On Oct 12, 7:50 am, Robert Clark wrote:
On Oct 12, 7:56 am, Craig Markwardt



wrote:
Robert Clark writes:
In researching the amount of energy required to ionize gas for ion
drives I was surprised by the total amounts of energy that would be
required to *fully* ionize the gas. This amount of energy is quite
large, actually huge, and so for actual ion drives the gas is only
minimally ionized.

... remainder deleted ... by typing in for example Xe 53 to get the last (54th) electron
ionization energy. However, not every ionization level for xenon is
given on this page either. After a web search, I found the total
amount of energy required to fully ionize xenon is about 200 keV.
Since 1 eV is about 100 kJ/mol , this is about, 2 x 10^10 J/mol. Since
the atomic weight of xenon is 130 this comes to 154 million joules per
gram, 154 billion joules per kilo.

...

This is not really a good method for storing energy. The losses are
too rapid.

In order to maintain (say) Xenon in a fully ionized state, a Boltzmann
equilibrium must be achieved. Removing the last two electrons of
Xenon takes ~40 keV each (ref. X-ray Data Book, xdb.lbl.gov). Thus,
the temperature must be (kB T 40 keV) where kB is the Boltzmann
constant. Let's say kB T = 100 keV, or a temperature of 1 billion Kelvins.
That is hot.

The Bremsstrahlung emissivity of a plasma scales approximately as,
W = 1.4e-34 ne nZ Z^2 T^(0.5) in Joule / s / cm^3
where ne is the electron density, nZ is the ion density, Z
is the atomic number of the ion. (ne and nZ must be in cm^{-3}).
Compare this to your quoted energy density of (200 keV / ion)
= (3.2e-14 Joules / ion), or an energy density of E = 3.2e-14 nZ Joules / cm^3.

Even your "low" density of 4e14 ions/cm^3, the ratio of W / E
is 1/(0.11 msec). In other words, all the internal (and ionization)
energy of the plasma will leak out in less than 1 millisecond by
bremsstrahlung radiation. This is radiation that can't be contained
by any magnetic field or trap, so it is unavoidable.

Not to mention the danger of carrying around a tank of 1 billion
degree plasma...

CM

Thanks for the informative response. Quite key here is that these are
*non-neutral* plasmas. That means the charges are all of the same
sign, all positive or all negative. In your formula you gave note this
would result in the Bremsstrahlung emissivity being zero since one of
the types of charge would be absent.
There has been alot of research on non neutral plasmas showing they
can be stored in magnetic/electrostatic traps for several days:

What is a nonneutral plasma?http://www-physics.ucsd.edu/~dhdpla/nnp.html

Bob Clark


Interesting.

Each galaxy emits plasma from its center at right-angles to
its disc (being 'blown off its accretion disc' NOT). The central
vortex separates infalling neutrons into negative and positive plasmas
and blows them out in opposite directions.

One kind of non-neutral plasma will be going out
one way, and the other will go out the other way.
The two will eventually re-combine into new stars.

John