Recently, for a class I teach, I used a question from an old exam which I took years ago. The question was: What is the ampacity of a #12 THHN copper conductor surface mounted on the wall of a walk-in cooler with an ambient temperature of 2 celsius? That is all of the information provided. No terminal temperature ratings were given and as no conduit fill was mentioned, no derating for that was implied or directed and no calculated load was given.
When I was solving the problem for the master I would use to grade the exams, I simply used the 90 degree column (THHN) from table 310.15 (B)(2)(a) and based on 110.14 (C)(1), the 60 degree column from 310.15(B) (16). I thus multiplied 20 by 1.15 which comes to 23 amps.
One of my students however, decided to divide 20 by 1.15 because I have taught a method for correcting for ambient temperature factors based on the equation: ampacity= calculated load/(adjustment factor), thus, by his reasoning, he took 20 and divided it by 1.15 and gave me 17.3.
It seems to me his mistake lies in the fact that no calculated load was given in the problem, only the size and insulation type.
Any thoughts?
I understand that, but he said his general formula was Ampacity...Not just Ampacity used below 86 degrees, Usually Ampacity is corrected for higher temps and or more than 3ccc, which is always less than 1. So I don't see how his "equation" works. I could be wrong, and if so would like to know the correct way.
Example, Lets look at a 20 AMP Load where there are 4CCC. The equation he gave would be Ampacity= 20/.8= 25....We all know Ampacity doesn't increase with 4CCC's..So ?? Who is doing what wrong ???
Again, the issue here is that you're assuming a load. What I was after is, given the conditions, how much would the allowable ampacity increase when the ambient temperature was so low. Clearly the allowable ampacity would increase. Mr Beck pointed out that the question was ambiguous.
In your example, like my student, you assumed a calculated load when there was none. The 20 amps is taken from the 60 degree column of table 310.15 (B)(16). It is not the calculated load however. No calculated load is given. I think that while the question was indeed ambiguous, it is important to read the question carefully and not insert information which is not present.
You have the wrong mathematical operators in your equations.
If you have 4 CCC's the conductor needs to be adjusted to 80% of it's normal capacity. 80% of 20 is 16 (20 x .80)
in the OP the 1.15 adjustment factor for a low ambient temp actually increases ampacity that 20 amp conductor is able to carry extra 15% because of low ambient temp and is good for 23 amps not 17.4. However in most circumstances it must still be protected by a 20 amp breaker, but if you have both low ambient temp with 1.15 adjustment and 4-6 conductors in raceway you have 20 x 1.15 x .80 = 18.4 for conductor available ampacity. Keep in mind this is for the 90 deg C column, a 12 AWG conductor would start derating from 30 amps not 20 amps in this instance so if we ran the same numbers there it would be good for 30 x 1.15 x .8 = 27.6 amps, terminations are likely only going to allow using this conductor though at the 75C level, and other then some specific loads (motors and air conditioners being a couple of the most popular) you still must protect this conductor with a 20 amp overcurrent device.
Now if you are adjusting the load value to find a conductor to match you work the other way.
Say you have a 15 amp load in the same ambient that allows for a 1.15 adjustment factor and put it in a raceway with 4-6 current carrying conductors in it you can go with the reciprocal functions of the adjustment factors and you want to know what minimum size conductor you need instead of starting with a 20 amp conductor (12 AWG) and adjusting it only to find it may be too small and then re-figuring adjustments with a 10 AWG just take the 15 amp load divide (instead of multiplying) by the adjustment factor of 1.15, then divide again (instead of multiplying) by the .80 adjustment for number of conductors and you get 15 / 1.15 / .80 = 16.3. You need to select a conductor that has an ampacity of at least 16.3 amps for this application. (note that you do need to multiply the 15 by 1.25 first if it is a continuous load)
The other confusion you have is with the adjustment factor higher then 1 for low ambient temps.
In my example with the 15 amp load above - if we disregard the number of current carrying conductors a moment we had 15 / 1.15 which is 13.04. That basically means the temp is cool enough that the insulation of a 13 amp conductor can handle the heat produced in the conductor carrying a 15 amp load. You will find that you still need a 14 AWG conductor in this instance for other reasons though, first because this 13 amp conductor is based on 90C insulation, you will still need a minimum of 15 amp conductor from either 60 or 75 degree column based on terminal ratings.
