Thermal relay calculation for single phase motor
Thermal relay calculation for single phase motor
(OP)
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
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





RE: Thermal relay calculation for single phase motor
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Thermal relay calculation for single phase motor
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.
RE: Thermal relay calculation for single phase motor
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.
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RE: Thermal relay calculation for single phase motor
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Thermal relay calculation for single phase motor
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
RE: Thermal relay calculation for single phase motor
For 230V single phase and neutral the current will be 0,48A * 66% = 0,8A.
Is it correct?
RE: Thermal relay calculation for single phase motor
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.
RE: Thermal relay calculation for single phase motor