Thermal relay calculation for single phase motor 2

[kallileo]

Hi,

I have a 3ph motor with the following specs:
230/400V3~ 50Hz
0.48/0.28A
cosΦ = 0,8

Some times we use permanent capacitor and use single phase voltage to run the motor.
Is it possible to calculate the current without measuring it in order to set the thermay relay accordingly?

Thanks
Leonidas

 

[waross]

The nameplate tells you what current the motor windings may carry without overloading. There may be less winding heating when all three windings are not fully energized. However when a stator is not energized equally there may be greater than normal rotor currents and rotor heating.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

As waross said the motor phase winding will not be -usually-symmetrically loaded. The phase[voltage] will be symmetrically only
if motor p.f.=0.5 .In order to create a symmetrical system for any pf an autotransformer [or transformer] could be used.
But anyway if pf will change-starting, no-load and so on-the current will not be symmetrical.

 

[jraef]

So if I get this right, you are using a 3 phase motor and running it from single phase source by starting with a phase shifting capacitor circuit, right?

First off, your motor, even with the capacitor to make it spin, cannot be run at full capacity, I hope you understand that. As the others mentioned, doing so creates a sever current imnbalance that creates negative sequence current in the rotor, which in turn creates torque that will oppose the normal torque direction, essentially making the motor fight itself. This results in higher than normal heating of the motor. If you want to protect it from damage, I would set the OL pick-up point at 58% of the nameplate FLC, 58% representing the equivalent lowering of the motor capability by the Sq. Rt. of 3 because the effective voltage is single phase, not 3 phase.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[waross]

Feeding two phases instead of three already results in a reduction to 66% How about an 85% setting?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

Since Rudolf Richter [Electrical Machines vol IV The Induction Machines ch.C 2c] Polyphase Machine with Condenser
consider -in case of single phase system supply and the same voltage-only 60% from the three phase rated power.
The pf is improved, indeed. So, the current would be:
Let's say efficiency=0.8[in both cases]
P3=sqrt(3)*0.48*230*.8^2/1000=0.122 kw[0.164 hp]
P1=0.6*P3=0.122*.6=0.073 KW
new pf=0.96
I1=0.073/0.96/.8/0.230=0.413 A
I1=0.86*I3

 

[kallileo]

So the FLC for 230V3~ is 0,48A as is stated on the nameplate.
For 230V single phase and neutral the current will be 0,48A * 66% = 0,8A.
Is it correct?

 

[7anoter4]

The problem is more complex a bit.
The motor has to develop the required power-the required torque at -more or less-the same rpm.
In order to overcome the current unbalance you have to reduce the required power to approx. 60% and the average current to 86% from rated.
The individual phase current could be 15-20% more [or less].
If you'll maintain the required torque it is possible to stop the motor as the maximum torque will be reduced by volt^2 that means at 1/3
from the rated break-down torque.
The slip will be -anyway-more than at rated voltage.

 

[1hagar]

how do you do the connections to this motor when running it on 230 volt? Is there any difference in speed and heat generated? Just had a problem the other day when the neutral was connected as a phase so motor ran with 2 phases and a neutral.(incorrect connection on supply side)