ampacity rating

[ohmboyz]

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?

 

[charlie b]

I have taught a method for correcting for ambient temperature factors based on the equation: ampacity= calculated load/(adjustment factor)

That can tell you much ampacity you need the wire to have. Your present question is asking what the ampacity of the wire actually is.

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.

No, his mistake was in thinking that "ampacity" is the same thing as "ampacity." If I may be allowed a criticism, I think you need to take some of the blame for that misconception. You are dealing with two different contexts, both of which have the word "ampacity" in their description. But a more complete description would clarify the situation. One context is "required ampacity," and the other is "available ampacity." There are other ways to phrase the to notions, but you need to make it clear that they are different, and that the same equation will not apply to both.

 

[ohmboyz]

thank you Mr Beck

thank you Mr Beck

Indeed, I agree. By not differentiating between required and available ampacity, I must take some responsibility for the confusion of my student. Nonetheless, I don't think he was correct in taking
the 20 amps from the 60 degree column and assuming that was the calculated load.

 

[charlie b]

I agree with you there. But in the absence of load information, we can assume that the load is no higher than 20 amps. Then his formula would tell you that you need a conductor with an uncorrected (i.e., tabulated) ampacity of 17.3.

I will say I do not understand why you would use a correction factor from the 90C column and an ampacity value from the 60C column. Neither is all that important anyway, from a practical perspective. You still have to keep the OCPD at 20 amps or lower.

 

[Carultch]

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?

If they do not tell you anything about the terminal ratings, then you have to assume that the terminals are rated at default temperatures per NEC110.14(C). For #12 wire, that would be the 60C column. Remember, this rule applies before you take in to consideration any ampacity adjustments or temperature correction. Your answer cannot exceed 25A, unless listed and marked otherwise.

Given that this is in a walk-in cooler, I'd consider it a DAMP LOCATION due to condensation. In a practical sense, most THHN is also rated as THWN-2, which allows you to use the 90C column anywhere for the wire ampacity. But when you are taking an exam, you can't assume that THHN is dual rated as THWN-2.

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.

They aren't asking you to size the wire to the load. They are asking you to take the wire that is there, and state what the ampacity is. It is an inverse of the problem you are usually solving.

 

[Henley]

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?

First, I agree with your Question/Answer. i am a little confused with what you meant by the other equation you said you use to teach Ampacity Correction. Shouldn't it be Ampacity(Corrected)= Calculated Load x Adjustment Factor?? I mean for most examples the correction factor is less than 1. If you divided by this it would increase the derated Ampacity ??. Maybe I'm just confused

 

[kwired]

If they do not tell you anything about the terminal ratings, then you have to assume that the terminals are rated at default temperatures per NEC110.14(C). For #12 wire, that would be the 60C column. Remember, this rule applies before you take in to consideration any ampacity adjustments or temperature correction. Your answer cannot exceed 25A, unless listed and marked otherwise.

Given that this is in a walk-in cooler, I'd consider it a DAMP LOCATION due to condensation. In a practical sense, most THHN is also rated as THWN-2, which allows you to use the 90C column anywhere for the wire ampacity. But when you are taking an exam, you can't assume that THHN is dual rated as THWN-2.




They aren't asking you to size the wire to the load. They are asking you to take the wire that is there, and state what the ampacity is. It is an inverse of the problem you are usually solving.

I agree most of the time we have a load and need to figure out what conductor ampacity is necessary to supply it. Not too often does one have a conductor and decide to match the load to the conductor.

I do think that default terminal temp these days can be safely assumed to be 75C if utilizing new equipment, but if using old equipment either pay attention to details or assume 60C to be safe - unless over a 100 amp circuit.

First, I agree with your Question/Answer. i am a little confused with what you meant by the other equation you said you use to teach Ampacity Correction. Shouldn't it be Ampacity(Corrected)= Calculated Load x Adjustment Factor?? I mean for most examples the correction factor is less than 1. If you divided by this it would increase the derated Ampacity ??. Maybe I'm just confused

Conductor in question was in a walk in cooler with a low ambient temp - adjustments for that situation do increase instead of decrease overall ampacity.

 

[Henley]

I agree most of the time we have a load and need to figure out what conductor ampacity is necessary to supply it. Not too often does one have a conductor and decide to match the load to the conductor.

I do think that default terminal temp these days can be safely assumed to be 75C if utilizing new equipment, but if using old equipment either pay attention to details or assume 60C to be safe - unless over a 100 amp circuit.

Conductor in question was in a walk in cooler with a low ambient temp - adjustments for that situation do increase instead of decrease overall ampacity.

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 ???

 

[ohmboyz]

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.

 

[ohmboyz]

If they do not tell you anything about the terminal ratings, then you have to assume that the terminals are rated at default temperatures per NEC110.14(C). For #12 wire, that would be the 60C column. Remember, this rule applies before you take in to consideration any ampacity adjustments or temperature correction. Your answer cannot exceed 25A, unless listed and marked otherwise.

Given that this is in a walk-in cooler, I'd consider it a DAMP LOCATION due to condensation. In a practical sense, most THHN is also rated as THWN-2, which allows you to use the 90C column anywhere for the wire ampacity. But when you are taking an exam, you can't assume that THHN is dual rated as THWN-2.




They aren't asking you to size the wire to the load. They are asking you to take the wire that is there, and state what the ampacity is. It is an inverse of the problem you are usually solving.

Indeed. Damp location, THWN. One more reason why the question was poorly worded. Thanks. Good catch

 

[david luchini]

Indeed. Damp location, THWN. One more reason why the question was poorly worded. Thanks. Good catch

THHN is rated 90deg for Dry and Damp locations. THWN would not be required for a damp location.

 

[Carultch]

THHN is rated 90deg for Dry and Damp locations. THWN would not be required for a damp location.

Ok, thanks for letting me know the subtle difference between damp and wet. I wasn't aware. To me, they mean the same, and I seldom ever specify anything other than dual rated wet/dry 90C wiring.

 

[Carultch]

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 ???

If it is up to me, I wouldn't take credit for an artificial ambient temperature in terms of wire ampacity. What if the cooling system fails, yet you still need to operate this circuit?

Same reason I'm uncomfortable taking credit for a heated roof in terms of snow loads.

 

[kwired]

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.