Voltage drop along one phase in a three phase circuit

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sf97ny

Member
Hello to all,
As an introduction, even though it is obvious to you, in three phase four wire 480V/277V wye system, we get the phase voltage of 277V by connecting one terminal of the load to the live end of a phase, and the other terminal to the neutral wire.
So for a distance L between tranformer/generator and load, we have the length L of the live conductor, and the length L of the neutral wire.
When I calculate the voltage drop between source and load, along the 2 wires, I am considering the distance L multiplied by 2.

Question: When the three circuit is perfectly balanced, the current through the neutral wire is 0 (zero). For the voltage drop along that perfectly balanced load, should I count only the live wire (distance L), or should I also add the length L of the neutral?
Personally, I think right now that I should count both wires - phase and neutral.
Thank you in advance for your answers.
Florin
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Florin, welcome to the forum! :smile:

If I understand what you're asking, you'd count the VD along each conductor, based on its current and impedance.

Just remember that there are three conductors, each with its own current and impedance (presumed equal here.)
 

Smart $

Esteemed Member
Location
Ohio
... For the voltage drop along that perfectly balanced load, should I count only the live wire (distance L), or should I also add the length L of the neutral?
Personally, I think right now that I should count both wires - phase and neutral.
Thank you in advance for your answers.
Florin
Depends on the type of loads being served by each component phase of the circuit. If it can reasonably be expected that all three component loads will be operating, and not, concurrently, then it is okay to use 1 ? L. Otherwise, use 2 ? L for the MWBC, or portion thereof.

An example would be three banks of store lights that are always on during business hours and always off overnight.
 
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sf97ny

Member
Thank you to all.
The perfect balanced load is the best case scenario, and a contractor or a designer sizing the wiring and conduit based on this best case scenario will end having in the real world more than the 3% maximum accepted by NEC.
 

raider1

Senior Member
Staff member
Location
Logan, Utah
Thank you to all.
The perfect balanced load is the best case scenario, and a contractor or a designer sizing the wiring and conduit based on this best case scenario will end having in the real world more than the 3% maximum accepted by NEC.

There is no maximum accepted voltage drop required by the NEC. The voltage drop recommendations in the NEC are located in Fine Print Notes and are not enforceable as code. (They are good recommendations though)

Chris
 

glene77is

Senior Member
Location
Memphis, TN
Question:
(1) When the three circuit is perfectly balanced, the current through the neutral wire is 0 (zero).
(2) For the voltage drop along that perfectly balanced load,
(a) should I count only the live wire distance "L" ),
or
(b) should I also add the length " L" of the neutral?

Florin,

Welcome to the forum.

As stated, assuming "ideal" condition, no harmonics, balanced, etc.

The bundle of energy in question
travels out on a Line wire and back on a Line wire.

The impedance involved will be Line Out and Line Back.

Further,
even if the loads are unbalanced,
then the current returns back to the source,
by way of the Line wires and/or the Neutral wire.

:smile:

The calculation for Voltage Drop will be along the lines of calculating the Neutral Unbalanced Current.
If the load gets off balanced, (and still no harmonics),
then the neutral current is a Pythagorean summation,
involving the square-root of
( the sum of Line Currents squared plus the (sum of each pair of Line Currents multiplied) ) ,
which I can't get keyed in this post. Jraef could key it in.

:smile:
 

drbond24

Senior Member
You are thinking about it too hard. :) The voltage drop equation for a single phase circuit already has a multiplier of '2' in it. You insert the one-way length of your circuit for 'L' and let the equation do the rest. It doesn't matter whether the current is returning on the neutral on another phase conductor. It's just 'plug and chug' equation as my college professors used to say.
 

mivey

Senior Member
For single phase, the multiplier is 2.

For three phase, the multiplier is sqrt(3) or 1.732
 

Smart $

Esteemed Member
Location
Ohio
For single phase, the multiplier is 2.

For three phase, the multiplier is sqrt(3) or 1.732
Of course!!!

But we are talking about a "hybrid"... three single phase which may or may not constitute a complete or balanced three phase circuit. Using 1 ? L per component circuit seems to me to be a fair compromise where likely to be in operation concurrently. One must remember that the single phase condition may still be possible under under out-of-the-ordinary conditions.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Can't you calculate the balanced 3ph loads using the 3ph multiplier, and the single phase loads using the 1ph multiplier, and add them together?
 

Smart $

Esteemed Member
Location
Ohio
Can't you calculate the balanced 3ph loads using the 3ph multiplier, and the single phase loads using the 1ph multiplier, and add them together?
I'm not sure we're talking about the same thing. I was referring to a balanced 3? MWBC. Sounds more like you are talking 3? 4-wire feeder conductors... but no matter. If you setup your calculation with cmil as the unknown, yes.

By the way, using 1 ? L for each of the three phase conductors is the same as using 1.732 ? L for the lot, because in each calculation you'd be using the respective voltages, where

VD1max = 3% ? V1 = I ? R
VD2max = 3% ? V2 = 1.732 ? I ? R
V2 = 1.732 ? V1

VD2max = 3% ? (1.732 ? V1)
V1 = VD1max ? 3%

VD2max = 3% ? 1.732 ? (VD1max ? 3%)
VD2max = 1.732 ? VD1max
 
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