Question 9.2: The Ideal Otto Cycle An ideal Otto cycle has a compression r...

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:

\begin{aligned} T_{1}=290 K \rightarrow u_{1} &=206.91 kJ / kg \\ v_{r 1} &=676.1 \end{aligned}

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

\begin{aligned} \frac{v_{r 2}}{v_{r 1}}=\frac{v_{2}}{v_{1}}=\frac{1}{r} \rightarrow v_{r 2}=\frac{v_{r 1}}{r}=\frac{676.1}{8}=84.51 \rightarrow T_{2} &=652.4 K \\ u_{2} &=475.11 kJ / kg \end{aligned}

\begin{aligned} \frac{P_{2} V _{2}}{T_{2}}=\frac{P_{1} V _{1}}{T_{1}} \rightarrow P_{2} &=P_{1}\left(\frac{T_{2}}{T_{1}}\right)\left(\frac{ V _{1}}{ v _{2}}\right) \\ &=(100 kPa )\left(\frac{652.4 K }{290 K }\right)(8)=1799.7 kPa\end{aligned}

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

\begin{aligned} q_{ in } &=u_{3}-u_{2} \\ 800 kJ / kg &=u_{3}-475.11 kJ / kg \\ u_{3} &=1275.11 kJ / kg \rightarrow T_{3}=1575.1 K \\ v_{r 3} &=6.108 \end{aligned}

\begin{aligned} \frac{P_{3} v_{3}}{T_{3}}=\frac{P_{2} v_{2}}{T_{2}} \rightarrow P_{3} &=P_{2}\left(\frac{T_{3}}{T_{2}}\right)\left(\frac{v_{2}}{v_{3}}\right) \\ &=(1.7997 MPa )\left(\frac{1575.1 K }{652.4 K }\right)(1)=4.345 MPa \end{aligned}

(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):

\begin{aligned} \frac{v_{r 4}}{v_{r 3}}=\frac{v_{4}}{v_{3}}=r \rightarrow v_{r 4}=r v_{r 3}=(8)(6.108)=48.864 \rightarrow & T_{4}=795.6 K \\ & u_{4}=588.74 kJ / kg \end{aligned}

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

\begin{aligned} -q_{ out } &=u_{1}-u_{4} \rightarrow q_{ out }=u_{4}-u_{1} \\ q_{ out } &=588.74-206.91=381.83 kJ / kg \end{aligned}

Thus,

w_{\text {net }}=q_{\text {net }}=q_{\text {in }}-q_{\text {out }}=800-381.83=418.17 kJ / kg

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

\eta_{ th }=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{418.17 kJ / kg }{800 kJ / kg }=0.523 \text { or } 52.3 \%

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

\eta_{\text {th,Otto }}=1-\frac{1}{r^{k-1}}                       (9–8)

\eta_{\text {th,Otto }}=1-\frac{1}{r^{k-1}}=1-r^{1-k}=1-(8)^{1-1.4}=0.565 \text { or } 56.5 \%

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:

MEP =\frac{W_{\text {net }}}{V_{\max }-V_{\min }}=\frac{w_{\text {net }}}{v_{\max }-v_{\min }}          (kPa)                   (9–4)

MEP =\frac{w_{\text {net }}}{v_{1}-v_{2}}=\frac{w_{\text {net }}}{v_{1}-v_{1} / r}=\frac{w_{\text {net }}}{v_{1}(1-1 / r)}

where

v_{1}=\frac{R T_{1}}{P_{1}}=\frac{\left(0.287 kPa \cdot m ^{3} / kg \cdot K \right)(290 K )}{100 kPa }=0.832 m ^{3} / kg

Thus,

MEP =\frac{418.17 kJ / kg }{\left(0.832 m ^{3} / kg \right)\left(1-\frac{1}{8}\right)}\left(\frac{1 kPa \cdot m ^{3}}{1 kJ }\right)=574 kPa

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