Back calculating Amps from Pump and Motor eff.

[mjpetrag]

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.



-Mike

 

[dabluffrat]

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...

 

[mjpetrag]

ok i got it to work. the hp curve on the pump curve matches up to the calculation. i was just trying to confirm the ammeter reading was accurate using the operating point on a pump curve.

-Mike

 

[mjpetrag]

just to be sure...

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

 

[BigInch]

The pump power curves include the efficiency of the pump curve given for the stated impeller size and pump speed, but DO NOT include the efficiency of the motor.

With a motor efficiency of 0.86, the power used by the motor is as you have calculated, 158.5 HP.

**********************
"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/

 

[CH5OH]

pump curve defines the correlation between head and flow
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)

 

[JLSeagull]

The motor efficiency is not constant; it depends upon the load.

 

[JJPellin]

There is another potential problem. When I have attempted similar exercises, I found that the actual voltage varied quite a lot from the nominal value. This can make a big difference in the result. Verify the actual voltage.

Johnny Pellin

 

[dabluffrat]

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...