Power Panel Loading Calculation

[7JLAman4]

I have the following 120V-3Ph-Delta panel that contains the following phase loading:
AB: 5.3A
CA: 13.0A
BC: 11.2A

How do I calculate the feeder ampacity? The drawing that I have shows a 21.0A load. Not certain how this number was calculated.

 

[DRWeig]

Can you post the drawing? Your terms are a little confusing, I'm not familiar with currents called out with two letters...

Good on ya,

Goober Dave

 

[waross]

For feeder calculation, I would assume equal loading on all three phases at the highest load. That would be 1.73 x 13 = 22.5 amps.
It looks as if the original designer vectorily added 11.2 and 13. That would be close to 21 amps.
AB, BC, CA means that the loads are connected line to line from "A" phase to "B" phase, from "B" phase to "C" phase and from "C" phase to "A" phase.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7JLAman4]

I've attached a copy of the drawing that contains the panel. Any ideas now? I believe the assumption is a 0.9pf for the branch circuits, however the calculation of the feeder is still puzzling!

 

[DRWeig]

Ah, thanks Waross. I missed the part where everything is 2-pole... I'm not used to 120V delta.

If the load were perfectly balanced, it would be a matter of summing the currents and dividing by 1.732. In your case, 29.5/1.732 = 17.0 amps, times 1.25 to size the feeder = 21.3.

However, with such a small number of 2-pole circuits, balance is hard to come by, so Waross's suggestion of assuming balance at the highest expected load is best. You should throw in some extra capacity for the spare breakers too.

Make sense?

Good on ya,

Goober Dave