Discussion
In an ideal Rankine cycle, steam enters the turbine at 10 MPa and 500 °C (h = 3375.1 kJ/kg, s = 6.5995 kJ/kg.K). It is cooled in the condenser at a pressure of 10 kPa. At 10 kPa, hf = 191.81 kJ/kg, sf = 0.6492 kJ/kg.K, hg = 2583.9 kJ/kg and sg = 8.1488 kJ/kg.K. The heat rejected in the condenser (in kJ/kg) is
Answer: A
The heat rejected is 1898 kJ/kg.
Expansion in the ideal turbine is isentropic, so the exit entropy is 6.5995 kJ/kg.K. At 10 kPa the dryness fraction and enthalpy are
\[ \begin{aligned} x &= \frac{6.5995 - 0.6492}{8.1488 - 0.6492} = 0.7934 \\ h_{2} &= 191.81 + 0.7934(2583.9 - 191.81) \\ &= 2089.7\ \text{kJ/kg} \end{aligned} \]
The condenser turns this wet steam into saturated liquid at \(h_{f}\), so it rejects \(h_{2} - h_{f} = 2089.7 - 191.8 = 1897.9 \approx 1898\) kJ/kg.
3796 is twice this value, 949 half of it and 2847 one and a half times it; none matches the isentropic exit state.
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