Feeder calc. w/cont. & derating

[greeny]

Hi all,
I would appreciate some input on a problem I have. Say I have 4 THHN current-carrying conductors feeding 100A of noncontinuous load and 80A continuous. What size wire do I need for the feeder at 75` terminals. Here's what I did:
80 amps x1 = 80
100 amps x1=+100
180 amps
Since I have to derate by .8 for the 4 current-carrying conductors, I do not need to derate for the continuous load per 210.19(A)(1). 3/0 at the 90` column gives me 225 amps. 225x.8(4 current-carrying conductors)=180...enough to carry the load. So I got 3/0 copper. Can anyone confirm this. The thing I find odd is that if there were only 3 current carrying conductors, the wires would need to carry 200 amps as the calculated load.
80x1.25+100= 200 When I have 4 current-carrying conductors, it only needs to carry 180 amps as the calculated load. Thank you for any and all input

 

[kingpb]

It is unclear as to why you have 4 current carrying conductors, please explain with more detail please.

 

[greeny]

Let's pretend it's powering single phase non-linear loads from a 3 phase supply.

 

[kingpb]

Ok, I confess, I don't undersatnd what your looking for.

You say feeders, which would indicate 3 phase, then you are talking about 4 current carrying conductors, and it is single phase loads.

Something just doesn't seem to add up!

 

[wirenut1980]

I don't see how 210.19(A)(1) allows you not to derate for continuous loads when derating for more than 3 current carrying conductors.

 

[charlie b]

I understand the situation. You have a three phase feeder going to a three phase panel that serves single phase loads. But you are mistaken in this statement:

Since I have to derate by .8 for the 4 current-carrying conductors, I do not need to derate for the continuous load per 210.19(A)(1).

That article tells us what ampacity you need. Separate from this, Table 310.16, along with table 310.15(B)(2)(a), tells us what ampacity the conductors have. You have to take both deratings (continuous load and 4 CCCs) into consideration, because they come from separate articles, and they serve different purposes.


100 times 1 plus 80 times 1.25 equals 200. That is the ampacity you need.

A 3/0 copper, at 90C, has an ampacity of 225. Derating this to 80% (for having 4 current-carrying conductors in a raceway) tells us that the ampacity of 3/0 copper is 180. That is not adequate for your application.

A 4/0 copper, at 90C, has an ampacity of 260. Derating this to 80% (for having 4 current-carrying conductors in a raceway) tells us that the ampacity of 4/0 copper is 208. That is adequate.

Finally, checking a spreadsheet for voltage drop calculations, I found that the VD for a 75 foot run of 4/0 is less than one volt. So VD will not be an issue.

(Edited to correct a typo, as pointed out by haskindm in post #8. Thanks for catching that.)

 

[PE-noy]

Feeder calc. w/cont. & derating

Kingpb,
Your question is is quite confusing. Answer the questions below and I'm sure someone will be able to give you a straight answer:
Load 1: Volts/Amps or kva/Phase/Type (lighting, motor, etc.)
Load 2: Volts/Amps or kva/Phase/Type (lighting, motor, etc.)

Best regards

 

[haskindm]

Charlie,
I agree with your calculation, but you also need to make sure that the temperature limitation of the terminal (the OP states 75 degree) is not exceeded. Since the 75 degree column ampacity is 230 amps for 4/0 copper this wire size will be OK. I assume that the "3/0" in your fourth sentence should be 4/0 to match the wire size at the beginning of the sentence.

 

[spsnyder]

As long as the conductor is 90' you can use the ampacity in that column for derating purposes. Charlie is correct in his calculation.

 

[iwire]

I also agree with Charlie.

 

[steve66]

On the off chance the 4th wire is a neutral wire, you may not have to count it per 310.15(B) (4).

 

[wirenut1980]

Sounds like the OP was asking an exam question of some sort. "Beware, real world and practicality may not apply."

 

[Smart $]

On the off chance the 4th wire is a neutral wire, you may not have to count it per 310.15(B) (4).

I believe greeny has already taken that into account. His second post stated...

Let's pretend it's powering single phase non-linear loads...

 

[greeny]

Okay everyone, thank you for all the responses. I had done the problem with the same steps as Charlie did and I got 4/0. But, the instructor at a recent seminar told me I was wrong. This guy is highly respected and his knowlege of the code is quite extensive. He said that the intention of 215.2a1 was not to "double tax" the conductor ampacity, and if any other form of adjustment or correction factors need to be applied, that we can omit the 125% for continuous loads. He said this is a widely misunderstood wording that appears in 210, 215, 230... Anyone else ever heard of this?

 

[don_resqcapt19]

greeny,

He said that the intention of 215.2a1 was not to "double tax" the conductor ampacity, and if any other form of adjustment or correction factors need to be applied, that we can omit the 125% for continuous loads.

I have heard that and it is supported by the code wording.

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.

Don

 

[steve66]

That is a new one for me. I have always thought "before" ment to multiply the continuous loads by 125%, then derate the wire.

Now I'm thinking that before just means to multiply the continuous loads by 125%, and use the unadjusted ampacity.

Steve

 

[wirenut1980]

I don't know if I buy this interpretation. Isn't the purpose of derating to offset conditions that would cause added heat to be generated on a conductor? And wouldn't we have to derate the circuit ampacity for the additional heat generated by the continuous loads and for the additional heat generated by having more than 3 current carrying conductors in a raceway?

 

[charlie b]

Charlie, I agree with your calculation, but . . . .

Charlie is correct in his calculation.

I also agree with Charlie.

I am not sure Charlie is right about this. Greeny’s instructor and Don bring up an interesting twist to this story. What does the code mean by the following words?

. . . before the application of any adjustment or correction factors. . . .

The problem I have with the phrase “before the application of” is that it does not say what happens after.

I am (or at least I used to be) certain that the code has us consider two questions separately, and in the following order: (1) How much ampacity does the conductor need to have, in order for the conductor to be adequate for this application? (2) What size of conductor do you need to select, in order that it have sufficient ampacity for this application? You use 220 to answer the first question, and (most often) you use 310.16 to answer the second. That phrase that Don quoted from 215.2(A)(1) follows a sentence that refers us to 220, so we could infer that that article is still addressing the first question. Here is one way to interpret that phrase. NOTE: For the moment, let us assume that we are going to use THHN, or another conductor that is rated for 90C.

• Step 1: You add 25% to the continuous load, then add in the non-continuous load. I see nothing that ever lets us not do this Step.
• Step 2: You take the results of Step 1, and look in Table 310.16 for a conductor that has at least that amount of ampacity. Please note that at this point, you can only use the 75C column, and not the 90C column, even though the conductor is rated for 90C.
• Step 3: Determine the ampacity that that size of conductor will have, under its conditions of use. To do this, you may need apply an “adjustment factor,” for more than 3 current-carrying conductors, and you may need to apply a “correction factor,” for temperature, but you can use the 90C column, as the starting point for this Step.
• Step 4: Compare the results of Step 3 to the 75C ampacity of the selected conductor size. Proceed to the next Step by selecting the smaller of these two numbers.
• Step 5: Compare the results of Step 4 to the “required ampacity,” as calculated per the instructions in Article 220. If the results of Step 4 are higher than the required ampacity, then you may use that size of conductor.
• Step 6: There is no Step 6! What I mean is that you do not have to compare the results of Step 5 (i.e., the final, calculated, adjusted and corrected ampacity of the selected conductor) against the results of Step 1. In other words, the final, calculated, adjusted and corrected ampacity does not need to be higher than the “125% continuous plus 100% non-continuous.” It only has to be higher than whatever 220 tells us is “required.”

Basically, I see no problem with this interpretation of 215.2(A)(1).

HOWEVER, please note that in Step 5, we are looking back to the “required ampacity,” a number that (presumably) we calculated earlier. The question now becomes, “When we calculated the ‘required ampacity’ earlier, did we have to add 25% to the continuous load?” I don’t know about anyone else, but I cannot find anything about “125% continuous plus 100% non-continuous” in Article 220.

So where does the “125% continuous plus 100% non-continuous” come into play? Well, it shows up in 215.2(A)(1), as has already been mentioned. It also shows up, in the exact same context, in 210.19(A)(1), and again in 230.42(A).

So what does this mean? It seems to mean that,

• IF a conductor’s ampacity, as listed in the table, exceeds the value of “125% continuous plus 100% non-continuous,”
• THEN you can still use that conductor, even if its “final calculated ampacity” is less than “125% continuous plus 100% non-continuous.”
• THE ONLY REQUIREMENT is that the “final calculated ampacity” be higher than the load calculated per 220, taking note that 220 never has us deal with “125% continuous plus 100% non-continuous.”

So let’s look at this specific example again.

• 100 times 1 plus 80 times 1.25 equals 200. You need to start with a conductor that has at least that amount of ampacity.
• A 3/0 copper, at 75C, has an ampacity of 200. That matches the “125% continuous plus 100% non-continuous” value of 200.
• Derating this 3/0 to 80% (for having 4 current-carrying conductors in a raceway), starting with its 90C rating of 225, tells us that the ampacity of 3/0 copper is 180. That is the same value as your calculated load (100 amps non-continuous plus 80 amps continuous). Therefore, a 3/0 is adequate for your application.

This appears to be the thought process of Greeny’s instructor. Anyone wish to comment?

 

[greeny]

Thank you, Don.
The code says "before the application of any adjustment or correction factors". BUT, it nevers says anything about AFTER the application of any adjustment or correction factors. Therefor, we can ignore that 125% factor when we have adjustment or correction factors. I think it should be read "without the application of any adjustment or correction factors".
I don't know how that one has slipped by me the past 5 years. I certainly was never taught it that way in school.
I called the instructor today, and he said it had been described to him that way at an international electrical inspectors convention. This is the real deal. He is going to be contacting a well-renowned senior electrical specialist at the NFPA to validate the interpretation and our state chief inspector to see if the state is on the same page(I'm taking my master's test tomorrow and I need to know how to interpret the wording in case there is a similar example on the test).
That still brings me back to my original problem. The conductor ampacity of a feeder with 3 current carrying conductors serving a continuous load is less than the conductor ampacity of a feeder with 4 current carrying conductors serving a continuous load. Can anyone confirm this?

 

[charlie b]

Therefore, we can ignore that 125% factor when we have adjustment or correction factors.

I disagree. You can never ignore the 125% factor. But perhaps you didn?t get a chance to read my previous post, before submitting yours.

That still brings me back to my original problem. The conductor ampacity of a feeder with 3 current carrying conductors serving a continuous load is less than the conductor ampacity of a feeder with 4 current carrying conductors serving a continuous load. Can anyone confirm this?

Not true at all. The ampacity of a conductor is not a function of continuous versus non-continuous. The fact that load is continuous will impact the ?required? ampacity, not the ?calculated ampacity.?

The ampacity is based on size, insulation rating, material (Cu versus Al), ambient temperature, number of current-carrying conductors in the same raceway, and method of installation (open air versus conduit versus underground ductbank).