Voltage Drop With A Twist

[1972Grady]

I have a unique question. In the field we are so use to saying single phase 208 or single phase 240. The issue I'm having is with single phase 208V. I know when you do a voltage drop equation for single phase 240V you use a multiplier of 2 because the phases are 180 degrees apart. I also know when you are doing voltage drop for 3 phase you use a multiplier of 1.732 because the phases are 120 degrees apart. But what multiplier should you use when using 2 legs of a 3 phase system?
I have a voltage drop calculator. I select 208 as my voltage. I select 3% for the the max desired voltage drop percentage. I select stranded Aluminum as the conductor. I select #1 for my wire size. I select 80A for the amps. There is another tab I can select, and that one is three phase or single phase. If I select 3 phase I can see the multiplier is 1.732050808, and the calculator says I can run a maximum distance of 178.00 feet to stay within my 3%. If I select single phase I can see the multiplier is 2, and the calculator says I can run a maximum distance of 154.15 feet to stay within my 3%. That's almost a 24 foot difference. I'm using 2 legs of a 3 phase 208V system for my feeder. So which one should I select? Single phase or 3 phase? Can I run 2 legs of a 3 phase 208V system using #1 AL at 80A 154.15 feet or can I run it 178 feet and still stay at or under my 3% VD?
I've googled single phase 208V voltage drop but I cant seem to find any formula that directly shows single phase 208V VD. I've found plenty that show the formula for 3 phase 208V VD. Same for single phase 480V VD formulas. I think the reason I cant find a single phase 208V or 480V voltage drop formula is because it doesn't exist and you have to use the three phase formula even when only using 2 legs of that three phase system.
I know you guys are experts and I would appreciate any and all feedback that can clear this up for me. Feel free to include math with your answers.

 

[GoldDigger]

I have a unique question. In the field we are so use to saying single phase 208 or single phase 240. The issue I'm having is with single phase 208V. I know when you do a voltage drop equation for single phase 240V you use a multiplier of 2 because the phases are 180 degrees apart. I also know when you are doing voltage drop for 3 phase you use a multiplier of 1.732 because the phases are 120 degrees apart. But what multiplier should you use when using 2 legs of a 3 phase system?
I have a voltage drop calculator. I select 208 as my voltage. I select 3% for the the max desired voltage drop percentage. I select stranded Aluminum as the conductor. I select #1 for my wire size. I select 80A for the amps. There is another tab I can select, and that one is three phase or single phase. If I select 3 phase I can see the multiplier is 1.732050808, and the calculator says I can run a maximum distance of 178.00 feet to stay within my 3%. If I select single phase I can see the multiplier is 2, and the calculator says I can run a maximum distance of 154.15 feet to stay within my 3%. That's almost a 24 foot difference. I'm using 2 legs of a 3 phase 208V system for my feeder. So which one should I select? Single phase or 3 phase? Can I run 2 legs of a 3 phase 208V system using #1 AL at 80A 154.15 feet or can I run it 178 feet and still stay at or under my 3% VD?
I've googled single phase 208V voltage drop but I cant seem to find any formula that directly shows single phase 208V VD. I've found plenty that show the formula for 3 phase 208V VD. Same for single phase 480V VD formulas. I think the reason I cant find a single phase 208V or 480V voltage drop formula is because it doesn't exist and you have to use the three phase formula even when only using 2 legs of that three phase system.
I know you guys are experts and I would appreciate any and all feedback that can clear this up for me. Feel free to include math with your answers.

It is simpler than you think.
Balanced three phase uses a multiplier of 1 since there is no current in the neutral. The voltage and current you use should be the line current, not the line to line (phase) voltage and current.
For 120 volt line to line the multiplier is 2, since there will be equal drop on the hot and the neutral wires.
For 240 line to line, the multiplier is also 2 since there will be a voltage drop on both not wires.
For 120V in balanced MWBC the multiplier is 1 since there will be no current in the neutral.
Note that the second and third situations, which seem identical but different in results really are identical, but the multiplier of 1 affects each side of the circuit independently and the sum of the two voltage drops is 2.
There is a problem of what voltage to use when calculating percent voltage drop, but the absolute value of the voltage drop is relatively straightforward.

When using an online or Excel VD calculator, you need to figure out just how that particular tool does the calculation based on your system type entry.

 

[1972Grady]

I'm not sure I follow you. Are you saying that if I have a sub feed from a 3 phase 208V system but I'm only using 2 phase conductors with a neutral to feed a sub panel then for the purpose of voltage drop my multiplayer would be 2 or 1.732?

 

[Ingenieur]

for a 3 ph ckt use sqrt3
for a single ph use 2
2 legs of a 3 ph ckt is single ph, so use 2

 

[Ingenieur]

It is simpler than you think.

Balanced three phase uses a multiplier of 1 since there is no current in the neutral. The voltage and current you use should be the line current, not the line to line (phase) voltage and current....

are you sure?
even though the current does not return via the neut, it does 'return' via a phase wire

so EACH ph line will see a vdrop = i line x line Z
so you will have drop on both lines, but due to ph angle not 2 times, but sqrt3 times
so balanced 3 ph vdrop = sqrt3 x i line x line Z
????

 

[gar]

180104-1723 EST

Ingenieur:

I am with GoldDigger.

If we have 2 wires from a voltage source, 208 or any other voltage, and the wire impedance for one wire is Z, and the load current is I, then voltage drop is 2*I*Z for the two wires assuming both wire impedances are the same. Where voltage drop means the voltage measured along the wires. This does not tell us the voltage across the load.

.

 

[Ingenieur]

????

 

[Ingenieur]

reread
he is saying l-n vdrop is (1 x line)/120
to me 3 phase means 208 in this case

but guess what?
no difference % wise
from my example
120 base
drop = 1 x 0.1 x 11.88 / 120 = 1%

208 base
drop = sqrt3 x 0.1 x 11.88 / 208 = 1%

 

[Grady_G]

I knew this one would be a good question. Perhaps this will help.
I'm running from a 3 phase 208V meter stack to single phase 208V dwelling units. I'm really trying to figure out how far I can run the feeds and stay at or under 3% voltage drop. If I use the multiplier of 2 then I have to upsize my wire at a shorter distance than I would if I use a multiplier of 1.732 so its important for me to know the proper multiplier. And like I said earlier google was no help so I figured one of you math wizzes might be able to shed some light on this.
I assumed the multiplier would be 1.732 because it being at 120 degrees phase angle with the other wire. When you measure 3 phase VD its line to line not line to line to line assuming its balanced. SO if its only line to line and you use the square root of three then why would it be any different if you only used 2 legs of a 3 phase system? It's still just line to line and its still 120 degrees phase angle.

 

[ActionDave]

Forget about the degrees, it's two wires. The 'lectrons are going out on one wire and coming back on the other, the multiplier is two.

 

[gar]

180104-1946 EST

Is voltage drop along a wire what you are really interested in or is it voltage across a load?

For any two terminal device, a wire with two ends is such a device, the voltage drop between those two terminals is I*Z.

.

 

[Ingenieur]

imo sqrt3
worse case is a balanced load, assume currents of:
L neut = L1 + L 2
L1 100/0 deg = 100 + 0j
L2 100/120 deg = -50 + 86.66j
Ln = 50 + 86.66j = 100/60 deg

assume a line Z of 0.05 Ohm for 1, 2 and n
L1 v drop = 5/0 deg
L2 v drop = 5/120 deg
Ln v drop = 5/60 deg

L1-Ln v drop = L1 + Ln drops
L1 = 5 + 0j +
Ln = 2.5 + 4.33j
total vdrop = 7.5 + 4.33j = 8.66/30 deg volt

using simple method
sqrt3 x 0.05 x 100 = 8.66 v
2 x 0.05 x 100 = 10 v

 

[50sThrowback]

when preforming voltage drop on feeders I find the terms "Single phase" and "three Phase" are irrelevant. what's important here is the line to line voltage of the feeders. for feeders that have a line to line voltage of 480 or 208 your multiplying factor will be sqrt(3). this is also true if your 208 (3 wire) feeder is derived from a 208 (4 wire) . The latter, line to line 240 volt feeder will have a multiplier of 2. Below I explain why.

as Grady_G explained a 240 volt line to line (3 wire) feeder is essentially "single phase" and as he described the two phase( conductor) voltages are 180 degrees out of phase respectively. this means using ohm's law ( shown below) our two phase voltage equations look like this:

ohm's law: V = I * R

V_P1 = I_p1 * R * sin(w*t - Φ)
V_P2 = I_p2 * R * sin(w*t - Φ -180)

Based off of Kirchoff's law of voltage we know that the line to line voltage of this feeder will be the absolute value of the difference of these two voltages, giving us the equation below

V_L-L = ABS[ I_p1 * R * sin(w*t - Φ) - I_p2 * R * sin(w*t - Φ -180) ]

Now assuming the load is balanced we know that I_p1 = I_p1 , which means I_p1 * R = I_p1 * R = Vp_max were V_pmax = 120
furthermore, for the sake of this calculation we can simplify assuming t = 0 and Φ = 90 giving us the equation below

V_L-L = ABS[ V_pmax* sin(90) - V_pmax * sin(-90) ]
V_L-L = ABS[ V_pmax* (1) - V_pmax * (-1) ]
V_L-L = V_pmax* 2

which is why 240 = 120* (2) for a 240 volt line to line (3 wire) feeder

Now for a "3 phase/ single phase " 480(4 wire) ,208 (4 wire) , and 208 (3 wire) feeder the same equation applies
However, this calculation differs in because the each phase voltage is 120 degrees out of phase respectively and a result your phase voltage equations will look like so

V_P1 = I_p1 * R * sin(w*t - Φ)
V_P2 = I_p2 * R * sin(w*t - Φ -120)
V_P3 = I_p3 * R * sin(w*t - Φ +120)

Now to simplify we can make the same assumption made before the exception of Φ, which, for the purposes of this calculation is Φ = 0 and we will take VP2 and VP3 to find our line to line voltage equation which is shown below.

V_L-L = ABS[ V_pmax* sin(120) - V_pmax * sin(-120) ]
V_L-L = ABS[ V_pmax* (sqrt(3)/2) - V_pmax * (-sqrt(3)/2) ]
V_L-L = V_pmax* sqrt(3)

which is why your multiplier for a 480(4 wire) ,208 (4 wire) , and 208 (3 wire) feeder is sqrt(3)

****IMPORTANT**** so what you might be thinking is "why am I attributing a 208 (3 wire) feeder with a multiplier of sqrt(3), its single phase! it should have a multiplier of (2)" well, that's why I believe the terms "single phase" and "three phase" are irrelevant in theory. This is because, what these terms represent in the trade, don't actually correlate to the physics. And in many scenarios like the one mentioned by Grady_G were are stuck thinking that a 208 (3 wire) feeder correlates to a multiplying factor of (2) when it does not. If I pull two hots and a neutral from a 208 (4) wire source though this feeder will be considered "single phase" each of the phases will be 120 degrees out of phase and as a result the math show above will give us a multiplying factor of sqrt(3).

Furthermore, this also applies to voltage drop. The equation I used is the same equation used to determine voltage across a resistor, assuming your conductor is the resistor. Using this equation engender voltage drop.

Please, if anyone believes my methodology is flaws, enlighten me. I have found that some, with higher pedigrees, disagree with my methodology> However, not a single one can mathematically show me why. If I am wrong show me why.

 

[gar]

180104-2137 EST

Read the first post.

But what multiplier should you use when using 2 legs of a 3 phase system?

Quite clearly he is describing a 2 wire voltage source. V = 208 V from two terminals that are floating out somewhere. That sentence references just two legs for the source voltage. And because it is just two wires it is a single phase source for his two wires.

Run other currents than his load thru either or both wires and you can get many different voltage drops.

Furthermore until you define the load impedance and the wire impedances you can not determine the load voltage. Load voltage is not necessarily Vsource - 2*Vdrop wiring.

.

 

[MyCleveland]

50sThrowback

Your discussion about the 3 phase voltages does NOT apply.
Disscussion on 208-1p VD, there is a single voltage across the two phase wires in question...
one phaser of mag 208 at a 30 den shift from the reference Y voltage angle.
Not sure why discussion drifted into multiple phasers, only one voltage.
The multiplier is 2.

 

[GoldDigger]

50sThrowback

Your discussion about the 3 phase voltages does NOT apply.
Disscussion on 208-1p VD, there is a single voltage across the two phase wires in question...
one phaser of mag 208 at a 30 den shift from the reference Y voltage angle.
Not sure why discussion drifted into multiple phasers, only one voltage.
The multiplier is 2.

A good analysis as far as it goes. When you load a single phase (line to line) of a 208Y/120 system the voltage drop will be in phase with the load current and will correspond to a two wire drop.

But when you put three such identical loads balanced across the three phases of the same source the voltage drop caused by one phase (say AB) will cause a partially opposing voltage drop on A to the voltage drop caused by the other phase sharing that wire (say AC).
This is where the root three comes in for a balanced delta load.

Previous posters have been trying to address all of the various scenarios mentioned by the OP, not just the two out of three situation he specifically asked about.

 

[1972Grady]

A good analysis as far as it goes. When you load a single phase (line to line) of a 208Y/120 system the voltage drop will be in phase with the load current and will correspond to a two wire drop.

But when you put three such identical loads balanced across the three phases of the same source the voltage drop caused by one phase (say AB) will cause a partially opposing voltage drop on A to the voltage drop caused by the other phase sharing that wire (say AC).
This is where the root three comes in for a balanced delta load.

Previous posters have been trying to address all of the various scenarios mentioned by the OP, not just the two out of three situation he specifically asked about.

Thank you for addressing my specific question in a direct way. I will use the multiplier of 2 in my scenario. But I still find it odd that I cant find not even one 208v or 480v single phase voltage drop formula anywhere on the net. Its very common for us electricians to run a 2 pole circuit from a 208V or 480V panel to a single phase load. You would think people would be using these voltages in examples of single phase voltage drop. But they only use 240v in the single phase examples. Any idea why that is or any idea where i can find some examples of 208V or 480V single phase voltage drop formulas?

 

[Ingenieur]

does the op's ckt have a neutral?
so he can serve 208 and 120 loads
if so, sqrt3
if not, 2

 

[winnie]

Very often equations are simplified to make them more useful/useable. But sometimes the _assumptions_ used to make these simplifications are not correct, in which case the simplifications are also incorrect.

The 'multiplier' in the voltage drop equation is a simplification, based on the assumption that all of the involved conductors for the circuit are the same length and carrying balanced current. If the conductors are different length (say because of a switch loop in the 'hot' conductor) or are carrying different current (say because the circuit in question is part of a MWBC and the neutral is carrying different current than the 'hot') then the equation that is using these multipliers will give a wrong answer.

The more basic approach (meaning lower level equations but more complex because you don't get to use the simplifying assumptions) you need to calculate the current and voltage drop in each section of wire, and then add up the voltage drops, using vector addition to account for the different possible phase angles involved.

I am going to take the OP's example, and fix the values to 1AWG aluminum at 75C, 180 feet, 80A, and calculate the voltage drop in various different situations.

According to NEC chapter 9 table 8, 1AWG aluminum has a resistance of 0.253 ohms per 1000 feet, and I am going to use these value without correction (yet another simplification). A _single_ conductor in the example has a resistance of 180/1000*0.253 = 0.04554 ohms. With 80A of current the voltage drop on a _single_ conductor is 0.04554*80 = 3.64V. How this 3.64V _per conductor_ changes the % voltage drop depends on the particular situation.

1) Single phase L-L load. You have 208V, and _2_ 3.64V drops, in single phase, so the total % drop is 2 * 3.64/208 = 3.5%

2) Unbalanced 120V loading with 80A on one leg and nothing on the other. You have _2_ 3.64V drops, in single phase, but with a 120V base voltage. The total % drop is 2 * 3.64 / 120 = 6.1% (relative to the 120V)

3) Balanced 120V loading with 80A on each leg but no L-L loading. You now have three 3.64V drops to consider, on each hot and on the neutral. The phase angle between the hot voltage drop and the neutral is 120 degrees, so you will see a 1.732 factor; the voltage drop seen by the 120V loads is 1.732 * 3.64 / 120 = 5.3%

Contrast this if the supply were true single phase:

1) Single phase L-L load. You have 240V, and _2_ 3.64V drops, in single phase, so the total % drop is 2 * 3.64/240 = 3.0%

2) Unbalanced 120V loading with 80A on one leg and nothing on the other. You have _2_ 3.64V drops, in single phase, but with a 120V base voltage. The total % drop is 2 * 3.64 / 120 = 6.1% (relative to the 120V) This is the same as the 2 of 3 phases case, since we just are pulling a single 120V load.

3) Balanced 120V loading with 80A on each leg but no L-L loading. You now have three drops to consider, 3.64V on each hot but 0V on the neutral. The voltage drop seen by the 120V loads is 3.64 / 120 = 3.0%

-Jon

 

[1972Grady]

does the op's ckt have a neutral?
so he can serve 208 and 120 loads
if so, sqrt3
if not, 2

The circuit in question is this. I'm pulling 2 legs and a neutral from a 208v 3 phase meter stack to a 100A dwelling unit panel. The breaker feeding the unit is 100A so I'm basing my voltage drop off of 80% of the breaker which would be 80A. So would I use the multiplier of 2 or 1.732 to determine the voltage drop for the sub feed feeding the unit panel?