I am trying to size some ladder cable tray with single conductors for a project and there is a debate in our office on how to calculate the sizes.
I am using 392.22(B)(1)(b) and 392.22(B)(1)(d) as the basis of my calculations.
We have a mix of conductors that are 250 to 500 and 2/0 to 4/0.
To calculate the size of tray for the 2/0 to 4/0 conductor portion, you simply add the diameters of the cable together as per 392.22(B)(1)(d).
Then I add the areas of the cables 250 and above together and multiply by 1.1 as per 392.22(B)(1)(b) to get the width for this portion, and add it to my 392.22(B)(1)(d) calculation.
The wording of 392.22(B)(1)(d) is the cause of the debate.
“(d) Where any of the single conductor cables are 1/0 through 4/0 AWG, the sum of the diameters of all single conductor cables shall not exceed the cable tray width.”
I contend that my method of doing both calculations for (b) and (d) and adding them together is correct.
The other side says that if there are any, even one cable that is 1/0 to 4/0, then all of the cables shall be calculated as side by side dimensions and not by area as for the 250-900 range. I say it is just for the 1/0-4/0 sizes.
There doesn't seem to be a rule for a mix of single conductors sizes 250-900 and 1/0-40.
This method causes the tray to be extra wide.
Which method is correct?
I am using 392.22(B)(1)(b) and 392.22(B)(1)(d) as the basis of my calculations.
We have a mix of conductors that are 250 to 500 and 2/0 to 4/0.
To calculate the size of tray for the 2/0 to 4/0 conductor portion, you simply add the diameters of the cable together as per 392.22(B)(1)(d).
Then I add the areas of the cables 250 and above together and multiply by 1.1 as per 392.22(B)(1)(b) to get the width for this portion, and add it to my 392.22(B)(1)(d) calculation.
The wording of 392.22(B)(1)(d) is the cause of the debate.
“(d) Where any of the single conductor cables are 1/0 through 4/0 AWG, the sum of the diameters of all single conductor cables shall not exceed the cable tray width.”
I contend that my method of doing both calculations for (b) and (d) and adding them together is correct.
The other side says that if there are any, even one cable that is 1/0 to 4/0, then all of the cables shall be calculated as side by side dimensions and not by area as for the 250-900 range. I say it is just for the 1/0-4/0 sizes.
There doesn't seem to be a rule for a mix of single conductors sizes 250-900 and 1/0-40.
This method causes the tray to be extra wide.
Which method is correct?
