120VAC Distribution panels

[EEStults7]

I have a 208Y/120 distribution panel that has 44, 15 amp
circuits (120VAC) on it and it is fed from a 3 phase, 480/208Y/120 -75 KVA transformer. I understand that full load would be ~361 amps for transformer. 44 X 15 = 660 amps at full load on the panel. How is this justifiable/OK?

Thanks

 

[augie47]

In a standard panel, the 44 circuits would be divided as evenly as possible between the 3 phases so each phase would have a potential load of 44/3 X 15 or 220 amps per phase.
In addition the number of breakers and size of those breakers is somewhat immaterial as the load diversity may well mean the supply is adequate.
Even if every one of the circuits was 100% loaded your load would only be 79kva (if I did the math correctly)

 

[EEStults7]

Thanks

Thanks

Augie,

Where would this type of load consideration be found in the NEC code? Maybe better said, where is derating found or addressed?


Thanks for your time.

 

[GoldDigger]

Augie,

Where would this type of load consideration be found in the NEC code? Maybe better said, where is derating found or addressed?


Thanks for your time.

There is no derating or diversity calculation based on the breaker sizes involved here at all
The rule for load calculation is that that you start with the actual connected load and its characteristics (motor, continuous, or non-continuous as well as type of appliance or load for residential) rather than just adding up the branch breaker numbers.

 

[augie47]

A common example of this might be a UPS feeding several computers.
You often find a 100 amp panel with 20 or so 20 amp branch breakers simply to provide individual protection to several outlets even though the total load might be well below 100 amps or, of course, your common residential 200 amp panel with 500 amps worth of breakers (adding up the individual breakers).

 

[suemarkp]

And if you have a bunch of general use receptacles on those 15A circuits, you consider each receptacle at 180VA per strap.

 

[charlie b]

I understand that full load would be ~361 amps for transformer.

How did you arrive at that value? My calculation says that 75,000 VA divided by 208 volts and then divided again by the square root of three yields a full load current of 208 amps.

44 X 15 = 660 amps at full load on the panel. How is this justifiable/OK?

It is justifiable because the calculation is irrelevant. The load on a 15 amp breaker is not counted at 15 amps. You need to know what is connected to that breaker. It may well be a couple table lamps that only draw 1 amp each.

 

[EEStults7]

PEr Charlie B. Questions & Answer

PEr Charlie B. Questions & Answer

1. 75K/(1.73*120)=361

2. Makes sense, I'm working on a project where the total load is still being established.

Thanks

 

[GoldDigger]

1. 75K/(1.73*120)=361

2. Makes sense, I'm working on a project where the total load is still being established.

Thanks

And that would be 361A total over three line to neutral loads if that is how they were wired.
And the 1.73 factor does not belong there, since the line current needs to be coupled with the line to neutral voltage.

So try 75K/(120 x 3) to get amps in any one line conductor.

 

[charlie b]

1. 75K/(1.73*120)=361

Not correct, as Golddigger has already pointed out.

2. Makes sense, I'm working on a project where the total load is still being established.

Then how do you know that that number of breakers of that rating will be appropriate for the project? On the surface, at least, it appears to me that you are taking the design process in the wrong sequence. First, you establish the load. Then you choose the electrical distribution equipment that would be needed to supply that load.