Voltage Drop Calculations w/ single and three phase loads

[Murloc]

VD = sqrt(3)*L*R*I , 3 phase
-------------
1000

VD = 2*L*R*I , 1 phase
-------------
1000
high leg delta system.
chiller MCA: 214 A, 3 phase
heater current: 10 A, 1 phase, A phase

A = 214 + 10 = 224 A
B = 214 = 214 A
C = 214 = 214 A
Neutral = 10 A

Is this the correct way to add currents per phase?
An engineer I asked said that you would just add the 3 phase and single phase loads directly.

Would it be ok for me to use the 3 phase voltage drop formula using single and 3 phase loads?

 

[wwhitney]

Under the approximation that the load current doesn't change with voltage drop, then the superposition principle would apply for calculating voltage drops. That's because V = I Z is a linear relationship between V and Z when I is fixed.

So you can separately calculate the 3-phase voltage drop for the 3-phase equipment using the 3-phase formula, and the single-phase voltage drop for the single-phase equipment using the single-phase formula, and add them up for the case that both loads are operating simultaneously. Note that when you add the voltage drops for the two loads, it will be a vector addition. As the 3-phase drop will move each of A, B, and C voltage towards their centroid, while the single phase voltage drop would move A and N towards their midpoint.

Cheers, Wayne

 

[Murloc]

Under the approximation that the load current doesn't change with voltage drop, then the superposition principle would apply for calculating voltage drops. That's because V = I Z is a linear relationship between V and Z when I is fixed.

So you can separately calculate the 3-phase voltage drop for the 3-phase equipment using the 3-phase formula, and the single-phase voltage drop for the single-phase equipment using the single-phase formula, and add them up for the case that both loads are operating simultaneously. Note that when you add the voltage drops for the two loads, it will be a vector addition. As the 3-phase drop will move each of A, B, and C voltage towards their centroid, while the single phase voltage drop would move A and N towards their midpoint.

Cheers, Wayne

right so you're saying that if you get

2% VD 120V single phase, A
2% VD 240V three phase, A

you add them together
4% VD @ 120V, single phase
4% VD @ 240V, three phase

 

[wwhitney]

Not quite. That would work if your single phase load were connected line to line, say A-B. Then the A-B VD would be 4%, with the A-C and B-C VDs being less.

If you have a 240V high leg delta, and you calculate the 3-phase VD for a balanced 3-phase load per the formula in the OP, that's the L-L VD. So if it's 2%, L-L decreases from 240V to 235.2V. The voltages A-B, B-C, and C-D are still equal to each other.

If you have a 2-wire A-N load on that system, and calculate its VD with the single phase formula, that's the VD A-N. The voltages of B and C don't change. And the A-N VD occurs half as voltage rise on N and half as voltage drop on A. So if it's 2%, then the A-N-B voltages would go from 120 / 0 / -120 to 118.8 / 1.2 / -120 from that load alone.

If you have both VDs at once, then the L-L voltages will no longer be balanced. You can say that the B-C VD is 2%, the A-B VD is 3% (not 4%, as half of the A-N VD occurs on N, half on A), and the A-C VD is somewhere between 2% and 3%, the exact value of which would require some vector calculation.

If you only care about the worst case L-L VD, then it would 3%.

Cheers, Wayne

 

[wwhitney]

P.S. There's a nice vector diagram that goes along with this question/answer, so if that's not clear to you, let me know and I'll figure out how to do a vector drawing and post it as an image.

Cheers, Wayne