Consider 2 kg of water vapor contained within a piston-cylinder. The steam under

Question

Consider 2 kg of water vapor contained within a piston-cylinder. The steam undergoes a isobaric expansion from state 1, where it is a saturated liquid with P=0.5 MPa to state 2, where u_2 = 1300 KJ/kg. During the process, there is heat transfer to the steam with magnitude 2000 KJ. Also a paddle wheel transfers energy to the steam by work with magnitude 1000 KJ. There is no significant change in the potential or kinetic energy of the steam. Determine the amount of energy transfer by work from the steam to the piston during the process, in KJ. Also determine the initial and final quality and the enthalpy change. [Selected Answer: W_piston = 1679.08 KJ, x_1 = 0, x_2 = 0.3438]

Explanation / Answer

State 1 is saturated liquid. Hence, x1 = 0.

P1 = 0.5 MPa.

From steam tables, at 0.5 MPa, u1 = uf = 639.68 kJ/kg, ug = 2561.2 kJ/kg, h1 = hf = 640.23 kJ/kg, hg = 2748.7 kJ/kg

Q - W = m(u2 - u1)

2000 - (-1000) - W = 2*(1300 - 639.68)

W = 1679.36 kJ

u2 = uf + x2*(ug - uf)

1300 = 639.68 + x2*(2561.2 - 639.68)

x2 = 0.3436

h2 = hf + x2*(hg - hf)

h2 = 640.23 + 0.3436*(2748.7 - 640.23)

h2 = 1364.79 kJ/kg

Enthalpy cange = m(h2 - h1) = 2*(1364.79 - 640.23) = 1449.13 kJ

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