Ampacity Question

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charlie b said:
Of course they have meaning. And we are bound to follow the rules they provide us.

But the amount of current you choose to run through a wire does not alter the ability of a wire to carry current. There are two very different and separate things going on here. Both are important. But they are different.

It is like saying you wish to buy a new TV, and your budget is $700. You go looking at stores, and you see many models at various prices. Question for you: if you go to a new store, and you see a different model, and it has different features than the others you have seen before, and if it is at a different price than the others you have see, does that fact cause your budget to become different?
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I don't follow the TV anology at all. I'd say a better analogy would be to limit yourself to one particular set. In this case, #4 THHN. Not #3, or #2, or THWN.
 
480sparky said:
Given the question states there are 4 CCCs, you derate to 80% based on 310.15(B)(2)(a).
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I agree.
480sparky said:
It also states it is a continuous load, so you need to derate again to 80% based on 210.19(A)(1).
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No, you don't derate the wire. You pick a wire that starts off with a higher ampacity, so that after it gets derated (for multiple CCCs), it will still have the minimum required ampacity.




Example:
  • Amount of load = 76 amps.
  • That load is continuous.
  • Therefore, I need a conductor with an ampacity of 76 * 1.25, or 95 amps.
  • I plan to put 4 CCCs in the conduit.
  • Therefore, I must derate whatever conductor I choose to use by 80% of the ampacity value given in the tables.
  • I try #4 THHN first. Tabulated value = 95. Derate by 80%, and you get an ampacity of 76. Compare required ampacity of 95 with available ampacity of 76, and conclude that I need a bigger wire.
    [*]I try #3 THHN next. Tabulated value = 110. Derate by 80%, and you get an ampacity of 88. Compare required ampacity of 95 with available ampacity of 88, and conclude that I need a bigger wire.
  • I try #2 THHN next. Tabulated value = 130. Derate by 80%, and you get an ampacity of 104. Compare required ampacity of 95 with available ampacity of 104, and conclude this this would work.
When I multiplied the tabulated ampacity by 80%, that was "derating" the conductor. When I multiplied the load by 125%, that was not "derating a conductor," but rather "calculating the required ampacity."
 
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charlie b said:
I agree.No, you don't derate the wire. You pick a wire that starts off with a higher ampacity, so that after it gets derated (for multiple CCCs), it will still have the minimum required ampacity.



Example:
  • Amount of load = 76 amps.
  • That load is continuous.
  • Therefore, I need a conductor with an ampacity of 76 * 1.25, or 95 amps.
  • I plan to put 54 CCCs in the conduit.
  • Therefore, I must derate whatever conductor I choose to use by 80% of the ampacity value given in the tables.
  • I try #4 THHN first. Tabulated value = 95. Derate by 80%, and you get an ampacity of 76. Compare required ampacity of 95 with available ampacity of 76, and conclude that I need a bigger wire.
  • I try #2 THHN next. Tabulated value = 130. Derate by 80%, and you get an ampacity of 104. Compare required ampacity of 95 with available ampacity of 104, and conclude this this would work.
When I multiplied the tabulated ampacity by 80%, that was "derating" the conductor. When I multiplied the load by 125%, that was not "derating a conductor," but rather "calculating the required ampacity."
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But we are not allowed, by the question itself, to calculate for a larger wire. We're stuck with #4. Your example is a real-world one, where the question Dennis is posing is more of a test question.

Besides, without a load, we can't even bother with changing wire sizes. For all we know, the load is only 5 amps and #4 was chosen for voltage drop.
 
480sparky said:
But we are not allowed, by the question itself, to calculate for a larger wire. We're stuck with #4.
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Correct. And the questions asks for the ampacity of that #4. And you get that answer by derating for the 4 CCCs, taking 95 * 80%, and getting 76. That is the available ampacity. The question of whether that is enough ampacity for the load was not asked, and it would not change the answer. If the required ampacity was 5 amps, then the available 76 ampacity would be sufficient. If the required ampacity was 85 amps, then the available 76 ampacity would not be sufficient. But the ampacity is still 76, regardless of what happens with the load.
 
Now there are two Becks with whom I agree,
I read too much into the question, I now agree "B".
In the past I have been assigned the project of preparing questions for tests. This accentuates what I learned in doing so: It ain't easy.
 
Well, you see why I posted this question. There are always a few of these on any exam. This particular question was just wrong from the get go because they used 70% instead of 80% for the fill.

This was a CD someone gave me that is based on the 2005 NEC but it is pretty good and valid for most question in 2008. It is by American Contractors Education Service-- whoever they are. :)

I am with Charlie on this perhaps for different reasons but I am very bad at taking tests simply because I tend to read too much into it.
 
augie47 said:
Now there are two Becks with whom I agree,
I read too much into the question, I now agree "B".
In the past I have been assigned the project of preparing questions for tests. This accentuates what I learned in doing so: It ain't easy.
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I think the intent was for an answer as Ken is doing simply because that was in their solution. I just don't see the continuous load applying.
 
I think the question is still a little broad. If you consider 310.15 B.4, Then it could change everything. As far as not having to count one of the wires as a ccc. If it's a neutral.
 
ggonza said:
I think the question is still a little broad. If you consider 310.15 B.4, Then it could change everything. As far as not having to count one of the wires as a ccc. If it's a neutral.
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A neutral is not stated, just the conductor is a CCC. This could be two 277v circuits, or it could be two 208v circuits, or it could be two 120v circuits, or it could be three 120v circuits in a MWBC or it could be three 277v circuits in a MWBC............
 
I believe that the disagreement here is the term 'ampacity' means something different from 'maximum load the circuit can serve'.

Ampacity is a feature of the wire itself. It is the amount of current that the wire can carry on a continuous basis without overheating. The ampacity of #4 THHN is 95A. However because we have 4 current carrying conductors (4 conductors serving non-linear loads) we are required to 'derate' the ampacity of the #4 THHN to 76A.

I agree with Charlie: the 'final derated ampacity' of the conductors is 76A.

But if you ask about the maximum current that a circuit can serve, a separate '80%' factor comes into play that you must apply.

Normal breakers are limited to supplying no more than 80% of their trip rating when supplying continuous loads. This means, for example, that an 80A breaker is limited to supplying no more than 64A on a continuous basis, _unless_ you have a '100% rated breaker'. On top of this, the breaker is required to protect the wire.

These requirements combine to reduce the maximum current that a circuit can serve. For example, if you have conductors with 80A ampacity, they must be protected by an 80A breaker, and that 80A breaker means that the circuit can only serve a 64A continuous load.

Now there is a hole in my reasoning: since I've made the claim that the first 80% factor is the ampacity change created by the number of current carrying conductors, and the second 80% factor is forced by a breaker limitation, it would seem that I could have conductors of 76A ampacity, protected by an 80A breaker (round up rule), and serve a continuous load of 64A. However if you observe the requirements of 210.19(A)(1) the maximum load allowed to be served is 60.8A unless you have a 100% rated breaker, in which case the maximum load allowed to be served is 76A.

-Jon
 
Dennis Alwon said:
Well, you see why I posted this question. There are always a few of these on any exam. This particular question was just wrong from the get go because they used 70% instead of 80% for the fill.

This was a CD someone gave me that is based on the 2005 NEC but it is pretty good and valid for most question in 2008. It is by American Contractors Education Service-- whoever they are. :)

I am with Charlie on this perhaps for different reasons but I am very bad at taking tests simply because I tend to read too much into it.
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I agree with Dennis and Charlie. The load that is supplied has no bearing on the ampacity of the wire. It is what it is. Derate for fill.
 
One-eyed Jack said:
The price of the TV has nothing to do with the amount of money you have to spend. Same as the load served has nothing to do with derating.
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I understand that part. But what I'm saying is we are limited to just one TV set (#4 THHN), not all that are available (16 to 2,000kcmil, and all types of insulation).
 
winnie said:
I believe that the disagreement here is the term 'ampacity' means something different from 'maximum load the circuit can serve'.

Ampacity is a feature of the wire itself. It is the amount of current that the wire can carry on a continuous basis without overheating. The ampacity of #4 THHN is 95A. However because we have 4 current carrying conductors (4 conductors serving non-linear loads) we are required to 'derate' the ampacity of the #4 THHN to 76A.

I agree with Charlie: the 'final derated ampacity' of the conductors is 76A.

But if you ask about the maximum current that a circuit can serve, a separate '80%' factor comes into play that you must apply.

Normal breakers are limited to supplying no more than 80% of their trip rating when supplying continuous loads. This means, for example, that an 80A breaker is limited to supplying no more than 64A on a continuous basis, _unless_ you have a '100% rated breaker'. On top of this, the breaker is required to protect the wire.

These requirements combine to reduce the maximum current that a circuit can serve. For example, if you have conductors with 80A ampacity, they must be protected by an 80A breaker, and that 80A breaker means that the circuit can only serve a 64A continuous load.

Now there is a hole in my reasoning: since I've made the claim that the first 80% factor is the ampacity change created by the number of current carrying conductors, and the second 80% factor is forced by a breaker limitation, it would seem that I could have conductors of 76A ampacity, protected by an 80A breaker (round up rule), and serve a continuous load of 64A. However if you observe the requirements of 210.19(A)(1) the maximum load allowed to be served is 60.8A unless you have a 100% rated breaker, in which case the maximum load allowed to be served is 76A.

-Jon
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This would not change the allowable ampacity of the wire. You must first determine the load ie 60.8 amps. If this load is continuous then you must apply the 125% which would then push the min req cir. ampacity to 76 which is what the #4 can safely carry. OCPD at 80 and it will not be loaded greater than 80% and it will carry a continuous load.
 
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