If i use 3c10awg MC cable for cabling 10 hp 460V motors, in accordance with table 310.15(B)(2)(a), i can make a bunch of 10 of this cable. My question is how many space it should have bettween this bunch of cables and another bunch of cables. (i mean in a ladder cable tray)
[This message has been edited by Bobby Mercier (edited 11-28-2002).]
Don has a very good point & I could see a fistfight or two breaking out over it. Maybe you need to space the bundles 1 bundle diameter apart.
I feel I correctly answered your question about space in the ordinary sense, but I didn't do so in the context of your question.
In any case, since you are dealing with #10 conductors, it appears that 310.15(B)(2)(a) would require a derating to 45%. Therefore, #10 would be OK since its final ampacity is more than 125% of the 14 amps the motor draws.
It would be interesting to apply Exception 5 to 310.15(B)(2)(a) by using #12 & supporting on bridle rings instead of cable tray. What do you think?
Few things are harder to put up with than the annoyance of a good example.
Don I believe that 392.11(A)(1) applies to the cable assembly. eg a 10-3 mc cable vs a 10-5 mc cable where you may have more than 3 current carring conductors in that one cable. I believe that 392.11(A)(3) applies to the total # of cable assemblies installed in the tray. Keeping cable assemblies seperated is a good practice.
Re: Table 310.15(B)(2)(a)#82582 12/06/0208:49 AM12/06/0208:49 AM