Talk:Fusion energy gain factor

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untitled

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)[reply]

Why isn't Q > 5 enough?

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 P_heat. If this power is taken as electricity from the grid, the amount of electricity needed for heating is P_{el_heat} = P_heat / ɳ_heat.

The electricity output would be, according to the article, P_elec = η_elec(1-f_ch)P_fus.

Now, what is wanted is that the output electricity is larger than the input electricity:

P_elec > P_{el_heat}

insert the definitions of P_elec and P_{el_heat}:

η_elec(1-f_ch)P_fus > P_heat / ɳ_heat

or

(Q = ) P_fus / P_heat > 1 / ( ɳ_heatη_elec(1-f_ch) )

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)[reply]

A fusion reactor will be successful if it produces electricity at a cost that is competitive with the alternatives. The Q needed to be successful in this sense will depend on the cost of the fusion plant and the cost of the alternatives. With the assumptions of the article and Q=22, f_recirc is 0.20 and you need to produce 25% more electricity than you plan to sell. With Q=10, f_recirc is 0.44 and you need to produce 78% more electricity than you sell. You're right that this could, in priciple, be OK. If a source of energy is cheap enough, it could tolerate a recirculating power fraction of even, say, 99%. The 20% figure is a soft limit that gives you an idea of when it starts to hurt. It is commonly used in studies, but other assumptions can be made. --Art Carlson 19:24, 14 May 2006 (UTC)[reply]

Whats included in input power

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)[reply]

It depends on the definition, but generally no, only heating sources are considered. However, it should be noted that on modern designs using superconducting magnets, the confinement system uses little energy anyway (after startup) and even the cooling load on the magnets would be low in relative terms. Maury Markowitz (talk) 11:11, 27 July 2017 (UTC)[reply]

\eta_heat

\eta_heat is used in the article but never defined.

Fixed. Maury Markowitz (talk) 13:47, 4 October 2018 (UTC)[reply]

Breakeven terminology

@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)[reply]

Maury, I appreciate your thoughts on this. Why do you say their (the institutions you listed) use of it might not be correct?

StevenBKrivit (talk) 00:40, 5 October 2018 (UTC)[reply]

I'm simply riffing on your comment in the checkin note; yes, there are other types of breakeven, and in theory, one could refer to any one of them. So in that respect simply saying "breakeven" is not technically perfect (the best kind of perfect!). But perfect is the enemy of good, and that applies here IMHO. Maury Markowitz (talk) 20:21, 5 October 2018 (UTC)[reply]

Thanks Maury. The cites you provided are absolutely rock-solid. The reasoning you provided - that of following the usage of those institutions - is sound. And if you ask any fusion expert, they will tell you, of course "breakeven" in the context of today's experimental reactors always refers to scientific breakeven. But that's not why I object to your addition. I object because Wikipedia's primary audience is the lay public. I can show you examples of journalists who a) lack your training in nuclear science, b) have gone to those Web sites, c) had no awareness of the different kinds of breakevens and d) written their fusion news stories using just the word "breakeven" and writing it as if it were engineering breakeven, and e) thereby misled the public. That's why I have a problem with your proposed addition. I think it's fair to say that the Wikipedia audience should be assumed to be the widest, most general audience in the world. We should make the article as clear as possible for this audience rather than simply repeating a bad practice by the fusion community. Your thoughts please?

StevenBKrivit (talk) 04:20, 9 October 2018 (UTC)[reply]

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)[reply]

@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)[reply]

Hi Maury. You are the one who has proposed to insert this new phrase. You jumped the gun by adding the phrase back in before our discussion was finished. Would you kindly remove it until we are done with this discussion? Second, would you please explain how your proposed addition improves the article? Thank you.

StevenBKrivit (talk) 06:51, 11 October 2018 (UTC)[reply]

Ha! I didn't even recall adding it in the first place, and it was only a year ago. I will clean all of this up. Maury Markowitz (talk) 20:30, 12 October 2018 (UTC)[reply]

@Maury Markowitz: You didn't clean it up. You didn't explain how using the word "breakeven" -- when it really means "scientific breakeven" -- improves the article. Then you went further, removed "scientific breakeven" from the lead entirely, removed it from the sub-heading, and moved the subject of "scientific breakeven" to the bottom of the article, when in fact, the top section is really all about "scientific breakeven." But you apparently don't want to make that clear to readers. You seem to be quite content with introducing and maintaining the ambiguous wording. Additionally, you assigned "scientific breakeven" to ICF when it applies to MCF as well as ICF. You've put in a lot of work to your latest round of edits which hide the term "scientific breakeven." You have shown several times that you are not amenable to discussing significant changes before making them. I do not have the time or energy to fight with you. I'm done. If you want another editor with whom to discuss additional edits on this page, you'll have to find someone else.

StevenBKrivit (talk) 21:58, 13 October 2018 (UTC)[reply]

Wut? I'm half way through the edit, that much should be obvious. Maury Markowitz (talk) 22:05, 13 October 2018 (UTC)[reply]

Edits are complete, now what is it you are concerned about? Maury Markowitz (talk) 23:37, 13 October 2018 (UTC)[reply]

@StevenBKrivit: Do you mind a re-read of the article in its current form? Maury Markowitz (talk) 13:52, 16 September 2019 (UTC)[reply]

eta_heat in denominator

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)[reply]

@Haraldmmueller: You only went astray on the meaning of Qe. That is the Q value you would need to break even after generation. Since there are losses along the way, Qe must be larger than Q. As you note, as the efficiency of that conversion drops, Qe grows. This is expected - the reactor needs to have a higher value of Q to offset greater losses in that case. I'll bet somewhere while reading this you said "ohhhhhhhh...."

To put that in perspective, Entler quotes a couple of studies that put the required value of Q between 5 and 8 depending on your assumptions. In contrast, LLNL reports put the required Q around 95! So, good luck with that.

Now all of that said, the section absolutely requires additional clarity. I have written to one author to understand the various terms, because different papers refer to all of these things using different subscripts and such. Entler develops a new Qe which is basically electrical power on the top and bottom, which reduces to a much simpler equation of (1-n)/n, but I can't replicate that and I think it's valuable. Maury Markowitz (talk) 16:12, 16 September 2019 (UTC)[reply]

Q

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)[reply]

"The fusion energy gain factor, usually expressed with the symbol Q, is the ratio of fusion power produced in a nuclear fusion reactor to the power required to maintain the plasma in steady state. The condition of Q = 1, when the power being released by the fusion reactions is equal to the required heating power, is referred to as breakeven, or in some sources, scientific breakeven."

The lead suffers from a bit of myopia. In physics, Q does not mean fusion energy gain factor. In physics, capital Q means energy release of a certain reaction. Ref: https://www.nndc.bnl.gov/qcalc/ In nuclear fusion research, Q *MAY* mean fusion energy gain factor. It also MAY mean other reaction calculations. One meaning, Q_fus, may account for injected thermal power into the plasma (used to calculate proximity to scientific breakeven). Another meaning, Q_eng, may account for full input power required to operate a fusion reactor (used to calculate proximity to engineering breakeven). Because Wikipedia should express science clearly for a layperson and not rely on ambiguities and assumptions that would be understood by those proficient in the science, the overly broad implication here of what "Q" means should be fixed.

Secondarily, the phrase "required to maintain the plasma in steady state" is simply wrong. In nuclear fusion science, Q, when it implicitly means fusion energy gain factor, or Q_fus when it explicitly denotes fusion energy gain factor, is the ratio of power produced by the fusion reactions to the injected thermal power required to produce those reactions. The Q_fus value required to maintain plasma in steady-state is predicted to be somewhere above Q_fus=10.

Does this help?

StevenBKrivit (talk) 21:18, 28 August 2020 (UTC)[reply]

Is the cost of tritium hindering experiments from using it?

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)[reply]

Hey ColaBear, I don't know the current price of tritium, but I'm sure its pretty damn expensive, so yeah, this is an impediment. But the greatest impediment to the use of tritium is that it just frikkin doesn't exist naturally on Earth. It's a man-made byproduct from, primarily, one type of nuclear fission reactor, the CANDU heavy-water reactors in Canada. And there isn't a whole lot of surplus every year, and the primary application for fresh tritium is to refresh the warheads in nuclear bombs because the half-life of tritium is so short. And yes, it will seriously activate the equipment, and it requires extreme caution in transportation and application.

StevenBKrivit (talk) 18:04, 21 May 2021 (UTC)[reply]

...fission reactor... :-) would be too nice, otherwise. --User:Haraldmmueller 18:40, 22 May 2021 (UTC)[reply]

thanks for pointing out the typo StevenBKrivit (talk) 14:33, 23 May 2021 (UTC)[reply]

Ivy Mike reference has no source

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)[reply]

Did the NIF actually claim "scientific breakeven" in 2013?

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)[reply]

Q for ICF is based on energy not power

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).

Guthrette (talk) 18:39, 25 July 2023 (UTC)[reply]

Wiki Education assignment: Engineering in the 21st Century_Section 1

This article was the subject of a Wiki Education Foundation-supported course assignment, between 20 August 2024 and 3 December 2024. Further details are available on the course page. Student editor(s): Treejump, DaFusionaters (article contribs). Peer reviewers: E102Group11, Team12E102F, 2024E102Group19.

— Assignment last updated by Treejump (talk) 20:52, 25 October 2024 (UTC)[reply]