Five kilograms of steam contained in 2m3 cylinder at 40 kPa

Five kilograms of steam contained in 2m³ cylinder at 40 kPa is compressed isentropically to 5000 kPa. What is the work needed?

TEXTBOOK ANSWER: 185 kJ

Solution

GIVEN DATA: Mass m= 5 Kg. Volume V₁=2m³ Pressure P₁= 40 kPa Pressure P₂= 5000 kPa. TO FIND: WORK NEEDED = WORK DONE. SOLUTION: WORKDONE = P₂V2 - P₁V₁ V₂ is unknown. FOR ISENTROPIC COMPRESSION PROCESS, P₁/P₂=(V₂/V₁)^k where k is ratio of specific heat of steam......k=1.33 for steam. SO 40/5000 = (V₂^k/2^1.33)........... V₂=(40/5000 × 2^1.33)^1/1.33 = 0.053 m³. So WORKDONE=(5000 × 1000 × 0.053) - (40 × 1000 × 2) = 265000 - 80000 = 185000 JOULES = 185000/1000 = 185 KILOJOULES (KJ).