Question

Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What are the temperature and the specific enthalpy of water at the compressor exit? use steam tables.

Answer 1

solution:

The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:

s= 8.2652 kJ/kgK

The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:

T₂ = T₁+ T₂ - T₁/ 8₂ - 8₁ ( s - s₁)

= (400 + 500 - 400/8.3271 - 8.0347 (8.2652 - 8))

= 478.83°C

The final enthalpy is determined in the same way:

h₂= h₁ + h₂ - h₁/s₂ - s₁ ( s - s₁)

= ( 3275.5+ 3486.6 - 3275.5/ 8.3271 - 8.0347) (8.265)

= 3441.91 kJ/kg