3 phase VD on floodlights

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Dickieboy

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How do you calculate voltage drop on a 3 phase ,480volt distribution system to a series of flood lights on poles?

Lets say you had 3 floods on each pole connected across the individual Phases AB, AC and BC.This of course keeps your loading across phases balanced and then you parallel off to the next pole and so on and so on.

Lets assume each floods ballast draw is 2 amps and your run length to the first pole is 500 feet,you have 5 poles with 3 fixtures each and the poles are spaced 100 feet apart.

What size wire is required from the source to the first pole with a max of 3% VD and then on to succesive poles with the wire size deceasing between each pole as you gain circuit length and the load decreases until you have reached the last pole.

dick
 
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In the question sited above I'm looking for the wire size's required at the 5 poles each would be somewhat smaller than the preceeding,,,,,,,,,,,,the VD is fixed max at 3%.

In the formula you posted I cannot use as I do not know the CM or K,,my math is baaaaaaad!

I think somewhere in the past I got into to this problem and had to deal with increasing values of amp feet and a moving center of load but it all has escaped me.

dick
 
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How about (Using the IEEE exact volt drop formula):

Section,amps,Type,Size,Section ft,Total ft,pf,Source,vd,Load,Section %,Total vd,Total %
1,10.0,Cu,10-1,500,500,1.0,480.0,10.39,469.6,2.17%,10.4,2.2%
2,8.0,Cu,10-1,100,600,1.0,469.6,1.66,467.9,0.35%,12.1,2.5%
3,6.0,Cu,10-1,100,700,1.0,467.9,1.25,466.7,0.27%,13.3,2.8%
4,4.0,Cu,10-1,100,800,1.0,466.7,0.83,465.9,0.18%,14.1,2.9%
5,2.0,Cu,10-1,100,900,1.0,465.9,0.42,465.5,0.09%,14.5,3.0%

Add:
For AC, the IEEE std 141 formula is :
Vd = V + IRcos@ + IXsin@ - sqrt[V^2-(IXcos@-IRsin@)^2]

where:
Vd = L-N volt drop
V = source voltage
I = current
R = AC Resistance from NEC Table 9 (Ohms to Neutral)
X = AC Reactance from NEC Table 9 (Ohms to Neutral)
@ = phase angle
 
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Dickieboy said:
What size wire is required from the source to the first pole with a max of 3% VD and then on to succesive poles with the wire size deceasing between each pole as you gain circuit length and the load decreases until you have reached the last pole.
Click to expand...
What's the load current of each luminaire?
 
Dickieboy said:
You lost me Mivey,,,,,,,,,,,,,

Larry 2 amps

dick
Click to expand...
#10 cu will work

The other vd calcs use an estimated calc. The IEEE std 141 is a more precise calculation that is done section by section so:

Section=the 5 sections
amps=amps flowing through that section
Type=cu or al conductor
Size=conductor size
Section ft=section length
Total ft=length to original source
pf=power factor
Source=voltage at source end of section
vd=section voltage drop
Load=voltage at load end of section
Section %=% vd in section
Total vd=vd back to original source
Total %=% vd back to original source
 
I just noticed the amps were line-line :roll: so the line current is 3.5 amps, not 2. You will need a #6 (6-7 means #6 with 7 strands) to the first pole, the #10 for the rest:

Section,amps,Type,Size,Section ft,Total ft,pf,Source,vd,Load,Section %,Total vd,Total %
1,17.3,Cu,6-7,500,500,1.0,480.0,7.35,472.6,1.53%,7.4,1.5%
2,13.9,Cu,10-7,100,600,1.0,472.6,2.88,469.8,0.61%,10.2,2.1%
3,10.4,Cu,10-7,100,700,1.0,469.8,2.16,467.6,0.46%,12.4,2.6%
4,6.9,Cu,10-7,100,800,1.0,467.6,1.44,466.2,0.31%,13.8,2.9%
5,3.5,Cu,10-7,100,900,1.0,466.2,0.72,465.4,0.15%,14.6,3.0%

ADD:
for #6: R=0.49 ohms /1000 ft and X = 0.051 ohms /1000 ft
for #10: R=1.2 ohms /1000 ft and X = 0.05 ohms /1000 ft
 
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In a nut shell does this make any sense?

Example follows? Using only the amps for one phase as its a balanced 3 phase system and calculating the load center for just the first two poles as a working example.

1st pole 2 amps at 500'=1000 ampft
2nd pole at 2 amps at 600'=1200 ampft

Load center for this would be, 1000 + 1200 divided by 2 for 1100' with a load of 4 amps,then calculate VD based on a 3 phase system with a 4 amp load at 1100 feet

dick
 
ooops I forgot the 1.73 in my example so it would be a 7 amp load at 1100 feet

Mivey I'm back on board with you .Thanks

dick
 
Mivey,

I keep coming up with #12 CU all the way through would be acceptable.
Using the standard voltage drop calculation for 3 phase:

CM = (1.732 X K X I X D)/VD

At the 1st pole CM = (1.732 X 12.9 X 10 X 100)/14.4 = 1551

At the 5th pole CM = (1.732 X 12.9 X 2 X 500)/14.4 = 1551

Since 1551 is so small, voltage drop will not be a problem and you can just go off of the ampacity tables. What am a doing wrong?
 
derek22r said:
Mivey,

I keep coming up with #12 CU all the way through would be acceptable.
Using the standard voltage drop calculation for 3 phase:

CM = (1.732 X K X I X D)/VD

At the 1st pole CM = (1.732 X 12.9 X 10 X 100)/14.4 = 1551

At the 5th pole CM = (1.732 X 12.9 X 2 X 500)/14.4 = 1551

Since 1551 is so small, voltage drop will not be a problem and you can just go off of the ampacity tables. What am a doing wrong?
Click to expand...
These poles are in series so every time you add a new pole, the total current goes up. Plus, it is 500 feet to the first pole.

Add: so the 5th pole would have 900 feet and a current of 5*2*1.73
 
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The cmil method with #6 & # 10 yields these section total vd:
1.53%
2.15%
2.61%
2.92%
3.07%

The cmil method with #12 only yields these section total vd:
6.40%
7.42%
8.19%
8.70%
8.96%
 
Whatcha mean they are in series????????

3 phase 480 bus(feeder) with taps to lights,parallel

dick
 
Sorry, I don't know where I got 100 ft to the first pole from. But I believe the OP stated that the poles would be in parallel. So the last pole would only see the current draw by it.

CM = (1.732 X 2 X 12.9 X 900) / 14.4 = 2792 = #14
 
Dickieboy said:
Whatcha mean they are in series????????

3 phase 480 bus(feeder) with taps to lights,parallel

dick
Click to expand...
Are they all tapped at the same point? I thought you meant:
Go out 500 feet, the hit the first light.
Continue on aonother 100 feet, then hit the second light.
...
This is series (or loop)

As opposed to:
Run a conductor 500 feet to the first light.
Run a conductor 600 feet to the second light.
...
This is radial
 
derek22r said:
Sorry, I don't know where I got 100 ft to the first pole from. But I believe the OP stated that the poles would be in parallel. So the last pole would only see the current draw by it.

CM = (1.732 X 2 X 12.9 X 900) / 14.4 = 2792 = #14
Click to expand...
It does not matter what the pole sees, it matters what the feeder sees all the way back to the source.

Dickieboy said:
What size wire is required from the source to the first pole with a max of 3% VD and then on to succesive poles with the wire size deceasing between each pole as you gain circuit length and the load decreases until you have reached the last pole.
Click to expand...
 
No its run 500 tap off 3 lights ,go another 100 tap off three more lights etc etc This is where I keep going back to the 3 phase load center and doing one calc for overall

dick
 
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
 
Dickieboy said:
No its run 500 tap off 3 lights ,go another 100 tap off three more lights etc etc This is where I keep going back to the 3 phase load center and doing one calc for overall

dick
Click to expand...
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.

[edit: I left out "pole" in the original description]
 
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