AC Motor Service Factor Rating

[aiki202]

Pertaining to the service factor rating on an AC 3PH motor: Correct me if I am wrong: the service factor rating value is a multiplier that is used to calculate the HP that the motor can provide above the rated (nameplate) HP. Before anyone explains why this rating should not be used when designing applications, keep in mind I have no intention of using this rating to get extra out of our motors. This is just for my own understanding.

We recently had an issue where an existing motor driving a fan and the fan shaft mount worn out. The fan was replaced but the new one had a more aggressive blade pitch and thus put more of a load on the motor. The motor is 460VAC, 3PH, 2HP, 60Hz, across the line contactor for control, and heater overloads for over-current protection (old system that will not be upgraded). The heater overloads continually trip out because of the increased load. One of our associates said why don't we just put in bigger heater overloads? This, of course, was not the correct solution, but it got me thinking.

Here is my question: If the service factor is a multiplier of the HP rating, is there a way to relate this multiplier to the motor current? If our 2 HP motor has a SF of 1.15, then it can produce 2.3 HP. What is the motor current at this rating? Correct me if I am wrong, but multiplying the FLA rating by the SF does not give a correct value. Or, is this one of those situations where there are too many variables for a generic relationship to be established, and each specific system would need to be analyzed individually? Or is there something I am missing totally?

Thanks for any feedback in advance.

 

[barclayd]

You are basically correct in your understanding of service factor. Operating a motor in the service factor 'window' will provide the speed and torque necessary for a load, but motor life is drastically reduced.

It is quite possible that the new fan requires more than 2 HP to operate.
The fan curves (pressure-volume) should be looked at to see what horsepower is required to develope the required static pressure and cfm.

Starter manufacturers have specific procedures to follow for selecting thermal units, based on nameplate FLA, Service Factor, etc.
Try to take current readings and compare them to the nameplate. That will give you an indication if the motor is too small.
If so, a bigger motor is necessary. If the existing motor is controlled by a NEMA Size 00 starter, you're screwed - you'll need a Size 0.

THEN you can install bigger heaters.

db

 

[Cold Fusion]

aiki -
Not necessarily in the order you asked:

... Or, is this one of those situations where there are too many variables for a generic relationship to be established, and each specific system would need to be analyzed individually? ...

There are a lot of variables:

NEMA MG-1 Condensed

9.15 Service Factor of Alternating-Current Motors
9.15.1 General-Purpose Alternating-Current Motors of the Open Type
When operated at rated voltage and frequency, general purpose alternating-current motors of the open
type shall have a service factor in accordance with Table 39. [MG 1-12.51.1 and MG 1-14.37]
When an induction motor is operated at any service factor greater than 1.0, it may have efficiency, power
factor, and speed different from those at rated load. Locked-rotor torque and current and breakdown
torque will remain unchanged. ...

... If our 2 HP motor has a SF of 1.15, then it can produce 2.3 HP. What is the motor current at this rating? Correct me if I am wrong, but multiplying the FLA rating by the SF does not give a correct value. ...

I dug out an electrical machinery reference. P(out) is closely porportional to I^2. So if the power output is up 15%, then expect the current to be up 32%. This is the simplified version, but perhaps is close enough for what you are after

... Or is there something I am missing totally?

I don't think so.

cf

 

[aiki202]

Thanks for the replies.

Cold Fusion, "I dug out an electrical machinery reference. P(out) is closely porportional to I^2. So if the power output is up 15%, then expect the current to be up 32%. This is the simplified version, but perhaps is close enough for what you are after"

:smile: This is basically what I was looking for. Thanks for the info.

 

[GeorgeB]

Thanks for the replies.

Cold Fusion, "I dug out an electrical machinery reference. P(out) is closely porportional to I^2. So if the power output is up 15%, then expect the current to be up 32%. This is the simplified version, but perhaps is close enough for what you are after"

:smile: This is basically what I was looking for. Thanks for the info.

ColdFusion, I'd like to see specifics on that reference, because I'm 99.9% sure it is wrong for a motor. Power output is approximately linearly proportional to current ... P=constant*E*I*PowerFactor*efficiency. The power factor drops a smidge and the efficiency drops a smidge. IME, current at 1.15 load is about 120% of current at 1.00 load.

Baldor's website has specific data on many motors including currents from no load to 125% load. Take a look at www.baldor.com.

Perhaps the information was on heaters (resistors) where P=constant*I^2*R.

(the constants vary with units and single/dc or 3 phase ... and probalby hundreds more)

 

[Cold Fusion]

(cut) I'd like to see specifics on that reference, because I'm 99.9% sure it is wrong for a motor. (cut)

George -
I wasn't ignoring you. I don't get to look at this site very often.
First, I agree P(O) ~ I^2 sounds really wrong. Paraphrasing what you said: Excepting minor changes in stator I^2R losses, and pf, P(O) = P(I) ~ I. I think you're exactly right, not 99.9%

The reference I looked at is Electric Machinery Fundamentals, edition 4, Induction Motors section 7.4, page 397. The current, I2, refered to in Eq. (7-31), appears to be (now anyway :-? ) the rotor current. (Also see Fig 7-12)

I don't know what I was thinking. Thanks for catching this. Color me emabarassed.

cf