Chapter 6

[dkfenger]

Rolls of boron wire. Each atom makes 6 electrons move at something like 1.3 million volts. Work the energy per ton out from there, but it is a bunch.

8.7 MeV per B11 consumed.
8.7 MeV * (6.022e23) / 11 in kWh => 21.2 kWh per gram of B11.

So a 100MW (fusion output) plant would require about 5kg of B11 per hour, and could run for over a week on a ton of fuel.

[ladajo]

Easy enough to figured what he charged to as it does note somewhere (I think without looking) the total number of caps in the bank.

Probably, we can also figure out something about the caps from the pics as well.

[jcoady]

"The coils will eventually be charged up to 25kV."

I implemented a Poisson Solver in MATLAB for a charged magnetic grid inside the vacuum chamber.

Nice.

Have you considered implementing a 1/8 sector rather than a 1/2? Might allow a small bit of extra resolution.

But nice none-the-less.

I actually implemented poisson solver for the whole device, and in order to visualize what the results were inside I split the mesh down the middle and plotted the result using MATLAB. I haven't tried just solving for a 1/8 sector yet. For the 3D tetrahedral mesh resolution I have an adaptive mesh that has a finer resolution around the magnetic grid and lower resolution as you move away from it.

MATLAB has a whole bunch of 3D visualization tools so there are many ways to look at the results. For instance you can slice up your 3D mesh to visualize along different planes like this.

http://www.mathworks.com/help/techdoc/ref/slice.html

Other visualization mechanisms like 3D plots, isosurfaces, contours are also available in MATLAB.

http://www.mathworks.com/help/techdoc/v ... liccy.html

I tried the my 3D poisson solver in matlab with a potential well added. This is the result I got.

http://dl.dropbox.com/u/5095342/images/polywell22.jpg

I also used the matlab function 'surfc' to print out the surface surface plot with a contour plot below which produced the following.

http://dl.dropbox.com/u/5095342/images/polywell20.jpg

http://dl.dropbox.com/u/5095342/images/polywell21.jpg


http://www.mathworks.com/help/techdoc/ref/surfc.html

[John Gallagher]

5kg of boron/hr for 100meg's ! I was wondering what pressure this thing may run at. Hydrogen and helium are real bears to pump. At 1 micron pressure you are looking at a million or two liters a second pump speed. One big diffusion pump! The backing pumps would be interesting also as turbo and diffusion pumps tend to have compression ratios under a thousand for these gases. If these were operating at a lower pressure they would have to be scaled by the same ratio. A diff pump would have backstreaming problems at 1 micron and a turbo would have cooling problems. Etc etc. This is not your father's vacuum system!

[ladajo]

Nice job on the well. Looks like your riggin' puts the center of the well at about 12.5KV.

Also remember that the machine is not 100% "well" efficient.

[ladajo]

5kg of boron/hr for 100meg's ! I was wondering what pressure this thing may run at. Hydrogen and helium are real bears to pump. At 1 micron pressure you are looking at a million or two liters a second pump speed. One big diffusion pump! The backing pumps would be interesting also as turbo and diffusion pumps tend to have compression ratios under a thousand for these gases. If these were operating at a lower pressure they would have to be scaled by the same ratio. A diff pump would have backstreaming problems at 1 micron and a turbo would have cooling problems. Etc etc. This is not your father's vacuum system!

I have had (disturbing) visions previously of converting existing ship steam turbines into super size vacuum pumps...
Wouldn't have to relocate them that way.

Maybe also be someway to re-rig a gas-turbine for the same purpose...

[MSimon]

Air is on the order of 1E19 molecules per cc.

[MSimon]

5kg of boron/hr for 100meg's ! I was wondering what pressure this thing may run at. Hydrogen and helium are real bears to pump. At 1 micron pressure you are looking at a million or two liters a second pump speed. One big diffusion pump! The backing pumps would be interesting also as turbo and diffusion pumps tend to have compression ratios under a thousand for these gases. If these were operating at a lower pressure they would have to be scaled by the same ratio. A diff pump would have backstreaming problems at 1 micron and a turbo would have cooling problems. Etc etc. This is not your father's vacuum system!

When I was running the numbers I always planned for two stages of turbo pumping. Ratioed at about 100 to 1 to give some slack. Then the backing pump.

I calculated the whole rig on being just this side of possible. But you know - half a order of magnitude in the wrong direction........

[D Tibbets]

Rolls of boron wire. Each atom makes 6 electrons move at something like 1.3 million volts. Work the energy per ton out from there, but it is a bunch.

8.7 MeV per B11 consumed.
8.7 MeV * (6.022e23) / 11 in kWh => 21.2 kWh per gram of B11.

So a 100MW (fusion output) plant would require about 5kg of B11 per hour, and could run for over a week on a ton of fuel.

Your numbers seem off.

http://en.wikipedia.org/wiki/Kilowatt_hour

One KWh = ~2 *10^22 eV.
Boron fusion yields >~8 MeV.
That means one P-B11 fusion yields ~ 4 *10-15 [EDIT should be in Wh, not KWh ]

This multiplied by Avogadros number yields ~10^8 KWh [EDIT 10^8 Wh or 100 MWh= my magnitude was correct, but the units were mixed up. the final results were ~ correct because I kept the units straight in my head, (Wh) though I wrote down the wrong unit (KWh)] per mole of boron and hydrogen, or ~ 12 grams to deliver ~ 100 MWh. This is ~ 500 times less than your calculation. You would need ~ 5 pounds of boron per week. To match this by chemically oxidizing boron (or carbon) would require ~ 5 * 30,000,000 (assuming energy yield of ~ 0.3 eV for chemical burning per atom) or ~ 150,000,000 pounds, or ~ 75,000 tons . Throw in about 30% thermal conversion efficience yields ~ 220,000 tons per week per 100 MWh
or ~ 50 railway cars per week for a 1 GW thermal plant. This comparison is consistant within a factor two to three.

Dan Tibbets

[John Gallagher]

Calculated by moles. 8E6 ev x9.5E4 coulombs/mole gives 760E9 joules/mole of Boron 11. So a100KW plant would need to burn about 5 gms/hr. This makes the vacuum pumps a bit smaller. Now I can get back to sleep.

[quixote]

I get different numbers than both of you.

Here's google calculator to show my work and so others can easily play with the numbers.

5.6042967 grams of 11b and p to generate 100 megawatt hours of energy. Of course this assumed that every atom of boron and every proton fused. I have no idea what the real ratios are, but if someone wants to plug them in there just multiply the 11b mass with the number of boron atoms it takes to generate a fusion on average, and also multiply the proton mass with the number of protons it takes.

Spelled out a little, here's what I did.

((100 megawatt hour) / (8 * (10^6) ev)) * ((11.00900 amu) + (1.00727638 amu))

100 megawatt hour / 8 MeV = 2.80867939 * 10^23 (note the units cancel out so the number is unitless).

So 2.80867939 * 10^23 is the number of fusions needed to create 100 megawatt hours of energy.

2.80867939 * 10^23 * ((11.009 amu) + (1.00727638 amu)) = 3.37498678 * 10^24 amu
11.009 amu is the mass of 11b (from memory)
1.00727638 amu is the mass of a proton (from wikipedia)

3.37498678 * 10^24 amu = 5.6042967 grams

It'd be fun and amusing if a few other people tried it out, too, to see how many different answers we can get for what should be a straight-forward calculation.

[mvanwink5]

In terms of boron, the proton weight is dropped (1 gm or 10% of weight). So, 5 gm/hr or 40 kg/year. (should fit in the laundry closet).

[ladajo]

So 2.80867939 * 10^23 is the number of fusions needed to create 100 megawatt hours of energy.

So we can cross this with the reported plasma density from chapter 6....

and then maybe estimate needed size using an alternative approach to the scaling calc.

[John Gallagher]

Now that the numbers for boron consumption have been 'refined' I was wondering what pressure range the plasma chamber can operate at. This should give some idea as to how a steady state system's vacuum system will have to be configured. It seems to me that electron confinement would be hampered by a high pressure/low mean free path. Charge shielding etc etc. So please direct me to the theory section that spells this out.

[D Tibbets]

Calculated by moles. 8E6 ev x9.5E4 coulombs/mole gives 760E9 joules/mole of Boron 11. So a100KW plant would need to burn about 5 gms/hr. This makes the vacuum pumps a bit smaller. Now I can get back to sleep.

I assume you made a typographical error like I did in the above post. The yield of ~ 5 grams of P-B11 would be near 100MWh, not 100KWh.

I rounded off more , resulting in a more approximate value. Also, as pointed out the burn up fraction is unknown. I generally operate on the assumption that much of the contained fuel is burned up before escape. A 50% to 80% burn up fraction may be reasonable (?). The actual fraction may be much less though, like 5-10 percent. A clever calculation of the losses from Bremsstrulung, etc. would establish the minimum burn up fraction that would be tolerable. Also the efficiency of the ion guns or gas puffers in delivering the ions into the Wiffleball is important. Because of these uncertainties the amount of boron (parent boron and derived alpha particles) transiting the machine per unit of time may be significantly more than the fusion numbers would indicate. That is an optimal condition.
Finally, the general density of hydrogens may be as much as ~10 times the boron density because of the intentional dilution used to control Bremmstrulung losses.

A simplistic estimate may be at least ~ 2 mg or ~ 0.02 moles of Boron per second and ~ 200 mg or 0.2 moles of hydrogen per second.This assumes almost perfect confinement and fusion burn up percentages. If no boron escapes, then the alpha numbers would be three times the boron numbers. The hydrogen numbers inside the Wiffleball would still be ~ 3 times greater than the alphas. But, the numbers of hydrogen ions escaping confinement may be significantly different (higher). If confinement is dependent on the number of passes before escape, the hydrogen would be escaping ~ 1.4 times as fast as the Boron loss rate,due to their higher speed.

Dan Tibbets

Dan Tibbets