Answer:
-In process 1 to 2, Work = -38.87 Btu and heat transfer = -300.5 Btu
-In process 2 to 3,Work = 0 Btu and heat transfer = -475.74 Btu
Explanation:
We are given the following ;
m = 2 lb
P1 = 100 lbf/in²
T1 = 400°F
T3 = 300°F
x3 = 0.6
And now, we need to plot the graph so let's find some values from the saturation tables which i have attached.
From the table A4E, at 300°F, we obtain
vg = vg3 = 6.4663 ft³/lbm
vf = vf3 = 0.01475 ft³/lbm
uf3 = 269.51 Btu/lbm
ug3 = 1099.8 Btu/lbm
V3 = V2 = vf3 + x3(vg3 - vf3)
= 0.01475 + 0.6(6.4663 - 0.01475) = 3.886 ft³/lbm
From table A-6E,at T1=400°F and P = 100 Psia, we have;
V1 = vg = 4.936 ft³/lbm and u1 = 1136.4 Btu/lbm
And from table A-5E attached and at 100 psia,
vf2 = 0.01774 ft³/lbm
Vg2 = 4.4327 ft³/lbm
uf2 = 298.19 Btu/lbm
ug2 = 1105.5 Btu/lbm
To find x2;
V2 = vf2 + x2(vg2 - vf2)
x2 = (v2 - vf2)/(vg2 - vf2)
x2 = (3.886 - 0.01774)/(4.4327 - 0.01774) = 0.8762
Now, let's find u2 and u3;
u2 = uf2 + x2(ug2 - uf2)
= 298.19 + 0.8762(1105.5 - 298.19) = 1005.56 Btu/lb
u3 = uf3 + x3(ug3 - uf3)
= 269.51 + 0.6(1099.8 - 269.51) = 767.69 Btu/lb
Work done for process 1-2 is given as;
W1-2 = mp(V2 - V1)
Where mp is the mass of the piston.
Now from the question mp = 2lbm,
Thus;
W1-2 = 2 x 100lbf/in²[(3.886 - 4.936) ft³/lb x (144in²/1ft²) x (1 Btu/778 ft.lbf) = -38.87 Btu
From energy balance;
Q1-2 = m(u2 - u1) + W1-2
Q1-2 = 2(1005.56 - 1136.4) + (-38.87)
Q1-2 = -300.5 Btu
Work done for process 2-3 will be zero because we have constant volume in this state.
Thus let's calculate Q2-3;
Q2-3 = m(u3 - u2) = 2(767.69 - 1005.56) = - 475.74 Btu
