arc flash energy question

[redwine47]

1 cal/cm2 equals ~4 joules. So does that mean an 8 cal incident energy equals the effect of a 32 cal incandescent light bulb? Doesn't seem like much heat.

 

[alehman]

No. It's 32 joules per cm^2.

 

[redwine47]

Yea, a typo either way it works out to 32 watts. Not much energy

 

[jghrist]

But 8 cal/cm² is energy over an area. If the arc were a point source radiating equally in all directions, the surface area of a sphere 24 inches (610 mm) away from the arc would be 46,700 cm². An arc would have to have a power output of 1.574 MW for one second to produce 8 cal/cm² at the 24 inch distance.

 

[itsmoked]

Keith Cress
kcress -

http://www.flaminsystems.com

 

[redwine47]

Thanks for the reply jghrist. I need to think your numbers. On a job today and tomorrow.

 

[Zogzog]

Well his numbers are exactly correct, it just dosent seem like a lot of heat because 1 cm2 is such a small area.


An arc flash can reach tempatures up to 36,000 degrees F, that about 20 time the tempature necessary to melt steel or about 5 times as hot as the surface of the sun.

 

[alehman]

Maximum sunlight intensity on earth is about 0.14 W/cm^2.
That means the sun's power output is 4*10^22 W.

 

[itsmoked]

Where's that really long thermocouple when you need it..

Keith Cress
kcress -

http://www.flaminsystems.com

 

[alehman]

It would just get all tangled up and pull the earth out of its orbit Keith.

I guess my point was that sunlight seems pretty bright, but it's only 0.14W/cm^2. The OP was commenting that 32W/cm^2 didn't seem like very much. Seems like more when you consider it's 230x stronger than sunlight.

 

[stevenal]

Mathcad says it is 33.5J/cm^2. Imagine the entire output of a 33.5 W filament (that normally emits in all directions) is concentrated on a square cm of skin for a full second.

 

[stevenal]

Upon further inspection, I see that Mathcad contains several different caloric units. It has the 15 degree cal, the 20 degree cal, the dietic cal, the mean cal, the cal, and the thermochemical cal. I used the second to the last, IEEE 1584 uses the last.

 

[jghrist]

The cal that Mathcad used in my calculation of the 1.564 MW is the one equal to 4.187 J.

Mathcad's 15 degree calorie cal15 equals 4.186 J. Mathcad's 20 degree calorie cal20 equals 4.182 J. Mathcad's Dietetic Calorie dcal equals 4186 J. Mathcad's kilocalorie kcal equals 4186 J. Mathcad's mean calorie mcal equals 4.19 J. Mathcad's Thermochemical Calorie tcal equals 4.184 J.

 

[alehman]

Why don't we just use joules? Everyone knows what that is.

 

[stevenal]

And in fact, IEEE S10-2002 lists the cal and the Cal as units that are not too be used. I'm already doing my part by refusing to read labels that mention the later. Seems strange that IEEE ignored their own rule with 1584.

 

[jghrist]

Why don't we just use joules? Everyone knows what that is.

Because NFPA 70E, while it uses joules as the main value in its clothing system table with calories parenthetically, uses nice even calorie values like 4, 8, 25, and 40.

 

[alehman]

Yeah, I know. I just question why they chose calories which is ambiguous.

 

[trevorarmco]

1.2cal/cm2 will cause a second degree burn. That's 5 joules/cm2, or 5watt-seconds/cm2.
My soldering iron is 25W, but the tip probably emits 10W of heat, the rest being lost elsewhere. Now- hold the tip- probably 0.5cm2- on your face for 0.25 seconds and see how it feels. Tell me when the blister finally heals.
BTW, a dietary calorie is actually a kilocalorie- so why don't I spontaneously ignite when I eat this 75 (k)calorie cookie?