Matt Donovan
Member
- Location
- United States
I have a condition with a mix of multi-conductor cables and single conductor cables in the same tray, including some single-conductor cables below 4/0. Section 392.22(A)(1) is titled "Ladder or Ventilated Cable Trays Containing Any Mixture of Cables", but its under the heading for Multiconductor cables, so I don't think this counts for a mix of multi- and single- cables. Do I calculate the required width for each cable type and add them together?
The other thing that's tripping me up is 392.22(B)(1)(d) says "where an of the single conductor cables are 4/0AWG or smaller, the sum of diameters of all single conductor sables shall not exceed the tray width". I have mostly larger cables, but a few 4/0 and smaller. Why does the presence of a couple smaller cables mean that all cables must follow that rule?
An example tray for my project is:
covered ladder style tray
(20) 400MCM XHHW-2
(12) 350MCM XHHW-2
(4) 300MCM XHHW-2
(12) 4/0 XHHW-2
(1) 1/0 THWN
(6) 4-Conductor #6 MC
I would be inclined to say:
- sum of conductor diameters for 4/0 and #6 MC cables is 7.4", so 8" is enough for this portion of cables.
- area of 250mcm and larger cables is 17 square inches. Per table 392.22(B)(1) col 1, 16" is enough for this portion of cables.
- the total tray then must be 8" + 16" = 24".
Does that make sense?
The other thing that's tripping me up is 392.22(B)(1)(d) says "where an of the single conductor cables are 4/0AWG or smaller, the sum of diameters of all single conductor sables shall not exceed the tray width". I have mostly larger cables, but a few 4/0 and smaller. Why does the presence of a couple smaller cables mean that all cables must follow that rule?
An example tray for my project is:
covered ladder style tray
(20) 400MCM XHHW-2
(12) 350MCM XHHW-2
(4) 300MCM XHHW-2
(12) 4/0 XHHW-2
(1) 1/0 THWN
(6) 4-Conductor #6 MC
I would be inclined to say:
- sum of conductor diameters for 4/0 and #6 MC cables is 7.4", so 8" is enough for this portion of cables.
- area of 250mcm and larger cables is 17 square inches. Per table 392.22(B)(1) col 1, 16" is enough for this portion of cables.
- the total tray then must be 8" + 16" = 24".
Does that make sense?