Rossi was asked about the resistance of the Quark-X, and he answered “zero.” When his fans, even, pointed out that this would make it a superconductor, he then “explained,”
Superconductivity is a completely different thing.
Obviously my “zero” was not absolute, it was jargon for good conductor ( otherwise I wouldn’t write it in letters). I just wanted to say that it is a good conductor, like copper, so that its resistance ( that cannot be R = 0 ) does not affect the circuit.
Rossi is lying. That is, he is making up a story for why he said “zero.” There is quite another reason.
Dr Andrea Rossi
You cannot disclose the voltage across the Ecat QX because there lies the core of your industrial secret, correct?
From Russia, with love
Rossi answered “Right.”
From the Gullstrom paper, we have a claim that the current through the QX is 100 mA. If we knew the resistance, we would know the voltage. Therefore, if he wants to keep the voltage secret, and he does, he cannot disclose the resistance, which is certainly not “zero,” it’s a finite value that he did not want to say. Never mind that this makes the Gullstrom report unintelligible.
So now we have ele tossing more flabber into the fire. (Flabber burns with intense heat. Great stuff.)
Yes plasma has extremely high conductivity.
That article actually has:
|Electrical conductivity||Very low: Air is an excellent insulator until it breaks down into plasma at electric field strengths above 30 kilovolts per centimeter.||Usually very high: For many purposes, the conductivity of a plasma may be treated as infinite.|
Glow discharge describes plasma devices. A plasma device will typically have high resistance until a breakdown voltage is reached. As current flows, the gas in the tube ionizes and conductivity increases, and the more the current, the lower the resistance, so it is said to exhibit “negative resistance.” However, that is only the change of resistance with current. The resistance is always positive.
Consider an ordinary neon lamp. I used to play with these for hours on end. The breakdown voltage of the common NE-2 lamp is about 90 volts. However, once the plasma has been created, it can be maintained with a voltage of about 50 volts. To use an NE-2 lamp with 120 VAC, one would have a 100KΩ resistor in series with it. That’s called a “ballast resistor.” If is not there and the voltage reaches breakdown and the plasma starts up, and if there is significant available current, the tube will burn out.
Specification for the operating current for an NE-2 appears to be 500 μA. The voltage drop across a 100K resistor would be, then, 50 V. That’s enough to protect the tube from burnout.
Rossi has such a habit of lying that he didn’t just say, “that is proprietary information” for the beginning. Instead, he imagined that he could just say “zero” and maybe the question would go away. The current is 100 mA. In spite of what Rossi has said, this gives us no information about the input power. We are told it’s DC.
However, if the QX is a plasma device, it would have a breakdown voltage higher than the operating voltage. So we may speculate that he starts the device with a voltage spike, then the 100 mA is operating current. (There may be a limiting resistor not shown, could be in the power supply circuit; the effect of that would be a voltage spike, because the drop through the limiting resistor would be low until current increases.) Another possibility is that the power supply is set for constant current. I was just looking at specs for a supply that had a constant current mode that was voltage-limited at 100 V. One could use it at up to 600 volts total by putting a voltage supply in series with it.
One could use such a supply with what is shown of the QX in the Gullstrom paper, just set the constant supply for 100 mA. The voltage would increase to 100 v; if that was enough to strike the plasma, then voltage would decline to keep the current constant and would then presumably be DC, constant current and voltage. If 100 V was not enough for the gas characteristics of the device, then up to 500 V more could be added, current control would be maintained.
The resistance of that NE-2 neon lamp under operating conditions is very roughly 100 KΩ. So the idea that a plasma is necessarily a very good conductor is … misleading. If the current were allowed to increase, it would become a better and better conductor, but would be getting hotter and hotter until it burns out.
When air breaks down, when the field strength is greater than 30KV/cm, what happens? If that voltage is from a very low-current source, like a van de Graaf generator, it makes a hot but very fast spark. I had one as a kid. What fun! Zap! But if the current source has much higher available current, the arc would burn through anything. That’s lightning! What is the conductance of a lightning strike?
From the Wikipedia article, the channel current in a lightning discharge is on the order of 40 KA, and the potential gradient is “hundreds of volts per meter.” (This is far less than the strike voltage, 3 MV/meter.). So the plasma resistance is on the order of 75 ohms per meter. Not close to “zero.”
The Wikipedia article is misleading, applied this way. Very hot plasmas have very high ionization, and thus very high conductivity. The power dissipated in that lightning channel is perhaps 10 MW per meter, if I’ve done the math right. It’s damn hot!
So Rossi’s comments about resistance are still misleading, unless he has plasma so hot in there that no material could survive contact with it. I don’t think so. Lightning is about 30,000 degrees K. It would have to be hotter than that.
No, it’s very simple: he did not want to disclose the voltage (or resistance, same thing), so it appears he dissembled.
Might he be telling the truth, i.e., that the resistance is very low? It’s not impossible, he might have something that is generating high ionization other than input voltage. I can’t imagine what it could be, but “I can’t imagine” is never a proof of anything.
Meanwhile, this is clearly misleading, again:
I’d say the fact that a plasma is very good at producing light and heat radiation is an good indication in most circumstances its a rather poor conductor of electrical current.
You have not read even the wikipedia page I linked.
Plasma is quite a perfect conductor. Any electrical discharge (lightning, sparks….. and many others) is the demonstration of that.
Actually, above, I showed that the resistance of a lightning channel was on the order of 75 ohms per meter. If there is an error in that, I’d appreciate correction. Lear was correct, as long as we also understand what “most circumstances” means. Lightning is a mere 30,000 K. Get it hotter, it would be a better conductor. The Wikipedia page was misleading. It happens, all kinds of people edit Wikipedia, and sometimes gross errors get through. The statement that ele may be relying on was unsourced, and actually vague. In a chart comparing the conductivity of gases and plasma, plasma conductivity is said to be:
Usually very high: For many purposes, the conductivity of a plasma may be treated as infinite.
For what purposes? And what is “usually”? What is the average plasma? I have two plasma devices above my desk. It’s a pair of 40W flourescent bulbs. This kind of device may be the most comment, the most “usual.”
If the current is not limited somehow, connected to a constant voltage source, those tubes would burn out, because resistance would keep decreasing with current until the power dissipation in the bulb was enough to destroy it. So, in a way, we might think of the resistance is zero, but doing so would drastically underestimate the dissipated power.
Rossi’s “zero” implied negligible power input and his continued explanation (likening the QX to a copper wire) did the same.
It is extremely unlikely that the Quark-X is hot enough or otherwise ionized enough to have very low resistance. But, to be sure, if somehow Rossi was making it so, he might well want to keep it secret, but, then, what is the point of the information about current in the Gullstrom article? It’s useless without voltage or resistance information.
How long will this be going on?
From JONP, yesterday:
Dr Andrea Rossi,
In the Gullstrom-Rossi paper the measurements of the voltage were made by a couple of voltmeters, but recently you said here that now you are using an oscilloscope. Why this modification of the measurement system?
August 29, 2017 at 5:31 PM
The oscilloscope gives us more information abut the nature of the electric energy that arrives to the E-Cat QX from the control box.
Rossi did not state why this “more information” was needed. As is typical, he doesn’t really answer questions. What is actually implied here is not something that Rossi wants to talk about, at all, and he’s avoided the whole issue. What is the power input to the device covered by the Gullstrom paper?
Rossi does not want anyone to know, but apparently wants people to assume that the COP is fantastic, huge, really big, epic, amazing!
Experiment with energy measurement done with a spectrometer
Description of the apparatus:
The circuit of the apparatus consists of a power source supplying direct
current, a 1-Ohm resistor load, and a reactor containing two nickel rods with LiAlH4 separated by 1.5 cm of space.
During the test, a direct current was switched on and off. When the current was switched on, a plasma was seen flowing between the two nickel rods.
Again, DIRECT CURRENT (merely switched on and off).
It takes substantial voltage to strike a plasma. Operating voltage may be much lower, but typically not anything approaching zero. Striking voltage for an NE-2 neon bulb is 90 V, and to maintain the plasma takes at least about 50 volts.
The current was running through the plasma but the plasma was found to be charge-neutral from a Van de Graaff test. This implies that the plasma has an equal amount of positive ions flying in the direction of the current and negative ions(electrons) in the opposite direction.
It would be completely astonishing to have a plasma with substantial current that was not charge-neutral. A plasma is ignited when the gas in the device is ionized, creating equal numbers of positive and negative ions. This, then, becomes a conductor, from the very high resistance before breakdown (and breakdown voltage will depend on the actual gas and pressure. I think hydrogen, the gas expected, has a higher striking voltage, but haven’t looked at the specific conditions here. So what Gullstrom says here is just ordinary expectation for a plasma device.
To run a Van de Graaf test, the best I could figure it out, one would apply a high voltage electric field across the plasma conductor and observe the behavior of the visible plasma discharge. However, to see the discharge at an operating temperature claimed (over 2600 K) would be very difficult. First of all, it would be incredibly bright, and the temperature claimed was external temperature, the surface of the device, not the internal temperature. Maybe with welding glasses one could look at it, but … seeing detail inside, I don’t expect that to be possible. Gullstrom provides no details of what was actually done, with an experiment that would be, if it is as he claims, of world significance. Naive. Students have lost their career over mistakes like this.
Input: 0.105 V of direct current over a 1 Ohm resistance.
Of course, the unit of current is the amp, not the volt. This was a clumsy expression that should not be excused in a PhD candidate, which Gullstrom is. What he means is that the direct current through the resistor creates a voltage drop of 105 millivolts.
Energy output: The wavelength of the radiations from the reactor was measured with a spectrometer ( Stellar Net spectrometer 350-1150 nm ) and was integrated with the value of 1100 nm ( 1.1 microns ).
Plasmas are hot, typically, but the radiation from them is not black-body radiation. It is not clear what Gullstrom actually saw, how much of this report is from his personal knowledge, and how much is simply what he was told.
The temperature of the surface of the reactor ( a perfect black body ) was
calculated with Wien’s equation: 2900/λ (micron) = 2900/1.1 = 2636 K
As per Boltzmann’s equation, the effect is: W = σ × × T
4 × A
A = 1.0 cm2
By substitution: W = 5, 67 × 1012 × 0.9 × 4.8 × 1013 = 244.9
That kind of calculation, based on assumptions of black-body radiation, based on a measurement in one band, if this is what was done, is fraught with hazards, as the Lugano test showed. For such a measurement to be reliable, it would need calibration and there is no hint of that. For an experimental report as part of a theory paper, to rely on and report experimental results, depending on what he is told by someone who is known for being deceptive (as can be seen by reading the ample documents in Rossi v. Darden, not just relying on what I say or what anyone else says), is very dangerous for his future. Does he realize what he’s diving into? What establishes the reactor as a perfect black body? A plasma will emit narrow-band radiation.
The melting point of nickel is 1728 K. What would happen to those nickel rods?
Experiment with energy measurement done with a heat exchanger
The system is displayed in figure 5. In the figure, the yellow thermometer
measures the temperature of the oil inside the heat exchanger. In the left in the figure there are two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance. In order to not burn the equipmet [sic], the experiment was set in a state with lower output power.
Bold above was added in v.3 of the paper. This comment is standard Rossi, there is no indication of what it actually means. Yes, too much input power, one could burn out any plasma device, and, as well, if there is XP, that could burn it out. But this experiment is showing what is purported (implied) to be steady state, though there is nowhere near enough data to evaluate it.
Calculations of the calorimetry made by the heat exchanger:
efficiency of the heat exchanger:90%
v.2 had “10%” Which made no sense, but which survived through two revisions.
Primary heat exchange fluid: lubricant oil ( Shell mineral oil )
Characteristics of the lubricant oil: D = 0.9 Specific Heat: 0.5
Calorimetric data of the fluid: 0,5 Kcal/h = 0.57 Wh/h
Flow heating: 1.58 C / 1.8″ x 11 g
Apparently ” was used for “seconds.” Very non-standard in a formal paper, but maybe in Sweden….
However, this is “flow heating,” with a heat rise in a weight of fluid. The physical arrangement of the “calorimeter” is unclear, and what is shown in the photo is a temperature gauge, an ordinary pocket digital type, manual read-out, There is no indication of flow or how flow was determined. Calling this a “heat exchanger” is, then, an unclear usage.
So maybe the pipe T that we see is full of oil. So how fast the oil temperature rises is then a measure of heat. A very crude measure, given what we see. Not flow calorimetry.
Basically, the claim here appears to be for a power of 0.57 W, based on heating the fluid 1.58 degrees C per 1.8 seconds. This is a very, very odd report. Trying to read that thermometer in 1.8 seconds, forget about it.
No, what would be done is to observe a higher temperature rise, for so many seconds, and then divide by the seconds, to get a per-second rise. but, then, why would this be reported “per 1.8 seconds, unless that were the actual measurement time.
I haven’t done the calculation, because without input power, this is all an exercise in meaninglessness.
Resulting rating: 20 Wh/h
Energy input: V=0.1 R=1 Ohm → W=0.01
That is not “energy input,” and this is totally obvious. That would be the energy input if the resistance of the plasma device were zero. However, plasmas are not that conductive, particularly at these relatively low temperatures (and, yes, these temperatures are low as plasmas go.
Instead, that power is the power dissipated in the 1 ohm resistor. Many have pointed that out. Yet this persists in the paper.
To strike the plasma might be taking 200 V. If that voltage continues across the device, it would probably burn out, plasma devices will always have current limit resistors (called “ballasts”). Or there is another possibility: The power supply is a constant-current supply. Such a supply commonly would provide high voltage until current begins to flow. Maybe this device depends on substantial nickel evaporation, takes time to warm up. Meanwhile, 100 mA at 200 V, why, that’s 20 W!
Back to the point here, an oscilloscope. That would show the time-behavior of the input voltage (or current if the scope is looking at the 1 ohm current shut, which is what that resistor is).
An oscilloscope, with direct current, what Gullstrom and Rossi claim, would show a straight line with much less precision than a meter. No improvement in information. So Rossi is essentially acknowledging that the power input is not “direct current,” not that simple. It varies, during the experiment.
Gullstrom was probably told DC and that’s what he reported. And looking at the meters in the photo, I was struck by the complete oddity of using two voltmeters to display the same voltage. I have never, ever, seen that done. Common voltmeters are very reliable devices, and if one is suspected of monkeying with one, well, one could monkey with two the same. It would prove nothing to use two. Instead, what is done if someone wants to check a test is to pull out their own voltmeter. Mats did that in some Rossi tests.
No, I find it likely that the second meter was being used to measure voltage, as one possibility, and then Rossi didn’t want to show that, so he changed the meter setting. As well, the meter might have been set to show AC current. Imagine this: input current, 0.1 A DC. A DC meter will generally show DC offset and not the AC component. So one could have as much AC in there as one wanted. Very simple. Of course, an oscilloscope would show what was really happening!