ENGR217 2025 Q1 (ENGR5003)

ENGR217 2025 Q1Past paperOld spec ENGR2172:220 marks24 min

ENGR217 Summer 2025 Q1[VALID] official answers

The compression ratio of a certain Diesel cycle (Figure Q1-1) is 19. The heat absorption $q_1 = 800\ \mathrm{kJ/kg}$ for 1 kg of air inputting. Given that the state at the beginning of compression is $t_1 = 25^\circ\mathrm{C}$ , $p_1 = 100\ \mathrm{kPa}$ , calculate:

Figure Q1-1: Schematic of a Diesel cycle (T-s diagram).

Formulas you may need

Ideal gas: $pv = R_g T$ , $R_g = 0.287\ \mathrm{kJ/(kg\,K)}$ for air (on the formula sheet)
$c_v = \dfrac{R_g}{\gamma - 1}$ , $c_p = \dfrac{\gamma R_g}{\gamma - 1}$ , $\gamma = 1.4$ (on the formula sheet)
Isentropic (adiabatic reversible) process: $pv^\gamma = \mathrm{const}$ , $Tv^{\gamma-1} = \mathrm{const}$ (on the formula sheet)
Constant-pressure heat addition: $q = c_p\,\Delta T$ (on the formula sheet)
Constant-volume heat rejection: $q = c_v\,\Delta T$ (on the formula sheet)
$\eta_{th} = \dfrac{w_{net}}{q_{in}} = 1 - \dfrac{q_{out}}{q_{in}}$ (on the formula sheet)
Diesel: $\eta = 1 - \dfrac{1}{r^{\gamma-1}}\left[\dfrac{r_c^\gamma - 1}{\gamma (r_c - 1)}\right]$ (on the formula sheet)
Cut-off ratio $r_c = v_3/v_2$ , compression ratio $r = v_1/v_2$ (learn this)
Carnot efficiency $\eta_C = 1 - \dfrac{T_L}{T_H}$ (temperatures in K) (learn this)
Mean effective pressure $\mathrm{MEP} = \dfrac{w_{net}}{v_{max} - v_{min}} = \dfrac{w_{net}}{v_1 - v_2}$ (learn this)

(a)
The pressure, temperature, and specific volume at each point in the cycle
v_1 =m^3/kg
p_2 =kPa
T_2 =K
v_2 =m^3/kg
p_3 =kPa
T_3 =K
v_3 =m^3/kg
p_4 =kPa
T_4 =K
v_4 =m^3/kg
[9]
(b)
The thermal efficiency of the cycle, and compare it with the Carnot cycle efficiency with the same temperature limits
eta_th =-
eta_Carnot =-
[6]
(c)
The mean effective pressure
MEP =kPa
[5]