Has anyone measured the light intensity given off in a fusion reactor?

[68658]

I was just wondering if, through all the fusion test done over the years, if anybody has ever mesaured the lux given off dureing a reaction?

 

[Drakkith]

Not sure. Of course, much of this is going to be given off by hot plasma that isn't fusing.

 

[68658]

well that's the thing, if you can more easly create a very bright low fusion per second reaction, than a high fusion per second rate, then why not go with the lesser and develope more sophiticated means of sapping energy off the reaction

i plan on trying something near this in an experiment but i don't want to do anything without knowing much about about the visible electromagnetic radiation given off

 

[Drakkith]

It's not a question of extracting the energy, it's being able to generate the energy in the first place. You cannot create a "very bright low fusion per second" that would generate enough power to sustain the reactions themselves. The more reactions per second you generate, the more energy you get, whether it is in photons or neutrons or whatever. IE you couldn't get a very bright low energy reaction, you would get a dim reaction. More reactions would be "brighter" to use your terms.

 

[68658]

Yes you are corect what i ment to say or should say is: the Lux given off will reach a plateau (if one where to graph the fusion rate to Lux) so the low reaction rate would be the fusions per second needed to sustain the Lux output needed, per the device used.

 

[Drakkith]

I don't know what you mean. The more reactions the more photons would be generated. Also, I don't understand your fascination with the light generated from the fusion. The energy from the fusion is almost entirely released in the reaction products, which you don't want to stay in the plasma. You want to capture them and generate heat or directly generate electricity from them. If they stayed and transferred all of their energy to the plasma, they would stay in the plasma (The charged particle would, the neutrons wouldn't) and poison the reaction.

 

[68658]

Essentialy my fassination is in shrinking a Dyson sphere (as an easy way to put it), the solar panels are an intermediary step in design. Also its not that i want all the particles to stay in the reaction they should escape naturaly because there's nothing strong holding them in, so most of them will get released. The idea is, is to lower the required reavtion rate of breakeven. Any and all ways to do this would idealy be built into a net gain reactor.

 

[Drakkith]

Ok. Am I correct in thinking that you think that gathering the light generated from the fursion lowers the breakeven point? If so, that only works in the sense that the reaction is already at point A in my other post.

For example, if you put 500 watts into the reaction AFTER all inefficiencies, and no matter the source of the power, and you only get 300 watts of power from the reaction, that isn't breakeven in any way. NOTE: It is important to understand what I mean by that. The 300 watts of power generated from the reaction is BEFORE you gather any of it. If you had 100% efficiencies in gathering the power from the reaction you would only get 300 watts of power. No matter what. That is all that is being generated.

In case B in my other post, if you "gather" the lost energy from the reactor from lost non fused particles, light, or whatever, you increase the Breakeven according to point B, but not point A. You need both.

 

[68658]

Ok so a build a highly effeicent fusion reactor, and i surround the reaction mass with solar cells that recharge an array of batteries. From the batteries i use an inverter from 12v to 120v (leaving out current type for simplicity) from 120v to a tesla coil (the 15kw transformer being apart of the 1.5Mv tesla coil). Now with this 1.5 Mv energy cappabilty i power the fuion reaction. Yet all i need from the fusion is the cappability to sustain the inverter, so directly the battery array.

 

[Drakkith]

You are skipping half the process lol.

Ok so a build a highly effeicent fusion reactor

This requires two things. Finding out how to sustain the reactions in a way that uses less inpute energy than it outputs. Second, generating enough power to overcome inefficiencies in the system. Again, note that the 1st step ignores everything related to the reactor itself, and only involves the amount of power finally input into the plasma, and then generated by the reactions. The breakeven for number 1 (or A) doesn't care if your reactor is 10% efficient at applying power to the reactions and gathering the produced power, of if it is 90% efficient.

 

[68658]

I agree that the reactor needs to be designed to be as eefient as possible at sustaining fusion at the lowest amount of input power. Yet what about the other end, the out put in over simplicated terms is it not possible through the use of electrical deffices to jump a set of batteries voltages up to megavolts of energy, hence lessing the input power before jumping it up to the megawatts of power I am discussing useing.

 

[russ_watters]

I agree that the reactor needs to be designed to be as eefient as possible at sustaining fusion at the lowest amount of input power. Yet what about the other end, the out put in over simplicated terms

"The other end" is unimportant. We can use 19th century technology to extract energy from the reactor if we ever get one that breaks even.

You're trying to solve a problem that has already been solved while ignoring the problem that hasn't!

is it not possible through the use of electrical deffices to jump a set of batteries voltages up to megavolts of energy, hence lessing the input power before jumping it up to the megawatts of power I am discussing useing.

Volts is not energy and you can't multiply energy with a transformer: conservation of energy applies. You are misunderstanding how basic electronic components work.

 

[Drakkith]

Lets say you input 500 watts from batteries. You get 300 watts out. (Making this number up, but the output is ALWAYS less than input currently) What now? Jumping the power up to 500 MW might give you 300 MW out. Still no gain and just a waste of power.

So you haven't done anything but waste power to burn a small amount of fusion fuel.

Edit: I don't think your knowledge on the basics of fusion power are sufficient. I would suggest reading more on breakeven and other basics on fusion to get an understanding on this.

 

[68658]

Ok so what exatly is the sigma value for breakeven again, its not something I've memorized. But anyway the fusion reactor keep only one spherical grid in the chamber positavaly charge the pressure vessel. Negativaly charge the inner grid place small torridial electromagnets on the grid through the center axis. Electrialy sheild the electromagnets from an outer cassing, neggativaly charge this casseing. On the outside of the vacuum chamber are a set of conainment electromagnets, placed in the shape of a truncated doedcahedron. Cover this with a faraday cage place a spherical outer vessel around all of this. When in operation deuterium gas will travel through the outer vessel, captureing any neutron that escape, forming small amounts of tritium, then from here into the main chamber where the gas becomes ionized. Fusion occureing will rapidaly expand hitting the cointment magnetic field, self "pressureising" itself, if the materials used are strong enough.

 

[russ_watters]

You're just fantasizing. There is no content in that post. It's just technobabble. Thread locked.