Summary, a Correction, and Further Comments

In our desire to make an opaque subject clearer to the nonexpert, we have focused on the essential parts of the nanophysics of fusion. The cross section for electron scattering from the ions, related to the resistivity of the plasma, it turns out, has been treated too simply, and needs a correction factor, which increases the plasma resistance. The correction factor known as “ln(A)” multiplies the strong scattering cross section p r22, which when multiplied by TGamow is the fusion cross section. In the scattering of electrons from ions, the cumulative effect of many small-angle scat­tering events make this substantial correction. This occurs because the Coulomb force has a long range. The correction does not apply to fusion, but only to the plasma resistance, because in that case the strong scattering is the only way to set up the fusion event, an accumulation of weak scattering events is of no use. The importance of the small scattering events is controlled by the Debye length,

Id = {квГє0/ж2)1/2, (3.71a)

Подпись: ln(L) = ln(21D/b) Summary, a Correction, and Further Comments Подпись: (4.40)

which is the maximum distance over which two particles in the plasma experience the Coulomb force. Beyond this distance, the small adjustments of positions of many electrons screen away the direct Coulomb force. In this case, the ratio of the Debye length to the classical turning point spacing r2, which is called “b” in the plasma literature, is important. Thus,

Taking 1D = 84.5 m and r2 = 111.3 f, we find ln(L) = 21.1, for the DT Tokamak plasma at 150 million K.

This means that the plasma resistivity and resistance values are too small by a factor 21.1, the electron mean free path reduced to 12.18 km, and the plasma resistivity becomes 3.0 x 10~10 V m. The voltage to produce the desired 10 MW heating rate has to be increased by a factor (21.1)1/2 = 4.6. So the needed dB/dt in the example above must be increased from 14.3 mT/s to 65.8 mT/s.

The design for ITER actually employs larger magnetic field values (http://en. wikipedia. org/wiki/ITER), 11.5 T for the toroidal field, leading to a maximum stored magnetic energy 41 GJ, a volume of 840 m3 and a central solenoid field up to 13.5 T (see Figure 4.10), which would limit the time of inductive heating at 10 MW to 13.5 T/

65.8 mT/s = 3.4min, assuming the central solenoid radius of 1 m.

Beyond this, there are several technical issues that in practice are important. The stability of the plasma has been assumed, which is a simplification. The toroidal magnetic field is approximately constant across the cross section, but more accurately decreases from the inner to outer walls of the torus. The induced current in the torus produces the usual circling magnetic field around the current, which is termed a “poloidal” field. In fact, an additional magnet, the “poloidal magnet,” produces a vertical magnetic field needed to keep the plasma from expanding outward in radius.

In Figure 4.9, we see that the high-temperature part of the plasma occupies a cross section less than the full vacuum vessel cross section, pb2, but let us assume b’ ~<b is the radius of the current density when the ohmic heating is being applied. (The plasma is actually diffuse and somewhat free to move in the mechanical cross section, and could only roughly be described as a torus of major radius a! and minor radius b’.) The resistance of the plasma Rp (4.36) when corrected for the small-angle scattering factor ln(L) = 21.1 is raised from 0.204 nV to 4.30 nV. At the chosen heating power 10 MW, the plasma current lp must satisfy

ip-Rp = 10 MW, (4.41)

which gives lp = 48.2 MA.

One aspect of the plasma equilibrium may be suggested by estimating the magnetic field produced by the plasma current lp itself. If the current were flowing in a straight solenoid of radius b, the magnetic field Bp(b) at its radius can be gotten from Amperes law,

B ■ dl — m0Ip. (4-42)

In this situation then, we estimate Bp(b) — m0Ip/2pb — 9.64 T, choosing b — 1m. If we imagine bending the solenoid to fit the major radius a, then the field B will become smaller on the outside and bigger on the inside and the relevant factor will be approximately (a ± b)/a. The difference in the plasma-current-induced magnetic fields Bp on the inside versus the outside is then DB « (2b/a) Bp(b). If a vertical field Bz — (b/a) Bp(b) is added (to increase the field on the outside), the current Ip will tend to be stabilized, with sum vertical fields equal on the inside and the outside. The required vertical field, in our rough analysis that neglects an outward force from the plasma pressure, is then

Bz — m0Ip/2pa — 1.56 T (for a — 6.2 m, Ip — 48.2 MA). (4.43)

The vector sum of this circling poloidal field and the toroidal field is then a helical field, which is the field that the ions and electrons in the plasma will follow, depending on Ip.

This situation is sketched in Figure 4.10. On the left, the magnetic field lines shown on a cut through the torus are distorted, stronger on the inside and weaker on the outside, estimated as DB above. The uniform vertical field Bz is shown added in the sense to strengthen the field on the outside of the torus. On the right, the resultant field is nearly symmetric as would occur for a straight solenoid. The detailed force balance has to include a component of the plasma pressure to the outside, which arises due to the curvature.

The difference DB is the origin of the tendency of the plasma to expand outward, which is corrected by the poloidal coil system. The vertical field Bz, estimated in Equation 4.43, should be applied to maintain the plasma centered in the physical cross section of the torus. A more accurate analysis, which includes the plasma pressure, gives

Bz — (M-0 Ip/4pa)[ln(8a/b) Tpplasma/pmagnetic T d], (4.43a)

where the correction d < 0 is given by Miyamoto [52]. Here pressure

Pplasma — 2NkBT — 2 x 1020 x 12.93 x 103 x 1.6 x 10~19 — 0.414MPa — 4.1 bar,

(4.44)

and pmagnetic is of the form (B’)2/2m0. If we interpret B’ — DB as defined above, then pmagnetic — (DB)2/2m0 — 3.84 MPa. It appears that our rough estimate is within a factor of two of the correct form (4.43a). A full analysis ofthe equilibrium is given in Section

6.3 of Miyamoto [52].

A variety of undesirable modes have been found to occur and to reduce power. The underlying nanophysics of the fusion power generation is quite clear, but the engineering design of a practical reactor to avoid the plasma instabilities is difficult. Furthermore, there are material degradation problems. We mentioned above the power loss from the plasma if highly charged nuclei such as tungsten are eroded into

the plasma. Wall erosion is also a maintenance problem, complicated by the fact that the first wall materials will become radioactive and require special handling.

An analysis of the availability in the sea of deuterium and lithium for a possible but probably unlikely age of fusion-produced electric power has been given by MacKay ([44], p. 172). Seawater contains 33 g of deuterium per ton, and this is a huge supply.

Each gram of deuterium represents 100,000 kWh, and the sea contains 197 million tons of seawater per person (at 7.0 billion humans). Lithium is also available in seawater at 0.17 ppm, which translates [44] to 3910 tons oflithium per person. If the lithium produces 2300 kWh per gram in the fusion reactors, and the energy usage per person is 105 kWh/day, then this energy source would last for over a million years.

The success ofthe ITER reactor is not guaranteed, and even so it is not a power reactor but a research reactor. In the words of David MacKay [44], “I think it is reckless to assume that the fusion problem will be cracked.”

Even if the problem is cracked, it is clear that a fusion reactor has to be large, a contributor to a power grid. While the fusion reactor seems quite safe, has no meltdown possibilities, and leaves few radioactive waste products, if indeed the engineering of the well-known science into a useful power reactor design succeeds, the end product will be very high technology only possible on a large scale, with high demands on an operator to make it work smoothly.

Updated: October 27, 2015 — 12:10 pm