Question 10.2: Problem: An engine operating on a cold air standard Diesel c...
Problem: An engine operating on a cold air standard Diesel cycle with a compression ratio of 20 and a cut‐off ratio of 2 compresses air that is at 100 kPa and 25 °C at the start of the cycle. Find the maximum temperature reached in the cycle and the heat added per kilogram of air.
Find: Maximum temperature T_3 reached during the cycle, heat added q per kilogram of air.
Known: Cold air diesel cycle, compression ratio r = 20, cut‐off ratio r_c = 2, intake air pressure P_1 = 100 kPa, intake temperature T_1 = 25 °C.
Diagram: Figure E10.2.
Assumptions: Constant specific heats evaluated at T_1 = 25 °C, air behaves as an ideal gas.
Governing Equations:
Compression ratio (isentropic, ideal gas) r = \frac{V_1}{V_2} = \left\lgroup \frac{P_2}{P_1}\right\rgroup ^{1/\gamma }
Cut‐off ratio r_c = \frac{V_3}{V_2} = \frac{T_3}{T_2}.
Heat added in cycle q_H = c_p (T_3 – T_2)
Properties: Specific heat ratio of air γ = 1.400 kJ / kgK (Appendix 1), specific heat of air at constant pressure c_p = 1.004 kJ / kg (Appendix 1).
The pressure after isentropic compression is
P_2 = P_1 r^γ = 0.1 \ MPa \times (20)^{1.400} = 6.62 \ 891 \ MPa.
The air temperature after isentropic compression is
T_2 = T_1 \left\lgroup\frac{V_1}{V_2}\right\rgroup ^{(\gamma -1)/\gamma } =T_1 r^{(\gamma -1)/\gamma } = 298.15 \ K \times (20 )^{(0.400)/1.400} = 701.710 \ K
The air temperature at the end of heat addition is
T_3 = T_2 r_c = 701.710 \ K \times 2 = 1403.42 \ K .
The heat added to the cycle, per unit mass of air, is then
q_H = c_p (T_3 – T_2) = 1.004 \ kJ/kgK \times (1403.42 \ K – 701.710 \ K) = 704.517 \ kJ / kg.
Answer: The maximum air temperature in the cycle is 1403 K and the heat added is 704.5 kJ / kg.