Is the handbook wrong?

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ryan_618

ryan_618

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Salt Lake City, Utah
I think there is an error in the handbook regarding conductor sizing and ampacity adjustment. Please review this example and let me know what you think.

We have a 25 amp continuous load. 60 degree terminals, THWN conductors, four current carrying conductors in teh raceway.

Here is the way I figure it to be done:

25*1.25 for continuous load=31.25
The conductors must have an ampacity of 31.25.

This would require an 8 AWG conductor. We then take the ampacity of 8 AWG and apply the adjustment required by 310.15(B)(2)(a), in this case, 80%. So,
40 (ampacity of at 60 degrees) * 0.80=32 amperes.

A 32 ampere conductor will carry this load, so we can use it. We would then select a 40 ampere OCPD to protect the conductor.

--------------------------------------------
Next example, same parameters:
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Determine the size of the overcurrent protective device (OCPD). Referring to 210.20(A), 125 percent of 25 amperes is 31.25 amperes. Thus, the minimum standard-size overcurrent device, according to 240.6(A), is 35 amperes.
Step 2.
Determine the minimum conductor size. The ampacity of the conductor must not be less than 125 percent of the 25-ampere continuous load, which results in 31.25 amperes. The conductor must have an allowable ampacity of not less than 31.25 amperes before any adjustment or correction factors are applied. Because there are four current-carrying conductors in the raceway, Table 310.15(B)(2)(a) applies. First calculate the ampacity of the conductor using the ampacity value calculated above:

Ampacity = computed load/0.80=39.06 amperes.
I think this is wrong
Because of the 60?C rating of the overcurrent device terminal, it is necessary to choose a conductor based on the ampacities in the 60? column of Table 310.16. The calculated load must not exceed the conductor ampacity. Therefore, an 8 AWG copper, Type TW conductor with an ampacity of 40 amperes is the minimum allowed computed load. However, conductors with a higher temperature rating such as THWN or THHN may be used, but only at their 60?C ampacity.

Which do you guys beleive to be correct...the first example, or the second?
 
Re: Is the handbook wrong?

I think you are using two different methods, but that you will come up with the same answer.

in this example both are saying to use #8
 
Re: Is the handbook wrong?

Maybe I am not understanding the question but I thought this section of the code...

110.14 (C) Temperature Limitations. The temperature rating associated with the ampacity of a conductor shall be selected and coordinated so as not to exceed the lowest temperature rating of any connected termination, conductor, or device. Conductors with temperature ratings higher than specified for terminations shall be permitted to be used for ampacity adjustment, correction, or both.
Click to expand...
....allowed us to make the calculations using the higher ampacity insulation as long as the final outcome did not exceed the terminal temp.

Am I missing something?
 
Re: Is the handbook wrong?

Both methods are the same till we get to the part about making the adjustment for 1-4 current carrying conductors.

In one we multiply the ampacity of the wire by 0.80
In the other we divide the circuit ampacity by 0.80

The net result should always give us the same wire size.
 
Re: Is the handbook wrong?

Why do you say 40 amp breaker and the example says 35 amp breaker ;)

I see where the wording of the example is inconsistant with the wording of the sections referenced. I may not be a code wiz, but I do know math.

You should always end up with the same answer, regardless of the method used.
 
Re: Is the handbook wrong?

Originally posted by ryan_618:
Why does the handbook say that I need a 39 amp conductor?
Click to expand...
Ryan, where in the handbook is this example?
 
Re: Is the handbook wrong?

Originally posted by ryan_618:
Why does the handbook say that I need a 39 amp conductor?
Click to expand...
Because the conductor needs to be adjusted for the 4 CCC's in the raceway. 39.06a(80%)=31.25a
 
Re: Is the handbook wrong?

Just to elaborate, the conductor needs to be a minimum of 31.25 amps to serve the continuos load. In order to achieve that ampacity with a derating of 80% the conductor, before derating, would need a minimum ampacity of 39.06 amps.

The example is correct. Hope this helps.


Trevor

[ January 29, 2006, 04:42 PM: Message edited by: infinity ]
 
Re: Is the handbook wrong?

Originally posted by ryan_618:
Anybody else?
Click to expand...
teacher.jpg


Anyone......anyone?
 
Re: Is the handbook wrong?

Originally posted by ryan_618:
Anyone? Badger?
Click to expand...
Me?

I agree with Trevor, the conductor must have at least a 39.06 rating before adjustments are made to result in a conductor ampacity of at least 31.25 after adjustments are made.

The OCP must have a rating of at least 31.25.
 
Re: Is the handbook wrong?

Ryan ;

Determine the size of the overcurrent protective device (OCPD). Referring to 210.20(A), 125 percent of 25 amperes is 31.25 amperes. Thus, the minimum standard-size overcurrent device, according to 240.6(A), is 35 amperes.
Click to expand...
i don't really get the idea why the NEC code book allow to round up to next size breaker for this type of conductor.

i dont know if i did make a mistake here or not but it should be sized down in case of distance ??
and if that refering to the wire size #8 i can understand it. but two diffrent conflect run into this and i did try to read this twice to make sure i am right track with this one.

but iam leaning to the first question there.

Thanks

Merci, Marc
 
Re: Is the handbook wrong?

Originally posted by ryan_618
Here is the way I figure it to be done:

25*1.25 for continuous load=31.25
The conductors must have an ampacity of 31.25.

This would require an 8 AWG conductor. We then take the ampacity of 8 AWG and apply the adjustment required by 310.15(B)(2)(a), in this case, 80%. So,
40 (ampacity of at 60 degrees) * 0.80=32 amperes.

A 32 ampere conductor will carry this load, so we can use it. We would then select a 40 ampere OCPD to protect the conductor.
Click to expand...
You have made a couple of mistakes in your solution here. First you did half of the equation chose a conductor and then did the second half of the equation.
The equation needs to be carried completely through before the conductor is chosen.

The second mistake you made is using your method of math where the final result would end with a rating of 32 amps the correct overcurrent device would be 35 amps.
Should this have been a classroom exercise I would have taken away extra points due to the double mistake you made.
:D
 
Re: Is the handbook wrong?

Mike, I almost hate to jump in here, but........
agree 31.25....in reference to the #8, would the derated value not be 44 amps? With the 60o rating of 40, I don't see the problem with the 40 amp ocp device. am i wrong ?
 
Re: Is the handbook wrong?

The original post had this problem
We have a 25 amp continuous load. 60 degree terminals, THWN conductors, four current carrying conductors in teh raceway.
Click to expand...
What parts of this question have any bearing on the answer? The 25 amp continuous load and four current carrying conductors in a raceway, and the conductors must be chosen from the 60 degree column. Every thing else is ?smoke? or just space that was filled in with words.

25 amps plus 25% because it is a continuous load. This can be arrived at with two different equations;
25 times 125% = 31.25
25 divided by 80% = 31.25

Now to get the 4 conductor derating of the conductors.
31.23 divided by 80% = 39.0375

The conductors that will be carrying the current to this 25 amp load are required to be no smaller that 39 amps.
The conductor is chosen from the 60 degree column and it is#8 of any type insulation that I choose as long as it is suitable for the location it is being installed in.
:)
 
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