Hello,

Maybe someone might help me:

I try to calculate a pump motor consumption in a cement plant. I can have the current by a software which measured it all the time: Average 21A.

Phase-phase voltage : 4160V (On the motor nameplate)
Power factor: 0.8

If I use this formula:
W = 1.732*21*4160*0.8 = 121 kWh

Is it correct?

Thank you for your help.

The formula looks correct to me.

P = sqrt(3)*I*VLL*p.f.
(assuming you don't have harmonics)

The numbers that you plug in deserve careful attention.

Actual voltage may vary from nameplate.

How did you come up with power factor of 0.8?

Note that when the motor when the motor operates at low load, the power factor is much lower than at full load. If power factor is just a guess (and you don't have any specific info from manufacturer), we might be able to take a guess about the power factor if you tell us more nameplate info... particularly horsepower and speed.

Another estimation:
shaft power = (slip / nominal slip) * nominal power rating.
Nominal values you find on nameplate.
slip = actual speed - synchronous speed
Arie Mol

Thanks.

The power factor is a guess. I'm waiting for info from the constructor.

Nameplate:
350HP
1186 rpm
4160V
46A

Measured Speed: 1180 rpm
Synchronous speed (6 poles): 1200 rpm

If the speed measurement is correct, then with an actual speed of 1180 rpm the motor would be overloaded 20/14 = 1.4 times, the amps would be much more than 46 Amps and the thermal overload relay would trip the motor.
If the amps reading of 21 Amps is correct then the actual speed should be halfway 1186 and 1200 rpm.
Or the nameplate 1186 rpm data is incorrect and should be well below 1180 provided 21 Amps / 1180 rpm reading is correct.
Pretty nice puzzle!

Of course, you meant to give the units as "kw" rather than "kwh"? And the division by "1000" is implied?

Power Rate : 0.853
Motor efficiency : 0.95

I will search for the speed.

Thanks.

From your nameplate data, knowing FLA, Voltage=4kv, (I assume the system is 4160,not the motor), horspower, power factor, the only unknown is efficiency and we can calculate it... comes out pretty close to 85%. I guess that's what you meant by "Power Rate"? Wow. 85% is a pretty darned low full- load efficiency for a 350hp motor. (is it a really old motor?)

Attached is be a rough graph of current vs load (the yellow graph) with some simplifying assumptions - nelgect leakage reactance and neglect losses. This leaves us with a magnetizing branch (reactive) and a load branch (real). You can see as you increase load to 70% and beyond, the current is almost linear with load. Down at low loads, it is far from linear, and also the errors associated with the assumed power factor and simplifying assumptions are bigger at low loads. From this graph we might estimate 21A corresponds to 35% load (but again this is the part of the graph that is not particularly accurate).

In general I trust current measurements more than speed measurements based on:

* Motor nameplate speed is allowed to deviate by up to 5 rpm
* actual slip vs power characteristics depend on rotor resistance which varies with temperautre
* depending on instrumentation used, errors in speed measurement can be significant. For example if you try to estimate it from a bin width 50cpm FFT.
* Frequency can vary from the nominal value- moreso overseas than in the US.

But it is good to have two different ways to estimate power for comparison. I agree with Arie in this case the comparison leads to a lot of questions.

Some possibilities to check to investigate
* - how accurate is your speed measurement.
* - Are you sure of your current measurement?
* - check for current balance
* - very very remote possibility - rotor bar problem increases rotor ressitance and causes lower speed for given load.
* Do you have a vfd?
* Do you have any other parameters that you might correlate with load to break the tie. That could include process parameters or motor winding temperature although neither of these is that exact.
* Maybe the actual loading changed between when the current was recorded and the speed was recorded?

This message has been edited. Last edited by: electricpete,

CurrentVsLoadR1.xls (17 KB, 18 downloads)

I wanted to say "power factor = 0.85" in my last message, not "power rate". The efficiency is 0.95. Moreover, an electrician told me that the current is 41 Amps, not 21.

I'm sorry for these mistakes.

Tomorrow I will have good speed measurement.

And I don't understand why the measured speed has to be higher than the nameplate (nominal) speed ?

Ok, PF = 0.85, Eff=0.95 makes more sense than PF=0.95, Eff = 0.85.
(either one would give FLA ~ 46A for 450hp 4kv motor)

I updated my graph with the new power factor. 21A would have been close to a no-load condition.

41A is close enough to full load amps that you can use a linear approximation.
Load level ~ 41A / 46A ~ 90%

quote:

And I don't understand why the measured speed has to be higher than the nameplate (nominal) speed ?

Load is roughly proportional to slip speed (sync speed minus actual speed)

P / Pnameplate ~ Slip / Snameplate
P /350hp = (1200 - rpm) / (1200-1186) = (1200 - rpm) / (14pm)

RPM = 1200 => P/350 = (1200 - 1200) / (14) ~ 0 (no load)
RPM = 1196.5 => P /350hp = (1200 - 1196.5) / (14) ~ 25% of full load
RPM = 1193 => P /350hp = (1200 - 1193) / (14) ~ 50% of full load
RPM = 1189.5 => P /350hp = (1200 - 1189.6) / (14) ~ 75% of full load
RPM = 1186 => P /350hp = (1200 - 1186) / (14) ~ 100% of full load
RPM = 1182.5 => P /350hp = (1200 - 1182.5) / (14) ~ 125% of full load

Again there are some cautions mentioned above regarding using speed to estimate load.

CurrentVsLoadR2.xls (18 KB, 15 downloads)

The measured speed is 1194 RPM.

I suppose I have to calculate the energy losses now.

Have you an estimate (%) for a motor like that?

I'm not sure I understand the question.

The motor rating is in output (mechanical) power.

The electrical formula is based on input power times efficiency

(efficiency = output / input)

You stated the efficiency ~ 0.95 which would be at full load. That would change slightly with load but not much at 41A which is ~ 90% load.

At 41A, Your losses would probably be around (1-0.95) * 90% * 350 HP. ~ 16HP ~ 12kw

This assumes you don't have severe voltage unbalance or other unusual conditions.

Hello everybody,

I have another problem:

This motor drives a screw (direct connection) which compress a powder. I would like to know the (nominal, current and minimum) torque of the screw, but I don’t know if I can with the data I have. To sum up quickly:

Nominal:
350 HP
46A
4160 V
1186 RPM
Power rate: 0.85
Efficiency: 0.95

Measured:
41 A (88% load)
1194 RPM
Power rate: 0.844
Efficiency: 0.951

I think the formula is: Power (W) = Speed (rad/s) * Torque (N.m) but I’m not sure.

With it I find:
Nominal Torque = 2 101 N.m
Current Torque = 1 841 N.m
Minimum Torque = ?
I have no idea if the values are corrects and not too big.

It is not really electricity, but how could I know the minimum torque to compress the powder?

Thanks again!

I double-checked your calculation for nominal (nameplate) torque and came up with the same result you did: 2100 N-m.

Minimum torque?.... Motor will match whatever torque is demanded by the driven equipment at the operating speed (up to rated torque continuously and breakdown torque momentarily). The minimum torque required by your particular driven equipment?.... I don't think we can tell you that.