Known A vapor power cycle operates with steam as the working fluid. The turbine and pump both have efficiencies of 85%.
Find Determine the thermal efficiency, the mass flow rate, in kg/h, the rate of heat transfer to the working fluid as it passes through the boiler, in MW, the heat transfer rate from the condensing steam as it passes through the condenser, in MW, and the mass flow rate of the condenser cooling water, in kg/h.
Schematic and Given Data:
Engineering Model
1. Each component of the cycle is analyzed as a control volume at steady state.
2. The working fluid passes through the boiler and condenser at constant pressure. Saturated vapor enters the turbine. The condensate is saturated at the condenser exit.
3. The turbine and pump each operate adiabatically with an efficiency of 85%.
4. Kinetic and potential energy effects are negligible.
Analysis Example 8.1Owing to the presence of irreversibilities during the expansion of the steam through the turbine, there is an increase in specific entropy from turbine inlet to exit, as shown on the accompanying T–s diagram. Similarly, there is an increase in specific entropy from pump inlet to exit. Let us begin the analysis by fixing each of the principal states. State 1 is the same as in , so h_{1}=2758.0 kJ / kg \text { and } s_{1}=5.7432 kJ / kg \cdot K.
The specific enthalpy at the turbine exit, state 2, can be determined using the isentropic turbine efficiency, Eq. 8.9,
\eta_{ t }=\frac{\left(\dot{W}_{ t } / \dot{m}\right)}{\left(\dot{W}_{ t } / \dot{m}\right)_{ s }}=\frac{h_{1}-h_{2}}{h_{1}-h_{2 s }} (8.9)
\eta_{ t }=\frac{\left(\dot{W}_{ t } / \dot{m}\right)}{\left(\dot{W}_{ t } / \dot{m}\right)_{ s }}=\frac{h_{1}-h_{2}}{h_{1}-h_{2 s }}
where h_{2 s } is the specific enthalpy at state 2s on the accompanying T–s diagram. From the solution to Example 8.1, h_{2 s }=1794.8 kJ/kg. Solving for h_{2} and inserting known values
\begin{aligned}h_{2} &=h_{1}-\eta_{ t }\left(h_{1}-h_{2 s }\right) \\&=2758-0.85(2758-1794.8)=1939.3 kJ / kg\end{aligned}
State 3 is the same as in Example 8.1, so h_{3}=173.88 kJ/kg.
To determine the specific enthalpy at the pump exit, state 4, reduce mass and energy rate balances for a control volume around the pump to obtain \dot{W}_{ p } / \dot{m}=h_{4}-h_{3}. On rearrangement, the specific enthalpy at state 4 is
h_{4}=h_{3}+\dot{W}_{ p } / \dot{m}
To determine h_{4} from this expression requires the pump work. Pump work can be evaluated using the isentropic pump efficiency in the form of Eq. 8.10b: Solving for \dot{W}_{ p } / \dot{m} results in
\eta_{ p }=\frac{\left(\dot{W}_{ p } / \dot{m}\right)_{ s }}{\left(\dot{W}_{ p } / \dot{m}\right)}=\frac{v_{3}\left(p_{4}-p_{3}\right)}{h_{4}-h_{3}} (8.10b)
\frac{\dot{W}_{ p }}{\dot{m}}=\frac{v_{3}\left(p_{4}-p_{3}\right)}{\eta_{ p }}
The numerator of this expression was determined in the solution to Example 8.1. Accordingly,
\frac{\dot{W}_{ p }}{\dot{m}}=\frac{8.06 kJ / kg }{0.85}=9.48 kJ / kg
The specific enthalpy at the pump exit is then
h_{4}=h_{3}+\dot{W}_{ p } / \dot{m}=173.88+9.48=183.36 kJ / kg
a. The net power developed by the cycle is
\dot{W}_{\text {cycle }}=\dot{W}_{ t }-\dot{W}_{ p }=\dot{m}\left[\left(h_{1}-h_{2}\right)-\left(h_{4}-h_{3}\right)\right]
The rate of heat transfer to the working fluid as it passes through the boiler is
\dot{Q}_{\text {in }}=\dot{m}\left(h_{1}-h_{4}\right)
Thus, the thermal efficiency is
\eta=\frac{\left(h_{1}-h_{2}\right)-\left(h_{4}-h_{3}\right)}{h_{1}-h_{4}}
Inserting values
\eta=\frac{(2758-1939.3)-9.48}{2758-183.36}=0.314(31.4 \%)
b. With the net power expression of part (a), the mass flow rate of the steam is
\dot{m}=\frac{\dot{W}_{\text {cycle }}}{\left(h_{1}-h_{2}\right)-\left(h_{4}-h_{3}\right)}
=\frac{(100 MW )|3600 s / h |\left|10^{3} kW / MW \right|}{(818.7-9.48) kJ / kg }=4.449 \times 10^{5} kg / h
c. With the expression for \dot{Q}_{ in } from part (a) and previously determined specific enthalpy values
\dot{Q}_{\text {in }}=\dot{m}\left(h_{1}-h_{4}\right)
=\frac{\left(4.449 \times 10^{5} kg / h \right)(2758-183.36) kJ / kg }{|3600 s / h |\left|10^{3} kW / MW \right|}=318.2 MW
d. The rate of heat transfer from the condensing steam to the cooling water is
\dot{Q}_{\text {out }}=\dot{m}\left(h_{2}-h_{3}\right)
=\frac{\left(4.449 \times 10^{5} kg / h \right)(1939.3-173.88) kJ / kg }{\left|3600 s / h \| 10^{3} kW / MW \right|}=218.2 MW
e. The mass flow rate of the cooling water can be determined from
\dot{m}_{ cw }=\frac{\dot{m}\left(h_{2}-h_{3}\right)}{\left(h_{ cw , \text { out }}-h_{ cw , in }\right)}
=\frac{(218.2 MW )\left|10^{3} kW / MW \| 3600 s / h \right|}{(146.68-62.99) kJ / kg }=9.39 \times 10^{6} kg / h
Skills Developed
Ability to…
• sketch the T–s diagram of the Rankine cycle with turbine and pump irreversibilities.
• fix each of the principal states and retrieve necessary property data.
• apply mass, energy, and entropy principles.
• calculate performance parameters for the cycle.
Quick Quiz
If the mass flow rate of steam were 150 kg/s, what would be the pump power required, in kW, and the back work ratio? Ans. 1422 kW, 0.0116.
