Arc fault.

[Reddy_kW]

We had an arc fault event here at work. A 13.8 kv motor faulted in the peckerhead. The relay techs that came out said it was a 20,000 amp fault and it lasted 6 cycles. It blew the box open, popped the heads off of about 50-60 5/16" bolts. My question is, how can I determine the cal/cm2 of the incident? I know I need to have more info, but I'm curious to know the energy created here.

 

[Haji]

You may find this link useful to find answer to your own question:
http://ecmweb.com/content/calculating-arc-flash-energy-levels

 

[Reddy_kW]

Thanks, but I believe it may be a little over my head. If I plug in 24" for working distance, after running the numbers I'm getting an answer in the neighborhood of 250 billion? My math doesn't seem to be telling me what I think it should after reading that article and plugging numbers.

 

[chris kennedy]

Check with these guys. Another free forum, good luck.

 

[ghostbuster]

With a trip time range from .051 to .111 sec (6 cycles is approx..096 sec) incident energy is approx. 1.8 cal. at 36" with 20 ka. of fault current (SKM arc fault program)

 

[G._S._Ohm]

Seems like this equation


E_MB = 1038.7 D_B^-1.4738 ? t_A [0.0093 F² 0.3453 F + 5.9673]

should be

= 1038.7 X DB^-1.4738 X tA X [(0.0093 X F^2) - (0.3453*F)+ 5.9675)]
The minus sign was missing in the first example.

You can cut and paste this second equation right into an Excel cell.

 

[G._S._Ohm]

Delete the last line of my prev. post.

36	=db
0.096	=ta
20	=f in kA
	1.410463342

I got 1.4 cals

 

[ATSman]

We had an arc fault event here at work. A 13.8 kv motor faulted in the peckerhead. The relay techs that came out said it was a 20,000 amp fault and it lasted 6 cycles. It blew the box open, popped the heads off of about 50-60 5/16" bolts. My question is, how can I determine the cal/cm2 of the incident? I know I need to have more info, but I'm curious to know the energy created here.

Nice Pics
What's the HP of the motor?

 

[Hugh_Hoagland_e-Hazard]

Arc Fault Energy

Arc Fault Energy

This event didn't have a lot of arc flash energy from a clothing perspective.

20kA for 6 cycles is about 8 cal/cm2 IF the arc is 12 inches long (which is unlikely looking at the equipment). This is at 12 inches from the arc.

This was likely in the 4-5 cal range but the door had a bit of pressure due to the current.

I'm basing this on my experience in doing arc flashes in testing. I have about 100,000 under my belt. The calculations would have added multipliers for arc in a box but they are often a little overkill in 15kV equipment like this.

Unlikely to have ignited clothing in my experience. Let me know if I can assist further.

 

[G._S._Ohm]

20kA for 6 cycles is about 8 cal/cm2 IF the arc is 12 inches long (which is unlikely looking at the equipment). This is at 12 inches from the arc.

Yes. From the formula above, I get

12 =db
0.1 =ta
20 =f in kA

7.4 cals/cm^2


But this "Doughty" equation is supposedly only valid for >= 18".

http://www.ibew702.org/uploads/Safety/Arc Flash Calc Spreadsheet 3-09.pdf