3 phase VD on floodlights

[mivey]

Yes, I understand that the first pole is going to require #10.

Is this calculation incorrect?

First Pole: CM = (1.732 X 10A X 12.9 X 500) / 14.4 = 7757 = #10
Last Pole: CM = (1.732 X 2A X 12.9 X 900) / 14.4 = 2792 = #14

[edit: I mean yes, that calculation is wrong]
The last pole has a feeder that has 3.5 amps on it in the last 100 foot section, but has accumulated amps from the other poles as it makes its way back to the source. The feeder section to the first pole has 17.32 amps on it.

 

[derek22r]

What you have just described is what I calculated.

This is the same as saying:
Go out 500 feet, the hit the first light pole.
Continue on another 100 feet, then hit the second light pole.

Isn't this a parallel lighting. A series circuit would be a single phase system in which only one conductor runs between each light. In the case of a series circuit all light would see the same current, but in a parallel circuit the current the conductors see would decrease from one light to the next.

 

[mivey]

Isn't this a parallel lighting. A series circuit would be a single phase system in which only one conductor runs between each light. In the case of a series circuit all light would see the same current, but in a parallel circuit the current the conductors see would decrease from one light to the next.

Maybe that is the confusion. I'm not talking about a series circuit. I'm talking about a series volt drop calculation.

Maybe I should have used "daisy chained" or something else instead.

Think of "I want to hook up a series of lights" as opposed to "I want to wire the lights in series".

 

[derek22r]

Okay, well this is new to me. I guess I've never done a series voltage drop calc. I would have used #10 all the way through.

I think I see what you are doing. When doing series VD you must take into the account the VD that already exists in the line. I tried this and my numbers are a little different, but still show that you have to increase the wire size to the last lights.

1st 2.24% #10
2nd 2.59% #10
3rd 2.867% #10
4th 3.046% #10
5th 3.13% #10

or

1st 2.24% #10
2nd 2.59% #10
3rd 2.867% #10
4th 2.878% #8
5th 2.883% #8

I understand the numbers are different than yours and I'm not saying that mine are right and yours are wrong. In fact I'm pretty sure that yours are more accurate.

 

[Dickieboy]

I extended and calculated for a 3 phase circuit at 3.5 amps on an equivalent load center distance of 2625' at 3% and it came up to 2.5% on a #8AWG.

I think we are all together

dick

 

[mivey]

This is where I keep going back to the 3 phase load center and doing one calc for overall

dick

FWIW, you can calculate the volt drop several ways: using the IEEE exact method (see post #4), the IEEE approximate method (V = IRcos@ + IXsin@), or the cmil method (see post #2 where K is 12.9 for cu and 21.2 for al and you can find the cm in the NEC table 8).

As for the load center method, you still have to make a series calculation until you have reached the allowable volt drop %.

The load center method works fine (except that you need to divide by the current, not the # segments), but it does not reduce the problem down to one calculation:
12124.36 amp-ft / 17.3 amp = 700 ft load center

700 ft @ 17.3 amp with #8 = 3.41% vd
700 ft @ 17.3 amp with #6 = 2.14% vd

As you can see, this is not a complete solution so you have more work to do. You will have to start the next calculation using the accumulated vdrop of the source sections. This is how you make a series vd calc.

The next step would be to calculate the vd in segment 1 with #6. The load center for the remaining segments is 250 ft, 13.9 amps, 3464.1 amp-ft and you will need #10 and it gets you a 3% vdrop. Since there is no remaining vd capacity, you are done (unless you want other wire combinations).

If using a computer, you get the same result making the series vd calculation by segment.

This is quite simple to put in a spreadsheet. Start all segments with the smallest wire needed to get to 3% or less, then use this size for segment 1. Then let segments 2+ have a size that gets to 3% or less, then use this size for segment 2. Then let segments 3+ have a size that gets to 3% or less, then use this size for segment 3...

You will find that once you get to segment #2, you have the same result as in the load center calc, meaning you have run out of vdrop allowance and you are done when you get to the #10 in segment 2.

 

[mivey]

I extended and calculated for a 3 phase circuit at 3.5 amps on an equivalent load center distance of 2625' at 3% and it came up to 2.5% on a #8AWG.

I think we are all together

dick

Not quite. I've got a different load center number. See my previous post #26.

[edit: spelling]

 

[Dickieboy]

Mivey I'm fine with the approach whether one of us is off on are math is not important the approach is what I'm looking for . The numbers will change in actualality

thanks for all the help in jogging me back to an old prob. I ran into this ampft calc as a requirement on a single phase design on a plant in California and the local code inspectors required it,

dick

 

[mivey]

Ok. I was just concerned that you had a 2625' load center which is a world of difference from a 700' load center. Just thought you might have a problem with your fundamental formula.

I'm guessing you've got it figured out now.

Glad to help.