Table 392.22(A) column 2 use for cable tray

[dm9289]

I have posted a similar post yesterday and got some help with a good resource, but im still confused about this calculation

Condition 392(A)(1)(c) 4/0 and smaller cables with 4/0 and larger cables combined and the use of column 2 Table 392(A) somehow im lost in the math or missing something. Specifically with the SD and its use. Any example problem would be great from anywhere not much luck on web with this condition. There is some online calculators but I'm trying to learn this the long way

 

[david luchini]

Sd is the sum of the diameters of all of the 4/0 and larger multiconductor cables.

If, for example, you had (3) 4/0 multiconductor cables with a diameter of 1.75" in a 12" wide tray.

Per column 2, you would have 14-1.2(1.75+1.75+1.75) or 7.7 sq. in. available for the smaller than 4/0 multiconductor cables.

 

[dm9289]

Sd is the sum of the diameters of all of the 4/0 and larger multiconductor cables.

If, for example, you had (3) 4/0 multiconductor cables with a diameter of 1.75" in a 12" wide tray.

Per column 2, you would have 14-1.2(1.75+1.75+1.75) or 7.7 sq. in. available for the smaller than 4/0 multiconductor cables.

i was not even in that neighborhood. So for small cables you will need spec sheet to find out area in sqin of the cable, or calculate by measured diameter, but big cables just diameter. I did not get that you were finding for area left for small cables. Your example is helpful

Thank you i get it much better

 

[m sleem]

Sd is the sum of the diameters of all of the 4/0 and larger multiconductor cables.

If, for example, you had (3) 4/0 multiconductor cables with a diameter of 1.75" in a 12" wide tray.

Per column 2, you would have 14-1.2(1.75+1.75+1.75) or 7.7 sq. in. available for the smaller than 4/0 multiconductor cables.

But, how initially we propose 12" wide tray for (3) 4/0 multiconductor cables ?