Conductor Ampacity In a duct bank?

[Mike01]

When installing conductors in an underground duct bank how is the correct ampacity selected and or calculated? After looking at different factors like load factor, dielectric loss, power factor of the cables, etc. and looking at the nec and their duct bank installations it refers to ?one? or ?two? circuits. If you had parallel feeders let?s say six sets of parallel conductors, two individual loads requiring six sets now you have a duct bank in the orientation of 2 rows high and 6 rows wide with that being the case is this considered a two circuit duct bank even though there are 12 used ducts? There are only two circuits routed through the duct bank but with parallel feeders? Or is each conduit considered a circuit? Therefore you have 12 ducts being used in the 2x6 configuration that does not exist in the nec details. The cable ampacity calculations I have found refer to load factor, depth, and temperature not necessarily duct bank configuration outside of what is indicated in the NEC?

 

[charlie b]

This is a very tough situation with at least one very simple answer. Whether you can take advantage of the simple answer depends on one thing: the basis of your ?required ampacity? value.

If you started by performing an Article 220 load calculation, and used the results of that calculation to size your feeders, then I believe you can simply take ampacity values straight out of Table 310.16. If instead you start with, ?I want to serve a 2000 amp switchboard and I want to be able to draw all of 2000 amps from that board,? then you need to calculate the ampacity of the underground conductors. Why the difference? Because the 220 calculation has built-in conservatisms, some of which will be ?expended? if you use the 310.16 ampacity values. In other words, we know that an underground ductbank does not meet the physical description at the top of Table 310.16 (i.e., not more than three current-carrying conductors in raceway). It fails to meet that description because the three conductors in one raceway are very close to three other conductors in raceway, and close to a bunch of others, so that the ability of any conductor to reject heat to its environment is hampered by the proximity of other nearby sources of heat. So in one sense using the 310.16 values is taking a step in the non-conservative direction. You compensate for that by using a ?required ampacity value? that was derived via an overly conservative process. Put another way, we know that if you tell the utility that a new building has a calculated load of 500 KVA, the utility is going to give you a 300 KVA (or smaller) service transformer. That is because the utility is fully aware that the actual load you will experience is going to be much lower than the calculated value.

By contrast, a board that really will serve 2000 amps worth of load needs conductors that have an ampacity of 2000 amps. And the conductor ampacity must take into account the physical circumstances of their installation.

To answer one of your questions, the notion of ?circuit,? as used in the NEC?s description of duct bank configurations, refers to some number of conductors within a single conduit. For example, a 2x2 array of conduits with three of the four conduits containing conductors would comprise ?three circuits?

Let me also say that you can forget about the 50% ?load factor? value that shows up in some of the NEC tables. You will never be able to take advantage of a tabulated ampacity that uses 50% LF, simply because no building is likely to have a load profile that gives a LF that low.

To answer your specific question about how to go about calculating ampacity, what I do is to use a computer software package designed for that purpose. The least expensive and easiest to use, in my limited experience with software options, is AmpCalc. www.calcware.com[FONT='Arial','sans-serif'][/FONT]

 

[bsh]

This is a tough one. I usually assume that the feeder must have the same capacity as the switchgear I'm feeding. That is worst case but for industial facilities it will all be needed. That said, soil acts like an insulator so the more conduits with current carrying conductors the more heat. This heat is contained in the earth and conduits adjacent to each other heat each other. All of this reduces the current capacity of the cable. The NEC has some good information but it is limited. The simplest way is to use a program to calculate the total ampacy of the conductors in the duct bank such as AmpCalc as Charlie suggested. I use EDSA. If you have access to the IEEE Brown Book (std 399), chapter 13 has some good tables. You will find that the ampacity of each cable goes down as more cables are added. Eventually you will find that you can't get enough copper in the ground. You might consider a ventillated trench with cable tray in it.

 

[Mike01]

IEEE Brown Book!!

IEEE Brown Book!!

A manufacturer list on their website that 750kcmil, 1/c, RHW-2, is rated in a duct for 535A (based on 1 ckt.) if I have a duct bank that is a 2x6 like stated above with 6-sets for two switchnboards through a ductbank, associated manholes the majority (90%) buried at approx. 30? below finished grade. According to the IEEE Brown Book Ch.13 Tabel 13-8 Fg: Grouping adjustment factor for 0-5000v 3/c, or triplexedin duct banks no spare ducts nonmetallic conduits 5? in diameter center to center of 7.5? if I select the cable size #750, No. of rows 2, No. of columns, 6 the adjustment factor indicates 0.522 so than this would be correct than:

535A (before adjustment)
535 * 0.522 = 279.27
279A per conductor
1600A sw.board
1600/279 = 5.73
Six sets would be = 1674A
So at 100% load factor a 1600A feeder would require 6-sets if 750kcmil.

Does this appear correct?

 

[jghrist]

A manufacturer list on their website that 750kcmil, 1/c, RHW-2, is rated in a duct for 535A (based on 1 ckt.) if I have a duct bank that is a 2x6 like stated above with 6-sets for two switchnboards through a ductbank, associated manholes the majority (90%) buried at approx. 30? below finished grade. According to the IEEE Brown Book Ch.13 Tabel 13-8 Fg: Grouping adjustment factor for 0-5000v 3/c, or triplexedin duct banks no spare ducts nonmetallic conduits 5? in diameter center to center of 7.5? if I select the cable size #750, No. of rows 2, No. of columns, 6 the adjustment factor indicates 0.522 so than this would be correct than:

535A (before adjustment)
535 * 0.522 = 279.27
279A per conductor
1600A sw.board
1600/279 = 5.73
Six sets would be = 1674A
So at 100% load factor a 1600A feeder would require 6-sets if 750kcmil.

Does this appear correct?

This appears to be correct if the ambient temperature does not exceed 30?C, the soil thermal resistivity (RHO) does not exceed 90?C-cm/W, and there are no other heat sources in the vicinity of the duct bank (like another duct bank). Because this adjustment factor is not covered in NEC tables, the calculation should be under the supervision of a registered engineer.