Calculating SFA

[friend81]

I tried to figure out the SFA of a motor using the following formula,

SFA = SF*HP/[SQRT(3)*Vll*EFF*PF]

Whereas,

SFA = Service factor Amps
SF = Service factor
SQRT(3)= 1.732
VII = line voltage
EFF = Efficiency
PF = Power Fator

Considering

SFA = 1.15*5HP / (1.732*220*0.75*0.81)
= 5.75 / 231.48
SFA = 0.02484

I would like to know how to arrive the SFA after this ?

 

[electricpete]

The equation you started with is not right since HP implies you plug numbers in without paying attention to the units. It would be right if you put P (power) instead of HP and wrote the units next to each value and then applied appropriate unit conversions (like converting hp to watts) until you arrive on amps.

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[friend81]

thanks electripete i arrived using watts instead of HP.

SFA = 1.15*3700 / (1.732*220V*0.75*0.81)
= 4255W / 231.48W
SFA = 18.38A

 

[LionelHutz]

Do you not know the rated current of the motor. Generally, multiply the rated current by the service factor and that would be the service factor current. At least I believe that is what you are trying to calculate.

 

[jraef]

Exactly, and in fact, HP ratings are meaningless from a standpoint of protection devices etc. The only valid numbers are nameplate FLA. All that calculating should only be used when the nameplate is missing. But then again if the nameplate is missing, how did you know the HP?

 

[friend81]

calculating directly using SF & FLA is also almost matching with the above arrived value,

SF = 1.15
FLA = 16.3A
SFA = 16.3*1.15
SFA = 18.745

but Nameplate SFA = 19.5A.

So i contacted the manufaturer & they replied that exact SFA cannot be arrived directly multiplying FLA & SF but the actual procedure is as follows,

SF is calculated based on the Motor power

eg: 5HP = 3700Watts
SF = 1.15 = 15%
3700 + 15% = 4255W
So the motor will be loaded to 4255W and current consumed in that load is declared as SFA.

 

[electricpete]

I doubt power factor is going down. Apparently the motor gets much less efficient above nameplate load, so the current increases more than the 15% increase in delivered mechanical power (?)

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[friend81]

Its true, PF comes down to 0.76. But is this small difference of 0.5 ( name plate rated 0.81) makes any significant difference ?

 

[LionelHutz]

If you have the motor data that tells you the power factor and efficiency at 115% load then you could calculate it. But, if you have that data you will also likely have the current at 115% load since the data sheets usually include a chart giving all 3 values for different motor loads.

If you do not have the info giving the efficiency, power factor and current at 115% load then you really don't have the info to calculate it anyways, no matter how you try to do it. The efficiency and power factor at 115% load will be different than the 100% load values as you have already pointed out.

As Jraef has pointed out, current is what heats the motor so it is fairly safe to use the service factor times the rated current to come up with the allowable current. However, NEMA MG-1 does define service factor as the allowable over power rating of the motor, ie you can operate at 1.15 times the rated power of the motor. But, NEMA MG-1 also is vague about what the service factor means and basically says you can operate in the service factor but it will reduce the life of the machine.

If you used nameplate values and got 18.4A calculating yet the manufacturer says it is 19.5A then the power factor and the efficiency do change and it does make a difference.

Maybe if you told us what you want this number for then we could tell you if it makes a difference or not. Lots of cheap overloads, for example, have errors in the current setting dial and measurement accuracy that are greater than the difference you are concerned with.

 

[friend81]

Actually i need this data for cable selection

 

[LionelHutz]

In that case, the electrical codes in North America say that the cables should be rated for at least 125% of the motor rated current. So, the cable should be good for at least 20.4A. You can probably use 10awg unless the wire run is long.

Can you elaborate on how the SFA is being used to select the cable?

 

[burnt2x]

LionelHutz,
Assuming the poster "truthfully answered what the numbers are for", I think we need to help this guy understand why he is wrong to think that way!

friend81,
I'm befriending you now! Get a copy of NEC or Google "cable sizing".

 

[friend81]

actually this pumpset is consuming 23A when installed in a borewell. The supply voltage is 224V & a cable of 10mm^2 * 160m is used to feed this motor.

The same pumpset when installed in testpit with same cable is consuming the SFA 19.5A.