GATE Papers >> Mechanical >> 2016 >> Question No 57

Question No. 57 Mechanical | GATE 2016

In a steam power plant operating on an ideal Rankine cycle, superheated steam enters the turbine at 3 MPa and $350^o\mathrm C$. The condenser pressure is 75 kPa. The thermal efficiency of the cycle is ________ percent.

Given data:

For saturated liquid, at $ P=75\;\mathrm{kPa},\;h_f=384.39\;\mathrm{kJ}/\mathrm{kg},v_f=0.001037\mathrm m^3/\mathrm{kg},s_f=1.213kJ/\mathrm{kg}-\mathrm K $
At $ 75\;\mathrm{kPa},\;h_{fg}=2278.6\mathrm{kJ}/\mathrm{kg},\;s_{fg}=6.2434\mathrm{kJ}/\mathrm{kg}-\mathrm K $
At $ P=3\;\mathrm{MPa}\; $ and $ T=350^o\mathrm C $(superheated steam), $h=3115.3\mathrm{kJ}/\mathrm{kg},s=6.7428\mathrm{kJ}/\mathrm{kg}-\mathrm K$


Answer : 25.8 : 26.1


Solution of Question No 57 of GATE 2016 Mechanical Paper

 

Given data:

$ \begin{array}{l}{\mathrm h}_3=384.39\mathrm{kJ}/\mathrm{kg}\\{\mathrm S}_3=1.213\mathrm{kJ}/\mathrm{kg}-\mathrm k\\{\mathrm h}_\mathrm{fg}=2278.6\mathrm{kJ}/\mathrm{kg}\\{\mathrm S}_\mathrm{fg}=6.2434\mathrm{kJ}/\mathrm{kg}-\mathrm K\\{\mathrm h}_1=3115.3\mathrm{kJ}/\mathrm{kg}\\{\mathrm S}_1=6.7428\mathrm{kJ}/\mathrm{kg}-\mathrm K\\{\mathrm S}_1={\mathrm S}_2\\6.7248={\mathrm S}_3+{\mathrm X}_2{\mathrm S}_\mathrm{fg}\\{\mathrm S}_1={\mathrm S}_2={\mathrm S}_3+{\mathrm X}_2{\mathrm S}_\mathrm{fg}\\6.7248=1.213+{\mathrm X}_2\left(6.2434\right)\\{\mathrm X}_2=0.8857\end{array} $

Calculating enthalpy:

$ \begin{array}{l}{\mathrm h}_1={\mathrm h}_\mathrm f=+{\mathrm x}_2\times{\mathrm h}_\mathrm{fg}\\{\mathrm h}_2=384.39+0.8857\times2278.6\\{\mathrm h}_2=2402.546\mathrm{kJ}/\mathrm{kg}\\{\mathrm h}_1=3115.3\mathrm{kJ}/\mathrm{kg}\\{\mathrm h}_4={\mathrm h}_3+\mathrm{vdP}\\{\mathrm h}_4=384.39+0.001037\times\left(3000-75\right)=387.423\mathrm{kJ}/\mathrm{kg}\end{array} $

Pump work $ =\mathrm{vdp}=0.001037\times\left(3000-75\right)=3.033\mathrm{kJ}/\mathrm{kg} $

Heat input $ ={\mathrm h}_1-{\mathrm h}_4=3115.3-387.423=2727.877 $

$ \begin{array}{l}\mathrm{Efficiency}\;\mathrm{of}\;\mathrm{cycle}=\frac{\mathrm{Net}\;\mathrm{work}}{\mathrm{Heat}\;\mathrm{input}}\\=\frac{\left({\mathrm h}_1-{\mathrm h}_2\right)-\mathrm{Pump}\;\mathrm{work}}{{\mathrm h}_1-{\mathrm h}_4}\\\frac{\left(3115.3-2402.546\right)-3.033}{2727.877}\\=\frac{709.721}{2727.877}=0.2601\\\mathrm\eta=26.01\%\end{array} $

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