Whether Service Factor of Induction motor is based on Power output or Full-load current drawn 2

[Soloten]

Is an Induction motor considered to be operating at Service Factor 1.0 if it is operating at rated kW output but at 90% rated voltage with 111% of the rated full-load current?

In page 2 in the second paragraph from the link, the author discusses this, but I am not clear as to what the author says with respect to the SF, especially when he mentions, 'If you have a SF of 1.15 for instance, but are operating at rated load, a voltage lower than rated load will not have much negative impact on the performance'. Is he referring to 'rated load' as rate kW output or rated Full load amps?

 

[jraef]

Your hypothetical isn't realistic. If referring to MECHANICAL output, that is a factor of torque and speed. Torque varies at the square of the applied voltage, so in your example a motor getting 90% of rated* voltage cannot deliver it's rated torque output, it is only going to deliver 81% of rated torque. If the LOAD only needed less than 81% of rated torque, which is often the case by the way, then you may never know the difference. But technically, it would not be operating at full rated kW output. Slip would increase and current would increase as the motor TRIES to get back to normal speed, so your 111% current might be expected, but again, the shaft power is not going to be "at rated".

What that author is saying basically is that "There is no free lunch"; you can't have it both ways. You can have the SF there to help ride through various anomalies, or you can use it to get a little more power from the motor, but NOT both things simultaneously, at least not without dire long term consequences. So when an OEM needs 22HP at the shaft and chooses to utilize a 20HP 1.15SF motor and run it into the SF, rather than selecting a 25HP motor and having it run cooler for longer, that OEM is sacrificing the "fudge factor" represented in that SF as an economic issue for HIS bottom line, while the burden will be borne by the end user who will have to replace that 20HP motor sooner rather than later (but most likely after the warranty period).

*Note that rated voltage of motors is always less than nominal line voltage, i.e. 460V motors are for operating on 480V systems. That is done on purpose for this very reason.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[waross]

The bottom line is the current.
DO NOT EXCEED RATED CURRENT.
Moderately reduced voltage reduces the maximum starting torque and the maximum pull out torque.
The motor will still develop full load torque but at increased current.
The basic formula for power is Volts times Amps = Watts.
At any given load in Watts or HP, if you reduce the Volts the Amps must increase to maintain that load.
The rated Amps is the maximum safe current the motor may carry without overheating.*
*Exception. With a service factor of 1.15 you can exceed the rated current by 15%
A question for the motor gurus.
I just tried to look compare SF 1.0 motors with SF 1.155 motors.
All I can find are SF 1.15 motors.
Have SF 1.0 motors been phased out or are my google skills lacking?

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

 

[Soloten]

Thanks Jraef and Waross!

1. Jraef, let me see if I understand this: In the hypothetical scenario, you are saying the rated MECHANICAL kW output is not going to be the rated output, but the current could possibly be 111% of rated amps if the voltage goes to 90% though. If my understanding is correct, the electrical power with 111% current and 90% voltage is close to the rated value, but isn't translating to Mechanical Output - So is the power factor poorer at that reduced voltage?

2. And with regard to Service Factor, sticking with the same example of 90% voltage and 111% current (or close), am I still considered to be operating at an SF of 1.11 or since as you mention, the motor is not reaching its rated kW MECHANICAL OUTPUT, am I considered to be operating at a lower SF based on the ratio of actual Mechanical output to the rated Mechanical output? Should the SF be re-evaluated if the motor is going to continually operate at a voltage less than the rated motor input voltage?

 

[waross]

Whoa, slow down.
Back to basics.
Heat kills motors.
Rated Amps plus the service factor is the safe limit for motor heating.
The mechanical output is determined by the load.
If the load demands more power than the motor is rated for, the motor will overheat.
Power equals Amps times Volts (times efficiency √3 x power factor) but you can't get away from Amps times Volts.
If the Volts go down the amps must go up to drive the same load.
Apply the service factor to the Amps.
The motor will produce the mechanical output demanded by the load.
If the load is too great, the motor will overheat and if not properly protected and disconnected will eventually burn out.
If the load exceeds the pullout torque the motor will stall and if not properly protected and disconnected will quickly burn out. (Probably in less than a minute.)
Somehow I am getting the impression that you are looking for a free lunch. Apologies if I am wrong.
Do what you want. Just don't exceed the service factor Amps.
Take a good look at the Cowern Papers.
They will provide you with all the information you want and answer all your questions.
You may also find the answer to some questions that you didn't know that you should ask.
Link

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