Hello,
My name is David and I have been a lurker on the fusor net for many years.
I have been following the advances in fusion be they somewhat small until now.
Initially after Dr. Rider’s paper noting how there was no possibility, I kind of gave up hope on the fusors. Now it would seem that this paper is in question concerning the polywell configuration. This was a major stumbling block – now gone? Maybe.
This new design seems to look promising. I do have a concern. It is with gathering energy from the p-b11 products - alpha particles. In the configuration of the fusion device the focal point is in the center of the polywell. Whether they fuse at the exact center does not matter. The ions go in and out and at some unkown moment collide and fuse. The direction of the products will also be unkown. To extract the energy of the particle the idea is to decelerate them with a high voltage grid. It would seem on the surface that the best configuration for this grid to obtain maximum products would be a sphere. This would not be the case. The electric field inside of a static charged sphere is zero. In fact any closed surface will be zero. Any curved surface will not have a uniform field. This is a fundamental problem that permeates the configuration of the machine for the idea of decelerating the alpha particles. Any Ideas?
David
decelerating grid
I see what you are saying. Based on the Faraday cage principle, the inside of a metal conductor should not have any e-field. However, you can get around that if you have an isolated charge inside the sphere. In fact, I think the MaGrid ought to do just about right.
It has to do with Gauss's law: the isolated charge will induce an opposite charge on the inside of the sphere so that the field within the metal surface itself is zero. So you have your alpha particles coming out of the MaGrid (where your criticism is correct, and there is no field) and suddenly the particles are in a field, and they travel through it to reach the container wall or collector grid or whatnot.
Does that make sense? Someone might want to check me on this; I'm actually not sure if they will use the Magrid that way or they'll want to set up another grid to create the field.
It has to do with Gauss's law: the isolated charge will induce an opposite charge on the inside of the sphere so that the field within the metal surface itself is zero. So you have your alpha particles coming out of the MaGrid (where your criticism is correct, and there is no field) and suddenly the particles are in a field, and they travel through it to reach the container wall or collector grid or whatnot.
Does that make sense? Someone might want to check me on this; I'm actually not sure if they will use the Magrid that way or they'll want to set up another grid to create the field.
Re: decelerating grid
zero relative to what?dbennett wrote:Hello,
My name is David and I have been a lurker on the fusor net for many years.
I have been following the advances in fusion be they somewhat small until now.
Initially after Dr. Rider’s paper noting how there was no possibility, I kind of gave up hope on the fusors. Now it would seem that this paper is in question concerning the polywell configuration. This was a major stumbling block – now gone? Maybe.
This new design seems to look promising. I do have a concern. It is with gathering energy from the p-b11 products - alpha particles. In the configuration of the fusion device the focal point is in the center of the polywell. Whether they fuse at the exact center does not matter. The ions go in and out and at some unkown moment collide and fuse. The direction of the products will also be unkown. To extract the energy of the particle the idea is to decelerate them with a high voltage grid. It would seem on the surface that the best configuration for this grid to obtain maximum products would be a sphere. This would not be the case. The electric field inside of a static charged sphere is zero. In fact any closed surface will be zero. Any curved surface will not have a uniform field. This is a fundamental problem that permeates the configuration of the machine for the idea of decelerating the alpha particles. Any Ideas?
David
If you're looking at it from outside the decel grids, perhaps. But even then, you have feedthrough connections to connect the magrid and e- guns, so it's not truly a closed surface.
Tom.Cuddihy
~~~~~~~~~~~~~~~~~~~~~
Faith is the foundation of reason.
~~~~~~~~~~~~~~~~~~~~~
Faith is the foundation of reason.
Solo, cuddihy
Thank you for your reply.
First let’s think about the spherical configuration. Yes, it will act as a Faraday cage.
Now let’s introduce the MaGrid into it.
Your idea that the electrical field will induce an opposite charge is correct in the wall of the sphere.
This is a dipole effect.
So let’s envision what's happening inside. First let's think of the MaGrid as a point charge. There is a dipole effect on a point on the inside of the sphere. So what does the field look like? Pick a point inside the sphere and sum or integrate to add the electric fields up from all points on the inside of the sphere. You will find the field cancels and again you are left with a zero electric field inside the sphere by the inside surface of the sphere (not the grid). There will be an electric field due to the grid. This would be calculated by using Gauss's Law (flux through a surface).
Now let’s complicate matters somewhat. The grid is not a point charge. There is a property of plasmas that they will shield electric fields. How much and of what strength is up to discussion or experiment. Let’s look geometrically to see how this might affect the E field inside the sphere. Again take a point on the inside of the sphere a near the surface. Everywhere that the Plasma is the field vector coming from the other side will be shielded somewhat like a shadow. The wiffel ball will certainly shield.
If we say that all the electric field vectors that are not in the "shadow region" will cancel then the only thing that will have a non canceling effect will be the shadow area. If we calculate the electric field due to the shadowed region at the diameter of the device then this will be effectively be the net gain that was not subtracted from the opposite side ..there will be a difference in electric field and thus a voltage and it will be the net voltage our alpha will see.
So how much voltage on the outer grid will be sufficient to stop an alpha particle? According to The Fusion Physics FAQ "p+B11 -> 3 He4 + 8.7 MeV". For sake of argument each He4 has 1/3 of the energy or 2.9Mev. I'm not sure if energy distributition is constant among the products. I would expect there is a conservation of momentum value to add on to the extra energy that is released in the fusion->fission process.
Now calculate how much charge will be needed on the sphere to produce a net voltage equal to the stopping voltage of 2.9Mev.
We will have to make some assumptions here as the final design is not done. Let’s say that the sphere is 15 ft in diameter, the wiffle ball is 8 in diameter. We must also assume that all products will be traveling outward with a vector line that if extended backward will pass through the wiffle ball.
By using similar triangles double 7.5ft and 8in to 15ft and 16" The diameter of the shadow is 16" from a point on the other side. Look at the size of the angle...inv sin(16"/180") = about 5 degrees. Since this is a small angle we will say that this is a disk of diameter 16". This may be a moot point anyway.
At a distance (the diameter) the voltage will be
V = 1/(4pi * EpsilonZero) * RO * 2pi * [sqrt(diameter_squared + radius_squared) - diameter]
2.9X10E6 = (1/(4*3.1416*8.8E-12)) *ro * 2 * 3.14168 *[sqrt(20.9 + .1643) - 4.572m]
ro = (.00032069/.11049) = 0.0029 coul/meter_squared = charge density
How much total charge on sphere..
area of sphere = 4pi r_squared.
A = 4 * 3.1415 * 5.225 meter_squared
A = 65.667 square meters
ro * A = coul
.190 coul
how much voltage will there need to be on the sphere?
V = q/(4pi * e0 * r)
V = .190 /(4*3.1415*8.8E-12 * 2.133meters)
V = 805305881 volts
or 805Mev
I have not subtracted the electric field from the MaGrid as this will have an attractive effect on the ion.
the MaGrid will be at 7 to 10 Kev ... this from other posts that I have read.
It would appear that this will be very small in comparison to what was just calculated.
I could have made errors
In fact this whole thing could be wrong, it would not surprise me.
Frankly I do hope I'm wrong as 800Mev seems insurmountable.
David
Thank you for your reply.
First let’s think about the spherical configuration. Yes, it will act as a Faraday cage.
Now let’s introduce the MaGrid into it.
Your idea that the electrical field will induce an opposite charge is correct in the wall of the sphere.
This is a dipole effect.
So let’s envision what's happening inside. First let's think of the MaGrid as a point charge. There is a dipole effect on a point on the inside of the sphere. So what does the field look like? Pick a point inside the sphere and sum or integrate to add the electric fields up from all points on the inside of the sphere. You will find the field cancels and again you are left with a zero electric field inside the sphere by the inside surface of the sphere (not the grid). There will be an electric field due to the grid. This would be calculated by using Gauss's Law (flux through a surface).
Now let’s complicate matters somewhat. The grid is not a point charge. There is a property of plasmas that they will shield electric fields. How much and of what strength is up to discussion or experiment. Let’s look geometrically to see how this might affect the E field inside the sphere. Again take a point on the inside of the sphere a near the surface. Everywhere that the Plasma is the field vector coming from the other side will be shielded somewhat like a shadow. The wiffel ball will certainly shield.
If we say that all the electric field vectors that are not in the "shadow region" will cancel then the only thing that will have a non canceling effect will be the shadow area. If we calculate the electric field due to the shadowed region at the diameter of the device then this will be effectively be the net gain that was not subtracted from the opposite side ..there will be a difference in electric field and thus a voltage and it will be the net voltage our alpha will see.
So how much voltage on the outer grid will be sufficient to stop an alpha particle? According to The Fusion Physics FAQ "p+B11 -> 3 He4 + 8.7 MeV". For sake of argument each He4 has 1/3 of the energy or 2.9Mev. I'm not sure if energy distributition is constant among the products. I would expect there is a conservation of momentum value to add on to the extra energy that is released in the fusion->fission process.
Now calculate how much charge will be needed on the sphere to produce a net voltage equal to the stopping voltage of 2.9Mev.
We will have to make some assumptions here as the final design is not done. Let’s say that the sphere is 15 ft in diameter, the wiffle ball is 8 in diameter. We must also assume that all products will be traveling outward with a vector line that if extended backward will pass through the wiffle ball.
By using similar triangles double 7.5ft and 8in to 15ft and 16" The diameter of the shadow is 16" from a point on the other side. Look at the size of the angle...inv sin(16"/180") = about 5 degrees. Since this is a small angle we will say that this is a disk of diameter 16". This may be a moot point anyway.
At a distance (the diameter) the voltage will be
V = 1/(4pi * EpsilonZero) * RO * 2pi * [sqrt(diameter_squared + radius_squared) - diameter]
2.9X10E6 = (1/(4*3.1416*8.8E-12)) *ro * 2 * 3.14168 *[sqrt(20.9 + .1643) - 4.572m]
ro = (.00032069/.11049) = 0.0029 coul/meter_squared = charge density
How much total charge on sphere..
area of sphere = 4pi r_squared.
A = 4 * 3.1415 * 5.225 meter_squared
A = 65.667 square meters
ro * A = coul
.190 coul
how much voltage will there need to be on the sphere?
V = q/(4pi * e0 * r)
V = .190 /(4*3.1415*8.8E-12 * 2.133meters)
V = 805305881 volts
or 805Mev
I have not subtracted the electric field from the MaGrid as this will have an attractive effect on the ion.
the MaGrid will be at 7 to 10 Kev ... this from other posts that I have read.
It would appear that this will be very small in comparison to what was just calculated.
I could have made errors
In fact this whole thing could be wrong, it would not surprise me.
Frankly I do hope I'm wrong as 800Mev seems insurmountable.
David
One error
dbennett,
The alphas are at two energies two at about 2.4 Mev another about 2.9 (IIRC).
However divide the voltage by 2 because the alpha has a charge of 2.
So you are talking grids of 1.2 MV and 1.5 MV (roughly).
The alphas are at two energies two at about 2.4 Mev another about 2.9 (IIRC).
However divide the voltage by 2 because the alpha has a charge of 2.
So you are talking grids of 1.2 MV and 1.5 MV (roughly).
If we fed protons and boron nuclei into the reactor in pulses, it could result in outward pulses of alpha particles. Induction (coils) could be used to impede the changes in magnetic field that would accompany the changes in alpha-current, and the coils would become a source of alternating current. This would have to reduce the alpha pulse velocity, wouldn't it?
Yes. If every thing was timed right.MisterX wrote:If we fed protons and boron nuclei into the reactor in pulses, it could result in outward pulses of alpha particles. Induction (coils) could be used to impede the changes in magnetic field that would accompany the changes in alpha-current, and the coils would become a source of alternating current. This would have to reduce the alpha pulse velocity, wouldn't it?
Collecting at DC has the advantage of not being frequency dependent. DC output or pulsed output. Same collection scheme.
I was thinking of having the alphas pass through a ring. Wire would be wound through the ring. The ring would concentrate the magnetic field from the alpha current, and the changing flux in the ring would induce pulsing current in the wire.MSimon wrote:Collecting at DC has the advantage of not being frequency dependent. DC output or pulsed output. Same collection scheme.
How could this scheme work to generate DC current from a constant stream of alphas when there would be no continual change in the magnetic field around the alphas?
No one has explained to me how the electrostatic deceleration of alphas with a grid would become a source of current.
Does the electrostatic acceleration of alphas require current? If not how is energy conserved?MisterX wrote:How could this scheme work to generate DC current from a constant stream of alphas when there would be no continual change in the magnetic field around the alphas?
No one has explained to me how the electrostatic deceleration of alphas with a grid would become a source of current.
Perhaps a book on particle accelerators would be a good start. Try the linear accelerator section.
how to convert
MSimon, MisterX
I will try to answer these 2 questions
Does the electrostatic acceleration of alphas require current? If not how is energy conserved?
Well yes and no.
Yes in the fact that you must first power up the grid and the wifle ball.
Lets say that we have an ideal machine. This means no leakage current from the decelerating grid or from the "wifle ball of electrons"
at the center. Nor are we talking keeping the magnets going. All of this would take allot of current.
Now we have a perfect machine sitting there and we drop a helium atom in it ... what happens. The nucleus sees a strong
electrostatic field and get accelerated toward the center. For a while the electrons in the 2s shell ride along until they see the field, repulsive for them, and strips them from the nucleus. Now we have an alpha particle rushing toward the center. I guess over time some alpha particles will becomes thermalized and capture electrons. Thus there will be a current to keep the wifle ball intact.
If not how is energy conserved?
We sometimes get confused because of our location here on earth. Let me explain.
This situation is like gravity and space.
If you were at a stationary position in space 200 miles from the surface you would have no kinetic energy and
only a possible potential energy. I say possible potential because if you were not near the earths gravity you would not have any forces on you. This is why we call it a potential well. Now if you fall into this well you would accelerate. You would convert potential energy to kinetic energy. Your potential would be negative hence the term "well". Coming down to earth one must expend this energy in our most archaic form i.e. heat. We sit at the bottom of a gravity well and must expend energy to climb out ..bummer. Sorry I digress….
Now our alpha particle is accelerating into a potential well. But it does not hit the surface it goes through and decelerates coming out the backside transforming kinetic to potential energy. Energy is conserved.
No one has explained to me how the electrostatic deceleration of alphas with a grid would become a source of current.
Here is how I would try to do it.
We have the outer grid a some extremely high voltage.
First we charge up the decelerating and decouple it from the system.
There will be a capacitor between the grid and the power supply that is held at a constant (high) voltage.
When an alpha particle approaches the grid it will add its positive voltage to the grid. This will be kinetic energy converted to potential energy. Small but there is power in numbers.
This will create a voltage across the capacitor. The capacitor voltage becomes an energy source that is small enough that we can deal with it with regular electronics. Now a circuit can be made to discharge this voltage through a coil and inductively picked up by another coil that is at or near ground potential. If the common mode potential
cannot be isolated then multiple coils can be used in steps to get to ground level.
There are concerns as to the voltage field outside the grid. It will have quite a gradient that will have to be taken into account as to where the physical placement of the electronics will be.
David
I will try to answer these 2 questions
Does the electrostatic acceleration of alphas require current? If not how is energy conserved?
Well yes and no.
Yes in the fact that you must first power up the grid and the wifle ball.
Lets say that we have an ideal machine. This means no leakage current from the decelerating grid or from the "wifle ball of electrons"
at the center. Nor are we talking keeping the magnets going. All of this would take allot of current.
Now we have a perfect machine sitting there and we drop a helium atom in it ... what happens. The nucleus sees a strong
electrostatic field and get accelerated toward the center. For a while the electrons in the 2s shell ride along until they see the field, repulsive for them, and strips them from the nucleus. Now we have an alpha particle rushing toward the center. I guess over time some alpha particles will becomes thermalized and capture electrons. Thus there will be a current to keep the wifle ball intact.
If not how is energy conserved?
We sometimes get confused because of our location here on earth. Let me explain.
This situation is like gravity and space.
If you were at a stationary position in space 200 miles from the surface you would have no kinetic energy and
only a possible potential energy. I say possible potential because if you were not near the earths gravity you would not have any forces on you. This is why we call it a potential well. Now if you fall into this well you would accelerate. You would convert potential energy to kinetic energy. Your potential would be negative hence the term "well". Coming down to earth one must expend this energy in our most archaic form i.e. heat. We sit at the bottom of a gravity well and must expend energy to climb out ..bummer. Sorry I digress….
Now our alpha particle is accelerating into a potential well. But it does not hit the surface it goes through and decelerates coming out the backside transforming kinetic to potential energy. Energy is conserved.
No one has explained to me how the electrostatic deceleration of alphas with a grid would become a source of current.
Here is how I would try to do it.
We have the outer grid a some extremely high voltage.
First we charge up the decelerating and decouple it from the system.
There will be a capacitor between the grid and the power supply that is held at a constant (high) voltage.
When an alpha particle approaches the grid it will add its positive voltage to the grid. This will be kinetic energy converted to potential energy. Small but there is power in numbers.
This will create a voltage across the capacitor. The capacitor voltage becomes an energy source that is small enough that we can deal with it with regular electronics. Now a circuit can be made to discharge this voltage through a coil and inductively picked up by another coil that is at or near ground potential. If the common mode potential
cannot be isolated then multiple coils can be used in steps to get to ground level.
There are concerns as to the voltage field outside the grid. It will have quite a gradient that will have to be taken into account as to where the physical placement of the electronics will be.
David
Regular electronics can handle 200 to 300 MW no problem.
===============================
Think of a linear particle accelerator.
Does it take current to accelerate the particles?If not where did the 1/2 mv^2 come from?
Now just operate it in reverse and use it to slow down particles. The current will reverse. Voila a power generator.
===============================
Think of a linear particle accelerator.
Does it take current to accelerate the particles?If not where did the 1/2 mv^2 come from?
Now just operate it in reverse and use it to slow down particles. The current will reverse. Voila a power generator.
Yes there will be a current associated with accelerating the alpha particle
toward the center.
There is an attractive force between the alpha partical and the wifle ball.
The wifle ball will be attracted to the alpha particle and the magnetic field
will need to be increased to keep the integrity of the wifle ball intact. Granted
for one particle it will be small. Many particles would cause a degradation in the integrity of the wifle ball, maybe.
I guess the question is a relative one. What is the magnitude of the current needed for the coils to keep the "ball" intact vs the current needed to accelerate the alphas at full power (max number of alphas)?
toward the center.
There is an attractive force between the alpha partical and the wifle ball.
The wifle ball will be attracted to the alpha particle and the magnetic field
will need to be increased to keep the integrity of the wifle ball intact. Granted
for one particle it will be small. Many particles would cause a degradation in the integrity of the wifle ball, maybe.
I guess the question is a relative one. What is the magnitude of the current needed for the coils to keep the "ball" intact vs the current needed to accelerate the alphas at full power (max number of alphas)?
With a liquid nitrogen cooled Cu magnet, the magnet power is about 1/100th the maximum grid current. This is for the test reactor and does not count the energy cost of cooling the liquid nitrogen.dbennett wrote:Yes there will be a current associated with accelerating the alpha particle
toward the center.
There is an attractive force between the alpha partical and the wifle ball.
The wifle ball will be attracted to the alpha particle and the magnetic field
will need to be increased to keep the integrity of the wifle ball intact. Granted
for one particle it will be small. Many particles would cause a degradation in the integrity of the wifle ball, maybe.
I guess the question is a relative one. What is the magnitude of the current needed for the coils to keep the "ball" intact vs the current needed to accelerate the alphas at full power (max number of alphas)?
For a fully operational reactor superconductors would be used for the coils so the energy requirements for the magnets would go way down.
I'm planning on about 2 MW of grid power (80KV @ 25A) for the test reactor. This may be far in excess of the requirements, but I wanted to be sure that current limiting on start-up or in operation was not a problem.
Surprisingly a power reactor might use less power because of fewer electron losses.