Here, I'm using the circular mills method as opposed to the resistive method:
To find the Voltage Drop of a single-phase ckt. I'd use: VD=2*R*L*I/CM
In order to size a conductor properly while staying within the 3% voltage drop recommendations I'd use: CM=2*K*L*I/VD
VD=voltage drop
L=length of conductor in ft.;one-way
CM=circular mills from Chap.9, Table 8
I=load current
R=the resistive value from Chap.9,Table 8 or 9
K=The D/C constant for the resistance of a 1000-mill conductor that's 1000 ft. long at 75 degrees C.

My question is with the K value. I've seen four different sets of values noted. The values are different for copper and aluminum.
The two sets that seem most accurate are:
cu-12.9
alu-21.2
...and
cu-12
alu-18

An electrician on a different forum stated that he uses cu-7 & alu-11. I'm not too sure about that.
I've been using the first set thinking that keeping the value slightly higher is playing it safe.

Any takers on this one?


I'd appreciate it

Dan

K=12.9 Copper

K=21.2 Alum.

Both at 75'C

Single Phase Two wire
Vd= 2K X L X XI over CM
CM= 2KxLxI over VD

Is what I use

I got it out of Ugly's 2002 edition page (48)

Thanks a lot,

You know, I've got an Ugly's--I should've checked it out there.
Another helpful person, like yourself, gave me this answer:

The value of K is direcly determined by temerature. The resistance of a conductor is R = (K/A) x 1000 ohms/1000ft.
K = volume of resistivity in ohms-cmil/ft
A = cross sectional area in cmil

K = 10.371 at 20C for CU
K = 17.002 at 20C for AL
K = 10.575 at 25C for CU
K = 17.345 at 25C for AL
K = 12.9 at 75C for CU
K = 21.2 at 75C for AL

Dan