# Thread: applying 110.14 (C)

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## applying 110.14 (C)

I am coaching one of my students who is preparing to sit for the California certification. She is using Tom and Tim Henry's 2014 exam preparation text. One of the chapter 2 (don't have it in front of me, but I believe it was this chapter) questions is what is the allowable ampacity for 4, 4/0 THWN copper conductors with an ambient temperature of 45 celsius. No terminal temperature rating is given. I told her that she had to use the 60 degree column as a starting point, even though THWN is listed in the 75 degree column. Accordingly, the calculation I suggested she use was (195)(0.80)(0.82). The result would then be 127.92, or 128. When we looked up the answer, the result was 150.88---151 amps. Clearly, the Henrys' starting point was 230 amps from the 75 degree column. In the absence of any information on terminal temperature rating, is it not true that one must proceed with the 60 degree column?

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## oops

Duh. Just realized that given the 4/0 size of the conductors, one is directed to use the 75 degree column. Apologies for wasting everyone's time.

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Originally Posted by ohmboyz
Duh. Just realized that given the 4/0 size of the conductors, one is directed to use the 75 degree column. Apologies for wasting everyone's time.

In the absence of any terminal rating specified, 60C is the default for 100A and less, while 75C is the default for everything over 100A. Even if the wire rating is higher. You may encounter a master statement at the beginning of the exam to default to the 75C column, since this represents what is most common in the field, but in the absence of such a statement, follow the defaults specified in 110.14(C).

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## 4/0

4. 4/0 THWN 45C IN CONDUIT = 197.6 AMPS.

5. Originally Posted by domnic
4. 4/0 THWN 45C IN CONDUIT = 197.6 AMPS.
The question assumes that the 4 conductors are all current carrying. (230*80%*.82=151)

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## please show me your math

Originally Posted by domnic
4. 4/0 THWN 45C IN CONDUIT = 197.6 AMPS.

Wondering how you got 197.6? Per the comment below, it was indeed four current carrying conductors. My bad I omitted that info. Accordingly, I believe the math goes something like (230) (.82)(80) (assumes 75 degree column---100 amps {110.14 (C)}given the size of the four 4/0's) which I believe leaves one with 150 +

7. Originally Posted by ohmboyz
Wondering how you got 197.6? Per the comment below, it was indeed four current carrying conductors. My bad I omitted that info. Accordingly, I believe the math goes something like (230) (.82)(80) (assumes 75 degree column---100 amps {110.14 (C)}given the size of the four 4/0's) which I believe leaves one with 150 +
Greetings,

The question gave you the size of the conductor which was clearly in the 75°C Column per 110.14(C)(1)(b) as well as the number of current carrying conductors which brings in Table 310.15(B)(3)(a) as well as the 45°C that brings in 310.15(B)(2)(a). As you concluded, you will start from the 75°C values due to those facts. Interesting note, we could have changed the THWN to THWN-2 and changed those values as well but alas it was only THWN being expressed.

The 230A x .80 x .82 = 150.88 (151 Amps) is the THWN's new de-rated ampacity.

But I see you already figured that out.

Many times on exams do not assume what the logical answer would be but simply what the question is asking. While it is important to teach the student the real world aspects of electrical work, in the exam environment we only focus on what the test question is asking and simply solve the question.

8. Originally Posted by MasterTheNEC
Greetings,

...........................................

Many times on exams do not assume what the logical answer would be but simply what the question is asking. While it is important to teach the student the real world aspects of electrical work, in the exam environment we only focus on what the test question is asking and simply solve the question.

Its one place where it seems experience may hurt. I've found experienced folks often take more into consideration in looking for an answer to a test question than is asked. For instance, taking the limitations of 240.4(D) into account when the question only asks about "ampacity"

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## 4/0

Originally Posted by domnic
4. 4/0 THWN 45C IN CONDUIT = 197.6 AMPS.
MY bad I used THWN-2.

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