Reply To: would nuclear energy really be accessible to all?

[Allan Brewer]

jamesr wrote: For the plasmoid at high temperature the dominant radiative cooling mechanism is via bremsstrahlung. The formula for which is approximated as:

P(br) = 1.7E-38*Z^2 *n_e * n_i *sqrt(Te) in W/m^3

where n_i, n_e are the ion & electron number densities in per m^3, and Te is the electron temperature in eV.

Taking the effective Z as 2 (combination of hydrogen & boron but with higher concentration of hydrogen) and ne =Z*n_i = 1e26/m^3 and Te= 1e9K = 80keV

Then I make P(br) ~ 4e17 W/m^3 for the plasmoid.

Multiplying this by the volume of the plasmoid (~10um diameter) an saying it lasts for 100ns gives only 4e-5J which seems way too small???

For the bulk plasma at a few thousand degrees the radiation comes from recombination and line radiation which is much harder to work out… I’ll try and find an estimate.

You also obviously have this hot, few thousand degree plasma in direct contact with the electrode surfaces – this I would expect to dominate everything.

I think for decaborane plasmoid n_e and n_i should be ~e29, derived as follows:
“…because of the very high density in the plasmoid, which is close to solid density…” (Eric L)
1cc water weighs 1g so 1m^3 water weighs 10^6 g
Relative density of decaborane is 0.9
so 1m^3 plasma weighs 0.9*10^6 g
Decaborane weighs 122g per mole
therefore 1m^3 contains 0.9*10^6 / 122 moles
each mole contains 6e23 molecules
each molecule gives 24 ions
so 1m^3 contains 6e23*24* 0.9*10^6 / 122 ions (= n_i = 1e29)
each molecule gives 64 electrons
so 1m^3 contains 6e23*64* 0.9*10^6 / 122 electrons (= n_e = 2.8e29)

This raises your P(br) to 2.24e24 W/m^3
10um cube gives 2.24e9W
100ns life gives 2.24e2W
notional 300 shots/s gives 70KWatts – how does that sound to you?