studentofthecode
Member
- Location
- New Jersey
I'm looking to find the maximum fill of MC cables for a project where the customer requests cable tray pathways to allow them easier retrofitting and expansion in the future in their proposed new building. We are designing a 24 inch wide open ladder/wire frame tray for distribution. In Article 392, section 392.22-(A)-1-(b) it states that "for cables #4/0 and smaller, the sum of the cross-sectional area of all cables shall not exceed the maximum allowable cable fill area in Column 1 of Table 392(A) for appropriate cable width". 24" wide tray (ventilated) is shown to be 28 sq. inches. My calculation is:
--12/2 MC in the Southwire cut sheets =.475 diameter.[12/3 is .505d]
--.475 squared=.2256 square inches
--28 square inches / .2256 square inches =125-12/2 MC's [12/3 is 109, 10/2 is 95, and 10/3 is 83]
This seems like a lot of cables for a 24 inch cable tray considering that they are not required to be in a single layer and de-rated [392.80(A)(1)(a)], because the cables have no more that 3 current carrying conductors.
Does this calculation appear correct?
Thanks
--12/2 MC in the Southwire cut sheets =.475 diameter.[12/3 is .505d]
--.475 squared=.2256 square inches
--28 square inches / .2256 square inches =125-12/2 MC's [12/3 is 109, 10/2 is 95, and 10/3 is 83]
This seems like a lot of cables for a 24 inch cable tray considering that they are not required to be in a single layer and de-rated [392.80(A)(1)(a)], because the cables have no more that 3 current carrying conductors.
Does this calculation appear correct?
Thanks