An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17°C, and 800 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Accounting for the variation of specific heats of air with temperature, determine

Question

An ideal Otto cycle

has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17°C, and 800 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure that occur during the cycle, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle.

Accepted Answer

An ideal Otto cycle is considered. The maximum temperature and pressure, the net work output, the thermal efficiency, and the mean effective pressure are to be determined.

Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 The variation of specific heats with temperature is to be accounted for.

Analysis The P-V diagram of the ideal Otto cycle described is shown in Fig. 9-19. We note that the air contained in the cylinder forms a closed system.

(a) The maximum temperature and pressure in an Otto cycle occur at the end of the constant-volume heat-addition process (state 3). But first we need to determine the temperature and pressure of air at the end of the isentropic compression process (state 2), using data from Table A-17:

[latex]T_{1}=290 \mathrm{~K} ightarrow u_{1} =206.91 \mathrm{~kJ} / \mathrm{kg} [/latex]

[latex]v_{r 1} =676.1[/latex]

Process 1-2 (isentropic compression of an ideal gas):

[latex]\begin{aligned}\frac{v_{r 2}}{v_{r 1}}=\frac{v_{2}}{v_{1}}=\frac{1}{r} ightarrow v_{r 2}=\frac{v_{r 1}}{r}=\frac{676.1}{8}=84.51 ightarrow T_{2} &=652.4 \mathrm{~K} \\quad\u_{2} &=475.11 \mathrm{~kJ} / \mathrm{kg}\end{aligned}[/latex]

[latex]\frac{P_{2} V_{2}}{T_{2}}=\frac{P_{1} V_{1}}{T_{1}} ightarrow P_{2} =P_{1}\left(\frac{T_{2}}{T_{1}} ight)\left(\frac{V_{1}}{V_{2}} ight) [/latex]

[latex]=(100 \mathrm{kPa})\left(\frac{652.4 \mathrm{~K}}{290 \mathrm{~K}} ight)(8)=1799.7 \mathrm{kPa}[/latex]

Process 2-3 (constant-volume heat addition):

[latex]q_{\mathrm{in}} =u_{3}-u_{2} [/latex]

[latex]800 \mathrm{~kJ} / \mathrm{kg} =u_{3}-475.11 \mathrm{~kJ} / \mathrm{kg} [/latex]

[latex]u_{3} =1275.11 \mathrm{~kJ} / \mathrm{kg} ightarrow T_{3}=1575.1 \mathrm{~K} [/latex]

[latex]\frac{P_{3} \mathrm{~V}_{3}}{T_{3}}=\frac{P_{2} \mathrm{~V}_{2}}{T_{2}} ightarrow P_{3} =P_{2}\left(\frac{T_{3}}{T_{2}} ight)\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{3}} ight) [/latex]

[latex]=(1.7997 \mathrm{MPa})\left(\frac{1575.1 \mathrm{~K}}{652.4 \mathrm{~K}} ight)(1)=4.345 \mathrm{MPa}[/latex]

(b) The net work output for the cycle is determined either by finding the boundary (P dV) work involved in each process by integration and adding them or by finding the net heat transfer that is equivalent to the net work done during the cycle. We take the latter approach. However, first we need to find the internal energy of the air at state 4 :

Process 3-4 (isentropic expansion of an ideal gas):

[latex]\begin{aligned}\frac{v_{r 4}}{v_{r 3}}=\frac{v_{4}}{v_{3}}=r ightarrow v_{r 4}=r v_{r 3}=(8)(6.108)=48.864 ightarrow T_{4} &=795.6 \mathrm{~K} \u_{4} &=588.74 \mathrm{~kJ} / \mathrm{kg}\end{aligned}[/latex]

Process 4-1 (constant-volume heat rejection):

[latex]-q_{ ext {out }} =u_{1}-u_{4} ightarrow q_{ ext {out }}=u_{4}-u_{1} [/latex]

[latex]q_{ ext {out }} =588.74-206.91=381.83 \mathrm{~kJ} / \mathrm{kg}[/latex]

Thus,

[latex]w_{ ext {net }}=q_{ ext {net }}=q_{ ext {in }}-q_{ ext {out }}=800-381.83=418.17 \mathrm{~kJ} / \mathrm{kg}[/latex]

(c) The thermal efficiency of the cycle is determined from its definition:

[latex]\eta_{\mathrm{th}}=\frac{w_{\mathrm{net}}}{q_{\mathrm{in}}}=\frac{418.17 \mathrm{~kJ} / \mathrm{kg}}{800 \mathrm{~kJ} / \mathrm{kg}}=0.523 ext { or } 52.3 \%[/latex]

Under the cold-air-standard assumptions (constant specific heat values at room temperature), the thermal efficiency would be (Eq. 9-8)

[latex]\eta_{ ext {th.Otto }}=1-\frac{1}{r^{k-1}}=1-r^{1-k}=1-(8)^{1-1.4}=0.565 ext { or } 56.5 \%[/latex]

which is considerably different from the value obtained above. Therefore, care should be exercised in utilizing the cold-air-standard assumptions.

(d) The mean effective pressure is determined from its definition, Eq. 9-4:

[latex]\mathrm{MEP}=\frac{w_{ ext {net }}}{v_{1}-v_{2}}=\frac{w_{ ext {net }}}{v_{1}-v_{1} / r}=\frac{w_{ ext {net }}}{v_{1}(1-1 / r)}[/latex]

where

[latex]v_{1}=\frac{R T_{1}}{P_{1}}=\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K} ight)(290 \mathrm{~K})}{100 \mathrm{kPa}}=0.832 \mathrm{~m}^{3} / \mathrm{kg}[/latex]

Thus,

[latex]\mathrm{MEP}=\frac{418.17 \mathrm{~kJ} / \mathrm{kg}}{\left(0.832 \mathrm{~m}^{3} / \mathrm{kg} ight)\left(1-\frac{1}{8} ight)}\left(\frac{1 \mathrm{kPa} \cdot \mathrm{m}^{3}}{1 \mathrm{~kJ}} ight)=574 \mathrm{kPa}[/latex]

Discussion Note that a constant pressure of 574 kPa during the power stroke would produce the same net work output as the entire cycle.

Question 9.2: An ideal Otto cycle has a compression ratio of 8. At the beg...

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