# Why do fusion reactors only work for a few seconds?

1. Aug 20, 2013

### rogerk8

Hi!

I wonder why it is so hard to make a fusion reaction last for more than a few seconds.

I have read a few elementary courses in Plasma Physics but do not remember so much.

What I do rememeber is that they said that whenever a fusion reaction starts within a Tokamak the plasma tends to approach the wall and when it does it of course cools off and that is when the fusion stops.

But do I remember correctly?

I mean, there are other ways to start a fusion reaction.

I am thinking of laser-beams. But the same time-problem seem to excist there.

I am very excited about the ITER-project which I think is about to start.

Reading some information about ITER says that they actually think the output energy will exceed the input energy to such an extent that they, if it wasn't a pure experimental site, could hook it up onto the elctrical grid.

Sounds amazing!

Hope they are right because there is quite much hydrogen here on the planet. Even if the type of hydrogen (Deuterium?) is very rare but yet available in our seas.

Further more, the only "exhaust" is harmless Helium.

Best regards, Roger

2. Aug 20, 2013

### Bill_K

It will be more impressive when the dollar value of the output energy exceeds what it costs to produce it.

Everything in the reactor's surroundings gets struck by the high-energy particles produced, becoming radioactive.

3. Aug 20, 2013

### rogerk8

The radioactivity is at least kept within the reactor.

4. Aug 20, 2013

### Astronuc

Staff Emeritus
Until the reactor is repaired and the activated part removed. Then it must be stored onsite, or shipped to a radioactive waste repository.

5. Aug 21, 2013

### Drakkith

Staff Emeritus
If I remember correctly it's something similar to the following:

The issue has to do with instabilities in the plasma. These instabilities make the reactor unable to contain part of the plasma, which is then able to touch the reactor walls. When the hot, ionized plasma touches the walls, it "scorches" the wall and heavy particles from the wall itself are thrown into the plasma. This introduction of heavy elements that cannot fuse into the plasma wrecks the reaction as a whole.

This is by design. The laser pulse needs to deliver its energy in an extremely short time or there simply wont be enough pressure on the fuel pellet to cause fusion. After fusion another pellet is shot into the chamber and ignited. Well, it's supposed to be that way at least. They weren't able to get the fusion to occur correctly, so it never really went anywhere.

Too bad we need tritium for it to work, which only occurs in very very very small amounts naturally, decaying with a half life of about 15 years I believe.

6. Aug 21, 2013

### the_wolfman

Lets break things up into magnetic (MFE) and inertial confinement fusion (ICF).

In ICF you use laser or particle beams to compress a fuel pellet to extremely large density. Once its compressed you quickly heat the pellet. Hopefully igniting fusion before the pellet expands and blows itself apart. The duration is the time between initial compression on the the deconstruction of the pellet. By their nature ICF experiments last only fractions of a second. An ICF power-plant will have to repeat this process at a rate of ~10 pellets a second.

MFE experiments are a different story. If fact there are a ton of different types MFE experiments Each one is a different story. In fact the Large Helical Devices, a stellarator in Japan) has run for hours. My understanding is that they could run even longer. But honestly there wasn't much a reason to do so.

Another type of MFE device is the tokamak. They're operation is usually limited to a few seconds or less. This because use induction to drive large (MA) toroidal currents. As such there operational life time is limited by the number of V-S stored in the capacitor banks to drive these currents. In fact, non-solenoidal (non-induction) start up and current drive is a hot area of active tokamak research.

Another thing that limits the operation of some experiments is heat generation. Even a modest experiment routinely creates plasmas with temperatures exceeding 1 million degrees. If you maintain a plasma for more than a few seconds, then you have to worry about the interior wall of you experiment heating up and melting. To avoid this you can actively cool your experiment. To save money most smaller experiments aren't not designed with cooling systems. Keep in mind that the characteristic time scale of a the plasma in most of these experiments is a few microseconds. So you can do a lot of meaningful experiments in a second.

This really isn't true. Impurity control is an issue and so is plasma stability. An instability can cause the plasma to disrupt ending the discharge. But over the years we've gotten pretty good at avoiding these (they still happen but less frequently). And then they happen most frequently when we are pushing the limits of our experiments (and understanding). Impurities radiate away energy much more efficiently than hydrogen, cooling the plasma. Actually fusion of DT isn't part of most modern experiments. If fact no operating tokamak currently uses DT fuel. Again this saves a ton of money. A ton. As discussed above, DT fusion releases a ton of neutrons. Before you can put DT in an experiment it has to be designed and build with adequate neutron shielding. And then once you put DT in the experiment, you have to perform most maintenance remotely. Both of these come at an enormous cost. Not to mention that tritium isn't cheep.

The D-T reaction is the easiest reaction to achieve, and our first generation fusion power plants will likely use D-T for this reason. But there are numerous other reactions/fuel cycles including D+He3, D+D catalysed, p+B, etc. Once we get fusion working its a small step from D-T to one of these other reactions. All of these reactions produce fewer neutrons than D+T. In fact p+B produces none. (Technically D-He3 doesn't either, but a burning D-He3 reactor will also fuse some D+D which produces neutrons). The neutrons are what activate the reactor walls. Using a D+D catalysed reaction, there is enough deuterium in the worlds oceans to meet our current energy demands for ~26 billion years. For comparison the universe is only ~14 billion years old.

7. Aug 21, 2013

### Drakkith

Staff Emeritus
Ah, so that's how it works.

Sure, but we have to get to the point where we can use something other than D-T first. Until then we're stuck with having to produce tritium and dealing with the activation of the reactor structure.

8. Aug 21, 2013

### rogerk8

Hi Wolfman!

This was extremely interesting to read!

Thank you very much for putting down so much time and energy into explaining it for me!

I'm especially interested in tokamaks (MFE).

Mega Amps, incredible!

I'm just playing around now...

Using

$$m\frac{dv}{dt}=qvXB$$

Solving the differential equation yields

$$w_c=\frac{|q|B}{m}$$

which is the cyclotron frequency i.e the frequency the charged (observe) particles spin around a perpendicular magnetic field B and with the Langmor radius of

$$r_L=\frac{mv_p}{|q|B}$$

I had to look this one up but I have "always" known that this type of equation is the key to understanding why tokamaks may work at all.

I do however not fully understand what the velocity constant vp (as in perpendicular to B) really mean.

Current drive research sounds very interesting. Where can I apply :)

Very interesting indeed.

I seem to remember that there was much talk about instabilities when I took the course (96). But impurity considerations is new to me.

You may gladly tell me more about how you have solved those problems today.

Would you mind explaining what a D+D catalysed reaction is?

Sounds very promising in the long run!

If you are up to it you may explain them all :)

In basic terms, of course.

It's a pity that D-T produces so many neutrons (which cannot be confined within the plasma due to no charge).

Take care!

Best regards, Roger
PS
I am a bit confused though. At one moment you say that DT isn't really a part of modern research but later on you say that D-T is the easiest to achieve. Is there a difference?

9. Aug 21, 2013

### rogerk8

10. Aug 22, 2013

### the_wolfman

When you fuse D+D there are two possible outcomes with equal probability.
D+D = T+ p
D+D = He-3 + n

Next you can fuse the T and He-3 each with another D. Using the reactions:
D + T = He-4 + n
D + He-3= He-4 + p

When you sum up these for reactions you get the simplified reaction
6D = 2 He-4 + 2n + 2p +43 MeV
We call this chain of reactions catalysed D-D.

A lot of MFE research is really understanding plasma physics related to confinement. In these experiments we use a surrogate plasma in lieu of DT (often a pure deuterium plasma). The basic underling physics is the same regardless of the "fuel" used. It is cheaper to not use tritium. It's like a engineer designing an air plane. If you want to study aerodynamics you could build a full scale plane and fly it or you could use a scaled replica and put it in a wind tunnel.

If you want to study a self heated burning plasma you have to use DT fuel. ITER which is designed to study just that is going to eventually use DT fuel. Like all other experiments, they are going to start with a surrogate fuel and work out all the kinks before they switch to DT.

This one is a little tricky. If you're looking at an individual particle its the velocity of particle perpendicular to B. However in plasma physics with often use a fluid model where the individual particles have a distribution of energies (often close to a Maxwellian). In these cases we care about some averaged gory-radius. We often use the average thermal velocity of the particles, but in some cases its more appropriate to use the ion sound speed.

11. Aug 22, 2013

### rogerk8

Hi Wolfman!

Please bear with me and treat me like your humble apprentice :)

Jesus, 43MeV!

Let's begin from the start by calculating the temperature and speed a proton would have at these energies.

$$E_{AV}=\frac{1}{4}mv^2=\frac{1}{2}kT$$

Solving for speed yields

$$v=\sqrt{\frac{2kT}{m}}$$

or

$$v=\sqrt{\frac{4E_{AV}}{m}}$$

Solving for temperature yields

$$T=\frac{2E_{AV}}{k}$$

I'm not sure of the details in the first equation because I think the kinetic energy should equal

$$Ek=\frac{mv^2}{2}$$

but according to Francis F. Cheng it doesn't (in the case of a Maxwellian distribution, anyway).

The first equation comes from the Maxwellian distribution of velocity or

$$f(v)=A\exp{(-\frac{mv^2/2}{kT})}$$

where, strangely enough, Ek is set different from kT.

Should they not always follow as has been described in the beginning?

Anyway, the number of particles per square meters is the velocity integration over f(v) (sorry, I do not know how to write integrals in tex)

$$n=\int_{-\infty,\infty}{f(v)dv}$$

Yielding

$$A=n\sqrt{\frac{m}{2\pi kT}}$$

But this means an integral of type

$$\int{\exp{(-ax^2)}dx}$$

will have to be calculated. And I do not know how!

Anyway, Beta tells me that the result should be as above.

The constant A then magically becomes

$$A=n\sqrt{\frac{m}{2\pi kT}}$$

Let's calculate the speed and temperature of a 43MeV "warm" proton (for simplicity).

$$k=1,38*10^{-23}$$

$$e=1,6*10^{-19}$$

$$m_p=1,67*10^{-27}$$

Using the above equations and Eav:

$$v=128*10^6 m/s$$

Which is relativistic....

And the temperature is

$$T=10^{12}K$$

This has to be wrong because we are not only talking millon degrees here, we are talking trillion degrees! :D

Finally, let's calculate the Magnetic Flux Density, B, in Teslas for confining such a fast particle while not touching the wall of the say 2 meters wide tokamak chamber:

The earlier formula is repeated here for convenience

$$R_{L}=\frac{mv}{|q|B}$$

which gives

$$B=\frac{mv}{|q|R_{L}}$$

Setting

$$q=+e$$

and

$$R_L=1$$

yields

$$B=1,34T$$

which however does not seem so much (ordinary transformer irons are often designed at some 1,6T).

But due to known istabilities we might not want the plasma to be so close to the tokamak wall.

Decreasing the Langmor radius some 10 times would yield a neccesary B of 16T.

How far from the truth am I?

So this means that you are experimenting with a plasma that you dont' really expect to ignite, so to speak?

This is very interesting.

I still don't understand the perpendicular velocity, vp as I have denoted it. How can there be a perpendicular velocity at all? Does it come from collisions or is it just a way to describe the x and y-components while gyrating?

The next bold part must say gyro-radius, right?

Finally, I tried Wikipedia regarding "ion sound speed" but found nothing.

So I would very much like an explanation of that.

Best regards, Roger

Last edited: Aug 22, 2013
12. Aug 22, 2013

### Staff: Mentor

Those 43 MeV are distributed on all reaction products (in 4 reactions!). You don't get a single proton with 43 MeV.
You are right that fusion products have an energy way above the thermal energy in the plasma - that's why they can heat the plasma.

Right. Most experiments are mainly interested in the plasma and its stability, you don't need fusion reactions to study that. With a pure deuterium plasma, you have nearly no neutron flux, so activation of the reactor materials is easier to handle.

The plasma is hot.

13. Aug 23, 2013

### Romulo Binuya

ITER, at a cost of 15 Billion Euros wants to fuse half a gram of hydrogen isotopes for at least 1000 seconds. Plasma confinement could be relatively easy to achieve, but I guess they will find it difficult to stabilize the quantum tunneling process to achieve the desired rate of proton fusions.

14. Aug 23, 2013

### Staff: Mentor

"stabilize the quantum tunneling process" does not make sense. If you have a plasma with good macroscopic parameters (density, temperature, composition), you get fusion.

15. Aug 23, 2013

### rogerk8

Let's try to calculate the current needed to generate 16T in vacuum (for simplicity):

Reusing the formula for an ordinary toroidal iron core

$$B=\mu_r\mu_0\frac{NI}{l_m}$$

where

$$\mu_r=1$$

and

$$\mu_0=4\pi*10^{-7}$$

Rearraging gives

$$I=\frac{l_mB}{N\mu_0}$$

Putting

$$B=16T$$

$$R_t=1+2m$$

which gives the mean magnetic length

$$l_m=2R_t*\pi=20m$$

which yields

$$I=255MA/N$$

This also sounds wrong.

Even though the turns (N) might be quite few to avoid resistive losses.

Best regards, Roger

16. Aug 23, 2013

### rogerk8

Summarizing the formulas:

$$m\frac{dv}{dt}=qvXB$$

Solving the differential equation yields

$$w_c=\frac{|q|B}{m}$$

which is the cyclotron frequency i.e the frequency the charged (observe) particles spin around a perpendicular magnetic field B and with the Langmor radius of

$$r_L=\frac{mv}{|q|B}$$

Using

$$E_{AV}=\frac{1}{4}mv^2=\frac{1}{2}kT$$

Solving for speed yields

$$v=\sqrt{\frac{2kT}{m}}$$

or

$$v=\sqrt{\frac{4E_{AV}}{m}}$$

Solving for temperature yields

$$T=\frac{2E_{AV}}{k}$$

I'm not sure of the details in the first equation because I think the kinetic energy should equal

$$Ek=\frac{mv^2}{2}$$

but according to Francis F. Cheng it doesn't (in the case of a Maxwellian distribution, anyway).

The first equation comes from the Maxwellian distribution of velocity or

$$f(v)=A\exp{(-\frac{mv^2/2}{kT})}$$

the number of particles per square meters is the velocity integration over f(v)

$$n=\int_{-\infty,\infty}{f(v)dv}$$

Yielding

$$A=n\sqrt{\frac{m}{2\pi kT}}$$

Let's calculate the speed and temperature of a 43MeV "warm" proton (for simplicity).

$$k=1,38*10^{-23}$$

$$e=1,6*10^{-19}$$

$$m_p=1,67*10^{-27}$$

Using the above equations and Eav:

$$v=128*10^6 m/s$$

Which is relativistic....

And the temperature is

$$T=10^{12}K$$

Let's calculate the Magnetic Flux Density, B, in Teslas for confining such a fast particle while not touching the wall of the say 2 meters wide tokamak chamber:

The earlier formula is repeated here for convenience

$$r_{L}=\frac{mv}{|q|B}$$

which gives

$$B=\frac{mv}{|q|r_{L}}$$

Setting

$$q=+e$$

and

$$r_L=1$$

yields

$$B=1,34T$$

But due to known istabilities we might not want the plasma to be so close to the tokamak wall.

Decreasing the Langmor radius some 10 times would yield a neccesary B of 16T.

Remembering that

$$w_c=\frac{|q|B}{m}$$

and

$$r_L=\frac{mv}{|q|B}$$

yields that

$$v=w_c*r_L=v$$

Trying to calculate the current for generating 16T in vacuum (for simplicity):

$$B=\mu_r\mu_0\frac{NI}{l_m}$$

where

$$\mu_r=1$$

and

$$\mu_0=4\pi*10^{-7}$$

Rearraging gives

$$I=\frac{l_mB}{N\mu_0}$$

Putting

$$B=16T$$

$$R_t=1+2m$$

which gives the mean magnetic length

$$l_m=2R_t*\pi=20m$$

which yields

$$I=255MA/N$$

17. Aug 24, 2013

### rogerk8

Hi Wolfman!

What do you think of this procedure:

Starting up a proton-based fusion reactor:

We know that a proton will gyrate around a perpendicular magnetic field, B.

So we will need a magnetic field to confine the plasma.

We cannot however use the basic "cyclotron-equation" to determine speed.

Just note that the proton will gyrate with the Langmor radius depending on speed (and B).

But we can aim at a certain average kinetic energy (and thus temperature).

With the use of the energy-equation above we can then calculate speed.

Putting this speed into the "cyclotron-equation" then gives the Langmor radius.

Which we must make sure is less than the tokamak chamber radius.

Realising this gives the minimum B required.

It is interesting to note that the amount of B does not affect speed at all!

Only the cyclotron frequency and the Langmor radius.

"All" we have to do now is to generate this B (while injecting our fuel).

Use of the toroidal formula for B gives an obscene amount of Amps :)

How far from the truth am I?

Best regards, Roger

Last edited: Aug 24, 2013
18. Aug 24, 2013

### Staff: Mentor

A magnetic field of 1T leads to ~3m radius per GeV momentum for hydrogen. Hydrogen with an energy of 10 MeV has 137 MeV of momentum, so it has a radius of ~40cm. Should be fine. The wall is cooled anyway (this is how the reactor extracts useful energy!)
Helium is even easier to confine, as it has twice the charge and gets the a similar momentum in the reactions (exactly the same in D+D->He+p).
A magnetic field of 1 T is not hard to produce - the LHC experiments have stronger fields.

@rogerk8: Why do you put all those empty lines in your posts? This just makes the posts longer than they really are.

19. Aug 24, 2013

### rogerk8

Hi mfb!

This is very fun!

Using

$$v=\sqrt{\frac{4E_{AV}}{m}}$$

and

$$r_L=\frac{mv}{|q|B}$$

I get approximatelly

$$v=600,000km/s (>c)$$

and

$$r_L=6m$$

What am I doing wrong?

Please explain to me like I was a child the diffence between the "energy" and the "momentum".

Because I really don't get it.

Actually, the Maxwellian distribution equation

$$f(v)=A\exp{(-\frac{mv^2/2}{kT})}$$

really puzzles me because there is obviously a difference between

$$mv^2/2$$

and

$$kT$$

While at the same time

$$E_{AV}=\frac{1}{4}mv^2=\frac{1}{2}kT$$

seem to hold.

Thanks for telling me. I have actually not thought so much about that "trivial" part! :)

Finally, regarding complaints about my writing style I just want to say that I am a Von Neumann machine of flesh and blood ;)

Best regards, Roger

Last edited: Aug 24, 2013
20. Aug 24, 2013

### Staff: Mentor

You are using nonrelativistic formulas for relativistic particles.

They are two different concepts, like mass and velocity. Their meaning cannot be shown by showing differences. "Where is the difference between mass and velocity?"
Non-relativisticly, $E=\frac{1}{2}mv^2$ and $p=mv$.

The Maxwell distribution is not relevant for fusion products. They are not in a thermal equilibrium.

21. Aug 24, 2013

### rogerk8

You are not that good of a teacher. :)

I fully understand the difference between E and p as you describe it.

But how is it relevant?

And how do I calculate the actual speed when it approaches or even "exceeds" the speed of light?

I find no clues at Wikipedia.

This is obviously wrong according to Wolfman above.

Best regards, Roger

22. Aug 24, 2013

### Staff: Mentor

I have no idea why you think some "difference" is relevant here.
The radius of charged particles in the magnetic field depends on the momentum of those particles (and the charge). I calculated the momentum of a proton with an energy of 10 MeV (=typical energy after a fission reaction).

Did you try special relativity?
$E=\gamma m c^2$, $p=\gamma m v$, $E^2=p^2c^2 + m^2 p^4$ with $\displaystyle \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
E=energy, p=momentum, m=mass, c=speed of light
This allows to convert between energy, momentum and velocity.

It is not. The plasma as a whole follows the distribution (well, approximately) with a temperature of the order of 100 million K, the high-energetic fusion products immediately after fusion do not.

23. Aug 24, 2013

### rogerk8

Now you can be called a teacher! :)

Thank you!

The correct expression for E should however be

$$E=\sqrt{(pc)^2+(mc^2)^2}$$

Which for

$$v=0$$

and

$$p=\gamma mv$$

reduces to

$$E=mc^2$$

Right?

It would however be nice to fully understand the expression for E above.

But you have given me a new insight.

Due to the term

$$(pc)^2=(\gamma mvc)^2$$

in the expression for energy above, E can be related to speed, v, in a relativistic way.

Desperatelly wanting a complete analytic expression:

$$E=\sqrt{(pc)^2+(mc^2)^2}=\sqrt{(\gamma mvc)^2+(mc^2)^2}=\sqrt{(mc\frac{v}{\sqrt{1-\frac{v^2}{c^2}}})^2+(mc^2)^2}$$

Which means that the speed we observe, v', is

$$v'=\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}$$

or

$$v=\frac{v'}{\sqrt{1+\frac{v'^2}{c^2}}}$$

So if I calculated 600,000km/s it really should be 268,000km/s?

Best regards, Roger

Last edited: Aug 24, 2013
24. Aug 25, 2013

### Staff: Mentor

That is just the square root on both sides. It is equivalent to the formula I posted.

Right (I compressed the quote a bit).

I would use $E=\gamma m c^2$ to do this.

I don't see why this should be true.

600,000km/s with non-relativistic formulas give Ekin=2mc2 or Etotal=3mc2 => γ=3. This corresponds to v=sqrt(8/9)c or v=283,000km/s.

25. Aug 26, 2013

### rogerk8

You don't give me much credit, do you? :)

If you scroll to your post, you can see that you happened to use p instead of c in the formula.

I just corrected that and added the square root for appearance.

This is interesting.

Above you say that you would use

$$E=\gamma mc^2$$

to calculate v from E in a relativistic way.

Then you thankfully use my example and simply put

$$\gamma=2$$

adding yet another gamma for the always present "rest energy", right?

But how can gamma equal 2 in the first place?

Looking at it from my point of view you seem to relate gamma to v/c in a linear way.

I can't see why this is possible.

By the way, I understand now that the masses above are "rest-masses" which gives the formulas.

But shouldn't the square root read

$$E=\sqrt{(pc)^2+(\gamma mc^2)^2}$$

Or is the second term always "rest energy" (gamma=1)

And where does the first term apply in our discussion?

I hope you don't feel that I'm asking too many questions.

Getting back to the Maxwellian distribution of speed I now think I understand what the distribution function actually means. Correct me if I'm wrong but doesn't it mean that the most probable speed is when

$$mv^2/2=kT$$

This while there are other speeds too, both higher and lower than "kT" and while their "temperature" could be the same, they are less probable to have that temperature.

Is something like this right?

Best regards, Roger
PS

$$E=\gamma mc^2$$

Is actually news to me because this means that energy approaches infinity when the speed approaches the speed of light.

I have heard people telling me that the speed of light is impossible to reach energy-wise, and now I know why.

Thank you!

Last edited: Aug 26, 2013