Understanding the National Electrical Code

[Smart $]

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.

I agree, "un-derating" is faster.

However, a bigger problem remains: most people are taught and still do the wire size determination incorrectly, just as you have. I am guilty myself. We compound the effect of the 125% factoring and derating.

I believe the part needing pointed out is the 125% factoring for continuous load is before the application of any adjustment or correction factors. Let's look at the requirements first...

Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.

The minimum feeder-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.

Restating the application example, we have a continuous load of 17A, with termination temperature limitation of 75?C, and we are going to use 90?C conductors, plus we have to derate them to 70% for 7-9 conductors in the homerun raceway.

1) 110.14(C): 17A results in 14 AWG Cu rated 20A@75?C

2) 210.19(A)(1): 17A?125%=21.25A results in 14 AWG Cu rated 25A@90?C

3) 310.15: 17A?70%=24.3A results in 14 AWG Cu rated 25A@90?C

 

[kwired]

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

:huh:

To a certain degree it is no different than working a mathmatical equation in reverse to check that you solved it correctly.

You are determining the ampacity needed instead of taking a conductor and then derating it only to find it is not what you needed, then picking a larger one and going through the process again. The missing number in the equation is the conductor size instead of the final ampacity

 

[kwired]

I agree, "un-derating" is faster.

However, a bigger problem remains: most people are taught and still do the wire size determination incorrectly, just as you have. I am guilty myself. We compound the effect of the 125% factoring and derating.

I believe the part needing pointed out is the 125% factoring for continuous load is before the application of any adjustment or correction factors. Let's look at the requirements first...





Restating the application example, we have a continuous load of 17A, with termination temperature limitation of 75?C, and we are going to use 90?C conductors, plus we have to derate them to 70% for 7-9 conductors in the homerun raceway.

1) 110.14(C): 17A results in 14 AWG Cu rated 20A@75?C

2) 210.19(A)(1): 17A?125%=21.25A results in 14 AWG Cu rated 25A@90?C

3) 310.15: 17A?70%=24.3A results in 14 AWG Cu rated 25A@90?C

I did increase it by 125% for continuos load in both examples. I did not figure out both 75 and 90 degree ampacities which must be done in many cases, sometimes even 60 degree ampacity needs figured as is the case with NM cable. For every conductor you will need to figure minimum needed ampacity for the termination temperature and minimum needed for insulation rating. Which ever results in the larger conductor is the minimum size needed. You do not need to derate for number in raceway for termination temperature - the termination is not in the raceway for starters.

The main goal of my post was pointing out the effectiveness of "un-derating" more so than giving a thouough explanation of sizing conductors. That really needs a thread of its own if we want to go there.

 

[George Stolz]

I dunno, we've about covered all the bases in this one thread. Send it anywhere it's been and it's still on topic.

I like the alternate method thrown out there, I've always gone the long way, it never occurred to me to take a shortcut.

 

[ggunn]

FWIW, that's the way I've always done it myself. Start with the needed ampacity and bump that number up by dividing by the appropriate derate factors, then find the correct conductor from the table. Compare with continuous use derating and use the more conservative (larger) conductor size.

 

[jaggedben]

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).

Notice that's the way I did it in my example. Except that I said 'divide by .58' instead of 'multiply by the reciprocal of .58'. 'Divide by' is probably easier for most electricians to understand.

 

[Smart $]

....

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. ...

I did increase it by 125% for continuos load ...

That's not the problem...

The problem is that you applied 70% derating to the 125% continuous value. These two factorings are completely separate.

17A?125%=21.25A

17A?70%=24.3A

NOT 17A?125%?70%=30.39A​

 

[Smart $]

Just remember there are three [separate] requirements for wire size determination, once minimum circuit ampacity is known. One each from Chapters 1, 2, and 3.

1. Termination Temperature Limitation
2. Continuous, Non-continuous Loads
3. Adjustment and Correction

 

[ggunn]

That's not the problem...

The problem is that you applied 70% derating to the 125% continuous value. These two factorings are completely separate.

17A?125%=21.25A

17A?70%=24.3A

NOT 17A?125%?70%=30.39A​

That's a mistake I made a few times when I was new at this. At least the error was in the right direction. Better to make the wire too big than too small.

 

[kwired]

That's not the problem...

The problem is that you applied 70% derating to the 125% continuous value. These two factorings are completely separate.

17A?125%=21.25A

17A?70%=24.3A

NOT 17A?125%?70%=30.39A​

You are correct, and I was aware the continuous value does not apply to insulation deration, yet overlooked it here, thanks for setting me straight. I probably have calculated conductor ampacity higher than needed before as this is easy to forget, but I guess that is better than having it too small.

 

[kwired]

Notice that's the way I did it in my example. Except that I said 'divide by .58' instead of 'multiply by the reciprocal of .58'. 'Divide by' is probably easier for most electricians to understand.

You are correct. I was taught multiply by the reciprocal, and that is what is stuck in my head for some reason. Because of this I will go .. ok we need the reciprocal of .58 and figure out what that is and then plug that value in the equation, but I need to train myself to think of just dividing by .58 because it will give the same result and really requires less thinking:slaphead:.

 

[George Stolz]

Just remember there are three [separate] requirements for wire size determination, once minimum circuit ampacity is known.

Smart, you have blown my mind. I have never even thought about holding the 125% as a separate requirement, but as a piece of the other two - thanks for clearing this up! I guess I had never looked at this from that angle before. :thumbsup:

For any skeptics (such as I was until I went looking):

The submitter?s change of the word ?before? to ?after? is technically incorrect. Making that change would require that the 125% sizing rule apply sequentially to the ampacity adjustment factors when, in fact, those ampacity adjustments can overlap the 125% calculation provided the conductor can carry the load and be protected by the overcurrent device. Applying the 125% after adjusting for ambient conditions or conduit fill would result in a conductor larger than necessary to comply with all of the rules.

 

[kwired]

Smart, you have blown my mind. I have never even thought about holding the 125% as a separate requirement, but as a piece of the other two - thanks for clearing this up! I guess I had never looked at this from that angle before. :thumbsup:

For any skeptics (such as I was until I went looking):

We don't add 25% for continuous loads because of the conductor or the insulation, we add it because of the overcurrent device. If the overcurrent device is rated 100% continuous then we don't have to add the 25% to conductor ampacity. Reason is standard overcurrent devices use conductor as a heat sink to dissipate heat away from device.

 

[George Stolz]

Yep.

So, let's take this a step further. I have a 20A continuous load on 12-2 NM and a 20A breaker. Legal?

 

[kwired]

Yep.

So, let's take this a step further. I have a 20A continuous load on 12-2 NM and a 20A breaker. Legal?

No.

Conductor would be acceptable, if not for 240.4(D).

The 20 amp continuous load x 125% would need a circuit with a minimum ampacity of 25 amps, unless on a 100% continuous rated overcurrent device. Then it should be legal. You probably won't find such an installation AFAIK.

 

[George Stolz]

Read it again. You sure about that?

 

[kwired]

Read it again. You sure about that?

So what is your loophole?

 

[George Stolz]

First, 334.80 limits the conductor to 60 degree rating for this application, so that wouldn't work.

However, MC cable or THHN would use 90 degree conductors. All 240.4(D) is stating is that a 12 not be protected with greater than a 20A breaker. In our application, we're doing that.

The terminals of the breaker are at 75 degrees.

Our ampacity of 90 rated 12 is 30. If we consider the terminals, then we are down to 25A, which is the minimum ampacity required for a 20A continuous load.

 

[kwired]

First, 334.80 limits the conductor to 60 degree rating for this application, so that wouldn't work.

However, MC cable or THHN would use 90 degree conductors. All 240.4(D) is stating is that a 12 not be protected with greater than a 20A breaker. In our application, we're doing that.

The terminals of the breaker are at 75 degrees.

Our ampacity of 90 rated 12 is 30. If we consider the terminals, then we are down to 25A, which is the minimum ampacity required for a 20A continuous load.

But 210.20(A) requires the overcurrent device to be not less than 100% of non-continuous load plus 125% of continuous load. So the branch circuit overcurrent device needs to be 25 amps to start with.

 

[George Stolz]

You're right, I blew it. Thanks for the gentle correction.