Panel Loading Question

[hanksmith]

Ok, I always have problems with this and I will try my best to explain my question.

First let me state that the values I am using are just made up, I need this to click in my head so I am trying to keep it dirt simple.

To start I will state a few things, I have a transformer, it's a delta-wye, 600V to 208/120, and 6kVA, just a small guy and it's connected to a small panel.

The panel has three loads connected to it, all the loads are 100W, the first is connected between phase B and neutral (120V single phase), the second is connected between phase A and C (208V single phase) and the third is connected 3-phase (208V three phase).

What would my over all phase currents be? I look at this and the single phase load will draw 0.83A from phase B (100W/120V), the 208V single phase load will draw 0.24A from phase A and 0.24 A from phase C (100W/208V/2), and the three phase load will draw 0.28A from each phase (100W/(208V*sqrt3), sum these up and you get 0.5A for A phase, 1.1A for B phase and 0.5A for C phase.

Does this look right?
Thanks

 

[tem1234]

quote: "the 208V single phase load will draw 0.24A from phase A and 0.24 A from phase C (100W/208V/2)"


it will be 100W/208W = 0.48A

everything else seems ok, if load are purely resistive (no VAR)

 

[rockman7892]

Dont you have to take the phase angles of the currents into consideration when adding load currents from different connections in order to compute the line currents?

 

[rbulsara]

Yes and No.

For the purpose of sizing feeder or service to meet NEC (assuming this in the USA), the "method" is correct and the correction pointed out by tem123 needs to be incorporated.

For engineering purposes and to be precise, the phase angles come in play. The 208V load (between two phases) current will be in phase with the line to line voltage (and hence 30 degree out phase with the other phase currents (single phase and balanced three phase).

However for precision is not warranted in all cases. Accuracy obtained by the simple method is good enough, for meeting code.



Rafiq Bulsara

http://www.srengineersct.com

 

[hanksmith]

For the single phase 208, I thought I would have to take the current 100/208=0.48 and divide this by 2 because you will essentially get half the current from one phase and half from the other.

 

[davidbeach]

Nope, you get all the current in on one phase and out on the other.