Re: micro tukomak

Reasoning regarding measurements

The rise in VSWR is the result of a change in the dynamic impedance of the
LCr circuit. The ionised gas forms part of the resistance. I use Thevethin's
theorem to regard it as a series resistor and if the plasma conducts better
that increases the load a reduces the dynamic resistance (this is an
increase in the effetive series resistance r) this is seen as a drop in the
S meter reading. This resukts in a mismatch thus increasing the VSWR. If the
ionised gas generates power (by fusion) then the dynamic resistance this is
because the effective series resistance goes down, if this become zero as
the part of the resistance due to the gas becomes negative then the dynamic
resistance (L/Cr) goes to infinity and if the effective series resistance
becomes negative the dynamic resistance becomes negative and the tuned
circuit will oscillate. If the ionised gas has resistance r then power
absorbed P=I^2R but if R goes negative then P goes negative too.

Negative resistance as a part of the effective series resistnce will change
the dynamic impedance and change the VSWR as well as the positive resistance
will. So my observation of a rise in reverse power shows a change in the
effective series resistance of the LCr circuit only this can be due to an
increase or a decrease of r. However the change in dynamic resistance for a
positive change of r is likely to be less than a negative change in r
because the negative change is approaching an asymtope that does not occur
with a positive change of r.

Conclusion - undecided.
"Zurikela" <zurabp@xxxxxxxxx> wrote in message
news:1d8350a9-f697-43b5-81da-99d81be72593@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Feb 4, 3:22 pm, "Chris" <ns_cjrs@xxxxxxxxxxxxxxxxxxxx> wrote:
And one more question: how does fusion energy transform in electrical
energy?

The plasma expands as it heats by the fusion reaction and this pushes the
flux out and causes more current to flow in the windings.

This increases the amplitude of the oscillations in the tuned circuit and
during the next part of the vibration the plasma in compressed and heated
more strongly than before, so the reaction gets hotter.

I use 2 Tor as the pressure now by the way. I found experimentally that
the
tube with hydrogen at a pressure less than 10^-4 Tor would not glow.

--
Chris.
London,
UK.

To reply remove ns_ from e-mail address

."Zurikela" <zur...@xxxxxxxxx> wrote in message

news:f4cf5a3c-b885-4bd1-a7ad-fb84ea22b52f@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Feb 3, 9:56 pm, "Chris" <ns_cjrs@xxxxxxxxxxxxxxxxxxxx> wrote:



Does "forward power" means 'input power" and ""reverse power" -
"output power"?

Yes.

--
Chris.
London,
UK.

To reply remove ns_ from e-mail address

."Zurikela" <zur...@xxxxxxxxx> wrote in message

news:0d09dd49-f5c4-4ac4-a541-3f4abb547637@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Feb 3, 1:54 pm, "Chris" <ns_cjrs@xxxxxxxxxxxxxxxxxxxx> wrote:

The input power backs off as the reverse power increases so cooling
the
plasma so with my present set up the power output cannot rise.

I had thought of using a linear power amplifier so heat and excite the
plasma. In this case I had intended to measure the reverse power as a
function of input power to extrapolate to see where the output power
equals
the input power at which point the system will self excite and give
out
net
power and then the exciter may be switched off and disconnected.

However I have been told by the manufacturer of the linear that the
reverse
power must not rise beyond the point where (forward power + reverse
power)/(forward power - reverse power) =2.5 or the amplifier will be
seriously damaged. Since the amplifier costs $3000 and I don't want to
risk
it, I cannot use one.

I have now designed a 50 Hz version that uses a battery to start it
and
there is no problem with the VSWR as it uses an oscillator to ignite
it.

I have sent an outline design to some local elecrical engineering
firms
that
are run by friends of mine and I have asked them to custom build one.

However if the design is successful they will probably hi-jack it.

--
Chris.
London,
UK.

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."David Kerber" <ns_dkerber@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in
message

news:MPG.23f135a119c61817989ce0@xxxxxxxxxxxxxxxxxxxxxx

In article <Gjdhl.5919$715.2...@xxxxxxxxxxxxx>,
ns_cjrs@xxxxxxxxxxxxxxxxxxxx says...
I think I am getting about 2 watts (continuous) of fusion power out
for
100
watts in to maintain the temperature and ionise the gas. I think so
because
if I tune the capacitor for a perfect match and power up the
reverse
power
starts to rise to 2 Watt after a delay as the tube gets hot. This
cannot
be
removed by adjusting the capacitor.

How do you plan to get a net power output from your device? 100W in
for
2W of fusion output doesn't strike me as a very good trade-off.

--
/~\ The ASCII
\ / Ribbon Campaign
X Against HTML
/ \ Email!

Remove the ns_ from if replying by e-mail (but keep posts in the
newsgroups if possible).

Does "forward power" means 'input power" and ""reverse power" -
"output power"?

From your recent experiments:" If my method is correct Poynting's
vector inside the coil will be 200/4.9E-4=407000 watt /m^2 so the E-
field will be given by:

<S>= (e0c/2)E^2 = 132.7x E^2 so E=SQR(407000/132.7)=55.38 volts/meter.

So for collision energy near windings,

40=m f p x 55.38 so m f p = 72 cm. That corresponds to a pressure of
10^-5 Tor.

I should be able to ionise the hydrogen in this tube.

Near the axis the field will be 13800 volts/meter so the collision
energy will be m f p x 13800 = 1 x 13800 or 13.8 KeV
This corresponds to a temperature of 2.2E11 K or 1100 times hotter
than the sun's core. Hot enough for proton fusion."
So iss your hypothesis?
And one more question: how does fusion energy transform in electrical
energy?

Thanks, now I'm fully informed, but many difficaltes in details: in
language and in physics.









i





.

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