Answer:
[tex]P_1 = 47.416 kPa[/tex]
[tex]Q_{12} = 2448.8 kJ[/tex]
Explanation:
Given data:
m = 1.5 kg
[tex]T_1 = 80[/tex] degree celcius
[tex]T_2 = 120[/tex] degree celcius
from saturated water table
at T = 120 degree C
[tex]v_2 = vg = 0.89133 m^3/kg[/tex]
Enthalapy hg = 2528.9 kj/kg
Initial state is 2 phase liquid
[tex]P_1 =P_{sat}[/tex] at 80 degree c
[tex]P_1 = 47.416 kPa[/tex]
[tex]v_1 =v_2 = 0.89133 m^3/kg[/tex]
[tex]v_1 = vf +x(vfg)[/tex]
At 80 degree C
vf = 0.001029 m^3/kg
vg = 3.4053 m^/kg
0.89133 = 0.001029 + x1 = ( 3.4053-0.001029)
[tex]x_1 = 0.261[/tex]
[tex]h_1 = hf + x_1 (hfg)[/tex] at 80 degree C
= 334.97 + 0.261 (2146.6)
[tex]h_1 = 896.35 kJ/kg[/tex]
[tex]Q_{12} = W_{12} + \Delta h[/tex]
= 0 + ( 2528.9 - 896.35)
[tex]Q_{12} = 1652.54 kJ/kg[/tex]
[tex] = 1632.54 \times 1.5 = 2448.8 kJ[/tex]
[tex]Q_{12} = 2448.8 kJ[/tex]
