Assume an ideal isentropic turbine and use the supplied table consistently. Its inlet specific entropy is s3=6.282kJkg−1K−1. At the condenser pressure the outlet mixture has s4=sf+χ(sg−sf), where χ is the vapour mass fraction. HenceThe outlet is approximately 71.5% vapour and 28.5% liquid by mass. Its specific enthalpy isThe source's common unit label omits the kelvin denominator for entropy: s must be interpreted in kJkg−1K−1, while h is in kJkg−1. A real irreversible turbine would increase entropy and require an efficiency correction; no such correction is supplied.
χ=8.667−0.2976.282−0.297=0.71505.
h4=84+0.71505(2538−84)=1838.74kJkg−1.
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