Building a WB-2 Polywell

Discuss the technical details of an "open source" community-driven design of a polywell reactor.

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D Tibbets
Posts: 2775
Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

Mattman, I do not think you need to be worried about neutron activation. The numbers are just to small to worry about. Assume you produce a billion neutrons per second and all of these manage to be trapped by the material of concern. You are activating at most ~ 10^-10 portion of the material. While this could cause a measurable activation. It could be controlled for. And it would have a miniscule effect on the strength and temperature of the material. The vastly more significant heat comes from the Ohimic heating of the magnet wires. With superconducting, the picture is somewhat different, but piratically you would have to increase the coolent fow very sightly. Using WB6 as a example of ~ 500 million neutrons per second, the neutron deposited heat would max out at under 1 milliwatt of heating. The contamination of a super conducted might be vulnerable, but again we are talking about less than 1 ppb levels.

This all changes when you are talking about commercial reactors that might be putting out many millions of Watts of fusion power, but for these feeble research efforts (perhaps 0.001 to 10 milliwatts of fusion output) the issue is minimal if not moot.

Dan Tibbets
To error is human... and I'm very human.

mattman
Posts: 459
Joined: Tue May 27, 2008 11:14 pm

Post by mattman »

That is tough to estimate.

Here is an attempt to estimate reactor burn out, from: http://thepolywellblog.blogspot.com/201 ... ation.html


The important points of this analysis are:

1. This is only a displacements per atom analysis - which is something ripped off from the fisson industry.
2. Boron Fusion is great for durability.
3. By nuetrons analysis only, the Polywell may have multi-year lifespan.
4. This is figured for a machine burning 1 Kg/Hour of fuel.
5. More work is needed.
6. LANL needs cross sections hot tritium, Helium-3 and Helium-4 interacting with other materials. This data was lacking so a worst case scenario was assumed.

Also, Sorry Guys, Excell Charts Do Not Translate Well on this forum.
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The Polywell will produce heat and neutrons, which will “burn out” the reactor core. Can we measure that effect into dollars and time? This burn out effect is actually several processes happening at the same time. There is embrittlement – where the neutrons knock holes in the surrounding materials. There transmutation – where alphas and neutrons hit the chamber walls and fuse with these materials. There is thermal stress – where the rapid heating and cooling causes problems in the walls. There may be other problems as well [31]. It is unclear which of these effects will have the biggest impact. This section will focus on the effect of bombardment. This is the “threat” - it is a mixture of nuclei, neutrons and protons hitting the walls. We do a simple analysis of this. We then try to connect that to an estimated price. That is why this calculation can only give us an estimate of how durable the reactor core will be.

Metals under attack from neutrons, protons and nuclei will crack. The degradation rate is hard to predict. Fortunately, how neutrons attack materials has already been studied. This is because existing reactors have to deal with neutrons. A U-235 nucleus, undergoing fission, releases an average of 2.4 fast neutrons [29]. The goal in fission plants is to slow down, but not absorb these neutrons. This is done using a moderator. The moderator also protects the reactor core is from the neutrons – giving the core a long durability. The goal is the same in a Polywell. Slow down these neutrons, so they can be used to heat a working fluid. But, these reactors will have no moderator. The core – the rings and chamber – will get blasted. As these materials get blasted by neutrons, the atoms are displaced and this leads to equipment failure. Cracks build. These cracks can swell with fusion products. The degradation by neutrons is measured by the rate of atom displacement [31]. The equation used to find the displacements per atom is included below.

Displacements Per Atom = Atoms Displaced/Orginal Atoms = Nuetron Flux * Time * Material Cross Section (Nuetron Energy)

A good rule of thumb is cracking becomes a problems at 1.0 displacement per atom [35]. As a reference point, parts of a 1300 megawatt PWR core can receive between 0 and 90 dpa [32]. Beyond this benchmark, degradation depends on many factors, including: thermal stress, local chemistry and material processing [31]. Also, hot protons and nuclei hitting the surface could degrade the walls as well. This is getting messy. So let us simplify. Let us apply the equation above. Will this equation predict cracks? The materials we use for the wall will be the ones considered elsewhere in this post, plus those common to todays’ fission reactors. By burning a fuel, it means that a specific byproduct would shoot out and hit the wall. That byproduct would be at a specific energy. Ideally, we need cross section data: for that specific byproduct, hitting that specific material at that specific energy. Cross sections measure how easily one thing hits something else. Below lists the cross sections we could find, for those energies, byproducts and materials from the Los Alamos’s nuclear information service [33].

Cross Sections Used: (eV)
Reaction: Byproduct: Energy: Carbon Tungsten Aluminum Iron Chromium Molybdenum Nickel Neodymium


D & T -> a + n Neutron 1.4E+07 1.3 5 1.8 2.5 2.4 3.6 3 4.8


D & T -> a + n Helium-4 3.5E+06 LANL lacks helium 4 data

D & D -> T + p Proton 3.0E+06 0 1.E-20 0.35 4.4E-03 0.02 unknown 5.9E-04 0


D & D -> T + p Tritium 1.0E+06 LANL only has tritium data for: H, He and Li


D & D -> He3 + n Neutron 2.5E+06 1.5 6.2 3 3.1 3.2 4.5 2.8 6.4


D & D -> He3 + n Helium-3 8.2E+05 LANL only has helium 3 data for: He & Li.


P & B11 -> 3a Helium-4 8.7E+06 LANL lacks helium 4 data

Lack of data! The ideal analysis will not work in this case because the numbers needed are unavailable. Of all the byproducts hitting the wall, the neutrons are the most damaging. They penetrate deepest and they can make the walls radioactive. If all the byproducts were neutrons; then the numbers would be available. We decided to go this route. That makes this analysis: a conservative, worst case scenario. Below is a list of the cross sections for neutrons at the hottest byproduct energy, when they hit that material.
Cross Sections Used (eV)
Reaction: Byproduct: Energy: Carbon Tungsten Aluminum Iron Chromium Molybdenum Nickel Neodymium

D & T -> a + n Neutron 1.4E+07 1.3 5 1.8 2.5 2.4 3.6 3 4.8

D & D -> T + p Neutron 3.0E+06 1.5 5.8 2.5 3.3 3.2 3.8 3 6.1

D & D -> He3 + n Neutron 2.5E+06 1.5 6.2 3 3.1 3.2 4.5 2.8 6.4

P & B11 -> 3a Neutron 8.7E+06 0.9 5.3 1.8 3.2 2.7 4.4 3.3 4.4


These cross sections were combined proportionally for each wall material. Steel was considered as 316 steel [34] and the proportions we used are shown below. It was assumed the reactor chamber was 15 feet in diameter with 3 inch wall thickness. Bussard had pushed for a wall with 1 cm tubes with coolant in them, to capture the heat [28]. The reactor was burning 1 kilogram per hour, at the aforementioned efficiencies. Example calculations for the worst, better and best case scenarios are worked out below.


D&T Nuetrons rate= 2.5E+16 Nuetrons/(cm^2*sec)=(1Kg/(1 hour))*(20%)*((1 mole)/1.0079g)*(1000g/1Kg)*(6.022e23/1mole)*(1/(656E5 〖cm〗^2 ))*(1/2)*((1 hour)/(3600 sec))

〖Stainless Steel〗_CS=17%*〖Chromium〗_CS+3%*〖Molybdenum〗_CS+.8%*〖Carbon〗_CS+63%*〖Iron〗_CS

(Tung.Carb.dpa)/Year= 1.99 dpa per year=2.5E16 Nuetrons/(cm*s)*(31449600s/(1 year))*(2.5 Barns)*((1E-24 〖cm〗^2)/(1 Barn))

Using the above equations, we can now compile a list of displacements per atom for each fuel and each material. This is included below:


Worst Case Scenario*:
(eV) Fluence: (dpa per year)
Reaction: Energy Convert N/(S cm^2) Tung Carb Steel Aluminum Molybdenum Neodymium

D & T -> a + n 3.5E+06 20% 2.5E+16 2.51 1.67 1.43 2.87 3.83

D & D -> T + p 3.0E+06 20% 1.2E+16 1.43 1.08 0.98 1.49 2.39

D & D -> He3 + n 2.5E+06 20% 1.2E+16 1.45 1.00 1.13 1.69 2.41





Better Scenario**:
(eV) Fluence: (dpa per year)
Reaction: Energy Convert N/(S cm^2) Tung Carb Steel Aluminum Molybdenum Neodymium

D & T -> a + n 3.5E+06 35% 4.4E+16 4.39 2.93 2.51 5.02 6.70

D & D -> T + p 3.0E+06 35% 2.2E+16 2.50 1.89 1.72 2.61 4.19

D & D -> He3 + n 2.5E+06 35% 2.1E+16 2.54 1.74 1.98 2.97 4.22

P & B11 -> 3a 8.7E+06 35% 4.6E+12 0.0005 0.0004 0.0003 0.0006 0.0006


P & B11 -> 3a 8.7E+06 20% 4.6E+12 0.0005 0.0004 0.0003 0.0006 0.0006



Best Scenario***:
(eV) Fluence: (dpa per year)
Reaction: Energy Convert N/(S cm^2) Tung Carb Steel Aluminum Molybdenum Neodymium

D & T -> a + n 3.5E+06 45% 5.7E+16 5.65 3.77 3.23 6.46 8.61

D & D -> T + p 3.0E+06 45% 2.8E+16 3.22 2.43 2.21 3.35 5.38

D & D -> He3 + n 2.5E+06 45% 2.7E+16 3.26 2.24 2.54 3.81 5.42

P & B11 -> 3a 8.7E+06 45% 4.6E+12 0.0005 0.0004
0.0003 0.0006 0.0006


This is very exciting. These dpa rates are very similar to todays’ PWR reactors. Bussards’ report also argued that the neutron flux experienced would be close to pressurized water reactors [28]. Does this mean a Polywell core would last as long as a typical fission core? A typical core today can last at least 20 years. This is an open question.


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* Worst Case: This assumes 20% fuel fused

** Better Case: This assumes 35% fuel fused

*** Best case: This assumes 45% fuel fused

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