Formula for kW -> amps

[drpeeb]

This is probably easy for you folks, just not me:

Could someone provide the formula to calculate the amperage rating for a given load when the power source is known?

I hope I've asked that right. My exact need is the amp rating for a 45 kW load on 208VAC 3-phase power.

Thank you

 

[RUWired]

The multiplier for 3-phase 208 volt is 360. (208x1.73). 45000/360=125

 

[Buck Parrish]

If it's a continuous load add 1.25% to what ruwired said.

 

[ATSman]

Power Formula

Power Formula

I googled it since my mind is gone..

KW = KV X Amps X PF X 1.73 so assuming 0.8 power factor:

45KW = 0.208KV X Amps X 0.8 X 1.73
Amps = 45 / .208 X .8 x 1.73
Amps = 45 / .2879
Amps = 156.3A

? someone pls check this

 

[roger]

? someone pls check this

See Ricks post.

Roger

 

[Jraef]

I googled it since my mind is gone..

KW = KV X Amps X PF X 1.73 so assuming 0.8 power factor:

45KW = 0.208KV X Amps X 0.8 X 1.73
Amps = 45 / .208 X .8 x 1.73
Amps = 45 / .2879
Amps = 156.3A

? someone pls check this

This is close enough if it is a motor or other inductive load, also a safe bet if it is a mixed load. If it is a purely resistive load, i.e. heating elements or incandescent lights, then the power factor is unnecessary. But few loads are purely resistive.

 

[jim dungar]

This is close enough if it is a motor or other inductive load, also a safe bet if it is a mixed load.

I agree, just don't convert HP straight to KW.

 

[drpeeb]

Thank you all.

The load is an FM broadcast tube-type transmitter. The mftr is calling for a 250A FUSED disconnect, which seemed high.

I guess that leads to this question then: Does the use of fuses require an increase in the disconnect's amp rating?

 

[Besoeker]

I agree, just don't convert HP straight to KW.

I don't follow you.
In what circumstances would the conversion be the wrong thing to do?

 

[jim dungar]

I don't follow you.
In what circumstances would the conversion be the wrong thing to do?

When many people try to determine the input KW of a motor which is rated in output HP, by simply converting HP to KW, they forget to account for the efficiency of the motor.

 

[Besoeker]

When many people try to determine the input KW of a motor which is rated in output HP, by simply converting HP to KW, they forget to account for the efficiency of the motor.

I see what you are getting at.
It doesn't make the conversion wrong. Just the interpretation of the result.