I have a client who has requested I go through a existing maintainance shop and determine what additional loads can be added to their emergency generator. Most of the existing loads are 277V lighting (approx. 12 circuits)and I took measurements on a few, but not all are on at the time I was there. We discussed using a "same as the other" to estimate the overall lighting loads. My quetion is this if I have 12 circuits at approx. 10 amps each can I add them for 120 amps of load then divide by three making the assumption a completely balanced system on the 480v panel thus loading at 40Amps on 480V?
Determining total 480V loading.
- Thread starter njb
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- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: Determining total 480V loading.
Please don't think I am making game of you on this one. What I would like you to understand is that you cannot, as a general rule, add amps to amps, and get amps. What you should do is to add VA, and then convert back to amps only at the very end. VA is VA, but amps is not amps.
Take your dozen 10 amp loads, and multiply each by 277 volts, and add them up. You get a total of 33,240 VA. Now divide by the phase to phase voltage of 480 volts, and divide again by the square root of 3. That gets you the 40 amps you had expected.
In this specific problem, you do get the same answer using your suggested method. But that was only because of the simple nature of this specific problem. In general, it won't work out correctly, unless you are adding VA, and not adding amps.
No. But the answer is 40 anyway.
Please don't think I am making game of you on this one. What I would like you to understand is that you cannot, as a general rule, add amps to amps, and get amps. What you should do is to add VA, and then convert back to amps only at the very end. VA is VA, but amps is not amps.
Take your dozen 10 amp loads, and multiply each by 277 volts, and add them up. You get a total of 33,240 VA. Now divide by the phase to phase voltage of 480 volts, and divide again by the square root of 3. That gets you the 40 amps you had expected.
In this specific problem, you do get the same answer using your suggested method. But that was only because of the simple nature of this specific problem. In general, it won't work out correctly, unless you are adding VA, and not adding amps.
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