5 lbm of saturated water vapor is to be condensed at a constant-pressure of 20 psia in a piston-cylinder device until all the water is a saturated liquid. Determine the work and heat transfer for the process.

Given: Water; m = 5 lbm; P = 20 psia; x1 = 1,0; x2 = 0.0
Assume: ?KE = ?PE = 0

What will be an ideal response?


Solution: The First Law for closed systems reduces to
Q – W = m (u2 – u1)
For a constant pressure process, the moving boundary work is W = P (V2 – V1) = mP(v2 – v1)
For water at 20 psia:
v1 = 20.09 ft3/lbm
u1 = 1082 Btu/lbm
v2 = 0.01683 ft3/lbm
u2 = 196.19 Btu/lbm
So, W = (5 lbm) (20 psia)(144 in2/ft2)(0.01683 - 20.09) ft3/lbm = -289,050 ft-lbf = -371 Btu
Q = -4800 Btu

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