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electrical submersible pump motor efficiency 1

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valdc

Industrial
Dec 14, 2009
8
I have a pump rated to deliver 75 gpm at a head of 740 feet
coupled to motor rated 20 Hp

The pump currently discharges 60 gpm and draws an average of 47.6 amps on the phases with 230 volts.

Question is how do i compute for the effeciency?

would it be-

[Pout = 14920 (from 20 Hp)] /
[Pin = A (from meter) * V (from meter) * 1.73 *.8 pf]

id get = 98%

or:

from the pump power curve, from the manufacturer, i've plotted the kw, against the current discharge of 60 gpm

ive got a value of 12.4kw = P2

P2 / P1 =
12.4kw / 15kw = 83% = p1 by the way is the pump motor requirement given from the manufacturers curve...

im confused. thanks!
 
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From 60 GPM and 740 Ft head and water at 8.35 lbs/gal you get 370740 ft/lbs/minute.
= 11.24 HP
= 8381 Watts (Work done per minute)

47.6 A x 230 V x 1.73 x .8pf = 15152 Watts. (Power consumed)

8381 Watts/15152 Watts = 55% efficiency.
Possible error sources:
Your dynamic head may be greater than 740 ft. Your voltage may be different. If your currents are not equal you may expect greater than normal heating losses in the motor. Your power factor may be lower than the assumed 0.8 Power factor often drops when a motor is less than fully loaded.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Efficiency of what? The motor, the pump, or the combination?
Your question suggests you want the efficiency of the motor alone. First method gave the efficiency of the motor.
 
One thing is pretty sure, the actual motor efficiency will be less than 98%.

Submersible motors are not known for high efficiency or good power factor.

Otherwise, as stevenal said, you need to decide what efficiency you are looking for.

David Castor
 
Like the second way for motor efficiency. The hp from the pump curve data for the present operating conditions gives you the motor output power. You then need to measure the input power with a 3-phase power meter.

Measuring voltage and current and assuming a power factor will make enough of an error that the rest of the calculation is pointless.
 
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