Double de-rating?

[david luchini]

We look at multi-outlet branch circuits differently you build your branch circuit conductors based on the load in this discussion a 10 amp load.
I build a multi-outlet circuit based on the rating of the circuit.

You seem to have changed the discussion from what the code allows you to do, to what would be the best way to design a circuit. This is two separate issues.

General purpose multi-outlet lighting circuit.

The one point you would most likely not agree is in 210.23 (A) it says a 15 amp circuit. That to me includes a circuit that has conductors rated at 15 amps Min. Without any initial consideration to what size load you put on it. After I build the circuit do not violated the rules , I build the circuit and limit the loads,

We do disagree on what a 15A circuit is. 210.3 says the "rating" of a branch circuit is "in accordance with the maximum permitted ampere rating or setting of the overcurrent device." There is nothing there about the ampacity of the conductors. A 15A c/b with 10A conductors, a 15A c/b with 15A conductors and a 15A c/b with 25A conductors are all 15A branch circuits.

You build the circuit to handle the load. If t is was an individual branch circuit I build it the way you do.

Because this is exactly what 210.19(A)(1) tells me to do.

After I build my 15 amp circuit with 12 CCC in conduit I will increase my conductors to 12AWG THHN And put your 10 amp load on it. Someone wants to add a lighting load to my circuit throw an amp probe on the circuit decide if you can add the lighting load to my 15 amp general purpose lighting branch circuit.

You built your circuit with 10 amp rated conductors my amp probe doesn?t tell me if i can add a load on your 15 amp general purpose branch circuit

Let's try another example and see how it fits into your method. Let's says we had six (6) 15A lighting circuits in the same raceway (12 ccc's.) The load on each circuit is 12Amps continuous.

210.19(A)(1) tells me that the minimum branch conductor size, before any adjustment factors, must have an ampacity that is not less than 125% of the continuous load: 12*1.25=15. Therefore, #14 awg is my minimum conductor size.

210.20(A)(1) say the OCPD shall be not less than 125% of the continuous load, so a 15A c/b is OK.

I plan on using #14awg THHN (25A at 90degc) as my conductor. T310.15(B)(2)(a) says that I must adjust the ampacity of my #14 THHN because I have 12 ccc's in the raceway. Therefore, the ampacity of my conductors is 12.5 (50% ajustment factor.)

Per 210.19(A)(1), the 12.5 ampacity of my #14 THHN (12.5) is not less than the maximum load to be served (12), so the #14 is acceptable. Per 240.4(B), the next higher standard overcurrent device is allowed, so the 15A c/b is acceptable for the #14 THHN. Per 240.4(D)(3), 15 Amp is the max OCPD for #14 copper, so we are compliant with this section as well.

By your definitions, we do not have a proper 15A circuit because the ampacity of the conductors is only 12.5A. But the circuit is already fully loaded, so no additional load can be added to the circuit. Why increase the conductors to #12 to get a 15A ampacity conductor when the #14 is properly sized, properly protected, and no additional load can be added?

 

[david]

Let's try another example and see how it fits into your method. Let's says we had six (6) 15A lighting circuits in the same raceway (12 ccc's.) The load on each circuit is 12Amps continuous.

210.19(A)(1) tells me that the minimum branch conductor size, before any adjustment factors, must have an ampacity that is not less than 125% of the continuous load: 12*1.25=15. Therefore, #14 awg is my minimum conductor size.

FPN: Table 310.16 through Table 310.19 are application tables for use in determining conductor sizes on loads calculated in accordance with Article 220.

Continuous Load. A load where the maximum current is expected to continue for 3 hours or more.

what is the total calculated lighting load in your example?

12 amp lighting load or 15 amp continues lighting (calculated) load?

 

[david luchini]

FPN: Table 310.16 through Table 310.19 are application tables for use in determining conductor sizes on loads calculated in accordance with Article 220.

Continuous Load. A load where the maximum current is expected to continue for 3 hours or more.

what is the total calculated lighting load in your example?

12 amp lighting load or 15 amp continues lighting (calculated) load?

The load on each circuit is 1440 Watts of lighting - 12 Amps @ 120Volts. The 12 Amp load is expected to continue for 3 hours or more, so it is a continuous load.

 

[david]

Let's try another example and see how it fits into your method. Let's says we had six (6) 15A lighting circuits in the same raceway (12 ccc's.) The load on each circuit is 12Amps continuous.

? 12 amps continuous loading
? 75 deg C. overcurrent device terminal rating
? Type THNN conductors
? Twelve current-carrying copper conductors in a raceway

210.19(A)(1) tells me that the minimum branch conductor size, before any adjustment factors, must have an ampacity that is not less than 125% of the continuous load: 12*1.25=15. Therefore, #14 awg is my minimum conductor size.

? Your calculated load for continuous loading is 15 amps
? Your calculated load in accordance with article 220 is 12 amps of lighting load
? FPN: Table 310.16 through Table 310.19 are application tables for use in determining conductor sizes on loads calculated in accordance with Article 220.

210.20(A)(1) say the OCPD shall be not less than 125% of the continuous load, so a 15A c/b is OK.

? Branch-circuit conductors shall have an ampacity not less than the maximum load to be served.
? 15 amp 75 deg. C. OCPD (Breaker)

I plan on using #14awg THHN (25A at 90degc) as my conductor. T310.15(B)(2)(a) says that I must adjust the ampacity of my #14 THHN because I have 12 ccc?s in the raceway. Therefore, the ampacity of my conductors is 12.5 (50% adjustment factor.)

? Your calculated load article 220 is 12 amps lighting
? Your calculated load continuous loading is 15 amps
You already took care of the min for continuous loading so lets make sure we do not double de-rate so lets not use the15 amp continuous calculated load
? lets look at the article 220, 12 amp lighting load
12 amp load 50% adjusted 12 / .50 = you have 24 amp adjusted conductors ampacity for application to table 310.16
? You need a 75 Deg C. conductor that can handle 24 amps

 

[david luchini]

? 12 amps continuous loading
? 75 deg C. overcurrent device terminal rating
? Type THNN conductors
? Twelve current-carrying copper conductors in a raceway
? Branch-circuit conductors shall have an ampacity not less than the maximum load to be served.
? lets look at the article 220, 12 amp lighting load
12 amp load 50% adjusted 12 / .50 = you have 24 amp adjusted conductors ampacity for application to table 310.16
? You need a 75 Deg C. conductor that can handle 24 amps

You may have overlooked that I am using THHN, which you will see in Table 310.16 is a 90degC conductor. I need a 90 deg C conductor that can handle 24 amps before applying adjustment factors. #14 Awg THHN fits the bill - it has an ampacity of 25.

But the final circuit will have a 15A c/b with a conductor having an ampacity of 12.5, which you previously stated would not be a proper 15A branch circuit. :?

BTW, you have referenced a Fine Print Note (FPN) in your last two posts. FPNs are "Explanatory Material" and are not enforceable as requirements of the Code. See 90.5(C).

 

[david]

I'm afraid I have to disagree with Charlie here (sorry Charlie - Go Irish.) Per 210.19(A)(1), your branch circuit conductor should have an ampacity not less than the maximum load to be served. You need a conductor with an ampacity of not less than 10. However, the minimum branch-circuit conductor size shall not be less than 125% of the continuous load (the load in this case is completely continuous.) So the minimum branch-circuit conductor size would need an ampacity of 12.5 - or #14.
If we had five circuits (10 CCCs) with a load of 10A each, and wanted to use a 75deg rated #14 awg conductor (20A ampacity) and applied the 50% adjustment factor from T310.15(B)(2)(a), then the #14 would have an ampacity of 10. Since the adjusted ampacity of the #14 is not less than the maximum load to be served (10amps), and the #14 is not smaller than the minimum required branch-conductor size, the ten (10) #14 CCCs would be acceptable. Example D3(a) in the Annex has a good example of this.

Example D3(a) in the Annex has a good example of this.

I never really agreed with your application of this example from the beginning

• Example D3(a) Industrial Feeders in a Common Raceway Ungrounded Feeder Conductors
The conductors must independently meet requirements for (1) terminations, and (2) conditions of use throughout the raceway run.
Minimum size conductor at the over-current device termination [see 110.14(C) and 215.2(A)(1), using
• 75?C ampacity column in Table 310.16]: 1/0 AWG.
Minimum size conductors in the raceway based on actual load [see Article 100, Ampacity, and 310.15(B)(2)(a) and correction factors to Table 310.16]:

For continuous loading consider the actual 100% load (article 220 load) and the additional 25% continuous loading as a calculated load for continuous loads.
Double de-rating is avoided when the actual article 220 calculated load in amps is adjusted for conductors in conduit rather than the calculated load for continuous loading

99,000 VA / 0.7 / 0.96 = 147,000 VA
[70% = 310.15(B)(2)(a)] & (0.96 = Correction factors to Table 310.16)
Conversion to amperes:
147,000 VA / (480V ? 3) = 177 A

Notice in this example the ampacity adjustment for raceway fill is applied to the article 220 calculated load. Adjusting the calculated 119.07 amps 70% (310.15(B)(2)(a) adjustment) to 170 adjusted amp to be applied to table 310.16

You may have overlooked that I am using THHN, which you will see in Table 310.16 is a 90degC conductor. I need a 90 deg C conductor that can handle 24 amps before applying adjustment factors. #14 Awg THHN fits the bill - it has an ampacity of 25.

You can use 90 deg C. rated conductors if you want to and I did not over look that you did, it really doesn’t matter unless your terminals are rated 100%. You have to consider the terminal rating of you over current device and it is rated at 75 deg. C. so your limited to the 75 deg ampacity coulomb even though you are using 90 deg C. rated conductors

But the final circuit will have a 15A c/b with a conductor having an ampacity of 12.5, which you previously stated would not be a proper 15A branch circuit. :?

BTW, you have referenced a Fine Print Note (FPN) in your last two posts. FPNs are "Explanatory Material" and are not enforceable as requirements of the Code. See 90.5(C).

Yes I understand that it is a fine print note or an informational note, but the information in the note is correct and what it tells you is the tables are tabulated to give you correct conductors sizes for the loads in amps as calculated in article 220 as adjusted for conditions of use as laid out for example in Table 310.15(B)(2)(a)
Your overcurrent protection device is a 15 amp breaker with 75% C. rated terminals
Your calculated article 220 load in amps is 12 your condition of use adjustment is 50%
Your adjusted condition of use amps is 24 amps to be applied to table 310.16 your overcurrent devise has terminals rated at 75 deg C. so your branch circuit conductor cannot be 14 AWG

Your 90 deg C insulation could handle the heat but you terminals cannot you are limited by the terminal rating

 

[david luchini]

I never really agreed with your application of this example from the beginning

• Example D3(a) Industrial Feeders in a Common Raceway Ungrounded Feeder Conductors
The conductors must independently meet requirements for (1) terminations, and (2) conditions of use throughout the raceway run.
Minimum size conductor at the over-current device termination [see 110.14(C) and 215.2(A)(1), using
• 75?C ampacity column in Table 310.16]: 1/0 AWG.
Minimum size conductors in the raceway based on actual load [see Article 100, Ampacity, and 310.15(B)(2)(a) and correction factors to Table 310.16]:

For continuous loading consider the actual 100% load (article 220 load) and the additional 25% continuous loading as a calculated load for continuous loads.
Double de-rating is avoided when the actual article 220 calculated load in amps is adjusted for conductors in conduit rather than the calculated load for continuous loading

Yes, this is what I said in my first post in this thread, and is exactly what I have been saying all along.

99,000 VA / 0.7 / 0.96 = 147,000 VA
[70% = 310.15(B)(2)(a)] & (0.96 = Correction factors to Table 310.16)
Conversion to amperes:
147,000 VA / (480V ? 3) = 177 A

Notice in this example the ampacity adjustment for raceway fill is applied to the article 220 calculated load. Adjusting the calculated 119.07 amps 70% (310.15(B)(2)(a) adjustment) to 170 adjusted amp to be applied to table 310.16

Yes, and I have done the same thing in the examples I have given. In the last example, the 220 calculated load of 12Amps is the basis for the adjustment for number of ccc's in the raceway, not 15 Amps.

You can use 90 deg C. rated conductors if you want to and I did not over look that you did, it really doesn’t matter unless your terminals are rated 100%. You have to consider the terminal rating of you over current device and it is rated at 75 deg. C. so your limited to the 75 deg ampacity coulomb even though you are using 90 deg C. rated conductors

This is not correct, and I'm not sure what your basis for this idea is. If you refer to 110.14(C) you will see that "Conductors with temperature ratings higher than specified for terminations shall be permitted to by used for ampacity adjustment, correction, or both." In other words, I can use the 90deg ampacity (of 25 for #14 THHN) for my adjustment for 12 current carrying conductors in the same raceway, on my circuit using 75 deg terminations. The adjusted ampacity is not allowed to be higher than the unadjusted ampacity of #14 @ 75deg, which is 20. But as you can see, the adjusted ampacity of 12.5 (25*0.5) is less then 20, so the conductor is acceptable.

Yes I understand that it is a fine print note or an informational note, but the information in the note is correct and what it tells you is the tables are tabulated to give you correct conductors sizes for the loads in amps as calculated in article 220 as adjusted for conditions of use as laid out for example in Table 310.15(B)(2)(a)
Your overcurrent protection device is a 15 amp breaker with 75% C. rated terminals
Your calculated article 220 load in amps is 12 your condition of use adjustment is 50%
Your adjusted condition of use amps is 24 amps to be applied to table 310.16 your overcurrent devise has terminals rated at 75 deg C. so your branch circuit conductor cannot be 14 AWG

Again, see 110.14(C). I need a conductor before adjustment that has an ampacity of at least 25. 110.14(C) permits me to use the ampacity of the higher temperature rating for my ampacity adjustment, so #14 THHN with an ampacity of 25 meets all the conditions for my circuit and can be used therein.

Your 90 deg C insulation could handle the heat but you terminals cannot you are limited by the terminal rating

My c/b terminals are rated to carry 15 Amps non-continuously without overheating, or 12 Amps continuously without overheating. The load I am putting on them is 12 Amps to be operated continuously. How will the terminals overheat? They clearly will not.

 

[david]

This is not correct, and I'm not sure what your basis for this idea is. If you refer to 110.14(C) you will see that "Conductors with temperature ratings higher than specified for terminations shall be permitted to by used for ampacity adjustment, correction, or both." In other words, I can use the 90deg ampacity (of 25 for #14 THHN) for my adjustment for 12 current carrying conductors in the same raceway, on my circuit using 75 deg terminations. The adjusted ampacity is not allowed to be higher than the unadjusted ampacity of #14 @ 75deg, which is 20. But as you can see, the adjusted ampacity of 12.5 (25*0.5) is less then 20, so the conductor is acceptable.

Example D3(a) Industrial Feeders in a Common Raceway

?If the utility corridor was at normal temperatures [(30?C (86?F)], and if the lighting at each building were supplied from the local separately derived system (thus requiring no neutrals in the supply feeders) the raceway result (99,000 VA / 0.8 = 124,000 VA; 124,000 VA / (480V ? 3) = 149 A, or a 1 AWG conductor @ 90?C) could not be used
? because the termination result (1/0 AWG based on the 75?C column of Table 310.16) would become the worst case, requiring the larger conductor.?

David
From the example D39(a) that you sent us too
In other word if the only adjustment was for current caring conductors in a race way and ambient temp adjustments where not a consideration in this example you could not use the 90 deg C rating of the XHHW-2 because of the 75 deg C rating of the equipment.

 

[david]

Yes, and I have done the same thing in the examples I have given. In the last example, the 220 calculated load of 12Amps is the basis for the adjustment for number of ccc's in the raceway, not 15 Amps.

What you are seeing is that the calculated article 220 lighting load before adjustment of current caring conductors in a raceway is 12 amps

What you are not seeing is the 12 amp article 220 calculated load is adjusted 50% because of the Twelve current caring conductors in a raceway

So the new article 220 calculated load is 24amps after adjustment

Your conductors must be sized to carry the total calculated load
24 amps

 

[david]

• 75 deg C rated Over current equipment
• 12 amp lighting load
• THHN Conductors
• 12 current caring conductors in a race way
• 15 amp rated branch circuit

15 amp rated circuit
• Continuous loading consideration ( 12 x 1.25 ) 15 amps
The 15 amp circuit can handle the continuous load
14 AWG ( 20 amp) 75 deg c conductors can handle the continuous load
Good so far, but that was just the first consideration

Now lets look at the worst case current caring conductors in a race way
(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.___________

You made the assumption that the underlined sentence takes you Table 310.15(B)(2)(a) in example D3(a) Industrial Feeders in a Common Raceway the 90 deg. C coulomb was only used to give a .96 adjustment for ambient temperature rather than a .94 ambient temperature adjustment had 75 deg c rated column been used

Note: the advandage of using a conductor with a 90 deg C. insulation in this example was for ambient temp. ampacity adjustment. It had no effect when it came to adjusting ampacity for current carrin conductors in a race way

• We need a Table 310.15(B)(2)(a) 50% adjustment 12 amp calculated lighting load adjusted to a 24 amp for table 310.16 application

For the purpose of adjusting ampacity for current caring conductors in a race way you need to consider the adjusted calculated ampacity 24 amps

So again take your 24 adjusted amps applicable to 310.16 consider the temperature limitation and choice a conductor that can carry 24 amps from the 75 deg C table .

 

[david]

Another example from the hand book for you consideration. The end result is going to be the same. In this example the use 60 deg rated equipment and 75 deg rate conductors.

Note: there is no ambient temp adjustment considerations There is no adjustment advantage from using THWN in this example for the 4 current caring conductors in a race way.

Is it becoming clearer now to you where this idea is being conceived from?


Calculation Example from the hand book
Determine the minimum-size overcurrent protective device and the minimum conductor size for the following circuit:
• 25 amperes of continuous load
• 60?C overcurrent device terminal rating
• Type THWN conductors
• Four current-carrying copper conductors in a raceway
Solution
STEP 1.
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 of the 60?C rating of the overcurrent device terminal, it is necessary to choose a conductor based on the ampacities in the 60?C column of Table 310.16.

The calculated load must not exceed the conductor ampacity. Therefore, an 8 AWG conductor with a 60?C allowable ampacity of 40 amperes is the minimum size permitted.

• Conductors with a higher allowable ampacity based on their insulation temperature rating may be used, but only at a 60?C allowable ampacity.

STEP 3.
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:

The conductor ampacity determined in Step 2 is not the calculated load that is used in this step. It is a calculation that is used to determine a minimum conductor ampacity and circular mil area to ensure sufficient dissipation of heat at the terminals of overcurrent protective devices supplying continuous loads. The calculated load is 25 amperes. The phrase “before the application of any adjustment or correction factors” has the effect of not requiring that the conductor be subjected to a “double-derating” where continuous loads are supplied by a conductor that is also subject to ampacity adjustment, because there are more than three current-carrying conductors in a cable or raceway or is subject to ampacity correction due to the installation being made where the ambient temperature exceeds 86?C

 

[david]

STEP 3.
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:

The conductor ampacity determined in Step 2 is not the calculated load that is used in this step. It is a calculation that is used to determine a minimum conductor ampacity and circular mil area to ensure sufficient dissipation of heat at the terminals of overcurrent protective devices supplying continuous loads. The calculated load is 25 amperes. The phrase ?before the application of any adjustment or correction factors? has the effect of not requiring that the conductor be subjected to a ?double-derating? where continuous loads are supplied by a conductor that is also subject to ampacity adjustment, because there are more than three current-carrying conductors in a cable or raceway or is subject to ampacity correction due to the installation being made where the ambient temperature exceeds 86?C

The math that did not show up when i copied and pasted this from the handbook was
the 25 amp load / .8 (80% correction factor for 4 current caring conductors in the racway)
31.25 amps so a 8AWG from the 60 deg C columb Table 310.16

 

[david luchini]

Example D3(a) Industrial Feeders in a Common Raceway

?If the utility corridor was at normal temperatures [(30?C (86?F)], and if the lighting at each building were supplied from the local separately derived system (thus requiring no neutrals in the supply feeders) the raceway result (99,000 VA / 0.8 = 124,000 VA; 124,000 VA / (480V ? 3) = 149 A, or a 1 AWG conductor @ 90?C) could not be used
? because the termination result (1/0 AWG based on the 75?C column of Table 310.16) would become the worst case, requiring the larger conductor.?

David
From the example D39(a) that you sent us too
In other word if the only adjustment was for current caring conductors in a race way and ambient temp adjustments where not a consideration in this example you could not use the 90 deg C rating of the XHHW-2 because of the 75 deg C rating of the equipment.

Congratulations...You read the sentence "If the utility corridor..." in the Annex D3(a) example, and completely misunderstood what it said. Go back and read the entire example. You will see "Determine the overcurrent protection and conductor size for the feeders in a common raceway, assuming the use of XHHW-2 insulation (90?C)." You will also see "This requires a 2/0 AWG conductor based on the 90?C Column of Table 310.16."

In other words, the example D3(a) gives a correct conductors size of 2/0 AWG from the 90?C Column even though the equipment has a rating of 75?C. This is in direct disagreement with what you stated in red above. What you have misunderstood in the section of the example that you have quoted is this: If there was no temperature correction required, and the adjustment for number of current carrying conductors was for 6 ccc's instead of 8 ccc's, then a #1AWG 90?C would have an ampacity that could handle the load (remember the load current in D3(a) is 119A, and #1AWG 90?C adjusted for 6 ccc's is 150*0.8=120A), but since #1 is smaller than the #1/0AWG required for the non-continuous load plus 125% of the continuous load before the application of any adjustment factors, #1 cannot be used.

What you are seeing is that the calculated article 220 lighting load before adjustment of current caring conductors in a raceway is 12 amps

What you are not seeing is the 12 amp article 220 calculated load is adjusted 50% because of the Twelve current caring conductors in a raceway

I suspect that I am not seeing this, because it is not correct. The load is not adjusted at all, the ampacity of the conductor is adjusted per 310.15(B)(2)(a)..."the allowable ampacity of each conductor shall be reduced..."

So the new article 220 calculated load is 24amps after adjustment

Your conductors must be sized to carry the total calculated load
24 amps

The load is 12Amps. It is 12Amps all day long, from the time I turn the lights on in the morning until the time I turn them off at night. If you could explain how 1440 Watts of lighting at 120V will give a load of 24 Amps, I'd love to hear it.

The conductor must have an ampacity of 12 AFTER being adjusted for number of ccc's in the raceway. I have done this in my example. The unadjusted ampacity of the conductor is 25. The adjusted ampacity of the conductor is 25*0.5=12.5. A conductor with an ampacity of 12.5 is large enough to carry a load of 12 Amps. I don't see that there is anything confusing about 12.5 being larger than 12.

 

[david luchini]

? 75 deg C rated Over current equipment
? 12 amp lighting load
? THHN Conductors
? 12 current caring conductors in a race way
? 15 amp rated branch circuit

15 amp rated circuit
? Continuous loading consideration ( 12 x 1.25 ) 15 amps
The 15 amp circuit can handle the continuous load
14 AWG ( 20 amp) 75 deg c conductors can handle the continuous load
Good so far, but that was just the first consideration

Yes, so far so good. This is the example that I laid out. The minimum allowed conductor size is #14 AWG per the second sentence of 210.19(A)(1). Just as I said in Post #61 where I proposed this example.

Now lets look at the worst case current caring conductors in a race way
(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.

You made the assumption that the underlined sentence takes you Table 310.15(B)(2)(a) in example D3(a) Industrial Feeders in a Common Raceway. The 90 deg. C coulomb was only used to give a .96 adjustment for ambient temperature rather than a .94 ambient temperature adjustment had 75 deg c rated column been used.

Note: the advandage of using a conductor with a 90 deg C. insulation in this example was for ambient temp. ampacity adjustment. It had no effect when it came to adjusting ampacity for current carrin conductors in a race way.

This is NOT CORRECT. Read the underlined sentence again. It says "...used for AMPACITY ADJUSTMENT, CORRECTION, or BOTH."

The AMPACITY CORRECTION is for the ambient temperature given in Table 310.16.
The AMPACITY ADJUSTMENT is for the number of current carrying conductors given in Table 310.15(B)(2)(a)
BOTH means that BOTH the ccc's adjustment factor AND the temperature correction factor can be based on the 90?C column.

If the only benefit of using the 90?C column in Example D3(a) was the smaller temperature correction factor (0.96 vs. 0.94) and not the higher rated ampacity of the #2/0AWG conductor (195 Amp vs. 175 Amp,) then the ampacity of the #2/0 conductor would have been 175A (2/0 @ 75?C) * 0.7 (adjustment factor for 8 ccc's) * 0.96 (temp. correction factor from 90?C column.)

......175A * 0.7 * 0.96 = 117.6A.

The load in example D3(a) was 119 Amps. A conductor with an ampacity of 117.6 would be too small for the load. But since the benefit of using the 90?C column in example D3(a) is BOTH the smaller temperature correction factor AND the higher conductor ampacity, the #2/0 AWG XHHW-2 works in the example.

......195A * 0.7 * 0.96 = 131.0A.

? We need a Table 310.15(B)(2)(a) 50% adjustment 12 amp calculated lighting load adjusted to a 24 amp for table 310.16 application

For the purpose of adjusting ampacity for current caring conductors in a race way you need to consider the adjusted calculated ampacity 24 amps

So again take your 24 adjusted amps applicable to 310.16 consider the temperature limitation and choice a conductor that can carry 24 amps from the 75 deg C table .

I don't know how many different ways I can come up with to say the same thing. You quoted the section above that allows you to use a conductor with HIGHER TEMPERATURE RATINGS than the terminations for adjustment factors yourself. But then you chose to ignore that section.

The load is 12 Amps,
the terminations of my equipment are rated for 75?C,
110.14(C) permits me to use conductors with higher temperature ratings for my ampacity adjustment,
#14 AWG THHN (rated 90?C) has an ampacity of 25 Amps,
applying the 0.5 adjustment factor to this gives me an ampacity of 12.5 Amps,
the 12.5 Amps adjusted ampacity of the #14 AWG THHN is higher than the maximum load it will carry (12 Amps),
therefore, the #14 AWG THHN conductor is acceptable for the load.

It's not anymore complicated than that.

 

[david luchini]

Another example from the hand book for you consideration. The end result is going to be the same. In this example the use 60 deg rated equipment and 75 deg rate conductors.

Note: there is no ambient temp adjustment considerations There is no adjustment advantage from using THWN in this example for the 4 current caring conductors in a race way.

Is it becoming clearer now to you where this idea is being conceived from?


Calculation Example from the hand book
Determine the minimum-size overcurrent protective device and the minimum conductor size for the following circuit:
• 25 amperes of continuous load
• 60?C overcurrent device terminal rating
• Type THWN conductors
• Four current-carrying copper conductors in a raceway

This example is WORTHLESS to our discussion.

The load is 25Amps. Since it is continuous, the second sentence of 210.19(A)(1) requires a minimum conductor size with an ampacity of 25*1.25 = 31.25. In the 60 deg column, this is #8 AWG. Therefore #8 is the smallest allowable conductor.

210.20 requires that the minimum OCPD for this circuit will be 25*1.25 = 31.25, or 35 Amp would be the smallest allowable circuit breaker or fuse.

There are 4 ccc's in your raceway, so you must apply an adjustment factor of 0.8. The ampacity of #8 at 60?C is 40. Adjusting the 40Amp ampacity for 4 ccc's gives you 40*0.8=32. The load is 25 - 32 is larger than 25, so #8 TW conductor would be acceptable for the application. Your example lists THWN (75?C) as the conductor. This is acceptable, but gives you no benefit, as the 60?C conductor is already acceptable for the load. Likewise, you could use THHN (90?C) as the conductor, but without any benefit.

A #10 AWG THHN conductor would give you the correct adjusted ampacity for the load (40*0.8=32), but #10 THHN would NOT be acceptable for the circuit because:
1) #10 is smaller than the minimum conductor size established by the second sentence of 210.19(A)(1), and
2) #10 connect be protected by the minimum required 35A OCPD, per 240.4(D).

Even if you changed the termination temp in this example to 75?C, to match the 75?C conductor, such that a #10 AWG THWN would be the minimum conductor size per 210.19(A)(1), you would still need a #8 AWG conductor because of 240.4(D).

So in this example, you start with a #8 conductor, and end with a #8 conductor.