Jim Lux wrote:
Most of the literature on sparks indicates that because the curve of conductivity vs temperature has a real steep step in it (around 6000-7000K) that you can model the spark as basically a uniform isothermal conductor. There's a sort of dynamic balance between power dissipation and spark diameter: more current dissipates more heat, which makes a larger channel, which then loses more heat because the radiating surface area goes as the diameter. The heat lost from the channel is by radiation, and you pick a convenient temperature (e.g. 7000K).
The current will follow the path of least resistance. So as long as there is a steep rise in conductivity due to temperature the arc will be as thin as possible since a thin arc will mean more conductivity. Or put in a different way: The energy spent in the arc is better put in to temperature than into width if you want to increase conductivity. The steep rise in conductivity will flatten out at some temperature, though. From that point on the width would increase. This kind of equilibrium would be in accordance with my assumption of 1/R ~ P^2 since an increase of power would mainly go into the width of the arc. Conductivity would rise with r^2 and power consumption proportional to r due to the surface area. I estimate the power dissipation of my 20kW 70cm arc to be about 600W/cm near the breakout point. At 6000K a black body radiator puts out around 100MW/m^2 (!). That would make my arc really thin, about 0.2 mm diameter. The arc certainly looks fatter. That makes me wonder, whether the Stefan-Boltzmann law really applies. Udo _______________________________________________ Tesla mailing list Tesla@xxxxxxxxxx http://www.pupman.com/mailman/listinfo/tesla