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Question from a layman:

In nuclear fission, you don't do all the work to split each nuclei every time. You set things up in a clever way so there is a chain reaction. Previously split nuclei cause other nuclei to split.

When generating hydroelectric power, you don't do all the work to move the water up high and let it fall every time, you set up the turbine in a place where water will fall on it naturally.

For nuclear fusion, it seems like you are doing all the work to make the atoms fuse each and every time. There is no chain reaction and you aren't letting just allowing nature to take its course.

Why would this process ever be energy net-positive? If you have to do all the work to make each nuclear fusion occur, and there are inefficiencies in converting forms of energy, won't you always get out less energy than you put in?

Is it theoretically possible just because you are converting mass into enormous amounts of energy?

Thanks.

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    $\begingroup$ Well, the Sun manages it.... What are you asking? $\endgroup$ Oct 4, 2022 at 21:15
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    $\begingroup$ Burning a hydrocarbon is not a chain reaction but it manages to produce net energy. $\endgroup$
    – Jon Custer
    Oct 4, 2022 at 21:29
  • $\begingroup$ When I was about 35 years younger we were supposedly 75 years away from a working system. I vaguely hoped to live to see one near the end of my days. Now we're about ... 75 years away from a working system and my life expectancy has not grown in proportion, so I getting less a bad vibe about this. $\endgroup$ Oct 4, 2022 at 22:06
  • $\begingroup$ In the sun, "the force of gravity eventually generates enough energy that atomic nuclei are moving fast enough so that they can fuse together." We don't have that gravitational energy available on earth so we have to do all the work to make the nuclei fuse. Will the process ever be energy net-positive on earth? $\endgroup$ Oct 5, 2022 at 1:30
  • $\begingroup$ Are you aware of the algebra of special relativity, that in the center of mass of a reaction of two particles, protons in the sun for example, the mass is converted to energy? hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html $\endgroup$
    – anna v
    Oct 5, 2022 at 3:52

3 Answers 3

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In nuclear fission, you don't do all the work to split each nuclei every time. You set things up in a clever way so there is a chain reaction. Previously split nuclei cause other nuclei to split.

Yup, very handy.

For nuclear fusion, it seems like you are doing all the work to make the atoms fuse each and every time. There is no chain reaction and you aren't letting just allowing nature to take its course.

No, there is something analogous to a chain reaction in fusion as well.

In the D-T reaction, and most of the others being studied for fusion reactors, an alpha particle is released along with a high-energy neutron. The neutron, being neutral, is basically invisible to the fuel and leaves, taking its energy with it. But the alpha has an electrical charge, which makes it very much visible to the fuel.

There are two main approaches to fusion energy, magnetic and inertial.

In the magnetic approach, the alpha is subject to the same magnetic fields that are confining the fuel. The alphas stay in the mix, bumping into the fuel ions and slowly depositing their energy into them. Thus the alphas heat the fuel up. As you increase the rate of fusion in the fuel, the rate of heating from these alphas also increases. There is a point where the alpha heating is equal to the energy losses from the neutrons and various other issues (radiation losses, leaking fuel, etc.)

This point is known as "ignition", and it is very much like the chain reaction in fission. As long as you can maintain those conditions, the plasma will continue to "burn" on its own without you having to add more energy. The difference is that fusion requires temperatures around 100 to 150 million K to work, whereas fission works fine at room temperature.

In the inertial approach, the mechanism is different but the outcome is the same. In this case, the fuel is compressed to super-high densities, about 100 times that of lead. In these conditions the alpha's charge interacts with the many ions around them, rapidly slowing them down and depositing their energy into the fuel. There is also a second mechanism working against you; the heat of these reactions causes the fuel to heat up and expand, which it does at very high speeds, so the fuel tends to blow itself apart.

NIF recently announced that it had achieved ignition. However, the total amount of energy released was small. This is due to that second effect, the fuel mass blew apart before the reaction got very far. But, for a brief instant, there was a fusion chain reaction going on. The goal is to make this time last longer, using more massive fuel capsules for instance, or using additional laser energy to keep it together longer. Time will tell.

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    $\begingroup$ "Time will tell." - time DID tell, they have now managed to increase the ignition to the point of Q > 1 $\endgroup$ Dec 14, 2022 at 15:07
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The difficulty is not in trying to discover a self-sustaining reaction. Nuclear fusion happens countless times every moment throughout the Universe, in the core of every hydrogen burning star. It is not only a theoretical possibility, it is a certainty all life on Earth relies on. Indeed, more energy is released by converting hydrogen to helium than the energy necessary to bring the positive charges together.

The primary difficulty of man-controlled nuclear fusion is attempting to replicate the conditions necessary for sustaining the reaction: temperatures many times hotter than the cores of stars in order to speed up the reaction so that we can benefit from it.

In other words, the problem is containment or "confinement": how do you hold a very high temperature high density plasma to reach conditions where we know a sustained reaction is possible, without the plasma cooling down (lower temperature) or spreading out (lower density).

Inertial confinement and magnetic confinement being two of the most promising avenues.

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  • $\begingroup$ agreed and edited $\endgroup$
    – Alwin
    Oct 5, 2022 at 0:30
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Imagine a boulder near the top of a large cliff. You have to exert (use up) some energy in order to push that boulder to the edge of the cliff, but once you've pushed it off the edge, the potential energy stored "in" the boulder is released as kinetic energy as the boulder falls to the ground. It can certainly be the case that the amount of energy that you had to put into the system in order to liberate that kinetic energy is smaller than the amount of kinetic energy that ends being liberated.

The point is that, sure, sometimes you have to put energy into a system in order to get the energy out that's already in there, but once you've done that, you've "uncorked the bottle" so to speak, and now the amount of energy that you get out is much larger than what you had to put in to get it started.

Note that it's not necessary to have a chain reaction here. Aside from the practicalities and the engineering and such (a very, very hard problem, as you know!), we just need to put enough energy into each pair of hydrogen atoms to get them to fuse and release more energy than we put in.

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  • $\begingroup$ Thanks for the answer. Without the gravitational energy of the sun available for us to use here on earth, I'm wondering if there even exists an analgous scenario to your anwser for fusion on earth. Are there any "rocks close enough to the cliff edge" for us at all? I guess the whole challenge is to come up with a clever and efficient way of putting enough energy in so that the nuclei fuse but we can't put in more than what will be released, if possible. $\endgroup$ Oct 5, 2022 at 1:56
  • $\begingroup$ @march It seems to me that the OP is thinking classically, has no concept of special relativity, i.e. that mass can be converted to energy, $\endgroup$
    – anna v
    Oct 5, 2022 at 3:48
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    $\begingroup$ @user2577361 Note that we aren't trying to replicate the stellar fusion reactions of plain hydrogen. That wouldn't be practical, as I explain here. Instead, we use deuterium & tritium. $\endgroup$
    – PM 2Ring
    Oct 6, 2022 at 14:43

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