equivalent impedance of cables in parallel 1

[odlanor]

we have 3 cables 4/0 AWG to calculate the equivalent impedance in parallel. Assuming current flow equal to three cables would have:
Req = R / 3 and Xeq = X / 3
Considering that in reality never has a distribution of current flow equal to 3 cables asks:
Is there a correction factor for the uneven distribution of current to be included in the calculation?

 

[dpc]

The only way to have a correction factor is to first determine what the actual current distribution will be. In general, this is not possible.

Cable impedances are always an approximation.

Also, if you are just trying to calculate the fault current, assuming equal impedances should give acceptable results.

David Castor

http://www.cvoes.com

 

[cuky2000]

The graph below from Okonite, based in the Carson's equation provides acceptable results to determine the average cable series inductive reactance to neutral as a function of the cable diameter and physical arrangement.

Resistance of the cable is more or less constant for the given temperature.

 

[7anoter4]

See also:

http://www.prysmianusa.com/export/s...ineering_Guide/General_Calculations_Rev_4.pdf

page 8/20.