Talk:Deuterium fusion

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Deuterium burning prevents this by acting as a thermostat that stops the central temperature rising above about one million degrees, which is not hot enough for hydrogen burning.[3][edit]

This is an interesting article.

The sentence "Deuterium burning prevents this by acting as a thermostat that stops the central temperature rising above about one million degrees, which is not hot enough for hydrogen burning" is obviously vital, but its is not entirely clear to me. Does this mean that excess heat above one million degrees is absorbed by the conversion of deuterium to helium? Is this like a phase change where adding heat energy to ice at the melting point will, for some period, cause melting without causing a further temperature rise? The fact that deuterium "burning" prevents a rise in temperature makes it sound as if the reaction is endothermic, but that doesn't seem right. Can this sentence be clarified?μηδείς (talk) 20:08, 10 July 2010 (UTC)[reply]

  • According to the Palla book the energy generation rate is proportional to (deuterium concentration) x (density) x (temperature)11.8. But the core is also more or less in a steady state ie. the energy generation should be constant. So if for some reason, say, the density increases then the deuterium concentration and/or the temperature need to decrease to keep the energy generation constant. But because of the 11.8 power of temperature you would need very large changes in density and deuterium concentration to cause even a moderate change in the temperature (for instance, doubling the temperature would mean you'd need to decrease D and rho to about 2^(-5.9)=1.7% of their original values to compensate). Thus, the temperature is prevented from rising very much above the deuterium ignition temperature. At least this is my understanding of the situation. I'm still looking for a source that explicitly explains it, because it takes a fair bit of cogitation and interpretation to extract that explanation from the sources I've found. Reyk YO! 22:12, 10 July 2010 (UTC)[reply]
Is it possible that what is meant is that while Deuterium is burning, the ambient temperature prevents the star's density from increasing, and the star from reaching a high enough temperature and density from the pressure of gravity to go on to burning hydrogen?μηδείς (talk) 23:20, 10 July 2010 (UTC)[reply]
This is not the same thing, but related. Hydrogen burning kicks in at about 10^7K and the thermostat effect keeps it at 10^6K. But once all the deuterium is gone the thermostat turns off and the star contracts, heating up as it goes. And once the temperature hits 10^7K, hydrogen burning starts. Reyk YO! 02:50, 11 July 2010 (UTC)[reply]
Yes, that is what I meant, does the Deuterium burning prevent further collapse and contraction and the heating which that would produce that would ignite hydrogen burning. Now, if you could find a way to say that in the article itself. . . Thanks. μηδείς (talk) 04:05, 11 July 2010 (UTC)[reply]
Does this mean that in the far future of the universe no large stars will be produced because deuterium will have become too scarce? XinaNicole (talk) 05:19, 25 April 2011 (UTC)[reply]

This is weird[edit]

The reason why Hydrogen Fusion requires the temperature of 10^7K is because the protons repel. But deuterium has essentially the same charge, so it shouldn't be easier to fuse, at a temperature of 10^6K. On top of that, the neutron in deuterium adds extra mass, which means more inertia, which makes it harder to accelerate. So why does deuterium fuse so easily? And finally, if it is so seemingly easy to accelerate the deuterium(the other requisite is still a proton, so it doesn't matter), why don't stars fuse 2 deuterium atoms and make helium-4? 32ieww (talk) 23:34, 28 January 2017 (UTC)[reply]

Energy produced by fusion of duterium.[edit]

I want to know that how much energy is released from 1 kg of duterium(fusion) as compared to 1 kg of coal . Siwan2002 (talk) 11:27, 24 December 2019 (UTC)[reply]

D+D fusion[edit]

Since D+D is the major reaction used in testing tokamaks including JET and ITER, I think it deserves a mention here... if not its own article.

It's not as easy as D+T of course, but it doesn't produce the high-energy neutron that is one of the main obstacles to use of D+T. Which sounded pretty good at one stage.

The problem with this is that there's a side reaction in which D+D produces Tritium. I'm not sure whether this was even predicted or its consequences appreciated before the D+D tests began at JET, but it was quickly demonstrated there when their supposedly pure Deuterium fuel began producing high-energy neutrons from D+T fusion!

Sad conclusion: No point in hoping that fusing pure Deuterium will avoid irradiating the PFM with these nasty high-energy neutrons, and thus avoid the need of expensive remote handling equipment to deal with the highly radioactive waste. Oh well, back to D+T.

I'm sure this can be sourced but don't have the sources to hand. Andrewa (talk) 17:06, 11 December 2020 (UTC)[reply]

  • I agree that D+D should be covered as a subsection of this article rather than a new one; I have never been a fan of diluting content over more pages than necessary. There's no huge rush though. Let me poke around for a bit and see what sources I can scrape up. Reyk YO! 17:35, 11 December 2020 (UTC)[reply]
    • Sounds good. The reaction currently described is important mainly as a stage of the stellar P+P chain (and that is very important indeed), and could be described as P+D but I'm not sure anybody does. But D+D is also important, and is an obvious meaning of Deuterium fusion. And not just for laypeople, even experts in fusion power are likely to use it in this sense too if they are unaware of its other meaning. I'm not such an expert but I do talk to those who are, and some of them know surprisingly little about how stars work. We live in an age of excessive specialisation IMO! Andrewa (talk) 23:44, 19 December 2020 (UTC)[reply]

Tritium vs deuterium fusion[edit]

Why is deuterium the most easily fused isotope available to protostars? Shouldn't it be tritium since it's got an extra neutron and occours naturally? Tritium's the most easily fused isotope on Earth. 2603:6000:8740:54B1:C941:7ADA:D88:9366 (talk) 16:34, 6 May 2023 (UTC)[reply]