Polywell Electricity Step-by-Step
|Near right is the electronic symbol for a battery, but I am using it as a symbol for any source of direct current: a battery, a charged capacitor or a power supply. For the following explanation, the particular type of DC source is not important.||Near left is the symbol for “ground.” Excess charge, positive or negative, eventually bleeds off to the ground. Voltages are given relative to ground. Strictly speaking, it can stand for a copper rod driven into the earth or the green wire in household wiring.|
First – Three small “up-front” electrical investments are required (right):
- “to Coils” current starts-up the magrid super-conducting coils. No outside current will be needed once super-conducting flow begins.
- "50 kV" charges the magrid covers. Once charged, additional current requirements are minimal.
- "1 kV" powers the electron guns which “fill-up” the Wiffle Ball. After that, the electron gun is only needed to replenish lost electrons.
Second – the Wiffle Ball forms:
- The Wiffle Ball is a spherical, magnetic confinement containing the electrons, which provide the negative potential well.
- The Wiffle Ball is formed as electron density increases in the well, pushing magrid’s magnetic field away, distorting it and closing off the electron escape routes (cusps).
- On the left is a “cut-away” view. Cross-sections of the magrid coils are shown in red. The spherical blue swarm of electrons in the center is pushing the magrid's magnetic field lines away, to form the Wiffle Ball
Third – Ion sources (right, in green) strip electrons from hydrogen and boron atoms, and drop the new H and B ions into the well. (671 volts is required to completely ionize a boron atom - another up-front investment.)
- The ions are attracted by the electron cloud, and accelerated into the center of the well.
- The electrons that were “stripped” from the H & B atoms are pumped into the ground. This turns out to be very important later!
Fourth – The now-energized H and B ions collide and fuse, producing:
- An alpha particle (He +2 ion) with an energy of 3,760,000 electron volts, and
- Two more alphas (He +2 ions), each with an energy of 2,460,000 electron volts,
...which all fly away from the center (shown on the left) carrying their millions of electron volts in energy.
Twenty percent of the alphas will strike the magrid, wasting their energy, heating and damaging the magrid. Prodigious design efforts will be required to address this problem.
Fifth - To convert the remaining energy to electricity, we need two concentric spherical grids outside of the magrid (right):
- A negatively charged Trap grid just outside of the magrid.
- A positively charged 1.22 million volt Collector shell outside of the Trap grid.
There is an electric field between the Trap grid and the collector shell, which slows down the escaping alphas (He +2 ions). The Trap grid also acts as a Faraday cage making the 1.22 million volt Collector invisible to the re-circulating electrons.
A highly mobile “sea” of electrons flows between the electron pair bonds that hold metal atoms together. A metal bond that is temporarily lacking an electron can be called a “hole.” Sometimes it is useful to think of these “holes” as positive charges, which can ripple rapidly across the bonding orbitals of the metal atoms. Even though the 1.22 million volt positive charge on the collector shell suggests that the metal has fewer electrons than protons, it still has many, many billions of electrons.
Sixth - The highly energetic positive alphas approach the positive 1.22 million volt collector shell, which is repelling them:
- They slow down, giving up their kinetic energy to positive “holes” in the collector shell, and pushing the “holes” down the wire that connects to it.
- Electrons rush up the same wire in the opposite direction, frantically seeking to compensate for the energetic positive “holes” rushing past them.
The repulsion of the 1.22 million volt collector nearly stops the 2.46 MeV alphas. (Don’t forget – the alphas have a charge of plus two. Multiply that 2 times the 1.22 million volt collector charge, and you get 2.44 MeV the work done in slowing the 2.46 MeV alpha.) The slowed alphas gently bump into the collector surface, immediately pick up two of the electrons, and turn into neutral helium atoms. The polywell’s electric current –the electricity- comes from all these electrons jumping out of the grids to neutralize the +2 alpha particles. The helium atoms are removed from the system with a vacuum pump to be recycled and sold.The 3.76 MeV alphas will have considerable excess kinetic energy -about 1.36 MeV total- when they smack into the collector grid surface, but they will still pick up two electrons and turn into helium atoms. The total energy for each fusion back in the well was 8.68 MeV (2.46 + 2.46 + 3.76). Of that, 1.36 MeV -almost 16% of each fusion- is wasted in the 3.76 MeV alpha collision with the collector. Also, about 20% of the original 100MW of fusion output is intercepted by the magrid. So 16% of the remaining 80% or another 12.5% (0.16 x .80 = 0.125) is lost to heat at the collector grid. A total of 32.5% of the original 100 MW -or 32.5 MW- will make heat instead of electricity. That is a LOT of heat: the two hydroelectric dams near my home together produced only 20 Mw before they were to be removed! Sadly, when combined with the inevitable down-time, this will lower the efficiency of the polywell to about 60%. Plus all of that wasted heat will have to be removed from the magrid and the collector shell by some sort of heavy-duty cooling system - certainly not a trivial task!
Seventh – the positive 1.23 MeV “holes” have a higher energy than the battery that is charging the collector, consequently:
- They are diverted (that is, they “overflow”) to the HVDC to AC inverter (high voltage direct current to alternating current inverter) which will be at the user end of the power line.
- The extra electrons that were earlier “stripped-off” in the ion source, rush out of the ground and through the inverter to fill the positive “holes.”
There is a net gain in power because:
a. For the ion source: V x I = P: 671 volts x 81.3 amps = 0.0546 million watts
b. For the inverter: V x I = P: 1.23 x 10^6 volts x 54.9 amps = 67.5 million watts
In other words, even though many of the same electrons were involved, their gain in energy was very close to 67.6 million watts! (The difference in amps is due to the 32.5% heat losses mentioned previously.)
The collector shell has a lot more surface area than the magrid, and it is much further away from the fusion reactions. Both the greater area and the greater distance work in favor of the collector shell. Nevertheless 12.5 megawatts of waste heat will need to be carried away from it. Four sections, similar to the one pictured left, would be constructed of standard 4” copper pipe, and bonded to the outside of the collector shell. Such a device should easily carry more than 9000 lbs of water per minute, providing a heat removal capacity of about 0.9 x 10^6 BTUs per minute or 15.8 megawatts, which is well over the 12.5 megawatt requirement.