Motor not rated for 208 volt 3 phase

[Strathead]

I am sure this has been asked and answered, but I can't find the right combination of key words to find it. I have a 3 phase motor rated 230 volts only with the appropriate amperage listed. The electrical system is 208 volt Wye. I believe you are allowed to run the motor, but what amperage do I use to size heaters?

 

[kwired]

I am sure this has been asked and answered, but I can't find the right combination of key words to find it. I have a 3 phase motor rated 230 volts only with the appropriate amperage listed. The electrical system is 208 volt Wye. I believe you are allowed to run the motor, but what amperage do I use to size heaters?

General rule is it will draw about 10% more at full load output rating then what the 230 volt rating is.


You may or may not want to increase overload setting by that much though depening on things like actual load, cooling, and duty cycles.

If motor is used hard and to capacity for long periods - it probably will fail eventually.

If motor is used for lighter duty then it's rating or not all that often, or even if there is extra cooling (like a fan with motor in it's own air stream) it could last a very long time.

 

[petersonra]

I am sure this has been asked and answered, but I can't find the right combination of key words to find it. I have a 3 phase motor rated 230 volts only with the appropriate amperage listed. The electrical system is 208 volt Wye. I believe you are allowed to run the motor, but what amperage do I use to size heaters?

If it is not rated for 208V, you can still try and run it at 208V but the motor may fail prematurely.

IMO your best options are to get the right motor for the voltage you have or to change the voltage to match the motor.

 

[ActionDave]

I agree with kwired about how well the motor may or may not like being run on 208V. I have seen many 230V motors run fine on 208.

The heaters would still be sized off the nameplate and the service factor would let you know how far you want to push them and still be code compliant.

 

[Ingenieur]

What is it driving? ie the load
is it constant or variable?
Connect it up and run it
measure the current
set the ol's at 115% of the measured or whatever code dictates based on sf
as long as the measured/running < nameplate it should be fine
most motor do not run at full load
I try to size mechanically for 90%

if the motor fla is rated at 230 would not i decrease at 208?
as would shaft power

 

[GoldDigger]

if the motor fla is rated at 230 would not i decrease at 208?
as would shaft power

The relationship is not that clear since a motor is not a resisitive linear load but rather a complex reactive load for which the resistive current at any given voltage rises with the shaft power the motor is delivering (and therefore the slip rate of the motor.)

For any given output (shaft) power the current will be higher at 208 than at 230 (240).
When you reach the maximum rated current, the motor will be delivering its maximum power output, and that power output will be lower at 208 than at 240.

The term FLA is hard to apply directly when running a motor at a lower voltage, since Fully Loaded is not as well defined. Delivering its full rated power at the rated voltage the motor will pull its FLA.
Delivering the maximum, lower, power it can safely deliver at 208 it will still be pulling its FLA.
The ohmic heating inside the motor depends only on the current, not the phase angle or the voltage.
There will also be some heating from magnetic losses, but I am ignoring those for now.

 

[Ingenieur]

I understand machines and reactive power/elements (I think lol)
but this is a machine wound for 1 HP AT 230 operating on 208

1 HP FLA from tables
230 4.2 A
208 4.6

but we are running a 230 rated/wound for 1HP on 208

v = i (R + j X) in which the cos(arctan X/R) is the pf
ignoring R which is basically a property of the windings and is constant
in this case X is primarily inductive = 2 Pi f L
all constants except L which can be consider constant/linear since poles/winding/etc remains the same and v delta is small with the only impact being the current delta

230 = 4.2 X or X = 54.76 ohm
208/54.76 = 3.9 A

to go to 4.6 A which is the rating for 1HP 208
208/4.6 = 45.2 a decrease of 17% of L despite more current
???

http://www.baldor.com/catalog/CNM20134#mobile-tab={"component":"tab","tab":"performance"}

small motor
current and torque drops proportional to v
230 0.86 fla
208 0.78 fla

0.78/0.86 = 0.907
208/230 = 0.904

torque, pull up
230 432 lb-in
208 344 lb-in

344/432 = 0.8 (sq of v ratio) since power~ torque ~ sq of V

should hold for a larger motor??
hmmmmm

 

[kwired]

I understand machines and reactive power/elements (I think lol)
but this is a machine wound for 1 HP AT 230 operating on 208

1 HP FLA from tables
230 4.2 A
208 4.6

but we are running a 230 rated/wound for 1HP on 208

v = i (R + j X) in which the cos(arctan X/R) is the pf
ignoring R which is basically a property of the windings and is constant
in this case X is primarily inductive = 2 Pi f L
all constants except L which can be consider constant/linear since poles/winding/etc remains the same and v delta is small with the only impact being the current delta

230 = 4.2 X or X = 54.76 ohm
208/54.76 = 3.9 A

to go to 4.6 A which is the rating for 1HP 208
208/4.6 = 45.2 a decrease of 17% of L despite more current
???

http://www.baldor.com/catalog/CNM20134#mobile-tab={"component":"tab","tab":"performance"}

small motor
current and torque drops proportional to v
230 0.86 fla
208 0.78 fla

0.78/0.86 = 0.907
208/230 = 0.904

torque, pull up
230 432 lb-in
208 344 lb-in

344/432 = 0.8 (sq of v ratio) since power~ torque ~ sq of V

should hold for a larger motor??
hmmmmm

current and torque may drop if there is a severe drop in voltage, but to drop from 240 to 208 isn't that severe.

Induction motors want to try to reach synchronous speed which is dependent on number of poles in the motor (which is fixed) and the frequency (which is also fixed if not using a VFD and the source Hz is stable - like the public grid is).

So if you run same motor on 208 instead of 240, in order to try to achieve same speed that motor is going to pull more current, as it is still going to try to do the same work. If you have limitations from the source you may limit current, otherwise it will still try to do same work.

Same principle works the other way, if you apply higher voltage current goes down. There is a point where core saturation is reached though and then destruction begins to occur.

 

[Strathead]

Follow up. The motor I referred to is a Baldor. Looking at the nameplate "attached" There is a line further down that indicates "usable at 208V 4.2 amps. Which is FAR greater than the 10% difference in voltage. What an odd way to create the nameplate. All that information should be in the same spot. How many times have you contorted yourself to read a dirty , nearly illegible nameplate on the back side of a motor? I hope the file attached.

 

[kwired]

Follow up. The motor I referred to is a Baldor. Looking at the nameplate "attached" There is a line further down that indicates "usable at 208V 4.2 amps. Which is FAR greater than the 10% difference in voltage. What an odd way to create the nameplate. All that information should be in the same spot. How many times have you contorted yourself to read a dirty , nearly illegible nameplate on the back side of a motor? I hope the file attached.

That motor also has a service factor or 1.4 - may have some impact on the 4.2 rating

It is more of a definite purpose motor though. Most general use motors are going to have a SF of 1.15 or 1.25.

 

[Strathead]

That motor also has a service factor or 1.4 - may have some impact on the 4.2 rating

It is more of a definite purpose motor though. Most general use motors are going to have a SF of 1.15 or 1.25.

So I only have to do my research here and I don't size heaters often...

Normally with a 1.15 SF is it correct that you use the actual amperage and size the heater? It accounts for the SF on its own. And if that assumption is correct, how do you size heaters for motors with a different SF?

 

[kwired]

So I only have to do my research here and I don't size heaters often...

Normally with a 1.15 SF is it correct that you use the actual amperage and size the heater? It accounts for the SF on its own. And if that assumption is correct, how do you size heaters for motors with a different SF?

That can get complex as the selection charts usually already have SF factored into them. You need to find some instructions that are not always on the selection chart itself. Many times if you have a 1.0 SF motor they tell you to multiply motor nameplate current by .90 then find that value in the chart.

In most cases if I had the motor you showed us I would size the overload as though it were a 1.15 SF motor, unless it is pushing that value from the start.

I guess NEC wise it still gets same overload protection anyway - this taken from 430.32(A)(1):

(1) Separate Overload Device.


A separate overload device that is responsive to motor current. This device shall be selected to trip or shall be rated at no more than the following percent of the motor nameplate full-load current rating:
Motors with a marked service factor 1.15 or greater 125%
Motors with a marked temperature rise 40°C or less 125%
All other motors 115%

 

[jap]

If it is not rated for 208V, you can still try and run it at 208V but the motor may fail prematurely.

IMO your best options are to get the right motor for the voltage you have or to change the voltage to match the motor.

I agree.

JAP>

 

[ActionDave]

So I only have to do my research here and I don't size heaters often...

Normally with a 1.15 SF is it correct that you use the actual amperage and size the heater? It accounts for the SF on its own. And if that assumption is correct, how do you size heaters for motors with a different SF?

Are you using actual old fashioned heaters or electronic overloads?

For heaters I would go 115% of the nameplate and no higher. Electronic overloads usually are set at nameplate but not always.

 

[Strathead]

Are you using actual old fashioned heaters or electronic overloads?

For heaters I would go 115% of the nameplate and no higher. Electronic overloads usually are set at nameplate but not always.

heaters because they are manual motor starters.

 

[ActionDave]

heaters because they are manual motor starters.

For what? Just curious. That looks like it must be a pump.

 

[Strathead]

For what? Just curious. That looks like it must be a pump.

It is for a "pad pump" for a greenhouse "roll up curtain" basically it circulates water that falls down through a perforated curtain on one end of the greenhouse while several wall mount fans draw air through it form the other end of the greenhouse.

 

[GoldDigger]

It is for a "pad pump" for a greenhouse "roll up curtain" basically it circulates water that falls down through a perforated curtain on one end of the greenhouse while several wall mount fans draw air through it form the other end of the greenhouse.

In that case it is likely that the pump/motor is running at less than full load power. If there is any problem with the motor overheating at 208V the load can be reduced by simply restricting the flow of water through the system.

 

[ActionDave]

It is for a "pad pump" for a greenhouse "roll up curtain" basically it circulates water that falls down through a perforated curtain on one end of the greenhouse while several wall mount fans draw air through it form the other end of the greenhouse.

I would use 115%/of the 230V nameplate amps and whatever size the chart of your brand of heaters says for this application.

 

[Ingenieur]

current and torque may drop if there is a severe drop in voltage, but to drop from 240 to 208 isn't that severe.

Induction motors want to try to reach synchronous speed which is dependent on number of poles in the motor (which is fixed) and the frequency (which is also fixed if not using a VFD and the source Hz is stable - like the public grid is).

So if you run same motor on 208 instead of 240, in order to try to achieve same speed that motor is going to pull more current, as it is still going to try to do the same work. If you have limitations from the source you may limit current, otherwise it will still try to do same work.

Same principle works the other way, if you apply higher voltage current goes down. There is a point where core saturation is reached though and then destruction begins to occur.

as you pointed out speed only depends on poles and freq
rpm = 120 x freq/poles iirc
no v or i
torque depends on v, i (or power) and slip
power = torque x n where n = 2 Pi rev/sec

at 208 it will turn the same speed but produce less torque
increase slip and current then torque goes up

but I've been wrong before, my wife will vouch