cable tray adjustment factor

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Jeff Q

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We are using 12 sets per phase including Neutral of 350 KCMIL cable on a 30" ladder type uncovered cable tray, need an amp rating more than 4000A to meet the spec.
350KCMIL =505A ampacity X12 sets =6060A . if multiplied by Cable tray Adjustment factor 0.65 per 392.80(A)(2)(b) we get 3939A and multiplied again by the 75c ambient temperature which is 0.88 per 310.15(B)(2)(a). Which gives us 3466A not meeting the 4000A criteria.
Can cable tray adjustment factor be 0.80 instead of 0.65 if we maintain a distance of 2.15*1.08(Cable OD)=2.32" between the bundle ? If so , The calculation looks like this. 505Ax12=6060A, x0.80(cable tray adjustment factor)= 4848A and then multiplied by ambient temp derating factor X 0.88=4226A meeting the criteria.
Please let me know if this is acceptable to get a free air rating of 80% if 2.322" of distance is maintained between each bundle.
 
75c (167F) ambient is no messing around. You are essentially fighting the 75c termination ratings on the equipment on either end.
What section does the 80% derate come from rather than the 65% shown in 392.80(A)(2)(b) ?
 
According NEC 392.80 Ampacity of Conductors, A) Ampacity of Cables, Rated 2000 Volts or Less, in Cable Trays,

(d) Where single conductors are installed in a triangular with a maintained diameter (2.15 × O.D.) of the largest conductor contained shall not exceed the allowable ampacities of two or three single insulated conductors rated 0 through 2000 volts supported on a messenger in accordance with 310.15(B)20.

For 75oC rated insulation it is 397 A. For 12 parallel cables 12*397=4764 A

If ambient air is 50oC then according to 310.15(B)(2) a derating factor of 0.89 it has to be employed 4764*0.89=4240 A
However, you need a 50" cable tray, in this case.
 
Julius Right said:
According NEC 392.80 Ampacity of Conductors, A) Ampacity of Cables, Rated 2000 Volts or Less, in Cable Trays,

(d) Where single conductors are installed in a triangular with a maintained diameter (2.15 × O.D.) of the largest conductor contained shall not exceed the allowable ampacities of two or three single insulated conductors rated 0 through 2000 volts supported on a messenger in accordance with 310.15(B)20.

For 75oC rated insulation it is 397 A. For 12 parallel cables 12*397=4764 A

If ambient air is 50oC then according to 310.15(B)(2) a derating factor of 0.89 it has to be employed 4764*0.89=4240 A
However, you need a 50" cable tray, in this case.
Click to expand...
This is the section/approach we could use if there was no neutral . We did try to achieve the ratings following this path. Unfortunately, the neutral creates the “4th” conductor (Reference NEC 2020 310.20(4)) which does not allow table 310.20 to be utilized as there are more than three insulated conductors. This is somewhat referenced in (d) indicated below where it talks about two or three insulated conductors.
Does neutral counts as a 4th Conductor?
NEC 392.80 (A)(d) starts with where single conductors are installed in triangular or "square" configuration, Does square configuration means including neutral or it counts it as 4th conductor ?
 
First, this article permits indeed to form a square. No need to derate the ampacity according to 310.15.(B)3a . Second, see:

310.15 Ampacities for Conductors Rated 0–2000 Volts.(B)(5) Neutral Conductor, a and c. If it is not about c case, you may neglect the neutral.

The problem remains in art 110.14. The maximum current has to be not more than 310 A. That means 12*310=3720 A only. If the cable insulation it is for 90oC then according to 110.14 and Table 310.15(B)(16) permissible is 350 A and then the total is 4200 A.
 
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