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INTRODUCTION:
The first step in realizing deuterium-tritium thermo nuclear fusion via the Plasma Impact Fusion (PIF)
process is generation of a suitable semi-stable plasma known as a spheromak. This web page reviews various
aspects of spheromak generation and spheromak lifetime.
SPHEROMAK GENERATOR:
As the name implies, a spheromak generator is an electrical apparatus used for forming a semi-stable plasma configuration known as a spheromak. Spheromak lifetime is the period of time during which a spheromak provides stable energy storage. This web site addresses issues related to spheromak generation and spheromak lifetime.
PLASMA SPHEROMAK GENERATION:
A plasma is a mixture of free electrons and ions. If a plasma is made by passing an electric current through a low pressure gas, due to the electron - ion mass ratio energy is preferentially transfered from the electric circuit to free electrons in the plasma. After some time, in accordance with the equipartition theorem, the energetic free electrons will interact with the ions and share their excess energy with the ions, thus forming a random plasma. However, if the geometry, ion density, local magnetic field and local electric field conditions are appropriate, when energy is initially transferred from the electric circuit to the plasma electrons, the plasma can form a semi-stable configuration known as a spheromak. A spheromak has the property that once formed it does not require an external magnetic field in order to continue to exist. The lifetime of a plasma spheromak is typically in the range 10 us to 2 ms. The lifetime of a plasma spheromak can be increased by reducing the ambient density of neutral gas molecules.
When there is an existing spheromak random movement of neutral gas atoms tends to eventually take these gas atoms into the spheromak where they are ionized and become part of the spheromak. Hence the spheromak formation process involves an injection of neutral gas and then ionizing the gas within the spheromak which traps the gas atoms. Hence a spheromak must sit in place at the spheromak generator to gradually absorb the randomly moving neutral gas atoms within the vacuum system.
On spheromak compression about 80% of these gas atoms recombine and raise the concentration of neutral gas molecules in the vacuum system. These neutral gas molecules will limit the spheromak lifetime unless the system volume is sufficient to absorb them.
The vacuum pump must be of sufficient quality that it pulls an equilibrium pressure of about 10^-10 bar.
SPHEROMAK SHAPE:
A representative spheromak photograph is shown on the General Fusion Inc. website. At first glance a spheromak appears to be a ball shaped plasma. However, closer inspection reveals that a spheromak is actually cylindrically symmetrical with important distinct geometrical feature dimensions including the core radius Ra, the equatorial radius Rs, the end radius Rf, the corner radius Rx and the overall length (2 Hf). The straight core length (2 Ha) is a geometrical feature dimension that is actually a function of other feature dimensions.
SPHEROMAK FORMATION:
Spheromaks are usually formed in a laboratory using a Marshall Gun. A Marshall Gun consists of two co-axial cylindrical electrodes that are insulated from each other and open at one end. A low pressure gas is introduced into the space between the two cylinders. A magnet inside the central cylinder and/or outside the outer cylinder is used to create an axial magnetic field between the two cylinders. The central cylinder is made highly positive (eg 20,000 volts) with respect to the outer cylinder. This voltage ionizes the gas and causes is a radial electron current between the two cylinders that interacts with the axial magnetic field to cause the free electrons in the plasma to revolve around the cylinder axis.
The graded axial current through the cylinders also causes a graded magnetic field between the two cylinders that points around the cylinder axis. This magnetic field is referred to herein as the plasma gun magnetic field.
The circulating electron current in the plasma has its own magnetic field that interacts with the plasma gun magnetic field. The forces created cause the plasma to be expelled out the open end of the plasma gun. If the amount of gas injection, plasma gun geometry, magnetic fields and electric fields are all appropriate a spheromak forms at the open end of the plasma gun.
PLASMA SHEET:
Under the circumstances of spheromak generation the electrons and ions form a Plasma Sheet. On separation from the Marshall Gun the plasma sheet encloses a volume the shape of a torus. Note that the torus cross section is not round. The axial core of the torus is almost straight. Inside the plasma sheet the magnetic field is toroidal. Outside the plasma sheet the magnetic field is poloidal.
The plasma sheet has a uniform net charge per unit area. Within the volume enclosed by the plasma sheet the electric field is cylindrically radial. Outside the plasma sheet, except in the straight spheromak core, the electric field is spherically radial. In the straight spheromak core the electric field is zero.
The free electrons move with constant total energy along a spiral path around the torus but remain within the plasma sheet. The ion and electron densities in the plasma sheet are almost equal which keeps the net positive charge of a spheromak relatively low. The position of the plasma sheet is determined by energy density balance between the adjacent electric and magnetic fields. The magnetic fields are a function of the number of free electrons and the free electron velocity. The electric fields are a function of the net charge on the plasma sheet.
Compression of a spheromak causes linear shrinkage of the plasma sheet. The net charge stays constant but the electric field energy, magnetic field energy and free electron kinetic energy all increase.
SPHEROMAK GEOMETRY:
The geometry of a spheromak can be characterized by the following parameters:
R = radius of a general point from the spheromak's axis of cylindrical symmetry;
Ra = spheromak's straight core radius;
Rs = spheromak's equatorial radius;
Rf = spheromak's end funnel radius;
Rx = spheromak surface radius of curvature at the outside corner;
Rc = radius of co-axial cylindrical enclosure;
H = distance of a general point from the spheromak's equatorial plane;
(2 Hf) = spheromak's overall length measured at R = Rf;
(2 Ha) = sphereomak's straight core length measured at R = Ra.
The spheromak ends are funnel shaped mirror images of each other. Radius Rf is the radius of the spheromaks end funnel at its widest point where the spheromak's overall length is longest.
At R = Rf the spheromak overall length is (2 Hf).
REPRESENTATIVE UNCOMPRESSED SPHEROMAK GEOMETRY:
The spheromak shown in the photograph on the General Fusion website has the following ratios:
Hfa = 2.7 Raa
Rxa = 0.3 Raa
Rfa = 3.0 Raa
Rsa = 4.2 Raa
where trailing subscript a designates an uncompressed spheromak value.
General Fusion experimentally measured the core magnetic field strength of this spheromak and found it to be:
Ba = 0.12 T
and found the corresponding free electron kinetic energy to be 20 eV to 25 eV.
INTERNAL FIELD ENERGY:
From the web page titled Spheromak Properties recall that:
Eo = (Qs)^2 / (8 Pi Epsilon Rs)
and
Ba = (Rs / Ra) (Es / C)
and
Es = Qs / (4 Pi Epsilon Rs^2)
Combining these equations gives:
Eo = (Qs)^2 / (8 Pi Epsilon Rs)
= (4 Pi Epsilon Rs^2 Es)^2 / (8 Pi Epsilon Rs)
= (2 Pi Epsilon Rs^3 Es^2)
= (2 Pi Epsilon Rs^3 (C Ba Ra / Rs)^2
= (2 Pi Epsilon Rs C^2 Ba^2 Ra^2)
= (4 Pi Rs Ra^2) (Ba^2 / 2 Mu)
This equation indicates that for a spheromak with core magnetic field strength Ba the energy Eo is given by:
Eo = (4 Pi Rs Ra^2) (Ba^2 / 2 Mu)
= (4 Pi Rs^3) (Ra / Rs)^2 (Ba^2 / 2 Mu)
The MFI model for an uncompressed spheromak, which numerically integrates the magnetic field energy over the spheromak volume, gives:
Em = (Ba^2 / 2 Mu)(Pi Rs^3) (.2306)
Thus:
Eo / Em = 4 (Ra / Rs)^2 / (.2306)
= 4 (1 / 4.2)^2 / (.2306)
= .9833
Thus the expressions for electric field energy and for magnetic field energy give identical results within measurement error. Hence:
Em = Eo
The importance of this relationship is that the magnetic field energy Em can be found by calculating Eo. Alternatively Eo can be calculated by measuring Ba and using the equation:
Eo = (4 Pi Rs^3) (Ra / Rs)^2 (Ba^2 / 2 Mu)
= (4 Pi Rc^3) (Rs / Rc)^3 (Ra / Rs)^2 (Ba^2 / 2 Mu)
This equation can be rearranged to give:
Ba^2 = 2 Mu (Rs / Ra)^2 (Rc / Rs)^3 (1 / 4 Pi Rc^3)
PLASMA INJECTOR DESIGN SEQUENCE:
A prototype General Fusion plasma injector, about (2 / 3) of the commercial size calculated by MFI herein, is shown in a photograph available on another web site.
1) Choose G = (Rca / Rcb) = 5
2)The external electric field energy around a sphere is given by:
Eep = [(Q Ne)^2 (Eke / Me)/ (4 Pi Rs C^2 Epsilon)]
For:
Ne = 3 X 10^16 free electrons
Eke = 20 eV
Rs = .5535 m
Numerical evaluation gives:
Eep = [(Q Ne)^2 (Eke / Me)/ (4 Pi Rs C^2 Epsilon)]
= [(1.602 X 10^-19 coul X 3 X 10^16)^2 (1.602 X 10^-19 J / eV X 20 eV) / (9.11 X 10^-31 kg)]
/ [4 X 3.14159 X .5535 m X 9 X 10^16 m^2 / s^2 X 8.85 X 10^-12 coul^2 s^2 / kg m^3]
= [9.026 X 9 X 10^6 coul^2 J / kg] / [415.5 X 10^4 coul^2 / kg]
= 19.53 J
Units check:
kg m / s^2 = (1 / Epsilon)coul^2 / m^2
or
Epsilon = coul^2 s^2/ m^3 kg
2) Vacuum capacitor manufacturers assume a maximum working electric field of about:
Ewm = 1.7 X 10^7 volts / m.
3) Assume that to meet the downstream fusion requirements the total energy Ett of each compressed spheromak at the instant of its injection into the reaction chamber is given by:
Ett = 200 J
Then:
Eob = Ett / 2
= 100 J
4) Express electric field Ewb as:
Ewb = Qsb / (Epsilon 4 Pi Rwb^2)
5) Express external electric field energy Eob in terms of other parameters:
Eob = Integral from R = Rsb to R = infinity of:
(Epsilon / 2) (Qsb / Epsilon 4 Pi R^2)^2 (4 Pi R^2 dR
= Qsb^2 / (8 Pi Epsilon Rsb)
6) Eliminate Qsb between these two equations to get:
Eob = [Ewb Epsilon 4 Pi Rwb^2]^2 / (8 Pi Epsilon Rsb)
7) Simplify this equation using spheromak property that:
(Rwb / Rsb) = Ex = 2.71828
to get:
Eob = [Ewb^2 Epsilon 2 Pi Rwb^3 Ex]
8)Choose Rwb = 0.3 m
9) Rearrange this equation to get:
Ewb^2 = (Eob) / [Rwb^3 Epsilon 2 Pi Ex]
or
Ewb = {(Eob) / [Rwb^3 Epsilon 2 Pi Ex]}^0.5
= {100 J / [(0.3 m)^3 X (8.85 X 10^-12 coul^2 / Nt-m^2) X (2 X 3.14159 X 2.71828)]}^0.5
= 4.95 X 10^6 {J Nt m^2 / m^3 coul^2}^0.5
= 4.95 X 10^6 {J^2 / m^2 Coul^2}^0.5
= 4.95 X 10^6 V / m
which is less than one third of the maximum electric field used by vacuum capacitor manufacturers.
10) Hence choose:
Rcb = 0.30 m
corresponding to a plasma injector neck diameter of:
2 X 0.30 m = 0.60 m
Note that the plasma injector has to be about 1.5X larger in linear dimensions than the current
prototype being constructed by General Fusion.
FIX FROM HERE ONWARDS
12) Use the spheromak property that:
Ba = (Rs / Ra) (Es / C)
to assert that:
Bab = (Rsb / Rab) (Esb / C)
= (Rsb / Rab) (Rcb / Rsb) (Ecb / C)
= (Rsb / Rab) (Ex) (Ecb / C)
which gives the axial magnetic field at state b for any specific value of:
(Rsb / Rab).
Numerical substitution for:
(Rs / Ra) = 4.2
and
Ecb = 3.5 X 10^7 volts / m
gives:
Ba = 4.2 X 2.71828 X 3.5 X 10^7 volts / m / (3 X 10^8 m / s)
= 1.33 T
General Fusion has measured a compressed spheromak axial magnetic field of about 2X this value but General Fusion experienced reduced compressed spheromak lifetime, probably due at least in part to electric field enhanced field emission within the plasma injector neck. Also it is possible that during the magnetic field measurements General Fusion was working with a spheromak with:
(Rs / Ra) > 4.2,
in which case the higher (Rs / Ra) value would at least partially explain General Fusion's compressed spheromak axial magnetic field measurement.
13) Use spheromak property that under compression:
Qsa = Qsb
Hence the spheromak net charge is given by:
Qsa = Ecb Epsilon 4 Pi Rcb^2
= Ecm Epsilon 4 Pi Rcb^2
= (3.5 X 10^7 volts / m) X (8.85 X 10^-12 coul^2 / Nt-M^2) X 4 X 3.14159 X (.30 m)^2
= 350.3 X 10^-6 coul
14) To prevent unwanted impact ionization of neutral gas molecules choose:
Ekea = 13.5 eV
= 13.5 eV X 1.602 X 10^-19 J / eV
= 21.627 X 10^-19 J
15) Recall that:
Vea = (2 Ekea / Me)^0.5
which gives the free electron velocity at state a.
16) Recall that:
(Vea / C) = Qsa / (Q Nea)
Hence:
Nea = (Qsa C) / (Q Vea)
= (Ecb Epsilon 4 Pi Rcb^2 C) / (Q (2 Ekea / Me)^0.5)
= [3.5 X 10^7 volts / m X 8.85 X 10^-12 coul^2 / Nt-m^2 X 4 X 3.14159 X (0.3 m)^2 X 3 X 10^8 m / s]
/[1.602 X 10^-19 coul (2 X 21.627 X 10^-19 J / 9.11 X 10^-31 kg)^0.5]
= [105.0956 X 10^3 (volts coul^2 m^2 m) / (m Nt m^2 s)]
/[1.602 X 10^-19 coul X (4.7479 X 10^12 J / kg)^0.5]
= [65.6027 x 10^22 (volts coul / Nt s)] / [2.179 X 10^6 m / s]
= 30.107 x 10^16 (volts coul s / Nt s m)
= 3.0107 X 10^17 free electrons
17) The corresponding value of Neb is given by:
Neb = Nea / G
= 3.0107 X 10^17 / 5 free electrons
= 6.0214 X 10^16 free electrons
18) The approximate volume Vol of the compressed spheromak is given by:
Vol = (4 / 3) Pi Rs^3
= (4 / 3) (3.14159) (0.3 m / 2.71828)^3
= .00563 m^3
19) Hence the average free electron/ion density in the compressed spheromak is:
Neb / Vol = 6.0214 X 10^16 / .00563 m^3
= 1069.52 X 10^16 / m^3
= 1.069 X 10^13 / Cm^3
Note that the free electron/ion density in the plasma sheet of the compressed spheromak can easily be two orders of magnitude higher than the average free electron/ion density in the compressed spheromak.
HYDROGEN ISOTOPE GAS INJECTION:
Note that Neb is almost four orders of magnitude below the number of hydrogen isotope ions that are required for the fusion energy pulse. Hence D-T must be injected into to the reaction chamber after the reaction chamber ports are closed. When these additional neutral hydrogen isotope molecules are injected into the reaction chamber the compressed spheromak lifetime will drop to virtually zero and the compressed spheromaks will immediately randomize.
SPHEROMAK MODEL:
We have a mathematical model that, with the exception of the ion density, accurately predicts the experimental observations of spheromak compression made by General Fusion. One of the most important aspects of this model is the spheromak field energy gain G through the conical plasma injector that is given by:
G = (Etb / Eta) = (Eob / Eoa) = (Rsa / Rsb)
UNCOMPRESSED SPHEROMAK:
Rca = G Rcb
= G (.30 m)
Choose G = 5.
Then:
Rca = 1.50 m
and
Rsa = Rca / Ex
= 1.5 m / 2.71828
= .552 m
Then:
Eoa = Eob / G
= 5.0 kJ / 5
= 1.0 kJ
Recall that:
Eo = (Ec^2 Epsilon 2 Pi Rc^3 Ex)
Rearrange this equation to get:
Ec^2 = Eo / Epsilon 2 Pi Rc^3 Ex
which at the upstream end of the plasma injector becomes:
Eca^2 = Eoa / [Epsilon 2 Pi Rca^3 Ex]
= (1000 J) / [(8.85 X 10^-12 coul^2 / Nt-m^2) x 2 X 3.14159 X (1.5 m)^3 X 2.71828]
= 5.3285 X 10^12 J Nt m^2 / coul^2 m^3
= 5.3285 X 10^12 (volt coul)^2 m / coul^2 m^3
= (5.3285 X 10^12 volt^2 / m^2
Hence the electric field at the inside surface of the enclosure at the upstream end of the plasma injector is:
Eca = 2.308 X 10^6 volts / m
which might be a limitation on the entire PIF concept if this field should prove unrealizable due to the spheromak generator construction.
If the main power supply is 23,080 volts, this electric field dictates a coaxial electrode spacing in the spheromak generator of:
23,080 volts / (2.308 X 10^6 volts / m) = 10^-2 m
= 1 cm
POWER SUPPLY REQUIREMENT:
The power supply for each spheromak generator / plasma injector must supply the total spheromak energy plus losses in about 10^-4 s while maintaining output voltage. Hence the power supply for each spheromak generator / plasma injector must have a pulse rating of over:
10^4 J / 10^-4 s = 10^8 W
= 100 MW
One of the constraints on the power supply will likely be development of the voltage Vsa at:
R = Rsa
while the spheromak is at the spheromak generator. This voltage is given by:
Vsa = Integral from R = Rsa to R = Rca of:
(Qsa / 4 Pi Epsilon R^2) dR
= [Qsa / (4 Pi Epsilon)] [(1 / Rsa) - (1 / Rca)]
Recall that:
Rsa = Rca / 2.71828
and
Qsa = (8 Pi Epsilon Rsa Eoa)^0.5
and
Rca = 1.50 m
Thus:
Vsa = [(8 Pi Epsilon Rsa Eoa)^0.5 / (4 Pi Epsilon)][1 / Rca][2.71828 - 1]
= [(2 Eoa Rsa) / (4 Pi Epsilon)]^0.5 [1 / Rca][2.71828 - 1]
= [(2 X 1000 J X 1.50 m / 2.71828) / (4 Pi X 8.85 X 10^-12 coul^2 / Nt-m^2)]^0.5 [1.71828 / 1.50 m]
= [9.9237 J m Nt m^2/ coul^2 X 10^12]^0.5 [1.71828 / 1.50 m]
= 3.608 X 10^6 J m / (coul m)
= 3,608,000 volts
The manner in which the spheromak spontaneously acquires this voltage while at the spheromak generator is not simple.
A high voltage can be realized for a short time (microseconds) by discharging a large capacitor bank through an inductor connected in series with the ion gun. Initially the gas breakdown voltage is much lower (eg 20,000 volts) but as the available supply of neutral gas in the ion gun is exhausted the ion gun impedance and hence the open circuit voltage across the ion gun rapidly rise. The corresponding rapid collapse of the inductor's magnetic field causes a short (microsecond) duration transient high voltage which provides the voltage, net charge, and electric field energy for the spheromak.
In order to realize spheromaks with free electron kinetic energies of ~ 13.5 eV this high voltage spike has to be carefully controlled and the appropriate number of deuterium gas molecules must be supplied to the spheromak generator. The required voltage gain Of:
3,608,000 / 20,000 ~ 180
is accompanied by a multiplicity of energy conversion inefficiencies.
MEASUREMENT OF SPHEROMAK NET CHARGE Qs:
Measurement of the spheromak's net charge Qs is a very useful parameter
because it can be used to calculate the spheromak's total field energy Et. Recall that:
Et ~ (Qs^2) / (4 Pi Epsilon Rs)
= [(Qs^2) (2.71828)] / (4 Pi Epsilon Rc)
The charge Qs on a spheromak can be found by measuring the change in image charge at
the enclosure wall as a spheromak passes. Consider an electrically isolated sense plate
of known area Ai that is located at the enclosure wall and hence at a known value
of Rc. The image charge Qi induced in this sense plate as a spheromak passes
by is given by:
(Qi / Qs) = [Ai / (4 Pi Rc^2)]
or
Qs = (Qi / Ai) 4 Pi Rc^2
The charge Qi is measured by integrating the current Ii that flows through
resistor Ri that is connected between the sense plate and the metal enclosure.
Qi = Integral from T = 0 to T= T of Ii dT
= Integral from T = 0 to T= T of (Vi / Ri) dT,
where:
Vi = voltage across resistor Ri.
Numerical Example:
Choose:
Ai = 10^-2 m^2
Rc = 1 m
Ri = 1000 ohms
Qs = 350.3 X 10^-6 coul
Then:
Qi = Qs Ai / [4 Pi Rc^2]
= 350.3 X 10^-6 coul X 10^-2 m^2 / [4 Pi m^2]
= .2787 X 10^-6 coul
Assume that the transistion from spheromak not present to spheromak present occurs in
10^-3 seconds. Then the average current Ii through the resistor during
this transition time interval is given by:
Ii = .2787 X 10^-6 coul / 10^-3 s
= .2787 X 10^-3 amps
and the average voltage Vi across the 1000 ohm resistor is:
Vi = .2787 volts.
Hence the peak value of the time integral of the voltage across resistor Ri gives
spheromak charge Qs according to the formula:
Qs = (Qi / Ai) 4 Pi Rc^2
= [4 Pi Rc^2 / Ai] Integral[(Vi / Ri) dT]
ELECTROSTATIC PRESSURE:
The high electric field surrounding ths spheromak will exert a negative mechanical pressure P on the inside walls of the enclosure. The value of P can be found from the conservation of energy relationship:
P dVol= (Epsilon / 2) Ec^2 dVol
or
P = (Epsilon / 2) Ec^2
Numerical evaluation for the plasma injector neck gives:
P = (8.85 / 2) X 10^-12 coul^2 / Nt-m^2 X (3.5 X 10^7 volts / m)^2
= 54.206 X 10^2 Pascals
= 5.42 kilopascals
= (5.42 / 101) bar
= .054 bar
If the plasma injector inside wall is rated as a vacuum pressure vessel to withstand 101 kilopascals external pressure, this change in pressure is not a significant increase in wall stress. However, the wall deflection caused by passage of a spheromak is sufficient to generate a noticeable acoustic thump.
This web page last updated April 8, 2013.
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