Question

At 250˚C a mixture of saturated steam and liquid water exists in equilibrium. If the
specific volume of the mixture is 0.04159 m³/kg, calculate the following: (a)percent moisture,
(b)enthalpy, and (c)entropy

Answer 1

Given:-

saturation temperature, $T_{s}=250^{0}C=523K$

specific volume of mixture, $v=0.04159m^{3}/kg$

To Find:-

dryness fraction, enthalpy and entropy

Explanation:-

$\text{From steam table, }\text{at }250^{0}C:-\\\\v_{g}=0.05037m^{3}/kg, v_{f}=0.001251m^{3}/kg, h_{f}=1085.8kJ/kg, h_{fg}=1714.6k/kg, s_{f}=2.794kJ/kg-K, s_{fg}=3.277kJ/kg-K\\\\\text{Using pure substance relation:-}\\\\v=v_{f}+x(v_{g}-v_{f})\\\\\Rightarrow0.04159=0.001251+x(0.05037-0.001251)\Rightarrow x=0.82\\\\\text{Now, }\\\\\text{enthalpy, }h=h_{f}+xh_{fg}=1085.8+(0.82\times1714.6)=2493.9kJ/kg\\\\\text{entropy, }s=s_{f}+xs_{fg}=2.794+(0.82\times3.277)=5.48114kJ/kg-K$