Talk:Lawson criterion
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I noticed an error in this article (as of March 12, 2007).
Lawson defined a criterion for a reactor that could either ignite or otherwise provide enough electrical energy to sustain itself. For a reaction with charged byproducts (the holy grail aneutronic reaction), this means that the Lawson Criterion is an ignition criterion. However, for the D-D reaction or the much easier D-T reaction, neutron thermal energy is captured, converted to electrical energy, and fed back into the plasma. When neutron energy is recaptured, the Lawson criterion is softened.
The current Wikipedia article says the Lawson Criterion is 1.5x10^20 s-m^-3. This applies to ignition in a D-T machine. The engineering break-even criterion that Lawson describes implies 6x10^19 s-m^-3.
Eric Meier U. Washington, Dept. Aeronautics and Astronautics 128.95.35.191 19:58, 13 March 2007 (UTC)
- As I read Lawson, he considers two cases, ignition with MCF and burn-up-fraction with ICF, which is just what this article does. The "See also" link to Fusion energy gain factor goes into more detail about the Q required for a practical reactor. Your number seems to be what is required to sustain a plasma with a conversion efficiency of 30%. If you were a real engineer, you would also want a bit left over to sell. So I disagree that the article is in error, but I encourage you to make some of these connections clearer. --Art Carlson 13:03, 14 March 2007 (UTC)
First, let me say that this is a wonderful and useful article (thanks to all of you who have put it together).
You said "he considers two cases, ignition with MCF and burn-up-fraction with ICF, which is just what this article does". I don't see any direct consideration of ignition. Lawson talks about engineering breakeven. Figure 2 of the original paper shows a dashed line for engineering breakeven requiring 6x10^19 s-m^-3. However, in the text, he does state:
"Conditions for a T-D-Li6 reactor (...) are easier though still severe. The corresponding values of temperature and nt are T=3x10^7 degrees, nt=10^14. To conclude we emphasise that these conditions, though necessary are far from sufficient."
So, formally, that's why the figure 10^20 is usually used as the Lawson Criterion. However, as you point out, a fusion reactor would not be useful without exceeding breakeven! (Except as a neutron source.)
Most references describe the Lawson Criterion as a condition for net energy from a fusion reaction, and not for a practical engineering reactor.
Cheers, Eric Meier 07:52, 18 March 2007 (UTC)
- Looking more closely at the article, I think you're right. He does consider "systems in which the [charged] reaction products are retained", but not in conjuction with general (conductive) heat losses. It is also true that the "softest" condition is when the charged products heat the plasma directly and the neutrons are converted to electricity for auxiliary heating. --Art Carlson 11:45, 18 March 2007 (UTC)
