Question

The pump of a water distribution system is powered by a 15-kW electric motor whose efficiency is 90 percent. The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the pressures at the inlet and outlet of the pump are measured to be 100 kPa and 300 kPa (absolute), respectively, determine the mechanical efficiency of the pump. (Page 99).

Answer 1

Answer:

Mechanical power of pump is 74.07%.    

Explanation:

Power of motor = 15 KW

Efficiency of motor= 90%

So the actual power(P) supplied by motor = 0.9 x 15 KW

P=13.5 KW

Water flow rate = 50 L/s

Volume flow rate = 50 L/s

We know that

$1000\ L/s=1\ m^3/s$

So

$Volume\ flow \rate =0.05\ m^3/s$

We know that pump is an open system and work input for open system can be calculated as

W=VΔP

ΔP is the pressure difference

V is the volume flow rate

So by putting the values

W=0.05 (300-100)            (here ΔP=300 - 100=200 KPa)

W=10 KW    

So mechanical power of pump

$\eta =\dfrac{W}{P}$        

$\eta =\dfrac{10}{13.5}$    

η =0.7407

Mechanical power of pump is 74.07%.