Voltage Drop on 3-wire Split Phase Circuit

[marti smith]

mivey

Pascal must have bored you to death. This is where the learning is. You are an inspiration.

 

[LarryFine]

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

Even then, I'd stil, stagger the conductor sizes as the loads diminish with circuit length.

 

[mivey]

Even then, I'd stil, stagger the conductor sizes as the loads diminish with circuit length.

Yes.

And I forgot to drop the load amps due to the voltage increase so you could use #10 for the 1st 2 160 ft sections and #12 for the remaining 160 ft sections.

 

[mivey]

mivey

Pascal must have bored you to death. This is where the learning is. You are an inspiration.

Thank you.

Well, I have programmed in many languages, Pascal being one of them. :grin:

I am often fascinated with these historic giants. They developed so many of the ideas we take for granted today, but made some of the most complex calculations you can find. Some of that stuff gives me a headache. I admire them for the leaps of faith they took, even when the scientific world at the time might have been ridiculing them.

 

[ramsy]

I don't think you can necessarily consider single phase loads as "balancing out" the neutral. ..may not be the best idea if you consider that you might lose some of the loads on one phase.

Agreements. My #8 neutral may need to be larger. Tungsten bulbs dim OK with voltage drop, but expired lamps on ckts w/ inductive ballast should not violate the other ballast-nameplate voltage / equipment listings (NEC 110.3b).

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.

There is an IEEE Red Book Std 141 formula for AC V-drop, illustrated after Table 9, in the 2005 NFPA 70 handbook. The last word on any formula tweaks may be found with member kingbp's "Validated & Verified" model called ETAP load flow. This apparently uses physical wires and loads for calibration.

I may have approximated conductor / terminal temperatures for checking NEC 110.14(C), but my results are usually stricter than NEC derating tables. My mileage also varies, since V-Drop standards and curricular convention seem to ignore industry Std 141 and AC power factors. No two answers are identical anyway, DC formulas may satisfy the conceptual exercise of raising Voltage-drop awareness, and the ball-park results may be conservative enough for most applications.

Re: my use of a #4 to satisfy a 3% VD, without an industry standard for "segmented" VD calcs it seemed to me isolating each segment of load (I) to panel (L) would find the highest impedance (Z) for the entire circuit.

A #14cu can handle 12.5A before a marked temperature rise, so if surrounding loads don't effect other wire segment impedance, total wire Length (Z) from panel to each load segment (Z) can find max circuit V-Drop. I admit, constant conductor temperature is not usually the case, and a poor assumption on my part.

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

Me too, so getting on the same page with everybody else I should sum each segment's load & length impedance between load points.

Summing V-drop segments on phase B now comes up #1cu ~ 3%.

 

[hardworkingstiff]

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.

I bet if one metered the neutral the data would support this to be a safe installation.

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

240-volt fixture is the way to go.

 

[mivey]

There is an IEEE Red Book Std 141 formula for AC V-drop, illustrated after Table 9, in the 2005 NFPA 70 handbook. The last word on any formula tweaks may be found with member kingbp's "Validated & Verified" model called ETAP load flow. This apparently uses physical wires and loads for calibration.

The formula you reference is indeed in the Red Book but is the approximate formula. The Red book also has the exact formula, which is the one I use.

The "formula tweak" I mention is a tweak to my own program which calculates line-line drops. The tweak I used gets me back to the line-neutral drop, which is the basic exact formula, and is correct, and is the last word according to IEEE.

The assumption here is that you know the resistance and reactance (easy enough), the current (also easy), and the load power factor (also easy). Doesn't really take rocket science to "validate and verify" the results.

There is no mysterious calculation that one program makes that another can't. Since the formulas are universally known, setting up a simple model or an exacting model, and making load assumptions are the remaining choices. That is true regardless of the software you use.

What is not calculated are the source and load vectors. I have derived those equations as well, but they are not really relevant for a simple voltage drop calculation.

 

[Wes G]

As I see it, you need to do a reverse calculation to allow for all loads and end up with appropriate voltage drop. To do this, start with the fartherst light and assume maximum amperage and work your way down.
At 2.5 A per lamp each side of the line would carry half the load, or 12.5A
1. 3.6V /12.5A =.288 max. ohms/1.6=max of .18ohms per m = #1wire to second light @ 320' (Odd numbers are listed after even numbers because last light has an even number)
2. 3.6V /12.5A =.288 max. ohms/1.44= max of .18ohms per m = #2wire to first light @ 160'
3. 3.6V /10A =.36 max. ohms /1.28=max of .28ohms per m = #3wire to fourth light @640'
4. 3.6V /10A =.36 max. ohms /1.12=max of .32 ohms per m = #3wire to third light @480'
5. 3.6V /7.5A =.48 max. ohms /.96=max of .6 ohms per m = #6wire to sixth light @960'
6. 3.6V /7.5A =.48 max. ohms /.8=max of .5 ohms per m = #6wire to fifth light @800'
7. 3.6V /5A =.72 max. ohms /.64=max of 1.12 ohms per m = #10wire to eighth light @1280'
8. 3.6V /5A =.72 max. ohms /.48=max of 1.5 ohms per m = #10wire to seventh light @1120'
9. 3.6V /2.5A =1.4 max. ohms /.32=max of 4.4 ohms per m = #14wire to tenth light @1660'
10. 3.6V /2.5A =1.4 max. ohms /.16=max of 8.75 ohms per m = #14wire to nineth light @ 1440'

 

[Wes G]

Ignore my previous post! 'Was using the wrong table etc. I know I should avoid the technical stuff that late at night!

 

[Smart $]

As I see it, you need to do a reverse calculation to allow for all loads and end up with appropriate voltage drop. To do this, start with the fartherst light and assume maximum amperage and work your way down.
At 2.5 A per lamp each side of the line would carry half the load, or 12.5A
1. 3.6V /12.5A =.288 max. ohms/1.6=max of .18ohms per m = #1wire to second light @ 320' (Odd numbers are listed after even numbers because last light has an even number)
2. 3.6V /12.5A =.288 max. ohms/1.44= max of .18ohms per m = #2wire to first light @ 160'
3. 3.6V /10A =.36 max. ohms /1.28=max of .28ohms per m = #3wire to fourth light @640'
4. 3.6V /10A =.36 max. ohms /1.12=max of .32 ohms per m = #3wire to third light @480'
5. 3.6V /7.5A =.48 max. ohms /.96=max of .6 ohms per m = #6wire to sixth light @960'
6. 3.6V /7.5A =.48 max. ohms /.8=max of .5 ohms per m = #6wire to fifth light @800'
7. 3.6V /5A =.72 max. ohms /.64=max of 1.12 ohms per m = #10wire to eighth light @1280'
8. 3.6V /5A =.72 max. ohms /.48=max of 1.5 ohms per m = #10wire to seventh light @1120'
9. 3.6V /2.5A =1.4 max. ohms /.32=max of 4.4 ohms per m = #14wire to tenth light @1660'
10. 3.6V /2.5A =1.4 max. ohms /.16=max of 8.75 ohms per m = #14wire to nineth light @ 1440'

Sorry to be so blunt, but your premise is flawed. It is impossible to have 3.6V drop at light #1 and also at light #10, with one exception, and that is if you use separate wires from the source to each load.

When operating under the 240V premise I posted a diagram for above, it is possible is to have:

0.72V drop for lights #1 and #2
another 0.72V drop (1.44V drop overall) for lights #3 and #4
another 0.72V drop (2.16V drop overall) for lights #5 and #6
another 0.72V drop (2.88V drop overall) for lights #7 and #8
another 0.72V drop (3.60V drop overall) for lights #9 and #10​

Note to mivey:
This is a network circuit with essentially two sources. The premise diagram I posted above is fairly accurate. However, there is a slight amount of voltage developed across the neutral wire sections I show as non-conducting. I considered it insignificant to the overall application.

Also, in the event of one fixture becoming disfunctional in the sense of non-conductive, the neutral conductor picks up the imbalance and voltage drop to the other working fixtures remains within nominal voltage tolerance. Of course that will change as the number of non-conductive fixtures powered by one leg increases.

I am of the opinion if one "bank" or major portion thereof is not working and not restored to working condition within a reasonable amount of time, the owner of the lighting array deserves the consequences.

 

[LarryFine]

Also, in the event of one fixture becoming disfunctional in the sense of non-conductive, the neutral conductor picks up the imbalance and voltage drop to the other working fixtures remains within nominal voltage tolerance. Of course that will change as the number of non-conductive fixtures powered by one leg increases.

If you follow this scenario to the extreme, you will end up with a single 120v circuit, with half of the power at half of the voltage, having the same current, resulting in the same voltage drop, so there's no increase in overall voltage drop.

This is one of the reasons we pro-MWBC guys like them; there's potentially half as much system-wide voltage drop as there would be with individual circuits; same with 3ph. It also points out why shared neutrals need not be counted as CCC's.

 

[Smart $]

If you follow this scenario to the extreme, you will end up with a single 120v circuit, with half of the power at half of the voltage, having the same current, resulting in the same voltage drop, so there's no increase in overall voltage drop.

Afraid not.

Say each balancing load tied into the neutral at the same location, say two fixtures every 320' rather than the OP's case of one every 160', all wiring of the same size. With one bank off in this scenario, the voltage drop at each load doubles... hence the "2" in the basic voltage drop formula... because the current running through this one bank of loads must travel through the neutral. When both banks are on, there is no neutral current, so the voltage drop per load is halved compared to the one bank off state.

This relationship is a bit more complex when balancing loads are not tied into the same point along the neutral run. For instance, in the OP's case and assuming 160' between the source and each subsequent load, where every other load is on a different leg, the balancing current (~2.5A) passes through every other section of the neutral. If the long bank is turned off, the current of the short bank adds up on it's return through the neutral conductor. There will be...

~12.5A on the section closest the source (whereas it was ~0A)
~10.0A on the next two (was ~2.5A and ~0A respectively)
~7.5A on the next two (was ~2.5A and ~0A respectively)
~5.0A on the next two (was ~2.5A and ~0A respectively), and
~2.5A on the last two in use (was ~2.5A and ~0A respectively).​

This will amount to much more voltage drop than while the long bank is on but less than double.

Note: the reason I'm using "~" to indicate current values greater than ~0A is because the current through each load becomes lesser with increased wire distance... i.e. lower voltage equates to lower current... something I see very few stepped VD drop calculations take into account. And my use of ~0A is an indication that the current in such sections of wire is not exactly 0A, but otherwise considered insignificant to the problem.

This is one of the reasons we pro-MWBC guys like them; there's potentially half as much system-wide voltage drop as there would be with individual circuits; same with 3ph. It also points out why shared neutrals need not be counted as CCC's.

This is roughly true as explained above, and only while all "banks" are on. Turn all but one off and and the one remaining on is no different than an individual branch circuit. To utilize the lesser voltage drop of a MWBC, it should be ascertained within reason that all banks will be on or off simultaneously.


FWIW, the following depicts my basic 3% VD calculation for the OP's case using Excel...

 

[mivey]

I am of the opinion if one "bank" or major portion thereof is not working and not restored to working condition within a reasonable amount of time, the owner of the lighting array deserves the consequences.

They do have a knack for finding the worst case scenario.

lower voltage equates to lower current... something I see very few stepped VD drop calculations take into account.

That's unfortunate. In addition to including adjustable parameters (like temperature), I think most should model a constant VA (like heavy industrial motors), constant impedance (typical for most commercial and residential loads), or constant current (essentially the same as 1/2 & 1/2 of the VA & impedance models). Some more sophisticated models use voltage & current measurements to define exponential load models or quadratic equations for the real & reactive power.

Afraid not...FWIW, the following depicts my basic 3% VD calculation for the OP's case using Excel...

Good answer.

 

[mivey]

lower voltage equates to lower current...

Good answer.

Ooops. I think we overlooked the fact that in the OP there was mention of ballasts being used, not incandescent.

If these are more like the universal electronic ballasts, these should be modeled more as constant VA loads instead of constant impedance loads, i.e., lower voltage equates to higher current.

 

[Smart $]

Ooops. I think we overlooked the fact that in the OP there was mention of ballasts being used, not incandescent.

Righto ye ar daer matey

If these are more like the universal electronic ballasts, these should be modeled more as constant VA loads instead of constant impedance loads, i.e., lower voltage equates to higher current.

I agree... and what of more traditional magnetic ballasts?

 

[mivey]

Righto ye ar daer matey


I agree... and what of more traditional magnetic ballasts?

For those, and even some electronic ballasts, the current drops with voltage.