Question

A cylinder fitted with a frictionless piston contains 4 lbm of superheated refrigerant R-134a vapor at 400 lbf/in.², 200 F. The cylinder is now cooled so the R-134a remains at constant pressure until it reaches a quality of 75%. Calculate the heat transfer in the process. 

Answer 1

C.V.: R-134a m₂ = m₁ = m; 

Energy Eq  m(u₂ - u₁) = ₁Q₂ - ₁W₂ 

Process: P = const. ⇒ ₁W2 = ⌡⌠PdV = P∆V = P(V₂ - V₁) = Pm(v₂ - v₁)

State 1: Table  h₁ = 192.92 Btu/lbm 

State 2:  h₂ = 140.62 + 0.75 × 43.74

= 173.425 Btu/lbm ₁Q₂ 

= m(u₂ - u₁) + ₁W₂ = m(u₂ - u₁) + Pm(v₂ - v₁) = m(h₂ - h₁)

= 4 × (173.425 – 192.92) = -77.98 Btu