Cable Tray Sizing

[Alley Cat Bad Dog]

Hi There,
I have question about an open ladder cable tray that I'm traying to size for one of the projects, and I'm confused between the following two methods, one it says a space between cables is required with at least the diameter of the cable as shown here:

or by the triangular conf. :



, and the other method shows sizing it based on adding the diameter of cables larger than 3/0 AWG + the (6in/7sq.in )* the cross sectional area of all cables equal or less than 3/0 AWG. also as shown bellow:


Thanks in advance!

 

[don_resqcapt19]

Look at what happens to the ampacity of the cables or conductors based on the different methods.

 

[Julius Right]

Nor NEC neither IEEE 835 indicates what is the ampacity of three [or four] conductor power cable in free air. But, it is reference to triplex. Triplex, of 3 single core cables, it is not a three conductors cable.
In NEC art.392.80 (1) Multiconductor Cables
(c) Where multiconductor cables are installed in a single layer in uncovered trays, with a maintained spacing of not less than one cable diameter between cables, ampacity for free air in accordance with 310.15(C).
Informative Annex B Application Information for Ampacity Calculation
Table B.310.15(B)(2)(3) Ampacities of Multiconductor Cables with Not More Than Three Insulated Conductors
For 75oC [limited by art.110.14C] 1/0 copper 160 A and for 500 kcml 415 A
In NEC art.392.80 (2) Single-Conductor Cables. (d) Where single conductors are installed in a triangular. See Table 310.15(B)(20).
For 75oC [limited by art.110.14C] 1/0 copper 183 A and for 500 kcml 496 A
Let’s take a 24” width cable tray.
According to Okonite catalogues 3*core cable diameters are:
3*1/0- 1.17” ;3*500 mcm-2.13 then:
11 cables 3*1/0 require 24.57 inch width [we take only 10, then]
6 cables 3*500 require 23.43 inch width
Single core cables:
1/0 -0.46; 500 mcm -0.87; 2.15*0.46= 0.989; 2.15*0.87= 1.8705
12 groups of 3*1/0 require 22.747inch width
6 groups of 3*500 require 20.58 inch width

 

[Julius Right]

For all cables, item 1 to 5, you need 30" width cable tray.
Sd=pi()/4*(3*2.26^2+4*1.55^2)=19.582 sq.in.
Total B cables cross-section area=20*0.17+20*0.2+4*0.8=10.6 sq.in.
According to Table 392.22A column 2 total B has to be not more than 35-1.2*19.582=11.5 sq.in.

 

[ron]

What kind of ampacity are you looking for? Free air 310.17 or standard 310.16.

 

[Alley Cat Bad Dog]

What kind of ampacity are you looking for? Free air 310.17 or standard 310.16.

Free air, I will be using a ladder cable trey run, from one building to another. and it will have all of the following cables:

36* 12AWG
10* 10AWG
6 * 8AWG
3* 6AWG
7 * 4AWG
2 *1AWG
3 * 1/0AWG
1 * 2/0AWG

4 * 350KCMIL

Note this is only an example of multiple cable trey runs

 

[Alley Cat Bad Dog]

Free air, I will be using a ladder cable trey run, from one building to another. and it will have all of the following cables:

36* 12AWG
10* 10AWG
6 * 8AWG
3* 6AWG
7 * 4AWG
2 *1AWG
3 * 1/0AWG
1 * 2/0AWG

4 * 350KCMIL

Note this is only an example of multiple cable trey runs

Tray **

 

[don_resqcapt19]

Free air, I will be using a ladder cable trey run, from one building to another. and it will have all of the following cables:

Then you have to use the rules that have the spacing between the cables or conductors. That is what I was leading to in post #2.