Questions Once Again

[Robthebob]

So I was wondering about the finer details and I came up with a list of energies that are required in preperation phase.

1. Obtaining boron 11, probably require chemical compounds to be broken down in order to get it.

2. Storing boron 11, since its never found free in nature, it may be rather reactive.

3. Obtaining Hydron.

4. Storing hydrgon, since its highly reactive.

5. Making the fuels into ions, hydron is probably going to be okay, wont be too bad, I dont know if you can strip all the electrons from the boron 11, even the inner level.

6. Starting the magnetic field.

7. Maintaining the Magnetic field, which may be free, because we can use super conductor coils, yay.

8. Maintaining the condition of the super conductivity.

9. When the three heliums form and fly out of the potenial well, it would cost the three heliums a lot of energy.

So under the right conditions after subtracting all these numbers, would we have enough energy left?

[TallDave]

Nebel or others can correct me if I'm wrong here, but I believe all of the costs above are trivial in terms of the net power equation (this is the case whenever you move from chemical to nuclear, thanks to E=mc**2; that's why fission fuels are a tiny fraction of the power cost while chemical fuels are generally the main cost driver). The big losses to overcome are the electron drive losses, which scale as r^2 according to Bussard, and the brem losses, which can hopefully be controlled by fuel mix and some other tricks.

The helium slowing down is where we gain energy. They trade kinetic energy for potential. It's sort of like a really strong guy who throws buckets of water up to the top of a mountain, which you then use to turn your water wheels on their way down. 93132 has a better explanation of this somewhere.

After that it's a question of how big you make it (with power scaling as roughly r^7), versus how much that costs, versus how much power density modern materials can withstand.

[blaisepascal]

So I was wondering about the finer details and I came up with a list of energies that are required in preperation phase.

1. Obtaining boron 11, probably require chemical compounds to be broken down in order to get it.

I don't have the figures handy, but Boron-11 is currently available on the open market from multiple suppliers. My assumption is that it's not too energy heavy to produce. Producing mixed-isotope boron is probably less energy intensive than producing aluminum. It's the separating out of the isotopes that's tricky, but even that's not too energy intense.

2. Storing boron 11, since its never found free in nature, it may be rather reactive.

Aluminum is reactive and never found free in nature. But it's easily stored in near-pure form. Based on its location in the periodic table, I'd expect it to be less reactive than carbon. The MSDS I've seen indicates it's not a fire or reactivity hazard.

3. Obtaining Hydron.

4. Storing hydrgon, since its highly reactive.[/quote]

Both are easy and long-solved problems.

5. Making the fuels into ions, hydron is probably going to be okay, wont be too bad, I dont know if you can strip all the electrons from the boron 11, even the inner level.

Again, easy and long-solved problems. Because ionization is more difficult than chemical changes, the ionization of boron is probably the most energy-intensive operation so far in the process. It's about 75MJ/mol, or 7GJ/kg.

6. Starting the magnetic field.

A startup cost. Amortized cost is minimal.

7. Maintaining the Magnetic field, which may be free, because we can use super conductor coils, yay.

8. Maintaining the condition of the super conductivity.

Both of these are lowish. Important, but lowish.

9. When the three heliums form and fly out of the potenial well, it would cost the three heliums a lot of energy.

The potential well they are flying out is about 50-100kV deep, and the energy "cost" to the heliums would therefore be about 600keV, out of the 8MeV of the heliums. It would give that energy back to the system, replacing the energy needed to accelerate the B and H to fusion energies.

So under the right conditions after subtracting all these numbers, would we have enough energy left?

It seems like "yes". Ionizing the boron would cost 7GJ/kg, and is probably the bulk of the energy costs involved in all the prep work and running the system. So let's call running the system. A 100MW reactor "burns" about 1kg of boron a day, thus generating about 9000 GJ/kg of fuel. If we assume 18GJ/kg for preprocessing and operating costs, we have a proposed gain of 50000%. There is a lot of room for badly estimated costs there.

Assuming we can get it to work at all, that is.

[MSimon]

Boron 11 (and its byproduct Boron 10) are manufactured to high purity for semiconductor mfg.

Initial supplies (at least for research purposes and possibly initial reactor use) will not be a problem. Since it is already done ramping up production will not be hard.

[Robthebob]

I guess in this case, the devil is not at all in the details.

One thing I thought would be a huge problem is maintaining the superconductivity of the system.

Wouldnt you need to keep the system's temperature at a very low temperature?

I still dont understand the whole, helium giving energy back into the well, what do you mean?

[blaisepascal]

One thing I thought would be a huge problem is maintaining the superconductivity of the system.

Wouldnt you need to keep the system's temperature at a very low temperature?

Technically, superconductivity isn't required for Polywell fusion -- just large magnetic fields. To generate those large magnetic fields you need big coils, which will need to be cooled so that (a) if they are not superconductors, they won't melt down due to resistive heating, or (b) if they are superconductor, they will stay within the superconductive range.

Cooling superconductor magnets of the field-strength needed is routinely done in MRI machines. There are some different engineering challenges for a Polywell -- on the plus side, the magnets are in a hard vacuum, so heating from conduction and convection are minimal; on the minus side, the coils and cooling have to conform to the field and be as small as possible, as well as have enough capacity to deal with the heating from the impinging alphas.

I still dont understand the whole, helium giving energy back into the well, what do you mean?

At the center of the reactor is a "virtual cathode" made of an electron plasma, held in place by the magnetic fields we are generating with the coils. The potential at the center of the cathode is significantly negative compared to ground. I don't know the exact target potentials, but they are in the range of -50kV.

A Boron ion sitting at the top of the potential well has 0 kinetic energy. When it falls through the center of the reactor, it has a kinetic energy of 250keV. It got this kinetic energy from the potential well, reducing the energy of the potential well by 250keV. If it fails to collide with a proton and fuse, the ion will continue through the center and climb back up the potential well until it has given up all its energy to the well, and has 0 kinetic energy again. At which point, it'll start to fall again.

A proton in the well does the same back-and-forth, exchanging 50keV between its kinetic energy and the potential well.

If it does fuse, then you'll have 3 alphas at the center of the well with a combined kinetic energy of 8.7MeV. To climb to the top of the potential well, those 3 alphas have to give up 300keV to the well, leaving a measly 8.4MeV to be recovered by the rest of the system.

One way to recover that remaining 8.4MeV is to set up another grid inside the vacuum chamber but far outside the magnets and reactor proper which is held at about +1MV. This creates another potential well the alphas have to climb out in order to escape. To do so, the alphas have to give up about 2MeV of kinetic energy to this well, which is recoverable through the system. This process is called "direct conversion", as it turns the kinetic energy of the alphas into directly usable electricity and theoretically high efficiencies.

[drmike]

It's positive. The alpha's push against the field and you transfer their kinetic energy into potential energy. If it was negative they would gain more kinetic energy, and you would have to supply it.

[Robthebob]

oh cool, I was wondering, how would the energy distribution work? Like, the reaction occurs, and three alphas have a combined 8.4 MeV of energy, but if you are looking at the energy of each alpha, would it just be 8.4 divided by 3? I guess?

Oh yeah, you answered my question very throughly, thank you. I first thought that when the alphas climb out of the well, the energy left would would be really small, I didnt know the energy generated by the fussion process was about 30 times more than the energy required to escape the well.

The other grid, why is it only +1MV? Because the alphas would only give up 2MeV of energy, shouldnt it be 4 times that, so we can get the most of what we have?

[blaisepascal]

oh cool, I was wondering, how would the energy distribution work? Like, the reaction occurs, and three alphas have a combined 8.4 MeV of energy, but if you are looking at the energy of each alpha, would it just be 8.4 divided by 3? I guess?

There are other posts which have the exact figures, but it's not 8.7MeV divided by 3. The full reaction is p+B11 --> C12* --> Be8 + alpha + energy --> 2alpha + energy. The initial C12 nucleus is not in a stable ground state, which is why it decays via alpha decay. That alpha has one energy level, and the two alphas from the fission of the Beryllium nucleus are at another energy level.

Oh yeah, you answered my question very throughly, thank you. I first thought that when the alphas climb out of the well, the energy left would would be really small, I didnt know the energy generated by the fussion process was about 30 times more than the energy required to escape the well.

If it wasn't, we probably wouldn't be looking too hard at that reaction. Nuclear energies tend to be large.

The other grid, why is it only +1MV? Because the alphas would only give up 2MeV of energy, shouldnt it be 4 times that, so we can get the most of what we have?

Keep in mind that I don't have the actual numbers in front of me, so I'm doing back-of-the-eyelid calculations, which are a bit rough.

If an alpha climbs a 1MV potential well, it gives up 2MeV of energy. If three alphas climb a 1MV potential well, they give up 2MeV each, or 6MeV total. That's why the grid needs to be 1MV, not 4MV.

With a 1MV potential well, we can easily recover 6MeV per reaction directly, and allow the remaining 2.7MeV be absorbed by the walls to generate power thermally. With a 1.25MV well, we can recover 2.5MeV from each alpha, and 7.5MeV per reaction.

A search turned up a reference that two of the alphas will have an an energy of about 2.46MeV each, which means we don't want a 1.25MV well, we want at most a 1.23MV well, otherwise the alphas will not reach the top and fall back in, a bad thing.

Other threads on this forum describe attempts at designing direct conversion mechanisms to capture as much energy from the alphas as possible, including multiple grids (at 1.2MV for the slow alphas, and 1.8MV for the fast alphas), shaped grids to attempt other separation methods, etc.

[drmike]

It's not exactly divided by 3, two have about the same energy and one is slightly less. I can't find the online reference right now, but it's out there. Dividing by 3 is close enough.

The reason you have 1 MV and 2 MeV is because the alpha has two charges. Each charge contributes to the particle's energy transfer,

Hey Blaise, you beat me to it! Can you post the reference to the energy distribution? We're going to need it again!

[blaisepascal]

Hey Blaise, you beat me to it! Can you post the reference to the energy distribution? We're going to need it again!

I found it half-way down the page at http://powerandcontrol.blogspot.com/200 ... usion.html, which cites a transcript of Bussard's Google talk. I know there have been a few postings on this forum which mention them (I just found viewtopic.php?t=629&highlight=direct+conversion, for instance). I'm sure I've seen some of the great p-B11 fusion animation that mentioned the energies as well.

[Robthebob]

I was actually talking with my professor about this a while ago, he said electrons are kinda fast and small, and may be harder to confine than protons and whatnot.

Is that right? I thought since electrons are a lot lighter, it would actually be easier to confine them.

[MSimon]

I was actually talking with my professor about this a while ago, he said electrons are kinda fast and small, and may be harder to confine than protons and whatnot.

Is that right? I thought since electrons are a lot lighter, it would actually be easier to confine them.

It is the charge to mass ratio that makes electrons easier to confine by about a factor of 40.

[Art Carlson]

In the following, I use the energies from the first ref of blaisepascal:

And yes, according to the transcript of the Google talk, there is the fact that the three alphas produced in p B11 reaction are not equal: the first one has 3.76 MeV, and the other two a median energy of 2.46 MeV.

p + B11 -> C12*
The total energy produced by this reaction is 8.68 MeV. Let's figure out how that energy is divided among the three alphas that are eventually produced.
C12* -> alpha + Be8*
The energy distribution of the alphas will have a spike at 3.76 MeV (with a width determined by the - relatively small - thermal spread of the reactants) since there is only one way that the C12* nucleus can decay. Conservation of momentum tells you that the Be8*, which has twice the mass of an alpha, will be moving in the opposite direction with half the velocity, and therefore half the kinetic energy of the alpha. That leaves 8.68 - 3.76 - (1/2)*3.76 = 3.04 MeV for the internal energy of the excited Be nucleus.
Be8* -> 2 alpha
The internal energy is divided equally to the two alphas (again, in order to conserve momentum), 3.04/2 = 1.52 MeV to each.

But remember that the velocities the alphas get from this decay must be added to the velocity of the Be nucleus they come from. Let's start over. Call the velocity of the first alpha v_0, the velocity of the excited Be nucleus v_1, and the velocity of the alphas produced when Be8* decays (relative to the Be8* velocity) v_2. We have, using only conservation of energy and momentum,
0.5 * m_alpha * v_0^2 = 3.76 MeV
(2*m_alpha) * v_1 = m_alpha * v_0
0.5 * m_alpha * v_2^2 = 1.52 MeV

It is easy to see that
v_1 = 0.5 * v_0, and
v_2 = sqrt(1.52/3.76) * v_0 = 0.64 v_0.
If the Be8* happens to decay so that the alphas fly apart along the direction of the Be8* velocity, then one of them will have velocity v_1+v_2 = 1.14 v_0 and the other will have velocity v_2-v_1 = 0.14 v_0. (Check on the math: 1^2+1.14^2+0.14^2 = 2.32 = 8.68/3.76) The energy of the faster alpha will be 1.14^2*3.76 = 4.89 MeV, and the slower one will have 0.14^2*3.76 = 0.07 MeV. If the Be8* happens to decay so that the alphas fly apart perpendicular to the direction of the Be8* velocity, then each will have velocity sqrt(v_1^2+v_2^2) = 0.81 v_0. (Check: 1^2+2*0.81^2 = 2.31) What results is an alpha energy distribution with a spike at 3.76 MeV containing 1/3 of the alphas, superimposed on a continuous distribution for the rest extending from essentially zero energy up to 4.9 MeV.

The continuum will make a gridded direct conversion system more complicated, less efficient, or both. The lowest energy alphas will not even be able to get out of the potential well holding the fuel ions, leading to a buildup of helium ash, which will exacerbate the bremstrahlung problem beyond the already extremely difficult level assuming pure fuel. This effect should be relatively easy to quantify. Rick, has this been published, do you know the result?