3 Phase 120/208v panel voltage drop calculation
- Thread starter Nycdb
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Carultch
Senior Member
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- Massachusetts
You could do either. There will be a factor in the numerator that will depend on which voltage you use.
Short answer:
Using 208V phase-to-phase voltage, there is a sqrt(3) factor in the numerator, in place of the 2 you use for single phase and DC.
Using 120V phase-to-neutral voltage, this factor in the numerator will drop out of the equation. The 2 that is there for single phase and DC becomes a 1.
To understand why, I think about in terms of "effective round trip length". For single phase 2-wire and DC 2-wire, it is the length along the circuit from the source to the load and back to the source. Twice the one way length (suppose 100 ft). Current travels 100 ft on the black wire, and 100 ft back on the white wire.
For three phase circuits, the effective round trip length is significantly less. And this is due to the fact that phases B and C, will be carrying the return current, while phase A is delivering the current to the load. Current only needs to travel out on phase A, and it will not create additional voltage drop on phases B and C. Phases B and C, already carry the return current of phase A's current when they carry the cycle of their own current, without imposing heating due to phase A current on any conductor other than phase A.
Using the phase-to-neutral voltage, the "effective round trip length", is the one-way length. Because the remaining two phases carry return current, which coincides with what they are already carrying.
Because the phase-to-phase voltage = sqrt(3)*phase-to-neutral voltage, we multiply by 1 in a fancy way, to make a formula that works for the phase-to-phase voltage. Multiply by sqrt(3)/sqrt(3), and replace (sqrt(3)*Vpn) with Vpp. The other sqrt(3) remains upstairs in the numerator. The effective round trip length is sqrt(3)*one-way-length, when using phase-to-phase voltage.
For split phase 3-wire systems carrying a balanced load, you can think of it as two opposite phase 120V sources sending current over an effective round trip length equal to the one-way-length. Or you can think of it as 240V sending current out on the black wire, and returning on the red wire, a round trip length equal to twice the one-way length. Either way, they give the same answer.
Thank you for the reply,
In the rule on not exceeding 3% voltage drop on the feeder and 5% total on the feeder plus the branch circuit.
Being that these are all single phase loads in a three phase panel with a neutral does that mean that 120v needs do be used to properly achieve the 3% or less on the feeder? Or does the fact the neutral wire brings back the few imbalanced amps across the 3 phase wires back to the service point allows you to use 208v?
I understand everything you said and of course you can use either voltage in the formula but the outcome of the % of the drop is not the same and I am specifically trying to size the feeder cable to not exceed 3% and I need to know which voltage to use (120 or 208) to properly increase the wire gauge to compensate for the drop to achieve 3% or less
Thank you.
This is of course understanding that 120v means the length of the phase wire plus the same length of the returning neutral to source and 208v as the distance from source to panel only.
winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
If the loads on the feeder are _perfectly balanced_ then there is no voltage drop on the neutral and you can use the 208V three phase calculation for that portion of the load.
Generally the loads are not perfectly balanced, but also the panel will probably never be running the full current predicted by the NEC load calcs. So for the most part people simply calculate the feeder voltage drop as if it is a balanced three phase load and call it good enough.
-Jon
Generally the loads are not perfectly balanced, but also the panel will probably never be running the full current predicted by the NEC load calcs. So for the most part people simply calculate the feeder voltage drop as if it is a balanced three phase load and call it good enough.
-Jon
Carultch
Senior Member
- Location
- Massachusetts
Consider the extreme worst case scenario of imbalance. All current is on phase A, and no current on phases B or C. All the current goes out on phase A, and returns on the neutral. It ends up being just like a single phase line to neutral load. The effective round trip length is twice the length, assuming neutral is the same size as phase A's wire. You couldn't be taking credit for reducing the effective round trip length if the remaining two phases aren't carrying their own current in the first place.
The three phase voltage drop formula depends on a balanced current among the phases, and most electricians/EE's will call it "good enough for our purposes". You usually design for balanced currents anyway, and if there is an imbalance, it is due to there being less load than your full calculated VA. In the interest of curiosity for how it really works when loads aren't balanced, I've produced a spreadsheet that calculates how voltage shifts due to the imbalance. Line resistance is exaggerated, so you can see the changes from source to load on the plot. Set it to 0.02 ohms for a code compliant example with a 100A load (~100 ft of #2 Cu wire), or set it to 0.2 ohms for an example you can see on the plot.
It should be close to the same. Here's a worked example.
Vpn = 120V
Vpp = 208V
I = 80A
L = 150ft (1 way)
r = 0.196 ohms/kft, #2 Cu
Using the phase-to-neutral voltage:
%Vd = I*r*L/(Vpn*1000 ft/kft) * 100%
%Vd = (80A)*(0.196)*(150ft)/((120V)*1000 ft/kft) * 100%
%Vd = (80)*(0.196)*(150)/((120)*1000) * 100%
%Vd = 1.96%
Using the phase-to-phase voltage:
%Vd = I*r*L*sqrt(3)/(Vpp*1000 ft/kft) * 100%
%Vd = (80A)*(0.196)*(150 ft)*sqrt(3)/((208V)*1000 ft/kft) * 100%
%Vd = (80)*(0.196)*(150)*sqrt(3)/((208)*1000) * 100%
%Vd = 1.959%
RD35
Senior Member
- Location
- Bloomington, Indiana
I'm surprised nobody commented on this statement. Maybe I am missing something. However, just for clarification, 210.19 (A) Informational Note No. 4 for branch circuits and 215.2 (A) (1) (b) Informational Note No. 2 for feeders are both merely suggestions for circuit efficiency and proper operation of equipment supplied. They are not rules. If a max. allowable voltage drop is a requirement on an installation, it usually is due to a requirement in the engineering specifications and/or a requirement of the AHJ for the particular location. The NEC does not set a requirement for this.
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RD35
Senior Member
- Location
- Bloomington, Indiana
Hey thanks Dennis! I'm sure I had come across that in the past....but I had completely forgotten about it! Nice catch! This is exactly why I love this forum. Keeps me on my toes and up to date! :thumbsup:
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