Answer:
W= -18.489 kJ
Q= -213.4 kJ
Explanation:
We are given following data for water:
[tex]V_{2} =\frac{V_{1} }{2} \\m=0.2kg\\P=400kPa[/tex]
From "Saturated Water Pressure Entry" table corresponding to pressure
P = 400kPa we can obtain specific volume and internal energy:
[tex]v_{f}=0.001084 m^3/kg\\ v_{1} =v_{g} =0.046246 m^3/kg\\\\u_{f}=604.29 kJ/kg\\\\u_{1} =u_{g} =2553.55 kJ/kg\\\\[/tex]
We can calculate final specific volume from given data in problem:
[tex]v_{2} =\frac{v_{1} }{2} =0.23123 m^3/kg[/tex]
Work done for constant pressure process is:
[tex]W=m.P.(v_{2} -v_{1} )\\W=-18.4984 kJ\\[/tex]
Find specific internal energy is given by:
[tex]u_{2} =u_{f} +x.(u_{g} -u_{f} )[/tex]
We must calculate first dryness fraction:
[tex]x=\frac{v_{2} }{v_{g}-v_{f} }-v_{f}\\ x=0.5[/tex]
Specific internal energy is:
[tex]u_{2} =u_{f} +x.(u_{g} -u_{f} )\\=1579 kJ/kg[/tex]
Heat transfer is given by:
[tex]Q=m(u_{2} -u_{1} )+W\\=-213.4kJ[/tex]
