text":"Row (a)","answer":[{"label":"T","value":120.21,"units":"C"},{"label":"u","value":1314.3,"units":"kJ/kg"}],"grade":"$undefined"},{"label":"b","text":"Row (b)","answer":[{"label":"p","value":476.16,"units":"kPa"},{"label":"x","value":0.6062,"units":"-"}],"grade":"$undefined"},{"label":"c","text":"Row (c)","answer":[{"label":"T","value":400,"units":"C"}],"grade":"$undefined"},{"label":"d","text":"Row (d)","answer":[{"label":"u","value":313.99,"units":"kJ/kg"}],"grade":"$undefined"},{"label":"e","text":"Row (e)","answer":[{"label":"T","value":172.94,"units":"C"},{"label":"u","value":731,"units":"kJ/kg"}],"grade":"$undefined"}],"solution":"$d7","confidence":"HIGH","solutionNote":"Numbers match the deck. Two descriptions differ: the deck labels (c) \"Dry steam\" with x = 1, but at 400 C and 1200 kPa the steam is superheated (x not defined), and it labels (e) \"Saturated liquid/vapour\" although x = 0 means saturated liquid. The deck also enters x = 0 for the compressed liquid in (d), where x is undefined."},{"id":"engr217-t1-p7","module":"ENGR5003","week":3,"label":"Old spec Tutorial 1 Problem 7","kind":"tutorial","level":1,"minutes":5,"grade":"third","tags":["steam-tables","superheated-steam","interpolation"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 7","paper":"ENGR217","question":"Problem 7","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"If 1 kg of steam has an enthalpy of 3300 kJ and a pressure of 2 MPa, use the steam tables to determine its\ntemperature.\n","formulaSheet":"- Linear interpolation: $\\dfrac{y - y_1}{y_2 - y_1} = \\dfrac{x - x_1}{x_2 - x_1}$ (learn this, Lecture 9)\n","answer":[{"label":"T","value":423.5,"units":"C","tol":0.01}],"solution":"$$h = 3300$ kJ/kg. At 2 MPa the saturated vapour has $h_g = 2798.3$ kJ/kg $< 3300$, so the steam is superheated.\nSuperheated table at 2 MPa: $h(400\\ \\text{C}) = 3248.4$, $h(500\\ \\text{C}) = 3468.3$ kJ/kg. Interpolating linearly:\n$$T = 400 + (500 - 400)\\frac{3300 - 3248.4}{3468.3 - 3248.4} = 400 + 100 \\times 0.2347 = 423.5\\ \\text{C}$$\nCheck: IAPWS-IF97 gives 423.6 C exactly; linear interpolation over a 100 C step is good to about 0.1 C here\nbecause $c_p$ hardly changes between 400 and 500 C.\n","confidence":"HIGH","solutionNote":"Matches the deck (423.47 C)."},{"id":"engr217-t1-p8","module":"ENGR5003","week":3,"label":"Old spec Tutorial 1 Problem 8","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["steam-tables","interpolation","absolute-pressure"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 8","paper":"ENGR217","question":"Problem 8","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"Water is heated in a vertical cylinder, closed at the top by a moveable piston. The piston has a mass of\n30 kg and a cross-sectional area of 0.006 $\\mathrm{m^2}$. If the local atmospheric pressure is 101.325 kPa, determine\nthe temperature at which the water starts boiling.\n","formulaSheet":"- Piston equilibrium: $p = p_{atm} + \\dfrac{mg}{A}$ (learn this)\n- Linear interpolation in the saturation (pressure) table (learn this, Lecture 9)\n","answer":[{"label":"pressure under the piston","value":150.3,"units":"kPa"},{"label":"boiling temperature","value":111.41,"units":"C","tol":0.005}],"solution":"The water boils when its temperature reaches the saturation temperature at the pressure it is under; it\nfirst boils just below the piston, where the pressure is lowest. Force balance on the (frictionless,\nstationary) piston:\n$$p = p_{atm} + \\frac{mg}{A} = 101.325 + \\frac{30 \\times 9.8}{0.006} \\times 10^{-3} = 101.325 + 49.0 = 150.3\\ \\mathrm{kPa}$$\nSaturation pressure table: $T_{sat}(150\\ \\text{kPa}) = 111.35$ C, $T_{sat}(175\\ \\text{kPa}) = 116.04$ C. Interpolating:\n$$T = 111.35 + (116.04 - 111.35)\\frac{150.325 - 150}{175 - 150} = 111.35 + 4.69 \\times 0.013 = 111.41\\ \\text{C}$$\nCheck: IAPWS-IF97 gives $T_{sat}(150.325\\ \\text{kPa}) = 111.415$ C; with $g = 9.81$ the pressure is 150.375 kPa and\n$T = 111.42$ C, so the choice of $g$ does not matter. Without the piston the water would boil at 100 C; the\nextra 49 kPa raises the boiling point by about 11 C.\n","confidence":"HIGH","solutionNote":"Matches the deck (150.325 kPa, 111.41 C)."},{"id":"engr217-t1-p9","module":"ENGR5003","week":3,"label":"Old spec Tutorial 1 Problem 9","kind":"tutorial","level":2,"minutes":15,"grade":"2:2","tags":["steam-tables","rigid-tank","dryness-fraction","interpolation","t-v-diagram","first-law"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 9","paper":"ENGR217","question":"Problem 9","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"A rigid tank of volume 1.5 $\\mathrm{m^3}$ contains 2 kg of wet steam at 75 C.\n","formulaSheet":"- Wet steam: $v = (1 - x)v_f + xv_g$, $u = (1 - x)u_f + xu_g$ (on the formula sheet)\n- First law, closed rigid system: $Q = \\Delta U = m(u_2 - u_1)$ (on the formula sheet)\n- Linear interpolation in the saturation tables (learn this, Lecture 9)\n","parts":[{"label":"a","text":"What is the dryness fraction?","answer":[{"label":"x","value":0.1815,"units":"-"}],"grade":"$undefined"},{"label":"b","text":"If the water is heated slowly until it is completely vaporised (becomes dry steam), determine the temperature and pressure at this point.","answer":[{"label":"T","value":125.95,"units":"C","tol":0.005},{"label":"p","value":239.2,"units":"kPa","tol":0.01}],"grade":"$undefined"},{"label":"c","text":"How much heat is required for the heating process?","answer":[{"label":"Q","value":3658,"units":"kJ"}],"grade":"$undefined"},{"label":"d","text":"Sketch the process on a T-v diagram, with respect to the saturation lines.","grade":"$undefined"}],"solution":"$d8","confidence":"HIGH","solutionNote":"Matches the deck (x = 0.1816 with v_f neglected, 125.97 C, 239.5 kPa, Q = 3657.6 kJ). The deck uses u1 = 706.5 kJ/kg from x = 0.1816; with v_f included x = 0.1814 and u1 = 706.1 kJ/kg, Q = 3658 kJ."},{"id":"engr217-t1-p10","module":"ENGR5003","week":3,"label":"Old spec Tutorial 1 Problem 10","kind":"tutorial","level":2,"minutes":15,"grade":"2:2","tags":["steam-tables","superheated-steam","dryness-fraction","interpolation","p-v-diagram"],"source":{"name":"ENGR217 (old spec) Thermo L6 Tutorial 1, Problem 10","paper":"ENGR217","question":"Problem 10","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 3 - Thermo L6 Tutorial 1 solutions (old spec ENGR217).pdf"},"text":"Wet steam is heated in a vertical cylinder, closed at the top by a moveable piston. Initially the cylinder\ncontains 0.01 $\\mathrm{m^3}$ of liquid and 0.19 $\\mathrm{m^3}$ of vapour in equilibrium at 700 kPa. Heat is transferred at\nconstant pressure until a temperature of 325 C is reached.\n","formulaSheet":"- Mass of each phase: $m_f = V_f/v_f$, $m_g = V_g/v_g$ (learn this)\n- Superheated steam table and linear interpolation (in $T$ and in $p$) (learn this, Lecture 9)\n","parts":[{"label":"a","text":"What is the initial temperature of the water?","answer":[{"label":"T1","value":164.95,"units":"C","tol":0.005}],"grade":"$undefined"},{"label":"b","text":"Determine the total mass of water.","answer":[{"label":"m","value":9.722,"units":"kg"}],"grade":"$undefined"},{"label":"c","text":"Determine the final volume.","answer":[{"label":"V2","value":3.78,"units":"m^3","tol":0.03}],"grade":"$undefined"},{"label":"d","text":"Show the process on a p-v diagram with respect to the saturation lines.","grade":"$undefined"}],"solution":"$d9","confidence":"HIGH","solutionNote":"Parts (a)-(b) match the deck (164.95 C, 9.7215 kg). For (c) the deck averages the four table values linearly in p, giving v = 0.39682 m^3/kg and V = 3.858 m^3; the correct value (IAPWS-IF97, or interpolating pv) is v = 0.3886 m^3/kg and V = 3.78 m^3, 2% lower. The answer box accepts both within tolerance."},{"id":"engr217-l18-q1","module":"ENGR5003","week":4,"label":"Old spec L18 Question 1","kind":"tutorial","level":2,"minutes":15,"grade":"2:2","tags":["ic-engines","otto-cycle","thermal-efficiency","net-work","p-v-diagram"],"source":{"name":"ENGR217 (old spec) Thermo L18 tutorial, Question 1","paper":"ENGR217","question":"Q1","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 6 - Thermo L18 Second law and gas turbine cycles (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 6 - Thermo L18 Second law and gas turbine cycles (old spec ENGR217).pdf"},"text":"Complete the following table for the ideal Otto cycle. The compression ratio is 5 and the specific heat\nsupplied is 750 kJ/kg. Assume 1 kg of air, $\\gamma = 1.4$, $c_v = 0.7175$ kJ/(kg K) and $R_g = 0.287$ kJ/(kg K).\nWhat is the net work output and the thermal efficiency?\n\n| State | p (bar) | V (m3) | T (K) |\n|---|---|---|---|\n| 1 | 1 | ? | 300 |\n| 2 | ? | ? | ? |\n| 3 | ? | ? | ? |\n| 4 | ? | ? | ? |\n\n(The same question is Question 1 in the current ENGR5003 Week 5 Lecture 15.)\n","formulaSheet":"- Ideal gas $pV = mR_gT$ (on the formula sheet)\n- Adiabatic process: $pV^\\gamma = \\text{const}$, $TV^{\\gamma-1} = \\text{const}$ (on the formula sheet)\n- Constant-volume heat addition $q_{in} = c_v(T_3 - T_2)$ (on the formula sheet)\n- Adiabatic work $W_{ab} = \\dfrac{p_bV_b - p_aV_a}{\\gamma - 1} = mc_v(T_b - T_a)$ (magnitude) (on the formula sheet)\n- Otto efficiency $\\eta = 1 - \\dfrac{1}{r_v^{\\gamma - 1}}$ (on the formula sheet)\n","parts":[{"label":"a","text":"Complete the table.","answer":[{"label":"V1","value":0.861,"units":"m^3"},{"label":"p2","value":9.518,"units":"bar"},{"label":"T2","value":571.1,"units":"K"},{"label":"p3","value":26.94,"units":"bar"},{"label":"T3","value":1616.4,"units":"K"},{"label":"p4","value":2.83,"units":"bar"},{"label":"T4","value":849.1,"units":"K"}],"grade":"$undefined"},{"label":"b","text":"What is the net work output and the thermal efficiency?","answer":[{"label":"W_net","value":356,"units":"kJ"},{"label":"thermal efficiency","value":47.47,"units":"%","tol":0.01}],"grade":"$undefined"}],"solution":"$da","confidence":"HIGH","solutionNote":"Matches the deck (table values; deck W_net = 355.8 kJ from rounded pressures, unrounded 356.0 kJ; eta = 47.4-47.5%)."},{"id":"engr217-l18-q2","module":"ENGR5003","week":4,"label":"Old spec L18 Question 2","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["ic-engines","otto-cycle","thermal-efficiency","mean-effective-pressure"],"source":{"name":"ENGR217 (old spec) Thermo L18 tutorial, Question 2","paper":"ENGR217","question":"Q2","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 6 - Thermo L18 Second law and gas turbine cycles (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 6 - Thermo L18 Second law and gas turbine cycles (old spec ENGR217).pdf"},"text":"In an ideal Otto cycle, the pressure and temperature of the air at the start of compression are 1 bar and\n330 K. The compression ratio is 8 and 1250 kJ/kg of heat is added at constant volume. Determine the maximum\npressure and temperature of the cycle, the thermal efficiency and the mean effective pressure.\nTake $\\gamma = 1.4$, $c_v = 0.7175$ kJ/(kg K), $R_g = 0.287$ kJ/(kg K).\n","formulaSheet":"- Adiabatic process: $pV^\\gamma = \\text{const}$, $TV^{\\gamma-1} = \\text{const}$ (on the formula sheet)\n- Constant-volume heat addition $q_{in} = c_v(T_3 - T_2)$; ideal gas $p/T$ const at constant $V$ (on the formula sheet)\n- Otto efficiency $\\eta = 1 - r_v^{1-\\gamma}$ (on the formula sheet)\n- Mean effective pressure $\\mathrm{MEP} = \\dfrac{w_{net}}{v_1 - v_2}$ (on the formula sheet)\n","answer":[{"label":"T_max","value":2500,"units":"K"},{"label":"p_max","value":60.6,"units":"bar"},{"label":"thermal efficiency","value":56.5,"units":"%","tol":0.01},{"label":"MEP","value":8.52,"units":"bar"}],"solution":"$db","confidence":"HIGH","solutionNote":"Matches the deck within rounding: the deck uses c_v = 0.718 for T3 (2499.1 K) and 0.7175 elsewhere, which gives w_net = 705.4 kJ/kg and eta = 56.4%; with a consistent c_v = 0.7175, T3 = 2500.3 K and eta = 56.5% = 1 - 8^-0.4. p_max 60.6 bar and MEP 8.52 bar agree."},{"id":"engr217-l18-q4","module":"ENGR5003","week":2,"label":"Old spec L18 Question 4","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["steady-flow","sfee","compressor","cooling-jacket","heat-exchanger-energy-balance"],"source":{"name":"ENGR217 (old spec) Thermo L18 tutorial, Question 4","paper":"ENGR217","question":"Q4","path":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 6 - Thermo L18 Second law and gas turbine cycles (old spec ENGR217).pdf","solutionsPath":"ENGR5003 - Thermodynamics and Heat Transfer/$ Old Spec/Week 6 - Thermo L18 Second law and gas turbine cycles (old spec ENGR217).pdf"},"text":"A compressor takes in air at the rate of 0.29 kg/s at 12 C and 100 kPa. It delivers air at 302 C and\n640 kPa. Conditions are steady and adiabatic. For air assume $c_p = 1.005$ kJ/(kg K); for water\n$c_p = 4.19$ kJ/(kg K). Neglect kinetic and potential energy changes.\n","formulaSheet":"- SFEE $\\dot Q - \\dot W = \\dot m\\left[(h_2 - h_1) + \\tfrac{1}{2}(C_2^2 - C_1^2) + g(z_2 - z_1)\\right]$ (on the formula sheet)\n- Ideal gas / liquid $\\Delta h = c_p\\Delta T$ (on the formula sheet)\n","parts":[{"label":"a","text":"What is the power requirement for the compressor?","answer":[{"label":"power (adiabatic)","value":84.52,"units":"kW"}],"grade":"$undefined"},{"label":"b","text":"Cooling water is now supplied around the compressor casing at 0.30 kg/s and 31 C, leaving at 72 C. The\nintake conditions and mass flow rate are unchanged, but the delivery temperature of the air falls to\n109 C. What is the new power requirement of the compressor?\n","answer":[{"label":"heat to the water","value":51.54,"units":"kW"},{"label":"power (cooled)","value":79.81,"units":"kW"}],"grade":"$undefined"}],"solution":"$dc","confidence":"HIGH","solutionNote":"Matches the deck (84.52 kW; 51.53 + 28.27 = 79.8 kW)."},{"id":"engr5004-ex-dyn-rect-01","module":"ENGR5004","week":1,"label":"Dynamics Session 1 Example 1","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["particle-kinematics","rectilinear-motion","differentiation","harmonic-motion"],"source":{"name":"ENGR5004 Dynamics Session 1 worked example 1 (slides 17-19)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 1.pdf"},"text":"A rotating crank drives a slotted yoke so that point P on the horizontal rod moves in a straight line. Its\nposition as a function of time is\n\n$$s = 0.4\\sin(2\\pi t)\\ \\text{m}$$\n\nwith $t$ in seconds and $s$ measured to the right.\n","figures":[{"src":"/figures/engr5004-ex-dyn-rect-01.png","caption":"Scotch-yoke mechanism: a crank rotating about a fixed pin drives a slotted yoke; point P at the end of the horizontal rod moves along a straight line, with position s measured to the right."}],"formulaSheet":"- Rectilinear motion: $v = \\dfrac{ds}{dt}$, $a = \\dfrac{dv}{dt} = \\dfrac{d^2s}{dt^2}$ (learn this)\n","parts":[{"label":"a","text":"Find the velocity and acceleration of P at $t = 0.375$ s.","answer":[{"label":"v","value":-1.78,"units":"m/s"},{"label":"a","value":-11.2,"units":"m/s^2"}],"grade":"$undefined"},{"label":"b","text":"What is the maximum magnitude of the velocity of P?","answer":[{"label":"v_max","value":2.51,"units":"m/s"}],"grade":"$undefined"},{"label":"c","text":"When the magnitude of the velocity of P is a maximum, what is the acceleration of P?","answer":[{"label":"a","value":0,"units":"m/s^2","tol":0.01}],"grade":"$undefined"}],"solution":"$dd","confidence":"HIGH","solutionNote":"Matches the slide answers (-1.78 m/s, -11.2 m/s^2, 2.51 m/s, 0)."},{"id":"engr5004-ex-dyn-rect-02","module":"ENGR5004","week":1,"label":"Dynamics Session 1 Example 2","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["particle-kinematics","rectilinear-motion","variable-acceleration","separation-of-variables"],"source":{"name":"ENGR5004 Dynamics Session 1 worked example 2 (slides 21-22)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 1.pdf"},"text":"After landing, an airplane deploys its drag parachute and travels along a straight runway with an\nacceleration\n\n$$a = -0.004\\,v^2\\ \\text{m/s}^2$$\n\nwhere $v$ is its speed in m/s.\n","formulaSheet":"- $a = \\dfrac{dv}{dt}$, so for $a = a(v)$: $\\displaystyle\\int_{v_0}^{v}\\frac{dv}{a(v)} = \\int_{t_0}^{t} dt$ (learn this)\n- $a\\,ds = v\\,dv$, so for $a = a(v)$: $\\displaystyle\\int_{v_0}^{v}\\frac{v\\,dv}{a(v)} = \\int_{s_0}^{s} ds$ (learn this)\n","parts":[{"label":"a","text":"Determine the time required for the velocity to decrease from 80 m/s to 10 m/s.","answer":[{"label":"t","value":21.9,"units":"s"}],"grade":"$undefined"},{"label":"b","text":"What distance does the plane cover during that time?","answer":[{"label":"s","value":520,"units":"m"}],"grade":"$undefined"}],"solution":"**Set-up.** Acceleration is given as a function of velocity, so separate variables. Take $t = 0$, $s = 0$\nwhen $v = 80$ m/s; positive direction is the direction of travel.\n\n**(a)** $\\dfrac{dv}{dt} = -0.004v^2$:\n\n$$\\int_{80}^{v}\\frac{dv}{v^2} = \\int_0^t -0.004\\,dt \\;\\Rightarrow\\; \\frac{1}{v} - \\frac{1}{80} = 0.004\\,t\n\\;\\Rightarrow\\; t = 250\\left(\\frac{1}{v} - \\frac{1}{80}\\right)$$\n\nAt $v = 10$ m/s: $t = 250(0.1 - 0.0125) = 21.9$ s (21.875 s).\n\n**(b)** Method 1 (slides): invert to get $v(t) = \\dfrac{2000}{8t + 25}$ m/s and integrate,\n\n$$s = \\int_0^{21.875}\\frac{2000}{8t + 25}\\,dt = 250\\ln\\left(\\frac{8(21.875) + 25}{25}\\right) = 250\\ln 8 = 520\\ \\text{m}$$\n\nMethod 2 (check, $v\\,dv = a\\,ds$):\n\n$$\\int_{80}^{10}\\frac{v\\,dv}{-0.004v^2} = s \\;\\Rightarrow\\; s = \\frac{1}{0.004}\\ln\\frac{80}{10} = 519.9\\ \\text{m}$$\n\nBoth give 519.9 m. Units: $0.004$ has units 1/m, so $1/0.004$ is in metres and $\\ln$ is dimensionless.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (t = 21.9 s, s = 519.9 m)."},{"id":"engr5004-ex-dyn-curv-01","module":"ENGR5004","week":1,"label":"Dynamics Session 2 Example 1","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["particle-kinematics","projectile-motion","constant-acceleration","cartesian-coordinates"],"source":{"name":"ENGR5004 Dynamics Session 2 worked example 1 (slides 13-15)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 2.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 2.pdf"},"text":"A projectile is launched at 10 m/s from a point on a plane surface that slopes downwards at $30^\\circ$ to the\nhorizontal. The launch velocity makes an angle $\\alpha = 80^\\circ$ with the downhill part of the slope\nsurface (so it points up and towards the downhill side, see the figure). The projectile lands further down\nthe slope. Assume negligible aerodynamic drag and take $g = 9.81$ m/s$^2$.\n\nFind the time of flight $t$ and the range $R$ measured along the slope.\n","figures":[{"src":"/figures/engr5004-ex-dyn-curv-01.png","caption":"Projectile launched at 10 m/s from a slope inclined at 30 degrees below the horizontal; the launch velocity makes angle alpha with the downhill slope; R is the distance along the slope to the landing point."}],"formulaSheet":"- Constant acceleration: $v = v_0 + a_c t$, $s = s_0 + v_0 t + \\tfrac12 a_c t^2$ (learn this)\n- Projectile (no drag): $a_x = 0$, $a_y = -g$ (learn this)\n","answer":[{"label":"t","value":2.32,"units":"s"},{"label":"R","value":17.2,"units":"m"}],"solution":"$de","confidence":"HIGH","solutionNote":"Matches the slide answers (R = 17.21 m, t = 2.32 s)."},{"id":"engr5004-ex-dyn-curv-02","module":"ENGR5004","week":1,"label":"Dynamics Session 2 Example 2","kind":"tutorial","level":1,"minutes":8,"grade":"third","tags":["particle-kinematics","cartesian-coordinates","variable-acceleration","integration"],"source":{"name":"ENGR5004 Dynamics Session 2 worked example 2 (slide 16)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 2.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR 5004 Dynamics Session 2.pdf"},"text":"A helicopter is instrumented with an $x$-$y$-$z$ accelerometer. During a test flight that starts from rest at\nthe origin at $t = 0$, the acceleration components recorded in the interval $t = 0$ to 10 s are\n\n$$a_x = 0.6t\\ \\text{m/s}^2, \\qquad a_y = 1.8 - 0.36t\\ \\text{m/s}^2, \\qquad a_z = 0$$\n\n($x$ horizontal, $y$ vertical).\n","formulaSheet":"- Cartesian components: $v_x = \\displaystyle\\int a_x\\,dt$, $x = \\displaystyle\\int v_x\\,dt$ (and the same for $y$, $z$) (learn this)\n","parts":[{"label":"a","text":"Find the helicopter's velocity components and position as functions of time.","grade":"$undefined"},{"label":"b","text":"Evaluate the position $(x, y)$ and the speed at $t = 10$ s, and sketch the flight path $y$ against $x$.\n","answer":[{"label":"x (t = 10 s)","value":100,"units":"m"},{"label":"y (t = 10 s)","value":30,"units":"m"},{"label":"speed (t = 10 s)","value":30,"units":"m/s"}],"grade":"$undefined"}],"solution":"**(a)** Starts from rest at the origin, so all constants of integration are zero.\n\n$$v_x = \\int_0^t 0.6t\\,dt = 0.3t^2\\ \\text{m/s}, \\qquad x = \\int_0^t 0.3t^2\\,dt = 0.1t^3\\ \\text{m}$$\n\n$$v_y = \\int_0^t (1.8 - 0.36t)\\,dt = 1.8t - 0.18t^2\\ \\text{m/s}, \\qquad\ny = \\int_0^t (1.8t - 0.18t^2)\\,dt = 0.9t^2 - 0.06t^3\\ \\text{m}$$\n\nand $v_z = 0$, $z = 0$ (planar flight). Check: differentiating $y$ twice returns $1.8 - 0.36t$.\n\n**(b)** At $t = 10$ s: $x = 0.1(1000) = 100$ m, $y = 0.9(100) - 0.06(1000) = 30$ m;\n$v_x = 30$ m/s, $v_y = 18 - 18 = 0$, so the speed is 30 m/s and the helicopter is flying horizontally.\nThe path rises steeply at first ($v_y/v_x = (1.8 - 0.18t)/(0.3t)$ is large for small $t$) and levels off at\n$y = 30$ m when $x = 100$ m ($v_y = 0$ at $t = 10$ s, the highest point of the interval).\n","confidence":"HIGH","solutionNote":"The slide's final line prints $y = 0.92t^2 - 0.06t^3$ m; integrating $1.8t$ gives $0.9t^2$, so the 0.92 is a\ntypo (the slide's own plot ends at y = 30 m, consistent with 0.9). Part (b) values were added here.\n"},{"id":"engr5004-ex-dyn-nt-01","module":"ENGR5004","week":2,"label":"Dynamics Session 3 Worked example 1","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["particle-kinematics","n-t-coordinates","circular-motion","tangential-acceleration","normal-acceleration"],"source":{"name":"ENGR5004 Dynamics Session 3 worked example 1 (slides 16-17)","path":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 3.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 3.pdf"},"text":"A boat travels around a circular path of radius $r = 40$ m at a speed that increases with time,\n$v = 0.0625\\,t^2$ m/s ($t$ in s). The boat starts from rest ($v = 0$ when $t = 0$).\n\nFind the magnitudes of the boat's velocity and acceleration at the instant $t = 10$ s.\n","formulaSheet":"- Normal-tangential components: $\\mathbf a = \\dot v\\,\\mathbf u_t + \\dfrac{v^2}{\\rho}\\mathbf u_n$, $a = \\sqrt{a_t^2 + a_n^2}$ (learn this)\n","answer":[{"label":"v","value":6.25,"units":"m/s"},{"label":"a_t","value":1.25,"units":"m/s^2"},{"label":"a_n","value":0.977,"units":"m/s^2"},{"label":"a","value":1.59,"units":"m/s^2"}],"solution":"**Set-up.** The path is known (a circle, $\\rho = r = 40$ m), so use n-t components. The velocity is always\ntangential, $\\mathbf v = v\\,\\mathbf u_t$.\n\n**Velocity** at $t = 10$ s: $v = 0.0625(10)^2 = 6.25$ m/s.\n\n**Acceleration.** Tangential component (rate of change of speed):\n$a_t = \\dot v = 0.125\\,t = 1.25$ m/s$^2$. Normal component (towards the centre):\n$a_n = v^2/\\rho = 6.25^2/40 = 0.977$ m/s$^2$. Magnitude:\n\n$$a = \\sqrt{1.25^2 + 0.977^2} = 1.59\\ \\text{m/s}^2$$\n\nCheck: units of $v^2/\\rho$ are $(\\text{m/s})^2/\\text{m} = \\text{m/s}^2$; at $t = 0$ both components vanish,\nas they should for a boat starting from rest with $a_t = 0.125t$.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (6.25 m/s, 1.25, 0.98 and 1.59 m/s^2)."},{"id":"engr5004-ex-dyn-nt-02","module":"ENGR5004","week":2,"label":"Dynamics Session 3 Worked example 2","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["particle-kinematics","n-t-coordinates","radius-of-curvature","normal-acceleration"],"source":{"name":"ENGR5004 Dynamics Session 3 worked example 2 (slides 18-21)","path":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 3.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 3.pdf"},"text":"A roller coaster travels along a vertical parabolic path $y = 0.01x^2$ ($x$, $y$ in m, origin at the bottom\nof the dip). At point B, where $x = 30$ m, it is moving down the track towards the origin with a speed of\n25 m/s, which is increasing at the rate of 3 m/s$^2$.\n\nFind the magnitude and direction of the roller coaster's acceleration at B.\n","figures":[{"src":"/figures/engr5004-ex-dyn-nt-02.png","caption":"Roller coaster track y = x^2/100; the car is at B, 30 m from the y axis, moving down towards the origin; the sketch shows a_t along the track (downhill) and a_n towards the centre of curvature."}],"formulaSheet":"- Normal-tangential components: $a_t = \\dot v$, $a_n = \\dfrac{v^2}{\\rho}$, $a = \\sqrt{a_t^2 + a_n^2}$ (learn this)\n- Radius of curvature of $y = f(x)$: $\\rho = \\dfrac{\\left[1 + (dy/dx)^2\\right]^{3/2}}{\\left|d^2y/dx^2\\right|}$ (learn this)\n","answer":[{"label":"rho","value":79.3,"units":"m"},{"label":"a_n","value":7.88,"units":"m/s^2"},{"label":"a","value":8.43,"units":"m/s^2"},{"label":"angle above the -x axis","value":38.2,"units":"deg"}],"solution":"$df","confidence":"HIGH","solutionNote":"Matches the slides (rho = 79.3 m, a_n = 7.88 m/s^2, a = 8.43 m/s^2 at 38.2 deg to the x axis, i.e. 141.8 deg\nfrom +x). The slide statement does not say which way the car moves; the slide's sketch (a_t pointing down the\ntrack) implies it is descending towards the origin, so that is stated in the question here.\n"},{"id":"engr5004-ex-dyn-polar-01","module":"ENGR5004","week":2,"label":"Dynamics Session 4 Worked example 1","kind":"tutorial","level":1,"minutes":5,"grade":"third","tags":["particle-kinematics","polar-coordinates","circular-motion"],"source":{"name":"ENGR5004 Dynamics Session 4 worked example 1 (slides 12-13)","path":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 4.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 4.pdf"},"text":"A car travels along a circular path of radius $r = 100$ m. At the instant considered, its angular position\nmeasured from the centre of the circle is changing at $\\dot\\theta = 0.4$ rad/s with $\\ddot\\theta = 0.2$ rad/s$^2$.\n\nUsing polar coordinates with origin at the centre, find the car's velocity and acceleration (components and\nmagnitudes).\n","formulaSheet":"- Polar velocity: $\\mathbf v = \\dot r\\,\\mathbf u_r + r\\dot\\theta\\,\\mathbf u_\\theta$ (learn this)\n- Polar acceleration: $\\mathbf a = (\\ddot r - r\\dot\\theta^2)\\mathbf u_r + (r\\ddot\\theta + 2\\dot r\\dot\\theta)\\mathbf u_\\theta$ (learn this)\n","answer":[{"label":"v","value":40,"units":"m/s"},{"label":"a_r","value":-16,"units":"m/s^2"},{"label":"a_theta","value":20,"units":"m/s^2"},{"label":"a","value":25.6,"units":"m/s^2"}],"solution":"**Set-up.** Origin at the centre of the circle, so $r = 100$ m is constant: $\\dot r = \\ddot r = 0$.\n\n**Velocity:** $\\mathbf v = \\dot r\\,\\mathbf u_r + r\\dot\\theta\\,\\mathbf u_\\theta = 0\\,\\mathbf u_r + 100(0.4)\\,\\mathbf u_\\theta$,\nso $v = 40$ m/s, entirely in the transverse direction (tangent to the circle).\n\n**Acceleration:**\n\n$$\\mathbf a = (0 - 100 \\times 0.4^2)\\mathbf u_r + (100 \\times 0.2 + 2 \\times 0 \\times 0.4)\\mathbf u_\\theta\n= -16\\,\\mathbf u_r + 20\\,\\mathbf u_\\theta\\ \\text{m/s}^2$$\n\n$$a = \\sqrt{16^2 + 20^2} = 25.6\\ \\text{m/s}^2$$\n\nCheck with n-t components: on a circle $a_n = v^2/\\rho = 40^2/100 = 16$ m/s$^2$ towards the centre ($-\\mathbf u_r$)\nand $a_t = \\dot v = r\\ddot\\theta = 20$ m/s$^2$. Same result.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (v = 40 m/s, a = -16 u_r + 20 u_theta, 25.6 m/s^2)."},{"id":"engr5004-ex-dyn-polar-02","module":"ENGR5004","week":2,"label":"Dynamics Session 4 Worked example 2","kind":"tutorial","level":3,"minutes":18,"grade":"2:1","tags":["particle-kinematics","polar-coordinates","n-t-coordinates","radar-tracking"],"source":{"name":"ENGR5004 Dynamics Session 4 worked example 2 (slides 14-16)","path":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 4.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 2/ENGR 5004 Dynamics Session 4.pdf"},"text":"A plane is flying along a vertical loop of radius 600 m at a constant speed of 400 km/h. When the plane is at\npoint A, its velocity is directed at $\\beta = 30^\\circ$ above the horizontal and the plane is entering the\nloop (the centre of the loop is above and behind the plane, on the concave side of its path).\n\nA radar at B on the ground tracks the plane. At this instant it gives the plane's position as $r = 800$ m and\n$\\theta = 30^\\circ$, where $\\theta$ is measured from the vertical through B towards the side the plane is\nflying to (so $\\mathbf u_r$ points $60^\\circ$ above the horizontal and $\\mathbf u_\\theta$ points $30^\\circ$\nbelow the horizontal, see the figure).\n\nFind the radial and transverse velocity components, and $\\dot\\theta$ and $\\ddot\\theta$ for this instant.\n","figures":[{"src":"/figures/engr5004-ex-dyn-polar-02.png","caption":"Radar at B tracks a plane at A: r = 800 m along a line at theta = 30 degrees from the vertical; the plane's velocity v is at beta = 30 degrees above the horizontal; the theta direction at A points down and to the right."}],"formulaSheet":"- Polar velocity: $v_r = \\dot r$, $v_\\theta = r\\dot\\theta$ (learn this)\n- Polar acceleration: $a_r = \\ddot r - r\\dot\\theta^2$, $a_\\theta = r\\ddot\\theta + 2\\dot r\\dot\\theta$ (learn this)\n- n-t components: $a_t = \\dot v$, $a_n = v^2/\\rho$ (learn this)\n","answer":[{"label":"v_r","value":96.2,"units":"m/s"},{"label":"v_theta","value":55.6,"units":"m/s"},{"label":"theta-dot","value":0.0694,"units":"rad/s"},{"label":"theta-double-dot","value":-0.039,"units":"rad/s^2"}],"solution":"$e0","confidence":"HIGH","solutionNote":"Matches the slide answers (96.2, 55.6 m/s, 0.0694 rad/s, -0.0390 rad/s^2). The slide prints\nv = 400 000/36000; the divisor should be 3600 (the result 111.1 m/s is correct). It also uses 111.2 in one line\n(a = 20.57 m/s^2); with v = 111.1 m/s, a = 20.58 m/s^2. The direction conventions for theta and the loop centre\nwere taken from the slide figures and are stated in the question.\n"},{"id":"engr5004-ex-dyn-relmot-01","module":"ENGR5004","week":3,"label":"Dynamics L5 Worked example 1","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["particle-kinematics","relative-velocity","translating-axes"],"source":{"name":"ENGR216 Dynamics L5 worked example 1 (slides 10-12)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L5 Relative motion analysis of two particles using translating axes (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L5 Relative motion analysis of two particles using translating axes (old spec ENGR216).pdf"},"text":"Two planes fly in the same horizontal plane. Plane A flies due east ($+x$) at $v_A = 650$ km/h. Plane B flies\nat $v_B = 800$ km/h along a line inclined at $60^\\circ$ to the east-west line, heading towards the south-west\n(its velocity points down and to the left in plan, at $60^\\circ$ below the $-x$ direction).\n\nFind the velocity of plane B relative to plane A, $\\mathbf v_{B/A}$ (magnitude and direction).\n","figures":[{"src":"/figures/engr5004-ex-dyn-relmot-01.png","caption":"Plan view: plane A flies to the right at 650 km/h; plane B flies at 800 km/h down and to the left along a line at 60 degrees to the horizontal."}],"formulaSheet":"- Relative velocity (translating axes): $\\mathbf v_B = \\mathbf v_A + \\mathbf v_{B/A}$ (learn this)\n- Law of cosines $c^2 = a^2 + b^2 - 2ab\\cos C$ and law of sines $\\dfrac{a}{\\sin A} = \\dfrac{b}{\\sin B}$ (learn this)\n","answer":[{"label":"v_B/A","value":1258,"units":"km/h"},{"label":"angle below the -x axis","value":33.4,"units":"deg"}],"solution":"**Vector method.** $x$ east, $y$ north.\n\n$$\\mathbf v_A = 650\\,\\mathbf i\\ \\text{km/h}, \\qquad\n\\mathbf v_B = -800\\cos 60^\\circ\\,\\mathbf i - 800\\sin 60^\\circ\\,\\mathbf j = (-400\\,\\mathbf i - 692.8\\,\\mathbf j)\\ \\text{km/h}$$\n\n$$\\mathbf v_{B/A} = \\mathbf v_B - \\mathbf v_A = (-1050\\,\\mathbf i - 692.8\\,\\mathbf j)\\ \\text{km/h}$$\n\n$$v_{B/A} = \\sqrt{1050^2 + 692.8^2} = 1258\\ \\text{km/h}, \\qquad\n\\theta = \\tan^{-1}\\frac{692.8}{1050} = 33.4^\\circ$$\n\nmeasured below the $-x$ axis (B appears to A to move west-south-west).\n\n**Check, graphical method.** Draw $\\mathbf v_A$ and $\\mathbf v_B$ from a common point; the angle between them\nis $120^\\circ$ and $\\mathbf v_{B/A}$ closes the triangle:\n$v_{B/A}^2 = 800^2 + 650^2 - 2(800)(650)\\cos 120^\\circ$ gives 1258 km/h, and the law of sines\n$\\sin\\theta = 800\\sin 120^\\circ/1258$ gives $33.4^\\circ$.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (1258 km/h at 33.4 deg)."},{"id":"engr5004-ex-dyn-relmot-02","module":"ENGR5004","week":3,"label":"Dynamics L5 Worked example 2","kind":"tutorial","level":2,"minutes":15,"grade":"2:2","tags":["particle-kinematics","relative-velocity","relative-acceleration","translating-axes","n-t-coordinates"],"source":{"name":"ENGR216 Dynamics L5 worked example 2 (slides 13-18)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L5 Relative motion analysis of two particles using translating axes (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L5 Relative motion analysis of two particles using translating axes (old spec ENGR216).pdf"},"text":"Car A travels west ($-x$) along a straight road at a steady 30 km/h. Car B is turning out of a junction on a\ncurve of radius 0.3 km. At the instant shown, the line from B to the centre of curvature of its path points\nup and to the right at $30^\\circ$ above the $+x$ axis, and B moves along its path (perpendicular to that line,\nup and to the left) at $v_B = 20$ km/h, with its speed increasing at $(a_t)_B = 1200$ km/h$^2$.\n\nFind the velocity $\\mathbf v_{B/A}$ and acceleration $\\mathbf a_{B/A}$ of car B relative to car A (magnitudes\nand directions). Axes: $x$ to the right, $y$ up the page.\n","figures":[{"src":"/figures/engr5004-ex-dyn-relmot-02.png","caption":"Car A moves left at 30 km/h; car B rounds a corner of radius 0.3 km, moving up and to the left at 20 km/h; the radius line from B to the centre of curvature is at 30 degrees to the horizontal."}],"formulaSheet":"- Relative motion (translating axes): $\\mathbf v_{B/A} = \\mathbf v_B - \\mathbf v_A$, $\\mathbf a_{B/A} = \\mathbf a_B - \\mathbf a_A$ (learn this)\n- n-t components: $a_t = \\dot v$, $a_n = v^2/\\rho$ towards the centre of curvature (learn this)\n","parts":[{"label":"a","text":"Find $\\mathbf v_{B/A}$.","answer":[{"label":"v_B/A","value":26.5,"units":"km/h"},{"label":"angle above +x","value":40.9,"units":"deg"}],"grade":"$undefined"},{"label":"b","text":"Find $\\mathbf a_{B/A}$.","answer":[{"label":"a_B/A","value":1794,"units":"km/h^2"},{"label":"angle above +x","value":72,"units":"deg"}],"grade":"$undefined"}],"solution":"$e1","confidence":"HIGH","solutionNote":"Agrees with the slides apart from typos: slide 17 prints $\\mathbf a_B = 544.7\\,\\mathbf i + 1706\\,\\mathbf j$ then uses\n554.7 (554.7 is correct), and writes $\\sqrt{554.7 + 1706^2} = 1790$; the correct magnitude is 1794 km/h^2\n(the graphical-method slide gives 1793). The slides give 40.8 and 40.9 deg for the velocity direction;\n40.9 deg is correct.\n"},{"id":"engr5004-ex-dyn-kin-01","module":"ENGR5004","week":3,"label":"Dynamics L6 Worked example 1","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["particle-kinetics","free-body-diagram","kinetic-diagram","spring","variable-acceleration"],"source":{"name":"ENGR216 Dynamics L6 worked example 1 (slides 12-14)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L6 Equation of motion for a particle and a system of particles (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L6 Equation of motion for a particle and a system of particles (old spec ENGR216).pdf"},"text":"A 10 kg block on a smooth horizontal surface ($\\mu_k = 0$) is pushed by a constant force $F = 500$ N whose\nline of action slopes downwards towards the block with a 3-4-5 slope (4 horizontal, towards the spring;\n3 vertical, downwards). A spring of stiffness $k = 500$ N/m is mounted between the block and a wall ahead of it.\nWhen $s = 0$ the block is at rest and the spring is uncompressed; $s$ is the block's displacement towards the\nwall. Take $g = 9.81$ m/s$^2$.\n","figures":[{"src":"/figures/engr5004-ex-dyn-kin-01.png","caption":"10 kg block pushed by F = 500 N along a 3-4-5 slope (down and to the right) towards a spring k = 500 N/m attached to a wall; s is the block displacement."}],"formulaSheet":"- Newton's second law: $\\sum F_x = m a_x$, $\\sum F_y = m a_y$ (learn this)\n- Linear spring: $F_s = k s$ (learn this)\n- $a\\,ds = v\\,dv$ when $a = a(s)$ (learn this)\n","parts":[{"label":"a","text":"Draw the free-body and kinetic diagrams of the block and find the normal reaction and $a(s)$.","answer":[{"label":"N","value":398.1,"units":"N"}],"grade":"$undefined"},{"label":"b","text":"Find the velocity of the block when $s = 0.5$ m.","answer":[{"label":"v","value":5.24,"units":"m/s"}],"grade":"$undefined"}],"solution":"$e2","confidence":"HIGH","solutionNote":"Matches the slide answer (v = 5.24 m/s); N was added (the slide writes the y equation but does not evaluate it)."},{"id":"engr5004-ex-dyn-kin-02","module":"ENGR5004","week":3,"label":"Dynamics L6 Worked example 2","kind":"tutorial","level":2,"minutes":8,"grade":"2:2","tags":["particle-kinetics","system-of-particles","connected-bodies","free-body-diagram","variable-force"],"source":{"name":"ENGR216 Dynamics L6 worked example 2 (slides 15-17)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L6 Equation of motion for a particle and a system of particles (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Dynamics L6 Equation of motion for a particle and a system of particles (old spec ENGR216).pdf"},"text":"A baggage truck A (360 kg) pulls cart B (220 kg) and cart C (130 kg) in a straight line on level ground. For a\nshort time, the driving friction force developed at the wheels of truck A is $F_A = 160t$ N ($t$ in s). Rolling\nresistance of the carts is negligible (treat the ground contact of B and C as smooth). The truck starts from rest.\n","parts":[{"label":"a","text":"Determine the speed of the train at $t = 2$ s.","answer":[{"label":"v","value":0.451,"units":"m/s"}],"grade":"$undefined"},{"label":"b","text":"What is the horizontal force in the coupling between truck A and cart B at $t = 2$ s?","answer":[{"label":"T","value":158,"units":"N"}],"grade":"$undefined"}],"formulaSheet":"- Equation of motion for a system of particles: $\\sum \\mathbf F_{ext} = m_{tot}\\mathbf a_G$ (internal forces cancel) (learn this)\n- $a = dv/dt$, so $v = \\int a\\,dt$ when $a = a(t)$ (learn this)\n","solution":"**(a)** Treat A, B and C as one system moving with a common acceleration $a$ (inextensible couplings). The\ncoupling forces are internal and cancel; the only external horizontal force is $F_A$:\n\n$$\\sum F_x = m a:\\quad 160t = (360 + 220 + 130)a \\;\\Rightarrow\\; a = \\frac{160t}{710} = 0.2254t\\ \\text{m/s}^2$$\n\n$$\\int_0^v dv = \\int_0^2 0.2254t\\,dt \\;\\Rightarrow\\; v = 0.1127(2)^2 = 0.451\\ \\text{m/s}$$\n\n**(b)** FBD of the truck alone at $t = 2$ s: driving force $160(2) = 320$ N forward, coupling force $T$ backward:\n\n$$320 - T = 360(0.2254 \\times 2) \\;\\Rightarrow\\; T = 320 - 162.3 = 158\\ \\text{N}$$\n\nCheck: FBD of carts B and C together, the only horizontal force is $T$ forward:\n$T = (220 + 130)(0.4507) = 158$ N. Same answer.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (v = 0.45 m/s, T = 158 N); the slide rounds a to 0.225t."},{"id":"engr5004-ex-dyn-kin-03","module":"ENGR5004","week":4,"label":"Dynamics L7 Worked example 1","kind":"tutorial","level":2,"minutes":8,"grade":"2:2","tags":["particle-kinetics","free-body-diagram","kinetic-diagram","cable-tension","reactions","cantilever"],"source":{"name":"ENGR216 Dynamics L7 worked example 1 (slides 9-10)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L7 Particle kinetics Rectangular coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L7 Particle kinetics Rectangular coordinate frames of reference (old spec ENGR216).pdf"},"text":"A uniform horizontal beam AB of length 2 m and mass $m_{AB} = 100$ kg is built into a wall at A (fixed support).\nA crate of mass $m_C = 15$ kg is lifted by a motor-driven cable: the cable runs vertically up from the crate to a\nsmall pulley at the free end B of the beam, then horizontally (at the level of the beam axis) to a motor on a\npier further to the right. The crate is accelerating upwards at 2 m/s$^2$. Neglect the size and mass of the\npulley; take $g = 9.81$ m/s$^2$.\n\nFind the cable tension and the reactions at the fixed support A.\n","figures":[{"src":"/figures/engr5004-ex-dyn-kin-03.png","caption":"Cantilever AB (2 m) fixed to a wall at A; the cable from the crate rises vertically to a pulley at B and runs horizontally to a motor on a brick pier to the right; the crate accelerates up at 2 m/s^2."}],"formulaSheet":"- Newton's second law for the crate: $\\sum F_y = m a_y$ (learn this)\n- Static equilibrium of the beam: $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M_A = 0$; a fixed support gives two forces and a moment (learn this)\n","answer":[{"label":"T","value":177,"units":"N"},{"label":"A_x (to the left)","value":177,"units":"N"},{"label":"A_y (up)","value":1158,"units":"N"},{"label":"M_A (anticlockwise)","value":1335,"units":"N m"}],"solution":"**Crate (dynamics).** FBD: tension $T$ up, weight $m_C g$ down. Kinetic diagram: $m_C a$ up.\n\n$$+\\uparrow\\ \\sum F_y = m_C a:\\quad T - 15(9.81) = 15(2) \\;\\Rightarrow\\; T = 15(11.81) = 177\\ \\text{N}$$\n\n**Beam (statics: the beam does not move).** The pulley at B carries both cable runs, so the beam feels $T$\ndownwards and $T$ horizontally to the right at B (both acting through B, on the beam axis). Weight\n$m_{AB}g = 981$ N acts at the mid-point (1 m from A). Fixed support: $A_x$, $A_y$, $M_A$.\n\n$$\\sum F_x = 0:\\quad A_x = T = 177\\ \\text{N}\\ (\\leftarrow)$$\n\n$$\\sum F_y = 0:\\quad A_y = 981 + 177 = 1158\\ \\text{N}\\ (\\uparrow)$$\n\n$$\\sum M_A = 0:\\quad M_A = 981(1) + 177(2) = 1335\\ \\text{N m (anticlockwise on the beam)}$$\n\nThe horizontal $T$ acts along the beam axis through A, so it has no moment about A.\n\nCheck: if the crate were at rest ($a = 0$), $T = 147$ N and $M_A = 1275$ N m; accelerating the crate upwards\nmust increase all three reactions, as found.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (T = 177 N, A_x = 177 N, A_y = 1158 N, M_A = 1335 N m). The slide's moment line\nwrites \"M_A - 9.81 - 354 = 0\"; the first term should be 981 N m (1 m x 100 kg x 9.81), which is what the\nslide's answer uses.\n"},{"id":"engr5004-ex-dyn-kin-04","module":"ENGR5004","week":4,"label":"Dynamics L7 Worked example 2","kind":"tutorial","level":3,"minutes":15,"grade":"2:1","tags":["particle-kinetics","dependent-motion","pulley","friction","connected-bodies","free-body-diagram"],"source":{"name":"ENGR216 Dynamics L7 worked example 2 (slides 11-13)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L7 Particle kinetics Rectangular coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L7 Particle kinetics Rectangular coordinate frames of reference (old spec ENGR216).pdf"},"text":"Block A ($m_A = 10$ kg) rests on a horizontal surface with kinetic friction coefficient $\\mu_k = 0.2$. A cord\nattached to A runs horizontally to a small pulley at the edge of the surface, down to a movable pulley from\nwhich cylinder B ($m_B = 20$ kg) hangs, and back up to a fixed hook on the ceiling (so B is supported by two\ncord segments). The cord is inextensible and the pulleys are frictionless and massless. At the instant\nconsidered, A is moving to the right (towards the edge) at 2 m/s and B is moving down.\n\nFind the speed of A when it has moved 4 m to the right. Take $g = 9.81$ m/s$^2$.\n","figures":[{"src":"/figures/engr5004-ex-dyn-kin-04.png","caption":"Block A on a rough horizontal surface; the cord runs from A over a pulley at the edge, down around a movable pulley carrying cylinder B, and up to a ceiling hook. Axes: +X right, +Y down."}],"formulaSheet":"- Newton's second law: $\\sum F = m a$ for each body (learn this)\n- Kinetic friction: $F = \\mu_k N$, opposing the sliding (learn this)\n- Dependent motion: write the cord length in position coordinates and differentiate (learn this)\n- Constant acceleration: $v^2 = v_0^2 + 2a_c(s - s_0)$ (learn this)\n","answer":[{"label":"T","value":71.9,"units":"N"},{"label":"a_A","value":5.23,"units":"m/s^2"},{"label":"v_A","value":6.77,"units":"m/s"}],"solution":"$e3","confidence":"HIGH","solutionNote":"The slide's T = 71.94 N, a_A = 5.232 m/s^2 and a_B = 2.616 m/s^2 are correct, but its final line\n$v_A^2 = 2^2 + 2(5.23)(4)$ gives $v_A = 6.77$ m/s, not the 6.15 m/s printed. Confirmed independently by\nwork-energy (6.77 m/s).\n"},{"id":"engr5004-ex-dyn-nt-kin-01","module":"ENGR5004","week":4,"label":"Dynamics L8 Worked example 1","kind":"tutorial","level":3,"minutes":12,"grade":"2:2","tags":["particle-kinetics","n-t-coordinates","free-body-diagram","kinetic-diagram","rotating-arm"],"source":{"name":"ENGR216 Dynamics L8 worked example 1 (slides 10-11)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf"},"text":"A fairground ride consists of an arm BC of length 10 m pinned to the ground at C; a seat at the end B is pin\nconnected to the arm so that it stays level. The arm swings upwards (anticlockwise), so the passenger at B moves\non a circle of radius 10 m about C, towards A at the top. At the instant when the arm is at $\\theta = 45^\\circ$\nabove the horizontal, the 75 kg passenger has a speed of 6 m/s, which is increasing at 0.5 m/s$^2$. Neglect the\npassenger's size and the mass of the seat; take $g = 9.81$ m/s$^2$.\n\nFind the horizontal and vertical components of the force of the seat on the passenger.\n","figures":[{"src":"/figures/engr5004-ex-dyn-nt-kin-01.png","caption":"Arm CB of length 10 m pinned at C, inclined at theta to the horizontal; the passenger at B moves along the circular arc towards A."}],"formulaSheet":"- Equations of motion in n-t components: $\\sum F_n = m\\dfrac{v^2}{\\rho}$, $\\sum F_t = m\\dot v$ (learn this)\n","answer":[{"label":"R_x (+ to the right)","value":-217,"units":"N"},{"label":"R_y (+ up)","value":571,"units":"N"}],"solution":"$e4","confidence":"HIGH","solutionNote":"Matches the slide answers (R_x = -217 N, R_y = 571 N)."},{"id":"engr5004-ex-dyn-nt-kin-02","module":"ENGR5004","week":4,"label":"Dynamics L8 Worked example 2","kind":"tutorial","level":1,"minutes":5,"grade":"third","tags":["particle-kinetics","n-t-coordinates","circular-motion","cable-tension"],"source":{"name":"ENGR216 Dynamics L8 worked example 2 (slides 12-13)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf"},"text":"A 2 kg block B slides on a smooth horizontal table. It is connected by a light cord, which passes through a\nhole at the centre of the table, to a 15 kg cylinder A hanging at rest below the table. Block B moves on a\ncircle about the hole at a constant speed of 10 m/s. Find the radius $r$ of the circular path. Take\n$g = 9.81$ m/s$^2$.\n","figures":[{"src":"/figures/engr5004-ex-dyn-nt-kin-02.png","caption":"Block B moves in a circle of radius r on a smooth table; the cord runs through the central hole to cylinder A hanging underneath."}],"formulaSheet":"- n-t equation of motion: $\\sum F_n = m\\dfrac{v^2}{\\rho}$ (learn this)\n","answer":[{"label":"r","value":1.36,"units":"m"}],"solution":"**Cylinder A** hangs at rest (its height does not change while B moves on a fixed circle), so\n$T = m_A g = 15(9.81) = 147.2$ N.\n\n**Block B.** FBD: weight and normal force (vertical, they cancel), cord tension $T$ towards the hole; no\nfriction (smooth table). With $+n$ towards the centre and constant speed ($a_t = 0$):\n\n$$\\sum F_n = m_B\\frac{v^2}{\\rho}:\\quad 147.2 = 2\\frac{10^2}{r} \\;\\Rightarrow\\; r = \\frac{200}{147.2} = 1.36\\ \\text{m}$$\n\nCheck: units $\\text{kg}\\,(\\text{m/s})^2/\\text{N} = \\text{m}$. If B went faster the required tension would exceed\n$m_Ag$ and B would spiral outwards (A rises), so 1.36 m is the only steady radius at 10 m/s.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (T = 147 N, r = 1.36 m)."},{"id":"engr5004-ex-dyn-nt-kin-03","module":"ENGR5004","week":4,"label":"Dynamics L8 Worked example 3","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["particle-kinetics","n-t-coordinates","banked-track","friction","normal-force"],"source":{"name":"ENGR216 Dynamics L8 worked example 3 (slides 14-15)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf"},"text":"A car of mass 1700 kg travels on a circular track of radius of curvature 100 m that is banked at\n$\\theta = 20^\\circ$ (the track rises away from the centre of curvature). The coefficient of static friction\nbetween tyres and road is $\\mu_s = 0.2$. Treating the car as a particle, find the maximum constant speed at which\nit can travel without slipping up the slope. Take $g = 9.81$ m/s$^2$.\n","figures":[{"src":"/figures/engr5004-ex-dyn-nt-kin-03.png","caption":"Cross-section of the banked track: the road surface is inclined at theta to the horizontal, rising away from the centre of curvature at distance rho."}],"formulaSheet":"- n-t equations of motion: $\\sum F_n = m\\dfrac{v^2}{\\rho}$, $\\sum F_b = 0$ (learn this)\n- Friction at impending slip: $F = \\mu_s N$, opposing the impending motion (learn this)\n","answer":[{"label":"N","value":19140,"units":"N"},{"label":"v_max","value":24.4,"units":"m/s"}],"solution":"**Set-up.** $n$ horizontal towards the centre of curvature, $b$ vertical, $t$ along the track. At the maximum\nspeed the car is on the point of sliding up the slope, so friction $F = \\mu_s N$ acts down the slope. The\nnormal force $N$ is perpendicular to the road, inclined at $\\theta$ to the vertical towards the centre.\n\n$$\\sum F_b = 0:\\quad N\\cos\\theta - F\\sin\\theta - mg = 0 \\;\\Rightarrow\\; N = \\frac{mg}{\\cos 20^\\circ - 0.2\\sin 20^\\circ} = 19\\,140\\ \\text{N}$$\n\n$$\\sum F_n = m\\frac{v^2}{\\rho}:\\quad N\\sin\\theta + F\\cos\\theta = 1700\\frac{v^2}{100}$$\n\n$$19\\,140(\\sin 20^\\circ + 0.2\\cos 20^\\circ) = 10\\,144\\ \\text{N} = 17v^2 \\;\\Rightarrow\\; v_{max} = 24.4\\ \\text{m/s}$$\n\nCheck (closed form): $v_{max}^2 = g\\rho\\,\\dfrac{\\sin\\theta + \\mu_s\\cos\\theta}{\\cos\\theta - \\mu_s\\sin\\theta}$ gives\n24.4 m/s (88 km/h); with $\\mu_s = 0$ it reduces to the ideal banking speed $\\sqrt{g\\rho\\tan\\theta} = 18.9$ m/s,\nwhich is lower, as it should be.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (N = 19141 N, v_max = 24.4 m/s)."},{"id":"engr5004-ex-dyn-nt-kin-04","module":"ENGR5004","week":4,"label":"Dynamics L8 Worked example 4","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["particle-kinetics","n-t-coordinates","normal-force","circular-motion"],"source":{"name":"ENGR216 Dynamics L8 worked example 4 (slides 16-17)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Dynamics L8 Particle kineticsn-t coordinate frames of reference (old spec ENGR216).pdf"},"text":"A tilting vehicle travels at a constant speed of 80 km/h around a horizontal circular track of radius 100 m.\nWhat tilt angle $\\theta$ of the vehicle (and seat) from the vertical will make the driver experience only a\nnormal reaction from the seat (no sideways friction)? Neglect the size of the car; take $g = 9.81$ m/s$^2$.\n","figures":[{"src":"/figures/engr5004-ex-dyn-nt-kin-04.png","caption":"Front view of a tilting three-wheeled vehicle; the body and driver lean by theta from the vertical towards the inside of the turn."}],"formulaSheet":"- n-t equations of motion: $\\sum F_n = m\\dfrac{v^2}{\\rho}$, $\\sum F_b = 0$ (learn this)\n","answer":[{"label":"a_n","value":4.94,"units":"m/s^2"},{"label":"theta","value":26.7,"units":"deg"}],"solution":"**Set-up.** Driver as a particle; $n$ horizontal towards the centre of the track, $b$ vertical. The only seat\nforce is the normal reaction $N$, perpendicular to the tilted seat, i.e. at $\\theta$ to the vertical, leaning\ntowards the centre.\n\n$v = 80/3.6 = 22.2$ m/s, $a_n = v^2/\\rho = 22.2^2/100 = 4.94$ m/s$^2$; constant speed so $a_t = 0$.\n\n$$\\sum F_b = 0:\\quad N\\cos\\theta = mg, \\qquad \\sum F_n = ma_n:\\quad N\\sin\\theta = m a_n$$\n\nDividing, $\\tan\\theta = a_n/g = 4.94/9.81 = 0.503$, so $\\theta = 26.7^\\circ$, independent of the driver's mass.\n\nCheck: $N = mg/\\cos 26.7^\\circ = 1.12\\,mg$; the driver feels 12 % heavier, pressed straight into the seat.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (a_n = 4.94 m/s^2, theta = 26.7 deg)."},{"id":"engr5004-ex-dyn-polar-kin-01","module":"ENGR5004","week":5,"label":"Dynamics L9 Worked example 1","kind":"tutorial","level":3,"minutes":18,"grade":"2:1","tags":["particle-kinetics","polar-coordinates","chain-rule","normal-force","rotating-arm"],"source":{"name":"ENGR216 Dynamics L9 worked example 1 (slides 10-12)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Dynamics L9 Particle kinetics Cylindrical (Polar) coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Dynamics L9 Particle kinetics Cylindrical (Polar) coordinate frames of reference (old spec ENGR216).pdf"},"text":"A ball P of mass 0.5 kg rests on the smooth curved surface of a semicircular block (vertical plane, radius\n$r_c = 0.4$ m). It is pushed along the surface by a smooth arm OA that rotates anticlockwise about the point O\nat the left-hand end of the block's flat base (O lies on the circle). The arm angle $\\theta$ is measured from the\nhorizontal base, so the ball's distance from O is $r = 2r_c\\cos\\theta$. At the instant $\\theta = 30^\\circ$,\n$\\dot\\theta = 0.4$ rad/s and $\\ddot\\theta = 0.8$ rad/s$^2$. Take $g = 9.81$ m/s$^2$.\n\nFind the force of the arm OA on the ball (and the normal force of the curved surface) at this instant.\n","figures":[{"src":"/figures/engr5004-ex-dyn-polar-kin-01.png","caption":"Semicircular block of radius r_c; arm OA pivoted at O (left end of the base) at angle theta; ball P sits on the curved surface against the arm, at distance r from O."}],"formulaSheet":"- Polar acceleration: $a_r = \\ddot r - r\\dot\\theta^2$, $a_\\theta = r\\ddot\\theta + 2\\dot r\\dot\\theta$ (learn this)\n- Equations of motion: $\\sum F_r = m a_r$, $\\sum F_\\theta = m a_\\theta$ (learn this)\n- Chain rule for $r = f(\\theta)$: $\\dot r = f'(\\theta)\\dot\\theta$, $\\ddot r = f''(\\theta)\\dot\\theta^2 + f'(\\theta)\\ddot\\theta$ (learn this)\n","answer":[{"label":"a_r","value":-0.542,"units":"m/s^2"},{"label":"a_theta","value":0.426,"units":"m/s^2"},{"label":"N_s (surface)","value":2.52,"units":"N"},{"label":"N_OA (arm)","value":3.2,"units":"N"}],"solution":"$e5","confidence":"HIGH","solutionNote":"Matches the slide answers (a_r = -0.542, a_theta = 0.426 m/s^2, N_s = 2.52 N, N_OA = 3.20 N)."},{"id":"engr5004-ex-dyn-polar-kin-02","module":"ENGR5004","week":5,"label":"Dynamics L9 Worked example 2","kind":"tutorial","level":3,"minutes":18,"grade":"2:1","tags":["particle-kinetics","polar-coordinates","chain-rule","spring","slotted-arm"],"source":{"name":"ENGR216 Dynamics L9 worked example 2 (slides 13-15)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Dynamics L9 Particle kinetics Cylindrical (Polar) coordinate frames of reference (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Dynamics L9 Particle kinetics Cylindrical (Polar) coordinate frames of reference (old spec ENGR216).pdf"},"text":"In a horizontal plane (neglect gravity), a smooth particle P of mass 0.08 kg is attached to an elastic cord from\nthe fixed point O to P (stiffness $k = 30$ N/m, unstretched length 0.25 m). A slotted arm rotating about O pushes\nP around the rim of a fixed circular disc of radius 0.4 m that passes through O, so that $r = 0.8\\sin\\theta$ m,\nwhere $\\theta$ is the arm angle. The arm rotates at a constant $\\dot\\theta = 5$ rad/s.\n\nFind the forces that the slotted arm and the disc rim exert on the particle when $\\theta = 60^\\circ$.\n","figures":[{"src":"/figures/engr5004-ex-dyn-polar-kin-02.png","caption":"Disc of radius 0.4 m with O on its rim; slotted arm through O at angle theta, rotating anticlockwise at 5 rad/s; particle P in the slot on the disc rim at distance r from O."}],"formulaSheet":"- Polar acceleration: $a_r = \\ddot r - r\\dot\\theta^2$, $a_\\theta = r\\ddot\\theta + 2\\dot r\\dot\\theta$ (learn this)\n- Equations of motion: $\\sum F_r = m a_r$, $\\sum F_\\theta = m a_\\theta$ (learn this)\n- Spring force: $F_s = k(\\ell - \\ell_0)$ (learn this)\n","answer":[{"label":"a_r","value":-34.6,"units":"m/s^2"},{"label":"a_theta","value":20,"units":"m/s^2"},{"label":"F_s","value":13.3,"units":"N"},{"label":"N (disc rim)","value":12.1,"units":"N"},{"label":"F (slotted arm)","value":7.67,"units":"N"}],"solution":"$e6","confidence":"HIGH","solutionNote":"Matches the slide answers (N = 12.1 N, F = 7.67 N). The slide writes the stretched length as 0.6938 m in the\nspring-force line; r = 0.6928 m (its result 13.28 N uses the correct value). The slide also labels the spring\nforce N_s in the r equation.\n"},{"id":"engr5004-ex-dyn-rot-01","module":"ENGR5004","week":5,"label":"Dynamics L10 Worked example","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["rigid-kinematics","fixed-axis-rotation","variable-acceleration","integration"],"source":{"name":"ENGR216 Dynamics L10 worked example (slides 18-19)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Dynamics L10 Planar rigid body motion Translation & Rotation (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Dynamics L10 Planar rigid body motion Translation & Rotation (old spec ENGR216).pdf"},"text":"Just after a fan is switched on, the motor gives the blade an angular acceleration $\\alpha = 20e^{-0.6t}$ rad/s$^2$,\nwhere $t$ is in seconds. At $t = 0$ the blade is at rest. The tip P of a blade is at $r = 0.5$ m from the axis.\n\nFind the speed and the magnitude of the acceleration of P at $t = 3$ s, and the number of revolutions the blade\nhas turned in the first 3 s.\n","figures":[{"src":"/figures/engr5004-ex-dyn-rot-01.png","caption":"Box fan; the blade tip P is 0.5 m from the axis of rotation."}],"formulaSheet":"- Angular motion: $\\omega = d\\theta/dt$, $\\alpha = d\\omega/dt$ (learn this)\n- Point on a body rotating about a fixed axis: $v = \\omega r$, $a_t = \\alpha r$, $a_n = \\omega^2 r$ (learn this)\n","answer":[{"label":"omega","value":27.8,"units":"rad/s"},{"label":"v_P","value":13.9,"units":"m/s"},{"label":"a_P","value":387,"units":"m/s^2"},{"label":"revolutions","value":8.54,"units":"rev"}],"solution":"**Angular velocity.** $\\alpha$ is a function of time, so integrate with $\\omega(0) = 0$:\n\n$$\\omega = \\int_0^t 20e^{-0.6t}\\,dt = 33.33\\left(1 - e^{-0.6t}\\right)\\ \\text{rad/s}$$\n\nAt $t = 3$ s: $\\omega = 33.33(1 - e^{-1.8}) = 27.82$ rad/s.\n\n**Velocity of P:** $v_P = \\omega r = 27.82(0.5) = 13.9$ m/s.\n\n**Acceleration of P.** $\\alpha(3) = 20e^{-1.8} = 3.306$ rad/s$^2$, so $a_t = \\alpha r = 1.65$ m/s$^2$ and\n$a_n = \\omega^2 r = 27.82^2(0.5) = 387.1$ m/s$^2$; $a_P = \\sqrt{1.65^2 + 387.1^2} = 387$ m/s$^2$ (almost entirely\ncentripetal).\n\n**Angle turned** ($\\theta(0) = 0$):\n\n$$\\theta = \\int_0^3 33.33\\left(1 - e^{-0.6t}\\right)dt = \\left[33.33t + 55.56e^{-0.6t}\\right]_0^3 = 100.0 - 46.4 = 53.6\\ \\text{rad}$$\n\ni.e. $53.6/2\\pi = 8.54$ revolutions.\n\nCheck: numerical integration in Python gives $\\omega(3) = 27.823$ rad/s and $\\theta(3) = 53.628$ rad. Limiting\ncase: as $t \\to \\infty$, $\\omega \\to 33.3$ rad/s, the fan's final speed.\n","confidence":"HIGH","solutionNote":"Matches the slide values (omega = 27.82 rad/s, v_P = 13.91 m/s, theta = 53.6 rad). The slide does not finish the\nacceleration of P or the number of revolutions; those were completed here (387 m/s^2, 8.54 rev). The slide's\ntheta(t) plot shows the antiderivative without subtracting its value at t = 0 (it starts at 55.6 rather than 0).\n"},{"id":"engr5004-ex-dyn-relvel-01","module":"ENGR5004","week":6,"label":"Dynamics L11 Worked example 1","kind":"tutorial","level":1,"minutes":8,"grade":"third","tags":["rigid-kinematics","relative-velocity","sliding-bar","general-plane-motion"],"source":{"name":"ENGR216 Dynamics L11 worked example 1 (slides 12-13)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L11 Rigid body relative motion analysis Velocity (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L11 Rigid body relative motion analysis Velocity (old spec ENGR216).pdf"},"text":"A bar AB of length 1.5 m has rollers at both ends: roller A runs on the horizontal floor and roller B runs on a\nvertical wall to the left of A. At the instant shown the bar makes $\\theta = 30^\\circ$ with the floor and roller A\nis moving to the right at 3 m/s.\n\nFind the angular velocity of the bar and the velocity of B.\n","figures":[{"src":"/figures/engr5004-ex-dyn-relvel-01.png","caption":"Bar AB of length 1.5 m; roller B on the vertical wall (upper left), roller A on the floor moving right at 3 m/s; the bar is at 30 degrees to the floor."}],"formulaSheet":"- Relative velocity on a rigid body: $\\mathbf v_B = \\mathbf v_A + \\boldsymbol\\omega \\times \\mathbf r_{B/A}$ (learn this)\n- $\\omega\\mathbf k \\times (x\\mathbf i + y\\mathbf j) = -\\omega y\\,\\mathbf i + \\omega x\\,\\mathbf j$ (learn this)\n","answer":[{"label":"omega (anticlockwise)","value":4,"units":"rad/s"},{"label":"v_B (down)","value":5.2,"units":"m/s"}],"solution":"**Set-up.** $x$ right, $y$ up, $\\boldsymbol\\omega = \\omega\\mathbf k$ (anticlockwise positive). Roller A moves\nhorizontally, roller B vertically: $\\mathbf v_A = 3\\mathbf i$ m/s, $\\mathbf v_B = -v_B\\mathbf j$ (assumed down).\n$\\mathbf r_{B/A} = -1.5\\cos 30^\\circ\\,\\mathbf i + 1.5\\sin 30^\\circ\\,\\mathbf j = (-1.299\\,\\mathbf i + 0.75\\,\\mathbf j)$ m.\n\n$$-v_B\\mathbf j = 3\\mathbf i + \\omega\\mathbf k \\times (-1.299\\,\\mathbf i + 0.75\\,\\mathbf j) = (3 - 0.75\\omega)\\mathbf i - 1.299\\omega\\,\\mathbf j$$\n\n$$\\mathbf i:\\ 0 = 3 - 0.75\\omega \\Rightarrow \\omega = 4\\ \\text{rad/s (anticlockwise)}; \\qquad\n\\mathbf j:\\ -v_B = -1.299(4) \\Rightarrow v_B = 5.20\\ \\text{m/s (down)}$$\n\n**Check (instantaneous centre).** The IC lies where the normals to $\\mathbf v_A$ and $\\mathbf v_B$ meet: vertically\nabove A and level with B, so $r_{A/IC} = 1.5\\sin 30^\\circ = 0.75$ m and $r_{B/IC} = 1.5\\cos 30^\\circ = 1.299$ m. Then\n$\\omega = 3/0.75 = 4$ rad/s and $v_B = 4(1.299) = 5.20$ m/s. Same answers.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (omega = 4 rad/s k, v_B = -5.2 j m/s)."},{"id":"engr5004-ex-dyn-relvel-02","module":"ENGR5004","week":6,"label":"Dynamics L11 Worked example 2","kind":"tutorial","level":2,"minutes":8,"grade":"2:2","tags":["rigid-kinematics","relative-velocity","slider-crank","general-plane-motion"],"source":{"name":"ENGR216 Dynamics L11 worked example 2 (slides 14-15)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L11 Rigid body relative motion analysis Velocity (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L11 Rigid body relative motion analysis Velocity (old spec ENGR216).pdf"},"text":"Crank OA (length 0.3 m) rotates anticlockwise about the fixed pin O at 12 rad/s. At the instant shown OA is\nvertical with A above O. Rod AB (length 0.6 m) connects A to a piston B that slides in a vertical guide; B is\nlevel with O, to the right, and AB makes $30^\\circ$ with the horizontal.\n\nFind the velocity of piston B and the angular velocity of rod AB.\n","figures":[{"src":"/figures/engr5004-ex-dyn-relvel-02.png","caption":"Vertical crank OA (0.3 m) rotating anticlockwise at 12 rad/s; rod AB (0.6 m) at 30 degrees to the horizontal down to piston B in a vertical guide, level with O."}],"formulaSheet":"- Fixed-axis rotation: $\\mathbf v_A = \\boldsymbol\\omega_{OA} \\times \\mathbf r_{A/O}$ (learn this)\n- Relative velocity: $\\mathbf v_B = \\mathbf v_A + \\boldsymbol\\omega_{AB} \\times \\mathbf r_{B/A}$ (learn this)\n","answer":[{"label":"omega_AB (anticlockwise)","value":12,"units":"rad/s"},{"label":"v_B (up)","value":6.24,"units":"m/s"}],"solution":"**Crank.** $\\mathbf v_A = 12\\mathbf k \\times 0.3\\mathbf j = -3.6\\,\\mathbf i$ m/s (A moves left).\n\n**Rod AB.** $\\mathbf r_{B/A} = 0.6\\cos 30^\\circ\\,\\mathbf i - 0.6\\sin 30^\\circ\\,\\mathbf j = (0.520\\,\\mathbf i - 0.3\\,\\mathbf j)$ m and\nB moves vertically, $\\mathbf v_B = v_B\\mathbf j$:\n\n$$v_B\\mathbf j = -3.6\\,\\mathbf i + \\omega_{AB}\\mathbf k \\times (0.520\\,\\mathbf i - 0.3\\,\\mathbf j)\n= (-3.6 + 0.3\\omega_{AB})\\mathbf i + 0.520\\omega_{AB}\\,\\mathbf j$$\n\n$$\\mathbf i:\\ \\omega_{AB} = 12\\ \\text{rad/s (anticlockwise)}; \\qquad \\mathbf j:\\ v_B = 0.520(12) = 6.24\\ \\text{m/s (up)}$$\n\n**Check (IC of AB).** Normal to $\\mathbf v_A$ (horizontal) is the vertical line through A; normal to $\\mathbf v_B$\n(vertical) is the horizontal line through B. They meet directly below A at the level of B, which is the pin O.\n$r_{A/IC} = 0.3$ m gives $\\omega_{AB} = 3.6/0.3 = 12$ rad/s, and $r_{B/IC} = 0.6\\cos 30^\\circ = 0.520$ m\ngives $v_B = 6.24$ m/s. Same answers.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (omega_AB = 12 rad/s, v_B = 6.24 m/s)."},{"id":"engr5004-ex-dyn-relvel-03","module":"ENGR5004","week":6,"label":"Dynamics L11 Worked example 3","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["rigid-kinematics","relative-velocity","gears","planetary-gears"],"source":{"name":"ENGR216 Dynamics L11 worked example 3 (slides 16-18)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L11 Rigid body relative motion analysis Velocity (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L11 Rigid body relative motion analysis Velocity (old spec ENGR216).pdf"},"text":"In a planetary gear train, the sun gear S (radius 80 mm) is held fixed. Planet gears P (radius 40 mm) mesh with\nS and with the internal ring gear R (pitch radius 160 mm). The planets are carried on pins of an arm attached to\nshaft A, which rotates about the common central axis. The ring gear rotates anticlockwise at $\\omega_R = 3$ rad/s.\n\nFind the angular velocity of each planet gear and of shaft A.\n","figures":[{"src":"/figures/engr5004-ex-dyn-relvel-03.png","caption":"Planetary gear set: fixed sun gear S (80 mm radius), planet gears P (40 mm radius) on an arm fixed to shaft A, and outer ring gear R rotating at omega_R."}],"formulaSheet":"- Relative velocity: $\\mathbf v_B = \\mathbf v_A + \\boldsymbol\\omega \\times \\mathbf r_{B/A}$ (learn this)\n- Meshing gears: the contact points have the same velocity (no slip) (learn this)\n","answer":[{"label":"omega_P","value":6,"units":"rad/s"},{"label":"omega_A","value":2,"units":"rad/s"}],"solution":"$e7","confidence":"HIGH","solutionNote":"Matches the slide answers (omega_P = 6 rad/s, omega_A = 2 rad/s anticlockwise). The slide line\n\"omega_A = -240/120 = -2 rad/s = 2k rad/s\" mixes a speed sign with the vector; the shaft turns anticlockwise\nat 2 rad/s, as the slide concludes.\n"},{"id":"engr5004-ex-dyn-ic-01","module":"ENGR5004","week":6,"label":"Dynamics L12 Worked example 1","kind":"tutorial","level":2,"minutes":8,"grade":"2:2","tags":["rigid-kinematics","instantaneous-centre","linkage","slider"],"source":{"name":"ENGR216 Dynamics L12 worked example 1 (slides 11-13)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L12 Planar rigid body motion Instantaneous centreof zero velocity (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L12 Planar rigid body motion Instantaneous centreof zero velocity (old spec ENGR216).pdf"},"text":"Link AB (0.4 m) is pinned to the ground at A and to link BD (0.4 m) at B; D is pinned to a block that slides\non a horizontal surface level with A. At the instant shown AB rises at $45^\\circ$ from A, BD falls at $45^\\circ$\nto D (the angle ABD is $90^\\circ$), and block D moves to the right at $v_D = 3$ m/s.\n\nFind the angular velocities of links AB and BD.\n","figures":[{"src":"/figures/engr5004-ex-dyn-ic-01.png","caption":"Links AB and BD, each 0.4 m, at 45 degrees to the horizontal with a right angle at B; A is a ground pin, D is pinned to a block moving right at 3 m/s."}],"formulaSheet":"- Instantaneous centre of zero velocity: intersection of the normals to two known velocity directions; $v = \\omega\\,r_{/IC}$ (learn this)\n- Fixed-axis rotation: $v_B = \\omega_{AB}\\,r_{B/A}$, perpendicular to AB (learn this)\n","answer":[{"label":"omega_BD (anticlockwise)","value":5.3,"units":"rad/s"},{"label":"omega_AB (clockwise)","value":5.3,"units":"rad/s"}],"solution":"$e8","confidence":"HIGH","solutionNote":"Matches the slide answers (omega_BD = 5.3 rad/s ccw, omega_AB = 5.3 rad/s cw)."},{"id":"engr5004-ex-dyn-ic-02","module":"ENGR5004","week":6,"label":"Dynamics L12 Worked example 2","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["rigid-kinematics","instantaneous-centre","rolling-wheel"],"source":{"name":"ENGR216 Dynamics L12 worked example 2 (slides 14-16)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L12 Planar rigid body motion Instantaneous centreof zero velocity (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L12 Planar rigid body motion Instantaneous centreof zero velocity (old spec ENGR216).pdf"},"text":"A wheel of outer radius 0.8 m has a hub of radius 0.3 m. The wheel rolls on its hub, without slipping, along a\nhorizontal rail (the outer rim overhangs the rail), and its centre C moves to the right at $v_C = 2$ m/s. Point A\nis on the hub at the same height as C, on the right-hand side; point B is at the top of the outer rim.\n\nFind the velocities of A and B at this instant.\n","figures":[{"src":"/figures/engr5004-ex-dyn-ic-02.png","caption":"Wheel (outer radius 0.8 m) rolling on its 0.3 m hub on a rail; C moves right at 2 m/s; A on the hub level with C, B at the top of the rim. A plan view shows the hub running in a slot."}],"formulaSheet":"- Rolling without slipping: the contact point is the IC, $v_C = \\omega r$ (learn this)\n- $\\mathbf v_P = \\boldsymbol\\omega \\times \\mathbf r_{P/IC}$ (learn this)\n","answer":[{"label":"omega (clockwise)","value":6.67,"units":"rad/s"},{"label":"v_A","value":2.83,"units":"m/s"},{"label":"v_B","value":7.33,"units":"m/s"}],"solution":"**IC.** The hub rolls without slipping, so its contact point with the rail has zero velocity: the IC is 0.3 m\ndirectly below C.\n\n$$\\omega = \\frac{v_C}{r_{C/IC}} = \\frac{2}{0.3} = 6.67\\ \\text{rad/s (clockwise)}, \\qquad \\boldsymbol\\omega = -6.67\\,\\mathbf k$$\n\n**Point A:** $\\mathbf r_{A/IC} = (0.3\\,\\mathbf i + 0.3\\,\\mathbf j)$ m, so\n\n$$\\mathbf v_A = -6.67\\mathbf k \\times (0.3\\,\\mathbf i + 0.3\\,\\mathbf j) = (2\\,\\mathbf i - 2\\,\\mathbf j)\\ \\text{m/s}, \\qquad v_A = 2.83\\ \\text{m/s}$$\n\ndirected at $45^\\circ$ below the horizontal (perpendicular to the line IC-A).\n\n**Point B:** $\\mathbf r_{B/IC} = (0.3 + 0.8)\\mathbf j = 1.1\\mathbf j$ m, so $\\mathbf v_B = -6.67\\mathbf k \\times 1.1\\mathbf j = 7.33\\,\\mathbf i$ m/s.\n\nCheck: $v_A = \\omega r_{A/IC} = 6.67(0.424) = 2.83$ m/s; relative velocity from C,\n$\\mathbf v_B = 2\\mathbf i + (-6.67\\mathbf k) \\times 0.8\\mathbf j = 7.33\\,\\mathbf i$ m/s.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (v_A = 2.83 m/s, v_B = 7.34 m/s from rounding 6.67 x 1.1; exact 7.33 m/s)."},{"id":"engr5004-ex-dyn-ic-03","module":"ENGR5004","week":6,"label":"Dynamics L12 Worked example 3","kind":"tutorial","level":3,"minutes":15,"grade":"2:1","tags":["rigid-kinematics","instantaneous-centre","four-bar-linkage","linkage"],"source":{"name":"ENGR216 Dynamics L12 worked example 3 (slides 17-20)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L12 Planar rigid body motion Instantaneous centreof zero velocity (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Dynamics L12 Planar rigid body motion Instantaneous centreof zero velocity (old spec ENGR216).pdf"},"text":"In a four-bar linkage, link CD (0.6 m) is vertical with D pinned to the ground and C above it; link BC (0.6 m)\nis horizontal with B to the left of C and E its midpoint; link AB runs from the ground pin A (level with D) up\nto B at $30^\\circ$ to the horizontal. CD rotates anticlockwise at $\\omega_{CD} = 6$ rad/s.\n\nFind the velocity of point E (magnitude and direction) and the angular velocity of link AB.\n","figures":[{"src":"/figures/engr5004-ex-dyn-ic-03.png","caption":"Four-bar linkage: AB at 30 degrees from ground pin A up to B; BC horizontal, 0.6 m, with midpoint E; CD vertical, 0.6 m, down to ground pin D; omega_CD = 6 rad/s anticlockwise."}],"formulaSheet":"- Instantaneous centre of zero velocity: intersection of the normals to $\\mathbf v_B$ and $\\mathbf v_C$; $v = \\omega\\,r_{/IC}$ (learn this)\n- Fixed-axis rotation: $v = \\omega r$, perpendicular to the link (learn this)\n","answer":[{"label":"omega_BC (clockwise)","value":10.39,"units":"rad/s"},{"label":"v_E","value":4.76,"units":"m/s"},{"label":"v_E angle above -x","value":40.9,"units":"deg"},{"label":"omega_AB (anticlockwise)","value":6,"units":"rad/s"}],"solution":"$e9","confidence":"HIGH","solutionNote":"Matches the slide answers (omega_BC = 10.39 rad/s cw, v_E = 4.76 m/s at 40.9 deg, omega_AB = 6 rad/s ccw)."},{"id":"engr5004-ex-dyn-relacc-01","module":"ENGR5004","week":7,"label":"Dynamics L13 Worked example 1","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["rigid-kinematics","relative-acceleration","instantaneous-centre","sliding-bar"],"source":{"name":"ENGR216 Dynamics L13 worked example 1 (slides 12-13)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L13 Planar rigid body relative motion analysis - acceleration (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L13 Planar rigid body relative motion analysis - acceleration (old spec ENGR216).pdf"},"text":"A 5 m rod AB has rollers at both ends: A runs down a vertical wall and B runs along the horizontal floor to the\nright of the wall. At the instant shown A is 4 m above the floor (so B is 3 m from the wall), and A is moving down\nat $v_A = 6$ m/s with a downward acceleration $a_A = 5$ m/s$^2$.\n\nFind the angular velocity and angular acceleration of the rod and the acceleration of B at this instant.\n","figures":[{"src":"/figures/engr5004-ex-dyn-relacc-01.png","caption":"Rod AB (5 m): roller A on the vertical wall, 4 m above roller B on the floor; A moves down with v_A = 6 m/s and a_A = 5 m/s^2."}],"formulaSheet":"- Instantaneous centre: $\\omega = v_A/r_{A/IC}$ (learn this)\n- Relative acceleration: $\\mathbf a_B = \\mathbf a_A + \\boldsymbol\\alpha \\times \\mathbf r_{B/A} - \\omega^2\\mathbf r_{B/A}$ (learn this)\n","answer":[{"label":"omega","value":2,"units":"rad/s"},{"label":"alpha (clockwise)","value":3.67,"units":"rad/s^2"},{"label":"a_B (to the left)","value":26.7,"units":"m/s^2"}],"solution":"$ea","confidence":"HIGH","solutionNote":"The accelerations match the slide (alpha = -3.67 rad/s^2, i.e. clockwise; a_B = 26.7 m/s^2 to the left). The\nslide labels omega = 2 rad/s as clockwise; with A sliding down the wall and B moving right, the rod turns\nanticlockwise (omega = +2k rad/s: v_B = v_A + omega x r_B/A gives +8i m/s only with the anticlockwise sense).\nThe omega^2 term does not depend on the sense, so the slide's accelerations are unaffected.\n"},{"id":"engr5004-ex-dyn-relacc-02","module":"ENGR5004","week":7,"label":"Dynamics L13 Worked example 2","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["rigid-kinematics","relative-acceleration","rolling-wheel","gears"],"source":{"name":"ENGR216 Dynamics L13 worked example 2 (slides 17-18)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L13 Planar rigid body relative motion analysis - acceleration (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L13 Planar rigid body relative motion analysis - acceleration (old spec ENGR216).pdf"},"text":"A gear of pitch radius 0.3 m rolls without slipping on a fixed horizontal rack. At the instant shown it has a\nclockwise angular velocity $\\omega = 12$ rad/s and a clockwise angular acceleration $\\alpha = 6$ rad/s$^2$. Point A is\nat the top of the gear, directly above the centre O.\n\nFind the acceleration of point A.\n","figures":[{"src":"/figures/engr5004-ex-dyn-relacc-02.png","caption":"Gear of radius 0.3 m rolling on a fixed rack, clockwise omega = 12 rad/s and alpha = 6 rad/s^2; A is the top point; a_O is horizontal."}],"formulaSheet":"- Rolling without slipping: $a_O = \\alpha r$ (centre moves on a straight line) (learn this)\n- Relative acceleration: $\\mathbf a_A = \\mathbf a_O + \\boldsymbol\\alpha \\times \\mathbf r_{A/O} - \\omega^2\\mathbf r_{A/O}$ (learn this)\n","answer":[{"label":"a_A x","value":3.6,"units":"m/s^2"},{"label":"a_A y","value":-43.2,"units":"m/s^2"},{"label":"a_A magnitude","value":43.3,"units":"m/s^2"}],"solution":"**Centre O.** Rolling without slip on a fixed rack: O moves horizontally with $a_O = \\alpha r = 6(0.3) = 1.8$ m/s$^2$,\nto the right (clockwise rolling moves the gear to the right).\n\n**Point A.** $\\boldsymbol\\omega = -12\\mathbf k$, $\\boldsymbol\\alpha = -6\\mathbf k$, $\\mathbf r_{A/O} = 0.3\\mathbf j$ m:\n\n$$\\mathbf a_A = 1.8\\mathbf i + (-6\\mathbf k) \\times 0.3\\mathbf j - 12^2(0.3\\mathbf j) = 1.8\\mathbf i + 1.8\\mathbf i - 43.2\\mathbf j\n= (3.6\\,\\mathbf i - 43.2\\,\\mathbf j)\\ \\text{m/s}^2$$\n\n$|\\mathbf a_A| = \\sqrt{3.6^2 + 43.2^2} = 43.3$ m/s$^2$, almost straight down (towards O): the centripetal part\n$\\omega^2 r$ dominates.\n\nCheck: the tangential part at the top is $a_O + \\alpha r = 2\\alpha r = 3.6$ m/s$^2$, consistent with the top of a rolling\nwheel moving at twice the centre's speed. (For comparison the contact point has $\\mathbf a = \\omega^2 r\\,\\mathbf j = 43.2\\mathbf j$ m/s$^2$,\nnot zero.)\n","confidence":"HIGH","solutionNote":"Matches the slide answer (a_A = 3.6 i - 43.2 j m/s^2); the magnitude was added."},{"id":"engr5004-ex-dyn-relacc-03","module":"ENGR5004","week":7,"label":"Dynamics L13 Worked example 3","kind":"tutorial","level":3,"minutes":18,"grade":"2:1","tags":["rigid-kinematics","relative-velocity","relative-acceleration","slider-crank","linkage"],"source":{"name":"ENGR216 Dynamics L13 worked example 3 (slides 19-22)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L13 Planar rigid body relative motion analysis - acceleration (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L13 Planar rigid body relative motion analysis - acceleration (old spec ENGR216).pdf"},"text":"Crank AB (0.7 m) is pinned to the ground at A and is vertical at the instant shown, with B above A. It rotates\nanticlockwise with $\\omega_{AB} = 3$ rad/s and $\\alpha_{AB} = 2$ rad/s$^2$ (anticlockwise). Link BC connects B to a\nslider block C that moves in a straight guide inclined on a 3-4-5 slope (rising to the right: 4 horizontal,\n3 vertical). At this instant C is 0.5 m to the left of A and 0.5 m below A.\n\nFind the velocity and acceleration of the slider block C (and the angular velocity and angular acceleration of BC).\n","figures":[{"src":"/figures/engr5004-ex-dyn-relacc-03.png","caption":"Vertical crank AB (0.7 m) about ground pin A, omega = 3 rad/s and alpha = 2 rad/s^2 anticlockwise; link BC down to slider C, 0.5 m left of and 0.5 m below A, in a guide on a 3-4-5 slope."}],"formulaSheet":"- Fixed-axis rotation: $v = \\omega r$, $a_t = \\alpha r$, $a_n = \\omega^2 r$ (learn this)\n- $\\mathbf v_C = \\mathbf v_B + \\boldsymbol\\omega_{BC} \\times \\mathbf r_{C/B}$, $\\mathbf a_C = \\mathbf a_B + \\boldsymbol\\alpha_{BC} \\times \\mathbf r_{C/B} - \\omega_{BC}^2\\mathbf r_{C/B}$ (learn this)\n","answer":[{"label":"omega_BC (anticlockwise)","value":1.125,"units":"rad/s"},{"label":"v_C (down the slope)","value":0.938,"units":"m/s"},{"label":"alpha_BC (clockwise)","value":3,"units":"rad/s^2"},{"label":"a_C (down the slope)","value":5.47,"units":"m/s^2"}],"solution":"$eb","confidence":"HIGH","solutionNote":"Matches the slide answers (omega_BC = 1.125 rad/s, v_C = 0.9375 m/s, alpha_BC = -3.0 rad/s^2, a_C = 5.47 m/s^2)."},{"id":"engr5004-ex-dyn-rbkin-01","module":"ENGR5004","week":7,"label":"Dynamics L14 Worked example 1","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["rigid-kinetics","translation","tipping","friction","free-body-diagram"],"source":{"name":"ENGR216 Dynamics L14 worked example 1 (slides 13-14)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L14 Mass moment of inertia and planar kinetics (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L14 Mass moment of inertia and planar kinetics (old spec ENGR216).pdf"},"text":"A uniform 50 kg crate, 1 m wide and 1 m high, rests on a horizontal surface with kinetic friction coefficient\n$\\mu_k = 0.2$. A horizontal force $P = 600$ N is applied to its left face at a height of 0.8 m above the floor.\nTake $g = 9.81$ m/s$^2$.\n\nFind the acceleration of the crate, and show whether it slides or tips.\n","figures":[{"src":"/figures/engr5004-ex-dyn-rbkin-01.png","caption":"1 m by 1 m crate on a rough floor; horizontal force P = 600 N applied to the left face 0.8 m above the floor."}],"formulaSheet":"- Rigid-body translation: $\\sum F_x = m(a_G)_x$, $\\sum F_y = m(a_G)_y$, $\\sum M_G = 0$ (learn this)\n- Kinetic friction $F = \\mu_k N$; tipping when the resultant normal force reaches the edge of the base (learn this)\n","answer":[{"label":"a_G","value":10,"units":"m/s^2"},{"label":"x (N_C from G)","value":0.467,"units":"m"}],"solution":"**Assumption.** The crate slides without tipping (pure translation, $\\alpha = 0$). Then the resultant normal force\n$N_C$ acts at an unknown distance $x$ to the right of the line through G, and friction $F = 0.2N_C$ acts to the left\nalong the floor. G is at the centre, 0.5 m above the floor; P acts 0.3 m above G.\n\n$$\\sum F_x = m a_G:\\quad 600 - 0.2N_C = 50a_G$$\n\n$$\\sum F_y = 0:\\quad N_C - 50(9.81) = 0 \\;\\Rightarrow\\; N_C = 490.5\\ \\text{N}$$\n\n$$\\sum M_G = 0\\ (\\text{anticlockwise}+):\\quad -600(0.3) + N_C x - 0.2N_C(0.5) = 0 \\;\\Rightarrow\\; x = 0.467\\ \\text{m}$$\n\n$$a_G = \\frac{600 - 98.1}{50} = 10.0\\ \\text{m/s}^2$$\n\n**Check the assumption.** $x = 0.467$ m $< 0.5$ m (half the base width), so the normal force stays within the base\nand the crate slides without tipping, as assumed. If $x$ had exceeded 0.5 m the problem would be re-solved assuming\ntipping about the front corner. Check: moments about the floor point under G give the same $x$, since\n$\\sum M_O = (m a_G)(0.5)$.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (N_C = 490 N, x = 0.467 m, a_G = 10.0 m/s^2)."},{"id":"engr5004-ex-dyn-rbkin-02","module":"ENGR5004","week":7,"label":"Dynamics L14 Worked example 2","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["rigid-kinetics","translation","normal-reactions","free-body-diagram"],"source":{"name":"ENGR216 Dynamics L14 worked example 2 (slides 15-16)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L14 Mass moment of inertia and planar kinetics (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Dynamics L14 Mass moment of inertia and planar kinetics (old spec ENGR216).pdf"},"text":"A 200 kg cart has its centre of mass G 0.2 m above the ground. Its wheels A (rear, left) and B (front) touch the\nground 0.3 m to the left and 0.2 m to the right of G respectively. A force $P = 50$ N is applied to the end of the\nhandle, which is 0.6 m to the right of G and 0.5 m above the ground; P is directed downwards and to the left at\n$60^\\circ$ to the horizontal, pushing the cart to the left. Neglect the mass of the wheels and rolling resistance;\ntake $g = 9.81$ m/s$^2$.\n\nFind the acceleration of the cart and the normal reactions at wheels A and B.\n","figures":[{"src":"/figures/engr5004-ex-dyn-rbkin-02.png","caption":"Cart with G 0.2 m above the ground; wheel A 0.3 m left of G, wheel B 0.2 m right of G; handle end 0.4 m beyond B and 0.5 m above the ground, where P = 50 N pushes down and to the left."}],"formulaSheet":"- Rigid-body translation: $\\sum F_x = m(a_G)_x$, $\\sum F_y = m(a_G)_y$, $\\sum M_G = 0$ (learn this)\n","answer":[{"label":"a_G (to the left)","value":0.125,"units":"m/s^2"},{"label":"N_A","value":765,"units":"N"},{"label":"N_B","value":1240,"units":"N"}],"solution":"**Set-up.** The cart translates ($\\alpha = 0$). Free wheels with negligible mass give vertical reactions only.\nComponents of P: $50\\cos 60^\\circ = 25$ N to the left, $50\\sin 60^\\circ = 43.3$ N downwards. Weight 1962 N at G.\n\n$$\\stackrel{+}{\\leftarrow}\\ \\sum F_x = m a_G:\\quad 25 = 200a_G \\;\\Rightarrow\\; a_G = 0.125\\ \\text{m/s}^2\\ (\\leftarrow)$$\n\n$$\\sum F_y = 0:\\quad N_A + N_B - 1962 - 43.3 = 0 \\;\\Rightarrow\\; N_A + N_B = 2005\\ \\text{N} \\qquad (1)$$\n\nMoments about G (anticlockwise +; the horizontal component acts 0.5 - 0.2 = 0.3 m above G, the vertical\ncomponent 0.6 m to the right of G):\n\n$$\\sum M_G = 0:\\quad -0.3N_A + 0.2N_B + 0.3(25) - 0.6(43.3) = 0 \\;\\Rightarrow\\; -0.3N_A + 0.2N_B = 18.48 \\qquad (2)$$\n\nSolving (1) and (2): $N_A = 765$ N, $N_B = 1240$ N.\n\n**Check (moments about A).** $\\sum M_A = (m a_G)\\,d$: $0.5N_B - 1962(0.3) - 43.3(0.9) + 25(0.5) = 5.0$ N m, and\n$m a_G \\times 0.2 = 200(0.125)(0.2) = 5.0$ N m. Consistent. The front wheel B carries more load because P pushes\ndown near it.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (a_G = 0.125 m/s^2, N_A = 765 N, N_B = 1240 N)."},{"id":"engr5004-ex-dyn-fixax-01","module":"ENGR5004","week":8,"label":"Dynamics L15 Worked example 1","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["rigid-kinetics","fixed-axis-rotation","moment-of-inertia","parallel-axis-theorem","free-body-diagram","kinetic-diagram"],"source":{"name":"ENGR216 Dynamics L15 worked example 1 (slides 10-13)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Dynamics L15 Kinetics and the equations of motion rigid body rotation about a fixed axis (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Dynamics L15 Kinetics and the equations of motion rigid body rotation about a fixed axis (old spec ENGR216).pdf"},"text":"A uniform slender horizontal rod of mass 15 kg and length 0.9 m is supported by a pin at O, 0.6 m from its left\nend (0.3 m from its right end), and by a vertical cord attached to its left end. The cord is cut. Take\n$g = 9.81$ m/s$^2$ and $I_G = ml^2/12$.\n\nFind the angular acceleration of the rod and the reactions at the pin O just after the cord is cut.\n","figures":[{"src":"/figures/engr5004-ex-dyn-fixax-01.png","caption":"Horizontal rod, 0.6 m to the left of pin O and 0.3 m to the right; G is 0.15 m left of O; a vertical cord at the left end is cut."}],"formulaSheet":"- Fixed-axis rotation: $\\sum F_n = m\\omega^2 r_G$, $\\sum F_t = m\\alpha r_G$, $\\sum M_O = I_O\\alpha$ (learn this)\n- Parallel-axis theorem: $I_O = I_G + m r_G^2$; slender rod $I_G = \\tfrac{1}{12}ml^2$ (learn this)\n","answer":[{"label":"alpha","value":16.4,"units":"rad/s^2"},{"label":"O_x","value":0,"units":"N","tol":0.01},{"label":"O_y (up)","value":110,"units":"N"}],"solution":"**Set-up.** G is at the rod's mid-point, 0.45 m from the left end, so $r_G = 0.6 - 0.45 = 0.15$ m to the left of O.\nJust after the cut the rod is still at rest: $\\omega = 0$, but $\\alpha \\neq 0$. G moves on a circle about O, so\n$a_n = \\omega^2 r_G = 0$ (along the rod, towards O) and $a_t = \\alpha r_G$ (vertically down).\n\n**Moment of inertia:** $I_O = \\tfrac{1}{12}(15)(0.9)^2 + 15(0.15)^2 = 1.0125 + 0.3375 = 1.35$ kg m$^2$.\n\n**Equations of motion** (FBD: weight 147.15 N at G, pin components $O_x$, $O_y$):\n\n$$\\sum F_n = m\\omega^2 r_G = 0 \\;\\Rightarrow\\; O_x = 0$$\n\n$$\\sum M_O = I_O\\alpha:\\quad 147.15(0.15) = 1.35\\alpha \\;\\Rightarrow\\; \\alpha = \\frac{22.07}{1.35} = 16.4\\ \\text{rad/s}^2$$\n\n(anticlockwise as drawn: the heavier left side drops).\n\n$$\\sum F_t = m\\alpha r_G:\\quad -O_y + 147.15 = 15(0.15)(16.35) \\;\\Rightarrow\\; O_y = 147.15 - 36.8 = 110\\ \\text{N (up)}$$\n\nCheck: the pin carries less than the weight (147 N) because G is accelerating downwards at $\\alpha r_G = 2.45$ m/s$^2$.\n","confidence":"HIGH","solutionNote":"Matches the slide answers (alpha = 16.4 rad/s^2, O_x = 0, O_y = 110 N)."},{"id":"engr5004-ex-dyn-fixax-02","module":"ENGR5004","week":8,"label":"Dynamics L15 Worked example 2","kind":"tutorial","level":3,"minutes":12,"grade":"2:2","tags":["rigid-kinetics","fixed-axis-rotation","moment-of-inertia","parallel-axis-theorem","composite-body"],"source":{"name":"ENGR216 Dynamics L15 worked example 2 (slides 14-16)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Dynamics L15 Kinetics and the equations of motion rigid body rotation about a fixed axis (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Dynamics L15 Kinetics and the equations of motion rigid body rotation about a fixed axis (old spec ENGR216).pdf"},"text":"A 30 kg solid sphere of radius 0.1 m is fixed to the end of a 10 kg uniform horizontal rod of length 0.2 m, whose\nother end is pinned to a wall at O (so the sphere's centre is 0.3 m from O). The assembly is held horizontal by a\nvertical cord AB attached to the top of the sphere. The cord is cut. Take $g = 9.81$ m/s$^2$,\n$I_G = \\tfrac25 mr^2$ for the sphere and $\\tfrac{1}{12}ml^2$ for the rod.\n\nFind the angular acceleration and the reaction at the pin O just after the cord is cut.\n","figures":[{"src":"/figures/engr5004-ex-dyn-fixax-02.png","caption":"Sphere of radius 0.1 m on the end of a 0.2 m horizontal rod pinned to a wall at O; cord AB from the top of the sphere to the ceiling is cut."}],"formulaSheet":"- Fixed-axis rotation: $\\sum F_n = m\\omega^2 r_G$, $\\sum F_t = m\\alpha r_G$, $\\sum M_O = I_O\\alpha$ (learn this)\n- Parallel-axis theorem $I_O = I_G + md^2$; sphere $I_G = \\tfrac25 mr^2$, slender rod $I_G = \\tfrac{1}{12}ml^2$ (learn this)\n","answer":[{"label":"I_O","value":2.95,"units":"kg m^2"},{"label":"alpha","value":33.2,"units":"rad/s^2"},{"label":"O_y (up)","value":60.2,"units":"N"}],"solution":"**Set-up.** Treat sphere and rod as one rigid body rotating about O. Just after the cut $\\omega = 0$, so there are\nno normal (centripetal) accelerations: $O_x = 0$.\n\n**Moment of inertia about O:**\n\n$$I_O = \\left[\\tfrac25(30)(0.1)^2 + 30(0.3)^2\\right] + \\left[\\tfrac{1}{12}(10)(0.2)^2 + 10(0.1)^2\\right] = 2.82 + 0.1333 = 2.953\\ \\text{kg m}^2$$\n\n**Moments about O:**\n\n$$\\sum M_O = I_O\\alpha:\\quad 30(9.81)(0.3) + 10(9.81)(0.1) = 98.1 = 2.953\\alpha \\;\\Rightarrow\\; \\alpha = 33.2\\ \\text{rad/s}^2$$\n\n**Vertical force** (each centre accelerates down at $\\alpha r$):\n\n$$-O_y + 40(9.81) = 30(0.3\\alpha) + 10(0.1\\alpha) = 10\\alpha \\;\\Rightarrow\\; O_y = 392.4 - 332.2 = 60.2\\ \\text{N (up)}$$\n\nCheck: the combined centre of mass is at $r_G = (30 \\times 0.3 + 10 \\times 0.1)/40 = 0.25$ m, and\n$392.4 - 40(0.25)(33.22) = 60.2$ N. The pin force is small because the body is almost in free fall at the\nsphere end.\n","confidence":"HIGH","solutionNote":"The slide rounds I_O to 2.95 kg m^2 and alpha to 33.3 rad/s^2, giving O_y = 59 N. Unrounded, I_O = 2.953 kg m^2,\nalpha = 33.2 rad/s^2 and O_y = 60.2 N (392.4 - 10 x 33.22); 60 N is the better value.\n"},{"id":"engr5004-ex-dyn-gpm-01","module":"ENGR5004","week":8,"label":"Dynamics L16 Worked example 1","kind":"tutorial","level":3,"minutes":15,"grade":"2:1","tags":["rigid-kinetics","general-plane-motion","radius-of-gyration","friction","free-body-diagram","kinetic-diagram"],"source":{"name":"ENGR216 Dynamics L16 worked example 1 (slides 9-10)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Dynamics L16 Kineticsof a rigid body moving in general plane motion (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Dynamics L16 Kineticsof a rigid body moving in general plane motion (old spec ENGR216).pdf"},"text":"A motorised spool of mass 200 kg and radius of gyration $k_G = 0.3$ m about its centre G stands on the ground on its\nouter rims (radius 0.6 m). A cable is wound round its hub (radius 0.4 m); it leaves the top of the hub at A and runs\nhorizontally to a fixed anchor B on a wall to the right. A motor applies an anticlockwise moment $M = 450$ N m to the\nspool. The coefficient of kinetic friction between the spool and the ground is $\\mu_k = 0.1$. Take $g = 9.81$ m/s$^2$.\n\nFind the angular acceleration of the spool and the tension in the cable.\n","figures":[{"src":"/figures/engr5004-ex-dyn-gpm-01.png","caption":"Spool with 0.6 m outer radius and 0.4 m hub on the ground; cable from the top of the hub (A) horizontally to anchor B on the right-hand wall; motor moment M = 450 N m anticlockwise."}],"formulaSheet":"- General plane motion: $\\sum F_x = m(a_G)_x$, $\\sum F_y = m(a_G)_y$, $\\sum M_G = I_G\\alpha$ (learn this)\n- Radius of gyration: $I_G = m k_G^2$ (learn this)\n- Kinetic friction: $F = \\mu_k N$, opposing the slip (learn this)\n","answer":[{"label":"alpha","value":5.08,"units":"rad/s^2"},{"label":"T","value":602,"units":"N"}],"solution":"$ec","confidence":"HIGH","solutionNote":"Matches the slide answers (alpha = 5.1 rad/s^2, T = 602 N)."},{"id":"engr5004-ex-dyn-we-01","module":"ENGR5004","week":9,"label":"Dynamics L17 Worked example 1","kind":"tutorial","level":3,"minutes":12,"grade":"2:2","tags":["work-energy","rigid-kinetics","rolling-wheel","spring","radius-of-gyration"],"source":{"name":"ENGR216 Dynamics L17 worked example 1 (slides 11-13)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Dynamics L17 Kinetic energy, work and the principle of work and energy (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Dynamics L17 Kinetic energy, work and the principle of work and energy (old spec ENGR216).pdf"},"text":"A 40 kg disc of radius 0.8 m and radius of gyration $k_G = 0.6$ m stands on a horizontal floor. A cord wrapped\naround its rim leaves the top of the disc (point A) horizontally and is connected to a spring of stiffness\n$k = 10$ N/m anchored to a wall on the left. A constant clockwise couple $M = 15$ N m is applied, so the disc rolls\nto the right, away from the wall, without slipping. It starts from rest with the spring unstretched.\n\nFind the angular velocity of the disc when its centre G has moved 0.5 m.\n","figures":[{"src":"/figures/engr5004-ex-dyn-we-01.png","caption":"Disc of radius 0.8 m rolling on the floor; a cord from the top point A runs left to a spring k = 10 N/m anchored to the wall; clockwise couple 15 N m."}],"formulaSheet":"- Principle of work and energy: $T_1 + \\sum U_{1-2} = T_2$ (learn this)\n- Kinetic energy in plane motion: $T = \\tfrac12 m v_G^2 + \\tfrac12 I_G\\omega^2$, $I_G = m k_G^2$ (learn this)\n- Work of a couple $U = M\\Delta\\theta$; work of a spring $U = -\\tfrac12 k(s_2^2 - s_1^2)$ (learn this)\n- Rolling without slipping: $v_G = \\omega r$, $\\Delta s_G = r\\Delta\\theta$ (learn this)\n","answer":[{"label":"theta turned","value":0.625,"units":"rad"},{"label":"U_1-2","value":4.375,"units":"J"},{"label":"omega","value":0.468,"units":"rad/s"}],"solution":"$ed","confidence":"HIGH","solutionNote":"Disagrees with the slide. The slide takes the rotation as arctan(0.5/0.8) = 0.559 rad; for rolling without\nslipping the angle turned is s/r = 0.5/0.8 = 0.625 rad, so U = 4.375 J (slide 3.37 J). The slide's final step also\nreports omega^2 as omega: sqrt(3.37/20.0) = 0.41 rad/s, not 0.17 rad/s. With the correct work, omega = 0.468 rad/s.\nThe figure shows the spring joined by a cord to the top of the disc; the 1 m stretch used on the slide (twice the\ntravel of G) assumes the cord is wrapped around the rim, which is stated in the question here.\n"},{"id":"engr5004-ex-dyn-we-02","module":"ENGR5004","week":9,"label":"Dynamics L17 Worked example 2","kind":"tutorial","level":2,"minutes":8,"grade":"third","tags":["work-energy","rigid-kinetics","fixed-axis-rotation","radius-of-gyration","pendulum"],"source":{"name":"ENGR216 Dynamics L17 worked example 2 (slides 14-15)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Dynamics L17 Kinetic energy, work and the principle of work and energy (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Dynamics L17 Kinetic energy, work and the principle of work and energy (old spec ENGR216).pdf"},"text":"The 50 kg pendulum of a Charpy impact machine rotates about a fixed pin A. Its centre of mass G is 1.25 m from A,\nand its radius of gyration about A is $k_A = 1.75$ m. It is released from rest with $\\theta = 0$, where $\\theta$ is\nthe angle of AG below the horizontal. Neglect friction; take $g = 9.81$ m/s$^2$.\n\nFind the angular velocity of the pendulum when $\\theta = 90^\\circ$ (AG vertical, G directly below A).\n","figures":[{"src":"/figures/engr5004-ex-dyn-we-02.png","caption":"Charpy impact machine: pendulum pivoted at A, G 1.25 m from A, theta measured downwards from the horizontal; the specimen S sits below A."}],"formulaSheet":"- Principle of work and energy: $T_1 + \\sum U_{1-2} = T_2$ (learn this)\n- Fixed-axis rotation: $T = \\tfrac12 I_A\\omega^2$, $I_A = m k_A^2$ (learn this)\n- Work of the weight: $U = mg\\,\\Delta y$ for a drop $\\Delta y$ of G (learn this)\n","answer":[{"label":"U_1-2","value":613,"units":"J"},{"label":"I_A","value":153.1,"units":"kg m^2"},{"label":"omega","value":2.83,"units":"rad/s"}],"solution":"**Work.** Only the weight does work (the pin reaction acts at the fixed point A). G drops by\n$\\Delta y = 1.25\\sin\\theta$, which is 1.25 m at $\\theta = 90^\\circ$:\n\n$$U_{1-2} = mg\\Delta y = 50(9.81)(1.25) = 613.1\\ \\text{J}$$\n\n**Kinetic energy.** Rotation about the fixed axis A: $I_A = mk_A^2 = 50(1.75)^2 = 153.1$ kg m$^2$ and\n$T_2 = \\tfrac12 I_A\\omega^2 = 76.56\\omega^2$; $T_1 = 0$.\n\n**Work-energy:** $0 + 613.1 = 76.56\\omega^2 \\Rightarrow \\omega = 2.83$ rad/s.\n\nCheck: G's speed is then $v_G = 1.25(2.83) = 3.54$ m/s, less than the free-fall value $\\sqrt{2g(1.25)} = 4.95$ m/s,\nbecause part of the energy goes into rotation about G ($k_A > r_{G/A}$).\n","confidence":"HIGH","solutionNote":"Matches the slide answers (U = 613.1 J, I_A = 153.1 kg m^2, omega = 2.83 rad/s)."},{"id":"engr216-pt2024-q1","module":"ENGR5004","week":4,"label":"Statics Progress Test 2024 Q1","kind":"past-paper","level":3,"minutes":35,"grade":"2:1","tags":["shear-bending-diagrams","bending-eccentric","frame","free-body-diagram","distributed-load","sinusoidal-load","combined-axial-bending","reactions"],"source":{"name":"ENGR216 Statics Progress Test 2024, Question 1 (solved in Statics session 21)","paper":"ENGR216","year":2024,"question":"Progress Test Q1","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 10 - Statics session 21 Revision Session (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 10 - Statics session 21 Revision Session (old spec ENGR216).pdf","tag":"VALID"},"text":"Consider the structure of Figure 1: a vertical member PQ of length $L$ rigidly joined at Q to a horizontal\nmember QR of length $2L$. P rests on a roller support (vertical reaction only) and R is pinned. Member PQ\nis loaded by the harmonic distributed normal load\n\n$$w(x) = w_0 \\sin\\left(\\frac{\\pi x}{L}\\right)$$\n\nacting horizontally to the right, with position $x$ measured along PQ with origin at P, and $w_0$ in N/m.\n","figures":[{"src":"/figures/engr216-pt2024-q1.png","caption":"Figure 1: frame PQR; roller at P, pin at R; half-sine load w(x) acting to the right on the vertical member PQ (length L); QR has length 2L."}],"formulaSheet":"- Resultant of a distributed load $P = \\int_0^{L} w(x)\\,dx$ and its position $x_P - x_0 = \\dfrac{\\int_0^{L} w(x)(x - x_0)\\,dx}{P}$ (on the formula sheet)\n- $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$, $V_D - V_C = -\\int_{x_C}^{x_D} w\\,dx$, $M_D - M_C = \\int_{x_C}^{x_D} V\\,dx$ (on the formula sheet)\n- Positive shear and positive bending sign convention (pictures) (on the formula sheet)\n- Combined axial force and bending (eccentric loading form) $\\sigma_x = \\dfrac{N}{A} - \\dfrac{My}{I}$ (on the formula sheet)\n- Solid circle $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n- $\\int x \\sin(ax)\\,dx = \\dfrac{\\sin(ax)}{a^2} - \\dfrac{x\\cos(ax)}{a}$ (learn this; integration by parts)\n- Static equilibrium $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$; roller gives one force, pin gives two (learn this)\n","parts":[{"label":"a","text":"Draw the free-body diagram of the entire structure, indicating all nonzero reactions.","grade":"2:2"},{"label":"b","text":"Determine all reactions acting on the structure as functions of $w_0$ and $L$.","grade":"2:2","answer":[{"label":"R_P (upward)","value":0.15915,"units":"w0*L"},{"label":"R_RH (to the left)","value":0.63662,"units":"w0*L"},{"label":"R_RV (downward)","value":0.15915,"units":"w0*L"}]},{"label":"c","text":"Calculate the axial load distribution $N(x)$, the shearing load distribution $V(x)$ and the bending\nmoment distribution $M(x)$ along members PQ and QR.\n","grade":"2:1","answer":[{"label":"M at Q (both members)","value":-0.31831,"units":"w0*L^2"},{"label":"V in PQ at Q","value":-0.63662,"units":"w0*L"}]},{"label":"d","text":"Draw the $N(x)$, $V(x)$ and $M(x)$ diagrams along members PQ and QR.","grade":"2:2"},{"label":"e","text":"Draw the free-body diagram of the vertical member PQ, indicating clearly all nonzero external and\ninternal loads, and give the expressions of all internal and external loads acting on it as functions\nof $w_0$ and $L$.\n","grade":"2:2","answer":[{"label":"F_QH on PQ (to the left)","value":0.63662,"units":"w0*L"},{"label":"F_QV on PQ (downward)","value":0.15915,"units":"w0*L"},{"label":"M_Q on PQ (clockwise)","value":0.31831,"units":"w0*L^2"}]},{"label":"f","text":"Calculate the magnitude of the maximum compressive stress, in kPa, in the cross section of member PQ at\n$0.75L$ from end P. Assume a circular cross section of radius $R = 200$ mm, and use $L = 1$ m and\n$w_0 = 30$ kN/m.\n","grade":"2:1","answer":[{"label":"M at x = 0.75L","value":-5.0126,"units":"kNm"},{"label":"max compressive stress","value":835.8,"units":"kPa"}]}],"solution":"$ee","confidence":"HIGH","solutionNote":"Reactions, N/V/M distributions and the member FBD match the deck. Part (f) does NOT: the deck evaluates sin(3 pi/4) as -sqrt(2)/2 (it is +sqrt(2)/2), giving M(0.75L) = -(w0 L^2/pi)(sqrt(2)/(2 pi) + 3/4) = -9.31 kN m and a maximum compressive stress of 1482 + 38 = 1520 kPa. The correct moment is -5.01 kN m (checked by direct numerical integration of the load) and the correct stress is 798 + 38 = 836 kPa."},{"id":"engr216-s14-np2","module":"ENGR5004","week":5,"label":"Statics new problem 2 (session 14)","kind":"tutorial","level":3,"minutes":45,"grade":"2:1","tags":["shear-bending-diagrams","frame","free-body-diagram","triangular-load","reactions","combined-axial-bending","i-section","beam-shear-stress","first-moment-of-area","design"],"source":{"name":"ENGR216 Statics \"new problem 2\" (set in session 11, solved in session 14)","paper":"ENGR216","question":"New problem 2","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Statics session 14 solution to new problem 2 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Statics session 14 solution to new problem 2 (old spec ENGR216).pdf"},"text":"The structure in Fig. 1 is made of columns AB and CD rigidly connected to beam BC. All three members have\nlength $L$. A is pinned and D rests on a roller (vertical reaction only). Column AB carries a triangular\ndistributed load acting horizontally to the right, growing linearly from zero at A to $w_0$ at B, i.e.\n$w(x) = w_0 x/L$ with $x$ measured along AB from A.\n\nAll members have the I cross section of Fig. 2: flange width $a = OP = ZW = 120$ mm, web height\n$SV = 120$ mm, flange thickness $b = OT = UZ = 20$ mm and web thickness $c = SR = 10$ mm (overall depth\n160 mm). Use $E = 200$ GPa and $w_0 = 30$ kN/m.\n","figures":[{"src":"/figures/engr216-np2-fig1.png","caption":"Fig. 1: portal frame ABCD (all members length L), pin at A, roller at D, triangular load on AB from 0 at A to w0 at B, acting to the right."},{"src":"/figures/engr216-np2-fig2.png","caption":"Fig. 2: I cross section; flanges OP and ZW of width a and thickness b, web SR of thickness c and height SV = a."},{"src":"/figures/engr216-np2-fig3.png","caption":"Fig. 3: the same frame with a uniform load w1 on beam BC."}],"formulaSheet":"- Resultant of a distributed load $P = \\int_0^{L} w(x)\\,dx$ and its position $x_P = \\dfrac{\\int_0^{L} x\\,w(x)\\,dx}{P}$ (on the formula sheet)\n- $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$ (on the formula sheet)\n- Combined axial force and bending $\\sigma_x = \\dfrac{N}{A} - \\dfrac{My}{I}$ (on the formula sheet)\n- Shear stress in a beam $\\tau = \\dfrac{VQ}{It}$, $Q = \\sum \\bar y_i A_i$ above the cut (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$ and the parallel axis theorem $I = \\bar I + Ad^2$ (on the formula sheet)\n- Static equilibrium; pin gives two forces, roller one (learn this)\n","parts":[{"label":"1","text":"Draw the free-body diagram of the loaded structure and calculate all reactions.","grade":"2:2","answer":[{"label":"R_Ax (to the left)","value":0.5,"units":"w0*L"},{"label":"R_Ay (downward)","value":0.33333,"units":"w0*L"},{"label":"R_D (upward)","value":0.33333,"units":"w0*L"}]},{"label":"2","text":"Calculate the normal load $N(x)$, shearing load $V(x)$ and bending moment $M(x)$ along all three members.","grade":"2:1","answer":[{"label":"M at B","value":0.33333,"units":"w0*L^2"},{"label":"V in BC","value":-0.33333,"units":"w0*L"}]},{"label":"3","text":"Draw $N(x)$, $V(x)$ and $M(x)$ in members AB, BC and CD.","grade":"2:2"},{"label":"4","text":"Draw free-body diagrams of members AB, BC and CD, clearly indicating all external and internal loads on each member.","grade":"2:2"},{"label":"5","text":"Calculate all internal and external loads acting on members AB, BC and CD.","grade":"2:2"},{"label":"6","text":"Calculate the maximum member length $L$ for which the maximum normal stress in the cross sections of\nmember AB does not exceed 200 MPa.\n","grade":"2:1","answer":[{"label":"I of the section","value":0.00002512,"units":"m^4"},{"label":"L max","value":2.48,"units":"m"}]},{"label":"7","text":"The structure is then subjected to a uniform distributed load $w_1 = 20$ kN/m on member BC (Fig. 3).\nVerify that the cross section is sufficient for the maximum shearing stress in the cross sections of\nmember BC not to exceed 100 MPa.\n","grade":"2:2","answer":[{"label":"V max in BC","value":24.8,"units":"kN"},{"label":"Q at the neutral axis","value":0.000186,"units":"m^3"},{"label":"tau max","value":18.36,"units":"MPa"}]}],"solution":"$ef","confidence":"HIGH","solutionNote":"Matches the deck (reactions, N/V/M, member loads, I = 2.5120e-5 m^4, A = 6.0e-3 m^2, L = 2.48 m, Q = 1.86e-4 m^3, tau max = 18.36 MPa). One slip in the deck: it writes V(x) = w1 L/2 - w1 x/2 for BC under w1; the correct expression is w1 L/2 - w1 x (its own sketch and V max = w1 L/2 are consistent with this), so the numbers are unaffected."},{"id":"engr273-2026-q1","module":"ENGR5004","week":4,"label":"ENGR273 2026 Q1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["shear-bending-diagrams","bending-eccentric","frame","free-body-diagram","distributed-load","bending-stress"],"source":{"name":"ENGR273 Summer 2026 Q1","paper":"ENGR273","year":2026,"question":"Q1","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL.pdf","solutionsPath":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL answers.pdf","tag":"VALID"},"text":"Figure Q1-1 shows a structure of negligible mass constrained by a fixed support at A, loaded by a\ndistributed triangular normal load $w(x)$ along beam DB and subjected to a concentrated moment\n$M_E = \\frac{1}{3} w_0 L^2$ at the end E of beam CE. Beam DB has length $2L$, and parts AB, BC and CE\nhave each length $L$. The expression of the distributed load is:\n\n$$w(x) = w_0 \\left(1 - \\frac{x}{2L}\\right) \\qquad \\text{(Q1-1)}$$\n\nwhere $x$ is the horizontal coordinate along beam DB with origin at D, and $w_0$ is the value of the\ndistributed load at end D.\n","figures":[{"src":"/figures/engr273-2026-q1.png","caption":"Figure Q1-1: frame fixed at A; column ABC, beam DB with triangular load w(x) (w0 at D), arm CE with moment M_E at E."}],"formulaSheet":"- Resultant of a distributed load $P = \\int_0^{L} w(x)\\,dx$ and its position $x_P - x_0 = \\dfrac{\\int_0^{L} w(x)(x - x_0)\\,dx}{P}$ (on the formula sheet)\n- Positive shear and positive bending sign convention (pictures) (on the formula sheet)\n- $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$, $V_D - V_C = -\\int_{x_C}^{x_D} w\\,dx$, $M_D - M_C = \\int_{x_C}^{x_D} V\\,dx$ (on the formula sheet)\n- Maximum bending stress $\\sigma_{max} = \\dfrac{|M| c}{I}$ (on the formula sheet)\n- Solid circle $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n- Static equilibrium $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$; a fixed support gives two forces and a couple (learn this)\n","parts":[{"label":"a","text":"Draw the free-body-diagram of the entire structure, indicating ONLY the nonzero reactions and applied loads.\n","marks":6,"grade":"2:2"},{"label":"b","text":"Calculate the reactions acting on the structure in figure Q1-1.\n","marks":6,"grade":"2:2","answer":[{"label":"A_x (right +)","value":0,"units":"w0*L"},{"label":"A_y (up +)","value":1,"units":"w0*L"},{"label":"M_A (CCW +)","value":-1,"units":"w0*L^2"}]},{"label":"c","text":"Calculate and draw as accurately as possible the shearing load diagram along beam DB, and the bending\nmoment diagram along column ABC.\n","marks":8,"grade":"2:1","answer":[{"label":"V at B in DB (sheet convention)","value":-1,"units":"w0*L"},{"label":"M in AB (D side compressed +)","value":1,"units":"w0*L^2"},{"label":"M in BC (D side compressed +)","value":-0.33333,"units":"w0*L^2"}]},{"label":"d","text":"Calculate the minimum diameter of the circular cross section of beam DB required for the maximum normal\nstress in any of the beam's cross sections not to exceed 150 MPa. Use $w_0 = 40$ kN/m and $L = 5$ m.\n","marks":5,"grade":"2:2","answer":[{"label":"|M|max in DB","value":1333.3,"units":"kNm"},{"label":"d_min","value":449,"units":"mm"}]}],"solution":"$f0","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers PDF on every result (A_y = w0 L, M_A = w0 L^2 clockwise,\nV(2L) = -w0 L, M_AB = +w0 L^2 and M_BC = -w0 L^2/3 with the same convention, d_min = 449 mm). Verified\nsymbolically by two routes (load integration and segment free bodies).\n"},{"id":"engr273-2026-q2","module":"ENGR5004","week":10,"label":"ENGR273 2026 Q2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["buckling","euler-load","effective-length","i-section","safety-factor"],"source":{"name":"ENGR273 Summer 2026 Q2","paper":"ENGR273","year":2026,"question":"Q2","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL.pdf","solutionsPath":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL answers.pdf","tag":"VALID"},"text":"Figure Q2-1 shows a rod of negligible mass of length $L$ subjected to a compressive axial load $F$. The rod\nis constrained by a fixed support at both end A and end B. The rod's cross section is depicted in\nFigure Q2-2, in which $b = 10a$.\n","figures":[{"src":"/figures/engr273-2026-q2.png","caption":"Figures Q2-1 and Q2-2: rod along z, fixed at A and B, load F at B; I-section with flanges b wide and a thick, web of thickness a and clear height b between the flanges (overall depth b + 2a)."}],"formulaSheet":"- Euler critical load $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L_e^2}$, $\\sigma_{cr} = \\dfrac{\\pi^2 E}{(L_e/\\rho_{min})^2}$, $\\rho = \\sqrt{I/A}$ (on the formula sheet)\n- Effective lengths: fixed-free $L_e = 2L$, pinned-pinned $L$, fixed-pinned $0.7L$, fixed-fixed $0.5L$ (on the formula sheet)\n- Rectangle $I_x = \\dfrac{1}{12} b h^3$, $I_y = \\dfrac{1}{12} h b^3$, $A = bh$ (on the formula sheet)\n- Parallel-axis theorem $I = I_c + A d^2$ (or build the I-section by subtracting rectangles) (learn this)\n- Axial stress $\\sigma = F/A \\le \\sigma_{all}$; buckling safety factor $P_{cr} \\ge FS \\cdot F$ (learn this)\n","parts":[{"label":"a","text":"Assuming that the compressive load $F$ exceeds by a small amount Euler's critical load, determine and\nclearly state if the rod will buckle in the $xz$ or $yz$ plane.\n","marks":6,"grade":"2:2","answer":[{"label":"I_x","value":690,"units":"a^4"},{"label":"I_y","value":167.5,"units":"a^4"}]},{"label":"b","text":"Draw as accurately as possible the shape of the buckled rod, labelling the two Cartesian axes defining the\nplane in which the rod buckled, and fulfilling the applied constraints in the drawn schematic.\n","marks":6,"grade":"2:2"},{"label":"c","text":"Using $F = 40$ kN, allowable normal stress of 100 MPa, Young's modulus of 200 GPa, $L = 1.6$ m and a\nsafety factor of 2 for the buckling critical load, calculate the minimum value of $a$ required for the rod\nnot to buckle and not to yield.\n","marks":7,"grade":"2:2","answer":[{"label":"a from buckling","value":3.528,"units":"mm"},{"label":"a from yield","value":3.651,"units":"mm"},{"label":"a_min","value":3.651,"units":"mm"}]},{"label":"d","text":"Assuming that the rod designed in part (c) is used with the fixed support at A but no constraint at B, and\nusing a safety factor of 2 in the buckling analysis, determine and state if the rod will buckle or yield\nunder the new constraint set-up.\n","marks":6,"grade":"2:2","answer":[{"label":"P_cr fixed-free","value":5.74,"units":"kN"},{"label":"P_cr / FS","value":2.87,"units":"kN"},{"label":"axial stress","value":100,"units":"MPa"}]}],"solution":"$f1","confidence":"HIGH","solutionNote":"Same method and conclusions as the unofficial worked answers PDF (buckles in the xz plane; yield governs in\n(c); buckles but does not yield in (d)), but the numbers differ: that PDF reads b as the OVERALL depth\n(A = 28a^2, I_y = 167.3a^4, a_min = 3.78 mm, P_cr = 6.58 kN). We read b as the clear web height between the\nflanges, as the dimension lines in Figure Q2-2 are drawn to the inner flange faces, and this reading\nreproduces the published answers of Tutorial Sheet 4 Problem 3 (same figure), so we trust a_min = 3.65 mm.\nUnder the other reading the answers would be 3.78 mm and 6.58 kN. Figure caption corrected from \"overall\nheight b\" to the clear web height.\n"},{"id":"engr273-2026-q3","module":"ENGR5004","week":8,"label":"ENGR273 2026 Q3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["rigid-kinematics","rigid-kinetics","crank-slider","relative-velocity","relative-acceleration","free-body-diagram"],"source":{"name":"ENGR273 Summer 2026 Q3","paper":"ENGR273","year":2026,"question":"Q3","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL.pdf","solutionsPath":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL answers.pdf","tag":"VALID"},"text":"The planar mechanism in Figure Q3-1 comprises a crank OA (uniform slender rod of length $a$, mass $m_1$)\npinned at O, and a coupler AB (uniform slender rod of length $b$, mass $m_2$) pinned at A and at the slider\npoint B. Point B is a frictionless slider of mass $m_3$ constrained to move along the vertical guide at\n$x = d$. At the instant shown, OA is horizontal along $+x$, so $A = (a, 0)$. The slider is at\n$B = (d, y_B)$ with $y_B > 0$. The slider velocity and acceleration are known and upward:\n\n$$\\mathbf{v}_B = v_B\\,\\mathbf{j}, \\qquad \\mathbf{a}_B = a_B\\,\\mathbf{j}.$$\n\nNeglect gravity and all frictions of the whole system. Positive angular sense is counterclockwise $+\\mathbf{k}$.\n","figures":[{"src":"/figures/engr273-2026-q3.png","caption":"Figure Q3-1: crank-coupler driving a vertical slider at x = d."}],"formulaSheet":"- Rotation about a fixed axis: $\\mathbf{v} = \\boldsymbol\\omega\\times\\mathbf{r}$, $\\mathbf{a} = \\boldsymbol\\alpha\\times\\mathbf{r} - \\omega^2\\mathbf{r}$ (learn this)\n- Relative velocity $\\mathbf{v}_B = \\mathbf{v}_A + \\boldsymbol\\omega\\times\\mathbf{r}_{B/A}$ (learn this)\n- Relative acceleration $\\mathbf{a}_B = \\mathbf{a}_A + \\boldsymbol\\alpha\\times\\mathbf{r}_{B/A} - \\omega^2\\mathbf{r}_{B/A}$ (learn this)\n- Equations of motion $\\sum\\mathbf{F} = m\\mathbf{a}_G$, $\\sum M_G = I_G\\alpha$ (or $\\sum M_O = I_O\\alpha$ about a fixed pin) (learn this)\n- Slender rod $I_G = \\dfrac{1}{12} m L^2$, $I_O = \\dfrac{1}{3} m L^2$ about an end (learn this)\n","parts":[{"label":"a","text":"Draw the free-body diagrams and kinetic diagrams for OA, AB, and the slider at B. Clearly state your axes\nand indicate all unknown reactions at pins/guide.\n","marks":8,"grade":"2:2"},{"label":"b","text":"Determine the instantaneous angular velocities $\\boldsymbol{\\omega}_{OA}$ and $\\boldsymbol{\\omega}_{AB}$,\nexpressed in terms of $a$, $b$, $d$, and $\\mathbf{v}_B$.\n","marks":8,"grade":"2:2","answer":[{"label":"omega_OA (CCW +)","value":1,"units":"v_B/a"},{"label":"omega_AB (CCW +)","value":0,"units":"rad/s"}]},{"label":"c","text":"Determine the instantaneous angular accelerations $\\boldsymbol{\\alpha}_{OA}$ and $\\boldsymbol{\\alpha}_{AB}$,\nexpressed in terms of $a$, $b$, $d$, $\\mathbf{v}_B$, and $\\mathbf{a}_B$.\n","marks":9,"grade":"2:1","answer":[{"label":"alpha_AB (CCW +)","value":-1,"units":"v_B^2/(a*y_B), y_B = sqrt(b^2-(d-a)^2)"}]}],"solution":"$f2","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers PDF for (b) and (c) (omega_AB = 0, omega_OA = v_B/a,\nalpha_AB = -v_B^2/(a y_B), alpha_OA = a_B/a + v_B^2 (d-a)/(a^2 y_B)). For (a) that PDF counts 7 unknowns\nagainst 8 equations \"minus one kinematic identity\"; we instead add an unknown driving couple M_O at O,\nbecause without a drive the frictionless, gravity-free mechanism cannot follow an arbitrary prescribed a_B\n(confirmed by solving the 8 equations symbolically: M_O is nonzero in general). The FBD content is otherwise\nthe same.\n"},{"id":"engr273-2026-q4","module":"ENGR5004","week":8,"label":"ENGR273 2026 Q4","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"first","tags":["rigid-kinematics","rigid-kinetics","instantaneous-centre","relative-acceleration","general-plane-motion"],"source":{"name":"ENGR273 Summer 2026 Q4","paper":"ENGR273","year":2026,"question":"Q4","path":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL.pdf","solutionsPath":"Second Year Past Exam Papers-20261003/Second Year Exam Papers 2026/ENGR273 in-person Main Exam Paper 2025-26 FINAL answers.pdf","tag":"VALID"},"text":"A uniform slender rod AB of mass $m$ has its ends constrained to move in two smooth guides (Figure Q4-1).\n\n- End A slides on the horizontal guide $y = 0$.\n- End B slides on a straight guide inclined at angle $\\beta$ to the $+x$ axis (guide line passes through the\n origin), i.e. $y = (\\tan\\beta)\\,x$, $0 < \\beta < 90^\\circ$.\n\nAt the instant shown:\n\n- $A = (x_A, 0)$ with known speed and acceleration to the right: $\\mathbf{v}_A = v_A\\,\\mathbf{i}$,\n $\\mathbf{a}_A = a_A\\,\\mathbf{i}$.\n- $B = (x_B, x_B\\tan\\beta)$ with known acceleration magnitude $a_B$ along the inclined guide as shown in\n figure Q4-1 (direction $+\\mathbf{t}$, where $\\mathbf{t}$ is the unit tangent of the guide).\n\nA driving actuator applies an unknown force $\\mathbf{P}$ at A along $+x$ (through the slider at A). Neglect\ngravity. Positive angular sense is counterclockwise $+\\mathbf{k}$. $L = AB$.\n","figures":[{"src":"/figures/engr273-2026-q4.png","caption":"Figure Q4-1: rod AB with A on the horizontal guide and B on the guide y = x tan(beta)."}],"formulaSheet":"- Instantaneous centre of zero velocity: intersection of the perpendiculars to two known velocity directions; $v = \\omega\\, r_{/IC}$ (learn this)\n- Relative velocity $\\mathbf{v}_B = \\mathbf{v}_A + \\boldsymbol\\omega\\times\\mathbf{r}_{B/A}$ (learn this)\n- Relative acceleration $\\mathbf{a}_B = \\mathbf{a}_A + \\boldsymbol\\alpha\\times\\mathbf{r}_{B/A} - \\omega^2\\mathbf{r}_{B/A}$ (learn this)\n- Equations of motion $\\sum\\mathbf{F} = m\\mathbf{a}_G$, $\\sum M_G = I_G\\alpha$, or about any point P: $\\sum M_P = I_G\\alpha + (\\mathbf{r}_{G/P}\\times m\\mathbf{a}_G)\\cdot\\mathbf{k}$ (learn this)\n- Slender rod $I_G = \\dfrac{1}{12} m L^2$ (given in the question)\n","parts":[{"label":"a","text":"Draw the free-body diagram and kinetic diagram of rod AB. Indicate guide reactions at A and B, the actuator\nforce $\\mathbf{P}$, mass centre G, and inertia terms $m\\mathbf{a}_G$ and $I_G$ (with\n$I_G = \\frac{1}{12} m L^2$).\n","marks":8,"grade":"2:2"},{"label":"b","text":"Determine the instantaneous centre of zero velocity (IC) of rod AB and locate it relative to points A and\nB. Hence determine the instantaneous angular velocity $\\boldsymbol{\\omega}_{AB}$ and the instantaneous\nvelocity of the mass centre G.\n","marks":10,"grade":"2:1"},{"label":"c","text":"Using planar rigid-body acceleration relations, determine the instantaneous angular acceleration\n$\\boldsymbol{\\alpha}_{AB}$ and the acceleration of the mass centre $\\mathbf{a}_G$. Hence determine the\nactuator force $\\mathbf{P}$ and the guide reactions at A and B required at this instant (in terms of the\ngiven variables).\n","marks":7,"grade":"first"}],"solution":"$f3","confidence":"MEDIUM","solutionNote":"Agrees with the unofficial worked answers PDF (same IC, omega_AB = v_A sin(beta) cos(beta)/(x_B - x_A cos^2 beta),\nboth routes for alpha_AB, a_G = (a_A + a_B)/2 and the same closed forms for N_B, P, N_A); its sample case\n(x_A = 0.8, x_B = 0.3, beta = 45 deg) was reproduced: omega = -5.00, alpha = -102.5, a_B = 61.9,\nP = 161.2, N_A = -73.2, N_B = 165.3. MEDIUM only because the question over-specifies the data (a_A and a_B\ncannot both be chosen freely); see the note in (c).\n"},{"id":"engr216-2025-a1","module":"ENGR5004","week":5,"label":"ENGR216 2025 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["shear-bending-diagrams","beam-shear-stress","frame","free-body-diagram","t-section"],"source":{"name":"ENGR216 Summer 2025 Q A1","paper":"ENGR216","year":2025,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25 answers.pdf","tag":"VALID"},"text":"Figure A1-a shows a structure of negligible mass constrained by a pin at end P and a roller at end R. Member\nPQ is loaded by a linear distributed load $w(x_v)$, which is 0 at end P and equals $w_0$ at end Q, with\n$x_v$ denoting the position along member PQ. The coordinate along member QR, with origin at Q, is denoted\nby $x_h$.\n","figures":[{"src":"/figures/engr216-2025-a1.png","caption":"Figure A1-a: frame with vertical member PQ (length L, pin at P, triangular load) and horizontal member QR (length 2L, roller at R). Figure A1-b: T cross-section."}],"formulaSheet":"- Resultant of a distributed load $P = \\int w\\,dx$ and its line of action $x_P = \\dfrac{\\int w\\,x\\,dx}{\\int w\\,dx}$ (on the formula sheet)\n- Equilibrium $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$; a pin gives two force components, a roller one force normal to its surface (learn this)\n- Load, shear and moment: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$, with the positive shear and positive bending pictures (on the formula sheet)\n- Transverse shear stress $\\tau = \\dfrac{VQ}{It}$ with $Q = \\bar y\\,A'$ (first moment about the neutral axis of the area on one side of the point) (learn this; on the 2025 sheet but not the 2026 one)\n- Radius of gyration $I = \\rho^2 A$ (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$ (on the formula sheet); parallel-axis theorem $I = I_c + A d^2$ and centroid of a composite area $\\bar y = \\dfrac{\\sum A_i \\bar y_i}{\\sum A_i}$ (learn this; only needed to check the given data)\n","parts":[{"label":"a","text":"Draw the free-body-diagram of the entire structure, indicating ONLY the nonzero reactions and applied loads.\n","marks":4,"grade":"2:2"},{"label":"b","text":"Calculate the reactions acting on the considered structure.\n","marks":6,"grade":"2:2","answer":[{"label":"P_x (to the left +)","value":0.5,"units":"w0*L"},{"label":"P_y (downward +)","value":0.16667,"units":"w0*L"},{"label":"R_y (upward +)","value":0.16667,"units":"w0*L"}]},{"label":"c","text":"Knowing that the expression of the bending moment $M(x_v)$ along member PQ is:\n\n$$M(x_v) = \\frac{1}{2} w_0 L^2 \\left[\\frac{x_v}{L} - \\frac{1}{3}\\left(\\frac{x_v}{L}\\right)^3\\right] \\qquad \\text{(Equation 1)}$$\n\nand the expression of the bending moment $M(x_h)$ along member QR is:\n\n$$M(x_h) = \\frac{1}{6} w_0 L^2 \\left[2 - \\frac{x_h}{L}\\right] \\qquad \\text{(Equation 2)}$$\n\ndetermine the expression of the sharing load $V(x_v)$ along PQ and the shearing load $V(x_h)$ along QR.\n","marks":4,"grade":"2:2","answer":[{"label":"V at P (xv = 0)","value":0.5,"units":"w0*L"},{"label":"V along QR","value":-0.16667,"units":"w0*L"}]},{"label":"d","text":"Draw as accurately as possible the shearing and the bending moment diagrams along PQ and QR, consistently\nwith the expressions given or obtained above.\n","marks":4,"grade":"2:2","answer":[{"label":"M at corner Q","value":0.33333,"units":"w0*L^2"}]},{"label":"e","text":"Calculate the minimum length $L$ required for the magnitude of the maximum shearing stress in the cross\nsection of member PQ at end P not to exceed 10 MPa. The cross section is reported in Figure A1-b. Use\n$a = 150$ mm, $b = 30$ mm and $w_0 = 40$ kN/m. The section is positioned so that the neutral axis of the\nbending moment is parallel to side AB. The centroid K of the section is at distance $d_K$ of 120 mm from\nside GH, and the radius of gyration past the neutral axis is 54.8 mm.\n","marks":7,"grade":"2:1","answer":[{"label":"L (limiting length)","value":1.875,"units":"m"}]}],"solution":"$f4","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers PDF in every part (reactions w0L/2 to the left, w0L/6 down, w0L/6 up;\nV(xv) = w0(L^2 - xv^2)/(2L), V(xh) = -w0L/6; L = 1.877 m with I from rho, 1.875 m with the exact I). The\nquestion says \"minimum length\" but the 10 MPa limit gives a maximum allowable L (see (e)).\n"},{"id":"engr216-2025-a2","module":"ENGR5004","week":6,"label":"ENGR216 2025 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["beam-deflection","bending-eccentric","propped-cantilever","statically-indeterminate","elastic-curve"],"source":{"name":"ENGR216 Summer 2025 Q A2","paper":"ENGR216","year":2025,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25 answers.pdf","tag":"VALID"},"text":"Figure A2 shows a beam of negligible mass with flexural rigidity $EI$ and length $2L$. The beam is constrained\nby a fixed support at end A and a roller at end B, and is subjected to a uniform distributed load $w_0$.\n","figures":[{"src":"/figures/engr216-2025-a2.png","caption":"Figure A2: beam of length 2L, fixed at A, roller at B, uniform load w0."}],"formulaSheet":"- Equilibrium $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$; a fixed support gives two forces and a couple, a roller one force; degree of indeterminacy = unknown reactions minus independent equilibrium equations (learn this)\n- Elastic curve $\\dfrac{d^2y}{dx^2} = \\dfrac{M(x)}{EI}$, slope $\\theta = dy/dx$ (learn this; on the 2025 sheet but not the 2026 one)\n- Load, shear and moment: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$, with the positive shear and positive bending pictures (on the formula sheet)\n- Slope is extreme where $d\\theta/dx = M/EI = 0$; deflection is extreme where $\\theta = 0$ (learn this)\n- Pure bending $\\sigma_{max} = \\dfrac{|M| c}{I}$ (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$, so a square of side $s$ has $I = s^4/12$, $c = s/2$ (on the formula sheet)\n","parts":[{"label":"a","text":"Draw the free-body-diagram of the entire structure, indicating ONLY the nonzero reactions and applied\nloads, and state if the structure is statically determinate or indeterminate, justifying your answer.\n","marks":4,"grade":"2:2"},{"label":"b","text":"Determine all nonzero reactions acting on the beam as functions of $w_0$ and $L$, knowing that the vertical\ndisplacement of its cross sections is:\n\n$$y(x) = \\frac{w_0 L^4}{EI}\\left[-\\frac{1}{4}\\left(\\frac{x}{L}\\right)^2 + \\frac{5}{24}\\left(\\frac{x}{L}\\right)^3 - \\frac{1}{24}\\left(\\frac{x}{L}\\right)^4\\right] \\qquad \\text{(Equation 3)}$$\n\nwhere $x$ denotes the position along the axis of the beam.\n","marks":6,"grade":"2:1","answer":[{"label":"A_y (upward +)","value":1.25,"units":"w0*L"},{"label":"B_y (upward +)","value":0.75,"units":"w0*L"},{"label":"M_A (anticlockwise +)","value":0.5,"units":"w0*L^2"}]},{"label":"c","text":"Draw as accurately as possible the bending moment diagram of the beam and the deformed shape of its mean\nline (elastic curve).\n","marks":4,"grade":"2:2","answer":[{"label":"M_max sagging","value":0.28125,"units":"w0*L^2"},{"label":"x at M_max","value":1.25,"units":"L"},{"label":"y_max (upward +)","value":-0.086658,"units":"w0*L^4/EI"}]},{"label":"d","text":"Calculate the positions along the beam at which the elastic curve has maximum slope as functions of $L$.\n","marks":4,"grade":"2:1","answer":[{"label":"x1 (steepest downward slope)","value":0.5,"units":"L"},{"label":"x2 (largest slope, at B)","value":2,"units":"L"}]},{"label":"e","text":"Knowing that the beam has square cross-section, determine the minimum side length of the cross-section such\nthat the maximum normal stress in the cross-section at end A does not exceed the material's allowable\nnormal stress of 200 MPa. Use $w_0 = 20$ kN/m and $L = 2.0$ m.\n","marks":7,"grade":"2:2","answer":[{"label":"s_min","value":106.3,"units":"mm"}]}],"solution":"$f5","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers PDF in every part (A_y = 5w0L/4, B_y = 3w0L/4, M_A = w0L^2/2\nanticlockwise; maximum slope at x = L/2 and x = 2L; s_min = 106 mm). Reactions also match the standard\npropped-cantilever result.\n"},{"id":"engr216-2025-b1","module":"ENGR5004","week":8,"label":"ENGR216 2025 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["rigid-kinetics","general-plane-motion","free-body-diagram","kinetic-diagram","smooth-surface"],"source":{"name":"ENGR216 Summer 2025 Q B1","paper":"ENGR216","year":2025,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25 answers.pdf","tag":"VALID"},"text":"A uniform rod OA with a mass ($m$) and length ($L$) is centrally affixed to a uniform block of mass ($m$).\n'O' is the centre of rotation for the rod OA, which is also the centroid of the block, as shown in Figure B1.\nThe block is positioned to move freely along a smooth horizontal surface. Initially, the system is at rest,\nand the rod OA is oriented horizontally with an initial counterclockwise angular velocity $\\omega_{OA}$.\n","figures":[{"src":"/figures/engr216-2025-b1.png","caption":"Figure B1: block of mass m on a smooth floor with rod OA (mass m, length L) pinned at the block centroid O."}],"formulaSheet":"- Relative acceleration (rigid body): $\\mathbf{a}_G = \\mathbf{a}_O + \\boldsymbol\\alpha\\times\\mathbf{r}_{G/O} - \\omega^2\\mathbf{r}_{G/O}$ (learn this)\n- Equations of motion: $\\sum\\mathbf{F} = m\\mathbf{a}_G$, $\\sum M_G = I_G\\alpha$; for a translating body $\\sum\\mathbf{F} = m\\mathbf{a}$ (learn this)\n- Slender rod about its centre: $I_G = \\dfrac{1}{12} m L^2$ (learn this)\n- Moments about another point P (check): $\\sum M_P = I_G\\alpha + (\\mathbf{r}_{G/P}\\times m\\mathbf{a}_G)_z$ (learn this)\n- Newton's third law at a pin: equal and opposite pin forces on the two bodies (learn this)\n- Friction: static $F \\le \\mu_s N$, kinetic $F = \\mu_k N$, opposing the (impending) relative sliding (learn this)\n","parts":[{"label":"a","text":"Draw the free-body and kinetic diagrams of the rod OA and the block. State appropriate coordinate systems.\n","marks":10,"grade":"2:1"},{"label":"b","text":"Determine the acceleration of point O on the initial instant.\n","marks":5,"grade":"2:1","answer":[{"label":"a_O (towards A +)","value":0.25,"units":"omega_OA^2*L"}]},{"label":"c","text":"Determine the angular acceleration of the rod OA at the instant of (b).\n","marks":5,"grade":"2:2","answer":[{"label":"alpha_OA (clockwise +)","value":1.5,"units":"g/L"}]},{"label":"d","text":"If there is friction between the block and the horizontal surface, will the acceleration of point O\nincrease or decrease? Give your reasoning.\n","marks":5,"grade":"2:1"}],"solution":"$f6","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers PDF (a_O = omega^2 L/4 towards A, alpha = 3g/(2L) clockwise,\nO_y = mg/4, N = 5mg/4, friction decreases a_O). The question's \"initially at rest\" with a nonzero omega_OA is\nread as \"block momentarily at rest\"; this does not change any result.\n"},{"id":"engr216-2025-b2","module":"ENGR5004","week":5,"label":"ENGR216 2025 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["particle-kinetics","particle-kinematics","rotating-frame","coriolis","friction"],"source":{"name":"ENGR216 Summer 2025 Q B2","paper":"ENGR216","year":2025,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2025/ENGR216 Main Exam 2024-25 answers.pdf","tag":"VALID"},"text":"A particle (P) of mass ($m$) slides in a tube with the relative velocity of $v_{rel}$. The tube and the rod\n(perpendicular to the tube) rotate counterclockwise about point O with angular velocity $\\omega$ and angular\nacceleration $\\alpha$. At the instant shown in Figure B2, the horizontal and vertical distance between O and P\nis $L$ and $h$, respectively. $\\mu_k$ and $\\mu_s$ are the kinetic and static friction coefficients between P\nand the tube.\n","figures":[{"src":"/figures/engr216-2025-b2.png","caption":"Figure B2: rod of length L pinned at O carrying a perpendicular tube; particle P at height h above the rod, moving up the tube at v_rel."}],"formulaSheet":"- Rotating-axes acceleration: $\\mathbf{a}_P = \\mathbf{a}_O + (\\mathbf{a}_{P/O})_{rel} + \\boldsymbol\\alpha\\times\\mathbf{r}_{P/O} + 2\\boldsymbol\\omega\\times(\\mathbf{v}_{P/O})_{rel} + \\boldsymbol\\omega\\times(\\boldsymbol\\omega\\times\\mathbf{r}_{P/O})$ (given in the question)\n- Unit-vector cross products: $\\mathbf{k}\\times\\mathbf{i} = \\mathbf{j}$, $\\mathbf{k}\\times\\mathbf{j} = -\\mathbf{i}$ (learn this)\n- Newton's second law for a particle: $\\sum\\mathbf{F} = m\\mathbf{a}$, in components (learn this)\n- Kinetic friction $F = \\mu_k N$, opposing the relative sliding; static $F \\le \\mu_s N$ only when there is no sliding (learn this)\n","parts":[{"label":"a","text":"Draw the free-body and kinetic diagrams of the particle P. State appropriate coordinate systems.\n","marks":10,"grade":"2:1"},{"label":"b","text":"Determine the acceleration of the particle P. Hint: use the following equation:\n\n$$a_P = a_O + (a_{P/O})_{rel} + \\alpha \\times r_{P/O} + 2\\omega \\times (v_{P/O})_{rel} + \\omega \\times (\\omega \\times r_{P/O}).$$\n","marks":10,"grade":"2:1"},{"label":"c","text":"Determine the normal force and friction force of the particle P.\n","marks":5,"grade":"2:2"}],"solution":"$f7","confidence":"MEDIUM","solutionNote":"Agrees with the unofficial worked answers PDF (a_P = -(omega^2 L + alpha h + 2 omega v_rel) i\n+ (a_rel + alpha L - omega^2 h) j; N = m(omega^2 L + alpha h + 2 omega v_rel); F = mu_k N). MEDIUM because the\nquestion does not say whether the plane is vertical or horizontal or whether gravity is ignored: N and F are\nthe same for \"no gravity\" and \"vertical plane\", but a_rel changes, and a horizontal plane with the weight on the\ntube would change F. Confirm the intended plane with the lecturer. No numerical answers (all results are\nsymbolic combinations, not single coefficients).\n"},{"id":"engr216-2024-a1-statics","module":"ENGR5004","week":6,"label":"ENGR216 2024 Statics Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["shear-bending-diagrams","bending-eccentric","beam-deflection","frame","i-section","combined-axial-bending"],"source":{"name":"ENGR216 Summer 2024 Statics Q A1","paper":"ENGR216","year":2024,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR216 Statics Online Exam 2024.pdf","tag":"VALID"},"text":"Figure A1 shows a constrained structure of negligible mass constrained by a roller at end P and a pin at end\nR. End P is loaded by a horizontal concentrated load $F$, as indicated in Figure A1.\n","figures":[{"src":"/figures/engr216-2024-a1-statics.png","caption":"Figure A1-a: frame with vertical PQ (length L, roller at P, horizontal load F at P) and horizontal QR (length 2L, pin at R). Figure A1-b: I cross-section."}],"formulaSheet":"- Equilibrium of a rigid body in the plane: $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$ (learn this)\n- Internal forces by the method of sections; sign convention for positive shear and positive bending as pictured, $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n- Pure bending $\\sigma_x = -\\dfrac{My}{I}$, $\\sigma_{max} = \\dfrac{|M|c}{I}$ (on the formula sheet)\n- Combined axial force and bending (eccentric loading form) $\\sigma_x = \\dfrac{N}{A} - \\dfrac{My}{I}$ (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$ (on the formula sheet); I-section by subtraction or by three rectangles with the parallel-axis theorem $I = I_c + Ad^2$ (learn this)\n- Elastic curve $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$ for the curvature argument in (d) (learn this; on the 2025 sheet but not the 2026 one)\n","parts":[{"label":"a","text":"Draw the free-body-diagram of the entire structure, indicating only nonzero reactions and applied loads.\n","marks":4,"grade":"third"},{"label":"b","text":"Calculate the reactions acting on the considered structure.\n","marks":4,"grade":"2:2","answer":[{"label":"P_y (up +)","value":0.5,"units":"F"},{"label":"R_x (right +)","value":-1,"units":"F"},{"label":"R_y (up +)","value":-0.5,"units":"F"}]},{"label":"c","text":"Calculate the axial load distribution $N(x)$, the sharing load distribution $V(x)$ and the bending moment\ndistribution $M(x)$ along members PQ and QR.\n","marks":6,"grade":"2:2","answer":[{"label":"N_PQ (tension +)","value":-0.5,"units":"F"},{"label":"|V_PQ|","value":1,"units":"F"},{"label":"N_QR (tension +)","value":-1,"units":"F"},{"label":"|V_QR|","value":0.5,"units":"F"},{"label":"|M| at Q","value":1,"units":"F*L"}]},{"label":"d","text":"Draw as accurately as possible the bending moment diagram $M(x)$ along PQ and QR and the elastic curve of\nthe whole structure, explaining why the curvature of the elastic curve is consistent with the diagram of\n$M(x)$ in both PQ and QR.\n","marks":6,"grade":"2:1"},{"label":"e","text":"Calculate the minimum length $L$ required for the magnitude of the maximum normal stress in the cross\nsection of PQ at distance $0.95L$ from P not to exceed 120 MPa. The beam cross section is reported in\nFigure A1-b. Use $a = t = 30$ mm, $b = 270$ mm and $F = 100$ kN. The section is positioned so that the\nneutral axis of the bending moment alone is parallel to side AB.\n","marks":5,"grade":"2:1","answer":[{"label":"A","value":22500,"units":"mm^2"},{"label":"I (axis parallel to AB)","value":257650000,"units":"mm^4"},{"label":"L_max","value":2.366,"units":"m"}]}],"solution":"$f8","confidence":"MEDIUM","solutionNote":"Agrees with the unofficial worked answers PDF (P_y = F/2 up, R_x = F left, R_y = F/2 down; same N, V, M;\nL = 2.37 m). MEDIUM only because of the paper's wording: (e) asks for a \"minimum\" length, but the stress rises\nwith L, so the 120 MPa limit gives a maximum L = 2.37 m; confirm the intended reading with the lecturer. The\npaper also prints \"N(x) (2,\" and \"sharing load\" in (c), read as shearing load.\n"},{"id":"engr216-2024-a2-statics","module":"ENGR5004","week":6,"label":"ENGR216 2024 Statics Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["beam-deflection","statically-indeterminate","elastic-curve","slope","sinusoidal-load"],"source":{"name":"ENGR216 Summer 2024 Statics Q A2","paper":"ENGR216","year":2024,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR216 Statics Online Exam 2024.pdf","tag":"VALID"},"text":"Figure A2 shows a beam of negligible mass with flexural rigidity $EI$ and length $L$. The beam is constrained\nby a roller at end A and a fixed support at end B, and is subjected to a distributed load $w(x)$ given by:\n\n$$w(x) = w_0 \\sin\\left(\\frac{\\pi x}{2L}\\right) \\qquad \\text{(Equation 1)}$$\n\nwhere $x$ denotes the distance from end A along the beam.\n","figures":[{"src":"/figures/engr216-2024-a2-statics.png","caption":"Figure A2: beam of length L, roller at A, fixed at B, load w(x) rising from zero at A."}],"formulaSheet":"- Resultant of a distributed load $P = \\int w\\,dx$ and its position $x_P = \\dfrac{\\int x\\,w\\,dx}{\\int w\\,dx}$ (on the formula sheet)\n- Equilibrium of a rigid body in the plane: $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$ (learn this)\n- Load, shear and moment: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$, positive shear and bending as pictured (on the formula sheet)\n- Elastic curve $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$, slope $\\theta = \\dfrac{dy}{dx}$ (learn this; on the 2025 sheet but not the 2026 one)\n- Solid circle $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n- Degrees to radians, $1^\\circ = \\pi/180\\ \\mathrm{rad}$ (learn this)\n","parts":[{"label":"a","text":"Draw the free-body-diagram of the entire structure, indicating only nonzero reactions and applied loads.\n","marks":4,"grade":"third"},{"label":"b","text":"Explain why the system is statically indeterminate (3 marks), and write the equations enforcing the static\nequilibrium of the system (3 marks).\n","marks":6,"grade":"2:2"},{"label":"c","text":"Knowing that the elastic curve of the beam is:\n\n$$y(x) = \\frac{w_0 L}{EI}\\left(\\frac{2}{\\pi}\\right)^4 \\left[-\\frac{1}{2}x^3 + \\frac{3}{2}L^2 x - L^3 \\sin\\left(\\frac{\\pi x}{2L}\\right)\\right] \\qquad \\text{(Equation 2)}$$\n\ncalculate all reactions acting on the beam as functions of $w_0$ and $L$.\n","marks":6,"grade":"2:1","answer":[{"label":"A_y (up +)","value":0.14385,"units":"w0*L"},{"label":"B_y (up +)","value":0.49277,"units":"w0*L"},{"label":"M_B (CCW +)","value":-0.087482,"units":"w0*L^2"}]},{"label":"d","text":"Provide a schematic of the elastic curve, making sure your drawing is consistent with the applied\nconstraints.\n","marks":4,"grade":"2:2"},{"label":"e","text":"The beam has a circular cross section. Determine the minimum value of the section radius such that the\nslope of the elastic curve at end A does not exceed 1 degree. Use $w_0 = 80$ kN/m, $L = 3.0$ m and\n$E = 200$ GPa.\n","marks":5,"grade":"2:2","answer":[{"label":"r_min","value":55.02,"units":"mm"}]}],"solution":"$f9","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers PDF (A_y = 0.144 w0 L up, B_y = 0.493 w0 L up, M_B = 0.0875 w0 L^2\nclockwise; theta_A = -0.01163 w0 L^3/EI; r_min = 55 mm). No official solution exists.\n"},{"id":"engr216-2024-a1-dynamics","module":"ENGR5004","week":5,"label":"ENGR216 2024 Dynamics Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["particle-kinetics","polar-coordinates","spring","normal-force","rotating-arm"],"source":{"name":"ENGR216 Summer 2024 Dynamics Q A1","paper":"ENGR216","year":2024,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR216 Dynamics Online Exam 2024.pdf","tag":"VALID"},"text":"The smooth particle $P$ moves in the slot of a rotating arm $OB$ and is attached to a spring extending from the\nfixed point $A$ to $P$. The mass of $P$ is $m$. The spring's stiffness is $k_{sp}$. The unstretched length of\nthe spring is $L_0$. The radial position and radial velocity of $P$ are $r$ and $\\dot{r}$, respectively. At the\ninstant shown in Figure A1: the angle $\\phi$ is known, and the length of the spring is $1.5L_0$. The arm $OB$\nis rotating with a constant angular speed $\\omega$ (clockwise) to the horizontal position. $\\mathbf{u}_r$ and\n$\\mathbf{u}_\\theta$ represent the unit vectors of radial and transverse directions, respectively.\n\nNote: $\\angle APB = \\phi = 60^\\circ$, $m = 3$ kg, $k_{sp} = 10$ N/m, $L_0 = 1.6$ m, $\\omega = 2$ rad/s.\n","figures":[{"src":"/figures/engr216-2024-a1-dynamics.png","caption":"Figure A1: particle P in the slot of arm OB (rotating clockwise about O), spring from fixed point A to P at angle phi."}],"formulaSheet":"- Polar velocity $\\mathbf{v} = \\dot r\\,\\mathbf{u}_r + r\\dot\\theta\\,\\mathbf{u}_\\theta$ (learn this)\n- Polar acceleration $\\mathbf{a} = (\\ddot r - r\\dot\\theta^2)\\,\\mathbf{u}_r + (r\\ddot\\theta + 2\\dot r\\dot\\theta)\\,\\mathbf{u}_\\theta$ (learn this)\n- Newton's second law in polar components $\\sum F_r = m a_r$, $\\sum F_\\theta = m a_\\theta$ (learn this)\n- Linear spring force $F_s = k_{sp}(L - L_0)$, directed along the spring (learn this)\n","parts":[{"label":"a","text":"Draw the free body diagram and the kinetic diagram of $P$.\n","marks":10,"grade":"2:2"},{"label":"b","text":"Determine the acceleration of the particle by using the variables: $\\phi$, $k_{sp}$, $\\omega$, $m$, $L_0$,\n$r$ and $\\dot{r}$.\n","marks":5,"grade":"2:1","answer":[{"label":"a_r (with data of (c))","value":1.3333,"units":"m/s^2"},{"label":"a_theta (with data of (c))","value":-20,"units":"m/s^2"}]},{"label":"c","text":"Given that the velocity of $P$ is $\\mathbf{v}_p = 5\\mathbf{u}_r - 2.4\\mathbf{u}_\\theta$ m/s at the instant,\ncalculate the numerical value for the normal force.\n","marks":5,"grade":"2:2","answer":[{"label":"N (u_theta +)","value":-66.93,"units":"N"}]},{"label":"d","text":"Under the conditions in (c), is the particle pressing against the lower part of the slot or the upper part?\nGive your reasoning.\n","marks":5,"grade":"2:1"}],"solution":"$fa","confidence":"MEDIUM","solutionNote":"Agrees with the unofficial worked answers PDF (a = 1.333 u_r - 20 u_theta m/s^2; N = -66.9 N; upper wall).\nMEDIUM because the paper does not say whether the arm moves in a horizontal or vertical plane. Gravity is\nneglected here (no g given); if the figure is a vertical plane, N = -37.5 N instead. Confirm which was intended;\nthe answer to (d) is the same either way.\n"},{"id":"engr216-2024-a2-dynamics","module":"ENGR5004","week":7,"label":"ENGR216 2024 Dynamics Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["rigid-kinematics","linkage","relative-velocity","relative-acceleration","slider"],"source":{"name":"ENGR216 Summer 2024 Dynamics Q A2","paper":"ENGR216","year":2024,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second year exam papers 2024/ENGR216 Dynamics Online Exam 2024.pdf","tag":"VALID"},"text":"Figure A2 shows a lightweight lever-type steel pushing machine. The crank $AB$ is rotating with\n$\\omega_{OA} = 0.5$ rad/s, connecting the crank $AB$ to make the rod $O_1B$ rotate around the axis $O_1$. Rod\n$DC$ is connected to slider block $C$ through a hinge, and the slider block $C$ can slide along rod $O_1B$.\n$OA = r$, $AB = \\sqrt{3}\\,r$, $O_1B = \\frac{2}{3}l$, $r = 0.2$ m, $l = 1$ m. At this instant,\n$BC = \\frac{4}{3}l$, $AB$ and $DC$ are both horizontal, and the induced angle between $CB$ and the horizontal\nline is $60^\\circ$. Neglect the tangential acceleration of point $A$.\n","figures":[{"src":"/figures/engr216-2024-a2-dynamics.png","caption":"Figure A2: crank OA, horizontal link AB, rod O1B pivoted at O1, slider C on O1B pushing horizontal rod DC."}],"formulaSheet":"- Rotation about a fixed axis: $\\mathbf{v} = \\boldsymbol\\omega\\times\\mathbf{r}$, $\\mathbf{a} = \\boldsymbol\\alpha\\times\\mathbf{r} - \\omega^2\\mathbf{r}$, $a_t = \\alpha r$, $a_n = \\omega^2 r$ (learn this)\n- Relative velocity on a rigid body $\\mathbf{v}_B = \\mathbf{v}_A + \\boldsymbol\\omega\\times\\mathbf{r}_{B/A}$ (learn this)\n- Relative acceleration on a rigid body $\\mathbf{a}_B = \\mathbf{a}_A + \\boldsymbol\\alpha\\times\\mathbf{r}_{B/A} - \\omega^2\\mathbf{r}_{B/A}$ (learn this)\n- Point sliding on a rotating body (rotating axes): $\\mathbf{v}_C = \\mathbf{v}_{C'} + \\mathbf{v}_{rel}$, $\\mathbf{a}_C = \\mathbf{a}_{C'} + \\mathbf{a}_{rel} + 2\\boldsymbol\\omega\\times\\mathbf{v}_{rel}$ (learn this)\n- Cross products in the plane: $\\mathbf{k}\\times\\mathbf{i} = \\mathbf{j}$, $\\mathbf{k}\\times\\mathbf{j} = -\\mathbf{i}$ (learn this)\n","parts":[{"label":"a","text":"Show that the velocity of point $B$ with respect to $A$, $v_{BA}$, is $\\frac{\\sqrt{3}}{30}\\,\\mathbf{j}$ m/s.\n","marks":5,"grade":"2:2","answer":[{"label":"v_B/A (j component)","value":0.057735,"units":"m/s"},{"label":"omega_O1B (CCW +)","value":0.17321,"units":"rad/s"},{"label":"omega_AB (CCW +)","value":-0.16667,"units":"rad/s"}]},{"label":"b","text":"Determine the tangential acceleration and angular acceleration of point $B$ with respect to $O_1$.\n","marks":10,"grade":"2:1","answer":[{"label":"(a_B)_t (along v_B +)","value":-0.022658,"units":"m/s^2"},{"label":"alpha_O1B (CCW +)","value":-0.033987,"units":"rad/s^2"}]},{"label":"c","text":"Determine the velocity of the slider block $C$.\n","marks":5,"grade":"2:2","answer":[{"label":"v_C (right +)","value":-0.4,"units":"m/s"}]},{"label":"d","text":"Determine the acceleration of the slider block $C$.\n","marks":5,"grade":"first","answer":[{"label":"a_C (right +)","value":0.15849,"units":"m/s^2"}]}],"solution":"$fb","confidence":"HIGH","solutionNote":"Agrees with the unofficial worked answers PDF (omega_O1B = 0.1732 rad/s CCW; alpha_O1B = 0.0340 rad/s^2 CW;\n(a_B)_t = 0.0227 m/s^2; v_C = 0.4 m/s left; a_C = 0.158 m/s^2 right). Every result was also confirmed by a\nnumerical simulation of the mechanism. No official solution exists.\n"},{"id":"engr216-2023-a1-statics","module":"ENGR5004","week":6,"label":"ENGR216 2023 Statics Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["beam-deflection","shear-bending-diagrams","bending-eccentric","frame","parabolic-load"],"source":{"name":"ENGR216 Summer 2023 Statics Q A1","paper":"ENGR216","year":2023,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR216 Exam Stat 2023.pdf","tag":"VALID"},"text":"Figure A1 shows a cantilevered structure of negligible mass. The vertical member AB of the structure is loaded\nby a parabolic distributed normal load $w(x)$, which is given by:\n\n$$w(x) = 4w_0\\left[-\\left(\\frac{x}{L}\\right)^2 + \\left(\\frac{x}{L}\\right)\\right] \\qquad \\text{(Equation 1)}$$\n\nwhere $x$ is the coordinate along the mean line of the vertical member AB (origin at A), which has length $L$.\n","figures":[{"src":"/figures/engr216-2023-a1-statics.png","caption":"Figure A1: vertical member AB (length L, fixed at A) with horizontal parabolic load, horizontal member BC (length L)."}],"formulaSheet":"- Distributed load resultant: $P = \\int w\\,dx$, acting at $x_P = \\dfrac{\\int x\\,w\\,dx}{\\int w\\,dx}$ (on the formula sheet)\n- Load-shear-moment relations: $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ and their integral forms (on the formula sheet)\n- Equilibrium of a rigid body: $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M_A = 0$ (learn this)\n- Elastic curve: $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$, slope $\\theta = \\dfrac{dy}{dx}$, integrate with the support conditions (learn this; on the 2025 sheet but not the 2026 one)\n- Small rotation of a straight member: transverse displacement $= L\\,\\theta$ (learn this)\n- Mean normal stress: $\\sigma_{mean} = N/A$ (learn this)\n- Pure bending: $\\sigma_x = -\\dfrac{My}{I}$, $\\sigma_{max} = \\dfrac{|M|c}{I}$ (on the formula sheet)\n- Square section: $I = \\dfrac{a^4}{12}$ (from $bh^3/12$) (on the formula sheet)\n","parts":[{"label":"a","text":"Draw the free-body-diagram of the entire structure, indicating only nonzero reactions and applied loads.\n","marks":5,"grade":"2:2"},{"label":"b","text":"Calculate the reactions acting on the considered structure (3 marks), and draw the elastic curve of the\nmember BC as accurately as possible, accounting for the suppressed degrees of freedom (2 marks).\n","marks":5,"grade":"2:2","answer":[{"label":"A_x (right +)","value":-0.66667,"units":"w0*L"},{"label":"A_y (up +)","value":0,"units":"w0*L"},{"label":"M_A (CCW +)","value":0.33333,"units":"w0*L^2"}]},{"label":"c","text":"Knowing that the bending moment $M(x)$ along the vertical member AB is\n\n$$M(x) = w_0 L^2\\left[\\frac{1}{3}\\left(\\frac{x}{L}\\right)^4 - \\frac{2}{3}\\left(\\frac{x}{L}\\right)^3 + \\frac{2}{3}\\left(\\frac{x}{L}\\right) - \\frac{1}{3}\\right] \\qquad \\text{(Equation 2)}$$\n\ndemonstrate that the magnitude $\\delta_C$ of the vertical displacement of point C is:\n\n$$\\delta_C = \\frac{1}{10}\\frac{w_0 L^4}{EI} \\qquad \\text{(Equation 3)}$$\n\nwhere $EI$ is the flexural rigidity (5 marks). State what is the mean normal stress in the cross sections of\nmembers AB and BC, and explain why (5 marks).\n","marks":10,"grade":"2:1","answer":[{"label":"delta_C (down)","value":0.1,"units":"w0*L^4/EI"},{"label":"sigma_mean in AB and BC","value":0,"units":"MPa"}]},{"label":"d","text":"Member AB has length $L = 2$ m and features a square cross section, and $w_0 = 40$ kNm. Calculate the\nminimum length of the edge of the square cross section such that the normal stress at the midpoint of\nmember AB does not exceed 130 MPa.\n","marks":5,"grade":"2:2","answer":[{"label":"|M| at midpoint of AB","value":10,"units":"kN m"},{"label":"a_min","value":77.28,"units":"mm"}]}],"solution":"$fc","confidence":"HIGH","solutionNote":"No official solution. Agrees with the unofficial worked answers PDF on every number (A_x = 2w0L/3 left,\nM_A = w0L^2/3 CCW, theta_B = w0L^3/(10EI) clockwise, delta_C = w0L^4/(10EI), sigma_mean = 0, a_min = 77.3 mm).\nThat PDF says the right (loaded) face of AB is in tension; with the load pushing AB to the right the left face\nis in tension, as stated here (no number is affected).\n"},{"id":"engr216-2023-a2-statics","module":"ENGR5004","week":6,"label":"ENGR216 2023 Statics Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["beam-deflection","shear-bending-diagrams","point-moment","elastic-curve","boundary-conditions"],"source":{"name":"ENGR216 Summer 2023 Statics Q A2","paper":"ENGR216","year":2023,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR216 Exam Stat 2023.pdf","tag":"VALID"},"text":"Figure A2 shows a beam of negligible mass with flexural rigidity $EI$ and length $2L$. The beam is subjected to\na pointwise moment $M_0$ at point P at a distance of $1.5L$ from the left end A.\n","figures":[{"src":"/figures/engr216-2023-a2-statics.png","caption":"Figure A2: beam of length 2L, roller at A, pin at B, concentrated moment M0 at 1.5L from A. Figure A3: rectangular section w x h."}],"formulaSheet":"- Equilibrium of a rigid body: $\\sum F_y = 0$, $\\sum M_A = 0$; a couple has the same moment about every point (learn this)\n- Positive shear and bending sign convention pictures; $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n- Elastic curve: $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$, two constants per segment, boundary and continuity conditions (learn this; on the 2025 sheet but not the 2026 one)\n- Rectangle: $I = \\dfrac{wh^3}{12}$ about the horizontal centroidal axis (on the formula sheet)\n","parts":[{"label":"a","text":"Draw the free body diagram of the loaded beam, indicating only nonzero reactions and applied loads\n(3 marks). Explain if and why the reactions would change if the moment was applied at the midpoint of the\nbeam (2 marks).\n","marks":5,"grade":"2:2","answer":[{"label":"A_y (up +)","value":0.5,"units":"M0/L"},{"label":"B_y (up +)","value":-0.5,"units":"M0/L"}]},{"label":"b","text":"Draw the elastic curve of the beam as accurately as possible, accounting for the suppressed degrees of\nfreedom.\n","marks":4,"grade":"2:1"},{"label":"c","text":"Determine the equation of the bending moment $M(x)$ along the entire beam as a function of $L$ and $M_0$.\n","marks":5,"grade":"2:2","answer":[{"label":"M just left of P (sagging +)","value":0.75,"units":"M0"},{"label":"M just right of P (sagging +)","value":-0.25,"units":"M0"}]},{"label":"d","text":"Knowing that the vertical displacement along the beam axis is:\n\n$$y(x) = \\frac{1}{EI}\\left(\\frac{M_0}{12L}x^3 + c_1 x + c_2\\right) \\quad \\text{if } 0 < x < 1.5L \\qquad \\text{(Equation 4)}$$\n\n$$y(x) = \\frac{1}{EI}\\left(\\frac{M_0}{12L}x^3 - \\frac{M_0}{2L}x^2 + c_3 x + c_4\\right) \\quad \\text{if } 1.5L < x < 2L \\qquad \\text{(Equation 5)}$$\n\nstate what are the four conditions to determine the constants $c_1$, $c_2$, $c_3$, and $c_4$ (3 marks), and\ncalculate them (3 marks).\n","marks":6,"grade":"2:1","answer":[{"label":"c1","value":-0.270833,"units":"M0*L"},{"label":"c2","value":0,"units":"M0*L^2"},{"label":"c3","value":1.229167,"units":"M0*L"},{"label":"c4","value":-1.125,"units":"M0*L^2"}]},{"label":"e","text":"As shown in Figure 3, the beam has a rectangular cross section of width $w$ and height $h$. Determine the\nminimum value of $h$ such that the magnitude of the vertical displacement of the beam at $x/L = 1.04$ does\nnot exceed 1 mm. Use $M_0 = 60$ kNm, $w = 180$ mm, $L = 2.0$ m and $E = 200$ GPa.\n","marks":5,"grade":"2:2","answer":[{"label":"h_min","value":246.8,"units":"mm"}]}],"solution":"$fd","confidence":"HIGH","solutionNote":"No official solution. Agrees with the unofficial worked answers PDF (A_y = M0/(2L) up, B_y = M0/(2L) down;\nM = M0 x/(2L) and M0 x/(2L) - M0; c1 = -13/48 M0 L, c2 = 0, c3 = 59/48 M0 L, c4 = -9/8 M0 L^2; h_min = 247 mm).\nEquation 5 as printed has a misprint (-M0 x^2/(2L) should be -M0 x^2/2); the transcription keeps the paper's\ntext and the solution explains the correction. That PDF swaps the supports (pin at A, roller at B); Figure A2\nshows the roller at A and the pin at B, which changes no number.\n"},{"id":"engr216-2023-b1-dynamics","module":"ENGR5004","week":2,"label":"ENGR216 2023 Dynamics Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["particle-kinematics","polar-coordinates","radius-of-curvature","crane"],"source":{"name":"ENGR216 Summer 2023 Dynamics Q B1","paper":"ENGR216","year":2023,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR216 Exam Dynamics 2023.pdf","tag":"PARTIAL","leftOut":"Part (b) (estimating the boom mass from a photograph and its effect on safe working load) is open-ended engineering judgement, not covered in ENGR5004."},"text":"Figure B1 shows an extendable boom mobile crane, typical of the type that was used to construct the new\nEngineering Building. Before the boom is retracted, distance AB = 25 m and $\\theta = 60^\\circ$. To retract the\nboom, Point B is moved toward Point A with a constant radial acceleration of $-0.5\\ \\mathrm{m/s^2}$ and is\nlowered toward the vehicle deck with a constant angular acceleration of $-0.1\\ \\mathrm{rad/s^2}$. At the time\n$t = 2$ seconds after the boom begins to retract, determine:\n","figures":[{"src":"/figures/engr216-2023-b1-dynamics.png","caption":"Figure B1: mobile extendable boom crane, boom AB at angle theta to the horizontal (left), and the crane on site (right)."}],"formulaSheet":"- Constant acceleration: $\\dot r = \\dot r_0 + \\ddot r\\,t$, $r = r_0 + \\dot r_0 t + \\tfrac12\\ddot r\\,t^2$ (and the same for $\\theta$) (learn this)\n- Polar velocity: $\\mathbf{v} = \\dot r\\,\\mathbf{u}_r + r\\dot\\theta\\,\\mathbf{u}_\\theta$ (learn this)\n- Polar acceleration: $\\mathbf{a} = (\\ddot r - r\\dot\\theta^2)\\,\\mathbf{u}_r + (r\\ddot\\theta + 2\\dot r\\dot\\theta)\\,\\mathbf{u}_\\theta$ (learn this)\n- Normal-tangential components: $a_t = \\dfrac{\\mathbf{a}\\cdot\\mathbf{v}}{v}$, $a_n = \\sqrt{a^2 - a_t^2}$, $\\rho = \\dfrac{v^2}{a_n}$ (learn this)\n","parts":[{"label":"a(i)","text":"The velocity of Point B\n","marks":5,"grade":"2:2","answer":[{"label":"v_r (outwards +)","value":-1,"units":"m/s"},{"label":"v_theta (theta increasing +)","value":-4.8,"units":"m/s"},{"label":"|v_B|","value":4.903,"units":"m/s"}]},{"label":"a(ii)","text":"The acceleration of Point B\n","marks":5,"grade":"2:2","answer":[{"label":"a_r (outwards +)","value":-1.46,"units":"m/s^2"},{"label":"a_theta (theta increasing +)","value":-2,"units":"m/s^2"},{"label":"|a_B|","value":2.476,"units":"m/s^2"}]},{"label":"a(iii)","text":"The radius of curvature of the path that B follows.\n","marks":5,"grade":"2:1","answer":[{"label":"rho","value":23.54,"units":"m"}]},{"label":"b(i)","text":"Using the photograph of the crane on site shown in Figure B1, estimate the mass of the boom if length\nAB = 60 m when fully extended. The boom is manufactured from hollow box section steel. State any\nassumptions that you make.\n","marks":5,"grade":"$undefined"},{"label":"b(ii)","text":"How would the mass of the boom influence the designed safe working load of the crane?\n","marks":5,"grade":"$undefined"}],"solution":"$fe","confidence":"HIGH","solutionNote":"No official solution. Part (a) agrees with the unofficial worked answers PDF (v = -1.0 u_r - 4.8 u_theta m/s,\n4.90 m/s; a = -1.46 u_r - 2.0 u_theta m/s^2, 2.48 m/s^2; rho = 23.5 m). Part (a) assumes the boom starts from\nrest at t = 0. Part (b) is off-syllabus and only an indicative estimate is given (the PDF estimates about 15 t too).\n"},{"id":"engr216-2023-b2-dynamics","module":"ENGR5004","week":8,"label":"ENGR216 2023 Dynamics Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"first","tags":["rigid-kinematics","rigid-kinetics","linkage","relative-acceleration","pin-forces"],"source":{"name":"ENGR216 Summer 2023 Dynamics Q B2","paper":"ENGR216","year":2023,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2023/ENGR216 Exam Dynamics 2023.pdf","tag":"VALID"},"text":"Figure B2 shows a crank driven assembly that comprises three connected slender bars (AB, BC and CDE) that\nproduce oscillatory motion at point E. Each of the bars are of uniform section and have a mass moment of\ninertia for $z$-$z$ rotation about their centre of mass ($I_G$) of $\\frac{1}{12}ml^2$, where $m$ = mass and\n$l$ = length. Data: AB = 0.3 m, BC = 0.5 m, CD = 0.3 m, DE = 0.6 m, $\\dot{\\theta} = 10$ rad/s. For the\ninstant shown:\n","figures":[{"src":"/figures/engr216-2023-b2-dynamics.png","caption":"Figure B2: crank AB (vertical, pinned at A) driving link BC and bar CDE pivoted at D; CDE horizontal."}],"formulaSheet":"- Rigid-body relative velocity: $\\mathbf{v}_B = \\mathbf{v}_A + \\boldsymbol\\omega\\times\\mathbf{r}_{B/A}$ (learn this)\n- Rigid-body relative acceleration: $\\mathbf{a}_B = \\mathbf{a}_A + \\boldsymbol\\alpha\\times\\mathbf{r}_{B/A} - \\omega^2\\mathbf{r}_{B/A}$ (learn this)\n- Instantaneous centre of zero velocity (as a check) (learn this)\n- Slender bar: $I_G = \\tfrac{1}{12}ml^2$ (given in the question)\n- Parallel-axis theorem: $I_D = I_G + m d^2$ (learn this)\n- Planar kinetics: $\\sum\\mathbf{F} = m\\mathbf{a}_G$, $\\sum M_G = I_G\\alpha$; about a fixed pin $\\sum M_D = I_D\\alpha$ (learn this)\n","parts":[{"label":"i","text":"Show that the velocity of point B is $-3\\mathbf{i}$ m/s\n","marks":2,"grade":"third","answer":[{"label":"v_Bx","value":-3,"units":"m/s"}]},{"label":"ii","text":"Show that the velocity of point C is $4\\mathbf{j}$ m/s\n","marks":3,"grade":"2:2","answer":[{"label":"omega_BC (CCW +)","value":10,"units":"rad/s"},{"label":"v_Cy","value":4,"units":"m/s"}]},{"label":"iii","text":"Determine the angular velocity of bar CDE\n","marks":3,"grade":"third","answer":[{"label":"omega_CDE (CCW +)","value":-13.333,"units":"rad/s"}]},{"label":"iv","text":"Determine the angular accelerations of bars BC and CDE\n","marks":6,"grade":"2:1","answer":[{"label":"alpha_BC (CCW +)","value":311.1,"units":"rad/s^2"},{"label":"alpha_CDE (CCW +)","value":-414.8,"units":"rad/s^2"}]},{"label":"v","text":"Determine the acceleration of the centre of mass of BC\n","marks":5,"grade":"2:2","answer":[{"label":"a_Gx","value":26.67,"units":"m/s^2"},{"label":"a_Gy","value":47.22,"units":"m/s^2"},{"label":"|a_G|","value":54.23,"units":"m/s^2"}]},{"label":"vi","text":"Given that the mass of BC = 2 kg and the mass of CDE is 3 kg, determine the forces exerted on BC by the\npins at B and C.\n","marks":6,"grade":"first","answer":[{"label":"B_x on BC (right +)","value":-570.7,"units":"N","tol":0.05},{"label":"B_y on BC (up +)","value":472.7,"units":"N","tol":0.05},{"label":"C_x on BC (right +)","value":624.1,"units":"N","tol":0.05},{"label":"C_y on BC (up +)","value":-358.6,"units":"N","tol":0.05}]}],"solution":"$ff","confidence":"MEDIUM","solutionNote":"No official solution. Agrees with the unofficial worked answers PDF on every number (omega_BC = 10, omega_CDE =\n-13.33 rad/s; alpha_BC = 311.1, alpha_CDE = -414.8 rad/s^2; a_G = 26.67 i + 47.22 j m/s^2; F_B = -570.7 i +\n472.7 j N, F_C = 624.1 i - 358.6 j N). MEDIUM only for (vi): the paper does not say whether gravity acts in the\nplane of the figure or whether the crank speed is constant; we assumed a vertical plane (g included) and\nalpha_AB = 0. Without gravity the pin forces change by 1-4 percent (the answer tolerance accepts both).\nConfirm the intended assumption with the lecturer.\n"},{"id":"engr216-2022-a1-statics","module":"ENGR5004","week":6,"label":"ENGR216 2022 Statics Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["shear-bending-diagrams","beam-deflection","bending-eccentric","frame","parabolic-load","slope"],"source":{"name":"ENGR216 Summer 2022 Statics Q A1","paper":"ENGR216","year":2022,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR216 Exam Part A 2021-22.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2022/ENGR216_statics_num_ans_2022.pdf","tag":"VALID"},"text":"Consider the cantilevered structure of Figure A1, loaded by a parabolic distributed normal load $w(x)$, given by\n\n$$w(x) = w_0\\left[-\\left(\\frac{x}{L}\\right)^2 + 2\\left(\\frac{x}{L}\\right)\\right] \\qquad (1)$$\n\nwhere $x$ is the coordinate along the mean line of the horizontal member BC (origin at B), which has length $L$.\n","figures":[{"src":"/figures/engr216-2022-a1-statics.png","caption":"Figure A1: column AB fixed at A, horizontal member BC of length L with parabolic load reaching w0 at C."}],"formulaSheet":"- Equilibrium of a rigid body: $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M_A = 0$ (learn this)\n- Resultant of a distributed load $P = \\int w\\,dx$ and its position $x_P = \\dfrac{\\int x\\,w\\,dx}{\\int w\\,dx}$ (on the formula sheet)\n- Load, shear and moment: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$, positive shear and bending as pictured (on the formula sheet)\n- Elastic curve $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$, so $\\theta_C - \\theta_B = \\dfrac{1}{EI}\\displaystyle\\int_B^C M\\,dx$ (learn this; on the 2025 sheet but not the 2026 one)\n- Combined axial force and bending: $\\sigma_x = \\dfrac{N}{A} - \\dfrac{M y}{I}$, $|\\sigma|_{max} = \\dfrac{|N|}{A} + \\dfrac{|M| c}{I}$ (on the formula sheet, as eccentric axial loading)\n- Solid circle: $A = \\pi r^2$, $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n","parts":[{"label":"a","text":"Draw the free-body-diagram of the entire structure, indicating all nonzero reactions.\n","marks":7,"grade":"2:2","answer":[{"label":"A_y (up +)","value":0.66667,"units":"w0*L"},{"label":"A_x","value":0,"units":"w0*L"}]},{"label":"b","text":"Demonstrate that the bending moment exerted by the fixed support at A has magnitude\n\n$$M_A = \\frac{5}{12} w_0 L^2 \\qquad (2)$$\n\nand state if this moment is clockwise or counter-clockwise.\n","marks":6,"grade":"2:2","answer":[{"label":"M_A (CCW +)","value":0.41667,"units":"w0*L^2"}]},{"label":"c","text":"Knowing that the bending moment $M(x)$ along the horizontal member BC is\n\n$$M(x) = w_0 L^2\\left[\\frac{1}{12}\\left(\\frac{x}{L}\\right)^4 - \\frac{1}{3}\\left(\\frac{x}{L}\\right)^3 + \\frac{2}{3}\\left(\\frac{x}{L}\\right) - \\frac{5}{12}\\right] \\qquad (3)$$\n\ndetermine the rotation of section C relative to section B as a function of $w_0$, $L$, and the flexural\nrigidity $EI$.\n","marks":6,"grade":"2:1","answer":[{"label":"theta_C - theta_B (CCW +)","value":-0.15,"units":"w0*L^3/EI"}]},{"label":"d","text":"The cross section of members AB and BC are circular with radius of 50 mm, $L = 1$ m and the allowable\nnormal stress of the material is 150 MPa. Briefly explain which cross section(s) of the given structure have\nmaximum magnitude of the normal stress, and calculate the maximum value of $w_0$ in kN/m such that the\nnormal stress in all cross sections of AB and BC never exceeds the allowable normal stress.\n","marks":6,"grade":"2:1","answer":[{"label":"w0_max","value":34.65,"units":"kN/m"}]}],"solution":"$100","confidence":"HIGH","solutionNote":"Matches the official numerical answers: (c) 3/20 w0 L^3/EI clockwise, (d) (w0)max = 34.65 kN/m.\n"},{"id":"engr216-2022-a2-statics","module":"ENGR5004","week":6,"label":"ENGR216 2022 Statics Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["beam-deflection","cantilever","triangular-load","stiffness-design","statically-indeterminate"],"source":{"name":"ENGR216 Summer 2022 Statics Q A2","paper":"ENGR216","year":2022,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR216 Exam Part A 2021-22.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2022/ENGR216_statics_num_ans_2022.pdf","tag":"VALID"},"text":"Consider the statically determinate system given by the cantilevered beam with flexural rigidity $EI$ and\nlength $L$ in Figure A2, subjected to a triangular distributed normal load varying from 0 at end A to $w_0$ at\nend B. The abscissa $x$ is the coordinate along the horizontal axis depicted in the figure with origin at end A\nof the beam.\n","figures":[{"src":"/figures/engr216-2022-a2-statics.png","caption":"Figure A2: cantilever fixed at A, free at B, triangular load from 0 at A to w0 at B. Figure A3: rectangular section w x h."}],"formulaSheet":"- Equilibrium of a rigid body: $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M_A = 0$ (learn this)\n- Resultant of a distributed load $P = \\int w\\,dx$ and its position $x_P$ (triangle: $P = w_0L/2$ at $2L/3$ from the zero end) (on the formula sheet)\n- Load, shear and moment: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$ (on the formula sheet)\n- Elastic curve $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$, so $M = EI\\,y''$; clamped end: $y = 0$, $y' = 0$ (learn this; on the 2025 sheet but not the 2026 one)\n- Rectangle bending about the axis parallel to the width $w$: $I = \\dfrac{w h^3}{12}$ (on the formula sheet)\n- Statically indeterminate beams by compatibility/superposition; cantilever with end load $P$: $\\delta = \\dfrac{PL^3}{3EI}$ (learn this)\n","parts":[{"label":"a","text":"For the system in Figure A2, sketch the deformed shape of the beam longitudinal axis as accurately as\npossible, accounting for the action of the constraint at the end A.\n","marks":6,"grade":"2:2"},{"label":"b","text":"Sketch the free body diagram of the loaded beam in Figure A2, clearly indicating all reactions and loads.\n","marks":6,"grade":"third","answer":[{"label":"A_y (up +)","value":0.5,"units":"w0*L"},{"label":"M_A (CCW +)","value":0.33333,"units":"w0*L^2"}]},{"label":"c","text":"Knowing that the equation of the deformed longitudinal axis of the beam is\n\n$$y(x) = \\frac{w_0}{EI}\\left(-\\frac{L^2 x^2}{6} + \\frac{L x^3}{12} - \\frac{x^5}{120L}\\right) \\qquad (4)$$\n\ndetermine the magnitude of the bending moment at the fixed support as a function of $w_0$ and $L$.\n","marks":4,"grade":"third","answer":[{"label":"|M_A|","value":0.33333,"units":"w0*L^2"}]},{"label":"d","text":"The value of the maximum allowable vertical displacement of the beam free end is 1 mm, and the beam has a\nrectangular cross section of width $w$ and height $h$, as shown in Figure A3. Determine the minimum value of\n$h$ to fulfill the aforementioned rigidity constraint. Use $w_0 = 50$ kN/m, $w = 150$ mm, $L = 1.5$ m and\n$E = 200$ GPa.\n","marks":4,"grade":"2:2","answer":[{"label":"h_min","value":210.15,"units":"mm"}]},{"label":"e","text":"State one constraint which could be applied to the end B to make the beam statically indeterminate, and,\nfor the system thus obtained, list all nonzero reactions acting on the beam.\n","marks":5,"grade":"2:1"}],"solution":"$101","confidence":"HIGH","solutionNote":"Matches the official numerical answers: (c) M_A = w0 L^2/3 counter-clockwise, (d) h_min = 210.1 mm (we get 210.15 mm; the official value is truncated).\n"},{"id":"engr216-2022-a1-dynamics","module":"ENGR5004","week":5,"label":"ENGR216 2022 Dynamics Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["particle-kinematics","rigid-kinematics","polar-coordinates","n-t-coordinates","coriolis","mechanism"],"source":{"name":"ENGR216 Summer 2022 Dynamics Q A1","paper":"ENGR216","year":2022,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR216 Exam Part B 2021-22.pdf","tag":"VALID"},"text":"Figure A1-1 shows a single link, AB, rotating anticlockwise about a fixed pivot at A.\n","figures":[{"src":"/figures/engr216-2022-a1-dynamics.png","caption":"Figure A1-1: link AB pivoted at A at angle theta, angular velocity omega and angular acceleration alpha anticlockwise."},{"src":"/figures/engr216-2022-a1-dynamics-2.png","caption":"Figure A1-2: actuator mechanism; link AB pivoted at A, block C slides on AB, link CD in a horizontal slide at height y."}],"formulaSheet":"- Fixed-axis rotation, vector form: $\\mathbf{v}_B = \\boldsymbol\\omega\\times\\mathbf{r}_{B/A}$, $\\mathbf{a}_B = \\boldsymbol\\alpha\\times\\mathbf{r}_{B/A} - \\omega^2\\mathbf{r}_{B/A}$ (learn this)\n- Polar coordinates: $\\mathbf{v} = \\dot r\\,\\mathbf{u}_r + r\\dot\\theta\\,\\mathbf{u}_\\theta$, $\\mathbf{a} = (\\ddot r - r\\dot\\theta^2)\\,\\mathbf{u}_r + (r\\ddot\\theta + 2\\dot r\\dot\\theta)\\,\\mathbf{u}_\\theta$; Coriolis term $2\\dot r\\dot\\theta$ (learn this)\n- Normal-tangential: $\\mathbf{v} = v\\,\\mathbf{u}_t$, $a_t = \\dot v$, $a_n = v^2/\\rho$ toward the centre of curvature (learn this)\n- Constraint differentiation: $x = y\\tan\\theta$, $\\dfrac{d}{d\\theta}\\tan\\theta = \\sec^2\\theta$, $\\dfrac{d}{d\\theta}\\sec^2\\theta = 2\\sec^2\\theta\\tan\\theta$, chain rule $\\dfrac{d}{dt} = \\dot\\theta\\dfrac{d}{d\\theta}$ (learn this)\n","parts":[{"label":"a","text":"Figure A1-1 shows a link connected to fixed pivot at A and having angular acceleration $\\alpha$ and\ninstantaneous angular velocity $\\omega$. Define and show on a sketch the velocity and acceleration\ncomponents of point B when described in\n\ni. Cartesian coordinate system components.\nii. Polar coordinate system components.\niii. Normal-tangential coordinate system components.\n","marks":6,"grade":"2:2"},{"label":"b","text":"Which coordinate system would you choose for an in-plane kinematic analysis of Figure A1-1, and why?\n","marks":2,"grade":"third"},{"label":"c","text":"For the configuration shown in A1-1, would point B experience a Coriolis acceleration? Give your reasoning.\n","marks":3,"grade":"2:2"},{"label":"d","text":"Figure A1-2 shows an actuator mechanism. Link CD moves in a horizontal slide and the end C carries a block\nwhich slides along AB. Show that the linear velocity of CD is\n\n$$\\dot{x} = (y\\sec^2\\theta)\\,\\omega$$\n\nand also that the linear acceleration of CD is\n\n$$\\ddot{x} = \\omega^2 (2y\\sec^2\\theta\\tan\\theta) + \\alpha y\\sec^2\\theta$$\n","marks":6,"grade":"2:1"},{"label":"e","text":"Calculate the linear acceleration of CD when $\\theta = 50^\\circ$, $y = 2$ m, $\\omega = 1$ rad/s clockwise and\n$\\alpha = 5\\ \\mathrm{rad/s^2}$ clockwise.\n","marks":4,"grade":"2:2","answer":[{"label":"xddot_CD (right +)","value":35.74,"units":"m/s^2"}]},{"label":"f","text":"If CD moves with constant linear velocity, and the angular velocity of AB is 1 rad/s clockwise, find the\nangular acceleration of AB when $\\theta = 50^\\circ$ and $y = 2$ m.\n","marks":4,"grade":"2:2","answer":[{"label":"alpha_AB (CW +)","value":-2.384,"units":"rad/s^2"}]}],"solution":"$102","confidence":"HIGH","solutionNote":"No official 2022 dynamics numerical answers were found (the 2022 answers folder has statics only). Constraint differentiation, a polar-coordinate resolution and a finite-difference check agree.\n"},{"id":"engr216-2022-a2-dynamics","module":"ENGR5004","week":9,"label":"ENGR216 2022 Dynamics Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["rigid-kinetics","work-energy","moment-of-inertia","fixed-axis-rotation","pulley","spring"],"source":{"name":"ENGR216 Summer 2022 Dynamics Q A2","paper":"ENGR216","year":2022,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2022/ENGR216 Exam Part B 2021-22.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c).\n","figures":[{"src":"/figures/engr216-2022-a2-dynamics.png","caption":"Figure A2-1: pulley of radius R on a fixed pivot; rope from a spring k (anchored below) over the pulley to a hanging mass m."}],"formulaSheet":"- Rotation about a fixed axis through G: $\\sum M_G = I_G\\,\\alpha$ (learn this)\n- Angle-dependent integration: $\\alpha = \\omega\\,\\dfrac{d\\omega}{d\\theta}$, so $I\\,\\omega\\,\\dfrac{d\\omega}{d\\theta} = T(\\omega)$ (learn this)\n- $\\displaystyle\\int \\frac{\\omega\\,d\\omega}{a - b\\omega^2} = -\\frac{1}{2b}\\ln(a - b\\omega^2)$; 1 rev $= 2\\pi$ rad, rpm $= \\omega \\cdot 60/(2\\pi)$ (learn this)\n- Newton's second law for a particle: $\\sum F = m a$ (learn this)\n- Rope over a pulley without slip: $a = \\alpha R$; pulley with inertia: $(T_1 - T_2)R = I_G\\alpha$, so $m_{eq} = m + I_G/R^2$ (learn this)\n- Work-energy: $T_1 + \\sum U_{1\\to2} = T_2$, spring work $-\\tfrac{1}{2}k(s_2^2 - s_1^2)$, gravity work $mg\\,\\Delta s$ (learn this)\n","parts":[{"label":"a","text":"Why is a 'moment of inertia' useful when determining the kinetics of a rotating planar rigid body?\n","marks":4,"grade":"third"},{"label":"b(i)","text":"A rotating disc driven by a motor has a moment of inertia $I_G = 50\\ \\mathrm{kg\\,m^2}$ and begins turning from\nrest at $t = 0$. The torque $T$ applied to the disc by the drive motor is given by $T = 140 - 0.02\\,\\omega^2$\nwhere $\\omega$ is angular velocity measured in rad/s. What will be the angular velocity in rpm (revolutions\nper minute) of the disc when it has turned 500 revolutions?\n","marks":5,"grade":"2:1","answer":[{"label":"omega at 500 rev","value":765.9,"units":"rpm"}]},{"label":"b(ii)","text":"What is the maximum angular velocity (in rpm) that the disc reaches?\n","marks":3,"grade":"2:2","answer":[{"label":"omega_max","value":799,"units":"rpm"}]},{"label":"c(i)","text":"Figure A2-1 shows a pulley-spring-mass system. The pulley has radius $R = 150$ mm and moment of inertia\n$I_G = 0.15\\ \\mathrm{kg\\,m^2}$ and the spring has spring constant $k = 120$ N/m. The mass $m = 4$ kg. The\nsystem is released from rest with the spring in the unstretched position. At the point when the mass has\nfallen 0.2 m, what is the angular acceleration of the pulley?\n","marks":4,"grade":"2:1","answer":[{"label":"alpha_pulley","value":9.525,"units":"rad/s^2"}]},{"label":"c(ii)","text":"For the condition in (i), what will be the tension in the rope between the mass and the pulley?\n","marks":4,"grade":"2:2","answer":[{"label":"T_rope (mass side)","value":33.53,"units":"N"}]},{"label":"c(iii)","text":"Determine the maximum distance the mass falls before rebounding.\n","marks":5,"grade":"2:2","answer":[{"label":"s_max","value":0.654,"units":"m"}]}],"solution":"$103","confidence":"HIGH","solutionNote":"No official 2022 dynamics numerical answers were found (the 2022 answers folder has statics only). Closed-form results agree with numerical integration; g = 9.81 m/s^2 assumed.\n"},{"id":"engr216-2021-a1-statics","module":"ENGR5004","week":9,"label":"ENGR216 2021 Statics Q A1","kind":"past-paper","level":3,"marks":23,"minutes":28,"grade":"2:1","tags":["bending-eccentric","beam-shear-stress","stress-transformation","buckling","cantilever","design"],"source":{"name":"ENGR216 Summer 2021 Statics Q A1","paper":"ENGR216","year":2021,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR216 Statics Exam 2021a.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2021/ENGR216_statics_num_ans_2021.pdf","tag":"VALID"},"text":"Consider the cantilevered beam of length $L$ in Figure A1, subject to the transverse load $F$. The beam has\nrectangular cross section of width $w$ and height $h$, and the second area moment past the neutral axis of\nthis cross section is $I$. The beam's material has Young's modulus $E$. The line of action of $F$ is parallel to\nthe side of height $h$, as indicated in Figure A1.\n","figures":[{"src":"/figures/engr216-2021-a1-statics.png","caption":"Figure A1: cantilever AB of length L fixed at A with transverse load F at B; rectangular section w x h with point S at mid-height of a side and T at the top."}],"formulaSheet":"- Pure bending: $\\sigma_x = -\\dfrac{M y}{I}$, $\\sigma_{max} = \\dfrac{|M| c}{I}$ (on the formula sheet)\n- Rectangle: $I = \\dfrac{b h^3}{12}$, $A = bh$, $\\rho = \\sqrt{I/A}$ (on the formula sheet)\n- Transverse shear stress: $\\tau = \\dfrac{VQ}{It}$, $\\tau_{max} = \\dfrac{3V}{2A}$ for a rectangle (learn this; on the 2025 sheet but not the 2026 one)\n- Principal stresses: $\\sigma_{1,2} = \\dfrac{\\sigma_x + \\sigma_y}{2} \\pm \\sqrt{\\left(\\dfrac{\\sigma_x - \\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$, $\\tan 2\\theta_p = \\dfrac{2\\tau_{xy}}{\\sigma_x - \\sigma_y}$ (learn this)\n- Euler buckling: $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L_e^2}$ with $L_e = 2L$ for a cantilever (on the formula sheet; $L_e = 2L$ is also given in the question)\n- Safety factor and strength limit: $P_{allow} = P_{cr}/FS$, $P \\le \\sigma_{all} A$ (learn this)\n","parts":[{"label":"a","text":"Considering the beam cross section at the fixed support A and making use of the plane stress theory, show\nthat the maximum normal stress at point S of the beam surface is given by Equation 1\n\n$$\\sigma_{S,max} = \\frac{3F}{2wh} \\qquad (1)$$\n\nand state if this normal stress is parallel to the beam axis, justifying your answer.\n","marks":7,"grade":"2:2"},{"label":"b","text":"Assuming $L = 1$ m, $F = 40$ kN, $w = 50$ mm, determine the height $h$ of the beam cross section if the\nmaximum allowable normal stress of the material of 120 MPa is not to be exceeded in any part of the beam.\n","marks":9,"grade":"2:1","answer":[{"label":"h_min","value":200,"units":"mm"}]},{"label":"c","text":"The beam designed in part (b) is now subject only to a centric compressive axial load, and $E = 200$ GPa.\nAssuming maximum allowable normal stress of 120 MPa and buckling safety factor of 2, determine the maximum\ncentric compressible axial load applicable to the beam fulfilling the strength and buckling stability\nrequirements. Please note that the equivalent length $L_{eq}$ of a cantilevered beam subject to axial load\nis twice its geometric length.\n","marks":7,"grade":"2:2","answer":[{"label":"P_max","value":514,"units":"kN"}]}],"solution":"$104","confidence":"HIGH","solutionNote":"Matches the official numerical answers ((b) h about 0.2 m; (c) maximum centric axial load 514 kN). Part (a) is\na show-that, so it has no answer box.\n"},{"id":"engr216-2021-a2-statics","module":"ENGR5004","week":6,"label":"ENGR216 2021 Statics Q A2","kind":"past-paper","level":3,"marks":27,"minutes":32,"grade":"2:2","tags":["beam-deflection","shear-bending-diagrams","propped-cantilever","statically-indeterminate","udl"],"source":{"name":"ENGR216 Summer 2021 Statics Q A2","paper":"ENGR216","year":2021,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR216 Statics Exam 2021a.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2021/ENGR216_statics_num_ans_2021.pdf","tag":"VALID"},"text":"Consider the statically indeterminate beam with Young's modulus $E$, cross sectional second area moment $I$\npast the neutral axis, and length $L$ depicted in Figure A2, subject to the constant distributed normal load\n$w_0$ along its entire length. The coordinate along the beam axis, denoted by $x$, has origin at end A of the\nbeam.\n","figures":[{"src":"/figures/engr216-2021-a2-statics.png","caption":"Figure A2: beam of length L fixed at A, roller at B, uniform load w0."}],"formulaSheet":"- Elastic curve: $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$ (learn this; on the 2025 sheet but not the 2026 one)\n- Load, shear and moment: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$ (on the formula sheet)\n- Distributed load resultant $P = \\int w\\,dx$ at its centroid (on the formula sheet)\n- Equilibrium of a plane rigid body: $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$ (learn this)\n- Fixed-end conditions $y = 0$, $y' = 0$; roller $y = 0$, $M = 0$ (learn this)\n","parts":[{"label":"a","text":"Sketch accurately the deformed longitudinal axis of the beam subject to the indicated load highlighting the\ngeometric conditions imposed by the constraints, and draw the free body diagram of the loaded beam.\n","marks":7,"grade":"2:2"},{"label":"b","text":"List and count all unknown reactions, and the equations available to calculate all such unknown reactions.\n","marks":6,"grade":"third"},{"label":"c","text":"Knowing that the equation of the elastic curve of the beam subject to the given load is:\n\n$$y(x) = \\frac{w_0}{EI}\\left(-\\frac{L^2 x^2}{16} + \\frac{5 L x^3}{48} - \\frac{x^4}{24}\\right) \\qquad (2)$$\n\ndetermine the magnitude of the maximum bending moment along the beam.\n","marks":8,"grade":"2:2","answer":[{"label":"|M|max","value":0.125,"units":"w0*L^2"}]},{"label":"d","text":"Determine at which position along the beam the maximum magnitude of the shearing load occurs.\n","marks":6,"grade":"2:2","answer":[{"label":"x at |V|max","value":0,"units":"L"},{"label":"|V|max","value":0.625,"units":"w0*L"}]}],"solution":"$105","confidence":"HIGH","solutionNote":"Agrees with the official numerical answers on the magnitudes: |M|max = w0 L^2/8 and |V|max = 5 w0 L/8 at x = 0.\nThe official sheet says the maximum bending moment \"occurs at x = 5/8 L\"; that location is wrong for w0 L^2/8.\nAt x = 5L/8 the moment is the local sagging maximum 9 w0 L^2/128; the overall maximum magnitude w0 L^2/8 is the\nhogging moment at the fixed end x = 0 (checked by equilibrium and superposition).\n"},{"id":"engr216-2021-a1-dynamics","module":"ENGR5004","week":5,"label":"ENGR216 2021 Dynamics Q A1","kind":"past-paper","level":3,"marks":18,"minutes":22,"grade":"2:2","tags":["rigid-kinematics","fixed-axis-rotation","gears","variable-acceleration"],"source":{"name":"ENGR216 Summer 2021 Dynamics Q A1","paper":"ENGR216","year":2021,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR216 Dynamics Paper 2021.pdf","tag":"PARTIAL","leftOut":"Part (d) (static and dynamic balancing of rotating machinery) is not in the ENGR5004 syllabus."},"text":"Figure A1 shows the drive mechanism from an industrial machine, whereby motor M drives geared shaft D via\nrotation of gear S. For a short time, gear S rotates with an angular acceleration of\n$\\alpha_S = (250\\,t^2 + 50)\\ \\mathrm{rad/s^2}$, where $t$ is in seconds.\n","figures":[{"src":"/figures/engr216-2021-a1-dynamics.png","caption":"Figure A1: motor M drives gear S (radius 60 mm) meshing with the gear on shaft D (radius 150 mm)."}],"formulaSheet":"- Fixed-axis rotation: $\\omega = \\omega_0 + \\int_0^t \\alpha\\,dt$, $\\theta = \\theta_0 + \\int_0^t \\omega\\,dt$ (learn this)\n- Meshing gears (no slip at the pitch point): $r_S\\,\\omega_S = r_D\\,\\omega_D$, $r_S\\,\\theta_S = r_D\\,\\theta_D$, $r_S\\,\\alpha_S = r_D\\,\\alpha_D$ (learn this)\n- Point on a rotating body: $v = \\omega r$, $a_t = \\alpha r$ (learn this)\n","parts":[{"label":"a","text":"Determine the angular velocity of shaft D when $t = 2$ s, starting from rest.\n","marks":4,"grade":"2:2","answer":[{"label":"omega_D","value":306.67,"units":"rad/s"}]},{"label":"b","text":"Determine the angular displacement of shaft D when $t = 2$ s, starting from rest.\n","marks":4,"grade":"2:2","answer":[{"label":"theta_D","value":173.33,"units":"rad"}]},{"label":"c","text":"State any assumptions that you have made, explaining your reasoning\n","marks":4,"grade":"third"},{"label":"d","text":"Write brief notes explaining why dynamic and static balance should be considered when designing and\ncommissioning rotating machinery.\n","marks":6,"grade":"$undefined"}],"solution":"$106","confidence":"HIGH","solutionNote":"No official solution available. Parts (a) and (b) were checked by integrating alpha_D directly, by the\npitch-point speed, and by numerical integration. Part (d) is off-syllabus and only an indicative answer is given.\n"},{"id":"engr216-2021-a2-dynamics","module":"ENGR5004","week":4,"label":"ENGR216 2021 Dynamics Q A2","kind":"past-paper","level":3,"marks":16,"minutes":19,"grade":"2:2","tags":["particle-kinetics","n-t-coordinates","radius-of-curvature","normal-force"],"source":{"name":"ENGR216 Summer 2021 Dynamics Q A2","paper":"ENGR216","year":2021,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR216 Dynamics Paper 2021.pdf","tag":"VALID"},"text":"In a warehouse, crates of mass 100 kg move down a slope defined by the equation $y = 0.12x^2$ (see Figure A2).\nAt the instant $x = 7$ m the crate in Figure A2 has a speed of 6 m/s. At this point, determine:\n","figures":[{"src":"/figures/engr216-2021-a2-dynamics.png","caption":"Figure A2: crate moving down the curve y = 0.12 x^2 at x = 7 m with speed 6 m/s."}],"formulaSheet":"- Normal-tangential kinetics: $\\sum F_t = m a_t$, $\\sum F_n = m a_n$, $a_t = \\dot v$, $a_n = \\dfrac{v^2}{\\rho}$ (learn this)\n- Radius of curvature of $y(x)$: $\\rho = \\dfrac{\\left[1 + (dy/dx)^2\\right]^{3/2}}{|d^2y/dx^2|}$ (learn this)\n- Slope angle: $\\tan\\theta = dy/dx$; magnitude $a = \\sqrt{a_t^2 + a_n^2}$ (learn this)\n","parts":[{"label":"a","text":"The rate of speed increase of the crate\n","marks":4,"grade":"2:2","answer":[{"label":"a_t","value":8.43,"units":"m/s^2"}]},{"label":"b","text":"The force the slope exerts on the crate\n","marks":4,"grade":"2:2","answer":[{"label":"N","value":617.4,"units":"N"},{"label":"N direction (CCW from +x)","value":149.24,"units":"deg"}]},{"label":"c","text":"The magnitude and direction of the crate's acceleration\n","marks":4,"grade":"2:2","answer":[{"label":"|a|","value":8.51,"units":"m/s^2"},{"label":"a direction (below -x axis)","value":51.43,"units":"deg"}]},{"label":"d","text":"State any assumptions that you have made, with full justification\n","marks":4,"grade":"third"}],"solution":"$107","confidence":"HIGH","solutionNote":"No official solution available. Checked by a Cartesian Newton's-law calculation (same acceleration vector) and\nthe at-rest limit for N. The frictionless assumption is the one the question invites in part (d).\n"},{"id":"engr216-2021-a3-dynamics","module":"ENGR5004","week":7,"label":"ENGR216 2021 Dynamics Q A3","kind":"past-paper","level":3,"marks":16,"minutes":19,"grade":"2:1","tags":["rigid-kinematics","relative-velocity","relative-acceleration","slider-crank","rolling-wheel"],"source":{"name":"ENGR216 Summer 2021 Dynamics Q A3","paper":"ENGR216","year":2021,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2021/ENGR216 Dynamics Paper 2021.pdf","tag":"VALID"},"text":"Figure A3 shows a slider mechanism that connects a wheel crank to the slider block B via link arm AB. If slider\nblock B has a velocity $v_B = 1$ m/s to the right, and acceleration $a_B = 2\\ \\mathrm{m/s^2}$, positive to the\nright as shown in the figure, determine:\n","figures":[{"src":"/figures/engr216-2021-a3-dynamics.png","caption":"Figure A3: wheel of radius 0.1 m connected by link AB (0.5 m) to slider block B moving right."}],"formulaSheet":"- Relative velocity: $\\mathbf v_B = \\mathbf v_A + \\boldsymbol\\omega \\times \\mathbf r_{B/A}$ (learn this)\n- Relative acceleration: $\\mathbf a_B = \\mathbf a_A + \\boldsymbol\\alpha \\times \\mathbf r_{B/A} - \\omega^2 \\mathbf r_{B/A}$ (learn this)\n- Rolling without slipping: $v_O = \\omega r$, $a_O = \\alpha r$; contact point has zero velocity (learn this)\n- Instantaneous centre of zero velocity (IC) method (learn this)\n","parts":[{"label":"a","text":"The angular velocity of link arm AB\n","marks":3,"grade":"2:2","answer":[{"label":"omega_AB","value":0,"units":"rad/s"}]},{"label":"b","text":"The angular acceleration of link arm AB\n","marks":5,"grade":"2:1","answer":[{"label":"alpha_AB (CCW +)","value":5.455,"units":"rad/s^2"}]},{"label":"c","text":"The angular acceleration of the wheel\n","marks":5,"grade":"2:1","answer":[{"label":"alpha_wheel (CCW +)","value":-4.545,"units":"rad/s^2"}]},{"label":"d","text":"State any assumptions that you have made, with full justification\n","marks":3,"grade":"third"}],"solution":"$108","confidence":"MEDIUM","solutionNote":"No official solution available. The answers depend on reading the figure: we assume the wheel rolls without\nslipping on the drawn ground line, A is on the rim directly above the centre and B is level with the contact\npoint (A is 0.2 m above B). Velocities were checked with the IC method and accelerations with a position-analysis\ndifferentiation in Python. Confirm with a lecturer whether the wheel rolls or turns on a fixed axle; the\nfixed-axle results are given at the end of (d).\n"},{"id":"engr216-2019-a1","module":"ENGR5004","week":9,"label":"ENGR216 2019 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["multiaxial-hooke","stress-transformation","generalised-hooke","poisson-ratio","plane-stress","mohrs-circle","max-shear-stress","thin-plate"],"source":{"name":"ENGR216 Summer 2019 Q A1","paper":"ENGR216","year":2019,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR216_PAPER_2019_FINAL (2).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/numerical_answers_216Statics-6exam_2019.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (e)\n\nSection A Appendix (as printed with the paper):\n\nThe generalised Hooke's laws for multi-axial load of an isotropic homogeneous material are:\n\n$$\\epsilon_x = +\\frac{\\sigma_x}{E} - \\frac{\\nu\\sigma_y}{E} - \\frac{\\nu\\sigma_z}{E}$$\n$$\\epsilon_y = -\\frac{\\nu\\sigma_x}{E} + \\frac{\\sigma_y}{E} - \\frac{\\nu\\sigma_z}{E}$$\n$$\\epsilon_z = -\\frac{\\nu\\sigma_x}{E} - \\frac{\\nu\\sigma_y}{E} + \\frac{\\sigma_z}{E}$$\n\nwhere $\\sigma$ denotes a normal stress component, $\\epsilon$ denotes a normal strain component, $x$, $y$, and $z$\ndenote the axis of a three-dimensional Cartesian system, $E$ denotes Young's modulus and $\\nu$ denotes\nPoisson's ratio.\n\nThe expression of the maximum and minimum normal stresses in a plane stress state are:\n\n$$\\sigma_{max/min} = \\frac{\\sigma_x + \\sigma_y}{2} \\pm \\sqrt{\\left(\\frac{\\sigma_x - \\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$$\n","formulaSheet":"- Generalised Hooke's law $\\epsilon_x = (\\sigma_x - \\nu\\sigma_y - \\nu\\sigma_z)/E$ etc. (given in the question; learn this)\n- Principal stresses $\\sigma_{max/min} = \\frac{\\sigma_x+\\sigma_y}{2} \\pm \\sqrt{\\left(\\frac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$ (given in the question; learn this)\n- Maximum in-plane shear stress $\\tau_{max} = \\sqrt{\\left(\\frac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2} = \\frac{\\sigma_{max}-\\sigma_{min}}{2}$ (radius of Mohr's circle) (learn this)\n- Normal strain $\\epsilon = \\delta / L$ (learn this)\n- Plane stress: $\\sigma_z = \\tau_{xz} = \\tau_{yz} = 0$ (learn this)\n","parts":[{"label":"a","text":"Making use of the generalised Hooke's laws for isotropic homogeneous materials subject to multi-axial load\nprovided in the Appendix explain why a three-dimensional structure subject to a single axial load can\ndeform also in directions different from that along which above said axial load acts.\n","marks":5,"grade":"third"},{"label":"b","text":"Provide the definition of plane stress state and one example of this multi-dimensional loading condition.\n","marks":5,"grade":"third"},{"label":"c","text":"A rectangular thin plate of homogeneous isotropic material is subject to a biaxial loading state made up of\na tensile stress $\\sigma_x = 120\\ \\mathrm{MPa}$ in the $x$ direction and a tensile stress\n$\\sigma_y = 171\\ \\mathrm{MPa}$ in the $y$ direction. Determine the magnitude of maximum shear stress in the\nplate (HINT: the expression of the maximum shear stress can be obtained from that of the maximum and minimum\nnormal stresses provided in the Appendix noting that the difference between maximum and minimum normal\nstresses is the diameter of Mohr's circle).\n","marks":5,"grade":"third","answer":[{"label":"tau_max (in-plane)","value":25.5,"units":"MPa"}]},{"label":"d","text":"Before being loaded, the plate has length $l_x = 1571\\ \\mathrm{mm}$ along the $x$ axis and\n$l_y = 533\\ \\mathrm{mm}$ along the $y$ axis, and after being loaded it elongates by $0.6\\ \\mathrm{mm}$ in the\n$x$ direction and $0.4\\ \\mathrm{mm}$ in the $y$ direction. Determine Young's modulus and Poisson's ratio of\nthe plate material.\n","marks":5,"grade":"2:2","answer":[{"label":"E","value":179.9,"units":"GPa"},{"label":"nu","value":0.3,"units":"-"}]},{"label":"e","text":"Knowing that the plate has thickness $l_z = 20\\ \\mathrm{mm}$, determine the variation of its thickness in mm\ndue to the applied loads, indicating whether this is a contraction or an elongation.\n","marks":5,"grade":"2:2","answer":[{"label":"delta_lz (thickening +)","value":-0.0097,"units":"mm"}]}],"solution":"$109","confidence":"HIGH","solutionNote":"Matches the official numerical answers (25.5 MPa; 180 GPa and 0.3; 0.0097 mm contraction). The official sheet\nlabels the thickness answer \"c.\" by mistake; it is part (e). Part (c) answer is the maximum in-plane shear stress\nas the hint directs; the absolute 3D maximum (85.5 MPa) is noted but not asked.\n"},{"id":"engr216-2019-a2","module":"ENGR5004","week":7,"label":"ENGR216 2019 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["beam-deflection","shear-bending-diagrams","elastic-curve","distributed-load","reactions","simply-supported","free-body-diagram","boundary-conditions"],"source":{"name":"ENGR216 Summer 2019 Q A2","paper":"ENGR216","year":2019,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR216_PAPER_2019_FINAL (2).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/numerical_answers_216Statics-6exam_2019.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d)\n\nConsider the statically determinate beam with flexural rigidity $EI$ and length $L$ depicted in Figure A2,\nsubject to the distributed load $w(x)$, where $x$ denotes the coordinate along the horizontal axis depicted in\nthe figure with origin at end A of the beam.\n","figures":[{"src":"/figures/engr216-2019-a2.png","caption":"Figure A2: beam AB of length L, roller support at A and pin support at B, downward distributed load w(x) decreasing from w0 at A to zero at B; x axis along the beam from A, y axis upwards at A."}],"formulaSheet":"- Elastic curve $\\dfrac{d^2y}{dx^2} = \\dfrac{M}{EI}$ (given in the question; learn this; on the 2025 sheet but not the 2026 one)\n- Resultant of a distributed load $P = \\int w\\,dx$ and its position $x_P = \\dfrac{\\int x\\,w\\,dx}{\\int w\\,dx}$ (on the formula sheet)\n- $dV/dx = -w(x)$, $dM/dx = V(x)$ and the positive shear / positive bending sign convention (on the formula sheet)\n- Equilibrium $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$; roller gives one reaction normal to its surface, pin gives two (learn this)\n- Boundary conditions for a simply supported beam: $y = 0$ at each support (learn this)\n","parts":[{"label":"a","text":"Knowing that the bending moment $M(x)$ is positive over the entire beam, and noting that the relationship\nbetween $M(x)$ and the local deflection $y(x)$ is:\n\n$$\\frac{d^2y}{dx^2} = \\frac{M}{EI} \\qquad (1)$$\n\nwhere the $y$ axis is depicted in Figure A2 and has origin at end A of the beam, sketch the deformed shape\nof the beam mean line. For your sketch, use the same Cartesian axes $x$ and $y$ defined in Figure A2.\n","marks":5,"grade":"2:2"},{"label":"b","text":"Sketch the free body diagram of the loaded beam in Figure A2.\n","marks":5,"grade":"third"},{"label":"c","text":"Knowing that the distributed load $w(x)$ is given by:\n\n$$w(x) = w_0\\left[1 - \\frac{x^3}{L^3}\\right] \\qquad (2)$$\n\nwhere $w_0$ is a constant of dimension N/m, demonstrate that the vertical reactive forces $R_A$ and $R_B$ at\nthe beam ends A and B are respectively:\n\n$$R_A = \\frac{9}{20} w_0 L \\quad \\text{and} \\quad R_B = \\frac{3}{10} w_0 L$$\n","marks":10,"grade":"2:2","answer":[{"label":"R_A (up +)","value":0.45,"units":"w0*L"},{"label":"R_B (up +)","value":0.3,"units":"w0*L"}]},{"label":"d","text":"Knowing that the equation of the deformed mean line of the beam is:\n\n$$y(x) = \\frac{w_0}{EI}\\left(\\frac{x^7}{840L^3} - \\frac{x^4}{24} + \\frac{3Lx^3}{40}\\right) + C_1 x + C_2 \\qquad (3)$$\n\ndetermine the constants $C_1$ and $C_2$ by imposing the compatibility condition of the beam deformation\nexpressed by Equation (3) with the constraints acting on the beam depicted in Figure A2.\n","marks":5,"grade":"2:2","answer":[{"label":"C1","value":-0.034524,"units":"w0*L^3/EI"},{"label":"C2","value":0,"units":"w0*L^4/EI"}]}],"solution":"$10a","confidence":"HIGH","solutionNote":"Matches the official numerical answers ($C_1 = -\\frac{29}{840}\\frac{w_0L^3}{EI}$, $C_2 = 0$). Part (c) results are\ngiven in the question; all results were checked symbolically (sympy) including $EI\\,y'' = M(x)$ for Equation (3).\n"},{"id":"engr216-2019-a3","module":"ENGR5004","week":10,"label":"ENGR216 2019 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["buckling","euler-load","slenderness-ratio","hollow-section","safety-factor","allowable-load"],"source":{"name":"ENGR216 Summer 2019 Q A3","paper":"ENGR216","year":2019,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR216_PAPER_2019_FINAL (2).pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2019/numerical_answers_216Statics-6exam_2019.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d)\n\nSection A Appendix (as printed with the paper):\n\nEuler's critical load $P_{cr}$ for doubly pinned rods is:\n\n$$P_{cr} = \\frac{\\pi^2 E I}{L^2}$$\n\nwhere $I$ is the smallest second area moment of the rod cross section, $E$ is the elastic or Young's modulus and\n$L$ is the rod length.\n\nThe second area moment $I$ of a solid circular cross section of radius $r$ about any diameter is:\n\n$$I = \\frac14 \\pi r^4$$\n","formulaSheet":"- Euler load $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L_e^2}$; pinned-pinned $L_e = L$ (given in the question; also on the formula sheet)\n- Solid circle $I = \\pi r^4/4$ (given in the question; also on the formula sheet); hollow circle $I = \\pi(r_2^4 - r_1^4)/4$ (on the formula sheet)\n- Radius of gyration $\\rho = \\sqrt{I/A}$, slenderness ratio $L_e/\\rho$, critical stress $\\sigma_{cr} = \\dfrac{\\pi^2 E}{(L_e/\\rho)^2}$ (on the formula sheet)\n- Normal stress $\\sigma = P/A$ and allowable load with a safety factor $P_{all} = P_{cr}/FS$ (learn this)\n","parts":[{"label":"a","text":"Provide a brief general definition of buckling in the context of the three main structural design\nrequirements.\n","marks":5,"grade":"third"},{"label":"b","text":"Define the slenderness ratio of a doubly pinned beam, and briefly explain how the likelihood of buckling\noccurrence varies with the value of this parameter.\n","marks":6,"grade":"third"},{"label":"c","text":"A beam pinned at both ends has length of $1.3\\ \\mathrm{m}$ and consists of a solid steel rod with Young's\nmodulus $E = 120\\ \\mathrm{GPa}$. The beam has a circular cross section with diameter of $40\\ \\mathrm{mm}$.\nTo reduce the weight of the member, the solid rod is replaced by a hollow rod, obtained by cutting a\nconcentric circular hole with diameter of $24\\ \\mathrm{mm}$. Assume that the effective length of the beam is\nthe same in all planes through the beam axis.\n\nDetermine the value of Euler's critical load of both the solid and the hollow rods, and calculate the\npercentage reduction of the critical load of the hollow beam with respect to the solid beam. The general\nexpression of Euler's critical load and the second area moment of a solid circular section past any\ndiameter are provided in the Appendix.\n","marks":6,"grade":"2:2","answer":[{"label":"P_cr solid","value":88.07,"units":"kN"},{"label":"P_cr hollow","value":76.65,"units":"kN"},{"label":"reduction","value":12.96,"units":"%"}]},{"label":"d","text":"Determine the maximum compressive load applicable to the hollow beam if the maximum allowable normal stress\nis $60\\ \\mathrm{MPa}$ and a safety factor of 2 is used for computing the actual critical buckling load.\n","marks":8,"grade":"2:2","answer":[{"label":"P_max","value":38.33,"units":"kN"}]}],"solution":"$10b","confidence":"HIGH","solutionNote":"Matches the official numerical answers (88.06 kN, 76.65 kN, 12.96%, 38.325 kN). The official sheet prints\n$\\pi^3 EI/L^2$, a typo: its numbers come from $\\pi^2 EI/L^2$ ($\\pi^3$ would give 277 kN). Our 88.07 kN vs the\nofficial 88.06 kN is rounding only.\n"},{"id":"engr216-2019-b1","module":"ENGR5004","week":5,"label":"ENGR216 2019 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["particle-kinematics","particle-kinetics","polar-coordinates","weight-components","loader-boom"],"source":{"name":"ENGR216 Summer 2019 Q B1","paper":"ENGR216","year":2019,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR216_PAPER_2019_FINAL (2).pdf","tag":"VALID"},"text":"Figure B1 shows a farm loader with an extendable boom. The boom is intended to be operated with the vehicle\nstationary. A service engineer has taken measurements of the loader in use and determined that the polar\ncoordinates of point A on the bucket are functions of time (in seconds) given by $r = 2 + 0.1t^2$ (m) and\n$\\theta = 0.03t^2$ (rad).\n","figures":[{"src":"/figures/engr216-2019-b1.png","caption":"Figure B1: farm loader; boom pivots at the cab, point A on the bucket at radius r and angle theta from the horizontal x axis (y up)."}],"formulaSheet":"- Polar velocity: $v_r = \\dot r$, $v_\\theta = r\\dot\\theta$, $v = \\sqrt{v_r^2 + v_\\theta^2}$ (given in the 2019 Section B appendix; not on the 2026 sheet, so learn this)\n- Polar acceleration: $a_r = \\ddot r - r\\dot\\theta^2$, $a_\\theta = r\\ddot\\theta + 2\\dot r\\dot\\theta$, $a = \\sqrt{a_r^2 + a_\\theta^2}$ (given in the 2019 Section B appendix; not on the 2026 sheet, so learn this)\n- Angular velocity: $\\omega = \\dot\\theta$ (given in the 2019 Section B appendix; learn this)\n- Newton's second law in polar components: $\\sum F_r = m a_r$, $\\sum F_\\theta = m a_\\theta$ (learn this; the 2019 appendix only gives $\\sum \\mathbf{F}_G = m\\mathbf{a}_G$)\n- Weight resolved along $\\mathbf{u}_r$, $\\mathbf{u}_\\theta$: $W_r = -mg\\sin\\theta$, $W_\\theta = -mg\\cos\\theta$ (learn this)\n","parts":[{"label":"a","text":"For the time $t = 3$ s, determine the velocity of point A in terms of radial and transverse components.\n","marks":6,"answer":[{"label":"v_r","value":0.6,"units":"m/s"},{"label":"v_theta","value":0.522,"units":"m/s"},{"label":"|v|","value":0.7953,"units":"m/s"}],"grade":"$undefined"},{"label":"b","text":"For the time $t = 3$ s, determine the angular velocity of point A.\n","marks":3,"answer":[{"label":"theta_dot (CCW +)","value":0.18,"units":"rad/s"}],"grade":"$undefined"},{"label":"c","text":"For the time $t = 3$ s, determine the acceleration of point A in terms of radial and transverse components.\n","marks":8,"answer":[{"label":"a_r","value":0.10604,"units":"m/s^2"},{"label":"a_theta","value":0.39,"units":"m/s^2"},{"label":"|a|","value":0.4042,"units":"m/s^2"}],"grade":"$undefined"},{"label":"d","text":"If the load bucket weighs 120 kg and has a centre of mass located at point A, determine the radial and\ntransverse components of force that act on the bucket from its supports for the time $t = 3$ s.\n","marks":8,"grade":"2:1","answer":[{"label":"F_r (outward +)","value":326.7,"units":"N"},{"label":"F_theta (increasing theta +)","value":1181.4,"units":"N"}]}],"solution":"$10c","confidence":"HIGH","solutionNote":"No official solution: the 2019 numerical answers file covers Section A (statics) only. Solved independently;\npolar and Cartesian methods agree for every part. In (d) the weight is included (the question asks for the\nforce from the supports); \"weighs 120 kg\" is taken as a mass of 120 kg.\n"},{"id":"engr216-2019-b2","module":"ENGR5004","week":7,"label":"ENGR216 2019 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["rigid-kinematics","rolling-wheel","relative-velocity","relative-acceleration","instantaneous-centre","sliding-bar"],"source":{"name":"ENGR216 Summer 2019 Q B2","paper":"ENGR216","year":2019,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR216_PAPER_2019_FINAL (2).pdf","tag":"VALID"},"text":"Figure B2 shows a disc and bar assembly. As the disc rolls along the plane surface, the bar slides along the\nplane surface. The disc is 1 m diameter and the bar is 1.2 m in length.\n","figures":[{"src":"/figures/engr216-2019-b2.png","caption":"Figure B2: disc (centre O, contact point Q) with angular velocity 3 rad/s and angular acceleration 4 rad/s^2, both counter-clockwise; bar PR pinned to the disc at P (level with O) with end R on the surface."}],"formulaSheet":"- Rolling without slipping: $v_O = \\omega r$, $a_O = \\alpha r$, contact point has zero velocity (learn this)\n- Relative velocity: $\\mathbf{v}_B = \\mathbf{v}_A + \\boldsymbol\\omega\\times\\mathbf{r}_{B/A}$ (given in the 2019 Section B appendix; not on the 2026 sheet, so learn this)\n- Relative acceleration: $\\mathbf{a}_B = \\mathbf{a}_A + \\boldsymbol\\alpha\\times\\mathbf{r}_{B/A} - \\omega^2\\mathbf{r}_{B/A}$ (given in the 2019 Section B appendix; not on the 2026 sheet, so learn this)\n- Instantaneous centre: $\\omega = v_A/r_{A/IC} = v_B/r_{B/IC}$ (given in the 2019 Section B appendix; not on the 2026 sheet, so learn this)\n","parts":[{"label":"a","text":"Determine the velocity of point Q.\n","marks":3,"grade":"third","answer":[{"label":"v_Q","value":0,"units":"m/s"}]},{"label":"b","text":"Determine the velocity of point O.\n","marks":3,"grade":"third","answer":[{"label":"v_O (right +)","value":-1.5,"units":"m/s"}]},{"label":"c","text":"Determine the velocity of point P.\n","marks":3,"answer":[{"label":"v_Px (right +)","value":-1.5,"units":"m/s"},{"label":"v_Py (up +)","value":1.5,"units":"m/s"},{"label":"|v_P|","value":2.121,"units":"m/s"}],"grade":"$undefined"},{"label":"d","text":"Determine the angular velocity of the bar PR, and also the linear velocity of the bar PR.\n","marks":8,"answer":[{"label":"omega_PR (CCW +)","value":-1.375,"units":"rad/s"},{"label":"v_R (right +)","value":-2.1875,"units":"m/s"}],"grade":"$undefined"},{"label":"e","text":"Determine the acceleration of point R.\n","marks":8,"grade":"2:1","answer":[{"label":"alpha_PR (CCW +)","value":-2.7,"units":"rad/s^2"},{"label":"a_R (right +)","value":-9.913,"units":"m/s^2"}]}],"solution":"$10d","confidence":"MEDIUM","solutionNote":"No official solution: the 2019 numerical answers file covers Section A (statics) only. Relative-velocity,\nIC and direct differentiation of the geometry all agree. MEDIUM only because two readings were assumed:\nP is taken on the rim of the disc level with O (as drawn; the text gives no dimension for P), and \"linear\nvelocity of the bar PR\" in (d) is taken as the velocity of the sliding end R (mid-point value also given).\nConfirm both readings with the lecturer or a classmate.\n"},{"id":"engr216-2019-b3","module":"ENGR5004","week":8,"label":"ENGR216 2019 Q B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["rigid-kinetics","fixed-axis-rotation","moment-of-inertia","parallel-axis-theorem","free-body-diagram","kinetic-diagram"],"source":{"name":"ENGR216 Summer 2019 Q B3","paper":"ENGR216","year":2019,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2019/ENGR216_PAPER_2019_FINAL (2).pdf","tag":"VALID"},"text":"The uniform steel plate shown in Figure B3 rotates about point C in the $x$-$y$ plane due to the combined\neffects of gravity and the vertical force, P. At the instant shown in Figure B3, the plate has an angular\nvelocity, $\\omega = 20$ rad/s and an angular acceleration, $\\alpha = 15\\ \\mathrm{rad/s^2}$ both in the\ncounter-clockwise direction. The plate has a mass of $m = 10$ kg, and dimensions $b = 0.5$ m and\n$h = 0.24$ m.\n","figures":[{"src":"/figures/engr216-2019-b3.png","caption":"Figure B3: rectangular plate b x h pinned at C (0.10 m from the left edge, 0.12 m above the bottom edge); vertical force P upward at the top right corner; alpha and omega counter-clockwise."}],"formulaSheet":"- Plate about its centroid: $I_G = \\dfrac{m}{12}(b^2 + h^2)$ (given in the question)\n- Moment of inertia definition $I_O = \\int_m r^2\\,dm$ and radius of gyration $k_O = \\sqrt{I_O/m}$ (given in the 2019 Section B appendix; learn this)\n- Parallel-axis theorem: $I_C = I_G + m d^2$ (learn this)\n- Fixed-axis rotation of G: $a_n = \\omega^2 r_{G/C}$ (towards C), $a_t = \\alpha r_{G/C}$ (learn this)\n- Equations of motion: $\\sum\\mathbf{F} = m\\mathbf{a}_G$, $\\sum M_C = I_C\\alpha$ (or $\\sum M_G = I_G\\alpha$) (given in the 2019 Section B appendix; not on the 2026 sheet, so learn this)\n","parts":[{"label":"a","text":"For the plate shown in Figure B3, the moment of inertia for rotation in the $x$-$y$ plane about an axis\nlocated at the centroid is given by\n\n$$I_G = \\frac{m}{12}(b^2 + h^2)$$\n\nExplain what a moment of inertia is, and calculate $I_G$ for the plate.\n","marks":4,"grade":"third","answer":[{"label":"I_G","value":0.2563,"units":"kg m^2"}]},{"label":"b","text":"In your answer book, sketch a free body diagram of the plate showing the location and direction of external\nforces acting on the plate, and sketch the corresponding kinetic diagram showing the couple moments that are\nproduced by the dynamic forces.\n","marks":6,"grade":"$undefined"},{"label":"c","text":"Calculate the normal and transverse components of acceleration acting on the plate's centre of mass\n(centroid).\n","marks":6,"answer":[{"label":"a_n (towards C)","value":60,"units":"m/s^2"},{"label":"a_t (up +)","value":2.25,"units":"m/s^2"}],"grade":"$undefined"},{"label":"d","text":"Determine the magnitude of force P.\n","marks":4,"answer":[{"label":"P","value":54.84,"units":"N"}],"grade":"$undefined"},{"label":"e","text":"Determine the reactions in the x direction, $C_x$ and the y direction, $C_y$ that are produced at point C.\n","marks":5,"answer":[{"label":"C_x (right +)","value":-600,"units":"N"},{"label":"C_y (up +)","value":65.76,"units":"N"}],"grade":"$undefined"}],"solution":"$10e","confidence":"HIGH","solutionNote":"No official solution: the 2019 numerical answers file covers Section A (statics) only. Moments about C and\nabout G give consistent P, C_x and C_y. C_x is the pin force on the plate (600 N towards C, in -x).\n"},{"id":"engr216-2018-a1","module":"ENGR5004","week":10,"label":"ENGR216 2018 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["buckling","bending-eccentric","stress-transformation","euler-load","eccentric-loading","neutral-axis","principal-axes","circular-section"],"source":{"name":"ENGR216 Summer 2018 Q A1","paper":"ENGR216","year":2018,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR216_Paper_2018_v5_FigA3fixed.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR216.pdf","tag":"VALID"},"text":"A pin-ended metal post of length $L = 2\\ \\mathrm{m}$ with circular cross section of radius $R$ supports a\ncompressive axial load $P = 400\\ \\mathrm{kN}$ applied at distance $e$ from the section centre, as indicated in\nFigure A1. Use a Young's modulus $E$ of $70\\ \\mathrm{GPa}$.\n","figures":[{"src":"/figures/engr216-2018-a1.png","caption":"Figure A1: circular cross section with axes x and y through the centre 0; A at the top (y = R), B at the left (x = -R), C at the bottom (y = -R); the load acts on the y axis at distance e below 0."}],"formulaSheet":"- Euler critical load, pin-ended: $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L_e^2}$ with $L_e = L$ (on the formula sheet; also printed in this paper's appendix)\n- Circle: $A = \\pi R^2$, $I = \\dfrac{\\pi R^4}{4}$, $\\rho = \\sqrt{I/A} = R/2$ (on the formula sheet)\n- Eccentric axial loading: $F = P$, $M = Pd$, $\\sigma = \\dfrac{N}{A} - \\dfrac{M y}{I}$, $M > 0$ compresses fibres at $y > 0$ (on the formula sheet; also printed in this paper's appendix)\n- Safety factor: $P_{cr} = FS \\cdot P$; allowable stress $|\\sigma| \\le \\sigma_{all}$ (learn this)\n- Uniaxial stress: maximum shear $\\tau_{max} = |\\sigma|/2$ on planes at $45^\\circ$ (learn this)\n- Principal axes $\\tan 2\\theta_p = \\dfrac{2\\tau_{xy}}{\\sigma_x - \\sigma_y}$, max shear axes $\\tan 2\\theta_s = -\\dfrac{\\sigma_x - \\sigma_y}{2\\tau_{xy}}$ (learn this; printed in this paper's appendix but not on the 2026 sheet)\n","parts":[{"label":"a","text":"Assuming $e = 0$, an allowable normal stress of $80\\ \\mathrm{MPa}$ and a buckling safety factor of 2.5,\ndetermine the minimum radius of the cross section required for yield- and buckling-free operation.\n","marks":3,"grade":"2:2","answer":[{"label":"R_min","value":52.11,"units":"mm"}]},{"label":"b","text":"Assuming $e = 0.25R$ and using the value of $R$ obtained at (a):\n\ndetermine the distance (with sign) of the neutral axis from the $x$-axis. Briefly comment on this result\nwith regard to the sign of the normal stress in the section, and sketch the diagram of the resulting normal\nstress along the $y$ axis of the section clearly indicating the neutral axis;\n","marks":4,"grade":"2:2","answer":[{"label":"y_NA (up +)","value":52.11,"units":"mm"}]},{"label":"c","text":"at positions A, B and C determine the normal stress acting on the post cross section and the maximum shear\nstress on the post surface.\n","marks":6,"grade":"2:2","answer":[{"label":"sigma_A (tension +)","value":0,"units":"MPa"},{"label":"sigma_B (tension +)","value":-46.89,"units":"MPa"},{"label":"sigma_C (tension +)","value":-93.79,"units":"MPa"},{"label":"tau_max at A","value":0,"units":"MPa"},{"label":"tau_max at B","value":23.45,"units":"MPa"},{"label":"tau_max at C","value":46.89,"units":"MPa"}]},{"label":"d","text":"Consider three small squared surface elements with two sides parallel to the axis of the post (axis $z$)\nand centered at points A, B and C. At these three points, determine the orientation of the principal axes\nand the maximum shear axes with respect to the axis $z$.\n","marks":7,"grade":"2:1","answer":[{"label":"theta_p at B and C (from z)","value":0,"units":"deg"},{"label":"theta_s at B and C (from z)","value":45,"units":"deg"}]},{"label":"e","text":"Briefly comment on how the stress variations due to the axial load eccentricity $e = 0.25R$ may impact on\nthe strength and the buckling stability of the post using the value of $R$ determined at (a).\n","marks":5,"grade":"first"}],"solution":"$10f","confidence":"HIGH","solutionNote":"Matches the official numerical answers (R = 52 mm; neutral axis y = 52 mm, compressive everywhere; sigma = 0,\n-46.89, -93.78 MPa and tau = 0, 23.44, 46.89 MPa). The official stresses use the unrounded R = 52.11 mm; with\nR = 52 mm exactly they become 47.1 and 94.2 MPa. Parts (d) and (e) have no official numbers.\n"},{"id":"engr216-2018-a2","module":"ENGR5004","week":9,"label":"ENGR216 2018 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["shear-bending-diagrams","stress-transformation","torsion","combined-loading","mohrs-circle","principal-stresses","axle"],"source":{"name":"ENGR216 Summer 2018 Q A2","paper":"ENGR216","year":2018,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR216_Paper_2018_v5_FigA3fixed.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR216.pdf","tag":"VALID"},"text":"Consider the free-body-diagram of the axle AD depicted in the left plot of Figure A2, with dimensions\n$AB = CD = 250\\ \\mathrm{mm}$ and $BM = MC = 760\\ \\mathrm{mm}$. The axle cross section, depicted in the right plot\nof Figure A2, has radius $R = 100\\ \\mathrm{mm}$. The axle is subject to an engine torque $T_0 = 150\\ \\mathrm{kNm}$\napplied at M, two resistive torques $T_0/2$ applied at A and D, two vertical loads $P_0 = 400\\ \\mathrm{kN}$\napplied at A and D, and two vertical reactions $P_0$ applied at B and C.\n","figures":[{"src":"/figures/engr216-2018-a2.png","caption":"Figure A2: left, axle AD along x with z up: downward loads P0 and resistive torques T0/2 at A and D, upward reactions P0 at B and C, engine torque T0 at the midpoint M. Right, circular cross section of radius R in the y-z plane with surface point K at the top (on the z axis)."}],"formulaSheet":"- Sign convention for positive shear and bending, $dV/dx = -w$, $dM/dx = V$ (on the formula sheet)\n- Bending stress $\\sigma_x = -\\dfrac{M y}{I}$ ($y$ measured from the neutral axis; here the vertical coordinate is $z$), $I = \\dfrac{\\pi R^4}{4}$ (on the formula sheet; also printed in this paper's appendix)\n- Torsion $\\tau = \\dfrac{T c}{J}$, $J = \\dfrac{\\pi R^4}{2}$ (learn this; printed in this paper's appendix but not on the 2026 sheet)\n- Principal stresses $\\sigma_{1,2} = \\dfrac{\\sigma_x + \\sigma_y}{2} \\pm \\sqrt{\\left(\\dfrac{\\sigma_x - \\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$, $\\tan 2\\theta_p = \\dfrac{2\\tau_{xy}}{\\sigma_x - \\sigma_y}$ (learn this; printed in this paper's appendix but not on the 2026 sheet)\n- Maximum in-plane shear $\\tau_{max} = \\sqrt{\\left(\\dfrac{\\sigma_x - \\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$; Mohr's circle centre $\\left(\\dfrac{\\sigma_x + \\sigma_y}{2}, 0\\right)$ and radius $\\tau_{max}$ (learn this)\n","parts":[{"label":"a","text":"Determine the expression of the bending moment $M(x)$, shear load $V(x)$ and torsional load $T(x)$ along the\nentire axle.\n","marks":4,"grade":"2:2","answer":[{"label":"M between B and C (sagging +)","value":-100,"units":"kNm"},{"label":"V in AB (formula-sheet +)","value":-400,"units":"kN"},{"label":"|T| in each half","value":75,"units":"kNm"}]},{"label":"b","text":"Sketch $M(x)$, $V(x)$ and $T(x)$ along the entire axle.\n","marks":3,"grade":"2:2"},{"label":"c","text":"State in which portion ($x$ interval) of the axle one has a plane stress state on the axle surface and\nexplain why.\n","marks":4,"grade":"2:1"},{"label":"d","text":"Determine the magnitude of the maximum normal stress at the point K indicated in the right plot of Figure A2\nacting on the axle surface at any axial position between B and M, and state if this stress is tensile or\ncompressive.\n","marks":7,"grade":"2:1","answer":[{"label":"sigma_max at K (tensile)","value":143.2,"units":"MPa"}]},{"label":"e","text":"Determine the magnitude of the maximum shear stress at the point K indicated in the right plot of Figure A2\nacting on the axle surface at any axial position between B and M.\n","marks":3,"grade":"2:2","answer":[{"label":"tau_max at K","value":79.58,"units":"MPa"}]},{"label":"f","text":"Calculate the centre coordinates and the radius of Mohr's circle referring to the axle surface point K at\nany position between B and M, and draw as accurately as possible Mohr's circle in the $\\sigma$-$\\tau$ plane,\nindicating in such plane the points corresponding to principal stresses and maximum shear stress.\n","marks":4,"grade":"2:2","answer":[{"label":"Mohr centre sigma_C","value":63.66,"units":"MPa"},{"label":"Mohr radius","value":79.58,"units":"MPa"}]}],"solution":"$110","confidence":"MEDIUM","solutionNote":"Matches the official numerical answers (M, V, T expressions; sigma_max = 143.2 MPa tensile; tau_max = 79.6 MPa;\nMohr centre (63.66, 0), radius 79.6 MPa). MEDIUM only because part (c) is worded ambiguously: strictly the whole\ntraction-free surface is in plane stress; the answer given (B to C, where V = 0 and only bending plus torsion act)\nis our reading of the examiner's intent and should be confirmed with a lecturer. The sign of T follows the\nofficial answer; the figure's torque arrows are hard to read, but only magnitudes enter (d) to (f).\n"},{"id":"engr216-2018-a3","module":"ENGR5004","week":7,"label":"ENGR216 2018 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["shear-bending-diagrams","beam-shear-stress","beam-deflection","cantilever","triangular-load","elastic-curve","rectangular-section"],"source":{"name":"ENGR216 Summer 2018 Q A3","paper":"ENGR216","year":2018,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2018/ENGR216_Paper_2018_v5_FigA3fixed.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2018/ENGR216.pdf","tag":"VALID"},"text":"Consider the cantilever beam of length $L$ depicted in the left plot of Figure A3. The beam is subject to a\nlinear distributed load which is zero at $x = 0$ (point A) and is $w_0$ N/m at $x = L$ (point B), has Young's\nmodulus $E$, and the second area moment of its cross section about the neutral axis is $I$.\n","figures":[{"src":"/figures/engr216-2018-a3.png","caption":"Figure A3: left, cantilever AB of length L built in at A (x = 0), free at B, with a downward linearly varying load from zero at A to w0 at B; right, rectangular cross section of width w and height h."}],"formulaSheet":"- Distributed load resultant $P = \\int w\\,dx$ and its position $x_P$ (on the formula sheet)\n- Sign convention for positive shear and bending, $dV/dx = -w(x)$, $dM/dx = V(x)$ (on the formula sheet; also printed in this paper's appendix)\n- Transverse shear stress $\\tau = \\dfrac{VQ}{It}$, $\\tau_{max} = \\dfrac{3V}{2A}$ for a rectangle (learn this; on the 2025 sheet but not the 2026 one; $VQ/It$ also printed in this paper's appendix)\n- Rectangle $I = \\dfrac{w h^3}{12}$ (on the formula sheet)\n- Elastic curve $EI\\,\\dfrac{d^2y}{dx^2} = M(x)$, integrated twice with boundary conditions (learn this; on the 2025 sheet but not the 2026 one; also printed in this paper's appendix)\n","parts":[{"label":"a","text":"Determine the magnitude and the orientation of the reactions acting on the beam.\n","marks":6,"grade":"2:2","answer":[{"label":"R_A (up +)","value":0.5,"units":"w0*L"},{"label":"M_A (CCW +)","value":0.33333,"units":"w0*L^2"}]},{"label":"b","text":"Demonstrate that the shear load $V(x)$ along the beam is given by:\n\n$$V(x) = \\frac{w_0 L}{2} - \\frac{w_0 x^2}{2L}$$\n\nand sketch as accurately as possible the function $V(x)$ along the beam axis.\n","marks":4,"grade":"2:2"},{"label":"c","text":"Assuming the beam has a rectangular cross section with sides of width $w$ and height $h$, shown in the right\nplot of Figure A3, and considering the shear stress variability in the cross section, indicate the position\nalong the beam ($x$ value) where the maximum shear stress occurs, and, at such position, determine the\nmaximum shear stress and the shear stress at $y = h/4$ as functions of $w_0$, $L$, $w$ and $h$.\n","marks":7,"grade":"2:2","answer":[{"label":"tau_max (at x = 0, y = 0)","value":0.75,"units":"w0*L/(w*h)"},{"label":"tau at y = h/4","value":0.5625,"units":"w0*L/(w*h)"}]},{"label":"d","text":"Knowing that the bending moment $M(x)$ along the beam is:\n\n$$M(x) = -\\frac{w_0 L^2}{3} + \\frac{w_0 L x}{2} - \\frac{w_0 x^3}{6L}$$\n\ndetermine the equation of the deformed mean line of the beam.\n","marks":5,"grade":"2:2"},{"label":"e","text":"Determine the rotation (magnitude and direction) of the end B of the beam as a function of $w_0$, $E$, $I$\nand $L$.\n","marks":3,"grade":"2:2","answer":[{"label":"theta_B (CCW +)","value":-0.125,"units":"w0*L^3/(E*I)"}]}],"solution":"$111","confidence":"HIGH","solutionNote":"Matches the official numerical answers (R_A = w0 L/2 up, M_A = w0 L^2/3 counter-clockwise; tau_max = (3/4) w0 L/(wh)\nat x = 0, tau(h/4) = (9/16) w0 L/(wh); the same elastic curve; theta_B = -w0 L^3/(8EI), clockwise).\n"},{"id":"engr216-2017-a1","module":"ENGR5004","week":1,"label":"ENGR216 2017 Q A1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["axial-loading","statically-indeterminate","compatibility","superposition","axial-deformation","reactions","displacement","stepped-bar"],"source":{"name":"ENGR216 Summer 2017 Q A1","paper":"ENGR216","year":2017,"question":"A1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR216_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR216 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (e)\n","figures":[{"src":"/figures/engr216-2017-a1.png","caption":"Figure A1 (dimensions in mm): horizontal stepped bar between rigid walls at A (left) and E (right). Rod (1), the larger diameter, runs from A to C; rod (2), the smaller diameter, runs from C to E. Axial loads P1 at B and P2 at D both act to the right. Lengths AB = 180, BC = 135, CD = 90, DE = 90."}],"formulaSheet":"- Axial deformation of a uniform segment $\\delta = \\dfrac{NL}{AE}$, summed over segments $\\delta = \\sum_i \\dfrac{N_i L_i}{A_i E_i}$ (learn this)\n- Normal stress $\\sigma = N/A$ and Hooke's law $\\sigma = E\\epsilon$ (learn this)\n- Area of a solid circle $A = \\pi r^2 = \\pi d^2/4$ (on the formula sheet)\n- Equilibrium along the bar $\\sum F_x = 0$ (learn this)\n- Compatibility for a bar between rigid supports: total change in length $= 0$; method of superposition (release a redundant support, then restore it) (learn this)\n","parts":[{"label":"a","text":"Explain what the difference between statically determinate and statically indeterminate structures is,\nand explain what is the standard procedure for determining all reactions for these two types of systems.\n","marks":4,"grade":"third"},{"label":"b","text":"Consider the structure depicted in Figure A1. The structure is made up of two cylindrical rods denoted by\n(1) and (2) joined at C and restrained by rigid supports at A and E.\n\nState whether the structure of Figure A1, constrained to maintain the same overall lengths between\npositions A and E before and after applications of the concentrated loads $P_1$ and $P_2$, is statically\ndeterminate or indeterminate, and justify your answer.\n","marks":3,"grade":"third"},{"label":"c","text":"Explain which conditions can be used to determine the reactions at the rigid supports A and E.\n","marks":3,"grade":"third"},{"label":"d","text":"The cross section diameter and Young's modulus of rod (1) are respectively $E_1 = 70\\ \\mathrm{GPa}$ and\n$d_1 = 40\\ \\mathrm{mm}$, and the corresponding parameters of rod (2) are respectively\n$E_2 = 200\\ \\mathrm{GPa}$ and $d_2 = 30\\ \\mathrm{mm}$. The values of the axial loads $P_1$ and $P_2$\nindicated in Figure A.1 are $P_1 = 90\\ \\mathrm{kN}$ and $P_2 = 120\\ \\mathrm{kN}$, and all four lengths\nindicated in the figure are in millimetres.\n\nDetermine the reactions at the rigid supports A and E, providing their magnitude and orientation (left or\nright).\n","marks":8,"grade":"2:2","answer":[{"label":"R_A (left +)","value":67.8,"units":"kN"},{"label":"R_E (left +)","value":142.2,"units":"kN"}]},{"label":"e","text":"Determine the magnitude and direction (left or right) of the displacement of point c.\n","marks":7,"grade":"2:2","answer":[{"label":"u_C (right +)","value":0.1047,"units":"mm"}]}],"solution":"$112","confidence":"HIGH","solutionNote":"The official 2017 numerical answers sheet covers Section B (dynamics) only, so there are no official numbers\nfor Section A. Results verified by three methods (compatibility, superposition, stiffness method) in Python.\nThe paper prints \"KN\"; written here as kN.\n"},{"id":"engr216-2017-a2","module":"ENGR5004","week":7,"label":"ENGR216 2017 Q A2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["beam-deflection","shear-bending-diagrams","elastic-curve","distributed-load","reactions","simply-supported","free-body-diagram","boundary-conditions","slope"],"source":{"name":"ENGR216 Summer 2017 Q A2","paper":"ENGR216","year":2017,"question":"A2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR216_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR216 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (e)\n\nSection A Appendix (as printed with the paper):\n\nThe differential relationship between shear load $V(x)$ and distributed load $w(x)$ acting on a beam is:\n\n$$\\frac{dV}{dx} = -w(x)$$\n\nConsider the statically determinate beam with flexural rigidity $EI$ and length $L$ depicted in Figure A2,\nsubject to the distributed load $w(x)$, where $x$ denotes the coordinate along the horizontal axis depicted in\nthe figure with origin at end A of the beam.\n","figures":[{"src":"/figures/engr216-2017-a2.png","caption":"Figure A2: beam AB of length L with a roller support at A and a pin support at B. A downward distributed load w(x) rises from zero at A to its largest value at B (concave-down curve, flat at B). x axis along the beam from A to the right, y axis upwards with origin at A."}],"formulaSheet":"- Elastic curve $\\dfrac{d^2y}{dx^2} = \\dfrac{M}{EI}$ (given in the question; learn this; on the 2025 sheet but not the 2026 one)\n- $dV/dx = -w(x)$ (given in the question; also on the formula sheet); $dM/dx = V(x)$ and the positive shear / positive bending sign convention (on the formula sheet)\n- Resultant of a distributed load $P = \\int w\\,dx$ and its position $x_P = \\dfrac{\\int x\\,w\\,dx}{\\int w\\,dx}$ (on the formula sheet)\n- Equilibrium $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$; a roller gives one reaction normal to its surface, a pin gives two (learn this)\n- Boundary conditions at simple supports: $y = 0$; slope $\\theta = dy/dx$ (rotation of the section, small angles) (learn this)\n","parts":[{"label":"a","text":"Knowing that the bending moment $M(x)$ is positive over the entire beam, and noting that the relationship\nbetween $M(x)$ and the local deflection $y(x)$ is:\n\n$$\\frac{d^2y}{dx^2} = \\frac{M}{EI} \\qquad (1)$$\n\nwhere the y axis is depicted in Figure A2 and has origin at end A of the beam, sketch the deformed shape of\nthe beam mean line. For your sketch, use the same Cartesian (xy) axes system as defined in Figure A2.\n","marks":3,"grade":"2:2"},{"label":"b","text":"Sketch the free body diagram of the loaded beam and determine the magnitude and orientation of the reactions\nacting on its two ends if the distributed load $w(x)$ is given by:\n\n$$w(x) = w_0\\left(\\frac{2x}{L} - \\frac{x^2}{L^2}\\right) \\qquad (2)$$\n\nwhere $w_0$ is a constant distributed load.\n","marks":5,"grade":"2:2","answer":[{"label":"R_A (up +)","value":0.25,"units":"w0*L"},{"label":"R_B (up +)","value":0.41667,"units":"w0*L"}]},{"label":"c","text":"Using the differential relationship between shear load $V(x)$ and distributed load $w(x)$ provided in the\nAppendix, demonstrate that, for the considered problem, the shear load along the beam is:\n\n$$V(x) = w_0\\left(\\frac{L}{4} - \\frac{x^2}{L} + \\frac{x^3}{3L^2}\\right) \\qquad (3)$$\n","marks":5,"grade":"2:2"},{"label":"d","text":"Knowing that the equation of the deformed mean line of the beam is:\n\n$$y(x) = \\frac{w_0}{EI}\\left(\\frac{x^6}{360L^2} - \\frac{x^5}{60L} + \\frac{Lx^3}{24}\\right) + C_1 x + C_2 \\qquad (4)$$\n\ndetermine the constants $C_1$ and $C_2$ by imposing the compatibility condition of the beam deformation\nexpressed by Equation (4) with the constraints acting on the beam depicted in Figure A2.\n","marks":6,"grade":"2:2","answer":[{"label":"C1","value":-0.027778,"units":"w0*L^3/EI"},{"label":"C2","value":0,"units":"w0*L^4/EI"}]},{"label":"e","text":"Determine the rotation (magnitude and direction) of the beam sections at ends A and B as functions of\n$w_0$, $EI$ and $L$.\n","marks":6,"grade":"2:1","answer":[{"label":"theta_A (CCW +)","value":-0.027778,"units":"w0*L^3/EI"},{"label":"theta_B (CCW +)","value":0.030556,"units":"w0*L^3/EI"}]}],"solution":"$113","confidence":"HIGH","solutionNote":"The official 2017 numerical answers sheet covers Section B (dynamics) only, so there are no official numbers\nfor Section A. All results checked symbolically (sympy): $EI\\,y'' = M(x)$ for Equation (4), $V(x)$ by two\nroutes, reactions by moments about A and B, end slopes by two methods. Same structure as ENGR216 2019 Q A2.\n"},{"id":"engr216-2017-a3","module":"ENGR5004","week":10,"label":"ENGR216 2017 Q A3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["buckling","axial-loading","euler-load","safety-factor","hollow-section","pin-jointed-truss","two-force-member","design","axial-deformation"],"source":{"name":"ENGR216 Summer 2017 Q A3","paper":"ENGR216","year":2017,"question":"A3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR216_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR216 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (e)\n\nSection A Appendix (as printed with the paper):\n\nEuler's critical load $P_{cr}$ for doubly pinned rods is:\n\n$$P_{cr} = \\frac{\\pi^2 E I}{L^2}$$\n\nwhere $I$ is the smallest second area moment of the rod cross section, $E$ is the elastic or Young's modulus and\n$L$ is the rod length.\n\nThe second area moment $I$ of a solid circular cross section of radius $r$ about any diameter is:\n\n$$I = \\frac14 \\pi r^4$$\n","figures":[{"src":"/figures/engr216-2017-a3.png","caption":"Figure A3: symmetric two-bar frame. Rods AB and CB, each of length L, are pinned to the ground at A (left) and C (right) and pinned together at the apex B. Each rod makes angle theta with the horizontal (marked at A). A vertical load P acts downwards at B; a dashed vertical line through B marks the axis of symmetry."}],"formulaSheet":"- Euler load for a pinned-pinned column $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L^2}$ ($L_e = L$) (given in the question; also on the formula sheet)\n- Solid circle $I = \\pi r^4/4$, $A = \\pi r^2$ (given in the question; also on the formula sheet); hollow circle $I = \\pi(r_2^4 - r_1^4)/4$, $A = \\pi(r_2^2 - r_1^2)$ (on the formula sheet)\n- Strength requirement $\\sigma = F/A \\le \\sigma_{all}$; buckling with a safety factor $P_{cr} \\ge FS \\times F$ (learn this)\n- Axial deformation $\\delta = \\dfrac{FL}{AE}$ (learn this)\n- Equilibrium of a pin joint, $\\sum F_x = 0$, $\\sum F_y = 0$, with two-force members carrying force along their axes (learn this)\n","parts":[{"label":"a","text":"Provide a brief general definition of buckling in the context of the three main structural design\nrequirements.\n","marks":3,"grade":"third"},{"label":"b","text":"Consider the statically determinate symmetric structure depicted in Figure A3, made up of 2 pinned rods\nsubject to a vertical load $P$ of $50\\ \\mathrm{kN}$ acting on the joint B, as indicated in the figure. Both\nrods have length $L$ of $500\\ \\mathrm{mm}$ and $\\theta = 60^\\circ$. Their material has Young's modulus\n$E = 200\\ \\mathrm{GPa}$, and allowable normal stress of $150\\ \\mathrm{MPa}$. Assume a buckling safety factor\nof 2, and also that the Euler critical load (axial force) of the rods is that of doubly pinned columns.\n\nDetermine the internal force in each rod.\n","marks":3,"grade":"third","answer":[{"label":"F_AB (compression +)","value":28.87,"units":"kN"},{"label":"F_BC (compression +)","value":28.87,"units":"kN"}]},{"label":"c","text":"Assuming that both rods have circular cross section, determine the minimum radius of the cross section\nrequired for the structure to safely support the load $P$, fulfilling buckling and strength requirements.\nEuler's critical load for doubly pinned rods and the second area moment of solid circular cross section\nabout any diameter are given in the Appendix.\n","marks":5,"grade":"2:2","answer":[{"label":"r_min","value":9.82,"units":"mm"}]},{"label":"d","text":"Assuming that both rods have hollow circular cross section with inner radius of $10\\ \\mathrm{mm}$,\ndetermine the minimum outer diameter of the cross section to safely support the load $P$.\n","marks":7,"grade":"2:1","answer":[{"label":"D_o min","value":25.4,"units":"mm"}]},{"label":"e","text":"Starting from the load specifications and final geometry as given in question (d) above, determine the\nouter diameter of the rods' cross section if the length of each rod must not decrease by more than\n$400\\ \\mu\\mathrm{m}$.\n","marks":7,"grade":"2:1","answer":[{"label":"D_o (shortening limit alone)","value":25.09,"units":"mm"},{"label":"D_o (all requirements)","value":25.4,"units":"mm"}]}],"solution":"$114","confidence":"MEDIUM","solutionNote":"The official 2017 numerical answers sheet covers Section B (dynamics) only, so there are no official numbers\nfor Section A. All calculations were done in Python and back-substituted; (b) checked by a linear solve,\n(c)-(e) by back-substitution and the strain route. MEDIUM only because (e) is ambiguous: \"starting from ...\nquestion (d)\" could mean the stiffness limit alone (D = 25.09 mm, our main reading) or the final design meeting\nall requirements (D = 25.40 mm, unchanged from (d) since strength still governs). Both are given; confirm with\nthe lecturer which one the marking scheme expects. The paper prints \"KN\"; written here as kN.\n"},{"id":"engr216-2017-b1","module":"ENGR5004","week":8,"label":"ENGR216 2017 Q B1","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["rigid-kinetics","translation","tipping","friction","normal-reactions","free-body-diagram"],"source":{"name":"ENGR216 Summer 2017 Q B1","paper":"ENGR216","year":2017,"question":"B1","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR216_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR216 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (c)\n\nThe storage cupboard shown in Figure B1 has a mass of 30 kg and contacts the floor via rollers at the points\nshown. The location of the centre of gravity (G) and the point of application of an externally applied force\n(200 N) are also shown in the Figure.\n","figures":[{"src":"/figures/engr216-2017-b1.png","caption":"Figure B1: storage cupboard on two floor rollers 0.9 m apart; centre of gravity G midway between the rollers, 1.2 m above the floor; a horizontal 200 N force pushes on the left face at height h above the floor."}],"formulaSheet":"- Force equation for the mass centre: $\\sum \\mathbf{F} = m\\mathbf{a}_G$ (given in the 2017 Section B formula sheet; not on the 2026 sheet, so learn this)\n- Rigid-body translation: $\\alpha = 0$, so $\\sum M_G = 0$, or about any point P, $\\sum M_P = (m a_G)\\,d$ where $d$ is the moment arm of $m\\mathbf{a}_G$ about P (learn this)\n- Kinetic (sliding) friction $F = \\mu N$, opposing the sliding motion (learn this)\n- Tipping condition: a roller is about to lift when its normal reaction falls to zero (learn this)\n","parts":[{"label":"a","text":"For the loading condition shown in Figure B1, draw free body and kinetic diagrams of the cabinet, making\nthe assumption that there is no friction acting between the floor and the rollers. Calculate the\nacceleration of the cupboard when acted on by the 200 N force.\n","marks":8,"grade":"third","answer":[{"label":"a_G (right +)","value":6.667,"units":"m/s^2"}]},{"label":"b","text":"If there is no friction between the floor and the rollers, calculate the values of dimension h for which\nthe cupboard is not predicted to tip over when acted on by the 200 N force.\n","marks":9,"grade":"2:1","answer":[{"label":"h_min","value":0.538,"units":"m","tol":0.05},{"label":"h_max","value":1.862,"units":"m"}]},{"label":"c","text":"If the rollers are locked into position such that the coefficient of sliding friction between the rollers\nand the floor takes a value of $\\mu = 0.3$, determine the acceleration of the cupboard when acted on by\nthe 200 N force.\n","marks":8,"grade":"2:2","answer":[{"label":"a_G (right +)","value":3.724,"units":"m/s^2"}]}],"solution":"$115","confidence":"HIGH","solutionNote":"Matches the official numerical answers: (a) 6.7 m/s^2, (b) 0.55 < h < 1.87 m, (c) 3.7 m/s^2. The official lower\nlimit 0.55 m comes from the rounded acceleration 6.7 m/s^2; the exact value is 0.538 m (tolerance 5% on h_min).\nG is assumed midway between the rollers, as drawn; the official answers confirm this.\n"},{"id":"engr216-2017-b2","module":"ENGR5004","week":8,"label":"ENGR216 2017 Q B2","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:1","tags":["rigid-kinematics","rigid-kinetics","slider-crank","relative-velocity","relative-acceleration","instantaneous-centre","two-force-member"],"source":{"name":"ENGR216 Summer 2017 Q B2","paper":"ENGR216","year":2017,"question":"B2","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR216_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR216 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d).\n\nIn a food factory, a machine for processing confectionery consists of two rollers mounted on a carriage which\nmoves back and forth over a base plate. Confectionery mixture is placed on the base plate and movement of the\ncarriage rolls the confectionery material out until the desired thickness is achieved. A crank OA and\nconnecting rod AC are used to move the carriage back and forth (Figure B2). The crank is driven by a geared\nelectric motor.\n\nThe following particulars are given:\n\n- Crank radius a: $1.2\\ \\mathrm{m}$\n- Connecting rod length b: $4.5\\ \\mathrm{m}$\n- Crank rotation speed (constant): $30\\ \\mathrm{rev/min}$\n- Mass of carriage, including rollers, m: $55\\ \\mathrm{kg}$\n","figures":[{"src":"/figures/engr216-2017-b2.png","caption":"Figure B2: crank OA (length a) pivoted at the fixed point O, at angle theta above the horizontal, turning counter-clockwise (curved 'Direction of cranking' arrow); connecting rod AC (length b) runs down to the right to pin C on the wheeled carriage, at angle beta from the vertical through C; the carriage rolls horizontally on a flat base plate."}],"formulaSheet":"- Angular speed conversion $\\omega = 2\\pi N/60$ (rev/min to rad/s) (learn this)\n- Relative velocity $\\mathbf{v}_B = \\mathbf{v}_A + \\boldsymbol\\omega \\times \\mathbf{r}_{B/A}$ (given in the 2017 Section B formula sheet; not on the 2026 sheet, so learn this)\n- Relative acceleration $\\mathbf{a}_B = \\mathbf{a}_A + \\boldsymbol\\alpha \\times \\mathbf{r}_{B/A} - \\omega^2 \\mathbf{r}_{B/A}$ (given in the 2017 Section B formula sheet; learn this)\n- Instantaneous centre $\\omega = v_A/r_{A/IC} = v_C/r_{C/IC}$ (given in the 2017 Section B formula sheet; learn this)\n- Fixed-axis rotation at constant speed: $\\mathbf{a}_A = -\\omega^2 \\mathbf{r}_{A/O}$ (learn this)\n- $\\sum \\mathbf{F} = m\\mathbf{a}_G$ for the translating carriage; a light pinned rod is a two-force member (given in the 2017 Section B formula sheet / learn this)\n","parts":[{"label":"a","text":"When the crankshaft is in the position shown, angle $\\theta$ is $30^\\circ$ and angle $\\beta$ is $60^\\circ$\ninstantaneously. Find the velocity of the carriage when in this position and the angular velocity of link\nAC.\n","marks":10,"grade":"2:2","answer":[{"label":"v_C (speed, carriage moving left)","value":3.77,"units":"m/s"},{"label":"omega_AC (CW)","value":0.8378,"units":"rad/s"}]},{"label":"b","text":"Find the acceleration of the carriage for the same position.\n","marks":7,"grade":"2:1","answer":[{"label":"a_C (magnitude, directed left)","value":10.48,"units":"m/s^2"}]},{"label":"c","text":"Find the angular acceleration of link AC for the same position.\n","marks":3,"grade":"2:2","answer":[{"label":"alpha_AC (CCW +)","value":1.114,"units":"rad/s^2"}]},{"label":"d","text":"When there is no confectionery mixture on the base plate, the carriage rolls with very little friction.\nFor this case, find the force in the connecting rod for the position shown in Figure B2.\n","marks":5,"grade":"2:1","answer":[{"label":"F_AC (tension)","value":665.9,"units":"N"}]}],"solution":"$116","confidence":"HIGH","solutionNote":"Matches the official numerical answers: v_C 3.8 m/s, omega_AC 0.84 rad/s, a_C 10.4 m/s^2, alpha_AC 1.1 rad/s^2,\nrod force 660 N. Our a_C = 10.48 m/s^2 and F = 666 N differ from the official 10.4 and 660 only through\nrounding (660 N follows exactly from a_C = 10.4). The rod is assumed light (mass not given). The crank\ndirection (counter-clockwise) only affects the sign of the velocities, not the accelerations or the force.\n"},{"id":"engr216-2017-b3","module":"ENGR5004","week":8,"label":"ENGR216 2017 Q B3","kind":"past-paper","level":3,"marks":25,"minutes":30,"grade":"2:2","tags":["rigid-kinetics","fixed-axis-rotation","moment-of-inertia","radius-of-gyration","connected-bodies","bearing-reaction"],"source":{"name":"ENGR216 Summer 2017 Q B3","paper":"ENGR216","year":2017,"question":"B3","path":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Exam Papers 2017/ENGR216_Exam_2017rev.pdf","solutionsPath":"Pre-Spec Change/Mechanical Engineering Year 2/Second Year Past Exam Papers/Second Year Numerical Answers 2017/ENGR216 Numerical Answers 2017.pdf","tag":"VALID"},"text":"Answer ALL parts (a) - (d)\n\nFigure B3 shows a load lifting mechanism. The lift cage is raised and lowered by rotation of the drum unit\nwhich is cable driven via the electric drive unit. The lift load capacity is 350 kg, the drum unit has a mass\nof 95 kg and the radius of gyration of the drum unit about its mounting at C is 0.4 m. When raising the lift\ncage, the drive unit produces a constant tension in the cable of 2000 N and the cable contacts the drum\nwithout slipping.\n","figures":[{"src":"/figures/engr216-2017-b3.png","caption":"Figure B3: drum unit mounted on a bearing at C, with an outer drum of radius 0.7 m and an inner hub of radius 0.35 m. The drive cable leaves the top-left of the outer drum and runs down to the left to the drive unit at 40 degrees to the horizontal; the lift cable hangs vertically from the right side of the 0.35 m hub and carries the lift cage."}],"formulaSheet":"- Moment of inertia $I_O = \\int_m r^2\\,dm$ and radius of gyration $k_O = \\sqrt{I_O/m}$, so $I_O = m k_O^2$ (given in the 2017 Section B formula sheet; not on the 2026 sheet, so learn this)\n- Fixed-axis rotation $\\sum M_O = I_O\\alpha$ (given in the 2017 Section B formula sheet; learn this)\n- $\\sum \\mathbf{F} = m\\mathbf{a}_G$ (given in the 2017 Section B formula sheet; learn this)\n- No-slip cable on a drum: $a = r\\alpha$ (tangential acceleration of the drum surface equals the cable acceleration) (learn this)\n","parts":[{"label":"a","text":"Explain what is meant by a moment of inertia of a solid.\n","marks":5,"grade":"third"},{"label":"b","text":"Draw free body and kinetic diagrams for the cage and the drum unit, and from these derive equations of\nmotion for the drum unit and the lift cage for the case when the lift cage is being raised by the drive\nunit.\n","marks":8,"grade":"2:2"},{"label":"c","text":"For the case when the lift is loaded to full capacity and being raised, calculate the tension in the lift\ncable, the acceleration of the lift cage and the angular acceleration of the drum unit.\n","marks":6,"grade":"2:2","answer":[{"label":"T (lift cable)","value":3851.7,"units":"N"},{"label":"a_cage (up +)","value":1.195,"units":"m/s^2"},{"label":"alpha_drum (CCW +)","value":3.414,"units":"rad/s^2"}]},{"label":"d","text":"For the case when the lift cage is loaded to full capacity and being raised, calculate the force in the\nbearing at C and its direction of action.\n","marks":6,"grade":"2:2","answer":[{"label":"C_x (right +)","value":1532.1,"units":"N"},{"label":"C_y (up +)","value":6069.3,"units":"N"},{"label":"|C|","value":6259.6,"units":"N"},{"label":"angle above horizontal","value":75.83,"units":"deg"}]}],"solution":"$117","confidence":"HIGH","solutionNote":"Matches the official numerical answers: T = 3851 N, a = 1.2 m/s^2, F_x = 1532 N, F_y = 6069 N ((a) and (b)\nare descriptive, N/A in the official sheet). The official answers also take the 350 kg capacity as the whole\nmoving mass of the cage (cage mass not given). The official sheet does not list alpha; ours is 3.41 rad/s^2.\n"},{"id":"engr5004-ts1-p1","module":"ENGR5004","week":1,"label":"Tutorial 1 Problem 1","kind":"exercise","level":2,"minutes":10,"grade":"2:2","tags":["axial-loading","statically-indeterminate","compatibility","rod-in-tube"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 1, Problem 1","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 10 tutorial 1 (old spec ENGR216).pdf"},"text":"A rod of length $L$, cross-sectional area $A_1$, and modulus of elasticity $E_1$ has been placed inside a tube\nof the same length $L$, but of cross-sectional area $A_2$ and modulus of elasticity $E_2$. A force $P$ is applied\non a rigid plate attached to both tube and rod, as shown in the sketch below. Determine:\n\nHINT: deformation of tube and rod is constrained to be the same.\n","figures":[{"src":"/figures/engr5004-ts1-p1.png","caption":"Rod (A1, E1) inside a tube (A2, E2), both fixed to the wall at the left and to a rigid end plate at the right; force P pushes the plate towards the wall."}],"formulaSheet":"- Axial deformation: $\\delta = \\dfrac{PL}{AE}$ (learn this)\n- Equilibrium of the end plate: $P_1 + P_2 = P$ (learn this: free-body diagram)\n- Compatibility (rigid plate): $\\delta_1 = \\delta_2$ (learn this)\n","parts":[{"label":"a","text":"the horizontal displacement $\\delta$ of the rigid plate as a function of $P$, $L$, $A_1$, $A_2$, $E_1$ and $E_2$;\n","grade":"$undefined"},{"label":"b","text":"the fixed support reactions acting on the rod ($P_1$) and tube ($P_2$) when $E_1 = E_2$;\n","grade":"$undefined"},{"label":"c","text":"the fixed support reactions acting on the rod ($P_1$) and tube ($P_2$) when $E_1 = 2E_2$;\n","grade":"$undefined"}],"solution":"$118","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet and with the worked solution in old-spec Statics session 10. The sheet\nprints \"rod (P2) and tube (P2)\" in part (c); this is a typo for rod ($P_1$) and tube ($P_2$) and has been corrected.\nSymbolic answers only, so no answer boxes.\n"},{"id":"engr5004-ts1-p2","module":"ENGR5004","week":2,"label":"Tutorial 1 Problem 2","kind":"exercise","level":3,"minutes":20,"grade":"2:1","tags":["multiaxial-hooke","axial-loading","poisson-ratio","dilatation","bulk-modulus"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 1, Problem 2","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 10 tutorial 1 (old spec ENGR216).pdf"},"text":"A steel beam has a rectangular cross section of height $l_x = 20$ mm and width $l_y = 30$ mm, and length\n$l_z = 1$ m (lengths $l_x$, $l_y$ and $l_z$ are measured respectively along $x$, $y$ and $z$ axes of a Cartesian\nsystem). The material of the beam has Young modulus $E = 200$ GPa, Poisson ratio $\\nu = 0.29$, and maximum\nallowable normal stress of 175 MPa. The beam is subject to a compressive centric axial load $P_z$ of 80 kN\napplied at its ends (load acts along $z$ axis).\n","formulaSheet":"- Normal stress under centric axial load: $\\sigma_z = P_z / A$ (learn this)\n- Generalised Hooke's law: $\\epsilon_x = \\dfrac{\\sigma_x}{E} - \\dfrac{\\nu\\sigma_y}{E} - \\dfrac{\\nu\\sigma_z}{E}$ (and cyclic) (learn this)\n- Maximum shear stress under axial load, on planes at $45^\\circ$: $\\tau_{max} = \\dfrac{P}{2A}$ (learn this)\n- Bulk modulus: $k = \\dfrac{E}{3(1-2\\nu)}$ (learn this)\n- Dilatation: $e = \\epsilon_x + \\epsilon_y + \\epsilon_z = \\dfrac{1-2\\nu}{E}(\\sigma_x + \\sigma_y + \\sigma_z)$ (learn this)\n","parts":[{"label":"a","text":"State whether the area of the cross section of the beam will increase or decrease under the effect of the\napplied centric axial load and explain why.\n","grade":"$undefined"},{"label":"b","text":"Determine the variation $\\delta_x$ of the section height $l_x$ in mm, indicating if such variation is a\ncontraction or an elongation.\n","answer":[{"label":"delta_x (elongation +)","value":0.00387,"units":"mm","tol":0.05}],"grade":"$undefined"},{"label":"c","text":"Determine the maximum axial load $(P_z)_{max}$ applicable to the beam and the maximum mean shear stress in\nthese conditions.\n","answer":[{"label":"(P_z)max","value":105,"units":"kN"},{"label":"tau_max","value":87.5,"units":"MPa"}],"grade":"$undefined"},{"label":"d","text":"In the loading condition (c), state whether the uniformly distributed normal load to be applied on the beam\nfaces normal to the $x$ axis leading to a zero variation of the section height $l_x$ is compressive or\ntensile and justify your answer.\n","grade":"$undefined"},{"label":"e","text":"In the loading condition (c), determine the magnitude of the uniformly distributed normal load $\\sigma_x$ to\nbe applied on the beam faces normal to the $x$ axis resulting in zero variation of the section height $l_x$.\n","answer":[{"label":"sigma_x (tension +)","value":-50.75,"units":"MPa"}],"grade":"$undefined"},{"label":"f","text":"Calculate the bulk modulus $k$ of the beam material, and the dilatation $e$ of the beam due to the axial\nload $(P_z)_{max}$ and the distributed load $\\sigma_x$, stating if the applied loads result in an increment\nof reduction of the beam volume.\n","answer":[{"label":"k","value":158.7,"units":"GPa"},{"label":"e","value":-0.0004741,"units":"- (dimensionless)"}],"grade":"$undefined"}],"solution":"$119","confidence":"HIGH","solutionNote":"Agrees with the old-spec Statics session 10 worked solution on every value, including $\\delta_x = +0.0038$ mm\n(an elongation). The answer key on the 2026-27 sheet prints $\\delta_x = -0.0038$ mm; the minus sign is a typo,\nsince the question's own part (a) logic and the session solution both give an elongation.\n"},{"id":"engr5004-ts1-p3","module":"ENGR5004","week":3,"label":"Tutorial 1 Problem 3","kind":"exercise","level":2,"minutes":12,"grade":"2:2","tags":["bending-eccentric","eccentric-loading","neutral-axis","circular-section"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 1, Problem 3","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 10 tutorial 1 (old spec ENGR216).pdf"},"text":"A beam has a constant circular cross section of radius 20 mm, and is subject to a tensile axial load of 4 kN.\n","formulaSheet":"- Eccentric axial loading: $F = P$, $M = Pd$, $\\sigma_x = \\dfrac{P}{A} - \\dfrac{My}{I}$ (on the formula sheet)\n- Solid circle: $A = \\pi r^2$, $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n","parts":[{"label":"a","text":"Determine the magnitude of the maximum stress in the cross section when the axial load is applied at the\ncentre of the section.\n","answer":[{"label":"sigma_max","value":3.18,"units":"MPa"}],"grade":"$undefined"},{"label":"b","text":"In the loading condition (a), state whether a neutral axis exists or not, and explain why.\n","grade":"$undefined"},{"label":"c","text":"State whether the maximum normal stress in the cross section when the axial load is applied at 10 mm from the\ncentre of the section is compressive or tensile and explain why.\n","grade":"$undefined"},{"label":"d","text":"In the loading condition (c), determine the magnitude of the maximum compressive and tensile stresses in the\ncross section.\n","answer":[{"label":"sigma_max tensile","value":9.55,"units":"MPa"},{"label":"sigma_max compressive (magnitude)","value":3.18,"units":"MPa"}],"grade":"$undefined"},{"label":"e","text":"In the loading condition (c), determine the distance of the neutral axis from the centre of the cross section.\n","answer":[{"label":"y_NA","value":10,"units":"mm"}],"grade":"$undefined"}],"solution":"$11a","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet and with the old-spec Statics session 10 worked solution."},{"id":"engr5004-ts1-p4","module":"ENGR5004","week":5,"label":"Tutorial 1 Problem 4","kind":"exercise","level":3,"minutes":20,"grade":"2:1","tags":["shear-bending-diagrams","bending-eccentric","beam-shear-stress","triangular-load","max-bending-stress"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 1, Problem 4","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 10 tutorial 1 (old spec ENGR216).pdf"},"text":"Consider a simply supported beam subject to the linear distributed load sketched below.\n","figures":[{"src":"/figures/engr5004-ts1-p4.png","caption":"Beam AB of length L, roller at A (left), pin at B (right), triangular load rising from zero at A to w0 at B; x along the beam from A, y up."}],"formulaSheet":"- Distributed load resultant: $P = \\int w\\,dx$, position $x_P = \\dfrac{\\int x\\,w\\,dx}{\\int w\\,dx}$ (on the formula sheet)\n- Load-shear-moment relations: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$ (on the formula sheet)\n- Maximum bending stress: $\\sigma_{max} = \\dfrac{|M|c}{I}$, rectangle $I = \\dfrac{bh^3}{12}$ (on the formula sheet)\n- Maximum transverse shear stress, rectangle: $\\tau_{max} = \\dfrac{3V}{2A}$ (learn this; on the 2025 sheet but not the 2026 one)\n","parts":[{"label":"a","text":"Determine the equations of shearing load $V(x)$ and bending moment $M(x)$ as a function of the position $x$,\nthe value of the distributed load $w_0$ at the right end of the beam, and the beam length $L$;\n","grade":"$undefined"},{"label":"b","text":"plot $V(x)$ and $M(x)$ along the beam axis;\n","grade":"$undefined"},{"label":"c","text":"determine the position along the beam where the maximum normal stress occurs and, knowing that the cross\nsection is square and has length $a$, calculate the value of such maximum normal stress as a function of\n$w_0$, $a$ and $L$;\n","answer":[{"label":"x at sigma_max","value":0.5774,"units":"L"},{"label":"sigma_max","value":0.3849,"units":"w0*L^2/a^3"}],"grade":"$undefined"},{"label":"d","text":"determine the position along the beam where the maximum shear stress occurs and the value of such maximum\nshear stress as a function of $w_0$, $a$ and $L$.\n","answer":[{"label":"x at tau_max","value":1,"units":"L"},{"label":"tau_max","value":0.5,"units":"w0*L/a^2"}],"grade":"$undefined"}],"solution":"$11b","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet and with the old-spec Statics session 10 worked solution."},{"id":"engr5004-ts2-p1","module":"ENGR5004","week":5,"label":"Tutorial 2 Problem 1","kind":"exercise","level":3,"minutes":20,"grade":"2:1","tags":["beam-shear-stress","shear-bending-diagrams","bending-eccentric","hollow-section","first-moment-of-area"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 2, Problem 1","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_2.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Statics session 13 tutorial 2 (old spec ENGR216).pdf"},"text":"The beam depicted in the left figure below is subject to two concentrated loads $P_1 = 60$ kN and $P_2 = 45$ kN,\nwith points of application defined by the dimensions $a = 0.4$ m, $b = 0.3$ m. The beam features the hollow\nsquare cross section depicted in the right figure below. Consider section n-n, the position of which is defined\nby $d = 0.2$ m. Determine:\n","figures":[{"src":"/figures/engr5004-ts2-p1.png","caption":"Left: beam AB, pin at A, roller at B, P1 at a from A, P2 at a + b from A, span 2a + b, section n-n at d from A. Right: hollow square section 100 mm x 100 mm, wall 12 mm, point p at the inside top-left corner."}],"formulaSheet":"- Load-shear-moment relations and sign convention: $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n- Rectangle second moment $I = \\dfrac{bh^3}{12}$ (hollow section by subtraction) (on the formula sheet)\n- Parallel-axis theorem: $I = I_c + Ad^2$ (learn this)\n- Transverse shear stress: $\\tau = \\dfrac{VQ}{It}$, $Q = \\bar y' A'$ (learn this; on the 2025 sheet but not the 2026 one)\n- Bending stress: $\\sigma_{max} = \\dfrac{|M|c}{I}$ (on the formula sheet)\n","parts":[{"label":"a","text":"the support reactions and the shearing load on section n-n;\n","answer":[{"label":"R_Ay (up +)","value":54.55,"units":"kN"},{"label":"R_By (up +)","value":50.45,"units":"kN"},{"label":"V_nn","value":54.55,"units":"kN"}],"grade":"$undefined"},{"label":"b","text":"the largest shear stress acting on section n-n;\n","answer":[{"label":"tau_max","value":28.7,"units":"MPa"}],"grade":"$undefined"},{"label":"c","text":"the shear stress at point p on section n-n;\n","answer":[{"label":"tau_p","value":21.6,"units":"MPa"}],"grade":"$undefined"},{"label":"d","text":"the largest normal stress on section n-n.\n","answer":[{"label":"sigma_max","value":98.2,"units":"MPa"}],"grade":"$undefined"}],"solution":"$11c","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet and with the old-spec Statics session 13 worked solution."},{"id":"engr5004-ts2-p2","module":"ENGR5004","week":7,"label":"Tutorial 2 Problem 2","kind":"exercise","level":3,"minutes":25,"grade":"2:1","tags":["beam-deflection","shear-bending-diagrams","statically-indeterminate","propped-cantilever","elastic-curve","parabolic-load"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 2, Problem 2","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_2.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Statics session 13 tutorial 2 (old spec ENGR216).pdf"},"text":"Consider the loaded beam shown below, subject to a parabolic distributed load, constrained by a fixed support at\nits left end, and a roller at its right end.\n","figures":[{"src":"/figures/engr5004-ts2-p2.png","caption":"Beam AB, L = 3 m, fixed at A (left), roller at B (right), parabolic load w = w0 (x/L)^2 rising from zero at A to w0 at B."}],"formulaSheet":"- Distributed load resultant $P = \\int w\\,dx$ and its position $x_P$ (on the formula sheet)\n- Load-shear-moment relations: $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n- Elastic curve: $EI\\dfrac{d^2y}{dx^2} = M(x)$, $\\theta = \\dfrac{dy}{dx}$ (learn this; on the 2025 sheet but not the 2026 one)\n","parts":[{"label":"a","text":"State if this structure is statically determinate or indeterminate and justify your answer;\n","grade":"$undefined"},{"label":"b","text":"knowing that the distributed load per unit length at the right end (end B) is $w_0 = 15$ kN/m, and the beam\nlength is $L = 3$ m, determine the reactions at the fixed support and the roller.\n","answer":[{"label":"R_Ay (up +)","value":5.25,"units":"kN"},{"label":"M_A (CCW on beam +)","value":4.5,"units":"kN m"},{"label":"R_By (up +)","value":9.75,"units":"kN"}],"grade":"$undefined"}],"solution":"$11d","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet ($R_{Ay} = 5.25$ kN, $M_A = 4.5$ kN m, $R_{By} = 9.75$ kN) and with the\nold-spec Statics session 13 worked solution, which uses the same $M(x) = -M_A + R_Ax - w_0x^4/(12L^2)$.\n"},{"id":"engr5004-ts2-p3","module":"ENGR5004","week":7,"label":"Tutorial 2 Problem 3","kind":"exercise","level":3,"minutes":25,"grade":"2:1","tags":["beam-deflection","elastic-curve","maximum-deflection","triangular-load","simply-supported"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 2, Problem 3","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_2.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 7 - Statics session 13 tutorial 2 (old spec ENGR216).pdf"},"text":"Consider the loaded beam shown below, subject to a triangular distributed load, constrained by a roller at its\nleft end, and pinned at its right end.\n\nNOTE: the beam of problem 3 is that for which we computed shearing load $V(x)$ and bending load $M(x)$ in\nproblem 4 of Tutorial 1. Please start the solution of problem 3 (tutorial 2) by using the expression of $M(x)$\ndetermined in problem 4 of Tutorial 1.\n","figures":[{"src":"/figures/engr5004-ts2-p3.png","caption":"Beam AB of length L, roller at A (left), pin at B (right), triangular load rising from zero at A to w0 at B; x along the beam from A, y up."}],"formulaSheet":"- Load-shear-moment relations: $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n- Elastic curve: $EI\\dfrac{d^2y}{dx^2} = M(x)$, maximum deflection where $\\dfrac{dy}{dx} = 0$ (learn this; on the 2025 sheet but not the 2026 one)\n","parts":[{"label":"a","text":"Calculate the position $x^*$ along the beam axis where the maximum deflection of the beam occurs, expressing it\nas a fraction of the beam length $L$;\n","answer":[{"label":"x*","value":0.5193,"units":"L"}],"grade":"$undefined"},{"label":"b","text":"calculate the absolute value of the maximum deflection of the beam $\\delta = |y(x^*)|$, expressing it as a\nfunction of the flexural rigidity $EI$, the beam length $L$, and the distributed load $w_0$ at beam end B\n(note that $E$ is Young modulus and $I$ is the second area moment past the cross section neutral axis), and\nstate if the deflection is upwards or downwards.\n","answer":[{"label":"delta","value":0.006522,"units":"w0*L^4/(EI)"}],"grade":"$undefined"},{"label":"c","text":"calculate the numerical value of the maximum deflection $\\delta$, assuming $I = 0.333\\times10^{-3}\\ \\mathrm{m^4}$,\n$w_0 = 66$ kN/m, $L = 6$ m, and $E = 200$ GPa.\n","answer":[{"label":"x*","value":3.116,"units":"m"},{"label":"delta","value":8.38,"units":"mm"}],"grade":"$undefined"}],"solution":"$11e","confidence":"HIGH","solutionNote":"Agrees with the old-spec Statics session 13 worked solution ($x^* = 0.5193L$, $y(x^*) = -0.006522\\,w_0L^4/(EI)$,\n$x^* = 3.116$ m, $\\delta = 8.4$ mm). The answer key on the 2026-27 sheet prints $x^* = 0.5293L$; that is a typo\n(its own $x^* = 3.116$ m with $L = 6$ m gives 0.5193).\n"},{"id":"engr5004-ts3-p1","module":"ENGR5004","week":9,"label":"Tutorial 3 Problem 1","kind":"exercise","level":2,"minutes":20,"grade":"2:2","tags":["stress-transformation","plane-stress","principal-stresses","maximum-shear-stress","mohr-circle"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 3, Problem 1","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_3.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Statics session 17 tutorial 3 (old spec ENGR216).pdf"},"text":"Part (a) refers to the state of plane stress sketched in fig. 1-a; parts (b) to (e) refer to the state of plane\nstress sketched in fig. 1-b.\n","figures":[{"src":"/figures/engr5004-ts3-p1.png","caption":"Fig. 1-a: square element ABDE with 27 MPa tension on the vertical faces, 45 MPa compression on the horizontal faces and 18 MPa shear (downward on the right face); oblique face AC from corner A, at 75 degrees to side AE. Fig. 1-b: element with 10 MPa tension on the vertical faces, 50 MPa tension on the horizontal faces and 15 MPa shear (downward on the right face)."}],"formulaSheet":"- Stress transformation: $\\sigma_{x'} = \\dfrac{\\sigma_x+\\sigma_y}{2} + \\dfrac{\\sigma_x-\\sigma_y}{2}\\cos2\\theta + \\tau_{xy}\\sin2\\theta$, $\\tau_{x'y'} = -\\dfrac{\\sigma_x-\\sigma_y}{2}\\sin2\\theta + \\tau_{xy}\\cos2\\theta$ (learn this)\n- Principal directions: $\\tan2\\theta_p = \\dfrac{2\\tau_{xy}}{\\sigma_x-\\sigma_y}$; principal stresses $\\sigma_{max,min} = \\dfrac{\\sigma_x+\\sigma_y}{2} \\pm \\sqrt{\\left(\\dfrac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$ (learn this)\n- Maximum shear: $\\tan2\\theta_s = -\\dfrac{\\sigma_x-\\sigma_y}{2\\tau_{xy}}$, $\\tau_{max} = \\sqrt{\\left(\\dfrac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$, normal stress $\\sigma' = \\dfrac{\\sigma_x+\\sigma_y}{2}$ (learn this)\n","parts":[{"label":"a","text":"determine normal and shear stresses acting on oblique face AC of shaded triangular element shown.\n","answer":[{"label":"sigma_AC","value":-49.2,"units":"MPa"},{"label":"tau_AC (magnitude)","value":2.41,"units":"MPa"}],"grade":"$undefined"},{"label":"b","text":"determine orientation of principal axes and sketch them in $xy$ Cartesian system indicated in figure,\n","answer":[{"label":"theta_p1","value":18.43,"units":"deg"},{"label":"theta_p2","value":108.43,"units":"deg"}],"grade":"$undefined"},{"label":"c","text":"determine principal stresses and shear stress on principal planes,\n","answer":[{"label":"sigma_max","value":55,"units":"MPa"},{"label":"sigma_min","value":5,"units":"MPa"}],"grade":"$undefined"},{"label":"d","text":"determine orientation of maximum shear stress axes (i.e. axes orthogonal to planes of maximum shear stress),\n","answer":[{"label":"theta_s1","value":-26.57,"units":"deg"},{"label":"theta_s2","value":63.43,"units":"deg"}],"grade":"$undefined"},{"label":"e","text":"determine maximum shear stress and corresponding normal stress\n","answer":[{"label":"tau_max","value":25,"units":"MPa"},{"label":"sigma' on max-shear planes","value":30,"units":"MPa"}],"grade":"$undefined"}],"solution":"$11f","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet and with the old-spec Statics session 17 worked solution on every\nnumber. Note: the sheet lists $\\theta_{p1} = 18.4^\\circ$ next to $\\sigma_{p1} = 55$ MPa, but substitution shows\n55 MPa acts on the plane whose normal is at $108.4^\\circ$ and 5 MPa at $18.4^\\circ$; do not pair them in order.\n"},{"id":"engr5004-ts3-p2","module":"ENGR5004","week":9,"label":"Tutorial 3 Problem 2","kind":"exercise","level":3,"minutes":25,"grade":"2:1","tags":["stress-transformation","shear-bending-diagrams","combined-loading","torsion","principal-stresses","axle"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 3, Problem 2","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_3.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Statics session 17 tutorial 3 (old spec ENGR216).pdf"},"text":"Consider the free-body diagram of the axle AD depicted in the left plot below, with dimensions\nAB = CD = 500 mm and BM = MC = 1000 mm. The axle cross section, depicted in the right plot below, has radius\n$R = 150$ mm. The axle is subject to an engine torque $T_0 = 350$ kNm applied at position M, two resistive\ntorques $T_0/2$ applied at A and D, two vertical loads $P_0 = 700$ kN applied at B and C, and two vertical\nreactions $P_0$ applied at A and D.\n","figures":[{"src":"/figures/engr5004-ts3-p2.png","caption":"Left: axle AD along x with z up; P0 down at B and C, P0 up at A and D, torque T0 at M, torques T0/2 at A and D. Right: circular cross section of radius R in the yz plane with point K on the surface at the top (on the z axis)."}],"formulaSheet":"- Load-shear-moment relations and sign convention: $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n- Bending stress $\\sigma_x = -\\dfrac{Mz}{I}$ (formula sheet: $\\sigma_x = -My/I$ with the vertical coordinate), solid circle $I = \\dfrac{\\pi R^4}{4}$ (on the formula sheet)\n- Torsion: $\\tau = \\dfrac{Tc}{J}$, $J = \\dfrac{\\pi R^4}{2}$ (learn this)\n- Principal stresses and maximum in-plane shear: $\\sigma_{1,2} = \\dfrac{\\sigma_x}{2} \\pm \\sqrt{\\left(\\dfrac{\\sigma_x}{2}\\right)^2 + \\tau^2}$, $\\tau_{max} = \\sqrt{\\left(\\dfrac{\\sigma_x}{2}\\right)^2 + \\tau^2}$ (learn this)\n","parts":[{"label":"a","text":"Determine the expression of the bending moment $M(x)$, shearing load $V(x)$ and torsional load $T(x)$ along the\nentire axle.\n","answer":[{"label":"M between B and C","value":350,"units":"kN m"},{"label":"|V| on AB and CD","value":700,"units":"kN"},{"label":"|T|","value":175,"units":"kN m"}],"grade":"$undefined"},{"label":"b","text":"Sketch $M(x)$, $V(x)$ and $T(x)$ along the entire axle.\n","grade":"$undefined"},{"label":"c","text":"State at which portions ($x$ interval) of the axle, one has a plane stress state on the axle surface and\njustify your answer.\n","grade":"$undefined"},{"label":"d","text":"Determine the magnitude of the maximum normal stress at point K indicated in the right plot below acting on the\naxle surface at any axial position between B and M, and state if this stress is tensile or compressive.\n","answer":[{"label":"sigma_K max (magnitude, compressive)","value":139.8,"units":"MPa"}],"grade":"$undefined"},{"label":"e","text":"Determine the magnitude of the maximum shear stress at the point K indicated in the right plot below acting on\nthe axle surface at any axial position between B and M.\n","answer":[{"label":"tau_K max","value":73.8,"units":"MPa"}],"grade":"$undefined"}],"solution":"$120","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet ($\\sigma_{max} = 139.8$ MPa compressive, $\\tau_{max} = 73.8$ MPa) and with\nthe old-spec Statics session 17 worked solution, including its piecewise $M$, $V$, $T$ and its answer to (c)\n(plane stress everywhere on the surface except at the points of concentrated load application).\n"},{"id":"engr5004-ts4-p1","module":"ENGR5004","week":10,"label":"Tutorial 4 Problem 1","kind":"exercise","level":2,"minutes":8,"grade":"2:2","tags":["buckling","euler-load","hollow-section","weight-reduction"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 4, Problem 1","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_4.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 10 - Statics session 20 tutorial 4 (old spec ENGR216).pdf"},"text":"A compression member of 510 mm effective length consists of a solid aluminium rod of diameter $D = 24$ mm. To\nreduce the weight of the member by 25%, the solid rod is replaced by a hollow rod of internal diameter\n$d = 12$ mm. Determine\n","figures":[{"src":"/figures/engr5004-ts4-p1.png","caption":"Solid circular section of diameter D and hollow circular section of outer diameter D and inner diameter d."}],"formulaSheet":"- Euler critical load: $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L_e^2}$ (on the formula sheet)\n- Solid circle $I = \\dfrac{\\pi r^4}{4}$, hollow circle $I = \\dfrac{\\pi(r_2^4 - r_1^4)}{4}$ (on the formula sheet)\n","parts":[{"label":"a","text":"the percent reduction in the critical load,\n","answer":[{"label":"reduction","value":6.25,"units":"%"}],"grade":"$undefined"},{"label":"b","text":"the value of the critical load for the hollow rod. Use $E = 70$ GPa.\n","answer":[{"label":"P_cr hollow","value":40.55,"units":"kN"}],"grade":"$undefined"}],"solution":"$121","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet (6.25%, 40.5 kN) and with the old-spec Statics session 20 worked solution."},{"id":"engr5004-ts4-p2","module":"ENGR5004","week":10,"label":"Tutorial 4 Problem 2","kind":"exercise","level":3,"minutes":30,"grade":"first","tags":["buckling","bending-eccentric","stress-transformation","column-design","eccentric-loading","safety-factor"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 4, Problem 2","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_4.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 10 - Statics session 20 tutorial 4 (old spec ENGR216).pdf"},"text":"A cantilevered metal beam has length $L = 0.6$ m, circular cross section of radius $R$, and is subject to a\ncompressive axial load $P = 650$ kN applied at distance $e$ from the section centre, as indicated in the figure\nbelow. The material of the beam has Young's modulus $E = 200$ GPa and allowable normal stress of 150 MPa.\n\nParts (b) to (e): assuming $e = 0.40R$ and using the value of $R$ obtained at (a).\n","figures":[{"src":"/figures/engr5004-ts4-p2.png","caption":"Circular cross section with centre 0, x axis horizontal and y axis vertical; the load acts at distance e below 0 on the y axis; A at the top (y = R), B at the left end of the x axis, C at the bottom (y = -R)."}],"formulaSheet":"- Euler critical load $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L_e^2}$; fixed-free $L_e = 2L$ (on the formula sheet)\n- Solid circle $A = \\pi R^2$, $I = \\dfrac{\\pi R^4}{4}$ (on the formula sheet)\n- Eccentric axial loading: $F = P$, $M = Pd$, $\\sigma = \\dfrac{P}{A} - \\dfrac{My}{I}$ (on the formula sheet)\n- Maximum in-plane shear for a uniaxial surface element: $\\tau_{max} = \\dfrac{|\\sigma|}{2}$; principal axes along the stress directions, maximum-shear axes at $45^\\circ$ (learn this)\n","parts":[{"label":"a","text":"Assuming $e = 0$, and a buckling safety factor of 2.0, determine the minimum radius $R$ of the cross section\nrequired for yield- and buckling-free operation.\n","answer":[{"label":"R_min","value":37.1,"units":"mm"}],"grade":"$undefined"},{"label":"b","text":"determine distance (with sign) of the neutral axis from the $x$-axis. Briefly comment on this result with\nregard to the sign of the normal stress in the section, and sketch the diagram of the resulting normal stress\nalong the $y$ axis of the section clearly indicating the neutral axis;\n","answer":[{"label":"y_NA","value":23.2,"units":"mm"}],"grade":"$undefined"},{"label":"c","text":"at positions A, B and C determine the normal stress acting on the beam cross section, and the maximum shear\nstress on beam surface.\n","answer":[{"label":"sigma_A (tension +)","value":90,"units":"MPa"},{"label":"sigma_B (tension +)","value":-150,"units":"MPa"},{"label":"sigma_C (tension +)","value":-390,"units":"MPa"},{"label":"tau_max at A","value":45,"units":"MPa"},{"label":"tau_max at B","value":75,"units":"MPa"},{"label":"tau_max at C","value":195,"units":"MPa"}],"grade":"$undefined"},{"label":"d","text":"Briefly comment on results of (c) with regard to capability of the designed beam to withstand the considered\neccentric axial load.\n","grade":"$undefined"},{"label":"e","text":"At points A, B and C determine the orientation of the principal axes and the maximum shear axes with respect\nto the axis $z$ coinciding with the axis of the beam.\n","answer":[{"label":"theta_p2 (from z)","value":90,"units":"deg"},{"label":"theta_s1 (from z)","value":45,"units":"deg"},{"label":"theta_s2 (from z)","value":135,"units":"deg"}],"grade":"$undefined"}],"solution":"$122","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet and with the old-spec Statics session 20 worked solution\n($R_{min} = 37.1$ mm, $y_{NA} = 0.625R = 23.2$ mm, $\\sigma_A = 90$, $\\sigma_B = -150$, $\\sigma_C = -390$ MPa,\n$\\tau_{max} = 45$, 75, 195 MPa, principal axes along $z$ and the tangent direction). The sign of $y_{NA}$ assumes the\nload is below the centre as drawn in the figure.\n"},{"id":"engr5004-ts4-p3","module":"ENGR5004","week":10,"label":"Tutorial 4 Problem 3","kind":"exercise","level":3,"minutes":25,"grade":"2:1","tags":["buckling","euler-load","effective-length","i-section","buckling-plane","safety-factor"],"source":{"name":"ENGR5004 Statics Tutorial Sheet 4, Problem 3","path":"ENGR5004 - Engineering Mechanics/$ General/TUTORIAL_SHEET_4.pdf"},"text":"The bottom left figure below shows a rod of negligible mass of length $L$ subjected to a compressive axial load\n$F$. The rod is constrained by a fixed support at end A. The rod's cross section is depicted in the bottom right\nfigure below, in which $b = 10a$.\n","figures":[{"src":"/figures/engr5004-ts4-p3.png","caption":"Left: rod AB along z, fixed at A, compressive load F at the free end B; x and y axes at A. Right: I-section with flanges b wide and a thick, web 2a thick, clear height b between the flanges."}],"formulaSheet":"- Euler critical load $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L_e^2}$; effective lengths fixed-free $2L$, fixed-fixed $0.5L$ (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$, $\\dfrac{hb^3}{12}$ (composite section by addition/subtraction) (on the formula sheet)\n- Parallel-axis theorem $I = I_c + Ad^2$ (alternative route for $I_x$) (learn this)\n- Axial stress $\\sigma = F/A$ (learn this)\n","parts":[{"label":"a","text":"Assuming that the compressive load $F$ exceeds by a small amount Euler's critical load, determine and clearly\nstate if the rod will buckle in the $xz$ or $yz$ plane.\n","answer":[{"label":"I_x","value":773.33,"units":"a^4"},{"label":"I_y","value":173.33,"units":"a^4"}],"grade":"$undefined"},{"label":"b","text":"Draw as accurately as possible the shape of the buckled rod, labelling the two Cartesian axes defining the plane\nin which the rod buckled, and fulfilling the applied constraints in the drawn schematic.\n","grade":"$undefined"},{"label":"c","text":"Using $F = 50$ kN, allowable normal stress of 90 MPa, Young's modulus of 180 GPa, $L = 1.4$ m and a safety\nfactor of 2 for the buckling critical load, calculate the value of $a$ required for the rod not to buckle and\nnot to yield.\n","answer":[{"label":"a","value":7.1,"units":"mm"}],"grade":"$undefined"},{"label":"d","text":"Assuming that the rod designed in question c) is constrained by a fixed support at both end A and end B, and\nusing a safety factor of 2 in the buckling analysis, determine the maximum axial load the rod can bear without\nbuckling and yielding.\n","answer":[{"label":"F_max","value":181.6,"units":"kN"}],"grade":"$undefined"}],"solution":"$123","confidence":"HIGH","solutionNote":"Agrees with the answer key on the sheet (buckles in the $zx$ plane with $I_x = 2320a^4/3$, $I_y = 520a^4/3$;\n$a = 7.1$ mm; $F_{max} = 181.6$ kN). The old-spec session 20 covers only Problems 1 and 2, so there is no\nlecturer worked solution for this problem; it closely resembles ENGR273 2026 Q2.\n"},{"id":"engr5004-ex-stat-axial-01","module":"ENGR5004","week":1,"label":"Statics Session 2 Problem 1","kind":"tutorial","level":1,"minutes":8,"grade":"2:2","tags":["axial-loading","statically-indeterminate","compatibility","reactions"],"source":{"name":"ENGR5004 Statics session 2 worked example 1 (Problem 1, slides 7-9)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf"},"text":"Bar AB of length $L$ and uniform cross section $A$ is attached to rigid supports at A (top) and B (bottom)\nbefore being loaded. A downward axial load $P$ is then applied at point C, a distance $L_1$ below A and\n$L_2$ above B ($L_1 + L_2 = L$). The bar is linear elastic with Young's modulus $E$.\n","figures":[{"src":"/figures/engr5004-ex-stat-axial-01.png","caption":"Left: vertical bar AB between rigid supports at A (top) and B (bottom), with a downward load P at C; AC = L1, CB = L2, total length L. Right: free-body diagram with upward reactions R_A at A and R_B at B."}],"formulaSheet":"- Equilibrium of the bar: $\\sum F = 0$ (learn this)\n- Axial deformation of a uniform segment $\\delta_i = \\dfrac{P_i L_i}{A_i E_i}$, summed over segments $\\delta = \\sum_i \\dfrac{P_i L_i}{A_i E_i}$ (learn this)\n- Compatibility for a bar between rigid supports: total change in length $= 0$ (learn this)\n- Normal stress $\\sigma = P/A$ (learn this)\n","parts":[{"label":"a","text":"Draw the free-body diagram of the bar and explain why the reactions cannot be found from equilibrium alone.\n","grade":"$undefined"},{"label":"b","text":"Find the reactions $R_A$ and $R_B$ in terms of $P$, $L_1$, $L_2$ and $L$.\n","grade":"$undefined"},{"label":"c","text":"Find the normal stresses in portions AC and BC, stating whether each is tensile or compressive.\n","grade":"$undefined"}],"solution":"$124","confidence":"HIGH","solutionNote":"Agrees with the slide solution. Symbolic answers only, so no answer boxes."},{"id":"engr5004-ex-stat-axial-02","module":"ENGR5004","week":1,"label":"Statics Session 2 Problem 2","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["axial-loading","statically-indeterminate","superposition","stepped-bar","compatibility","reactions"],"source":{"name":"ENGR5004 Statics session 2 worked example 2 (Problem 2, slides 11-13 and Appendix 1 slide 16)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf"},"text":"A vertical steel bar ($E = 200$ GPa) is fixed to rigid supports at A (top) and B (bottom), with a close fit at\nboth supports before the loads are applied. Going down from A the bar passes points D, C and K:\n$l_{AD} = l_{DC} = l_{CK} = l_{KB} = 0.15$ m. Portion AC has area $A_{AD} = A_{DC} = 250\\ \\mathrm{mm^2}$ and\nportion CB has area $A_{CK} = A_{KB} = 400\\ \\mathrm{mm^2}$. Downward loads $P_D = 300$ kN at D and\n$P_K = 600$ kN at K are applied.\n\nDetermine the reactions at A and B using the superposition method, treating the reaction at B as redundant.\n","figures":[{"src":"/figures/engr5004-ex-stat-axial-02.png","caption":"Stepped vertical bar between rigid supports at A (top) and B (bottom). Points D, C, K divide it into four 0.15 m lengths; area A_AC above C, larger area A_CB below C; downward loads P_D at D and P_K at K."}],"formulaSheet":"- Axial deformation of a stepped bar $\\delta = \\sum_i \\dfrac{P_i l_i}{A_i E}$ (learn this)\n- Superposition: release the redundant support, find the deformation due to the loads $\\delta_L$ and due to the redundant reaction $\\delta_R$, impose $\\delta_L + \\delta_R = 0$ (learn this)\n- Equilibrium $\\sum F_y = 0$ (learn this)\n","parts":[{"label":"a","text":"With support B removed, find the elongation $\\delta_L$ of the bar due to $P_D$ and $P_K$.\n","answer":[{"label":"delta_L","value":5.625,"units":"mm"}],"grade":"$undefined"},{"label":"b","text":"Find the elongation $\\delta_R$ due to an upward force $R_B$ alone at B (as a multiple of $R_B$), and hence\n$R_B$ and $R_A$.\n","answer":[{"label":"R_B","value":577,"units":"kN"},{"label":"R_A","value":323,"units":"kN"}],"grade":"$undefined"}],"solution":"$125","confidence":"HIGH","solutionNote":"Agrees with the slides ($R_B = 577$ kN, $R_A = 323$ kN). The data give very high stresses (for example\n$323\\ \\mathrm{kN}/250\\ \\mathrm{mm^2} = 1292$ MPa in AD), far above the yield stress of ordinary steel, so stresses\nare not asked here; the example is about the method.\n"},{"id":"engr5004-ex-stat-axial-03","module":"ENGR5004","week":1,"label":"Statics Session 2 Practice Problem A","kind":"tutorial","level":1,"minutes":6,"grade":"third","tags":["axial-loading","axial-deformation","stepped-bar","free-body-diagram"],"source":{"name":"ENGR5004 Statics session 2 worked example 3 (Appendix 2 practice problem A, slides 17-18)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf"},"text":"A horizontal steel rod ($E = 200$ GPa) is fixed to a wall at A and is free at D. Going right from A it passes\nB and C. Axial loads act at three points: $P_B = 350$ kN at B to the right, $P_C = 200$ kN at C to the left and\n$P_D = 130$ kN at D to the right. Lengths: $l_{AB} = l_{BC} = 0.3$ m, $l_{CD} = 0.4$ m. Areas:\n$A_{AB} = A_{BC} = 5.8\\ \\mathrm{cm^2}$, $A_{CD} = 1.9\\ \\mathrm{cm^2}$.\n\nDetermine the internal force in each portion and the overall deformation of the rod.\n","figures":[{"src":"/figures/engr5004-ex-stat-axial-03.png","caption":"Horizontal rod fixed to a wall at A; larger section from A to C (through B), smaller section from C to D. P_B to the right at B, P_C to the left at C, P_D to the right at the free end D."}],"formulaSheet":"- Axial deformation of a stepped bar $\\delta = \\sum_i \\dfrac{P_i L_i}{A_i E_i}$ (learn this)\n- Internal force from a section FBD (take the free part, tension positive) (learn this)\n","answer":[{"label":"P_AB","value":280,"units":"kN"},{"label":"P_BC","value":-70,"units":"kN"},{"label":"P_CD","value":130,"units":"kN"},{"label":"delta","value":1.91,"units":"mm"}],"solution":"$126","confidence":"HIGH","solutionNote":"Agrees with the slides (1.9 mm)."},{"id":"engr5004-ex-stat-axial-04","module":"ENGR5004","week":1,"label":"Statics Session 2 Practice Problem B","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["axial-loading","axial-deformation","two-force-member","free-body-diagram","displacement"],"source":{"name":"ENGR5004 Statics session 2 worked example 4 (Appendix 3 practice problem B, slides 19-21)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf"},"text":"The horizontal rigid bar BDE is supported by two vertical pin-ended links. Link AB (aluminium, $E = 70$ GPa,\narea $500\\ \\mathrm{mm^2}$, length 0.3 m) hangs from a fixed pin A down to B; link CD (steel, $E = 200$ GPa, area\n$600\\ \\mathrm{mm^2}$, length 0.4 m) hangs from a fixed pin C down to D. $BD = 0.2$ m and $DE = 0.4$ m, and a\ndownward force of 30 kN acts at E. Treat bar BDE as rigid.\n\nDetermine the vertical deflection of (a) B, (b) D and (c) E.\n","figures":[{"src":"/figures/engr5004-ex-stat-axial-04.png","caption":"Rigid bar BDE hung from two vertical links: AB (0.3 m) above B and CD (0.4 m) above D, both pinned to a fixed support. BD = 0.2 m, DE = 0.4 m; 30 kN downward at E."}],"formulaSheet":"- Equilibrium of a rigid body: $\\sum M_B = 0$, $\\sum M_D = 0$ (learn this)\n- Axial deformation $\\delta = \\dfrac{FL}{AE}$ (learn this)\n- Rigid bar: displaced points stay on a straight line (similar triangles) (learn this)\n","parts":[{"label":"a","text":"Deflection of B (give the direction).","answer":[{"label":"delta_B (upwards)","value":0.514,"units":"mm"}],"grade":"$undefined"},{"label":"b","text":"Deflection of D (give the direction).","answer":[{"label":"delta_D (downwards)","value":0.3,"units":"mm"}],"grade":"$undefined"},{"label":"c","text":"Deflection of E (give the direction).","answer":[{"label":"delta_E (downwards)","value":1.93,"units":"mm"}],"grade":"$undefined"}],"solution":"$127","confidence":"HIGH","solutionNote":"Agrees with the slides (0.514 mm up, 0.300 mm down, 1.928 mm down)."},{"id":"engr5004-ex-stat-axial-05","module":"ENGR5004","week":3,"label":"Statics Session 2 Problem (rods CE and DF)","kind":"tutorial","level":2,"minutes":18,"grade":"2:1","tags":["axial-loading","statically-indeterminate","compatibility","displacement","shear-bending-diagrams"],"source":{"name":"ENGR5004 Statics session 2 problem set for week 4 (slide 14, with answers), worked in old-spec Statics session 7 Problem 2 (slides 9-13)","path":"ENGR5004 - Engineering Mechanics/Week 1/ENGR5004_Statics_session_02_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Statics session 07 (old spec ENGR216).pdf"},"text":"Rod CE and rod DF are attached to the rigid horizontal bar ABCD, which is pinned at B. Distances along the bar:\n$AB = 338$ mm, $BC = 225$ mm, $CD = 150$ mm. Both rods hang vertically down from the bar to fixed supports\nat E and F: rod CE has length 450 mm and diameter 10 mm; rod DF has length 562 mm and diameter 15 mm. Both\nrods are steel with $E = 200$ GPa. A vertical downward load of 100 kN is applied at end A of the bar.\n\nDetermine:\n","figures":[{"src":"/figures/engr5004-ex-stat-axial-05.png","caption":"Rigid bar ABCD pinned at B; 100 kN downward at A, 338 mm left of B. Rod CE (450 mm long) and rod DF (562 mm long) run down from C and D to fixed supports E and F; BC = 225 mm, CD = 150 mm."}],"formulaSheet":"- Equilibrium of a rigid body: $\\sum M_B = 0$, $\\sum F_y = 0$ (learn this)\n- Axial deformation $\\delta = \\dfrac{FL}{AE}$, circle $A = \\pi d^2/4$ (learn this; circle area on the formula sheet)\n- Rigid bar rotating about B: displacements proportional to distance from B (similar triangles) (learn this)\n- Bending moment by the method of sections; $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n","parts":[{"label":"a","text":"the internal force in each rod;","answer":[{"label":"F_CE","value":25.02,"units":"kN"},{"label":"F_DF","value":75.12,"units":"kN"}],"grade":"$undefined"},{"label":"b","text":"the vertical displacements of bar ends A and D;","answer":[{"label":"delta_D (up)","value":1.19,"units":"mm"},{"label":"delta_A (down)","value":1.08,"units":"mm"}],"grade":"$undefined"},{"label":"c","text":"the maximum absolute value of the bending moment in the bar, and where it occurs.","answer":[{"label":"|M|max","value":33.8,"units":"kN m"}],"grade":"$undefined"}],"solution":"$128","confidence":"HIGH","solutionNote":"Agrees with the answers on the session 2 slide and the worked solution in old-spec session 7 ($F_{DF} = 75.12$ kN,\n$F_{CE} = 25.02$ kN, $\\delta_D = 1.19$ mm, $\\delta_A = 1.08$ mm, $|M|_{max} = 33.8$ kN m at B).\n"},{"id":"engr5004-ex-stat-multiaxial-01","module":"ENGR5004","week":2,"label":"Statics Session 3 Example 1","kind":"tutorial","level":1,"minutes":5,"grade":"third","tags":["multiaxial-hooke","poisson-ratio","axial-loading"],"source":{"name":"ENGR5004 Statics session 3 worked example 1 (slide 5)","path":"ENGR5004 - Engineering Mechanics/Week 2/ENGR273_Statics_session_03_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 2/ENGR273_Statics_session_03_v1.pdf"},"text":"A 500 mm long, 16 mm diameter rod made of a homogeneous isotropic material is subjected to an axial tensile load\nof 12 kN. Its length increases by 300 $\\mu$m and its diameter decreases by 2.4 $\\mu$m.\n\nDetermine the modulus of elasticity $E$ and Poisson's ratio $\\nu$ of the material.\n","figures":[{"src":"/figures/engr5004-ex-stat-multiaxial-01.png","caption":"Rod of length L = 500 mm and diameter d = 16 mm fixed to a wall, pulled along x by 12 kN; delta_x = 300 micrometres, delta_y = -2.4 micrometres."}],"formulaSheet":"- Normal stress $\\sigma_x = P/A$, circle $A = \\pi d^2/4$ (learn this; circle area on the formula sheet)\n- Normal strain $\\epsilon = \\delta/L$, Hooke's law $\\sigma = E\\epsilon$ (learn this)\n- Poisson's ratio $\\nu = -\\dfrac{\\epsilon_y}{\\epsilon_x}$ (uniaxial load along $x$) (learn this)\n","answer":[{"label":"E","value":99.5,"units":"GPa"},{"label":"nu","value":0.25,"units":"-"}],"solution":"**Set-up.** Uniaxial tension along $x$; lateral strain measured on the diameter. Tension and elongation positive.\n\n$$A = \\frac{\\pi d^2}{4} = \\frac{\\pi (0.016)^2}{4} = 2.011\\times10^{-4}\\ \\mathrm{m^2}, \\qquad \\sigma_x = \\frac{P}{A} = \\frac{12\\times10^3}{2.011\\times10^{-4}} = 59.7\\ \\mathrm{MPa}$$\n\n$$\\epsilon_x = \\frac{\\delta_x}{L} = \\frac{300\\times10^{-6}}{0.500} = 6.0\\times10^{-4}, \\qquad \\epsilon_y = \\frac{\\delta_y}{d} = \\frac{-2.4\\times10^{-6}}{0.016} = -1.5\\times10^{-4}$$\n\n$$E = \\frac{\\sigma_x}{\\epsilon_x} = \\frac{59.7\\times10^6}{6.0\\times10^{-4}} = 99.5\\ \\mathrm{GPa}, \\qquad \\nu = -\\frac{\\epsilon_y}{\\epsilon_x} = \\frac{1.5\\times10^{-4}}{6.0\\times10^{-4}} = 0.25$$\n\nCheck: $\\nu$ is dimensionless and lies in the physical range $0 < \\nu < 0.5$; $E \\approx 100$ GPa is typical\nof a copper or titanium alloy. Note the lateral strain uses the diameter (16 mm), not the length.\n\nMarks (likely): stress 1, strains 2, $E$ and $\\nu$ 2.\n","confidence":"HIGH","solutionNote":"Agrees with the slide (E = 99.5 GPa, nu = 0.25)."},{"id":"engr5004-ex-stat-multiaxial-02","module":"ENGR5004","week":2,"label":"Statics Session 3 Example 2","kind":"tutorial","level":2,"minutes":8,"grade":"2:2","tags":["multiaxial-hooke","generalised-hooke","bulk-modulus","dilatation"],"source":{"name":"ENGR5004 Statics session 3 worked example 2 (slides 13-14)","path":"ENGR5004 - Engineering Mechanics/Week 2/ENGR273_Statics_session_03_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 2/ENGR273_Statics_session_03_v1.pdf"},"text":"A steel block ($E = 200$ GPa, $\\nu = 0.29$) is subjected to a uniform pressure $p$ on all its faces. Before the\nload is applied the edges of the block in the $x$, $y$ and $z$ directions are $l_x = 100$ mm, $l_y = 50$ mm and\n$l_z = 75$ mm. Under the load, $l_x$ decreases by $\\delta_x = 0.04$ mm.\n","formulaSheet":"- Generalised Hooke's law: $\\epsilon_x = \\dfrac{\\sigma_x}{E} - \\dfrac{\\nu\\sigma_y}{E} - \\dfrac{\\nu\\sigma_z}{E}$ (and cyclic) (learn this)\n- Bulk modulus: $k = \\dfrac{E}{3(1-2\\nu)}$; dilatation $e = \\epsilon_x + \\epsilon_y + \\epsilon_z = -\\dfrac{p}{k}$ under hydrostatic pressure (learn this)\n- Volume change $\\Delta V = eV$ (learn this)\n","parts":[{"label":"a","text":"Determine the change in length of the other two edges.","answer":[{"label":"delta_y","value":-0.02,"units":"mm"},{"label":"delta_z","value":-0.03,"units":"mm"}],"grade":"$undefined"},{"label":"b","text":"Determine the pressure $p$ applied to the block faces.","answer":[{"label":"p","value":190.5,"units":"MPa"}],"grade":"$undefined"},{"label":"c","text":"Determine the overall volume change of the block.","answer":[{"label":"Delta V","value":-450,"units":"mm^3"}],"grade":"$undefined"}],"solution":"$129","confidence":"HIGH","solutionNote":"Agrees with the slides (p = 1904.8 bar = 190.5 MPa, k = 158.7 GPa, Delta V = -450 mm^3)."},{"id":"engr5004-ex-stat-multiaxial-03","module":"ENGR5004","week":2,"label":"Statics Session 4 Example","kind":"tutorial","level":1,"minutes":4,"grade":"third","tags":["multiaxial-hooke","shear-strain","generalised-hooke"],"source":{"name":"ENGR5004 Statics session 4 worked example (slide 13)","path":"ENGR5004 - Engineering Mechanics/Week 2/ENGR273_Statics_session_04_v1.pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/Week 2/ENGR273_Statics_session_04_v1.pdf"},"text":"A rectangular block of material with shear modulus $G = 620$ MPa is bonded to two rigid horizontal plates. The\nblock is 50 mm high, 200 mm long (in the direction of the load) and 62 mm wide. The lower plate is fixed and the\nupper plate is subjected to a horizontal force $P$ along the 200 mm length, under which the upper plate moves\nby 1 mm.\n\nDetermine:\n","figures":[{"src":"/figures/engr5004-ex-stat-multiaxial-03.png","caption":"Block 50 mm high, 200 mm long and 62 mm wide bonded between a fixed lower plate and an upper plate pulled horizontally by P along the 200 mm direction."}],"formulaSheet":"- Average shearing strain $\\gamma_{xy} \\approx \\tan\\gamma_{xy} = \\dfrac{\\text{displacement}}{\\text{height}}$ (learn this)\n- Hooke's law in shear $\\tau_{xy} = G\\gamma_{xy}$, average shear stress $\\tau = P/A$ (learn this)\n","parts":[{"label":"a","text":"the average shearing strain in the block;","answer":[{"label":"gamma_xy","value":0.02,"units":"rad"}],"grade":"$undefined"},{"label":"b","text":"the force $P$ exerted on the upper plate.","answer":[{"label":"P","value":153.8,"units":"kN"}],"grade":"$undefined"}],"solution":"**(a)** The top face slides 1 mm relative to the bottom face across a height of 50 mm. For small angles\n\n$$\\gamma_{xy} \\approx \\tan\\gamma_{xy} = \\frac{1}{50} = 0.020\\ \\mathrm{rad}$$\n\n(exact $\\arctan(0.02) = 0.019997$ rad, so the small-angle approximation is excellent).\n\n**(b)** Hooke's law in shear, then force = average shear stress times the bonded area ($200 \\times 62$ mm):\n\n$$\\tau_{xy} = G\\gamma_{xy} = 620\\times10^6 \\times 0.020 = 12.4\\ \\mathrm{MPa}$$\n\n$$P = \\tau_{xy}A = 12.4\\times10^6 \\times 0.200 \\times 0.062 = 1.538\\times10^5\\ \\mathrm{N} = 153.8\\ \\mathrm{kN}$$\n\nCheck: units Pa m$^2$ = N. The shear stress acts on the horizontal faces, so the area is length times width,\nnot the height.\n\nMarks (likely): strain 2, stress 1, force 2.\n","confidence":"HIGH","solutionNote":"Agrees with the slide (0.020 rad, 153.8 kN)."},{"id":"engr5004-ex-stat-bending-01","module":"ENGR5004","week":3,"label":"Old spec Statics Session 5 Example 1","kind":"tutorial","level":1,"minutes":4,"grade":"third","tags":["bending-eccentric","bending-stress","rectangular-section","max-bending-stress"],"source":{"name":"ENGR216 (old spec) Statics session 5 worked example 1 (slide 8)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 05 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 05 (old spec ENGR216).pdf"},"text":"A steel bar of rectangular cross section, 20 mm wide and 64 mm deep, is subjected to two equal and opposite\ncouples $M$ acting in the vertical plane of symmetry of the bar (pure bending about the horizontal centroidal\naxis). Determine the value of $M$ that makes the maximum stress equal to the allowable stress of 250 MPa.\n","figures":[{"src":"/figures/engr5004-ex-stat-bending-01.png","caption":"Bar loaded by equal and opposite end couples M and M' in its vertical plane of symmetry; cross section 20 mm wide by 64 mm deep."}],"formulaSheet":"- Pure bending: $\\sigma_x = -\\dfrac{My}{I}$, $\\sigma_{max} = \\dfrac{|M|c}{I}$ (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$ (on the formula sheet)\n","answer":[{"label":"M","value":3.41,"units":"kN m"}],"solution":"**Set-up.** The neutral axis is the horizontal centroidal axis (perpendicular to the plane of symmetry), so the\nextreme fibres are at $c = 64/2 = 32$ mm.\n\n$$I = \\frac{bh^3}{12} = \\frac{0.020 \\times 0.064^3}{12} = 4.369\\times10^{-7}\\ \\mathrm{m^4}$$\n\n$$\\sigma_{max} = \\frac{Mc}{I} = \\sigma_{all} \\;\\Rightarrow\\; M = \\frac{I\\sigma_{all}}{c} = \\frac{4.369\\times10^{-7} \\times 250\\times10^6}{0.032} = 3.41\\times10^3\\ \\mathrm{N\\,m} = 3.41\\ \\mathrm{kN\\,m}$$\n\nCheck: section modulus $S = bh^2/6 = 0.02 \\times 0.064^2/6 = 1.365\\times10^{-5}\\ \\mathrm{m^3}$, and\n$S\\sigma_{all} = 3.41$ kN m, the same. Using $h = 64$ mm as the depth matters: bending the bar the other way\n($h = 20$ mm) would give only 1.07 kN m.\n\nMarks (likely): $I$ and $c$ 2, $M$ 2.\n","confidence":"HIGH","solutionNote":"Agrees with the slide (3.413 kN m)."},{"id":"engr5004-ex-stat-bending-02","module":"ENGR5004","week":3,"label":"Old spec Statics Session 5 Example 2","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["bending-eccentric","eccentric-loading","combined-axial-bending","neutral-axis","circular-section"],"source":{"name":"ENGR216 (old spec) Statics session 5 worked example 2 (slides 12-14)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 05 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 05 (old spec ENGR216).pdf"},"text":"An open-link chain is made by bending low-carbon steel rod of diameter $2c = 13$ mm into the shape shown. The\nchain carries a load of 712 N, whose line of action passes through the centres of the curved ends, at a\ndistance $d = 16$ mm from the centroid of the cross section of each straight portion.\n\nDetermine:\n","figures":[{"src":"/figures/engr5004-ex-stat-bending-02.png","caption":"Open chain link made from 13 mm diameter rod; 712 N pulls on each curved end; the load line is 16 mm from the centreline of each straight side."}],"formulaSheet":"- Eccentric axial loading: $F = P$, $M = Pd$, $\\sigma_x = \\dfrac{P}{A} - \\dfrac{My}{I}$ (on the formula sheet)\n- Solid circle: $A = \\pi r^2$, $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n- Maximum bending stress $\\sigma_{max} = \\dfrac{|M|c}{I}$ (on the formula sheet)\n","parts":[{"label":"a","text":"the largest tensile and compressive stresses in the straight portion of a link;","answer":[{"label":"sigma_t,max","value":58.2,"units":"MPa"},{"label":"sigma_c,max","value":-47.5,"units":"MPa"}],"grade":"$undefined"},{"label":"b","text":"the distance between the centroid and the neutral axis of the cross section.","answer":[{"label":"y_0","value":0.66,"units":"mm"}],"grade":"$undefined"}],"solution":"$12a","confidence":"HIGH","solutionNote":"Agrees with the slides (58.17 and -47.45 MPa, 0.66 mm); 58.18 MPa with unrounded intermediate values."},{"id":"engr5004-ex-stat-sfbm-01","module":"ENGR5004","week":3,"label":"Old spec Statics Session 6 Problem","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["shear-bending-diagrams","sinusoidal-load","distributed-load","simply-supported","max-bending-stress"],"source":{"name":"ENGR216 (old spec) Statics session 6 worked example (Problem, slides 13-15)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 06 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 06 (old spec ENGR216).pdf"},"text":"A simply supported beam AB of length $L$ (pin at A, roller at B) carries the downward distributed load\n\n$$w(x) = w_0 \\sin\\frac{\\pi x}{L}$$\n\nwith $x$ measured from A.\n","figures":[{"src":"/figures/engr5004-ex-stat-sfbm-01.png","caption":"Simply supported beam AB of length L (pin at A, roller at B) under a half-sine downward load w = w0 sin(pi x / L), zero at both ends and w0 at midspan."}],"formulaSheet":"- Resultant of a distributed load $P = \\int_0^L w\\,dx$ at its centroid (on the formula sheet)\n- Load, shear and moment: $\\dfrac{dV}{dx} = -w(x)$, $\\dfrac{dM}{dx} = V(x)$, $V(x) - V_A = -\\int_{x_A}^{x} w\\,dx$, $M(x) - M_A = \\int_{x_A}^{x} V\\,dx$ (on the formula sheet)\n- Maximum bending stress $\\sigma_{max} = \\dfrac{|M|c}{I}$; square of side $a$: $I = a^4/12$, $c = a/2$ (on the formula sheet)\n","parts":[{"label":"a","text":"Determine the equations of the shear load $V(x)$ and the bending moment $M(x)$.","grade":"$undefined"},{"label":"b","text":"Plot $V(x)$ and $M(x)$ along the beam axis.","grade":"$undefined"},{"label":"c","text":"Assuming the cross section is square with side length $a$, determine the position along the beam where the\nmaximum normal stress occurs and the value of that stress.\n","grade":"$undefined"}],"solution":"$12b","confidence":"HIGH","solutionNote":"Agrees with the slides. Symbolic answers only, so no answer boxes."},{"id":"engr5004-ex-stat-sfbm-02","module":"ENGR5004","week":3,"label":"Old spec Statics Session 6 Review Problem","kind":"tutorial","level":1,"minutes":10,"grade":"third","tags":["shear-bending-diagrams","superposition","udl","simply-supported","reactions"],"source":{"name":"ENGR216 (old spec) Statics session 6 worked example 2 (Appendix review problem, slides 17-23)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 06 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 3 - Statics session 06 (old spec ENGR216).pdf"},"text":"A simply supported beam AB of length $L$ (pin at A, roller at B) carries a uniform distributed load $W$ (N/m)\nover its whole length and a concentrated load $P$ (N) at midspan. Using superposition (uniform load alone plus\nconcentrated load alone), determine the shear force and bending moment along the beam and sketch both diagrams.\n","figures":[{"src":"/figures/engr5004-ex-stat-sfbm-02.png","caption":"Simply supported beam of length L (pin at A on the left, roller at B on the right) with a uniform load W along the whole span and a concentrated load P at midspan; x measured from A."}],"formulaSheet":"- Positive shear and positive bending sign convention (pictures); $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n- Equilibrium $\\sum F_y = 0$, $\\sum M = 0$ for the reactions; method of sections (learn this)\n- Superposition of load cases for a linear structure (learn this)\n","solution":"$12c","confidence":"HIGH","solutionNote":"Agrees with the slides. The slide sketch does not mark where $P$ acts, but the solution places it at midspan, so the\nstatement says so. Symbolic answers only, so no answer boxes.\n"},{"id":"engr5004-ex-stat-sfbm-03","module":"ENGR5004","week":3,"label":"Old spec Statics Session 7 Problem 1","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["shear-bending-diagrams","udl","simply-supported","max-bending-stress","reactions"],"source":{"name":"ENGR216 (old spec) Statics session 7 worked example 1 (Problem 1, slides 5-8)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Statics session 07 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Statics session 07 (old spec ENGR216).pdf"},"text":"Beam AC is simply supported (pin at A, roller at C) with span 9 m. It carries a uniformly distributed load of\n20 kN/m over the 6 m length AB; the remaining 3 m, BC, is unloaded.\n","figures":[{"src":"/figures/engr5004-ex-stat-sfbm-03.png","caption":"Simply supported beam AC, pin at A, roller at C; 20 kN/m downward over AB = 6 m, BC = 3 m unloaded."}],"formulaSheet":"- Load, shear and moment: $\\dfrac{dV}{dx} = -w$, $\\dfrac{dM}{dx} = V$ and their integral forms (on the formula sheet)\n- Maximum bending stress $\\sigma_{max} = \\dfrac{|M|c}{I} = \\dfrac{|M|}{S}$ with section modulus $S = I/c$ (on the formula sheet: $|M|c/I$; $S$ learn this)\n","parts":[{"label":"a","text":"Draw the shear and bending moment diagrams of the beam.","answer":[{"label":"R_A","value":80,"units":"kN"},{"label":"R_C","value":40,"units":"kN"},{"label":"M at B","value":120,"units":"kN m"}],"grade":"$undefined"},{"label":"b","text":"Determine the location of the section where the maximum normal stress due to bending occurs and, assuming an\nelastic section modulus $S = 1270\\ \\mathrm{cm^3}$, the magnitude of that stress.\n","answer":[{"label":"x from A","value":4,"units":"m"},{"label":"M_max","value":160,"units":"kN m"},{"label":"sigma_max","value":126,"units":"MPa"}],"grade":"$undefined"}],"solution":"$12d","confidence":"HIGH","solutionNote":"Agrees with the slides (R_A = 80 kN, R_C = 40 kN, M_max = 160 kN m at x = 4 m, 126 MPa)."},{"id":"engr5004-ex-stat-shear-01","module":"ENGR5004","week":4,"label":"Old spec Statics Session 8 Example","kind":"tutorial","level":2,"minutes":8,"grade":"2:2","tags":["beam-shear-stress","shear-flow","first-moment-of-area","parallel-axis-theorem","i-section"],"source":{"name":"ENGR216 (old spec) Statics session 8 worked example (slides 8-9)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Statics session 08 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 4 - Statics session 08 (old spec ENGR216).pdf"},"text":"A beam is made of three equal planks, each of cross section $a \\times b$ with $a = 100$ mm and $b = 20$ mm,\nnailed together into an I section: two horizontal flanges ($a$ wide, $b$ thick) and a vertical web ($b$ thick,\n$a$ deep) between them. The nails joining each flange to the web are spaced at $z = 25$ mm along the beam and\nthe vertical shear in the beam is $V = 500$ N. Determine the shearing force in each nail.\n","figures":[{"src":"/figures/engr5004-ex-stat-shear-01.png","caption":"I section of three planks: flanges a wide and b thick, web b thick and a deep between them, nailed through the flanges into the web."}],"formulaSheet":"- Shear flow (horizontal force per unit length) $q = \\dfrac{VQ}{I}$; nail force $F = qz$ (learn this; $\\tau = VQ/(It)$ is on the formula sheet)\n- First moment of area $Q = \\bar y A$ of the area beyond the cut, about the neutral axis (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$ and the parallel axis theorem $I = \\bar I + Ad^2$ (on the formula sheet)\n","answer":[{"label":"q","value":3704,"units":"N/m"},{"label":"F per nail","value":92.6,"units":"N"}],"solution":"$12e","confidence":"HIGH","solutionNote":"Agrees with the slides (q = 3704 N/m, F = 92.6 N)."},{"id":"engr5004-ex-stat-shear-02","module":"ENGR5004","week":4,"label":"Old spec Statics Session 9 Problem 1","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["beam-shear-stress","first-moment-of-area","i-section","reactions"],"source":{"name":"ENGR216 (old spec) Statics session 9 worked example 1 (Problem 1, slides 3-5)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 09 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 09 (old spec ENGR216).pdf"},"text":"Beam AB is simply supported (pin at A, roller at B) and made of three planks glued together into an I section:\na top flange 100 mm wide and 20 mm thick, a web 20 mm thick and 80 mm deep, and a bottom flange 60 mm wide and\n20 mm thick (total depth 120 mm). The centroid C is 68.3 mm above the bottom face and the centroidal second\nmoment of area about the neutral axis is $8.63\\times10^{-6}\\ \\mathrm{m^4}$. Two downward loads\n$P_1 = P_2 = 3.0$ kN act at distances $a = 0.6$ m and $a + b = 1.0$ m from A, and the span is\n$2a + b = 1.6$ m. Each glued joint is 20 mm wide.\n\nDetermine the shear stress in each glued joint (joint u, between the top flange and the web, and joint l,\nbetween the web and the bottom flange) at section n-n, located between A and $P_1$.\n","figures":[{"src":"/figures/engr5004-ex-stat-shear-02.png","caption":"Left: simply supported beam AB with loads P1 and P2 at a and a + b from A (span a + b + a); section n-n between A and P1. Right: I section, top flange 100 x 20 mm, web 20 x 80 mm, bottom flange 60 x 20 mm, centroid C 68.3 mm above the bottom; glued joints u (top) and l (bottom)."}],"formulaSheet":"- Shear stress in a beam $\\tau = \\dfrac{VQ}{It}$, $Q = \\bar y A$ of the area beyond the cut (on the formula sheet)\n- Centroid of a composite area $\\bar y = \\dfrac{\\sum A_i \\bar y_i}{\\sum A_i}$ (learn this; only needed to check the given data)\n","answer":[{"label":"tau_u","value":1450,"units":"kPa"},{"label":"tau_l","value":1216,"units":"kPa"}],"solution":"$12f","confidence":"HIGH","solutionNote":"Agrees with the slides (1450 kPa and 1216 kPa). With the unrounded centroid ($\\bar y_1 = 41.67$ mm) $\\tau_u = 1448$ kPa;\nthe answer box tolerance accepts both.\n"},{"id":"engr5004-ex-stat-shear-03","module":"ENGR5004","week":4,"label":"Old spec Statics Session 9 Problem 2","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["beam-shear-stress","bending-stress","rectangular-section","design","allowable-load"],"source":{"name":"ENGR216 (old spec) Statics session 9 worked example 2 (Problem 2, slides 6-9)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 09 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 5 - Statics session 09 (old spec ENGR216).pdf"},"text":"A simply supported timber beam AB of length $L$ and rectangular cross section (width $b$, height $h$) carries a\nsingle concentrated load $P$ at its midpoint C.\n","figures":[{"src":"/figures/engr5004-ex-stat-shear-03.png","caption":"Simply supported timber beam AB with load P at midpoint C (AC = CB = L/2); rectangular section of width b and height h."}],"formulaSheet":"- Maximum transverse shear stress, rectangle: $\\tau_{max} = \\dfrac{3V}{2A}$ (learn this)\n- Maximum bending stress $\\sigma_{max} = \\dfrac{|M|c}{I}$, rectangle $I = \\dfrac{bh^3}{12}$ (on the formula sheet)\n- Simply supported beam with central load: $V_{max} = P/2$, $M_{max} = PL/4$ (learn this: from the shear and moment diagrams)\n","parts":[{"label":"a","text":"Show that the ratio $\\tau_{max}/\\sigma_{max}$ of the maximum shear and normal stresses in the beam is equal to\n$h/(2L)$.\n","grade":"$undefined"},{"label":"b","text":"Determine the cross section height $h$ and width $b$, knowing that $L = 3.5$ m, $P = 60$ kN,\n$\\tau_{max} = 1100$ kPa and $\\sigma_{max} = 31.0$ MPa.\n","answer":[{"label":"h","value":0.248,"units":"m"},{"label":"b","value":0.165,"units":"m"}],"grade":"$undefined"},{"label":"c","text":"For the beam just designed, determine the maximum load $P$ that can be applied if the maximum allowable shear\nstress is 1600 kPa.\n","answer":[{"label":"P_max","value":87.3,"units":"kN"}],"grade":"$undefined"}],"solution":"$130","confidence":"HIGH","solutionNote":"Agrees with the slides for h = 0.248 m and b = 0.165 m. The slide writes the final answer as \"87.2 KPa\": the\nunits should be kN (it is a force), and with unrounded h and b the value is 87.3 kN.\n"},{"id":"engr5004-ex-stat-defl-01","module":"ENGR5004","week":6,"label":"Old spec Statics Session 11 Example 1","kind":"tutorial","level":1,"minutes":8,"grade":"third","tags":["beam-deflection","elastic-curve","cantilever","boundary-conditions","maximum-deflection"],"source":{"name":"ENGR216 (old spec) Statics session 11 worked example 1 (slides 7-8)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 11 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 11 (old spec ENGR216).pdf"},"text":"A cantilever AB of length $L$ and flexural rigidity $EI$, fixed at A, carries a downward concentrated load $P$\nat its free end B. Take $x$ from A along the beam and $y$ upwards.\n","figures":[{"src":"/figures/engr5004-ex-stat-defl-01.png","caption":"Cantilever AB of length L, fixed at A (left), with a downward load P at the free end B."}],"formulaSheet":"- Elastic curve: $\\dfrac{d^2y}{dx^2} = \\dfrac{M(x)}{EI}$, slope $\\theta = \\dfrac{dy}{dx}$ (learn this; on the 2025 sheet but not the 2026 one)\n- Fixed end: $y = 0$ and $\\theta = 0$ (learn this)\n- Bending moment by the method of sections, positive sagging (on the formula sheet)\n","parts":[{"label":"a","text":"Sketch the elastic curve of the beam.","grade":"$undefined"},{"label":"b","text":"Calculate the slope $\\theta(x)$ and deflection $y(x)$ of the elastic curve.","grade":"$undefined"},{"label":"c","text":"Calculate the maximum deflection of the beam.","grade":"$undefined"}],"solution":"$131","confidence":"HIGH","solutionNote":"Agrees with the slides. Symbolic answers only, so no answer boxes."},{"id":"engr5004-ex-stat-defl-02","module":"ENGR5004","week":6,"label":"Old spec Statics Session 11 Example 2","kind":"tutorial","level":2,"minutes":8,"grade":"2:2","tags":["beam-deflection","elastic-curve","cantilever","maximum-deflection","slope"],"source":{"name":"ENGR216 (old spec) Statics session 11 worked example 2 (slides 9-10)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 11 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 11 (old spec ENGR216).pdf"},"text":"A cantilever AB of length $2L$ and flexural rigidity $EI$, fixed at A, carries a downward concentrated load $P$\nat its midpoint C ($AC = CB = L$).\n\nYou may use the result for a cantilever of length $L$ with an end load $P$: $\\theta(x) = \\dfrac{1}{EI}\\left[-PLx + \\dfrac{Px^2}{2}\\right]$,\n$y(x) = \\dfrac{1}{EI}\\left[-\\dfrac{PLx^2}{2} + \\dfrac{Px^3}{6}\\right]$.\n","figures":[{"src":"/figures/engr5004-ex-stat-defl-02.png","caption":"Cantilever AB of length 2L fixed at A; downward load P at the midpoint C, AC = CB = L."}],"formulaSheet":"- Elastic curve: $\\dfrac{d^2y}{dx^2} = \\dfrac{M(x)}{EI}$, slope $\\theta = \\dfrac{dy}{dx}$ (learn this; on the 2025 sheet but not the 2026 one)\n- Unloaded segment: $M = 0$, so it stays straight (rigid rotation), small angles $y_B \\approx y_C + L\\theta_C$ (learn this)\n","parts":[{"label":"a","text":"Sketch the deformed shape of the beam.","grade":"$undefined"},{"label":"b","text":"Calculate the maximum deflection of the beam.","grade":"$undefined"}],"solution":"$132","confidence":"HIGH","solutionNote":"Agrees with the slides. Slide 9 says the left half deforms like \"example 2\"; it means example 1 (the end-loaded\ncantilever). Symbolic answer only, so no answer box.\n"},{"id":"engr5004-ex-stat-defl-03","module":"ENGR5004","week":6,"label":"Old spec Statics Session 11 Example 3","kind":"tutorial","level":1,"minutes":10,"grade":"2:2","tags":["beam-deflection","elastic-curve","simply-supported","udl","maximum-deflection","slope"],"source":{"name":"ENGR216 (old spec) Statics session 11 worked example 3 (slides 11-12)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 11 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 11 (old spec ENGR216).pdf"},"text":"A simply supported beam AB of length $L$ and flexural rigidity $EI$ (pin at A, roller at B) carries a uniform\ndistributed load $w$ over its whole length. Take $x$ from A and $y$ upwards.\n","figures":[{"src":"/figures/engr5004-ex-stat-defl-03.png","caption":"Simply supported beam AB of length L (pin at A, roller at B) under a uniform downward load w."}],"formulaSheet":"- Elastic curve: $EI\\dfrac{d^2y}{dx^2} = M(x)$, slope $\\theta = \\dfrac{dy}{dx}$ (learn this; on the 2025 sheet but not the 2026 one)\n- Boundary conditions for a simply supported beam: $y = 0$ at each support (learn this)\n","parts":[{"label":"a","text":"Sketch the deformed shape.","grade":"$undefined"},{"label":"b","text":"Calculate the slope $\\theta(x)$ and deflection $y(x)$ of the elastic curve.","grade":"$undefined"},{"label":"c","text":"Calculate the maximum deflection and the rotation of end A.","grade":"$undefined"}],"solution":"$133","confidence":"HIGH","solutionNote":"Agrees with the slides. Symbolic answers only, so no answer boxes."},{"id":"engr5004-ex-stat-defl-04","module":"ENGR5004","week":6,"label":"Old spec Statics Session 12 Example 1","kind":"tutorial","level":3,"minutes":20,"grade":"2:1","tags":["beam-deflection","elastic-curve","cantilever","distributed-load","boundary-conditions","maximum-deflection"],"source":{"name":"ENGR216 (old spec) Statics session 12 worked example 1 (slides 6-9)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 12 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 12 (old spec ENGR216).pdf"},"text":"A cantilever ABC of length $3a$ and flexural rigidity $EI$ is fixed at A. Portion AB ($0 \\le x \\le 2a$) carries a\nuniform distributed load $w$ acting UPWARDS; portion BC ($2a \\le x \\le 3a$) is unloaded, and a downward\nconcentrated load $P = \\tfrac{2}{3}wa$ acts at the free end C. Take $x$ from A and $y$ upwards.\n","figures":[{"src":"/figures/engr5004-ex-stat-defl-04.png","caption":"Cantilever fixed at A, length 3a: upward uniform load w on AB (length 2a), downward load P = (2/3)wa at the free end C (BC = a)."}],"formulaSheet":"- Elastic curve: $EI\\dfrac{d^2y}{dx^2} = M(x)$, slope $\\theta = \\dfrac{dy}{dx}$, two constants per segment; boundary and continuity conditions (learn this; on the 2025 sheet but not the 2026 one)\n- Equilibrium $\\sum F_y = 0$, $\\sum M_A = 0$; fixed support gives a force and a couple (learn this)\n- Bending moment by the method of sections; $\\dfrac{dM}{dx} = V$ (on the formula sheet)\n","parts":[{"label":"a","text":"Calculate the deflection and rotation of the free end C.","grade":"$undefined"},{"label":"b","text":"Sketch $M(x)$ and $y(x)$.","grade":"$undefined"},{"label":"c","text":"Calculate the maximum deflection of the beam.","grade":"$undefined"}],"solution":"$134","confidence":"HIGH","solutionNote":"Agrees with the slides ($\\theta_C = -5wa^3/(3EI)$, $y_C = -8wa^4/(3EI)$, maximum at C). Symbolic answers only, so no answer boxes."},{"id":"engr5004-ex-stat-defl-05","module":"ENGR5004","week":6,"label":"Old spec Statics Session 12 Example 2","kind":"tutorial","level":3,"minutes":18,"grade":"2:1","tags":["beam-deflection","statically-indeterminate","propped-cantilever","elastic-curve","udl","maximum-deflection"],"source":{"name":"ENGR216 (old spec) Statics session 12 worked example 2 (slides 10-16)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 12 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 6 - Statics session 12 (old spec ENGR216).pdf"},"text":"A beam AB of length $L$ and flexural rigidity $EI$ is fixed at A and supported by a roller at B. It carries a\nuniform downward load $w$ over its whole length. Take $x$ from A and $y$ upwards.\n","figures":[{"src":"/figures/engr5004-ex-stat-defl-05.png","caption":"Propped cantilever: beam AB of length L, fixed at A (left), roller at B (right), uniform downward load w."}],"formulaSheet":"- Equilibrium $\\sum F_x = 0$, $\\sum F_y = 0$, $\\sum M = 0$; a fixed support gives two forces and a couple, a roller one force; degree of indeterminacy = unknown reactions minus independent equilibrium equations (learn this)\n- Elastic curve: $EI\\dfrac{d^2y}{dx^2} = M(x)$, slope $\\theta = \\dfrac{dy}{dx}$ (learn this; on the 2025 sheet but not the 2026 one)\n- Fixed end $y = 0$, $\\theta = 0$; roller $y = 0$ (learn this)\n","parts":[{"label":"a","text":"Determine the reactions of the supports.","grade":"$undefined"},{"label":"b","text":"Sketch the elastic curve.","grade":"$undefined"},{"label":"c","text":"Determine the position of the maximum deflection.","answer":[{"label":"x_delta / L","value":0.5785,"units":"-"}],"grade":"$undefined"}],"solution":"$135","confidence":"HIGH","solutionNote":"Agrees with the slides (5wL/8, 3wL/8, wL^2/8). The slide gives $x_\\delta = 0.5784L$; the exact root\n$(15 - \\sqrt{33})/16 = 0.57846$ rounds to 0.5785. Same set-up as ENGR216 2021 A2 (in the bank), but here the\nelastic curve is derived rather than given.\n"},{"id":"engr5004-ex-stat-stress-01","module":"ENGR5004","week":8,"label":"Old spec Statics Session 15 Example","kind":"tutorial","level":1,"minutes":5,"grade":"third","tags":["stress-transformation","plane-stress"],"source":{"name":"ENGR216 (old spec) Statics session 15 worked example (slide 11)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Statics session 15 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Statics session 15 (old spec ENGR216).pdf"},"text":"A square element is in plane stress with $\\sigma_x = \\sigma_y = 100$ MPa (tension) and $\\tau_{xy} = 50$ MPa\n(positive, i.e. upwards on the right face). Calculate the normal and shear stresses on the plane whose normal\nmakes an angle $\\theta = 30^\\circ$ (counterclockwise) with the $x$ axis.\n","figures":[{"src":"/figures/engr5004-ex-stat-stress-01.png","caption":"Square element with sigma_x on the vertical faces, sigma_y on the horizontal faces and positive shear tau_xy (upwards on the right face, rightwards on the top face); an oblique cut whose normal is at angle theta to x."}],"formulaSheet":"- Stress transformation: $\\sigma_{x'} = \\dfrac{\\sigma_x+\\sigma_y}{2} + \\dfrac{\\sigma_x-\\sigma_y}{2}\\cos2\\theta + \\tau_{xy}\\sin2\\theta$, $\\tau_{x'y'} = -\\dfrac{\\sigma_x-\\sigma_y}{2}\\sin2\\theta + \\tau_{xy}\\cos2\\theta$ (learn this)\n","answer":[{"label":"sigma_x'","value":143.3,"units":"MPa"},{"label":"tau_x'y'","value":25,"units":"MPa"}],"solution":"**Set-up.** Sign convention of the slides: tension positive, $\\tau_{xy} > 0$ upwards on the positive-$x$ face,\n$\\theta$ counterclockwise from $x$ to the normal $x'$.\n\n**Normal stress.** Since $\\sigma_x = \\sigma_y$, the cosine term vanishes:\n\n$$\\sigma_{x'} = \\frac{100 + 100}{2} + 0 + 50\\sin(60^\\circ) = 100 + 25\\sqrt3 = 143.3\\ \\mathrm{MPa}$$\n\n**Shear stress.**\n\n$$\\tau_{x'y'} = -0 \\cdot \\sin(60^\\circ) + 50\\cos(60^\\circ) = 25.0\\ \\mathrm{MPa}$$\n\nCheck (Mohr's circle): centre $(100, 0)$, radius $R = \\sqrt{0^2 + 50^2} = 50$ MPa; the point for the $x'$ face is\n$(100 + 50\\sin60^\\circ,\\ 50\\cos60^\\circ) = (143.3, 25)$, which lies on the circle:\n$(143.3 - 100)^2 + 25^2 = 50^2$. Also $\\sigma_{x'} + \\sigma_{y'} = 143.3 + 56.7 = 200 = \\sigma_x + \\sigma_y$\n(invariant).\n\nMarks (likely): formula 1, each stress 2.\n","confidence":"HIGH","solutionNote":"Agrees with the slide (143.3 MPa and 25 MPa)."},{"id":"engr5004-ex-stat-stress-02","module":"ENGR5004","week":8,"label":"Old spec Statics Session 16 Example","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["stress-transformation","plane-stress","principal-stresses","maximum-shear-stress","principal-axes"],"source":{"name":"ENGR216 (old spec) Statics session 16 worked example (slides 9-12)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Statics session 16 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Statics session 16 (old spec ENGR216).pdf"},"text":"An element is in plane stress with $\\sigma_x = 100$ MPa (tension), $\\sigma_y = 20$ MPa (compression) and a\nshear stress of 80 MPa acting upwards on the right face and rightwards on the top face ($\\tau_{xy} = +80$ MPa).\nDetermine:\n","figures":[{"src":"/figures/engr5004-ex-stat-stress-02.png","caption":"Element ABCD: 100 MPa tension on the vertical faces, 20 MPa compression on the horizontal faces, 80 MPa shear (upwards on the right face BC, rightwards on the top face AB)."}],"formulaSheet":"- Principal directions: $\\tan2\\theta_p = \\dfrac{2\\tau_{xy}}{\\sigma_x-\\sigma_y}$; principal stresses $\\sigma_{max,min} = \\dfrac{\\sigma_x+\\sigma_y}{2} \\pm \\sqrt{\\left(\\dfrac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$ (learn this)\n- Maximum shear: $\\tan2\\theta_s = -\\dfrac{\\sigma_x-\\sigma_y}{2\\tau_{xy}}$, $\\tau_{max} = \\sqrt{\\left(\\dfrac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$, normal stress $\\sigma' = \\dfrac{\\sigma_x+\\sigma_y}{2}$ (learn this)\n- Stress transformation $\\sigma_{x'}$, $\\tau_{x'y'}$ (to decide which angle gives which stress) (learn this)\n","parts":[{"label":"a","text":"the principal planes;","answer":[{"label":"theta_p1","value":26.6,"units":"deg"},{"label":"theta_p2","value":116.6,"units":"deg"}],"grade":"$undefined"},{"label":"b","text":"the principal stresses, and which principal plane carries which;","answer":[{"label":"sigma_max","value":140,"units":"MPa"},{"label":"sigma_min","value":-60,"units":"MPa"}],"grade":"$undefined"},{"label":"c","text":"the planes of maximum in-plane shear, the maximum shear stress, and the normal stress on those planes.","answer":[{"label":"theta_s1","value":-18.4,"units":"deg"},{"label":"tau_max","value":100,"units":"MPa"},{"label":"sigma'","value":40,"units":"MPa"}],"grade":"$undefined"}],"solution":"$136","confidence":"HIGH","solutionNote":"Agrees with the slides except one slip: slide 10 writes $\\theta_{p2} = \\theta_{p1} + 90^\\circ = 111.6^\\circ$; it should\nbe $116.6^\\circ$. The slide leaves \"which angle gives $\\sigma_{max}$\" and \"which gives positive $\\tau_{max}$\" as questions;\nthey are answered above.\n"},{"id":"engr5004-ex-stat-stress-03","module":"ENGR5004","week":8,"label":"Old spec Statics Session 16 Problem","kind":"tutorial","level":3,"minutes":15,"grade":"2:1","tags":["stress-transformation","combined-loading","axle","torsion","principal-stresses","max-shear-stress"],"source":{"name":"ENGR216 (old spec) Statics session 16 worked example 2 (Problem, slides 13-16)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Statics session 16 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 8 - Statics session 16 (old spec ENGR216).pdf"},"text":"The solid axle of an automobile, radius $r = 16$ mm, is acted upon by the forces and torque shown: an upward\n3 kN reaction at the wheel, a downward 3 kN load from the car body applied 0.15 m inboard of the wheel, and a\ntorque of 350 N m. Point H is on the top of the axle, 0.2 m further inboard than the body load, in the\ncentral part of the (symmetric) axle where there is no shear force. Use axes with $x$ along the axle from the\nwheel towards H and $y$ normal to $x$ in the surface plane at H; with these axes the torque produces a\nnegative shear stress $\\tau_{xy}$ at H.\n\nDetermine:\n","figures":[{"src":"/figures/engr5004-ex-stat-stress-03.png","caption":"Axle with the wheel at the left: 3 kN up at the wheel, 3 kN down 0.15 m inboard, torque 350 N m; point H on the top of the axle a further 0.2 m inboard."}],"formulaSheet":"- Bending stress $\\sigma_{max} = \\dfrac{|M|c}{I}$, solid circle $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n- Torsion $\\tau_{max} = \\dfrac{Tr}{J}$, $J = \\dfrac{\\pi r^4}{2}$ (learn this)\n- Principal stresses $\\sigma_{max,min} = \\dfrac{\\sigma_x+\\sigma_y}{2} \\pm \\sqrt{\\left(\\dfrac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$, $\\tan2\\theta_p = \\dfrac{2\\tau_{xy}}{\\sigma_x-\\sigma_y}$, $\\tau_{max} = \\sqrt{\\left(\\dfrac{\\sigma_x-\\sigma_y}{2}\\right)^2 + \\tau_{xy}^2}$ (learn this)\n","parts":[{"label":"a","text":"the principal axes and principal stresses at point H on top of the axle;","answer":[{"label":"theta_p (sigma_min face)","value":18.9,"units":"deg"},{"label":"sigma_max","value":18.7,"units":"MPa"},{"label":"sigma_min","value":-158.5,"units":"MPa"}],"grade":"$undefined"},{"label":"b","text":"the maximum shear stress at the same point.","answer":[{"label":"tau_max","value":88.6,"units":"MPa"}],"grade":"$undefined"}],"solution":"$137","confidence":"HIGH","solutionNote":"Agrees with the slides ($\\sigma_x$ = 140 MPa compressive, $\\tau_{xy} = -54.4$ MPa, principal stresses 18.65 and\n-158.6 MPa, $\\tau_{max}$ = 88.65 MPa; small differences are rounding of 139.9 to 140). The slide lists\n$\\theta_{p1} = 18.9^\\circ$ next to $\\sigma_{max}$; substitution shows $18.9^\\circ$ is the $\\sigma_{min}$ direction and\n$108.9^\\circ$ the $\\sigma_{max}$ direction.\n"},{"id":"engr5004-ex-stat-buckling-01","module":"ENGR5004","week":9,"label":"Old spec Statics Session 18 Problem","kind":"tutorial","level":2,"minutes":12,"grade":"2:2","tags":["buckling","euler-load","column-design","safety-factor","circular-section"],"source":{"name":"ENGR216 (old spec) Statics session 18 worked example (Problem, slides 10-12)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Statics session 18 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 9 - Statics session 18 (old spec ENGR216).pdf"},"text":"A 2 m long pin-ended steel column of solid circular cross section has $E = 200$ GPa and allowable normal stress\n$\\sigma_{all} = 100$ MPa. Using a factor of safety $k = 2.5$ on Euler's critical load for buckling, determine the\ncross-section radius needed if the column is to safely support:\n","formulaSheet":"- Euler load for a pinned-pinned column $P_{cr} = \\dfrac{\\pi^2 E I_{min}}{L^2}$ ($L_e = L$) (on the formula sheet)\n- Solid circle $A = \\pi r^2$, $I = \\dfrac{\\pi r^4}{4}$ (on the formula sheet)\n- Design: $P_{cr} \\ge kP$ and $\\sigma = P/A \\le \\sigma_{all}$; the larger radius governs (learn this)\n","parts":[{"label":"a","text":"a 380 kN load;","answer":[{"label":"r","value":39.6,"units":"mm"}],"grade":"$undefined"},{"label":"b","text":"an 800 kN load.","answer":[{"label":"r","value":50.5,"units":"mm"}],"grade":"$undefined"}],"solution":"$138","confidence":"HIGH","solutionNote":"The slides round the radii DOWN: (a) r = 39 mm (exact 39.57 mm; 39 mm would give a safety factor of only 2.36,\nbelow 2.5) and (b) r = 50 mm from the stress condition (exact 50.46 mm; 50 mm gives 101.9 MPa, just over\n100 MPa). The answers here are the exact minimum radii; in design round up. The method and conclusions\n(buckling governs in (a), stress governs in (b)) agree with the slides.\n"},{"id":"engr5004-ex-stat-buckling-02","module":"ENGR5004","week":10,"label":"Old spec Statics Session 19 Example 1","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["buckling","effective-length","slenderness-ratio","radius-of-gyration","safety-factor","allowable-load"],"source":{"name":"ENGR216 (old spec) Statics session 19 worked example 1 (slides 8-10)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 10 - Statics session 19 (old spec ENGR216).pdf","solutionsPath":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 10 - Statics session 19 (old spec ENGR216).pdf"},"text":"A 2 m long steel column of solid circular cross section with radius $r = 40$ mm has $E = 200$ GPa and allowable\nnormal stress $\\sigma_{all} = 100$ MPa. Using a factor of safety $k = 2.5$ on Euler's critical load for buckling,\ndetermine the maximum compressive axial load the column can carry if:\n","formulaSheet":"- Euler critical stress $\\sigma_{cr} = \\dfrac{\\pi^2 E}{(L_e/\\rho)^2}$, $\\rho = \\sqrt{I/A}$ (on the formula sheet)\n- Effective lengths: pinned-pinned $L_e = L$, fixed-fixed $L_e = 0.5L$ (on the formula sheet)\n- Solid circle $\\rho = r/2$ (on the formula sheet: $I = \\pi r^4/4$, $A = \\pi r^2$)\n- Allowable load $P_{max} = \\min(\\sigma_{cr}/k,\\ \\sigma_{all})\\,A$ (learn this)\n","parts":[{"label":"a","text":"the column is pinned at both ends;","answer":[{"label":"P_max","value":396.9,"units":"kN"}],"grade":"$undefined"},{"label":"b","text":"the column is fixed at both ends.","answer":[{"label":"P_max","value":502.7,"units":"kN"}],"grade":"$undefined"}],"solution":"$139","confidence":"HIGH","solutionNote":"Agrees with the slides (396.8 kN and 502.6 kN; differences are rounding)."},{"id":"engr5004-ex-stat-buckling-03","module":"ENGR5004","week":10,"label":"Old spec Statics Session 19 Example 2","kind":"tutorial","level":2,"minutes":10,"grade":"2:2","tags":["buckling","buckling-plane","effective-length","slenderness-ratio","rectangular-section","design"],"source":{"name":"ENGR216 (old spec) Statics session 19 example 2 (slide 15, statement only)","paper":"ENGR216","path":"ENGR5004 - Engineering Mechanics/$ Old Spec/Week 10 - Statics session 19 (old spec ENGR216).pdf"},"text":"A steel column of length $L$ and rectangular cross section (sides $a$ and $b$) is fixed at its base B and\nsupports a centric compressive load $P$ at its top A. Two smooth, rounded fixed plates at A stop end A from\nmoving in the $y$ direction (so in the vertical $xy$ plane of symmetry the top is effectively pinned) but allow\nit to move freely in the $z$ direction (so in the $xz$ plane the top is free). Side $a$ is measured along $y$\nand side $b$ along $z$; $x$ is the column axis.\n","figures":[{"src":"/figures/engr5004-ex-stat-buckling-03.png","caption":"Column of length L fixed at base B, centric load P at the top A between two smooth rounded plates; rectangular section with sides a and b; x along the column axis, y and z horizontal."}],"formulaSheet":"- Effective lengths: fixed-free $L_e = 2L$, fixed-pinned $L_e = 0.7L$ (on the formula sheet)\n- Radius of gyration $\\rho = \\sqrt{I/A}$, slenderness ratio $L_e/\\rho$, $\\sigma_{cr} = \\dfrac{\\pi^2E}{(L_e/\\rho)^2}$ (on the formula sheet)\n- Rectangle $I = \\dfrac{bh^3}{12}$ about the axis parallel to $b$ (on the formula sheet)\n","parts":[{"label":"a","text":"Determine the effective slenderness ratio for buckling in the $xy$ plane.","grade":"$undefined"},{"label":"b","text":"Determine the effective slenderness ratio for buckling in the $xz$ plane.","grade":"$undefined"},{"label":"c","text":"Determine the ratio $a/b$ of the two sides of the cross section that gives the most efficient design against\nbuckling, and justify this choice.\n","answer":[{"label":"a/b","value":0.35,"units":"-"}],"grade":"$undefined"}],"solution":"$13a","confidence":"HIGH","solutionNote":"The deck gives only the statement (no worked solution); solved here from the effective lengths on the formula\nsheet. The slide text does not say which plane the plates restrain; the statement here fixes it as the $xy$\nplane with side $a$ along $y$. If the roles of the planes were swapped the answer would be $b/a = 0.35$.\n"}],"modules":[{"code":"ENGR5001","title":"Control and Robotics","short":"Control","folder":"ENGR5001 - Control and Robotics","weekOneStart":"2026-10-05","weeksTaught":10,"assessment":[{"name":"Exam","weight":100}],"weeks":{"1":"Part 1 - lectures 1-2"}},{"code":"ENGR5002","title":"Fluid Mechanics and Mass Transfer","short":"Fluids","lecturer":"Dr L. Capozzi","folder":"ENGR5002 - Fluid Mechanics and Mass Transfer","weekOneStart":"2026-10-05","weeksTaught":10,"assessment":[{"name":"In-person exam (2 h)","weight":70},{"name":"CFD group report","weight":30}],"predecessor":"ENGR271 (2025-26 paper is the closest match, same title and formula-sheet style)","weeks":{"1":"Fluids at rest - properties, pressure, hydrostatic forces","2":"Fluids at rest - buoyancy and stability","3":"Fluids in motion - continuity, Bernoulli","4":"Fluids in motion - pipe flow, losses, momentum","5":"Fluid machinery","6":"Fluid machinery","7":"Rheology","8":"Rheology","9":"Mass transfer","10":"Mass transfer"}},{"code":"ENGR5003","title":"Thermodynamics and Heat Transfer","short":"Thermo","folder":"ENGR5003 - Thermodynamics and Heat Transfer","weekOneStart":"2026-10-05","weeksTaught":10,"assessment":[{"name":"Summer exam","weight":70},{"name":"Lab report (jet engine data + heat transfer design)","weight":30}],"predecessor":"ENGR272","weeks":{"1":"Thermodynamics","2":"Thermodynamics","3":"Thermodynamics","4":"Thermodynamics","5":"Thermodynamics","6":"Thermodynamics (practical session)","7":"Heat transfer","8":"Heat transfer (practical session)","9":"Heat transfer","10":"Heat transfer"}},{"code":"ENGR5004","title":"Engineering Mechanics","short":"Mechanics","folder":"ENGR5004 - Engineering Mechanics","weekOneStart":"2026-10-05","weeksTaught":10,"assessment":[{"name":"Exam","weight":100}],"predecessor":"ENGR273 (statics) and ENGR216 (dynamics)","weeks":{"1":"Statics sessions 1-2, Dynamics sessions 1-2","2":"Statics sessions 3-4, Dynamics sessions 3-4"}}],"ctx":{"glossary":{"continuum":{"id":"continuum","term":"Continuum hypothesis","short":"Treat the fluid as a continuous medium so that density, velocity, pressure and temperature are defined at every point, even though it is really made of molecules. Valid when the Knudsen number is below about 0.001.","lecture":"engr5002-l1","slide":28},"mean-free-path":{"id":"mean-free-path","term":"Mean free path","symbol":"\\lambda","units":"m","short":"Average distance a molecule travels between collisions. About 68 nm for air at 1 atm and 20 deg C; inversely proportional to pressure at fixed temperature.","lecture":"engr5002-l1","slide":29},"characteristic-length":{"id":"characteristic-length","term":"Characteristic length","symbol":"L","units":"m","short":"The size of the system that the flow passes through, normally the smallest relevant dimension (pipe diameter, channel height, body length).","lecture":"engr5002-l1","slide":29},"knudsen":{"id":"knudsen","term":"Knudsen number","symbol":"Kn","units":"dimensionless","short":"Kn = mean free path / characteristic length. Kn below 0.001 means the continuum hypothesis holds; 0.001 to 0.1 slip flow; 0.1 to 10 transition; above 10 free molecular flow.","lecture":"engr5002-l1","slide":29,"formula":"knudsen"},"fluid":{"id":"fluid","term":"Fluid","short":"A substance that deforms continuously under the action of a shear stress, however small. Consequence - a fluid at rest carries no shear stress, only pressure.","lecture":"engr5002-l1","slide":31},"shear-stress":{"id":"shear-stress","term":"Shear stress","symbol":"\\tau","units":"Pa (N m^-2)","short":"Tangential force per unit area. In a moving fluid it is set by the rate of deformation (viscosity); in a fluid at rest it is zero.","lecture":"engr5002-l1","slide":31},"density":{"id":"density","term":"Density","symbol":"\\rho","units":"kg m^-3","short":"Mass per unit volume. Water is taken as 1000 kg/m^3 in this module's exams.","lecture":"engr5002-l1","slide":32,"formula":"density"},"specific-weight":{"id":"specific-weight","term":"Specific weight","symbol":"\\gamma","units":"N m^-3","short":"Weight per unit volume, gamma = rho g. Water is 9.81 kN/m^3. It is what sets hydrostatic pressure, p = gamma h.","lecture":"engr5002-l1","slide":32,"formula":"specific-weight"},"relative-density":{"id":"relative-density","term":"Relative density","symbol":"RD","units":"dimensionless","short":"Density of the substance divided by the density of water (1000 kg/m^3). Old papers call it specific gravity, SG or S. First step in nearly every hydrostatics question is rho = RD x 1000.","lecture":"engr5002-l1","slide":32,"formula":"relative-density"},"molar-mass":{"id":"molar-mass","term":"Molar mass","symbol":"M","units":"kg mol^-1","short":"Mass of one mole. Given in g/mol on data sheets; divide by 1000 before using in rho = pM/(RT).","lecture":"engr5002-l1"},"ideal-gas-density":{"id":"ideal-gas-density","term":"Ideal gas density","symbol":"\\rho = pM/(RT)","units":"kg m^-3","short":"Density of an ideal gas from its absolute pressure, molar mass and absolute temperature. R = 8.314 J/(mol K).","lecture":"engr5002-l1","formula":"ideal-gas-density"},"hydrostatics":{"id":"hydrostatics","term":"Hydrostatics","short":"Fluids at rest. Only pressure acts on any surface, and pressure increases with depth as p = p0 + rho g h.","lecture":"engr5002-l2"},"no-slip":{"id":"no-slip","term":"No-slip condition","short":"A fluid in contact with a solid wall moves at the wall's velocity. Holds in the continuum regime; fails once Kn exceeds about 0.001 (slip flow).","lecture":"engr5002-l1","slide":30}},"formulas":{"knudsen":{"type":"formula","id":"knudsen","name":"Knudsen number","latex":"Kn = \\frac{\\lambda}{L}","meaning":"The ratio of how far a molecule travels between collisions to the size of the flow system. Small Kn means many collisions inside the system, so continuum averages work.\n","symbols":[{"sym":"Kn","name":"Knudsen number","meaning":"Rarefaction parameter","units":"dimensionless"},{"sym":"\\lambda","name":"Mean free path","meaning":"Average distance between molecular collisions","units":"m"},{"sym":"L","name":"Characteristic length","meaning":"Size of the system (pipe diameter, channel height, body length)","units":"m"}],"unitCheck":"$$[Kn] = \\dfrac{\\mathrm{m}}{\\mathrm{m}} = 1$, dimensionless as required.","when":"Whenever you must justify treating a flow as a continuum (an M2 \"justify your assumption\" step), or decide whether standard results (no-slip, Hagen-Poiseuille, Bernoulli) apply.\n","assumptions":"L must be the relevant size, normally the smallest dimension the flow passes through. The boundaries 0.001, 0.1 and 10 are order-of-magnitude guides, not sharp lines.\n","mistakes":["Inverting the ratio (L / lambda).","Forgetting unit conversion (nm, um, mm).","Using a large overall length when the flow passes through a tiny gap."],"sheet":{"status":"NOT_GIVEN_MEMORISE","note":"Not on the ENGR271 2026 formula sheet or the Week 1 formula sheet. Trivial to remember, but you must also know the regime boundaries, especially Kn < 0.001."},"tags":["knudsen","continuum"],"drills":["engr5002-l1-k1","engr5002-l1-k2","engr5002-l1-k3"],"module":"ENGR5002","lecture":"engr5002-l1"},"density":{"type":"formula","id":"density","name":"Mass density","latex":"\\rho = \\frac{m}{V}","meaning":"Mass per unit volume. At a point, rho is the limit of delta m / delta V as delta V shrinks (see the continuum section).\n","symbols":[{"sym":"\\rho","name":"Density","meaning":"Mass per unit volume","units":"kg m^-3"},{"sym":"m","name":"Mass","meaning":"Mass of the sample","units":"kg"},{"sym":"V","name":"Volume","meaning":"Volume occupied by that mass","units":"m^3"}],"unitCheck":"$$\\dfrac{\\mathrm{kg}}{\\mathrm{m^3}} = \\mathrm{kg\\,m^{-3}}$. Correct.","when":"Converting between mass and volume of fluid; every pressure, flow and force calculation starts with rho.\n","assumptions":"rho = m/V is an average; it equals the point density only if the fluid is uniform (homogeneous). Liquids are treated as incompressible (constant rho). Gas density changes with pressure and temperature; for an ideal gas rho = pM/(RT), which you meet in Week 1 Exercise 6.\n","mistakes":["Litres to m^3 (1 L = 10^-3 m^3).","g/cm^3 to kg/m^3 (multiply by 1000).","Molar mass in g/mol must become kg/mol in pM/(RT)."],"external":"Typical values: water 1000 kg/m^3 (exam value); seawater about 1025; oils 800-920; mercury 13 600; air at 20 deg C and 1 atm about 1.2.","sheet":{"status":"NOT_GIVEN_MEMORISE","note":"Not written on the ENGR271 2026 sheet as a definition (it appears only inside m_dot = rho A v). Basic definition."},"tags":["density"],"drills":["engr5002-l1-d1","engr5002-l1-d2","engr5002-l1-d3"],"module":"ENGR5002","lecture":"engr5002-l1"},"specific-weight":{"type":"formula","id":"specific-weight","name":"Specific weight","latex":"\\gamma = \\frac{m g}{V} = \\rho g","meaning":"Weight per unit volume, i.e. how heavy each cubic metre of fluid is.","symbols":[{"sym":"\\gamma","name":"Specific weight","meaning":"Weight per unit volume","units":"N m^-3"},{"sym":"\\rho","name":"Density","meaning":"Mass per unit volume","units":"kg m^-3"},{"sym":"g","name":"Gravitational acceleration","meaning":"9.81 m/s^2 unless told otherwise","units":"m s^-2"}],"unitCheck":"$$[\\rho g] = \\mathrm{kg\\,m^{-3}} \\times \\mathrm{m\\,s^{-2}} = (\\mathrm{kg\\,m\\,s^{-2}})\\,\\mathrm{m^{-3}} = \\mathrm{N\\,m^{-3}}$. Matches the left-hand side.","when":"Whenever weight matters - hydrostatic pressure (p = p0 + gamma z in Lecture 2 is the same as rho g z), buoyancy, weight of fluid in a tank. Some questions give gamma instead of rho (Week 1 Exercise 9 gives rho g = 12.36 kN/m^3), so you must convert both ways.\n","assumptions":"Depends on g, so unlike rho it is not purely a material property (it would differ on the Moon). Use g = 9.81 m/s^2 unless told otherwise.\n","mistakes":["Quoting gamma in kg/m^3.","Mixing kN and N (water: gamma = 9.81 kN/m^3 = 9810 N/m^3)."],"sheet":{"status":"NOT_GIVEN_MEMORISE","note":"Not on the ENGR271 2026 sheet or the current Week 1 sheet (the old ENGR217 2024 sheet defined rho g = gamma in its pump section)."},"tags":["specific-weight"],"drills":["engr5002-l1-s1","engr5002-l1-s2","engr5002-l1-s3"],"module":"ENGR5002","lecture":"engr5002-l1"},"relative-density":{"type":"formula","id":"relative-density","name":"Relative density","latex":"RD = \\frac{\\rho_{fluid}}{\\rho_{water}}","meaning":"How many times denser a substance is than water (water conventionally taken at 4 deg C, where rho_w = 1000 kg/m^3).\n","symbols":[{"sym":"RD","name":"Relative density (old papers: specific gravity, SG or S)","meaning":"Density ratio","units":"dimensionless"},{"sym":"\\rho_{fluid}","name":"Density of the substance","meaning":"","units":"kg m^-3"},{"sym":"\\rho_{water}","name":"Density of water","meaning":"1000 kg/m^3 in this module's exams","units":"kg m^-3"}],"unitCheck":"$$\\dfrac{\\mathrm{kg\\,m^{-3}}}{\\mathrm{kg\\,m^{-3}}}$ = dimensionless. Correct.","when":"Nearly every hydrostatics question gives fluids as an RD (Week 1 Exercises 1, 2, 5, 8; ENGR271 2026 Q1). Step one is always rho = RD x 1000 kg/m^3.\n","assumptions":"Works for liquids and solids; for gases RD relative to water is tiny (about 0.001) and rarely used. RD < 1 means lighter than water (floats on it, if immiscible); RD > 1 means heavier.\n","mistakes":["Giving RD units.","Dividing the wrong way.","Forgetting to multiply by 1000."],"sheet":{"status":"GIVEN","note":"On the ENGR271 2026 formula sheet (\"Relative density RD = rho_A / rho_w\"). Not on the current Week 1 sheet, so treat it as given but know it anyway."},"tags":["relative-density"],"drills":["engr5002-l1-r1","engr5002-l1-r2","engr5002-l1-r3"],"module":"ENGR5002","lecture":"engr5002-l1"},"ideal-gas-density":{"type":"formula","id":"ideal-gas-density","name":"Ideal gas density","latex":"\\rho = \\frac{p M}{R T}","meaning":"Density of an ideal gas from its absolute pressure, molar mass and absolute temperature.","symbols":[{"sym":"\\rho","name":"Density","meaning":"","units":"kg m^-3"},{"sym":"p","name":"Absolute pressure","meaning":"Not gauge pressure","units":"Pa"},{"sym":"M","name":"Molar mass","meaning":"Convert g/mol to kg/mol first. For a mixture use the mole-fraction weighted average, M = sum(x_i M_i)","units":"kg mol^-1"},{"sym":"R","name":"Universal gas constant","meaning":"8.314","units":"J mol^-1 K^-1"},{"sym":"T","name":"Absolute temperature","meaning":"deg C + 273.15","units":"K"}],"unitCheck":"$$\\dfrac{\\mathrm{Pa}\\cdot\\mathrm{kg\\,mol^{-1}}}{\\mathrm{J\\,mol^{-1}\\,K^{-1}}\\cdot\\mathrm{K}} = \\dfrac{\\mathrm{N\\,m^{-2}}\\cdot\\mathrm{kg}}{\\mathrm{N\\,m}} = \\mathrm{kg\\,m^{-3}}$. Correct.","when":"Any gas density question, and the atmosphere problem in Week 1 Exercise 6.","assumptions":"Ideal-gas behaviour; uniform composition and temperature.","mistakes":["Using gauge pressure or deg C.","Leaving M in g/mol (answer 1000 times too big)."],"sheet":{"status":"UNKNOWN","note":"Not on the Week 1 sheet. Given in the exercise sheet. Treat as memorise until the papers show otherwise."},"tags":["ideal-gas","density"],"drills":[],"module":"ENGR5002","lecture":"engr5002-l1"}}}}]}]],["$L13b","$L13c","$L13d"],"$L13e"]}],{},null,false,null]},null,false,"$13f"]},null,false,null],"$L140",false]],"m":"$undefined","G":["$141",["$L142"]],"S":true,"h":null,"r":"$undefined","s":"$undefined","a":"$undefined","l":"$undefined","p":"$undefined","d":"$undefined","b":"XwsFlVSIM7fzpkmRIGTCG"}