Radiation heat transfer?

[Ryan]

what mode of heat transfer would be the most in case of a fusor, between the plasma and outer environment? searching through the earlier post i could find not much reference to radiation heat transfer...wouldnt the temperature gradient be large enough for this to be significant..

[Doug Coulter]

You're actually asking a fairly complex question, which also contains a probably incorrect assumption.
The latter is that this is a plasma. It's not neutral by quite a long way according to my Faraday probe measurements here. You are also assuming most of the energy is thermal, when perhaps quite a lot of it is not (anymore than a bullet is hot because it's going fast - it might *also* be hot, but that's not where most of the energy is).

In the grid, the main heat source is ions hitting it, and the main heat transfer out is usually radiative, with some conduction through the HV feed through stalk. My own fusor is designed to enhance that latter by using a very fat rod there, and the former by using graphite end caps on my grid.

Most of the heat in the outer shell is from electrons, emitted from the grid when ions strike it, being accelerated to the supply voltage (more or less) and then striking the tank walls.

There's very little gas in a fusor in operation. No matter how you define its temperature, it doesn't transfer much to anything, in or out. There are for sure some randomly oriented neutral atoms in there, moving fast - no one has a great handle on what fraction under what operating conditions and how much of the total heat energy is contained in those. But it can't be much, there's just not much there to be hot. We also don't see any huge pressure rise/fall with power on/off, which would indicate that it's not a huge effect. I've only seen that in fairly rotten vacuum conditions with a lot of contamination on the tank walls that gets boiled off quickly when power is first applied. After its gone, so is the effect of pressure rising noticeably with power input.

[Frank Sanns]

If the question is: Where does the input energy go? The answer is that it ends up as heat.

There are not two solid bodies contacting so Conduction is ruled out quickly.

There is only a very small surface area glowing and energy is radiation at Temperature^4. Thermal Radiation alone can not account for the kind of energy transfers going on.

The last is Convection. Yes there is a non solid medium of atoms, electrons, and ions. They are not in thermal equilibrium but instead are in non uniform velocity distributions. Still, they could be considered Forced Convection or mass transfer and are responsible for the lion's share of energy transfer to the fusor shell.

Frank Sanns

[John Futter]

Frank
I like your reasoning
But why not e=mv^2 as the energy transferral as aposed to your term of convection
the m above taking into account the mass of electrons and also the mass of the ions (note: -quite different). Then the V term again a known due to knowing the voltage applied electrons moving faster because of their smaller mass etc etc.

Then of course the plasma is hot very hot so the radiative losses are not insignificant the T^4 term rises very rapidly with temperature differential..

i'll have another beer

[Richard Hull]

Rest fully assured that 99.999999% of the input energy, ultimately, heats the shell by many processes. Energy goes into the fusor and less than .0000000001% ever leaves it that is not pure, low grade heat. It is effectively, a room space heater when talking about the input energy expenditure and its resultant expression as work done.

As a heater, the fusor is 99.999999% efficient.

Richard Hull

[Chris Bradley]

I suspect that, actually, there is very little heat transfer from the plasma to the environment. The plasma may have a high 'temperature' but is very low density, so the total 'heat' in it is negligible.

It is also very low ionisation, meaning only a few fractions of a percent of it are actually ionised. I'd not hazard a guess, but if you take a look at the Debye length, for electric fields to be present the ionisation density looks like it'll be in the ~10^9/cm^3 range, while the background is at 10^14/cm^3 range. Looks like 'more than' 999 in 1000 nucleii in there are part of neutral gas molecules.

So where does all the energy go that you pump in? I'd say it is the same answer as to where the energy goes in a resistor. The electrons bump into stuff at the shell and dissipate their electrical potential there.

I've never really thought why a flow of electrons causes a resistor to get hot, but I'd say a fusor gets hot for pretty much the same reason (well, 99% of the heating, there is a trickle from ion impacts and the neutral gas getting hot). I guess there is some sort of quantum explanation for why an electron deposits its electrical potential as heat energy in a conductor. Can anyone spread light on how a current heats its conductor?

[Frank Sanns]

Energy is energy and 1/2 mv^2 is good. Solving for a 1 inch inner grid at 2,000 degrees gives only watt or two watts of radiant energy so the rest of the input power (i.e 600 watts) is mass transfer or electrically forced convection depending on definition.

Frank Sanns.

[David Geer]

That's simple Electrical Principles... a current of voltage A moves/vibrates at a specified frequency. When going through an object the vibrations cause the particles to "bump" into the container thus transferring energy as it bounces off. This leads to the "bumped" particle gaining frequency and producing heat.

Easiest explanation is to rub your hands together really fast and you'll understand the particle interactions for where the heat comes from.

[Chris Bradley]

David Geer wrote:
> When going through an object the vibrations cause the particles to "bump" into the container thus transferring energy as it bounces off.

I was looking for more than such a trivial and insubstantial answer. What bumps into what, exactly, and how is energy transferred? Do you mean the electron bumps into the nucleii?

Bear in mind that you only get conduction if there are 'free' electrons already bumping around in a sea of free electrons (like if you pour a bucket of water in one side of a swimming pool filled to just overflowing, then you'll get a bucket of water out the other side).

We know, experimentally, that if an electron of e Coulombs changes from a voltage -V to 0 Volts, then Ve Joules of energy is dissipated. Wherever, and however it changes its voltage, it will dissipate Ve somehow. Thus, an electron borne with -V potential will deposit its energy, Ve, in the shell once it gets there.

I was hoping there is a more fundamental answer to this, but if 'bumping around' works for people then let me not disabuse them of that idea.

[Ryan]

To Chris bradley:
i found this on:http://www.asmcommunity.net/board/index ... ic=14134.0

"Resistive heating is not a mechanical phenomena, so it is not a frictional mechanism. At the microscopic level, it is a quantum physics event involving speed distribution of conduction electrons, number of collisions with the ionic cores of the material, the average distance a electron travels before colliding with something, changes in the drift speed of the electrons, and a whole lot more. Moving the charge carriers (electrons in this case) in one direction cause more electron collisions to occur with the material lattice than would happen if no movement were present. This causes the electrons to lose some of their kinetic energy, which is transferred to the ionic core of the material lattice by the collisions. This makes the lattice vibrate faster, which by quantum theory, raises the temperature of the resistive material. If the voltage remains constant, the loss of electron kinetic energy causes a slower electron drift velocity, which in turn means less charge entering and leaving the resistor per unit time, so the current will be less. The charge flow response is almost instantaneous, and the charge flow per unit time (current) is decreased. That is why resistance decreases current and dissipates heat when current is present."

[Ryan]

Thanks for all your replies....but i was seeking an explanation as to how plasma in the fusor core could result with only 130 degrees on chamber wall. (specially when the radiation equation involves T^4)

[Chris Bradley]

You are now asking about plasma-wall interactions, which is a sideways question to the one you asked. There is plenty of literature on the web on this. Enter the following seach term "langmuir debye sheath plasma wall" and you'll be able to pick a listing to satisfy the depth of analysis you want on the subject. Might be better to start a new thread if you have follow-up questions.

Thanks for looking for an answer to my question. Still unsure what this means in simpleton+1-level physics, which is where I am floating around. I note wikipedia discusses electrical resistance by saying heating in the conductor arises from "destructive interference of free electron waves on non-correlating potentials of ions". I get a feeling I understand that - somewhat!

[David Geer]

Doesn't matter if they are free electrons or not. Particle interactions will share the vibrational energies until everything reaches an even balance. So if, particle A and Particle B are of the energy frequencies 3MeV and 5MeV, constant interactions will eventually lead to both levelling out at 4MeV. The transfer of motion is what causes the heat. This is why a capacitor or feedstock gets hot. Electrons flow through, innumerable interactions occer between the medium and the free electrons and produces heat. This is neither insubstantial or trivial because it is a basic principle. The only other variables to look at are resistivity and conductance for the time it takes for the exchange to occur.

-David

[Chris Bradley]

David Geer wrote:
> Doesn't matter if they are free electrons or not. Particle interactions will share the vibrational energies until everything reaches an even balance. So if, particle A and Particle B are of the energy frequencies 3MeV and 5MeV, constant interactions will eventually lead to both levelling out at 4MeV. The transfer of motion is what causes the heat.

So, your interpretation seems to be that the electrons in a conductor at a higher potential than in another [separate] conductor are thermal in nature. That is, the electrons in an isolated wire at 100V are at a higher temperature than those in a 0V wire, hence their energy would flow into the 0V wire because that energy would be thermalising with the cooler electrons there (were the conductors to come into contact).

Yet, oddly, the 100V wire, with all these hotter electrons, is no hotter?

Your explanation also seem to be implying that they do not even need to be conductors for the electrons to flow, because you mention they do not need to be free electrons.

Your explanation doesn't seem quite correct, on the face of it, does it? Especially given that your explanation does not seem to require any differential charge to get a current moving.