Question 10.4: The Ideal Reheat Rankine Cycle Consider a steam power plant ...

A steam power plant operating on the ideal reheat Rankine cycle is considered. For a specified moisture content at the turbine exit, the reheat pressure and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig. 10–13. We note that the power plant operates on the ideal reheat Rankine cycle. Therefore, the pump and the turbines are isentropic, there are no pressure drops in the boiler and condenser, and steam leaves the condenser and enters the pump as saturated liquid at the condenser pressure.
(a) The reheat pressure is determined from the requirement that the entropies at states 5 and 6 be the same:

State 6:      \begin{aligned}&P_{6}=10 kPa \\&x_{6}=0.896 \quad \text { (sat. mixture) } \\&s_{6}=s_{f}+x_{6} s_{f g}=0.6492+0.896(7.4996)=7.3688 kJ / kg \cdot K\end{aligned}

Also,

h_{6}=h_{f}+x_{6} h_{f g}=191.81+0.896(2392.1)=2335.1 kJ / kg

Thus,

State 5:    \left.\begin{array}{l}T_{5}=600^{\circ} C \\s_{5}=s_{6}\end{array}\right\} \quad \begin{aligned}&P_{5}= 4 . 0 M P a \\&h_{5}=3674.9 kJ / kg\end{aligned}

Therefore, steam should be reheated at a pressure of 4 MPa or lower to prevent a moisture content above 10.4 percent.

(b) To determine the thermal efficiency, we need to know the enthalpies at all other states:

State 1:    \left.\begin{array}{l}\begin{array}{l}P_{1}=10 kPa \\\text { Sat. liquid }\end{array}\end{array}\right\} \quad \begin{aligned}&h_{1}=h_{f @ 10 kPa }=191.81 kJ / kg \\&v_{1}=v_{f @ 10 kPa }=0.00101 m ^{3} / kg\end{aligned}

State 2:      \begin{aligned}P_{2}=& 15 MPa \\s_{2}=& s_{1} \\w_{\text {pump,in }}=& v_{1}\left(P_{2}-P_{1}\right)=\left(0.00101 m ^{3} / kg \right) \\& \times[(15,000-10) kPa ]\left(\frac{1 kJ }{1 kPa \cdot m ^{3}}\right) \\=& 15.14 kJ / kg \\h_{2}=& h_{1}+w_{\text {pump,in }}=(191.81+15.14) kJ / kg =206.95 kJ / kg\end{aligned}

State 3:            \left.\begin{array}{l}\begin{array}{l}P_{3}=15 MPa \\T_{3}=600{ }^{\circ} C\end{array}\end{array}\right\} \begin{aligned}&h_{3}=3583.1 kJ / kg \\&s_{3}=6.6796 kJ / kg \cdot K\end{aligned}

State 4:              \left.\begin{array}{l}P_{4}=4 MPa \\s_{4}=s_{3}\end{array}\right\} \begin{aligned}&h_{4}=3155.0 kJ / kg \\&\left(T_{4}=375.5^{\circ} C \right)\end{aligned}

Thus

\begin{aligned}q_{\text {in }} &=\left(h_{3}-h_{2}\right)+\left(h_{5}-h_{4}\right) \\&=(3583.1-206.95) kJ / kg +(3674.9-3155.0) kJ / kg \\&=3896.1 kJ / kg \\q_{\text {out }} &=h_{6}-h_{1}=(2335.1-191.81) kJ / kg \\&=2143.3 kJ / kg\end{aligned}

and

\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{2143.3 kJ / kg }{3896.1 kJ / kg }=0.450 \text { or } 45.0 \% 

Discussion This problem was solved in Example 10–3c for the same pressure and temperature limits but without the reheat process. A comparison of the two results reveals that reheating reduces the moisture content from 19.6 to 10.4 percent while increasing the thermal efficiency from 43.0 to 45.0 percent.