Parallel Conductors

[jimmac49]

When conductors are paralleled the NEC requires that they are the same length among other things. We have a 2000 amp bus that calls for 5 sets of cables to feed it. It can be difficult or impossible to get them all exactly the same length due to space constraints. What is considerd a safe approach to this problem. I think the length of the run would be a big factor. Say a 20' run verses 200' run, I would think the longer run requires a more exact length. Any thoughts would be appreciated.

 

[LarryFine]

Say a 20' run verses 200' run, I would think the longer run requires a more exact length.

On the contrary, I'd say the shorter the run, the more critical the match. It's all about the division of current, which is based on relative resistances; the lowest resistance carries the greatest share.

It's more important to maintain length (and thus, resistance) equality within a given percentage than a specific number of inches, so the longer the run, the less the relative resistance difference will matter.

Also, the closer to the limit of conductor ampacity your current ends up being, the more likely a small current difference can overload the conductor with the lowest resistance (including terminals, etc.).

Added: Plus, remember that the matching applies to the paralleled conductors of one given phase; phases need not match each other, although they likely will be close anyway.

I don't know a specific percentage to attain, but there are plenty of engineering types (but, we like them anyway ) on this forum who may be able to give you a figure.

 

[jimmac49]

Thanks Larry I appreciate your help

Jim

 

[petersonra]

My inclination would be that the longest one should be something like no more than 5% longer than the shortest one. That would keep the currents pretty close.

 

[don_resqcapt19]

You can get a close approximation of the current in each set using the following:

set 1 current = total current / ((1/L1) + (1/L2) + (1/L3) + (1/L4) + (1/L5)) * (1/L1)

L1 through L5 are the lengths of the conductors

Use *(1/L2) at the end for set two and so on for the other sets.

 

[rcwilson]

The reactance of the cables also affects the load sharing. The relative impedance of the individual paralleled cables determines the distribution of current, not just the relative resistance.

Based on my old GE Short Circuit Calculation handbook, the impedance of three, single-conductor 500 kcmil cables in non-magnetic conduit is 0.0276 +j.0373 = 0.0464 ohms/1000 ft. (75C conductor temperature, tinned copper, RHHW, triplexed). Total impedance at 60 Hz is almost twice the resistance. (0.046 versus 0.027)

Cable impedance changes with the installation method. Cables in steel conduit have higher impedance due to the magnetic effect (0.0551 Ohms/1000'). If the cables are routed spaced apart in open air, the impedance is higher because the A phase current's magnetic field is not cancelled out by the B&C phase currents as well as it is when the wires are twisted together in a conduit.

If you parallel cables and bunch all of the A phase cables together, some mutual impedance effects will affect the load sharing. (I measured 20% difference in loading on a 5/phase 1200 amp circuit where all cable lengths were within 1%. The difference was due to the way the cables were laid in the tray.)

Bottom line - keep the cable lengths close to each other (+/-5%) but don't worry about it much. If possible, route the wires in bundles of A,B,C,N to minimize inductive effects. If your load is really 2,000 Amps, throw in an extra 10% to cover unequal sharing of load between paralleled cables.

 

[don_resqcapt19]

Bob,
Yes, of course it is not just the resistance that determines load sharing, but assuming the conductors are in raceways and that the installation is in compliance with 310.4, other than length, can't we assume that the resistance and the reactance are virtually the same for each set and do a load sharing calculation based only on length? That is what the formula I gave does.

If the installation is not in raceways or not full sets (a,b,c,n) bundled together in tray, I can see where you would have to use some other method.

 

[rcwilson]

Don - Good observation. If all wiring is in conduit with A, B, C, & N in each conduit, the parallel impedances will be essentially equal and proportional to the length of each conduit run.

IMHO - doing a lot of work to make each wire the same resistance is not productive. Keep each parallel circuit essentially the same and there won't be problems. We only get in trouble when we throw 8 per phase 500 kcmil in a tray or on a rack in random order and claim we have a 3,000 A circuit.

 

[LarryFine]

The reactance of the cables also affects the load sharing. The relative impedance of the individual paralleled cables determines the distribution of current, not just the relative resistance.

I used the term 'resistance' to make it easier. The NEC requires similar conductor, insulation, and raceway characteristics for just those reasons.

Based on my old GE Short Circuit Calculation handbook, the impedance of three, single-conductor 500 kcmil cables in non-magnetic conduit is 0.0276 +j.0373 = 0.0464 ohms/1000 ft. (75C conductor temperature, tinned copper, RHHW, triplexed). Total impedance at 60 Hz is almost twice the resistance. (0.046 versus 0.027)

But, the paralleled sets should still be of equal impedance, right, presuming equal installation characteristics? So, that's moot.

Cable impedance changes with the installation method. Cables in steel conduit have higher impedance due to the magnetic effect (0.0551 Ohms/1000'). If the cables are routed spaced apart in open air, the impedance is higher because the A phase current's magnetic field is not cancelled out by the B&C phase currents as well as it is when the wires are twisted together in a conduit.

Again, same = same.

If you parallel cables and bunch all of the A phase cables together, some mutual impedance effects will affect the load sharing. (I measured 20% difference in loading on a 5/phase 1200 amp circuit where all cable lengths were within 1%. The difference was due to the way the cables were laid in the tray.)

Well, that qualifies as a non-similar installation, methinks. Did relaying the cables alleviate the difference, with no other changes such as remaking any connections?

Bottom line - keep the cable lengths close to each other (+/-5%) but don't worry about it much. If possible, route the wires in bundles of A,B,C,N to minimize inductive effects. If your load is really 2,000 Amps, throw in an extra 10% to cover unequal sharing of load between paralleled cables.

All good advice. I wasn't busting chops.


I do like to remember that terminations also play a part in load division. They all have to be tight. We've seen IR images and pics of overheated connections. It's more important with paralleled conductors, because a failure under high load can quickly cascade.

 

[hardworkingstiff]

Since most conductors we use are 90C and the terminations 75C, how does that affect anything? Let's say you have one conductor a little shorter and it's carrying a bit more current. When it starts to go over the 75C ampacity (let's say less than 10%), the conductor will still be fine but the termination starts to become an issue. Since the other conductors on the same termination are below the 75C rating, would that heat sink the termination point enough to keep problems from developing?

 

[don_resqcapt19]

If you have a short conduit run with flat (horizontal) 90s, it would be easy to gain a foot of length for each of the runs. if you have a 2000 amp load using five sets of 600 kcmil, and the lengths are 20, 21, 22, 23 and 24', the 20' length will be slightly overloaded. It would run at 438 amps and the 24' one would run 365 amps. (note: the currents listed here are based on the formula in my earlier post)

 

[petersonra]

Since most conductors we use are 90C and the terminations 75C, how does that affect anything? Let's say you have one conductor a little shorter and it's carrying a bit more current. When it starts to go over the 75C ampacity (let's say less than 10%), the conductor will still be fine but the termination starts to become an issue. Since the other conductors on the same termination are below the 75C rating, would that heat sink the termination point enough to keep problems from developing?

There is a fair amount of slop built into the NEC. A lot of what is allowed is pretty conservative.

 

[jimmac49]

Thanks for the information guys I appreciate it.

Jim