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> Fulfilling the Lawson criterion for a burning plasma doesn’t yet mean that we can produce useful energy with fusion power plants. The next step toward that go
by im3w1l 4y ago
> Fulfilling the Lawson criterion for a burning plasma doesn’t yet mean that we can produce useful energy with fusion power plants. The next step toward that goal would be to demonstrate a fusion scheme that produces as much energy as that contained in the laser pulses driving the reaction. In other words, the scheme should have a net gain, G, of 1.
This is what I had always kinda sorta thought, but seeing someone put it in words made me realize that it actually doesn't make sense at all.
If 1 unit of laser energy gets you 0.2 units of fusion, that means you end up with 1.2 units of heat energy total. Okay so 1.2 units of heat just sounds like a crappy heat pump. But given that the heat is so damn hot (meaning ~0 entropy) you could in theory convert it all to electricity and come out ahead.
- sacred_numbers 4y agoThe main problem is that -- especially with laser fusion-- you are not inputting 1 unit of electricity and getting out 1.2 units of high grade heat. You are inputting 100 units of electricity, most of which is lost as low grade heat through wires, cooling devices, etc to power a laser that outputs 1 unit of photons. The fusion adds a bit of energy, but not enough to surpass the 100 units you put in. I don't know the exact efficiency of these lasers, so 100x may be wrong, but I think I'm in the right order of magnitude.
- im3w1l 4y agoAh that makes sense. I had no idea the lasers were that inefficient. Still it's not an inherent feature of the problem. But maybe getting the fusion reaction to self-sustain is the easier thing to solve.
- nine_k 4y agoLasers are likely 10% efficient, not 1%, but it's still an order of magnitude less than needed to break even.
- fsh 4y agoThe laser at NIF has a 0.5% efficiency [1]. Some lasers are highly efficient (for example diode lasers). High-energy pulsed lasers are usually not. [1] https://physics.aps.org/articles/v14/168 https://physics.aps.org/articles/v14/168
- zmgsabst 4y agoHow would you convert it into energy without losses? For your example of 0.2 increase, you need 83% efficiency — or 91% efficiency on both input and output. Steam turbines may not hit that level of efficiency [1]. If you assume that input and output are each 80% efficient, you need to generate 0.5 units per unit put into the system to break even; if they’re 70% efficient, you need to generate 1 unit per unit put into the system. [1] — https://www.epa.gov/sites/default/files/2015-07/documents/catalog_of_chp_technologies_section_4._technology_characterization_-_steam_turbines.pdf https://www.epa.gov/sites/default/files/2015-07/documents/ca...