Concrete encasement for duct banks

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erniep

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I have a large commercial project served by 3 - 3000A service sections for 277/480 3ph lighting and HVAC equipment. The common duct bank from the service yard to the building has up to 20 - 4" conduit(incl spares)in five groups of 4's with 24" spacing in between and 18" cover. The typical conduit has 4 - 750 KCMIL and 1 - #4/0 GND. Plans call for a "river" of concrete slurry to encase the conduit w/6" of slurry all around. Heat dissipation has become an issue - but NEC '99 Article 310 does not offer any guidelines as to when heat build-up becomes a problem. Are there guidelines for the max number of conduit at "X" voltage that can be placed in a duct bank before heat becomes a problem?

I apologize for trying to be succinct with my first post. Here's the actual condition by bank.
bank 1: 1400A, 4-4"C w/4-750 KCMIL, 1-#4/0 GND.
bank 2: 2 - 3"C and 2 -4"C all spares
bank 3: 1000A, 3-4"C w/4-500 KCMIL, 1-#2/0 GND., 1-4"C spare
bank 4: 1000A, 3-4"C w/4-500 KCMIL, 1-#2/0 GND., 1-4"C spare
bank 5: 600A, 1-4"C, 1-2 1/2"C wire size unknown
Distance from SES to elec. room - 500ft. Is heat buildup and the loss of ampacity a concern? Is filling the void between banks w/concrete beneficial, or should the banks be individually formed?

[ January 16, 2004, 01:50 PM: Message edited by: erniep ]
 
Re: Concrete encasement for duct banks

Annex B goes into more depth. But, this is under engineering supervision

Roger
 
Re: Concrete encasement for duct banks

First, this is not a code issue. It is a design issue. There are table in the 2002 code in Annex B
that may be of assistance. From your post it appears to me that you have 5 duct runs, each consisting of 4 4??" conduits for a total of 20 conduits. Each conduit has 3 750 phase conductor,
a 750 neutral conductor and a 4/0 GND conductor.
Since the duct runs are 24" apart, I do not think you will have mutual heating between the ducts.
Assuming that is correct, you can use table 310.16
for conductor ampacity. Making another assumption that the load will not produce harmonic currents,
an assumption that is likely incorrect, then the
ampacity of the 750 at 75 degree equals 475 amps
per conductor. 20 runs of 750 at 475 amps equals a total of 9500 amps. However the posts states that there are some spares in the duct runs, therefore the total runs of 750 must be less than the assumes 20 and the ampacity is less that the 9500 amps.
If the assumptions that there is no harmonic load is incorrect, then 310.15.B(4)c requires that the neutral may be required to be counted as a current carrying conductor. If so, then a 0.80 derating factor must abe applied to the ampacity of the 750 conductor. The 750 derated at 0.80 x 475 amps = 380 amps x 20 runs = 7600 amps.
 
Re: Concrete encasement for duct banks

Sorry Bob, the title to Table 310.16 is "Table 310.16 Allowable Ampacities of Insulated Conductors Rated 0 Through 2000 Volts, 60?C Through 90?C (140?F Through 194?F), Not More Than Three Current-Carrying Conductors in Raceway, Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30?C (86?F). This table allows only Three Current-Carrying Conductors in Raceway, Cable, or Earth (Directly Buried). There is no mention of a concrete duct or multiple raceways. IMO you need Annex B.
 
Re: Concrete encasement for duct banks

Take a closer look at Detail 2 in Annex B, Table B.310.2. It shows four conduits within a single ductbank. At first glance, it might appear to be similar to your description of the ?common duct bank.? But in Detail 2, one of the conduits is empty. If you want to use the ampacities of Table B.310.7, and to use the ?Detail 2? columns of that Table, then you have to leave one conduit empty in each set of 4. If your installation is different, then you would be best advised to turn this over to one of those ?engineering supervision? type persons.
 
Re: Concrete encasement for duct banks

Charlie
I agree with you statement regarding using Annex B
for cable ratings. I used a Rho of 90 and a LF of
75. I don't think a LF of 100 is applicable. The post mentioned that there were spare ducts in the
duct banks so I used table B310.6 and Figure B310.1 to interpolate a new cable capacity. If I did it correctly, the capacity for 750 came out as 473 amps. That was close the the value in table 310.16. If there is one spare in each 4x4 duct run, then there are 15 runs of 750. This amounts to a capacity of 7125 amps. There does not seem to be enough wire for this service. We need Ernie to add some additional information.
 
Re: Concrete encasement for duct banks

And just to complicate matters in one more small way, the person signing the calculation would have to justify the use of a 75% load factor. But that is no easy matter. First of all, the term ?Load Factor? is not defined anywhere in the NEC. We must then revert to standard industry definitions: ?Load Factor? is the ratio of average demand divided by peak demand. If you had a facility that was running at 80% power for 16 hours each day, and dropped back to 40% for the other 8 hours, the average would be 67%, the peak would be 80%, and the LF would be 67% / 80%, or 0.83. To get an LF of 75%, you would need a higher peak or a lower average. That sort of manipulation is the reason for ?engineering supervision.?
 
Re: Concrete encasement for duct banks

Charlie B
Since we do not have the information regarding this load at this time, we can only guess. If this load is typical commercial we could estimate the peak for 8 hours at 80% and drops to 40% for 16 hour. If this is so, then the average would be
0.53. LF would equal 0.53/0.80 = 0.66. If you assume the peak occurs for 12 hrs and drops to 40% for 12 hrs the average would equal 0.60 and the LF would equal to 0.75. I hope Ernie will take time to add some additional information.
 
Re: Concrete encasement for duct banks

I tried posting this just as the topic came out, but my computer at the office wouldn't allow that. I'm sorry it it is not as timely as it should have been, but technology stuck it to me and I'm having to post it from home. After all this, I hope it's worth it.

Just something that I found that may speak to your situation.

Neher-McGrath Calculations

The Neher-McGrath Calculations provide a method for calculating underground cable temperatures or ampacity ratings and are derived from the following technical paper:

J. H. Neher and M. H. McGrath,"The Calculation of the Temperature Rise and Load Capability of Cable Systems", AIEE Transactions, Part III, Volume 76, pp 752-772, October, 1957.

This paper considers the complicated heat transfer issues associated with the determination of underground system ampacities. The paper cites the following basic equation for calculation of a cable ampacity:

nmcalc.gif


However, this single equation masks the great complexity involved in these procedures. There are scores of complicated equations involved in developing the terms in this equation and those required for temperature calculations. (The paper defines over 80 variables and contains in excess of 70 formulas excluding appendices.) To solve for unique ampacities or temperatures at each cable position, a multiple set of equations must be developed to take into account interference heating from every position in the system, and a matrix solution technique for simultaneous equations utilized.
 
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