Analyzing the Otto Cycle
The temperature at the beginning of the compression process of an air-standard Otto cycle with a compression ratio of 8 is 540°R, the pressure is 1 atm, and the cylinder volume is 0.02 ft ^{3}. The maximum temperature during the cycle is 3600°R. Determine (a) the temperature and pressure at the end of each process of the cycle, (b) the thermal efficiency, and (c) the mean effective pressure, in atm.
Known An air-standard Otto cycle with a given value of compression ratio is executed with specified conditions at the beginning of the compression stroke and a specified maximum temperature during the cycle.
Find Determine the temperature and pressure at the end of each process, the thermal efficiency, and mean effective pressure, in atm.
Schematic and Given Data:
Engineering Model
1. The air in the piston–cylinder assembly is the closed system.
2. The compression and expansion processes are adiabatic.
3. All processes are internally reversible.
4. The air is modeled as an ideal gas.
5. Kinetic and potential energy effects are negligible.
Analysis
a. The analysis begins by determining the temperature, pressure, and specific internal energy at each principal state of the cycle. At T_{1}=540^{\circ} R, Table A-22E gives u_{1}=92.04 Btu/lb and v_{ r 1}=144.32.
For the isentropic compression process 1–2
v_{ r 2}=\frac{V_{2}}{V_{1}} v_{ r 1}=\frac{v_{ r 1}}{r}=\frac{144.32}{8}=18.04
Interpolating with v_{ r 2} in Table A-22E, we get T_{2}=1212^{\circ} R and u_{2}=211.3 Btu/lb. With the ideal gas equation of state
p_{2}=p_{1} \frac{T_{2}}{T_{1}} \frac{V_{1}}{V_{2}}=(1 atm )\left(\frac{1212^{\circ} R }{540^{\circ} R }\right) 8=17.96 atm
The pressure at state 2 can be evaluated alternatively by using the isentropic relationship, p_{2}=p_{1}\left(p_{ r 2} / p_{ r 1}\right).
Since Process 2–3 occurs at constant volume, the ideal gas equation of state gives
p_{3}=p_{2} \frac{T_{3}}{T_{2}}=(17.96 atm )\left(\frac{3600^{\circ} R }{1212^{\circ} R }\right)=53.3 atm
At T_{3}=3600^{\circ} R, Table A-22E gives u_{3}=721.44 Btu/lb and v_{ r 3}=0.6449.
For the isentropic expansion process 3–4
v_{ r 4}=v_{ r 3} \frac{V_{4}}{V_{3}}=v_{ r 3} \frac{V_{1}}{V_{2}}=0.6449(8)=5.16
Interpolating in Table A-22E with v_{ r 4} \text { gives } T_{4}=1878^{\circ} R , u_{4}=342.2 Btu/lb. The pressure at state 4 can be found using the isentropic relationship p_{4}=p_{3}\left(p_{ r 4} / p_{ r 3}\right) or the ideal gas equation of state applied at states 1 and 4. With V_{4}=V_{1}, the ideal gas equation of state gives
p_{4}=p_{1} \frac{T_{4}}{T_{1}}=(1 atm )\left(\frac{1878^{\circ} R }{540^{\circ} R }\right)=3.48 atm
b. The thermal efficiency is
\eta=1-\frac{Q_{41} / m}{Q_{23} / m}=1-\frac{u_{4}-u_{1}}{u_{3}-u_{2}}
=1-\frac{342.2-92.04}{721.44-211.3}=0.51(51 \%)
c. To evaluate the mean effective pressure requires the net work per cycle. That is,
W_{\text {cycle }}=m\left[\left(u_{3}-u_{4}\right)-\left(u_{2}-u_{1}\right)\right]
where m is the mass of the air, evaluated from the ideal gas equation of state as follows:
m=\frac{p_{1} V_{1}}{(\bar{R} / M) T_{1}}
=\frac{\left(14.696 lbf / in .^{2}\right)\left|144 in .^{2} / ft ^{2}\right|\left(0.02 ft ^{3}\right)}{\left(\frac{1545}{28.97} \frac{ ft \cdot lbf }{ lb \cdot{ }^{\circ} R }\right)\left(540^{\circ} R \right)}
=1.47 \times 10^{-3} lb
Inserting values into the expression for W_{\text {cycle }}
W_{\text {cycle }}=\left(1.47 \times 10^{-3} lb \right)[(721.44-342.2)-(211.3-92.04)] Btu / lb
= 0.382 Btu
The displacement volume is V_{1}-V_{2}, so the mean effective pressure is given by
\text { mep }=\frac{W_{\text {cycle }}}{V_{1}-V_{2}}=\frac{W_{\text {cycle }}}{V_{1}\left(1-V_{2} / V_{1}\right)}
1 =\frac{0.382 Btu }{\left(0.02 ft ^{3}\right)(1-1 / 8)}\left|\frac{778 ft \cdot lbf }{1 Btu }\right| \left| \frac{1 ft ^{2}}{144 in.^{2}}\right|
=118 lbf / in .^{2}=8.03 atm
1 This solution utilizes Table A-22E for air, which accounts explicitly for the variation of the specific heats with temperature. A solution also can be developed on a cold airstandard basis in which constant specific heats are assumed. This solution is left as an exercise, but for comparison the results are presented for the case k = 1.4 in the following table:
| Parameter | Air-Standard Analysis | Cold Air-Standard Analysis, k = 1.4 |
| T_{2} | 1212°R | 1241°R |
| T_{3} | 3600°R | 3600°R |
| T_{4} | 1878°R | 1567°R |
| \eta | 0.51 (51%) | 0.565 (56.5%) |
| mep | 8.03 atm | 7.05 atm |
Skills Developed
Ability to…
• sketch the Otto cycle p–v and T–s diagrams.
• evaluate temperatures and pressures at each principal state and retrieve necessary property data.
• calculate thermal efficiency and mean effective pressure.
Quick Quiz
Determine the heat addition and the heat rejection for the cycle, each in Btu. Ans. Q_{23}=0.750 \text { Btu, } Q_{41}=0.368 \text { Btu }.
