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capacitor energy vs pwr factor
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From: Wysock, William C. [SMTP:Wysock-at-courier8.aero-dot-org]
Sent: Tuesday, February 10, 1998 12:57 PM
To: Tesla List
Cc: ttr
Subject: RE: capacitor energy vs pwr factor
To all working with NST's as a power source.
With any NST (containing a magnetic leakage shunt in its core;
whether it be "stock" or "modified" for higher current) and
irrespective of using external a.c. electrolytic power factor
correction capacitors, there seems to be something
fundamental missing in everyone's calculations. Remember,
you can hard-wire short the secondary terminals on an NST
without burning the thing up. If your NST was rated for example,
at 60 ma., you will have 60 ma. flowing in the short circuit. BUT,
you will have "0" volts. The voltage output of an NST is
inversely proportional to the amount of current being drawn.
The only time you have say, 15 KV rms at the output terminals
is when NO current is being drawn. The bottom line is that
at no point do you have 15 KV *and* 60 ma., at the same time.
The load voltage and current will be less (for the point where
the current sine wave phase angle crosses the voltage sine
wave phase angle,) to yield maximum usable power.
Bill Wysock
-------------------------------------------
Tesla Technology Research
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From: Tesla List
To: Tesla List
Subject: capacitor energy vs pwr factor
Date: Tuesday, February 10, 1998 8:51AM
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From: bmack [SMTP:bmack-at-frontiernet-dot-net]
Sent: Monday, February 09, 1998 8:50 PM
To: Tesla List
Subject: Re: capcitor energy vs pwr factor
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> From: Tesla List <tesla-at-pupman-dot-com>
> To: 'Tesla List' <tesla-at-pupman-dot-com>
> Subject: capcitor energy vs pwr factor
> Date: Sunday, February 08, 1998 10:09 PM
>
>
> ----------
> From: Antonio Carlos M. de Queiroz [SMTP:acmq-at-compuland-dot-com.br]
> Sent: Saturday, February 07, 1998 2:14 PM
> To: Tesla List
> Subject: Re: capcitor energy vs pwr factor
>
Antonio et all,
I should have stated my point in a more concise way (something to work on).
What I was tring to bring out in the previous post, was the physics of the
transformer itself provides E and I approximately in phase at the
secondary.
<snip>
As far as power is concerned, there is no doubt that the cap disapates all
it's energy when the gap fires, but I was quetioning how much of the
AVAIlABLE power is sucks from the transformer.
If current was really cosinsoidal and voltage was sine as previously
suggested,
it would only absorb about a third of the peak energy comparded to high
power
factor scenario. Do the integration for the hypothetical 15kv-at-30ma NST ,
multiply by 120, you will see what I mean.
Comments?
Muddleing and modeling in NY
Jim McVey
> Jim McVey wrote:
>
> > Some say that E/I is the inductive reactance of the secondary and that
> > matching this with capacitive reactance -at-60 HZ produces resonance.
>
> True, if the inductive reactance of the secondary is used for current
> limiting, as in a typical NST. E is the open-circuit voltage and I is
> the short-circuit current. Assuming that all the current limitation is
> due to the inductance, L=E/(I*2*PI*60), and a capacitance C=I/(E*2*PI*60)
> results in resonance at 60 Hz. (The transformer insulation doesn't like
> this...)
<snip>
> > Thus I propose that we consider the secondary resistive, which yeilds a
> > 45 degree phase angle when the capacitive reactance is equal to the
> > E/I "resistance" and this approaches zero as the leakage and choke
> > reactances are considered.
>
> This can happen if you use a normal transformer with low series
inductance
> and a resistive ballast as current limiter at the primary side. This is
> very inefficient, however, as the ballast dissipates great part of the
energy.
>
> Antonio Carlos M. de Queiroz
> mailto:acmq-at-compuland-dot-com.br
> http://www.coe.ufrj.br/~acmq
>
>