The Ideal Regenerative Rankine Cycle
Consider a steam power plant operating on the ideal regenerative Rankine cycle with one open feedwater heater. Steam enters the turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some steam leaves the turbine at a pressure of 1.2 MPa and enters the open feedwater heater. Determine the fraction of steam extracted from the turbine and the thermal efficiency of the cycle.
A steam power plant operates on the ideal regenerative Rankine cycle with one open feedwater heater. The fraction of steam extracted from the turbine 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. 10–19. We note that the power plant operates on the ideal regenerative Rankine cycle. Therefore, the pumps and the turbines are isentropic; there are no pressure drops in the boiler, condenser, and feedwater heater; and steam leaves the condenser and the feedwater heater as saturated liquid. First, we determine the enthalpies at various states:
\left.\begin{aligned}\text{State 1:}\quad\quad P_{1}=10 \mathrm{kPa} \\\text { Sat. liquid }\end{aligned}\right\} \begin{aligned}h_{1}&=h_{ f @ 10 \mathrm{kPa}}=191.81 \mathrm{~kJ} / \mathrm{kg} \\v_{1}&=v_{ f @ 10 \mathrm{kPa}}=0.00101 \mathrm{~m}^{3} / \mathrm{kg}\end{aligned}\begin{aligned}\text{State 2:}\quad\quad P_{2}&=1.2 \mathrm{MPa} \\s_{2}&=s_{1}\\w_{\text {pump,in }} & =v_{1}\left(P_{2}-P_{1}\right)=\left(0.00101 \mathrm{~m}^{3} / \mathrm{kg}\right)[(1200-10) \mathrm{kPa}]\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^{3}}\right) \\& =1.20 \mathrm{~kJ} / \mathrm{kg} \\h_{2} & =h_{1}+w_{\text {pump I,in }}=(191.81+1.20) \mathrm{kJ} / \mathrm{kg}=193.01 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
\left.\begin{aligned}\text{State 3:}\quad\quad P_{3}=1.2 \mathrm{MPa} \\\text { Sat. liquid }\end{aligned}\right\} \begin{aligned}v_{3}&=v_{ f @ 1.2 \mathrm{MPa}}=0.001138 \mathrm{~m}^{3} / \mathrm{kg} \\h_{3}&=h_{ f @ 1.2 \mathrm{MPa}}=798.33 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
\begin{aligned}\text{State 4:}\quad\quad P_{4}&=15 \mathrm{MPa} \\s_{4}&=s_{3}\\w_{\text {pump II,in }} & =v_{3}\left(P_{4}-P_{3}\right)=\left(0.001138 \mathrm{~m}^{3} / \mathrm{kg}\right)[(15,000-1200) \mathrm{kPa}]\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^{3}}\right) \\& =15.70 \mathrm{~kJ} / \mathrm{kg} \\h_{4} & =h_{3}+w_{\text {pump II,in }}=(798.33+15.70) \mathrm{kJ} / \mathrm{kg}=814.031 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
\left.\begin{aligned}\text{State 5:}\quad\quad P_{5}&=15 \mathrm{MPa} \\T_{5}&=600^{\circ} \mathrm{C}\end{aligned}\right\} \begin{aligned}h_{5}&=3583.1 \mathrm{~kJ} / \mathrm{kg} \\s_{5}&=6.6796 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\end{aligned}
\left.\begin{aligned}\text{State 6:}\quad\quad P_{6}&=1.2 \mathrm{MPa} \\s_{6}&=s_{5}\end{aligned}\right\} \begin{aligned}h_{6}&=2860.2 \mathrm{~kJ} / \mathrm{kg} \\(T_{6}&=218.4^{\circ} \mathrm{C})\end{aligned}
\begin{aligned}\text{State 7:}\quad\quad P_{7}&=10 \mathrm{kPa} \\s_{7}&=s_{5} x_{7}=\frac{s_{7}-s_{f}}{s_{f g}}=\frac{6.6796-0.6492}{7.4996}=0.8041 \\h_{7}&=h_{f}+x_{7} h_{f g}=191.81+0.8041(2392.1)=2115.3 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
The energy analysis of open feedwater heaters is identical to the energy analysis of mixing chambers. The feedwater heaters are generally well insulated ( \dot{Q} = 0), and they do not involve any work interactions (\dot{W} = 0). By neglecting the kinetic and potential energies of the streams, the energy balance reduces for a feedwater heater to
\dot{E}_{\text {in }}=\dot{E}_{\text {out }} \longrightarrow \sum\limits_{\text {in }}{\dot{m} h} =\sum\limits_{\text {out }}{\dot{m} h}
or
y h_{6}+(1-y) h_{2}=1\left(h_{3}\right)
where y is the fraction of steam extracted from the turbine \left(=\dot{m}_{6} / \dot{m}_{5}\right). Solving for y and substituting the enthalpy values, we find
y=\frac{h_{3}-h_{2}}{h_{6}-h_{2}}=\frac{798.33-193.01}{2860.2-193.01}=0.2270
Thus,
\begin{aligned}q_{\text {in }}&=h_{5}-h_{4}=(3583.1-814.03) \mathrm{kJ} / \mathrm{kg}=2769.1 \mathrm{~kJ} / \mathrm{kg} \\q_{\text {out }}&=(1-y)\left(h_{7}-h_{1}\right)=(1-0.2270)(2115.3-191.81) \mathrm{kJ} / \mathrm{kg} \\&=1486.9 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
and
\eta_{\mathrm{th}}=1-\frac{q_{\mathrm{out}}}{q_{\mathrm{in}}}=1-\frac{1486.9 \mathrm{~kJ} / \mathrm{kg}}{2769.1 \mathrm{~kJ} / \mathrm{kg}}=0.463 \text { or } 46.3 \%
Discussion This problem was worked out in Example 10-3c for the same pressure and temperature limits but without the regeneration process. A comparison of the two results reveals that the thermal efficiency of the cycle has increased from 43.0 to 46.3 percent as a result of regeneration. The net work output decreased by 171 \mathrm{~kJ} / \mathrm{kg}, but the heat input decreased by 607 \mathrm{~kJ} / \mathrm{kg}, which results in a net increase in the thermal efficiency.
