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Joined: Nov 2003
Posts: 36
M
Member
I am going to a electrical tech school. We are now working on calculating wire sizes and such. I want to make sure that I am doing this correctly so I am going to post some sample problems and my answers so you all can check my work. I apprieciate any tips.

1) 25 #12AWG copper cunductors type THHN, in conduit. The ambient temp is 95*F What is the ampacity of each conductor?
A:the chart lists #12 copper THHN @30
the correction factor is 0.96 for 95*
so 30(0.96)=28.8A
then 50% correction for 25 in conduit
so 0.96(0.50)=14.4A

2) a single phase load is 2800' from the source and draws a current of 86A, and operates at 480V. Maximum voltage drop must not exceed 3%. What size of Aluminum conductor should be used?
A: 480V(3%)=14.4V

R=14.4V/86A
R=0.167

CM=(K)L/R
CM=(17)2800/0.167
CM=285,030
the closest size is 300Kcmil

3) A 3 phase motor operates on a 480V and is located 1800' from the source. Current is at 235A. Using copper voltage drop not to exceed 6%
A: 480V(0.06)=28.8V
L= 1800(1.732)=3118

CM=10.4(3118)/0.123
CM= 263,636
all 3 shall be 300Kcmil

Sorry if these formulas are unclear.

Thanks for any help
MW

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Joined: Sep 2003
Posts: 650
W
Member
Double check on #2; remember that a circuit requires _two_ conductors, and you will have voltage drop in _each_. In other words, the length of wire over which you need a maximum of 3% voltage drop will be 2*2800'.

For #3 it appears that the equations compensate for the return path, by multiplying the length over which you calculate the voltage drop by 1.732 (the square root of 3).

I've never worked with k in units of ohm*foot/cmil, but when I calculated it for aluminium from k given in ohm-cm, I got 16.7 at 20C, which agrees with 17. I didn't double

-Jon

Joined: Nov 2003
Posts: 36
M
Member
thanks winnie, my instructor pointed out the footage problem in number 2 to me. I understand now.


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