Respuesta :
Solution :
Given
Volume, [tex]$V_1 = 1 \ m^3$[/tex]
Temperature, [tex]$T_1=20 \ ^\circ C[/tex]
[tex]$x_1=0.5$[/tex]
From the saturated water table, corresponding to [tex]$T_1=20 \ ^\circ C[/tex], we get the saturated liquid, vapor specific and the entropy.
[tex]$v_f=1.0010 \ m^3/kg$[/tex]
[tex]$v_g=57.791 \ m^3/kg$[/tex]
[tex]$s_f=0.2966 \ kJ/kg-K$[/tex]
[tex]$s_g=8.6672 \ kJ/kg-K$[/tex]
Now calculating the initial specific volume
[tex]$v_1=v_f+x_1 \cdot(v_g-v_f)$[/tex]
[tex]$=1.0018+05 \cdot(57.791-1.0018)$[/tex]
[tex]$= 29.8973 \ m^3/kg$[/tex]
Calculating the initial specific entropy:
[tex]$s_1=s_f+x+1 \cdot (s_g-s_f)$[/tex]
[tex]$=0.2966+0.5 \cdot (8.6673 - 0.2966)$[/tex]
[tex]$= 4.48 \ kJ/kg-K$[/tex]
So final volume of the vessel is two times bigger as the initial volume
[tex]$V_2=2 .V_1$[/tex]
[tex]$= 2 \times 29.8973 = 59.8 \ m^3/kg$[/tex]
If we interpolate the values from tables between [tex]$v_g=57.791 \ m^3/kg$[/tex] and [tex]$v_g=61.293 \ m^3/kg$[/tex], we can get final temperature and specific entropy corresponding to value of [tex]$v_2$[/tex] :
Final temperature, [tex]$T_2= 19.6 ^\circ C$[/tex]
and [tex]$s_2 = 8.68 \ kJ/kg-K$[/tex]
Calculating change in entropy
[tex]$\Delta s = s_2-s_1$[/tex]
[tex]$=8.68-4.48 = 4.2 \ kJ/kg-K$[/tex]
