Mike Dolt Enterprises
Mike Dolt Enterprises
User: morgan3520, Location: Dallas, TX
Asked on: Monday, March 31, 2025 at 12:43 pm EDT
Answered on: Monday, March 31, 2025 at 12:46 pm EDT
Question: I need help deciphering Table 392.22(A)(1) Column 2, for cable tray fill ratio calcs. I am the engineer on the project, the contractor want to use multi-conductor cable for all load being ran in cable tray. Checking my math: I have 6 cable 4/0 and larger in 48" cable tray (Max. Allowable Fill - 56in2). Sd, sum of diameters (see attachment). Allowable fill will be (56-13.8)in2. So when adding up the diameter of the cables, are the 4/0 and larger cables area (19in2) included or excluded?
Okay, let's break down NEC 392.22(A)(1) Column 2 and apply it to your situation with multi-conductor cables in a 48" cable tray. This is a common point of confusion.
Understanding NEC 392.22(A)(1) Column 2
Column 2 of Table 392.22(A)(1) addresses the allowable fill area for multi-conductor cables when the cable tray is 6 inches wide or wider. Here's the crucial part: "Where all of the cables are 4/0 AWG or larger, the sum of the diameters of all cables shall not exceed the cable tray width..." "...Where cables are 4/0 AWG or larger are installed in the cable tray with cables smaller than 4/0 AWG, the sum of the cross-sectional areas of all cables smaller than 4/0 AWG shall not exceed the value derived from the following equation:Allowable Fill Area = [0.20 x cable tray width (in.)] x cable tray width (in.) - (sum of the diameters of all cables 4/0 AWG or larger, in.)"Your Specific Scenario
You have: A 48" wide cable tray. 6 multi-conductor cables that are 4/0 AWG or larger. The contractor wants to use only multi-conductor cables. Max. allowable fill - 56in2 The sum of all diameters cables 4/0 AWG or larger is 13.8in2Applying the Code
Since you have both 4/0 AWG or larger cables as well as cables smaller than 4/0 AWG, you must use the Allowable Fill Area equation. Allowable Fill Area = [0.20 x cable tray width (in.)] x cable tray width (in.) - (sum of the diameters of all cables 4/0 AWG or larger, in.)" Allowable Fill Area = [0.20 x 48"] x 48" - 13.8" Allowable Fill Area = 446.08 - 13.8"* Allowable Fill Area = 443.08 in2The correct formula that needs to be use to determine the allowable fill area is, (Sd) sum of the diameters of all cables 4/0 AWG or larger. The diameter of each cable needs to be added together to arrive at the (Sd). This is the number that will be used in the formula provided above.
Answer
Your understanding is on the right track, but it's important to carefully apply the language of the code. In your calculation, the sum of the cross-sectional areas of all cables smaller than 4/0 AWG (19")is not included. Flag for review