# Thread: Formula for kW -> amps

1. Junior Member
Join Date
Sep 2008
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2

## Formula for kW -> amps

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

2. Senior Member
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Feb 2006
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Pa.
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The multiplier for 3-phase 208 volt is 360. (208x1.73). 45000/360=125

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

4. ## 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

5. Originally Posted by tonytonon

? someone pls check this
See Ricks post.

Roger

6. Originally Posted by tonytonon
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.

7. Moderator
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Originally Posted by Jraef
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.

8. Junior Member
Join Date
Sep 2008
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2
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?

9. Originally Posted by jim dungar
I agree, just don't convert HP straight to KW.
In what circumstances would the conversion be the wrong thing to do?

10. Moderator
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Originally Posted by Besoeker
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.

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