I have a ground faulted 1750mcm cable that is part of a 2000 amp feeder circuit. The existing circuit uses two 1750mcm cables per phase. We will probably replace all 6 cables due to the fact that they are 20 years old and we will probably damage them if we just try to replace the single bad cable. This cable isn't readily available and has very long manufacture lead times, not to mention a 10,000' minimum order. I am considering using 4 runs of 1000mcm per phase to replace the original circuit. 1000mcm is common and can be delivered right away at the lengths we need. My question is, how do I properly calculate the ampacity of this proposed circuit? It runs in cable tray for 600', then goes under ground in conduit for 25' where it enters the main sub station. I know it will require additional conduit and termination space. What other items do I need to consider or be concerned about? I don't want the electrical contractor to be my only source of council on this. Thanks!
ampacity calculation for feeder repair
- Thread starter Tfret
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- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
I am confused 2 runs of 1750 KCM copper is only good for 1300 amps and then you need to consider voltage drop.
I should have mentioned this is a 2.3kv system. 392.12 deals with the capacity of the tray. It doesn't look that is a problem. 4 conductors in a bundle per phase all in a single layer seems to fit this code. The max continous load is never over 1000 amps. The system has a 2000 amp breaker.
Cow
Senior Member
- Location
- Eastern Oregon
- Occupation
- Electrician
Doesn't the copper to ampacity ratio get a little out of hand when you get over 500mcm or so? I just can't see 1750mcm being practical from a cost and labor standpoint. Two 400mcm conductors is more ampacity than that and less than half the copper?
- Location
- Illinois
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- retired electrician
Dennis,
Using the rule in 392.11(B)(1) and a 75?C rated conductor, I come up with 1920 amps for paralleled 1750 kcmil copper.
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
That is for cable tray but the wire goes in conduit underground does that not change thing?
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
That's the point I am curious about. Does this not change the amp. or does310.15(A)(2) come into play.
I said look at 392.11. 392.11.B1 refers to the ampacity of cable installed in a cable tray. Since you have less that 5% of the cable in conduit, I believe you
can use 310.15(A)(2). If your max load is 1000 amps why not consider a 1200
amp service? 1000 amps is a large load on a 2.4 kv system. Is this a primary service from a utility? Are you sure its 1000 amps a 2.4 kv?
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Cold Fusion
Senior Member
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- way north
The existing is likely MV-90, unshielded. Shouldn't the tray part be using 392.13 (current code)?
Also, 310.15 doesn't apply - shouldn't it be 310.60?
cf
Also, 310.15 doesn't apply - shouldn't it be 310.60?
cf
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Art 392 states for MC cables, >2000 volts Tabl3 310.69 and 310.70 apply.
In conduit it appears it might be 310.70, but I find nothing on 1750.
Bob mentioned a 5% rule and I am curious as to where that is in the Code.
In conduit it appears it might be 310.70, but I find nothing on 1750.
Bob mentioned a 5% rule and I am curious as to where that is in the Code.
- Location
- Illinois
- Occupation
- retired electrician
Dennis,
I looked right past the conduit part.
I believe you are correct regarding 392.13. Are you referring to the duct bank rating in 310.60. I still think that you could use the logic in 310.15 A2 exception.
Augie
310.15(A)2 exception says 10% but 310.15 says for cable rated 0 - 2000 volts. The question is does this apply in this installation. I am not aware of another section that applies to 5 kv cables. It seems to me that the logic of
310.15(A)2 exception could be used.
310.15(A)2 exception says 10% but 310.15 says for cable rated 0 - 2000 volts. The question is does this apply in this installation. I am not aware of another section that applies to 5 kv cables. It seems to me that the logic of
310.15(A)2 exception could be used.
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
I don't think 310.15(A)(2) applies
To me this states 10% of the 600 feet in the cable tray or 10' whichever is less. 10% of 600 is 60 feet. Thus 10 feet is the max. allowed of the conduit. In the OP's example there is 25 feet of conduit therefore it does not apply.
To me this states 10% of the 600 feet in the cable tray or 10' whichever is less. 10% of 600 is 60 feet. Thus 10 feet is the max. allowed of the conduit. In the OP's example there is 25 feet of conduit therefore it does not apply.
Cold Fusion
Senior Member
- Location
- way north
I wouldn't think so. The only regulation that applies is 310.60.D. This is a design problem. I don't see it being solved by applying regulation that says it doesn't fit. My guess is the code panels realized that a 2000A, 4160V feeder just is not within the cookie cutter design approach of the NEC.
cf
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