cooling a WB and maximum size of WB's

[BenTC]

When the gyroradius of the 6 MeV (maximum energy) alphas is below the size of the donut hole they don't hit the coil casings. Cooling rqmts go way down. That happens at around 1T for a 1 m "hole". The coils are where you have the maximum heat problem. Out at the wall you have lots of area.

That would seem to indicate a significant change in the dynamics of the Polywell as scale increases past that point.

[MSimon]

When the gyroradius of the 6 MeV (maximum energy) alphas is below the size of the donut hole they don't hit the coil casings. Cooling rqmts go way down. That happens at around 1T for a 1 m "hole". The coils are where you have the maximum heat problem. Out at the wall you have lots of area.

That would seem to indicate a significant change in the dynamics of the Polywell as scale increases past that point.

It is not scale per se. It is field strength vs coil opening. The crossover point is about 1 m T. i.e .5 m "hole" 2 T. 2 m "hole" .5 T.

[KitemanSA]

It is not scale per se. It is field strength vs coil opening. The crossover point is about 1 m T. i.e .5 m "hole" 2 T. 2 m "hole" .5 T.

Correct me if I am wrong, but your contention is that this cross over is the condition wherein most of the alphas from pB&j make it out the cusps rather than hitting the MaGrid, no?

[D Tibbets]

The surface area of the direct conversion system should be much larger than the basic spherical surface area. Unfortunately it's also harder to cool...

There are two advantages that the collection could have. One is that they may, at least in part, be in the shadow of the magrid- less x-rays, slightly less gamma rays. . Also, they are further away, so the radiation and nonfocused charged partical flux will be more diffuse. If the collection grid is a venitian blind arrangement, the firs (cooled?) layer would again shield more distantal layers. . Ideally, the charged particals would have minimal heating energy once they reach the collection grid. But if efficiency tops out at 85%, then there would be significanlt residual heat. Would this impact the collection grid, or the more distant wall? Does the 85% efficiency refer to an actual very good conversion of the charged particle energies, but reconizes the percentage of the heating energy from x-rays, etc?

Dan Tibbets

[MSimon]

It is not scale per se. It is field strength vs coil opening. The crossover point is about 1 m T. i.e .5 m "hole" 2 T. 2 m "hole" .5 T.

Correct me if I am wrong, but your contention is that this cross over is the condition wherein most of the alphas from pB&j make it out the cusps rather than hitting the MaGrid, no?

Yes.

[agricola1964]

When the gyroradius of the 6 MeV (maximum energy) alphas is below the size of the donut hole they don't hit the coil casings. Cooling rqmts go way down. That happens at around 1T for a 1 m "hole". The coils are where you have the maximum heat problem. Out at the wall you have lots of area.

Figure 3T for a 1 m "hole". Scale up and down accordingly. i.e. 6T for a .5 m "hole". And of course the field gets larger closer to the coils. So the estimate is very conservative.

even if the electrons do never hit the magnets, the 15% power that is NOT converted has to show up somewhere - and the only places it can show up is either the magnets or the chamber wall ..

the exact transfer mode is not that relevant - but you have to take those 15% (or whatever - based on the total effciency) out of the system and that means cooling ..

what i an trying to figure out what the gross cooling demands are for a unit of size X

[93143]

the only places it can show up is either the magnets or the chamber wall

No, it can show up on the alpha collector plates. A bit of it might also show up on the electron guns, ion guns, and trap grid, but hopefully they will be small enough and well protected enough to not cause big problems...

If you assume that the outer wall area is a reasonable approximation of the area that has to be cooled in a large aneutronic Polywell, you can calculate the maximum power for a given size of reactor.

At 1 MW/m², with a 10 m diameter chamber, you get 314 MW of cooling power, or about 2.1 GW of core power if you assume 85% efficiency.

At 10 MW/m², which is more difficult but not necessarily impossible, you get 3.14 GW of cooling power, or about 21 GW of fusion power.

For a 5 m diameter chamber (which could achieve the same power output with weaker magnets, due to field concentration), with 10 MW/m² you get 785 MW of cooling power, or 5.2 GW of fusion power.

These are not hard numbers, and design specifics could easily modify them substantially. For one thing, the collector plates will probably have a higher surface area than the outer wall, but they will be harder to plumb...

The nozzle of the RS-25 (a.k.a. the SSME) is an encouraging data point...

[MSimon]

When the gyroradius of the 6 MeV (maximum energy) alphas is below the size of the donut hole they don't hit the coil casings. Cooling rqmts go way down. That happens at around 1T for a 1 m "hole". The coils are where you have the maximum heat problem. Out at the wall you have lots of area.

Figure 3T for a 1 m "hole". Scale up and down accordingly. i.e. 6T for a .5 m "hole". And of course the field gets larger closer to the coils. So the estimate is very conservative.

even if the electrons do never hit the magnets, the 15% power that is NOT converted has to show up somewhere - and the only places it can show up is either the magnets or the chamber wall ..

the exact transfer mode is not that relevant - but you have to take those 15% (or whatever - based on the total effciency) out of the system and that means cooling ..

what i an trying to figure out what the gross cooling demands are for a unit of size X

The 15% of ions that would hit the grid in a lower field device will get converted to electrical energy. It adds some to the heat load. It adds more to the energy out.

[TallDave]

even if the electrons do never hit the magnets, the 15% power that is NOT converted has to show up somewhere

Well, keep in mind, "waste heat" probably mostly ends up lost in a Carnot cycle somewhere in a steam turbine (even in a p-B11, I imagine you push your heat from cooling through a turbine if you can; Simon can tell me if this is crazy). It's mostly keeping the surfaces and superconductors healthy that matters.

[MSimon]

even if the electrons do never hit the magnets, the 15% power that is NOT converted has to show up somewhere

Well, keep in mind, "waste heat" probably mostly ends up lost in a Carnot cycle somewhere in a steam turbine (even in a p-B11, I imagine you push your heat from cooling through a turbine if you can; Simon can tell me if this is crazy). It's mostly keeping the surfaces and superconductors healthy that matters.

It depends on how much efficiency you are willing to give up. If you are willing to reject the heat at 80C or 90C everything gets smaller.

[agricola1964]

even if the electrons do never hit the magnets, the 15% power that is NOT converted has to show up somewhere

Well, keep in mind, "waste heat" probably mostly ends up lost in a Carnot cycle somewhere in a steam turbine (even in a p-B11, I imagine you push your heat from cooling through a turbine if you can; Simon can tell me if this is crazy). It's mostly keeping the surfaces and superconductors healthy that matters.

PRECISELY ..

the problem is that in order to feed the carnot cycle you first need to move that "waste heat" from the surfaces it first impinges on (which are either the magnet rings and/or the inner chamber surface) and move it where it does some good

however there is an practical upper limit how much energy you can move away (among other things you need to find the physical space for all that plumbing and you do not want to have to resort to strange cooling methods or mediums)

which returns to my orginial question - since the power out put (and thus unconverted "waste" power) climbs much faster then the avaialble surface to be cooled there will be a point, where you simply can no longer move the "waste heat" away fast enough ... this marks the upper power lmit of a practical polywell reactor

now - if i uunderstood what i have read so far the "r" in the scaling laws refers to the radius of the fusion region, or roughly the radius of the "hole" in one of the magnets of the cubical magnet arrangment?

what would be a rough guestimate for the "r" of a 100 or 1000 MW unit?

[TallDave]

OK, so we're on the same page there. I think 1MW/m^2 was the practical limit. Simon did a calc on that and probably remembers better. It turned out to be the limiting factor.

After that limit (around 100MW) you can grow your power only at the square of your radius (inverse square law), which is very bad for economics -- cost = r^3.

The precise meaning of r has been debated a bit (do we measure form center bore? is it the space between casings?). Strictly speaking, the r in r^3 is the ball of fusing plasma, but the machine has a radius too and I think we just assume the two Rs scale the same (i.e. 10x bigger machine = 10x bigger plasma ball). Bussard's estimate was a machine radius of 1.5m would be sufficient for 100MW. I think that was at 5T.

There are some more calculations floating around here, if you search, or you can start from ~.001W from WB-6 at .15m and .1T and go from there.

[ladajo]

I would think that the size of the plasma spikey ball is dependant on not only the size of the coils but the field applied. And that relationship changes as relative surface areas and distances change. Not linear.

I do not think that we can look at it as simple as 10X bigger coil = 10X bigger Plasma. Applied field and impact on the plasma with the resulting non-linear changes must be considered as you grow it.

[TallDave]

Well, that was Bussard's approach. I'll welcome any attempt to give us a more precise one.

[D Tibbets]

I read in science fiction books that superconductor always maintains the same temperature across itself. Is it true?
If so then all the cooling problems can be solved by having a superconducting heat pipe.

Nobody has chimed in, so I will. In my ignorance I'll assume that the constant temperature of a superconducter is due to the enviorment it is in, not due to superconducting. A superconductor by definition does not have any resistance, so no heatingdue to current occurs. But, a system cannot be completely isolated. There is heat leackage through the walls from the outside. This will heat the liquid that is used for cooling. As that liquid (helium of nitrogen)warms, it will start to boil/ vaporize rapidly. This consumes alot of heat energy. so the liquid will not rise in temperature till all of it is converted to vapor. Consider boiling water. It maintains it's temperature at 100 degrees C, despite large heating inputs.
It takes ~ 100 times as much heat energy to convert water to steam, as it does to heat liquid water one degree C (while below boiling point). This is called the heat of fission(?). Similar things happens at the freezing point- heat of fusion. I don't know how liquid helium and liquid nitrogen compares to water in this regard. Also, I don't know if this occurs in the phase transition between a gas and a plasma.

Dan Tibbets