I am new to the forum and I'm studying for the California electrical exam. I was wondering if any could explain when to divide and when to multiple the correction factor by the amperage? I have been doing either or witch ever gives me the correct answer but in the real world I won't have possible answers in front of me.
Ambient Temperature Correction Factor
- Thread starter James14
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packersparky
Senior Member
- Location
- Wisconsin
- Occupation
- Inspector
You always multiply by the value given in table 310.15(B)(2)(b) for ambient temperature correction. For example: 2 AWG copper XHHW has an ampacity of 115. You are going to use it in an ambient temperature of 115 degrees F. 115 X .85 (the value in the table) = 97.75 amperes. The ampacity of 2 AWG XHHW copper used when the ambient temp is 115 degrees is 97.75 amps.
jeremy.zinkofsky
Senior Member
- Location
- nj
I think you're missing the point of the question.
If it is too hot or cold, the wire is no longer good for it's rating in 310.15. The correction factor is a percentage. Table 310.15 tells you what each conductor is rated for at an ambient temperature of 60C, 75C, and 90C. If the temperature changes, the ampacity will change as a result.
So, a #6 in a 35C environment is only rated for 88% of the value listed in the 75C column. 88% of 65A is 57.2A (0.88*65 = 57.2A).
What this ultimately means is that a #6 in a 35C environment can only carry 57.2A. If you need 60A protection, you are no longer allowed to use a #6 because it is now rated for 57.2A. In a 35C environment, a 60A load would need #4 feeders.
Another thing to note; you have to apply this adjustment twice if the wire is on a roof and at a different temperature than what is tabulated in 310.15.
Adamjamma
Senior Member
- Location
- Jamaica and london
I thought you had to figure which temperature situation gave you the biggest difference factor, or worst case, and set your correction figures for that? You might calculate losses such that one part of cable is 57 amps and another part of cable is 59 amps but you use the 57 amps because that is lowest figure? Or am I reading the calculations hints wrong? Instructor was saying you could, due to conditions, have fice calculations on different factors, but then you choose worst case and adapt for that one.
jeremy.zinkofsky
Senior Member
- Location
- nj
That might be over analyzing a bit. I have always looked at historical temperatures for the area and applied the closest temp correction factor.
Carultch
Senior Member
- Location
- Massachusetts
The NEC specifies the calculation problem of "given a certain wire at conditions of use, what is the ampacity?". If you plug-and-chug an equation directly according to the way the NEC specifies how to calculate ampacity, you would multiply by a temperature correction factor. So given #6 Cu THWN-2 wire with an 0.96 temperature correction factor, the wire ampacity is 75A*0.96.
In a design problem, your calculation is instead, "given an operating current and conditions of use, what wire size should be used?". This is the inverse problem of what the NEC directly prescribes, and therefore when you rearrange it with algebra, you get division instead of multiplication.
You will get both of these questions on an exam. So it is a matter of which direction your calculation goes.
As an example, 56A operating current at 32C ambient. One calculation I would do, is 56A/0.96 = 58.3A. Then take 58.3A, and look it up in the 90C column of the ampacity table. It is between the value for #8 and #6, and thus we round up to #6. There are more calculations that govern whether this is a sufficient size for a specific situation (terminal ratings, 240.4(B) ocpd, etc), but leave that for another topic.
- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
See if this helps. A conductor's ampacity is based on its ability to endure the heat that will be created as current passes through the wire, without sustaining damage to its insulation system. A bigger wire has a higher ampacity because the same amount of current releases less heat (i.e., given that the larger wire has lower resistance). But to the present point, the hotter a wire gets, the greater the potential degradation of the insulation. If you run 40 amps through a wire in an ambient temperature environment of 30C, the insulation will reach some temperature. If you run the same current though the same wire in a hotter environment, the insulation will reach a higher temperature. To reduce the chances of damage to the insulation when the ambient temperature is higher, you have to reduce the amount of current you run through the wire. The temperature adjustment factors are all below 1.0 for temperatures above 30C. You multiply the ampacity in the table by the adjustment factor, and the resulting ampacity is lower than the tabulated value.
In a design problem, your calculation is instead, "given an operating current and conditions of use, what wire size should be used?". This is the inverse problem of what the NEC directly prescribes, and therefore when you rearrange it with algebra, you get division instead of multiplication.
I think that is the answer I was looking for. Thanks for the help. I have been studying without having a instructor and some of my coworkers are not to knowledgeable on the test.
I think that is the answer I was looking for. Thanks for the help. I have been studying without having a instructor and some of my coworkers are not to knowledgeable on the test.
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