20A 450' & VD

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Alwayslearningelec

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Feeding a (3) 20A loads from 20A breakers with distance of 420'. There will be (3) hots, (1) neutral & (1) ground in one conduit( not sure size yet)
I'm coming up with #2 wire taking into account VD. Is this correct?
 
If you have 3 un-coordinated single phase loads then voltage drop will vary depending upon what is on.

Have a single load on and all your current is through one hot and neutral, and you need #2 or #3. (The Southwire calculator says #2, but that is assuming the full 20A and also assuming conductors at 75C...but cooler conductors will have less resistance and less drop.)

Have two single loads and you still have full current on the neutral, but the voltage drop angle on the neutral is changed and you get slightly less drop.

Have all three single loads running and now your neutral is invisible for voltage drop purposes. Now #6 is just fine. Possibly #8 if your '20A load' is really a '20A circuit with a max 16A continuous load'.

-Jon
 
Not sure how any of these computations are being done without specifying the system voltage.

Obviously you need to confirm that the loads are 20A loads, not just loads that require a 20A circuit, since the actual current should be used for VD calculations.

winnie said:
If you have 3 un-coordinated single phase loads then voltage drop will vary depending upon what is on.
Click to expand...

Seems like based on your comments it would be economical to size the neutral larger than the 3 ungrounded conductors. E.g. something like a 1/0 neutral and (3) #4 ungrounded conductors (a guess, would need to confirm the VD in the various cases). That would be less copper than (4) #2s, not sure if it's less $.

Cheers, Wayne
 
You are correct, you need to know the system voltage to do voltage drop calculations.

However based upon the lengths and conductor sizes one can pretty well rule out 480/277V and 416/240V. 240/139V is very unlikely to have L-N loads, so that pretty much leaves 208/120V.

Interesting thought about using the smaller ungrounded conductors with a larger neutral. When the neutral is loaded the larger size reduces voltage drop, when the loads are balanced the system is 3 phase with reduced voltage drop. Plus increasing the ungrounded conductors means increasing the EGC, but not the neutral.

3 #6 hots, 1 #1 neutral, 1 #6 EGC on a 208/120V network in copper would have a voltage drop of 3.3% if only one 'circuit' were loaded at 20A, and 2.6% if all 'circuits' were evenly loaded at 20A.

-Jon
 
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On the 120 vd calc you used #6 ungrounded and #1 grounded 20 amp load. Care to share your math with me. I'm interested in how you did it. Never done it that way. Thanks in advance
 
Tulsa Electrician said:
On the 120 vd calc you used #6 ungrounded and #1 grounded 20 amp load. Care to share your math with me. I'm interested in how you did it. Never done it that way. Thanks in advance
Click to expand...
Using formulas, just omit the "2" in the single phase formula, repeat the calculation for each size, and add the voltage drops.

Using a VD calculator, just do the calculation twice, once for #6 and once for #1, then average the voltage drops.

Cheers, Wayne
 
Let say you have an MWBC with p ungrounded conductors (p = 2 or 3 for single split phase or for 3 phase, respectively) serving L-N loads and you need to satisfy voltage drop requirements, but you want to minimize the total copper required. What ratio of size should you use for the ungrounded conductors vs the neutral?

So call A the area (in unspecified units) of the ungrounded conductors and N the area of the neutral. The voltage drop requirements are going to be of the following form:

1/A + 1/N <= 1 (most unbalanced case, only one ungrounded conductor has current).
1/A <= 1 (most balanced case, all ungrounded conductor currents are equal).

where I've set a bunch of constants to 1 for simplicity (the optimal ratio of A/N won't depend on the constants, length of the run, etc). The former inequality implies the latter, as N >0, so we can ignore the latter. And we want to minimize pA + N while satisfying the inequality. That minimum will occur at the boundary when 1/A + 1/N = 1, i.e. N = A/(A-1).

So we need to find the value of A that minimizes pA + A/(A-1). Some algebra and calculus gives us that A = (p + sqrt(p))/p, and then N = (p+sqrt(p))/sqrt(p). Whence N/A = sqrt(p).

The upshot, if I did the math correctly, is that for a MWBC with 2 ungrounded conductors (single phase or 2 of 3 phases), the optimal ratio of neutral area to ungrounded area would be sqrt(2) =1.414, and for an MWBC with 3 ungrounded conductors (3 phase), the optimal ratio would be sqrt(3) = 1.732. If conductors came in arbitrary sizes.

Of course, as conductors come in fixed sizes, for AWG that would be 1-2 AWG sizes larger for the the former case, and 2-3 AWG sizes larger for the latter case. And there may be a chance to further shrinking one of the conductor types due to the oversizing necessitated by the discrete sizes available.

Cheers, Wayne
 
Ok I have to admit something new to consider.
I plan on doing the math. I see how this would work not sure I would. The voltage drop being greater on the ungrounded conductor. Once I do that that goes against informational note on VD. Then it allows an impact in the EGC since it is based off of ratio of ungrounded conductor. Another item for me to consider is conductor size terminated on the OCD. Next item for me to consider is the grounded conductor not having over current. protection.
Again new to me so I'm rationalizing the method pros and cons.

So what I will look at is 20 amp load (s). Voltage 208/120 three phase 4 wire multiwire branch circuit with 120 volt loads. Max unbalanced current on netutal 20 amps.
Distance of 420' one way.
I will look at not only the VD of the circuit load. I will look at the VD on the EGC and impact on the EGC on conductor withstand rating if the smaller phase conductor is a bolted ground fault at the load.
Not looking for a code violation. Looking to understand the concept and impact of the installation.

Food for thought, what if they use a #10 ungrounded and larger grounded ( netutal) conductor. Then they use a #10 EGC.

See ya in a few days. My math skills not that good.
 
Tulsa Electrician said:
On the 120 vd calc you used #6 ungrounded and #1 grounded 20 amp load. Care to share your math with me. I'm interested in how you did it. Never done it that way. Thanks in advance
Click to expand...

For openers I got the calculation wrong. Let me describe how to do it correctly.

Voltage drop is caused by the resistance and inductance of the wires. For loads on relatively small conductors at 60Hz you can often approximate things by ignoring inductance, and I did that here. Resistances simply add in series.

I looked up the resistance of #6 wire (NEC table 9) as 0.491 ohm/kFT, and the resistance of #1/0 wire as 0.122 ohm/kFT.

0.491 * 420/1000 + 0.122 * 420/1000 = 0.257 ohm

A key point here is that the resistance given at 75C, but since these conductors will be very lightly loaded they will be cool, so I recalculated the resistance at 25C. See the formula at the bottom of table 9. R(25) = R(75) * ( 1 + 0.00323 * (25 - 75)) Correct R = 0.215 ohm

Then you use Ohm's Law V = I R to get V(drop) = 0.215 * 20 = 4.3V = 3.6%

Sorry for rushing and missing the numbers by a bit.

-Jon
 
wwhitney said:
Some algebra and calculus gives us that A = (p + sqrt(p))/p, and then N = (p+sqrt(p))/sqrt(p). Whence N/A = sqrt(p).
Click to expand...

With the proper substitution, the algebra and calculus is simple enough to post. So again the goal is to find the minimum of pA + N on the curve 1/A + 1/N = 1.

Let b = N/A. Then 1/A + 1/bA = 1, and A = 1 + 1/b with N = b + 1. The target pA + N = p + p/b + b + 1. Differentiating with respect to b gives 1 - p / b2. Which is 0 when b2 = p.

Cheers, Wayne
 
Ok, I understand some of this pretty straight forward.
What I am struggling a touch with is where we're allowed to do this or better it does not. Then we reduce the C based on what.
Now the other calculus stuff this may take me a while.
Thanks for sharing this.
Interesting the way other do the math.
 
Now what if bolted ground fault.
We have a #6 which done the above way is all required. What impact if any on the circuit s ability to deliver clearing fault current or open OCD in time with out conductor damage.
Once I'm past that I would be inclined to go, ok.
 
Well it looks old dog new trick.
Thank you for sharing.
Put this one in the tool box of tricks.
When I share this with other I will also give the credit to you.
I see also in 2020 code on upside EGC how this could be applied.

Thank you
 
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