Cable Tray Width

[timm333]

I am trying to figure out how to calculate the minimum required width (in inches) of cable tray as per NEC 392.22(A)(1)(c); when both larger (>= 4/O AWG) and smaller (< 4/O AWG) cables are present in the same cable tray. This minimum width (in inches) would be the sum of two parts.

PART 1: Minimum width required (in inches) for larger cables. This part is easy as it would be the sum of outer diameters of all larger cables. They have denoted it as “Sd” in column 2 of NEC Table 392.22(A)(1).

PART 2: Minimum width required (in inches) for smaller cables. For this part the fill area of smaller cables (in square inches) can be calculated from column 2 of NEC Table 392.22(A)(1). My question is that for this part, how to calculate the minimum required width (in inches) after calculating the fill area (in square inches)?

Thanks

 

[JoeStillman]

To convert the square inches of a bundle of small cables to width, just divide by the depth. If the tray is 2" deep and you want to fill it to 50%, then your square inches are equal to your width in inches.

 

[Julius Right]

First calculate the part of cable tray for large cables and then the remaining part of small cables has to be calculated.
Let's say we have 5 cables of 500 kcmil of 0.943 inches diameter and 20 cables of 2/0 of 0.528 inches dia.
The cable tray width first part is 5*0.943=4.715"
Now, the cross-section area of the small cables 0.524^2/4*pi=0.21565*20= 4.313 in^2-total.
Then 4.313+5.658=9.971".Using column 2[A1c], the closer to this cable tray width is 10.5.

 

[Julius Right]

Mr. JoeStillman's post reminds me of the ICEA P-54-440 standard although this standard refers to solid bottom tray only [according to Stolpe].
ANSI/NEMA WC 51 ICEA P-54-440 AMPACITIES OF CABLES INSTALLED IN CABLE TRAYS
First we have to calculate the apparent fill depth [d]
The depth of cables in trays should be calculated as follows:

calculated Depth,inches=(n1*d1^2+n2*d2^2+...nn*dn^2)/w

where: d1,d2..dn =diameter of cables, inches
n1,n2,...nn number of cables of diameter d1,d2,..dn respectively. w=width of tray, inches.
That means we don't have to separate the cables according to the conductor cross-section area.
Let's take the same example as above.
n1=5 d1=0.943 and n2=20 d2=0.528
d=(5*0.943^2+20*0.528^2)/10.5=0.95447[close to 1”]
From Table 5-2 for single core 600 V insulated non-jacketed cables:
for 500 kcmils 450 A and for 2/0 awg 137 A.

Now from NEC art 392.22 A(3)c and 392.80A(2)b: 65% of Table 310.15(B)(17):
2/0 awg 265 for 75oC rated and 65%=172.25A
500 kmils 620 for 75oC rated and 65%=403 A

 

[Julius Right]

There are some correction factors in ICEA P-54-440 which I neglected.
First, since the rated temperature is here 90, for 75oC a factor of 0.86 has to be employed.
Second, the cable diameter- for 2/0 in the Table 5-2 is 0.58 and for 500 kcmils 1".
Then, after corrections applied, for 2/0 the ampacity remains 107.26 A and for 500 mcm 365 A.