Re: Load Diversity
Originally posted by sherwin: I see that if 'E' is twice 'N' then the square root term goes to one and the solution refers back to B310.11.
I think you mean if "N" is twice "E," or if "N" is bigger than twice "E."
Your question has led me to suspect that the article is using the phrase "load diversity" in the context of "not all wires will be carrying current at the same time." In other words, if you have two pumps, and if all wires are in the same conduit, and if the switch is set up to turn on one or the other (but never both), then only half the wires will be carrying current at any one moment. That would give you (in the apparent context of this formula) a load diversity of 50%.
That is not a standard use of the phrase "load diversity," as you may have seen in my earlier posts on this topic. So the rest is a guess on my part.
Count how many conductors are in the raceway. That is "N." Now suppose that you have control switches to prevent all of the wires from carrying current at the same time. So count up how many might be carrying current at the same time. Please note that it does not matter how much current, or does the current-carrying capability of each wire matter. If it carries current at the same time as any other wire, then count it. This total count is "E." Here is what you do with the information:
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- <font size="2" face="Verdana, Helvetica, sans-serif">If "E" is half of "N," then you have 50% load diversity (again, in this non-standard context). That means that A2 is the same as A1, as you have already pointed out.</font>
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- <font size="2" face="Verdana, Helvetica, sans-serif">If "E" is less than half of "N" (let us say that "E" is one eighth of "N"), then the load diversity is lower than 50% (in this case, 12.5%), and A2 would be equal to 2 times A1. The rule tells you to use A1, if this happens.</font>
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- <font size="2" face="Verdana, Helvetica, sans-serif">If "E" is more than half of "N" (let us say that "E" is 3/4 of "N"), then the load diversity is higher than 50%. You would expect to not be able to use the same factors that show up in the table, since they are based on 50% LD. In this example, A2 would be about 81% of A1, so you would have to derate by 81% of Table B310.11.</font>
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- <font size="2" face="Verdana, Helvetica, sans-serif">Finally, if "E" is equal to "N" (it cannot be higher, or you have mis-counted) then you have 100% load diversity. That means that A2 is about 71% of A1, so you would have to derate by 71% of Table B310.11.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">If this makes sense to you, I would be happy, for that would mean that it made sense to at least one of us.
Good luck.