Talk:Decay heat: Difference between revisions

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Performed a partial rewrite. Since decay heat is a very significant issue for nuclear reactor safety, I will probably be revisiting this article, as well as others that should know about decay heat.--Burzum 07:09, 31 December 2005 (UTC)[reply]

Good! I shall read with interest (though I'm no great expert in this area). Just to warn you that the last 'Nuclear Technology' topic I contributed to, got a 'too technical' tag slapped on it Bob aka Linuxlad 16:44, 31 December 2005 (UTC)[reply]

Edit request: "This results in 13 MeV" --> "This results in 23 MeV"

I disagree. 10 MeV of neutrinos (of the 23 MeV total for beta decay) are not deposited in the core. This results in only 13 MeV being deposited in the core.--Burzum 00:08, 16 June 2006 (UTC)[reply]

Even the 13 MeV may be high - does this include decays with long halflife that are unlikely to happen while the fuel is still in the reactor? --JWB (talk) 18:00, 20 May 2008 (UTC)[reply]

Yes. But this isn't an issue. Decays of nuclides with exceptionally long half-lives do not deposit a significant amount of energy. And the long and steady power history ensures that the 13 MeV number is reasonable since these nuclides will build up to a steady state inside the reactor. Before the reactor shuts down, it is still producing 6.5% of its power from decay heat. The rest comes from induced fission. After it shuts down and the induced fission rate is made negligible, the only heat production will be from the same decay heat. This previously steady state value will then decrease because no more radioactive nuclides are being produced as fission products. Cheers.--Burzum (talk) 03:53, 9 September 2008 (UTC)[reply]

I changed the definition of T_subscript_S to the correct definition as stated in reference [3]. The equation is number (3) of reference [3]. 68.35.150.245 (talk) 02:13, 9 January 2011 (UTC) Mike McNaughton, mcnaught@unm.edu[reply]

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 :::Yes.  But this isn't an issue.  Decays of nuclides with exceptionally long half-lives do not deposit a significant amount of energy.  And the long and steady power history ensures that the 13 MeV number is reasonable since these nuclides will build up to a steady state inside the reactor.  Before the reactor shuts down, it is still producing 6.5% of its power from decay heat.  The rest comes from induced fission.  After it shuts down and the induced fission rate is made negligible, the only heat production will be from the same decay heat.  This previously steady state value will then decrease because no more radioactive nuclides are being produced as fission products.  Cheers.--[[User:Burzum|Burzum]] ([[User talk:Burzum|talk]]) 03:53, 9 September 2008 (UTC)
+I changed the definition of T_subscript_S to the correct definition as stated in reference [3]. The equation is number (3) of reference [3].
+[[Special:Contributions/68.35.150.245|68.35.150.245]] ([[User talk:68.35.150.245|talk]]) 02:13, 9 January 2011 (UTC) Mike McNaughton, mcnaught@unm.edu