Motor FLA

Can someone please explain how the FLA in the attached table came to be 14 AMPS instead of....


(1000*0.75kW)/(1Pf * 115vac) = 6.5AMPS


assuming a power factor of 1?


Thanks

I think it depends on the efficiency of the motor.
From Table 430-148 & 430-150 of the NEC the FLA of 1 HP single phase motor is 16A.
The actual motor amps may be higher or lower than the
average values listed .
For more reliable motor protection, use the actual motor current as listed on the motor nameplate.
this table should be used as a guide only.

What you measure is the Apparent Intensity, which is the vector sum of Active Intensity and Reactive Intensity.

In alternate currents the thing is not as easy as applying Ohm's law.

The relationship between Active and Apparent is the power factor.

The power factor in a motor is always less than 1, in the motor plate is usually indicated the full load factor but when it works with less load is even smaller, being close to 0 when the motor rotates without load

As stated, power factor is never 1 and the NEC tables (I'm assuming that's where those numbers were sourced) are always very conservative. I can't recall seeing a motor whose nameplate FLA was the same or greater than the NEC table's.

I am looking at a manual for a vacuum pump. Motor full load amp nor horsepower is specified in manual for the pump motor. All it shows is motor power consumption. this is a single phase motor 115VAC 60Hz 0.90kW. Based on this limited information, can I calculate FLA from this info and be certain that it is correct or at least on the high end? I am trying to select a Contactor with appropriate thermal overload protection to energize it. Without knowing power factor nor motor efficiency, if I calculate FLA as shown below;

(1000 x .90kW) / 115VAC = 7.82 AMPS

will the above value be reasonably close?


Thanks.

I would include power factor and efficiency too.

For single phase:
FLA = (kW * 1000) / (V x pf x eff)

My rule of thumbs are 0.8pf and 0.9eff.

skyfox said:

(1000 x .90kW) / 115VAC = 7.82 AMPS

will the above value be reasonably close?


Thanks.

Click to expand...

No! For calculate the size of contactor, motor proteccion, cables etc. you have to use the Apparent power (kVA) and not the Active power (kW). Use the intensity specified in the motor plate or the manufacturer intensity tables.

lfe said:

No! For calculate the size of contactor, motor protection, cables etc. you have to use the Apparent power (kVA) and not the Active power (kW). Use the intensity specified in the motor plate or the manufacturer intensity tables.

Click to expand...

Can you elaborate on this a bit?

We are not talking about the coil that energize the contractor correct? Contactor that I am looking at is rated for 24VDC Coil operation and contacts carrying AC line voltages are rated for for 15AMPS. Thermal overload relay add on for that same contactor is adjustable between ~7 to 20 AMPS if I recall correctly. There is nothing in the contractor selection options asking for a kVA rating.

So if I calculate my FLA using the rule of thumb values HJTRBO provided (.8pF and .9 Efficiency) FLA is around 11 AMPS. So Why Can't I use a contactor rated for say 15 AMPS with a thermal overload relay rated for 7 ~ 15Amps adjustable and set it to say for an example 12 to 13 amps? Where does the kVA figure in to this? What am I missing or overlooking?

This Vacuum pump is designed for laboratory use and comes with a 3 prong power cord 3m in length attached to it and is meant to be plugged in to a wall outlet. I have no other motor information from the Manufacturer other than the Pump kW rating of .90kW at 115V 60 Hz.


Thanks

A 15amp ac3 rating contactor will be an excellent choice. Overload as you describe will also be a good too.

Also, the kVa figure is taken into account with the formula I use with the pf.

For single phase
kVa = kW / pf

skyfox said:

Can you elaborate on this a bit?


Thanks

Click to expand...

For sure the wikipedia article will explain it better than me : https://en.wikipedia.org/wiki/AC_power

It says :
Apparent power is the product of the rms values of voltage and current. Apparent power is taken into account when designing and operating power systems, because although the current associated with reactive power does no work at the load, it still must be supplied by the power source. Conductors, transformers and generators must be sized to carry the total current, not just the current that does useful work ...


As you can see, the thing is much more complex than the Ohm's law that you try to apply wrongly.

Most designers simply use an approximate formula to calculate the Apparent Intensity, as some proposed, or consult the manufacturer's tables that in your case indicate 15A.

Thanks.

Best illustration of this:

I was hoping another member would have chimed in by now.

The national electric code is the US standard and they set the requirements based on testing (I think), regardless, they set the rules for motor fla.

it has been way to long since I had motors class at UT Martin.

you might post this question on www.mikeholt.com which is an electrical web site for a better answer. lots of electrical professionals there.
note, I am a member of the website only.

james

You must by code follow the NEC tables unless a local authority supersedes the NEC. Some large cities, have their own code. By me states adopt a certain year for the NEC.
The NEC tables are conservative for a reason. A few years from now, after you hit the lottery, a new guy might put in a poorly rewound motor that is very inefficient, the wiring would still be protected by the NEC tables, you must also size your wire to carry the loads listed in the table. It gives you a defensible position in case something catastrophic were to occur.

JordanCClark said:

Best illustration of this:

Click to expand...

Thanks for this. I'm stealing it.