VOLTAGE DROP

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I am feeding roadway light poles about 200 feet on center with 277 volt and using a common neutral. The first pole will be on phase A, the second pole on phase B, the third pole on phase C, the fourth pole on phase A. the fifth pole on phase B and so on. Since I am using a common neutral I think Current Flow gets involved and Vector Analysis gets involved. Is there a computer program that I can use and if not is there a long hand formula that I can use. Some of the circuits are long (1500 to 1800) feet from the panel source. I need to make sure I am not exceeding a total of 5% VD.--If I use the conventional VD formula I think the answer will be larger than it really is with using a common neutral. I need to minimize my wire size so I am trying to find the correct computer program or long hand formula.-- I would appreciate any help I can get. Thanks. My e-mail address is --

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THANKS

[ January 09, 2006, 06:06 PM: Message edited by: charlie b ]
 
Re: VOLTAGE DROP

Why not use 480 and feed them all with one circuit?
Either way its just a standard 2 wire application, no complicated vectors required.
Compute each section seperatly, then add the load to the next section to get its volt drop
Make sure to use the actual VA of the ballast, not lamp wattage.
Some ballasts are sensitive to voltage dips and, what type are you using?
 
Re: VOLTAGE DROP

this is inserting question but few thing i will like to know before we can go on with this question here

1) what type of luminaire head you will be using [Ex Metal Halide or HPS or CFL ]

2) total foot/meter of the run of lumiaires from begin to the end of run.

3) numbers of lumaires total per circuit will each it have own photoelectric cell switching device or will use common photocell device to switch all lumiaires at the same time


those question i will like to know before we can give you the answer here


Merci Marc

P.S. if very long run i will try to suggest to check the lumiaire manufacter to see if they have 480 ballast for it.
 
Re: VOLTAGE DROP

Yesterday I posted a question about volt drop and in return received questions about my question. Answers are: 1. We are using 277 volt with common neutral from a 277/480 volt-3 phase-4 wire control panel because the Client asked for 277 thinking that if they lose phase A they will still have lights operating on phases B & C. 2. We are using Cobra Heads with HPF AutoReg Ballasts. 3. We are using 250 watt HPS lamps and 400 watt HPS lamps and I have the EOL amps for these lamps. 4. All lighting circuits will go back to same control cabinet with one lighting contactor controlled by one photocell. 5. A four conductor cable duct (Phase A,B,C and Common Neutral) will feed the lights on one side of the roadway. Every third light pole will be on phase A. So on phase A the first pole will be 300 feet from the control panel, second pole 500 feet from control panel, third pole 700 feet from control panel, fourth pole 900 feet from control panel, fifth pole 1100 feet from control panel and sixth pole 1300 feet from control panel. So the first pole will be 300 feet from the control panel and then the other light poles will be spaced 200 feet from each other. 6. This is a DOT project and the DOT tells me they have a formula to determine the volt drop but they legally can't give me their formula. 7. I previously worked on a project where we had a similar situation with light poles along a roadway and we fed the lights from a 240/480 volt single phase delta panel. In that case we were feeding the lights with 240 and using a common neutral. Every other light pole was on phase A. So in a continuous row the light poles were wired as phase A, phase B, phase A, phase B, etc. This was also a DOT project and the DOT gave us a diagram showing how to calculate the volt drop because with a common neutral there was a change in direction of current flows. The difference here is we have 3 phase instead of 2 and its a wye distribution and not a delta distribution. 8. I think we are going to have change in current flows with the 277 volt circuits but I can't find a method of calculating the volt drop. H E L P !!!
My e-mail is bwalldroff@jmt.com --THANKS
 
Re: VOLTAGE DROP

I'm not an E.E, but it sounds as though you are complicating this matter. Three phase systems are meant to be set-up using all three phases and one neutral, since the phases are 120 degrees apart. A standard method of computing vd should be fine in my opinion but I would assume this would give you more vd computed than you actually would have. Now I understand your issue, you are concerned with oversizing the conductors?

[ January 10, 2006, 03:55 PM: Message edited by: BruceH ]
 
Re: VOLTAGE DROP

If I'm reading your post right, it's just a 3-wire circuit with a shared neutral, not a 3-phase circuit. Kind of a risky setup in my humble opinion. If the neutral gets broken somewhere along the line you could potentially backfeed 480 into your ballasts. I think the technical term is - KA-BLOOM.

Someone correct me if I'm wrong. I only learn when I make mistakes.
 
Re: VOLTAGE DROP

I agree that the voltage drop equations work for whatever it is you're doing. They don't discriminate based on application type.

I think we are going to have change in current flows with the 277 volt circuits but I can't find a method of calculating the volt drop. H E L P !!!
You're going to be using this neutral in the same way it would be used for single phase. All of the current is going to be present on it. There are no special calculations. You'll need a bigger neutral than the phase conductors.

DOT. In case they drop a phase. :D
 
Re: VOLTAGE DROP

If the neutral gets broken somewhere along the line you could potentially backfeed 480 into your ballasts. I think the technical term is - KA-BLOOM.

Someone correct me if I'm wrong. I only learn when I make mistakes.

Unless "I" have it wrong, there would be no KA-BLOOM because there's only two wires connected to the ballasts. A lost neutral would only leave an open circuit.

Edit: Hey, what's the DOT gonna do to protect themselves if they drop the neutral? :D

[ January 10, 2006, 04:35 PM: Message edited by: physis ]
 
Re: VOLTAGE DROP

It is true that these three phase currents add vectorially. In this case any two load currents would add as follow:

Iload @0 + Iload @-120 = 2xIloadxcos(60) = Iload @-60

Add Iload @-240 which cancels Iload @-60

I would start at the far end and work backwards. If all lights are on, the neutral currents would be Iload, Iload, Zero. Then the pattern would repeat.

PM me if you want more detail.

Rattus
 
Re: VOLTAGE DROP

This is nothing to do with VD, but I would use the 480 set up as AB, BC, CA ect and just fuse each pole at the base. That way, if one ballast goes bad you dont loose the entire load (pleasing the customer) and you save money on the install.
 
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