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Re: Thermal imaging of fusor grid / plasma

Posted: Thu Jan 24, 2019 11:15 pm
by ian_krase
You can likely get away with a smaller window.

Re: Thermal imaging of fusor grid / plasma

Posted: Fri Jan 25, 2019 4:31 am
by Andrew Seltzman
Hi Ben,

Seek thermal makes a usb thermal camera for a little over $100 ... :rk:2:pf:0

For the window look for ZnSe laser cutter lenses on ebay, you can get them pretty cheap. II-VI also makes 2" diameter lenses for laser cutters; these can easily be mounted in a rotatable 2.75" CF ring: ... rk:43:pf:0

A measurement of the plasma temperature would require a langmuir probe of thomson scattering system.

Re: Thermal imaging of fusor grid / plasma

Posted: Fri Jan 25, 2019 9:26 pm
by Ben_Barnett
That gives me some direction (and makes me grateful for Ebay once again).

With the ZnSe glass on the viewport, if I attach the thermal camera on the outside, I should be able to see the inner grid? I'm curious how the idea to use ZnSe glass came about. Is it reducing the luminosity of the light, or changing the wavelength?

Thanks for the help.

Re: Thermal imaging of fusor grid / plasma

Posted: Sat Jan 26, 2019 4:19 am
by Andrew Seltzman
The Zinc Selenide (ZnSe) lens will replace the glass viewport and seal to the conflat flange with a viton o-ring. ZnSe is transparent in the IR wavelengths, so it will let the thermal camera see into the fusor.

You can also use a germanium viewport like I am currently using on the latest upgrade:

Re: Thermal imaging of fusor grid / plasma

Posted: Fri Apr 05, 2019 4:39 am
by Rich Feldman
Kind of late to this party.
It's my understanding that thermal-infrared cameras and exotic optical materials, like Andrew has been talking about, are for target temperatures in the sub-red-hot range. Like the narrow liquid-cooled tubes in Andrew's grid.

Ben, have you noticed that many fusioneers use grid wire made from refractory metals, like tungsten or molybdenum?
That's 'cause they have enough high voltage power that a stainless steel grid would melt. Implies operating temperatures so incandescent that you'd need a pretty dark filter to safely look at with your eyes. Luminous enough to easily get the brightness temperature with ordinary visible-light cameras and optical materials.

You could make a science-fairish optical pyrometer, by arranging for camera to simultaneously view the fusor grid and a reference piece of the same kind of wire. Little mirrors might be useful. Reference wire is heated with ordinary electric current to make its brightness, thus its temperature, match the incandescent grid. The rocket science in imaging thermometers is their ability to focus and detect radiant energy, at wavelengths where glass is opaque and silicon is transparent or insensitive.

Here's a factoid from incandescent lamp research 100 years ago -- maybe I learned it from a CRC handbook. A typical tungsten filament's luminance is a bit lower than blackbody at the same temperature, because the emissivity is less than 1.0. Its color temperature is higher than the actual temperature, because the emissivity is spectrally tilted in the blue direction. I bet if the incandescent wire were hollow, with a tiny hole offering a view to the interior, that spot in a closeup image would fairly represent the blackbody color and luminance.

Thermal imaging of fusor grid / plasma

Posted: Wed Jun 19, 2019 11:34 am
by PatrickRothschild
Great Information sharing .. I am very happy to read this article .. thanks for giving us go through info. Fantastic post. I appreciate this post. This will consist of an array of small copper segments mounted to the grid with a standoff of known thermal conductivity, the collector to grid temperature difference will them be proportional to the heat flux flowing through the standoff.

Re: Thermal imaging of fusor grid / plasma

Posted: Wed Jun 19, 2019 7:11 pm
by John Futter
it is even easier than you make out all you have to know is the current flowing into the grid times the accel voltage = heat input
being in vacuum we are neglecting the neutron production as it is an impossibly small fraction of the power in and we are ignoring the conduction losses through the grid stalk (these are usually stainless a bad heat conductor

q = σ T4 A
q= power input
σ = 5.6703 10-8 Stefan Boltzmans constant
A=radiative area in sq meters
T4 = temp in kelvins raised to the forth power

edit to makeup for room temp T4 is modified to (T1^4 - t2^4) t1 = grid t2 = room temp

Re: Thermal imaging of fusor grid / plasma

Posted: Thu Jun 20, 2019 12:04 am
by Rich Feldman
I think you didn't factor in the emissivity of grid metal, which isn't even close to blackbody.
e.g. if e=0.4, absolute T needs to be about 25% higher to match the radiant power density of blackbody.
Let's not talk about what fraction of HV power is deposited on the current-sourcing surfaces, as opposed to the current-sinking inner grid.

Been meaning to follow up on my previous speculation about measuring the spectral emissivity of tungsten.
Discovered a marvelous and authoritative paper by Larrabee, at MIT in 1957. ... 9ec6c7.pdf

His monochromator and PMT were focused at a spot on a long straight tube of pure tungsten, with 0.125" OD and 0.001" wall thickness. Whose fabrication makes a good story by itself.
Blackbody reference, at nearly the same temperature, is a 0.013" round hole offering a view to the interior.
A detail I didn't think of is a zero-luminance reference, so scattered light reaching the detector can be subtracted out.
That's another 13 mil hole, that goes through both front and back sides of the luminous cylinder. The glass vacuum enclosure and filament placement are carefully designed to minimize reflected light in the all-the-way-through view.

Re: Thermal imaging of fusor grid / plasma

Posted: Thu Jun 20, 2019 1:23 am
by John Futter
Twas only the grid current I was concerned with as it is what heats the grid minus what is lost due to secondary electron emission no i did not factor in Emisivity of Tungsten now we can 0.471 ahead as a constant. Just need a handle of secondaries or suppress them to get an accurate count and you will get an arm wavey answer that is closer than an educated guess