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Joined: Sep 2019
Posts: 21
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Hello,

I have this question, "For a ladder tray; with ten single 500 KCM THHN conductors , ( diameter of .949 from table 5 page 685) what is the smallest width of tray?"
The answer given is, "12 392.22.B.1.B"

I determined it to be 10.55" using 90%. ((0.949 x 10) / 90%) but missed using table 392.22(B)(1) because I only referenced 392.20(D) and missed 392.22(B)(1)(b). So I was wrong with a fill in answer, but might have gotten multiple choice.

Once I looked at the answer, I read the code in 392.22(B)(1)(b) which needs the conductor cross-sectional area, which brings in A=πR².
That math leads me to 7.85524" using 90% (((3.14*(.949/2)²)*10)/90%)
Using the table 392.22(B)(1), that leads to 8" width.

But the answer given was 12". So Why is the answer using diameter but the code requires cross-sectional area?
What am I missing here?
Which is the correct way to determine the smallest width for a Ladder Cable Tray?

Thanks,
Phil


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2017 / 2014 NEC & Related Books and Study Guides
Joined: Apr 2002
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I use the diameter of the conductor from page 685 table. By experience, 12" wide would be the smallets width.

The key word above is 'experience'. As to going to the 'Code' MINIMUM size calcs, in the real world that I work within, that does not happen. Six data centers, and the minimum width is 24" to 36" tray. Looks sort of 'dumb' sometimes with only two (2) fiber within inner ducts.

Also, I wonder if the person that created the question looked at any of the mfg. documentation, and saw that the smallest tray for power use is 12", based on weight of the 500KCMil.


John
Joined: Sep 2019
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Thanks John.

That diameter of a cable listed in Ch.9, Table5, is that total diameter including insulation, or only internal conductor diameter? Must be with insulation, huh?
Can that column be used in calculations if it is total including insulation?

Kindly let me introduce another, similar question related to cable tray minimum width.

Question:
"What is the minimum width for a cable tray with 12 cables of 3 conductor 4/0AWG-CU type MC cable?"

Answer Given:
18.84 Annex D13(a)

Take note when I got this question (exactly as given above), I was not given the Cable O.D. nor Annex D13(a).
So what I did was go to Ch.9-Table 8 for the 4/0 measurements of 211600cm, 0.219in2, & 0.528" diameter.
There's a big difference between a 0.528" diameter and a 1.57" Cable O.D. (from Annex D13a) so my answer was way off.

How do you deal with this on the test, or is it unlikely to get a question without the cable OD given?

Thanks,
Phil

PS: I test tomorrow, 2019-09-18, Master, Minnesota. So my final cram is going on today <:


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It's likely questions on the test may come from the recommended study materials, so if you get these questions you know what they're looking for.

Good Luck!

Let us know what happens.


Bill
Joined: Apr 2002
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Ditto to what Bill said above.

I'll kick around this query when I have a few minutes, but I'm up to my eyebrows right now.


John
Joined: Sep 2019
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Took my test on Wednesday (Sep18th).
Still awaiting results.
Phil


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It sounds like you prepared and I bet you are getting your card.


Greg Fretwell
Joined: Sep 2019
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Originally Posted by gfretwell
It sounds like you prepared and I bet you are getting your card.

Good News!!

I passed!!

Whew, tough test!

Thanks for all the support.
This forum helped alot <:
Phil


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Good deal.


Greg Fretwell

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