Question

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Compute the thermal efficiency ofa Rankine cycle with isentropic compression and expansion for which steam leaves the boiler as saturated vapor at 6 MPa and is condense? at 50 kPa

Answer 1

solution: First calculate the pump work. The specific gravity is 1.03 and the change in pressure is 5.95 MPa, so the change in enthalpy is (1.03)*(5.95)=6.13. Then the value of h₂ is h₁ + w_p =34.5+6.1 = 346.6.

Next, at 50 kPa, the saturated liquid entropy is 1.0910 and that of saturated vapor is 7.5939, and the entropy of the vapor before and after expansion in the turbine is s₃ =s₄ =5.8892, so the quality x is calculated as follows:

x= (5.8892-1.0910)/(7.5939-1.0910) = 0.738

Since h₁ = 340.5 and at 50 kPa the enthalpy of saturated vapor is 2645.9, the value of h₄ is calculated as follows:

h₄ = h₁ + x(h_g-h₁) = 340.5 - 0.738 (2645.9 - 340.5) =2041.6

The values of h and s for 1 to 4

State h s

1 340.5 Not needed

2 346.6 Not needed

3 2784.3 5.8892

4 2041.6 5.8892

Heat input into the cycle is h₃ - h₂ =2437.7. Work output is h₃ - h₄ = 742.7. After deducing pump work, net work output is 736.6. Thermal efficiency is therefore

   $\eta$ th = W_net /Q_net =736.6 /2437.7 = 0.302 =30.2%