COPPER LOSSES OF MOTOR

[MIK2000]

Dear Sirs
I have some question about to determine losses of a small three-phase induction motors with sqiruell cage rotor. Motor is 2kw - 2 poles, nominal current is 4.8A, Y connection, 400V/50 Hz.


I locked the rotor and voltage is zero, than the voltage is increased, till nominal current obtain.

question : at this point, Does the input power(Pk) taken from supply give to me the copper losses (rotor and stator)?
As i know yes it does. (Pk is 338W -three-phase)

Now i need to seperate rotor and stator copper losses i mean how many power does belong to rotor and stator?

Stator opper losses =3 * Io^2 * R (cold)
=3 * 2,2^2 * 2,5
=36,1 W
R is resistance of each phase , and i mesaured it by using an ohmmeter.

So Copper losses = rotor copper loss + stator copper loss
338w = rotor copper losses +36,1
rotor copper losses=338-36,1
=310,7w

Question: There is something wrong Rotor copper losses is to much also i calculating rotor copper losses by using formula = slip * Rotor input power
=(slip) * (Input power-stator losses)
=(0.037) *(2688- core losses - stator copper loss)
=(0.037)*(2688-128-36,1)
=93w is rotor copper losses
93+36,1= 129W total copper losses

Does this value suppose to equal Pk? but it does not.

Do you have an idea where am i doing a mistake by calculating copper losses?

Thank you very much

 

[ScottyUK]

Your formula (slip) * (Input power-stator losses) includes windage and rotor iron losses, not just copper losses. Windage is probably higher than you think, especially if the rotor has a fan impeller cast into the ends.


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If we learn from our mistakes I'm getting a great education!

 

[MIK2000]

Thank you ScottUK
So you mean that Stator losses in that formula shuold be included windage (friction+wind), and rotor iron losses ,
but those losses all about rotor? in this formula i need to find stator losses (stator iron + stator copper ) Am i wrong?

 

[edison123]

Your locked rotor test loss is high @ 17% of the motor rated power. The power factor in locked rotor (and in no-load run) is very small in the range of 0.05 (to 0.2). You need special metering to measure these losses correctly.

Also, you need to use a Kelvin' Bridge or micro-ohmmeter to measure the winding resistance correctly.

 

[MikiBg]

What is Io in stator copper losses calculation? It should be nominal current (4,8A), as you described locked rotor test. Then you'll have 172,8W stator, and 165,2W rotor copper losses.
How did u get core losses?

 

[MIK2000]

Thank you edison123,
power factor is at no load is 0,12 (400v) and locked rotor pf= 0,62 (68v) i think they are not very small. ohmmeter is already micro ohmmeter i didnt write down here 0,001 . But Pk is high you are right. I test locked rotor at 50Hz does it affect the Pk?

 

[edison123]

As MikiBg says, the stator copper loss should be calculated for a current of 4.8 A due to wye connection. Stator resistance should be per phase. If you measured stator resistance across lines (say U to V), then per phase resistance is half of that.

The pf during locked rotor test is not that high. You need to recheck.

50 hz is correct frequency to do the test.

What does the OEM test certificate say about the losses ?

 

[MIK2000]

I have found one my mistake by calculating stator copper losses, Io is no-load current and i must take In (nominal current)as you say.
Ressistance is per phase, no problem here.

Iron losses (Pfe) =Po(no-load input power)-Pcu(172,8w)-Pwf(friction and windage)

Pfe=209-172,8-28
Pfe=8,2W (is it normal?)

Pfw=friction and windage losses. I get Pfw by using no-load test. voltage is applied from 400v to 100V, and i draw a curve power and voltage than, i try to get power at zero voltage by copmlating the curve . So the power at zero voltage is friction and windage losses.

 

[MikiBg]

During no load test u have Io as stator current, therefore 36,1W stator copper loss and Pfe=144,9W according to your data.

 

[MIK2000]

Thank you MikiBG, i confused stator copper losses, when i calculate Pfe i must take Io in stator copper losses, but when i calculate total stator cooper losses In must be taken.

 

[jraef]

I got this information from EPRI a few years ago from an paper discussing improvements in designing energy efficient motors.

Of the TOTAL losses in a motor, the division works out like this;
Stator copper losses (35%).
Rotor copper losses (25%).
Core losses (25%) from hysteresis and eddy currents in the lamentations.
Windage and bearing losses (5%).
The last 10% of losses are stray load losses caused by leakage flux induced by load current fluctuations.

I though that might prove useful for you once your are able to determine your true actual losses.


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