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Re: Neonics wonderland!!!!
Terry -
Your calcs assume that the primary capacitor is fully charged and
discharged at each bang. I believe this could be in error because the
secondary output spark varies in length at each bang indicating the primary
capacitor joules and charging is varying.
Assuming the primary capacitor does not fully discharge how would you test
to determine the number of joules put into the primary capacitor at each
bang when the syatem is operating at 120 BPS? Or do you think that the
NUMBER of primary capacitor charges varies and varies the length of the
secondary spark output?
The amount of primary capacitor joules per bang would affect the input
watts per ft of spark rating.
John Couture
----------------------------------
-----Original Message-----
From: Tesla List <tesla-at-pupman-dot-com>
To: tesla-at-pupman-dot-com <tesla-at-pupman-dot-com>
Date: Friday, June 11, 1999 2:08 PM
Subject: Re: Neonics wonderland!!!!
>Original Poster: Terry Fritz <twftesla-at-uswest-dot-net>
>
>At 09:28 PM 6/10/99 -0700, you wrote:
>>
>> Dr Resonance, All -
>>
>> Would you hazard a guess as to what wattage the 15 KV 60 ma NST's would
>>produce in Tesla coil operation? This is not a trick question. I have been
>>contemplating this question for some time but so far have not come up with
a
>>reasonable answer.
>>
>> John Couture
>>
>
>
>Hi John,
>
> There is a simple way of finding this that gets around all the distorted
>waveforms and phase angle problems associated with trying to integrate the
>real wave forms. Simply find the firing voltage and calculate the energy
>in joules stored in the primary cap and multiply by the break rate.
>
>P = 0.5 x Vfire^2 x Cpri x BPS
>
>In my LTR coil:
>
>Vfire = 20500 volts
>Cpri = 28 nF
>BPS = 120
>so the delivered power to the tank circuit (the transformer and protection
>filter burns off some power) is:
>
>0.5 x 20500^2 x 28e-9 x 120 == 706 watts
>
> I use a typical 15kV / 60mA transformer. I should point out that in this
>configuration the RMS current from the transformer is around 85mA and the
>peak voltage is 21.8kV. So the voltage is close to nominal while the
>current from the transformer is about 41% over the rating. However, my
>runs are not real long and it is a new transformer run at room temperature
>so it should do fine even though the windings dissipate 41% more heat than
>they are really designed for. Neons are, of course, over designed to
>withstand the harsh environments they normally work in. However, old
>second hand devices my be at the end of their life anyway...
>
> It should be noted that the system runs that 85mA (actually it is 94mA
>because of the filter caps which adds another 16 watts...) from the
>transformer through two 5K ohm filter resistors that burn off another 72
>watts bringing the total delivered power "from" the transformer up to 778
>watts. The windings (secondary) in the transformer are dissipating a total
>of 35 watts. With 10.6 amps RMS going into the primary (which dissipates
>another 27 watts) the total power into the whole thing totals 805 watts
>(really 821 watts...).
>
> If one simply multiplies the RMS voltage by the RMS current you get 1272
>watts (10.6 x 120 = 1272). However, that does not take into account the
>phase angle and the distortion in the current waveform. If one assumes the
>voltage and current are pure sine waves, the input phase angle could be
>assumed to be 50 degrees. However, the 900VA nameplate rating of the
>transformer is exceeded by 41%. I guess you could say its running with
>"enhanced" performance. :-)
>
>At least, that's my "guess" ;-)
>
>Cheers,
>
> Terry
>
>