calculating X-sectonal Area to handle 150A in Aluminum.

Discussion in 'General Electronics Chat' started by cjphelps, Aug 22, 2008.

  1. cjphelps

    Thread Starter New Member

    Aug 21, 2008
    Hi Everyone ,

    I am doing what I though was a simple calulation and getting a funny result. Maybe someone could help me out with this:

    I need to calculate the cross section of solid Aluminum needed to handle 150Amps @ 20V, used by my group for a 75 cm long piece of solid square aluminium. Assume Temperature K = 298

    What Ive done so far:

    1.V=IR 20 = 150*R R= 20/150 R=0.1333

    p= 2.709X10^-8 OHM*m
    p(cm) = 2.709X10^-6OHM*cm

    I substitute R from Ohms Law into equation and solve for A.

    A=pl/R A= 1.5698X10^-3 cm*cm

    This seems to small.

    I compare this to the awg of Aluminum required to handle 150A = Size 0
    n = −(m−1) where m=1, n = 0, plugging into d = .005"*92^(36-n/39)
    AREA=(PI*D^2)/4 = 82.88X10^-3"^2 = cross section for 1/0 wire which handles 150Amps.

    this is much bigger then my calc.
    Why is my calculation so off?

    Thank you in advance!
  2. Ratch

    New Member

    Mar 20, 2007

    So you want to dissipate 3 kW in a 75 cm square bar of aluminum? You got a problem. You are specifying a low resistance material, so the diameter has to be very small in order to raise the resistance high enough. The aluminum bar will instantly melt at that high wattage and small diameter, because aluminum has a fairly low melting point compared to other metals, and it cannot dissipate that much power in that shape. You are comparing it the the AWG which tries to set a minimum of resistance/length in order to not unduly raise the temperature of the wire. What are you trying to do, make a heater or transfer electrical energy?

  3. cjphelps

    Thread Starter New Member

    Aug 21, 2008
    Sorry I am being very confusing...! I want to dissipate nothing in the bar, without over enigneering it! I want to deliver all 3000Watts to a load down the line, while using the smallest possible x-section of aluminum . The constraint is that the aluminum has to be 75cm long.

    Sorry to start off so confusing:)
  4. kubeek

    AAC Fanatic!

    Sep 20, 2005
    Are you sure the voltage difference between the two ends of the alu bar will be 20V?
    That means you will dissipate 3KW in the bar.

    You say you want to use it as a conductor, so there ideally should be 0V difference. Your calculation is so way off because it calculates some completely different case.
    I would stick with the 1/0 AWG wire or bigger.

    BTW I am really glad that Imperial system doesn´t involve voltage, current etc. I personally have problems with any other units than length, so using a special Brittish Ampere would definitely complete the mess.
  5. Externet

    AAC Fanatic!

    Nov 29, 2005
    3 cm^2 does not seem too small. It is a 17.3 x 17.3 mm bar, = 300mm^2 section

    0.324" diameter round wire is 8.23 mm diameter; is 53 mm^2 section ; and 1/0 is not for 150 Amperes. That would be 3/0 = 85mm^2

    Will some day these stupid AWG numbering systems be abolished, inches too ? :rolleyes:
    Last edited: Aug 22, 2008
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    Is this AC, DC, pulse, or ????