# Help on a sizing conductor question

#### Bravepotato

##### Member
Guys, after studying Mike's Motors, conductor sizing and voltage drop DVD's and book for the last 3 days almost 10 hours a day I am going back to test my newfound knowledge on Ray Holders cheap Practice exam test, which is what I studied for a week the first time I took the Masters last March (passed NEC general knowledge section, failed calculations by 3 points). I am doing MUCH better on these questions now. Mikes stuff is worth every penny.

I cannot make sense of this question. I found my FLC in the 3 phase table, (74.8A), multiplied it by the obligatory 125% for continuous duty motors, (section 430.22) and then I multiplied that by the .75% I found in Table 310.15(B)(2)(a) for the given ambient temp rise of 50 degrees C (with 75 degree terminals) in the question, and came up with 70.12 amps. I realize that is a lesser amperage then what the 3 phase table shows, and definitely lesser than the 125% rule, and I realize if I sized my conductor by this number (70 amps) I would end up with a gauge that wouldn't support the startup of the motor.

However, I cannot for the life of me find where it tells you to "divide" the correction factor given in table 310. 15b2a....it clearly says "Multiply" (yellow highlighter). In the notes or otherwise.

Can anyone give some insight because Id like to know the correct procedure before I continue on with trying to answer these types of questions. Thanks!!

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#### david luchini

##### Moderator
Staff member
Guys, after studying Mike's Motors, conductor sizing and voltage drop DVD's and book for the last 3 days almost 10 hours a day I am going back to test my newfound knowledge on Ray Holders cheap Practice exam test, which is what I studied for a week the first time I took the Masters last March (passed NEC general knowledge section, failed calculations by 3 points). I am doing MUCH better on these questions now. Mikes stuff is worth every penny.

I cannot make sense of this question. I found my FLC in the 3 phase table, (74.8A), multiplied it by the obligatory 125% for continuous duty motors, (section 430.22) and then I multiplied that by the .75% I found in Table 310.15(B)(2)(a) for the given ambient temp rise of 50 degrees C (with 75 degree terminals) in the question, and came up with 70.12 amps. I realize that is a lesser amperage then what the 3 phase table shows, and definitely lesser than the 125% rule, and I realize if I sized my conductor by this number (70 amps) I would end up with a gauge that wouldn't support the startup of the motor.

However, I cannot for the life of me find where it tells you to "divide" the correction factor given in table 310. 15b2a....it clearly says "Multiply" (yellow highlighter). In the notes or otherwise.

Can anyone give some insight because Id like to know the correct procedure before I continue on with trying to answer these types of questions. Thanks!!
Multiply 150 by 0.75...you get 112.5.
Divide 112.5 by 0.75...you get 150. These are inverse operations.

So multiply the #1 THWN allowable ampacity of 130 by 0.75, you get 97.5 which is larger than the required 93.5. A #1 THWN has sufficient ampacity.
Or divide 93.5A by 0.75, you get 124.6A. A #1 TWHN has sufficient ampacity. These are inverse operations.

You are getting the same result by working backwards from the required ampacity.

#### winnie

##### Senior Member
Perhaps another way of saying this:

The ambient correction factors in table 310.15(B)(2)(a) are the change in conductor ampacity because of ambient temperature. You start with a conductor 'normally rated' (at 20C) at say 100A; but because of the elevated temperature you multiply by 0.75 to get the ampacity at the elevated temperature of 50C.

But you are asking the inverse question, so you need to do the inverse operation as David Luchini describes. You know the ampacity that you need, and what to find the conductor which will provide it. So you take the number that you want to get (93.5A) and divide by the correction factor and get '125A'. Then you look for a conductor 'normally rated' for this. 1AWG copper with a 75C rating has an ampacity of 130A, and 130 * 0.75 = 97.5 , which is greater than the 93.5 that you need.

-Jon

#### Bravepotato

##### Member
Perhaps another way of saying this:

The ambient correction factors in table 310.15(B)(2)(a) are the change in conductor ampacity because of ambient temperature. You start with a conductor 'normally rated' (at 20C) at say 100A; but because of the elevated temperature you multiply by 0.75 to get the ampacity at the elevated temperature of 50C.

But you are asking the inverse question, so you need to do the inverse operation as David Luchini describes. You know the ampacity that you need, and what to find the conductor which will provide it. So you take the number that you want to get (93.5A) and divide by the correction factor and get '125A'. Then you look for a conductor 'normally rated' for this. 1AWG copper with a 75C rating has an ampacity of 130A, and 130 * 0.75 = 97.5 , which is greater than the 93.5 that you need.

-Jon
Thanks Jon, that's basically what I figured out throughout the 10 hours I spent studying today. I guess I over thought it....When I finally realized WHY I was using the correction factor to begin with, it hit me like a brick, and I knew then that I was to multiply to increase the temp rating and divide to lower. It should've come natural but I've been taking in a lot of info. Spent 2 full days going over Mikes Motor calc dvd and section in the book, and then another 2 days just going through the test questions, making sure I have it 100% before I moved on to other subjects. After 13 years in the trade, 5 year apprenticeship, 7 years as a licensed journeyman and 2 years as a union contractor going from spending my last 500 dollars on materials for my first job, to having done 2.5 million in work 2 years later, (renting a masters) I can honestly say that after spending the last 5 days on this study material I have never been more deadly with a code book and especially calculations. I've just got to pass the Masters Calculations portion on Nov 26th and I'm home free

#### Bravepotato

##### Member
Thanks to Dave as well. I now know what you were trying to tell me lol.

#### Judy L Venegas

##### New User
Multiply 150 by 0.75...you get 112.5.
Divide 112.5 by 0.75...you get 150. These are inverse operations.

So multiply the #1 THWN allowable ampacity of 130 by 0.75, you get 97.5 which is larger than the required 93.5. A #1 THWN has sufficient ampacity.
Or divide 93.5A by 0.75, you get 124.6A. A #1 TWHN has sufficient ampacity. These are inverse operations.

You are getting the same result by working backwards from the required ampacity.