Steam occupies a piston-cylinder device at 600 kPa and 250oC, at a volume of 0.5 m3. The steam is cooled at constant pressure until the volume is 0.1 m3.

Question

(a) Determine the work done and the heat transfer for the process.
(b) Using your computer model for a piston-cylinder assembly, plot the work and heat transfer for the process for final volumes ranging between 0.05 m3 and 0.40 m3.
Given: Water; T1 = 250oC; P1 = P2 = P = 600 kPa; V1 = 0.5 m3; V2 = 0.1 m3
Assume: ?KE = ?PE = 0.
What will be an ideal response?

Answer 1

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)

(a) W = (600 kPa)(0.1 m3 – 0.5 m3) = -240 kJ

For water: v1 = 0.3939 m3/kg

So m = V1/v1 = 1.27 kg

u1 = 2721.31 kJ/kg

v2 = V2/m = 0.0787 m3/kg

x2 = 0.247; u2 = 1138.39 kJ/kg

So, Q = -2250 kJ



(b) For the range of final volumes: