Voltage drop

[Jpflex]

well, it was "free" afterall

You’re right and I’ve considered that too so my bad. I was just mad because their error make me pull a lot of hair out

 

[Tulsa Electrician]

I did a quick run at it.
Re doing the math later.
It said to use chapter 9 table 8.
I used old school R=VD/I.

 

[Jpflex]

I did a quick run at it.
Re doing the math later.
It said to use chapter 9 table 8.
I used old school R=VD/I.

Yes this is for circular mill size of wires and I used this as well

 

[wwhitney]

If current I=110 (full load, no 125% factor), and distance D = 2 * 245', with allowable voltage drop Vd = 3% * 240 = 7.2V, then allowable total resistance is R = 7.2 / 110 = 65.5 milliohms, and allowable resistance per kft = 65.5 / 0.49 = 134 milliohms/kft.

Then if you want to use the DC resistances from Table 8, #1/0 Cu is the smallest size with a lower resistance/kft, of 122 milliohms/kft. Or Table 9 tells you the effective Z for the conditions assumed in the Table for #1/0 Cu is 130 milliohms/kft, same answer.

Cheers, Wayne

 

[Tulsa Electrician]

I got 1/0 as well. Did the math a little diff.
Thank you, Wayne

I like to show first year the old way for field lengths on a simple 20 amp circuit. Using VD= I/R #12 @ .002
It really opens there's eyes to voltage drop

The R= VD/I is just a variation of that. Then move them into 2k.
So they understand K

Thanks for sharing JP flex