Kindly advise 215.2 (A) (1)

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m sleem

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I'm confused regarding 215.2 (A) (1). How to size feeders that less than 600v based on the continuity of the load (1.25*continous+continous) regardless the correction factor (catalog ampacity only). It means when we have correction factor 0.5 & according to continuity rule the conrinuity is calculated after demand, so the actual ampacity will be less than the demand. Is it ok?




 

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Smart $

Esteemed Member
Location
Ohio
I'm confused regarding 215.2 (A) (1). How to size feeders that less than 600v based on the continuity of the load (1.25*continous+continous) regardless the correction factor (catalog ampacity only). It means when we have correction factor 0.5 & according to continuity rule the conrinuity is calculated after demand, so the actual ampacity will be less than the demand. Is it ok?
Generally (with emphasis), conductor sizing is a four-step process...

1) Minimum size of conductor for terminal temperature limitations (using calculated load value; does not include 125% factoring for continuous loads).

2) Minimum ampacity of conductor before adjustment and correction must equal or exceed 125% continuous plus noncontinuous sum of calculated load.

3) Minimum ampacity of conductor after adjustment and correction must equal or exceed calculated load (does not include 125% factoring for continuous loads).

4) The overcurrent protection device must have a rating not less than 125% continuous plus noncontinuous sum of calculated load. The adjusted and corrected conductor ampacity must at a minimum be greater than next smaller standard overcurrent device rating [240.4(B); does not apply in all instances]. Where 240.4(B) does not apply, the adjusted and corrected conductor ampacity must equal or exceed overcurrent protection rating.

Ultimately, the chosen conductor must meet all four requirements.


EDIT to add: Calculated load includes demand factoring.
 
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m sleem

Senior Member
Location
Cairo
Generally (with emphasis), conductor sizing is a four-step process...

1) Minimum size of conductor for terminal temperature limitations (using calculated load value; does not include 125% factoring for continuous loads).

2) Minimum ampacity of conductor before adjustment and correction must equal or exceed 125% continuous plus noncontinuous sum of calculated load.

3) Minimum ampacity of conductor after adjustment and correction must equal or exceed calculated load (does not include 125% factoring for continuous loads).

4) The overcurrent protection device must have a rating not less than 125% continuous plus noncontinuous sum of calculated load. The adjusted and corrected conductor ampacity must at a minimum be greater than next smaller standard overcurrent device rating [240.4(B); does not apply in all instances]. Where 240.4(B) does not apply, the adjusted and corrected conductor ampacity must equal or exceed overcurrent protection rating.

Ultimately, the chosen conductor must meet all four requirements.


EDIT to add: Calculated load includes demand factoring.
Is there any difference between 1 & 3 ?
 

m sleem

Senior Member
Location
Cairo
Please check this example:

Continous load is 400A
Non-continous load is 250A

thus:
1-min ampacity (before correction) will be 750A

2-min ampacity (after correction) will be 600A, where: correction factor is 0.92
So, next higher size will be 700A

3-OCPD will be 800A

We shall select between 1&2, which one is higher

So, the designed cable ampacity will be 750A, where 240.4(B) is applicable.
 

Smart $

Esteemed Member
Location
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Please check this example:

Continous load is 400A
Non-continous load is 250A

thus:
1-min ampacity (before correction) will be 750A

2-min ampacity (after correction) will be 600A, where: correction factor is 0.92
So, next higher size will be 700A

3-OCPD will be 800A

We shall select between 1&2, which one is higher

So, the designed cable ampacity will be 750A, where 240.4(B) is applicable.
Assuming 600V or less here, and no requirements or design issues regarding oversizing of OCPD for issues such as motor starting currents...

1) 75?C rated terminations, minimum copper conductor size for 650A is 1750kcmil (or 2 sets of 400kcmil at 335A each).

2) 125% ? 400A + 250A = 750A. 90?C-rated copper conductor minimum size is 2000kcmil (or 2 sets of 400kcmil at 380A each)

3) 650A ? 0.92 = 707A. 90?C-rated copper conductor minimum size is 1750kcmil (or 2 sets of 400kcmil at 380A each).

4) 800A OCPD. 240.4(B) applies. Adjusted and corrected ampacity can be as low as 701A. 701 ? 0.92 = 762A. 90?C-rated copper conductor minimum size exceeds 2000kcmil. Will have to use 500kcmil parallel 90?C copper conductors at a combined pre-adjusted/corrected ampacity of 430A each.

The minimum size 90?C-rated copper conductor satisfying all four requirements is 2 sets of 500kcmil. Increasing number of sets will reduce the minimum required size.
 
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david luchini

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Assuming 600V or less here, and no requirements or design issues regarding oversizing of OCPD for issues such as motor starting currents...

1) 75?C rated terminations, minimum copper conductor size for 650A is 1750kcmil (or 2 sets of 400kcmil at 335A each).

2) 125% ? 400A + 250A = 750A. 90?C-rated copper conductor minimum size is 2000kcmil (or 2 sets of 400kcmil at 380A each)

3) 650A ? 0.92 = 707A. 90?C-rated copper conductor minimum size is 1750kcmil (or 2 sets of 400kcmil at 380A each).

4) 800A OCPD. 240.4(B) applies. Adjusted and corrected ampacity can be as low as 701A. 701 ? 0.92 = 762A. 90?C-rated copper conductor minimum size exceeds 2000kcmil. Will have to use 500kcmil parallel 90?C copper conductors at a combined pre-adjusted/corrected ampacity of 430A each.

The minimum size 90?C-rated copper conductor satisfying all four requirements is 2 sets of 500kcmil. Increasing number of sets will reduce the minimum required size.

I'm of the opinion, from 110.14(C), that Item #2 should be based on the 75?C rating, 2 sets of 500kcmil would be the minimum allowable (Cu) conductor size.
 

Smart $

Esteemed Member
Location
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I'm of the opinion, from 110.14(C), that Item #2 should be based on the 75?C rating, 2 sets of 500kcmil would be the minimum allowable (Cu) conductor size.
Item #2 is regarding compliance with 215.2(A)(1), as pictured in the OP. What in 110.14(C) makes you believe you have to use 75?C rating for 215.2 compliance when the conductor is 90?C rated?
 

david luchini

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Item #2 is regarding compliance with 215.2(A)(1), as pictured in the OP. What in 110.14(C) makes you believe you have to use 75?C rating for 215.2 compliance when the conductor is 90?C rated?

110.14(C)(1) (effectively) says that for a 75?C termination rating, you should use conductors rated 75?C, or conductors with higher temperature ratings, provided that the ampacity of such conductors does not exceed the 75?C ampacity of the conductor size used.

The 75?C ampacity of 2 sets of 400mcm Cu is only 670A, but 215.2(A)(1) requires the minimum conductor size, before any adjustment or correction factor, to have an ampacity of 750, in this case.

110.14(C) permits the conductors with higher temperature ratings than the termination to be used for ampacity adjustment, correction, or both. As the minimum conductor size requirement from 215.2(A)(1) is being done before any adjustment or correction factor, it should be based on the 75?C rating.

The Annex Example D3(a) reflects this approach. It has a total load of 99,000VA (119A at 480V). The minimum conductor size based on 125% of continuous plus 100% of non-continuous must have an ampacity of not less than 136A. They tell us that this is 1/0AWG. (Based on 90?C, the min. conductor would be #1AWG.) The example then goes on to apply the adjustment and correction factors to the 90?C ampacity of the XHHW-2 conductors that they are planning to use.
 

augie47

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Maybe I'm looking at it wrong, or just differently.
To me, per 215.2 the minimum amapcity, before adjustment would be 750 amps (125% continuous + non continuous).
Assuming 75? termination the final conductor would need to be no less than 750 amps at it's 75? rating.
So for 750 amps of calculated load, the minimum conductor would be 375 amps for a two conductor run, in this case (2)
500s Cu,
If we have a derting factor of .92 that could be applied to the 90? rating provided the conductor has 90? insulation. In the case of 500 Cu we have 90? of 430 with a derated ampacity of 395.6, still above the 375 limitation.

I thinik this is along with what David is stating.
 

david luchini

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If we have a derting factor of .92 that could be applied to the 90? rating provided the conductor has 90? insulation. In the case of 500 Cu we have 90? of 430 with a derated ampacity of 395.6, still above the 375 limitation.

I thinik this is along with what David is stating.

Yes, that's basically what I'm saying, except that the adjusted ampacity from the 90? rating wouldn't need to be above the 375 limitation. It would need to be above the 325A limitation from Smart$'s line #3 (based on the actual load amperage) AND above the 350.5A limitation from Smart$'s line #4 (based on protecting the conductor with the next size up OCPD.) So for this example, the minimum conductor size would be 500mcm, and the minimum adjusted ampacity would be 350.5A per parallel set.
 

Smart $

Esteemed Member
Location
Ohio
110.14(C)(1) (effectively) says that for a 75?C termination rating, you should use conductors rated 75?C, or conductors with higher temperature ratings, provided that the ampacity of such conductors does not exceed the 75?C ampacity of the conductor size used.

The 75?C ampacity of 2 sets of 400mcm Cu is only 670A, but 215.2(A)(1) requires the minimum conductor size, before any adjustment or correction factor, to have an ampacity of 750, in this case.

110.14(C) permits the conductors with higher temperature ratings than the termination to be used for ampacity adjustment, correction, or both. As the minimum conductor size requirement from 215.2(A)(1) is being done before any adjustment or correction factor, it should be based on the 75?C rating.

The Annex Example D3(a) reflects this approach. It has a total load of 99,000VA (119A at 480V). The minimum conductor size based on 125% of continuous plus 100% of non-continuous must have an ampacity of not less than 136A. They tell us that this is 1/0AWG. (Based on 90?C, the min. conductor would be #1AWG.) The example then goes on to apply the adjustment and correction factors to the 90?C ampacity of the XHHW-2 conductors that they are planning to use.
Well the D3(a) Example certainly makes me question my approach... but in re-reading 110.14(C)(1)(b)(2) and 215.2(A)(1) several times, I see nothing that says the D3(a) Example is implemented correctly. It seems the Example combines two requirements into one, when they are independent of each other.

Nothing in 110.14(C) says its determination must use noncontinuous plus 125% continuous current in establishing ampacity values. The general statement specifically says
Conductors with temperature ratings higher than specified for terminations shall be permitted to be used for ampacity adjustment, correction, or both.
215.2(A)(1) specifically says
The minimum feeder-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.
What advantage would there be to establishing a minimum conductor size before the application of any adjustment or correction factors if the size after the application of adjustment and correction must still be not less than the noncontinuous load plus 125 percent of the continuous load???
 

Smart $

Esteemed Member
Location
Ohio
...
The Annex Example D3(a) reflects this approach. It has a total load of 99,000VA (119A at 480V). The minimum conductor size based on 125% of continuous plus 100% of non-continuous must have an ampacity of not less than 136A. They tell us that this is 1/0AWG. (Based on 90?C, the min. conductor would be #1AWG.) The example then goes on to apply the adjustment and correction factors to the 90?C ampacity of the XHHW-2 conductors that they are planning to use.
Round 2... :p

I think the Example combines 110.14(C) and 215.2(A)(1) requirements simply because it usually works out to making no difference. I'll work the Example using my steps but out of order and cascade the determined current values from preceding steps rather than establishing a minimum conductor size for each step...

4) OCP minimum calculated 136A in Example. Next standard size 150A. Next lower standard size OCPD is 125A. Minimum conductor ampacity afforded protection is 126A [240.4(B)].

3) Minimum conductor ampacity based on actual load must be 177A before adjustment and correction per Example. Using XHHW-2 copper conductors, the minimum size must be 2/0 AWG (table ampacity 195A @ 90?C; next smaller size is 1/0 AWG rated 170A, so too small).

2) & 1) Determine compliance with 110.14(C) and 215.2(A)(1)... 75?C column ampacity of 2/0 AWG is 175A,which is greater than noncontinuous load plus 125% continuous load (136A; see minimum value in Step 4). 2/0 AWG is in compliance.

:thumbsup:
 
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Smart $

Esteemed Member
Location
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4) OCP minimum calculated 136A in Example. Next standard size 150A. Next lower standard size OCPD is 125A. Minimum conductor ampacity afforded protection is 126A [240.4(B)].

3) Minimum conductor ampacity based on actual load must be 177A before adjustment and correction per Example. Using XHHW-2 copper conductors, the minimum size must be 2/0 AWG (table ampacity 195A @ 90?C; next smaller size is 1/0 AWG rated 170A, so too small).

...
Note if we did it completely my way backwards, we would actually use the 126A value from Step 4, and back figure the before adjustment and correction ampacity from that value rather than from the 119A/99000VA used in the Example.

Example: 99,000 VA/0.7/0.96/(480V ? √3)= 177 A

Mine: 126A/0.7/0.96= 188A

2/0 AWG still good!!!
 

david luchini

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Nothing in 110.14(C) says its determination must use noncontinuous plus 125% continuous current in establishing ampacity values.

Of course not, that requirement is specific to 215.2(A)(1), and isn't about establishing an ampacity value, but establishing a minimum conductor size.

215.2(A)(1) specifically says

The minimum feeder-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.

Yes, and 110.14(C) says that 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.

The lowest temperature rating in the example circuit is 75?, so the temperature rating associated ampacity of the (400mcm in your example) conductors should be 75?, which gives the two sets of 400mcm an ampacity of 670 instead of 760 (before the application of any adjustment or correction factor.)

What advantage would there be to establishing a minimum conductor size before the application of any adjustment or correction factors if the size after the application of adjustment and correction must still be not less than the noncontinuous load plus 125 percent of the continuous load???

It needn't still be "not less than the noncontinuous load plus 125 percent of the continuous load". After the application of any adjustment or correction factor (for which 110.14(C) permits to use the ampacity associated with the higher temperature rating), 215.2(A)(1) tells us that the ampacity must only be large enough to supply the load (or 650A in the OP.) And 240.4 tells us that the conductor must be protected in accordance with their ampacities (or next size up in this case) so 701A would be the minimum allowable ampacity for an 800A OCPD.
 

david luchini

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I think the Example combines 110.14(C) and 215.2(A)(1) requirements simply because it usually works out to making no difference.

It may usually work out...but lets make a couple of changes to the original post and apply your method.

Lets go with 300A non-continuous and 320A continuous load, and the same 0.92 correction factor. This would give us...

1) 75?C rated terminations, minimum copper conductor size for 620A is 2 sets of 350kcmil at 310A each.

2) 125% ? 320A + 300A = 700A. 90?C-rated copper conductor minimum size is 2 sets of 350kcmil at 350A each.

3) 620A ? 0.92 = 674A. 90?C-rated copper conductor minimum size is 2 sets of 350kcmil at 350A each.

4) 700A OCPD. 240.4(B) applies. Adjusted and corrected ampacity can be as low as 601A. 601 ? 0.92 = 653A. 90?C-rated copper conductor minimum size is 350kcmil parallel at a combined pre-adjusted/corrected ampacity of 350A each.

The 90?C-rated copper conductor satisfying all four requirements is 2 sets of 350kcmil. But the 75?C rated terminations will overheat with the continuous load. The 125% factor for the continuous load should have the effect of oversizing the terminations so that they can handle the extra heat from the continuous loading.

In the example above, however, we would allow the 350kcmil lugs, negating the effects of adding in the 25% of continuous load. [/QUOTE]
 
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Smart $

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Location
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Of course not, that requirement is specific to 215.2(A)(1), and isn't about establishing an ampacity value, but establishing a minimum conductor size.
I agree.

Yes, and 110.14(C) says that 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.

The lowest temperature rating in the example circuit is 75?, so the temperature rating associated ampacity of the (400mcm in your example) conductors should be 75?, which gives the two sets of 400mcm an ampacity of 670 instead of 760 (before the application of any adjustment or correction factor.)
And the load is 650A...
...
3) 650A ? 0.92 = 707A. 90?C-rated copper conductor minimum size is 1750kcmil (or 2 sets of 400kcmil at 380A each).
...
I just showed in my Step 3 the reverse of 110.14(C) permitting use a higher rated conductor for adjustment and correction. Unnecessary for 110.14(C) compliance, but necessary for 215.2(A)(1) first sentence requirement, which is what Step 3 is concerning.

Nevertheless, 2 sets of 400kcmil (335A each at 75?C rating) meets 110.14(C) requirement:

2 ? 335A = 670A

670A
> 650A
...as stated in Step 1
 
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david luchini

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I agree.


And the load is 650A...

But the second part of 215.2(A)(1) doesn't say anything about figuring the ampacity based on the load. It says the minimum conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125% of the continuous load. That is 750A. The minimum size conductor shall have an allowable ampacity not less than 750A selected and coordinated so as not to exceed the lowest temperature rating of any connected termination, conductor or device (in this case 75?C.) The conductor which does that is two sets of 500mcm.
 
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Smart $

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Location
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But the second part of 215.2(A)(1) doesn't say anything about figuring the ampacity based on the load. It says the minimum conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125% of the continuous load. That is 750A. The conductor shall have an allowable ampacity not less than 750A selected and coordinated so as not to exceed the lowest temperature rating of any connected termination, conductor or device (in this case 75?C.) The conductor which does that is two sets of 500mcm.
But where does 215.2(A)(1) say the second sentence applies to 110.14(C) determination? The subject of the first sentence is ampacity. The subject of the second sentence is size. 110.14(C) requirements are regarding ampacity... and says nothing regarding size.

215.2 Minimum Rating and Size.

(A) Feeders Not More Than 600 Volts.

(1) General.
Feeder conductors shall have an ampacity not
less than required to supply the load as calculated in Parts
III, IV, and V of Article 220.
The minimum feeder-circuit
conductor size, before the application of any adjustment or
correction factors, shall have an allowable ampacity not
less than the noncontinuous load plus 125 percent of the
continuous load.

Compliance with the second sentence of 215.2(A)(1) is accomplished in Step 2. Perhaps I should have reversed Steps 2 and 3... ???
 
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