1
The best that I determined is 500 KCmil Al.
So you are shooting for 5%?
2
given a 700' run of 500 AL I get the following results
11.87% drop @ 200 amps
I'm confused- do you mean 11.87 volts/ just under 5% of 240V?
3
If you load is truly 200 amps, running 1000 MCM aluminum would get you down to about 5.4% voltage drop along the 1000 MCM conductors.
Same there, do you mean 5.4 volts, and not 5.4% of 240V which is 12.96V? 5.4 volts would be a 2.25% drop on 240V.
I'm using Mike Holt's equation here.
I=(CM×VD)÷(2×KxD)
K= 21.2 for aluminum.
I=(CM×VD)÷(2 ×21.2 ohms×700ft)=XXX Amps
VD= (240*.05) = 5% is 12V.
#1 - So (500,000 x 12V) / 29680 = 202 amps.
#2 - And 500,000 x 11.87V / 29680 = 199.96 amps.
#3- While 5.4 % of 240V is 12.96V.
So 1,000,000 *12.96 /29680 = 436 amps, that seems like double what is needed.
Am I right here?
Say 500 feet of 2AWG copper, 480 3 phase, if you want 3% VD, 3% = 83 amps?
(2AWG * 3%VD) / (1.732*12.9*500)= 83A
or
(61631 * 14.4) / (1.732*12.9*500)= 83A
Don't Let Voltage Drop Get Your System Down
http://ecmweb.com/code-basics/dont-let-voltage-drop-get-your-system-down