ECN Forum Forums Photo Discussion Forums Photos for General Discussion Meter Fire

Very impressive, I am curious to find out what the fault current was. There is a formula for that in my ohm's law book, that uses length of the wire, size and the voltage to give you potential fault current. I do believe that there is a resistance table for lengths of wire that is needed also.

I used to know that of the top of my head, that's why this site is so great, gives me that continuing education that I need. Sure can't remember it all!

And people wonder why the service disconnect has to be nearest the point of entrance. Think about that service raceway burning up like that inside of the building.
Don

The motel owner is really lucky this time, if the gas service let go, it will go up in a big fireball. By the way, did the cutout on the pole trip?

That sounds rather suspicious!

No...all there would be is a big torch and if the FD applies water to protect the building there would be little additional damage.
Probably not. I watched 30' of 4" rigid with 4 500 kcmil copper inside vaporize and the primay fuse never opened.
Don

[This message has been edited by resqcapt19 (edited 09-15-2006).]

To get the exact fault current available use a loop impedance tester like the MIT 310 from AVO Megger.
It will impose a brief testload on the 230 or 115 Volts mains and calculates the available short circuit current.
Depending upon length and size of cable, condition of connectors etc. it should be less than 0.4 Ohms (in NZ case).
Say 230 volts / 0.4 Ohms = 575 Amps.

It is amazing fire works but quite often the overall resistance may be a few ohms and the fault current may be 200 Amps or so when arcing.

A low resistance fault say 0.02 Ohms will give a large current but not always a hughe arc 11500 Amps initially to blow the fault clear with a lot of damage, then the after arcing will take place with a lower current.

The formula is the standard one

U
____

I * R

Where U or V or E is voltage
I is current
R is total resistance

Powerfactor can be neglected in this case.

Thanks for the refresher RODALCO. Simple ohm's law, less resistance = more current flow. Oh and another thing, those cutouts, fuses...up on the poles. Sounds like they aren't much good for their intended purpose.

Their intended purpose is to protect the utility distribution sytem, not to protect the transformer and its secondary conductors.
Don

Thank you for the information on the cutouts Don. That was something I was never aware of until now. I just "assumed" that the intened purpose was o.c.p for the drop and pole trannie.

I know that old saying about assuming things, I should know better. Maybe if I was a lineman instead of an inside wireman I would have known this? Thanks anyway though.