Adjustment and correction factors

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Brandon002

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Location
Tennessee
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I want to be an electrician
I’ve see people multiple and divide it
I was wondering if anyone could help clear it up for me


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1 divided by1.25 = .8
1.25 times .8 = 1

100 divided by 1.25 = 80
80 times 1.25 = 100

A motor that draws 80 amps requires a conductor of 1.25 times 80 = 100.
What's the Max load I can put on a 100-amp breaker? I can divide by 1.25 or multiply by .8.

Way over simplified.


What exactly is your question?
 
Another way to state Toms examples is; one way is the "inverse" of the other.
 
ptonsparky said:
1 divided by1.25 = .8
1.25 times .8 = 1

100 divided by 1.25 = 80
80 times 1.25 = 100

A motor that draws 80 amps requires a conductor of 1.25 times 80 = 100.
What's the Max load I can put on a 100-amp breaker? I can divide by 1.25 or multiply by .8.

Way over simplified.


What exactly is your question?
Click to expand...

If we 15 amp continuous load and an ambient temperature 65 and four current carrying conductors
would you 15*1.25=18.75
18*.65 * .80=9.36
I was wondering if this is the correct way


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Brandon002 said:
If we 15 amp continuous load and an ambient temperature 65 and four current carrying conductors
would you 15*1.25=18.75
18*.65 * .80=9.36
I was wondering if this is the correct way


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Click to expand...
No, you would use 15 for both equations
15 x 1.25 =18.75
15 / .80 = 18.75
 
ptonsparky said:
You need a conductor that has a final ampacity of 18.75.
18.75/.65=28.8
or
18.75*1.53
Click to expand...

I think I figured it out

So if you had a known load
Load /(adj*cf)=ampacity

But if it was an unknown load
Would you pick an ampacity and multiply it by the ambient temperature factor and ccc factor?


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Brandon002 said:
If we 15 amp continuous load and an ambient temperature 65 and four current carrying conductors
would you 15*1.25=18.75
18*.65 * .80=9.36
I was wondering if this is the correct way


Sent from my iPhone using Tapatalk
Click to expand...
15 amp continuous load and non 100% rated overcurrent device (which is the majority of cases) needs additional 25% for conductor ampacity

So 15 amps x 1.25 = 18.75 amps is the minimum ampacity needed for the termination rating. If a 75C termination then you can not have less than 14 AWG conductor even if you can use 90 C ampacity table for further adjustments when needed.

Next is the four current carrying conductors adjustment with a .80 adjustment. Inverse of that is 1.25. Assuming you have 90C conductor insulation you can either look at 310.16 and multiply the 25 ampacity of a 14 AWG by .80 giving you 20 amps of max allowable for said 14 AWG conductor or you can multiply your 15 amps (actual load) by 1.25 giving you the 18.75 as what size you can use for minimum ampacity needed for the 90C conductor insulation, 14 AWG still fits all the requirements here.

Then you mentioned ambient temperature, but you did not mention whether that was 65F or 65C which could be a huge difference in the next calculation. For that you already have an adjusted minimum ampacity for number of conductors of 18.75. If your ambient is 65F you actually increase allowable ampacity with an adjustment factor of 1.08 which would give your 25 amp @ 90C 14 AWG an increased capacity of (25x1.08=27) 27 amps. Or you can do the reciprocal thing again to the minimum ampacity required which would be 1 / 1.08= .9259 x the so far determined minimum ampacity of 18.75 = 17.36. The final minimum ampacity needed for a 90C conductor here is 17.36, which still allows a 14 AWG conductor. Had the adjustments been enough to allow less than 14 AWG @ 90C then you still limited to the 75C termination rating that requires a 14 AWG conductor.

If your ambient was 65C the adjustment for that would have been .65 or with the reciprocal method an adjustment of 1.54 which would have given you the 18.75 adjusted for number of conductors x 1.54 = 28.88 minimum conductor ampacity @ 90C which would mean you need at least 12 AWG conductor.

Keep in mind for general loads the small conductor rule in 240.4(D) would require 15 amps max overcurrent protection on 14 AWG conductor and since your adjustments that come into play here put you at a minimum load and overcurrent protection that is over 15 amps you would need 12 AWG and 20 amp overcurrent device in most situations. Motors and air conditioning equipment is one place where 240.4(D) does not apply though so you possibly could use 14 AWG there and because of other rules on overcurrent protection for those applications the protection could be higher as well.
 
Just wanted to point out that for a continuous load with adjustment and correction factors, you don't use all 3 effects at once.

The termination ampacity is determined by the uncorrected and unadjusted ampacity at the termination temperature rating (typically 75C), and it must be at least 125% * the load. The wire ampacity is determined by the corrected and adjusted ampacity based on the wire insulation rating (typically 90C), and it must be at least 100% * the load.

Cheers, Wayne
 
I bet your head is spinning after reading kwired's excellent explanation. It is all there, but you may need to take it a piece at a time. Regarding your fist post, I have had discussions with other instructors about the 1.25 vs. .8 question and I firmly believe we should not confuse you more than the code already does. The codes states that continuous loads shall be calculated at 125% of their full load. Until you are completely comfortable with this, you shouldn't concern yourself with the reciprocal. Then the code says that 4-6 conductors requires and 80% adjustment. This is a completely separate calculation and should be viewed as such. Do one calculation at a time.
 
So formula wise, if L is the load, and the correction/adjustment factors multiplied together are F, and A75 is the table 75C ampacity, and A90 is the table 90C ampacity, we typically have the requirements:

1.25*L <= A75
L <= F*A90

But A75 = sqrt(45/60) * A90 = 0.866 * A90 (although the tables may deviate somewhat), so the equations may be written as:

L <= 0.693 * A90 (continuous load)
L <= F * A90

Meaning whichever is smaller, F or 0.693, will control the final selection. As the tables deviate somewhat, if the two values are close, we'll need to double check with the tables. But at least this gives you a rule of thumb for when termination rating controls and when ampacity controls.

Cheers, Wayne
 
Strathead said:
I bet your head is spinning after reading kwired's excellent explanation. It is all there, but you may need to take it a piece at a time. Regarding your fist post, I have had discussions with other instructors about the 1.25 vs. .8 question and I firmly believe we should not confuse you more than the code already does. The codes states that continuous loads shall be calculated at 125% of their full load. Until you are completely comfortable with this, you shouldn't concern yourself with the reciprocal. Then the code says that 4-6 conductors requires and 80% adjustment. This is a completely separate calculation and should be viewed as such. Do one calculation at a time.
Click to expand...

I’m pretty sure I’ve figured it out but one more question
Do you apply for continuous 1.25 first and then the ambient factor and the ccc factors


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Brandon002 said:
I’m pretty sure I’ve figured it out but one more question
Do you apply for continuous 1.25 first and then the ambient factor and the ccc factors
Click to expand...
No, you never apply them both at the same time. There are two separate checks, one using the continuous 1.25 and the termination temperature rating, no adjustment/correction factors, the other using the insulation temperature rating and adjustment and correction factors, no continuous 1.25.

Cheers, Wayne
 
wwhitney said:
No, you never apply them both at the same time. There are two separate checks, one using the continuous 1.25 and the termination temperature rating, no adjustment/correction factors, the other using the insulation temperature rating and adjustment and correction factors, no continuous 1.25.

Cheers, Wayne
Click to expand...

Could you show me an example please


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Brandon002 said:
Could you show me an example please
Click to expand...
100A continuous load, 3 CCCs, 75C termination temperature, 90C insulation rating. Ambient temperature in part of the run is 60C, which controls correction and adjustment; temperature correction factor for 90C insulated conductors in 60C ambient is sqrt((90-60)/(90-30)) = 0.707 (or check the table).

First check: At termination, the 75C table ampacity must be at least 100A * 125%. So check Table 310.16 to find conductors in the 75C column that are at least 125A. For Cu, #1 is the smallest allowable size. [Note that for a 100A 100% rated breaker, the conductor termination ampacity at the breaker need only be 100A in the 75C column. But the utilization equipment is very likely to require the 125% factor.]

Second check: For the run of the wire in the 60C ambient, the 90C table ampacity must be at least 100A/0.707 = 141A. So find conductors in the 90C column that are at least 141A. For Cu, #1 is the smallest allowable size.

Third check (only when some of the loading is continuous, or sometimes when 240.4(B) does not apply): if using a normal (non-100% rated) breaker, then the OCPD for the 100A continuous load must be at least 125A. 240.4 requires that the wire have an ampacity of at least 111A, via subsection (B) which allows rounding up to the next higher OCPD if the conditions are met. The ampacity is the value in the second check. So we actually need a conductor whose 90C column entry is at least 111/0.707 = 157A. For Cu, #1/0 is the smallest allowable size.

So for this load, we could use a 125A regular OCPD with #1/0 Cu conductors. Or if we were able to get a 100A 100% rated OCPD (not so likely), we could use it with #1 Cu conductors.

Cheers, Wayne
 
Last edited:
Brandon002 said:
Could you show me an example please
Click to expand...
Here's the same example with the possibility of 100%-rated breakers removed for clarity:

100A continuous load, 3 CCCs, 75C termination temperature, 90C insulation rating. Breaker is regular (non-100% rated) 125A. Ambient temperature in part of the run is 60C, which controls correction and adjustment; the temperature correction factor for 90C insulated conductors in 60C ambient is sqrt((90-60)/(90-30)) = 0.707 (or check the table).

First check: At termination, the 75C table ampacity must be at least 100A * 125%. So check Table 310.16 to find conductors in the 75C column that are at least 125A. For Cu, #1 is the smallest allowable size.

Second check: For the run of the wire in the 60C ambient, the 90C table ampacity must be at least 100A/0.707 = 141A. So find conductors in the 90C column that are at least 141A. For Cu, #1 is the smallest allowable size.

Third check: 240.4(B) says our wire must have an ampacity of at least 111A, which is 1A more than the next smaller breaker size. The ampacity is the value in the second check. So we actually need a conductor whose 90C column entry is at least 111/0.707 = 157A. For Cu, #1/0 is the smallest allowable size.

So for this load, we could use a 125A regular OCPD with #1/0 Cu conductors, assuming voltage drop doesn't require further upsizing.

Cheers, Wayne
 
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