An engine operating on the ideal Otto cycle is considered. For specific source and sink temperatures, the exergy destruction associated with this cycle and the second-law efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) We take the engine bordering the heat source at temperature T_{H} and the environment at temperature T_{0} as the system. This cycle was analyzed in Example 9-3, and various quantities were given or determined to be
\begin{aligned}r &=8 & P_{2} &=1.7997 \mathrm{MPa} \\T_{0} &=290 \mathrm{~K} & P_{3} &=4.345 \mathrm{MPa} \\T_{1} &=290 \mathrm{~K} & q_{\text {in }} &=800 \mathrm{~kJ} / \mathrm{kg} \\T_{2} &=652.4 \mathrm{~K} & q_{\text {out }} &=381.83 \mathrm{~kJ} / \mathrm{kg} \\T_{3} &=1575.1 \mathrm{~K} & w_{\text {net }} &=418.17 \mathrm{~kJ} / \mathrm{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_{\text {dest }}=T_{0} s_{\text {gen }}=T_{0}\left(s_{e}-s_{i}-\frac{q_{\text {in }}}{T_{b, \text { in }}}+\frac{q_{\text {out }}}{T_{b, \text { out }}}\right) ( kJ / kg ) (9-32)
\begin{aligned}s_{3}-s_{2} &=s_{3}^{\circ}-s_{2}^{\circ}-R \ln \frac{P_{3}}{P_{2}} \\&=(3.5045-2.4975) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}-(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \frac{4.345 \mathrm{MPa}}{1.7997 \mathrm{MPa}} \\&=0.7540 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\end{aligned}
Also,
q_{\text {in }}=800 \mathrm{~kJ} / \mathrm{kg} \quad\text { and }\quad T_{\text {source }}=1700 \mathrm{~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 \mathrm{~K})\left[0.7540 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}-\frac{800 \mathrm{~kJ} / \mathrm{kg}}{1700 \mathrm{~K}}\right] \\&=82.2 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
For process 4-1, s_{1}-s_{4}=s_{2}-s_{3}=-0.7540 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, q_{41}=q_{\text {out }}=381.83 \mathrm{~kJ} / \mathrm{kg}, and
T_{\text {sink }}=290 \mathrm{~K}. Thus,
\begin{aligned}x_{\text {dest, } 41} &=T_{0}\left[\left(s_{1}-s_{4}\right)_{\text {sys }}+\frac{q_{\text {out }}}{T_{\text {sink }}}\right] \\&=(290 \mathrm{~K})\left(-0.7540 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}+\frac{381.83 \mathrm{~kJ} / \mathrm{kg}}{290} \frac{\mathrm{K}}{\mathrm{K}}\right) \\&=\mathbf{1 6 3 . 2} \mathrm{kJ} / \mathrm{kg}\end{aligned}
Therefore, the irreversibility of the cycle is
\begin{aligned}x_{\text {dest, cycle }} &=x_{\text {dest }, 12}+x_{\text {dest }, 23}+x_{\text {dest }, 34}+x_{\text {dest }, 41} \\&=0+82.2 \mathrm{~kJ} / \mathrm{kg}+0+163.2 \mathrm{~kJ} / \mathrm{kg} \\&=245.4 \mathrm{~kJ} / \mathrm{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.
x_{\mathrm{dest}}=T_{0}\left(\frac{q_{\mathrm{out}}}{T_{L}}-\frac{q_{\mathrm{in}}}{T_{H}}\right) ( kJ / kg ) (9-34)
(b) The second-law efficiency is defined as
\eta_{\mathrm{II}}=\frac{\text { Exergy recovered }}{\text { Exergy expended }}=\frac{x_{\text {recovered }}}{x_{\text {expended }}}=1-\frac{x_{\text {destroyed }}}{x_{\text {expended }}}
Here the expended energy is the energy content of the heat supplied to the air in the engine (which is its work potential) and the energy recovered is the net work output:
\begin{aligned}x_{\text {expended }}=& x_{\text {heat,in }}=\left(1-\frac{T_{0}}{T_{H}}\right) q_{\text {in }} \\=&\left(1-\frac{2900 \mathrm{~K}}{1700 \mathrm{~K}}\right)(800 \mathrm{~kJ} / \mathrm{kg})=663.5 \mathrm{~kJ} / \mathrm{kg} \\& x_{\text {recovered }}=w_{\text {net,out }}=418.17 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
Substituting, the second-law efficiency of this cycle is determined to be
\eta_{\mathrm{II}}=\frac{x_{\text {recovered }}}{x_{\text {expended }}}=\frac{418.17 \mathrm{~kJ} / \mathrm{kg}}{663.5 \mathrm{~kJ} / \mathrm{kg}}=0.630 \text { or } \mathbf{6 3 . 0 \%}
Discussion The second-law efficiency can also be determined using the exergy destruction data,
\eta_{\mathrm{II}}=1-\frac{x_{\text {destroyed }}}{x_{\text {expended }}}=1-\frac{245.4 \mathrm{~kJ} / \mathrm{kg}}{663.5 \mathrm{~kJ} / \mathrm{kg}}=0.630 \text { or } 63.0 \%
Note that the exergy destruction associated with heat transfer involving both the heat source and the environment are accounted for in the results.
