An ideal Otto cycle is considered. The maximum temperature and pressure, the net work output, the thermal efficiency, the mean effective pressure, and the power output for a given engine speed 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. 9-20. 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 \mathrm{~K} \rightarrow u_{1} &=206.91 \mathrm{~kJ} / \mathrm{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 \quad T_{2}&=652.4 \mathrm{~K} \\u_{2}&=475.11 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
\begin{aligned}\frac{P_{2} \mathrm{v}_{2}}{T_{2}}=\frac{P_{1} \mathrm{v}_{1}}{T_{1}} \rightarrow P_{2} &=P_{1}\left(\frac{T_{2}}{T_{1}}\right)\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right) \\&=(100 \mathrm{kPa})\left(\frac{652.4 \mathrm{~K}}{290 \mathrm{~K}}\right)(8)=1799.7 \mathrm{kPa}\end{aligned}
Process 2-3 (constant-volume heat addition):
\begin{aligned}q_{\mathrm{in}}&= u_{3}-u_{2} \\800 \mathrm{~kJ} / \mathrm{kg} &=u_{3}-475.11 \mathrm{~kJ} / \mathrm{kg} \end{aligned}
\begin{aligned}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad u_{3}=1275.11 \mathrm{~kJ} / \mathrm{kg}\quad \rightarrow \quad T_{3}&=1575.1 \mathrm{~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 \mathrm{MPa})\left(\frac{1575.1 \mathrm{~K}}{652.4 \mathrm{~K}}\right)(1)=4.345 \mathrm{MPa}\end{aligned}
(b) The net work output for the cycle is determined either by finding the boundary (P d V) 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 \quad T_{4} &=795.6 \mathrm{~K} \\u_{4} &=588.74 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
Process 4-1 (constant-volume heat rejection):
\begin{aligned}-q_{\mathrm{out}} &=u_{1}-u_{4} \rightarrow q_{\mathrm{out}}=u_{4}-u_{1} \\q_{\mathrm{out}} &=588.74-206.91=381.83 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
Thus,
w_{\text {net }}=q_{\text {net }}=q_{\text {in }}-q_{\text {out }}=800-381.83=418.17 \mathrm{k} J / \mathrm{kg}
(c) The thermal efficiency of the cycle is determined from its definition:
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{418.17 \mathrm{~kJ} / \mathrm{kg}}{800 \mathrm{~kJ} / \mathrm{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, } O \text { tto }}=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:
\mathrm{MEP}=\frac{W_{\mathrm{net}}}{V_{\max }-V_{\min }}=\frac{w_{\mathrm{net}}}{v_{\max }-v_{\min }} ( kPa ) (9-4)
\mathrm{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 \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K}\right)(290 \mathrm{~K})}{100 \mathrm{kPa}}=0.8323 \mathrm{~m}^{3} / \mathrm{kg}
Thus,
\mathrm{MEP}=\frac{418.17 \mathrm{~kJ} / \mathrm{kg}}{\left(0.8323 \mathrm{~m}^{3} / \mathrm{kg}\right)\left(1-\frac{1}{8}\right)}\left(\frac{1 \mathrm{kPa} \cdot \mathrm{m}^{3}}{1 \mathrm{~kJ}}\right)=574 \mathrm{kPa}
(e) The total air mass taken by all four cylinders when they are charged is
m=\frac{V_{d}}{v_{1}}=\frac{0.0016 \mathrm{~m}^{3}}{0.8323 \mathrm{~m}^{3} / \mathrm{kg}}=0.001922 \mathrm{~kg}
The net work produced by the cycle is
W_{\text {net }}=m w_{\text {net }}=(0.001922 \mathrm{~kg})(418.17 \mathrm{~kJ} / \mathrm{kg})=0.8037 \mathrm{~kJ}
That is, the net work produced per thermodynamic cycle is 0.8037 \mathrm{~kJ} / \mathrm{cycle}. Noting that there are two revolutions per thermodynamic cycle \left(n_{\mathrm{rev}}=2\right. rev/cycle) in a fourstroke engine (or in the ideal Otto cycle including intake and exhaust strokes), the power produced by the engine is determined from
\dot{W}_{\text {net }}=\frac{W_{\text {net }} \dot{n}}{n_{\text {rev }}}=\frac{(0.8037 \mathrm{~kJ} / \text { cycle })(4000 \mathrm{rev} / \mathrm{min})}{2 \text { rev/cycle }}\left(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)=26.8 \mathrm{~kW}
Discussion If we analyzed a two-stroke engine operating on an ideal Otto cycle with the same values, the power output would be calculated as
\dot{W}_{\text {net }}=\frac{W_{\text {net }} \dot{n}}{n_{\text {rev }}}=\frac{(0.8037 \mathrm{~kJ} / \text { cycle })(4000 \mathrm{rev} / \mathrm{min})}{1 \text { rev/cycle }}\left(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)=53.6 \mathrm{~kW}
Note that there is one revolution in one thermodynamic cycle in two-stroke engines.
