NEC and cable loading in duct banks 1

[BJC]

Does anyone else find the interpolation chart B.310.1 in the NEC confusing?
Has anyone ran accross a good artical with a monkey see monkey do explanation?
I want to use a 100% load factor with a Rho of 60 ( preferably 55 -which is concrete ). The tables use a Rho of 90 and 120 with 100% load factor but not Rho 60.
I'll keep trying, just wanted to see in the meantime if I was getting dumber or the code is written as badly ( English seems to a second language to many engineers ) as I alway new it was.
Thanks

 

[kenvlach]

I’m not an electrician, but occasionally buy busbar & cables, & use easy-to-understand websites like Okonite’s.
If I understand you correctly, the concrete should carry away heat at a faster rate than normal ductwork, so the allowable ampacity of your cable should be higher (or else a bigger safety factor for the same ampacity). I don’t whether you are referring to cables in buried concrete, but the following may be helpful.

Okonite has some tables with correction factors for various ambient air and soil temperatures which “are derived from AIEE-IPCEA “Power-Cable Ampacities”, joint publication S-135-1 and P-46-426 which includes more complete tables covering additional earth resistivities and load factors.”

“The following tables relate to insulated cables in underground ducts, in free air, in conduit in air, and directly buried in earth. The values are based on 90°C and 105°C conductor temperatures and an ambient temperature of 20°C for all cables in underground duct or directly buried in the ground and 40°C for all cables in air.”

They do use the values Earth thermal resistivity: RHO 90, 100% load factor, for both duct & direct burial. I tried pasting the tables, but formatting was lost, so please see the link below:

http://www.okonite.com/engineering/ampacity-correction-factors.html

 

[DougMSOE]

If memory serves correctly, look in the color book series of IEEE. I think it is the brown book. Numerous tables addressing your problem.

 

[cuky2000]

Chart B.310.1 should be used in conjunction with table B-310-7.

EXAMPLE: For Rho=55, determine the ampacity of 4/0 AWG type THHW copper cables in 3 electrical ducts configuration similar to detail 2 on fig B-310-2

a- From table B-310-7, 3 electrical ducts select current as follow:

I₁= 280 A ( Rho= 60 & LF=50%)
I₂= 201 A (Rho=120 & LF=100%)

b- Calculate the ratio of I₂/I₁= 201/280 = 0.72

c- From the bottom portion of Chart B.310.1, select the intersection point of Rho=55 on the horizontal axe and LF=100% on the vertical axe. _{(Only for LF= 100% the bottom horizontal axe have same value of 55 as the value on the intermediate horizontal line between the bottom and upper portion of the Chart.)}

d- In the upper portion of the Chart B.310.1, interpolate a curve for I₂/I₁= 0.72 between 0.70 & 0.75

e- Determine the intersection point between the interpolated curve for I₂/I₁= 0.72 and a vertical line from 0.55 starting on the horizontal axe between the upper and bottom portion of the Chart.

f- Determine the factor F=0.84 (approx.) on the right side of vertical axe on the upper portion of the Chart.


g- Multiply FxI₁=0.84x280=235A. This is the adjusted ampacity for 4/0 Cu cable for a thermal resistivity, Rho =55 ^oC-cm/watt.

 

[jbartos]

Also, there is no direct definition of "LF stands for Load Factor" in the whole NEC. I guess it is expected easy to guess. Also, this kind of guessing appears to be a part of the engineering profession.

 

[cuky2000]

Dear Jbarto;

I should agree with you that there is not extensive explanation on the NEC about the Load factor (LF). However, there are more details explanation in sources referred by the NEC such as Neher-McGrath, IEEE, ICEA Standard, cable manufacturer literature and other sources.

The IEEE Std. 100 defines the Load Factor (LF) as the ratio of the average load over a designated period of time to the peak load occurring in that period.

The load factor (LF) take in account the allowance to increase the non continuous cable current rating for variable cable loading since there is a delay of temperature rise do to the heat absorption by the surrounding duct system elements (conduit, cable, concrete, earth, air, etc).

For example, consider estimate the load factor for the following loading cycle:
8AM to 8PM (12 hr)…………………… 500 A (peak)
8PM to 12M (4 hr)…………………… 300 A
12M to 8AM (8 hr)……………………… 100 A

Average Load = (12hx500A +4hx300A + 8hx100A)/24h = 350 A
Load Factor (LF)= (Average Load/Peak)*100 = (350/500)*100 = 70%

I hope this could help to clarify the concept on load factor.


NOTE:
_{The NEC provides a conservative approach and allow under qualified engineering supervision to do more aggressive and economic design. The price for those interested to go beyond the present NEC table is to study other sources. Please the enclose article "Underground Ampacity: Rho Makes A Difference" site and additional comment in this matter.}

<

http://www.southwire.com/tech/pubs/pcu/95/pcu1095.htm#rho>

 

[BJC]

Thanks Cuky2000. Very helpful Your explanation was the way I was reading it. Has a discussion here about ductbanks. Utilites can build a duct bank and get 3,000 amps out of a 2,500 KVA transformer, but not the rest of us ( if we have the local inspector looking over our shoulder.).

The last time I engineered a 2,500 KVA padmount transformer I wanted to use a busduct. The utility objected ( I wanted to cut a hole in their transformer) and the architect didn't like the ugly thing ( his opinion). I used precast trench with cable tray and used the tray rating of the cable. The installation had two transfomrers feeding the same switchgear. Both were supposedly sized to cary 67% of the facility load and in the event one transformer failed the other would carry 133% for the duration.

Does anyone know of a duct bank using forced air cooling? I could use 2,3 or 4 ducts to blow air into and out of the ductbank. If the exiting air was below 68 degrees the cables could be loaded to the single duct rating. It would take a PLC to monitor current, air temperature and turn the fan on an off.

 

[dpc]

I think the forced cooling would work, but I would be worried about getting approval (variance) from local inspectors.

I have done trench installations using special thermal sand backfill and cable spacers for some large 15 kV feeders. Okonite provided calculations on ampacity of the configuration based on Neher-McGrath, and everyone was happy, except probably the poor electricians who had to build it. I believe that this was 6 or 8 conductors per phase.

The trench was covered on top and was accessible, if necessary, in an emergency.

You might also talk to the Calvert Company. They make an engineered &quot;cable bus&quot; system that is a hybrid between rigid bus and cable systems.

Good luck.

 

[BJC]

Cuky2000

Check your math above. I get a load factor of 66.6% Your method is correct but you may have hit the wrong key on the figuring machine.

 

[beghtesady]

You also need to keep in mind that the Rho depends upon the medium(s) surrounding the cable. In a duct bank there are typically four layers of medium to consider. Each medium has a different rate of heat transfer due to thickness, material and wire size and hence different Rhos. Assuming that we have a continues load (instead of a variable load) over a long period of time and that the medium around the conductors is evenly proportioned, you need to develop the equivalent Rho value (this is a heat transfer equation)using the Rho values of the air, conduit, concrete and finally the soil and consider the heat contribution of the other conductors in the adjacent ducts (or conduits in the bank). Only after you have established this quantity then you can utilize the tables listed in the NEC to find the current carrying ampacity of the conductors.