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

The Ideal Reheat Rankine Cycle

Consider a steam power plant that operates on the ideal reheat Rankine cycle. The plant maintains the inlet of the high-pressure turbine at 600 psia and 600°F, the inlet of the low-pressure turbine at 200 psia and 600°F, and the condenser at 10 psia. The net power produced by this plant is 5000 kW. Determine the rate of heat addition and rejection and the thermal efficiency of the cycle.

Is there any advantage to operating the reheat section of the boiler at 100 psia rather than 200 psia while maintaining the same low-pressure turbine inlet temperature?

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An ideal reheat steam Rankine cycle produces 5000 kW of power. The rates of heat addition and rejection and the thermal efficiency of the cycle are to be determined. Also, the effect of changing reheat pressure is to be investigated.

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. 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. From the steam tables (Tables A-4E, A-5E, and A-6E),

\begin{aligned}h_{1}&=h_{ f  @  10  \mathrm{psia}}=161.25  \mathrm{Btu} / \mathrm{lbm} \\v_{1}&=v_{ f  @  10  \mathrm{psia}}=0.01659  \mathrm{ft}^{3} / \mathrm{lbm}\end{aligned}

\begin{aligned}w_{\text {pump,in }} &=v_{1}\left(P_{2}-P_{1}\right) \\&=\left(0.01659  \mathrm{ft}^{3} / \mathrm{lbm}\right)[(600-100)  \mathrm{psia}]\left(\frac{1  \mathrm{~B} \mathrm{u}}{5.404  \mathrm{psia} \cdot \mathrm{ft}^{3}}\right) \\&=1.81  \mathrm{Btu} / \mathrm{lbm}\end{aligned}

h_{2}=h_{1}+w_{\text {pump,in }}=161.25+1.81=163.06  \mathrm{Btu} / \mathrm{lbm}

\left.\begin{aligned}P_{3}&=600  \mathrm{psia} \\T_{3}&=600^{\circ} \mathrm{F}\end{aligned}\right\} \begin{aligned}h_{3}&=1289.9   \mathrm{Btu} / \mathrm{lbm} \\s_{3}&=1.5325   \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\end{aligned}

\left.\begin{aligned}P_{4}&=200  \mathrm{psia} \\s_{4}&=s_{3}\end{aligned}\right\} \begin{aligned}x_{4}&=\frac{s_{4}-s_{f}}{s_{f g}}=\frac{1.5325-0.54379}{1.00219}=0.9865 \\h_{4}&=h_{f}+x_{4} h_{f g}=355.46+(0.9865)(843.33)=1187.5  \mathrm{Btu} / \mathrm{lbm}\end{aligned}

\left.\begin{aligned}P_{5}&=200  \mathrm{psia} \\T_{5}&=600^{\circ} \mathrm{F}\end{aligned}\right\} \begin{aligned}h_{5}&=1322.3 \mathrm{Btu} / \mathrm{lbm} \\s_{5}&=1.6771 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\end{aligned}

\left.\begin{aligned}P_{6}&=10  \mathrm{psia} \\s_{6}&=s_{5}\end{aligned}\right\} \begin{aligned}x_{6}&=\frac{s_{6}-s_{f}}{s_{f g}}=\frac{1.6771-0.28362}{1.50391}=0.9266 \\h_{6}&=h_{f}+x_{6} h_{f g}=161.25+(0.9266)(981.82)=1071.0  \mathrm{Btu} / \mathrm{lbm}\end{aligned}

Thus,

q_{\text {in }}=\left(h_{3}-h_{2}\right)+\left(h_{5}-h_{4}\right)=1289.9-163.06+1322.3-1187.5=1261.7  \mathrm{Btu} / \mathrm{lbm}

\begin{array}{l}q_{\mathrm{out}}=h_{6}-h_{1}=1071.0-161.25=909.7  \mathrm{Btu} / \mathrm{lbm} \\w_{\mathrm{net}}=q_{\mathrm{in}}-q_{\mathrm{out}}=1261.7-909.8=352.0  \mathrm{Btu} / \mathrm{lbm}\end{array}

The mass flow rate of steam in the cycle is determined from

\dot{W}_{\text {net }}=\dot{m} w_{\text {net }} \rightarrow \dot{m}=\frac{\dot{W}_{\text {net }}}{w_{\text {net }}}=\frac{5000  \mathrm{~kJ} / \mathrm{s}}{352.0  \mathrm{Btu} / \mathrm{lbm}}\left(\frac{0.94782  \mathrm{Btu}}{1  \mathrm{~kJ}}\right)=13.47  \mathrm{lbm} / \mathrm{s}

The rates of heat addition and rejection are

\begin{array}{l}\dot{Q}_{\text {in }}=\dot{m} q_{\text {in }}=(13.47  \mathrm{lbm} / \mathrm{s})(1261.7  \mathrm{Btu} / \mathrm{lbm})=16,995  \mathrm{Btu} / \mathrm{s} \\\dot{Q}_{\text {out }}=\dot{m} q_{\text {out }}=(13.47  \mathrm{lbm} / \mathrm{s})(909.7  \mathrm{Btu} / \mathrm{lbm})=12,250  \mathrm{Btu} / \mathrm{s}\end{array}

and the thermal efficiency of the cycle is

\eta_{\mathrm{th}}=\frac{\dot{W}_{\text {net }}}{\dot{Q}_{\text {in }}}=\frac{5000  \mathrm{~kJ} / \mathrm{s}}{16,995  \mathrm{Btu} / \mathrm{s}}\left(\frac{0.94782  \mathrm{Btu}}{1  \mathrm{~kJ}}\right)=0.279 \text { or } 27.9 \%

If we repeat the analysis for a reheat pressure of 100 psia at the same reheat temperature, we obtain a thermal efficiency of 27.3 percent. Thus, operating the reheater at 100 psia causes a slight decrease in the thermal efficiency.

Discussion     Now we try to address this question: At what reheat pressure will the thermal efficiency be maximum? We repeat the analysis at various reheat pressures using appropriate software. The results are plotted in Fig. 10-14. The thermal efficiency reaches a maximum value of 28.1 percent at an optimum reheat pressure of about 325 psia.

 

10.13
10.13

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