## Nuclear Fusion Power

Reasoning

I have named it the Tomahawk reactor after my Red Indian girl friend as she is completely different from all the others.

A short talk about it http://youtu.be/PwUGlGxiJ_g

I have the notion that a low pressure hydrogen tube inside the coil of a tuned circuit will produce power after the fusion reactions are ignited by a pulse of electrical energy supplied to the coil.

The plasma is coupled to the coil because it is a conductor inside the coil and when fusion reactions take place the plasma heats an so expands and changes the inductance of the coil. This adds to the current in the coil by changing a parameter (the inductance) and so maintains oscillations and power may be taken from the tuned circuit.

The plasma is maintained and contained near the axis by a restoring central force due to the repulsion between the exciter current in the coil and the induced current in the ionised gas. This exciter current also heats the gas.

Further Theory

The reasoning is as follows: This is a diagram of forces inside a coil
with a coil inside the outer coil both carrying current

This theory assumes that the windings of the coil act as an antenna sending electromagnetic radiation towards the axis and concentrating the energy as it approaches the axis.

It is like the interior of a cylindrical mirror focusing the radiation at the centre.

Fusion of hydrogen nuclei occurs when two high energy protons collide. Fusion only occurs when there is sufficient energy in the collision.

In the set up the energy of particles in the tube is = mean free path x field strength.

Under the influence of the alternating current in the coil the electrons and ions follow circular paths thus having unlimited path length.

The electrons and ions travel very fast as they are very light and make thousands of orbits in one 1/4 cycle of the alternating current in the coil.

The acceleration is due to the circular Hertzian waves emanating from the electrons oscillating in the winding of the coil. These circular electromagnetic waves move towards the axis of the coil in concentric circles. As you know the electric a magnetic fields are at right angles and in phase. The electric field lines from concentric circles round the axis getting stronger towards the centre. The magnetic field is along the coil but getting stronger towards the centre.

The electrons and ions are moved in orbits by these fields which get tighter and more energetic as you move towards the axis.

The electrons are accelerated by the electric vector and deflected towards the axis by the longitudinal magnetic field of the electromagnetic wave.

As their velocity increases the magnetic field gets stronger confining them to their orbit. Ions close to the axis experience a stronger electromagnetic field and acquire higher energy.

The condition that the centripetal force=force due to movement through magnetic field will be met for some orbits as the electromagnetic field changes during the oscillation of the current amplitude.

The accelerated ions hit unionised atoms that stay put. The gas contains both ionised and unionised hydrogen atoms.

This has the effect of heating the gas near the axis and since there is so little gas and the collisions are so energetic the temperature of the region near the axis goes up very significantly. There will be practically no conduction or convection away from the central region and the radiation is trapped by the surrounding ionised gas and this radiation pressure is the mechanism where the heat of the fusion reactions are transferred to movement of the plasma against the magnetic field thus inducing current in the coil which maintains oscillation and the transfer of power.

As the amplitude of the wave increases over the 1/4 cycle this condition will be met for increasingly smaller radii orbits and the trons will tend to be compressed towards the axis since there is a force of repulsion from the windings on the trons forcing them towards the axis.

I’ve been considering mean free path and field strength and the energy of collision as my basis for the selection of pressure.

I selected the pressure in the tubes to be 1E-7 Tor. The mean free path at
this pressure is about a Km so the field strength of 500 volts/meter will
give rise to energy of collision of 500 KeV. This will be enough to ionise
the hydrogen easily and near the axis the field strength will be R/r times
higher. If R=25 mm and r=0.001 mm R/r = 25000 so the energy of collision
will be 1.2 GeV. This is enough for the fusion of Protons. (P + P>D: D +
D>He plus loads of energy). .

There are 10^9 nuclei per cc at this pressure so the small volume of 0.001 cc near the axis will have 10^6 nuclei. The energy given off per collision is 2.61 x 10^15 joules per mole and 1 mole is 10^27 nuclei so since there are only 10^6 nuclei there is 2.61 x 10^15 x 10^ (6-27) = 2.61 x 10^-6 joules to be given off in this small volume. So if there is a cross section of 0.1 and each batch of ions arrives every 5 x 10^-7 seconds (2 MHz oscillator) then the power will be 2.61 x 10^-7/5 x 10^-7 joules every second that is 0.5 watt.

This high energy may be high enough for mass to energy conversion of the protons. For 1 mole of hydrogen nuclei this is 9 x 10^13 Joule for 10^6 nuclei this is 9 x 10-8 Joule giving a power of 5KW

There are 10^9 nuclei in the tube so if 10^5 react every 5 x 10^-7 seconds it will last 10^9/10^5 x 5 x 10^-7 seconds. That is 5 x 10^-3 seconds.

And give out 2.5 x 10^-3 Joules

So for reasonable powers of 20 KW the number of reactions per second is calculated thus: number of reactions per second shall be = power/energy per reaction.

20x10^3/2.61^15/10^27 = 7.66 x 10^15 at 0.1 cross section the number of ions needed is 7.66 x 10^16. If the volume of reaction is 0.1 cc the number of nuclei/cc should be 7.66 x 10^17. This corresponds to a pressure of 1 mbar or 0.75 Tor. The mean free path at this pressure is 0.1 mm. To reach the energy required for fusion a field strength of 400,000 volts/meter is required to ionise the hydrogen and near the axis the field strength rises to 1 x 10^10 volt/meter and with a mean free path of 0.1 mm the energy reached on collision will be 1 MeV. This would be enough for D + D reactions. The equivalent temperature would be 1.16 x 10^13 K or 11 billion degrees K. This is enough for P + P reactions and is over a million times hotter than the sun.

So for a 26 mm coil a potential of 12000 volts per turn is required so for a 10 turn coil an anode voltage of 1.2 million volts is required.

Experiment says differently: I ionised and heated to white heat hydrogen gas at 0.001 Tor (10^-3 Tor) in an 850 watt microwave oven. So I believe a power of 1 KW would be sufficient for this pressure.

The power developed depends on the number of nuclei per meter cubed which depends on the pressure and the field strength required to give sufficient energy to ionise the gas and to accelerate the ions depends on the mean free path and this depends on the pressure. So the pressure to use depends on this trade off. So for high power, high pressures are needed and thus very high potentials but for low power, low pressure and low potentials may be used.

The energy given off by the fusion reaction is so high that even with very low pressures the power is quite high.

The initial stage is to ionise the hydrogen, here the energy has to reach about 40 eV.

In order to execute fusion the energy of the ions has to exceed 200 MeV.

In a coil the field due to the acceleration of electrons in the wire of the coil is the potential per turn divided by the circumference at the radius of measurement. This means that the field strength increases as 1/r towards the axis.

During the cycle of the exciting potential the field will increase from zero to a maximum and then return to zero over 1/2 a cycle. During the cycle ions in the gas will be forced into spiral paths towards the axis with ever increasing energy.

Once near the axis their energy will be so high that fusion will occur and then during the next 1/4 cycle the extra energy will be acting against the magnetic field and thus increase the current so as to maintain oscillations.

My oscillator has a field strength of 500 volts/meter around each turn. To ionise hydrogen an energy of 40 eV is required this requires the electron to fall through 40/500 meters = 0.08 meters or 8 cm. Suppose we select a pressure for a mean free path of 10 cm. That corresponds to 10^-3 mbar or 7.5 x 10^-5 Tor.

So near the axis the field strength will be R/r=25/0.001 x 500 = 1.25 x 10^7 volts/meter so an ion would collide with an energy of 1.2 MeV. This is equivalent to 20 billion degrees K a million times hotter than the sun and high enough for P + P >D reactions to occur.

Actual Values From experiments.

The tube I have at the moment has a pressure of 10^-3 Tor and did not ionise as the mean free path at this pressure is 10 cm. With a field strength of 500 volts per meter the energy on collision would be 50 volts and this is not enough. The power of my oscillator is not very high but when I put the same tube in an 850 watt microwave oven it did ionise. In a 650 Watt oven it did not ionise so it is very critical.

I do not know the field strength of radiation at 850 watt but you should be able to work it out from Poytings vector.

I'll look up the formula S=E x H

<S>=(e0c/2)(E^2) so if S=850 =132.75 E^2  so E^2=850/132.75 = so E=2.53 volts/meter.

This means that the m f p is effectively given by m f p x field strength= 40 so m f p=40/2.53 = 15 meters

For the 650 watt oven,

<650>=132.75 E^2 E=sqr(650/132.75) = 2.2 volts/meter

m f p= 40/2.2 =18 meters

So 15 meters is the path the electron takes to ionisation this is more than the m f p of 10 cm.

So the path it needs is 150 times the mean free path.

So for a path of 200 meters the mean free path will be 1.3 meter making a pressure of 10^-5 Tor.

I think it may be a bit more complicated because the energy S is per square meter. The microwave oven is about 300 mm x 400 mm so is only 0.12 square meters and my coil is only 25 mm diameter so about 1.963 x 10^-3 square meter

So the power density in the microwave oven is about 850/0.12 = 7083 watt/meter^2 So the E-field will be 7.3 volts/meter and for the 650 watt oven the power density is 5416 watt/meter^2 and the E field will be 6.3 volts/meter

So the path to ionise is 40/5.25 meter = 7.6 meter this is 76 times the m f p.

For my oscillator the power was measured at 4 watt,

so:

If there is only 4 watt available then the field strength is worked out:

energy/turn 4/33 = 0.12 watt <S>=0.12/4.9 x 10 ^-4=244.46 watts/m^2

E = Sqr(244.46/132.5)=1.36 volts/meter.

So to compute the path we use path = 40/1.36 = 29 meters so by experiment:

So the mean free path comes out at 29/76 = 0.38 meter or 380 cm. I estimate that the pressure should be about 10^-6 Tor.

If the power travels down the coil then Poynting's vector is 4/4.9 x 10^-4 = 8163 watt/meter^2. So E=SQR(8163/132)=7.86 V/m.

So the path=40/7.86 = 5 meters so m f p = 66 cm. That makes the pressure 10^-5 Tor.

So I should get ionisation on either my 10^-5 Tor tube or my 10^-6 Tor tube.

The tube with a pressure of 10^-7 Tor has a m f p of 100 meters so the path will be 7.6 Km so the ionisation potential will be p x field strength=40 field strength= 40/7600 = 5.26 E-3 volts per meter so the power to ionise will be given by

<S>=132.75 x (5.26E-3) ^2 = 3.5 E -3 watts per meter squared so for a coil 3 cm x 3 cm the power required will be 3.2 E-6 watt.

And with a collision path length of 7.6 Km the energy of the collision with an accelerating field due to a power of 100 watt in 6.25 x 10^-4 meter square will be sqr(160000/132.5) x 7600 = 9.4 MeV. This is equivalent to 11 trillion K. (11x10^13). Hot enough for fusion.
(using 1 KeV == 1.16 x 10^7 K).

With an accelerating potential of 500 volts/meter the energy would be 3.8 MeV so the temperature equivalent is 44.08 trillion K. (4.408 x 10^13).

This requires a radio frequency power of 132.5 x 500 x 500 = 32 million watt/meter squared. In a small container of 6.25E-4 power would be 20 KW.

This would probably sufficient for matter into energy conversion as in a black hole.

We could drive quite a lot a machines from power like that. The power out would be very low unless the hydrogen could be replenished. So the unit would need the hydrogen drawn slowly through it. E=mc^2 so if 1 g of hydrogen was consumed every second the power would be 10^-3 x (3x10^8)^2 = 90,000 GW.

For fusion the power would be less.

4P>He plus 2.61 x 10^15/1000 Joules so for 1 g/second the power would be 6.5 x 10^11W. This is 652.5 GW.

Inductance of coil in presence of ionised gas inside.

The ionised gas is a conductor and to a first approximation screens out the magnetic flux from its interior.

The inductance of a coil is given by:

µAn2/l where A is the area of the coil, n in the number of turns and l the length.

Where A is the area of the coil.

When the plasma shuts off some of the area the total area is reduced to Ac-Ap so the inductance will be reduced to (Ac-Ap)/Ac = Ac(1-Ap/Ac)

If the inductance without the plasma is L1 then it will be L1(1-Ap/Ac)

The frequency of oscillation is 1/(2л√(LC)) and so is proportional to 1/√(1-Ap/Ac).

So as the plasma changes its area with temperature the inductance changes this tends to maintain oscillations. (See theory of parametric amplifier).

The tube will have a resonant frequency which is due to an acoustic standing wave inside the tube. This will be the radius of the tube/velocity of sound.

The expansion and contraction resonant frequency would be a sound wave propagating through the gas and so the resonant frequency will depend on the velocity of sound in the rarefied gas and the dimensions of the tube.

The tube is 25 mm in diameter and the velocity of sound in hydrogen is: 1.3 Km/sec so the resonant frequency will be 104 KHz.

If the hydrogen becomes monatomic in the tube then the velocity becomes: 1.4 Km/sec and the frequency becomes  114 KHz.

So the exciting current should be at this frequency or a harmonic.

The 20th harmonic is 2 MHz.

The new oscillator with a 50 volt PSU (36 watt).

I think I need 1000 watt at 10^-3 Tor but I'm trying a lower pressure and
lower power. My oscillator is only 9 watt but this PSU should bring it up to
36 watt. The potential across the tube will rise to 1442 volts RF. That is
14,425 volt/meter. It takes 50,000 volts/meter to break down hydrogen so this is not enough..

Inside the coil Poynting's vector will be 36/4.9E-3 = 7346 watt/m^2 so the
E-field of the waves propagated towards the axis will be given by
<S>=e0c2E^2 so E^2=<S>/e0c/2 =<S>/132.75 so E=sqr(7346/132.75)=7.43
volt/meter. m f px7.43=10000x7.43 = 7430 eV. It might ionise hydrogen at
10^-7 Tor

These collisions may cause fusion.

I'm considering using a 600 watt oscillator, this will give 5888 volt across the coil and 58,889 volts/meter that should be sufficient to break down the hydrogen. Then the circular waves propagating towards the axis will compress it and contain it and because they are focussed on the axis the power rises to a maximum in the vicinity of the axis and so the pressure is highest here and the collision energy will rise to very high values here.

Inside the coil Poynting's vector will be 600/4.9E-3 = 122448 watt/m^2 so the E field propagating inwards will be 30.4 volts/meter near the axis it will be 1200 volts/meter. If the mean free path is 1 meter than the mean collision energy will be 1.2KeV this is equivalent to 19 million degrees K. So the gas needs to be at a pressure of: 10^-6 Tor.

The spot of glue turned out to be a red herring. The mechanism of ionisation involves the microwave energy heating the gas first and then it ionises in the electric field.

The energy density in the oven is 850/(0.2x0.28x0.28) = 54209 watts/m^3. In my oscillator the energy density is 12/(4.9E-3x0.1) = 24489 Watt/m^3 so it will take 2.2 times as long to heat up. If the energy of the oven is concentrated in the tube then the power is 850/12 or 70. So it will take 70 times as long to heat up = 70x 40 seconds = 47 minutes.

note: from "Introduction to Plasma Physics and
Controlled Fusion" by Francis F. Chen:

ni/nn ~= 2.4E21 * T**(3/2.) * e**(-Ui/(K*T))/ni

ni: number of ionized atoms per m**3
nn: number of neutral atoms per m**3
T: gas temperature in degrees Kelvin
K: Boltzmann's constant
Ui: Ionization energy of the gas in ergs
e: Euler's number

It seems that a Townsend discharge starts first and the current heats the gas then at a particular temperature more current flows and the glow discharge starts.

In the coil the repulsion between the windings and the plasma current will keep the ionised gas away from the wall of the tube and compress it towards the axis. The focusing of the energy will mean that the region near the axis gets a lot hotter.

A coil is a length of wire.

If we take a length of straight wire and pass a current through it the effect of the current is greatest near the wire and falls away as you go away from the wire.

The effect of current is only measurable by its effect on another current carrying wire.

The force between the wires is mediated by virtual photons, one of which connects one electron with another electron on a one on one basis. The force mediated by one photon does not fall off with distance. The inverse square law is the effect of the density of photons falling off according to the inverse square law because they spread out from the source.

With two parallel wires a short distance apart the area the photons occupy at distance r is equal to 2лr so the same number of photons leaving the wire has to be spread out, so the number of photons falls as 1/r and thus the force falls off as 1/r.

In a circular loop each wire has photons spreading out falling off in density as 1/r in a coil the next loop also has this property and so the force falls of as 1/r from each loop towards the axis.

If you use the law of Biot-Savart the force from an element of a wire in another element of wire = (I*L * I*L /r**2) Sin (theta) where theta is the angle between the conductors and r is the distance between them. With a loop of wire inside another loop of wire both carrying current I the way to calculate the force between the conductors is by integrating round each loop. A coil is a set of loops.

If you do this you find that the force is a maximum when the wires are close and smaller when they are further away. The force varies as 1/r towards the axis where it turns round. It is not zero at the axis.

For alternating current flowing in the outer winding then a current is induced in the inner winding these current repel each other.

With a transformer there is a force of repulsion between the inner and outer winding known as the bursting force. If the inner coil is displaced from the central position one side is closer to the wire than the other so there is an overall force toward the central position.

If the inner coil is a conducting gas then it will be compressed towards the axis and be perfectly stable as a pillar of ionized gas round the axis.

With hydrogen, which is difficult to ionise, the exciting current causes a Townsend discharge first that does not glow, then as the gas heats up the number of conducting electrons and ions increases and a glowing ionised conducting gas forms. The ionised gas is forced away from the walls and heats up through ohmic heating. This is greater near the axis because the photons from the windings of the coil are focused on the axis so the gas gets very hot here.

Hot enough for proton-proton fusion at 200 MK.

The heat of the fusion reactions makes the gas expand against the photon field that confines it and this causes more current in the windings thus transferring power to the external circuit.

The oscillations then become self sustaining and power may be extracted.

It may be necessary to control the reaction with a shunt regulator.

It has just occurred to me that the power of the reactor may be controlled using a spark gap as a shunt regulator.

Temperature of the Inductor

The heating effect on the coil depends on the power but the cooling rate is proportional to T^4 (Stephan's law of cooling).

So 100 W=k * (273+60)^4 so k=8.13E-9. So with 1000 watt the temperature will be given by:

1000=8.13E-9 *T^4 so

T^4=1000/8.13E-9 so

T=(1.23E11)^1/4=592K=319.16C

The melting point of copper is 1083C so it won't melt. It will just get very hot. The melting point of solder is 188 C so the soldered joints might melt. The enamel coating will probably burn off.

Pressure in my prototype reactor. About 6 amp flows in the coil of 13 turns so 13x6 amp flows in the plasma this is 78 amp. The force near the windings is e0 x length of turn x 78 x 78= 4.23E-4 N. This is over the length of the coil of 6 cm so the pressure is 8.9E-2 Pa. Near the axis at 0.1 mm it is 250 times higher or 22.25 Pa. Atmospheric pressure is 10E5 Pa so the pressure is 0.16 Tor.

If the power is raised to 1000 watt from 100 A the current will rise by 3 times so the pressure will rise to 1.4 Tor.

That means that the pressure in the tube is not exceeded indicating that there will be a lot of unionised gas as an atmosphere to the plasma. If no convection takes place it will help to insulate it.

I am unsure if this reasoning is correct as with all my reasoning.

Estimate of potential across coil.

The power is 100 watt into 50 ohms so since Power=V^2/R Voltage=SQR(100x50)=70 volt so the potential across the coil of 13 turns is 910 volts.

When I increase the power to 1000 watt the current will be 18 amp, potential 2730 volt.

I=V/R so the current in is 1.4 amp. So the Q is 4.2.

The current indicator read 1.4 with the tube but 1.9 without.

So the resistance of the tube robbed the coil.

1.9/1.4 x 6 = 8 amp.

The current in the coil without the tube is therefore 8 amp so the dynamic resistance without tube is 910/8 = 113 ohm but with the tube is 910/6 = 151 so the tube is a resistance of 38 ohm. It acts as a series resistor.

Actually the current goes down when you put in the tube so it a series resistance of 38 ohms. So as the resistance of the tube falls the current will rise and when the resistance goes negative when it develops power the current will rise still further.

If the tube develops power the resistance of the tube will fall and then go negative so the indicator will rise.

The hydrogen ionises in a two stage process, at first a Townsend discharge takes place then when the gas is hot enough more electrons are released from the hydrogen atoms and what is called an anomalous discharge occurs and the gas becomes ionised and glows.

The containment occurs because the ionised gas shuts out flux from the plasma making the flux higher between the coil windings and the plasma and this compresses the gas. The force on the plasma is proportional to I^2.

Because the region near the axis has a higher plasma current density than near the winding the heating effect is higher. The ration of current density near the axis to near the windings is about 2000:1 so the heating effect is 4E6 times higher. The gas insulates the inside from the outside and the lower pressure region near the winding (the gas is forced out of this area by the force of repulsion of unlike currents) this acts as a vacuum insulation too. The radiation from the axial region is reflected back by the ionised gas outside thus increasing the insulation.

That is why fusion will start near the axis.

When I run 100 watt in I get 0.5 Watt out. This cannot be removed by tuning the capacitor. So is this power generated by the plasma?

If power is being generated by the plasma the ratio power out/power in will be a increase. To do measurement I need a circulator. This will also stop power getting back to the transmitter enabling it to maintain full power. It would also mean that the receiver would not see the power generated by the plasma when I key the transmitter off. Although some power would get through to the receiver it would not be enough to damage it.

I have now discovered that circulators for the frequency I use are not possible owing to their high cost.

I have been thinking about the mismatch. My circuit is a tuned circuit inductively coupled to a transmitter with an impedance of 50 ohms. The best match was found by trial and error to be two coils each of two turns in parallel at each end loosely coupled to the inductor coil of 16.5 turns. The impedance of the tuned circuit at resonance is its dynamic impedance. If this changes the impedance is still purely resistive as this does not change the frequency of tuning it just makes it broader.

The plasma is inductively linked to a coil carrying RF. The plasma itself is equivalent to a loop of wire which has resistance. The circuit is therefore a transformer with the one turn and series resistor being the plasma. So the plasma is in series. The primary of the transformer is the coil of the tuned circuit.

I would suggest that the effective series resistor of the tuned circuit is in series with the load of the ionised gas (I’m using Thevethin’s Theorem). That would mean that the ionised gas will not change the centre frequency of the LC circuit.

Because P=I2R if the some of the ionised gas is making power then that part of R is effectively negative that means that the effective series resistance of the LC circuit is reduced thus increasing the dynamic impedance and so creating a mismatch.

When the gas first ionises the extra load means that the effective series resistance in increased thus reducing the dynamic impedance. This also creates a mismatch.

I cannot tell the difference in my set up. See below.

Dynamic Resistance:

Rd=L/Cr where r is the effective series resistance.

For a change in the effective series resistance:

Rd+ДRd=L/C(r+Дr)

So if the transmitter sees 50 ohms initially then an increase in the effective series resistance would result in a reduction of the input impedance of the reactor and a reduction of the effective series resistance would result in an increase of the dynamic impedance.

So if the effective resistance changes then the transmitter would see a change in impedance. Both an increase and a decrease in the input impedance would result in a mismatch. So a change in the effective series resistance would result in a mismatch.

As I have no means of measuring the input impedance of the test cell I only have the VSWR to go on. This is (forward power-reverse power)/ (forward power + reverse power).

I think that as the VSWR is always positive so I think the effect o a change in the dynamic resistance always increases the VSWR. So there is no way of showing that the cell is generating power from the VSWR.

I think that if the gas was generating power the effective series resistance (r I equation 1) would go down. I’m thinking of a resistor made up of two resistors one is the resistance of the ionised gas, when this generates power the effect is that of a negative series resistor. (Power absorbed=I2R so if R goes negative the power is emitted).

The effect of a positive effective series is to reduce the dynamic impedance. If the effective series resistance goes down then the dynamic resistance goes up if the effective resistance is zero then the dynamic resistance goes to infinity. If effective series resistance goes negative the dynamic resistance goes negative and the circuit will self oscillate.

This is seen on the VSWR as an increase from zero when the circuit is matched to a higher and higher mismatch as the dynamic resistance increases and when the dynamic resistance goes to infinity the VSWR rises to infinity.

After that the forward power may be reduced to zero and there will still be a reverse power showing.

In fact the ionised gas shows a series resistance that decreases the dynamic impedance that creates a small mismatch. If the gas conducts better this reduces the extra series resistance and will increase the dynamic resistance, making this smaller or zero will not change the dynamic resistance as the effective series resistance approaches its value that it was without the tube. The only way the dynamic resistance can increase is if the effective series resistance decreases and this can only occur if the plasma resistance goes negative meaning that power is being produced and this can only come from fusion in the gas.

The mismatch was a VSWR of 1.2 with Forward Power=100 and reverse power = 2

(VSWR-1)/(VSWR+1)=(ZL-ZS)/(ZL+ZS)

0.2/2.2 = (Z-50)/(50+Z)

0.167*(50+ZS)=Z-50

8.35 +50 + 0.167 Z=Z

58.35=Z*0.833

Z=69.62 OHMS

Since the turns ratio is 15:1 the impedance ratio is 225:1 so the impedance of the tank circuit for the input impedance of 50 ohms is 11250 but for the state where the reverse power is 2 watt then the impedance of the 15664.5

Rd=L/Cr =2.3E-6/8E-10r=2875/r

with the input impedance of 59 ohms r=2875/15664.5 = 0.18 ohms when the input impedance is 50 ohms r=0.26 ohms so the resistance of the plasma is 0.18-.26 = - 0.08 so the plasma has a negative resistance of 0.08 ohms so 2 watt of fusion is going on.

This represents  burning1E-10 g of hydrogen per second.

You wrote:

Note of error:

There are 10^9 nuclei per cc at this pressure so the small volume of 0.001 cc near the axis will have 10^6 nuclei. The energy given off per collision is 2.61 x 10^15 joules per mole and 1 mole is 10^27 nuclei so since there are only 10^6 nuclei there is 2.61 x 10^15 x 10^ (6-27) = 2.61 x 10^-6 joules to be given off in this small volume. So if there is a cross section of 0.1 and each batch of ions arrives every 5 x 10^-7 seconds (2 MHz oscillator) then the power will be 2.61 x 10^-7/5 x 10^-7
joules every second that is 0.5 watt.

Correction of power generation by Larry.

You would be closer to 2.61 x 10^15 x 10^6 / (6.022 x 10^23) = 0.00433 joules = 0.00433 watt second for the volume. So you would have 0.00433 x .2 x 10^7 = 0.000866 x10^7 = 866 watts. Seems like you will be able to heat a small room.

Larry

I have an LC circuit with a tube of hydrogen at a pressure of 2 Torr inside the coil. I set the tuned circuit capacitor so the circuit is tuned to the transmitter frequency and I selected a matching coupling coil so that the reflected power is zero when the coil is hot. Then I put the tube of hydrogen inside the inductor and bring the transmitter power up again. At first the reflected power is zero but after a few seconds this rises to about 6 watt.

This rise must be due to the plasma generated in the tube as it glows.

The cold gas does not conduct electricity and so the tube has no effect when it is cold.

The hot has contains ionised gas that conducts electricity. Now the gas inside the coil is effectively a single turn of a conductor with resistance. This resistance is reflected back to the primary of the tank circuit as in a transformer so it appears to the primary as a small resistance.

In the tuned circuit the dynamic impedance is L/Cr where r is the effective resistance. In the case of no tube the dynamic impedance is L/Cr but in the hot gas case the dynamic impedance is L/C(r+R) where R is the resistance of the gas reflected back to the primary. When the gas is hot R is quite low and does not effect the dynamic impedance much but if power is being generated by the gas as it would if a small amount of fusion is taking place then R will go negative and the sum r+R would go down. This increases the dynamic resistance and so the transmitter would see an increased VSWR.

Once r+R=0 then the dynamic impedance goes to infinity and the VSWR would rise to infinity. If r+R<0 then the LC circuit will oscillate and the reverse power would rise above the forward power. At that point the transmitter can be turned off and disconnected and the LC circuit with the tube of low pressure hydrogen would become an oscillator and a source of power.

I think I have demonstrated containment by this method and heating. It is possible that I have demonstrated a small amount of fusion power and the generation of electrical power using this method.

The method I am using is not mentioned in the literature. The confinement time is high, approaching infinity, and the nuclei density is high too and the gas is at a pressure of 2 tor, the temperature is actually calculated in terms of the energy of the ions. And the hydrogen is ionised and a current passes through it induced by the rf current in the coil. I assert that the radio energy density near the axis is 2000 times that near the winding because it is focussed. I did work this out and I got a collision energy equilvalent to 200 million K. This is high enough for proton fusion. (4P+4e>He+gamma). At present the loss from the plasma prevents the temperature building up but if I use higher power this will be greater than the loss and so the interior will heat up. That means the lawson criteria is close to being achieved.

The RF power near the winding is 100 watt so near the axis it will be 200KW. This is absorbed by the ions. If I recall correctly P=E^2/275 where 275 is the impedance of free space so 2E5=E^2/275 so E^2=55E6 sp E=7416 volts/meter. The mean free path at 2 Tor is about 5 mm but the tail of the distribution of paths before colision bring a 100 fold increase to 0.5 meters giving high energy ions a collision energy of 3.7KeV. As the excitation power increases to 1KW this increases to 37 KeV. This means that at low excitation energies the number of fusion events is low but much higher than room temperature and at 1KW excitation the number will greatly increase.

According to my textbook on "controlled nuclear fusion" 1KeV is equivalent to 1.16E7K so 3.7 KeV > 42.9 million K and 37 KeV > 429 Million K. Both of these figures is hotter than the suns interior and 429 MK is hotter than the estimated 200 MK at the sun's core.

That would imply that the prototype fusion reactor can reach 400 Million K near the axis and therefore generate fusion reactions.

The reaction rate will increase rapidly with increasing energy of excitation as both the pressure (remember the repulsion of unlike currents compresses the plasma) and the proportion of ions energetic enough to execute fusion will increase.

I think the 4P>He is a two stage reaction, P+P+2e>D+e+ +e- (e+ +e- > gamma) then D+D+2e>He4. No neutrons produced. There will be other side reactions that will yeild neutrons.

The neutrons are stopped by the ionised gas around the core and decay N> P+e- and take part in further P+P reactions.

Here is a correction:

The power near the windings is not 100 W/meter it is 100 W/area of coil. This is 0.06 m x 0.025 x 0.024 x 3.14/4 = 2.945E-5 so the power density near the windings is 3E6 Watts/meter^2. So the power density near the axis will be 6.79E9 watts/m^2 so the E field will be SQR (6.79E9 x 275) = 1.3MV/meter for a mean free path of 5 mm the collision energy will be 6.8KeV. This is equivalent to a temperature of 1.16E7x6.8 K= 78MK. This temperature is as hot as the sun and will support P+P reactions. The more energetic ions will have energies up to 680KeV giving a temperature of 7.8 Billion K. This is hotter than the core of the sun.

However the temperature reached also depends on the losses from the centre and although the central region is thermally insulated by ionised gas and low pressures giving low radiative and convection transport there is a loss of heat from the tube as seen because the tube heats up.

With higher input power this loss will be overcome and the temperature of the central region will rise. As fusion reactions become more frequent the heat content will rise and this further increases the fusion reaction rate.

The reaction rate depends on both the temperature and the pressure. The pressure in the sun is thousands of tonnes force per meter squared. The temperature is about 200 million K.
The reaction rate at the pressure in the tube is much lower than in the pressure in sun at the same time. However the force on the ions from the current in the windings compresses them into the axis. And because of focussing the pressure near the axis is higher than near the windings.

The pressure near the windings is the force/area and the force is the current x number of turns x e0 so the pressure is INe0/(length of coil x circumference of coil) this is about 6x15x8.85E-7/0.06 x 0.0785 = 0.0168 Pa but near the windings it will be 0.0168 x 2000 Pa = 33.82 Pa.
This is additional to the pressure of 2 torr of the gas in the tube. one atmosphere is about 10^5 Pa so 2 Tor is 263 Pa so the total pressure is 296 Pa.
The reaction rate is the probability of a fusion x collision frequency and this may be calculated. The probability of fusion is known as the cross section. The collision frequency depends on the particle density and therefore the pressure.
I looked up the power density from "controlled thermonuclear reactions" (ISBN 0-88275-326-6) where P+P reactions are not listed as the cross section is too low but for D-D at 40 KeV is 4 watt /cm^3 so my 29 cm^2 tube with only 0.1 of the volume being active is therefore 11 watt. I cannot see from the graph how to estimate the effect of pressure.

However at 2 Tor the particle density is 10^16 particles per cm^3 and if the chances of a collision is given by the mean free path of 5 mm then the number of fusion reactions in a cc is 1/8 x 10^16 per cc so if each collision makes a fusion reaction then there will be 2.61E15 x 1/8 x 10^16 /10^23 per cc 32E6 joules. The collision frequency is 1/transit time of 5 mm this comes to 2x10^11 collisions per second.
So the power will be 32E6 x 2x10^11 or 7.2 x 10^18 watt.
I would suggest that experiments show that the power level obtained is between the two values.

The cross section is given in barns. I do not understand Barns. However the probability of a collision resulting in fusion depends on the energy. So with a higher power exciter the cross section will increase thus increasing the reaction rate and thus the power of the reaction. At 1KW exciter the collision energy will rise to 400 eV where the D+D reaction rate is astronomical. The P+P cross section is smaller than D+D.

When a P approaches another P the protons become polarised. This is quatised so a polaron is exchanged (also known as a meson). This makes the force between the two protons change from repulsion to attraction. There is a probablity which rises with collision energy that the polaron which is a virtual particle will obtain enough energy to become real and becomes a positron which is ejected and takes away the collision energy. In which case one of the Protons becomes a neutron and becomes bound to the proton by the exchange force as the N and P change places as P<>N exchange takes place. This is then a Deuteron.

I could not find a reference to the cross section for this reaction.

This is the first interaction. The next stage is D+D>He4. This has a low cross section, however the reaction D+D>He3+n has a cross section of about 80 mBarn at 400KeV.

My previous experiments showed no signs of neutrons so the neutrons must have been mopped up by He3+n>He4 because of the high particle density.

The rapid removal of the D formed by the P+P reaction means that the reaction P+P>D is favoured over D>P+P and so the reaction will go.

This is similar to the reaction kinetics of wet chemistry.

The other reaction that is occuring is e+ + e->2gamma. In previous experiments these gammas were detected at 0.51 MeV. The energy of positron-electron anhilliation.

Most of these gammas are stopped by the ionised gas outside the active region.

Looking at astrophysics I see the P+P reaction chain takes place at 20MK and the cycle takes 10^9 years to complete. This is for a single reaction chain. However a 400 KeV collision is equivalent to 4640MK so using the theory of exponential increase where the reaction rate doubles for every 40 KeV increase the rate at this temperature will be 994 times faster than at 20 MK (1.7KeV) so the reaction cycle will complete in 1E6 years or 3.17E13 seconds. If the collision frequency is 2E11 collisions per second the reaction rate will be 6.3E-3 fusions per second making a power of 32E6x6.3E-3 per cc =2MW per cc. My reactor has an active volume of 2.3E-2 cc so the expected power is 4.7KW.

It has occurred to me that when there is no tube I could not get a balance because the resistance of the windings was causing a mismatch but when the tube was in place power generated by the ionised gas neutralised the loss by the resistance and thus allowed a match to be made.

The spectrograph result indicated that line broadening had taken place. The factor is SQR (kT/mc^2) where k is boltzman's constant and m the mass of the molecule and c the velocity of light.

Comparison with the published spectrum of hydrogen shows that the red line had split into two and looked about 10 x broader than normal.

c=3E8: k=1.3E-23 m=1/6E23 so T=line broadening^2 mc^2/k=10^2 x (1/6E23 x 3E8^2)/1.3E-23 =1.1E18K assuming the spectral lines had been broadened by a factor of 10. That is much too hot so my measurements are wrong.

I measured the line again and the hot red line taking the left hand line was 0.3 wide and the reference line was 0.2 wide making the broadening 1.5

So T=2.25 x factor =2.475E18, if I use 0.3-0.2/0.2 then the broadening becomes 0.5 and T=5.5E15 K.

This is all much to hot so I must be making a bad measurement.

The torus was difficult to wind a commercial unit needs the coil to come in two halves that clamp together. The linear version was easier to ionise as I saw very little ionisation in the torus at 200 watt. Also in the torus the current peak is outside the coil near the soldered joints causing them to overheat and melt. So I will concentrate on the linear version now. Because the tube has been used for some time I need some new ones filled.

I did some calculations and I do not think the effect would work at 50 Hz unless the coil was very big.

When the plasma current flows since it is a radio frequency alternating current there will be an associated electric field. Since this is part of the electromagnetic wave travelling inwards then its magnitude will be given by the ratio of E to H in the wave. The H-field is along the coil and is related to the current flowing locally. I suppose you could take the wire in the coil as an antenna with waves propagating inwards. I suppose a coil makes a spiral wave propagating through the coil.

A coil is a length of wire wrapped up into a helix. So if you take it as a length of wire and the concept of current making radio waves around it the wrapping up process makes a radiating wire putting radio waves into the axis where they form a standing wave.

We can use the concept of ampere-turns per meter to get the near H field and then use the impedance of free space to get the electric field in volts per meter.

Z0=376.7 so E/H=376.7 so E=376.7 x H but in my coil with 24 amp flowing in the coil H=24 x 15 / 0.06 A-T / m = 6000 so the E field is 2260200 volt per meter. That is 2.2E6 V/meter. So an ion in hydrogen at a pressure of 2 Tor which has a mean free path of 2E-4 meters will get an energy of 400eV for an average collision. That corresponds to a temperature of 6.4 MK thus explaining the spectrum.

For a temperature of 200E6K a collision energy of 17 KeV is required

However if I can get a 6 KV air vane variable capacitor at 1000pF then I have the capacity to get 2000 ampere turns at 130 amp. But my transmitter is only 1000 watt so the current I can deliver is only 24 x SQR (1000/100) = 75 amp or 1130 ampere turns so the plasma current will be 1130 amp. This means I will be able to reach a collision energy of 1.4KeV this is equivalent to 22MK.

I should get some fusion events with this energy.

To get 200 MK a transmitter power of 100 KW is required.

This is from collision energies not from Ohmic heating. The losses from the surface go up at T^4 but the ionised gas insulates the interior and so the outer layers will be much cooler than inside.

When I tested it and found 400 watt thermal power, I was unable to understand why no reverse power was seen. I now realise that reverse power is only seen for a change of phase caused by a reflection. The power generated in the coil is in the same phase as the incident power so merely adds to it.

I therefore need an e-field meter to measure the radiated Hertzian power.

The spectrum shows line broadening indicating ion velocities along the line of sight. The broadening is greater along the long axis of the tube than the transverse axis. This indicates that the velocity if higher along the long axis of the tube than in the transverse direction. So it looks like the ions spriral up and down the tube at very high velocities.

The longitudinal broadening is very pronounced 2.5/0.1 = 25 and the transverse broadening is 1/0.1=10 so the ions are moving at 2.5 times the velocity in the longitudinal direction than the transverse direction.

If you make this a kinetic temperature we get a temperature given by delta f=fo SQR (kT/mc^2) so

SQR (kT/mc^) = f/fo = lo/l

kT/mc^2=(lo/l)^2

T=(lo/l)^2 x mc^2/k =(25 x 650x10^-9)^2 x ((1 / (6 x10^23)) x (3 x 10^8)^2) / 1.3 x 10^-23 = 3x 10^6

So the kinetic temperature longitudinally is 3 million K and transversely the temperature is 1.9 million K.

The density of hydrogen at STP is 0.00008988 g/cc so at 10^-8 tor it will be 8.988E-5 x 1E-8/760 = 1.18E-15 g/cc. The phial was 10 cm long by 2.5 cm in diameter so its volume would be 49 cc so the mass of hydrogen present would be 5.79E-14 gram or 2.896E-13 mole. The reaction overall is 4P>He so (1.0079 x 4 – 4.0026) = 0.029 so the energy released by the fusion of 2 moles of H2 would be 2.61E12 Joules that is 1.305E12 per mole. So if all the hydrogen in the tube was converted, 0.3779 Joule was released.

The conversion took 180 seconds so the power level was therefore 2 mW..

In order to obtain this transmutation I supplied 350 watt for about 180 seconds so the energy input was 63000 Joule

To obtain useful power levels higher pressures are needed and therefore higher exciter power to excite the gas. At some point the system will become self exciting and then we can generate useful power. I guess this occurs at 3 Tor and 1000 watt exciter power.

I think it generated power because the reverse power rose to 200 watt and the forward power rose from 300 watt to about 700 watt before I used the tuner to reduce the reverse power signal to zero and the forward power fell back to 300 watt.

The unit got very hot so I tuned it off by cutting off the exciter power.

I would like to try my new design for a toroidal core and try tubes of higher pressure hydrogen, 10, 50 and 100 Tor to see if I can generate power as the power generated depends on the pressure of hydrogen.

I would also like to try the test I did with hydrogen at 10-8 Tor again.

But I think I am getting close to the danger zone so it needs to be taken up with the bigger labs. I would like to be there to supervise although I have given out full details on my web page.

The fact that transmutation occurred at low pressure is a reflection of the mean free path at 10-8 Tor is higher than at higher pressures.

 Vacuum range Pressure in hPa (mbar) Molecules / cm3 Molecules / m3 mean free path Ambient pressure 1013 2.7*1019 2.7*1025 68 nm Low vacuum 300 - 1 1019 - 1016 1025 - 1022 0.1 - 100 μm Medium vacuum 1 - 10-3 1016 - 1013 1022 - 1019 0.1 - 100 mm High vacuum 10-3 - 10-7 1013 - 109 1019 - 1015 10 cm - 1 km Ultra high vacuum 10-7 - 10-12 109 - 104 1015 - 1010 1 km - 105 km Extremely high vacuum <10-12 <104 <1010 >105 km

From Wikipedia

The mean free path at 10-8 Tor is about 10000 meter so the energy acquired by ions before collision in a radiation field of 500 watt is from S=E x H: E/H=376 so |S|=E2 x 376 Therefore E=√(|S|/376)

The area is π(0.032/2)2 so |S|=6.2E5 watt/meter squared

so E=40.6 volts/meter so the energy acquired for the average collision is given by mean free path (10Km) x accelerating field this comes to 4E5 eV or 400 KeV. This corresponds to a temperature of 1.16E7 x 400 K= 420 MK.

The field along the tube is about 5000 volts so 5 KeV is acquired longwise so about 5.8 MK.

So the temperature equivalent is 425.8 MK.

This is hot enough for the higher energy process of 4P>He to take place.

The length is about 0.06 meter.

For higher pressures such as 3 Tor the temperature acquired by the gas by Ohmic heating is important. If no energy comes out because of the good insulation the temperature rise depends on the specific heat of the gas, the mass and the power input.

The specific heat capacity of hydrogen is about 15 KJ/Kg/K. so for a volume of 2 cc at a density 8.9E-5 g/cc at STP. This at 3 Tor its density will be 3.5E-7 g/cc so the mass of 2 cc will be 7E-7 g or 7E-10 Kg the heat capacity of 2 cc will be 1E-8 KJ/K so at 1000 watt the temperature rise / sec will be  1 x 1E8 degree/sec at 500 watt it will be 5E-1xE8 (=5E7) degree/sec so to reach  200E6 K will take 4 sec.

Only the very low pressure tube showed transmutation the high pressure tube take a higher activation energy and so convert more slowly.

I have had a break and so I have had time to consider. The basic idea is a tube of hydrogen inside a Tesla coil. I am using a Tesla coil but without a tuned primary. Tesla coils generate 100,000 volts or more so the ion energy would be much higher resulting in a higher collision energy. My reactor only has a pd generated of 5KV.

With higher pressure hydrogen the accelerating voltage has to be higher to achieve the same collision energy but the higher pressure means that more energy is released because the number of collisions is higher.

If the mean free path is 0.1 mm then to get a collision energy of 10 KeV the accelerating potential difference needs to be 100 million volts/meter. With a mean free path of 10 mm the pd has to be 1 million volts/meter and for a mean free path of 10 cm the PD needs to be 100,000 volts/meter.

Depressing Calculation:

I calculated that I got 1 microwatt out of a tube with a pressure of 10^-8 Tor

That would imply that the tube I am using has hydrogen at a pressure of 10^-4 Tor. (30 Watt)

This fits the curve given in my book on fusion reactions for D-D reactions at 100 KeV. My collision energy is about 800 KeV and I am using the P-P reaction.

With a pressure of 1 Tor a power of 100000 watt would be expected and to obtain collision energy of 800 KeV at 1 Tor the power of the exciter would need to be: 1000000 x 100 watt = 100 megawatt.

At atmospheric pressure the power would be 1E9 and the exciter power 1470 x 100 MW = 1.47 E11.

It looks like we won’t get ignition at all. Why do H-bombs work? How does the sun work?

I tried a 1 Tor tube and there was no excess RF power with exciter power of 100 watt.

It seems that the exciter power density has to be proportional to the pressure and the power density generated is proportional to the pressure so to make the thing self exciting you need to change something else. Ignition is not possible like that.

In fact the power density is proportional to the square of the pressure so ignition is possible and the exciter power density is linearly proportional to the pressure.

Pressure......power out......power in
1E-8 Tor....1E-6 watt
1E-4 Tor.....30 watt.   .... 100 watt
1 Tor...........1000 watt ... 1E6 Watt
10 Tor ........100,000 watt ... 1E7 Watt
100 Tor ......1E7 Watt ..........1E8 Watt
1000 Tor ....1E9 Watt ..........1E9 Watt.

My reaction chamber is about 20 mm by 30 mm diameter so the volume is piR^2l=3.14x(15/1000)^2x20/1000 = 2.8E-7meter^3 so the power density for ignition is 3.48 Watt/meter^3.

This is impractical for me.

Deuterons fuse at a lower energy than Protons and so my guess is that 10 Watt exciter would yield 30 watt and so would start up at 10 watt.

Electric field strength misconception.

I made a mistake over the energy acquired by ions in an electric field.

So I need a higher potential difference. I'll mull over that.

I looked up the conversion: 1KeV ~ 1.16E7K so 7KeV~ 78 million K. From other readings I believe this is hot enough for fusion to take place.

The general Idea that I am trying is an induction coil with low pressure hygrogen inside.

By observation I can see that the hottest part of the gas is in the region of the mid point of the coil.

Measurement with an oscilloscope reveals that at 20 Watt exciter power the radio frequency potential is 800 Volt but the probe reduces the Q thus reducing the RF potential.

I used a loop of wire thus making a 16:1 transformer and the measured RF potential (peak to peak) was 100 volt across the loop thus making the total potential across the coil as 1600 volt at 100 watt exciter power so to reach 17000 volt would require 12KW

At 20 Watt the voltage is 40 volt across the loop making the RF potential across the coil of 640 volt.

At current meter reading of 6 amp the dial reading of the little meter was 1.

At 70 watt exciter power the current was 6 amp at the end connection to the coil and the RF potential was 1280 volt so the energy in the circuit was 7680 Joule this is 109 times the exciter power so the Q is 109. This means that at 800 watt exciter power the energy content would be 87200 Joule so the RF potential would rise to 4300 volt and the current to 20.2 amp. That means I would require 12.5 KW to reach 17KeV. The current flowing would be 78.8 Amp so I would need thicker wire.

However at 4.3 KeV the equivalent temperature is 50 MK so some fusion would be going on.

At 1.28 KeV the equivalent temperature is 15 MK so I would still expect fusion to take place.

That is with hydrogen. With deuterium I would expect it to take off at 50 watt exciter power.

With the exciter power of 100 watt the Geiger counter registered only background.

The direct measurement gave 800 volt at 20 watt but using the transformer gave 640 meaning the true voltage requires a correction of 800/640 = 1.25. This slightly alters my estimates.

At lower pressure the mean free path is higher. There would be an optimum pressure for maximum power.

There is a threshold of exciter power for each value of n.

Suppose it is 1 KeV (temperature equivalent 16 MK) then the threshold exciter power is given by:

100 W > 5000 > 8x10^6 volts/meter > 800 volt/mm >

800 x mean free path (mm)=1000 eV so mfp=1.25 mm > 10^-2 Tor this is at 10^13 particles/cc so the energy released/cycle will be

2.61x10^15x10^13x10^-27 = 2.61 x 10^-2 so the power at 3.7 MHz will be

96MW.

At 10^-4 Tor the mfp is about 10 cm so the potential for 1000 eV comes to 100 volt so the power comes to 35 watt and the power out will be 96KW

At 1 Tor the mfp goes down to 0.1 mm and n=10^16 so to reach 1000 eV the power needs to be 156 times as much or 15.6KW and the power out will be 96000 x 10^6 W = 96 GW.

These calculations assume all collisions result in fusion. Experiment shows the power out to be less than this.

There are four regions in the reactor: Region A is the winding. Region D is the free radiation region where there is no ionised gas, region B where the radiation penetrates, heats and exerts radiation pressure on the ionised gas and region C the reaction region where the radiation does not penetrate. There regions overlap.

In the reaction region the gas it hottest and is compressed by radiation pressure by radiation that it shuts out by being a good conductor. In this region fusion reactions take place. As the wave passes over the ionised gas is compressed and then expands as the pressure rises due to fusion reactions thus acts against the wave making electricity in the coil by induction (Le Chattilier's effect).

In the transition region the ions are driven inwards and longwise by the radiation and the gas is more ionised towards region C, the gas also gets hotter as region C is approached. There is no boundary between region C and region B.

In region D the gas is cool and unionised but there is no boundary between region B and D.

If Q is the heat of the fusion reaction over the compression cycle and the gas is both heated and compressed cyclically by the radiation we have a hysteresis curve where the area of the curve is the energy converted to electrical power. This will be shown by the product potential curve and the current curve over one complete cycle. The integral (IV) over one cycle = Q.

For no energy generated the integral =0.

The engine diagram manifests itself in a phase lag between the current and potential wave. As the potential increases so current flows against the ion pressure thus lagging behind the potential wave then as the potential wave fades the current is maintained by the fusion reactions. Simple phase calculations show that since the system is losing power at the moment there is calculated to be a phase lead of the current. As the phase increases negatively the power generated is calculated to increase linearly with the phase lag.

I think the pressure 10-4 Tor may be too low. Maximum power is developed with pressure between 3 Tor and 10-4 Tor.  More pressure more power, probably 10 tor is best as at higher powers the reaction becomes uncontrollable leading to a nuclear explosion.

I suggest that the deuterium in the hydrogen reacts producing helium - 3 and a neutron. This neutron reacts with a proton making another deuteron thus screening out neutron emission and making more deuterium.

There will be a best pressure for each exciter power as the generation of power depends on both the pressure and the exciter power with the generated power increasing with exciter power and decreasing with pressure but the maximum power is higher as the pressure rises but the exciter power has to increase too.

Power = A*exp (B*exciter power), Power = C*(Pressure)^2. But the threshold exciter power for each Pressure is higher as the pressure increases.

So Power = D*(pressure)^2*Exp (B*((exciter Power)-K)) where K is a function of pressure that includes the mean free path and the local accelerating electric field strength.

I would have thought that a fusion reactor could be made from a high inductance coil terminated by split copper rings placed at the physical ends of the coil. Inside the coil is a glass tube of low pressure hydrogen. This arrangement would start up by itself if the Q was nigh enough.

I have a coil of 18,000 T and with a current of 20 ma the plasma current will be 360 A. The chamber is 40 mm x 20 mm diameter so the plasma current density is about 900000 A/m^2.

The Tesla power reactor took its power from deuterium fusion, this made helium-3 and the neutron reacted with a proton in the hydrogen gas so generating more deuterium.

The cycle starts with high voltage along the axis of the coil that moves the ions (once ionised) that makes the gas hot and some fusion, then the current compresses the ionised gas with more fusion and the heat generated expands the gas and as the current in the windings goes down this expansion generates more current that maintaining oscillations.

Normal Science.

My reactor was reverse engineered from the description given to me by my mother before she died recently at 100.

I am now working out a lower frequency unit for domestic mains use.

The reactor tube is now ceramic with it sitting in the middle of a bigger coil former that supports it. I need a bigger winding and capacitor for 50 Hz. The primary is on a bigger former and the secondary is mounted inside.

20/12/2012 18:10 Primary with 4 windings. The exciter coil. the sense coil, the shunt regulator coil and the power out coil. The secondary high voltage coil former.

The central holes are to mount a 11 mm ceramic tube to take the mixture of deuterium and hydrogen.

Windings.

I have yet to calculate the winding data.

`Here is the data for the secondary:`
```Find number of turns from inductance 1=yes
Do you know desired resonant frequency 1=yes 1
frequency 50
capacitance 40000E-12
Inductance 253.302968
enter length of coil (meters) 60E-3
enter diameter of coil (meters) 60E-3
Area 0.28274333e-2
number of turns 20682.1007
diameter of wire in mm 0.175
Turns per layer 342.857143
Number of Layers 60.3227937
Diameter of coil 0.81112978e-1
Length of Wire 4287.97008 Meters
Resistance of Coil 29564.6754 Ohms
dynamic impedance 214193.936
Resonant frequency is 50.0
Q= 2.69164034
end? 1=yes ```
` `
` `
`Computer Program that gave this output. It is badly written. It uses a Basic interpreter.`
```[Main]
Let Flag1=0
Let Flag2=0
let Pi=3.14215926
Let ResistivityOfCopper=1.68E-8
input "Find number of turns from inductance 1=yes "; OK
if OK=1 then goto [FindNumberOfTurns]
input "Do you know desired resonant frequency 1=yes ";OK
if OK=1 then goto [ResonantFrequencyKnown]```
```[one]
'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer```
```[Loop1]
input "diameter of wire in mm " ; DiameterOfWire
let DiameterOfWire=DiameterOfWire/1000
If Flag1=1 then [TurnsKnown]
input "diameter of former in meters ";DiameterOfFormer
input "length of former in meters " ; LengthOfFormer
input "number of turns "; NumberOfTurns
goto [TurnsUnknown]
[TurnsKnown]
Let DiameterOfFormer=CoilDiameter
Let LengthOfFormer=CoilLength
[TurnsUnknown]
let TurnsPerLayer=LengthOfFormer/DiameterOfWire
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*DiameterOfWire
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*DiameterOfWire
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((DiameterOfWire/2)^2)*Pi
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"```
```'Input "1 for end 2 for loop ";OK
'if OK=2 then goto [Loop1]
```
```if Flag1=1 or Flag2=1 then goto [InductanceKnown]
'Program to calculate inductance from number of turns```
```'  inductance is = mu Area of coil face times number of turns squared over length
'mu = 4.pi.10E-7
let Pi=3.1415926
Let mu=4*Pi*10E-7
[loop2]
'input "enter coil diameter in meters "; CoilDiameter
Let CoilDiameter=DiameterOfFormer```
```'input "enter coil lengh in meters "; CoilLength
let CoilLength=LengthOfFormer
'input "enter number of turns ";NumberOfTurns```
`Let inductance=mu*Pi*(CoilDiameter/2)^2*NumberOfTurns^2/CoilLength`
```Print "coil inductance is "; inductance
```
```'input "1 for exit 2 for loop "; ok
'if ok=2 then goto [loop2]
```
```[InductanceKnown]
'to calculate dynamic impedance```
```'dynamic impedance is L/Cr where r is the resitance of the coil.
[loop3]
'input "inductance (H) "; inductance
If Flag2=0 then input "capacitance (F) "; capacitance
'input "resistance of winding (ohms) "; resistance
let resistance=ResistanceOfCoil
let impedance = inductance/(capacitance*resistance)
print "dynamic impedance "; impedance
'input "1 exit 2 loop "; ok
'if ok=2 then goto [loop3]
' ******************************
'stop```
```'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))```
```let pi=3.1415926
[loop4]
'input "inductance of coil "; inductance
'input "capacitance of capacitor "; capacitance
let ResonantFrequency=1/(2*pi*(sqr(inductance*capacitance)))
print "Resonant frequency is "; ResonantFrequency
'input "1 for exit 2 for loop "; ok```
```'if ok=2 then goto [loop4]
'Q=dynamic impedance . 2.pi.f.L
Let Q=impedance/(2*pi*ResonantFrequency*inductance)
Print "Q= ";Q```
```input "end? 1=yes ";ok
if ok=1 then goto [Finish]
Goto [Main]
[FindNumberOfTurns]
Let Flag1=0
'program to calculate number of turns from inductance
let pi=3.1415926
let mu=4*pi*10E-7
'inductance is muAN^2/l so N=sqr(length*inductance/mu*Area)
[loop5]
if Flag2=0 then input "enter inductactance (henrys) ";inductance```
`input "enter length of coil (meters) ";CoilLength`
`input "enter diameter of coil (meters) ";CoilDiameter`
`let Area=pi*(CoilDiameter/2)^2:print "Area ";Area`
`Let NumberOfTurns=sqr(CoilLength*inductance/(mu*Area))`
```Print "number of turns ";NumberOfTurns
let Flag1=1
'input "1 for exit 2 for loop "; ok
'if ok=2 then goto [loop5]
goto [one]```
```[ResonantFrequencyKnown]
'program to calculate Inductance from resonant frequency
'f=1/(2.pi.sqr(l.c)) so f^2=1/4.pi^2.l.c so 4.pi^2.l.c.f^2=1 so l=1/(4.pi^2..c.f^2)
let pi=3.1415926
[loop6]
Flag2=0
input "frequency ";frequency
input "capacitance ";capacitance
let inductance=1/(4*pi^2*capacitance*frequency^2)```
```Print "Inductance ";inductance; "Henry"
'input "1 for loop";ok
'if ok=1 then goto [loop6]
Flag2=1
goto [FindNumberOfTurns]```
```[Finish]
end

```
`END OF LISTING`
` `

The sense coil is about 15 turns, the shunt coil is about 38 T the primary is 10 turns and the power coil is 791 T. The output power is about 300 w at 250 v at the frequency of 50 Hz.

The output power depends on the diameter of the wire used in the windings. 2 mm wire in the secondary would give about 60 kW at 250 V output.

I have not dealt with the regulator but I would think it would be a triac shunting the regulator coil controlled by the sense coil through a diode and capacitor smoother.

Shunt Regulator Shunt Regulator

The sense winding of 15T 1mm wire has an ac potential across if from the secondary that is rectified by the diode and smoothed by the 100 MFD capacitor. This current passes through the 10 K pot with a tap that is applied to the triac gate via a 2V zenner diode and 1K resistor. When the secondary potential reaches 6000 volt the triac switches on and shunts the inductor through the inductively coupled shunt coil of 40 T, 2mm wire.

The primary winding has ten turns so to achieve 600 AT at the tuned secondary with a Q of 3 the current needs to be just 20 amp pulse.

I now realise that the triac needs AC on the gate so here is the revised circuit. Triac Shunt Regulator  revised version with AC on the gate.

The sense coil has 6 T of 1 mm wire and the shunt coil has 40T of 2 mm wire. When the AC potential of the secondary rises above 6000 V the triac conducts thus controlling the power by shunting the winding.

Inductance of a multilayer coil. Each layer adds inductance L plus the mutual inductance of each layer on the other except one. I make the multiplier n/(n-1) where n is the number of layers.

see program below:

```'Program to find the inductance of a multilayer coil by summing the inductance
'of each layer then multiplying by nl^2 where nl is the number of layers..
'Inductance = 4.piE-7.A.n^2/l```
```'Program to find the inductance of a multilayer coil by summing the inductance of each layer and the effect of mutual inductance
'Inductance = 4.piE-7.A.n^2/l```
```'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
let Pi=3.14215926
Let ResistivityOfCopper=1.68E-8
Let mu=4E-7*Pi
[Loop]
input "diameter of wire in mm " ; DiameterOfWire
let DiameterOfWire=DiameterOfWire/1000
input "diameter of former in meters ";DiameterOfFormer
input "length of former in meters " ; LengthOfFormer
input "number of turns "; NumberOfTurns
let TurnsPerLayer=LengthOfFormer/DiameterOfWire
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*DiameterOfWire
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
Let TotalInductance=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*DiameterOfWire
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
let LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
Let TotalInductance=LayerInductance+TotalInductance
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((DiameterOfWire/2)^2)*Pi
Let TotalInductance=TotalInductance*NumberOfLayers^2
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"
Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"
Input "1 for end 2 for loop ";OK
if OK=2 then goto [Loop]

Another Program to help with coil calculations```
```'Program to find the inductance of a multilayer coil by summing the inductance of each layer and the effect of mutual inductance
'Inductance = 4.piE-7.A.n^2/l```
```'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
let Pi=3.14215926
Let ResistivityOfCopper=1.68E-8
Let mu=4E-7*Pi
[InductanceLoop]
input "diameter of wire in mm " ; DiameterOfWire
let DiameterOfWire=DiameterOfWire/1000
input "diameter of former in meters ";DiameterOfFormer
input "length of former in meters " ; LengthOfFormer
input "number of turns "; NumberOfTurns
let TurnsPerLayer=LengthOfFormer/DiameterOfWire
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*DiameterOfWire
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
Let TotalInductance=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*DiameterOfWire
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
let LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
Let TotalInductance=LayerInductance+TotalInductance
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((DiameterOfWire/2)^2)*Pi
Let TotalInductance=TotalInductance*NumberOfLayers^2
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"
Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"
input "1=exit 2=InductanceLoop ";OK
if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance```
```'dynamic impedance is L/Cr where r is the resitance of the coil.

```
```[ResonanceLoop]
input "Capacitance (F) "; Capacitance
let impedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "dynamic impedance "; impedance```
```'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))

```
```'input "capacitance of capacitor "; Capacitance
'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency```
`input "1 for exit 2 for loop "; ok`
```if ok=2 then goto [ResonanceLoop]
```
```'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil```
```Print "Q " ;Q

```
```input "1 exit 2 loop "; ok
if ok=2 then goto [InductanceLoop]
```
`End`
`End of listing.`
`Data Result of above program.`
```diameter of wire in mm 0.175
diameter of former in meters 60E-3
length of former in meters 60E-3
number of turns 20000
Turns per layer 342.857143
Number of Layers 58.3333333
Diameter of coil 0.80416667e-1 Meter
Length of Wire 4136.25765 Meters
Resistance of Coil 28518.6493 Ohms
Inductance of Multilayer Coil 1647.99297 Henry
1=exit 2=InductanceLoop 1
Capacitance (F) 6.258E-9
dynamic impedance 9234020.67 Ohm
Resonant frequency is 49.5503076 Hz
1 for exit 2 for loop 2
Capacitance (F) 6.25E-9
dynamic impedance 9245840.22 Ohms
Resonant frequency is 49.5820096 Hz
1 for exit 2 for loop 2
Capacitance (F) 6.26E-9
dynamic impedance 9231070.51 Ohms
Resonant frequency is 49.5423915 Hz
1 for exit 2 for loop 2
Capacitance (F) 6.2E-9
dynamic impedance 9320403.45
Resonant frequency is 49.7815356
1 for exit 2 for loop 2
Capacitance (F) 6.15E-9
dynamic impedance 9396179.09
Resonant frequency is 49.9834899
1 for exit 2 for loop 2
Capacitance (F) 6.1E-9
dynamic impedance 9473196.95
Resonant frequency is 50.1879222
1 for exit 2 for loop 2
Capacitance (F) 6.12E-9
dynamic impedance 9442238.79
Resonant frequency is 50.1058487
1 for exit 2 for loop 2
Capacitance (F) 6.115E-9
dynamic impedance 9449959.34
Resonant frequency is 50.1263293
1 for exit 2 for loop 2
Capacitance (F) 6.13E-8
dynamic impedance 942683.546
Resonant frequency is 15.8319313
1 for exit 2 for loop 2
Capacitance (F) 6.13E-9
dynamic impedance 9426835.46 Ohms
Resonant frequency is 50.0649627 Hz
1 for exit 2 for loop 1
Q 18.1810302
1 exit 2 loop ```
```So to make the tuned circuit resonant at 50 Hz, use a 20000T coil with 0.175mm wire on former 60mm dia by 60 mm long
tuned with a 6.13nF capacitor. ```
```
```

A  20000T, 1647 H winding taking 15 mA at 50 Hz will develop 7757 volt. With the Q of 18 the primary of 10 T will require  1.6 mA and need 215 mV to drive it. This has an impedance of  0.1 ohm to make it 8 ohm 77 T are needed. Here the current of 2mA will be driven by 1.66V.

When there is ionised gas in the tube it acts as a parallel resistor and takes current from the circuit, when it develops power its resistance goes negative.

```'Program to find the inductance of a multilayer coil by summing the inductance of each layer and the effect of mutual inductance
'Inductance = 4.piE-7.A.n^2/l```
```'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
let Pi=3.14215926
Let ResistivityOfCopper=1.68E-8
Let DensityOfCopper=8.94E3
'Kg per m^3
Let mu=4E-7*Pi
[InductanceLoop]
input "diameter of wire in mm " ; DiameterOfWire
let DiameterOfWire=DiameterOfWire/1000
input "diameter of former in meters ";DiameterOfFormer
input "length of former in meters " ; LengthOfFormer
input "number of turns "; NumberOfTurns
let TurnsPerLayer=LengthOfFormer/DiameterOfWire
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*DiameterOfWire
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
Let TotalInductance=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*DiameterOfWire
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
let LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
Let TotalInductance=LayerInductance+TotalInductance
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((DiameterOfWire/2)^2)*Pi
Let TotalInductance=TotalInductance*NumberOfLayers^2
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"
Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"
input "1=exit 2=InductanceLoop ";OK
if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance```
```'dynamic impedance is L/Cr where r is the resitance of the coil.

```
```[ResonanceLoop]
input "Capacitance (F) "; Capacitance
let impedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "dynamic impedance "; impedance```
```'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))

```
```'input "capacitance of capacitor "; Capacitance
'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency```
`input "1 for exit 2 for loop "; ok`
```if ok=2 then goto [ResonanceLoop]
```
```'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil```
`Print "Q " ;Q`
`'to calculate the total mass of copper in the winding Area times lengh times density`
```let TotalWindingMass=Pi*((DiameterOfWire/2)^2)*LengthOfWire*DensityOfCopper
Print "Total Winding Mass ";TotalWindingMass

```
```input "1 exit 2 loop "; ok
if ok=2 then goto [InductanceLoop]
```
`End`

Number of Turns from Inductance of a multilayer winding.

```'To find number of turns from inductance in a multilayer coil
'Program to find the inductance of a multilayer coil by summing the inductance of each layer and the effect of mutual inductance
'Inductance = 4.piE-7.A.n^2/l```
```'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
let Pi=3.1415926
Let ResistivityOfCopper=1.68E-8
Let DensityOfCopper=8.94E3
'Kg per m^3
Let mu=4E-7*Pi
input "Desired inductance Henry ";DesiredInductance
'to find initial inductance
'Program to calculate inductance from number of turns```
```'  inductance is = mu Area of coil face times number of turns squared over length
'mu = 4.pi.10E-7
[EstimateLoop]

```
```input "diameter of former in meters ";DiameterOfFormer
input "length of former in meters " ; LengthOfFormer```
`Inductance=DesiredInductance`
`let NumberOfTurns=sqr(LengthOfFormer*Inductance/(mu*Pi*(DiameterOfFormer/2)^2))`
```Print "Number of Turns "; NumberOfTurns
```
```'input "1=exit 2=loop "; ok
if ok=2 then goto [EstimateLoop]
'end of estimate section
input "diameter of wire in mm " ; DiameterOfWire
let DiameterOfWire=DiameterOfWire/1000
Let TempLow=0
Let TempHigh=NumberOfTurns
[InductanceLoop]
```
```'input "diameter of former in meters ";DiameterOfFormer
'input "length of former in meters " ; LengthOfFormer
'input "number of turns "; NumberOfTurns
let TurnsPerLayer=LengthOfFormer/DiameterOfWire
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*DiameterOfWire
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
Let TotalInductance=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*DiameterOfWire
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
let LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
Let TotalInductance=LayerInductance+TotalInductance
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((DiameterOfWire/2)^2)*Pi
Let TotalInductance=TotalInductance*NumberOfLayers^2```
```if abs(TotalInductance-DesiredInductance)<DesiredInductance/100 then goto [Finish]
'Print "TempHigh ";TempHigh;" TempLow ";TempLow;" NumberOfTurns ";NumberOfTurns;" TotalInductance ";TotalInductance;" DesiredInductance ";DesiredInductance
'input "ok "; ok
If (TotalInductance<DesiredInductance) then goto [Less]
'Print "More"
Let TempHigh=NumberOfTurns
Let NumberOfTurns=(TempHigh+TempLow)/2```
```goto [InductanceLoop]
[Less]
'Print "less"
Let TempLow=NumberOfTurns
Let NumberOfTurns=(TempHigh+TempLow)/2

```
```Goto [InductanceLoop]
[Finish]
'Print "Inductance " ;TotalInductance
'Print "Temp "; Temp
Print "NumberOfTurns ";NumberOfTurns
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"
Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"
'input "ok" ;ok```
```Print "Number of Turns is "; NumberOfTurns
Print "End of calculation"
end```

End of Listing

31-12-2012 6:50

```'To start the calculation from the capacitance and resonant frequency
'To start the calculation from the capacitance and resonant frequency
'program to calculate Inductance from resonant frequency
'f=1/(2.pi.sqr(l.c)) so f^2=1/4.pi^2.l.c so 4.pi^2.l.c.f^2=1 so l=1/(4.pi^2..c.f^2)
let Pi=3.1415926
[ResonanceLoop]
input "Resonant Frequency Hz ";ResonantFrequency
input "Capacitance ";Capacitance
let DesiredInductance=1/(4*Pi^2*Capacitance*ResonantFrequency^2)```
```Print "Inductance ";DesiredInductance
'input "1 to exit 2 for loop ";ok
'if ok=2 then goto [ResonanceLoop]

```

'To start the calculation from the capacitance and resonant frequency
'To start the calculation from the capacitance and resonant frequency
'program to calculate Inductance from resonant frequency
'f=1/(2.pi.sqr(l.c)) so f^2=1/4.pi^2.l.c so 4.pi^2.l.c.f^2=1 so l=1/(4.pi^2..c.f^2)
[start]
let Pi=3.1415926
[ResonanceLoop]
input "Resonant Frequency Hz ";ResonantFrequency
input "Capacitance ";Capacitance
let DesiredInductance=1/(4*Pi^2*Capacitance*ResonantFrequency^2)

Print "Inductance ";DesiredInductance
'input "1 to exit 2 for loop ";ok
'if ok=2 then goto [ResonanceLoop]

'To find number of turns from inductance in a multilayer coil
'Program to find the inductance of a multilayer coil by summing the inductance of each layer and the effect of mutual
inductance
'Inductance = 4.piE-7.A.n^2/l

'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
let Pi=3.1415926
Let ResistivityOfCopper=1.68E-8
Let DensityOfCopper=8.94E3
'Kg per m^3
Let mu=4E-7*Pi
'input "Desired inductance Henry ";DesiredInductance
'to find initial inductance
'Program to calculate inductance from number of turns

' inductance is = mu Area of coil face times number of turns squared over length
'mu = 4.pi.10E-7
[EstimateLoop]

input "diameter of former in meters ";DiameterOfFormer
input "length of former in meters " ; LengthOfFormer

Inductance=DesiredInductance

let NumberOfTurns=sqr(LengthOfFormer*Inductance/(mu*Pi*(DiameterOfFormer/2)^2))

Print "Estimated Number of Turns "; NumberOfTurns

'input "1=exit 2=loop "; ok
'if ok=2 then goto [EstimateLoop]
'end of estimate section
input "diameter of wire in mm " ; DiameterOfWire
let DiameterOfWire=DiameterOfWire/1000
Let TempLow=0
Let TempHigh=NumberOfTurns
[InductanceLoop]

'input "diameter of former in meters ";DiameterOfFormer
'input "length of former in meters " ; LengthOfFormer
'input "number of turns "; NumberOfTurns
let TurnsPerLayer=LengthOfFormer/DiameterOfWire
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*DiameterOfWire
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
Let TotalInductance=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*DiameterOfWire
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
let LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
Let TotalInductance=LayerInductance+TotalInductance
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((DiameterOfWire/2)^2)*Pi
Let TotalInductance=TotalInductance*NumberOfLayers^2

if abs(TotalInductance-DesiredInductance)<DesiredInductance/100 then goto [Finish]
'Print "TempHigh ";TempHigh;" TempLow ";TempLow;" NumberOfTurns ";NumberOfTurns;" TotalInductance ";TotalInductance;"
DesiredInductance ";DesiredInductance
'input "ok "; ok
If (TotalInductance<DesiredInductance) then goto [Less]
'Print "More"
Let TempHigh=NumberOfTurns
Let NumberOfTurns=(TempHigh+TempLow)/2

goto [InductanceLoop]
[Less]
'Print "less"
Let TempLow=NumberOfTurns
Let NumberOfTurns=(TempHigh+TempLow)/2

Goto [InductanceLoop]
[Finish]
'Print "Inductance " ;TotalInductance
'Print "Temp "; Temp
'Print "NumberOfTurns ";NumberOfTurns
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"
Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"
'input "ok" ;ok

Print "Number of Turns is "; NumberOfTurns
'to find dynamic impedance

'dynamic impedance is L/Cr where r is the resitance of the coil.
'[ImpedanceLoop]
'input "capacitance (F) "; Capacitance
'input "Inductance Henry ";TotalInductance
'input "Resistance ";ResistanceOfCoil

let DynamicImpedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "dynamic impedance "; DynamicImpedance
'input "1 exit 2 loop "; ok
'To find mass of winding
'Pi(D/2)^2.lengthofwire.density
Let MassOfWinding=Pi*((DiameterOfWire/2)^2)*LengthOfWire*DensityOfCopper
Print "Mass of winding ";MassOfWinding;" Kg"
'if ok=2 then goto [ImpedanceLoop]

'To find Q
'Q=2.Pi.f.L/R

Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil

Print "Q=";Q

Print "End of calculation"
input "Run again =2 ";ok
if ok=2 then goto [start]
end
End Of Listing

7-1-2012 1:19 am Outline reactor design sketch 2012 Shunt Regulator Design 2013

11-5-2012 11:28 pm

Hi, I met a friend who is another university graduate, badly damaged by Christians, who looked at my description of the mechanism of the reactor and has found a sound theoretical basis. His derivation is based on Maxwell’s equations and he says energy will be produced and turned into electrical power. Because of psychiatric attack I cannot do it myself although I passed exams in Maxwell’s equation.

I am hoping he will post them to me.

12-1-2013 6:19

He did not and I looked up "Coaxial Waveguide" and found little mention but it was some help.

The coil makes the wave spiral round and so slows it down the wave velocity is approximately c/n where c is the velocity of light and n is turns per meter.

Now the proposed coil is 17000 T and its length is 0.06 m so n=283E3 so the slow wave velocity is c/283E3 or 1060 m/s at 50 Hz the wavelength. 21.2 m and half wave is 10 m however there is capacitance as well. So we have a capacitance of 3nF and an inductance of 870 H so the propagation velocity is: given in terms of the inductance per meter and the capacitance per meter. Now the resonant frequency of this system is 32 Hz so a wave sits in the winding with a current peak in the centre and the potential peak at each terminal so by taking the ends of the winding to the ends of the reaction chamber the E-field is applied along the tube. This moves the ions with 6KV. The gas is heated by a Townsend discharge and then ionises with an anomolous discharge and glows and the gas gets hot, the current in the gas and the current in the winding are opposite so they repel. This compresses the ionised gas and fusion occurs. This heats the gas and pushes the ionised gas against the current in the winding and this increases the current by le-chatiliers principle. I suppose the power produced is connected with the engine cycle. I suppose an oscilloscope is needed.

14-1-2013 9:21 am

The power from fusion probably follows an exponential law with temperature like chemical reactions with the law being of the form A*Exp(-activation energy/RT) where A is a chamber constant and R is the gas constant, T is the absolute temperature.

The activation energy is the value of the energy required to bring the nuclei close enough for the strong nuclear force to be greater than the electrostatic repulsion. This is about 6000KeV

The mean free path in gas is a function of pressure.

So the power fro fusion increases exponentially with temperature.

The condition for ignition in a reactor is where the power generated is equal to the power to heat the hydrogen.

I used electrical induction to heat the gas and with a chamber of 12 cc it required 800 watt for ignition, in my new one the chamber is 0.3 cc so the power input for ignition will be 18 watt.

After that the power will rapidly increase until the shunt regulator limits the power by robbing the chamber of its energy by shorting the shunt coil. The power level at which this happens is set by the sense coil and the potentiometer.

The output power is limited by the current capacity of the windings. For the 40 mA coil the limit is about 4 mA at 6000V or 24 W 18 is required for ignition so the output power is about 6W

22-01-2013 2:11 am

Graphire: a new invention seen on TV.

There was a program on TV recently (CLICK BBC NEWS) that told that a person had invented a new substance “Graphire” which is a monatomic layer of graphite deposited in a vacuum oven. I do not know how. It is liquid proof and can be used to waterproof almost anything but it is also a two dimensional room temperature super conductor.

If the substance was deposited on threads like nylon it would make a superconducting wire.

One use would be my reactor. There are plenty of others.

I have now had confirmation from Jos that the theory of my reactor is correct.

29-1-2013 10:56 am

Since the reaction graph of rate against temperature the reaction has a maximum it reaches this temperature and tends to sit there.

5-02-2013 0:27

New generating algorithm

```'Program to find the inductance of a multilayer coil by summing the inductance of each layer and the effect of mutual inductance
'Inductance = 4.piE-7.A.n^2/l```
```'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
let Pi=3.14215926
'for room temp
Let ResistivityOfCopper=1.68E-8
Let DensityOfCopper=8.94E3
'Kg per m^3
'for superconductor
Let ResistovityOfCopper=0
'for LN temp
'Let ResistivityOfCopper=2.647E-9
Let mu=4E-7*Pi
Let e0=8.85E-12
[InductanceLoop]```
```input "diameter of wire in mm " ; DiameterOfWire
Input "Diameter Of Lumen mm "; DiameterOfLumen
input "Thickness of Insulation mm ";ThicknessOfInsulation
input "diameter of former in mm ";DiameterOfFormer
input "length of former in mm " ;LengthOfFormer
'Input "Resonant Frequency Hz ";ResonantFrequency```
```let DiameterOfWire=DiameterOfWire/1000
Let DiameterOfLumen=DiameterOfLumen/1000
Let ThicknessOfInsulation=ThicknessOfInsulation/1000
Let DiameterOfFormer=DiameterOfFormer/1000
Let LengthOfFormer=LengthOfFormer/1000
input "number of turns "; NumberOfTurns
let TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulation)
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
Let TotalInductance=0
Let TotalCapacitance=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*(DiameterOfWire+2*ThicknessOfInsulation)
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
let LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
Let LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsulation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer
Let InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
Let TotalInductance=LayerInductance+TotalInductance
Let TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/(((DiameterOfWire/2)^2)-((DiameterOfLumen/2)^2))*Pi
Let TotalInductance=TotalInductance*NumberOfLayers^2```
```Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"
Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"
Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; " microfarads"
'input "1=exit 2=InductanceLoop ";OK
'if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance```
```'dynamic impedance is L/Cr where r is the resitance of the coil.

```
```[ResonanceLoop]
'input "Resonant Frequency Hz "; ResonantFrequency

```
```input "capacitance of external capacitor "; ExternalCapacitance
Let Capacitance=TotalCapacitance+ExternalCapacitance
'input "Capacitance (F) "; Capacitance
let impedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "dynamic impedance "; impedance
'print "dynamic impedance - Infinite"```
```'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))

```
```'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency```
`input "1 for exit 2 for loop "; ok`
```if ok=2 then goto [ResonanceLoop]
```
```'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil```
`Print "Q " ;Q`
```'Print "Q infinite"
'to calculate the total mass of copper in the winding Area times lengh times density```
```let TotalWindingMass=Pi*(((DiameterOfWire/2)^2)-((DiameterOfLumen/2)^2))*LengthOfWire*DensityOfCopper
Print "Winding Mass ";TotalWindingMass;" Kg"
[PotentialLoop]
Input "Rms Potential ";RmsPotential
'to calculate slow wave```
```Print "Half Wavelength "; LengthOfFormer/2
Print "Velocity "; LengthOfFormer*ResonantFrequency
Print "Peak Potential "; 1.4*RmsPotential
Let PeakCurrent=1.4*RmsPotential/(TotalInductance*2*Pi*ResonantFrequency+ResistanceOfCoil)
Print "Peak Current ";PeakCurrent
Let PeakAmpereTurns=PeakCurrent*NumberOfTurns
Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)
Print "Peak Ampere Turns "; PeakAmpereTurns
Print "Plasma Current "; PeakAmpereTurns
Let PlasmaCurrentDensity=PeakAmpereTurns/((1.5E-3)*LengthOfFormer/3)
Print "Plasma Current Density ";3*PlasmaCurrentDensity; " Amp per meter^2, near centre"
Let PressureOnPlasma= ((3*PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer-(6E-3))/2)
Print "Additional Pressure of Plasma ";PressureOnPlasma; " Pa"
Print "Power lost through resistance "; ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"
'Power = 0.5 watt/m^3/kPa^2 10torr=133 Pa
Let PowerGenerated=0.5*(LengthOfFormer/3)*3^2*2*Pi*((133+PressureOnPlasma)/1000)^2
Print "Power Generated ";PowerGenerated;" Watt"
input "1 exit 2 loop "; ok
if ok=2 then goto [PotentialLoop]
```
```End
```

6-2-2013 11:11

I am considering using silica tubes, vacuum sealed with plugs, containing low pressure hydrogen (10Tor), 100 mm long by 10 mm diameter with a 6 mm lumen to slide into the reactor as the energy source.

9-02-2013 13:58

New generating algorithm. only cosmetic changes

'Program to find the inductance of a multilayer coil by summing the inductance of each layer and the effect of mutual inductance
'Inductance = 4.piE-7.A.n^2/l

'to find the final diametere of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
[Start]
let Pi=3.14215926
'for room temp
Let ResistivityOfCopper=1.68E-8
Let DensityOfCopper=8.94E3
'Kg per m^3
'for superconductor
'Let ResistovityOfCopper=0
'for LN temp
'Let ResistivityOfCopper=2.647E-9
Let mu=4E-7*Pi
Let e0=8.85E-12
[InductanceLoop]

input "diameter of wire in mm " ; DiameterOfWire
Input "Diameter Of Lumen mm "; DiameterOfLumen
input "Thickness of Insulation mm ";ThicknessOfInsulation
input "diameter of former in mm ";DiameterOfFormer
input "length of former in mm " ;LengthOfFormer
'Input "Resonant Frequency Hz ";ResonantFrequency

let DiameterOfWire=DiameterOfWire/1000
Let DiameterOfLumen=DiameterOfLumen/1000
Let ThicknessOfInsulation=ThicknessOfInsulation/1000
Let DiameterOfFormer=DiameterOfFormer/1000
Let LengthOfFormer=LengthOfFormer/1000
[TurnsLoop]
input "number of turns "; NumberOfTurns
let TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulation)
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
Let TotalInductance=0
Let TotalCapacitance=0
[ForLoop]
LayerN=LayerN+1
Let LayerDiameter=DiameterOfFormer+(LayerN-1)*(DiameterOfWire+2*ThicknessOfInsulation)
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
let LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
Let LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsulation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer
Let InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
Let TotalInductance=LayerInductance+TotalInductance
Let TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance
if LayerN<NumberOfLayers then goto [ForLoop]
Let ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/(((DiameterOfWire/2)^2)-((DiameterOfLumen/2)^2))*Pi
Let TotalInductance=TotalInductance*NumberOfLayers^2

Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding
Print "Length of Wire ";LengthOfWire; " Meters"
Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"
Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"
Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; " microfarads"
'input "1=exit 2=InductanceLoop ";OK
'if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance

'dynamic impedance is L/Cr where r is the resitance of the coil.

[ResonanceLoop]
'input "Resonant Frequency Hz "; ResonantFrequency

input "capacitance of external capacitor "; ExternalCapacitance
Let Capacitance=TotalCapacitance+ExternalCapacitance
'input "Capacitance (F) "; Capacitance
let impedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "dynamic impedance "; impedance
'print "dynamic impedance - Infinite"

'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))

'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency

input "1 for exit 2 for loop for another capacitance "; ok

if ok=2 then goto [ResonanceLoop]
input "1 for exit 2 for new inductance ";ok
if ok=2 then goto [TurnsLoop]

'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil

Print "Q " ;Q

'Print "Q infinite"
'to calculate the total mass of copper in the winding Area times lengh times density

let TotalWindingMass=Pi*(((DiameterOfWire/2)^2)-((DiameterOfLumen/2)^2))*LengthOfWire*DensityOfCopper
Print "Winding Mass ";TotalWindingMass;" Kg"
[PotentialLoop]
Input "Rms Potential ";RmsPotential
'to calculate slow wave

Print "Half Wavelength "; LengthOfFormer/2
Print "Velocity "; LengthOfFormer*ResonantFrequency
Print "Peak Potential "; 1.4*RmsPotential
Let PeakCurrent=1.4*RmsPotential/(TotalInductance*2*Pi*ResonantFrequency+ResistanceOfCoil)
Print "Peak Current ";PeakCurrent
Let PeakAmpereTurns=PeakCurrent*NumberOfTurns
Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)
Print "Peak Ampere Turns "; PeakAmpereTurns
Print "Plasma Current "; PeakAmpereTurns
Let PlasmaCurrentDensity=PeakAmpereTurns/((1.5E-3)*LengthOfFormer/3)
Print "Plasma Current Density ";3*PlasmaCurrentDensity; " Amp per meter^2, near centre"
Let PressureOnPlasma= ((3*PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer-(6E-3))/2)
Print "Additional Pressure of Plasma ";PressureOnPlasma; " Pa"
Print "Power lost through resistance "; ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"
'Power = 0.5 watt/m^3/kPa^2 10torr=133 Pa
Let PowerGenerated=0.5*(LengthOfFormer/3)*3^2*2*Pi*((133+PressureOnPlasma)/1000)^2
Print "Power Generated ";PowerGenerated;" Watt"
input "1 exit 2 loop "; ok
if ok=2 then goto [PotentialLoop]
input "1 exit 2 again " ;ok
if ok=2 then goto [Start]

End

I have discovered silica tubes are very expensive so I may try borosilicate tubes filled with low pressure hydrogen.

This calculation is for deuterium at initial pressure of 10 Tor. I am not going to use deuterium but hydrogen gas with deuterium in it. However I have found that it is very difficult and expensive to use this gas and so I will be using reduced pressure air. Air does have some deuterium. The power will be lower as the partial pressure is very low. This means that the rms potential across the coil must me high.

04/03/2013 07:18

My calculations of inductance did not match the experimental result and the calculations for self inductance were also wrong. So I used experimental result for the inductance with a multimeter.

My calculations for winding resistance were right after correcting a simple mistake.

The calculations for power are based on the published result of 0.5W/KPa^2/cc at 6KeV (the peak of the published curve for deuterium). I calculated the pressure from the ampere turns in the winding (600) and the repulsion between this current sheet and the induced ion current in the ionised gas in the reaction chamber. 600x600*mu/(distance between current sheets)/(length of interaction)/(2*Pi*(radius of reaction chamber)^2)/1000 KPa.

The reaction chamber is 6mm Diameter and the hot region is about 20 mm long and this is where the ion pressure occurs.

The resonant frequency is calculated from the inductance and capacitance using published formula and the current is from the published formula for the reactance of the inductor and the resistance at resonance.

Since I intend to use natural hydrogen the deuterium present is the reactive species so the power will be much less. It is a second order reaction so the power will be the calculated result for pure deuterium*(Fraction of deuterium present)^2

Looking this up in Wikipedia I see that the fractional content of Deuterium is 1/6400 so the power will be 24E-9 x 5E14 or 12E6 with an RMS potential across the reactor coil of 2000 v

See the computer output below:

Diameter of Wire 0.17

Diameter Of Former mm 60

Length Of Former mm 60

Number of Turns 13000

Turns per layer 352.941176

Number of Layers 36.8333333

Diameter of coil 0.72523333e-1

Length of Wire 2727.04675 Meters

resistance of winding 2018.06811

inductance by measurement 7.6

varactor capacitance (uF)12

Capacitance 0.13333333e-5

dynamic impedance 2824.48346

Extermal Capacitance 0.0000015

Resonant frequency is 49.9879868

1 for exit 2 for another turns value 1

Q 1.18304593

Winding Mass 0.55347263 Kg

Rms Potential 2000

Half Wavelength 0.03

Velocity 2.99927921

Peak Potential 2800.0

Peak Current 0.89568024

Peak Ampere Turns 11643.8431

Plasma Current 11643.8431

Plasma Current Density 1.16438431e9 Amp per meter^2, near centre

Ion Pressure of Plasma 3.15563757e10 Pa

Power lost through resistance 1156.41515 Watt

Power Generated if Deuterium 5.63215935e14 Watt

Power Generated by natural hydrogen 13517182.4 Watt

Potential Across Varactor 237.600808 Volt Actually a series capacitor of 12 uF to adjust the frequency to nearly 50 Hz

Potential Across External Capacitor 1900.80646 Volt

1 exit 2 loop 1

1 exit 2 again 1

Correction for reaction chamber volume calculation.

```Diameter of Wire 0.17
Diameter Of Former mm 60
Length Of Former mm 60
Number of Turns 13000
Turns per layer 352.941176
Number of Layers 36.8333333
Diameter of coil 0.72523333e-1
Length of Wire 2727.04675 Meters
resistance of winding 2018.06811
inductance by measurement 7.6
Capacitance 0.13326914e-5
dynamic impedance 2825.84404
Extermal Capacitance 0.0000015
Resonant frequency is 50.0000252
1 for exit 2 for another turns value 1
Q 1.18333084
Winding Mass 0.55347263 Kg
Rms Potential 2000
Half Wavelength 0.03
Velocity 3.00000151
Peak Potential 2800.0
Peak Current 0.89555444
Peak Ampere Turns 11642.2077
Plasma Current 11642.2077
Plasma Current Density 1.16422077e9 Amp per meter^2, near centre
Ion Pressure of Plasma 3.1547512e10 Pa
Power lost through resistance 1156.09033 Watt
Power Generated if Deuterium 5.62899582e14 Watt
Power Generated by natural hydrogen 13509590.0 Watt
Potential Across additional series capacitor 238.539935 Volt
Potential Across External Capacitor 1900.0819 Volt
1 exit 2 loop 1
1 exit 2 again 1``` This is the ignition circuit.

I realized that the longitudinal field from the ends of the winding only ionise the gas so if I supply this ionisation field from another source the winding does not need to be taken out at the ends to put across the reaction chamber so the shunt regulator will operate properly. The spark gap limiter must go inside but be replaced by a high voltage diac or high voltage neon.

The ignition driver is now a commercial 12v 3 A AC PSU and the regulator is updated for the simpler one outer coil design. I also put a shorting switch across A1 and A2 as an emergency cut off.

Here is the updated regulator circuit with diac. On looking at the print-out for current in the reactor coil at various potentials, the current is 63 mA at 200V.  In the standard wire tables the current capacity for 0.17mm wire is 54 mA. The power level for natural hydrogen at that current is given as 1000 W. With deuterium the power level is 5E10 W.

That means I cannot run it at any higher exciter power.

I think using air is bad because of the endothermic reaction of oxygen with nitrogen. So we must use a mixture of deuterium and hydrogen.

The exciter winding got hot as I ran it so it needs to be made of thicker wire, wire tables show that I need 22 SWG (0.71mm) wire.

I suggest a winding 20mm wide on my former of 300T.

So I need a new outer former to wind the exciter coil. Once the reactor ignites then the exciter winding becomes the output winding by throwing a switch.

I realised I could make a major simplification by connecting a high impedance source at 250 V AC directly across the reactor coil thus the outer coils can be dispensed with. The current taken by the tuned circuit is the driver potential/dynamic impedance.  The dynamic impedance is about 2800Ohms so the current will be 89mA.

This simplification loads the resonant circuit so it is wrong.

I think the reactor would be better if the reactor coil was smaller in diameter because it will be closer to the reaction chamber so the ion pressure is higher for the same ampere turns.

I used a smaller former and using 0.250mm wire with a current capacity of 140mA I wound 9629T with an inductance of 1.176H.

```diameter of wire in mm 0.25
diameter of former in mm 20
length of former in mm 55
number of turns 9629
Turns per layer 220
Number of Layers 43.7681818
Diameter of coil 0.41884091e-1
Length of Wire 950.503176 Meters
resistance of winding 325.248
Inductance by measurement 1.176
capacitance of Main capacitor (uF) 10
Capacitance 0.86111111e-5
Dynamic impedance 419.88803
Resonant frequency is 50.0044383
1 for exit 2 for loop for another capacitance 1
Q 1.13621215
Winding Mass 0.41719521 Kg
Rms Potential 100
Half Wavelength 0.0275
Velocity 2.75024411
Peak Potential 140.0
Peak Current 0.28438255
Peak Ampere Turns 2738.31953
Plasma Current 2738.31953
Plasma Current Density 2.98725767e8 Amp per meter^2, near centre
Ion Pressure of Plasma 8.01134653e9 Pa
Power lost through resistance 18.7885157 Watt
Power Generated if Deuterium 3.32753914e13 Watt
Power Generated by natural hydrogen 798609.393 Watt
Potential Across Additional Series Capacitor 14.5963579 Volt
Potential Across Main Capacitor 90.4974192 Volt
1 exit 2 loop 1
1 exit 2 again 1 ```
`The RMS reactor current is 200 mA and the plasma current 2738. The power is: 798KW with natural hydrogen.`

With the thinner wire with more turns the print-out is:

```diameter of wire in mm 0.150
diameter of former in mm 20
length of former in mm 55
number of turns 19407
Turns per layer 366.666667
Number of Layers 52.9281818
Diameter of coil 0.35878455e-1
Length of Wire 1715.85985 Meters
resistance of winding 1630.94898
Inductance by measurement 4.76
capacitance of Main capacitor (uF) 3
Capacitance 0.29126214e-5
Dynamic impedance 1002.03421
Resonant frequency is 42.736241
1 for exit 2 for loop for another capacitance 2
capacitance of Main capacitor (uF) 1
Capacitance 0.99009901e-6
Dynamic impedance 2947.7317
Resonant frequency is 73.299164
1 for exit 2 for loop for another capacitance 2
capacitance of Main capacitor (uF) 1.5
Capacitance 0.14778325e-5
Dynamic impedance 1974.88295
Resonant frequency is 59.9964738
1 for exit 2 for loop for another capacitance 2
capacitance of Main capacitor (uF) 2
Capacitance 0.19607843e-5
Dynamic impedance 1488.45858
Resonant frequency is 52.0862898
1 for exit 2 for loop for another capacitance 2
capacitance of Main capacitor (uF) 2.2
Capacitance 0.21526419e-5
Dynamic impedance 1355.79739
Resonant frequency is 49.7109934
1 for exit 2 for loop for another capacitance 2
capacitance of Main capacitor (uF) 2.2
Capacitance 0.21072797e-5
Dynamic impedance 1384.98285
Resonant frequency is 50.2431944
1 for exit 2 for loop for another capacitance 2
capacitance of Main capacitor (uF) 2.2
Capacitance 0.20496894e-5
Dynamic impedance 1423.8968
Resonant frequency is 50.9441479
1 for exit 2 for loop for another capacitance 2
capacitance of Main capacitor (uF) 2.2
Capacitance 0.21246106e-5
Dynamic impedance 1373.68525
Resonant frequency is 50.0378526
1 for exit 2 for loop for another capacitance 1
Q 0.91774797
Winding Mass 0.2711253 Kg
Rms Potential 100
Half Wavelength 0.0275
Velocity 2.75208189
Peak Potential 140.0
Peak Current 0.63242939e-1
Peak Ampere Turns 1227.35571
Plasma Current 1227.35571
Plasma Current Density 1.33893351e8 Amp per meter^2, near centre
Ion Pressure of Plasma 1.60945252e9 Pa
Power lost through resistance 4.65946898 Watt
Power Generated if Deuterium 1.34297669e12 Watt
Power Generated by natural hydrogen 32231.4406 Watt
Potential Across Additional Series Capacitor 3.24387045 Volt
Potential Across Main Capacitor 91.4181671 Volt
1 exit 2 loop 1
1 exit 2 again 1```

The RMS reactor current is 44 mA and the plasma current is 1227A and the power using natural hydrogen is 32KW.

It does mean the primary needs to be 2000AT so 2000T at 1 Amp or 400T at 5Amp.

I do not know how to calculate the impedance.

Former Design:  Resonance system calculation:

```diameter of wire in mm 0.170
diameter of former in mm 20
length of former in mm 55
number of turns 19407
Turns per layer 323.529412
Number of Layers 59.9852727
Diameter of coil 0.40394993e-1
Length of Wire 1852.41378 Meters
resistance of winding 1370.82255
Inductance by measurement 4.74
Resonant Frequency Hz 50
Capacitance 2.13680687 uF
Dynamic impedance 1618.19863
Resonant frequency is 50
1 for exit 2 for loop for another capacitance 1
Q 1.0864889
Winding Mass 0.37595994 Kg
Rms Potential 100
Half Wavelength 0.0275
Velocity 2.75
Peak Potential 140.0
Peak Current 0.69162856e-1
Rms Current 0.4940204e-1
Peak Ampere Turns 1342.24355
Plasma Current 1342.24355
Plasma Current Density 1.46426569e8 Amp per meter^2, near centre
Ion Pressure of Plasma 1.92486342e9 Pa
Power lost through resistance 4.68380755 Watt
Power Generated if Deuterium 1.92093194e12 Watt
Power Generated by natural hydrogen 46102.3666 Watt
Potential Across Main Capacitor 103.010016 Volt
1 exit 2 loop 1
1 exit 2 again 1```

The capacitor may be assembled from cheap capacitors (because the potentials are not high) in a parallel and series combination. Capacitor combination for reactor 2. Total capacitance 2.1369µF

After discovering an error in the value of π I recalculated and found the capacitance was 2.1375µF. The revised circuit is below: Revised circuit for capacitor board for reactor 2 Revised outer former design for reactor 2.

I have yet to calculate the turns and wire for the single winding. It is likely to be 400T of 0.9mm wire. At 2 Amp this will be 800 AT. A changeover switch would make the exciter winding into the output winding.

A 600V DIAC could be used as the upper limiting shunt.

Modified design for Reactor 2 to take a bigger internal winding. I have changed the discs at each end of the inner former to be 8 mm think.

I have also altered the regulator design to remove the transformer because the triac takes 700 v. I am a bit worried about the 10K resistor. The transformer must have loaded the resonant circuit and reduced the power. I think a 1 M ohm linear pot would be better. The regulator triac shunts the resonant tuned circuit across the winding so shutting off the power the triac then goes open circuit so the power remains constant at the level set by the pot.

I have a small 1 amp toggle switch connected across the reactor coil to turn it off. This simply shorts the coil and stops the reactor. This is built into the regulator box and is connected across A1 and A2 on the regulator triac.

It means "turns times winding current" if 40 mA is flowing in a 20000T winding then the current sheet is 800 A so the plasma curent is 800 A. The plasma is generated by the ionisation wire which is excited by 20KV 60KHz power supply.

The ion pressure is 800.800.mu/D/length of winding. But since the current is actually a standing wave them there is a maximum current in the middle and zero at the ends making the pressure a bit quite a bit bigger near the middle. The pressure I calculated as about 10^8Pa and the plasma heats by ohmic heating with the maximum near the middle and near the axis. When the temperature reaches 6KeV (that is about 6E6 K the power reaches its maximum. This is only near the middle and near the axis the insulation is because the ionised gas does not diffuse outward because of the ion pressure and is kept near the maximum because it is the maximum current. The fusion reactions futher heat the gas and the ions exert pressure on the containment current this inducing more current in the winding. The current density near the centre is very high because the area over which it flows is about 1mmx6mm so the current density is 800/6E-6 or 8E8A/meter^2.
The copper loss is the resistance of the winding (2000Ohm) and the current (40mA) which is 80 Watt.
The power generated is 0.5 Watt/cc/kPa^2 and with a reaction volume of 18E-2 so the power is 18E-2x0.5x(10E8/1000)^2 or 9E10Watt (90GW). It cannot be run at that power because the current generated is too high for the windings and since this extra current increases the pressure the power simply rises exponentially until the unit explodes.
The power is controlled by the shunt regulator by shunting the windings when the potential rises to the preset level.
To make a high power unit the reaction chamber remains the same but the resonance winding is made of insulated copper tube of the same inductance. This will have a smaller number of turns because the winding has a bigger diameter. I do not know how to work the inductance out from the number of turns and dimensions of the winding. You can estimate from the gaussian formula for a single turn coil and after winding, measure the inductance. Then calculate the capacitance for resonance at 50 Hz. Calculate the potential reached at maximum power to estimate the voltage rating of the capacitor and similarly the current rating. To get 50KW at 400 V the current output will be 125A. So the tank circuit must run at about 1250A so only 10% is taken to load. The output winding will have about 400T of 125A wire with the output voltage of 400 volt. To calculate the potential across the tank circuit the turns of the resonance winding need be known. So build the resonance circuit first and then find the potential it develops at low power (the exciter across the resonance coil) and then measure work out the turns for the output winding.
To reduce the copper loss and cool the winding the copper tube winding needs liquid nitrogen flowing in the tube.

Correction:

I think I made an error in the winding current. If you take the turns ratio and work out the current taken by the output coil (125A) and divide by the turns in the resonance winding and multiply by the output winding turns and then multiply by ten you get the winding current in the resonance coil. With a big number of turns like 20,000 then the winding current will be 25A. So we need a 30 A wire for the winding. AHH that’s better!

I make the resonance winding potential 20,000 volt. I am told a 20KV 30 A 50 Hz capacitor of a few microfarads could be made.

I make the plasma current 600,000A

The plasma is a single turn of low potential high current and experiment indicates that the temperature is less than ambient. The exciter generates current after the ionisation wire makes the gas in the reaction chamber conduct by causing ionisation. The pickup coil is in parallel with the inner winding. The inner winding is the resonance winding and the phenomena depends on resonance. The regulator keeps the current in the inner winding constant by shunting the winding if the potential reaches a predetermined level set by the regulator. When the pickup winding is loaded it takes power from the inner winding thus dropping the potential since the fusion reactions are very powerful and are only kept in check by the regulator the potential of the inner winding rises to meet the load.

The plasma loop is electrically equivalent to a negative resistance because of the fusion reactions it contains.

The number of turns in the transformer formula is replaced by the length of wire and the electromagnetic model is replaced by photons and special relativity.

Scintillation counter to detect neutrons.

A suitable neutron detector may be made from a sheet a 8 mm acrylic 4cmx7cm with the last 3 cm as a triangle with the apex cut off to form a flat it. A phototransistor is placed in a hole drilled in the flat bit. The output of the photo transistor is amplified and gated to be counted. The acrylic sheet is wrapped in aluminium foil as used for cooking and then in black PVC adhesive tape to keep out alpha and beta particles. The neutrons make a flash of light, I think it is Cherenkov radiation, whose amplitude depends on its energy. The flash is picked up by the phototransistor. Update:

The resonance coil winding ends do not need to be brought to each end of the coil and wound round the reaction chamber as the ionisation potential is provided by a separate high voltage, high frequency power supply.

The winding ends must be terminated on a terminal block as is normal practice.  Reactor 2 v 5 inner coil.

In this the hydrogen from the reaction is recycled and acts as a blanket to absorb neutrons and more hydrogen is added. Elsewhere a helium filter extracts the helium formed in the reaction.

There has to be a terminal block to take the winding ends to a stable point. This is not shown.

Hydrogen cycle.

The hydrogen cycles and passes over the reaction tube to form a hydrogen blanket that picks up neutrons mitted by the reaction and thus creates more deuterium. The natural hydrogen (it contains some deuterium) passes through the quartz reaction tube at low pressure (50Tor) where the reaction takes place. After reaction the gas passes through a ceramic tube where the helium which is super diffusive passes through the wall and into a steel enclosure to be pumped away for storage and export. The hydrogen exits this tube depleted of the helium product of the rectant.

See below: Hydrogen Cycle

The plasma generator:

In order that current will flow in the reactor chamber, the gas has to be ionised. this is done with a plasma generator with electrodes placed on the outside of the quartz tube on each end and capacatively coupled to the gas. This is 20KV 40 KHz.

This is turned off when the ionised gas gets hot inside.

The hydrogen is generated by the electrolysis of tap water by a pulsed doc high voltage source. The hydrogen has to be dried and he deuterium in it gives of its energy during nuclear fusion and the exhaust to the reactor is helium and oxygen. The helium may be separated and sold and the oxygen released to the atmosphere.

A 20KW unit at 50 HZ would weigh about 700 Kg. 2Kw about 70Kg. The 7 Kg unit would only genersate 200 W. You could probably do with one coil rather than two. This would be simpler but more difficult to regulate as it would need more power capacity triac. In fact the inner coil could be a red herring! With one coil the measured inductance was 8 µH so the capacity for 50 Hz will be: 1266514.84  and so impractical and that is why I chose the inner coil design.

Tiny reactor: Tiny fusion reactor. Designed to operate at 50 Hz.

The reactor must be commercially viable so the fuel must only be deuterium as this has to be manufactured so it becomes a way of making money on the fuel/ The "energy tubes" would slide into the reaction chamber and replaced when the deuterium is mostly transmuted into helium. Thee mus be a way of detecting when this has happed to safely change the energy tube. This tube contains deuterium with hydrogen at 20 Tor. The power of the tube is determined by the ratio of deuterium to hydrogen V:V.

20/10/2016 22:53:53