A piston-cylinder assembly contains 2 lb of water, initially at 100 lbf/in.^2 and 400 degree F. The water undergoes two processes in series: a constant-pressure process followed by a constant-volume process. At the end of the constant-volume process, the temperature is 300 degree F and the water is a two-phase liquid-vapor mixture with a quality of 60%. Neglect kinetic and potential energy effects. Sketch T-v and p-diagrams showing the key states and the processes. Determine the work and heat transfer for each process, all in Btu.

-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

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