Motor Nameplate speed 4

[gronk2003]

We are having a discussion here with regarding the load on a 800 1786 RPM 4000 volt motor.
We did an online test on this motor and it indicated that it was running at 94.1% FLA and at a speed of 1789.46. The nameplate (RPM 1786 105 FLA) matched up pretty close to indicating that in fact this motor is running at 100%...or so I thought. We are being told that, since the nameplate is 1786 RPM and the motor is running at 1789.46 that the motor is running only at 70% load. This conclusion was arrived at by assuming that at 100% full load the motor would run at 1786 or 14 RPM slip. Since it is running at 1789 that gives him 4 RPM slip difference which he calculated out to be 71.42% load 4 RPM slip / 14 RPM
slip *100.
How accurate is the nameplate speed on a motor and can the above formaula be used in determining the %load?
Also the measured voltage was 4162 average at 60hZ.

 

[BigLe]

With small slip, load is proportional to square of the voltage and proportional to slip itself. for norminal case: slip =14/1800=0.0077777. Under your case slip=10.5/1800=0.0058333. See if you have the same voltage as rated, 0.0058333/0.007777 = 75%. But when you are running at 4162Volts instead of 4000. you should do 0.75*(4162/4000)^2=0.812. So you are running at 81% load.

 

[electricpete]

Great point by BigLe - slip is roughly proportional to voltage squared.

Also current is roughly inversely proportional to voltage.

Both these factors will push estimated load by current higher than estimated load by slip in the case where voltage below nameplate.

Also consider your speed measurement may be off by perhaps 1 or 2 rpm and even a 0.05hz change in grid frequency would cause a 3 rpm change in syncronous speed of a 4-pole motor. If based on measurement error and grid frequency deviation you have estimated slip speed 2 rpm below actual it would explain your question.

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[Marke]

Hello gronk2003

The bottom line is that both slip and current at rated load are dependent on the voltage supplied to the motor.
It is difficult to determine accurately from the slip or the current, if the motor is operating at rated load if the voltage is not exactly as specified by the manufacturer.
As Pete points out, the frequency of the supply also has a major bearing on the apparent slip.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[aolalde]

The nameplate speed in motors manufactured under NEMA MG1 is not exact. The standard allows +/- 10% slip variations when all the power supply parameters ( Volts, Hertz, form factor, harmonics content) are into the nameplate and standard values.
That means the slip is not the best way to estimate the load power consumption. If the motor parameters are known the best approach is based on the electric input.

HPload = SQRT(3)*kV*Iline*PF*EFF/0.746

PF can be measured but the efficiency (EFF) has to be estimated. As an approximate exercise for the data you provided, I will get the product EFF*PF nominal based on the nameplate data and I will assume their product stays very close, in spite of the voltage deviation and possible frequency deviation too.

EFF*PF = 800*.746 / (1.732*4*105) = 0.8204 (Nameplate)

And the actual load power (HP)

HP = 1.732*4.162* 98.8*0.8204/0.746 = 783.24 HP

The load is approx 97.9%

 

[BigLe]

To get a thorough understanding of the load needs more tests. As aolalde pointed out, power factor and efficiency are to be provided.But from my personal experience.. efficiency may not change much within +/-20% full load but power factor may. If it's hard to measure these two quantities in your facility, a closer look and supply frequency may help. Again, as pete and mark said, voltage/frequency flucturation on the line, rpm readings errors plus inacuracy of the nameplate data make it only an approximation to get the actual number.

But it really doesn't matter for your application as long as the pf and efficiency are within the spec and the motor is running happily. any ways the motor was not designed to match exact customer load, which may subject to change at any time.