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Arc Blast Pressure 1

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mykh

Electrical
Mar 6, 2005
55
I am trying to calculate pressures developed by electrical fault. I 've come across formula published in from Ralph Lee's paper, "Pressures Developed by Arcs":

Pounds/ft^2=(11.5×kA arc)÷(distance from arc in feet)^0.9

There are also other publications referring to the same source. My concern is that arc pressure in the formula above does not depend on system voltage. Arcing energy is a function of voltage and arcing current. Does anybody know why the pressure does not depend on voltage? Please correct me if I am missing something.
 
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I don't see it. The ability of the arc to occur is driven by voltage but the actual arc is a cloud of metal and plasma. The plasma is mostly electrons, that is, current. Less current less arc.

You are saying you think the arc is more related to power than then to current? I would argue, "why are arc welders all about current?"

Keith Cress
kcress -
 
So, would the pressure effect of say 208V 100kA arc be equivalent to 5kV 100ka arc, for example? Hard to believe.
 
Pressure effect would be the same, if you could get the 208V arc started. Voltage has a lot to do with whether or not the arc will start or if started will sustain. But once you have an arc the voltage has very little to do with it.
 
Hi mykh. I see your point. One is essentially 20MW and one is 500MW. It is my understanding that there just isn't a lot of voltage across an established arc because its conductivity is so high. I think you need zilch voltage to continue an arc.

During an event a lot of the voltage drop would be in the supply network I doubt you would see the 5kV across an actual arc. But that is really a guess on my part.

Again its the current that is providing the media of the arc. Not the voltage. If you have high voltage and less current you just have a spark.

Keith Cress
kcress -
 
My opinion is the higher the system voltage the higher voltage drop in the supply network AND gap between conductors. Hence, higher voltage produces higher arcing current (consider IEEE 1584 formula for arcing current). Thermal effect and heat rise are proportional to product of current and voltage OR square of current ( assuming arc resistance depends mostly on equipment geometry and electrode material ), and would be different depend on system voltage. I thought more thermal energy means more pressure. I am still having little discomfort using Ralph Lee's formula but your inputs really help. Thanks.
 
So if you assume 100 A across a 100 ohm resistor, you can compute the power loss without knowing the voltage. It's the same concept at work for the energy created from an arc.

 
dpc, you are absolutely right, the things that bothers me is that in your example the energy would be proportional to Iarc^2, not to Iarc alone as in Ralph Lee's equation. This is empirical formula and I would have more confidence if I knew the voltage range they used to take pressure measurements. Any ideas?
 
Ralph Lee's equation is completely theoretical. If you want empirical, take a look at IEEE 1584. Test data was used up through 15 kV. There is actual pressure data available from testing on dummies. It can be amazingly high.

 
The arc has "negative resistance" - or at least negative slope, once the arc has formed. That means that the more current there is, the more ions to conduct and hence a lower arc voltage. That's why the old carbon arc lamps needed a series impedance, plus a gap regulator. And that is also why a flourescent light can't work without a ballast.



Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
"a lower arc voltage" read "lower apparent resistance"

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
Thanks Gunnar. You have supplied the "missing link" in this discussion.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
This is a very interesting thread.!

Thinking back to some of the more theory related stuff in school, and about static electricity, a couple things come to mind: static electricity has an exceptionally high voltage (electric field) but very little capacity to cause high pressure in an 'arcing' condition.

If you looked at the electric field in a static electricity (putting shirt on, etc) situation, you would find the electric field would be high magnitude in a very thin line from one surface to another. So, there's only so many electrons that can flow in this relatively small cross section of the electric field that is capable of moving electrons in an arc. Additionally, there's only so many electrons concentrated on the surface of your material before they flow over to your other surface, and then your electric field drops down because there's quickly equipotential on both surfaces.

In an arc flash scenario, however, once you breakdown the air or whatever is between your two surfaces of different electric potential, there's going to be substantially more cross sectional area of electrical field lines that can carry current. And a greater supply of electrons for that matter. Your generator or whatever is supplying your system makes sure of that until your fault is cleaered.

Skogs, although I think I see what you are getting at, I think you should expand upon your 'negative resistance' idea. Are you saying that an arc causes a negative slope in the I-V characteristic? I think that's what you're saying.

 
Yes jimgineer. Exactly.

You can see this in any device that uses an arc. An arc welder, for instance. If there weren't any current limiting, it would not be possible to weld at all. The current would increase untill a breaker trips or a fuse blows.

Once the arc has started, the voltage across it is relatively constant. Independent of arc current. Actually, the voltage rises if you reduce the current down to the arc stability limit where the arc snaps off.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
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