Voltage Drop on 3-wire Split Phase Circuit

[ronlow00]

OK we are having a debate and I was wondering if anyone has some input. We have a street lighting project which utilizes 120V ballasts (2.5A) and using a split wire system (L-N-L). Say we have 10 lights; 5 lights on one hot and 5 on the other, alternating the connected circuit as you progress down the street. Any thoughts on what the voltage drop should be? If you want to assume a distance lets assume 160' between lights (320' between lights being feed from the same hot conductor).

 

[Buck Parrish]

Are you sharing a neutral? We would need to know the wire size befor we can make a correct calculation. Their are guys on this forum. That probably can do it in their heads. How ever I am not one of them.
See here http://www.csgnetwork.com/voltagedropcalc.html I hope this does not take all the fun of the debate out of it.

 

[hardworkingstiff]

OK we are having a debate and I was wondering if anyone has some input. We have a street lighting project which utilizes 120V ballasts (2.5A) and using a split wire system (L-N-L). Say we have 10 lights; 5 lights on one hot and 5 on the other, alternating the connected circuit as you progress down the street. Any thoughts on what the voltage drop should be? If you want to assume a distance lets assume 160' between lights (320' between lights being feed from the same hot conductor).

The answer will be different depending on whether this is 120/240 or 208Y/120 "single-phase".

You didn't give us a wire size/type.

 

[Smart $]

OK we are having a debate and I was wondering if anyone has some input. We have a street lighting project which utilizes 120V ballasts (2.5A) and using a split wire system (L-N-L). Say we have 10 lights; 5 lights on one hot and 5 on the other, alternating the connected circuit as you progress down the street. Any thoughts on what the voltage drop should be? If you want to assume a distance lets assume 160' between lights (320' between lights being feed from the same hot conductor).

Not enough info to answer your question. Additionally, you may not be asking the right question. Typical recommendation is 3% or less voltage drop... but not required. For greater voltage drop, you typically sacrifice the length of operation of both ballasts and lamps in addition to reduced lumen output per lamp.

So we need to know...

• Do you want to limit the voltage drop to 3%? If not, what is the maximum percentage drop you wish to apply? OR
• If a size and type of wire has been predetermined, what is it?
• What is the distance from the power source to the first fixture?
• Are both hots to be energized and deenergized concurrently at all times?
• Are the ballasts standard or electronic?
• What wiring method is to be used?

 

[Smart $]

The answer will be different depending on whether this is 120/240 or 208Y/120 "single-phase".

You didn't give us a wire size/type.

For many, including myself, split phase is taken to mean 120/240.

 

[Smart $]

Are you sharing a neutral? We would need to know the wire size befor we can make a correct calculation. Their are guys on this forum. That probably can do it in their heads. How ever I am not one of them.
See here http://www.csgnetwork.com/voltagedropcalc.html I hope this does not take all the fun of the debate out of it.

This one is a little tricky to enter into the typical one load at "x" distance calculator... but it can be done using a little logic coupled with some basic math.

 

[hardworkingstiff]

For many, including myself, split phase is taken to mean 120/240.

DUH on me.

 

[Buck Parrish]

This one is a little tricky to enter into the typical one load at "x" distance calculator... but it can be done using a little logic coupled with some basic math.

That little dot beside his name, if you hover your curser over it . It reads "He's gone"
But yeah your right we need the wire size and it takes math.
My wife has a masters in math and ststistics. She knows nothing about electrical work, yet she can solve electrical calculation practice test questions very easily.

 

[ronlow00]

For this system it is a 120/240V system.
Yes the neutral is being shared.
Assume each ballast has a current rating of 2.5A.

We are shooting for for 3% voltage drop. Depending on how it is calculated we get from #8 -#4 AWG conductor sizes.

The debate is if the neutral aids in lowering the voltage drop by carying the unbalanced current.

Also let us know how you are calculating this.

Thanks

 

[mivey]

My first observation would be to use a higher voltage. But:

Just taking one leg: unity pf, 20 ft from supply to first light (180 for 2nd leg), 320 ft to next light...1300 ft to farthest light (1460 for 2nd leg)

You would need #8 CU to get <10% volt drop at the farthest light on 1st leg and #6 CU for 2nd leg.

You would need #4 CU to get <5% volt drop at the farthest light on both legs.

You would need #2 CU to get <3% volt drop at the farthest light on both legs.

 

[mivey]

The debate is if the neutral aids in lowering the voltage drop by carying the unbalanced current.

Also let us know how you are calculating this.

Thanks

I think the neutral would reduce the volt drop but I calculated each sequence of lights without a shared neutral.

I used a simple sequence calculation using the exact volt drop formula.

PS: I can't see how you get to a #8. Even at 240 volts on a sequence set you would need #6 on the shortest run.

 

[ronlow00]

So from the math I take it you took each circuit as a single phase circuit.

Is this the correct method or does the shared neutral and balancing loads need to considered when sharing the neutral?

That is how one of us calculated the voltage drop but a Contractor is telling us we are way off and the conductors are about double the size of what would be actually needed.

 

[ronlow00]

I streched out some of the distances to make them all the same and easier for discussions

 

[mivey]

So from the math I take it you took each circuit as a single phase circuit.

I did and it would be a conservative approach.

Is this the correct method or does the shared neutral and balancing loads need to considered when sharing the neutral?

That would be more precise.

That is how one of us calculated the voltage drop but a Contractor is telling us we are way off and the conductors are about double the size of what would be actually needed.

I doubt it is double but I would have to think about it. Maybe I'll get froggy and modify my spreadsheet.

 

[hardworkingstiff]

As a contractor, I will say a #4 would give reasonable service. I would also warn that they would need to run both circuits to keep voltage drop % lower than running just one circuit.

 

[iMuse97]

#4AWG sounds about right to me, off hand. On a project with a little less distance than mentioned by OP, we ran #6 for 9 light poles, on two circuits, 120/240.

 

[Smart $]

...Also let us know how you are calculating this.

The attached pdf shows the premise on which I determine the numbers to enter into a one load, one distance, one size wire calculator.

View attachment 3715

Note I assumed 160' to the first fixture. Works out to a #1 for 2.6% VD... but that will depend a little on actual versus assumed variables.

 

[winnie]

You only have voltage drop if you actually have current flowing through the conductor. Since the shared neutral carries less current, you get less total voltage drop. In the limit of perfectly balanced loading, you have no current in the neutral, and no voltage drop on the neutral.

Consider a 120V circuit, 1000 feet long in 8ga wire, with 12.5A of load right at the very end. This is vastly simplified from your original calculation question, but uses similar distances and loads.

8ga is about 0.6 ohms per 1000 feet, so on the 'hot' conductor we have 0.6 ohms and 12.5A, for 7.5V of drop. We also have 0.6 ohms on the 'neutral' conductor and 12.5A, so again we have 7.5V of drop. The total drop seen by this 120V circuit is 15V, or 13%.

Now consider a 120/240V multiwire circuit, again 1000 feet long in 8ga wire, where we have _added_ another 12.5A of load, so that now we have two 120V 12.5A loads. For the first load (the 'black' load) we have 1000 feet of hot conductor, with a resistance of 0.6 ohms, and the same 7.5V drop we had above. But now there is _no_ current flowing on the neutral. Since there is no current flow on the neutral, there is no voltage drop. Thus the 'black' load has only 7.5V of drop, or 6.5%. The same applied to the 'red' load.

Another way to look at this is that you still have the same 15V of drop, but now it is 15V of 240V, or 6.5%, rather than 15V of 120V.

If there is any way you can reasonably operate at higher voltage, then you will significantly reduce your voltage drop for a given conductor size.

-Jon

 

[ramsy]

Avoiding plans or engineering with Voltage Drop calcs like this is always fun. Many engineers have trouble with some VD calcs, unless using a verified modeling tool.

Forum member "kingpb" used to share his corporate-purchased, modeling-systems results on this forum, which some tried to emulate with spreadsheets. The problem here breaks my spreadsheet, without some agreement on neutral current thru each section of the circuit.

Jon beat me to the point, but here's a similar way neutral current could affect wire size, if AC power factor = 1.

Assuming all lights are working on a 3-wire 120/240v ckt, one obvious place neutral current wont equal zero (cancel between phases) is End Of Line, 2.5A, 1600 ft total, shown on Smart'$ diagram.

A #8cu neutral is under 3% VD when plugging in 2.5A, 800ft for calculators that need 1-way distance.

Using Smart'$ diagram, size phase B by checking max VD at each load.

Wire between panel & Load1
320 ft. 1-way with 12.5A from all fixture loads = #6cu < 3%

Wire between panel & Load2
640 ft. 1-way with 10.0A from 4 fixture loads = #4cu < 3%

Wire between panel & Load3
960 ft. 1-way with 07.5A from 3 fixture loads = #4cu ~ 3%

Wire between panel & Load4
1280 ft. 1-way with 05.0A from 2 fixture loads = #4cu < 3%

Wire between panel & Load5
1600 ft. 1-way with 02.5A from 1 fixture load = #6cu < 3%

If this method is Ok for < 3% VD, then the smaller wire works.
#8 neutral,
#4 hot on phase B,
#6 hot on Phase A.

 

[mivey]

Avoiding plans or engineering with Voltage Drop calcs like this is always fun. Many engineers have trouble with some VD calcs, unless using a verified modeling tool.

Forum member "kingpb" used to share his corporate-purchased, modeling-systems results on this forum, which some tried to emulate with spreadsheets. The problem here breaks my spreadsheet, without some agreement on neutral current thru each section of the circuit.

I find it entertaining as well. It keeps the old cobwebs cleaned out.

Modeling tools verified by whom? We have found many errors in canned software. It always pays to make your own verification. I wrote my own software as well as the software for the company where I worked. I guess since they paid me you could say it was corporate-purchased.

Jon beat me to the point, but here's a similar way neutral current could affect wire size, if AC power factor = 1.

Assuming all lights are working on a 3-wire 120/240v ckt, one obvious place neutral current wont equal zero (cancel between phases) is End Of Line, 2.5A, 1600 ft total, shown on Smart'$ diagram.

A #8cu neutral is under 3% VD when plugging in 2.5A, 800ft for calculators that need 1-way distance.

Using Smart'$ diagram, size phase B by checking max VD at each load.

Wire between panel & Load1
320 ft. 1-way with 12.5A from all fixture loads = #6cu < 3%

Wire between panel & Load2
640 ft. 1-way with 10.0A from 4 fixture loads = #4cu < 3%

Wire between panel & Load3
960 ft. 1-way with 07.5A from 3 fixture loads = #4cu ~ 3%

Wire between panel & Load4
1280 ft. 1-way with 05.0A from 2 fixture loads = #4cu < 3%

Wire between panel & Load5
1600 ft. 1-way with 02.5A from 1 fixture load = #6cu < 3%

If this method is Ok for < 3% VD, then the smaller wire works.
#8 neutral,
#4 hot on phase B,
#6 hot on Phase A.

The best way would be to model the system impedances and loads.

I have written a program for this and also have some canned software for such things, but where is the fun in that? Besides, my program is not graphically driven and is a little more cumbersome to set up the data than in Excel. It was written back in the Fortran/Basic days. While I have modified it from time to time, I just have not felt like converting it to take full advantage of Visual. The exercises I do here are strictly for my enjoyment.

I like to take the chance to walk back through the calcs. Using canned software and not walking back through the calcs sometimes can come up and bite you because you might forget some of the under-lying assumptions that were made for the program.

Now consider this exercise for fun:

I made a quick and dirty modification of my spreadsheet. To keep from re-writing the whole thing, I made 3 calcs: one for phase A, one for phase B and one for the neutral, with the correct amps in each line section. I assumed the loads were simple so that equal currents from A & B will cancel at the neutral. This means that 1/2 of the neutral line segments have no current and the other 1/2 have 2.5 amps.

So, my thought was that the normal single-phase calc assumes that the neutral is the same as the phase and uses a multiplier of "2" (3-phase calcs use a 1.732 multiplier). So if I take 1/2 of the % drop from the A or B calc, and add it to 1/2 the % drop from the neutral calc, I would get the total drop. I may add that this is approximate for this circuit, but much closer than using no shared neutral.

See if you agree with this logic or have I over-thunk it?

Anyway, using Smart $'s lengths, allowing for about a 3% drop, with <1.5% on the neutral, the end results for the short run, long run, and neutral are: 100% pf: #3|#2|#8 = 1.90%|1.65%|1.13% => short=3.03% , long=2.77% .... 90% pf: #3|#2|#8 = 1.87%|1.65%|1.05% => short=2.92% , long=2.70%

Using equal sized wire yields: 100% pf with #4 = 2.36%|2.68%|0.45% => short=2.81% , long=3.13% .... 90% pf with #4 = 2.29%|2.60%|0.45% => short=2.72% , long=3.03%

You can compare this with the conservative (non-shared neutral) results of: 100% pf: short run with #2 = 2.74%, long run with #1 = 2.60% .... 90% pf: short run with #2 = 2.75%, long run with #1 = 2.69%

This means a #2 or #1 in the conservative case vs a #4 for the more precise case (but maybe not so smart...more later): 1.6 times the cmil in one run and 2 times the cmil in the other run for an average of 1.8 times. That is pretty close to doubling the wire size so the contractor was correct along that line of thinking.

I was wrong in thinking the neutral current would be reduced by a fraction when it is actually reduced down to one fixture load for 1/2 the segments and zero for the other.

Now for the rub: I don't know that a total size reduction is such a good idea. Your worst case scenario would be losing one leg (or all the lights on one leg) and the neutral would then carry full current from the other leg and you would have a full 2-wire voltage drop (the conservative case).

I don't think you can necessarily consider single phase loads as "balancing out" the neutral. Otherwise we could say we will balance our panels and use a very small neutral. Three-phase loads can be excluded from the neutral of course.

Using ramsy's conductors yields:

100% pf: #6|#4|#8 = 3.73%|2.68%|1.13% => short=4.86% , long=3.81% .... 90% pf: #6|#4|#8 = 3.53%|2.60%|1.05% => short=4.58% , long=3.64%

In conclusion, you could reduce the wire sizes for volt drop with the shared neutral, but it may not be the best idea if you consider that you might lose some of the loads on one phase.


A final point: You could just use 240 volt fixtures and you could just use #8 all the way with no neutral.