# Ampacity and deratings



## pnk7285 (Oct 28, 2009)

Hello, folks:

I must take an examination on Friday and am struggling a bit with problems pertaining to ampacities and to the proper procedure for derating ampacities. I show below a sample problem and my proposed solution. I would like to know whether or not the proposed solution is correct, and, if so, whether the underlying logic is correct. If you can help, I would be much obliged!


*Problem*

A 240V/120V single-phase, three-wire feeder circuit supplies a continuous load of 180 A and a non-continuous load of 100 A. Given that:

(1) the terminations for the feeder are rated for 75 degrees C,

(2) the ambient temperature is 40 degrees C, 

calculate the size of THHN conductors that are required for this application. What size overcurrent protection device is required?


*Proposed solution*

Step #1: Calculate the true load: 180 A + 100 A = 280 A

Step #2: Adjust true load for presence of continuous load: adjusted load = 1.25 X 180 A + 100 A = 325 A

Step #3: Use result of step #2 to establish minimal size of THHN conductors:

Using 90 degree C column of Table 310.16, see that 350 kcmil THHN conductors are the smallest that have an uncorrected ampacity, 350 A, that is equal to or greater than 325 A. 

Step #4: Adjust uncorrected ampacity for 40 degree C temperature

Using the 90 degree C column of Table 310.16, I see that the correct factor is 0.91.
Therefore, adjusted ampacity = 350 A X 0.91 = 318.5 A

Step #5: Compare adjusted ampacity with 75 degree ampacity of 350 kcmil conductors:

318.5 A is greater than 310 A, which is the 75 degree C ampacity of 350 kcmil conductors

Therefore use 310 A rather than 318.5 A as being the allowable ampacity of the conductors
Allowable ampacity = 310 A

Step #6: Compare allowable ampacity with true load:

310 A is greater than 280 A, so conductors will carry the load

Step #7: The conductors must be protected according to their allowable ampacity, which is 310 A. In addition the results of step #2 establishes a minimal rating for the overcurrent protection device, namely 325 A, that actually exceeds 310 A. Unfortunately 325 A does not correspond to a standard size, so we must select the next largest standard size, which is 350 A. 


CONCLUSIONS: 350 KCMIL THHN CONDUCTORS SATISFY ALL REQUIREMENTS OF THE NEC PROVIDED THAT THE OVERCURRENT PROTECTION DEVICE IS RATED AT 350 A. 


I have a few specific questions: 

(1) Is step #3 correct? Approximately 50% of my classmates insist that we must use the 75 degree column even at this early stage of the calculation because the termination devices are rated at 75 degrees C. I do not see how to justify this point of view for circuits in excess of 100 A. 

(2) Is the use of the correction factor for 90 degree conductors in step #4 correct?

(3) Is the comparison with the true load in step #6 correct? I note that if the proper comparison should be with the adjusted load, then 350 kcmil conductors would not be acceptable. 

If there should be anyone out there in cyberspace with the time and inclination to be of assistance I would greatly appreciate hearing from you.


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## nolabama (Oct 3, 2007)

good luck !! if you are taking the icc test your thinking about this waaay to much :thumbup:


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## nolabama (Oct 3, 2007)

wait a minute - that looks like a union exam question and not a license exam question - am i right?


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## user5941 (Mar 16, 2009)

Size your OCP device frst based on 215.3 then select your conductor using the 74C column based on the calculated load now apply your derating factors if the selected conductor after derating cannot be protected by the OCP device then select the next size wire at this point you can use the 90c table for derating.You size the OCP device then find the wire size based on 75c


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## pnk7285 (Oct 28, 2009)

*reply to nolabama*



nolabama said:


> wait a minute - that looks like a union exam question and not a license exam question - am i right?


Actually neither. This is merely a question from my textbook. I am a student at a nearby technical college. I must say, however, that your response cheers me up a bit. I'm glad to know that I won't run into too many problems like this.


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## pnk7285 (Oct 28, 2009)

rewire said:


> Size your OCP device frst based on 215.3 then select your conductor using the 74C column based on the calculated load now apply your derating factors if the selected conductor after derating cannot be protected by the OCP device then select the next size wire at this point you can use the 90c table for derating.You size the OCP device then find the wire size based on 75c


***********

Thanks a million, Rewire. I must say that your explanation is far clearer than is either my textbook or the NEC. According to your recipe, then, the correct solution is:

step #1: OCPD = 350 A based upon next standard size above the calculated load

step #2: using 75 degree C column in conjunction with calculated load, 325 A, I observe that 400 kcmil conductors are required.

step #3: derate using the 90 degree ampacity:

allowable ampacity = 380 A X 0.91 = 345.8 A, which is substantially greater than the true load.

This conductor may, in fact, be protected by a 350 A OCPD, so the problem is solved. 

The correct answer is therefore 400 kcmil, as 50% of my classmates insisted. 


Thanks again for a great explanation.


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## user5941 (Mar 16, 2009)

welcome to the forum


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