utilizing a lowly loaded high leg

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dhsvcs

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miami
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electrician
I have a job 120/240 v high leg delta. there are 4 panels, only 1 is 3 phase for air conditioning. The 3 panels are 1 phase 120/240 v, drawing under 200 a each. The 4th panel is 120/240 delta with a high leg ( B at panel ) with a 88 amp draw on the B phase. The other phases, A & C, draw around 200 A each. There are only 2 circuits, 1 A-N & 1 C-N
presently being used at about 10 A each. most of the loads are 1 phase 240 v ac's connected between A & C phase. Can I move some of the single phase ac's to A-B & B-C
which would give me more available current to work with on my A & C legs ? The neutral is full size to the panel. The high leg coming in from the utility company has only 88 amp load at present and is extremely under loaded, as it is 500 mcm when it could be reduced down to 4/0 safely..
Thanx, dhsvcs
 
Yes, as long as the load does not require a neutral.

dhsvcs said:
how do you calculate the load per phase ?
Click to expand...
Uh-oh. That's a no-no here. :eek: (j/k)

It's easier to use Kw or Kva, and then add them. Think load between lines and not per line.
 
So
1. add up the kva per item, single phase between a-c, a-b, b-c and add them to the single phase kva / 240
2 add up the kva per item on the 3 phase items and divide by 4.1714 or whatever that is
3. add the 2 totals and try to balance out the loads as close as possible ?
4. then add up the actual 1 phase loads per leg to get a minimum neutral load ? The single phase ac's have no neutral so I cant see them adding load on the neutral unless there is some weird harmonic or wye phase triangulation....Even with an equation, it should be a really low load on the neutral..
 
You should probably check out this other similar thread.

208V Wild Leg Question

Hello everyone. I have 3 phase service 120/240/208. 2 legs @ 120V and a wild leg @ 208V. Any combination of 2 = 240V. My question is this: if my PTAC unit specs allow for a range of 193V - 253V, could I just use the 208V leg and N for the PTAC? I realize my current draw would go up. Any...
forums.mikeholt.com
 
I did,,thanx....I have some studying ahead of me....
Started as an electrician in 1974 and still learning...
 
dhsvcs said:
I have a job 120/240 v high leg delta. there are 4 panels, only 1 is 3 phase for air conditioning. The 3 panels are 1 phase 120/240 v, drawing under 200 a each. The 4th panel is 120/240 delta with a high leg ( B at panel ) with a 88 amp draw on the B phase. The other phases, A & C, draw around 200 A each. There are only 2 circuits, 1 A-N & 1 C-N presently being used at about 10 A each. most of the loads are 1 phase 240 v ac's connected between A & C phase. Can I move some of the single phase ac's to A-B & B-C which would give me more available current to work with on my A & C legs ?
Click to expand...
This could be figured out if we had all loads in KVA. But using the info provided we can still calculate the load currents so that they can be distributed between the phases more evenly:

The current through the load across phases A and B (call it IA-B) must be equal to the load current IB-C across phases B and C. This is because of the "mirror image" symmetry of the equal line currents IA = IC = 200A, and equal load currents IA-N = IC-N = 10A.
Because of this symmetry IA-B = IB-C = 88A /√3 = 50.8A just like in a regular delta, because the two load currents from both sides of the delta through phase B are equal. However, the two load currents IA-B and IA-C are not balanced at phase A, and neither are IB-C and IA-C at phase C and so you can't just divide by √3 to get the load current IA-C.

I came up with the following equation using vectors is:
IA-C = [ √ (4 IA² - 3 IA-B ² ) - IA-B ] / 2
I can give more detail how I got this but it's kind of a pain to type it all out.
Calculating this for the OP's situation of IA = 200 and IA-B = 50.8 we get IA-C = 170A
The total of all load currents is 50.8A + 50.8A + 170A = 271.6A. If we distributed this evenly there would be a 271.6A /3 = 90.5A load across each pair of phases. So we need to transfer 90.5A - 50.8A = 39.7A from the A-C load over to A-B, and also transfer the same 39.7A load from A-C to B-C. Then A-C would be left with 90.5A - 10A = 80.5A of 240V loads because the 10A through the A-N and C-N loads is still on A-C.

By the way we can multiply the total of all load currents across 240V to get: 240V x 271.6A = 65.18 KVA total for all loads.
 
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