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Author Topic: Voltage drop question  (Read 9479 times)

Mark Cadwallader

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Voltage drop question
« on: April 13, 2017, 07:21:58 pm »

I'm looking at bidding an outdoor event in a local city park. If they put the stage in the usual location, I figure that I'll need to run about 200 feet from the 14-50R to a spider box. The local rental house has 6/4 cable with CS connectors. My question is when I'm figuring the voltage drop, do I use the 240 volt table, or do I use the 120 volt table because each "hot" conductor is at 120 volts?

On a related note, what loss should I assume for each time I join another length of the 6/4 cable?  I assume that two 100' cables are better than four 50' lengths, but I haven't seen any data on the question.
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Tim McCulloch

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Re: Voltage drop question
« Reply #1 on: April 13, 2017, 07:29:10 pm »

I'm looking at bidding an outdoor event in a local city park. If they put the stage in the usual location, I figure that I'll need to run about 200 feet from the 14-50R to a spider box. The local rental house has 6/4 cable with CS connectors. My question is when I'm figuring the voltage drop, do I use the 240 volt table, or do I use the 120 volt table because each "hot" conductor is at 120 volts?

On a related note, what loss should I assume for each time I join another length of the 6/4 cable?  I assume that two 100' cables are better than four 50' lengths, but I haven't seen any data on the question.

You should use the 120v table as that is the voltage your load(s) operate at.

You are correct that additional connections increase the resistance of the circuit but I'd argue that it's not enough to worry about... However (you knew that was coming, right?) there is a Code limit to how may connections may be utilized to make your feeder run.  The connection at the line end (service) does not count, nor does the connection at your distro, but the limit is 2 connections IIRC.  The example I'm recalling was for a 250 ft run you'd need 2x 100 ft feeders and a 50 ft feeder; using 1x 100ft and 3x 50 ft were a no-go.  I'm not on the computer with the NEC or Id look it up.

For 200 ft, I'd have them rent a generator.
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Stephen Swaffer

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Re: Voltage drop question
« Reply #2 on: April 13, 2017, 10:04:04 pm »

I'm looking at bidding an outdoor event in a local city park. If they put the stage in the usual location, I figure that I'll need to run about 200 feet from the 14-50R to a spider box. The local rental house has 6/4 cable with CS connectors. My question is when I'm figuring the voltage drop, do I use the 240 volt table, or do I use the 120 volt table because each "hot" conductor is at 120 volts?

On a related note, what loss should I assume for each time I join another length of the 6/4 cable?  I assume that two 100' cables are better than four 50' lengths, but I haven't seen any data on the question.

What "table" are you referring to?  More out of curiousity than anything.  Voltage drop is calculated using ohms law, E=IR, the voltage you are running at doesn't really matter-until you try to calculate the "percentage" of voltage drop.  What is important to know is the current flowing through each conductor.  If you are running 120 V equipment from a 14-50R spider box, it  is presumably reasonably well balanced between the 2 phases-the voltage drop you would be concerned with is the more heavily loaded side (and remember-you are really concerned with peak load because that is when voltage drop will be the largest.)
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Steve Swaffer

Mark Cadwallader

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Re: Voltage drop question
« Reply #3 on: April 14, 2017, 12:53:27 am »

What "table" are you referring to?  More out of curiousity than anything.  Voltage drop is calculated using ohms law, E=IR, the voltage you are running at doesn't really matter-until you try to calculate the "percentage" of voltage drop.  What is important to know is the current flowing through each conductor.  If you are running 120 V equipment from a 14-50R spider box, it  is presumably reasonably well balanced between the 2 phases-the voltage drop you would be concerned with is the more heavily loaded side (and remember-you are really concerned with peak load because that is when voltage drop will be the largest.)

Pocket Ref, 3rd Ed., by Thomas J Glover. Page 149; "Wire Size vs. Voltage Drop"; maximum footage @ 120 volts, 1 phase, 2% max voltage drop. Copper wire, solid, 2-conductor, K=11 (77*-121*F).   I recognize that I'm using stranded wire, but K=11 for stranded as well.  There is a similar table for 240 volts on the same page.

I have thought about being sure to balance the load between the two legs, and it "should be" fairly close.  Thank you, gents.
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TJ (Tom) Cornish

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Re: Voltage drop question
« Reply #4 on: April 14, 2017, 07:27:19 am »

Pocket Ref, 3rd Ed., by Thomas J Glover. Page 149; "Wire Size vs. Voltage Drop"; maximum footage @ 120 volts, 1 phase, 2% max voltage drop. Copper wire, solid, 2-conductor, K=11 (77*-121*F).   I recognize that I'm using stranded wire, but K=11 for stranded as well.  There is a similar table for 240 volts on the same page.

I have thought about being sure to balance the load between the two legs, and it "should be" fairly close.  Thank you, gents.
Make sure you know if your calculator requires using the whole path length or just the simple distance - I.e. A 100' cord has 200' of circuit length.
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Mike Sokol

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Re: Voltage drop question
« Reply #5 on: April 14, 2017, 09:17:19 am »

Make sure you know if your calculator requires using the whole path length or just the simple distance - I.e. A 100' cord has 200' of circuit length.

And the interesting thing about voltage drop and overheated extension cords is that they really don't care about the actual voltage applied to them. It's the current that actaully creates the voltage drop and heating. Here's a video I made for the RV industry using a 3-volt Glo-Melt transformer to apply 30 amperes of current though a piece of 16 gauge extension cord. My transformer is only drawing around 100 watts (less than 1 amp) on the 120-volt side, but making 30 amps at 3 volts on the secondary side. The cable doesn't care if the one end of the wire has 120 volts and the other end has 117 volts, or the one end has 3 volts and the other end has 0 volts. It's all about the differential voltage which in this case is 3 volts. Of course in a actual hookup that would be 2 times 3 volts for both the hot and neutral run in the same extension cord, which totals 6 volts. And my short length of cable (maybe 10 feet) would have to multiplied to the length of the actual extension cord run. So I'm guessing that if you dead shorted the far end of a 100 ft, 16 gauge extension cord it might draw 30 amps of current. What's dangerous in the RV industry is that you can buy a 30-amp to 15-amp dogbone adapter that will actually let you load a skinny extension cord with 30 amperes. As you can see, in just 5 minutes or so it will approach the boiling point of water. https://www.youtube.com/watch?v=WZznobYGF_c
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Mark Cadwallader

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Re: Voltage drop question
« Reply #6 on: April 14, 2017, 09:44:57 am »

The book provides the Ohm's Law formula for voltage drop on the prior page, which shows the double length of the run, and walks through the calculation.  The tables are built for showing what is the max length of different gauges of line that will not exceed a 2% drop.

I'm now wondering what the other providers have done for other events with the stage in the usual place. I think I'll suggest the stage be placed in a different location, nearer the 14-50R. 
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Jay Barracato

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Re: Voltage drop question
« Reply #7 on: April 14, 2017, 10:47:16 am »

And the interesting thing about voltage drop and overheated extension cords is that they really don't care about the actual voltage applied to them. It's the current that actaully creates the voltage drop and heating. Here's a video I made for the RV industry using a 3-volt Glo-Melt transformer to apply 30 amperes of current though a piece of 16 gauge extension cord. My transformer is only drawing around 100 watts (less than 1 amp) on the 120-volt side, but making 30 amps at 3 volts on the secondary side. The cable doesn't care if the one end of the wire has 120 volts and the other end has 117 volts, or the one end has 3 volts and the other end has 0 volts. It's all about the differential voltage which in this case is 3 volts. Of course in a actual hookup that would be 2 times 3 volts for both the hot and neutral run in the same extension cord, which totals 6 volts. And my short length of cable (maybe 10 feet) would have to multiplied to the length of the actual extension cord run. So I'm guessing that if you dead shorted the far end of a 100 ft, 16 gauge extension cord it might draw 30 amps of current. What's dangerous in the RV industry is that you can buy a 30-amp to 15-amp dogbone adapter that will actually let you load a skinny extension cord with 30 amperes. As you can see, in just 5 minutes or so it will approach the boiling point of water. https://www.youtube.com/watch?v=WZznobYGF_c
We are currently doing simple resistive circuits in physics. I like to over emphasize that while we analyze in voltage (kirchhoff's loop law) the fundamental principle is the conservation of charge, I.e. current.

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Jay Barracato

Stephen Swaffer

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Re: Voltage drop question
« Reply #8 on: April 14, 2017, 12:50:36 pm »

In Mike's case, a 3 volt drop on a 3 volt supply = 100% voltage drop, on the 120 volt supply a 3 volt drop is 2.5%-usually considered acceptable-the chart is using a conservative 2%-but that would be the only difference between the charts is on the percentage calculation.

On the 3 volt circuit, you have 0% effeciency in your transmission of power, on the 120 volt 97.5%.  If you used the same current on the same wire at 240 volts you would still have 3 volts dropped, but now the voltage drop as a percentage is 1.25% resulting in an effeciency of 98.75 %.

Normally, the load will remain constant, not the current-so if you double the voltage, you halve the current resulting in half the voltage drop.  Now, in the example above you only have .625% voltage drop.  Which shows the advantage of running at a higher voltage.
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Steve Swaffer

Mike Sokol

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Re: Voltage drop question
« Reply #9 on: April 14, 2017, 07:28:34 pm »

In Mike's case, a 3 volt drop on a 3 volt supply = 100% voltage drop, on the 120 volt supply a 3 volt drop is 2.5%-usually considered acceptable-the chart is using a conservative 2%-but that would be the only difference between the charts is on the percentage calculation.

Exactly. However, because the heating effect is uniform across the length of the extension cord (resistor), my little 10 ft piece of test cable should exhibit the same localized temperature rise as a 100 ft cable. So if you take a 100 ft of 16 gauge extension cord and make pass 30 amperes of current, it should heat up similarly to my short short cable with 3 volts passing 30 amperes of current. I've simply created a small sample from the middle of a larger sample.
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Mike Sokol
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Re: Voltage drop question
« Reply #9 on: April 14, 2017, 07:28:34 pm »


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