Hello everyone i been trying to make sence of this thermodynamics problem but ca

Question

Hello everyone i been trying to make sence of this thermodynamics problem but can't seem to get the right answer that's posted on the bottom of the question help with this would be great. here it is...........

Water at 60 psia and 800 F is confined by a piston to a closed cylinder with stops. The initial volume is twice as great as that which exists within the cylinder when the piston rests on the stops. The water is cooled until the piston rests on the stops. Determine (a) the temperature (if superheated) or quality (if saturated) at the end of the stops, (b) the temperature at which the water is in the saturated-vapor state, (c) the work per pound mass during this process, and (d) the heat transfer per pound mass during the process.

Answers: (a) 0.868, (b) 292.7 F, (c) -69 Btu/lbm, (d) -374 Btu/lbm

Explanation / Answer

Given

P1 = 60 psia

T1 = 800 F

Saturation pressure at T1 = 800 F is

Psat = 3204 psia

Since P1 < Psat The state is Superheated vapor

Now Properties of water at P1 and T1 are

v1 = 12.45 ft^3/lbm

u1 = 1293 Btu/lbm

h1 = 1431 Btu/lbm

Also Given

v2 = v1/2 = 6.225 ft^3/lbm

P2 = P1 = 60 psia

The saturation properies of steam at P2 = 60 psia are

Tsat = 292.7 F

vf = 0.01738 and vg = 7.18 (Units ft^3/lbm)

uf = 262 and ug = 1098 (Units Btu/lbm)

hf = 262.2 and hg= 1178 (Units Btu/lbm)

Since

vf < v2 < vg The state is LIQUID-VAPOR MIXTURE
a)

v2 = (1-x)*vf + x*vg

6.225 = (1-x)*0.01738 +x*7.18

x = 0.867

u2 = (1-x)*uf + x*ug

u2 = 986.7 Btu/lbm

h2 = (1-x)*hf + x*hg

h2 = 1056 Btu/lbm

b)

Since it is liq-vapor mix

T2 = Tsat = 292.7 F

c)

W = P*(v2-v1) = 60*(6.225-12.45) = -60*6.225 psia*ft^3/lbm

W = (-373.2/0.0833^2)*0.001285 Btu/lbm

W = -69.17 Btu/lbm = (dQ-dU ) (h2-h1-u2+u1)

d)

dQ = (h2-h1) = (1056-1431)

dQ = -375 Btu/lbm

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