We need to tabulate Service Factor Amps for various motors we are considering matching with drive. Can't find the SFA published in any materials either in Baldor Motors catalog or on their website. Does anyone know if this information is published anywhere? I'm still waiting for Baldor Motors to contact me back.
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Baldor Motors/SFA 1
- Thread starter tmo42
- Start date
Maybe I have misunderstood the need, but I can’t find the term “service factor amps” or “service factor current” in 1999 NEC Article 430 or 240. There are a number of references to “motor nameplate full-load current”. I think “service factor amps” could be a phrase cooked up as a marketing term by the low-budget motor trade, and subject to possibly some significant abuse and maybe borderline shady huckstering. By Article 430, one does not size motor-overload protection based on “service factor amps”—but uses “motor nameplate full-load current.” This is possibly why a manufacturer may understandably hesitate using or publishing data based on the term.
- Thread starter
- #4
The reason I ask, is that Baldor lists the Amps at service factor on the nameplate of the motor. But they don't seem to publish it. We're in the practice of sizing to run at as much amps as possible without causing damage to the motor. We use the SFA to gauge about how much we can do this safely, so it's of reasonable importance.
It just seems odd that they'd list something on a nameplate, but not publish it anywhere. I tried to formulate how SFA would related to FLA, but so far haven't managed to arrive at any plated values.
It just seems odd that they'd list something on a nameplate, but not publish it anywhere. I tried to formulate how SFA would related to FLA, but so far haven't managed to arrive at any plated values.
electricpete
Electrical
"I tried to formulate how SFA would related to FLA, but so far haven't managed to arrive at any plated values"
I would think that to a close approximation:
SFA ~ FLA *SF
There is some error in the above due to the fact that the load-component increases by SF but the magnetizing current does not increase. So SFA will be slightly less than FLA*SF.
If no-load current were available, a better approximation could be made.
NLA = No-load amps
FLA = Full load amps
LA= sqrt(FLA^2-NLA^2) = load component of the current.
Your approximation for SFA will be
SFA = sqrt(NLA^2+SF^2*LA^2)
If you don't have NLA, try using 20%*FLA for 2-pole motors, 25% for 4-pole, 30% for 6-pole.
I would think that to a close approximation:
SFA ~ FLA *SF
There is some error in the above due to the fact that the load-component increases by SF but the magnetizing current does not increase. So SFA will be slightly less than FLA*SF.
If no-load current were available, a better approximation could be made.
NLA = No-load amps
FLA = Full load amps
LA= sqrt(FLA^2-NLA^2) = load component of the current.
Your approximation for SFA will be
SFA = sqrt(NLA^2+SF^2*LA^2)
If you don't have NLA, try using 20%*FLA for 2-pole motors, 25% for 4-pole, 30% for 6-pole.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: The usage of SFA would be a deviation from industry standards since empirical approximations are applied. A question is how to justify usage of SFA, if the motor or equipment experiences a short life-cycle in comparison to the motor applications that use normally applied design margins.
I was thinking it could be possible for the motor manufacturer to work backwards, having access to motor design information, and calculate the amperes at rated conditions of voltage and ambient, that would produce the temperature rise of the stated SF. On the other hand, it could be a sales dept. calculation.
electricpete
Electrical
jbartos - NEMA MG-1 defines SFA in section 1.78. Further more section 10.41 for medium polyphase motors requires that SFA be marked on the nameplate (in addition to FLA) when SF exceeds 1.15
The condensed version of MG1-2002 references “service factor amps” as a label requirement in §7.10. I stand corrected. It is worth noting that the same document also states: A motor operating continuously at any service factor greater than 1.0 will have a reduced life expectancy compared to operating at its rated nameplate horsepower… In those applications requiring an overload capacity, the use of a higher horsepower rating is recommended to avoid exceeding the temperature rises for the class of insulation system used… [§9.15.1]
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1
- #10
ElectricPete,
People have been burning up motors for years because they have mistakenly
multiplied the motor FLA by the SF number and operated them in an overloaded state. While it is rare to see such a failure in the first thirty days of operation...failure does occur over time as the motor life is consequentially foreshortened.
The amps do not track the power because of the power factor; which
increases when the motor is loaded above rated. As a consequence, the
increased power (rated power x PF)....occurs at an amperage value that is
less than FLA x PF number). So, as any motor manufacturer will tell you,
operating at the SF x FLA will result in an overloaded motor.... and one
which will void the warranty.
I would be remiss if I did not also address the issue of efficiency as the motor is operated into the service factor. Efficiency does drop off some as the load is increased above 100%. So, the the increase in current due to loss of efficiency is somewhat offset by the increase in power factor. I leave it to others to decide the resultant consequence. But from tests made on a dynamometer at a major motor manufacturer, I can say that you get to the SF power at a value less than the SF x the rated amps.
Now then, to get back to tmo42's original post, operating the motor at the SF
on inverter power can cause the motor to overheat and/or operate
overloaded as it negates any margin for harmonic heating in the motor. Many
motor manufacturers who rate their motors for definitive use on inverter
power, will rate them as follows:
1.0 SF Class B rise on sinewave power
1.15 SF Class F rise on Sinewave power
1.0 SF Class F rise on inverter power
Such motors typically have class 'F' insulation.
For tvo42 I would suggest he have a conversation with Baldor to see
what FLA they sanction for the particular motors when operated on inverter
power. You may find that you cannot use the Service Factor while operating on inverter power or you may not be able to use the full measure.
And lets not lose sight of the fact that not all inverters are created equal.
Differences in modulation schema do have considerable impact upon the
harmonic content in the consequent output waveform. Tests run on various
manufactures that employ sine-coded or star modulation show that the rms
fundamental output voltage can be down by as much as 82% from the rms
fundamental of the applied mains voltage. Consequence is that the motor
operates voltage starved, having to slip more to produce a given value of
output torque. Thus, less torque per ampere means that more amperes are
required to produce the same value of torque.
Oh, and using an ammeter to observe the motor amps when on inverter power, can lead you astray too. The ammeter, even the true RMS types, will show you the total current; that is, the fundamental + the harmonic currents.
Only the fundamental produces useful torque.... some of the harmonic currents/torques actually subtract from the delivered shaft torque.
Interesting, isn't it, that what you see, is not necessarily what you get.
So the decision to operate into the service factor of a given motor should not be made lightly. There are many factors to consider before taking the leap.
People have been burning up motors for years because they have mistakenly
multiplied the motor FLA by the SF number and operated them in an overloaded state. While it is rare to see such a failure in the first thirty days of operation...failure does occur over time as the motor life is consequentially foreshortened.
The amps do not track the power because of the power factor; which
increases when the motor is loaded above rated. As a consequence, the
increased power (rated power x PF)....occurs at an amperage value that is
less than FLA x PF number). So, as any motor manufacturer will tell you,
operating at the SF x FLA will result in an overloaded motor.... and one
which will void the warranty.
I would be remiss if I did not also address the issue of efficiency as the motor is operated into the service factor. Efficiency does drop off some as the load is increased above 100%. So, the the increase in current due to loss of efficiency is somewhat offset by the increase in power factor. I leave it to others to decide the resultant consequence. But from tests made on a dynamometer at a major motor manufacturer, I can say that you get to the SF power at a value less than the SF x the rated amps.
Now then, to get back to tmo42's original post, operating the motor at the SF
on inverter power can cause the motor to overheat and/or operate
overloaded as it negates any margin for harmonic heating in the motor. Many
motor manufacturers who rate their motors for definitive use on inverter
power, will rate them as follows:
1.0 SF Class B rise on sinewave power
1.15 SF Class F rise on Sinewave power
1.0 SF Class F rise on inverter power
Such motors typically have class 'F' insulation.
For tvo42 I would suggest he have a conversation with Baldor to see
what FLA they sanction for the particular motors when operated on inverter
power. You may find that you cannot use the Service Factor while operating on inverter power or you may not be able to use the full measure.
And lets not lose sight of the fact that not all inverters are created equal.
Differences in modulation schema do have considerable impact upon the
harmonic content in the consequent output waveform. Tests run on various
manufactures that employ sine-coded or star modulation show that the rms
fundamental output voltage can be down by as much as 82% from the rms
fundamental of the applied mains voltage. Consequence is that the motor
operates voltage starved, having to slip more to produce a given value of
output torque. Thus, less torque per ampere means that more amperes are
required to produce the same value of torque.
Oh, and using an ammeter to observe the motor amps when on inverter power, can lead you astray too. The ammeter, even the true RMS types, will show you the total current; that is, the fundamental + the harmonic currents.
Only the fundamental produces useful torque.... some of the harmonic currents/torques actually subtract from the delivered shaft torque.
Interesting, isn't it, that what you see, is not necessarily what you get.
So the decision to operate into the service factor of a given motor should not be made lightly. There are many factors to consider before taking the leap.
electricpete
Electrical
Hi jomega - Did you read my whole message? I gave SF x FLA as a first approximation. I clearly stated SFA will be a little less than computed this way and the refinement is to calculate SFA = sqrt(NLA^2+SF^2*LA^2) which gets us closer
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion to electricpete (Electrical) May 3, 2003 ///\\jbartos - NEMA MG-1 defines SFA in section 1.78. Further more section 10.41 for medium polyphase motors requires that SFA be marked on the nameplate (in addition to FLA) when SF exceeds 1.15
///Thank you. It is good to know. I actually meant "The usage of SFA would be a deviation from industry standards when empirical approximations are applied." This implies when the SFA is not on the motor nameplate. Even, when it is on the nameplate and properly used, the motor may experience a shorter life expectancy, which is hard to justify; especially, if it results in a higher cost of motor procuring. Obviously, if one is irreplaceable or indispensable, there is nothing to be worry about.\\
///Thank you. It is good to know. I actually meant "The usage of SFA would be a deviation from industry standards when empirical approximations are applied." This implies when the SFA is not on the motor nameplate. Even, when it is on the nameplate and properly used, the motor may experience a shorter life expectancy, which is hard to justify; especially, if it results in a higher cost of motor procuring. Obviously, if one is irreplaceable or indispensable, there is nothing to be worry about.\\
electricpete
Electrical
I will mention that the method I suggested is not empirical but based upon the equivalent circuit, with simplifying assumptions (neglect leakage reactances, neglect change in efficiency).
An example calculation of the above method using a motor test data from the Reliance catalog:
50hp 2-pole 460v motor. FLA = 55.5A 125% current is 68.7A
Since current at 125% load is given, we will for purposes of this excercize assume that this is a 1.25 SF motor. (the objective of the method is to calculate current at some load above full load given FLA... it doesn't matter whether it is the actual motor SF or not).
> FLA:=55.5:
> NLA:=0.2*FLA; # For 2-pole motor
NLA := 11.10
> LA:=sqrt(FLA^2-NLA^2);
LA := 54.38
> SF:=1.25:
> SFA:=sqrt(NLA^2+SF^2*LA^2);
SFA := 68.88
> # actual amps at 125% load is 68.7 Remarkable agreement.
>
> # Note that actual NLA is higher than the estimate used. If actual NLA were used the SFA estimate in this case would not be as close to actual!... would be around 68.0A. Still very close to actual. The error in estimating NLA improved the accuracy (by luck).
An example calculation of the above method using a motor test data from the Reliance catalog:
50hp 2-pole 460v motor. FLA = 55.5A 125% current is 68.7A
Since current at 125% load is given, we will for purposes of this excercize assume that this is a 1.25 SF motor. (the objective of the method is to calculate current at some load above full load given FLA... it doesn't matter whether it is the actual motor SF or not).
> FLA:=55.5:
> NLA:=0.2*FLA; # For 2-pole motor
NLA := 11.10
> LA:=sqrt(FLA^2-NLA^2);
LA := 54.38
> SF:=1.25:
> SFA:=sqrt(NLA^2+SF^2*LA^2);
SFA := 68.88
> # actual amps at 125% load is 68.7 Remarkable agreement.
>
> # Note that actual NLA is higher than the estimate used. If actual NLA were used the SFA estimate in this case would not be as close to actual!... would be around 68.0A. Still very close to actual. The error in estimating NLA improved the accuracy (by luck).
electricpete
Electrical
I believe the error of neglecting change in efficiency in my method above is usually conservative (since usually efficiency decreases above full load), and will lead toward a low estimate of SFA. Not sure about the effect of neglecting leakage reactance.
Perhaps worth mentioning that if the above example were calculated using the simple SF*FLA = 55.5*1.25 = 69.375, the result is within 1% of the actual 125% current (68.7).
I agree with the other folks who have suggested that sharpening your pencil to get the last few percent out of a motor is usually penny-wise and pound foolish, in the long-term.
Perhaps worth mentioning that if the above example were calculated using the simple SF*FLA = 55.5*1.25 = 69.375, the result is within 1% of the actual 125% current (68.7).
I agree with the other folks who have suggested that sharpening your pencil to get the last few percent out of a motor is usually penny-wise and pound foolish, in the long-term.
tmo42, SF is simply a multipier of horse power. It is the acceptable limit that a motor can be loaded to without damage. I have been a technician for 20 years and I have never sized a motor to a load using it's SF or SFA. A motor with a 1.15 SF is a better motor than a 1.00 SF. It's just a higher duty motor. FLA is always used to size OL protection.
electricpete
Electrical
Typically the OL can be set higher for SF=1.15 than for SF=1.0.
Many times I have come onto the project after the mechanical engineer has purchased the package, which for example, may be load equal to 98% or 101% of motor HP, with a 1.15 SF. What do you do then? Probably set the motor relay at 1.15 pickup (for example with a motor protection relay).
Another case is the process control people, since current is available at the PLC, to adjust the process to maximize motor amperes, which they choose as either FLA or SF*FLA (regardless of voltage also). They measure amps, then adjust the process.
These are the real things I see.
Another case is the process control people, since current is available at the PLC, to adjust the process to maximize motor amperes, which they choose as either FLA or SF*FLA (regardless of voltage also). They measure amps, then adjust the process.
These are the real things I see.
Here is an interesting link for more debate. See the first sentence after "Important".
Keep in mind, this is unique to the pool and irrigation industry!
Keep in mind, this is unique to the pool and irrigation industry!
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