voltage drop and continuous load

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hamburgerpatty

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My application is a continuous lighting load of 78 amps at 240volt single phase.
My distance is 475' away from source.

My question is "Do I take in account my continuous load first before I calculate for voltage drop, or what?"

I come up with;

78 * 1.25 = 97.5 Amps for continuous
97.5 A *2*K*475/cm = 3/0 wire = 6.1 volt = 2.6% Vdrop

If I don't have to wory about the continuous load in my calcuation for voltage drop requirements.

I come up with
78 A *2*K*475/cm = 2/0 wire = 6.2 volt = 2.6% Vdrop
and the 2/0 wire would meet my requirements for continuous load @ 125% of FLA

What is the correct procedure for calcuating Voltage Drop while taking in a continuous load that is 475 feet away from the source?
 
Re: voltage drop and continuous load

The 125% is an NEC requirement that has nothing to do with the formula for voltage drop.

Use the 78 amps for your calculations. :)
 
Re: voltage drop and continuous load

I was going to be lazy and not do the math but I changed my mind.

Your formula is correct: 2KLI/cmil

this is what I'm using

K = 12.88 (for copper)
L = 475
I = 78

(2KLI) = 954,408

2/0 = 133100 cmil. 2KLI/133100 = 7.17v/240 = 2.99%

1/0 = 105600 cmil. 2KLI/105600 = 9.038v/240 = 3.77%

1 AWG = 83690 cmil. 2KLI/83690 = 11.404v/240 = 4.75%

1 AWG is good for 110 amps in the 60?C column.

I usually don't do loads like this so don't take my advice but I think because it's only lighting (no motors or other equipment) I would allow a little more voltage drop. I think I would use 1 gauge. :)
 
Re: voltage drop and continuous load

Keep in mind that HID lighting has a start up current that will exceed the normal running current of the fixture which can result in voltage starving the fixtures at startup.
What it is I havent a clue. :confused:
 
Re: voltage drop and continuous load

I got this from the site below that describes HID a little.

The ballast is an electronic module that has a circuit board lined with several small high current capacitors
Click to expand...
(I never heard of a capacitor being both small and high current but ok)

When current is first applied to a large discharged capacitance for a breif instant it acts like a short. It's very breif and as soon as the capacitance is charged it no longer draws current from the source. This is a discription of what happens with DC, AC is very different. I didn't read far enough to see if the ballast makes DC but I presume it does.

http://www.intellexual.net/hid.html#hid

:)
 
Re: voltage drop and continuous load

I just found that the link I posted and the information I sited is for automobile HID's.

It may or may not be relevant to AC HID's. :eek:
 
Re: voltage drop and continuous load

There was a similar thread on buck/boost transformer a while ago. Would the 3% total voltage drop for branch circuit to farthest load or 5% total voltage drop from feeder to farthest load apply in this case? I've heard about 10% rule but doesn't code dispute this? Section 210.19(A)1 FPN#4 In this case I'm assuming source is from circuit breaker so wouldn't you not want to go over 3% drop?

[ November 29, 2004, 04:43 PM: Message edited by: four ]
 
Re: voltage drop and continuous load

Originally posted by four:
There was a similar thread on buck/boost transformer a while ago. Would the 3% total voltage drop for branch circuit to farthest load or 5% total voltage drop from feeder to farthest load apply in this case? I've heard about 10% rule but doesn't code dispute this? Section 210.19(A)1 FPN#4
Click to expand...
Remember that the NEC does really care about voltage drop.

That aside the UL + or - 10% works perfectly with the NECs suggested + or - 5%.

The utility is also allowed (most times) + or - 5%.

The utility can be 5% low and the wiring in the building can drop it another 5% and you end up with 10% total drop. Still within the UL required operating range.
 
Re: voltage drop and continuous load

Thanks iwire that helps on the 10%. I've seen in the past people figure 10%from circuit breaker to motor and I didn't think that was right. I was wondering if his selection of #1 wire was correct because that had over 3% voltage drop and i was figuring no more than 3% from circuit breaker to furthest load?
 
Re: voltage drop and continuous load

My choise of 1 AWG is not correct under the 3% branch 2% feeder rule of thumb Four and I'm not trying to advise it but depending on the load I might still use it. I was actually hoping for some more educated opinions to be added to mine. :)

Editted for clarity.

[ November 29, 2004, 05:29 PM: Message edited by: physis ]
 
Re: voltage drop and continuous load

:)
Physis:
Thanks for the info I wasn't trying to correct anyone I was also trying to get more clarification in this matter because I see this problem often but it is hard to convince people there is problem when lights are on and motors are still turning.

[ November 29, 2004, 08:19 PM: Message edited by: four ]
 
Re: voltage drop and continuous load

We had a case one time a few years ago that we had several fixtures that would not ignite properly or not at all, After replacing everything else we notice that the voltage drop was fine when the lamp was lit but when it was trying to start it would drop to over 20%. We pulled in new wires and everything worked like it was suppose to. That's when we found out that the startup current was much higher than the running current.
 
Re: voltage drop and continuous load

Wayne, I thought your example sounded like something aside from or in addition to voltage drop. So I did some math to get an idea of the current difference you would have to have to go from 3% to 20% voltage drop at 240 volts. If you use the same 1 gauge at 475 feet you would have to go from 50 amps to get a 3.05% drop to 330 amps to get a 20.1%.

I think the situation you're describing has to be attributable to something other than or in addition to voltage drop.

In the senario I describe what should be a 50 amp load and have a 3% voltage drop would have to maintain a 330 amp draw to cause a 20% voltage drop.
 
Re: voltage drop and continuous load

Sam I think if I can remember the wire was a #8 with the load around 35-40 amps and the last light was about 300' all this was fed from a 208 volt circuit go figure? That's why the original electrician ran from the job as he didn't know what he was doing. And we didn't think to look at the wire size first. There were 5 4 head poles with 400 watt MH fixtures in a parking lot all fed from these two #8 running from one to the other. so with all these fixtures starting at the same time the start up current was high I don't remember how high. The first three in line from the building worked ok but it was the last two poles that were intermittent if they came on at all. And at the time we didn't even know how they were fed. We never thought someone would feed them like this.
Live and learn. :D

[ November 30, 2004, 01:28 AM: Message edited by: hurk27 ]
 
Re: voltage drop and continuous load

That arrangement's a little more delicate. I think you had too much voltage drop to start with and probably some bad connections to boot. :)
 
Re: voltage drop and continuous load

I appreciate all of your input.

My original question I believe was answered.
(What is the correct procedure for calcuating Voltage Drop while taking in a continuous load that is 475 feet away from the source?)
Hence the word "procedure"

If that were the case, my correct answer would be;
(78 A *2*K*475/cm = 2/0 wire = 6.2 volt = 2.6% Vdrop
and the 2/0 wire would meet my requirements for continuous load @ 125% of FLA)

By the way, the start up current for these fixtures out at that pole is 78 Amps @ 240 Volt
 
Re: voltage drop and continuous load

Hamburgerpatty, I'm still curious about one thing. I've checked my math a couple of times originally (standard for me) and I checked it again now. And I keep getting the numbers I posted. I'm not interested in trying to say you're doing something wrong but I am interested in knowing how you got your numbers. There are many ways to get an answer for this but I used the same formula you used and our numbers don't match.

Mostly just to make sure I'm not goofing it up somewhere (I have before) Can you post your math?

Edit: Sorry, I mean replace the variables with values. :)

[ December 02, 2004, 10:20 PM: Message edited by: physis ]
 
Re: voltage drop and continuous load

Physis, the magnetic ballast is essentially an inductor which limits the current in a fluorescent light. In some cases it provides a momentary overvoltage to get the light started.

The electronic ballast I presume does the same thing, and I presume it could provide DC which would eliminate the flicker in fluorescent lights, but there may be other considerations. Anyone know if electronic ballasts provide DC?
 
Re: voltage drop and continuous load

This is all I'm talking about Rattus :)

By me:
K = 12.88 (for copper)
L = 475
I = 78

(2KLI) = 954,408

2/0 = 133100 cmil. 2KLI/133100 = 7.17v/240 = 2.99%
Click to expand...
By Hamburgerpatty
(78 A *2*K*475/cm = 2/0 wire = 6.2 volt = 2.6% Vdrop
Click to expand...
Edit: I left out some math

[ December 03, 2004, 04:44 PM: Message edited by: physis ]
 
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