Parking Lot Vd.

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ronball

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Champaign Il.
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Electric Contractor
i'm not to good on voltage drop. I HAVE 7 PARKING LOT LITES ( 400WATT) EA. AT 208 VOLTS. TOTAL DISTANCE TO LAST LITE BEING ABOUT 1000 FEET.
WHAT SIZE WIRE WOULD I RUN. RON
 
Parking Lot Vd.

We Are Wiring 7- 400w Lites @ 208 V From Source. Total Distance To End Lite Being Around 1000ft. What Size Wires Do I Need To Run.
 
ronball said:
We Are Wiring 7- 400w Lites @ 208 V From Source. Total Distance To End Lite Being Around 1000ft. What Size Wires Do I Need To Run.
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Big ones. Gave you about the same info you gave us.
 
ronball said:
We Are Wiring 7- 400w Lites @ 208 V From Source. Total Distance To End Lite Being Around 1000ft. What Size Wires Do I Need To Run.
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when you say 208 volts from source does that mean 2 hot wires to each light? Or does that mean one hot with 208 (high leg) and a neutral?
 
Ron,

As others have said, you've not given enough information to answer the question.

1) How are the lights being connected? line-line or line-neutral? As a single phase circuit (2 hots) or as a three phase circuit (3 hots)?

2) What is the current draw of the lights? If you assume 208V and 400W and 'unity power factor' then you get about 2A...but lamps with ballasts have ballast losses and power factor, so the current flow is probably higher.

3) How much voltage drop can the lights tolerate? Some lighting systems will adjust for low voltage and work just fine; others will get dimmer and less efficient but otherwise be fine; still others won't function at all if the voltage is too low.

4) How is the circuit arranged? All lights evenly spaced in a row, all clustered 1000 feet away, or spread out over an area?

Generally you figure out how many amps each conductor is carrying, and how much voltage drop you can accept in that conductor. You then divide allowable voltage drop by current to get allowable resistance for that conductor.

-Jon
 
steelersman said:
Am I missing something here? Can't you use Ohm's law: I=P/E, I=400/208, I=1.923 Amps per light?
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if the op meant that the ballasts use 400W, then that's correct. But a lot of people tend to say its a 400W fixture when its really a ?W fixture with a 400W bulb, and in that case ohm's law wouldn't necessarily be close.
 
Without knowing the point to point distances I will just treat it like all seven fixtures are 1000' away.

The 400 watt ballast I linked to draws 2.2 amps at 208.

Thats 15.4 amps total, in my opinion that is to close to 16 amps to feel good about and would make the branch circuit protection 30 amps.

Back to voltage drop, I think you would need about a 3 AWG copper to stay under 3% drop.

Keep in mind that your EGC will have to be the same size as the circuit conductors, if you run 3 AWG hots you will have to run 3 AWG EGC. 250.122(B)
 
VD

VD

Lites Are Actually About 150 ' Apart, With The Last Lite Being About 1000' Away From Panel? Do I Not Need To Run A Neutral , And Only Run A Grd.
 
You don't need to run a neutral if the load doesn't need a neutral. 208V lights just need the two 'hots'; there is nothing to connect the neutral to. If you also have some other loads (such as control devices) that require the neutral, then run the neutral.

With lights spaced out as you suggest, an easy approximation is to treat the full load at the average distance; so take iwire's current number of 15.4A but at 575 feet rather than 1000 feet.

You start with the 208V. Now you multiply by the allowable voltage drop. Iwire used 3%, and that is a reasonable number, but you really need to make a design decision based upon the lights that you are using. If you can tolerate a 5% drop, then you can use smaller wire. Working with 3%: 208 * 0.03 = 6.24V

The allowed 6.24V is on two conductors together, so you only get 3.12V of allowed drop per conductor.

You take the allowed voltage drop and divide by current to get allowed resistance:
3.12 / 15.4 = 0.2 ohms.

From the allowed resistance you can calculate wire size. This is easy to do with tables. The tables normally list 'ohms per 1000 feet' or something similar.

The above calculation gives 0.2 ohms per 575 feet, or 0.2 / 575 * 1000 = 0.35 ohms per 1000 feet. Now you go to your table. The result is somewhere between a #5 and a #6 copper conductor; you can't buy #5 and #6 is slightly too small (given the above assumptions). If you accept a 3.5% drop then #6 works.

Note that the difference between the #3 that iwire suggested and the #6 that I mention is because of a change in assumptions going into the calculations. Iwire was calculating with the full load at 1000 feet, I was calculating with the load at an average of 575 feet.

Changing assumptions again: if you can run this as a 3 phase circuit, with 3 hots and the lights balanced between the 3 legs, then your current per leg goes down, and you can work with #8 conductors.

This is why the missing details are so key; just by picking different design parameters, the solution is anything from #3 conductors to #8 conductors.

-Jon
 
winnie said:
You don't need to run a neutral if the load doesn't need a neutral. 208V lights just need the two 'hots'; there is nothing to connect the neutral to. If you also have some other loads (such as control devices) that require the neutral, then run the neutral.

With lights spaced out as you suggest, an easy approximation is to treat the full load at the average distance; so take iwire's current number of 15.4A but at 575 feet rather than 1000 feet.

You start with the 208V. Now you multiply by the allowable voltage drop. Iwire used 3%, and that is a reasonable number, but you really need to make a design decision based upon the lights that you are using. If you can tolerate a 5% drop, then you can use smaller wire. Working with 3%: 208 * 0.03 = 6.24V

The allowed 6.24V is on two conductors together, so you only get 3.12V of allowed drop per conductor.

You take the allowed voltage drop and divide by current to get allowed resistance:
3.12 / 15.4 = 0.2 ohms.

From the allowed resistance you can calculate wire size. This is easy to do with tables. The tables normally list 'ohms per 1000 feet' or something similar.

The above calculation gives 0.2 ohms per 575 feet, or 0.2 / 575 * 1000 = 0.35 ohms per 1000 feet. Now you go to your table. The result is somewhere between a #5 and a #6 copper conductor; you can't buy #5 and #6 is slightly too small (given the above assumptions). If you accept a 3.5% drop then #6 works.

Note that the difference between the #3 that iwire suggested and the #6 that I mention is because of a change in assumptions going into the calculations. Iwire was calculating with the full load at 1000 feet, I was calculating with the load at an average of 575 feet.

Changing assumptions again: if you can run this as a 3 phase circuit, with 3 hots and the lights balanced between the 3 legs, then your current per leg goes down, and you can work with #8 conductors.

This is why the missing details are so key; just by picking different design parameters, the solution is anything from #3 conductors to #8 conductors.

-Jon
Click to expand...
Nice explanation Jon. Only question I have at this point is how can you use 3 phase for this application?
 
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