How much power is lost to the cathde

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Maciek Szymanski
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How much power is lost to the cathde

Post by Maciek Szymanski »

Continuing the plasma experiments I wanted to estimate how much power is lost to the fusor cathode.

Assuming that at the reduced pressures and temperatures of the cathode higher than 500ºC the convection can be neglected, the heat from the cathode can be lost only by radiation. At the equilibrium, when the cathode temperature is constant the heat lost by radiation must equal the head received from the ions impacting the cathode. Thus the power lost to the cathode is:

Code: Select all

                2
Pk = S⋅Φ = S⋅σ⋅T

where:
S total area of the cathode
σ Stefan Blozman constant 5.67e-8 W/m^2*K^4
T cathode temperature   
The cathode temperature may be easily measured with the optical pyrometer. Then the power may be calculated from the above formula and compared with the total electrical power consumed by the fusor.

The experimental setup is described in my previous posts, especially viewtopic.php?f=6&t=13921.
The pyrometer of Pyrolux II type with measurement range 700ºC to 2000ºC was installed on the tripod. A small flat mirror was installed above the fusor viewport to allow observation of the cathode through the pyrometer. The celling lights above the fusor were turned off during the measurements.


9E8CCE2C-53C2-45A8-8C8A-1D3174E247BE.jpeg

The pyrometer installed in front of the fusor.


B6F811FE-828C-4D75-AF34-D59990CC9EBE.jpeg
The detail of the observation mirror

The measurements were conducted for two types of the wire cathodes. Both were constructed from three 35 mm loops of the stainless stell wire. One was made of 0.8 mm wire and other from 0.6 mm. The radiating area of the cathode was calculated as:

Code: Select all

        2
Sk = 3⋅π⋅D⋅d

where:
D is the cathode diameter
d is the wire diameter
The calculated areas are 192e-6 m^2 for the 0.6 mm wire and 264e-6 m^2 for the 0.8 mm respectively. The transparency ratio of the 0.6mm cathode is 94.8% and 93.1% for the 0.8 mm.

First the measurements were made in air at 1.5e-2 Torr for different power settings with both cathodes. Then for the 0.8 mm cathode the measurement at the maximum power output of the power supply for different pressures were made. In all measurements the secondary side ballast resistor was not used. The primary side ballast was adjusted as needed. The voltage and current flowing through the fusor was also noted to allow calculation of the total power consumed.


A0B35FC8-B03E-494B-89E2-691A3979D6BB.jpeg
The precent of power lost to the cathode vs total power. Air, pressure 1.5e-2 Torr. Continuous line 0.6 mm wire, dashed line 0.8 wire.


3FD1861B-672A-4322-95DB-49EA72EBA61F.jpeg
The percent of power lost to the cathode and total power vs pressure. 0.8 mm wire cathode. Continuous line - relative power lost to the cathode, dashed line - total fusor power.

What can be interesting - the thiner wire leads to higher power lost to the cathode. It may be attributed to the stronger electric field around the wires allowing easier ion interception. Lower pressures and higher power also seems to reduce the loses to the grid. It may be suspected, that the loses are smaller not due to higher power but the higher voltage resulting in higher ion energies, so plotting the power loss vs voltage may be interesting.
Last edited by Maciek Szymanski on Tue Apr 06, 2021 3:25 am, edited 1 time in total.
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Maciek Szymanski
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Re: How much power is lost to the cathde

Post by Maciek Szymanski »

Here is the relation of the power lost to the cathode vs fusor voltage.


200C2CC2-5283-4C55-8D53-136F560DE3FC.jpeg
Power lost to the cathode as the function of voltage. Continuous line - power lot to the cathode as % of total power, dashed line - total power in watts, dash - dot line - pressure in torrs.

The reduction of grid loses at 1400V may be attributed to the effect of convection cooling. The measurement was taken at relatively high pressure of 0.1 Torr. At this pressure the assumption of purely radiative heat exchange may no longer apply.
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Richard Hull
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Re: How much power is lost to the cathde

Post by Richard Hull »

This chart is good, but doesn't show issues at fusor operational voltages. The shell is where all the power loss tends to occur, where your chart does, indeed, indicate as the grid losses dip at higher voltages and lower pressures. Shell losses are electron bombardment and neutral collisions.

Richard Hull
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Re: How much power is lost to the cathde

Post by Maciek Szymanski »

That seems logical. The numbers are relative - so if grid loses get lower (by percent) other loses are getting higher, assuming that 100% of fusor input are "loses". I suppose that most of the remaining 86% must be lost to the shell minus a small amount radiated by plasma. For the 0.6 mm grid when voltage is raised from 2 kV to 3 kV the power lost to the grid increases (in absolute numbers) by factor 2.2 while the total power by 3.4. So it means that at higher powers the grid loses gets less significant that the other loses. I have measured the grid loses as they are easier to measure than the shell loses.
Unfortunately I don't have a reliable way to measure the power lost to the solid cathode, as it doesn't heat to temperatures high enough to ne measured with the pyrometer. A thin walled sphere should do - that would be interesting if the the cathode transparency is needed at all?
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Re: How much power is lost to the cathde

Post by Andrew Seltzman »

Fantastic work. I have a liquid cooled grid on my fusor which can directly measure heat deposited in the grid, I was measuring 10-15% of the input power in star mode pressure regions, looks like your measurements are getting the same value. It's really cool that the two measurement techniques match up.

http://www.rtftechnologies.org/physics/ ... -index.htm
http://www.rtftechnologies.org/Design/A ... 5_phys.pdf
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Maciek Szymanski
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Re: How much power is lost to the cathde

Post by Maciek Szymanski »

That's gerat to see the the measurements agree, using different methods!
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Rich Feldman
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Re: How much power is lost to the cathde

Post by Rich Feldman »

Applause for the analog optical pyrometer! Can you get a picture of view through the eyepiece?
Hard to see how a more modern instrument could be much more accurate than manually matching the brightness of reference lamp filament and luminous target, when one partly eclipses the other in field of view.
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Maciek Szymanski
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Re: How much power is lost to the cathde

Post by Maciek Szymanski »

I like to realy on the simple instruments. They are easy to repair and calibrate with primary physical principles. I have a lot of bad experience with aging of plastics and electronic components as well as with interface compatibility problems related to the surplus bought modern (I mean 5-15 years old) equipment.

Here is the photo through the eyepiece of the pyrometer:

250D06A1-2C29-410F-90AB-F464609BE7C8.jpeg
0,8 mm wire cathode. 7e-2 Torr, 1800 V 55 mA, cathode temperaure 950ºC, with red filter, without ND filter

The field of view of the eyepiece is very limited, so it was very hard to make the photo. As I not used any camera attachment it was not possible to press the filament button, so the filament is cold. During the measurements I adjusted the position of the pyrometer so the as much as possible of the filament is coincident with the cathode wire. Again no way to do it with a hand held camera.
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Re: How much power is lost to the cathde

Post by Rich Feldman »

Thank you; beautiful picture & tells the story well.
I guess there's a wee risk of indicated temperature being increased by plasma luminance, but that would apply to any optical pyrometer (even one with an imaging sensor).

Sometimes I put too much effort into hands-free holding of camera, in place where human eye normally goes.
The fixturing in link below includes a popsicle stick and a walnut shell.
viewtopic.php?f=13&t=13363&p=86944#p86944
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Re: How much power is lost to the cathde

Post by Rich Feldman »

In calculation of radiant power emitted by cathode, I saw the familiar blackbody formula with constant 5.67e-8 in SI units.
Didn't see the emissivity factor, perhaps 0.6 for polished stainless steel at these temperatures,
and closer to 0.4 for tungsten (at ordinary lamp filament temperatures, and visible red wavelengths).

Could matter significantly when we figure the actual wattage radiated by hot cathode.

It's a smaller concern for the measurement of actual temperature using optical pyrometers. Indicated absolute temperature is perturbed by only the fourth root of the emissivity ratio between target and instrument filament, and fourth root of optical loss factor in windows and mirrors that aren't part of the instrument.
All models are wrong; some models are useful. -- George Box
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