Principle of Nuclear Fission Direct Energy Conversion

[Aakash Sunkari]

TL;DR Summary

A basic level question: In a basic Direct Energy Conversion scheme for nuclear fission, positively-charged fission fragments are slowed down by an electric potential between an anode and a cathode, with the fragment depositing it's charge at the anode. The question: how is the kinetic energy of the particle converted into electrical energy (other than by static charge buildup)?

Good day everyone,

I've recently been researching direct energy conversion schemes for nuclear fission, and I have a question on the basic physics behind the device (specifically on how it converts kinetic energy to electricity).

In essence, the "basic" scheme for fission DEC devices is that:

• The fissionable fuel is used as a cathode, and an anode surrounds the fuel-cathode. A potential of several million volts exists between the anode and the cathode.
• Fission fragments are ejected during fission and are decelerated by the electric potential. The fragment comes to a complete stop at the anode and the particle deposits its charge.
• An external circuit relieves the charge and produces a current.

Note that various methods are used to deal with the electrons in the atoms (which would otherwise offset the positive charges).

Diagram of a simple fission DEC:

My question is: how does decelerating the charged fission fragments allow us to convert their kinetic energy to electrical energy?
I understand how the buildup of charge in the anode can be released as a current, but that doesn't have anything to do with converting the the kinetic energy of the particles to electrical energy (or so I think?)
And what is the physics principle (name or concept) that explains how slowing down a charged particle will produce a current? I've heard this principle referenced many times in research papers but I've never seen it listed in any of my textbooks.

Thank you all for your time and help!
Additional resources: A review paper which summarizes the working principles of a basic fission DEC.

 

[hutchphd]

The charged particle could not get to the anode unless it expends some kinetic energy to get there. For the system held at voltage V additional charge q adds qV to the system. You choose V (the anode-cathode difference) to be as high as you can and still get charges to the Anode. It is the same as charging a capacitor "from the inside"!

 

[Astronuc]

Summary:: A basic level question: In a basic Direct Energy Conversion scheme for nuclear fission, positively-charged fission fragments are slowed down by an electric potential between an anode and a cathode, with the fragment depositing it's charge at the anode. The question: how is the kinetic energy of the particle converted into electrical energy (other than by static charge buildup)?

In essence, the "basic" scheme for fission DEC devices is that:

• The fissionable fuel is used as a cathode, and an anode surrounds the fuel-cathode. A potential of several million volts exists between the anode and the cathode.
• Fission fragments are ejected during fission and are decelerated by the electric potential. The fragment comes to a complete stop at the anode and the particle deposits its charge.
• An external circuit relieves the charge and produces a current.

Note that various methods are used to deal with the electrons in the atoms (which would otherwise offset the positive charges).

I presume the device would operate initially in a vacuum, however, one should realize that various fission products are isotopes of noble gases, Xe and Kr, and volatiles, e.g., Br, I, and low melting temperature metals like Rb, Cs. Some of the volatile elements form volatile compounds (TeI, CsI). The device would have to be cooled from the outside of the anode.

Furthermore, realize that for every fission product (fp) directed toward the anode, one stays with the cathode. The energy in fps is something like 170 MeV; the lighter fp will have more KE, e.g., ~0.6*170 = 102 Mev, and the heavier fp ~ 68 MeV; range in solids is about 6 microns for the light fp, and about 2-3 microns for the heavier fp. As they separate, they reconfigure electrons.

Realize also that there are 7 or 8 prompt gammas of average energy ~ 1MeV each (actually there is an energy distribution) and various beta particles. The gammas can induce free electrons from the photoelectric effect and Compton scattering, which will be initially in the 100s of keV range, and each will scatter producing more electrons.

I would expect such as device to be impractical. On the other the other hand, I've seen thermionic systems.

 

[shjacks45]

Any additional positively charged fission decay fragments are repelled by the anode and attracted to the grids negative charge. A mol wt 100 ion accelerated to 1 MeV gives the fragment 100 MeV of energy. So an anode at 1 MV would mean only a fraction of fissions' energy is captured, most of the fragments ending at the grid.
Also don't follow how discharging the anode-cathode generates power.

 

[artis]

@Aakash Sunkari well besides to what has already been said your intuition I think is in the right direction.
The charge of the particles released would constitute a current but their kinetic energy would simply be released as heat within the electrodes as they strike them.
So in such a scheme one would recover current from the charges and also need a lot of cooling for the electrodes. This removed heat would be used the same way as in every existing nuclear power plant, aka to drive a steam or gas (or whatever other coolant is used) cycle to convert heat energy to mechanical torque for driving a generator.I think @Astronuc will correct me if I'm wrong but direct energy conversion is more discussed with respect to nuclear fusion rather than fission because in fission most of the released energy is as kinetic energy of the fission fragments and neutrons which then due to interaction with matter release heat.
For example fast neutrons from fission are scattered within the coolant and become slow neutrons in this process the coolant is heated up but it also helps keeping the fuel and surroundings at a reasonable temperature.

 

[Astronuc]

A mol wt 100 ion accelerated to 1 MeV gives the fragment 100 MeV of energy.

This is a mis-statement. Accelerating a mass to 1 MeV gives 1 MeV of energy. Accelerating a charge of 100 e (charge on electron/proton) over a potential of 1 MV would yield an energy of 100 MeV. However, 1) one is not going to find a bare nucleus of 100 e, and 2) the dimensions are not provided, but I expect they would be on the order of cm, so a potential of 1 MV over 1 cm is a substantial electrical field, 100 MV/m. I suspect one would produce an arc between cathode and anode.

As soon as fission occurs, the electrons reconfigure rapidly within each of two (or three) nuclei. The ions maybe something like +8, +7, but briefly as electrons are attracted (from other neutral atoms in which the first ionization potential is at most a few eV) to them.

Other problems with the design include the fact that at most, half of the fission products will escape the U layer, and one still needs a source of neutrons in a configuration that maintains criticality while produceing copious neutrons.

I think @Astronuc will correct me if I'm wrong but direct energy conversion is more discussed with respect to nuclear fusion rather than fission because in fission most of the released energy is as kinetic energy of the fission fragments and neutrons which then due to interaction with matter release heat.

Fusion uses a plasma, or highly ionized atomic gas, so naturally, with ions and electrons separated, it would make sense to collect the ions and electrons, separately and have the electrons travel through a load (producing a current) to eventually neutralize the ions. In fission, the fission products do not travel far and are certainly not fully ionized (only partially), but become neutralized fairly quickly. In the OP system, most of the energy of the fission products would be dissipated as thermal energy, i.e., collisions and recombination (with low energy X-ray emissions).