Known A regenerative vapor power cycle operates with steam as the working fluid. Operating pressures and temperatures are specified; the isentropic efficiency of each turbine stage and the net power output are also given.
Find Determine the thermal efficiency and the mass flow rate into the turbine, in kg/h.
Schematic and Given Data:
Engineering Model
1. Each component in the cycle is analyzed as a steady-state control volume. The control volumes are shown in the accompanying sketch by dashed lines.
2. All processes of the working fluid are internally reversible, except for the expansions through the two turbine stages and mixing in the open feedwater heater.
3. The turbines, pumps, and feedwater heater operate adiabatically.
4. Kinetic and potential energy effects are negligible.
5. Saturated liquid exits the open feedwater heater, and saturated liquid exits the condenser.
Analysis The specific enthalpy at states 1 and 4 can be read from the steam tables. The specific enthalpy at state 2 is evaluated in the solution to Example 8.4. The specific entropy at state 2 can be obtained from the steam tables using the known values of enthalpy and pressure at this state. In summary, h_{1}=3348.4 kJ / kg , h_{2}=2832.8 kJ / kg , s_{2}=6.8606 kJ / kg \cdot K , h_{4}=173.88 kJ / kg.
The specific enthalpy at state 3 can be determined using the isentropic efficiency of the second-stage turbine
h_{3}=h_{2}-\eta_{ t }\left(h_{2}-h_{3 s }\right)
With s_{3 s }=s_{2}, the quality at state 3s is x_{3 s }=0.8208; using this, we get h_{3 s }=2146.3 kJ/kg. Hence,
h_{3}=2832.8-0.85(2832.8-2146.3)=2249.3 kJ / kg
State 6 is saturated liquid at 0.7 MPa. Thus, h_{6}=697.22 kJ/kg.
Since the pumps operate isentropically, the specific enthalpy values at states 5 and 7 can be determined as
h_{5}=h_{4}+v_{4}\left(p_{5}-p_{4}\right)
=173.88+\left(1.0084 \times 10^{-3}\right)\left( m ^{3} / kg \right)(0.7-0.008) MPa
\times\left|\frac{10^{6} N / m ^{2}}{1 MPa } \right| \left|\frac{1 kJ }{10^{3} N \cdot m }\right|=174.6 kJ / kg
h_{7}=h_{6}+v_{6}\left(p_{7}-p_{6}\right)
=697.22+\left(1.1080 \times 10^{-3}\right)(8.0-0.7)\left|10^{3}\right|=705.3 kJ / kg
Applying mass and energy rate balances to a control volume enclosing the open feedwater heater, we find the fraction y of the flow extracted at state 2 from
y=\frac{h_{6}-h_{5}}{h_{2}-h_{5}}=\frac{697.22-174.6}{2832.8-174.6}=0.1966
a. On the basis of a unit of mass passing through the first-stage turbine, the total turbine work output is
\begin{aligned}\frac{\dot{W}_{ t }}{\dot{m}_{1}} &=\left(h_{1}-h_{2}\right)+(1-y)\left(h_{2}-h_{3}\right) \\&=(3348.4-2832.8)+(0.8034)(2832.8-2249.3) \\&=984.4 kJ / kg\end{aligned}
The total pump work per unit of mass passing through the firststage turbine is
\begin{aligned}\frac{\dot{W}_{ p }}{\dot{m}_{1}} &=\left(h_{7}-h_{6}\right)+(1-y)\left(h_{5}-h_{4}\right) \\&=(705.3-697.22)+(0.8034)(174.6-173.88)=8.7 kJ / kg\end{aligned}
The heat added in the steam generator per unit of mass passing through the first-stage turbine is
\frac{\dot{Q}_{\text {in }}}{\dot{m}_{1}}=h_{1}-h_{7}=3348.4-705.3=2643.1 kJ / kg
The thermal efficiency is then
\eta=\frac{\dot{W}_{ t } / \dot{m}_{1}-\dot{W}_{ p } / \dot{m}_{1}}{\dot{Q}_{\text {in }} / \dot{m}_{1}}=\frac{984.4-8.7}{2643.1}=0.369
b. The mass flow rate of the steam entering the turbine, \dot{m}_{1}, can be determined using the given value for the net power output, 100 MW. Since
\dot{W}_{\text {cycle }}=\dot{W}_{ t }-\dot{W}_{ p }
and
\frac{\dot{W}_{ t }}{\dot{m}_{1}}=984.4 kJ / kg \quad \text { and } \quad \frac{\dot{W}_{ p }}{\dot{m}_{1}}=8.7 kJ / kg
it follows that
\dot{m}_{1}=\frac{(100 MW )|3600 s / h |}{(984.4-8.7) kJ / kg }\left|\frac{10^{3} kJ / s }{1 MW }\right|=3.69 \times 10^{5} kg / h
1 Note that the fractions of the total flow at various locations are labeled on the figure.
Skills Developed
Ability to…
• sketch the T–s diagram of the regenerative vapor power cycle with one open feedwater heater.
• 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 entering the first-stage turbine were 150 kg/s, what would be the net power, in MW, and the fraction of steam extracted, y? Ans. 146.4 MW, 0.1966.
