Bob Clukey
Member
I want to install 12, 500 kcmil single conductors in a cable tray. Please explain the steps to properly size the cable tray? The cable tray will be botton vented.
Cable Tray Sizing
- Thread starter Bob Clukey
- Start date
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ron
Senior Member
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- New York, 40.7514,-73.9925
Look in Article 392 for sizing and ampacity calculations. It is pretty intense based on lots of spacing requirements.
Julius Right
Senior Member
- Occupation
- Electrical Engineer Power Station Physical Design Retired
Let's say it is about 4*3*500 MCM CU/XLPE low-voltage-4 groups, 3 lives, no neutral cable. Cable O.D. =0.87 ".The overall cross-section area=0.685 inch^2.
In order to avoid current imbalance it is better to run these cables in equilateral triangle-and the best in an alternate order- A[C]B-B[C]A -C in the upper position.
1-In order to get 580 A/cable [Table 310.15(B)20 EPR/XLPE insulation] you have to follow
art.392.22(B)(1)(b) recommendations:
The clearance has to be 2.15*dia then the total width=(4*2+3*2.15)*.87= 12.57"
The next width [standard] it is 18".
There is a maximum permitted filling [From Table 392.22(B)(1)]:
12 cables total area=0.685*12 =8.2 inch^2 [maximum permitted: 19.5]
If there are 3 groups of 4 cables [3 lives 1 neutral] the total width= (3*2+2*2.15)*.87=8.96"
And for 12" cable tray width [From Table 392.22(B)(1)]: maximum permitted: 13 inch^2
2-One may avoid the clearance of 2.15 *o.d. using only 65% from Table 310.15(B)17=700*.65=455 A per one cable. See 392.17(2)(b).
Then the total width will be 3*2*.87=5.22"[6" standard]
However you have to choose 9" for permissible maximum 9.5 inch^2.
In order to avoid current imbalance it is better to run these cables in equilateral triangle-and the best in an alternate order- A[C]B-B[C]A -C in the upper position.
1-In order to get 580 A/cable [Table 310.15(B)20 EPR/XLPE insulation] you have to follow
art.392.22(B)(1)(b) recommendations:
The clearance has to be 2.15*dia then the total width=(4*2+3*2.15)*.87= 12.57"
The next width [standard] it is 18".
There is a maximum permitted filling [From Table 392.22(B)(1)]:
12 cables total area=0.685*12 =8.2 inch^2 [maximum permitted: 19.5]
If there are 3 groups of 4 cables [3 lives 1 neutral] the total width= (3*2+2*2.15)*.87=8.96"
And for 12" cable tray width [From Table 392.22(B)(1)]: maximum permitted: 13 inch^2
2-One may avoid the clearance of 2.15 *o.d. using only 65% from Table 310.15(B)17=700*.65=455 A per one cable. See 392.17(2)(b).
Then the total width will be 3*2*.87=5.22"[6" standard]
However you have to choose 9" for permissible maximum 9.5 inch^2.
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