Question 12.CSGP.114: The power plant shown in Fig. 12.21 combines a gas-turbine c......

The power plant shown in Fig. 12.21 combines a gas-turbine cycle and a steam- turbine cycle. The following data are known for the gas-turbine cycle. Air enters the compressor at 100 kPa, 25°C, the compressor pressure ratio is 14, the heater input rate is 60 MW; the turbine inlet temperature is 1250°C, the exhaust pressure is 100 kPa; the cycle exhaust temperature from the heat exchanger is 200°C. The following data are known for the steam-turbine cycle. The pump inlet state is saturated liquid at 10 kPa, the pump exit pressure is 12.5 MPa; turbine inlet temperature is 500°C. Determine
a. The mass flow rate of air in the gas-turbine cycle.
b. The mass flow rate of water in the steam cycle.
c. The overall thermal efficiency of the combined cycle.

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a) From Air Tables, A.7:  P _{ r 1}=1.0913, \quad h _1=298.66, \quad h _5=475.84 \,kJ / kg

\begin{aligned}& s _2= s _1 \Rightarrow P _{ r 2 S }= P _{ r 1}\left( P _2 / P _1\right)=1.0913 \times 14=15.2782 \\& T _2=629 K , h _2=634.48 \\& w _{ C }= h _1- h _2=298.66-634.48=-335.82 \,kJ / kg\end{aligned}

At  T _3=1523.2 \,K : P _{ r 3}=515.493, h _3=1663.91 \,kJ / kg

\dot{ m }_{ AIR }=\dot{ Q }_{ H } /\left( h _3- h _2\right)=\frac{60000}{1663.91-634.48}= 5 8 . 2 8 \,k g / s

b)

\begin{aligned}P _{ r 4 S } & = P _{ r 3}\left( P _4 / P _3\right)=515.493(1 / 14)=36.8209 \\& ⇒ T _4=791 \,K , \quad h _4=812.68 \,kJ / kg \\w _{ T \text { gas }} & = h _3- h _4=1663.91-812.68=851.23 \,kJ / kg\end{aligned}

Steam cycle:    – w _{ P } \approx 0.00101(12500-10)=12.615 \,kJ / kg

h _6= h _9- w _{ P }=191.83+12.615=204.45 \,kJ / kg

At 12.5 MPa, 500 °C:  h _7=3341.1 \,kJ / kg , \quad s _7=6.4704 \,kJ / kg K

\dot{ m }_{ H _2 O }=\dot{ m }_{ AIR } \frac{ h _4- h _5}{ h _7- h _6}=58.28 \frac{812.68-475.84}{3341.1-204.45}= 6 . 2 5 9 \,kg / s

c)

\begin{gathered}s _8= s _7=6.4704=0.6492+ x _8 \times 7.501, \quad x _8=0.7761 \\h _8=191.81+0.7761 \times 2392.8=2048.9 \,kJ / kg \\w _{\text {T steam }}= h _7- h _8=3341.1-2048.9=1292.2 \,kJ / kg\end{gathered}

\begin{aligned}\dot{ W }_{ NET } & =\left[\dot{ m }\left( w _{ T }+ w _{ C }\right)\right]_{ AIR }+\left[\dot{ m }\left( w _{ T }+ w _{ P }\right)\right]_{ H _2 O } \\& =58.28(851.23-335.82)+6.259(1292.2-12.615) \\& =30038+8009=38047 \,kW =38.05\, MW \\\eta_{ TH } & =\dot{ W }_{ NET } / \dot{ Q }_{ H }=38.047 / 60= 0 . 6 3 4\end{aligned}

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