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Sizing Conductors for Voltage Drop-K value #99497
05/12/06 08:11 AM
05/12/06 08:11 AM
workn26  Offline OP
Junior Member
Joined: Dec 2005
Posts: 8
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:

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


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Re: Sizing Conductors for Voltage Drop-K value #99498
05/14/06 08:48 PM
05/14/06 08:48 PM
Yoopersup  Offline
Joined: Mar 2003
Posts: 840
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)

Re: Sizing Conductors for Voltage Drop-K value #99499
05/14/06 08:56 PM
05/14/06 08:56 PM
workn26  Offline OP
Junior Member
Joined: Dec 2005
Posts: 8
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




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