MV Part of parallel circuits isolated, part not isolated

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jdcpe17

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Location
Austin, Texas
This is a bit of an odd ball situation. I am not really seeing any reason not to proceed in either way. Any thoughts? ....


I have four 5" conduits in a duct bank by themselves. I have a A-B-C phases and ground to run using 1000kcmil wire, medium voltage, each phase has four parallel runs, and run a full size 350 ground in each conduit. Both options gives me a fill % of 39.0% which is just under the 40% allowed.

Option 1: Run 1 parallel run A-B-C-g in each conduit, so four sets, side by side (A-B-C-g, A-B-C-g, A-B-C-g, A-B-C-g,)

Option 2: Run 3 parallel runs of each phase per conduit for 3 conduits (A-A-A-g, B-B-B-g, C-C-C-g) then the fourth conduit run A-B-C-g.

Running separately is allowed per 310.10(H)(3) as long as 2-6 are followed and I believe they are.

Thoughts?

Thank you for your time.
 

charlie b

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Location
Lockport, IL
Occupation
Retired Electrical Engineer
Option 2 is not allowed. It does not meet 310.10(H)(3) (2011 NEC), in that the electrical characteristics of three of the conduits (i.e., isophase) do not match those of the fourth. You could, under certain circumstances, have A-A-A-A-G, B-B-B-B-G, and C-C-C-C-G, but this would not work for your sizes of cables and conduits. See the exception to 300.(3)(B)(1).
 
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jdcpe17

Member
Location
Austin, Texas
thank you for your help.

I am not following why option 1 would not work per 300.3.b.1. I noticed you wrote A-A-A-A-g to indicate 4 parallel runs in one conduit, but my option 1 would have A-B-C-g so the conduit is not filled up. The 350 ground is per 250.122 for the 2500 ocd.

Are you trying to say it is not allowed to run parallel sets of A-B-C-g underground?
 

charlie b

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Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
I said Option 2 would not be allowed. I did not address Option 1. So now I will; Option 1 is OK.
 

rcwilson

Senior Member
Location
Redmond, WA
The parallel cables in option 2 do not have equal impedance. The ABCg group of cables will have a lower impedance than the other sets since the phase wires are much closer together. Spacing between phases has a large effect on the circuit impedance. At 1.5" spacing for tight triplex, the reactance for 1000 kcmil is about 0.03 ohms per 1000'. Put the phases in 5 " conduits with 2" between conduits, spacing = 7" and reactance = 0.068 ohms per 1000'.

The ABCg group will carry more current than the others. Instead of each wire carrying 25% of the total current the division could be three carrying 19.5% each and the fourth carrying 42%
 

iMuse97

Senior Member
Location
Chicagoland
The parallel cables in option 2 do not have equal impedance. The ABCg group of cables will have a lower impedance than the other sets since the phase wires are much closer together. Spacing between phases has a large effect on the circuit impedance. At 1.5" spacing for tight triplex, the reactance for 1000 kcmil is about 0.03 ohms per 1000'. Put the phases in 5 " conduits with 2" between conduits, spacing = 7" and reactance = 0.068 ohms per 1000'.

The ABCg group will carry more current than the others. Instead of each wire carrying 25% of the total current the division could be three carrying 19.5% each and the fourth carrying 42%

Just a small question: is the difference of reactance the only determining factor in the current that follows each path? And is that mathematically derived, such that I could have figured it out if I knew how to determine the reactance?
 

rcwilson

Senior Member
Location
Redmond, WA
Just a small question: is the difference of reactance the only determining factor in the current that follows each path? And is that mathematically derived, such that I could have figured it out if I knew how to determine the reactance?

For my calcs, I assumed the resistances were all equal and used a nomograph from Okonite to estimate the reactance based on the phase-to-phase conductor spacing. In real life, the resistances of the cables will not be equal, nor will the phase-to-phase distances be uniform. Current division among the calbes will be determined by each cables impedance relative to the total combined circuit impedance.

Go to Okonite's website and look in ther technical data section for the inductance calculations. Other reference books also list cable inductances.
 
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