E-Cat QX Picture Posted in New Rossi-Gullstrom Paper (COP of 2000 reported with Calorimetry)

Thanks to Observer for pointing out a new paper that was published has been added to the Rossi-Gullstrom paper that was published on July 18 on Arxiv.org here. The title of the article is “Nucleon polarizability and long range strong force from σI=2 meson exchange potential”

It is a very involved theoretical paper for the most part. Here is the abstract.

“We present a theory for how nucleon polarizability may be used to extract energy from nucleons by means of special electromagnetic conditions.
Also a new theory for a long-range strong force is introduced by enhancing the role of the σI=2 meson in nucleon-nucleon potential obtained
through isospin mixed σ mesons. The novelty in the idea is to let an imaginary mass exchange particle be enhanced by absorbing only
one particle in an entangled state of two particles. The imaginary mass particle is not intendent to be free and contravance the law of physic; it is merely included as a binding exchange particle in a system with total positive invariant mass. In order to validate part of the theory, we introduce an experiment that in many ways have motivated this study”

At the end of the paper is an appendix which reports on two experiments, the first has already been reported in a previous paper, the second is one that has used calorimetry as the measurement system.

“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 is two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance.

Calculations of the calorimetry made by the heat exchanger:
efficiency of the heat exchanger:10%
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
Resulting rating: 20 Wh/h
Energy input: V=0.1 R=1 Ohm → W=0.01
The COP of the system with the calorimetric measurement is substantially conciliable with the measurements made by the Wien’s equation and the Boltzmann equation.”

On the last page is an image of the experimental setups.

  • Engineer48

    COP 2,000!!!!!

    • Buck

      For me . . . these results taken as true, repeatable, and verifiable by a 3rd party simplify my evaluation of the now settled lawsuit.

      Also, I am like you . . . the notion of a COP = 22,000 simply boggles the mind.

      I have heard that some of the more efficient heat exchangers/transformers are in the range of 30-40%, pointing to optimized industrial versions of the E-Cat QX being installed with an projected COP in the expected range of 6600 to 8800.

      Can you imagine the size and efficiency of the power plant if you were driving a supercritical CO2 turbine.

      WOW ! ! ! !

      • Richard Hill

        Surely there is Defense interest in these figures. What better way to power a laser weapon?

        • Buck

          The DoD is already on record for recognizing that an operational LENR platform would change everything: international relations, international commerce, international stability, and of course their domestic analogues.

    • Observer

      So, what is the COP if you negate the power dissipated across the 1 ohm ballast resistor?

      Should we quote Buzz Lightyear at this point?

      • GiveADogABone

        Infinite, if the QX superconducts? The resistor and associated current is just fitted to stabilise the QX and is a parasitic loss?

  • Björn-Ola

    I don’t understand the calculation but it is good to have two different measurements supporting each other.

    • Engineer48

      COP = 2,000 was done via flowing water through a 10% efficient heat exchanger.

      • Björn-Ola

        Yes I understand that but I don’t understand the calculation.

        A typo, oil, not water.

        • Thomas Kaminski

          The calculation a simply that required to estimate the energy out (in joules), give a temperature rise over a period of time by using a fluid with a known heat capacity. The so-called “bomb calorimeter” enclose a reaction chamber surrounded by a fluid (usually water) in a well thermally insulated case and measure the temperature rise in the fluid as the reaction goes to completion. For caloric content of food, it is burned in the chamber along with an oxygen source, causing a known quantity of the fluid to rise in temperature. See this Wikipedia entry for a description:

          https://en.wikipedia.org/wiki/Calorimeter

          The basic principal is that the temperature rise in a fluid is proportional to the heat absorbed times the heat-capacity of the fluid. The heat capacity of the fluid is assumed to be a constant (per unit mass) times the mass of the fluid. Twice as much heat evolved will cause twice a much of a temperature change.

          The rate of temperature rise (delta-Temperature divided by time) is related to the power generated. A constant power produces a linear increase in temperature over time.

          Rossi is simply measuring the temperature change over a fixed time for a known amount of fluid (the oil) and using that to estimate the power generated.

          • Björn-Ola

            Thank you Thomas. Yes I understand that. I”m doing that all the time in my work.
            Multiplying the flow with the heat capacity and the delta t.
            But I didn’t understand their figures and calculation.
            But here it is:
            The heat capacity for mineral oil is 1.67 kJ/kg,K or 0,4 kcal/kg
            http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html
            Is the flow 11g/1,8sek? That is 0,0061kg/s
            1,56°C x 1670 x 0,0061= 16W

            Using heat capacity 0,5 kcal/ kg instead of 0,4: 1,56 x 2090 x 0,0061 = 20W
            Right on spot!

            I don’t know about the heat exchanger efficiency 10%
            Do they mean that the oul captures only 10% of the heat?

          • Thomas Kaminski

            That might be related to the ratio of heat capacity of the whole device relative to the heat capacity of the oil. I am not sure where it comes from. If the device plus oil heat capacity is considered, then the temperature rise will be less than that of the oil alone.

            One other question I have is how was the device operated. I am not familiar with the shown temperature probe, but my guess is that it has a time constant of at least several seconds. If so how could Rossi find “1.58 C” rise in 1.8 seconds? I suspect that they instead applied the inout DC current/voltage for 1.8 seconds and then waited a few seconds until the temperature stabilized. SO, my guess is that it is operated in a “pulse” mode. What do you think?

          • Björn-Ola

            I think they measured 11 g in 1,8 sek and they calculated the flow from that, just as I did. A bit odd. The
            temperature is constant. It has to be to make this calculation. 1,58C is the difference between the temperature in to and out from the heat exchanger, delta t.

          • Björn-Ola

            I’m sorry, I didn’t study the picture, silly me. There is no flow and only one termometer, so it is obviously that they meassured the speed of the rising temperature.
            My calculation is still valid though.

  • GiveADogABone
  • Engineer48
  • Thomas Kaminski

    I am confused by the calorimetric results. Is the fluid (oil) stationary or moving? If it is moving, what is the mass flow rate? Also, I do not see how the “plasma” is formed with a 0.1 volt applied source. When the paper talks of a 1 Ohm “brown” resistance, is this the resistance of the Ecat-QX or a separate resistor?

    It would be nice to have a schematic so I knew what is being applied and what electrical characteristics are being measured.

    • GiveADogABone
      • Thomas Kaminski

        I did — see above. As for the calorimetry, what the heck is the meaning of “Flow heating: 1.58 C / 1.8″ x 11 g”??? Mixing inches and grams confuses the heck out of me. Perhaps the ” is really supposed to be seconds. That is, the DeltaT is 1.58C in 1.8 seconds with 11 grams of oil…..

        • Engineer48
          • Thomas Kaminski

            No flow. I think I have it figured out. He has measured the temperature rise in Degrees C with a 1.8 second time period. Given 11 grams of oil with a heat capacity of 0.5 (Calories/(gram-DegC)). The result is:

            8.69 cal = 0.5*11*1.58
            36.4 joules = 8.68 cal * 4.184 joules/cal
            1 watt = 1 joule/sec
            So: 20.2 Watts = 36.4 joules/1.8 seconds

            I don’t know what to do with the “heat exchanger efficiency”

          • Engineer48

            Hi Thomas,

            Ok that seems to work but yea where is the 10% efficiency?

            Seems we need to follow the bread crumb trail.

          • Buck

            What about all the heat/energy being absorbed by the green plastic fitting and not by the oil?

          • Thomas Kaminski

            That might be it. If you are just measuring the oil, but the rest of the apparatus is also absorbing heat, that will lead to an error.

          • Buck

            Assuming they are just slapping their testing apparatus together to get a ball park measurement . . . it is easy to consider that they skipped the insulation. Further, when you are dealing with an original COP = 22,000, engineering for 100% precision and accuracy may not be your first priority.

          • Thomas Kaminski

            I really don’t know how you use “heat exchanger efficiency” in a calorimeter. Normally you assume that the calorimetry is well insulated and all of the heat is absorbed by the fluid.

          • GiveADogABone

            Heat exchanger efficiency is a Second Law of Thermodynamics concept and relates to dropping heat through a temperature difference. Entropy is a related concept.
            https://en.wikipedia.org/wiki/Entropy

            Thermodynamic Efficiency”[of a heat exchanger], e, is entirely dependent on terminal temperatures and nothing else:
            e = (t2 – t1) / (T1 – t2), where T1 is the temperature of the hot fluid at inlet, t1 is the temperature of the cold fluid at inlet, and t2 is the temperature of the cold fluid at outlet.
            Source: Process Heat Transfer by D.Q. Kern, pp170–171, and others.
            https://www.quora.com/How-do-I-calculate-the-efficiency-of-a-shell-and-tube-heat-exchanger
            Peter England, Heat transfer engineer in oil & gas for 25 yrs

    • Engineer48

      Hi Thomas,

      Seems the Brown resistor is in series with the PSU and QuarkX input. Is used to measure circuit current via voltage drop across it with one meter and the other measures the voltage between the 2 QuarkX rods.

      • Thomas Kaminski

        As GADAB suggested I went back and read the comments. The 1 ohm is the resistance of the QuarkX according to Rossi. I am still confused about the calorimetry. Also, there must have been a meter to read current into and voltage across the device. It is not clear from the picture how each of the two meters are connected.

  • Engineer48
  • Engineer48

    This has my interest.
    Is this a QuarkX reactor in the flesh?

    https://uploads.disquscdn.com/images/742660de20073442432fbc9e52748ff40e834886583ade570ab5199b050319da.png

    On a wooden board?

    If this is a raw QuarkX reactor with a COP 22,213, my GOD what has Rossi done?

    • Thomas Kaminski

      Hmmm. Nice observation. Looks like the test apparatus without the calorimetry. That must have been the one used to make the radiation measurement.

    • Stephen

      Yes nice! I wonder if it was the dummy unit or the device used for radiation measurement as Thomas mentions below….

      Can we work out the dimensions from this picture? I suppose the active element is the clear part between the two sections.

      It does look beautiful. It must be stunning in operation.

      The clamps look pretty strong and robust. I wonder if that is a requirement.

      • Observer

        Those aren’t clamps.

        • Stephen

          Ahh ok thanks… I was wondering.

          So I guess they are just connectors for the rods to the QX.

          It looks like really good well designed apparatus.

          Im really happy to see this I can hardly wait now for October.

      • LukeDC

        That is a 50Mhz Tektronix TBS1052b in the background so I suppose you could use its dimensions to work out the exact size of the QX

    • Stephen

      The bolts on the second grey box with the second device unit look like they could be mountings for the device too so it could be mounted like first one bieng tested.

      I wonder if the grey box contains all the electronics to control the device. If so that is a stunning piece of self contained kit. You could take one of those plonk it front of any scientist or engineer switch it on then just sit back and say there you go! … explain that!

    • Frank Acland

      Yes, this is the description from the first Rossi-Gullstrom paper:
      “Description of the apparatus The circuit of the apparatus is made by a power source to supply direct current, a load made a 1 Ohm resistance, a reactor containing two nickel rods with LiAlH4 separated by 1.5 cm of space.”

      From this angle we can see just one of the nickel rods.

    • US_Citizen71

      The oil filled “T” is sprinkler system parts as well. https://hydrorain.com/products/blu-lock/1-blu-lock-fittings It appears to be an older BL 402-130 in light green.

  • DNI

    I don’t understand how the Energy input is calculated. If it’s a 1 ohm resistor in series with the QuarkX and the voltage is measured over the resistor. Then It seems to me like it’s the input power to the resistor that is calculated. Not the input power to the QuarkX.

    • US_Citizen71

      You measure the voltage across the entire system and subtract the voltage across the resistor leaving you with the voltage across the QX. This was likely done so that the resistance of the resistor can be measured independent of the setup for verification. The reactor resistance will likely measure as an open circuit when it is unpowered and not up to temperature.

      • DNI

        But it seems like only the voltage across the resistor is measured. I agree it would be possible to calculate the power in to the QuarkX if we have the voltage over the resistor and the voltage over the QuarkX (or the total voltage). But it doesn’t seems like this is how it is done.

        • US_Citizen71

          It is difficult to tell what is being measured in the picture. It appears that both sets of leads from the multimeters are measuring across the the reactor on the right. It maybe a staged shot that is not wired as it would be for a test. Rossi is a master at obfuscation so I wouldn’t read into any image he puts out too much.

          • DNI

            It’s in the report “In the left in the figure there is two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance.”

          • US_Citizen71

            Which page?

            edit: never mind found it.

          • DNI

            In the Appendix, page 18.

          • US_Citizen71

            Your quote states “the mV of the current passing through the 1 Ohm brown resistance.” It appears he is using lamp cord or such to be his resistor for whatever reason. I can’t tell how the brown cord connects You can barely see a bit of red on the right, I initially assumed it was an alligator clip that should lead back to the meters, I also assumed it connect directly to the reactor. I may be wrong on both assumptions. There is likely a multimeter across the power source, which gives the voltage of the system. More than one way to skin a cat.

          • DNI

            If there is one voltmeter measuring across across the reactor V1 (or across the power source) and one measuring across a 1 ohm resistor (connected in series with the reactor) V2. Then it would be easy to calculate the power in to the reactor as:
            I = V2/R= V2
            P Reactor = V1*I = V1V2

            But for some reason this is not how it is described in the report. In the report they have calculated the power to the 1 ohm resistor.

            Of course if the resistance of the reactor also is 1 ohm then the value would be correct. But there is now way to know the resistance of the reactor with the setup described in the report.

          • DNI

            If there is one voltmeter measuring across the reactor V1 (or across the power source) and one measuring across a 1 ohm resistor (connected in series with the reactor) V2. Then it would be easy to calculate the power to the reactor as:

            I = V2/R= V2
            P Reactor = V1*I = V1V2

            But for some reason this is not how it is described in the report. In the report they have calculated the power to the 1 ohm resistor.

            Of course if the resistance of the reactor also is 1 ohm then the value would be correct. But there is no way to know the resistance of the reactor with the setup described in the report.

          • US_Citizen71

            There is also no way to know that the resistance of the reactor is not 1 Ohm.

          • Thomas Kaminski

            It sure would e a lot simpler if the current into the reactor and the voltage across it were both measured. As it is, we have to take Rossi’s work that the resistance is one ohm. The formula for electrical power in is V^2/R. For R=1 Ohm and v=0.10 volts, P = 0.01 Watts.

          • US_Citizen71

            Oh I completely agree!!!

    • Rossi has earlier clarified that 1 ohm is the input impedance of the device (= reactor).

      • DNI

        Do you mean that the calculation of the power in to the QuarkX is based entirely on Rossis claim that the input impedance is always 1 ohm. That sounds like a very stupid and risky way to calculate the power.

        • Not entirely on Rossi’s claim in this case, but on Carl-Oscar Gullström’s and Andrea Rossi’s report. They assume 1 ohm for their calculations. Everything is risky in this business. Even stupidities can appear. You are right.

          • DNI

            I can’t understand why there is no measurement for either the total input voltage or the voltage over QuarkX in the report. With any of those voltages it would be easy to verify the input power. As long as this is not corrected or explained I can not take this report seriously.

          • On page 19 of the report, you can see 100 mV reading on the voltmeter display, probably measuring the total input voltage over the QX. On page 18 it is explained: “Energy input: V=0.1 R=1 Ohm → W=0.01”. Further, 1 ohm is assumed as the input impedance of the reactor according to Rossi on JoNP:
            “March 24, 2017 at 6:54 PM
            Michel:
            It is the input impedance of the reactor.
            Warm Regards,
            A.R.”

          • DNI

            Page 17 in the report:
            “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.”
            “Input: 0.105 V of direct current over a 1 Ohm resistance.”
            Page 18 in the report:
            “In the left in the figure there is two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance.”

            To me there is no question that the report say that they measured the voltage over the resistor. And use this voltage to calculate the power in to the reactor. Which doesn’t make any sense.

          • Sorry to repeat myself, but if you think like this:
            “a 1-Ohm resistor load” = input impedance of the QX (as confirmed by Rossi) and
            the load is resistance-only for DC, then it makes full sense and matches the calculations on the report.

            But if you consider “a 1-Ohm resistor load” = a separate resistor (not confirmed), it doesn’t make any sense, as you said.

          • DNI

            Maybe im splitting words here. But it say a 1 ohm resistor load AND a reactor.

            But if you are right we don’t know how the resistance of the reactor is measured. It would be better to actually use a resistor in series and then measure the voltage across both reactor and resistor. Then we would have everything we need to calculate the power to the reactor. Without the need to trust that Rossi measured the resistance of the reactor undocumented but correctly.

          • Thomas Kaminski

            Agreed. The picture in this report, however, shows both meters reading the same voltage. I suspect that it is the voltage across the reactor, but can’t confirm that from the picture.

          • DrD

            Those concerns are very valid. Why ever can’t he be unambigously clear. There’s room for serious error here.
            I suppose the only saving grace is that he would need a massive error to negate such a positive COP.

          • Fully agree.

          • A separate brown resistor was found in the picture! You were right. Now it seems they approximating the input power based on the assumption that the reactor has a constant 1 ohm impedance as well. Anyway, this is a report about the theory, not COP. Maybe this is enough to study the behaviour of the reactor.

          • DNI

            Hard to understand why they didn’t measure the voltage across the reactor or the total input voltage then. Maybe they did but then why not put in the report?

            Reminds me a bit of the demos where all water was assumed to be vaporized because Rossi said so. Here the resistance of the reactor is assumed to be 1 ohm because Rossi said so.

        • US_Citizen71

          If…
          Vtotal = 200mV
          Rbrown = 1 Ohm
          Vbrown = 100mV

          Vtotal=Vbrown + Vreactor
          200mv=100mV + Vreactor
          100mV=Vreactor

          Prove Vtotal 200mV

          If you can’t than you are simply assuming Rossi/Gullstrom lie when they state the QX has a resistance of 1 Ohm. The paper needs more detail obviously, but don’t you think it a bit premature to call them liars?

          • Thomas Kaminski

            What you show is one way to use a known resistor (Rbrown) of 1 ohm to and a voltmeter calculate the current. However, the picture shows both meters connected to the same terminals. Perhaps they were just cross-checking the meters to see if they agreed on the same measurement voltage.

          • US_Citizen71

            Double measurement was my assumption.

          • US_Citizen71

            What I showed was 2 known measurements and a WAG at another, which is exactly what DNI has been doing up and down the thread. My math shows how to calculate an unknown voltage drop of an unknown resistor with the known total (my WAG) and a known resistor in series with the unknown resistor.

          • Engineer48

            Hi Thomas,

            I suspect you are correct and the parallel meters was just a check of the meters.

          • Thomas Kaminski

            And they differ by 0.4 millivolt out of roughly 100. That is typical of the accuracy of a good hand-held meter.

          • DNI

            I have not called them liars.

        • If the input is 10mW then it will be a simple matter of breaking the continuity and running an ampmeter measuring current.

    • Andreas Moraitis

      I would expect the resistance of the QX to be negligible as long as there is plasma between the electrodes. In this case, the voltage across the whole system would be very close to the voltage across the resistor. However, before the plasma state is reached, the resistance of the QX will be orders of magnitude higher. Does the plasma break down between the DC pulses? Maybe it doesn’t if the gaps are very short, but that would be information of interest.

  • Hhiram

    Arxiv.org is an open-review pre-publication platform, so I would suggest that the community here try to compile a list of questions and criticisms, and then convey those directly to Rossi and/or Gullstrom so they can revise the paper accordingly.

    Frank: you might invite Gullstrom to participate directly in the forum discussion here as well.

  • Guru Khalsa

    Let me see if I got this right. We have 2 Quark Xs, one free standing and one connected to a heat exchanger. the free standing one is metered. But what is the big box it is sitting on, the control and battery? Hopefully the demo will be more transparent. Still its nice to see pictures

    • Guru Khalsa

      On second thought it looks like both QXs are metered hence the two readings.

      • Engineer48

        Hi Guru,

        QX reactor above the 2 meters is not powered.

    • Guru Khalsa

      Another thought: since the QXs are suppose to both create heat and DC can we say that OHM’s law applies?

  • Rene

    Yes, if the 0.1V is an average. A 10,000V pulse duration of 1/100,000th of a sec, once a second. Or, 10KV duration 1 microsecond 10x/sec.

    • CWatters

      In which case their choice of meter to measure the average voltage is dubious.

      • Rene

        Once the plasma arc is struck its resistance drops to near zero. So long as sufficient current (or other reactions) keep the plasma hot, no further starter pulses are necessary. My WAG is that the high voltage starter pulses are all of a tens of milliseconds to get the plasma started.

  • Karl Venter

    Are we going to have secret boxes with wires coming out like in the photo in the demo?
    I would like to see the demo of the qx heating a drum of x amount of litre with a delta t from beginning of test to end
    Flow is difficult to measure accurately at low rates let alone 100mV
    I have a Fluke and a cheapie voltmeter and they are 10% different at 1 volt?

  • Cars will be run on Stirling cycle engines if this turns out to be real.

  • Engineer48

    The 1 Ohm Brown Resistor has been found:

    https://uploads.disquscdn.com/images/1d5a6e9fb45a8c0a6b4b9a4b4596c5cd06d0106a2c85ac8aac0ebf520200120f.png

    Note the bare wire that connects it to the 2 pin female connector.

    Would now assume it is in series with the QX reactor input and that one of the meters measured the voltage drop across the Brown resistor to determine the current passing through the QX reactor.

    • Great! And if we dare to trust that the reactor input impedance is 1 ohm as well, the gauge reading will allow us to calculate the power input as done in the report.

    • Thomas Kaminski

      Nice find. I guess my “old man eyes” can’t hold a candle to yours.

      • GiveADogABone

        Get some good graphic image handling software for your computer and a hi-res screen will help?

        • Engineer48

          Hi GADAB,

          For sure that helps my old engineers eyes.

          Have a 32 inch 1,920 x 1,080 monitor.

          • GiveADogABone

            There is a Matrox VDU card with six outputs, so you could get to 11,520×1080 by putting the screens side by side or 5760×2160 for a squarer look. Thin edged VDUs would help as well.

          • US_Citizen71

            70″ UHD TVs work quite well too!

        • Thomas Kaminski

          I suspect that it is more the mental state that finds things hidden in the image. Engineer48: Did you previously work for the Australian equivalent of the Defense Intelligence Agency as an image analyst?

    • Chapman

      Yes. It is a simple Ballast Resistor.

      But that means we are likely misinterpreting the data given…

      • Engineer48
        • Chapman

          Of course. BUT, I believe he is driving it with a DC Constant Current Source.

          He WANTS 100mA through the reactor. We do not know the initial cold resistance of the LAH. The monitored 100mV on the ballast verifies the series current, but who knows what the initial voltage was. The resistor will ALWAYS be 100 mV, but that would not be the applied voltage.

          My point is that we do not know the drop on the reactor – and therefore do not know the power, or the COP. We only know the series current, as attested to by the Ballast voltage drop.

          You show the second meter on the reactor, but the wiring in the pic is sketchy.

          • Engineer48

            Hi Chapman,

            We don’t know what Rossi wants through the QX reactor.

            What I do know is using 2 meters to me screams a standard measurement technique, one meter measured the series current via the voltage drop across the 1 ohm resistor and the other meter measured the voltage across the QX reactor.

            Also remember that neither author is a native English speaker nor writer, so their English may not be a clear as what we may write.

            However I’m very sure both authors know how to measure the input power consumed by the QX reactor, so I’ll continue to support their claims.

          • Chapman

            I am not disputing their claims in any way. I am wondering what the triggering voltage threshold is.

            The device is clearly current dependant. The required voltage is just a secondary requirement, but one needs to know it to determine input power. There is NO way that the reactor resistance is one ohm, hot and cold, constant….

            Rossi knows his stuff. But he is also not divulging everything. AND GOOD FOR HIM. I am not criticizing. It is up to us to fill in the blanks and make certain logical deductions as to what he is up to.

          • GiveADogABone

            Checking out the meter leads in the photo, they join at two three-way points on the table top to the right of the meters. Two leads then go away. The black lead can be seen and goes to a terminal on the QX and the red lead is hidden but is close to the hidden end of the 1 Ohm resistor.

  • Engineer48

    Nice image of the inside of the QX reactor.

    Note clear housing and what appears to be an round electrode with a flattish end.

    https://uploads.disquscdn.com/images/ed3ff2ffc65a075aa14b0f9785eb49365c4bdc40470cf6b5256aa7b8492d007f.png

    • artefact

      On the “flattish” end there appears to be a black flat dot. Fuel of the QX?
      I always thought that the electricity trigger needs to go directly through the fuel.

    • Stephen

      Hi Engineer48.

      Elsewhere on another forum some weeks ago someone indicated that plasma lamps might be related in some way.

      These lamps use electrodes or microwaves to excite and control the plasma to emit light. They typically use a mixture of Metal Halides that are ionized and heated to very high temperatures to ensure light is emitted from the released and excited halide ions at various colour s.

      https://en.m.wikipedia.org/wiki/Plasma_lamp

      https://en.m.wikipedia.org/wiki/Sulfur_lamp

      https://en.m.wikipedia.org/wiki/Electrodeless_lamp

      I wonder if the E-cat QX is a synergy between this technology and his previous LENR technology. Both at engineering level in the electrodes and materials but also using Metal Hydrides in the plasma instead of Metal Halides.

      Incidentally I think these types of lamps some times need a resistor load. Could this be another reason for the presence of the resistor in the E-Cat QX apparatus? or is it something else like just for measurements.

    • Ged

      A dusty plasma LENR reactor?

      • Engineer48

        Hi Ged,

        For those willing to follow them and learn, Rossi drops breadcrumbs.

      • Stephen

        Yup I think this is close to axils earlier ideas and they look good possibilities to me.

        Or alternatively if we consider the plasma lamp anology. maybe when in operation instead of using external applied high voltages or microwaves to generate the plasma the ions are generated directly or indirectly by LENR in the Rods and the heat and light comes from of the plasma like in a normal plasma lamp?

        • Chapman

          You know what REALLY irks me? If you read the paper carefully, it sure looks like what they are proposing is REAL close to one of Axil’s way-back posts suggesting teleportation of nucleons between adjacent atoms…

          Oh crap…

          And here I gave Axil all kinds of crap about it!

          Now, Rossi is suggesting that a dramatic stimulated increase of the range of the strong force allows the sorting and transfer of nucleons between adjacent atoms, and as such there is never any true emitted radiation, but rather a set of targeted and direct transfers. But that is still close enough to qualify as being basically the same as Axil’s musings.

          I hate this!!! It burns me to the core to have to do it. But…

          SORRY AXIL!!!! (wherever you are)

          [I hope he never reads this, because he will rub my nose in it every time I criticize him from here on…]

          • Stephen

            Hehe you know. Axil is often one step ahead of us. I’ve seen this time and time again. I respect his out of the box ideas more and more these days.

          • Chapman

            Yeah, I know… But it still BURNS!!!

            So, for HONOR’S sake, I will admit to YOU that he was right, but you have to promise to keep it between us, ok? I would not want him to get all puffed up, and getting too big for his britches…

            Besides, the theory may be bogus. It looks a little bit too much like grasping at straws, and reeks of desperately seeking ANY valid explanation. Way too many IF’s in the equation. But it still proves that Axil was thinking along the same lines as Rossi, and that counts for a lot. So, his idea may yet turn out to be wrong, but his THINKING appears to have been valid, so I give him kudos… (now I have to go find some pepto-bismol, cause just sayin that outloud gave me heartburn).

          • Stephen

            I understand. I will keep quiet for honors sake 😉

            Thanks by the way for bringing clarity in this thread. I appreciated your post about what is known facts as opposed to what we are guessing.

          • Chapman

            I would not have spoken up about it, but I kept seeing Mats asking a simple question, and he was dead right, but he kept being put off and dismissed.

          • Bob Greenyer

            Piantelli – “and then there is a reorganisation of nuclei”

            If the clear device in the photo is the QX, then it looks a lot like shoulders plasma focus device “Single EVO Capable of >50 KeV Output From 1KeV Input” from ICCF10

            See page 22
            https://goo.gl/M9tayN

            Shoulders said that EVOs mix all the nucleons up and they settle out.

          • Chapman

            Well I’ll be damned!

            Thanks for the heads-Up!!!

            I must admit that I love the idea that it is not random radiation causing a cascade of secondary effects, but instead a very finite point-to-point interaction under the control of a defined force. It makes things cleaner, and once fully understood will lead to a much more predictable and CONTROLLABLE reaction.

  • georgehants

    All very interesting but for genuine Cold Fusion followers an open, repeatable, confirmable demonstration of a device showing a COP above one is what is needed.
    This just gives more things to talk around and around ad nauseam.
    The World waits and suffers while Rossi plays his silly little games.
    He is like a person mumbling on for seven years that they have found a cure for all cancers and releasing nothing for those that need it.

    • Buck

      George,

      your opinion, especially that of the last sentence, seems to come from an inability to empathize and imagine the challenge of discovering a new view of physics and the consequent technology, discovering how to refine and gain control of this new form of physics that does not come with a manual, contesting investors who arguably are looking to take control of your revolutionary technology with selfish greedy intent.

      You honorably argue for compassion for those billions in need, for a level of charity where Rossi simply give his technology away, and for ignoring the simple reality of the forces arrayed against a ~ 1x / 5000 year revolutionary energy source. You ignore an apparent truth . . . only Rossi has refined and gained control of a new physics to the point where a highly evolved stable reactor attains a COP = 22,000 and the same reactor harnessed as a heat engine attains a COP = 2000.

      You argue for the picture of Rossi as a self-centered mumbler with no compassion.

      I argue that the very speck you claim is in Rossi’s eye may also be in yours.

      Please show compassion, empathy, and imagination in expanding your understanding of Rossi and what it is like to walk in his shoes.

      • georgehants

        Buck, thank you for your reply, I have never said Rossi etc. should not be fairly rewarded for their work just like everybody else in the World.
        I am very satisfied with my definition of Empathy which is to take every opportunity to understand and help others in contrast to greedily caring about oneself.
        As I have said many times with Rossi’s original knowledge from seven years ago, had it been freely released, then Cold Fusion could possibly now be helping and saving many lives.
        I see no other way of describing Empathy.

        • Buck

          “I see no other way of describing Empathy” . . . . and that is the problem.

    • Richard Hill

      GH: How much would you pay for a black box with a COP above one?

  • So, the latest Planet Zero theory is that Rossi (& company) are mismeasuring input power by a factor of 20,000?

  • Engineer48

    Hi CW,

    All LENR reactors are thermal amplifiers. Their thermal gain does vary but like dogs, different dog, same leg action.

    So with a thermal gain / amplification of 22,213 what do your expect the output watts will be with an input of 1 watt?

  • Ged

    Direct electrical activation perhaps, ala plasma? Or just such a small mass that 10 mW of electrical resistance power is enough to heat past the activation barrier?

  • Greg Leonard

    What was the voltage drop across the plasma? Surely that should have been used in determining the power input.

  • Gerard McEk

    It was not clear to me why the Gullström/Rossi documents provide such a poor information of the input power.
    In the up-loaded picture I show you why I believe it is so poor. The shown data does not provide any information about the real input power, still they seem to use that.
    First I thought it is done deliberately to hide the real COP.
    What is needed, is the DC source voltage or the voltage over the QX, to enable you to calculate the input power.

    I know some of you would say that the voltage over the plasma is zero, but that would make the COP infinite.

    However, another possibility is that the QX produces a voltage and Andrea does not want to publish this yet. If that voltage is 200 V negative when the plasma is on and the DC source provides a positive current of 0.1 A at zero voltage (the required voltage is then supplied by the plasma), then the calculation is right!
    If the QX is off and needs to be started, the DC source needs to supply 200V during a short time.

    What do you think?

    https://uploads.disquscdn.com/images/a381c703b1efab4f59292afd874b7be448616764b5650ab36061187c9e2b8e43.png

    • Andreas Moraitis

      If the plasma is permanently maintained, there should be no principal problem with this setup (see below). However, if the DC is pulsed you would anyway have to take very short samples and integrate the data subsequently. A PCE-830 could do that if the pulse frequency is not too high (interrupted plasma might cause other problems, though). Did anyone identify the meters? Are they capable of integrating power with a high resolution?

      • Gerard McEk

        If you have a DC chopper using a choke and a high enough frequency the current could be near to DC and if the plasma is stable also the voltage should not change so much. Nevertheless, the easiest way is to measure the power at the mains side of the power supply.

        • Andreas Moraitis

          AR just answered that the input is plain DC. So we have one problem less 🙂

    • sam

      I think it’s great that we finally
      have a picture of the QX.

    • Engineer48
      • Gerard McEk

        Yes, my diagram is based on what I read in the document. As you know it is totally inconceivable that the input power can be measured over this 1 ohm resistor, unless you assume that the reactor produces a voltage and the DC source is able to control a 100 mA current and also able to supply a high enough voltage to start the QX. Once started the QX produces a voltage which the DC source controller can use to maintain the 100 mA to the reactor. The voltage over the DC source will then be near to zero.

    • Dr. Mike

      Gerard,
      A schematic diagram certainly should be included in a published paper to show how measurements were made. However, where is the control system in your diagram? A control system could be providing many watts of rf power to the device (unless there is no control system). One other meter needed in your diagram is an oscilloscope across the dc source (or the reactor) to verify that it really is just a dc source. Rossi claims the E-Cat QX can be run from a dc source, but this doesn’t mean that he is only supplying the device with only dc.

      • Gerard McEk

        The control system is part of the DC source. It is able to supply the required voltage to start the QX. (I suggest 200 V).
        Once started it will maintain a 100 mA current through the reactor, using a voltage that is generated by the reactor. The voltage of the source will be near to zero when the plasma in the QX is on. The controller in the DC source is probably a chopper using a choke to maintain the current to 100 mA. Maybe the 1 ohm resistor is being used for feed-back to the controller.

  • GiveADogABone

    Al is adjacent to Si in the periodic table. Plenty of Al in Lithium Aluminium Hydride. Just transfer a proton? V to Cr is a bit more enigmatic.

  • Frank Acland

    Andrea Rossi explained this to me in an interview I had with him yesterday. I am still transcribing it, but here is an excerpt, since it is relevant to the discussion here:

    “We have measured only that resistance [the 1 Ohm resistor] because that is the only resistance we have in the circuit. If the E-Cat has a resistance, that makes our calculations more conservative, because, as you well know, the resistance goes in the denominator when you make the calculus of the amps. You have volts as the numerator, and the resistance as the denominator. So the bigger the resistance, the smaller is the amount of amps.

    “To be conservative, since the datum of the resistance of the E-Cat QX is confidential, we just do not consider the resistance. Because correctly we should have to make the sum of the resistance of the resistor that has been put in the circuit, and the resistance of the E-Cat. So we should have amps = volts/R1 (the resistor)+R2 (the resistance of the E-Cat). But we do not consider the resistance of the E-Cat, we consider it as if it is a perfect conductor, and we only consider the one 1 Ohm to make the calculation of the amps.”

    • Chapman

      Bingo!

      I knew there was no way the reactor resistance was 1 ohm.

      This begins to make more sense now…

    • GiveADogABone

      ‘Energy input: V=0.1 R=1 Ohm → W=0.01’
      Seems to me that this statement is no longer true.
      The real value of V is 0.1 + something and the something drops the CoP.

      • Thomas Kaminski

        Actually, the way I interpret Frank’s comment from Rossi is that the resistance of the QuarkX is in series with the 1-ohm resistor, but they are measuring the voltage across the combined load. That would drop the power estimate, because any additional resistance will lead to a smaller current. The voltage is the voltage across both.

        • Frank Acland

          That’s what Rossi is saying, I think.

        • GiveADogABone

          Now agreed. It is the circuit diagrams in the thread that fooled me.

      • Engineer48

        HI GADAB,

        I think what Rossi is saying is he has a accurate digital DC power supply that outputs 0.1 vdc and a 1 ohm series resistor that has a voltage drop of 0.1 volts meaning the total series circuit current, including the resistance of the QX, is 0.1 amps. Then total circuit power consumed (resistor + QX) is 0.01W which is higher than the actural QX input power as the 1 ohm resistor dissipation is included. So the QX power input is less than total circuit power and is conservative.

        • DrD

          Hi Eng,
          I agree, that makes sense.

        • Thomas Kaminski

          If what you say is true, the actual voltage across the device is quite small. Could it be acting like a superconductor? If so, how do you get an input power with zero volts?

          • Engineer48

            Hi Thomas,

            Rossi did say he consider the QX to be a perfect conductor. Even so there is always some resistance.

            What does that tell us?

          • Frank Acland

            I don’t think he meant that it IS a perfect conductor. But they are just using that assumption for the purpose of making a conservative calculation. He says that the E-Cat QX does have a resistance, but they don’t want to disclose what it is.

          • Engineer48

            Hi Frank,

            which is what I said. There is some resistance.

            What all this means is the real COP is through the roof. MUCH higher than 22,213.

          • Thomas Kaminski

            Interesting. Perhaps there is an AC component to the measurement, sort of like an inductor. It has zero resistance for DC, but has a reactance that stores energy. I think, based on the calorimetry, that there is actually a pulse applied for 1.8 seconds during which the effect happens and heat is evolved. Then they wait until the temperature measurement stabilizes to get the calorimetric number.

          • DrD

            It isn’t surprising that the QK doesnt behave as a simple resistor. It would be interesting to record the power output of his supply = [V(across the supply) times I] and then subtract the power dissipated in the resistor and the difference is the power consumed (or even produced) by the QK. No doubt a very small (variable) value.
            That’s a very useful interview post by Frank, looking forward to reading the rest.

          • Engineer48

            Hi DrD,

            Maybe Rossi has found a way to self power, from with-in the reaction, the QX, always in SSM, but is not ready to release it?

          • Andreas Moraitis

            If there is a current, there must be a voltage, even in a superconductor.

          • Gerard McEk

            Please read my assumption below Thomas. I believer there is a good chance that the reactor produces a negative voltage, while having a positive current.

          • Thomas Kaminski

            If that is the case, then the direction of power flow changes and QuarkX becomes a generator, not a load. However, the sum of the voltages around the loop still has to be zero, so the “power supply” (not specifically shown) has to absorb power.

    • Thomas Kaminski

      Frank: Did he explain the calorimetry setup? I do not recognize the temperature probe, but from the paper, I concluded he must have made a difference in temperature reading of 1.58 C in 1.8 seconds. Based on typical time constants for a temperature probe (several seconds), I conclude that he pulsed the power for 1.8 seconds then waited for a “peak” temperature reading and used that for the calorimetry estimate.

      • Frank Acland

        No, sorry Tom. I didn’t ask about that.

      • US_Citizen71

        It is a kitchen thermometer, I use a different model from the same company to check my chicken on the barbeque.

    • Gerard McEk

      That makes my assumption stronger. I assume that the reactor, once in operation produces a negative voltage. The DC source is able to use that voltage to control the current of 0.1 A. Once running the voltage over the DC source will probably be near to zero. Still the DC source needs some external power for starting up the QX and controlling the current. Please read it below.

    • GiveADogABone

      “correctly we should have to make the sum of the resistance of the resistor that has been put in the circuit, and the resistance of the E-Cat. So we should have amps = volts/R1 (the resistor)+R2 (the resistance of the E-Cat).”

      [As written, this equation is wrong and dimensionally incorrect on the right hand side. It should have brackets around I=V / (R1+R2) ]

      “But we do not consider the resistance of the E-Cat, we consider it as if it is a perfect conductor, and we only consider the one 1 Ohm to make the calculation of the amps.”

      [So R2 is set to zero when it is not. That makes I bigger than real and therefore I^2R1 losses are bigger than real in the resistor.

      What about the I2R losses in the QX that are driven by the real I from the power supply? Are they set to zero by R2=0? That is not conservative.

      Are the conservative resistive losses more than offset by the non-conservative QX resistive losses?]

      ————————-

      Is the voltage V in the equation :-
      1: over the 1 Ohm resistor, or
      2: over the reactor, or
      3: over the 1 Ohm resistor and the reactor?

      It needs to be over the 1Ohm resistor AND the reactor to be conservative for the CoP calculation. V(overall)^2/R1 is bigger than V(overall)^2/(R1+R2)

      There are/were circuit drawings in the thread that do not conform to the necessity of that analysis.

      • Frank Acland

        Yes, my mistake. I was just transcribing what AR said, not thinking about the correct syntax.

        And the answer is 1 — the voltage in the equation is just over the 1 Ohm resistor.

        • SG

          Are you sure about that? I think it is 3. Worth double-checking with Rossi?

          • Frank Acland

            The way I understand it is that Rossi says he is ignoring the resistance of the E-Cat in the calculation he did in the paper, and that is a conservative calculation because if the combined resistance (E-Cat + 1 Ohm resistor) is higher, the amps would be lower and the overall COP of the system would increase. That’s why I picked answer 1.

          • SG

            Right, but the V must be across the e-Cat and the 1 Ohm resistor, and I=V/(R1 + R2)

          • Frank Acland

            I see what you mean, and yes, I think in reality the voltage reading was across the combined load (E-Cat and 1 Ohm). But the calculation in the paper ignores the internal resistance of the E-Cat.

          • GiveADogABone

            Surely, that takes us to answer 3? V(overall), R2 set to zero and a high current that hides the real value of R2.

          • Frank Acland

            Yes, I agree, in reality it is 3.

          • GiveADogABone

            It would be good to get the authors to remove the circuit diagrams that show otherwise, IMO. They had me fooled for quite a while.

        • GiveADogABone

          Frank,

          Frank Acland ECW Admin GiveADogABone • an hour ago
          And the answer is 1 — the voltage in the equation is just over the 1 Ohm resistor.

          Thomas Kaminski GiveADogABone • 3 hours ago
          Actually, the way I interpret Frank’s comment from Rossi is that the resistance of the QuarkX is in series with the 1-ohm resistor, but they are measuring the voltage across the combined load. …
          Frank Acland ECW Admin Thomas Kaminski • 3 hours ago
          That’s what Rossi is saying, I think.

          I am still confused. You say the answer is 1 – just the 1 Ohm resistor,
          but before in response to Thomas you agreed ‘measuring the voltage across the combined load’.

          Answer 1 is not demonstrably conservative V(over R1)/R1 gives a realistic I but I^2R1 leaves no margin for the electrical power dissipation in R2.
          Answer 3 V(overall)/(R1+0) is greater than V(overall)/(R1+R2) and is conservative for the power dissipation in R1+R2 whatever the value of R2.

    • SD

      Can you ask Rossi if he ran a control experiment?

      • artefact

        He said on JONP some time ago that he created a dummy which gave a COP below 1.

    • This is wrong. The current is the same in the whole circuit. If the reactor has a resistance, the voltage over the resistance will be decreased since some of the voltage will be across the reactor. Consequently the current decreases. You simply don’t need to know the resistance of the reactor to calculate the correct current. But you need to know the voltage over the reactor, or the voltage over the total circuit and over the resistance, to calculate the input power in the reactor.

      • Chapman

        You are 100% dead-on correct SIR!!! (not surprising) 🙂

  • Engineer48

    In addition to the TexTronix digital scope,

    https://uploads.disquscdn.com/images/f60c2ed3800b910a8c2275baacaa473bcb074f43efa353611fa2c1a5bd3ac350.png

    there are 2 scope probes that one would assume are used with the scope to monitor the QX reactor voltage and current waveforms.

    https://uploads.disquscdn.com/images/ad0f9d283779db0d51282cbdd6550b034461b88e6a83707768d9f3a152ae9d46.png

  • Frank Acland

    AR also told me that “we use two voltmeters to make a double check. The difference of the measurement is the margin of error of two different voltmeters ( several mV )”

    • Engineer48

      Hi Frank,

      OK that makes sense as knowing the power consumed by the 1 ohm resistor and having an accurate digital reference DC power supply, the power consumed by the QX reactor can be easily calculated without needing a meter across the QX reactor input.

      Many ways to bake this cake.

      • I still don’t y understand well how the measurement was made. To measure the voltage over the 1 ohm resistance to calculate input power in the reactor is of course wrong since you need to know the voltage over the reactor. If I remember right, this was the way it was described in the first paper.
        So is there another description now — substracting the power consumed by the resistor from the overall consumed power (resistor + reactor), by knowing the voltage over the whole circuit (and the current — through the resistance, and thus in the whole circuit)?
        Was this what Rossi told you Frank? So why wasn’t it in the paper?

        • Frank Acland

          I asked him about the resistance of the reactor, and he gave the answer I transcribed below. I am not sure why that was not mentioned in the paper, though.

          • Right. But he provides V(resistor), right? Does he provide V(overall) too? Then V(reactor) is the difference and we can calculate its impedance from the current, which is known if we have V(resistor). Or?

          • Frank Acland

            He just provides one Voltage reading, so i don’t think we can make that deduction.

        • GiveADogABone

          Mats,
          I believe Rossi is trying to hide the internal impedance of the QX; its confidential.

          The real current is calculated by V(overall)/(R1+R2) but R2 must be hidden. Set R2 to zero, so your calculated I becomes V(overall)/R1.
          Now the useful property of V(overall)/R1 is that it is always greater than V(overall)/(R1+R2) and therefore conservative for the CoP calculation. V(overall)/R1 gives no information about R2 but allows a conservative CoP claim.

          If you know the real voltage over the resistor and its resistance, you have the real current. If you have the real voltage overall, the real current and the resistor resistance, then you can calculate the internal impedance of the QX.

          We are playing games with data hiding. You state, ‘and the current ‘. That is the conservative current and not the real current.

          • Frank Acland

            Yes, he admits he wants to keep the resistance of the QX confidential.

          • Chapman

            Yeah… That is the impression I was getting too.

            My questioning the interpretation HERE of the data he presented was not accusing him of being stupid or lying, but rather observing that the numbers alone do not add up to anything he would be likely to do. There were a number of ways to stack the numbers and have them add up, but the application was pointless! Rossi is not stupid. If he was conducting a test, I am betting he was harvesting some useful data. I think he was throwing us a bone and giving us a peek, but certainly not drawing back the curtain and inviting us all to behold all his secrets. It is a fine line he is walking.

            God bless him for sharing what he did!

            Also, did you happen to notice that the rig he used is NOT an experimental apparatus? What he is showing there is a DEMONSTRATION unit, and the half finished second unit beside it shows he is making more.

            You know what that means???

            Rossi is on the move!!!

        • AdrianAshfield

          Mats,
          You do know the current as the resistor and reactor are in series.
          Rossi stated today that the current was simple DC, not pulsed.
          The resistance of the reactor is probably well below 1 ohm, So knowing the output voltage of the power supply Rossi can determine the power exactly.
          If the reactor resistance is less than 1 ohm, as implied, it uses even less power than the resistor. That is to say very little indeed.

        • Gerard McEk

          Mats, I assume that the reactor produces a DC voltage when in operation (see below). If that is the case, the DC source is only required to control the current at about zero voltage. I guess AR does not want to detail this and produced the halfway house details in this document. I hope to detail that tomorrow in another diagram.

        • Andreas Moraitis

          If the resistance of the reactor is much lower than the resistance of the resistor (which one could expect if there is plasma between the nickel electrodes), there will be no significant difference in the voltages. The reactor would have the same effect as an additional piece of wire. Thus, that measurement method might be ok for practical purposes. But I agree that in a presentation it would be better to document all possibly relevant parameters.

      • Steve H

        I2R could also be used to deduce the power consumed. If memory serves correctly, it is also the heat generating capability of a DC circuit.
        I2R ( current squared x resistance) for those not familiar with the nomenclature.

  • Buck

    If memory serves me, you will find multiple sources at LENRProof.com. This is a good starting point. I should note that in older versions of LENRProof, it also points to the EU as a knowledgeable agency aware of LENR and its implications.

    Finally, you should do a search of this blog site. It has been discussed here as well.

  • Andreas Moraitis

    You must apply a voltage at least once to obtain a current. If you loop the superconductor subsequently, the current may continue to flow without the applied voltage. But I think in a linear system it would break down immediately as soon as the voltage is removed.

    • Thomas Kaminski

      Andreas: Typically a superconducting magnet is charged by using the inductance of the coil. The voltage across the inductor is proportional to L, the inductance in Henries, and di/dt the rate of change in the current. To initially charge the magnet, a portion of the wire is heated above the superconducting state and a big power supply is applied to start the charging. With di/dt equal to the supply voltage, the current builds slowly. When it reaches the operating current, the shunt is cooled to superconducting levels, closing the loop and locking in the current. The supply can then be removed.

      It is interesting that the equation for the time constant of an RL (resistance/inductance) series circuit is L/R. Since R=0, the “time constant” is infinity and the current never decays.

      • Andreas Moraitis

        Thanks Thomas, I have always been wondering how this is technically realized.

        • Thomas Kaminski

          Your are welcome Andreas. One other point: the leads to charge the magnet constitute a thermal leakage path down to the liquid helium bath, so they are usually made as an assembly that can be removed after charging.

  • CWatters

    My background is in electronics. I’m with Mats, as far as I can see we do not have enough information to calculate the input power.

  • Chapman

    I see a lot of folks that are asking the right questions, only to be shut down by a few well meaning folks insisting on some pretty wrong facts.

    Ignore all the diagrams, the speculations, the guesses. Look at the facts we know, and the actual statements from Rossi.

    1. There is a 1 Ohm resistor in series with the QuarkX reactor.
    2. There are two meters hooked up in parallel reading the voltage across the resistor, but nothing is stated regarding the applied power source.
    3. The use of two redundant meters is an issue of protocol, and protects the test from bad data due to meter failure or inacuracy, which is also the likely reason they are two completely different makes and models.
    4. During the test, in the frame sample provided, we see that the resistor voltage drop is 100 mV. From this we can calculate that the series current at that time is 100 mA.
    5. After powered operation for a duration of 1.8 seconds, the oil bath surrounding the reactor showed a temperature increase that calculates out to 20 watts (per second) of generated heat.
    6. Rossi says that the reactor has low/negligible/nonexistent resistance. It is not stated if that assumption pertains only to the operational state, or even when “cold”.
    7. When taken as a whole, if the voltage drop on the reactor really is zero, the total power consumed by the ASSEMBLY (including ballast resistor) is just the resistors 10 mW. This results in an operational COP of 2,000.
    8. If all the above is TRUE AND ACCURATE, then the COP is actually orders of magnitude greater, because the resistor is actually producing 10 mW of heat directly, and ALL of the heat from the reactor is FREE, and has no mathematical connection to input power level.
    9. That means that the Quark is entirely current dependant, and the circuit could just as well have used a .5 ohm ballast, reduced the voltage to 50 mV, maintained the exact same 100 mA current through the reactor, and exhibited an operational COP of 4,000. Go to a .25 ballast and you get a COP of 8,000. Because the voltage drop on the ballast has no direct bearing on the reactor, but it simply sets the current passing THROUGH the reactor. This is basic electronics…

    I have seen so many people talking in circles and convincing each other that Rossi said something OTHER than what was printed right there before our eyes. And folks are changing their minds and saying “yeah, I guess that’s right”. WRONG. Read it again. It says what it says.

    Now, if someone has additional actual FACTS to throw in, fine. But do not just have a group hug and decide that the paragraphs in the intro suddenly transformed or mutated. Rossi said very specific things. Stick with what we know. We can theorize about the MISSING facts, but we can’t just decide we do not like, and will abandon, the actual facts given.

    Either there is something fundamental Rossi excluded, or the Quark is a current dependant reaction chamber that utilizes the presence of a 100 mA current to stimulate the release of nuclear binding energy from a small reserve of an as yet not fully disclosed amalgam of Li dust, LiAH, and ???. The reaction is singularly dependent on the current passing through the reactor, and yet exhibits little to no independant electrical resistance, making it susceptible to unstable runaway conditions, which require an external driver/ballast resistor to clamp max current to within safe levels. The size of that ballast is dependant upon tolerance factors and power handling ability of the selected ballast. The smaller the ballast, the higher the overall COP of the circuit as a whole.

    And you want to know what is REALLY wrong with this picture? There is ONE fact that makes me doubt these numbers, and the zero resistance of the reactor… If the reactor has zero resistance, then there is NOTHING keeping him from daisy chaining 100 of them in series with a single ballast resistor fixing a 100 mA series current flowing through ALL of them, and delivering a 2KW reactor running off a single AAA battery!!! That’s technically right, but at the same time SOOOOO wrong that there MUST be something missing.

    • Nice.

    • Thomas Kaminski

      Good summary. However, I think that one thing is missing. It is not at all clear that there is not a “controller” or “power supply” also in the loop providing the forcing function for the 100 mA current. If there is none, then the only conclusion is that the QX must also be a generator. You cannot have 100 mA of current flow without something supplying current.

      • Chapman

        I agree with you 100 percent. My point is that taking the data at face value, ignoring the fact that the most crucial info regarding the power supply is missing, leads to a circuit approximation that looks like a miracle device with truly impossible properties. Yet we can not selectively choose WHICH of the stated facts we will toss and maintain any reasonable validity.

        There is more here than meets the eye, and much we simply do not know. But he DID give us a peek at the form factor, as well as a clear idea of “scale” and versatility. Obviously, there is more to the power story, as well as something curious going on regarding the QX resistance and POSSIBLE active generation. But we can only guess. But if those details are important enough for him to have so blatantly hid them, they must be real whoppers!

        And, once more, I am not criticizing him for withholding info. His presentation is in no way misleading, it simply has some omitted facts that he needs to keep secret at this time. But the arguing sure proves why he does not bother with public demonstrations! Nothing is ever good enough to satisfy folks… even his fans!

        We begged to see it – well, now we have! We wanted to know it’s power scale – well, he has shown us, generally, the voltage and current conditions for its basic operation.

        I am overjoyed! I thank him for sharing all that he did. I will take it and be grateful, and will make no demands of “Where’s the Gravy???” I only point out the importance of the missing data because folks are jumping to conclusions refusing to recognize that the missing parts are the most important to what they are trying to figure out.

        • Thomas Kaminski

          Agree whole heartedly. The fact that we have an image of the QX with dimensions viewable is a great gift. I eagerly await the details. ….but, in the mean time, conjecture is fun.

    • Leonard Weinstein

      The voltage source was not described, and if it was a battery or higher voltage source, where is the source of the remaining voltage drop? This is not a complete story.

      • Chapman

        Exactly my point.

  • GiveADogABone

    CoP (Coefficient of Performance) is really a concept related to Carnot machines (e.g. heat engines and heat pumps).
    https://en.wikipedia.org/wiki/Carnot%27s_theorem_(thermodynamics)
    The formula for this maximum efficiency is eta(max) = 1-Tc/Th

    The QX is not a Carnot heat engine/pump.

    The energy efficiency of the QX, calculated as the (Energy Out)/(Energy In) could be taken overall. The electrical power dissipation takes place in the power supply, the ballast resistor and maybe the QX itself. I believe Gerard McEk has it right :
    http://e-catworld.com/2017/03/21/new-paper-by-rossi-and-gullstrom-reports-quarkx-experiment-calculated-cop-22000/
    Gerard McEk Omega Z • 4 months ago
    It is clear that the calculated input power for the plasma is wrongly calculated. It is the power over the input resistor, not over the plasma. The COP (22,223) cannot be calculated, based on the provided data. It can be any value.

    The most meaningful measure of overall efficiency in the commercial world would be the measured heat output divided by the power flow at the mains input to the power supply (see https://en.wikipedia.org/wiki/Heat_rate_(efficiency)). In the scientific world, the provided data is meaningless and the QX impedance confidential.

    What all this says about the future demo is a question. My advice would be to drop all mention of CoP and call it a heat rate test, then plead confidentiality about the QX apart from its measured heat output.

    • Right GADAB. I would argue for the same definition of any power plant — output power / overall input power.
      So in the QX case we need to know the total power consumed. Using the power consumed by the resistor and assuming the power dissipation in the QX is almost zero without proving it will not do.
      (Maybe this has already been said now — I haven’t read all comments).
      Interesting, however, the hypothesis of electrical power being generated in the QX, complicating the circuit analysis a bit.

      • Thomas Kaminski

        In general the COP is defined as “Q/W” where Q is the heat output and W is the work in for a heat pump. Since the QX is really a heat generator with electrical power in, COP is appropriate. However, if the device does generate electrical power (maybe! — we will see), COP would be meaningless. W and Q are both assumed to be positive quantities. The QX would have “infinite” COP, since the work in is “less than zero”.

        • Omega Z

          Rossi has indicated that about 10% of QX output is DC voltage. I think he is quite some way off from economically harvesting it. At least according to his postings on JONP.

    • Omega Z

      Yes, Rossi has pointed out in the past and I have posted it in my comments on occasin that “COP” isn’t the coreect term. The correct terminology is to give power in verses power out.

      I believe Rossi continues to use the term COP because it’s easier then constantly correcting every one on the blogs.

  • Gerard McEk

    Hi all,
    I think I might have a solution for all the questions about measuring the input power, which seems to be wrongly written in the AR/G paper. I have written my thoughts below in previous comments, but here is a new diagram that might better explain it:

    https://uploads.disquscdn.com/images/9a89524ac102a9e3c3791e727c7d3a640882e982ae0e0e81066f1823c00ef6ec.png

    When in operation the reactor (right) produces a DC voltage (depicted by a DC source in the reactor). The reactor controller controlling the 100 mA uses this voltage to maintain the current. The voltage over the controller drops to about zero when in operation. The internal resistor (impedance of the reactor) is such that the voltage over the reactor also drops to zero, when the current is 100 mA.

    This is all in accordance with the Rossi/Gullström paper.

    When the reactor is off, its impedance is infinite. To start the reactor a relatively high voltage will be needed. I have suggested 200 V max, but could be more or less. This voltage must be generated by the controller. At the moment the reactor plasma starts the controller must react very quickly, otherwise the current will be far to large. This will be one of the difficulties for controlling the reactor. The controller will have a mains connection to guarantee proper operation under all circumstances.

    I have asked Andrea Rossi if this assumed design of the E-cat QX is right and he said:

    Andrea Rossi
    July 21, 2017 at 5:50 PM

    Gerard McEk:
    I am not going to release more information before the presentation.
    Warm Regards,
    A.R.

    He didn’t say ‘no’, so perhaps I am on the right way! 🙂

    • DNI

      You can’t have 0,1 V over 1 ohm resistor and at the same time 0 V across both the Reactor controller and the Reactor. The sum of the voltages must be zero (Kirchhoff’s voltage law).

      • Gerard McEk

        That’s right, that is why I said about zero. At other locations in circuitry there will be some voltage drops, especially over the controller. These are minor though in comparison to the voltage generated by the QX (I assume).

        It could be that the the QX is a current source and that no active control is needed while in operation. The contoller is then just needed to start the QX with a short positive pulse and a short negative pulse to shut it down.

        • DNI

          I’m not sure you understand my objection. If you have 0,1 V over the 1 ohm resistor. Then the sum of the voltages across Reactor controller and Reactor must also be 0,1 V (positive or negative depending on how you define the polarity). Or do you mean that 0,1 V is about zero.

          • Gerard McEk

            Yes, that’s what I meant. The sum of the voltages will obviously be zero in a closed circuit. I believe the voltage of the DC source in the reactor is a lot higher than the 0.1 V over the resistor. The resistor is in fact an electrical load on the reactor. From the past we know that the E-catX could deliver maximal 10% of the total power electrically: 2 W. Assuming the E-cat QX is a 100 mA current source, the resistor can have a maximum value of 200 Ohm, causing a voltage of 20 V over the resistor.

          • DNI

            When you say resistor above I’m not sure if you mean the resistance in the reactor or the 1 ohm resistor.

            Maybe you could give an example with the voltages across Reactor controller, resistor, internal resistance inside Reactor and DC source inside Reactor.

          • Gerard McEk

            If is assumed that the reactor has a current source, and the controller has a zero impedance and output voltage, the voltage over the external resistor is 0.1 Volt and equal to the voltage over the reactor. The voltage over the internal impedance of the reactor depends on the voltage over the current source Ucs and is Ucs- 0.1V.

          • DNI

            Thanks, then I think I understand how you mean.

      • US_Citizen71

        What if the reactor has a negative resistance like a metal halide light?

        • DNI

          I cant see how this will make any difference. The sum of the voltages in a closed circuit must still be zero.

        • CWatters

          Rossi says same conductivity as silver.

          • US_Citizen71

            Which is essentially saying the same thing. The resistance of an arc light is all but zero once it is struck.

  • Thomas Kaminski

    The fact that the voltage across the QX is not being measured (or at least, not being reported) makes it difficult to guess what the actual input power is . We only have Rossi’s statement that the resistance of the QX is negligible.

    • Chapman

      Thomas,

      You are without doubt the most consistent shooter on the firing line!

      For a long time, thread after thread, post after post, you have been nailing the most fundamental facts and ideas with laser precision right from the get-go. You are not the loudest, or the flashiest, or the most bombastic. You are simply the rightestest… 🙂

      • Thomas Kaminski

        Gosh, gee, I don’t know what to say. I guess it is the lifelong career in engineering that has taught me to look for an answer that fits the data…

        • Chapman

          I am not calling you a genius, so do not think I am going overboard, but the difference between high intelligence AND genius is that intelligence finds the right answer to a given question rapidly, while GENIUS is a matter of spotting the right question to be asking in the first place. Lots of intelligent people waste their time spinning their wheels following pointless paths. Genius comes into play by focusing the intellect on what matters. Any answer can be chased down given a little discipline and determination, but being able to spot what is relevant is the key to real insight.

          Your comments are always focused on the core issues. You spot what matters quickly and seem to ignore distractions, regardless of how amusing or interesting they may be. And you do not get lost on tangents.

          A very useful trait for an engineer! I, on the other hand, go chasing after every blinking light that crosses my vision! I am as easily distracted as a cat being toyed with by a laser pointer…

          • Thomas Kaminski

            I just returned from a morning coffee klatch with old friends, one of whom is “Chapman”. You are welcome to join us if you are ever in Madison Wisconsin.

            I appreciate you humor as well as you knowledge of those things that I cannot understand. As Clark said: “Any sufficiently advanced technology is indistinguishable from magic.”

            Your knowledge of quantum physics is something to be admired. As far as I am concerned, the mathematics required to understand Rossi’s paper is so far beyond my understanding, that it might as well be magic. I have to take it as “faith” rather than “knowledge”.

            You also have the ability to distill it all down to my level, especially when it comes to the legal aspects of the contracts/lawsuits. Thanks.

          • clovis ray

            I agree, and some folks has the ability to pass info on with ease and make it so understandable. thanks guys for using a few extra words so we might understand.

      • sam

        I agree.Thomas says his sight is not what it used to be but his brain sure is not
        failing.

  • Dieter von Holten

    Hi all,

    i just want to point out that a COP of 2000 is NOT reported ! Rossi drops some numbers and leaves it to the readers to combine them in any way they like and conclude a COP….
    DvH

  • Ged

    Well… no, there is. It depends what you mean though I guess? It is a complicated subject… Unlike the other three fundamental forces, the strong force grows stronger with distance. And since gluons, the mediators of the strong force are massless, they technically have “infinite” range to act, just like photons (and gravitons if they are actually a thing, which many theories say they aren’t and many say they are).

    The thing is, gluons carry charge and a different anti-charge (strong charge called “color”, not electrical charge), and thus they don’t end up zooming through space like a chargless photon, since they find something to interact with before they do. But they could in the right circumstances, I guess.

    I hope I explained that somewhat understandably. For a more in-depth description, here is the answer from Fermi lab scientists science/inquiring/questions/strong_force.html

  • Thomas Kaminski

    Steve: The thought of a tunnel diode did occur to me. Also, a microwave magnetron where an applied DC current sets up an oscillating electrical field in the GigaHertz range. Maybe there is really an oscillation in the TeraHertz range (long-wave IR) that is also powered by the thermal excitation.

  • sam

    JPR

    July 22, 2017 at 5:09 AM

    Update

    Andrea Rossi

    July 22, 2017 at 5:51 AM

    JPR:
    Today great internal test.
    Warm Regards,
    A.R.

  • Chapman

    Try telling that to Yoda! He goes ON and ON about it incessantly…

    I am tired, and confused, and wet in uncomfortable places that are starting to itch.

    I am tired of this endless swamp.

    I want to go back to Tatooine! I never thought I would be saying this, but I miss the old moisture farm….

  • GiveADogABone

    https://en.wikipedia.org/wiki/Negative_resistance
    negative differential resistance (NDR) (dv/di<0)
    Examples of devices with negative differential resistance are … and gas discharge tubes
    The power output of the QuarkX is AC that is produced by NDR.

    The Wikipedia page explains it all.

    • Chapman

      Now WHY did you have to go and open THAT can of worms? 🙂

      Yes, if the Quark has an NDR kink, it could be used to amplify a pilot AC signal. And IF that amplification taps the internally liberated energy rather than drawing additional supply voltage then it would be a means of sucking some of the generated energy directly out as a generated electrical product. Tapping the QX energy would then reduce the portion that is thermalized, and you could tune the pilot signal to select how much electricity vs how much heat you wanted to produce.

      Running the system purely in a DC mode would preclude all of that, but the bottom of the kink WOULD represent a minimum static resistance point, and would be the “sweet spot” that Rossi would want to idle the QX at in such a DC system as he presented, in order to achieve the highest COP and efficiency. He would not be oscillating up and down the curve, but rather just surfing along down in the trough.

      • GiveADogABone

        In the photo of the QX there seem to be only two electrical terminals.
        How do you get the output power of the QX out while putting in the battery power?
        If the battery is steady DC and the output power is AC there is no problem; just a simple capactitive filter will separate them.

        Also, the wiring to the QX is not a high voltage, well protected/insulated affair.
        How do you start the QX?
        Generate the HV impulse internally via the usual ballast/starter circuits of a fluorescent tube running from the DC.
        The 1 Ohm resistor is part of the ballast.
        https://en.wikipedia.org/wiki/Arc_lamp

        • Chapman

          Great ideas!

          • GiveADogABone

            The NDR curve of the QX would, of course, be of great interest.

        • Thomas Kaminski

          I would concur with your idea. I suspect that there is a buck/boost switching supply in the grey box that used an inductor in the “boost” mode to start the plasma ignition and then switches to a current-limited “buck” mode to apply the “DC” current once ignited.

          The picture shows a thin, red clip lead going off behind the grey box. The right end of the QX with Calorimeter does not seem to be connected to anything. Note that the other grey box on the left has the four screws tied together with copper braided wire. These seem to be the place where the clamp would be connected, implying that the screws are electrically connecting to the clamps that somehow connect the QX device. The circuits inside of the grey box must be using those screws to electrically connect to the screws/QX. There is also a hole in the grey box on the right in a similar position to the grey box on the left (after rotating it 90 degrees). Is this where the power comes in?
          https://uploads.disquscdn.com/images/09da01f23125f9fd66ec08e7baaae90adb729e31706338df768b871a07a03afb.jpg

          • Goodrice

            I wonder if power to the gray (control?) box which likely supplies the necessary HV and waveform comes from this:

            http://i.imgur.com/xPHkHNC.png

          • Thomas Kaminski

            Good find. It is very likely the same power supply, but it does not seem to be connected to the other boxes.

          • Thomas Kaminski

            Also, it is simply an adjustable DC supply. It does not seem to have any Ac component. Rossi just confirmed that the DC voltage should not be confused with what is applied to the QX by the control system:

            Andrea Rossi
            July 23, 2017 at 1:57 PM
            Oystein Lande:
            Our power source can be either 120 or 220 V AC, or we can use 24V DC batteries.
            Obviously your calculation is wrong, because one thing is the voltage at the power source, a totally different thing is the voltage that goes to the E-Cat through the circuitry of the control system.
            In the same Gullstroem-Rossi paper you can read the voltage measured by the 2 voltmeters.
            Warm Regards,
            A.R.

          • Chapman

            As GADAB pointed out, the DC control supply that biases the system and maintains the optimum operating range is separate from any AC component, and that secondary signal can easily be capacitively coupled directly to the reactor, and the power product can be harvested the same way. The AC component would effectively piggy back the DC. It would be a totally separate, and OPTIONAL, circuit added in parallel to the existing “heat only” DC control circuitry, bridging across the reactor contacts. You could connect it with alligator clips and never break the existing circuit!

            An absolutely beautiful design arrangement!

          • Thomas Kaminski

            I think this is entirely possible. If you look at the grey box, the end clamps seem to be electrically connected to something inside. The AC pulse generator could be inside the grey box. It could be capacitively connected, or even even connected via an electrical switch, such as a diode. You could use an inductor “kickback” such as what is used in spark coils to boost 12 volt car batteries to several hundred volts.

          • Chapman

            Yeah. And it SIMPLIFIES the mysteries, rather than compound them. And I always work with confidence in the fact that truth identifies itself by simplifying complexity.

        • GiveADogABone

          http://www.cmsim.org/images/1_CHAOS2012_Proceedings_Papers_A-B.pdf
          In plasma physics it is well known that the S-type NDR is related to the appearance and disruption of a complex space charge structure (e.g. double layers, multiple double layers, etc.) [1,2,3], whereas the N-type NDR is related to the spatio-temporal dynamics of such a structure [4,5], or to the onset of lowfrequency instabilities [6,7,8].

          A negative resistance requires an active component in the electrical circuit able to act as a source of energy. In plasma systems, this component could be a selfconsistent double layer existing at the border of a fireball. The double layer works as a nonlinear element of circuit able to convert thermal energy into electrical energy, creating all the conditions necessary for the appearance of the S-type NDR in the current-voltage characteristic of a plasma conductor.

          Here, we report on experimental results and theoretical modeling of the S-type NDR effect in plasma conductors. The proposed theoretical model explains the relation between the negative differential resistance and the self-structuring of plasma as double layer, as well as the shape of the current-voltage characteristic of an electrode immersed into plasma, in conditions in which a double layer structure appears in front of it.

  • GiveADogABone

    A key kink?
    https://uploads.disquscdn.com/images/b7c9254277a98a1f8f0363a681e48ca2999e9ba22ca100f8de51b0cf0cd12201.png

    Gas discharge devices are CCNR(current controlled negative resistance) or ‘S’ type, so require a bias current to operate in the NDR(Negative DifferentialResistance mode.
    https://en.wikipedia.org/wiki/Negative_resistance

  • CWatters

    What do people make of the claim by Rossi that the conductivity is the same as silver? Silver is the best conductor, better than copper, gold or nickel. How can that be possible unless it’s either made of silver or a super conductor of some sort?

    • GiveADogABone

      Hot plasma?