From the steam table, the following values are obtained h i n , l = 640.1 k J / k g {\displaystyle h_{in,l}=640.1kJ/kg} h i n , v = 2747.5 k J / k g {\displaystyle h_{in,v}=2747.5kJ/kg} h o u t = 3065 k J / k g {\displaystyle h_{out}=3065kJ/kg}
Let the fraction of liquid water be w l = mass of liquid water total mass of water {\displaystyle w_{l}={\frac {\text{mass of liquid water}}{\text{total mass of water}}}}
Using energy balance Q = Δ H = ∑ i m ˙ o u t , i ⋅ h o u t , i − ∑ i m ˙ i n , i ⋅ h i n , i = m t o t a l ⋅ h o u t − m i n , l ⋅ h i n , l − m i n , v ⋅ h i n , v = h o u t − w l ⋅ h i n , l − ( 1 − w l ) ⋅ h i n , v = − 1.93 × 10 6 k J / h m t o t a l (From Question 3c) {\displaystyle {\begin{aligned}Q&=\Delta H\\&=\sum _{i}{\dot {m}}_{out,i}\cdot h_{out,i}-\sum _{i}{\dot {m}}_{in,i}\cdot h_{in,i}\\&=m_{total}\cdot h_{out}-m_{in,l}\cdot h_{in,l}-m_{in,v}\cdot h_{in,v}\\&=h_{out}-w_{l}\cdot h_{in,l}-(1-w_{l})\cdot h_{in,v}\\&={\frac {-1.93\times 10^{6}kJ/h}{m_{total}}}{\text{ (From Question 3c)}}\\\end{aligned}}}
Rearranging the expression and solving for w l {\displaystyle w_{l}} w l = 0.765 {\displaystyle w_{l}=0.765}
