Thread: parallel conductor formulas/question

1. Junior Member
Join Date
Apr 2006
Posts
2

parallel conductor formulas/question

I have a hypothetical question to ask.

Details
480v, 3 phase, 60 Hz system
(10) 500 mcm conductors per phase and neutral installed in wire way extending form the generator output bus to a piece of switchgear.

Brief Theory
All of the conductors comprising a phase need to be the same length etc. to ensure that they each carry the same amount of current.

Question
What formula can be used to predict what impact a variance in conductor resistance will have? Specifically what impact a variance in length of one conductor would have?

Assume no abnormal effects from inductive reactance exist.

Also what is the formula used to determine the impact inductive reactance would have on the individual conductors?

Why am I asking?
We were researching a purported variance in individual conductors of a parallel feeder and I was hearing anecdotal evidence about the effects of conductor length, impedance, and inductive reactance. The testimony was all over the place and I really found myself wanting to be able to predict the actual outcomes using math.
Thanks
Rick

2. Use the formula for parallel resistors to find the resistance of all the legs in parallel. Use this resistance along with the load to find the voltage drop for the parallel sets of conductors. Use the voltage drop and the resistance of each individual conductor to to find the current on that conductor.
It is my understanding that the inductance of an installation like this is often the biggest problem. To minimize its effect, you should group the conductors in the wireway with all 3 phases and the grounded conductor, if used, in each group.
Don

3. The impedance, Z, adds by the reciprocal of the sum of the reciprocals. Therefore given 10 conductors of equal length, and equal impedance, the impedance of the 10 conductors in parallel will be equivalent to 1/10th the impedance of a single conductor. Pretty straight forward math.

Now, if you vary the impedance of 1 set of conductors by increasing the length, then you have the reciprocals of 9 conductors of equal impedance added to the reciprocal of the other conductor, and invert.

Example:
Assume 100ft of (10) 500 KCMIL all same length-

4. Moderator
Join Date
Jun 2003
Posts
5,501
Don has the exact answer, but in a nutshell, here is what I think King is trying to say:

All the conductors have the same voltage across them. If you start with all of them the same length, they each have equal currents.

If you shorten one cable by 10%, its resistance is reduced by 10%. With the same voltage across it, you are going to get a 10% increase in current in that one cable.

This will hold as long as the cable resistance is small compared to the load resistance.

Similar effects will hold with changes in length and inductance, again, as long as the inductance of the cable is small compared to the load inductance.

Steve

5. King,
While it may not change the total impedance by much, the circuit is a current divider and the current in the conductors will change. Using your numbers, the longer conductor will carry ~350 amps and the other 9 will each carry ~361 amps assuming a 3600 amp total load. If you have a design where the parallel conductors are loaded to the maximum permitted by the code, it is very easy to have one or more of the conductors operating above it permitted ampacity as the result of a small difference in impedance.
Don

edited for spelling
Last edited by don_resqcapt19; 11-30-07 at 07:48 PM.

6. Senior Member
Join Date
May 2005
Location
Redmond, WA
Posts
755
Based on my experience in measuring the actual current in paralleled conductors, the relative impedance of the conductors is what determines the load sharing among conductors. A cable's reactance is the sum of its self-reactance and the mutual reactance from the magnetic fields created by the flow of current in all of the other conductors in the circuit. The mutual inductance is a function of the cable-to-cable spacing and orientation.

If the cables are tightly bundled in ABC groups, the cables' mutual reactance’s will be similar because the magnetic fields of the three phase currents add to cancel the magnetic field outside each ABC group and the cable spacing in each group will be very similar. The mutual reactance of the A-B, B-C and C-A cables will be similar in each group and between groups, If the conductors lay in the wireway in a random pattern, the magnetic fields cannot cancel as well, so each cable's current creates an impedance in the other cables. Depending on the curren, the length of the run, and the cable configuration, there may be significant discrepancies in the current distribution.

Unless the cables are held in a rigid position for most of the run it is almost impossible to calculate the impedance.

One solution is to use cable bus which is like cable in tray, but the cables are held in supports that maintain cable spacing for cooling purposes to improve the ampacity rating. Suppliers use computer programs to calculate the current distribution and suggest the best cable configuration. By using cable bus, you get a UL certified cable ampacity and do not have to make educated guesses about the load sharing among cables.

If I don’t use cable bus, I assume at least a 10% variation in amps from inductive effects.

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