Why do fusion reactors only work for a few seconds?

Click For Summary

Discussion Overview

The discussion centers on the challenges of sustaining fusion reactions for extended periods, particularly in the context of magnetic confinement fusion (MFE) and inertial confinement fusion (ICF). Participants explore various factors that limit the duration of fusion reactions, including plasma stability, reactor design, and the nature of the fuels used.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants note that in a Tokamak, plasma tends to approach the walls, leading to cooling and cessation of fusion.
  • Others discuss the role of instabilities in the plasma that prevent containment, causing hot plasma to contact the walls and introduce heavy particles that disrupt the reaction.
  • One participant mentions that ICF relies on compressing fuel pellets with laser or particle beams, which limits the duration of the reaction to fractions of a second.
  • Another viewpoint highlights that MFE experiments vary widely, with some devices like the Large Helical Device capable of running for hours, while Tokamaks typically operate for only a few seconds due to limitations in current drive and heat generation.
  • Concerns are raised about the radioactivity produced in fusion reactors, with some participants arguing that while the exhaust is harmless helium, the reactor surroundings become radioactive due to high-energy particles.
  • There is a discussion about the need for tritium in fusion reactions, with one participant noting its scarcity and the challenges it presents for reactor design and maintenance.
  • Some participants propose alternative fusion reactions, such as D+He3 or D+D catalyzed reactions, which may produce fewer neutrons and have different implications for reactor design.

Areas of Agreement / Disagreement

Participants express a range of views on the limitations of fusion reactors, with no consensus on the best approach or the most significant challenges. Disagreements exist regarding the specifics of plasma behavior, the implications of radioactivity, and the feasibility of different fusion reactions.

Contextual Notes

Participants mention various technical limitations, including the need for cooling systems in experiments, the impact of impurities on plasma stability, and the economic considerations of using different fuels. The discussion reflects a complex interplay of factors affecting fusion reactor performance.

  • #31
Some kA is similar to the current in the LHC magnets, they are designed to provide a field of ~8T.
 
Engineering news on Phys.org
  • #32
the_wolfman said:
Roger if you want to do some back of the envelope calculations for ITER I recommend using a magnetic field of ~5T and a bulk plasma temperature of 10keV.

For most fusion applications the energies are low enough that non-relativistic formulas for the speed and gyroradius are ok (You can check this of course). There are exceptions but they're not important for the current discussion.

Also a word of caution. The idea that a charged particle gyrates around a "magnetic field line" is the basis for magnetic confinement but it is far from the end all be all. It is often important that the gyroradius be much smaller than the minor radius of a device (if you want you can check this for ITER which has a minor radius of 2m), but the size of a fusion reactor is ultimately determined by radial transport and stability. Remember we want to reach 100 million degrees in the core of the plasma. However stability and transport limit how steep of a pressure gradient we can have. A very crude way to determine the minimum size of a reactor would be to divide 100 million degrees by the maximum gradient.

Hi Wolfman!

Thank you for these data.

I am much obliged!

First, my calculations regarding B(I) is somewhat wrong. A formula between

B=\mu_0\frac{NI}{lm}

inside a long solenoid or toroid and

B=\mu_0\frac{NIA}{2\pi r^3}=\mu_0\frac{NIR^2}{2r^3}=\mu_0\frac{NI}{2R}

in the center of a short solenoid where the distance from top of coil and onto z-axis, r, has been set to R (which might not be possible).

Anyway, usinq R=2m and my estimated lm=0,5m gives

\frac{B(long solenoid)}{B(short solenoid)}=8

Let's say that the constant is 4. Then I(5T) would equal

I=\frac{B4l_m}{\mu_0N}=8MA/N

Now I will leave this part once and for all (maybe except for calculating an exact formula).

Plasma pressure, p, is defined by

p=nkT

where n is the particle density.

For an isoterm (whatever that means) plasma we then have

\nabla p=kT\nabla n

More of this later...

kT seems related to eV according to Cheng:

eV=kT

So calculating the temperature of a 10keV hot plasma yields:

T=\frac{10^3*1,6^{-19}}{1,38^{-23}}

or 116MK~100MK

Returning to the Maxwellian distribution the average energy however relates to kT and v:

E_{av}=mv^2/4=kT/2

where there is a kT/2 for each degree of freedom (three in my amateur book).

And the most probable speed for two protons (~deuterium) is

v=\sqrt{\frac{2kT}{2m_p}}=\sqrt{\frac{kT}{m_p}}

or 980000m/s~1000km/s

Which indeed is non-relativistic...

The Larmor radius is

r_L=\frac{2m_pv}{|2q|B}=\frac{m_pv}{|q|B}

or 2mm(?)

And the cyclotron frequency is

w_c=\frac{|2q|B}{2m_p}=\frac{|q|B}{m_p}

or 480Mrad/s.

And now I will have to go to bed :)

Tomorrow I will wright about drifts in a plasma (=transport?)

Roger
 
  • #33
My calculations regarding B(I) is still somewhat wrong while using a formula for a long solenoid or toroid (where lm is included).

The below formula is for a short solenoid where r indicates the distance from top of coil to z-axis

B=\mu_0\frac{NIA}{2\pi r^3}=\mu_0\frac{NIR^2}{2r^3}=\mu_0\frac{NI}{2R}

the latest term comes from the assumption that r can be set to R in the middle of the short solenoid (which might not be possible).

Anyway, usinq R=2m I(5T) would equal

I=\frac{2RB}{\mu_0N}=16MA/N

As far as I have seen in pictures the coils are stacked some 1m apart around the tokamak and supposing each coil do not have more turns than some 10 (due to resistive losses) the current through them all connected in series would be some 2MA.

However, this is not the "important" part. The important part is that the B-field will be curved and non-uniform along the

\phi-axis

Which introduces a gradient in the B-field. I will get back to this later on.

Now I will continue to work :)

Roger
PS
It was too late to edit the above...
 
Last edited:
  • #34
Now I wish to somewhat prove the above formula.

Consider a current carrying loop in vacuum then these four equations apply:

\nabla \cdot B=0
B=\nabla XA
A=\frac{\mu_0}{4\pi}\int_v{\frac{J}{R}dv}

and while

Jdv=JSdl=Idl

B=\frac{\mu_0I}{4\pi}\oint_c\frac{dlXa_R}{R^2}

where the first is from Maxwell's Equations, the second is a consequence while using an arbitrary vector A, the third is a definition of the vector magnetic potential A and the fourth is the Biot-Savart Law.

Using b as the loop radius and R as the distance from dl to point we get

dl=bd\phi a_{\phi}
R=a_zz-a_rb
R=\sqrt{z^2+b^2}

then

dlXR=a_{\phi}bd\phi X (a_zz-a_rb)=a_rbzd\phi + a_zb^2d\phi

and while the r-components cancel out we get using the Biot-Savart Law above

B=\frac{\mu_0I}{4\pi}\int_{0 2\pi}a_z\frac{b^2d\phi}{(z^2+b^2)^{3/2}}

or

B=\frac{\mu_0I}{2}\frac{b^2}{(z^2+b^2)^{3/2}}=\frac{\mu_0I}{2}\frac{b^2}{R^3}

It is interesting to note that z=0 is valid.

Roger
PS
I have used my book's (Field and Wave Electromagnetics by David K. Cheng) convention due to simplicity which means I thereby hopefully state the correct equations.
 
  • #35
This chapter of my interest is about different drift mechanisms in a plasma.

Using

m\frac{dv}{dt}=q(E+vXB)

and the fact that

F=qE

we get the common expression

v_{force}=\frac{1}{q}\frac{FXB}{B^2}

Where we can put

F_E=qE

or

F_g=mg

or

F_{cf}=a_r\frac{mv_{//}^2}{R_c}

Where the first is the force from a electric field upon a charge and the second is the force on the particle due to gravity and the third is the centrifugal force as the particles move along a curved line of force, B.

Therefore

v_E=\frac{EXB}{B^2}
and
v_g=\frac{m}{q}\frac{gXB}{B^2}
and
v_R=\frac{1}{q}\frac{F_{cf}XB}{B^2}=\frac{mv_{//}^2}{qB^2}\frac{R_cXB}{R_c^2}

where Rc is the curvature radius of the lines of force.

It is interesting to note that

|v_E|=|\frac{E}{B}|

It is harder to explain why

F_{\nabla B}=-/+\frac{qv_pr_L}{2}\frac{dB}{dy}a_x

where B is along the z-axis and grad B is along the y-axis and the sign denotes ion drift direction.

Which put in the vforce equation above gives

v_{\nabla B}=-/+\frac{v_pr_L}{2}\frac{\nabla BXB}{B^2}

p is here denoting perpendicular (to B).

In a curved vacuum field we may add

v_R

to

v_{\nabla B}

thus

v_{cv}=v_R+v_{\nabla B}=\frac{m}{qB^2}\frac{RcXB}{Rc^2}(v_{//}^2+\frac{1}{2}v_p^2)

Now I quote Francis F. Chen:

"It is unfortunate that these drifts add. This means that if one bends a magnetic field line into a torus for the purpose of confining a thermonuclear plasma, the particles will drift out of the plasma no matter how one juggles the temperature and magnetic fields".

Finally we have the diamagnetic drift (considering the plasma as a fluid)

v_D=-\frac{\nabla pXB}{qnB^2}

where

F_D=-\frac{\nabla p}{n}

where p denotes pressure and n volume particle density.

For an isoterm plasma

\nabla p=kT\nabla n

this gives

v_D=-\frac{kT}{qn}\frac{\nabla nXB}{B^2}

But the total drift perpendicular to B is actually

v_{ED}=v_E+v_D

Roger
PS
Most of the time I am using the book Plasma Physics and Controlled Fusion by Francis F. Chen. Please note that there is a certain Cheng here inside this thread also...
 
  • #36
This is no longer a discussion thread, it looks more like your private blog. If you want to continue this, please use our blog function and not discussion threads.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 28 ·
Replies
28
Views
27K
  • Poll Poll
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 18 ·
Replies
18
Views
8K