New Circuit Load - 208/3 Phase with single phase loads

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Jerrodlk

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I am installing a 208/3 phase circuit to feed 2 welders, 1 blower motor, and a 5kva transformer.

Welder 1 - 208/1phase - 56 amp input current - non-continuous load
Welder 2 - 208/1 phase - 56 amp input current - non-continuous load
Blower motor - 208/3 phase - 5hp (16.7amp) - continuous load
Transformer - 5kva - 208/1 primary; 120v/1 secondary - non-continuous load

I assume feeding all these loads through a distribution block and fused individually. A am assuming 90 amp feeder wire.

The 2 welders will share 1 leg. Do we assume that leg then goes up to 112amp at full load?

How do you calculate the wire size to feed these loads with 208/3?

Thanks,
 
I could be way wrong but here is what I have.

Welder 1 = 56amp
Welder 2 = 56amp
Blower Motor = 16.7 amp x 125% = 21 amp
5kva transformer connected at 208v = 24amp

56+56+21+24 = 157amp

157amps / 1.732 = 90.64 amps

I am asking for help because I could be very wrong.
 
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Slightly off question but why are you using a transformer to obtain 120 off a 208Y/120 system ?
 
Sorry to say, not correct math.

You need to treat each phase separately, and add up the loads connected to that phase. But you can't use a simple sum, since the different loads will have different phase angles.

To really do this accurately you need to know the power phase angle of each load. You can approximate by treating everything as having the same power phase angle.

Doing this I get 1.732 * 56 + 21 = 118A on the most heavily loaded phase (yes, I skipped a bunch of steps). The key is that 3 phase loads just add up, and the two welder currents add up (2 *56) then get reduced because of the different phase angles ( * 1.732/2)

Jon
 
No. If there is no neutral available you have one solution.
Unbalanced single phase loading gets into math I have not used in ages, I'll step aside and let younger minds address your particular situaion but it is not as simple as you have calculated. :)
 
I'd be inclined to put one welder A to b, the second welder b to c, and the blower and transformer c to A. Then your worst case is 56 amps on any wire.
 
petersonra said:
I'd be inclined to put one welder A to b, the second welder b to c, and the blower and transformer c to A. Then your worst case is 56 amps on any wire.
Click to expand...

Unfortunately not quite.

That is exactly the connection approach that I would use as well, but when you have welder 1 from A to B and welder 2 from B to C, _both_ welders are connected to B, placing (56+56)*(1.732/2) on B. Then you need to add the blower.

Jon
 
winnie said:
Unfortunately not quite.

That is exactly the connection approach that I would use as well, but when you have welder 1 from A to B and welder 2 from B to C, _both_ welders are connected to B, placing (56+56)*(1.732/2) on B. Then you need to add the blower.

Jon
Click to expand...
Oops
 
Ok. This job consists of more than 2 welders and blower. 14 welders total.
Assuming I can connect the blowers and transformer elsewhere, Can I connect 3 welders (208/1phase/ 56 amp) to a single 100amp circuit?
 
Jerrodlk said:
Ok. This job consists of more than 2 welders and blower. 14 welders total.
Assuming I can connect the blowers and transformer elsewhere, Can I connect 3 welders (208/1phase/ 56 amp) to a single 100amp circuit?
Click to expand...
Three equivalent 56A single phase loads connected in delta creates a 97A 3 phase load.

I won't answer your question because it depends upon the details of continuous vs non-continuous and circuit sizing for welders that I do not fully remember. Hope my handling the math part of how the loads will combine helps.

-Jon
 
Jerrodlk said:
Ok. This job consists of more than 2 welders and blower. 14 welders total.
Assuming I can connect the blowers and transformer elsewhere, Can I connect 3 welders (208/1phase/ 56 amp) to a single 100amp circuit?
Click to expand...
You could. But if you do and they happen to run nine welders at a time, you would end up with about 100 amps per wire, as it is non-continuous I would suggest this is code legal, without looking it up. what are you going to do with the other 5 welders?

I think your best bet is a 200 Amp panelboard, and appropriately sized conductors for a 200 Amp circuit. 4/0 AL would probably work. That will give you enough to wire up everything. You might want to run the numbers first because you will be close to 180 Amps and I only did this in my head and as my previous "oops" indicates I might be in error.
 
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It’s for a welding school, not a production application. The odds are pretty low that we will see full amps on 2 welders at the same time.
I’m thinking 5 circuits total for the welders with #3 feeder wire. Then distribute through lower amperage fuses to each welder.
I will need separate circuits for the blowers, etc.
Any further input would be appreciated especially if there are separate rules for welders.
 
I don't remember if there are any special rules for sizing over current protection devices for welders. I think you are allowed some dispensation on conductor size. But I would have to look it up and it's not worth the effort to me. I am pretty sure that you can take advantage of some kind of factor to reduce the calculated load but it don't think that changes the overcurrent protection device size. But I could be wrong.

Just curious why you are so set on trying to run smaller wires. Number three copper and 4/0 aluminum are probably about the same cost
 
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