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Re: Capacitor C/Peek
>>From pierson-at-msd26.enet.dec-dot-comWed Oct 23 22:20:11 1996
>Date: Wed, 23 Oct 96 20:54:06 EDT
>From: pierson-at-msd26.enet.dec-dot-com
>To: mail11: ;
>Cc: pierson-at-msd26.enet.dec-dot-com
>Subject: Capacitor C/Peek
>I was pondering something exceedingly basic, and getting phunny numbers.
>Specifically, what energy is in the cap in the primary tank?
>I keep getting a number which seems small, someone check my math:
> Joules == watt-seconds == (C(Ve2))/2
>right?
> C= 0.025 uFd (for the CP caps, just to pick a number...
> V=10,000 V (to pick another)
>Rolling this together:
> ((2.5)*(10e-8)*(10e4)e2))/2= 1.25 Joules
>Izzat right?
>I was expecting a larger number....
>Am i dropping a decimal somewhere???
<snip>
> regards
> dwp
dwp,
Your math is fine. However, if that 10 kV was transformer RMS you
can redo your calculation to the peak (X 1.414) value as this is what
it could charge your cap to if allowed to by the gap. With this view you
get 2.5 Joules. If you can 'pop' your 2.5 Joules in one half
squarewave at 100 kHz (5 microseconds), your reward is a 500 kW pulse. : )
Create a nice half sinewave instead and you get a _peak_ power of
1.414 x RMS or a 707 kilowatt pulse.
Now take a 12000 volt RMS xfmer and charge a 0.125 mfd cap to 16968
volts. That's 17.99 Joules. Bang this in 5 microseconds and your
sinewave peak power is 5.088 megawatts! That's approximately the
spec of R. Hull's Nemesis coil and exactly those of my MTC system. My
first half sinewave is 6.25 microseconds though (80 kHz), so my theoretical
peak power is lower at merely 4.07 megawatts, and then taking K=0.2 and gap
losses into account I'm down to maybe 15% of that (a measely 610 kilowatts)
out the top?
I realize that the above is fast and simplified 'figurin' which
ignores a lotta stuff so flames are welcomed if I'm out more than 10 db. : )
rwstephens