Calculations For Voltage Drop Of Wire - Electrical - DIY Chatroom Home Improvement Forum

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05-25-2011, 03:40 PM   #1
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## Calculations for voltage drop of wire

I was just talking with a design tech that I know and was comparing voltage drop data. I was surprised how far apart we were and it seems he uses a different method for the calculation. Mine seem to arrive at the same number as about any calculator online but his were up to 3X as much voltage drop.

Just curious what formula and tables references you use? Let's leave the actual application and and loads and such out of it and concentrate on the actual value. I would also like to know if you guys look mostly at 90c for service entrance wire or if you build for 75c?

Thanks

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05-25-2011, 04:34 PM   #2
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Were you using the total length of the circuit which is the out and back distance or only out? Were you using the same design load?

Your wire needs to be sized based on the lowest rated part of the system. You will not find terminations rated 90C, but the wire can be derated from the 90C ratings.

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Answers based on the National Electrical Code. Local amendments may apply. Check with your local building officials.

 05-25-2011, 04:48 PM #3 Member   Join Date: Jul 2009 Posts: 244 Rewards Points: 242 Because of the length of run, wire heat is not as much an issue as Vdrop. Loads are being calculated phase - phase with equal Vdrop between them so of course it would double. I am more or less looking for hard calculations here. As for the temp rating, I thought one could use a higher current rating but lower temp rating to offset operating higher temp wire? Is that not the case? Obviously the NEC will limit the current in the conductor to a safe level anyway and right now our wires are looking to run about 55C but Vdrop is a concern. I am working back through the physics of it and going to compare the NEC book to actual periodic table data just too see if there are engineered margins there. Sometimes too many compounded margins ends up getting ridiculous IMO. Last edited by viper; 05-25-2011 at 04:51 PM.

 05-25-2011, 06:57 PM #4 Lic Elect/Inspector/CPO   Join Date: Apr 2011 Location: NJ Posts: 369 Rewards Points: 250 See 210.19 A1 FPN #4 The note says that the 3% is just for a branch circuit. For the Feeder and the Branch circuit its a total of 5% If you supply the information, I will do a calc and we could compare results.
 05-25-2011, 07:35 PM #5 Member   Join Date: Jul 2009 Posts: 244 Rewards Points: 242 Thanks! Example would be: 4/0 Aluminum 240V 1ph 500ft from transformer to load 150A load Another is 500mcm Al, 240V 1ph 500ft as well 300A load Oh, these would both be underground in conduit installations so I am figuring 15-20C ambient.
 05-25-2011, 07:50 PM #6 Member   Join Date: Nov 2007 Location: Nashua, NH, USA Posts: 7,971 Rewards Points: 1,548 We do not use temperature when figuring voltage drop. However when choosing the wire size needed to avoid temperature problems (in conduit, in free air, kind of insulation, etc) we might already come up with a wire size large enough to not suffer excessive voltage drop. __________________ The good conscientious technician or serviceperson will carry extra oils and lubricants in case the new pump did not come with oil or the oil was accidentally spilled, so the service call can be completed without an extra visit.
05-25-2011, 07:54 PM   #7
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Quote:
 Originally Posted by viper Thanks! Example would be: 4/0 Aluminum 240V 1ph 500ft from transformer to load 150A load Another is 500mcm Al, 240V 1ph 500ft as well 300A load Oh, these would both be underground in conduit installations so I am figuring 15-20C ambient.

#1 i get -15V(6.25%), to get under 5% i show needing 300kcmil AL, for 3% i show 500kcmil AL

#2 i get -12.72V(5.3%)to get under 5% i show needing 600kcmil AL, for 3% i show 1 run of 900kcmil AL, 2 runs of 500kcmil AL, 3 runs of 300kcmil or 4 runs of 250kcmil AL

 05-25-2011, 09:24 PM #8 Member   Join Date: Jul 2009 Posts: 244 Rewards Points: 242 Huh, that is yet another value. do you happen to have the equation that was used to calculate that? One that seems to be floating around the net that I am not sure I agree with is (2*L*R*I)/1000 That produces 12.06% for the 4/0 using .0804/K for resistance. The NEC books I think is .1 ohms for 4/0 so there is a little discrepancy there.
 05-25-2011, 09:39 PM #9 Member     Join Date: Apr 2011 Location: Almost Arkansas Posts: 2,764 Rewards Points: 2,000 My turn....300 mcm copper....................500 mcm alum.
 05-25-2011, 10:19 PM #10 Sparky   Join Date: Mar 2011 Location: Central Florida Posts: 706 Rewards Points: 510
 05-26-2011, 05:49 AM #11 Lic Elect/Inspector/CPO   Join Date: Apr 2011 Location: NJ Posts: 369 Rewards Points: 250 #1 500 kcm #2 900 kcm
 05-26-2011, 11:20 AM #12 Member   Join Date: Jun 2007 Posts: 3,790 Rewards Points: 240
05-26-2011, 04:07 PM   #13
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http://www.csgnetwork.com/wiresizecalc.html
I use this often.

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