Fusion, Politics and Thermodedgummits!

[davidtrimmell]

dabbler, I am not sure about that patent.

But even if it works I don't know where it gets us. Here is an example:
Ok 1 RAD = 100 ergs/gram. So let us assume we have a source of 10,000 RADs/hour (pretty hot source). In that field of X or gamma (>50kev) you will be depositing 3.6 watts into one gram of matter every hour. Now lets go further and say we are using 1000 grams of our miracle semiconductor to extract electricity from our x-ray field, that comes out to 3.6KW/Hr. Now we need to determine the efficiency, from memory (hey, this is all from memory so hopefully I am not making a total fool of myself!) the HPGe (high purity Germanium) detectors are usually less than 10% (now this is *detection* efficiency, not efficiency in producing electrons), but we will be generous here and assume 10%. So we now have 360 watts/hour. I don’t know about you but that isn’t going to get us vary far. Really, the only way that has any hope to me is the possibility of extracting the energy from the *charged* particles…

David Trimmell

PS. I may have goofed my math, but I know I am with a order of magnitude or so. This is also the reason why the electric generators in spy satellites, etc. using Pu-238 are only utilizing the thermal heat, converting that to electricity using thermo-electric generators.

[dabbler]

Just for the sake of exercise, what's the yield/gram on silicon PV's at
the expected fifteen percent efficiency using visible light? Drat! I gotta
go find the wattage/m^2 for sunlight. I'll be back.

dabbler.

[Adam Szendrey]

1,4 kW peak, must be around 1 kW average.

Adam

[dabbler]

The figures I found were all over the map, but the site that seemed to
actually understand math says that it runs 930 watts / m^2. I make
that out to be about 140 watts/m^2 converted power in ordinary PV
cells, assuming that they are very good ones.

A running fusion plant would operate 7 days/wk, 24 hrs/day. You don't
get that from sunlight. I also think that PV tuned for x-rays will turn out
to be far more efficient than 10%. The reason being is that you get
more than one chance with x-rays. In other words, the geometry is as
important as the material. Check out how Chandra works.

dabbler.

[davidtrimmell]

Dabbler, I am not sure you get my point. Just try and back calculate how much energy is being produced by *FUSION* to get the x-ray flux I imagine?! I have been quite the Liberal here, but you will be spending dollars to save a couple cents of wasted energy. Worry about getting a real net even energy consumption before trying to salvage the wasted x-ray photons!

Personally I do not think Fusion will ever break even in a homocentric sense. We cannot emulate the stars, but we can be more creative. Perhaps we will find a "sustainable" way to push the cart up the hill, but we will still have to push it up there...

Regards,

David Trimmell

[dabbler]

How much energy would that be, David? You see, I don't know. All I
know is what I've read and what I've read says that nearly fifteen
percent of the energy shows up as x-ray or gamma radiation. That's
not trivial.

Perhaps what I read was in error. I don't disagree with you that the
real gain comes from the charged particles, but fifteen percent is
fifteen percent and the more energy we are talking about, the larger
that fifteen percent is. Is the fifteen percent figure in error?

I am, of course, talking about ground applications here, not aircraft or
spacecraft. In a spacecraft, you'd accept that fifteen percent loss
without blinking provided you were well over break-even on other
outputs. You would do that to save on reaction mass.

[Richard Hull]

What form of fusion are you discussing? I assume D-D?

If done well, there are no gamma rays or x-rays in D-D fusion. The fusor has a high waste of X-rays in D-D, but that is not due to fusion itself, just lost electrons slamming into the outer shell. The x-rays are isotropic and as such, 100% are automatically spent as heat in the outer shell up to about 50kv. Above that, they radiate outward. To capture those over 50kev would be a job in the sense of converting them back to electricity directly.

Compton scattering and the photoelectric effect would get electrons back in the mix which could be scooped up but conversion efficiency is below 10% as power efficiency goes.

Dave was right, Nothing beats thermal to electric conversion even to this date for compact, inexpensive and simple systems that are actually installable and in use. Most radiant energy to electricity systems are not happy with the nasty high energy photons. The devices that convert such energy are usually very expense or exotic and are made only in the small to keep costs down.

Thermal conversions work so well due to the thick absorber which can be dirt if need be. This absorber warrants that 100% of the photon's energy will be given up with zero losses as heat in the medium. Converting that heat to electricity, unfortunately is not so sweet, as thermoelectric conversion is crappy.

Richard Hull

[dabbler]

Well, like you said, Richard, "Reseach is what I do when I don't know
what I'm doing."

I'm learning a lot around here, thanks.

dabbler.

[JoelMoses]

Question woud a high pressure, high temperature plasma stop X or Gamma Photons.

[3l]

Hi Dabbler:

It only hurts for a minute....just scream and go on.

Happy Fusoring!
Larry Leins
Fusor Tech