wire derating for ac units.

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stew

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I have to install an additional circuit 208 v 3 phase to a new rooftop ac unit. The min circuit amps are 26.5 so I will use a #10 thhn to feed. The max breaker size is 50 so I will use that for a breaker. Hers the rub. I want to put this circuit in an existing conduit with other loads to hit the roof. There are 4 # 10 And 6 # 12 in the pipe now for a total of 10 current carring conductors. How then do I apply the derating of this to correspond to the breaker when the breaker is already larger than the required size for a normal load?
 
Re: wire derating for ac units.

Just divide the required circuit ampacity by the derating factor to find the new wire size. In this case the required ampacity is 26.5/.5 = 53 amps. Now just pick a wire with an ampacity of at least 53 amps in the 90?C column and at least 26.5 amps in the 60?C column. What about the other conductors in the conduit? They only had a derating factor of 70% before you added your conductors. It is likely that they are now too small.
Don
 
Re: wire derating for ac units.

I'm with Don on this one. Now you have to go back a recalculate for all of the existing conductors in the conduit. You may need to re-pull all new conductors even for the existing circuits.
 
Re: wire derating for ac units.

your logic escapes me on the calculation.
Click to expand...
I'm assuming you mean this:

ampacity is 26.5/.5 = 53 amps
Click to expand...
Don's just turned the equation around.

53 x .5 = 26.5

Is the same thing as:

26.5 / .5 = 53

I know people don't like algebra but I think this is easier with letters.

A x B = C

Is the same thing as:

C / B = A

or

C / A = B

That's how Ohm's law works too.

____________________________________________________________________________________

I look at it like this.

C is made out of A and B. (A x B = C)

If you want B, Then you can take A out of C (C / A = B)

Same thing to get A. (C / B = A)

Beacause you get C by putting A and B together. (A x B = C)

Sometimes I don't explain things very well. I think now is one of those times. :( :D
 
Re: wire derating for ac units.

the math is ok i just didnt really get the correlation he showed as i am used to making sure the derated amps meet or exceed the breaker size. as this is one of those exception to the rule the same as electric motor loads where the breaker is normally always larger than the wire size to start wtih i did not see the reasoning behind the math. also i dont understand why with thhn we need to use the 60 deg table.
 
Re: wire derating for ac units.

Stew,
You might be able to use the 75?C ampacities if the equipment terminations are marked as suitable for such use.
As far as the wire size, all I did was to provide you with a conductor that has an ampacity that equals or exceeds the required conductor ampacity under the conditions in your installation. It has nothing to do with the breaker size. You circuit requires a minimum ampacity of 26.5 amps. Your installation conditions require a derating of at least 50% as a result of having 10 current carrying conductors in the raceway. It may also require additional derating as a result of high ambient temperatures, but I did not calculate for that condition. Under the conditions specified, the smallest permitted copper conductor is #6. In this case it does not make any difference as to the use of the 60? or 75? column as #8 in the 75?C column only has an ampacity of 25 amps when there are 10 current carrying conductors in conduit. #6 copper has a 60?C ampacity of 27.5 amps and 32.5 amps at 75?C, again under the conditions specified for your installation.
Don
 
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