Supply voltage effects on motor current

[noel0589]

Hi all,

I'm trying to fill out a table for 3 phase ac induction motors.
A 50HP motor has ratings of 148 full load amps and 808 locked rotor current at 200 volts.
If power supply is 208 volts, I use a proportional equation which results in a higher locked rotor current (about 840)due to increased voltage. Also, for full load amps i use the equation:
current = (HP*746)/(1.73*Appliedvoltage*Efficiency*powerfactor)
This says full load amps will drop (to about 142A) due to an increased voltage.
However, I have read that applying increased voltage above ratings can result in saturation of the magnetizing part which would result in an increase of drawn current, not a decrease as is stated in the equation. Is it because this saturation occurs when supplied voltage is well above the rated voltage?

Thanks!

 

[buzzp]

As a rule of thumb and a generic explanation of what you are seeing:
Above about 5% of nameplate voltage, the current will drop (assuming fully loaded), above 5% nameplate voltage, the current draw increases above full load amps. This is why you see a small reduction in current when increasing the applied voltage to 208 (only 4% increase in nameplate voltage).
I am sure others will give some detailed explanation of what is happening within the motor to make this happen.

 

[aolalde]

noel0589

Your equation is valid for any condition, but the Power factor (PF) and Efficiency (EFF) both change with the magnetic saturation and motor losses.
For instance if the voltage is increased 10% the flux densities increase 10% too, but depending on the saturation of the magnetic circuit the magnetizing current will increase more than 10%, reducing the power factor. By the other hand the magnetic losses will increase approx with the squared ratio of the flux density which most probably will reduce some points of the efficiency too.

Based on a generic curve published by EASA, "Effect of Voltage Variation on Induction Motor Characteristics", with 10 % voltage increase the full load current will increase around 3.5%, the PF will be reduced 10% and the EFF will reduce 1%. Note that these are average generic performances and could change from motor to motor.

 

[buzzp]

Yep, thats where I got my info. If you look at the curve you will see the current drops slightly above nameplate voltage then it rises above about 5% nameplate voltage.

 

[noel0589]

Thanks for responses,

I've seen the voltage variation vs. motor performance curves.
One thing is a little confusing to me, however.
For a 230/460 V motor, which are the same motor,
when voltage is increased from 230 to 460, there is a drop in locked rotor and full load current. However, according to voltage variation graphs, if voltage is increased 10 percent, Locked rotor and full load currents increase. I understand this probably has to do with motor windings being in series or parallel depending on the application. However, the same goes when looking at 200V motors vs. 230V motors. The 200v motor draws higher currents.
In general, the higher the voltage rating the lower the currents drawn. But the higher the voltage on a particular motor, the higher the currents drawn. Can someone explain this or correct me?
Many Thanks!

 

[Skogsgurra]

noel,

When you think about your last post: "In general, the higher the voltage rating the lower the currents drawn. But the higher the voltage on a particular motor, the higher the currents drawn. Can someone explain this or correct me?" you will probably find out why this is so.

No? Let's say that you have a 1 HP motor that is wound for one voltage (U). This motor consumes a certain current (I). Efficiency (eff) and power factor (PF) are given at full load. The power will then be (for a three-phase motor):

power = U x I x sqrt(3) x eff x PF

Now, for a 1 HP motor, power is constant even if you change the winding. Let's say you have a 200 V motor and then change the winding to work with 230 V (more turns and thinner wire). The U in the above equation goes up (from 200 to 230 V) and then the I required to keep power constant is reduced to Ix200/230.

It is quite natural - the P=UxI formula in disguise - actually.

The other effect (having one motor and changing the applied voltage) is something totally different. And those effects have been discussed in earlier postings.

 

[aolalde]

noel0589:

"In general, the higher the voltage rating the lower the currents drawn. But the higher the voltage on a particular motor, the higher the currents drawn. Can someone explain this or correct me?"

Please read my post above.

 

[noel0589]

I have been looking around online and have come across a lot of contradicting information concerning motor characteristics under variations in voltage.
According to G.E. if voltage is increased say 10% on a particular motor power factor will reduce by 4%, and full load current will decrease by 6-7%.
But I have also seen reports describing an increase of full load current under a voltage increase of 10%.
I am leaning towards EASA reports but my boss believes in a decrease of FLA under increasing voltage and with all the contradicting information I am not exactly sure how to prove one right or wrong without actual testing, which I don't have the means to. Can anyone add any advice?
Thanks!

 

[buzzp]

Not sure what proof to give you other than doing the tests himself.
In general, and as previously stated, if a motor is fully loaded and the supply voltage is increased up to say 3%, the running current will drop slightly. Anything above about 5% of nameplate voltage will cause an increase in current. I believe this is due to the saturation of the core but someone else can tell you why. In any case, this will happen with any squirrel cage induction motor which is FULLY loaded.

 

[DickDV]

I have found a large variation between manufacturers on this subject.

Reliance, for example, typically builds motors with a relatively large capacity for utilizing higher voltages without higher currents, in my experience.

On the other hand, I have had some experience with Weg motors under the same circumstances and it seems that the slightest increase in voltage over nameplate increased current rather than decreasing it.

I'm no expert on what goes on inside the motor but I suspect that the difference is in the magnetic circuit design.

 

[aolalde]

I see it this simple: IL = HP*.746/ (1.732*kV*EFF*PF); for whatever condition.

What changes in the motor? Only the EFF and the PF! (And the voltage or frequency by the power supply)

EFF is function of the looses (Stator I^2*R, Rotor I^2*R, Core Loss, Stray load loss, Friction and windage loss)

PF is function of the magnetizing current Im (magnetic circuit saturation) and the phase current Iw (power delivered (HP) at the shaft + losses).
PF = Iw / SQRT (Iw^2 + Im^2)

PF and EFF change dynamically for each motor condition; voltage, load, temperature, frequency, wave form, etc. And remember “ All motors are created different”.