voltage drop

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Re: voltage drop

There is always voltage drop along a length of wire with current flowing. It is the product of current times resistance, or IR.
 
Re: voltage drop

Originally posted by bbvolt:
would there be a drop on a 480 volt line at 220
feet at a max of 40 amps using # 8 wire
3 phase of course 4 wire
Click to expand...
Almost 2.5% or 12 volts of drop using the info you provided.
 
Re: voltage drop

I used the impedance of #8 from Table 9 of the NEC, .69 ohms/1000'. For a balanced three phase load the calculation would be

220'x(.69/1000)x 40 amps = 6.07 volts drop line to neutral or 6.07/277 = 2.19% drop.

Because the load is assumed to be balanced there is no voltage drop in the neutral return current, and no need for multiplying by 2 as in a single phase calculation.
 
Re: voltage drop

For three phase voltage drop Vd = (R/ft.)(L)(I)(1.732)

(.00069)(220)(40)(1.732) = 10.5v

For three phase instead of length times 2 it's length times the square root of 3.

10.5/480 = .0219 or 2.19%

Jtester, I have no idea how you still came out with 2.19%. Maybe your method works too. :)
 
Re: voltage drop

Sam

Your equations and mine are exactly the same. You approached it from a line to line perspective and I used line to neutral. Notice your equation has 1.732 in the first part of the calculation, and then you have 1.732 in the denominator when you divided by 480 volts. The two 1.732's cancelled each other out. I didn't include the square root of 3 in the first part, and then I divided by 277 volts in the percentage part.

Same equation, just looking at it from L-n vs L-L.

Thanks

Jim
 
Re: voltage drop

Jim, The first time I looked at your math with only the one wire length I thought you were goofing it up. I see what you're saying now, that's actually pretty cool using the neutral with no voltage drop, it haddent occured to me. :)
 
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