Jim, to answer this question, you also need to know the total length of the wire run.
The resistivity (rho) of any material is a temperature dependent variable, but the variance is small, and values for many materials at 20 degrees C are available on the internet. A quick search provides the resistivity value for copper as 1.724E-8 ohm*meters.
Resistance, resistivity, length and cross-sectional area are related by the equation
R = (rho * l )/A
Voltage, current and resistance are related by the equation
V=I*R
Combining the two, and rearranging for A gives
A = (I * rho * l ) / V
Now, it is important to note that the V above is not your nominal operating voltage, but rather a voltage corresponding to the acceptable voltage drop in the wire. Many electrical codes give this as around 2% of the nominal voltage, so for a 36V system that would be 0.72 V. You didn't provide a wire length, but for the purpose of discussion, for a 10 meter wire, you have
A = (150 * 1.724E-8 * 10) / 0.72 = 3.6E-5 m^2, or 36 mm^2. The corresponding AWG wire gauge is about 2 AWG.
This is large, but this would be necessary if you were running at full power continuously, and could not tolerate voltage drops higher than 0.72V. Provided you stay within safety limits (i.e. the isolation voltage limit of the wire insulation), and can tolerate more heat, and more voltage drop in the wire at maximum current, you can derate this to be a more appropriate fit to the expected continuous load.
-Sean
On Jan 15, 2009, kocpnt@tds.net wrote:
Hi All,I am looking for advice. I am looking for the proper size of copper wire to power my sub.This wire will go from the batteries and at maximum draw will need to handle about 150 amps total load at 36 volts DC. I realize that there places to look for this, but felt that the experienced people in the community might have some good insights!Thanks in advance,Jim K