3-phase AC motor efficiency

[isutechman]

Ok this may seam like a basic question, but my background is mechanical engineering and I seam to be failing to remember what I learned in EE class.

I have a 250 hp motor that is driving an air compressor and it is rated at 440v and 262.5 amps and has a PF 1. I want to find the efficiency of this motor and I was using this equation
I=(0.746*hp)/(1.73*E*eff*PF)
I = amps
hp = horsepower
E = line to line voltage
eff = efficiency
PF = power factor

I run this equation and I keep getting an efficiency of like 0.09%

Any ideas what I'm doing wrong?
Thanks

 

[ozmosis]

are you multiplying by 440(volts). It should be 0.44 (kV)
and I suspect your PF would be slightly less than 1, depending on the motor used.

 

[DickDV]

Switch the position of I and .746hp in the equation so you have

.746*hp = I*E*PF*1.73*Eff

I/m a bit rusty on that but I think that's correct. Of course, the motor isn't at a PF of 1.00 either. Depending on load, it could be anywhere from .92 to .30.

 

[Skogsgurra]

The factor 0.746 is wrong. Use 746 instead. That will give you 93.2 percent efficiency.

Gunnar Englund

http://www.gke.org

 

[isutechman]

Thanks for the help. I got the PF=1 from the nameplate on the motor, how would I go about calculating the power factor for my load?

 

[Skogsgurra]

Is this a synchronous motor? You can't have P.F = 1 on an asynchronous motor.

Gunnar Englund

http://www.gke.org

 

[isutechman]

Yes it is a synchrous motor.

 

[Skogsgurra]

OK. Then your P.F. is determined by the excitation. Can be set to any value - at least as long as you stay inside the capability curve.

Gunnar Englund

http://www.gke.org

 

[aolalde]

isutechman

If you do not have an instrument to measure the power factor (PF), you can adjust the motor excitation to fit PF = 1.0.
The procedure is simple, under the desired operating load, move the field excitation up slowly and watch the line currents, increase excitation until you notice that the line currents increase too, now reduce gradually the excitation; the line current must decrease up to a point when reducing the field increases the line current again. The point of minimum line current is when the motor has unit power factor (PF=1.0). Then you can calculate the efficiency corresponding to PF=1.0 with your formula and the recommendations above .

 

[LionelHutz]

FYI, using 0.746 as the hp conversion factor gives you motor power in kW (1000's of watts or 0.746kW/hp). This makes your answer off by a factor of 1000. So, use plain old W (watts) or convert either the voltage or current values to k. In other words, either use 746 or use one of 0.440kV or 0.2625kA.