3 phase current, two 208V loads, how much current?

[turboii]

so i have two pieces of equipment that require 208V single phase, that i want to power off a 3-phase supply that has 110V Line to neutral or 208V line-line, with leads A, B, C, and N. So there is 208V between any 2 phases, and 110V between any phase and neutral.

I have the first 208V outlet wired across phases A and B and the second outlet wired across phases B and C. If the two loads draw 30A each, this means there are 30A on lines A and C, but how much current is drawn on the line B? would it be 60A, or do the differing phases cancel some of that out? how do i calculate the current on line B?

 

[mivey]

so i have two pieces of equipment that require 208V single phase, that i want to power off a 3-phase supply that has 110V Line to neutral or 208V line-line, with leads A, B, C, and N. So there is 208V between any 2 phases, and 110V between any phase and neutral.

I have the first 208V outlet wired across phases A and B and the second outlet wired across phases B and C. If the two loads draw 30A each, this means there are 30A on lines A and C, but how much current is drawn on the line B? would it be 60A, or do the differing phases cancel some of that out? how do i calculate the current on line B?

30 * sqrt(3) = 52 amps

 

[winnie]

Do a search on 'vector addition'.

The phase B to A current has a different phase angle than the B to C current, and so to add the two currents you need to account for the phase angle difference. If you represent the current flow as a vector, where the length of the vector corresponds to the amount of the current flowing, and the direction of the vector corresponds to the phase angle, then you can simply connect the vectors 'head to tail' to add them. The resultant vector (going from the origin to the last 'head' in the chain of added vectors) is the net current flow.

In the case that you described, if we _assume_ unity power factor, then we know the phase angle difference of the current flow from the phase angle difference of the associated voltages...The B to A current is 60 degrees away from the B to C current. In the case where both of these currents have the same magnitude, then the vector addition is particularly simple, and the answer is sqrt(3) * 30 = 52A.

-Jon

 

[LarryFine]

In the case where both of these currents have the same magnitude, then the vector addition is particularly simple, and the answer is sqrt(3) * 30 = 52A.

And if the two loads are not equal?

 

[winnie]

And if the two loads are not equal?

Pick your vector addition technique of choice.

I'm partial to decomposing the vectors into X and Y components, adding separately, and then converting back to a single vector...but now a days I usually cheat and fire up a cad package and just draw out the vectors

-Jon

 

[Besoeker]

so i have two pieces of equipment that require 208V single phase, that i want to power off a 3-phase supply that has 110V Line to neutral or 208V line-line

Doesn't 110V line to neutral give 190V line to line?

 

[mivey]

Doesn't 110V line to neutral give 190V line to line?

Me thinks you are correct.

 

[LarryFine]

Doesn't 110V line to neutral give 190V line to line?

Back when I was a helper, the high leg was 190v. I guess it became 208v when 110v "became" 120v.

 

[Besoeker]

Back when I was a helper, the high leg was 190v. I guess it became 208v when 110v "became" 120v.

I suppose I'm just used to what seems to me the simpler system used in UK and elsewhere.
LV is 400/230V
All single phase domestic and commercial kit is single phase 230V.
Three phase LV is 400V from which the 230V l-n is derived.
No high leg, no centre tapped single phase, no mixture of 120, 240, 277, 208....stuff.
I like simple.