If a motor runs on 230V at 47A, how many amps will it draw at 208V?

[kwired]

I'm sure the essential questions are "How much circuit capacity needs to be provided?" and "What's the appropriate OCPD size?"
If the motor nameplate doesn't specify the full-load current at 200 volts, I'd base the motor-starter overload setting on the 230-volt full-load current, and specify at least 125% of that (60 amps) for the circuit ampacity.

If this is a new installation, (and maybe even if it's not -- Do the Math) I would return the 230-volt motor and install a 200-volt motor. This is a big enough consumer that its operating efficiency shouldn't be ignored.

Might depend on what it is used for. As has been mentioned for centrifugal fans and pumps if it happens to slow down, the load demand from the fan/pump will decrease as well.

Then when it comes to overall energy efficiency something that runs 24/7 at lesser efficiency cost more in energy losses than something that only gets occasional use. That occasional use could be a few minutes a day, couple hours but only once a month, or completely random and maybe sits for a year or more but runs for several hours when it is used.

 

[Besoeker3]

If it''s the number he has to work with for 230/240v, the corresponding number at 200/208v would be use in whatever calculations he's doing.

Possibly so. But if the FLC is greater/much greater than 47A , a reduction in voltage would likely result in a reduction in current drawn.

 

[junkhound]

First, the manufacturer said it will run fine at the lower voltage. Next, I have tried the "is / of" ratio formula and that doesn't seem to work (the result is a lower amperage which I know to be false). I also tried the "percentage" formula which I believe gave me the same answer (45A and some change). Will anyone share the formula or at least how to derive it? Thanks!

Since you have gotten so many non-consistent answers, to start with there is NOT 'the formula', there are sets of formula since you need to take the load into account.

Here is a simple experiment for you to perform. on a small motor if you wish.
1. non-synchronous induction motor and fan with a variac - at 240 Vac the fan will cause the motor to pull X amps. As you decrease the voltage to you will see the fan slow down due to increase in slip and the current will DROP since fan power is related to cube of speed.
2. same motor, connect to a dynamometer (say a generator with load) at CONSTANT dyno power. As you decrease the voltage, the current will INCREASE near proportionally.

Put a tach on the motor and a wattmeter in front of your variac and it will help you understand the relationships.

PS: did not look, but probably a Utube video on similar type experiment?

 

[Besoeker3]

1. non-synchronous induction motor and fan with a variac - at 240 Vac the fan will cause the motor to pull X amps. As you decrease the voltage to you will see the fan slow down due to increase in slip and the current will DROP since fan power is related to cube of speed.

With the same frequency there may be a slight drop in speed and an attendant increase in slip. More slip generally means more current.
Probably rather more than the cube law reduction due to slight reduction in speed.

 

[kwired]

Since you have gotten so many non-consistent answers, to start with there is NOT 'the formula', there are sets of formula since you need to take the load into account.

Here is a simple experiment for you to perform. on a small motor if you wish.
1. non-synchronous induction motor and fan with a variac - at 240 Vac the fan will cause the motor to pull X amps. As you decrease the voltage to you will see the fan slow down due to increase in slip and the current will DROP since fan power is related to cube of speed.
2. same motor, connect to a dynamometer (say a generator with load) at CONSTANT dyno power. As you decrease the voltage, the current will INCREASE near proportionally.

Put a tach on the motor and a wattmeter in front of your variac and it will help you understand the relationships.

PS: did not look, but probably a Utube video on similar type experiment?

With the same frequency there may be a slight drop in speed and an attendant increase in slip. More slip generally means more current.
Probably rather more than the cube law reduction due to slight reduction in speed.

Depends on motor design also. PSC and shaded pole motors have low torque rating and are easily speed controlled simply by lowering voltage. In those cases applying 208 instead of 240 likely does slow the motor down and if driving a centrifugal fan or pump lowers the mechanical load as well.

Most other induction motors have enough torque they are going to try to maintain rated speed, if voltage drops then it will take more current to get to speed. With centrifugal fans and pumps, mechanical load stays the same if speed stays the same.

If the supply source can't deliver enough current then you may see increased slip.

 

[Besoeker3]

Depends on motor design also. PSC and shaded pole motors have low torque rating and are easily speed controlled simply by lowering voltage. In those cases applying 208 instead of 240 likely does slow the motor down and if driving a centrifugal fan or pump lowers the mechanical load as well.

Most other induction motors have enough torque they are going to try to maintain rated speed, if voltage drops then it will take more current to get to speed. With centrifugal fans and pumps, mechanical load stays the same if speed stays the same.

Depends on how heavily the motors were loaded. On light loads the current may well reduce.

 

[kwired]

Depends on how heavily the motors were loaded. On light loads the current may well reduce.

If load remains the same and speed remains the same supplying it with 208 instead of 240 should result in more current being drawn at 208 vs 240.

In most such cases the current will be approximately 10% higher if the mechanical load remains the same.

 

[Besoeker3]

If load remains the same and speed remains the same supplying it with 208 instead of 240 should result in more current being drawn at 208 vs 240.

In most such cases the current will be approximately 10% higher if the mechanical load remains the same.

What if the motor was close to idling as it sometimes is on cyclic loads such as accelerating a bench grinder after it has got up to speed?

 

[kwired]

What if the motor was close to idling as it sometimes is on cyclic loads such as accelerating a bench grinder after it has got up to speed?

I'd guess average current is 10% higher at 208 than at 240.
Either way it is probably well below full load rated current if it is at/near idle, and the power factor is probably low and a good portion of current is reactive current.

 

[drcampbell]

What if the motor was close to idling ... ?

Who cares? The circuit supplying it needs to be designed & installed based on full-load current.

 

[Besoeker3]

Who cares?

The OP who asked the question.

 

[kwired]

Who cares? The circuit supplying it needs to be designed & installed based on full-load current.

The OP who asked the question.

We don't really know if OP's 47 amp motor is full load rated for 47 amps or if that is what it typically draws when in use. I don't think it matters a lot other than for a couple specific motor types, and I doubt you find many shaded pole or PSC motors that draw 47 amps either so those are sort of out of the question anyway.

 

[junkhound]

Here is a 230V PSC 184 frame motor draws 49 A driving 4000psi 4GPM pressure washer.

Nameplate of only 24.6A, 5 HP. Does not overheat.

Does have supplemental cooling, just thought to throw this out there to confuse matters

 

[EEEC]

Wow! There is a lot of information to process... Most of which goes well over my head, but I am "getting the drift". Thank you!

A friend is asking for my help in upgrading his Brewery. The piece of equipment in question is a chiller that the mfg said will run on 208V 3ph with no issue. I thought we might need to use a buck/booster... The spec sheet says 230V 3ph: FLA 41, MCA 47, MOC 74. The purpose for my question was indeed for calculating his entire load increase to compare to the existing service amperage. Purely for planning at this stage...

 

[LarryFine]

For planning, adding 10% to each amperage value should suffice: 45, 52, and 82.

Not 100% sure you can increase the MOC, but likely; if 70 won't work, I'd try 80.

 

[EEEC]

Copy that! Thanks!

 

[kwired]

Wow! There is a lot of information to process... Most of which goes well over my head, but I am "getting the drift". Thank you!

A friend is asking for my help in upgrading his Brewery. The piece of equipment in question is a chiller that the mfg said will run on 208V 3ph with no issue. I thought we might need to use a buck/booster... The spec sheet says 230V 3ph: FLA 41, MCA 47, MOC 74. The purpose for my question was indeed for calculating his entire load increase to compare to the existing service amperage. Purely for planning at this stage...

I'd bet your chiller seldom if ever draws full rated amps (when supplied with 240 volts), but still something to take into consideration.

 

[Besoeker3]

We don't really know if OP's 47 amp motor is full load rated for 47 amps or if that is what it typically draws when in use.

exameninto

 

[jeremy.zinkofsky]

First, the manufacturer said it will run fine at the lower voltage. Next, I have tried the "is / of" ratio formula and that doesn't seem to work (the result is a lower amperage which I know to be false). I also tried the "percentage" formula which I believe gave me the same answer (45A and some change). Will anyone share the formula or at least how to derive it? Thanks!

Calculate the VA at 230V then use that value to calculate amps at 208V. VA stays the same no matter what voltage.

 

[Besoeker3]

Calculate the VA at 230V then use that value to calculate amps at 208V. VA stays the same no matter what voltage.

That assumes no difference in PF