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15. Confinement of positive ions and electrons with a static electric and magnetic field
Experiments with the simulation program: simplified SEM fusor design with only 2 rings
Fig.1. Exp. 15.1 Only two negatively charged rings and a magnetic field of 0,3 T. Cross section with the electric field displayed: arrow is direction, color is strenght (the redder, the stronger). Fig.2. Exp. 15.1
The magnetic field is relatively weak. The particles stay more or
less confined. The maximum vertical speed of the D+ ions is about
2,2 E6 m/s (I checked this in the simulation). This is about the
speed necessary for fusion. This configuration with only two rings is quite simpler, but the one with three rings and two spheres does confine the particles better. ----------------- Fig.3. Exp. 15.2
Less voltage, a little more magnetic field. The D+ ions are injected a little below the lower ring, the
electrons are injected a little above the lower ring, chcc:=1; When the D+ ions are more or less cross the centre region, their vertical speeds are a little bit larger than their initial speeds (about 5,5E+5). ---------------------- Fig.4. Exp. 15.3
The speed of the electrons is slowly reduced in the simulation (to simulate Bremsstrahlung). if
DecreaseSpeedElectrons=true then It is curious that the speed of the electrons in the vertical
direction (z) is a lot more reduced that the speed in the
Fig.5. Exp. 15.3
The same experiment, but a little later in time. The electrons have spread out a little more in the x-y plane.
Fig.6. Exp. 15.3
The same experiment, again a little later in time. The electrons
are even more spread out in the B = m.v / ( r . q) With a horizontal speed of 3.103 m/s -> r = m.v / (B .q) = 9.10 Ũ10−31 . 3.103 / (0,5 . 1.6021Ũ10−19 ) = 3,38 .10−8 m
Fig.6. Exp. 15.3
The same experiment, again a little later in time. The electrons
have even more spread out in the Exp. 15.3 fig. 7 screenshot.png (part of the screen is made grey, to see the field lines better) When the electrons escape: as seen in the screenshot, at some
distance from the vertical centre axis, the
Fig. 8. Exp. 15.4
The speeds of the electrons is reduced, but only as long no one
of the vx,
vy and vz components are not smaller But why in 15.4 the electrons do now not spread out, and in experiment 15.3 yes they do? And why the vz component of the speed of the electrons in experiment 15.3 becomes very small (almost zero), and in experiment 15.2 (without Bremsstrahlung) not ? When we keep reducing the speeds in the program, the vz
component becomes almost zero, but the vx and vy components keep
some small value (about 1E2, 1E3 m/s):
Fig. 9. Exp. 15.5
if DecreaseSpeedElectrons=true then The electrons keep well confined in the centre area. But there
are only few (not 1020 ). Again the speed component vz is a lot more reduced than the components vx and vy. I tried out: if (vx>1E2) and (vy>1E2) and (vz>1E2) and the component vz did not change. So the vz is reduced by the program, and not because vx and vy are reduced.
Fig. 10. Exp. 15.6
The mass and the charge of the electrons is multiplied with 108
in the above experiment.
The speed of the electrons is decreased: The electrons escape.
Fig. 11 Charging the middle ring with a negative charge, will not stop
the electrons from escaping, see the image above
Fig.12 Exp. 15.7.
Fig. 12. Exp. 15.7 screenshot1.png The mass and the charge of the electrons is multiplied with 108 in the above experiment.
The speed of the electrons is decreased:
The electrons spread out, due to the augmented charge (x 108
). Interactions between the particles is on. Example that demonstrates that we can not simply represent a group of particles by a few particles. ------------------------- Letīs create a negative charge in the centre region. I was thinking to generate the D+ ions random, but then many of them will colide with the negative charged rings up and down, and this will cause (huge?) losses. It seems to be that the best is to inject both the D+ and the electrons.
In the link below an electron gun
is sold with an energy range: 1 eV to 100 keV and a beam current
range: 1 nA to 20 mA (or even 50 mA)
(1 A = 6.242Ũ1018 electrons/s)
Letīs take 1 mA -> 1.10-3 A = 1.10-3 C/s -> 1.10-3 C/s / (1,6.10-19 C) = 6,3. 1015 electrons/s It seems therefore plausible to have an electron gun that produces about 6.3. 1015 electrons/ s ( 1 electron/1,59 10-16 s)
If the energy of an electron is 1 eV, then:
― meVe2 =
1 eV = 1,602.10-19 J
=> Ve = √ (2
. 1,602.10-19 /
(9,1095 . 10-31)
) = 5,931. 105 m/s It seems to be possible to give them speeds of 1 . 106 m/s (as in the simulation program).
How much charge we have after one minute:
60 x 6,3. 1015
electrons
= 3,8. 1017
electrons
In the simulation program we will use the middle ring to simulate, quite approximated, a negative charge in the centre region. How much negative voltage we must give it to be equivalent with 6,0 10-2 C ?
See also here for some calculations
According the simulation program, for a charged ring, if V = 100 kV then Q = 8,93.10-7 C
6,0 . 10-2 C / ( 8,93. 10-7 C ) = 6,8 . 104 6,8 104 x 100 kV = 6,8 . 106 V = 6800 kV
That would be a huge acumulation of charge.
Letīs take in the simulation program a middle ring with -20 kV.
With how many electrons this will correspond? 100 kV -> 8,93.10-7 C 20 kV -> 1,8 . 10-7 C -> 1,12 1012 electrons
To have an electric equilibrium we should have the same amount of D+ ions.
How much fusion energy we could get from 1,12 1012 D+ ions? (see also Droom11.html# )
As 2 D+ ions fuse,
1,12 1012 D+ ions correspond to 1,12 1012 x 2,9E-13 J = 0,33 J.
8,78E10 J =2283 ltr diesel
https://www.convert-measurement-units.com/conversion-calculator.php?type=oelaequivalent 1 m3 diesel = 41 Gigajoule 1 joule = 2,4 .10-8 ltr diesel
Letīs say we want to produce in a Sem Fusor (0,5 m diameter, 1 m high + magnet and equipment) the amount of energy that is the equivalent of 1000 ltr diesel in one hour. 1000 ltr diesel = 4,17 . 1010 joule -> 4,17 . 1010 / 3600) = 11,6 . 106 joule/s = 11,6 MW 4,17 . 1010 joule / ( 2,9E-13 J ) = 1,4 . 1023 D+ ions. massa deuterium ion D+ = 2 . 1,67E-27 kg = 3,34E-27 kg 1,4 . 1023 D+ ions x 3,34E-27 kg = 4,8 . 10-4 kg D+ ion = 0,48 gram deuterium gas
Suppose we inject this quantity of D+ ions into the vacuum chamber.
1,4 . 1023 electrons . 1,6.10-19 C = 22400 C / hour = 6,2 C / s = 6,2 A
There are telephone chargers that deliver about 3 A, a micro wave about 8 A.
To get the energy of 1000 ltr diesel in one hour we need 0,48 gram deuterium gas (if it fuses completely). This is not so much.
Control: massa deuterium ion = 2 . 1,67E-27 kg = 3,34 E-27 kg 0,48 gram deuterium -> 0,48 E-3 / 3,34 E-27 = 1,44 E23 ions -> charge = 1,44 E23 . 1,602.10-19 = 2,3 E 4 C 2,3 E 4 C / 3600 = 6,4 A
Just for curiosity.
Fig. 13. F = 9E9 1,15 E4 . 1,15 E4 / 0,52 = 4,8 E18 N,
which is huge!! (4,8 E14 ton force)
Letīs charge the middle ring with 1 V.
I have simulated this, 1 V is very little, but almost no effect on the D+ ions.
Fig. 14. Exp. 15.8.
The middle ring is charged with -10 kV in the simulation. This should be our cloud of electrons.
hydrogen[i].z:= 0.50 -dr+r2r+0.05; -> the D+ ions are injected a little bit above the lower ring.
-10 kV corresponds with 1,12 1012 / 2 electrons = 0,6. 1012 electrons ( see #Voltage to amount of electrons )
With a 1 mA electron gun (1 E-3 C/s) this will produce: ( 1 . 10-3 C/s) / (1,6.10-19 C/electron) = 6,25 . 1015 electrons/s We could use an electron gun with a lot less capacity, if we want to create an electron cloud with a charge equal, more or less, to the charged middle ring in the simulation.
The time it takes to produce this amount of electrones with this electron gun: 0,6. 1012 / 6,25 . 1015 = 9,6 . 10-5 s
If we fuse an equal amount of D+ ions in this
time, weīll get: 0.6E12 .
2,9E-13 J = 0,174 J
(see
Droom11.html# )
----------------------
We want to inject the deuterium ions. To maintain electric equilibrium, we also have to inject electrons. If all injected deuterium (0,5 gram/hrs) fuse, we would obtain a power of 11,6 MW with a beam of D+ ions and electrons corresponding with a current of 6,2 A.
If we generate random the D+ ions (not inject them, but just a little D2 gas inside the vacuum chamber), probably many D+ ions will collide against the negatively charged rings and cause losses. But if some D+ ions keep trapped and fuse, then this would be also quite an achievement!
Doubts:
How much of the injected deuterium will fuse? Can an electro gun and a "deuterium ion gun" be made with a capacity of about 6 A?
It should be interesting to do real systematic experiments.
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20 February
2014 ..
by
Rinze
Joustra www.valgetal.com