## Paul Nicholson on Energy Transfer of a Capacitive Reactance

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• Been wanting to make a reply to tesla list regarding Pauls statements, but I figured I better mull things over first... I may have a conceptial difference of
Message 1 of 1 , Jun 1, 2002
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Been wanting to make a reply to tesla list regarding Pauls
statements, but I figured I better mull things over first... I may
have a conceptial difference of opinion with Paul, but perhaps I
dont, which is the reason for mulling it over.

The first problem I have is with the way he views energy transfer. I
view it as the amount of energy times the number of transfers divided
by that time period. Paul however uses pi times that amount, and
finally does give a good explanation as to why this is so. Even
though he has explained why pi times more energy transfer takes place
with a capacitive reanctance, I am unsatisfied with that explanation,
for the simple fact that the actual energy transfer can be any
quantity with respect to the energy input, and this is only
negotiated by the q of the coil. In my thinking the VAR input then
does not have to equal the actual energy transfer, and for the case
of a capacitive reactance this VAR quantity actually is pi times
more than the noted energy transfer, but to make the VAR equation
equal, the pi multiplier is made in place by equation.

Here is Paul's Reply from tesla list;
Date:Thu, 30 May 2002 17:27:49 -0600
From:"Tesla list" <tesla@...>
To:tesla@...
Subject:Re: VA and stored energy in capacitors

Jolyon wrote:
Since relationship between apparent power (VA), frequency (F) and
stored energy (E) in a capacitor in an AC circuit is VA=4pFE, where
it is assumed that F is the line frequency (50 to 60 Hz) during the
period when the capacitor is being charged by the transformer prior
to the firing of the spark-gap, ...
I had wrote back asking why this was 4 pi *f*E where E is the energy
storage,(.5CV^2), where in my thinking it should be 2*f*E. Pauls
reply shows that I did not factor in Vpeak, and instead used Vrms,
and he also shows where this pi comes from...
In general for a capacitor carrying a sinusoidal current, the stored
energy is 0.5 * C * Vp^2 where Vp is the peak voltage. This peak
voltage is sqrt(2) * Vrms, so in terms of RMS voltage, E = C *
Vrms^2. The associated RMS current is 2 * pi * F * C * Vrms, so we
have
VA(R)
= Vrms * Irms
= V * 2 * pi * F * C * Vrms
= 2 * pi * F * C * Vrms^2
= 2 * pi * F * E
I put the (R) in VA(R) to remind that we're talking reactive VA, not
real power - the energy delivered to the C in one quarter-cycle is
returned to the source during the next so the net energy transfer is
zero. For example, a resonating LC behaves as if the stored energy E
shuffles back and forth 2 * pi * F times per second. The above
doesn't quite apply to an idealised TC charging circuit, because the
discharge of the C is via another path. The line-side RMS ammeter is
only registering the charging current, not the two-way charge-
discharge current. To a first approximation the RMS ammeter reading
is halved to pi * F * C * Vrms. So now VA = Vrms * Irms = V * pi * F
* C * Vrms = pi * F * C * Vrms^2 = pi * F * E where I've dropped the
(R) because now some portion of the VA is real and some reactive.
This wouldn't be expected to apply to a real charging circuit as it
stands because the currents and voltages are no longer sinusoidal -
form factors other than sqrt(2) relate RMS readings to peak values,
readings depend on just where in the circuit you put the metering,
and the effect of chokes and ballasts all come into play. Best thing
to do is to model your particular circuit to estimate how your actual
meter readings relate to the peak voltages and stored energy.
> doesn't the same equation apply when the gap fires when F becomes
> the resonant frequency of the Tesla coil primary?
After the gap fires the low resistance of the arc effectively
decouples the charging circuit from the primary LC. Things get a lot
simpler, the LC currents and voltages are sinusoidal, and RMS meter
readings in the LC circuit would obey VA(R) = 2 * pi * F * E during
each cycle of the ringdown. Of course, real meters report an RMS
value averaged over the response time of the meter's pointer which is
typically much longer than the whole firing event, so again some
circuit modeling would have to be applied to make sense of real
measurements
. > Does this not go some way into accounting for high peak powers
> observed in TC discharges?
It certainly describes the 'power compression' taking place in the
charging/firing/primary circuit. Roughly speaking, energy stored in C
over a quarter-cycle of line frequency is released over a quarter-
cycle of the resonant frequency, so you're getting a 'compression
ratio' in the order of F_res/F_line. However, the discharges from the
secondary are compressed further still, because once the bulk of your
stored energy has made its way to the topload, it can all discharge
to ground in nano-seconds. Again, the formula VA = 2 * pi * F * E
could be said to apply during the arc discharge, but this time the F
is the frequency of oscillation of the discharge itself, which
involves the topload C and the self-inductance of the arc and its
ground return path - a very high frequency. Therefore ultimately, the
peak power in the discharge could be compressed by up to
F_discharge/F_line. -- Paul Nicholson --
Heres a reply I never sent because maybe I dont know what I"m talking
Thank You Paul for this great explanation of energy tranfer! It
solves a paradox that has bothered me for some time. We know that for
a LC series resonance as the C quantity approaches resonance I^2R as
real power input approaches VI as apparent power input, and that at
resonance they become equal for ideal resonators. The
misunderstanding I had was that VI exceeded I^2R in reactive
measurements, because this accounts for the "extra" energy taken by
the fields in oscillation, and that apparent power measurements
exceed the true power as heat loss because of this. If we think like
that, we end up wondering how more energy can be going into the the
resonant fields exchanging energy, when no portion of energy is
allocated for that situation because VI=I^2R in ideal resonance. You
have answered this question very well with the realization that the
net higher amounts of energy transfer found in resonance actually
consists of that made by that "borrowed and returned" net zero
energy transfer of the source.

What this then implies to me is that while we have always considered
a reactance a case where energy is borrowed and returned from the
source, leading to the observation that VI > I^2R, the implication
that this ceases to happen in resonance is wrong!

Essentially in this correct understanding, in resonance we are only
enabling the reactive components to borrow and return energy Q times
the former levels. VI can = I^2R at resonance, but this does not
cancel the energy going into the magnetic and electric fields,
instead it maximizes it, where for very large L values generally the
amount of energy transfer at resonance can exceed the actual perfect
resonance energy transfer as heat.

Paul has noted that for a capacitive reactance the energy transfer in
VARS is given as 2 pi * E(rms)* f

But logically I see the actual energy transfer as 2*f*(CV^2), where
(CV^2) is the correction of the normal Vrms energy storage as .5CV^2,
for the peak voltage value which will contain twice the energy,
although only 40% more voltage, but because of the exponential term
V^2, this is so. So my estimate is pi times less (ACTUAL) energy
transfer than what the VAR input indicates, and again this is easily
possible since the actual energy transfer can be ANY quantity, and
that this is negotiated to be within practical parameters by how the
acting Q will differ from the real predicted Q actions, and it is
this predicted Q that determines the ratio between the apparent power
and that of the real. So by X(L)/R, since concievably this can be
made to any ratio, then any ratio of energy transfer vs power inputed
can then be obtained,(within practical limits).

To see whether this is actually so let us find the condition where
the VAR input actually equals the energy transfer,(according to my
hypothesis). Since we know that by my definition the capacitive
reactance is only allowing a current that shows pi times less energy
transfer than the VAR readings indicate, again to show what Paul has
done is the following;

First he has taken the formula for capacitive reactance X(C)= 1/(2
pi*f*C),
Where Z= X(C), and solved for I(rms)
Since V=IZ, I = V/Z = V(rms)/X(C) = V(rms)*2 pi*f*C
or,
I(rms) =V(rms)*2 pi*f*C
which is a nice formula for determining the current in a capacity
without having to resort to finding the capacitive reactance of the
capacity to determine the conduction at the impressed voltage.

Since apparent power = V(rms)*I(rms), by replacing I(rms) by the
above substitution we have;
Apparent Power = V(rms)* V(rms)*2 pi*f*C, or V(rms)^2*2 pi*f*C, or
2 pi*f*(CV(rms)^2)

Now then this is again showing that VAR is pi times the actual
energy transfer for the capacitive reactance. Let us first find out
if this is true for an inductive reactance, where now we can specify
the inductor to have a q of pi.

Since 3.14 =X(L)/R, for a 1 ohm inductor to simplify calculations of
I^2R, we now need to find the inductance for 60 hz. X(L)= 2pi*F*L
3.14 = 6.28*60*L where L = 8.3 mH

Now we need to specify an input voltage to make comparisons of
reactive power vs real, and again convenience will allow us to use 1
volt(rms) We also know that for capacity the true energy storage is
not .5CV^2 but CV^2 to account for Vpeak, so the same routine should
be employed to account for the energy storage of .5LI^2 becoming
double at I(peak). Thus the energy to be moved 120 times per second
becomes LI^2. Now we need to find Z for the conduction at 1 volt.
Z = sq rt{ 3.14^2*1^2} = 3.295, where the inverse of this becomes the
conduction of .303 A at 1 volt

We then have a VAR of .303
LI^2 =.0083*(.303)^2 = 7.64 *10^-4, where moving this 2 f times
is .0917 joules/sec, and moving it 2 pi f times is .288 joules/sec,
clearly under the VAR input of .303

Now let us find the real power input by I^2R as .0918 watts, so that
3.3 times the real power is shown by the VI of the apparent.

Apparently,(no pun intended!), for a inductor with a q of pi, this
will cause the the I^2R heating losses to be approximately
equivalent to the energy transfer, since R=1 and also (L)* 2f = .0083
* 120 = .996 If we then selected the inductor to have a higher q
than pi, say 2 pi for example, in one example this would cause R to
be reduced to .5 ohms, so for the VAR input, the contribution of
resistance to the impedance Z would be minimalized, and Z becomes
closer to the sq rt of Pi squared, or pi itself, or the value of X(L)
itself. We are not changing the inductive value to gain an increased
q, only the resistance of that inductor is being changed, so while R
is varied to lower resistances to gain a higher Q for the coil, the
inductance L or inductive reactance X(L) remains unchanged. Since the
conduction of the reactance determines the I in the VI apparent
power, as the resistance is minimized, the current is not maximized
and only approaches the highest possible value of 1 volt across pi
ohms of impedance, or it approaches a current of the inverse of pi
or pi^-1 = .318 A Thus as the resistance is lowered the apparent
power input approaches .318 VAR, but the I^2R quantity as real power
does get lowered proportionally to the decrease in resistance. Thus
the ratio of energy transfer vs heat losses was equal for a inductor
with a q of pi, so for an inductor with a q of 2 pi, the ratio of
energy transfer to heat loss must also ~ double. Thus we now have a
reliable mechanism to estimate the ratio of energy transfer to that
of heat losses, by making a ratio of the coils Q factor to that of Pi.

Now let us return to the case of X(C), where the energy transfer has
according to my hypothesis shown itself to be pi times less then the
VAR input. It seems easy to speculate that by giving X(C) a
cancelling reactance X(L): then the VAR input will change on account
of the rising I as a consequence of ohms law conduction value being
made, but V input remains identical so the left side of the equation
VI =2f(CV^2) will have its value increased Q or pi fold. However the
right term (CV^2) will have its value increased exponentially pi
squared, leading to the observation that the energy transfer should
now equal the VAR input, since originally it was thought to be pi
times less then that inputed, and the situation has been corrected by
now allowing pi times more multiplication for the right term in the
expression.

Thus for the original inductor of 1 ohm with 1 volt impressed
resulting in 1 amp conduction at resonance we have a simple VAR
input of 1, also equalling the heat losses of I^2R, another
tripartide unity of triple ones, and if the inductor has a Q of Pi,
the energy transfer is then 2f*C*pi^2. To find out if this also
equals one we must assign a value to C given these circumstances of
using a 8.3 mh inductor of 1 ohm @ 60 hz resonance.

Since both the conduction and impressed voltage are known, we can try
Paul's derived formula to find the unknown C, where I(rms) =V(rms)*2
pi*f*C
1= 1*6.28*60*C, C =(6.28*60)^-1 = 2654 uf, or .00265 F, thus in
actuality we CANNOT use that formula to determine an unknown capacity
in resonance, we can only use it to find an unknown capacity in
actual reactance current tests.

Finding X(L) for 8.3 mh; this has already been noted to be Pi, thus
we must also make an X(C)= Pi
Pi =1/2 Pi*f*C , Pi squared=1/2fC
2fC= 1/pi^2, C=1/2f pi^2= 8.44 *10^-4 Farad

By using Thompson Resonance formula
R(F)= 1/(2 pi*sq rt {LC}) we find that the needed C value is 848 uf
where the energy storage is then .0083 joules which transfering 120
times /sec yeilding 1 watt of energy transfer, thus at resonance with
an inductor of pi q factor the apparent power, the real power. and
the energy transfer all equal one watt.

To find the situation where the energy transfer is twice as great as
the ohmic losses, and also the apparent power, we need only make the
inductor have a q of 2 pi. The amperage term in VI will be doubled by
using half the resistance, for a new q of 2 pi. Formerly the energy
transfer was equal to the apparent power with an inductor with a q of
pi, so with twice the q factor, and twice the resonant rise of
voltage, we have 4 times the energy storage by V^2, and also I^2 in
each of the terms CV^2 and LI^2 as those field energies in
transition. Since the left term became doubled, but the right term
quadrupled this then indicates twice the energy in transition than is
being inputed. Thus it seems to make better sense to assume that the
VAR input can enable any energy transfer, according to the q of the
inductor being used.

Sincerely HDN
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