Unsure of Wire Size

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Re: Unsure of Wire Size

Originally posted by itasca_mn:
The results are the same, except that we are approaching it from opposite directions. Using your approach you must use an educated guess at a conductor size, and derate its ampacity.
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The results are not the same, I come up with a 6 awg for the application and you came up with 8 awg.

I do not have to make an educated guess.

If I have to deal with a 50% downward adjustment I can just add 200% (the inverse of 50% down)

I needed a conductor that was permissible to use on a 30 amp circuit, 30 + 200% = 60 amps that means a 6 awg.

Except for general purpose branch circuits where exact loads are impossible to pin down, you always base conductor size on the load being served.
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Yes and you also have to figure in your breaker size.

To take from a recent thread here what size conductors and OCP does a fixed load of 900 amps need. ;)

[ December 01, 2003, 03:27 PM: Message edited by: iwire ]
 
Re: Unsure of Wire Size

Originally posted by itasca_mn:
Article 210.19 states that the conductor must equal 125% of the continuous load plus the noncontinuous load before the application of any adjustment factors. A #8 THHN is good for 55 amps before the application of adjustment factors. Well above the 30 amps required for the 24 amp continuous load.
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That gives me something to think about, I was basing my opinion on 240.4(A) and 310.15

This is interesting. :)
 
Re: Unsure of Wire Size

iwire,

Please understand that I'm not trying to create a huge issue here, only trying to straighten this whole thing out. My company is paying a high octane instructor to teach us our 16 hour continuing ed class every two years for license renewal here in Minnesota. This instructor is probably the most knowlegeable person regarding the NEC that I have ever met. His classes put others that I have attended to shame. I have sent him off an email requesting that he shed some more light on this issue. He has taught this method for years, and is adamant that this method is letter per letter as the NEC requires us to size conductors. He is very highly respected in the electrical industry, and is one of only a few fulltime code instructors in the USA.
 
Re: Unsure of Wire Size

Originally posted by itasca_mn:
iwire,

Please understand that I'm not trying to create a huge issue here, only trying to straighten this whole thing out.
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That is what it is all about here :) like I said this is interesting.

Heres what I think, your method for determining the conductor size for the load served is correct.

Lets go with your result of 27.5 amp conductor size needed.

There is no 27.5 amp breaker available, and if the this circuit could not have 240.4(B) applied you could not put it on a 30 amp breaker.

You can not put this load on a 25 amp breaker so you are stuck with the 30 amp breaker which means bumping up the conductor size to meet or exceed the breaker.

That is why I asked you for the conductor size and OCPD for a 900 amp load.

What you would find is you can find a set of conductors to carry the load 4 sets of 4/0s = 920 amps but no 920 amp breaker is available.

The next available breaker would be 1000 amps but you can not put 920 amp rated conductors on the 1000 amp breaker.

You will need to bump up the conductor size to cover 1000 amps. This is over and above the continuous and non-continuous loads, but you have no choice you must use (most times) conductors rated higher than the breaker.

Bob
 
Re: Unsure of Wire Size

iwire,

I'm not sure I fully understand your last response.

The 900 amp load is a horse of a different color because of 240.4(C) for loads over 800 amps. 240.4(C) requires the ampacity of the conductors to be equal to or greater than the overcurrent device that protects them where the overcurrent device is rated at over 800 amps. In this instance you would indeed need to bump up the wire size.

But for for a conductor sized at 27.5 amps, 240.4(B) allows us to use the next higher standard sized OCPD without bumping up the wire size, as long as all of the conditions outlined in (1), (2), and (3) are met. A 30 amp breaker would be fine in this case. Isn't this the whole intent of 240.4(B)? Or did I miss something?
 
Re: Unsure of Wire Size

No you did not miss a thing :) here is my first post in this thread.

This will require 6 awg for each circuit as Dereck has said or you could apply 240.4(B) and use 8 awg.
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What would you say to the wire size if there was two outlets on the circuit?

With two outlets on the circuit 240.4(B) could not be used.

So my point was that while your method will get us the wire size needed for the load, you may have to bump it up to comply with 240.4

240.4 Protection of Conductors.
Conductors, other than flexible cords, flexible cables, and fixture wires, shall be protected against overcurrent in accordance with their ampacities specified in 310.15, unless otherwise permitted or required in 240.4(A) through (G).
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Bob
 
Re: Unsure of Wire Size

I don't think the code really means what it says in 210.19(A) where the words "before the application of any adjustment or correction factors" appear. I say this because of the substantiation for the proposal that added these words to the code. These words first appeared in the '99 code as a result of proposal 2-5. This proposal was submitted by the "NEC TCC Task Group on the Useability of the NEC". The proposal involved a reorganization and rewrite of Articles 210, 215,and 220. Part of the substantiation for this rewrite said: " ... The objective of this proposal is to eliminate redundancy in present Articles 210, 215, and 220 and to provide a logical layout of the rules for the user of the Code. It is not intended that any changes in the technical requirements be made by this proposal ... " The added words, when applied as written, do make a change in the technical requirements of the code.
Don
 
Re: Unsure of Wire Size

Okay guys let me try to clarify this.

Step 1. Size the wire to carry 22A continuously.
22A x 1.25 = 27.5A [210.19(A)(1)]

Step 2. Go to Table 310.16, select the wire at 60/75C. Lets assume 60C. 10 AWG rated 30A

Step 3. Size the circuit protection device, 22A x 1.25 = 27.5A = 30A [210.20(A) and 240.6(A)].

Step 4a. Adjust the conductor ampacity based on 10-20 conductors 310.15(B)(2)(a) at 90C.40A x .5 = 20A. Since conductor does not have the ampacity to be protected by a 30A protection device, we need to try the next larger conductor.

Step 4b. Lets try 8 AWG = 55A x 0.50 = 27.5A

Since this is supplying a single receptacle, 240.4(B) permits 27.5A wire to be protected by a 30A protection device.

There are multiple ways to do this calculation to get the same results.

Minimum wire required 10 AWG 210.19(A)(1)
Minimum protection 30A 210.20(A)
Minimum wire after adjustment required: 8 AWG = 27.5A
 
Re: Unsure of Wire Size

Don, iwire, etal,

Here is the response I got from my continuing ed instructor. His name at the end is purposely omitted for privacy. His response was regarding the question Don posed in one of his posts, not the original question. First the question from Don in quotes, then my instructor's reply.

Itasca,
My point is that your step 3 is not correct. You must adjust from the required conductor ampacity and not the actual load. Look what happens if the continuous load is 24 amps. 125% of 24 amps is 30 amps and a #10 is suitable if no adjustment factors are required. If you apply a 50% adjustment factor to 24 amps, you get 48 amps requiring a #8, but when the 50% adjustment factor is applied to the required conductor ampacity of 30 amps, you get 60 amps requiring a #6.

Don
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Matt,
I'm not in the office, but I'll try to answer your question.

The basic rule in 210.19 is that the conductor be sized for 125% of continuous load plus 100% of the noncontinuous load.
Steps 1 and 2 satisfy the basic rule.

What many people overlook is that conductors are rated to carry 100% of their rated ampacity continuously - period. The reason for increasing conductors to 125% of the continuous load is to make them a better heat sink for the breaker or fuse termination point. You will see evidence of this in the code wherein if an overcurrent device is rated for 100% operation, the conductors are selected for 100% of the load.

It is for this reason that step 3 and 4 are based on 100% of the load. Steps 3 and 4 are related to heat either from adjacent conductors or ambient temperature. In the example given the 6 amperes added to the 24 ampere load does not generate heat, so it's permitted to be ignored in Steps 3 and 4.

It's easy to lose sight of the fact that the conductor is only going to carry 24 amperes. The #8 AWG 90 deg C rated conductor satisfies all requirements because it is more than 125% of the continuous load and is rated at 55 amperes x .5 = 27.5 amperes continuous load.

There isn't anything in the Code that requires derating (Steps 3 and 4) to be done at 125% of the continuous load.

Hope this helps.

[ December 02, 2003, 10:10 AM: Message edited by: itasca_mn ]
 
Re: Unsure of Wire Size

Don,

I consider you and iwire to be a couple of the most knowlegeable people I've met online. Now we all know even more! :)
 
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