Adjustment factors & Load diversity

[HEAVY DUTY]

Can anyone explain what "load diversity" is as referenced in appendix B, Table B-310-11.

I would like to put 22(current carrying) - #12 THHN CU wires in a 1" conduit and install on 20 A breakers. These circuits all have minimal loads and are not loaded to their maximum load(no diversity) as referenced in Table 310-15(B)(2)(A). Average load is approx. 6 A.

I am being told that I must derate to 45%.

I feel that I can derate to 70% of my allowable ampacity according to Appendix B, Table B-310-11.

Am I not understanding what is meant by load diversity vs no diversity?

I would be grateful for any input.

 

[jumpinjohnny]

This is a good question. Does this mean 1 amp on one ckt and 5amp on another and 12 amp on another. If so then apply Table B.310.11. But if you look at Table 310.15 (B) (2)a, this is where your 45% is from.

 

[LarryFine]

Load diversity means it can be assured that not all conductors will be loaded simultaneously. I believe you'll be stuck with going to #10 wire just to be able to use 15a breakers.

30a (@90 deg) at 45% = 13.5a, too small for even 15a breakers.

40a at 45% = 18a. = #10cu on 15a breakers and 1.25" EMT.

55a at 45% = 24.75a = #8cu on 20a breakers and 1.5" EMT.

If you still want 20a breakers, you'll need to go to #8 wire and 1.5" conduit.

 

[charlie b]

If all 20 conductors can be carrying current at the same time, regardless of whether each carries 20 amps or 1 amp, then you do not have “load diversity.” If you want to use a single conduit for 22 conductors, and if you wish the ampacity to be at least 20 amps for a #12 THHN, then you need to prevent more than 12 conductors from carrying current at the same time.

I worked out the value of 12 from the formula located under Table B.310.11.

 

[jumpinjohnny]

If all 20 conductors can be carrying current at the same time, regardless of whether each carries 20 amps or 1 amp, then you do not have ?load diversity.? If you want to use a single conduit for 22 conductors, and if you wish the ampacity to be at least 20 amps for a #12 THHN, then you need to prevent more than 12 conductors from carrying current at the same time.

I worked out the value of 12 from the formula located under Table B.310.11.

Now I am really intrigued. Never had to use this formula before. How did you determine What N is equal? We know that E is equal to 22 (the number of conductors in the conduit. And if I understand A1 is equal to conductor amps from 310.16 multiply by the appropriate factor from B.310.11. Am I on the right track or am I missing something.