310.14(B) Engineering Supervision and total heat in conduit calculation

[brycenesbitt]

Question:

Under engineering supervision, what latitude if any does the engineer to count
heat rather than wires? Anyone dealt with a situation that required it, and how did it go?

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The situation:
An electrical room with a 2" existing rigid conduit and pull string, empty.
It leads roughly 30" through a post tension slab to a garage below.
The game is to get as many circuits as possible through that conduit without having to core the slab again.
They have to be individual circuits.

Wires will be landed on terminal blocks such as Eaton 14002-4, then on to eventual OCPDs.
But they're of mixed types.
Some are continuous loads, some not.
Some are low voltage control cables.
Some are at different capacities with different OCPDs.

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Assuming they were all the same type, do I have it about right?

1. #8 wire at 90C, start at 55A per 310.16
2. Apply 50% factor, so down to 27.5 amp.
3. 36 total #8 conductors for the 2" conduit ("Jam probability 9%" per south wire). 37.64% fill OK.

Total capacity of 12 circuits * 27.5 amps = 330 amps continuous.
But can better be done?

1. #10 wire at 90C, start at 40A per 310.16
2. Apply 45% factor, so down to 18 amp.
3. Now we're good to 66 #10 conductors for the 2" conduit ("Jam probability 13%" per south wire).

Total capacity of 22 circuits * 18 amps = 396 amps.
Or maybe

1. #12 wire at 90C, start at 30A per 310.16
2. Apply 35% factor, so down to 10.5 amp.
3. Now we're good to 102 #12 conductors for the 2" conduit ("Jam probability 16%" per south wire). 38.52% fill.

Total capacity of 34 circuits * 10.5 amps = 357 amps.


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Can the distance to the terminal block be considered to reduce the actual terminal temperature:

75C Terminal block --------------20 feet of wire at 70F ambient --------- 30 inch of conduit ------------ 20 feet of wire at 90F ambient ----- 75C terminal



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So apparently the #10's are the best for this formula, no?

But the question was about 310.14(B). The way I'm reading it the engineer discretion is only to apply a formula on an individual
wire, not a total load calculation on a bundle of wires. Is that a correct reading? Could an engineer look at this entire system of wires (continuous loads and non-coincident loads) and do a temperature calculation instead for the inside of the 30 inch constrained conduit ? Or a measurement?

 

[Julius Right]

If you intend to use the formula shown on 310.14(B) article, then you need to know some data about cables and environment.
Let's say we have multi-core cables of 3 live conductors and 1 neutral of 4*1/0 copper conductors XLPE [90oC] insulation. Let’s take from Okonite catalogue approx." O.D.[ 1.307 in2], that means only 1 cable may run through 2" RMC [see NEC ch.9 Article 344- Rigid Metal Conduit (RMC)]-for 40% 1.363 in^2.

If the cable is 4*10 awg then the O.D. is 0.56"[0.56^2/4*pi()=0.246 in^2] 5 cables may run here. Then Tc=90 Ta=25 [the concrete temperature]. The problems are Yc-skin and proximity effect-and Rca.Yc may be calculated according to Neher&McGrath [or IEEE 835/94].
I’d prefer IEC 60287-1-1 [it is simpler, in my opinion].
Let's say all 5 cables are the same type and the load will be maximum permissible.
The Rca -thermal resistance- it is a chain of Ti,Tj,Td and Te resistance i=insulation for individual core;j=jacket for all 4 [3 loaded] of a cable ;d=between cables and conduit and e=from conduit to concrete.
The idea is to consider the heat [here considered in watts] it is like a current and the temperature drop in each thermal resistance envelope is like a voltage drop.
If you have all the cable the same type, then you'll take 1 part through insulation 3 through jacket 15 between cable bunch and conduit and the same number through conduit to concrete thermal resistance.
If the conductor is different then, the electrical resistance and the current will be different, so an index must be introduced as a ratio between maximum heat produce in a main cable and each of other cables.
Let's say we have 3 of 10 awg copper and other of 12 awg .The electrical resistance of 10 awg
will be 1000 µΩ/ft [dc 20oC] and 12 awg 1600.
The current for 10 awg is 10 A and 6.6 A for 12.
Then for one conductor of 10 awg copper we have 10^2*1000/10^6=0.1 W/ft and 0.07 for 12.
If we take 1 per 10 awg will be 0.7 for 12.
Now when calculating Rca we shall take 1*Ti+3*Tj+(3*3+3*2*0.7)(Td+Te)

 

[wwhitney]

An electrical room with a 2" existing rigid conduit and pull string, empty.
It leads roughly 30" through a post tension slab to a garage below.
The game is to get as many circuits as possible through that conduit without having to core the slab again.
They have to be individual circuits.

If the constraint is just 30", i.e. you can set boxes at both ends and then fan in/out to multiple raceways/cables, then (2017) NEC 310.15(A)(2) Exception basically means you can ignore derating within that 2.5' long conduit for any circuit at least 25' long.

If you can figure out how to modify the conduit so it is just 24" long, then per Chapter 9 Table 1 you can use up to 60% fill.

Cheers, Wayne

 

[brycenesbitt]

If the constraint is just 30", i.e. you can set boxes at both ends

It's got a box on the garage ceiling, then 2" rigid extending up 3 feet or so.
Maybe could hack that off.

I rather add a heat sink to the wires above and below the slab. Let it be 90C for the 30 inches sure, it's not going to be that
hot for much more distance.....

 

[Julius Right]

If there are only single core cables of the same type then, you may use table Table C.9 Maximum Number of Conductors or Fixture Wires in Rigid Metal Conduit (RMC) (Based on Chapter 9: Table 1, Table 4, and Table 5).
So, in 2" RMC 40% you may run 64 insulated conductors of 10 awg or 37 of 8 awg.
To calculate the ampacity -in case of 100% full load-the conduit details are required:
you must know if it is totally sink in concrete slab and if it is so what is the distance from surface up to conduit centre-line.
However, calculated as per IEC 60287 the result is close to NEC [NFPA 70/23] using Table 310.16 and 310.15(C)1.
You are right that the formula 310.14(B) returns the current flowing through a single conductor but, it takes into consideration the presence of all other loaded conductors in the definite space.

 

[WD40]

However, calculated as per IEC 60287 the result is close to NEC [NFPA 70/23] using Table 310.16 and 310.15(C)1.

My boss was a foreman on a huge project with the same situation, the PM had the EE do some ampacity calculations without telling my boss. My Boss knew the coring contractor well and texted him. guy came drilled two more holes he ran the conduits and it was done.
Days later the EE came back to the PM and said his ampacity calc came out to less than the NEC allowed not sure how that was possible but he tells that story anytime someone brings up that section of code.