Calculations Explain Destruction of FilamentsDecember 8, 2016
It’s clear that the existence or non-existence of filaments—dense vortices of current and plasma—in the current sheet has a huge impact on the functioning of FF-1, because the filaments are the first stage of compressing the plasma. The disruption of the filaments is a key reason for lower-than-expected fusion yields in past experiments. Yet, while many researchers have studied filamentation, it has been difficult to get a clear understanding of when they form and how they are destroyed. New calculations by Lerner have used the concept of minimum dissipation of energy to figure out this problem. In general, processes will minimize the amount of energy they dissipate or turn into heat—Mother Nature is not lazy, but she is efficient. If hydrodynamic friction dominates this energy dissipation, the Rayleigh-Taylor instability will conserve or produce filaments because filaments dissipate less energy moving through the background gas than does a solid sheet. But if electrical resistance dissipates more energy, then the electric current tends to spread out to minimize this resistance. This spreading can disrupt filaments, or prevent their generation.
Impurities, such as tungsten, can greatly increase electrical resistance, because when such heavy ions lose several electrons, their greater charge makes collisions with electrons in the current much more likely. The new calculations show that filaments with large radii tend to be more stable than those with small ones. But with tungsten impurities of even 4% by mass, filaments would have to be so large they would merge with their neighbors, and so be disrupted. Since in the last experiment, we had about 20% tungsten impurities by mass, that indicates that we can avoid disruption of the filament only if we can reduce the impurities by ten-fold or so. This gives us a good quantitative goal to aim for in the elimination of oxygen. To achieve this level of impurity, initial oxygen in the chamber has to be less than about 10 micrograms, an achievable level.
Lerner also made more detailed calculation of the damage that can be expected from the recombination radiation for both filaments and a non-filamented current sheet. Further research showed that the most damaging form of radiation is “dielectronic recombination radiation,” which occurs when a free electron recombines with an ion, and the energy is absorbed by a second, bound electron, which jumps to a higher energy level. The second electron then radiates the energy when it jumps back down to lower energy. The calculation showed that with even small tungsten impurities, some 30 MW/cm2 of radiation will hit the end of the anode, enough to melt tungsten and cause heavy erosion. So Lerner does not expect this source of erosion to improve greatly in the next experiment. However, these impurities will enter the plasma too late to affect the fusion yield. In addition, once the experiments switch to beryllium as the electrode material, recombination radiation will be practically eliminated. The beryllium ions, with only 4 electrons, will be completely stripped. In this case, the calculation indicates a 40-fold decrease in radiation, falling below that needed to melt beryllium.