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capcitor energy vs pwr factor
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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.
By definition, an ac device that maintains a zero power phase angle is a
RESISTANCE. The inductance of secondary is effectively cancelled by
by "transformation" , leaving only the residual leakage reactance.
Think of it as a sine source in series with the E/I "limiting resistor
(which
IS the transformer), leakage reactance, choke reactance, and capacitor.
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...)
>
> > Wouldn't the resonance actually be due to the sum of the transformer
> > secondary leakage reactance and the series choke? The transformer
> > presents an in phase and therefore resistive component to the model
which
> > limits Q. Right?
>
> Assuming E=15 kV and I = 30 mA, for example: E/I=500 kOhm and L=1326 H.
> L is much larger than any reasonable filter choke inductance, and the
> secondary wire resistance is certainly much smaller that 500 kOhm. It
> limits the Q, yes.
>
> > Others say that the current is cosine with sine voltage and the T/4
> > instantaneous
> > energy is the integral of these function's product.
>
> This assumes steady-state conditions, and is exactly what happens with
> the primary capacitor in this situation. Actually, this would happen
> exactly only if there is no firing of the spark gap for several cycles.
>
> > If this is the case,
> > the absorbed power is absolutely dismal compared to an "in phase
> > system".
>
> No. The spark gap fires and all the energy in the capacitor is put to
> use. The system operates very far from steady-state sinusoidal
conditions,
> and so any consideration about phase angles, power factors, etc. is
> not exactly correct. The voltage and the current in the primary of the
> power transformer can be considered as approximately sinusoidal in most
> cases, because of the heavy filtering caused by the large series
inductance
> of a NST secondary, and measurements of simulations of this would show a
> high power factor (See the ROTJIT program), with efficient energy
transfer
> from the power line. Tesla was aware of this, certainly.
>
> > 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
>
>