Aluminum vs Copper? [RE-wrenches]

Hugh Piggott hugh at scoraigwind.co.uk
Mon Sep 4 14:38:24 PDT 2006


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>Thanks Hugh, Jeff, Chris and others for leading me down this path. 
>If I am understanding this correctly Hugh, and don't get too worried 
>about greater than 5% line losses at max output, and use aluminum 
>with proper precautions, then I can, say, use a 2/0 aluminum cable 
>with a line loss of about 8% (based on the derated 40 amps instead 
>of 52) or a #2 copper, and not be too concerned?

Bear in mind that there are 3 wires and so the loss is increased by 
the heating of all three.  I find it simpler to use the 52 amp 
current that is probably flowing in one pair of wires at any instant 
(well not quite because of some sharing but broadly speaking...)  I 
therefore size it like a DC circuit using the resistance of the two 
wires and the current at 52 amps.   This exaggerates the loss 
slightly.

>  And if economics demanded I could even go to a higher line loss 
>bearing in mind that: "As loss increases, the situation becomes much 
>more complex, since more than 2 wires will start to conduct at once 
>during the changeover." as you state in your answer to Paul Gipe on 
>http://www.scoraigwind.com/CABLE/index.htm?

Yes indeed.  More on that follows
At 10:49 am -0700 15/10/99, Brian Acker wrote:
>
>The latter case of a high-impedance alternator with sinusoidal phase
>currents yields somewhat different results. The rms phase current is:
>
>         Irms = [pi/(3*sqrt(2))] * IDC = 0.74 * IDC
>
>Total power loss is obtained the same way as in the former case:
>
>         Ploss = 3 * Irms^2 * R = 3 * (0.74 * IDC)^2 * R
>
>where R is again the one-way (150') conductor resistance. I hope this
>discussion
>sheds some more light on the wiring loss problem.


So in the best case the loss is about 19% lower.  (thanks to Brian)



>Honestly I don't pretend to know what that really means but if I can 
>get away with 8-10% without seriously compromising charging 
>potential then the wire cost savings will very pleasant.

Yes and wire resistance depends on temperature, which I am guessing 
will be low in your case, whereas the resistance tables usually 
assume about 70 degrees C.

-- 
Hugh

Scoraig Wind Electric
http://www.scoraigwind.co.uk/


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