Question 10.1: Problem: An engine operating on a cold air standard Otto cyc...
Problem: An engine operating on a cold air standard Otto cycle takes air at 100 kPa and 25 °C and compresses it isentropically to 2.2 MPa. The work output from the cycle is 200 kJ / kg of air. Find the efficiency of the cycle, and the maximum temperature reached in the cycle.
Find: Efficiency η_{th,Otto} of the Otto cycle, maximum temperature T_3 reached.
Known: Cold air standard Otto cycle, intake air pressure P_1 = 100 kPa, intake temperature T_1 = 25 °C, pressure after compression P_2 = 2.2 MPa, work output w_{net} = 200 kJ / kg.
Diagram: Figure E10.1.
Assumptions: Constant specific heats evaluated at T_1 = 25 °C, air behaves as an ideal gas.
Governing Equations:
Otto cycle efficiency (isentropic, ideal gas) \eta _{th,Otto}=\frac{w_{net}}{q_H} =1-\frac{1}{r^{\gamma -1}}
Heat added in cycle q_H=c_v (T_3-T_2)
Properties: Specific heat ratio of air \gamma = 1.400 kJ / kgK (Appendix 1), specific heat of air at constant volume c_v = 0.717 kJ / kg (Appendix 1).
The compression ratio of the cycle can be found, since the process uses an ideal gas and is isentropic:
r=\frac{V_1}{V_2} =\left\lgroup\frac{P_2}{P_1} \right\rgroup ^{1/\gamma }=\left\lgroup\frac{2.2 \ MPa}{0.1 \ MPa} \right\rgroup ^{1/1.400}=9.096 \ 48.
The cycle efficiency can then be found,
\eta _{th,Otto}=1-\frac{1}{r^{\gamma -1}} =1-\frac{1}{9.096 \ 48^{0.400}} =0.586 \ 524.
The air temperature at the end of compression is
T_2=T_1\left\lgroup\frac{P_2}{P_1} \right\rgroup ^{(\gamma -1)/\gamma }=298.15 \ K \times \left\lgroup\frac{2.2 \ MPa}{0.1 \ MPa} \right\rgroup ^{0.400/1.400}=721.081 \ K.
The heat added to the cycle, per unit mass of air is
q_H=\frac{w_{net}}{\eta _{th,Otto}} =\frac{200 \ kJ/kg}{0.586 \ 524} =340.992 \ kJ/kg.
Then the maximum temperature in the cycle is
T_3=T_2+\frac{q_H}{c_v} =721.081 \ K+\frac{340.992 \ kJ/kg}{0.717 \ kJ/kgK} =1196.66 \ K.
Answer: The efficiency of the cycle is 58.7% and the maximum air temperature in the cycle is 1197 K.