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electure #170917 11/15/07 12:32 AM
Joined: Mar 2004
Posts: 814
B
Member
Ever try Martindale's Calculator site? It is amazing. Here's a link to the electrical ones: http://www.martindalecenter.com/Calculators4A.html

electure #170942 11/15/07 04:51 PM
Joined: Jun 2007
Posts: 301
J
Member
electure,
He asked for the formula, he got the calculators.
Whats your beef. I read all the posts.

If you require lenghthy responses on this forum just let me know.

JValdes #170950 11/15/07 06:58 PM
Joined: Dec 2000
Posts: 4,294
Member
"He asked for the formula, he got the calculators."

The sentence should be self explanatory. Some members are here to learn, (my aged self included). In depth education is sound education. Unlike calculators, it explains the hows and whys.

As a matter of fact, until Greg was kind enough to post his calculator, SparkyNC got 1. (The other one wouldn't open on SparkyNC's computer.)



"I" don't require anything.
A moderator is just a member who's been around the site and carefully read/responded to a lot of threads, nothing more.


electure #170953 11/15/07 07:59 PM
Joined: Nov 2002
Posts: 174
K
Member
It may be to your advantage to check into using a buck-boost transformer to reduce your wire size at the far end.

kale #170986 11/16/07 04:52 PM
Joined: Nov 2006
Posts: 348
I
ITO Offline
Member
Originally Posted by sparkync
I have a 50 amp 120/240 camper power cord going 250 ft. I've been trying to find what the voltage drop is. Can anyone help me with the formula? Thanks Steve
I will have to assume the amperage at 50 amps max.


Wow, all that talk and still no formula?

VD = (2LIK)/CM

Or in English:

VD – Voltage Drop
L- Lengths
I- Amps
K- Constant for Copper (use 12.9 for CU and 21 for AL)
CM- Circular Mills (page 635 Table 8 of the NEC-2005)

Or you can back into it by:
CM = (2LIK)/ VD

So if 5% of 240 = 12v then…

(2 x 250f x 50A x 12.9)/12v = 26,875 cm

Go to page 635 Table 8 of the NEC-2005: Looks like # 6 is (26,240) so the next size up is #4 (41,740)



Last edited by ITO; 11/16/07 04:54 PM.

101° Rx = + /_\
ITO #170999 11/16/07 07:43 PM
Joined: Dec 2000
Posts: 4,294
Member
The formula is in the link I gave, but not the answer wink

electure #171000 11/16/07 08:12 PM
Joined: Apr 2002
Posts: 7,381
Likes: 7
Member
Time to type.....
Voltage drop is a FPN in the NEC, 2% feeders, 3% branch for a total of 5%. But, it's a FPN, not 'Code'.

That said...thanks to Ito for doing that typing.

Kale:
The buck/boost solution is only recomended (IMHO) with a constant load. If you size the B/B for a 50 amp load, and it drops to 10 amps, you can/may get an overvoltage situation at the 'load'.

Sounds similar to my class Wed PM



John
HotLine1 #171005 11/16/07 08:49 PM
Joined: Jul 2004
Posts: 9,928
Likes: 34
G
Member
If I was going to go with a transformer, I would go with 2. Step that up to 480 and back down to 120/240 but at that point bigger wire would probably be cheaper. Trying to boost the drop up at the far end is just paying extra to hide the waste heat you are losing along the way.
#1 al would get you in under 3% and probably not break the bank. #2al gets you up to 3.3% but that assumes 50a load and that would have to be a peak. I bet 40a would be a better working number. If you do that you can get in around 2.5% with #4 copper or the #2al.


Greg Fretwell
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