4000 amp main

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bob

Senior Member
Location
Alabama
This was a question on another forum I thought might stir some great answers. The question did not say but I assume that it is and underground service.


On a project with a 4000 amp service, the engineer
says that we must install 15 parallel 500 kcmil copper conductors to meet the code. The local utility company requires that we install 11 parallel 500 kcmil conductors. The engineer is using table B310.7(Annex B) as backup for his position. However 310.15A1 states " Ampacities for conductors shall be permitted to be determined by tables or under engineering supervision, as provided in 310.15B and C. We believe, as does the local utility company, that Table 310.16 would apply if we opt for the table method of sizing the conductors. What is your opinion?
 

don_resqcapt19

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Location
Illinois
Occupation
retired electrician
Re: 4000 amp main

Bob,
There is a problem with mutual heating in underground duct banks and if you install cables in an underground bank and load them to their Table 310.16 ampacities, they will self destruct. The only reason that this problem does not show up more often is because the calculations in Article 220 result in a load that is much higher than the actual load will be.
Don
 

charlie

Senior Member
Location
Indianapolis
Re: 4000 amp main

Don, to back up your point, I wrote the following to be used by our engineers to size our transformers. This is for an electric utiliy to use, not for general use. If we undersize a transformer, we have to go back and replace it, even if it is 20 years later. :)

For commercial buildings, a reasonable rule is to use a demand factor of 50% for all connected load. Another rule, although somewhat shaky, is to take 80% of the switch size and multiply the resultant by 50%.
 

bob

Senior Member
Location
Alabama
Re: 4000 amp main

Bennie what I have listed is all the info I know. This was not my question.

Don I agree with your statement. The NEC caculation provide a lot of fat for service and feeder loads. Thats good I suppose.

Charlie when I used to do this, for commercial buildings I gave 100% for lighting, 100% for the A/C units and threw in a little extra for the other load. Here on the gulf coast there is no diversity for A/C units. Temp is 96 today.

To the question I think the engineer is correct in his efforts to use the table of the UG duct banks. I do have a problem if he used table B310.7
as there are only 6 ducts shown. He apparently chose 273 amps per conductor with Rho = 90 and LF = 100%. I would like to see the table with LF= 75
and Rho = 90.

[ August 28, 2003, 02:45 PM: Message edited by: bob ]
 

sc57ford

Member
Location
South Carolina
Re: 4000 amp main

Did a project recently very similar. Had a 2500 kVA cast coil transfromer with fans the rating becomes 3136kVA. That is 3772 Amps at 480 volts. My guess is that the engineer is trying to cover the fan rating of the transformer. Therefore, load factor would not enter in just because of Don's earlier post. The 15 500kCMil cables rated at 273 amps gives him 4095 amps, about 9 percent over the transformer output. (And of course we all know transformers can run higher than rated output for some time).
 

charlie b

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Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: 4000 amp main

If the engineer wants to use Table B310.7 for a 15-duct configuration, then I would say that that engineer needs more supervision. The number 273 is clearly related to a 6-duct configuration, and cannot be applied to the problem at hand.

If the NEC 220 method was used to determine that you needed a 4000 amp service, then there would be conservatisms built into the ?calculated load.? Using Table 310.16 for the service ampacities will expend some of this conservatism. But this would be less severe on the cables than ignoring the load calculation and simply running 380 amps through each of 15 parallel underground 500 MCM cables. To this extent, I agree with Don and the Other Charlie. But I think the right thing to do is to perform a Neher-McGrath calculation, using one of several commercially-available software packages.

I agree with Bob, that Table B310.7 would be far more useful if it employed an LF of 75, instead of 50. The Load Factor is the ratio of average to peak. To get a 50% ratio, you would have to run the facility at full load for 12 hours a day, and turn it completely off (main breaker open!) for the other 12 hours. To get a 75% ratio, you would have to run the facility at full load for 12 hours a day, and reduce the load to 50% for the other 12 hours. This is a far more realistic load profile.
 
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