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.
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