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Alwayslearningelec

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If you has this scenario where you had l6ight poles, 1st @ 100' from source, 2nd 300', 4th 400' and so on to the last on @ 700'

1. If you had to run say #1's due to voltage drop and all those lights were on the same circuit you would have to bring the #1's throughout the entire run all the way to the end correct? You would then splice down to a smaller conductor at the fixture.
2. What would be the best way to down size the #1 conductor to terminate on a 1P-20A breaker?

Thanks.

And maybe I'm not figuring this correct. I'm plugging into VD drop calculator.
Each light draws ~1amp but for the load rating input I'm putting in 16A( 80% of the 20A breaker).

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To get the actual Vd at the end of the run you would have to run the calculation several times and add them up because each time you pass a branch point going away from the source the current drops by the amount of current that is diverted into that branch, and you need to consider the Vd up to each branch point as well.
 
ggunn said:
To get the actual Vd at the end of the run you would have to run the calculation several times and add them up because each time you pass a branch point going away from the source the current drops by the amount of current that is diverted into that branch, and you need to consider the Vd up to each branch point as well.
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Right so essentially it wouldn't be 16 amps needed at the end of the run. But that's easy to figure I guess of I 'm putting 16 light poles on a 20A circuit as they draw 1 amp each.

Your saying I'd have a 16A to the first light, then 15A at the second etc. dropping 1A at each fixture???
 
horsegoer said:
If you has this scenario where you had l6ight poles, 1st @ 100' from source, 2nd 300', 4th 400' and so on to the last on @ 700'
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I assume you mean 7 light poles, spaced 100' apart, with the last one at 700'. Note that voltage drop calculations should always be done using the actual circuit current, not anything based on the circuit or OCPD rating. If you are planning for future expansion, you should just include all the future loads now.

If this is a 2-wire circuit of constant wire size, and each light is a constant current load, independent of voltage (at least up to the variation that VD will cause), then all that matters is the sum of current * distance over all the lights. I.e. if each light draws 1A (input current to its driver), then you can sum 1A * 100' + 1A * 200' + 1A * 300' . . . = 2800 A-ft. You get the same answer if you look at the current on each segment individually: 7A * 100' + 6A * 100' + 5A * 100' . . . = 2800 A-ft.

Now you can use a VD calculator and enter any combination of amps and distance that multiply to 2800A-ft. E.g. 7A and 400 ft (which represents the total current and the average distance to a lamp). That should give you your minimum wire size for the case that all segments are the same.

If you want to reduce the wire size as you go along, the math gets more complex. If you're willing to change wire size for each of the 7 segments, then there is some choice for each segment that will minimize the total copper while keeping the VD at any specific limit, but that would be a lot of work to determine, and (possibly) 7 different sizes isn't practical. More to follow.

Cheers, Wayne
 
A reasonable one round optimization might go like this: determine the minimum single size wiring that would work for your VD limit, per the above. Say your VD limit works out to 7.2V (which I chose just because that's 3% of 240V, both of which are plausible values). Since wire sizes are discrete, the smallest size that will give you under 7.2V will actually provide some lower VD. To make up a number, maybe it's 6.5V. That means you have an extra 0.7V of headroom to possibly take advantage of.

In which case you could say "okay, if we downsize starting after the first 3 lights, what size wire could we use?" You'd calculate the VD for the first 400' per the above method (1A * 100' + 1A * 200' + 5A * 300' = 7A * 100' + 6A * 100' + 5A * 100' = 1800 A-ft) for the given wire size, which would be just 6.5V * 1800/2800 = 4.2V. Then you'd look at the remaining 400' (1A * 100' + 1A + 200' + 1A * 300' + 1A * 400' = 4A * 100' + 3A * 100' + 2A * 100' + 1A * 100' = 1000 A-ft = 2800 A-ft - 1800 A-ft), and calculate the minimum wire size for your remaining VD budget (which would be 7.2V - 4.2V = 3.0V).

Your first segment is dropping 6.5 / 2800 = 2.3 mV / A-ft, while your second segment is then allowed to drop 3.0 / 1000 = 3.0 mV / A-ft. For the assumption that impedance is inversely proportional to area, that means you second segment could have 2.3/3 = 77% of the area. That means the second segment could be 1 AWG smaller (which is a reduction to 79% of the area).

The upshot for this example, where my made up numbers meant that the actual VD given the wire size worked out to be 90% of the allowable VD, is that mixing wires sizes doesn't seem worth it.

Cheers, Wayne
 
One might consider the use of a MWBC since you have nearly equal loading of the circuits should cut voltage drop in half.
 
wwhitney said:
A reasonable one round optimization might go like this: determine the minimum single size wiring that would work for your VD limit, per the above. Say your VD limit works out to 7.2V (which I chose just because that's 3% of 240V, both of which are plausible values). Since wire sizes are discrete, the smallest size that will give you under 7.2V will actually provide some lower VD. To make up a number, maybe it's 6.5V. That means you have an extra 0.7V of headroom to possibly take advantage of.

In which case you could say "okay, if we downsize starting after the first 3 lights, what size wire could we use?" You'd calculate the VD for the first 400' per the above method (1A * 100' + 1A * 200' + 5A * 300' = 7A * 100' + 6A * 100' + 5A * 100' = 1800 A-ft) for the given wire size, which would be just 6.5V * 1800/2800 = 4.2V. Then you'd look at the remaining 400' (1A * 100' + 1A + 200' + 1A * 300' + 1A * 400' = 4A * 100' + 3A * 100' + 2A * 100' + 1A * 100' = 1000 A-ft = 2800 A-ft - 1800 A-ft), and calculate the minimum wire size for your remaining VD budget (which would be 7.2V - 4.2V = 3.0V).

Your first segment is dropping 6.5 / 2800 = 2.3 mV / A-ft, while your second segment is then allowed to drop 3.0 / 1000 = 3.0 mV / A-ft. For the assumption that impedance is inversely proportional to area, that means you second segment could have 2.3/3 = 77% of the area. That means the second segment could be 1 AWG smaller (which is a reduction to 79% of the area).

The upshot for this example, where my made up numbers meant that the actual VD given the wire size worked out to be 90% of the allowable VD, is that mixing wires sizes doesn't seem worth it.

Cheers, Wayne
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Thank you very much for that response . So actually your wire might get smaller as you go further and have less load? I definitely need to digest everything you wrote because I’d like to make sure I’m figuring it properly. What size wire would you run for the situation I gave ?
 
horsegoer said:
Thank you very much for that response . So actually your wire might get smaller as you go further and have less load? I definitely need to digest everything you wrote because I’d like to make sure I’m figuring it properly. What size wire would you run for the situation I gave ?
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You didn't specify the voltage, or confirm that it's 7 lamps. And is 1A definitely the full input current to the fixture?

At first glance, it's probably worth downsizing the wire, although in a longer example it might be.

Cheers, Wayne
 
wwhitney said:
You didn't specify the voltage, or confirm that it's 7 lamps. And is 1A definitely the full input current to the fixture?

At first glance, it's probably worth downsizing the wire, although in a longer example it might be.

Cheers, Wayne
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Thank you very much Wayne.
I have a few different scenarios.
One with the 7 lamps I posted and another with 15 light posts about 120' apart.
Yes current draw is 1A.
Voltage is 120V.
 
Some preliminary comments: depending on the fixture details, they might work fine on a voltage well under 120V, in which case you could use smaller wires and accept the resulting voltage drop and power waste. Also, if this is still in the design phase, you'd be better off with higher voltage fixtures, 208V or 240V.

So let's do one example (which will be the worst case mentioned so far), you can modify for other cases. 120V 2-wire supply, 15 fixtures drawing 1A each, spaced 120' apart regularly, 1st one at 120' from the source, last one at 15*120' = 1800' from the source. 2% allowable VD, so 2.4V. One size conductor throughout.

We saw earlier that the voltage drop at the end will be the same as a single 15A load located at the average distance away, which is 960'. 2.4V / 15A = 0.16 ohms, that's the allowable impedance of the circuit. And the circuit length is 2*960' = 1920'. So we need a conductor impedance of at most 0.083 ohms / kft.

If we use (2017) NEC Chapter 9 Table 8 DC Resistance at 75C, then we need a 3/0 Cu or 300 kcmil Al conductor. If we increase the allowable VD to 5% (6V), then our maximum conductor impedance rises to 5/2 * 0.083 = 0.208 ohms/kft, which would allow #2 Cu or 1/0 Al.

There's also Table 9, which gives impedances for circuits with 0.85 PF in different conduit types. Since we don't have the PF, not sure if it's appropriate to use. But for comparison, if we assume PVC conduit, for the 2% VD case we need 4/0 Cu or 300 kcmil Al; for the 5% VD case, we need #2 Cu or 1/0 Al. So almost the same answers, except for the 2% VD, copper conductor case.

Cheers, Wayne
 
To look at the benefit of an MWBC, let's suppose the supply is 208Y/120 and we run a 4-wire MWBC. Wire the 15 fixtures in rotation, so the first is A-N, the second is B-N, the third is C-N, etc. A slightly non-conservative simplifying assumption is that there's no VD on the neutral. That would correspond to zero current on the neutral, while for the idealized case (no harmonics) the neutral current will be either 0A (between the sets of 3 fixtures) or 1A (within a set of 3).

With that assumption, now the current on the ungrounded conductors is only 5A each at the supply end, and the mean circuit length is only 960'. 2.4V / 5A = 0.48 ohms, and for 960', the allowable impedance is up to 0.500 ohms/k-ft. Using Table 9 PVC conduit that gives #6 Cu or #4 Al.

The upshot is that for 2% VD, PVC conduit, and Al conductors, you could use a 2-wire circuit and need (3) 300 kcmil Al conductors in the conduit (since for a 20A circuit, the EGC has to match the ungrounded conductors in size). Or you could use a 4-wire MWBC and need (5) #4 Al conductors.

Cheers, Wayne
 
horsegoer said:
Why is that ?
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Assuming your using 120volt circuit, the perks of utilizing a MWBC is you only have to pull one additional wire to basically have 2 circuits (although it’s considered 1 circuit). And equally loading both ungrounded conductors of the circuit you will have basically 0 neutral current if the lights all operate simultaneously therefore working as a 240v circuit and cutting your voltage drop down by half.
 
horsegoer said:
Right so essentially it wouldn't be 16 amps needed at the end of the run. But that's easy to figure I guess of I 'm putting 16 light poles on a 20A circuit as they draw 1 amp each.

Your saying I'd have a 16A to the first light, then 15A at the second etc. dropping 1A at each fixture???
Click to expand...
Yes. Also the voltage will be less at each node (fixture) due to the voltage drop up to that point..
 
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