3-Phase vs. Balanced MWBC Voltage Drop

[JDBrown]

If you have a balanced 3-phase MWBC, do you calculate the voltage drop the same way you would a 3-phase circuit with no neutral? My initial reaction was that of course you would calculate them the same way, but the more I think about it, the more I doubt that reaction.

When calculating single-phase voltage drop, I use:
VD = 2IR

For 3-phase I use:
VD = IR*sqrt(3)

where R is the one-way resistance of a conductor.

However, the more I think about it, the more I think I could get away with using VD = IR for a balanced MWBC.

Consider the following contrived example: Let's say there are 3 identical light fixtures installed so that each one is 10' away from a common junction box. A 100' homerun goes from the panel to the junction box, and from there it branches off to the 3 fixtures, with one fixture connected to each phase.



I would clearly use VD = 2IR to calculate the voltage drop from each fixture to the common junction box. I understand that the 2 in this formula comes from the fact that the current must travel up the phase conductor and back down the neutral conductor. However, in the case of the homerun, my neutral now carries the vector sum of all three phase currents. Since they are equal in magnitude, the total current on the neutral conductor is zero. Since the current on the homerun neutral is zero, the voltage drop is also zero. Therefore, it seems that the voltage drop on my homerun could be calculated as VD = IR.

Is this correct, or am I missing something? I expected to find a reason that I would need to include that sqrt(3) in the formula, as is required for a straight 3-phase circuit, but I'm not seeing it. Am I missing something here, or is the voltage drop really less for a balanced MWBC than for an equivalently sized 3-phase circuit?

 

[GoldDigger]

If you have a balanced 3-phase MWBC, do you calculate the voltage drop the same way you would a 3-phase circuit with no neutral? My initial reaction was that of course you would calculate them the same way, but the more I think about it, the more I doubt that reaction.

When calculating single-phase voltage drop, I use:
VD = 2IR

For 3-phase I use:
VD = IR*sqrt(3)

where R is the one-way resistance of a conductor.

However, the more I think about it, the more I think I could get away with using VD = IR for a balanced MWBC.

Consider the following contrived example: Let's say there are 3 identical light fixtures installed so that each one is 10' away from a common junction box. A 100' homerun goes from the panel to the junction box, and from there it branches off to the 3 fixtures, with one fixture connected to each phase.

View attachment 9883

I would clearly use VD = 2IR to calculate the voltage drop from each fixture to the common junction box. I understand that the 2 in this formula comes from the fact that the current must travel up the phase conductor and back down the neutral conductor. However, in the case of the homerun, my neutral now carries the vector sum of all three phase currents. Since they are equal in magnitude, the total current on the neutral conductor is zero. Since the current on the homerun neutral is zero, the voltage drop is also zero. Therefore, it seems that the voltage drop on my homerun could be calculated as VD = IR.

Is this correct, or am I missing something? I expected to find a reason that I would need to include that sqrt(3) in the formula, as is required for a straight 3-phase circuit, but I'm not seeing it. Am I missing something here, or is the voltage drop really less for a balanced MWBC than for an equivalently sized 3-phase circuit?

I understand your confusion, and it took me awhile to get past it. Maybe this will help you.

For single phase, the current, I, is the current in each of the wires and is also the current going through the load.
For a balanced pair of 120 volt loads, the voltage drop will be IR where R is the one way wire resistance. And the % VD will be 100 x (IR /120)
For exactly the same resistive loads but without a neutral attached, the voltage drop will be 2 x IR, since we have to count the voltage drop in both wires.
And the %VD will be 100 x (2IR/240) which when you simplify it becomes just 100 x (IR/120). So either way you work it out the %VD stays the same, giving some reassurance that the method is right.

Now for a four wire wye, say 208Y/120 for example, the current in each of the phase lines is identical to the current in an attached wye load. Result is that the %VD is 100 x (IR/120). No current in the neutral means no voltage drop.
So far, still so good.

Now look at the same circuit with the loads wired in delta. The current we use in the formula can be either the line current or the line-to-line current.
and the voltage can be either the line to neutral voltage or the line to line voltage. We can start out hoping that the %VD will come out the same in both methods.
Start with the current, I, being the line current. Now the voltage drop on each line is going to be IR. But the voltage drop across each load is going to be the vector sum of the voltage drops at each end, which works out to IR x sqrt(3). And the voltage across each load is going to be 208V which is 120 x sqrt(3).
Now we find that the %VD is 100 x (IRsqrt(3)/120sqrt(3). That simplifies to 100 (IR/120). Hooray!
So the formula VD = IRsqrt(3) that you mention is using the line current, I and telling you what the voltage drop seen by each line-to-line load is. Exactly the same thing as we calculated using the dropped voltage on each line. And the %VD which results is exactly the same as we calculated for the wye connection.

If you are going to use VD instead of %VD, you just have to realize what the practical result will be.
In the case of wye loads, a motor or resistive element will see a voltage drop of exactly IR. So a 120 volt nominal load will see instead 120 - VD = 120 -IR.
In the case of delta loads, a motor or resistive element designed for 208 volts will see a voltage drop of exactly IR sqrt(3), for an effective voltage of 208 - IR sqrt(3).

My work here is done....

 

[JDBrown]

...
And the voltage across each load is going to be 208V which is 120 x sqrt(3).
...

:slaphead: D'oh! That's what I get for working everything out symbolically, without plugging in actual numerical values. Thanks for pointing out what should have been obvious: it's notated differently, but in reality it's the same thing. :dunce: