calculating amps from a give kW load

[mull982]

I am trying to calculate the FLA for a motor given the motor kW rating (this is a German motor and they like to give their ratings in kW rather than hp.)but my numbers are not matching up with those given by the manufacturer. I am trying to calculate the FLA as an exercise for myself to try to figure out how to derive them from a given kW.

This is a 480V, 60hz motor rated at 13kW. The motor has a power factor of 75%. I am taking the given 13kW and converting it to a kVA value by dividing it by its power factor of .75. This gives me the following results:

13kW / .75 = 17.3kVA

Then dividing the kVA by the voltage and the factor of 1.73 I get the following:

17.3kVA / .48kV / 1.73 = 20.87A

As you can see the FLA I calculated is 20.87A compared to the 25A that the manufacturer is saying. Did I go about the calculation right for calculating the Amps from a given kW value?

Thanks.

 

[charlie b]

I get the same results, but I am skeptical about the input information. For example, given the country of origin, are you certain it is not 230 volt, 50 hz? Or is this motor made in Germany for use in the US? Also, is the motor data given in terms of actual load, or in terms of minimum circuit ampacity? Finally, it is possible that the manufacturer simply rounded the value up, so as to make sure it is conservative.

 

[mayanees]

efficiency??

efficiency??

mull,
I'd suggest using an efficiency of ~ 0.85, and that will get you there.
JM

 

[jim dungar]

I am trying to calculate the FLA for a motor given the motor kW rating (this is a German motor and they like to give their ratings in kW rather than hp.)

The value of motors is always the OUTPUT whether in KW or in HP.

As mayanees, suggested you need to include the efficiency of the motor when converting output KW to input KW.

 

[weressl]

The value of motors is always the OUTPUT whether in KW or in HP.

As mayanees, suggested you need to include the efficiency of the motor when converting output KW to input KW.

There is mechanical KiloWatt and then there is electrical KiloWatt.....

 

[iwire]

There is mechanical KiloWatt and then there is electrical KiloWatt.....

Would I be correct in assuming a 13 KW motor is roughly equivalent to 17.5 HP motor?

 

[jim dungar]

There is mechanical KiloWatt and then there is electrical KiloWatt.....

Yes, but a motor is rated in mechanical output. When converting between mechanical kW and an electrical kW you need to consider the efficiency of your transfer device(s).

 

[mull982]

I get the same results, but I am skeptical about the input information. For example, given the country of origin, are you certain it is not 230 volt, 50 hz? Or is this motor made in Germany for use in the US?

The motor spec sheet has the motor listed with the voltage at 460V and the frequencey at 60hz. The motor was made in Germany with its intended use to be in the US.

mull,
I'd suggest using an efficiency of ~ 0.85, and that will get you there.
JM

The efficiency listed for the motor is 87.5. I never thought of using the efficiency in the calculation, do you typically use this in this kind of calculation? I assume since the 13kW is given in output power, once you figure out this current associate with the output power you can divide it by the efficiency of the motor to figure out how much current the motor will draw? (or convert output power to input power first and then perform calculation)

Using the 87.5 efficiency and the 460V rating, I get an input current of 24.89 which is darn close to the number they are giving. (I'm sure the rounded) Maybe this is how they got their number. Does the NEC table 430.250 take into account the efficiency of motors when giving the FLA for a given horse power?

 

[jim dungar]

Does the NEC table 430.250 take into account the efficiency of motors when giving the FLA for a given horse power?

Kind of.

No motor built to NEMA standards (MG-1) can have an FLA higher than those shown in the NEC. That is why the NEC table values can be used prior to have the actual motor on hand. And, why the NEC values are not to be use for sizing OLRs.

 

[kingpb]

Yes, but a motor is rated in mechanical output. When converting between mechanical kW and an electrical kW you need to consider the efficiency of your transfer device(s).

Huh? Mechanical output in Hp is equivalent to 746W @ 100% efficiency. The motor rating, built to IEC 60034 or NEMA MG-1, is based on output power whether in Hp or kW.

Power = Work/time, therefore Hp and KW are not different, they are simply expressed in different terms or units.

The current listed in the NEC Tables is the current to the motor, or input for a given output Hp. Using the rated Hp you cannot derive the rated input current without also knowing the power factor, and efficiency. Certain reasonable assumptions can be made to get close. This is why it was difficult for the OP to determine current from output kW.

For general application and planning purposes, and prior to actually knowing the motor nameplate ratings, is why the NEC provides the Tables. But, people forget the Table is only valid for normal belted motors and typical torque characteristics, which means there are many cases were the do not apply.

Rule to Remember - Motor current is related to input power, Hp rating is output power, the bridge between input current and output Hp is power factor and efficiency.

 

[LarryFine]

The current listed in the NEC Tables is the current to the motor, or input for a given output Hp. Using the rated Hp you cannot derive the rated input current without also knowing the power factor, and efficiency.

Rule to Remember - Motor current is related to input power, Hp rating is output power, the bridge between input current and output Hp is power factor and efficiency.

Don't forget that we also have to take voltage at the terminals into account. Motor input power is related to output HP. Motors attempt to be constant-power devices, which is why current will attempt to rise as terminal voltage sags.

When heavily loaded, a motor will slow a bit, current will rise as the motor attempts to keep torgue up, the current exacerbates the voltage drop, causing a snowball of current (and heat) rise. This is why we say that motors abhor voltage drop.

 

[Wirenuts1]

I agree with the calculation.

Formula for Finding AMP's, with a 3 phase

Amp= kilowatts X 1000
-----------------
Volts X PF X 1.73


HP = volts X AMPS X EFFICENCY X PF X 1.73
-------------------------------------
746


With this formula i got
20.87 amps
17.42 HP

Can you all compare notes and double check