CABLE DAMAGE CURVE AND NEC 240.4 (B)

[mbrooke]

I don't respect your DIY engineering enough to consider changing my mind, which is why I have asked for real examples.

I have over 40 years of experience as a licensed engineer, so I am comfortable with my ability to apply engineering fundamentals.

You don't know all the fundamentals based on what you've revealed thus far.

Few take pics categorizing every single failure they've encountered (assuming the client allows it), and still then I feel like you'd find some alternate meaning to explain away the context implied.

Hiding behind the code doesn't make you right. The code is a minimum standard predominantly driven by litigation numbers, funding more so today. I can design a system following the code to a T while failing to meet its intent of life safety in several regards.

 

[kwired]

Back feed or no back feed 100,000 amps of fault current can damage smaller cables before the breaker opens.

There is a curve, high fault currents lower final temps of the conductor, but beyond a certain point you see a sharp rise once you hit the maximum clearing time.

Can you give more info as to how time domain plays into this?

will the resistance of the conductor even allow 100kA to flow in a small conductor? Maybe if really short length it can become like a fuse but for at least 10 feet I think the conductor resistance becomes a current limiter.

 

[mbrooke]

will the resistance of the conductor even allow 100kA to flow in a small conductor? Maybe if really short length it can become like a fuse but for at least 10 feet I think the conductor resistance becomes a current limiter.

#12 cu, 277/480 volt system.

2.0 ohms per 1000 feet at 75*C, ignoring X=

0.002 ohms per foot

Assuming 10 feet = 0.02 ohms= 13,850 amps

In Jim Dungar's defense sputtering will reduce the amount of RMS current but will still trip the breaker magnetically. So this does help.

The thing is 3,836,450 watts of power would be dissipated in one second across those 10 feet, 64,000 watts in one cycle.

 

[kwired]

#12 cu, 277/480 volt system.

2.0 ohms per 1000 feet at 75*C, ignoring X=

0.002 ohms per foot

Assuming 10 feet = 0.02 ohms= 13,850 amps

In Jim Dungar's defense sputtering will reduce the amount of RMS current but will still trip the breaker magnetically. So this does help.

The thing is 3,836,450 watts of power would be dissipated in one second across those 10 feet, 64,000 watts in one cycle.

How much does it take to start melting said #12 copper?

We must also remember the OCPD is supposed to protect not just the conductor but the conductor insulation, but as I was trying to get at is there is no way you are pushing 100kA through that small of a conductor with much length to it too much resistance. I knew this just from doing available fault current calculations at services. Most cases with smaller circuits (up to 100 amps) you generally are under 10kA if you have at least 25' of conductor to the source no matter how stout that source is.

 

[mbrooke]

How much does it take to start melting said #12 copper?

We must also remember the OCPD is supposed to protect not just the conductor but the conductor insulation, but as I was trying to get at is there is no way you are pushing 100kA through that small of a conductor with much length to it too much resistance. I knew this just from doing available fault current calculations at services. Most cases with smaller circuits (up to 100 amps) you generally are under 10kA if you have at least 25' of conductor to the source no matter how stout that source is.

Big difference to be honest, however you do not want to exceed a final temperature of 150*C-160*C.

 

[jim dungar]

You don't know all the fundamentals based on what you've revealed thus far.

I am still waiting for your real world examples.

It would not be hard to find people to vouch that I have more than a fair grip on fundamentals, but I feel no need to prove my qualifications to you.

 

[mbrooke]

I am still waiting for your real world examples.

It would not be hard to find people to vouch that I have more than a fair grip on fundamentals, but I feel no need to prove my qualifications to you.

On black and white you insinuate sub cycle clearing time of breakers, or absence of adiabatic limits. That speaks for itself, no matter how many NEC believers exist on this forum.

If I find a real world example on my hard drive, I will post it. But I'm not putting all that much effort into it.

 

[mbrooke]

I am still waiting for your real world examples.

It would not be hard to find people to vouch that I have more than a fair grip on fundamentals, but I feel no need to prove my qualifications to you.

Try this, its free, and yes I'm well aware it applies to sizing EGCs- but it goes to show you just how big a conductor must be to be considered fully protected:

https://myelectrical.com/tools/cpc-sizing-calculator

 

[jim dungar]

On black and white you insinuate sub cycle clearing time of breakers, or absence of adiabatic limits.

I have been stating that UL testing proves that breakers can clear in enough time to protect properly applied cables.

 

[mbrooke]

I have been stating that UL testing proves that breakers can clear in enough time to protect properly applied cables.

Can you describe the testing more in depth? Cable lengths, fault current and temp measurements?

 

[mbrooke]

Here is a simple version in determining energy let through vs time through for a given wire size:

myCableEngineering.com > The adiabatic equation

Electrical cable sizing software. Current capacity to BS 7671, ERA 69-30 and IEC 60502. Impedance and voltage drop to IEC 60909 and CENELEC CLC/TR 50480. Cloud based - any device, anywhere.

mycableengineering.com

 

[don_resqcapt19]

Can you describe the testing more in depth? Cable lengths, fault current and temp measurements?

Read UL 489. You can read it for free online. Just not a very user friendly system, exactly the same system that NFPA uses to let you read their standards for free.
I do know that there is a section that says the insulation on the conductors connected to the breaker for all of the tests, including the interrupting rating test, cannot be damaged. I do not know what size conductors are used for all of the tests.

 

[mbrooke]

Read UL 489. You can read it for free online. Just not a very user friendly system, exactly the same system that NFPA uses to let you read their standards for free.
I do know that there is a section that says the insulation on the conductors connected to the breaker for all of the tests, including the interrupting rating test, cannot be damaged. I do not know what size conductors are used for all of the tests.

Alright, Can you link it?

 

[don_resqcapt19]

Alright, Can you link it?

Just do a search for UL 489...like with the NFPA you have to register with an email address, but it is free. Once you find it, you click on digital view.

 

[mbrooke]

Just do a search for UL 489...like with the NFPA you have to register with an email address, but it is free. Once you find it, you click on digital view.

Lemme try.

Is it the same for UL1699?

 

[don_resqcapt19]

Lemme try.

Is it the same for UL1699?

All of the UL standards are viewable.

I believe that any organization that creates ANSI standards must have some method for the public to view those standards for free.

 

[mbrooke]

All of the UL standards are viewable.

I believe that any organization that creates ANSI standards must have some method for the public to view those standards for free.

Neat. Was this recent? I lost my PDFs years back and found the hard way that I couldn't view them for free.

 

[don_resqcapt19]

Neat. Was this recent? I lost my PDFs years back and found the hard way that I couldn't view them for free.

I found about about it a year or so ago...not sure how long they have been providing free access.

 

[mbrooke]

I found about about it a year or so ago...not sure how long they have been providing free access.

I'd say thats when they must've gone free, but then again I tend to miss the obvious.