Question

5. A pump operating at steady state receives 1.2 kg/s of liquid water at 50o C, 1.5 MPa. The pressure of the water at the pump exit is 15 MPa. The magnitude of the work required by the pump is 21 kW. Stray heat transfer and changes in kinetic and potential energy are negligible. Determine the work required by a reversible pump operating with the same conditions, in kW, and the isentropic pump efficiency. Determine the work required by a reversible pump operating with the same conditions, in kW. Determine the isentropic pump efficiency.

Answer 1

Answer:

the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW

the isentropic pump efficiency is 78%

Explanation:

Given that;

m = 1.2 kg/sec

T = 50 degree Celsius   { Vf = 0.001012 m^3/kg}

P1 = 1.5 Mpa

P2 = 15  Mpa

W-actual = 21 kw  

W reversible = m*Vf (p2 - p1)

= 1.2 * 0.001012 * ( 15*10^3 - 1.5*10^3)

= 1.2 * 0.001012 * 13500

= 16.39 kW

the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW

Isentropic Pump efficiency = W-reversible / W-actual

= 16.39 / 21 = 0.78

= 78%

the isentropic pump efficiency is 78%