Back calculating Amps from Pump and Motor eff.
Back calculating Amps from Pump and Motor eff.
(OP)
If I have a pump running at 160 hp (119.3 kW) and I am trying to find amps I used the following. (3 phase 480V motor).
119300W/480V = 248.5 A
3 phase correction = 248.5/sqrt(3) = 143.5 A
power factor and motor efficiency correction = 143.5/(.95*.88) = 171.7 A
Now do I divide that 171.7 A by the pump efficiency at the operating point? I'm not 100% sure if 171.7 is correct or not.
119300W/480V = 248.5 A
3 phase correction = 248.5/sqrt(3) = 143.5 A
power factor and motor efficiency correction = 143.5/(.95*.88) = 171.7 A
Now do I divide that 171.7 A by the pump efficiency at the operating point? I'm not 100% sure if 171.7 is correct or not.
-Mike





RE: Back calculating Amps from Pump and Motor eff.
Pump Brake Horse power (BHP) = (Head * Flow * SG) / (3960* Pump Effy)
Motor Input Power (HP) = BHP / Motor Effy
Convert motor input HP to KW then back out your Amps as you were doing.
A more practical approach.... If you have a NEMA motor driving the pump, the full load Amp (FLA) rating of the motor should be published. Likely a 200 hp motor... so the motor will only be 80% full load HP. This will affect where you're at on the motor efficiency curve.
If this a proposed application, I would select the motor I intend to use and ask a motor mfg to quote it. Also ask for a typical performance curve (test curve) for the motor. You should be able to pick off from there the approx. Amp draw at 80% load.
If this is an existing application, you should be able to read motor amps at the power source.
Did you know that 76.4% of all statistics are made up...
RE: Back calculating Amps from Pump and Motor eff.
-Mike
RE: Back calculating Amps from Pump and Motor eff.
does the power curve on the pump curve assume motor efficiency to be 100%? From the spec sheet it says
TDH required = 220 FT
Capacity = 1800 GPM
SG = 1.05
HP = 137 HP
Efficiency = 77%
When I do the calculation...
P = (220'*1800 GPM*1.05)/(3960*.77) = 136.3 HP
Now that sounds right, but if I take into account the motor efficiency
P = 136.3/.86 = 158.5 HP
Does the pump curve use the 136.3 HP or the 158.5 HP to determine the horsepower line?
-Mike
RE: Back calculating Amps from Pump and Motor eff.
With a motor efficiency of 0.86, the power used by the motor is as you have calculated, 158.5 HP.
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"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/
RE: Back calculating Amps from Pump and Motor eff.
pressure(pascal)=head(meter) x density (kg/m3)x g (9.81m/s2)
hydraulic power (kW)=pressure(pascal) x flow (m3/s)
efficiency curve defines correlation between hydraulic power and pump shaft power
efficiency(%)=hydraulic power (kW) x 100/shaft power (kW)
electric power (kW)=
shaft power (kW) x 100 / efficiency motor (%)
electric current,3ph (A)=
electric power (kW)/(3root3 x voltage x cos phi)
RE: Back calculating Amps from Pump and Motor eff.
RE: Back calculating Amps from Pump and Motor eff.
Johnny Pellin
RE: Back calculating Amps from Pump and Motor eff.
Pump curve uses the 136.3 HP this is the hydraulic HP independent of the motor driving it.
The 158.5 HP you calculate is the INPUT power to the motor (Motor Efficiency = Output power / Input Power). Convert to KW... 158.5*.746 = 118.2 KW line power
Did you know that 76.4% of all statistics are made up...