Understanding the National Electrical Code

[thortin simpson]

I have a problem understanding the ampacity tables 310.16,and310.17.
My problem is how do I know which coloumn I should be looking in for example in Mike's book Understanding The National Electrical Code on page 334 has a question:

What size Type NM cable is required to supply a 10kw,240v,single phase fixed heater with a 3a blower motor. THE TERMINALS ARE RATED 75 C

Well the way i worked it out was 10,000 /240=42+3=45x1.25=56A.....Since it says 75C the smallest size to be able to hold 56A is a No.#6 awg
So where am I going wrong.

 

[Ken9876]

NM cable is only rated at 60 c , you can use the 90 coloumn for derating.

 

[kwired]

I think this may be part of why it is common to see 9.6 kW heat strips for air handlers instead of 10kW. By the time you add a blower you have more than a 6 awg @ 60 deg can handle, and would need multiple circuits. 424.22(B) limits these elements to 60 amp overcurrent protection.

 

[buddhakii]

You can only use the column that contains the lowest rating of the wire or terminations for ampacity. For instance ifyou have 90 degree wire but 60 degree lugs then you would use the 60 degree column. Most lugs are rated for 75 degrees so that is the column you will use the most. If you have to derate for anything such as more than 3 current carrying conductors then you can use the column with the temp rating of the wire, however the final result cannot exceed the 75 degree column if terminating to 75 degree lugs. Hope that makes more sense.

 

[augie47]

If you are like me you may never fully "understand" the NEC.
The more you learn about it, it seems the less you know.

As ken pointed out, the "trick" here is the use of NM. 334.80 points out that the final terminating temperature of NM can not exceed it's 60? rating.
In this case you need to take into account 310.16, 110.14, and 334.80.
#6 THHN would be correct, but with an NM you would need a #4 due to the 60? limitation.

 

[Dennis Alwon]

I think this may be part of why it is common to see 9.6 kW heat strips for air handlers instead of 10kW. By the time you add a blower you have more than a 6 awg @ 60 deg can handle, and would need multiple circuits. 424.22(B) limits these elements to 60 amp overcurrent protection.

I have seen many newer 10 kw units that call for 56 or 57 amps. It really angers me because what normally was used is #6 because the amps was less than 55amps, and now it is not compliant.

 

[kwired]

I have seen many newer 10 kw units that call for 56 or 57 amps. It really angers me because what normally was used is #6 because the amps was less than 55amps, and now it is not compliant.

I have never seen a heat strip that was an even multiple of 5 kW for air handlers, They are almost always multiples of 4.8 kW. Unit heaters yes, I have seen and they almost always are in multiples of 5 kW, on ones that are larger than 5kW.

 

[iwire]

I have seen many newer 10 kw units that call for 56 or 57 amps. It really angers me because what normally was used is #6 because the amps was less than 55amps, and now it is not compliant.

I don't understand why that would anger an EC, if the job costs more you should make more not less.

 

[jaggedben]

You can only use the column that contains the lowest rating of the wire or terminations for ampacity. For instance ifyou have 90 degree wire but 60 degree lugs then you would use the 60 degree column. Most lugs are rated for 75 degrees so that is the column you will use the most. If you have to derate for anything such as more than 3 current carrying conductors then you can use the column with the temp rating of the wire, however the final result cannot exceed the 75 degree column if terminating to 75 degree lugs. Hope that makes more sense.

I have been struggling a bit to clearly understand this as well, but I think your post just helped me a bunch.

Say I am using 90C wire with 75C terminals, and I also need to derate for temperature and/or fill. I must size the wire for the required ampacity either as if it were rated at 75C, or as derated from 90C, whichever size is larger. Is that correct?

Do others agree?

 

[George Stolz]

Right process, but you use the smaller result.

 

[kwired]

Right process, but you use the smaller result.

smaller ampacity, larger conductor (I think when he said larger he was refering to the conductor size)

Add: Thinking about it more I think I mean to say smaller as in temperature rating not ampacity.

 

[George Stolz]

Agreed.

Simplest: Figure out your ampacity required, then turn to look at your lugs to see what degree terminals/conductor to worry about. One step at a time.

 

[Smart $]

Agreed.

Simplest: Figure out your ampacity required, then turn to look at your lugs to see what degree terminals/conductor to worry about. One step at a time.

The smallest size wire is determined by termination temperature limit. Once that size is determined, a higher temperature-rated conductor can only be the same size or larger. So while using a higher temperature-rated conductor allows for derating, it either verifies the size is appropriate or needs to be upsized. Hence, it would be the larger.

 

[jaggedben]

The smallest size wire is determined by termination temperature limit. Once that size is determined, a higher temperature-rated conductor can only be the same size or larger. So while using a higher temperature-rated conductor allows for derating, it either verifies the size is appropriate or needs to be upsized. Hence, it would be the larger.

Now you see why people get confused!

When I said 'larger' I meant the conductor size.

So, let's say I have a 15amp circuit, using 90C wire, 75C lugs, operating at an ambient temperature up to 70C.

Table 310.16, in the 75C column, says I need 14awg wire. So based on my lugs, I can't use a smaller size than 14awg.

Now for my temperature derating, I need to look at 310.15(B)(2)(a), and I get a correction factor of 0.58 for 90C wire at 70C. Divide 15amps by .58 and I need an ampacity of 25.9 amps. Go back to 310.16 and I see that I need 12awg to meet that ampacity.

12awg is bigger than 14awg, so 12awg is my result.

Correct?

 

[Smart $]

Now you see why people get confused!

When I said 'larger' I meant the conductor size.

So, let's say I have a 15amp circuit, using 90C wire, 75C lugs, operating at an ambient temperature up to 70C.

Table 310.16, in the 75C column, says I need 14awg wire. So based on my lugs, I can't use a smaller size than 14awg.

Now for my temperature derating, I need to look at 310.15(B)(2)(a), and I get a correction factor of 0.58 for 90C wire at 70C. Divide 15amps by .58 and I need an ampacity of 25.9 amps. Go back to 310.16 and I see that I need 12awg to meet that ampacity.

12awg is bigger than 14awg, so 12awg is my result.

Correct?

Yes.

 

[jaggedben]

Yes.

Thanks!

 

[George Stolz]

Correct?

Pretty much, but your "ampacity" is 17.4. The conductor is good for 17.4 amps under the conditions specified.

Ampacity. The maximum current, in amperes, that a conductor can carry continuously under the conditions of use without exceeding its temperature rating.

 

[jaggedben]

Pretty much, but your "ampacity" is 17.4. The conductor is good for 17.4 amps under the conditions specified.

Right. The completely thorough way to have said it would be "I need a wire that has an ampacity of at least 25.9amps according to table 310.16, so that when a correction factor of .58 is applied, the ampacity exceeds the 15amp current of the circuit."

 

[kwired]

One way I was taught was to find the ampacity needed instead of selecting a conductor, derate it, find out it is too small, then do it all over with the next size larger.

Lets say you have a continuous load of 17 amps. For continuous loads you need to multiply the load by 1.25, so at this point you need a conductor with an ampacity of 21.25 amps. Now lets say you need a deration of 70% because of number of conductors in the raceway. Instead of picking a conductor, derating, then seeing if it is still large enough just multiply your needed ampacity of 21.25 by the recriprocal of 70% (1/.70 = 1.43).

21.25 x 1.43 = 30.39. At this point we need a conductor with an ampacity of 30.39. If we still need to consider ambient temperature we can still multiply this value by the reciprocal of whatever adjustment is required, and that result is minimum ampacity required of our conductor. Select the one that is equal or greater than the result.

Most of us (myself included) were taught to do this the way that can involve more steps. With the 17 amp load above we will add 125% for continuous load and pick a 12 AWG @75 deg conductor because it is more than 21.25 ampacity. We will then multiply that by 70% for conductors in raceway plus derate for ambient only to find out the 12 AWG is too small and then we do it all over again with 10 AWG.

 

[cadpoint]

One way .....

Lets say you have a continuous load of 17 amps. For continuous loads you need to multiply the load by 1.25, so at this point you need a conductor with an ampacity of 21.25 amps. Now lets say you need a deration of 70% because of number of conductors in the raceway. Instead of picking a conductor, derating, then seeing if it is still large enough just multiply your needed ampacity of 21.25 by the recriprocal of 70% (1/.70 = 1.43).

21.25 x 1.43 = 30.39. At this point we need a conductor with an ampacity of 30.39. If we still need to consider ambient temperature we can still multiply this value by the reciprocal of whatever adjustment is required, and that result is minimum ampacity required of our conductor. Select the one that is equal or greater than the result.

....

Well is this true, does this way work every time? Is this Test Proof and Rock solid ?

:huh: