Wire sizing

A

AndrewGia93

Member
Location
PA
Occupation
Electrician
Running a new service to a pole barn about 500 feet from a ct cabinet. 200 amp 208 3 phase and the load calc for the building is a max of about 140 amps. Not really getting too much help from engineers or township to for my calculations on wire sizing. Trying to run aluminum wire. Thoughts?
 
AndrewGia93 said:
Appreciate it! In this situation would you calculating according to the full 200 amps or estimated load
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Your choice. Depends on how much you want to allow for future loads.
Once you run the numbers whoever is paying will likely lean toward estimated or calculated load and not the full 200.
 
3* 250 kcmils aluminum conductor 75oC, in a 4" conduit, underground [ρe120 W.cm/oC-in NEC it is only 90], may be loaded with 200 A. However, if you limit 3% voltage drop, then you need 750 kcmils for 500' length.
 
According NEC 2023 Annex B- B.2 Typical Applications Covered by Tables
Typical values of thermal resistivity (Rho) are as follows:
Very dry soil (rocky or sandy) = 120
[Rho=oC.cm/W]
If it is about unknown location the worst case is preferred to be taken into consideration, in my opinion.
 
infinity said:
When you have a minute please update you profile to include your location.
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Decided to run parallels to bring the size down and for some future proofing. The south wire voltage drop app is convenient but is it overkill at all? I’m assuming when you input the amps it’s accounting for continuous?
 
AndrewGia93 said:
Decided to run parallels to bring the size down and for some future proofing. The south wire voltage drop app is convenient but is it overkill at all? I’m assuming when you input the amps it’s accounting for continuous?
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For voltage drop calculations it doesn't matter if it's a continuous or non-continuous load. As far as overkill it just tells you the numbers based on what parameters you input. The key number being the current value you use. Some see a feeder that is say 200 amps and they use that in their calculation when the actual load is much less. That will result in conductors much larger than what is needed for the smaller actual load.
 
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