11.16
Confinement of positive ions and electrons with a static electromagnetic field
Trying
to calculate the possibility of fusion

Let's take for the dimensions of the SEM fusor's vacuum
chamber 1 mtr x 45 cm diameter = 1 .
π.(0,45/2)^{2 }= 0,16 m^{3}
Suppose we made a good vacuum (1.10^{6} Torr) and filled the
vacuum chamber again with deuterium gas up to a pressure of 30 microns
(3.10^{2} Torr).
1 Torr = 1000 microns, 1
micron = 1e3 Torr = 1E3 Torr =
10^{3} Torr
1 atmosphere=1013 mbar=760 Torr (mm of Hg) = 760 000 micron (millitorr,
mTorr ) =
101325 pascal (N/m^{2}) 1 bar = 100 000 pascal
e2 = E2 =10^{2}
1 J = 2,6e8 ltr diesel
(
hypertextbook.com
)
(see:
http://www.fusor.net/board/viewtopic.php?f=42&t=10776
)
How many D atoms we have now ? (with a
pressure of
3.10^{2} Torr)
Ideal gas law:
p.V = nRT
p = pressure of the gas (Pa or N/m^{2 )
}
(1 torr = 1/760 atm = 133,3 Pa)
> 3E2 torr = 4,0 N/m^{2
}V = volume of the gas ( m^{3})
n = amount of substance of gas (in moles)
R = ideal gas constant =
8.3144598
J.mol^{1}.K^{1}
Let's take for the temperature T = 293
ºK
> n = p.V / R.T = 4,0 .0,16 / (8,3145 . 293
) = 2,63E4 mol (1 mol = 6,022 E23 molecules D_{2})
2,63E4 . 6,022 E23 = 1,58 E20 molecules
D_{2 }(3,16E20 ions D+)_{
} => density = 1,58 E20
molecules/0,16 m3 = 9,88E20 molecules/m3 = 1,98E21 D ions /m3
In
Wendelstein7X a
plasma density of
of 3.10^{20} particles
per cubic metre is aimed.
massa
deuterium ion
D^{+} = 2 . 1,67E27 kg = 3,34E27 kg
Total massa deuterium in the vaccum
chamber = 1,58 E20 . 2. 3,34 E27 kg = 1,06E6 kg (1 mg molecules D_{2 }
)
100 kWh = 9,4 ltr diesel =
fusion energy of 4,1 E3 gram of deuterium (see
Droom11.html#
)
1,06E6 kg deuterium is equivalent with
(1,06E6) / (4,1 E6) . 9,4 = 2,43 ltr diesel.
This is some amount of energy, not so
much, but at least our vacuum chamber will not be a potential "nuclear bomb"
, which is an advantage.

The
nuclear force is powerfully attractive between
nucleons at distances of about 1 femtometre (fm,
or 1.0 × 10^{−15} metres),
but it rapidly decreases to insignificance at distances beyond about 2,5 fm.
How much energy does it cost
to bring two deuterium ions at a distance of 1 femtometre?
Z_{ }
= 1 (amount of positive protons in a
deuterium ion)_{
} k is the
Coulomb's constant = 9,0×10^{9}
N m² C^{−2}
1 eV = 1,6 . 10^{ 19 }
J
= >
U_{coul} = k . Z_{ }. Z_{ }. e^{ 2 } / r = 9.0 . 10^{ 9
}. 1 . 1 . (1,6.10^{ 19}
)^{ 2 }
/ (1.10^{ 15})^{
}= 2,3 .10^{ 13
}Nm = 1,44 MeV
(see:
wiki/Coulomb_barrier )
The kinetic energy of one D+ ion
(with massa m and speed v) = 0,5.m.v^{2 }= 0,5. (2 .
1,67 . 10^{27}) . v^{2 }=
2,3 .10^{
13 }J
=> v = 11,8 . 10^{ 6} m/s
(T = 1,1.10^{10
}K
)
So to overcome the Coulomb barrier and one D+ ion has no
speed, the other one must have a speed of 11,8 . 10^{ 6}
m/s (collision head on, according this simplified calculation).
If the two D+ ions have
opposite speeds: 0,5.m.v^{2 }+^{ }0,5.m.v^{2 }
= (2 . 1,67 . 10^{27}) . v^{2 }
=^{ }2,3 .10^{
13 }J => they must colide with 8,3 .10^{6} m/s .
(1/2).m.v^{2 }
= (3/2) . k_{B} .T
k_{B} =
1,38.10^{23} J.K^{1})
> T =
1/3 . m_{(D+ ion)} .
(8,3
.10^{6)}^{2 . / }k_{B }=_{ }
5,6.10^{9 }K
http://www.splung.com/content/sid/5/page/fusion

In
alpha.web.cern.ch/penningtrap
a device a little bit simular to the SEM fusor
is used, but not for fusion. Here the goal is to trap antiprotons and
positrons in order to produce antihydrogen. According to them the electrons
cool down due to
cyclotron_radiation .
"Antiprotons,
however, are far more massive. It would take antiprotons over 300 years of
sitting in a 1 Tesla magnetic field to cool through cyclotron radiation
alone!"
That the electrons cool down in the SEM fusor is only favorable, I think. So
they form even better a negative cloud: a virtual cathode (I suppose). The protons
(D+ ions) however should be hot, move with high speeds in order to be able to fuse.

When any charged particle (such as an
electron, a proton, or an ion) accelerates, it radiates away energy in the form
of electromagnetic waves. For velocities that are small relative to the speed of
light, the total power radiated is given by the Larmor formula:
(see:
/hyperphysics.phyastr.gsu.edu/hbase/particles/synchrotron.html
and
Larmor Formula
)
P = 2Kq^{2}a^{2
}/ (3c^{3})
K = Coulomb's constant = 8.987.10^{9 }Nm^{2/}C^{2
} a = the accelaration of the particle
q = the charge of the particle
c = the speed of light = 3.10^{8 }m/s
The charge for both the electrons and
the deuterium ions is 1,60E19 C
Let´s take for the mean speed 2E6 m/s
(more or less the speed necessary for fusion, see
Droom11.html#
)
B = 1 tesla (strength of the magnetic
field)
B.v.q = m.v^{2} /
r = m .a (see:
Droom11.html#Fig._4
)
v = speed
a = acceleration (centripetal)
m = mass of the particle
Energy loss of an electron
due to the magnetic field
a _{electron }= B.v.q /
m = 1. 2E6 . 1,6E19 /(9,1E31) = 3,52E17 m/s^{2 }
=> P _{electron }=
2 . 9E9 . (1,6E19 )^{2} . (3,52E17)^{2 }/ (3(3E8)^{3})
= 57,095E6 / 8,1E25 = 7,049E19
J/s
kinetic energy of an electron =
0,5.m.v^{2 } = 0,5 . 9,1E31 . (2E6)^{2}
= 1,82E18 J
Due to the centripetal
acceleration caused by a magnetic field of 1 tesla, an electron with a
speed of 2E6 m/s would lose in about 2,6 s its kinetic energy.

Energy loss of a deuterium
ion due to the magnetic field
a_{D+}
= 1 . 2E6 . 1,6E19 _{ }/ (2.1,67E27 ) = 9,58E13 m/s^{2}
=> P_{D+
}= 2 . 9E9 . (1,6E19 )^{2} . (9,58E13)^{2 }/
(3(3E8)^{3}) = 4,229 / 8,1E25 = 5,22E26 J/s
kinetic energy of a D+ ion
= 0,5.m.v^{2 } = 0,5 .(2 . 1,67E27 )
. (2E6)^{2} = 6,68E15 J
Due to the centripetal
acceleration caused by a magnetic field of 1 tesla, a deuterium ion with
a speed of 2E6 m/s would lose in 1,3 E11 s (4058 year!) its
kinetic energy. Negligible
Energy loss of a deuterium
ion due to the electrostatic field
Let's consider the configuration
of #Fig.7a
.
The deuterium ions move
vertically up and down covering a distance of about 70 cm.
Suppose a D+ ion has in the
centre a speed of 2E6 m/s and when reached a vertical distance of 35 cm
from the centre, its speed is zero.
Formulas of motion:
0,35 = 2E6 . t + 0,5.a.t^{2
}0 = 2E6 + a .t
a = 2E6 / t
0,35 = 2E6 . t + 0,5.2E6 . t
=> t = 0,35 / (1E6 ) = 3,5E7 s and
a = 5,7E12 m/s^{2}
This is a mean (de)acceleration.
In reality it will not be constant, but for simplicity let's assume the
(de)acceleration is constant, only switching from direction (sign) when
passing the centre.
The mean acceleration due to the
electric field is smaller than the one due to the magnetic field, and
therefore the energy loss due to this acceleration is also negligible..
Conclusion
=> The
energy loss (Larmor formula) of the deuterion ions due to the
accelerations caused by the magnetic and electric field of the SEM fusor
is neglibible ☺
=>
The electrons loose quickly their kinetic
energy according the Larmor formula.

Lawson's criterion
Confinement time is the time the plasma is maintained at a temperature
above the critical ignition temperature.
For deuteriumdeuterium fusion:
n.
τ
≥ 10^{16 } (s/cm^{3})
( see the link above)
n = ion density in the plasma
We already calculated before that in the vacuum
chamber of our SEMfusor
(1mtr x 45 cm
diameter ) with a pressure of 3E2 torr there would be 1,58 E20
molecules D_{2. }Suppose they are all ionized, then we would
have 3,16E20 D^{+} ions.
Suppose this is our ion density
in the plasma and the temperature is above the critical ignition
temperarure
=> n
= 3,16E20 / 0,16 = 1,975E21 ions/m^{3 }= 1,975E15 ions/cm^{3}
=>
τ
≥ 10^{16 / }
(1,975E15)
= 5,1 s

https://en.wikipedia.org/wiki/Migma#Problems
Two possible energy losses:
Electrons and D+ ions collide and neutralize,
meaning they are no longer subject to the elecric and magnetic field and
leave the reaction chamber.
Collisions between D+ ions and electrons will give
energy to the electrons which will be released by
bremsstrahlung/synchroton radiation.

In
https://en.wikipedia.org/wiki/Solar_core
"At the center of the Sun, fusion power is estimated by
models to be about 276.5 watts/m^{3}. Despite
its intense temperature, the peak power generating density of the core
overall is similar to an active compost heap,
and is lower than the power density produced by the metabolism of an adult
human."
And here:
coldfusionnow.org/powerequivalenttothesunwealreadyhaveit/

How much energy is stored in an operational SEM
fusor?
Suppose the mean speed of the D^{+}
ions is v = 2E6 m/s (speed needed for fusion, see
Droom11.html#
)
The vacuum chamber of 1mtr x 45 cm diameter with a pressure of 3E2
torr contains 3,16E20 D^{+} ions (already
calculated before).
m_{(D+ ion)} = 2 . 1,67E27 kg
Total kinetic energy = amount of D^{+} ions .
½ m_{(}_{D+
ion)} v^{2}
= 3,16E20 .
½ .(
2 . 1,67E27). (2E6 )^{2}
= 21,1E5 J (1 J = 2,61E8 ltr diesel. see
energy_conversion_calculator
=> 55. 10^{3} ltr diesel
, 47 gram)
The kinetic energy of the electrons is neglible because of their small
mass.
The energy per unit volume in a region of space of permeability, μ_{0},
containing magnetic field, B,
is: u = ½
. B^{2} /
μ_{0},
( see
wikipedia.org/wiki/Magnetic_energy )
B = 1 tesla => u = 0,5 . 1 / (4
π
10^{7
) }
=^{ }4,0.10^{5 }J/m^{3}
Vaccuum chamber SEM fusor = 0,16 m^{3 }
=> magnetic energy = 6,4 .10^{4} J
(1,7.10^{3 }ltr diesel , 1,5 gram)
The electric energy stored in one of the charged
spheres (see
Droom11.html#Fig._7a ) :
E
=
½ QV
V = 250 kV = 9E9.Q /r
r = 6 cm
=> Q = 1,67E6 coulomb
=>
E
=
½
. 1,67E6 . 250E3 = 0,21 J ( 5,4E9 ltr diesel)
A system of point charges also has a potential
electric energy, see
experiment9.html.
Let's calculate how much energy
it costs to bring one of the spheres of the Sem fusor from infinity to a
distance of 1 m from the other one (distance between their centres).
E = Q . 9E9 . Q / 1 =
(1,67E6) .9E9 . (1,67E6)
= 0,025 J , which is neglectable.
The potential energy of the rings and spheres
together will also be neglectable (the energy it would take to bring
them from infinity together into the Sem fusor configuration).
We have totally 2 charged
spheres and 3 charged rings, so the estimated total electrical energy
will be approximately 5 x 0,21 J = 1,05 J (2,6E 8 diesel).
=> Most of the energy is stored in the
kinetic movement of the D+ ions. The electrical energy of the
charged spheres and rings and the magnetic field is relatively small.
28 January
2018
by
Rinze
Joustra www.valgetal.com


