Question 10.7: Second-Law Analysis of an Ideal Rankine Cycle Determine the ...

The Rankine cycle analyzed in Example 10–1 is reconsidered. For specified source and sink temperatures, the exergy destruction associated with the cycle and exergy of the steam at turbine exit are to be determined.
Analysis In Example 10–1, the heat input was determined to be 2728.6 kJ/kg, and the heat rejected to be 2018.6 kJ/kg.
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{aligned} &s_{2}=s_{1}=s_{f @ 75 kPa }=1.2132 kJ / kg \cdot K \\ &s_{4}=s_{3}=6.7450 kJ / kg \cdot K \quad \text { (at } 3 MPa , 350^{\circ} C \text { ) } \end{aligned}

Thus,

\begin{aligned} x_{\text {dest }, 23} &=T_{0}\left(s_{3}-s_{2}-\frac{q_{\text {in }, 23}}{T_{\text {source }}}\right) \\ &=(290 K )\left[(6.7450-1.2132) kJ / kg \cdot K -\frac{2728.6 kJ / kg }{1600 K }\right] \\ &=1110 kJ / kg \end{aligned}

\begin{aligned}x_{\text {dest }, 41} &=T_{0}\left(s_{1}-s_{4}+\frac{q_{\text {out }, 41}}{T_{\text {sink }}}\right) \\&=(290 K )\left[(1.2132-6.7450) kJ / kg \cdot K +\frac{2018.6 kJ / kg }{290 K }\right] \\&=414 kJ / 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+1110 kJ / kg +0+414 kJ / kg \\ &= 1 5 2 4 k J / 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.
The exergy (work potential) of the steam leaving the turbine is determined from Eq. 10–22. Disregarding the kinetic and potential energies, it reduces to

x_{ dest }=T_{0}\left(\frac{q_{ out }}{T_{L}}-\frac{q_{ in }}{T_{ H }}\right) \quad ( kJ / kg )                      (10–21)

\psi=\left(h-h_{0}\right)-T_{0}\left(s-s_{0}\right)+\frac{V^{2}}{2}+g z \quad ( kJ / kg )                     (10–22)

\begin{aligned} \psi_{4} &=\left(h_{4}-h_{0}\right)-T_{0}\left(s_{4}-s_{0}\right)+\frac{{V_{4}^{2}}^{\nearrow ^0} }{2}+ {g z_{4}}^{\nearrow ^0}  \\ &=\left(h_{4}-h_{0}\right)-T_{0}\left(s_{4}-s_{0}\right) \end{aligned}

where

\begin{aligned} &h_{0}=h_{@ 290 K , 100 kPa } \cong h_{f @ 290 K }=71.355 kJ / kg \\ &s_{0}=s_{@ 290 K , 100 kPa } \cong s_{f @ 290 K }=0.2533 kJ / kg \cdot K \end{aligned}

Thus,

\begin{aligned} \psi_{4} &=(2403.0-71.355) kJ / kg -(290 K )[(6.7450-0.2533) kJ / kg \cdot K ] \\ &=449 kJ / kg \end{aligned}

Discussion Note that 449 kJ/kg of work could be obtained from the steam leaving the turbine if it is brought to the state of the surroundings in a reversible manner.