Reply To: Branson Prize: $25M for removing 1 Gigaton CO2/year

[Ferret]

The functioning of a DPF device is controlled by the so-called “drive factor”: D = I / ( a sqrt(p) ), where I is the peak current, a the anode radius and p is the pressure inside the device. For optimum neutron yield in deuterium, D = 78.46 kA / ( cm sqrt(mbar) ). In order to use the normal atmospheric pressure p = 760 Torr instead of 7 Torr, p has to increase 100 times. Thus, either I increases 10 times too or a decreases 10 times. I would go for a decrease of a. Say a = 0.2 cm (anode diameter of 0.4 cm), a catode 0.5 cm in radius (1 cm diameter) and p = 1000 mbar, it follows I = 0.5 MA. Up to now it should work with deuterium fuel at normal atmospheric pressure. For decaborane, maybe a higher I is necessary, say 1-2 MA.

Of course, there are problems with such a device. For one thing, the anode has to be a solid bar, i.e. no cooling hole in the center. That way it could take some of the mechanical stress from the plasmoid. Next, it may not last long. As Mr. Olsen found out, the current passing through it and the X-rays will vaporize a layer at its surface each time the device is fired. But it should take a few pinches.

One advantage of having a smaller anode is that the device needs far less energy. If I’m correct, the energy scales as a^3, thus a 10 times decrease of a means a 1000 times decrease for the energy. Instead of, say, 5-10 kJ used in other devices, only 5-10 J would be required. Thus, the capacitor bank may be reduced to just one capacitor.

That is, if the pinch does occur. Why would it not occur? This is a real question, not rethorical.

There is one possible reason: the flow of plasma along the anode may be disrupted by the higher pressure, somehow. Is it the case? Or maybe other reason?

PS: The capacitor fires at about 10 kV in air, quite a high voltage. Its capacity: about 100 nF or 0.1 uF.