Sonntag, Borgnakke and van Wylen
9.123 An insulated piston/cylinder contains R-22 at 20°C, 85% quality, at a cylinder volume of 50 L. A valve at the closed end of the cylinder is connected to a line flowing R-22 at 2 MPa, 60°C. The valve is now opened, allowing R-22 to flow in, and at the same time the external force on the piston is decreased, and the piston moves. When the valve is closed, the cylinder contents are at 800 kPa, 20°C, and a positive work of 50 kJ has been done against the external force. What is the final volume of the cylinder? Does this process violate the second law of thermodynamics? Solution: C.V. Cylinder volume. A transient problem. Continuity Eq.: m2 - m1 = mi Energy Eq.: m2u2 - m1u1 = 1Q2 + mihi - 1W2 Entropy Eq.: m2s2 - m1s1 = 1Q2/T + misi + 1S2 gen Process: 1Q2 = 0, 1W2 = 50 kJ State 1: T1 = 20oC, x1 = 0.85, V1 = 50 L = 0.05 m3 P1 = Pg = 909.9 kPa, u1 = uf + x1ufg = 208.1 kJ/kg v1 = vf + x1vfg = 0.000824 + 0.85×0.02518 = 0.022226 m3/kg, s1 = sf + x1sfg = 0.259 + 0.85×0.6407 = 0.8036 kJ/kg K m1 = V1/v1 = 2.25 kg State 2: T2 = 20oC, P2 = 800 kPa, superheated, v2 = 0.03037 m3/kg, u2 = 234.44 kJ/kg, s2 = 0.9179 kJ/kg K Inlet: Ti = 60oC, Pi = 2 MPa,
hi = 271.56 kJ/kg, si = 0.8873 kJ/kg K Solve for the mass m2 from the energy equation (the only unknown) m2 = [m1u1 - 1W2 - m1hi] / [u2 - hi] =
2.25 × 208.1 – 50 – 2.25 × 271.56 = 5.194 kg 234.44 – 271.56
V2 = m2v2 = 0.158 m3 Now check the second law 1S2 gen = m2s2 - m1s1 - 1Q2/T - misi = 5.194 ×0.9179 – 2.25 × 0.8036 – 0 – (5.194 – 2.25) 0.8873 = 0.347 kJ/K > 0,
Satisfies 2nd Law