Double de-rating?

[aelectricalman]

If the correction factor for conduit fill and continuous load exists, do you have to correct them both? Does the NEC make provisions so you dont have correct twice?
The two correction factors are both aimed at reducing heat around conductors so it seems only the most stringent of the two would apply. How do you handle this instance.

My example is parking lot lighting. I have voltage drop, conductor deration for more than 3 current carrying conductors and continuos load deration.

Happy new year. Hope you all had a good Christmas.

 

[kwired]

Deration for continuous load in general is going to effect the minimum size conductor at 60 or 75 deg for the terminations.

Deration for number of conductors in the raceway is going to effect the minimum size conductor for insulation purposes at 90 deg for the insulation.

You need to find minimum size needed for terminations and minimum size needed for insulation - which ever results in a larger conductor is the minimum size you need.

Do not take any deration into consideration for voltage drop.

Find out what minimum size conductor is needed and if that allows too much voltage drop then increase conductor size accordingly. No minimum ampacity calculations are needed for voltage drop considerations you are already larger than minimum as far as overcurrent protection is concerned.

 

[kwired]

Deration for continuous load in general is going to effect the minimum size conductor at 60 or 75 deg for the terminations.

Deration for number of conductors in the raceway is going to effect the minimum size conductor for insulation purposes at 90 deg for the insulation.

You need to find minimum size needed for terminations and minimum size needed for insulation - which ever results in a larger conductor is the minimum size you need.

Do not take any deration into consideration for voltage drop.

Find out what minimum size conductor is needed and if that allows too much voltage drop then increase conductor size accordingly. No minimum ampacity calculations are needed for voltage drop considerations you are already larger than minimum as far as overcurrent protection is concerned.

 

[kwired]

I did not intentionally post same thing twice:?

 

[aelectricalman]

Thanks for input.

Sure you didn't.

Sure you didn't.

 

[aelectricalman]

I need a sanity check, please.

I need a sanity check, please.

If I have the following:

4 208 V lighting circuits in one PVC scedule 40 conduit
120/208 3phase 4 wire system
550 ft one directional distance
10 amps utilization continuous load
8 current carrying conductors ( two on each circuit)
30deg C AT
Copper
THHN 90 deg
Terminals 75 deg
max 5 % VD

What would your conductor size be?

I have my answer but I want to check my math given the above situation. Any help would be awesome!

 

[Smart $]

If the correction factor for conduit fill and continuous load exists, do you have to correct them both? Does the NEC make provisions so you dont have correct twice?
The two correction factors are both aimed at reducing heat around conductors so it seems only the most stringent of the two would apply. How do you handle this instance.

My example is parking lot lighting. I have voltage drop, conductor deration for more than 3 current carrying conductors and continuos load deration.

Happy new year. Hope you all had a good Christmas.

A very common mistake... but they apply independently. Refer to 210.19(A) and 215.2(A)(1). Allowable ampacity of conductors shall be not less than noncontinuous plus 125% continuous load, before the application of any adjustment or correction factors.

 

[kwired]

I come up with min conductor of 12.5 amps needed 10 amps x 1.25 for continuous load.

I then come up with deration for 8 ccc (90 deg conductors) requiring about 14.25 amp conductor.

14.25 is larger of two and is minimum size needed before considering voltage drop.

14 AWG copper is all that is needed.

If you want to maintain 5% max voltage drop @ 10 amps of load then I came up with needing 8 AWG copper.

 

[aelectricalman]

A very common mistake... but they apply independently. Refer to 210.19(A) and 215.2(A)(1). Allowable ampacity of conductors shall be not less than noncontinuous plus 125% continuous load, before the application of any adjustment or correction factors.

By using both in conjunction I would need #6 wires versus, using the larger of the two and having #8. It just seems a little excess to run #6 550 just to get a 10amps draw under 5% VD.

 

[aelectricalman]

Thats perfect. Exactly what I was thinking. I like to trust my mind but its good to verify every once in a while. Thank you all for the great help.

 

[Smart $]

If I have the following:

4 208 V lighting circuits in one PVC scedule 40 conduit
120/208 3phase 4 wire system
550 ft one directional distance
10 amps utilization continuous load
8 current carrying conductors ( two on each circuit)
30deg C AT
Copper
THHN 90 deg
Terminals 75 deg
max 5 % VD

What would your conductor size be?

I have my answer but I want to check my math given the above situation. Any help would be awesome!

Is it possible to put the lighting on a 3? circuit? A 3? circuit is 15% more efficient compared to individual 1? circuits... and you wouldn't have to derate for excess of 3 conductors.

Edit to modify: I should have said two (2) 3? circuits... 6 conductors at 11.55A each if completely balanced.

 

[kwired]

Is it possible to put the lighting on a 3? circuit? A 3? circuit is 15% more efficient compared to individual 1? circuits... and you wouldn't have to derate for excess of 3 conductors.

Edit to modify: I should have said two (2) 3? circuits... 6 conductors at 11.55A each if completely balanced.

15% more efficient? Outside of line losses the load will consume the same amount of energy.

How did you come up with 11.55 Amps? 3 balanced 10 amp loads should draw 17.32 amps per phase.

 

[Smart $]

15% more efficient? Outside of line losses the load will consume the same amount of energy.

How did you come up with 11.55 Amps? 3 balanced 10 amp loads should draw 17.32 amps per phase.

I'm talking line current efficiency.

For example:

(3) 10A 208V 1? ckts you need 6 wires. 6240VA total
(2) 8.66A 208V 3? ckts you need 6 wires. 6240VA total

15% more efficient... which can substantially offset voltage drop issues compared to running 1? circuits.

However, in this case we have (4) 10A 208V 1? ckts which is 8320VA...

8320VA?208V?√3?2ckts=11.55A

Check voltage drop for these 3? circuits? I'm getting 3.6% using #8.

 

[charlie b]

I come up with min conductor of 12.5 amps needed 10 amps x 1.25 for continuous load. I then come up with deration for 8 ccc (90 deg conductors) requiring about 14.25 amp conductor. 14.25 is larger of two and is minimum size needed before considering voltage drop.

I do not agree with this approach, and believe instead that both factors (not just the larger of the two) must be taken into account. You need a wire that has an ampacity of 12.5. With 8 CCCs in a conduit, whatever wire you pick will have to be derated to 70% of its tabulated ampacity. So if 70% of the tabulated value must exceed 12.5, then the tabulated value must exceed 12.5 divided by 70%, or 17.8. The smallest wire that fits that description is indeed #14. So I agree with your result, just not the approach that brought you there.

If you want to maintain 5% max voltage drop @ 10 amps of load then I came up with needing 8 AWG copper.

I agree with this result.

 

[charlie b]

If the correction factor for conduit fill and continuous load exists, do you have to correct them both? Does the NEC make provisions so you dont have correct twice?

You had me confused initially, but then I think I understood your question. So please allow me to split hairs for a moment. The derating that is related to the number of CCCs in a raceway is an ?adjustment factor,? not a ?correction factor,? and it is not the same as the issue of ?conduit fill.? Secondly, adding 25% to the current taken up by a continuous load is also not at ?correction factor.? Nor for that matter is it an ?adjustment factor,? and nor does it constitute ?derating.? It has no name; it?s just part of the calculation process. I know these are just a minor language issues, but sometimes it is important to be precise in our use of terms, especially when we are dealing with complex questions.

Finally, we do ?correct? twice in this situation, and in others. If you had thrown a higher ambient temperature into the mix, then we would have to correct three times.

 

[kwired]

I'm talking line current efficiency.

For example:

(3) 10A 208V 1? ckts you need 6 wires. 6240VA total
(2) 8.66A 208V 3? ckts you need 6 wires. 6240VA total

15% more efficient... which can substantially offset voltage drop issues compared to running 1? circuits.

However, in this case we have (4) 10A 208V 1? ckts which is 8320VA...

8320VA?208V?√3?2ckts=11.55A

Check voltage drop for these 3? circuits? I'm getting 3.6% using #8.

???


We have (4) 10 amp 208 volt circuits. That is 2080 VA each. You can balance three of them across all three phases. Balanced VA is 6240. 6240/208/1.732 = 17.32 A.

That portion of the total will draw 17.32 amps per phase.

You will have 10 more amps on the two phases you connect the fourth load to.

The feeder to all of this carries that no matter what as long as you balance as much as possible.

8320VA?208V?√3?2ckts=11.55A

You can not divide total VA by voltage when load is not balanced and get amps per phase. I have no idea where the "?2ckts" is coming from. I have no idea where you got this from and am assuming you made some kind of mistake here. I know you better than that. Or maybe I made a mistake - set me straight if so.

 

[kwired]

I do not agree with this approach, and believe instead that both factors (not just the larger of the two) must be taken into account. You need a wire that has an ampacity of 12.5. With 8 CCCs in a conduit, whatever wire you pick will have to be derated to 70% of its tabulated ampacity. So if 70% of the tabulated value must exceed 12.5, then the tabulated value must exceed 12.5 divided by 70%, or 17.8. The smallest wire that fits that description is indeed #14. So I agree with your result, just not the approach that brought you there.
I agree with this result.

My mistake. I have a bad habit of not applying 125% of continuous load when applying derating factors, and I don't know where I got that from. It clearly states in 210.19, 215.2, and maybe elsewhere that this applies before application of adjustments or correction factors. It works out a lot of the time anyway as long as you don't get into the 50% or more deration factors. So yes 12.5 amps min needed. For 8 CCC 90 deg conductors they would need to be 17.8 amps min not 14.25.

My main point was to emphasize that you must figure out what size conductor is needed to satisify the 60 or 75 deg termination, and you also need to figure out what size conductor is needed for the 90 degree insulation value of the conductors in the raceway. (Unless you are not using 90 degree conductors.) Which ever result gives you a larger conductor is the minimum size conductor needed.

 

[Smart $]

???


We have (4) 10 amp 208 volt circuits. That is 2080 VA each. You can balance three of them across all three phases. Balanced VA is 6240. 6240/208/1.732 = 17.32 A.

That portion of the total will draw 17.32 amps per phase.

You will have 10 more amps on the two phases you connect the fourth load to.

The feeder to all of this carries that no matter what as long as you balance as much as possible.


You can not divide total VA by voltage when load is not balanced and get amps per phase. I have no idea where the "?2ckts" is coming from. I have no idea where you got this from and am assuming you made some kind of mistake here. I know you better than that. Or maybe I made a mistake - set me straight if so.

The OP says these are lighting circuits. It could be that each individual luminaire is a 10A load. However, it could also be several to many luminaires combined to make the 10A load per circuit. That is why I asked if putting on 3? cirucits were possible. What if these four proposed circuits actually powered 36 ballasts? You could arrange these 6-6-6 on two (2) 3? circuits.

Regarding the "?2ckts" ,,,,,,, the efficiency comparison is made using two 3? circuits because it has the same number of wires as three 1? circuits. And the line current of two balanced 3? circuits powering the same loads is half that of a one balanced 3? circuit. :slaphead:

Keep working on the math and you'll eventually figure it out...

 

[david luchini]

I do not agree with this approach, and believe instead that both factors (not just the larger of the two) must be taken into account. You need a wire that has an ampacity of 12.5. With 8 CCCs in a conduit, whatever wire you pick will have to be derated to 70% of its tabulated ampacity. So if 70% of the tabulated value must exceed 12.5, then the tabulated value must exceed 12.5 divided by 70%, or 17.8. The smallest wire that fits that description is indeed #14.

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.

 

[Smart $]

You had me confused initially, but then I think I understood your question. So please allow me to split hairs for a moment. The derating that is related to the number of CCCs in a raceway is an ?adjustment factor,? not a ?correction factor,? and it is not the same as the issue of ?conduit fill.? Secondly, adding 25% to the current taken up by a continuous load is also not at ?correction factor.? Nor for that matter is it an ?adjustment factor,? and nor does it constitute ?derating.? It has no name; it?s just part of the calculation process. I know these are just a minor language issues, but sometimes it is important to be precise in our use of terms, especially when we are dealing with complex questions.

Finally, we do ?correct? twice in this situation, and in others. If you had thrown a higher ambient temperature into the mix, then we would have to correct three times.

If I'm understanding you correctly here, you are making the very common mistake I referred to earlier. Article 210 and 215 say the 125% factoring for continuous loads applies to the allowable ampacity of the conductor BEFORE the application of any adjustment or correction.

So here is all we do is make that assessment: A 10A continuous load requires a conductor having a table ampacity of 12.5A. If we use a 90?C conductor, any size listed in the Table 310.16, for example, meets that requirement... even #18 copper. That ends the compliance assessment.

Now when making the 310.15 adjustment for more than 3 ccc's you use the 10A value, not the 12.5A value. Same goes for making the 110.14(C) compliance assessment, the 10A value is used.