​New E-Cat QX Picture and New Rossi-Gullstrom Paper (Very high COP reported with Calorimetry)​

  • So this paper contains details of a new Rossi experiment. I say details, but the word is really innappropiate:


    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 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.


    That, and the photo, is all we have.


    Even with this information, if these results spur theoretical physicists into speculation it just shows that those who do theory can't be trusted with experimental stuff.


    The first problem is the way the input power is measured. As previously, but now more specifically, Rossi measures the voltage across a known brown resistance of 1 ohm as being 100mV, and from that deduces (V^2/R) that the power is 10mW! Note how his calculation, with one V and one R, leaves no room for ambiguity or the possibility that there are here two voltages, one across the cell and one across the resistor.


    The previous paper could perhaps have been mis-stating what Rossi was doiung, but this time it is quite clear. He is measuring the input power through a resistor! What does this have to do with the input power through the plasma tube? Nothing.


    If we guess that this resistor is in series with the plasma tube we get a plasma current of 100mA. At a plasma voltage of 200V that would give input power of 20W matching the claimed output power. Those are plausible values, but I'm not confident of them, and of course the measurement we are given say nothing about the plasma voltage and hence the input power. Rossi's determination here of that output power is even weirder. So strange that I can't decode it.


    He has a fluid (11g worth) with a known spec heat cap and density. No units given for these, but we can math with typical values for lubricant oil and get the answer.


    He then details as calorimetric data for the fluid 0.5kCal/h. What does this mean? The only interpretation I can put on it is that he calculates the energy input to the oil from its temperature rise. Unfortunately we are not given the temperature rise anywhere. there is so much lost from this data that it is clear Rossi, unconscously or not, is wanting not to attract the attention of guys like TC & Paradigmnoia who like looking at details and decoding them.


    Finally heat exchanger efficiency: 10%. Seems impossibly low for anything that could correspond to that setup (with the oil recirculating?). But I don't know, and since the output-side data is not given it is irrelevant.


    Bottom line: input power. Certainly measured wrong. Real input power depends on voltage across the cell which we do not know. Output power: your guess is as good as mine!


    Worth noting. Rossi has discovered a clever way of getting bad experimental data into Arxiv. Publish it as a one paragraph summary in a theoretical paper written by somone else.

  • Would you mind expanding a little on the problems with:


    Rossi measures the voltage across a known brown resistance of 1 ohm as being 100mV, and from that deduces (V^2/R) that the power is 10mW!


    (It mentions a DC input).


    Stephan, I'd assume they calibrated the calorimeter ("efficiency of the heat exchanger:10%") by just running it with the resistance described above?


    Finally heat exchanger efficiency: 10%. Seems impossibly low for anything that could correspond to that setup (with the oil recirculating?).


    I agree. Are they definitely recirculating?

  • One would normally measure the voltage across the resistor to compute the current and then use the voltage across the tube to measure the voltage. Then the input power is V*I if the voltage and current are DC or in phase. If the input voltage is pulsed DC, it may be difficult to measure depending on the pulse duration. It is all measurable if someone knows what they are doing.

  • The first problem is the way the input power is measured. As previously, but now more specifically, Rossi measures the voltage across a known brown resistance of 1 ohm as being 100mV, and from that deduces (V^2/R) that the power is 10mW! Note how his calculation, with one V and one R, leaves no room for ambiguity or the possibility that there are here two voltages, one across the cell and one across the resistor.


    You're making things up. He is (most likely) using two voltmeters in a redundant configuration to confirm that the current corresponds to the 100mV voltage measurement, and sure enough, they essentially match, as expected. To take your conjecture as true, one would have to assume that Rossi made his trick right out there in the open for all to see! That is why your conjecture is most likely nonsense.

  • One would normally measure the voltage across the resistor to compute the current and then use the voltage across the tube to measure the voltage. Then the input power is V*I if the voltage and current are DC or in phase. If the input voltage is pulsed DC, it may be difficult to measure depending on the pulse duration. It is all measurable if someone knows what they are doing.


    Rossi has stated on his blog that the QuarkX uses DC. This is about as simple as a measurement gets. THH is grasping.

  • You're making things up. He is (most likely) using two voltmeters in a redundant configuration to confirm that the current is about 100mV, and sure enough, they essentially match, as expected. To take your conjecture as true, one would have to assume that Rossi made his trick right out there in the open for all to see! That is why your conjecture is most likely nonsense.



    IHFB - stick to non-tech stuff or learn that current is not measured in mV?


    How can you make sense of his power calculation - explicitly the power dissipated in 1 ohm (the brown resistor)?


    I'm calling you on this, ante up. What does Rossi's own calculation mean?

  • Zeus:


    If they put a resistor inside the heat exchanger with a resistance that yielded the same output power and documented it I would be much more pleased than

    with what we have now, does he mean that the 1 ohm resistance is a separate resistance or that the cell represent the resistance. Knowing that they really

    calibrated it with the dummy at the range of the reactor would be helpful assuming an honest experiment of cause.

  • THH,


    It was a mis-type, and I have a massive headache right now. He is measuring the voltage across the known resistance, redundantly. He has stated it is DC. This is a simple calculation, and you know it.


    Facts are tough, are they not? It is a simple calculation, which is why in this case Rossi is bang to rights.

  • From the paper:

    W=5,67×10^12×0.9×4.8×10^13=244.9

    I'm only a cable guy, used to use volts, amps, ohms and watts in kilo or milli...

    Is that outcame above really correct?

    I mean 244.99 is the correct value, but multiplying something with 10 to the power of 12 and then again with 10 to the power of 13, I get much more zeros added to 244.9 ....

    244.944.000.000.000.000.000.000.000

    Can someone please enlight me?!

  • If the brown resistor is a resistor in series of the reactor you will have the same current U/R = 100mA going through both systems.

    Now the power is I^2R which means that the power can be anything in the reactor depending of the resistance of it. So this setup

    cannot be used to calculate input power. This is a pretty silly mistake and relly wondering if it represent any truth. It's hard to see

    from the picture where the plus side of the volt meters goes and atm we cannot say much about the setup.

  • If the brown resistor is a resistor in series of the reactor you will have the same current U/R = 100mA going through both systems.

    Now the power is I^2R which means that the power can be anything in the reactor depending of the resistance of it. So this setup

    cannot be used to calculate input power. This is a pretty silly mistake and relly wondering if it represent any truth. It's hard to see

    from the picture where the plus side of the volt meters goes and atm we cannot say much about the setup.


    Correct. The picture proves noting, and Rossi's stated calculations are wrong.


    so what's new?

  • So,


    Mineral oil on shell side, water in tube side,

    Overall heat transfer coefficient is close to

    100 btu/hr/ft2/f?

    Water comes in at say 55f out at?

    Oil in at? Out at ?


    To calculate heat exchanger efficiency he should have this data, no?


    Would be nice to KNOW both fluid flows,

    Delta T's, specific heats and not have to mke assumptions.


    With oil start temp before Quark x,

    after Quark x and oil flow, energy input can be calculated.

    Then, if this quantity is as large as AR suggests, back calculate how much electrical energy is needed to heat the oil to the quartz temp.

    Is it even possible to get that much electrical power/energy thru the wire size used?

  • Certainly measured wrong.

    This is just your prejudice ! THH This is a repetition of Lugano with a better technology.

    From the paper:


    W=5,67×10^12×0.9×4.8×10^13=244.9


    I'm only a cable guy, used to use volts, amps, ohms and watts in kilo or milli...

    Is that outcame above really correct?


    There is a mistype in the exponent of Stefen Boltzmann constant. Because the unit used is cm^2 it should be 10^-12.

    Is just a mistype the result is correct.

  • Is just a mistype the result is correct.


    @RB: Nobody measures power over a voltage, if the resistance is not known...(and do not tell us that the resistence is constant and not complex...)


    I always recommend to use a small accumulator, charged with a given wattage! Most Russian experimenters use this method, which avoids unwanted cheats.

  • is I^2R which means that the power can be anything in the reactor depending of the resistance of it.

    A Plasma is conductive. Almost 0 resistance. The power is limited by the internal resistance of the power supply and the resistor in series.

    The power measured in this way is just a higher limit of the power spent in the reactor chamber.

    Correct. The picture proves noting, and Rossi's stated calculations are wrong.


    so what's new?

    The Rossi stated calculations are not wrong.

    And this is not new. They were correct even before.

  • Nobody measures power over a voltage, if the resistance is not known...(and do not tell us that the resistence is constant and not complex...)

    Is a different type of experiment. Charge a capacitor with a known energy and discharge it in the device.

    We use it for Plasma Focus devices.


    The measure of Rossi is a higher limit. But correct to demonstrate the effect.

  • A Plasma is conductive. Almost 0 resistance. The power is limited by the internal resistance of the power supply and the resistor in series.

    The power measured in this way is just a higher limit of the power spent in the reactor chamber.

    The Rossi stated calculations are not wrong.

    And this is not new. They were correct even before.

    Sure, if the resistance of the reactor is less than 1ohm, you are correct, but that is not stated in the article which shows that one need hidden facts to interpret it correctly, which indicate that the report needs to be reworked. A voltage over the reactor would be nice to have as well as possible, as supplement to really underwrite the stated resistance proprerties of the reactor.

  • randombit0 : THX for the quick clarification.

    Another question:

    The first picture of the early E-Cat QuarkX showed mostly bright blue light and paler blue light in the centre.

    From the paper I read: integrated spectrum emission wavelength is 1100nm -> that's invisible infra red.

    How can the bright blue light be explained?

    (Black body emission about 2700k isn't infra red, from all I know)

  • Correct. The picture proves noting, and Rossi's stated calculations are wrong.


    so what's new?

    First Jed maintains the E-Cat QX doesn't even exist, so the photo is apparently a figment of one's imagination.

    Then THH can't resist speculating in his usual negative way, that one can't determine the power by measuring the voltage drop across a known resistor in series with the reactor, to get the current, and measure the output voltage pf the power supply. As it it supposed to be DC it should be fairly simple.


    Also curious is where my last post on the settlement thread disappeared to