Answer:93 %
Explanation:
Given
Volume(V)=200 L
entropy=25 KJ/K
At [tex]T_{sat}=250 ^{\circ}[/tex]
Properties are
[tex]s_f=2790 J/kg-K[/tex]
[tex]s_g=6070 J/kg-K[/tex]
[tex]\rho _g=20 kg/m^3[/tex]
[tex]\rho _f=799 kg/m^3[/tex]
Entropy at \eta quality is given by
[tex]\frac{S}{m}=s_f+\chi s_{fg}[/tex]
[tex]\frac{25}{m}=2790+\chi \left ( 6070-2790\right )[/tex]
[tex]\frac{25}{m}=2790+\chi \left ( 3280\right )----1[/tex]
Also for mass
[tex]\left ( \frac{m}{V}\right )^{-1}=\rho _f^{-1}+\chi \left ( \rho _g^{-1}-\rho _f^{-1}\right )[/tex]
[tex]\left ( \frac{m}{0.2}\right )^{-1}=799^{-1}+\chi \left ( 20^{-1}-799^{-1}\right )[/tex]
[tex]\left ( \frac{0.2}{m}\right )=1.25\times 10^{-3}+\chi \left ( 0.04875\right )---2[/tex]
From 1 and 2
we get
m=4.3 kg
[tex]\chi =0.93[/tex] or 93%
