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%
Step-by-step 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%