Answer:
The correct solution will be "0.0344 m³/kg". The further explanation is given below.
Explanation:
The given values are:
[tex]P_1=700 \ Kpa[/tex]
[tex]=0.7 \ Kpa[/tex]
From table,
[tex]T_1=26.69^{\circ} C[/tex]
[tex]h_f=88.82 \ kJ/kg[/tex]
[tex]P_2=160 \ Kpa[/tex]
[tex]h_f=31.21 \ kJ/kg[/tex]
[tex]T_{sat} = -15.6^{\circ}C[/tex]
Now,
[tex]\Delta T=T_{in}-T{out}[/tex]
[tex]=-15.6-26.69[/tex]
[tex]=-42.29^{\circ}C[/tex]
[tex]V=V_f+x(V_g-V_f)[/tex]
[tex]=V_f+\frac{h-h_f}{h_g-h_f}(V_g-V_f)[/tex]
On putting the value, we get
[tex]=(0.0007437)+(\frac{88.82-31.21}{241.11-31.21})(0.12348-0.0007437)[/tex]
[tex]=0.0344 \ m^3/Kg[/tex]
