Wire loss calculation

[edy252]

hi all....

i've been trying to calculate the wire loss in a 3-phase wind generator...i expect 62 amps as maximum...the wires that i'll use are (each wire): 11 m of 2, 4mm wires joint together as to act as one wire..they have apprx. 1.5 ohms/1000m resistance, so for two wires joint together it will be 1.5/2 = 0.75 ohms/1000 m

i've seen Hugh's notes about line losses.....he explains how to calculate this loss and finally got to a formula:

loss (in watts) = (total amps)(total amps)(resistance in one wire)*2 (as i understood it)

so, concerning my machine:

resistance in one wire = (11/1000)*0.75 = 0.00825 ohms

so the loss = 62*62*0.00825*2 = 63.42 W

while somebody else ( i dont remember who ) said that we calculate the loss by:

loss (in W) = (2/3 * total amps)(2/3 * total amps)(resistance in one wire)

i  get : 41.3*41.3*0.00825 = 14 W loss!

so which way is right? or are they both wrong?!!! if so, how to calculate this loss

thx in advance

[baggo]

Hi edy252,

At a guess the first formula applies to DC current after the output is rectified.

The AC current in the alternator windings is not constant as the output is a sine wave.

Although I am no expert on AC theory I presume the average AC current in the windings is 2/3 of the peak current, hence the second formula.

John

[Flux]

Hugh is right

For normal rectifiers feeding a battery, current is carried in 2 wires, which swap round in sequence.  The loss is the same as DC but the heating is less as it is shared between 3 wires.

The rms AC current is 0.82 times the DC current.

Flux

[wooferhound]

Here is a bunch of "Voltage Drop Calculators" . . .

http://www.electrician.com/vd_calculator.html

http://www.stanselectric.com/vdrop.html

http://www.elec-saver.com/freetool06.htm

[edy252]

hi...and thank you all for your replies.

what i didnt mention is that the rectifiers are 11 m AWAY from the genny, so the calculation of loss is for AC current going through these wires