Re: Power in a TC System

• To: tesla-at-pupman-dot-com
• Subject: Re: Power in a TC System
• From: "R.E.Burnett" <R.E.Burnett-at-newcastle.ac.uk> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)
• Date: Mon, 15 May 2000 11:38:50 -0600
• Approved: twftesla-at-uswest-dot-net
• Delivered-To: fixup-tesla-at-pupman-dot-com-at-fixme

Hi Gavin,

The interactions between the tank capacitor and leakage L of the NST are
quite complex which makes analysis a little tricky.  By definition the
P=0.5*BPS*C*V^2 formula is correct for the amount of power passing through
the tank capacitor.

The power drawn from the wall will be slightly higher than this figure
because of I^2R losses and such in the NST and wiring.  The supply current
will then be higher again due to poor power factor,  so supply current
alone is not a good indicator of real power.

I don't think you can just treat the primary cap as an impedance across
the HV lines,  because of the firing of the gap etc.  With a static gap
the firing is actually quite chaotic,  and power drawn can fluctuate
dramatically.

Hope this helps,
						Cheers,

						-Richie,

On Mon, 15 May 2000, Tesla List wrote:

> Original Poster: "Gavin Dingley" <gavin.dingley-at-astra.ukf-dot-net> 
> 
> Hi all,
> Up until now, I have been using the primary tank capacitor reactance to
> judge the power used by my TC system. That is, I calculated the
> capacitors reactance at the mains frequency (50Hz) and divided the
> square of the kV output of my NST by it. P=V^2 / X. Does this give a
> true indication of system power?
> 
> The alternative would be to multiply the number of brakes per second in
> the SG by the energy stored in the capacitor, that is,
> 
> ((C V^2) / 2) * Bps
> 
> Which is correct?
> 
> Regards,
> 
> Gavin, U.K.
> 
> 
> 
> 

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