current at 50Hz on modified 60Hz motor

[foretmark]

I am taking a single phase 5HP 60Hz motor running at 220v and changing the centrifugal switch spring so it will start at 50Hz (for outside of US). I'm told by the manufacturer that this is all that need be done.
The V/Hz has gone up from 3.7 to 4.4, but I cannot drop the voltage down any. It will not be running 100% duty cycle, although the duty cycle is unknown. I do know that worse case it would be around 50%.
What is the best way to accurately calculate the FLA at 50Hz? The FLA of this motor at 220v is around 26A. Could multiplying this by 1.2 (60/50) be an accurate way to determine a new FLA? Seems too simple to me.

 

[waross]

No.
The full load amps stays the same. The limit of full load amps is the I²R heating of the windings.
You will be running at 120% voltage. That may put you into magnetic saturation of the iron. The current will rise rapidly with only a small increase in voltage at the saturation point. The lifetime of a motor with a magnetically saturated stator may be measured in minutes. (In severe cases in seconds.)
Try a small auto transformer to buck the voltage down 20%.

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

 

[foretmark]

This may not be the place to ask, but is it code-safe to throw an autotransformer inline with a motor circuit like that? I don't want to drop the voltage for the control circuit as well...

Incidentally, and I apologize if this has been mentioned, I didn't see it in FAQ....but what is the acceptable deviation on a V/Hz ratio, or is there such a thing?

Also, you mention that the FLA stays the same...I meant to say actual current draw. I assume it would increase due to the 'lower' voltage.

 

[waross]

Current increase may be estimated as 6/5 for a constant load.
Over the optimum V/H ratio quickly puts you into saturation.
Under the V/H ratio and torque drops off quickly. V/H is an optimum design value for each motor.
If your controls are 50Hz rated run at rated voltage. If the frequency of control coils is being changed run from the auto transformer to maintain the V/H ratio.
I have used a lot of auto transformers to convert 480 VAC to 600 VAC and 600 VAC to 480 VAC. Be careful that you have your protection installed correctly.

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

 

[ScottyUK]

Rated FLA of the motor will drop because the shaft driven fan will move much less air at the lower speed and the cooling will be reduced, unless you're planning to replace the fan too. You could always do the furriners a favour and install an IEC motor designed for 50Hz and in an IEC frame size.


----------------------------------

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

 

[foretmark]

furriners...had to run that one through a few times..

I think that's the plan....of course the task of testing it here in the US reverses my problem!

 

[itsmoked]

foretmark; Your last comment is the one to go with. Test it. Regardless of the voltage and frequency if the current remains below the FLA you should be good to go. As an additional check use a temp gun on the motor after it has reached thermal equilibrium and see if it is reasonably below the allowed running temperature.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[ScottyUK]

Hi Keith,

Rated FLA relies on rated cooling air flow which in turns relaies on the fan spinning at rated speed: you ain't got rated cooling flow at 5/6 rated speed, you've got about 70% of rated flow. I forgot to mention in my previous post that the internal fan on the rotor has the same problem reduced speed too. The chances are the motor manufacturer will have a derating curve intended for VSD applications, but likely you could use the curve as the basis for a guess at the true derating. The volt-Hz problem will make the curve a little optimistic. Or maybe a lot optimistic, depending on how near the margins the design is to start with.


----------------------------------

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

 

[foretmark]

ok, we got 50Hz motor....however, being in the US, they're testing at 60Hz and they say the current increases as pressure builds....I believe at the goal pressure (when it stopped appropriately per the system), the current reading was roughly 60% higher than FLA. (fla ~13, acutal ~23). I'm told the time between starting and stopping was about 10 seconds.

So this would be due to the core being saturated? Other than measuring the winding resistance, is there any way to verify the integrity of the windings?

 

[itsmoked]

This is confusing.. Please state again why you have.


A 50Hz motor you are running with 60Hz and you are having over current? (this is what I'm reading and the opposite of the original question)

Keith Cress
kcress -

http://www.flaminsystems.com

 

[waross]

I think I may have been a little confusing in my second answer.

Current increase may be estimated as 6/5 for a constant load.

This applies to a motor developing the same horse power at reduced frequency and voltage. IT ONLY APPLIES to a motor that is less than 83% loaded at the higher voltage and frequency.
The nameplate current is the maximum current allowed, regardless of frequency and/or voltage.



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

 

[ScottyUK]

"The nameplate current is the maximum current allowed, regardless of frequency and/or voltage."

We're gonna have to disagree on this Bill! The current is an indirect measure of the temperature of the motor windings, which is the parameter which we are really concerned about. The current rating is valid only at stated conditions: ambient temperature, nominal elevation (or not exceeding some elevation), running at rated speed, etc, etc. All these factors affect the ability of the motor to reject heat to the atmosphere, and if any of them change then the current rating also changes. Slowing down the cooling fan is definitely going to change the current rating.


----------------------------------

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

 

[foretmark]

ok, sorry for the confusion. My original question was about using a 60Hz motor with a new starting spring in the start cap circuit at 50Hz. I did not like the sound of that, and said we needed to get a motor designed to run at 50Hz....and we did.

Obviously, we don't have 50Hz, so we tested it at 60Hz. (nameplate 220v 50Hz FLA 12.4, Tested 208(211)v 60Hz). The design is such that the motor kicks on when the pressure drops to a setpoint...I noticed that when the motor kicked on, it immediately hit it's FLA (give or take a few amps-I could only measure at incoming to unit, not just motor) and rose roughly 60% before shutting off (per design).

I guess my question here, is my V/Hz is 3.5. The spec V/Hz is 4.4. Could the fact that the motor is pulling more current be a direct result of the lower ratio, because of the torque being lower and will be more if not completely within spec with the proper power supply?

 

[ScottyUK]

Assuming this motor is conncted to its load, the load is being driven at 20% overspeed and that's almost certainly where the extra energy is going. The lower V/Hz would tend to reduce heating in the stator iron, although the iron losses from hysteresis would tend to increase it.


----------------------------------

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

 

[itsmoked]

Thanks for the re-state that cleared things up well.

I'm with Scotty there unless you have done nothing mechanical to compensate for the standard nameplate speed of a 50Hz motor.

Example: If your original motor was spec'd to run 3450RPM (60Hz) the same motor type motor made for 50Hz operation would run at about 5/6 x 3450 or 2875RPM. Knowing this you re-geared the machine for this motor speed. Now you actually run the motor at 60Hz you would be feeding the gearing at 3450RPM again. This would represent running the original machine at something like 4100RPM.

I would expect a serious increase in current as HP = T x speed. HP is delivered by the V x I. V isn't going anywhere so I must go up.
To add to this problem torque requirements often increase with a machine's speed.

Example 2: By running this motor name plated to be the same HP as your original 60Hz motor you are actually converting it to a 6/5ths higher HP! Of course it will read a higher current draw, it has to get the added HP from somewhere.

Furthermore dropping its voltage down from 220V to 208V exacerbates this by requiring even more I to make up the voltage loss in the HP~VxI equation.

Keith Cress
kcress -

http://www.flaminsystems.com