 Solution

Assume an ideal isentropic turbine and use the supplied table consistently. Its inlet specific entropy is $s_3=6.282\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. At the condenser pressure the outlet mixture has $s_4=s_f+\chi (s_g-s_f)$, where $\chi$ 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 $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, while $h$ is in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. A real irreversible turbine would increase entropy and require an efficiency correction; no such correction is supplied.