Lecture 1: Introduction

A short conceptual lecture - module admin (slides 1-24), then the physical foundations (slides 25-33) - the continuum hypothesis, the Knudsen number, the definition of a fluid, and density, specific weight and relative density. These ideas are used in every later lecture.

Exam relevance: what the papers show

What the papers show. The most useful paper is ENGR271 (2025-26), the predecessor of this module with the same title and the same lecturer's style (slide 27 even refers to ENGR271). Every old-spec fluids paper (ENGR217 / ENGR266, 2017-2025) was searched as well.

  • No past question tests the continuum hypothesis, Knudsen number or the fluid/solid definition directly. These are flagged [CHECK]: they could appear as short "explain" parts (the 2026 paper has 5-mark theory parts like Q1(b)), but there is no evidence yet.
  • Relative densityRDRDUnits: dimensionlessDensity 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.Open in lectureQuestions is examined every year, embedded in hydrostatics questions (ENGR271 2026 Q1; ENGR217 2024 Q A1(b); ENGR217 2017, 2018 and 2021 ship-stability questions). The full hydrostatics questions belong to Lectures 2 and 3; here you only practise the L1 step.
  • Old-spec papers say "specific gravity" (SG or S); the current module says relative density (RD). Same quantity. Use RD in your answers.

Exam format (slides 14-16 and the ENGR271 2026 paper). 2-hour in-person exam, 70 percent of the module; 30 percent is the CFD group report (due 12:00, Wednesday 2 December 2026). The 2026 paper had three compulsory questions (hydrostatics 30 marks, pipe flow and momentum 40 marks, mass transfer 30 marks) with a formula sheet at the end.

1. Module overview

Nothing on slides 1-24 is examinable as fluid mechanics, but four facts matter for planning.

ItemDetail
Weekly plan
W1-2 fluids at rest; W3-4 fluids in motion; W5-6 fluid machinery; W7-8 rheology; W9-10 mass transfer (slide 9)
Labs
CFD Lab 1 (laminar and turbulent pipe flow), CFD Lab 2 (flat-plate boundary layer), two supporting sessions (slide 19). Attendance is mandatory and monitored (slide 22)
Coursework
Group report, max 5 pages main body, plus references, appendices, an AI-use reflective statement (max 1 page) and a contribution statement (slides 20-22)
Exam
2 hours, in person; assesses AHEP outcomes M1 (apply principles), M2 (problem analysis, justify assumptions), M3 (analytical techniques such as Moody charts, correlations) (slide 16)

2. A multi-scale view of fluids

The three scales at which a fluid can be described.

Intuition. Water pouring from a pipe looks smooth (macroscopic). Zoom into a tiny cube of it and you can still talk about "the pressure on this face" (microscopic). Zoom further and it is a crowd of molecules moving randomly and colliding (molecular).

Physical meaning of each scale.

  • Molecular: about Avogadro's number of molecules (6.02×10236.02 \times 10^{23} per mole) in constant random thermal motion. Collisions carry momentum, energy and mass from place to place. This is the origin of viscosity, heat conduction and diffusion.
  • Microscopic (from molecules to continua): the properties you use in engineering are averages over many molecules - temperature is average molecular kinetic energy; Densityρ\rhoUnits: kg m^-3Mass per unit volume. Water is taken as 1000 kg/m^3 in this module's exams.Open in lectureQuestions is mass per unit volume; viscosity is momentum transport between layers by molecules; diffusivity is the rate of molecular mixing (the link to the mass-transfer half of the module, Fick's law).
  • Macroscopic: pipe flow, drag, lift, pumps and turbines; the systems you design.

The key idea for the whole module: we never track molecules; we use averaged properties defined at every point. The continuum section explains when that is allowed.

3. From phenomena to equations

Slide 27 previews the three conservation laws the module is built on. You are not expected to use these yet; they are introduced properly in later weeks. What you should take from the slide is the structure: each physical phenomenon has a balance equation.

PhenomenonConserved quantityEngineering (integral) form on slide 27Taught in
Fluids accelerate and exchange momentum with their surroundings
Momentum
∑F=m˙ (vout−vin)\sum F = \dot m\,(v_{out} - v_{in})
Fluids in motion (W3-4)
Energy is transported and transformed
Energy
pρg+v22g+z=constant\dfrac{p}{\rho g} + \dfrac{v^2}{2g} + z = \text{constant} (Bernoulli)
Fluids in motion (W3-4)
Species move and mix
Mass of species i
ddt(m ci)=m˙inci,in−m˙outci,out\dfrac{d}{dt}(m\,c_i) = \dot m_{in} c_{i,in} - \dot m_{out} c_{i,out}
Mass transfer (W9-10)

The second row of equations on the slide (with D/DtD/Dt and ∇\nabla) are the same laws written at a point, which is only possible because of the Continuum hypothesisTreat 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.Open in lecture. Those differential forms are what CFD solves (the coursework).

4. The continuum concept

Defining density at point C by shrinking an elementary volume, and the density measured as the volume shrinks.

Intuition. Put a small box around point C in a glass of water and measure the mass inside divided by the box volume. Shrink the box. For a while the answer stays the same, but once the box is so small that it holds only a handful of molecules, the answer jumps around wildly depending on whether a molecule happens to be inside at that instant.

Reading the graph (right of the slide).

  • Large δV\delta V (right side): the density is a steady value. This is the continuum region.
  • Very small δV\delta V (left side): density fluctuates because of individual molecules. This is the non-continuum region.
  • The dashed line marks the smallest volume, call it δVmin\delta V_{min}, that still contains enough molecules for a stable average.

The definition (slides 28 and 32).

ρ=lim⁡δV→δVminδmδV\rho = \lim_{\delta V \to \delta V_{min}} \frac{\delta m}{\delta V}

The slides write the limit as δV→0\delta V \to 0. Read the "0" as "small compared with the flow, but still large compared with molecular spacing". Mathematically we then treat δVmin\delta V_{min} as zero, which is what lets us write density, velocity, pressure and temperature as continuous functions ρ(x,y,z,t)\rho(x,y,z,t), v(x,y,z,t)v(x,y,z,t) and so on, and take derivatives of them.

What the continuum hypothesis lets you do (slide 30):

  1. Each fluid element (a small control volume) contains a very large number of molecules.
  2. Microscopic fluctuations average out.
  3. Density, velocity, pressure and temperature become continuous functions of space and time.

5. When does the continuum hold? The Knudsen number

Intuition. The model fails when the molecules are so far apart, relative to the size of the thing they flow through, that a molecule may cross the whole device without hitting another. Then "the velocity at a point" stops meaning anything. This happens in very low-pressure (rarefied) gases, such as high-altitude flight, vacuum systems and some micro- and nano-devices. It almost never happens in liquids.

The Mean free pathλ\lambdaUnits: mAverage 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.Open in lecture λ\lambda is the average distance a molecule travels between collisions. Compare it with the size of the system LL:

Formula

Knudsen number

Not given - memorise
What each term means
KnKnKnudsen numberdimensionless
λ\lambdaMean free pathm
LLCharacteristic lengthm
Click to pin

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.

SymbolNameMeaningSI units
KnKnKnudsen numberRarefaction parameterdimensionless
λ\lambdaMean free pathAverage distance between molecular collisionsm
LLCharacteristic lengthSize of the system (pipe diameter, channel height, body length)m
Unit check
[Kn]=mm=1[Kn] = \dfrac{\mathrm{m}}{\mathrm{m}} = 1, dimensionless as required.
When to use it
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.
Assumptions and limits
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.
Common 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.
Formula sheetNot 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.

Drills

Plug in, then rearrange, then use it in context. Attempt before revealing.

K1DrillCurrent specThird3 min

Air at atmospheric conditions (λ=68 nm\lambda = 68\ \mathrm{nm}) flows in a pipe of diameter 50 mm. Calculate KnKn and state the regime.

K2DrillCurrent specThird4 min

For the same air (λ=68 nm\lambda = 68\ \mathrm{nm}), what is the smallest channel dimension for which the continuum hypothesis is valid?

K3DrillCurrent specThird4 min

At about 120 km altitude the mean free path of air is roughly 3 m. A satellite of characteristic length 2 m passes through. Find KnKn, state the regime, and say whether the methods of this module could predict its drag.

Flow regimes by Knudsen number. This module works entirely in the continuum regime.
Kn rangeRegimeWhat it means
Kn < 0.001
Continuum
Standard fluid mechanics (this module), no-slip at walls
0.001 - 0.1
Slip flow
Continuum equations still roughly usable but the gas slips at walls; the no-slip condition fails
0.1 - 10
Transition
Neither model works well; needs molecular methods
Kn > 10
Molecular (free molecular)
Molecules mostly hit walls, not each other

6. What is a fluid?

Molecular and continuum views of fluids and solids under shear.

Molecular view (left of the slide). In a solid, molecules are tightly packed in fixed positions and strongly bonded. In a liquid they are close together but free to move past each other. In a gas they are far apart and move almost independently. This is why liquids and gases are both fluids: neither has fixed molecular positions.

Continuum view (right of the slide): the definition. Apply the same tangential (shear) force FF to the top of a block of height hh:

  • Steel (solid): the block deforms by a finite displacement δx\delta x and then stops. The internal Shear stressτ\tauUnits: Pa (N m^-2)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.Open in lecture balances FF. Remove FF and it springs back (elastic). A solid resists shear by deforming its shape by a fixed amount.
  • Water (fluid): the top surface keeps moving, at a velocity δu\delta u. The deformation grows continuously for as long as the shear stress acts. A fluid cannot resist a shear stress at rest; it flows.

That is why the slide labels the solid with a displacement δx\delta x but the fluid with a velocity δu\delta u. For a solid, shear stress is related to the amount of deformation (strain); for a fluid, shear stress is related to the rate of deformation (strain rate, roughly δu/h\delta u / h).

Two consequences you will use in Lecture 2.

  1. A fluid at rest has no shear stress. If there were any, it would be flowing. So in a static fluid the only surface stress is pressure, which acts normal to any surface. This is the foundation of HydrostaticsFluids at rest. Only pressure acts on any surface, and pressure increases with depth as p = p0 + rho g h.Open in lecture.
  2. Shear stress needs motion. The fluid's resistance to the rate of deformation is viscosity. For a Newtonian fluid this is Newton's law of viscosity, τ=μ dvdy\tau = \mu\,\dfrac{dv}{dy}, which is on the ENGR271 2026 formula sheet (rheology weeks). You do not need it yet; just see where it comes from.

7. Density, specific weight and relative density

Intuition. Density tells you how much mass is packed into a space. Specific weight tells you how heavy that space is, which is what matters in hydrostatics because pressure comes from the weight of fluid above. Relative density is just density compared with water, a quick number for "heavier or lighter than water".

Formula

Mass density

Not given - memorise
What each term means
ρ\rhoDensitykg m^-3
mmMasskg
VVVolumem^3
Click to pin

Mass per unit volume. At a point, rho is the limit of delta m / delta V as delta V shrinks (see the continuum section).

SymbolNameMeaningSI units
ρ\rhoDensityMass per unit volumekg m^-3
mmMassMass of the samplekg
VVVolumeVolume occupied by that massm^3
Unit check
kgm3=kg m−3\dfrac{\mathrm{kg}}{\mathrm{m^3}} = \mathrm{kg\,m^{-3}}. Correct.
When to use it
Converting between mass and volume of fluid; every pressure, flow and force calculation starts with rho.
Assumptions and limits
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.
Common 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).
Useful facts [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.
Formula sheetNot written on the ENGR271 2026 sheet as a definition (it appears only inside m_dot = rho A v). Basic definition.

Drills

Plug in, then rearrange, then use it in context. Attempt before revealing.

D1DrillCurrent specThird2 min

A sample of 0.25 m30.25\ \mathrm{m^3} of oil has a mass of 215 kg. Find its density.

D2DrillCurrent specThird3 min

Find the mass of 50 L of mercury (RD = 13.6).

D3DrillCurrent specThird4 min

A rectangular tank 2.0 m×1.5 m×1.0 m2.0\ \mathrm{m} \times 1.5\ \mathrm{m} \times 1.0\ \mathrm{m} is full of diesel (ρ=850 kg m−3\rho = 850\ \mathrm{kg\,m^{-3}}). Find the mass and the weight of the diesel.

Formula

Specific weight

Not given - memorise
What each term means
γ\gammaSpecific weightN m^-3
ρ\rhoDensitykg m^-3
ggGravitational accelerationm s^-2
Click to pin

Weight per unit volume, i.e. how heavy each cubic metre of fluid is.

SymbolNameMeaningSI units
γ\gammaSpecific weightWeight per unit volumeN m^-3
ρ\rhoDensityMass per unit volumekg m^-3
ggGravitational acceleration9.81 m/s^2 unless told otherwisem s^-2
Unit check
[ρg]=kg m−3×m s−2=(kg m s−2) m−3=N m−3[\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 to use it
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.
Assumptions and limits
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.
Common mistakes
  • Quoting gamma in kg/m^3.
  • Mixing kN and N (water: gamma = 9.81 kN/m^3 = 9810 N/m^3).
Formula sheetNot on the ENGR271 2026 sheet or the current Week 1 sheet (the old ENGR217 2024 sheet defined rho g = gamma in its pump section).

Drills

Plug in, then rearrange, then use it in context. Attempt before revealing.

S1DrillCurrent specThird2 min

Find the specific weight of seawater, ρ=1025 kg m−3\rho = 1025\ \mathrm{kg\,m^{-3}}.

S2DrillCurrent specThird3 min

Glycerine has γ=12.36 kN m−3\gamma = 12.36\ \mathrm{kN\,m^{-3}} (Week 1 Exercise 9 data). Find its density and RD.

S3DrillCurrent specThird4 min

A vertical column of liquid with γ=8.0 kN m−3\gamma = 8.0\ \mathrm{kN\,m^{-3}} is 3.0 m tall. What is the weight of liquid resting on each square metre of the column's base? (This is your first hydrostatic pressure calculation; Lecture 2 builds on it.)

Formula

Relative density

On the formula sheet
What each term means
RDRDRelative density (old papers: specific gravity, SG or S)dimensionless
ρfluid\rho_{fluid}Density of the substancekg m^-3
ρwater\rho_{water}Density of waterkg m^-3
Click to pin

How many times denser a substance is than water (water conventionally taken at 4 deg C, where rho_w = 1000 kg/m^3).

SymbolNameMeaningSI units
RDRDRelative density (old papers: specific gravity, SG or S)Density ratiodimensionless
ρfluid\rho_{fluid}Density of the substancekg m^-3
ρwater\rho_{water}Density of water1000 kg/m^3 in this module's examskg m^-3
Unit check
kg m−3kg m−3\dfrac{\mathrm{kg\,m^{-3}}}{\mathrm{kg\,m^{-3}}} = dimensionless. Correct.
When to use it
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.
Assumptions and limits
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.
Common mistakes
  • Giving RD units.
  • Dividing the wrong way.
  • Forgetting to multiply by 1000.
Formula sheetOn 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.

Drills

Plug in, then rearrange, then use it in context. Attempt before revealing.

R1DrillCurrent specThird1 min

Crude oil has RD = 0.90 (Week 1 Exercise 1). Find its density.

R2DrillCurrent specThird2 min

A 1.00 L bottle holds 1.36 kg of a liquid. Find its RD.

R3DrillCurrent specThird3 min

Oil (RD = 0.8) and water are poured into the same open tank and do not mix. Which liquid ends up on top, and why? Is this consistent with the arrangement in ENGR271 2026 Q1?

8. From Week 1 Exercises: air as a mixture

The Week 1 exercise sheet mostly covers Lectures 2 and 3 (pressure, manometers, forces on surfaces, buoyancy); those exercises are solved in the L2 and L3 packs. Exercise 6 part 1 is a density calculation and fits here.

Formula

Ideal gas density

Formula sheet status unknown
What each term means
ρ\rhoDensitykg m^-3
ppAbsolute pressurePa
MMMolar masskg mol^-1
RRUniversal gas constantJ mol^-1 K^-1
TTAbsolute temperatureK
Click to pin

Density of an ideal gas from its absolute pressure, molar mass and absolute temperature.

SymbolNameMeaningSI units
ρ\rhoDensitykg m^-3
ppAbsolute pressureNot gauge pressurePa
MMMolar massConvert g/mol to kg/mol first. For a mixture use the mole-fraction weighted average, M = sum(x_i M_i)kg mol^-1
RRUniversal gas constant8.314J mol^-1 K^-1
TTAbsolute temperaturedeg C + 273.15K
Unit check
Pa⋅kg mol−1J mol−1 K−1⋅K=N m−2⋅kgN m=kg m−3\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 to use it
Any gas density question, and the atmosphere problem in Week 1 Exercise 6.
Assumptions and limits
Ideal-gas behaviour; uniform composition and temperature.
Common mistakes
  • Using gauge pressure or deg C.
  • Leaving M in g/mol (answer 1000 times too big).
Formula sheetNot on the Week 1 sheet. Given in the exercise sheet. Treat as memorise until the papers show otherwise.

From Week 1 Exercises

E6.1Exercise sheetCurrent specThird8 min

Week 1 Exercises, Exercise 6 part 1

Air at sea level has p0=101 325 Pap_0 = 101\,325\ \mathrm{Pa} and uniform temperature T=300 KT = 300\ \mathrm{K}; R=8.314 J mol−1 K−1R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}. Its molar composition is 78% N2\mathrm{N_2} (M=28.02 g mol−1M = 28.02\ \mathrm{g\,mol^{-1}}), 21% O2\mathrm{O_2} (32.00 g mol−132.00\ \mathrm{g\,mol^{-1}}) and 1% Ar (39.95 g mol−139.95\ \mathrm{g\,mol^{-1}}).

  1. (a)
    Calculate the average molar mass of the mixture.
  2. (b)
    [Added, needed for Exercise 6 parts 2-4] Using ρ=pMw/(RT)\rho = p M_w / (R T), calculate the density of air at sea level.

9. Exam questions on this lecture

Past paper questions (adapted to the L1 step)

X1Adapted past paperCurrent specThird5 marks6 min

ENGR271 2025-26 Q1 (data and RD step only)[PARTIAL] Left out: Parts (a)-(d) (absolute and gauge pressure, interface condition, piezometer heights) test Lecture 2 hydrostatics and are in that pack.

An open vertical tank contains a 1.0 m layer of water (ρw=1000 kg m−3\rho_w = 1000\ \mathrm{kg\,m^{-3}}) with a 1.0 m layer of oil of relative density RDoil=0.8RD_{oil} = 0.8 floating on top. Take g=9.81 m s−2g = 9.81\ \mathrm{m\,s^{-2}}.

  1. (i)
    Calculate the density and the specific weight of the oil and of the water.
    [3]
  2. (ii)
    Hence calculate the weight of each layer resting on 1 m21\ \mathrm{m^2} of horizontal area.
    [2]
X2Adapted past paperOld spec ENGR217Third3 marks4 min

ENGR217 2024 Q A1(b) (data only)[PARTIAL] Left out: The pump head-rise calculation (manometer plus Bernoulli, Lectures 2 and W3-4).

A pumped liquid has specific gravity S1=0.82S_1 = 0.82 and the manometer fluid is mercury with S2=13.6S_2 = 13.6. State what "specific gravity" means in the current module's notation, and find the density and specific weight of both fluids.

No real exam question tests the continuum hypothesis, the Knudsen number or the definition of a fluid, so the following are generated in the style of the ENGR271 2026 paper (short theory parts worth 5 marks, calculation parts worth 5-10 marks, data listed in the stem, g=9.81 m/s2g = 9.81\ \mathrm{m/s^2}, ρw=1000 kg/m3\rho_w = 1000\ \mathrm{kg/m^3}).

Generated exam-style questions

G1GeneratedCurrent specThird5 marks6 min

Generated, based on ENGR271 2026 Q1(b) "explain" style

State the continuum hypothesis and explain, with reference to the Knudsen number, when it is valid. Explain why the density at a point is defined as the limit of δm/δV\delta m / \delta V as δV\delta V becomes small, and why that volume cannot literally shrink to zero.

G2GeneratedCurrent spec2:210 marks12 min

Generated, based on ENGR271 2026 Q2(c) "calculate and justify" style

Air at 101.3 kPa and 20 deg C has a mean free path λ=68 nm\lambda = 68\ \mathrm{nm}. At constant temperature, λ\lambda is inversely proportional to pressure. Air flows through a tube of internal diameter 10 mm connected to a vacuum system.

  1. (a)
    Calculate the Knudsen number at atmospheric pressure and state the flow regime.
    [3]
  2. (b)
    Determine the pressure below which the continuum hypothesis is no longer valid for this tube.
    [4]
  3. (c)
    The system is pumped down to 10 Pa. Calculate KnKn, state the regime, and explain whether the no-slip condition and the standard pipe-flow results of this module can be used.
    [3]
G3GeneratedCurrent specThird5 marks6 min

Generated, based on ENGR271 2026 Q1(b) style

A block of steel and a layer of water, each of height hh, are fixed at their base and subjected to the same constant tangential force FF on their top surface. With the aid of sketches, describe how each responds, and hence give a definition of a fluid. State what this implies about the stresses acting in a fluid at rest.

G4GeneratedCurrent spec2:210 marks12 min

Generated, based on ENGR271 2026 Q1(a) calculation style

A sealed cylindrical drum has internal diameter 0.58 m and internal height 0.85 m. It weighs 0.25 kN when empty and 2.20 kN when completely full of an oil.

  1. (a)
    Calculate the density of the oil.
    [4]
  2. (b)
    Calculate its specific weight.
    [2]
  3. (c)
    Calculate its relative density, and state whether it would sit above or below water in a tank if the two did not mix.
    [2]
  4. (d)
    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.
    [2]
G5GeneratedCurrent spec2:210 marks12 min

Generated, based on Week 1 Exercise 6 and ENGR271 2026 Q3 mixture style

Natural gas in a storage vessel is modelled as an ideal gas mixture of 90% methane (M=16.04 g mol−1M = 16.04\ \mathrm{g\,mol^{-1}}) and 10% ethane (M=30.07 g mol−1M = 30.07\ \mathrm{g\,mol^{-1}}) by mole, at 5.0 bar absolute and 15 deg C. Take R=8.314 J mol−1 K−1R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}.

  1. (a)
    Calculate the average molar mass of the gas.
    [3]
  2. (b)
    Calculate its density using ρ=pM/(RT)\rho = pM/(RT).
    [3]
  3. (c)
    Calculate its relative density (with respect to water) and specific weight.
    [2]
  4. (d)
    Explain why the density of a gas cannot always be treated as constant in hydrostatic calculations, whereas the density of a liquid usually can.
    [2]

10. What to memorise from Lecture 1

ItemStatusNotes
Definition of a fluid (deforms continuously under any shear stress)
MUST MEMORISE
Consequence: no shear stress in a fluid at rest
Continuum hypothesis (three bullet points, slide 30)
MUST MEMORISE
Justify it with Kn
Kn=λ/LKn = \lambda / L, continuum if Kn<0.001Kn < 0.001; regimes 0.001 / 0.1 / 10
NOT GIVEN - memorise
Not on the 2026 sheet
ρ=m/V\rho = m/V, rho = limit of δm/δV\delta m / \delta V
NOT GIVEN - memorise
γ=ρg\gamma = \rho g (N/m^3)
NOT GIVEN - memorise
RD=ρ/ρwRD = \rho / \rho_w
GIVEN (ENGR271 2026 sheet)
RD = SG in old papers