An Actual Steam Power Cycle
A steam power plant operates on the cycle shown in Fig. 10–5. If the isentropic efficiency of the turbine is 87 percent and the isentropic efficiency of the pump is 85 percent, determine (a) the thermal efficiency of the cycle and (b) the net power output of the plant for a mass flow rate of 15 kg/s.
A steam power cycle with specified turbine and pump efficiencies is considered. The thermal efficiency and the net power output 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. 10–5. The temperatures and pressures of steam at various points are also indicated on the figure. We note that the power plant involves steady-flow components and operates on the Rankine cycle, but the imperfections at various components are accounted for.
(a) The thermal efficiency of a cycle is the ratio of the net work output to the heat input, and it is determined as follows:
Pump work input:
\begin{aligned}w_{\text {pump,in }} &=\frac{w_{s, \text { pump,in }}}{\eta_{P}}=\frac{v_{1}\left(P_{2}-P_{1}\right)}{\eta_{P}} \\&=\frac{\left(0.001009 \mathrm{~m}^{3} / \mathrm{kg}\right)[(16,000-9) \mathrm{kPa}]}{0.85}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^{3}}\right) \\&=19.0 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
Turbine work output:
\begin{aligned}w_{\text {turb,out }} &=\eta_{T} w_{s, \text { trrb,out }} \\&=\eta_{T}\left(h_{5}-h_{6 s}\right)=0.87(3583.1-2115.3) \mathrm{kJ} / \mathrm{kg} \\&=1277.0 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
\text{Boiler heat input:} \quad\quad q_{\text {in }}=h_{4}-h_{3}=(3647.6-160.1) \mathrm{kJ} / \mathrm{kg}=3487.5 \mathrm{~kJ} / \mathrm{kg}Thus,
\begin{array}{c}w_{\text {net }}=w_{\text {turb,out }}-w_{\text {pump,in }}=(1277.0-19.0) \mathrm{kJ} / \mathrm{kg}=1258.0 \mathrm{~kJ} / \mathrm{kg} \\\\\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{1258.0 \mathrm{~kJ} / \mathrm{kg}}{3487.5 \mathrm{~kJ} / \mathrm{kg}}=0.361 \text { or } 36.1 \%\end{array}
(b) The power produced by this power plant is
\dot{W}_{\text {net }}=\dot{m} w_{\text {net }}=(15 \mathrm{~kg} / \mathrm{s})(1258.0 \mathrm{~kJ} / \mathrm{kg})=18.9 \mathrm{MW}
Discussion Without the irreversibilities, the thermal efficiency of this cycle would be 43.0 percent (see Example 10-3c).