Rule of thumb

[Red Wiggler]

Is there a rule of thumb or some sort of percentage that can be used when dealing with Parallel feeder runs and the issues within the code.

My concern is with multiple parallel runs and how they are terminated involving relatively short runs... 6 parallel runs of less than 100'

Is there a rule of thumb or percentage that can work to ensure there will not be one common phase conductor is pulling more of the load than the other common phase conductors with in the parallel.

Someone mentioned 1%.... if that is the case, than 1% of the 100' run would be 12". Doesn't that sound like too much of a difference? (5 A phase conductors equal 100' and the 6th phase conductor is 99'). Is this acceptable?

I guess to truely find the answer, I will be forced to do a resistance test on all of the common phases to ensure they are all equal (resistance wise) to meet the requirments of the code, because it is impossible to ensure they are all the "same" length as they are installed in a ductbank.

 

[qcroanoke]

Morepower posted this a while back:

Taken from Ferm?s Fast Finder Index?, play with this formula to see the effects of different lengths of parallels:

Lp*It/Lt = Ip

Where: Lp = Length of a parallel conductor

It = total current of the load

Lt = Total length (sum of parallels)

Ip = Current on a parallel conductor

I run across this problem an awful lot and it seems that few people understand the importance of the rule.

 

[G._S._Ohm]

Two conductors joined have a non-zero contact resistance. For a badly installed wirenut this can be 50 milliohms or more.

If the contact resistance between conductors at the terminations is negligible compared to conductor resistance
then the conductor resistance should give you a good idea of how equally the conductor current shares.

There are ways to measure conductor or contact resistance without a micro-ohmmeter.

 

[don_resqcapt19]

With the 6 sets and the lengths given, assuming a 2000 amp load, my estimate is that there would be ~ 336 amps on the 99' one and ~ 332 amps on the 100' ones.

 

[don_resqcapt19]

Morepower posted this a while back:

Taken from Ferm?s Fast Finder Index?, play with this formula to see the effects of different lengths of parallels:

Lp*It/Lt = Ip

Where: Lp = Length of a parallel conductor

It = total current of the load

Lt = Total length (sum of parallels)

Ip = Current on a parallel conductor

I run across this problem an awful lot and it seems that few people understand the importance of the rule.

That would show more current on the longer conductor and that is not the case. The most current is on the shortest conductor.

 

[iceworm]

That would show (Ferm's) more current on the longer conductor and that is not the case. The most current is on the shortest conductor.

Thanks Don. That that is what i saw as well. (But then I never liked Ferm anyway)

For six cables, one different from the rest, I get:
I6/It = 1/(5a+1)
where
I6 is the current in the different one,
It is the total current for all six.
a is the ratio difference of the different one - for example, if 5 are 100' and one is 99' then a = .99

For all random lengths the algebra gets a bit messier (more messy?)

RW -
To answer your question: For parallel run conduits/raceways, and minor differences in termination lengths, the minor difference in current sharing rarely matters. The few times I have been questioned, I didn't measure the resistance. I just showed the excess current the short one could carry was still under the allowed ampacity of the cable.

Of course if you have sized the cables to within 1% of their full load and the difference is 2% - - - It will cost you and should.

ice

 

[don_resqcapt19]

The method I use is to assume that the lengths are the "resistance" and solve it as a parallel set of resistors to get the total "voltage drop" and then solve for the currents on each length. The resistance is directly proportional to the length so this works assuming that everything else is identical.

As someone else posted you can have differences in the contact resistance of the connections that could exceed that of a few feet of wire.

 

[defears]

I've always thought that once a conductor of a parallel runs gets a few amps above the other than the path of least resistance is the other wire so more current goes to it to keep everything within reason. I've never seen maybe more than 5 amps diffrence between any set I've ever worked on. Maybe I just got lucky.

 

[G._S._Ohm]

I've always thought that once a conductor of a parallel runs gets a few amps above the other than the path of least resistance is the other wire so more current goes to it to keep everything within reason. I've never seen maybe more than 5 amps diffrence between any set I've ever worked on. Maybe I just got lucky.

Yes, as the current goes up in a conductor the resistance of that conductor goes up slightly at "+0.393 percent per degree C" due to I squared R heating
and so the other cooler conductors take more of the current.
The final current allocation could be solved for by using graphs.

". . .5 amps diffrence between any set. . ." out of how many amps in one conductor?