Question

Determine the compressor inlet work required to compress isentropically 150 kPa water to 2 MPa, assuming that the water exists as a) saturated liquid and b) saturated vapor in the initial state.

Answer 1

Compresser work input is generally calculated as change in enthalpy between inlet and exit states. Since it is isentropic compression, entropy remains constant.

(a) Inlet state conditions : P1 = 150 kPa and saturated liquid

Using steam tables at above conditions,

h1 = 467.1 kJ/kg and s1 = 1.433 kJ/kg k

Exit conditions : P2 = 2 MPa and s2 =s1 =1.433 kJ/kgk

At above values, h2 = 469 kJ/kg

Hence inlet work required, W = h2 - h1 = 469 - 467.1 = 1.9 kJ/kg

(b) Inlet conditions : P1 = 150 kPa and saturated vapor

Using steam tables, h1 = 2693.4 kJ/kg, s1 =7.223 kJ/kgk

Exit conditions : P2 = 2 MPa and s2 = 7.223 kJ/kg k

So, at above conditions, h2 = 3310 kJ/kg

Hence inlet work required, W = 3310 - 2693.4 = 616.6 kJ/kg