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Re: Calculating streamer breakout of top-loads



Original poster: "Malcolm Watts" <m.j.watts-at-massey.ac.nz> 

On 8 Sep 2003, at 8:00, Tesla list wrote:

 > Original poster: "Rikard Titus" <rikard_titus-at-hotmail-dot-com>
 >
 >
 >
 >
 > Hi Gerry,
 >
 > >Hi Rikard,
 > >
 > >Do you mean:  1/2 * (0.05uf x 13KV ^^2)  = 5.6 joules
 > >5.6J x 345 pps = 1940 W
 >
 > Correct.Forgot about the square.
 > One way or another,some 1000 W is missing,(see dr Resonance's post).
 >
 >
 > >John's empirical spark length formula of 1.7 sqrt (power) is based on
 > >xformer input power not the power delivered thru the spark gap. Is this
 > >right???
 >
 > I am not familiar with this formula.Provide me with link/more info .
 >
 >
 >
 > If so, seems like1940 watts at the spark gap could mean
 > >3000W
 > >xformer input power
 >
 >
 > Hey wait a minute!
 > If PT losses 34 % of a real power,than it means it is mature for garbage  !
 >
 > -Rik

Not necessarily. The degree of (over)loading of a transformer plays a
not-insignificant part in the efficiency stakes. Some power supplies
are terrible losers. Iron-cored ballasts such as welders for example
cannot be wonderful performers with a high harmonic content from the
gap generating eddy currents in the core. To some extent that are
being asked to behave like SMPS inductors.

       For this reason, whenever I have been considering efficiency,
how well you can do with such-and-such a pri-sec combo, etc. it has
always been useful to specify whether one is dealing with wallplug
power or primary power (E.BPS) since the two are obviously not the
same. For me, the parameter of interest has always been primary power
since running conditions can be set such that it is easily and
repeatably quantifiable, particularly on a per-shot basis. It also
eliminates poorly defined losses which can vary from one transformer
to the next from the mix.

Malcolm