Tap rules for parallel conductors

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GoldDigger said:
For small percentage variations the % difference in current will be roughly equal and opposite to the % difference in length (assuming that we can ignore the termination resistance. For short runs of parallel wire the termination resistance and its variation may be more important than the actual wire resistance.) The impedance of the rest of the circuit has no effect on the current division.
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I know all that. I'm asking how much are we talking about with this scenario?
 
GoldDigger said:
Different people will give different answers to the basic question of whether putting a tap which does not break the original conductor (insulation piercing or split bolt) counts as terminating the conductor for the purpose of defining the extent of parallel wires. On the other hand, it arguably would still be the origin point of a new set of parallel tap connectors.
A larger number will say that cutting the original wire to make the tap connection does definitely terminate the original parallel run and requires bringing the parallel conductors back together in a Polaris connector, on a bus bar, or other means.
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parallel tap.jpg

Here is an example I've put together to show why you have to tap both sets, and preserve the symmetry. If you can preserve the symmetry by matching tap conductors to feeder conductors one-to-one, that will also preserve the symmetry, as long as they all tap at equal positions on the circuit.

Given:
R1 = 0.0520 Ohms
R2 = 0.0001 Ohms
V2 = 480V
It = 200A
IL = 500A

Find:
Iu
V1
Ia1
Ia2
Ib

Each parallel path resistance R1 represents 500 ft effective round trip length of 250 kcmil wire. The final couple of feet is represented by R2. Each path is rated for 255A on this 500A circuit, therefore this circuit overloads due to current Ia2. Had you distributed the tap current It among both paths, the current through both R2's would be within the ampacity.
 
Thanks for the detailed example of why asymmetric taps are not acceptable.

I was discussing only the question of whether symmetric taps (one for one) could be implemented as one tap per original conductor or whether all of the original set would have to be joined back together (Polaris, bus, etc) in order to add the tap conductors at that new common point.
 
Carultch said:
View attachment 18536

Here is an example I've put together to show why you have to tap both sets, and preserve the symmetry. If you can preserve the symmetry by matching tap conductors to feeder conductors one-to-one, that will also preserve the symmetry, as long as they all tap at equal positions on the circuit.

Given:
R1 = 0.0520 Ohms
R2 = 0.0001 Ohms
V2 = 480V
It = 200A
IL = 500A

Find:
Iu
V1
Ia1
Ia2
Ib

Each parallel path resistance R1 represents 500 ft effective round trip length of 250 kcmil wire. The final couple of feet is represented by R2. Each path is rated for 255A on this 500A circuit, therefore this circuit overloads due to current Ia2. Had you distributed the tap current It among both paths, the current through both R2's would be within the ampacity.
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Why do you pick V2 as 480V instead of V1?
 
ggunn said:
Why do you pick V2 as 480V instead of V1?
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To keep the numbers round when calculating the load resistance. Specify ILOADas 200 and VLOAD as 480 and RLOADcomes out even.
The difference in the result is small, the difference in the arithmetic chore is larger.
 
Smart $ said:
I agree... but how much difference in length are we talking. Code provides no tolerance and making certain each set has an exactly equal length is impossible.
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No tolerance- but technically a few engineering calcs can substantiate a violation of 310.15 B 16.
 
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