[hardworkingstiff]

I'm looking at feeding three 480-volt single-phase loads. The loads are not even, so it will not be possible to balance the loads.

A-B = 155A
B-C = 320A
A-C = 210A

What would the total load be on the 3-phase feeder with all 3 single-phase loads on?

I came up with

A = 189A
B = 299A
C = 292A

Of course the amperage on B phase would go up if the A-B load was turned off.

 

[HARRY304E]

I'm looking at feeding three 480-volt single-phase loads. The loads are not even, so it will not be possible to balance the loads.

A-B = 155A
B-C = 320A
A-C = 210A

What would the total load be on the 3-phase feeder with all 3 single-phase loads on?

I came up with

A = 189A
B = 299A
C = 292A

Of course the amperage on B phase would go up if the A-B load was turned off.

I think you've got to base it off of your largest load.

 

[hardworkingstiff]

I think you've got to base it off of your largest load.

I understand that I need to size the circuit to handle the largest load by itself, but I'm trying to understand what the current in the 3-phase feeder will be with all three single-phase loads turned on.

I was of the understanding that the line current is .865 of the load current that is shared between two other phases. In other words if A-B load is 100A and the B-C load is 100A, the current on B is 100*.865 or 86.5A.

Since my loads are not balanced I took the shared amperage and multiplied the shared load by .865 then added in the rest of the higher load to get the current on that phase. (I hope that makes sense).

I'm looking for confirmation that I understand this correctly or if I'm wrong, please explain where and why.

Thanks

 

[BBQ]

Lou, when I have seen this one come up before the engineers seemed to indicate it is very long and drawn out to do these calculations. I don't believe it is anywhere near as simple as you have done it.

 

[KDC]

Rough calcs, assuming all the loads are pure resistive and in phase with their respective voltages, I get:
A 317 A
B 419 A
C 462 A`

Your line currents in a delta system (which is what this is, effectively) are going to be larger than your currents through the individual loads.

In the case of your example of 100A, if it were a balanced system (3 100 Amp loads AB, BC,CA), I'd expect to see 173A on B phase, not 86.

 

[ohmontherange]

A - 105a
b - 137a
c - 153a

 

[HARRY304E]

Lou, when I have seen this one come up before the engineers seemed to indicate it is very long and drawn out to do these calculations. I don't believe it is anywhere near as simple as you have done it.

True that,I think the annexes have some good examples.