As a welder and electrician I understand the importance of current carrying capabilities of wire.
"The photosphere, then, is plasma in the 'arc' mode. We say this because the Sun emits power at a rate of over 63 million watts/sq meter from its photospheric surface. This is equivalent to a power output of 40 kW from each square inch of that surface. Some have questioned whether the photosphere's relatively low temperature (~5800K) disqualifies it from being in arc mode. In 1944 C.E.R. Bruce of England's Electrical Research Institute proposed that the "photosphere has the appearance, the temperature, and the spectrum of an electric arc; it has arc characteristics because it is an electric arc, or a large number of arcs in parallel." And, it is difficult to imagine a plasma discharge in anything other than arc mode that could radiate 40 kW of power from each square inch of its surface area. Can you imagine the light from forty 1000 watt light bulbs coming out of a one square inch area?
40Kw per square inch is huge current. For a 40 volt arc welder it would be about 1000 amps which is the max that a 1 inch wire can carry.
Xenon short arc lamps, which are the brightest common single point light source, seem to be available up to about 10kW, which makes sense given their electrode size. With the xenon short arc lamps, once a plasma arc is struck the voltage goes down. So the sun had to have a higher initial voltage to light up?
Can plasma carry as much current per square inch as solid wire? Can someone point me to the equation to figure that out?
The plasma tube experiments that I have done I have only ever seen a few milliamps.
Capacitor discharge experiment are higher current but a flux tube forms... Am I correct in thinking that a flux tube is because the thin plasma cannot carry the current?
Brant
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