A steam power plant operating on the simple ideal Rankine cycle is considered. For specified 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 We take the power plant bordering the furnace at temperature T_{H} and the environment at temperature T_{0} as the control volume. This cycle was analyzed in Example 10-1, and various quantities were determined to be q_{\text {in }}=2729 \mathrm{~kJ} / \mathrm{kg}, w_{\text {pump,in }}=3.0 \mathrm{~kJ} / \mathrm{kg}, w_{\text {turt }, \text { out }}=713 \mathrm{~kJ} / \mathrm{kg}, q_{\text {out }}=2019 \mathrm{~kJ} / \mathrm{kg}, and \eta_{\text {th }}=26.0 percent.
(a) 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\quad \text{and} \quad x_{\text {dest , 34}}=0
Processes 2-3 and 4-1 are constant-pressure heat-addition and heat-rejection processes, respectively, and they are internally reversible. But 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 irreversibility associated with each process is determined from Eq. 10-19. The entropy of the steam at each state is determined from the steam tables:
x_{\text {dest }}=T_{0} s_{\text {gen }}=T_{0}\left(s_{e}-s_{i}+\frac{q_{\text {out }}}{T_{b, \text { out }}}-\frac{q_{\text {in }}}{T_{b, \text { in }}}\right) (kJ/kg) (10-19)
\begin{array}{l}s_{2}=s_{1}=s_{ f @ 75 \mathrm{kPa}}=1.2132 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\s_{4}=s_{3}=6.7450 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \quad\left(\text { at } 3 \mathrm{MPa}, 350^{\circ} \mathrm{C}\right)\end{array}
Thus,
\begin{aligned}x_{\text {dest, 23} } &=T_{0}\left(s_{3}-s_{2}-\frac{q_{\text {in,23 } }}{T_{\text {source }}}\right) \\&=(300 \mathrm{~K})\left[(6.7450-1.2132) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}-\frac{2729 \mathrm{~kJ} / \mathrm{kg}}{800 \mathrm{~K}}\right] \\&=\mathbf{6 3 6} \mathrm{k} \mathrm{J} / \mathrm{kg}\\\\x_{\text {dest, 41} } &=T_{0}\left(s_{1}-s_{4}+\frac{q_{\mathrm{out}, 41}}{T_{\text {sink }}}\right) \\&=(300 \mathrm{~K})\left[(1.2132-6.7450) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}+\frac{2019 \mathrm{~kJ} / \mathrm{kg}}{300 \mathrm{~K}}\right] \\&=360 \mathrm{k} \mathrm{J} / \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,3 } 4}+x_{\text {dest, 41 }} \\&=0+636 \mathrm{~kJ} / \mathrm{kg}+0+360 \mathrm{~kJ} / \mathrm{kg} \\&=996 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
The total exergy destroyed during the cycle could also be determined from Eq. 10-21. Notice that the largest exergy destruction in the cycle occurs during the heat-addition process. Therefore, any attempt to reduce the exergy destruction should start with this process. Raising the turbine inlet temperature of the steam, for example, would reduce the temperature difference and thus the exergy destruction.
x_{\text {dest }}=T_{0}\left(\frac{q_{\text {out }}}{T_{L}}-\frac{q_{\text {in }}}{T_{H}}\right) ( kJ/kg ) (10-21)
(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 exergy is the exergy content of the heat supplied to steam in boiler (which is its work potential) and the pump input, and the exergy recovered is the work output of the turbine:
\begin{array}{c}x_{\text {heat,in }}=\left(1-\frac{T_{0}}{T_{H}}\right) q_{\text {in }}=\left(1-\frac{300 \mathrm{~K}}{800 \mathrm{~K}}\right)(2729 \mathrm{~kJ} / \mathrm{kg})=1706 \mathrm{~kJ} / \mathrm{kg} \\x_{\text {expended }}=x_{\text {heat,in }}+x_{\text {pump,in }}=1706+3.0=1709 \mathrm{~kJ} / \mathrm{kg} \\x_{\text {recovered }}=w_{\text {turbine }, \text { out }}=713 \mathrm{~kJ} / \mathrm{kg}\end{array}
Substituting, the second-law efficiency of this power plant is determined to be
\eta_{\mathrm{II}}=\frac{x_{\text {recovered }}}{x_{\text {expended }}}=\frac{713 \mathrm{~kJ} / \mathrm{kg}}{1709 \mathrm{~kJ} / \mathrm{kg}}=0.417 \quad \text { or } \quad 41.7 \%
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{996 \mathrm{~kJ} / \mathrm{kg}}{1709 \mathrm{~kJ} / \mathrm{kg}}=0.417 \text { or } 41.7 \%
Also, the system considered contains both the furnace and the condenser, and thus the exergy destruction associated with heat transfer involving both the furnace and the condenser are accounted for.
