Answer:
[tex]V=0.972m^2[/tex]
Explanation:
the energy balance is :
[tex]E_{in} -E_{out} =[/tex]Δ[tex]E_{system}[/tex]
0=ΔU=[tex]mm(u_{2} -u_1)[/tex]
[tex]u_1=u_2[/tex]
the properties of water are from table
[tex]P_1=500 kpa\\T_1=60 degree C\\v_1=v_f_{-->60} =251.6 Kj/Kg[/tex]
verified if we get a quantity between 0 and 1
[tex]P_2=10 kpa\\u_2=u_1[/tex]
we have,
[tex]v_f=0.001\\v=14.670 m^3/kg\\u_f=191.79\\v_{fg}=2245.4 Kj/Kg\\[/tex]
therefore,
[tex]x_2=\frac{u_2-u_f}{u_{fg} }[/tex]
we replace the values in this equation:
[tex]x_2=\frac{251.6-191.79}{2245.4}\\ x_2=0.026[/tex]
so,
[tex]T_2=T_{sat--10kpa} =45.81\\v_2=v_f-x_2v_{fg}[/tex]
we replace the values in this equation:
[tex]v_2=0.38886m^3/kg[/tex]
we have,
[tex]V=mv_2[/tex]
we replace the values in this equation:
[tex]V=0.972m^2[/tex]