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Re: Streamer Voltage



Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

At 07:44 AM 5/7/2003 -0600, you wrote:
>Original poster: "Harold Weiss by way of Terry Fritz 
><teslalist-at-qwest-dot-net>" <hweiss-at-new.rr-dot-com>

Distance is only proportional to voltage for uniform fields.  The more 
non-uniform the field, the more it deviates from some nice cm/kV 
metric.  Once you get into non-uniform fields and spark discharge... 
loosely, anything where the radius of curvature is much less than 1 cm per 
30 kV, things like stored energy and the RF properties of the source start 
to play a big effect.  The pulsed/RF nature of tesla coils is also significant.


For a DC source (i.e. something like a Van deGraaff generator), the voltage 
will stabilize at roughly that where the surface field is 31 kV/cm... A 6" 
sphere (15 cm diameter) has a radius of 7.5 cm, so when the voltage  gets 
to around 230 kV, the corona losses will start to radically increase as the 
air breaks down at the surface.


The voltage drop along a streamer is a different matter, and depends very 
much on the current flowing in the streamer.  A typical lightning stroke at 
10 kA has a voltage drop of about 1000 V/meter.  To a first order, the area 
of the spark channel and hence the resistance, is proportional to current.



>Hi All,
>
>This has been bugging me for a while. Is the streamer voltage measured by 
>the combination of topload geometry plus streamer length, or just by 
>streamer length?  With my 4" coils, the 6" sphere had a breakdown voltage 
>of 571KV by Sd=((MV/7.5)*2)*100.  Sd=sphere dia in cm.  MV= voltage in 
>megavolts.  It's 18" streamer would come to 502.92KV at 1.1KV/mm.  If the 
>two add together it would make the total voltage 1MV+.  If it were just 
>streamer length the 502KV would not even produce corona on the sphere.  Or 
>could it be that, the voltage peaks enough to produce corona, and then 
>drops, as the streamer is formed.
>
>David E Weiss