Question 9.10: Second-Law Analysis of an Otto Cycle Determine the exergy de...

The Otto cycle analyzed in Example 9–2 is reconsidered. For specified source and sink temperatures, the exergy destruction associated with the cycle and the exergy purged with the exhaust gases are to be determined.
Analysis In Example 9–2, various quantities of interest were given or determined to be

\begin{aligned}r &=8 & P_{2} &=1.7997 MPa \\T_{0} &=290 K & P_{3} &=4.345 MPa \\T_{1} &=290 K & q_{\text {in }} &=800 kJ / kg \\T_{2} &=652.4 K & q_{\text {out }} &=381.83 kJ / kg \\T_{3} &=1575.1 K & w_{\text {net }} &=418.17 kJ / kg\end{aligned}

Processes 1-2 and 3-4 are isentropic \left(s_{1}=s_{2}, s_{3}=s_{4}\right) and therefore do not involve any internal or external irreversibilities; that is, X_{\text {dest, } 12}=0 and X_{\text {dest }, 34}=0 .
Processes 2-3 and 4-1 are constant-volume heat-addition and heat-rejection processes, respectively, and are internally reversible. However, the heat transfer between the working fluid and the source or the sink takes place through a finite temperature difference, rendering both processes irreversible. The exergy destruction associated with each process is determined from Eq. 9–32.
However, first we need to determine the entropy change of air during these processes:

X_{ dest }=T_{0} s_{ gen }=T_{0}\left(s_{e}-s_{i}-\frac{q_{ in }}{T_{b, in }}+\frac{q_{ out }}{T_{b, out }}\right) \quad( kJ / kg )

\begin{aligned} s_{3}-s_{2} &=s_{3}^{\circ}-s_{2}^{\circ}-R \ln \frac{P_{3}}{P_{2}} \\ &=(3.5045-2.4975) kJ / kg \cdot K -(0.287 kJ / kg \cdot K ) \ln \frac{4.345 MPa }{1.7997 MPa } \\ &=0.7540 kJ / kg \cdot K \end{aligned}

Also,

q_{\text {in }}=800 kJ / kg \quad \text { and } \quad T_{\text {source }}=1700 K

Thus,

\begin{aligned} x_{\text {dest }, 23} &=T_{0}\left[\left(s_{3}-s_{2}\right)_{\text {sys }}-\frac{q_{\text {in }}}{T_{\text {source }}}\right] \\ &=(290 K )\left[0.7540 kJ / kg \cdot K -\frac{800 kJ / kg }{1700 K }\right] \\ &=82.2 kJ / kg \end{aligned}

For process 4-1, s_{1}-s_{4}=s_{2}-s_{3}=-0.7540 kJ / kg \cdot K , q_{R, 41}=q_{\text {out }}= 381.83 kJ / kg \text {, and } T_{\text {sink }}=290 K . Thus,

x_{\text {dest }, 41}=T_{0}\left[\left(s_{1}-s_{4}\right)_{\text {sys }}+\frac{q_{\text {out }}}{T_{\text {sink }}}\right]

\begin{aligned} &=(290 K )\left[-0.7540 kJ / kg \cdot K +\frac{381.83 kJ / kg }{290 K }\right] \\ &=163.2 kJ / kg \end{aligned}

Therefore, the irreversibility of the cycle is

\begin{aligned} x_{\text {dest }, \text { cycle }} &=x_{\text {dest }, 12}+x_{\text {dest }, 23}+x_{\text {dest }, 34}+x_{\text {dest }, 41} \\ &=0+82.2 kJ / kg +0+163.2 kJ / kg \\ &=245.4 kJ / kg \end{aligned}

The exergy destruction of the cycle could also be determined from Eq. 9–34. Notice that the largest exergy destruction in the cycle occurs during the heat-rejection process. Therefore, any attempt to reduce the exergy destruction should start with this process.
Disregarding any kinetic and potential energies, the exergy (work potential) of the working fluid before it is purged (state 4) is determined from Eq. 9–35:

x_{\text {dest }}=T_{0}\left(\frac{q_{\text {out }}}{T_{L}}-\frac{q_{\text {in }}}{T_{H}}\right) \quad( kJ / kg )               (9–34)

\phi=\left(u_{4}-u_{0}\right)-T_{0}\left(s_{4}-s_{0}\right)+P_{0}\left(v_{4}-v_{0}\right) +\frac{V^2}{2} +gz           (kJ/kg)              (9–35)

\phi_{4}=\left(u_{4}-u_{0}\right)-T_{0}\left(s_{4}-s_{0}\right)+P_{0}\left(v_{4}-v_{0}\right)

where

\begin{aligned} &s_{4}-s_{0}=s_{4}-s_{1}=0.7540 kJ / kg \cdot K \\ &u_{4}-u_{0}=u_{4}-u_{1}=q_{\text {out }}=381.83 kJ / kg \\ &v_{4}-v_{0}=v_{4}-v_{1}=0 \end{aligned}

Thus,

\phi_{4}=381.83 kJ / kg -(290 K )(0.7540 kJ / kg \cdot K )+0=163.2 kJ / kg

which is equivalent to the exergy destruction for process 4-1. (Why?) Discussion Note that 163.2 kJ/kg of work could be obtained from the exhaust gases if they were brought to the state of the surroundings in a reversible manner.