Ductbank - Increasing Ampacity

[Keri_WW]

Info:
2000A service at 208V 3-phase, 5 sets 4#500 kcmil, 1#250 kcmil GND. I was planning on using a 6-way ductbank with one being a spare (all at 4"). I'm also using the rho value of 90 which is pretty typical for the area.

Question:
Table B.310.7 is showing me that I won't have the ampacity to meet the demand. Is adding another parallel set the best option that I have? I understand that the numbers might be off a bit because there isn't a chart for 5 ducts....

Thanks for any help and info you guys can provide. This is my first ductbank design from scratch and I want to make sure I understand it correctly before I get too far along.

Keri
:grin: :grin:

 

[charlie b]

You need to start with a load calculation. A 500 MCM is only good for 380 amps, per Table 310.16, and five of them are good for 1900 amps. You can't use that for a 2000 amp service, unless the calculated load is under 1900. If you do get a calculated load under 1900, then you don't need to use Table B.310.7.

On the other hand, if you just say that you want 2000 amps of conductor capacity, in an underground configuration, that I would say you have two choices. One is to use Table B.310.7, give the 500 MCM conductors credit for 249 amps each, and install 8.03 of them. The other is to have the ampacity of the conductors calculated by a Professional Engineer, using the "Neher McGrath Method." You would probably find, however, that that method gives you a lower number than that shown in Table B.310.7.

This, of course, brings us back to your best next step being to perform a load calculation.

 

[QES]

charlie mention about 500MCM with 249amp, this value should be a little bit higher because you are running parallel set.

 

[bob]

charlie mention about 500MCM with 249amp, this value should be a little bit higher because you are running parallel set.

The ampacity is 249 amps for each set of 3 #500 kcm. Actually if Rho is 90 the rating is 273 amps for the 6 duct configuration. However 273 amps x 6 only equals 1638 amps. Even using 750 kcm only gets you 1980 amps. You won't be able to use these tables since there are only 6 ducts installed.
If you install 2 runs of 3 ducts separated at least 24 inches, the ampacity of 500 kcm is 341 amps x 6 equals 2046 amps. Need to verify the ampacity at 24 inch separation. Keri, Is this economically feasible?

 

[Keri_WW]

Thanks for the information. Currently I'm at about 2100A connected / 1400A demand load. These numbers are without complete HVAC loads and pumps (only guesstimates at this point). Hopefully the numbers won't increase by a whole lot.

Edit: I did a quick extrapolation, and for the 5 duct set that I want to run, I'll have 297A (1485A total)... looks like I'll need 6 sets after all..... Charlie, I'm a little confused on why you say if I'm not at 100% load that I don't have to use Table B.310.7.... can you enlighten me?

Thanks again!!

Keri
:grin:

 

[charlie b]

Charlie, I'm a little confused on why you say if I'm not at 100% load that I don't have to use Table B.310.7.... can you enlighten me?

I was thinking two or more steps beyond what I had written. Sorry for the confusion.

It is not a matter of being, or not being, at 100% load. The NEC load calculation process is known to be conservative. As an illustration, you will note that utility companies tend to give you a transformer that is rated between 40% to 50% of your calculated load. They are well aware that the actual load will be far lower than the calculated load.

Some jurisdictions, and my state is among them, will allow you to use Table 310.16 for underground ductbank configurations. But they only allow this if the amount of load is based on an Article 220 load calculation. They are essentially allowing us to ?expend? some of the conservatism built into the calculation process, by allowing us to use a higher value of ampacity than that shown in B.310.7.

Does that make sense?

 

[Keri_WW]

Sure does! Thanks for clarifying.