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An anonymous editor has slapped three templates on this article, Expert, expansion, and confusing. It would help to have a little more detail on the problems perceived. I already am an expert on the subject, and I find the presentation clear albeit succinct. I'm not sure where expanding it would make it much clearer. Please help me here. -- Art Carlson 10:05, 30 April 2006 (UTC)
I don't quite understand the high Q value necessary. Wouldn't a fusion reactor be successful if it could deliver more electricity to the net than what it is consuming?
Let's assume that most of the energy that the reactor needs is used for the heating of the plasma. Let's call the required power Pheat. If this power is taken as electricity from the grid, the amount of electricity needed for heating is Pel_heat = Pheat / ɳheat.
The electricity output would be, according to the article, Pelec = ηelec(1-fch)Pfus.
Now, what is wanted is that the output electricity is larger than the input electricity:
Pelec > Pel_heat
insert the definitions of Pelec and Pel_heat:
ηelec(1-fch)Pfus > Pheat / ɳheat
or
(Q = ) Pfus / Pheat > 1 / ( ɳheatηelec(1-fch) )
which is the same expression used in the article save for the recirculation of electricity.
Put in the numbers in the article, and the result is Q > 1 / (0.7*0.4*(1-0.2)) = 4.46 ≈ 5. Won't this mean that a fusion reactor with Q > 5 will deliver more power than it uses? Of course you need to deliver more than that to have some energy to sell, but wouldn't, say, Q=10 be enough? -- Maccer83 12:07, 14 May 2006 (UTC)
Intro says "...the power required to maintain the plasma in steady state." - Does this include any of the power used in toroidal or poloidal magnets ? I guess it only includes the magnet drive power used to induce ohmic heating ? - Rod57 ( talk) 10:27, 9 December 2015 (UTC)
\eta_heat is used in the article but never defined.
@ StevenBKrivit: I disagree with the concern that "breakeven" should not be tied to "scientific breakeven". I understand it, but disagree with it.
Simply put, if you see the term "breakeven" in a work, it is almost always referring to scientific breakeven. And I'm not talking about "layman" sources, this is the way the term is used by the labs themselves in almost every review. For instance, that's what LLNL calls it, its what Los Alamos calls it, it's what Culham calls it, its what Frascati calls it and it's what ITER calls it.
On the contrary, it's pretty difficult to find source that includes the term that isn't talking about scientific breakeven. One can indeed find examples, but without fail they always include a clarifying term. I cannot find a single paper that uses the term "breakeven" to refer to, say, engineering breakeven. If a paper is referring to engineering breakeven, it clearly makes that distinction. And even in those papers, they always use "breakeven" to refer to scientific breakeven even in their description of the terms.
Simply put, if the term "breakeven" appears in an article about fusion, without a clarifying statement to the contrary, it is talking about scientific breakeven. That might not be "correct", and it might not be what you want, but it is what it is. Our job is to report that, not fix it.
Maury Markowitz ( talk) 13:43, 4 October 2018 (UTC)
Our mission is to make a great encyclopedia, not fix the entire planet. The clarity offered in the middle of the article covers this. I'll expand the lede slightly, but I really think we're well outside the lines now. Maury Markowitz ( talk) 11:10, 9 October 2018 (UTC)
@ StevenBKrivit: Actually, on further study the only use of "scientific breakeven" I can find in use is the somewhat dubious definition used by NIF in 2013. Do you have any examples of the term being used outside the context of NIF? Maury Markowitz ( talk) 17:49, 9 October 2018 (UTC)
Wut? I'm half way through the edit, that much should be obvious. Maury Markowitz ( talk) 22:05, 13 October 2018 (UTC)
@ StevenBKrivit: Do you mind a re-read of the article in its current form? Maury Markowitz ( talk) 13:52, 16 September 2019 (UTC)
I get lost: In Q_E = ..., eta_heat is in the denominator. So, a very small eta_heat (say 0,015) would produce a huge Q_E. Already this puzzles me - although I cannot see what would be wrong with the Q_E equation.
But then, in the final section, we are told "ICF devices have extremely low Q. This is because the laser is extremely inefficient; ... eta _heat [for] lasers [is] on the order of 1.5%" ... so a low eta_heat results in a low Q - but this directly contradicts the Q_E equation.
Where do I go astray? -- User:Haraldmmueller 10:27, 22 March 2019 (UTC)
Hi @ Maury Markowitz: Hi I'd like to invite you to discuss this. My understanding of Q value is that is a very broad denominator of energy released in a nuclear reaction. Any kind of nuclear reaction. Are we in agreement on this?
StevenBKrivit ( talk) 18:04, 29 July 2020 (UTC)
Hi,
'Additionally, fusion fuels, especially tritium, are very expensive, so many experiments run on various test gasses like hydrogen or deuterium', says the article. But is it really the cost of tritium? Isn't it the fact that experiments with tritium seriously activate all the equipment, blankets etc., making these experiments so costly? Thanks, ColaBear ( talk) 17:17, 21 May 2021 (UTC)
The description for the image for this page reads.
"The explosion of the Ivy Mike hydrogen bomb. The hydrogen bomb is the only device currently able to achieve fusion energy gain factor significantly larger than 1."
This is not explained at all in the page and none of the references cover it as far as I can see. Should the caption or the image be changed? Xzpx ( talk) 08:28, 3 July 2023 (UTC)
This article says that the NIF claims to have achieve "scientific breakeven" in 2013 but the sources don't provide any mention that they claimed "scientific breakeven." The NIF's publish research shows a breakeven of the "total fuel energy gain" (the ratio of fusion energy to energy absorbed into D-T fuel) but no mention of "scientific breakeven." Guthrette ( talk) 20:59, 24 July 2023 (UTC)
Even though the units given for the NIF's experiments are given in joules, the article still states that the Q is based of P instead of E. The article is even titled "Fusion Energy Gain Factor." I feel like we need to rewrite parts of this article to note the difference between Q in MCF(given in power) and ICF(given in energy).
![]() | This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||||||||||||
|
An anonymous editor has slapped three templates on this article, Expert, expansion, and confusing. It would help to have a little more detail on the problems perceived. I already am an expert on the subject, and I find the presentation clear albeit succinct. I'm not sure where expanding it would make it much clearer. Please help me here. -- Art Carlson 10:05, 30 April 2006 (UTC)
I don't quite understand the high Q value necessary. Wouldn't a fusion reactor be successful if it could deliver more electricity to the net than what it is consuming?
Let's assume that most of the energy that the reactor needs is used for the heating of the plasma. Let's call the required power Pheat. If this power is taken as electricity from the grid, the amount of electricity needed for heating is Pel_heat = Pheat / ɳheat.
The electricity output would be, according to the article, Pelec = ηelec(1-fch)Pfus.
Now, what is wanted is that the output electricity is larger than the input electricity:
Pelec > Pel_heat
insert the definitions of Pelec and Pel_heat:
ηelec(1-fch)Pfus > Pheat / ɳheat
or
(Q = ) Pfus / Pheat > 1 / ( ɳheatηelec(1-fch) )
which is the same expression used in the article save for the recirculation of electricity.
Put in the numbers in the article, and the result is Q > 1 / (0.7*0.4*(1-0.2)) = 4.46 ≈ 5. Won't this mean that a fusion reactor with Q > 5 will deliver more power than it uses? Of course you need to deliver more than that to have some energy to sell, but wouldn't, say, Q=10 be enough? -- Maccer83 12:07, 14 May 2006 (UTC)
Intro says "...the power required to maintain the plasma in steady state." - Does this include any of the power used in toroidal or poloidal magnets ? I guess it only includes the magnet drive power used to induce ohmic heating ? - Rod57 ( talk) 10:27, 9 December 2015 (UTC)
\eta_heat is used in the article but never defined.
@ StevenBKrivit: I disagree with the concern that "breakeven" should not be tied to "scientific breakeven". I understand it, but disagree with it.
Simply put, if you see the term "breakeven" in a work, it is almost always referring to scientific breakeven. And I'm not talking about "layman" sources, this is the way the term is used by the labs themselves in almost every review. For instance, that's what LLNL calls it, its what Los Alamos calls it, it's what Culham calls it, its what Frascati calls it and it's what ITER calls it.
On the contrary, it's pretty difficult to find source that includes the term that isn't talking about scientific breakeven. One can indeed find examples, but without fail they always include a clarifying term. I cannot find a single paper that uses the term "breakeven" to refer to, say, engineering breakeven. If a paper is referring to engineering breakeven, it clearly makes that distinction. And even in those papers, they always use "breakeven" to refer to scientific breakeven even in their description of the terms.
Simply put, if the term "breakeven" appears in an article about fusion, without a clarifying statement to the contrary, it is talking about scientific breakeven. That might not be "correct", and it might not be what you want, but it is what it is. Our job is to report that, not fix it.
Maury Markowitz ( talk) 13:43, 4 October 2018 (UTC)
Our mission is to make a great encyclopedia, not fix the entire planet. The clarity offered in the middle of the article covers this. I'll expand the lede slightly, but I really think we're well outside the lines now. Maury Markowitz ( talk) 11:10, 9 October 2018 (UTC)
@ StevenBKrivit: Actually, on further study the only use of "scientific breakeven" I can find in use is the somewhat dubious definition used by NIF in 2013. Do you have any examples of the term being used outside the context of NIF? Maury Markowitz ( talk) 17:49, 9 October 2018 (UTC)
Wut? I'm half way through the edit, that much should be obvious. Maury Markowitz ( talk) 22:05, 13 October 2018 (UTC)
@ StevenBKrivit: Do you mind a re-read of the article in its current form? Maury Markowitz ( talk) 13:52, 16 September 2019 (UTC)
I get lost: In Q_E = ..., eta_heat is in the denominator. So, a very small eta_heat (say 0,015) would produce a huge Q_E. Already this puzzles me - although I cannot see what would be wrong with the Q_E equation.
But then, in the final section, we are told "ICF devices have extremely low Q. This is because the laser is extremely inefficient; ... eta _heat [for] lasers [is] on the order of 1.5%" ... so a low eta_heat results in a low Q - but this directly contradicts the Q_E equation.
Where do I go astray? -- User:Haraldmmueller 10:27, 22 March 2019 (UTC)
Hi @ Maury Markowitz: Hi I'd like to invite you to discuss this. My understanding of Q value is that is a very broad denominator of energy released in a nuclear reaction. Any kind of nuclear reaction. Are we in agreement on this?
StevenBKrivit ( talk) 18:04, 29 July 2020 (UTC)
Hi,
'Additionally, fusion fuels, especially tritium, are very expensive, so many experiments run on various test gasses like hydrogen or deuterium', says the article. But is it really the cost of tritium? Isn't it the fact that experiments with tritium seriously activate all the equipment, blankets etc., making these experiments so costly? Thanks, ColaBear ( talk) 17:17, 21 May 2021 (UTC)
The description for the image for this page reads.
"The explosion of the Ivy Mike hydrogen bomb. The hydrogen bomb is the only device currently able to achieve fusion energy gain factor significantly larger than 1."
This is not explained at all in the page and none of the references cover it as far as I can see. Should the caption or the image be changed? Xzpx ( talk) 08:28, 3 July 2023 (UTC)
This article says that the NIF claims to have achieve "scientific breakeven" in 2013 but the sources don't provide any mention that they claimed "scientific breakeven." The NIF's publish research shows a breakeven of the "total fuel energy gain" (the ratio of fusion energy to energy absorbed into D-T fuel) but no mention of "scientific breakeven." Guthrette ( talk) 20:59, 24 July 2023 (UTC)
Even though the units given for the NIF's experiments are given in joules, the article still states that the Q is based of P instead of E. The article is even titled "Fusion Energy Gain Factor." I feel like we need to rewrite parts of this article to note the difference between Q in MCF(given in power) and ICF(given in energy).