"Soft enough to excavate with a diamond drill"
Oxymoron?
Sorry to drill you Axil..I sort of know what you are getting at
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any idea how such a little "glow stick" with 1mm diameter and 30 mm long can be manufactured / hand made? Not to think of several layers, quartz, ultra hard boron nitride, filled with oil (which kind of oil btw can be used at 2700°C ??), connected to line power and heat exchangers?
I don't know what the final dimensions are. He started with 100 Watts and something bigger. He told me earlier this year when I asked what the current dimensions were now, that they were 25 mm x 6 mm if I recall.
There is no point in speculating further. Boron nitride is one of the few materials that will put up with those temperatures. Quartz won't. I'm guessing there will be a low temperature version that is liquid cooled and a high temperature version with a tungsten sheath that isn't cooled except by radiation/convection. None of this is known for sure. (Except by the trolls who make up stuff about which they are certain,) Machining techniques for these materials are fairly well known.
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No.
Would you please elaborate on your answer a bit? Nothing too complicated. A simple example would suffice.
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Would you please elaborate on your answer a bit? Nothing too complicated. A simple example would suffice.
The voltage drop across the resistor only tells you the current going through the reactor, not the applied voltage across it.
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@Adrian Ashfield ,
As I thought I understood things, the total voltage in a series circuit is equal to the sum of the voltages across all the resistances.
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I don't know what the final dimensions are. He started with 100 Watts and something bigger. He told me earlier this year when I asked what the current dimensions were now, that they were 25 mm x 6 mm if I recall.
There is no point in speculating further. Boron nitride is one of the few materials that will put up with those temperatures. Quartz won't. I'm guessing there will be a low temperature version that is liquid cooled and a high temperature version with a tungsten sheath that isn't cooled except by radiation/convection. None of this is known for sure. (Except by the trolls who make up stuff about which they are certain,) Machining techniques for these materials are fairly well known.
It looks to me that Rossi found in his development that the smaller the reactor became, the more reliable that it became. So Rossi downsized the reactor as much as he could. The power density of a very small reactor is easier on the tube material.
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Kirchoff's Law: Sum of the voltages around =0
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Kirchoff's Law
Would covering that circuit with a 100 kiloVolt electrostatic field change the behavior of the circuit any?
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My impression from the photo was that the light came from a small hole (window) in the sheath. It wasn't clear enough to tell for sure.
From a window in the shielding they put in front of it. The light can be any colour required. (Yeah, me too!) The tube is AFAIK sapphire and very small and slender, the electrodes bright silver in colour, the max duty cycle without a heat exchange fluid is 10%. That's all I know.
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Would covering that circuit with a 100 kiloVolt electrostatic field change the behavior of the circuit any?
I don't think so
But a varying magnetic field would affect Kirchoffs assumptions.
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From a window in the shielding they put in front of it. The light can be any colour required. (Yeah, me too!) The tube is AFAIK sapphire and very small and slender, the electrodes bright silver in colour, the max duty cycle without a heat exchange fluid is 10%. That's all I know.
2,050 °C
Melting point 2,030–2,050 °C It is not Sapphire.
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As I thought I understood things, the total voltage in a series circuit is equal to the sum of the voltages across all the resistances.
I didn't think Rossi gave the total voltage. I would also like to see the voltage across the reactor: if it is a plasma and also generates electricity I suspect it is a little more complicated.
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So maybe I worded something wrong along the way...
0.1 V total volts, plus 0.1 V drop across the resistor, = zero volts, no?
Where voltage drop is -0.1 V.
Alternately, adding the voltages measured across the series resistors, when using a positive value, results in the positive voltage value of the supply.
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Why not? 10% duty cycle remember.
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@Adrian Ashfield ,
Rossi used the voltage drop (reported as a positive voltage) across the 1 ohm resistor to calculate the power consumed by the Quark. Of course it just calculates the power dissipated by the resistor, although it should give the correct current to the device (or for Axil's sake, maybe the current coming out of it).
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Why not? 10% duty cycle remember.
Where does that come from?
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A plasma is not the same as a simple resistor. It seems that the resistance of the reactor is at least "low." Did Rossi say the total voltage was exactly 0.1v? I would still like to see a direct measurement.
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Why not? 10% duty cycle remember.
Rossi said that the max temperature was 2700K or was it 3000K regardless of the duty cycle. But you can get that Info from the horse's mouth soon enough.
- Steven N. Karels
July 7, 2017 at 3:27 AM
Dear Andrea Rossi,
The posted operating point of the QuarkX reactor is about 2700 degC. This above the melting point of almost all metals. How does one heat the QuarkX and sustain that temperature?
a. Use of tungsten wire for resistive heating? If so, oxidation is of a concern.
b. Inductive heating? If so, then it requires the interior (fuel) container to be transparent to RF waves.
c. A special ceramic (or similar) material with the tungsten wires embedded within the material?
d. Other technique? - Translate
Andrea Rossi
July 7, 2017 at 7:08 AM
Steven N. Karels:
confidential
Warm Regards,
A.R.
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Andrea Rossi
May 9, 2017 at 8:16 PMSvein Henrik:
We have a temperature in the primary about 2700 C, so we can convert to any P and T of steam we want.
Any fluid can be used, also for example oil .
COP: enough.
Warm Regards,
A.R. - Steven N. Karels
July 7, 2017 at 3:27 AM
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I would still like to see a direct measurement.
On the surface it looks like the 1 ohm resistor is simply to prevent the power pack from seeing a dead short.
Seems to me it is possible to control the current output directly by the power pack.
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A plasma is not the same as a simple resistor. It seems that the resistance of the reactor is at least "low." Did Rossi say the total voltage was exactly 0.1v? I would still like to see a direct measurement.
There is a photo of two parallel voltmeters on the reactor in the Gullstrom-Rossi paper (v3).
