Requirements of H-Ni LENR Devices and Their Implications on the Lugano Test

The following guest post was submitted by Michael Lammert. 

Requirements of H-Ni LENR Devices and Their Implications on the Lugano Test

by Michael Lammert (AKA Dr. Mike)

October 31, 2014

I’ve continued to think about the results of the Lugano test and Rossi’s E-Cat devices. Two questions that I’ve been considering are: What factors are required to produce an optimum H-Ni LENR device (ala Rossi’s E-Cat) and are the active “consumption” powers in the Lugano test reasonable and correct?

As I see it, key requirements for a useful E-Cat type device include:

  1. Ability to generate enough heat within the Ni powder to produce a useful output.
  2. Ability to control the low energy nuclear reactions sufficiently to provide stable operation without thermal runaway.
  3. Ability to deliver the heat to a load upon demand.
  4. Ability to dissipate the heat from the Ni to its surroundings fast enough to prevent the Ni from melting.

Although we don’t know yet what exact combination of Ni powder, hydrogen, catalysts, and electromagnetic pulses are needed to produce a working LENR device, there doesn’t appear to be any question that an E-cat device can produce sufficient heat to be useful. Not only is Rossi currently working hard to get his 1 MW plant installed, others have independently demonstrated useful heat from LENR devices.

The control of an LENR device for stable operation is going to be a difficult, but not impossible task, at least until the theory of operation is well understood. The ability to deliver heat to a load upon demand should just be a straight forward engineering problem, except for the control function. Therefore, both items #2 and #3 will benefit greatly when a solid theory of LENR is established.

I believe that getting the heat generated in the Ni dissipated to its surroundings will be the primary limiting factor for how much power can be generated from a given size device. Ni has a specific heat of 0.44J/gr/oK, which means that 0.44 Joules of energy will raise the temperature of 1 gram of thermally isolated Ni by 1 oK. It would only take about 520 Joules (1 Joule = 1W-sec) to raise that 1 gram of thermally isolated Ni from room temperature to its melting point (0.44*(1455-273) = 520). Luckily the Ni powder in the E-Cats is not thermally isolated. In the Hot-Cat, the Ni is in contact with both other Ni particles and the alumina reactor. Both Ni and alumina have reasonable thermal conductivities (91 and 30 W/m/oK, respectively). The Ni will also radiate heat proportionally to T4, and also have heat transferred via convection to the mostly hydrogen gas within the reactor. (Although there is no data on what the partial pressure of hydrogen is during operation, cooling through transfer of energy to the hydrogen might be made somewhat more effective by removing most of the air within the reactor prior to start-up since hydrogen has about a 7 times higher heat conductivity than air.)

How fast can heat be dissipated from the Ni powder through conduction, radiation and convection? For the answer to this question I would request a thermal engineer to make a calculation based on assumptions that are appropriate to the operating conditions of the Hot-Cat. The closest electrical analogy that I found to this thermal dissipation problem is data on the internet for temperature vs. current data for NiCr 60 wire (60%Ni). A 1 gram piece of 18 gauge (1 mm diameter) NiCr 60 wire would be about 15.5cm long and have a high temperature resistance of about 0.24 ohms. The NiCr would melt at a current of about 33 amps or at a power of about 260W (332*.024). (This is for a straight wire in air.)   I would expect a gram of Ni powder sitting on alumina in a hydrogen environment to have a somewhat higher maximum dissipation power than the 260W for a piece of NiCr wire. However, if the Ni was clumped in a pile, it might be expected to melt at even a lower power level than the 260W. The solution to the dissipation problem is to add more Ni powder, distribute the powder in a uniform layer over the entire reactor internal surface, and run the reactor at a lower power generation rate per gram of Ni powder.

Now let’s look at the numbers from the Lugano test. It was claimed that 0.55 grams of Ni was used in the reactor with average output powers of about 1660W for the first part of the run and about 2320W for the second part. Assuming all of this excess power is generated uniformly in the 0.55 grams of Ni powder, dissipation in the Ni would have to be 3018W/gr in the first run and 4018W/gr in the second portion. Although I will wait to see a calculation of the maximum possible dissipation per gram of Ni powder from a thermal engineer before making a final conclusion, I believe these dissipation numbers are about an order of magnitude too high for the 0.55 gram of Ni powder not to melt.

If it is assumed that it is not probable that the measured excess power seen in the Lugano test could have come from heat generation in the small amount of Ni, where did the heat generation come from? In my previous post I pointed out the problem with the “Joule heating” calculations, reprinted here:

  1. Problem with the “Joule Heating” Calculation

         The “Joule heating” calculation for the Cu wire for the dummy run on pages 13-14 seems to be fairly straight forward. The “Joule heating” is simply the resistance of the wire times the current squared flowing through that wire. Sum the Joule heating in the 3 Cu wires from the controller and the 6 Cu wires to the device and you have the power that comes out of controller, but doesn’t participate in heating the Inconel coils. This is such a simple calculation, that it seems unlikely that an error would be made in other calculations of Joule heating. However, the “Joule heating” in the Cu wires for the active run has been calculated in Table 7, page 22 as about 37W for the input power at 800W and about 42W for the operation at 920W. These “Joule heating” calculations imply that the current in the Cu wires was 2.35 times as high in the 800W active run as it was in the dummy run (SQRT(37/6.7) = 2.35). The only way for this to be possible is for the Inconel resistors to have a very large negative temperature coefficient of resistance. Although the report did not specify what type of Inconel was used in the coils, the data sheets for various Inconels show well less than 10% variation in resistivity over a wide temperature range. For example, Inconel 625 has a resistivity of 135.9 micro-ohm-cm at 427 oC and 133.9 micro-ohm-cm at 1093 oC. Other Inconels have a slightly increasing resistivity as the temperature increases. Also it should be pointed out that if the Inconel used in the coils in this experiment had a large negative TCR, then the Joule heating as calculated in Table 7 would have been much higher than 42W for the 900W portion of the test. The calculated “Joule heating” powers are directly proportional to the “consumption” powers, indicating no change in resistivity of the Inconel coils as the temperature increases from about 1260 oC to 1400 oC in the two portions of the active runs. Questions for the authors: 1. What is the source of the error in the “Joule heating” calculation for the active run?   2. What type of Inconel was used in the resistor coils? 3. What was the current flowing through the resistors for each of the active power levels?

 

In re-reading the Lugano report I found that the authors actually answered my third question on the current levels on the active power: On page 3 they say that the potentiometer on the TRIAC power regulator was used to set the operating point, “normally 40-50 Amps”. Using the data in Table 7, page 22 for the Joule heating numbers, the average current for File #2 can be calculated as I2 = 19.5* SQRT(36.98/6.7) = 45.81A (where 19.5 is the current in the dummy run and 6.7 is the Joule heating in the dummy run), and the average current for File #7 can be calculated as I7 = 19.5* SQRT(42.18/6.7) = 48.93A. Indeed both of these currents are within the 40-50A range quoted by the authors on page 3. Since the resistance of the Inconel coils is known to change by less than +/- 5% over the operating temperature range (look this up if you don’t believe it) the input power for the active runs will equal to the dummy power times the ratio of the Joule heating powers. For File #2 the actual power supplied by the TRIAC regulator (“consumption” power in Table 7) is P2= 486*36.98/6.7 = 2682W (+/-5%) (486 is the power for the dummy run), and for File #7 the actual power supplied by the TRIAC regulator is P7 = 486*42.18/6.7 =3060W (+/-5%). (The “Joule heating” powers must be subtracted from these values to get the actual power delivered to the Inconel wire resistors.) Using these calculated “consumption” powers results in File #2 having a net power production of -217W with a COP of .918 and File #7 having a net power production of 180W with a COP of 1.06. If these numbers are correct, one can see that the power output due to the LENR effect is in the noise of the measurement technique. However, the analysis of the “fuel” and the “ash” clearly show that nuclear reactions took place within the reactor during the active runs, but perhaps at a power level in the low 100’s of Watts.

My speculation as to a possible source of an error in the set-up is that the connection of the power source supplying the “specific electromagnetic pulses” mentioned on page 1 of the report is somehow interfering with getting an accurate power measurement from the PCE 830 meter on the output of the TRIAC power regulator. There is no electrical diagram showing how this power source is wired to the reactor and the TRIAC power regulator, and the authors did not state whether this power supply was turned on during the dummy run. I would further speculate that the authors measured the higher current and relatively low power during the active runs and assumed that Inconel resistor coils just had a large negative temperature coefficient of resistance. (This is what I assumed until I looked up the TCR for the various types of Inconel.)

It would certainly be disappointing if the results of the Lugano test are clouded by an “instrumentation” error; however, it wouldn’t be the first time and won’t be the last time that the measurement technique caused an error in reported scientific results. In 2011 the scientists working on the OPERA project at Italy’s INFN Gran Sasso Laboratory reported that they had measured neutrinos that were traveling slightly faster than the speed of light. It took about 6 months with help from scientists outside their laboratory to finally confirm that they had an error in their measurement technique.

Recommendations

   My recommendations to the Lugano authors and others working on E-Cat experiments are as follows:

  1. Have a thermal engineer estimate the maximum power dissipation rate for 1 gram of Ni in the Hot-Cat reactor based on assumptions that match the operating conditions for the Lugano test.
  2. Have the electrical set-up of the Lugano test reviewed by an electrical engineer to determine if the power source supplying the “specific electromagnetic pulses” could have interfered with making an accurate power measurement during the active runs.
  3. Re-examine the radial uniformity of the reactor temperature data. If the Ni powder was mostly at the bottom of the reactor, I would expect to see at least a 100 oC temperature difference between the bottom of the reactor and the top of the reactor (needs verified by a thermal engineer’s calculation). If the radial temperature uniformity is about the same in the active runs as the dummy run, then there is a good chance that most of the observed output power is coming from input power to the Inconel coils, not from LENR in the Ni powder.
  4. For future Hot-Cat experiments it might be beneficial to supply the “specific electromagnetic pulses” via a fourth coil wound on the reactor.
  5. The MFMP group would have a much higher chance of success if they tried to duplicate the original E-Cat, rather than the Hot-Cat. If they decide to go ahead and build a copy of the Hot-Cat, they should probably follow my recommendation #4 above and load the reactor with a minimum of 5-10 grams of “fuel”.

Michael Lammert.

  • Ged

    More misunderstandings of how joule heating and input power work. The input power is virtually independent of joule heating and vice versa (I.e. this is why high tension lines don’t burst into flames).

    -Joule heating is directly, geometrically, dependent on amperage.-

    Input power is dependent linearly on amperage And voltage. You can up the input power by upping the voltage without significantly affecting joule heating. Again, this is exactly why we use high voltage lines to carry MW of electricity, not high amperage.

    The input power is measured and reported directly. Stop fallaciously trying to conflate joule heating and input power and try to rewrite their actual measurements from the mains. Remember, they measured the power going to the controller, as well as from it. You cannot invent power with a controller, or we’ve just discovered a whole new power source and law of physics.

    The joule heating was -subtracted out- of a the input power, as it is line loss seen by each wire and is not a splitting of input power as each wire sees the same load, so if there is any error here regarding joule heating the affect on the COP calculations is minor.

    • Ged

      Also, joule heating is -not a measured quantity-. It is a calculated one based on assumptions. It is not fit nor useful for back calculations, especially when we -do- have the measured quantities we care about.

      • ivanc

        Where did your studied electricity?
        This is based in V=RI, and Power=VI, so Power=R I I (or I^2)
        as the feeding cables are copper R is constant, and mesurable. if you take to readings at different set voltages.
        now you could calculate I, and I is in series with the device, so is the input current,.
        that will cause the input power.

        I will add a question, What power is spent creating the electromagnetic signal?
        Or is just an harmonic result of the use of TRIACS.

        A serious report will have show an oscilloscope with the wave form.

        • Ged

          I don’t think so.

          We aren’t talking about the heating of the reactor by the resistors, that’s not the “Joule heating” the authors are reporting on or putting in their table. We’re talking about loss of power by transmission down the lines which they are subtracting out from the heating consumption power of the reactor. Or on page 13: “For this reason, it is expedient to evaluate what portion of the current, fed to the system by the power mains, is dissipated by the cables as Joule heat.”

          If you believe what you say, then the “Joule heating” they report is the input power, which means they were only inputting 7 W to the dummy reactor (page 14), or 41 W to the actually E-cat experimental run (Table 7)? Or is their input power the 900 W they supply?

          Everyone seems to be conflating the heating of the reactor (the 900 W) with the calculated -power loss- that never makes it to the reactor for the purposes of heating (the 41 W).

        • Dr. Mike

          ivanc,
          I believe the “specific electromagnetic pulses” are supplied by a pulse generator- probably the brown box seen in Figure 3, page 4 of the Lugano report. As far as I can tell both the power needed to create the electromagnetic pulses and the power they deliver to the reactor are not included in the COP calculations. I assume that no oscilloscope waveform was shown because it is proprietary. Mentioning the “specific electromagnetic pulses” in the introduction on page 1 and then not discussing them anywhere else in the report certainly did not enhance the report. I would have been satisfied with the report stating that the nature of these pulses was proprietary, and therefore no oscilloscope waveforms could be shown. If the report had just shown how the pulse generator was hooked up in the Figure 4 wiring diagram, someone might have been able to determine if the pulse generator was affecting the PCE-830 power measurements.
          Dr. Mike

      • Dr. Mike

        Ged,
        One key point of my post is that I believe the the PCE-830 is giving an erroneous reading of the power because of the way the pulse generator was hooked up to the TRIAC regulator and heater coils. I can’t even imagine how the pulse generator was hooked up. Even though the Joule heating is a calculated quantity, it ir comes from a single calculation of the resistance of the Cu wire and a measured quantity the current. If you read the Lugano report carefully you will see that the current was measured by both the PCE-830 and clamp-on ammeters. It can be assumed that this is an accurate measurement, because a problem would have been noted if the two measurements did not agree. Finally note that even though the Cu wire resistance was a calculated value, by using a ratio of powers to currents squared, the value of the calculated resistance of the Cu wire, which is assumed to remain constant, the calculated resistance cancels out of the ratios. (P1/P2 =I1*I1*R/I2/I2/R =I1*I1/I2/I2) These same equations were used to calculate input powers to the Inconel coils without actually calculating he Inconel resistance.
        Dr. Mike

        • forget it,
          you cannot fool those powermeter with triacs. this kind of devices is designed to measures tricky losses at high frequency because of switching artifact on IBGT, Mosfet…
          the bandwidth in not infinit, but a rule of electricity is that if the voltage have no noticable harmonics, then the current whatever it is transfort few power…

          as said before the only way to smuggle energy, is a fraud with huge DC or HF energy injected as voltage, with current harmonics or DC matching the voltage.

          the power is the sum of power hy harmonics. if no voltage above some HF or at DC, no power forget it… even a magician cannot change that.

          note that in that case the front PCE830 is the one to follow.

          the second hone works without problem, but there is huge voltage and current harmonics and the computation is above the level of a skeptic.

          note that even on the intermediate PCE830, measuring DC voltage OR current (yes OR… if one is null, no power can pass) and HF voltage OR current…

          note that for high frequency, a way to measure HF current, above PCE830 (100harminics, 5kHz) is to measure RF in the room…
          few amp at 5kHz+ cannot be ignored by simply an osciloscope with a tiny wire as antenna.

          It tested it as kid.

          the bottom line is that only a conspiracy of fraud, is a credible explanation, beside the evidente hypothesis that it works.

    • Dr. Mike

      Ged,
      What you say about power lines and why they use high voltage is correct; however I disagree with your statement : “You cannot calculate input power from joule heating unless you exactly know the voltage at all times.” If you know the resistances of both the Cu wire and the Inconel wire, and you know the current through both wires, you can calculate the joule heating in both the Cu wire and the Inconel wire. The circuit schematic for one phase of the heater would just be a resistor for the input Cu wire, a resistor for the Inconel wire, and a resistor for the the output Cu wire connected in series across the output of TRIAC regulator power supply. It might have been clearer if I had just calculated the resistance for each of the 3 heating coils and used this number in my input power calculations. Using the same assumption as the authors,that is, the current is about the same in all three heater coils and their resistance is about the same, the resistance of one Inconel heater wire can be calculated from R=P/I/I. or R=(486-6.7)/3/9.85/9.85 = 1.647 ohms. One of my main point is that this resistance changes by less than +/-5% over the entire temperature range of operation. From the Joule heating in the Cu wire for the active runs, one can calculate the current that was flowing (through both the Cu wire and the Inconel wire). For File #2, the current can be calculated from as I2 =9.85*SQRT(36.98/6.7) = 23.14A, and for File #7 the current can be calculated as I7 = 9.85*SQRT(42.18/6.7) = 24.71A. Now the power to a single heater wire can be calculated as P=I*I*R, so P2 = 23.14*23.14*1.647= 881.9W and P7 = 24.71*24.71*1.647 = 1005.6W. The total power to all 3 heater coils is just these numbers multiplied by 3 or 2646W for File #2 and 3017W for File #7. These numbers are can be compared to the powers I calculated in my post for the power delivered to the coils: for File #2: 2682-37 = 2645W and for File #7: 3060-42 = 3018W.
      Please note that we do know the voltage across the coils: V=I*R or for File #2 the RMS voltage is V2 = 23.14*1.647 = 38.11V and for File #7 the RMS voltage is V7 = 24.71*1.647 = 40.70V. Also note that I misquoted the current coming out of the TRIAC supply as 19.5A in my post- it should have been 19.7A. However, since i didn’t use this current in my calculations I believe the power and COP numbers are correct.

      I hope this is a better explanation for you on how input powers to the Inconel coils can be calculated from the data provided by the authors.
      Dr. Mike

      • Ged

        The problem with your calculations and even theirs is you are assuming to know the resistance. You calculate it, but they do too and their numbers are wildly different from yours, by three fold. Such as on page 14: “one may easily deduce that the electrical resistance of the three cables exiting the regulator (Circuit 1, C1) is = R1 = 4.375*10-3 Ω, whereas that of the cables splitting off from these (Circuit 2, C2) = R2 = 2.811*10-3 Ω”. And then “For each of the six 50 cm lengths of copper cable, the relevant resistance is 7.028*10-4 Ω”.

        Their joule heating values are a summation across all the circuits.

        But if we use the resistance they used in their calculations from which you’re back calculating, then the voltage is actually in the millivolts by the calculations and assumptions you make here.

        The authors don’t actually supply data on the amperage during operation. What they say on page 13 is: “In the present run of the E-Cat the current flow may actually be higher than 40 A. For this reason, it is expedient to evaluate what portion of the current, fed to the system by the power mains”. They only say “normally 40-50 A” on page 3, but normally doesn’t mean that’s what we’re seeing, and they admit as such in page 13.

        The authors never actually supply the amps or volts measured as far as I have been able to find in the report linked here by E-cat World. Without that knowledge, you can’t back calculate correctly, it’s impossible. And the divergence of their values from yours shows that in spades.

        Instead of believing the author’s table 7 showing consumption in watts, you’re focusing on joule heating while having no reliable information to exactly calculate it. And you come to a completely different resistance value for their wires than they have, while their resistance values are what were used to give the joule heating they report — meaning your results are immediately divergent from them.

        The fact one can drop input power but still raise joule heating by raising the amps while lowering the volts more, shows that in the absence of amperage or voltage data directly supplied, we can’t guess and back calculate input power. They -do- supply the input power consumed though.

        So, I have to side solidly with the authors in the absence of any more available data. I cannot agree with your analysis at all due to divergence from the reported resistance for the wires and the reported input power. Better data and or assumptions than currently presented will have to be supplied to convince me otherwise.

        • AlbertNN

          The analysis by Gullstrom is not compatible with the amount of fuel, 1g, that was used in the latest test.

  • Ged

    More misunderstandings of how joule heating and input power work. The input power is virtually independent of joule heating and vice versa (I.e. this is why high tension lines don’t burst into flames).

    -Joule heating is directly, geometrically, dependent on amperage.-

    Input power is dependent linearly on amperage And voltage. You can up the input power by upping the voltage without significantly affecting joule heating. Again, this is exactly why we use high voltage lines to carry MW of electricity, not high amperage.

    The input power is measured and reported directly. Stop fallaciously trying to conflate joule heating and input power and try to rewrite their actual measurements from the mains. Remember, they measured the power going to the controller, as well as from it. You cannot invent power with a controller, or we’ve just discovered a whole new power source and law of physics.

    The joule heating was -subtracted out- of the input power, as it is line loss seen by each wire and is not a splitting of input power as each wire sees the same load, so if there is any error here regarding joule heating the affect on the COP calculations is minor.

    You cannot calculate input power from joule heating unless you exactly know the voltage at all times.

    • Ged

      Also, joule heating is -not a measured quantity-. It is a calculated one based on assumptions. It is not fit nor useful for back calculations, especially when we -do- have the measured quantities we care about.

      • ivanc

        Where did your studied electricity?
        This is based in V=RI, and Power=VI, so Power=R I I (or I^2)
        as the feeding cables are copper R is constant, and mesurable. if you take to readings at different set voltages.
        now you could calculate I, and I is in series with the device, so is the input current,.
        that will cause the input power.

        I will add a question, What power is spent creating the electromagnetic signal?
        Or is just an harmonic result of the use of TRIACS.

        A serious report will have show an oscilloscope with the wave form.

        • Ged

          I don’t think so.

          We aren’t talking about the heating of the reactor by the resistors, that’s not the “Joule heating” the authors are reporting on or putting in their table. We’re talking about loss of power by transmission down the lines which they are subtracting out from the heating consumption power of the reactor. Or on page 13: “For this reason, it is expedient to evaluate what portion of the current, fed to the system by the power mains, is dissipated by the cables as Joule heat.”

          If you believe what you say, then the “Joule heating” they report is the input power, which means they were only inputting 7 W to the dummy reactor (page 14), or 41 W to the actually E-cat experimental run (Table 7)? Or is their input power the 900 W they supply?

          Everyone seems to be conflating the heating of the reactor (the 900 W) with the calculated -power loss- that never makes it to the reactor for the purposes of heating (the 41 W).

          • fact police

            Ged> Everyone seems to be conflating the heating of the reactor (the 900 W) with the calculated -power loss- that never makes it to the reactor for the purposes of heating (the 41 W).

            No one is doing this. Everyone understands the joule heating refers to the heat dissipated by the wires outside the reactor. That has nothing to do with the inconsistency that has been identified.

            It’s very simple. If you increase the current in the wires, the current increases in the coils by the same factor.

            Therefore, if the power dissipated in the wires is increased, the power dissipated in the coils increases by the same factor. Unless the resistances change.

            The authors report a 6-fold increase in the power dissipated in the wires (joule heating), but only a 2-fold increase in the power dissipated by the coils. One of those must be wrong, unless the resistance of the coils reduces by a factor of 3, which is not consistent with its behavior at day 10.

          • Ged

            They don’t report only a 2-fold increase in power dissipated by the coils. I do not see where you get that from, so if I’m mistaken there, please show me the reference.

            What they do show, according to page 7 for instance, is the input power directly detected going to the reactor — this is not the dissipation by the coils, but the electric draw of the system. Therefore, you can have a 2-fold increase in input power and a 6 fold (50% more than 4 fold, and if you look at page 3, you’ll notice their I^2 is varying by at least 56%) in joule heating, simply by having their feedback controller drop voltage to keep current pumped. We see them report having to modulate this controller-keeping-the-power-input-the-same-by-changing-current during the actual run.

          • fact police

            Ged> They don’t report only a 2-fold increase in power dissipated by the coils.

            It’s implicit. The input power is dissipated in the wires and the coils, so the amount dissipated in the coils is the input power minus the joule heating.

            In the dummy run that’s 486-7 = 479 W. In the live run, from table 6, file 6 (e.g.) it’s 923 – 42 = 881 W, for a factor of 1.84.

            Ged> What they do show, according to page 7 for instance, is the input power directly detected going to the reactor — this is not the dissipation by the coils, but the electric draw of the system.

            Yea, I get that. The same argument could have been made using the total input: If the joule heating increases by some factor, the total input must increase by the same factor unless there’s a change in resistance. But the total input is pretty close to the coil dissipation, so it doesn’t change things very much.

            Ged> Therefore, you can have a 2-fold increase in input power and a 6 fold (50% more than 4 fold, and if you look at page 3, you’ll notice their I^2 is varying by at least 56%) in joule heating,

            No. That’s completely wrong. A 2-fold increase in input power should produce a 2-fold increase in joule heating. So, even a 4-fold increase is wrong by a factor of 2. And no, their I^2 is not varying by 56%. It’s in that range. They claim the overall power measurement is accurate to 10%, so there is no way the I^2 can vary by that amount. You’re not reading the report correctly.

            Ged> simply by having their feedback controller drop voltage to keep current pumped.

            Ged, voltage and current are not independent in this circuit. You can’t increase the voltage without increasing the current. The current determines both the power in the wires and in the coils, and the current is the same.

          • ivanc

            I did not said the joule heating is the input power, the joule heating is the loss of power in the transmission line, but carry a current, and this current is eaxtly the one that causes the input power, following p=vi, so if you have the instantaneous current and instantaneous voltage you could integrate in the cicle, or use and RMS metter, but please note RMS metters depends on the wave form, so if you using a metter design for sinusiodal wave on a different wave the device that makes the measurment could be wrong, this is why you need the wave with an osciloscope, to integrate the power in function of time.
            I doublt the phisiists will know the details of electrical measurements. they used a distorted wave and used clamp amperimeters, this sound like disaster to me.
            in any case if they need to control the voltage/or current, they should use an autotransformer and not TRIACs to avoid the problems with instruments
            for you to understand this better, the current will be the same but the voltage different, the drop of voltage in the cu wire will be small, and the one on the load will be large. but this values related, as total voltage = voltage in cu + voltage in inconel resitors

        • Dr. Mike

          ivanc,
          I believe the “specific electromagnetic pulses” are supplied by a pulse generator- probably the brown box seen in Figure 3, page 4 of the Lugano report. As far as I can tell both the power needed to create the electromagnetic pulses and the power they deliver to the reactor are not included in the COP calculations. I assume that no oscilloscope waveform was shown because it is proprietary. Mentioning the “specific electromagnetic pulses” in the introduction on page 1 and then not discussing them anywhere else in the report certainly did not enhance the report. I would have been satisfied with the report stating that the nature of these pulses was proprietary, and therefore no oscilloscope waveforms could be shown. If the report had just shown how the pulse generator was hooked up in the Figure 4 wiring diagram, someone might have been able to determine if the pulse generator was affecting the PCE-830 power measurements.
          Dr. Mike

      • Dr. Mike

        Ged,
        One key point of my post is that I believe the the PCE-830 is giving an erroneous reading of the power because of the way the pulse generator was hooked up to the TRIAC regulator and heater coils. I can’t even imagine how the pulse generator was hooked up. Even though the Joule heating is a calculated quantity, it ir comes from a single calculation of the resistance of the Cu wire and a measured quantity the current. If you read the Lugano report carefully you will see that the current was measured by both the PCE-830 and clamp-on ammeters. It can be assumed that this is an accurate measurement, because a problem would have been noted if the two measurements did not agree. Finally note that even though the Cu wire resistance was a calculated value, by using a ratio of powers to currents squared, the value of the calculated resistance of the Cu wire, which is assumed to remain constant, the calculated resistance cancels out of the ratios. (P1/P2 =I1*I1*R/I2/I2/R =I1*I1/I2/I2) These same equations were used to calculate input powers to the Inconel coils without actually calculating he Inconel resistance.
        Dr. Mike

        • forget it,
          you cannot fool those powermeter with triacs. this kind of devices is designed to measures tricky losses at high frequency because of switching artifact on IBGT, Mosfet, PWM switched power…
          the bandwidth in not infinite, but a rule of electricity is that if the voltage have no noticeable harmonics, then the current whatever it is transport few power…

          as said before the only way to smuggle energy, is a fraud with huge DC or HF energy injected as voltage, with current harmonics or DC matching the voltage.

          the power is the sum of power hy harmonics. if no voltage above some HF or at DC, no power forget it… even a magician cannot change that.

          in that case the front PCE830 is the one to follow.

          the second one works without problem, but there is huge voltage and current harmonics and the computation is above the level of a skeptic.

          even on the intermediate PCE830, measuring DC voltage OR current (yes OR… if one is null, no power can pass) and HF voltage OR current…

          note that for high frequency, a way to measure HF current, above PCE830 (100 harmonics, 5kHz) is to measure RF in the room…
          few amp at 5kHz+ cannot be ignored by simply an oscilloscope with a tiny wire as antenna.

          I tested it as kid. even 50Hz is detected by an antenna.

          the bottom line is that only a conspiracy of fraud, is a credible explanation, beside the evidente hypothesis that it works.

    • fact police

      Ged wrote:

      The input power is virtually independent of joule heating and vice versa

      Not if the cables are simply connected to the load resistor as in this case. In this case the input power is proportional to the joule heating, to the extent that the resistance remains constant.

      Ged> -Joule heating is directly, geometrically, dependent on amperage.-

      For constant resistance, the joule heating is proportional to the square of the current.

      Ged> Input power is dependent linearly on amperage And voltage.

      But for a constant resistance, the current is proportional to the voltage according to Ohm’s law. So the input power is equal to I^2*R or V^2/R.

      Ged> You can up the input power by upping the voltage without significantly affecting joule heating.

      Not in the circuit in question. If you up the voltage, the current increases proportionally, and the power goes up by the square.

      Ged> Again, this is exactly why we use high voltage lines to carry MW of electricity, not high amperage.

      But then there are transformers to reduce the voltage before it gets used. There are no transformers involved in the circuit in question.

      Ged> The input power is measured and reported directly.

      Well the device measures voltage and current and then calculates the input power. The joule heating is determined from the current measurement, and the resistance calculation.

      Ged> Stop fallaciously trying to conflate joule heating and input power

      It’s not fallacious. In the given circuit, input power is proportional to joule heating unless the resistance changes. This has been acknowledged frequently.

      Ged> Remember, they measured the power going to the controller, as well as from it.

      Right. But it was measured with the same kind of device, which, experts tell us, is vulnerable to a simple reversal of a current clamp.

      Ged> You cannot invent power with a controller, or we’ve just discovered a whole new power source and law of physics.

      No one is claiming that. The claim is that the 6-fold increase in joule heating is not consistent with a 2-fold increase in reactor heating, unless there is a 3-fold reduction in coil resistance. One of these was determined from an independent current measurement. The other from a power measurement with the PCE. Unless there really is a 3-fold resistance reduction, one of those measurements is wrong. (And the 3-fold reduction in resistance is not consistent with what happens when the power is increased at day 10.)

      In any case, if the ecat works as claimed, *then* you have a whole new power source and law of physics, and you don’t seem to have a problem with that.

      • Ged

        What? You use Joule’s first law, I^2*R, used for calculating joule heating, and use it to calculate input power? I think we are talking about very different “joule heating”s then.

        Look at what the report says on page 14: “In the calculations that follow, relevant to the dummy reactor and the E-Cat’s power production and consumption, the watts dissipated by Joule heating will be subtracted from the power supply values.”

        We aren’t talking about the heating being done to the reactor by the resistors, we’re talking about the line loss due to joule heating from the resistance of the wires to the amperage flowing down them. Exactly like any transmission wire. The input power consumed is supplied by the authors in their table 7, and the joule heating they supply is the “loss” that is subtracted from the power consumption being used to heat the reactor; and this loss has been calculated by them, not directly measured like the power consumption has been.

        That’s the subject I’m talking about.

        • fact police

          Ged> What? You use Joule’s first law, I^2*R, used for calculating joule heating, and use it to calculate input power?

          No, I’m using the law to show how the power depends on the resistance and the current. The current through the wires is the same as the current through the heating coils. If you double the input current, the current doubles through all the wires and all the coils. So the power dissipated by the wires has to increase by the same factor as the power dissipated in the coils. That’s all that’s needed for this argument.

          The paper itself claims the power lost in the wires increases 6-fold. If that’s true, and the resistances don’t change, then the power in the coils must also increase 6-fold. But according to the power meter, it is only 2-fold. Something is wrong.

          Ged> Look at what the report says on page 14: “In the calculations that follow, relevant to the dummy reactor and the E-Cat’s power production and consumption, the watts dissipated by Joule heating will be subtracted from the power supply values.”

          That doesn’t change the discrepancy. The authors used the measured current to calculate the joule heating in the wires. If that current were used to calculate the coil heating, and if the dummy run was right, then there should be a 6-fold increase in the coil power. If the power meter gives a different value, either the current measurement is wrong, or the power measurement is wrong. Or the resistance of the coils decreases 3-fold between the dummy run and the live run.

          Ged> We aren’t talking about the heating being done to the reactor by the resistors, we’re talking about the line loss due to joule heating from the resistance of the wires to the amperage flowing down them. Exactly like any transmission wire. The input power consumed is supplied by the authors in their table 7, and the joule heating they supply is the “loss” that is subtracted from the power consumption being used to heat the reactor.

          That’s the subject I’m talking about.

          We’re all talking about that subject. But the authors claim a 6-fold increase in the loss in the wires, but a 2-fold increase in the coil heating. That’s only possible if the resistance of the coils changes 3-fold.

          • Ged

            Resistance need not change, the controller need only drop the voltage and voila. We already have an amperage range of 40-50 going on here, so is resistance constantly changing to drive these amp changes?

            I have tried to find where they use current to calculate input power, but I haven’t located it. I see they measured it with the PCE directly, not calculating it just from current (which wouldn’t work without voltage anyways).

            Also, we see the control system varying the amperage to hold input power steady in the paper itself, invalidating your claims that that can only change by resistance changes. To wit (page 7): “After this initial period, we noticed that the feedback system had gradually cut back the input current, which was yielding about 790 W. We therefore decided to increase the power, and set it slightly above 900 W”

            Even further, we see on page 3, that the power supply itself would not be able to handle the kilowatts you guys are trying to claim, such as : “its maximum nominal power consumption is 360 W” (I am assuming that’s for each of the three channels independently, so 1080 W total).

            Your claims that voltage and amperage to the system can’t fluctuate while holding input power the same without changing resistance is demonstrably mistaken according to the paper itself when they saw current being cut back and had to up the power to counter it. I guess that is your Waterloo?

          • fact police

            Ged> Resistance need not change, the controller need only drop the voltage and voila.

            If the voltage drops, the current drops, and the power drops, and it’s by the same factor in the wires and the coils.

            Ged> We already have an amperage range of 40-50 going on here, so is resistance constantly changing to drive these amp changes?

            The operating point is sent normally to 40 – 50 amps does not mean it varies by that amount. It means it is somewhere in that range. And even if it were changing by that range, it would change the same in the wires and the coils. And if it is changing, then the voltage has to be changing too, if the resistors are constant.

            Ged> I have tried to find where they use current to calculate input power, but I haven’t located it. I see they measured it with the PCE directly, not calculating it just from current (which wouldn’t work without voltage anyways).

            They use the current to calculate the joule heating. It’s described in detail for the dummy run. So, when they report the joule heating in the live run, you can get the current. They don’t use the current to get the total input power.

            But the point is that joule heating scales with total input power, and we know how the joule heating scales (x6), so the input power should scale the same way.

            Ged> Also, we see the control system varying the amperage to hold input power steady in the paper itself, invalidating your claims that that can only change by resistance changes.

            I didn’t say that. I said current and voltage have to change proportionally if resistance is constant. So, in the above case, the change in the current is produced by a change in voltage. The triac varies the voltage directly, not the current.

            Ged> Even further, we see on page 3, that the power supply itself would not be able to handle the kilowatts you guys are trying to claim, such as : “its maximum nominal power consumption is 360 W”

            That’s the power dissipated by the controller itself; not the output power. This is clear from the previous sentence, where they say “furthermore, [the two PCEs, one upstream and one downstream of the control unit] enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be the same as the nominal value declared by the manufacturer.”

            Ged> Your claims that voltage and amperage to the system can’t fluctuate while holding input power the same without changing resistance is demonstrably mistaken according to the paper itself when they saw current being cut back and had to up the power to counter it.

            My claim is a simple statement of Ohm’s law. For a given resistance, the ratio of voltage to current is fixed, and a single power corresponds to a single current (or voltage). If the ecat works it will be a revolution in nuclear physics. You are now claiming it is a revolution in electrical engineering as well.

            Ged> I guess that is your Waterloo?

            No. Still yours. For a simple circuit under consideration, voltage and current cannot be independently changed, and nothing in the report claims they are. And the power in the wires scales with the power in the coils. As long as resistance doesn’t change.

          • ivanc

            Why the resistance will change? a small change is possible but not a large, any drop in voltage will cause a drop in amperage.
            the resistor are used to heat the device they are not the active component. so we should not expect any special behavior on them, and treat them as the resistence of any electrical heating device.

      • Dr. Mike

        fact police,
        Thanks for the line by line reply to Ged. I believe you really helped him understand what then problem is in the data.
        Dr. Mike

        • Ged

          The problem is we don’t know what data is used to calculate joule heating in table 7, we’re making crazy assumptions to try to claim their, actually measured, input power is somehow wrong based on a -sum total- of a -calculated- joule heating power loss from the copper wires feeding the reactor. All the wires. All the circuits. Maybe more than fully accounted for in their description. We don’t know the actual amps, and we don’t know the actual volts. If volts go down, and amps go up, their measured (again, -measured-) input power would stay the same while their joule heating could go way up. 40^2 is 1600 while 50^2 is 2500, that’s a 56% difference, while for input power to stay the same at 900 W, voltage simply has to vary from 22.5 V to 18 V, or a 20% change in voltage affecting joule heating by 56% for the same input power. That’s how huge the -minimum- variation is we are dealing with, and the variation could be so much larger. Notice how 56% is within the difference of 4 versus 6 (50%) for the claim that the joule heating increase for the double input power increase should have been 4x yet it was a 6x. That’s just a 50% larger increase than expected which is explained by a 56% variance space for the joule heating based on the “normal” amperage range given — and they say it could be higher without presenting a cap on how high it could go.

          So, I find it completely bizarre to make huge sweeping claims that their reported input power is extraordinarily wrong when 1) that’s what they measured, 2) we don’t know the actual amps during the test for those values in the table, 3) we don’t know the actual volts during the test for those values in the table, 4) we don’t know all the wiring completely, just what they share with the dummy, and 5) joule heating is a calculated line loss value (thus could be calculated wrong) where a small voltage change can lead to disproportionate change in joule heating at the same input power.

          We don’t have remotely enough facts to use joule heating they report for anything, while we have the -actual input power- reported by them. Why ignore that for something we can’t accurately use? Explain that.

          • fact police

            Ged> The problem is we don’t know what data is used to calculate joule heating in table 7,

            Yes, we do. They explain it in detail in the dummy run. They use the measured line current, and the calculated resistance. If they use the same calculated resistance in both cases, and get a 6-fold increase in the joule heating, then they had to have measured a current root(6) times higher.

            Ged> we’re making crazy assumptions to try to claim their, actually measured, input power is somehow wrong based on a -sum total- of a -calculated- joule heating power loss from the copper wires feeding the reactor.

            The only assumption is that they use the same resistance as they used in the dummy run, and that they can measure the current. And what it shows is that the two determinations of the input power don’t agree. One must be wrong.

            Ged> All the wires. All the circuits. Maybe more than fully accounted for in their description. We don’t know the actual amps, and we don’t know the actual volts.

            It only matters that they used the same wires in both cases. They only need the current and the resistance to get the power. They report the power, so we can calculate the current.

            Ged> If volts go down, and amps go up,

            That can only happen if the resistance goes down.

            Ged> their measured (again, -measured-) input power would stay the same while their joule heating could go way up.

            For a 6-fold increase in joule heating with a 2-fold increase in reactor heating, you have to assume a 3-fold decrease in the coil resistance. This is not plausible.

            Ged> 40^2 is 1600 while 50^2 is 2500, that’s a 56% difference, while for input power to stay the same at 900 W, voltage simply has to vary from 22.5 V to 18 V, or a 20% change in voltage affecting joule heating by 56% for the same input power.

            Unless the resistance changes, whatever change in voltage produces an increase in joule heating, will also increase the heating in the coils.

            Please address this point. Everything hinges on it, and your latest dalliance with variance does nothing to resolve the problem.

            Notice how 56% is within the difference of 4 versus 6 (50%) for the claim that the joule heating increase for the double input power increase should have been 4x yet it was a 6x.

            No. You’re mistaken. If the joule heating is 4x, then the input power has to be 4x as well. (It’s the current that’s only 2x.) So, you still have a factor of 2 to account for, even if this 56 % made any sense, and it doesn’t. (The authors did not say the current was 45+/-5A; they said it was in the range of 40 to 50 A. And from the input current measurement in the dummy run given as 19.7A suggests much higher accuracy. What kind of scientist measures current with a 25% error bar?)

            Ged> So, I find it completely bizarre to make huge sweeping claims that their reported input power is extraordinarily wrong when 1) that’s what they measured,

            They also measured the current to calculate the joule heating.

            Ged> 2) we don’t know the actual amps during the test for those values in the table,

            Yes, we do. They report the power, show the procedure and the resistance for the dummy run, allowing current to be calculated.

            Ged> 3) we don’t know the actual volts during the test for those values in the table,

            Only the current and resistance are needed to calculate the power.

            Ged> 4) we don’t know all the wiring completely, just what they share with the dummy,

            But we know what they are accounting for from the dummy run. If they changed things between the dummy and the live run, then it’s not a dummy run.

            Ged> and 5) joule heating is a calculated line loss value (thus could be calculated wrong) where a small voltage change can lead to disproportionate change in joule heating at the same input power.

            It doesn’t matter if it’s calculated wrong, as long as it’s calculated the same way in the dummy and live runs. And you’re wrong about the disproportionate change in joule heating. Unless the resistances change, the joule heating and power input are proportionate.

            Ged> We don’t have remotely enough facts to use joule heating they report for anything, while we have the -actual input power- reported by them.

            That’s wrong. We have enough facts. Based on the circuit shown, any increase in the joule heating should be the same as the increase in the coil heating. If the factor they increase by is different by a factor of 3, there is a mistake somewhere. It saps confidence from the paper.

            Ged> Why ignore that for something we can’t accurately use? Explain that.

            The possibility for error on the power meter are at least as plausible as errors on the input current measurement.

          • ivanc

            The other thing we need to undestand is we not dealing with large voltages or currents, this is only 900watts. so if we use 50 amps

            p=vi , so v=p/i, then 900/50 = 18v.
            so we only putting 18v. and using large currents???

    • Mark Szl

      OMG, we are now having articles and post where the blind are leading and reading the blind. Can a certified electrical engineer which specializes in power measures and calculations speak up and steer us straight?

      • Thomas Clarke

        Several have done so. You might like to check out andrea.s for detailed knowledge of the PCE-830 on Mats Lewan’s comment thread [url=https://matslew.wordpress.com/2014/10/09/interview-on-radio-show-free-energy-quest-tonight/]here[/url] where the whole matter is debated at great length.

        The power calculations needed here require somone with basic knowledge of ohms law, KCL, KVL and phasors. That many have. On this thread you will find that fact police has been entirely correct.

    • Dr. Mike

      Ged,
      What you say about power lines and why they use high voltage is correct; however I disagree with your statement : “You cannot calculate input power from joule heating unless you exactly know the voltage at all times.” If you know the resistances of both the Cu wire and the Inconel wire, and you know the current through both wires, you can calculate the joule heating in both the Cu wire and the Inconel wire. The circuit schematic for one phase of the heater would just be a resistor for the input Cu wire, a resistor for the Inconel wire, and a resistor for the the output Cu wire connected in series across the output of TRIAC regulator power supply. It might have been clearer if I had just calculated the resistance for each of the 3 heating coils and used this number in my input power calculations. Using the same assumption as the authors,that is, the current is about the same in all three heater coils and their resistance is about the same, the resistance of one Inconel heater wire can be calculated from R=P/I/I. or R=(486-6.7)/3/9.85/9.85 = 1.647 ohms. One of my main point is that this resistance changes by less than +/-5% over the entire temperature range of operation. From the Joule heating in the Cu wire for the active runs, one can calculate the current that was flowing (through both the Cu wire and the Inconel wire). For File #2, the current can be calculated from as I2 =9.85*SQRT(36.98/6.7) = 23.14A, and for File #7 the current can be calculated as I7 = 9.85*SQRT(42.18/6.7) = 24.71A. Now the power to a single heater wire can be calculated as P=I*I*R, so P2 = 23.14*23.14*1.647= 881.9W and P7 = 24.71*24.71*1.647 = 1005.6W. The total power to all 3 heater coils is just these numbers multiplied by 3 or 2646W for File #2 and 3017W for File #7. These numbers are can be compared to the powers I calculated in my post for the power delivered to the coils: for File #2: 2682-37 = 2645W and for File #7: 3060-42 = 3018W.
      Please note that we do know the voltage across the coils: V=I*R or for File #2 the RMS voltage is V2 = 23.14*1.647 = 38.11V and for File #7 the RMS voltage is V7 = 24.71*1.647 = 40.70V. Also note that I misquoted the current coming out of the TRIAC supply as 19.5A in my post- it should have been 19.7A. However, since i didn’t use this current in my calculations I believe the power and COP numbers are correct.

      I hope this is a better explanation for you on how input powers to the Inconel coils can be calculated from the data provided by the authors.
      Dr. Mike

      • Ged

        The problem with your calculations and even theirs is you are assuming to know the resistance. You calculate it, but they do too and their numbers are wildly different from yours, by three fold. Such as on page 14: “one may easily deduce that the electrical resistance of the three cables exiting the regulator (Circuit 1, C1) is = R1 = 4.375*10-3 Ω, whereas that of the cables splitting off from these (Circuit 2, C2) = R2 = 2.811*10-3 Ω”. And then “For each of the six 50 cm lengths of copper cable, the relevant resistance is 7.028*10-4 Ω”.

        Their joule heating values are a summation across all the circuits.

        But if we use the resistance they used in their calculations from which you’re back calculating, then the voltage is actually in the millivolts by the calculations and assumptions you make here.

        The authors don’t actually supply data on the amperage during operation. What they say on page 13 is: “In the present run of the E-Cat the current flow may actually be higher than 40 A. For this reason, it is expedient to evaluate what portion of the current, fed to the system by the power mains”. They only say “normally 40-50 A” on page 3, but normally doesn’t mean that’s what we’re seeing, and they admit as such in page 13.

        The authors never actually supply the amps or volts measured as far as I have been able to find in the report linked here by E-cat World. Without that knowledge, you can’t back calculate correctly, it’s impossible. And the divergence of their values from yours shows that in spades.

        Instead of believing the author’s table 7 showing consumption in watts, you’re focusing on joule heating while having no reliable information to exactly calculate it. And you come to a completely different resistance value for their wires than they have, while their resistance values are what were used to give the joule heating they report — meaning your results are immediately divergent from them.

        The fact one can drop input power but still raise joule heating by raising the amps while lowering the volts more, shows that in the absence of amperage or voltage data directly supplied, we can’t guess and back calculate input power. They -do- supply the input power consumed though.

        So, I have to side solidly with the authors in the absence of any more available data. I cannot agree with your analysis at all due to divergence from the reported resistance for the wires and the reported input power. Better data and or assumptions than currently presented will have to be supplied to convince me otherwise.

        • fact police

          Ged> The problem with your calculations and even theirs is you are assuming to know the resistance.

          Actually, that’s not a problem at all. The discrepancy is completely independent of the resistance. It only depends on the temperature independence of the resistors, and for copper and inconel it’s certainly close enough. If so, the resistance cancels.

          Regardless of the values of the resistors, if they stay the same, or to the extent they stay the same, the power dissipated in the wires and the coils will remain in the same proportion.

          Therefore, if the power dissipated in the wires increases 6-fold, the power dissipated in the coils has to increase 6-fold as well.

          Ged> You calculate it, but they do too and their numbers are wildly different from yours, by three fold.

          Actually Lammert doesn’t calculate the resistance, and if fact, it’s not necessary, as explained above.

          Ged> The authors don’t actually supply data on the amperage during operation. What they say on page 13 is: “In the present run of the E-Cat the current flow may actually be higher than 40 A. For this reason, it is expedient to evaluate what portion of the current, fed to the system by the power mains”. They only say “normally 40-50 A” on page 3, but normally doesn’t mean that’s what we’re seeing, and they admit as such in page 13. The authors never actually supply the amps or volts measured as far as I have been able to find in the report linked here by E-cat World. Without that knowledge, you can’t back calculate correctly, it’s impossible.

          But they give the current in the dummy run as 19.6 A. And then they say the joule heating is 6-fold higher in the live run, which they can only determine from the current. If the power is 6-fold higher, then the current is root(6) times higher (because the power goes as the square of the current), which is 48 A. So, if you trust the authors’ report of the joule heating, then you can back calculate the current. And 48A, is gonna produce 6 times higher power in the coils compared to the dummy run.

          Ged> Instead of believing the author’s table 7 showing consumption in watts, you’re focusing on joule heating while having no reliable information to exactly calculate it.

          I don’t think you’re following the argument. The authors give the joule heating and the coil heating in the dummy run and the live run. The ratios should be the same if the resistances don’t change. But the ratios are different by a factor of 3. So something is wrong.

          Ged> The fact one can drop input power but still raise joule heating by raising the amps while lowering the volts more,

          If the resistances are constant, then you can’t get more current with lower voltage. That’s your Waterloo.

          • Ged

            I’m sure their control system is perfectly capable of dropping the voltage and increasing the amps to maintain a set power rating, without any change in resistance having anything to do with it. Just ask any burned out electric motor or electronic device that was supplied too little voltage by its controller and thus drew too many amps and overheated. Looking at their wiring diagram, we know their total resistance is lower than the resistance of any one particular resistor segment since it’s in parallel; we also know the amps get divided but the voltage doesn’t. If you want to push more amps across each individual resistor, while keeping power steady, you have to drop the voltage. We also know their controller is giving out at least a 40-50 amp range (page 3), and they plainly state their amperage does not stay steady (page 14): so is their resistance in the wires constantly varying enough to draw 40-50 amps randomly? That would invalidate your claim resistance doesn’t change. But no, it’s all the controller playing with voltage.

            This can easily change the ratios, without changing the resistance at all.

          • Obvious

            I was thinking about dropping the voltage. A lot. Maybe ~25 V. That would put a glow on a wire alright, if the resistor wires can manage the current.

          • Ged

            Their control system can take it at least, and we know that they saw the current climb above 40 A during the run which they talked about in vague terms (and that it had a “normal” range it was fluctuating in of at least 40-50 A); they even talk on page 7 of the controller actually dropping the current to hold power steady early in the run, and then they bumped up the power to compensate. But we don’t know what happened after that, and we don’t know the actual amperage or voltage values for any of the table 7 values, just that the controller was changing those values dynamically by the feedback system.

            The controller also can only take a certain input power before it’d blow itself, and that looks to be around 1 kW from what I read in the specs (but I could be misinterpreting them, and it could be able to handle a lot more), so if it really did draw 2 or 3 kW, it should probably have melted down their controller, or possibly blown a fuse on the main’s circuit (other machinery was on the circuit upstream, and suddenly drawing 2-3 kW is no small matter).

            All these claims that somehow amps can’t change without changing resistance is absurd to me, as we actually see it occurring in the report, and in electric motors, and in electronic devices, and in any system where you change voltage but have a controller trying to hold power steady. Or maybe I really am crazy? You tell me.

          • fact police

            Ged> Their control system can take it at least, and we know that they saw the current climb above 40 A during the run which they talked about in vague terms (and that it had a “normal” range it was fluctuating in of at least 40-50 A);

            They do *not* say fluctuating. They say it is in the range of 40 to 50 A. That level of “fluctuation” in the current is completely inconsistent with the fact that the power was consistent within a few per cent in the power graphs.

            Not that it affects the inconsistency at all.

            Ged> they even talk on page 7 of the controller actually dropping the current to hold power steady early in the run, and then they bumped up the power to compensate.

            It simply says the current was reduced, which they noticed because the power decreased. The assumption is that the power depends directly on the current.

            Ged> But we don’t know what happened after that, and we don’t know the actual amperage or voltage values for any of the table 7 values,

            Only the current is needed to determine the power, and that can be deduced from their reporting of the joule heating as explained elsewhere.

            Ged> The controller also can only take a certain input power before it’d blow itself, and that looks to be around 1 kW from what I read in the specs (but I could be misinterpreting them, and it could be able to handle a lot more), so if it really did draw 2 or 3 kW, it should probably have melted down their controller, or possibly blown a fuse on the main’s circuit (other machinery was on the circuit upstream, and suddenly drawing 2-3 kW is no small matter).

            Yes, you are misinterpreting the specs. The 360 W is the power dissipated in the controller as is clear from the text. So the controller would not have melted down.

            3-phase circuits can handle much more than 2 to 3 kW. An ordinary household, single-phase circuit can handle up to about 2 kW, so this would have been nothing for a 3-phase circuit, and may be why they used a 3-phase circuit.

            The paper says unrelated equipment was connected to the same panel, not on the same circuit. If Rossi’s ntention was to draw up to 3 kW, he could easily have ensured there would be no problem.

            Ged> All these claims that somehow amps can’t change without changing resistance is absurd to me,

            The current can change without changing the resistance, but the voltage-current ratio can’t. Ohm’s law is not absurd.

            Ged> as we actually see it occurring in the report,

            As you say, we don’t know the voltage, so you can’t say that the voltage-current ratio changed.

            and in electric motors,

            This is not an electric motor. There is no back emf with a resistor network.

            Ged> and in any system where you change voltage but have a controller trying to hold power steady.

            The controller controls the voltage. That’s all. It increases it to increase the current and the power, and decreases the voltage to decrease the current and power. You can’t increase the voltage and decrease the current in a resistor network.

            Ged> Or maybe I really am crazy? You tell me.

            I’m sure you’re not crazy. But I’m fairly certain you’re wrong about this.

          • Ged

            Also, as I pointed out elsewhere, you only have to drop the volts by around 20% to drive a 50% increase in joule heating while holding power steady, which would completely account for what was reported in the paper. And there’s a 56% increase in the joule heating from 40 A to 50 A, the range they claim they were “normally” bouncing within. Hmm.

          • fact police

            Ged> Also, as I pointed out elsewhere, you only have to drop the volts by around 20% to drive a 50% increase in joule heating while holding power steady,

            I don’t follow. Joule heating of a resistive load can be expressed V^2/2. So if you drop the voltage by 20%, the power drops by 35% and the joule heating drops by 35%.

            Ged> which would completely account for what was reported in the paper.

            No, it doesn’t account for anything. A drop in the voltage results in a drop in the current through the wires and the coils. So they continue to scale. And if one increases 6-fold, the other must also increase 6-fold. If a different measurement gives 2-fold, one of them is wrong by a factor of 3. Nothing you’ve said about fluctuations changes that.

            Ged> And there’s a 56% increase in the joule heating from 40 A to 50 A, the range they claim they were “normally” bouncing within. Hmm.

            You’re adding words like fluctuate and bouncing which are not present in the paper. Such a fluctuation is completely contrary to the relatively flat power, which varies by only a few per cent, within each of the two regimes.

            And again, that fluctuation would appear in both the wires and the coils, and therefore would not affect the ratio.

          • Thomas Clarke

            Just to add. Uncetrainty in these numbers due to changing powers is not posssible. The values are integrated by the PCE-830 and stored every 0.5s. The numbers in the report are averages from this which take into account, precisely, all fluctuations.

            You should be glad that in many ways these measurements were well done. It does mean there is no wiggle-room in this area.

          • fact police

            Ged> I’m sure their control system is perfectly capable of dropping the voltage and increasing the amps to maintain a set power rating, without any change in resistance having anything to do with it.

            How does that work? Has someone repealed Ohm’s law for the ecat?

            You said yourself that the only thing the power dissipation depends on in a simple resistor is the current. If you increase the current with the same resistance, how does that not increase the power? If the joule heating increases, the coil heating must necessarily increase unless it changes resistance.

            Ged> Just ask any burned out electric motor or electronic device that was supplied too little voltage by its controller and thus drew too many amps and overheated.

            A motor is not a resistive load. It represents an inductor and a generator (generating back emf). I already admitted when I mentioned transformers that it’s possible to change voltage and current independently of power. But not in the kind of circuit we’re considering here.

            Ged> Looking at their wiring diagram, we know their total resistance is lower than the resistance of any one particular resistor segment since it’s in parallel; we also know the amps get divided but the voltage doesn’t. If you want to push more amps across each individual resistor, while keeping power steady, you have to drop the voltage.

            The first part of that was ok. The second part makes no sense. The current depends directly on the voltage. Look at the analysis in wikipedia under 3-phase power. The currents follow Ohm’s law. There is no way in that circuit to increase the current with lower voltage.

            And again, whatever change in current in the wires, it’s the same changes in the coils.

            Ged> We also know their controller is giving out at least a 40-50 amp range (page 3),

            No. We know it is *in* that range.

            Ged> and they plainly state their amperage does not stay steady (page 14):

            Right. The plots of the power show that as well. The average current “changed from day to day”. So?

            Ged> so is their resistance in the wires constantly varying enough to draw 40-50 amps randomly?

            They don’t say it varies between 40 and 50 amps, but even if it did, if the resistance is constant, then the current change is accompanied by a voltage change.

            Ged> That would invalidate your claim resistance doesn’t change. But no, it’s all the controller playing with voltage.

            What’s your point here. I said the voltage and current could not be adjusted independently *if* the resistance was constant.

            Ged> This can easily change the ratios, without changing the resistance at all.

            Only if it violates Ohm’s law.

          • Thomas Clarke

            Ged: voltage, current, power and resistance are exactly related by equations from which you get that the Input power / Joule power ratio is the same as the total resistance / wire resistance ratio.

            I can lead you through every step if you have time.

          • Thomas Clarke

            Correct. If you want to split hairs:

            A nonlinear component – e.g. a diode, or a varistor, could in principle change the ratio at different input powers. But a X3 change is asking a lot and there is no conceivable nonlinear component that could operate as the heater does.

            The beauty of the total power/joule heating power ratio is that it is insensitive to almost everything! And the data from which it is derived is securely stored by the testers direct from the PCE-830. So any possible mistake here can be resolved.

  • georgehants

    Is anybody else going to mention that if the University’s and tax funded scientific establishment were putting in the money and manpower that this proven phenomenon deserves, then all these continual questions may have been answered long ago, or at least we would be receiving reports from many places of the successes and problems involved now.
    Would it be out of order for just one scientist to write a topic page asking these questions and demanding answers NOW?
    Who the hell is ment to do Research and science in this World, single little guys like Mr. Rossi et al, a few volunteers like MFMP etc.with a handful of dollars between them.
    What are these academic comedians doing with their time, beyond living in ivory towers and giving out crazy dictates of what is allowed and what is forbidden to be investigated by the lowly scientific frightened underlings.
    Billions on hot fusion without any results, virtually zero on Cold Fusion and these pages are putting out just an almost deafening silence on the situation.
    Please do not get annoyed about my seeming repetition etc. it is to me unbelievable the wasted repetition and speculation on these pages regarding what Mr.Rossi et al are doing.
    I once asked Mr Rossi on JONP if he drank tea or coffee to illustrate the inane behaviour of supposedly grown people.

    • Dr. Mike

      georgehants,
      I agree. (See my comment to Ophelia above.) I am still hoping the Lugano report gets enough coverage to get university physics departments working on LENR. I assume that key to getting public money flowing into LENR is get those holding the purse-strings interested. No physicist is going to waste time writing a proposal to secure funds for research if he/she believes the proposal will be turned down.
      Dr. Mike

      • georgehants

        Dr. Mike I understand what you are saying, do you agree that is a terrible situation for science to be in?
        Do you agree that it is up to scientists to put that faulty situation right?
        Why are they not on strike or protesting at the White House or Parliament?
        From these pages it appears very few of them have any sort of care or concern for the terrible life harming crimes being committed by science.

        • Dr. Mike

          georgepants,
          I think that science has always been in this terrible situation. New ideas always are hard to accept. It certainly is up to scientists to spread the word on LENR, but I think it will continue to progress slowly until results get published in journals that have a large audience. Perhaps a successful installation of Rossi’s 1MW plant will be the catalyst to finally get LENR moving forward.
          Dr. Mike

      • AlbertNN

        I am sorry to say that the Lugano report is of to low quality to be used for rationalising spending time and money on a replication of the proposed effect at a university or research institution.

        • georgehants

          AlbertNN, you appear to be saying these academic clowns only look at something after it has been fully proven by other people and never do their own Research etc.
          As I ask above, then what are these wasters there for, just to spend taxes on there entertainments and refreshment.
          Maybe you are saying on unproven rubbish like GW or hot fusion.

          • AlbertNN

            Hot fusion is proven, as it has been demonstrated in experiments.

            There is to date no demonstration of the E-cat that lives up to the standards necessary to get funding for materials and time at a university. I had large hopes for the last report, but was disappointed of what I read.

            We should fund research where it has benefit to the society and a reasonable probability of success. The latter can’t be said for any investigations into the E-cat at this stage, as too much is secret, and the demonstrations this far has been very unsatisfactory.

    • Thomas Clarke

      as a non-lurking skeptopath I can answer that by saying that what from your point of view looks like proof, from the point of view of the scientific establishment or even the LENR establishment (some of it) looks like nothing of the kind.

      You can only resolve that apparent difference by understanding in detail what those with views different from you use to form judgements.

      I’d also point out that no-one is stopping Rossi from raising money (he seems good at that) and spending it on his research. If he really has 100s of working e-cats then it would seem well spent and even one of those, in the hands of NASA who have offered, would make the world take notice.

      • georgehants

        Thomas, I am unable to understand what your comment has to do with what I have written.
        Please answer, who is meant to Research and investigate new possible phenomenon in this World if not main-line academic science etc.
        Or should they just sit on their arses and wait for that phenomenon to be researched and proven by small people like Mr. Rossi et al.
        If you think that, then what the devil are these wasters being paid for?

      • Omega Z

        Rossi doesn’t need to raise money. That is Industrial Heats department now that they are the owners of the technology.

        As to how convincing the test was, I don’t’ think it is Rossi’s intent to provide indisputable results to the gallery. Not in best interests of a business development.

        Most of the complaints & questions about the test would be valid if this were strictly a scientific endeavor, But it is not. We need to keep this in mind in the discussion. There is a ton of data that we Don’t, and Wont have access to. It will be provided only to a few with a need to know.

        I have no doubt that If Rossi/IH made every detail available, Dozens of large entities would setup a small R&D lab to replicate it, But not a single one of them would build any products for the market.

        They would focus on developing an alternate technique that could be patented to protect their Billions it will cost to bring such a product to market. This would delay LENR introduction to market by many years.

  • Buck

    Dr. Mike,

    I find it very easy to express that your time spent, writing down and sharing your framing of the problem (perspective, observations, questions, and suggestions), is greatly appreciated. It honors Rossi/IH and all others fighting the tide. And it is a blast to read a coherent depiction.

    Thank you.

    Oh . . . and Happy Halloween . . . it think your posting will only scare the lurking skeptopaths.

    • Dr. Mike

      Buck.
      Thanks for your response. We all should be honoring Rossi for his efforts to make LENR useful for mankind! I’m not so sure this post will scare the lurking skeptopaths since I am pointing out a possible problem with the input power measurements. However, even if I’m correct that there is some error in the input power measurement, science and knowledge will have advanced from the efforts of the Lugano scientists. Many thanks to them for their work on this project!
      Dr. Mike

  • Buck

    Dr. Mike,

    I find it very easy to express that your time spent, writing down and sharing your framing of the problem (perspective, observations, questions, and suggestions), is greatly appreciated. It honors Rossi/IH and all others fighting the tide. And it is a blast to read a coherent depiction.

    Thank you.

    Oh . . . and Happy Halloween . . . it think your posting will only scare the lurking skeptopaths.

    • Dr. Mike

      Buck.
      Thanks for your response. We all should be honoring Rossi for his efforts to make LENR useful for mankind! I’m not so sure this post will scare the lurking skeptopaths since I am pointing out a possible problem with the input power measurements. However, even if I’m correct that there is some error in the input power measurement, science and knowledge will have advanced from the efforts of the Lugano scientists. Many thanks to them for their work on this project!
      Dr. Mike

      • Thomas Clarke

        I’m a non-lurking skeptopath (what a word! are we allowed credopaths too?) but not scared.

        I’ve followed all of the Rossi tests and have been awaiting one that would be rigorous. I have some hope that this one will be, mainly because the PCE-830 was measuring all the inputs, was up to the job, and all its data is stored.

  • Ophelia Rump

    As Rossi said, the time of testing is over. If tests had any merit we would have seen that merit demonstrated by now.

    People will see what they want to see. There are enough people with open minds and open eyes for the market to bring us magic if institutions will not bring us science.

    • Bernie777

      Ophelia…..I wish I could hit the up arrow ten times. Well said, bring on the customers!!

    • Dr. Mike

      Ophelia,
      I have to disagree with you that there is no benefit in second guessing test reports. Science only progresses when results are published and others learn from what is published. Often someone finds an error (could be a minor error or might be a major error) in a published result and getting this error corrected is what really keeps science moving forward. I certainly won’t claim that I’m sure everything that I said in my post is correct. (Actually I know it isn’t- I misquoted the dummy run current as 19.5A when the report said it was 19.7A) However, what if as a result of this post Rossi finds out that his measurement of active input power is not correct for the Hot-Cat? What if this post results in some calculations that shows how much Ni should be used in the Hot-Cat?
      I believe the testing is far from over. I would hope that the physics departments in every university in the world would look at the results of the Lugano tests and decide they had to start running their own LENR experiments.
      Do you have any comments on the key points of my post?
      Dr. Mike

    • LuFong

      Actually if the results aren’t analyzed and the test repeated, it’s just a demonstration and I would agree with you that there is no merit to this. I think that a thorough analysis of the test and test results will either strengthen the results or diminish them. It’s also true that many of the posters on this blog (including me) aren’t very capable of analyzing the test results but that is no reason to stop those that can.

      All of us want to know as much as possible about the E-Cat and there will be precious little additional news coming from IH in the next year. Let us enjoy this report as long as possible!

  • Ophelia Rump

    As Rossi said, the time of testing is over. If tests had any merit we would have seen that merit demonstrated by now.

    People will see what they want to see. There are enough people with open minds and open eyes for the market to bring us magic if institutions will not bring us science.

    There is no benefit in second guessing test reports. It is a waste of time when the reports themselves get no recognition. A laundry list could be made of why the results should have been made higher because the testing was designed conservatively.

    • Bernie Koppenhofer

      Ophelia…..I wish I could hit the up arrow ten times. Well said, bring on the customers!!

      • Mark E Kitiman

        At the time of writing this he only has 7 up-votes… but using the clamp reversal thingy equation then he really has 21 !!!
        But if he is blushing from your praise, then I would have to question the IR measurements, surely his pink cheeks are less than incandescent.

    • kasom

      I agree and I am full of wishful thinking like all of us.
      But why is IH back to the low temp 1MW?

      1MW is a fart in the wind when it comes to heat of real common industrial purposes.

      The dogbone-Cat glowsticks should been tested under load in between.

      pls. Tom Darden explain to the public, why is there still no HOT-CAT in operation worldwide?

      It can not depend on IP issues, this would be ridiculous IMHO, because IH should be abke to sell HEAT only and keep control of the equipment as well.

      • Dr. Mike

        kasom,
        I would guess IH is concentrating on the 1MW plant because getting this plant running is a contractual obligation. I believe that everyone involved would rather be working on the development of the Hot-Cat.
        Dr. Mike

        • Omega Z

          Focus on the 1Mw Lt-Cat is just the logical next step.
          Both from a business and R&D point of view.

          Focusing on the Ht-Cat is just an observers wish point of view…
          It is a lot of R&D away from useful work.

    • Dr. Mike

      Ophelia,
      I have to disagree with you that there is no benefit in second guessing test reports. Science only progresses when results are published and others learn from what is published. Often someone finds an error (could be a minor error or might be a major error) in a published result and getting this error corrected is what really keeps science moving forward. I certainly won’t claim that I’m sure everything that I said in my post is correct. (Actually I know it isn’t- I misquoted the dummy run current as 19.5A when the report said it was 19.7A) However, what if as a result of this post Rossi finds out that his measurement of active input power is not correct for the Hot-Cat? What if this post results in some calculations that shows how much Ni should be used in the Hot-Cat?
      I believe the testing is far from over. I would hope that the physics departments in every university in the world would look at the results of the Lugano tests and decide they had to start running their own LENR experiments.
      Do you have any comments on the key points of my post?
      Dr. Mike

      • Thomas Clarke

        To my knowledge this is the first test with safely stored independent data good enough to provide a definite answer to the question: “How was the apparent COP produced”. If as seems likely now that data shows that a clamp was reversed on the active run but not the dummy, or that the power in measurements were wrong, we still have clear evidence of a carefully measured e-cat with a definite COP of 1.

        That result is more informative than previous tests, where there have been possible measurement issues that could result in the observed COP but no proof that the COP is 1.

        So actually, one way or the other, the details of this test are significant. I would expect the external world to pay attention to a test that convincingly showed COP=1 because in spite of the lack of independence there is no conceivable motive for these professors to rig that results, and the exactness of the result (confirming expected science to within say 10%) makes experiment error unlikely.

        It may seem strange that a negative result should be more powerful than a positive one, but there is no LENR theory that predicts the precise COP observed, whereas there is a standard theory that predicts the precise COP. So an accurate confirmation of the standard theory has more weight than any non-standard and non-predicted result.

        Also – when the testers clearly have a vested emotional interest in a positive result it is only human nature that positive mistakes are more likely than negative ones.

  • Obvious

    The actual power supplied by the triac is reported in column 1, table seven. Calculating it obtusely and getting a result that is three times higher means that those latter calculations are grossly wrong.

    • Dr. Mike

      Obvious,
      I would agree with your second sentence if you would add: those latter calculations are grossly wrong OR THE POWER MEASUREMENT IS INCORRECT! The Cu wire Joule heating measurements are based on a calculation of resistance and a measurement of current. The report states that the current measurements were verified with independent clamp-on ammeters so I have to assume that if a difference were ever seen between the clamp-on ammeters and the PCE-830, it would have been investigated immediately. If you make calculations using only the ratios of the Joule heating numbers, that ratio represents the square of the current ratio, a parameter that was measured by two independent meters. So do you believe a parameter that was measured with two independent meters, or a number that was just read off a single meter.
      If the experimental set-up had really been wired as shown in Figure 4, it would have been hard to imagine how the PCE-830 could have had an erroneous power reading. However, the authors say on page 1 that “specific electromagnetic pulses” were added to the coils, and the picture in Figure 3, page 4 shows another brown box hooked up to the reactor (presumably a pulse generator). My hypotheses is that the connection of this pulse generator resulted in an erroneous power readings on the PCE-830 for the active runs.
      Please continue to look over the data and post your future thoughts.
      Dr. Mike

      • Obvious

        Sorry if my post came across with attitude. I have been dragging my brain through all the calculations and inserting various possibilities and tracking the changes through the math.
        I think the specific pulses are after the control box, so they should be included in the mean power consumption. I don’t think the power consumption measurements are wrong. They are the easiest to do and verify.
        The pulses are not likely to be high powered, but the pulse box could use a highly variable amount of power. There is no way to determine the characteristics of the pulse box.
        The calculated values are prone to various assumptions, many of which are neither falsifiable nor verifiable, unless the math jives.

        • Dr. Mike

          Obvious,
          You are entitled to post with attitude! If the repetition rate of the pulses is low (no data on this) the pulses won’t be adding much power to the reactor. It would have been good for the authors to state that the “specific electromagnetic pulses” contributed very little power to the reactor if this really is true.
          The power consumption measurements should have been easy, but the reported values for the active runs do not “jive” with the other data in the report. What is needed is further explanation from the authors on why the Joule heating in the Cu wires does not correlate with the Joule heating in the Inconel wires for the active runs.
          Dr.Mike

          • Obvious

            Thanks for sticking around to defend your theory. We will all learn something from this discussion, eventually.

      • Ged

        Or their power measurement is correct and we are missing something about their table 7 -calculation-. I find it far easier to assume a math error or lack of our knowledge of all their math parameters, than an instrumentation error; especially since our example is the dummy run and we are given only a vague “may actually be higher than 40” for amperage and nothing on voltage.

        We’re making an awful lot of assumptions to scream that their reported input power is wrong when just a slight increase in amps and drop in volts could explain everything. Also recall their “joule heating” is not simply the Inconel heater wires, but the power line loss of the copper wires feeding the system.

        That’s the point I’ve been trying to make — especially since they are measuring from the main line, as well as after the controller, so this isn’t subject to calculation error like calculating joule heating is, and 2 or 3 KW is a little high for their set up to handle (based on watt densities I’m finding for similar wires and inconel).

        • fact police

          Ged> Or their power measurement is correct and we are missing something about their table 7 -calculation-.

          I agree. Those are the options.

          Ged> I find it far easier to assume a math error or lack of our knowledge of all their math parameters, than an instrumentation error; especially since our example is the dummy run and we are given only a vague “may actually be higher than 40” for amperage and nothing on voltage.

          The current can be determined from the power, so it’s only a question if we trust their calculation of the power. And even that only depends on them using the *same* value for the wire resistance in both calculations, and a measure of the current. Using the same resistance seems reliable enough, so this also comes down to an instrumental question of their measurement of the current.

          And others have shown that a very simple reversal of a current clamp can produce a factor of 3 error in the power, so that is not that difficult to imagine.

          Ged> We’re making an awful lot of assumptions to scream that their reported input power is wrong when just a slight increase in amps and drop in volts could explain everything.

          I disagree. The only assumption is that they measure the current correctly, and use the same value for the wire resistance in both cases. Trusting the authors on that requires no more confidence than trusting they didn’t miss an inverted current clamp. And the latter involves a very rare scientific revolution.

        • Obvious

          Hmmm.

          • Obvious

            Then the total current into the dummy must be 34.12 A.

          • AlbertNN

            The sum of the currents going into the E-cat at any one time is always zero. It is not possible to add RMS-currents, as the RMS value only give you the magnitude, and not the phase. It is possible to add the three time-variant currents in phasor representation, and then the sum would still be zero.

          • Obvious

            19.7 amps average current is flowing towards the reactor through each of the C1 wires. Presumably this current returns down one or two of the other two C1 wires, depending on phasing. What is the total current flowing into the wiring (heater and all) by which heat is made? (IE: what amount of current is the electrical company going to charge you for? Certainly not the instantaneous zero current (that would be nice).)

          • AlbertNN

            The sum of the currents in the three wires going to the E-cat must be zero. This is a consequence of the law of charge conservation, and expressed by Kirchoff’s laws.

            The electrical company is not charging for current, they are charging for power, J or Ws. Also commonly expressed as Wh.

          • Obvious

            The 3 phase electrical line has constant voltage.

            The electrical company cannot charge for energy if they cannot measure the total current moving through 3 phases and add them together and get a number bigger than zero.

            The three phases can be treated as a single phase.

          • AlbertNN

            Please study the theory of electrical engineering and three phase systems. None of your statements this time are correct.

          • Obvious

            I’m obviously not describing my point or question clearly, so I’m not getting any help here. Clearly I do not deal with 3 phase power often.
            Is the total current consumption of the entire dummy assembly, including cables:
            A) 19.7 A (the average of three phases, which are balanced)
            B) a factor of 19.7 A due to three phases (times 1.73)
            C) a sum of 19.7 A due to three phases
            D) a factored sum of some sort

          • AlbertNN

            Current is never consumed. It is talked about like that sometimes, but what you actually are talking about, and what you should add, is power, as in Watts.

          • Obvious

            Summing the power in each phase will give the total power for the system.
            But that requires knowing either U or R. If I’m trying to solve for R or U (V), I would like to know what the total I is. We have a P value.
            Is the 486 W for the dummy run the instantaneous Power, or the total power?

  • Obvious

    The actual measured power supplied by the triac is reported in column 1, table seven. Calculating it obtusely and getting a result that is three times higher means that those latter calculations are grossly wrong.
    The approximate increase in power supplied after day five is about 10%, which is about the same as the increase in Joule heating, (but these values are calculated, and may have a minor error).

    EDIT: I think there IS something weird about the Joule heating curve, but I’m still trying to work it out.

    • fact police

      Obvious> The approximate increase in power supplied after day five is about 10%, which is about the same as the increase in Joule heating.

      Right. And that indicates that the resistance of the heater is largely independent of temperature.

      However, compared to the dummy run, the joule heating increases 6-fold, whereas the claimed heater power increases only 2-fold. For a constant resistance the heater power and the joule heating should be proportional. So something clearly changes from the dummy run to the real run. That’s the conundrum.

      • Obvious

        A doubling of the current input should result in a 4 times increase in Joule heating, due to I^2. Since the highest input is nearly double the dummy run, then normally I would expect to see something like ~28 W, rather than something in the 40 W range for the Joule Heating values.

        The possibility remains that a very skinny pulse (modulated) of AC is used for the Joule heating of the dummy (not at full power). This skinny pulse would not conduct power through all the C2 leads, since the full 120° is not used, and the normal Kirchhoff law applies, as though it were DC.
        Then (possibly) when higher power levels are delivered, then the AC square root of three rule applies, when the width of the pulse is wide enough to use all three phases through the C2 wires, increasing the current load, and therefore Joule heating. There is no indication of this calculation methodology in the report, however.

        • Ged

          I think they just miscalculated the joule heating for table 7, or we’re missing that the joule heating is a huge summation of all the wires that are conducting electricity anywhere along the path, and something is being added in that wasn’t in the dummy example. But I think they just misreported since the input power is an actual measured quantity but the joule heating is not. And we do not know the data used to calculate the joule heating during the live run, only their example on the dummy (i.e. we get no amperage or voltage data for the live run, at least that I can find).

          • Obvious

            The Compact Fusion Triac uses 1.3W per amp load per phase. Maybe that got added in to the Joule heating for the fueled run? (wild guess)

          • Thomas Clarke

            They clearly state that the dummy calculations are done to show the methodology, and therefore must use the same for the real run.

            If they are so incompetent as to make a X3 mistake in the Joule heating power then it could just as easily be a X3 mistake in the input power.

            Actually the in power is the more vulnerable measurement. Either of the following things could reduce the power measurement by a factor of 3 but not current, from which the Joule heating power is derived:
            Current Clamp reversed (see andrea.s elsewhere on the waveform evidence for this)
            Display set to line power instead of total power

            The clamps were taken off and replaced between the dummy and active runs, because the reactor was removed, so the clamp reversal option looks possible.

          • Mark E Kitiman

            Is this your assumption?

            ‘They clearly state that the dummy calculations are done to show the
            methodology, and therefore must use the same for the real run.’

            Where do they state this ? Why must they use the same for the real run?

            And…

            ‘The clamps were taken off and replaced between the dummy and active runs, because the reactor was removed’

            Please expand… the reactor was removed to where?

          • Dr. Mike

            Ged,
            The Cu wire Joule heating in table 7 actually agrees with the statement on page 3 that the normal operating current was 40-50 Amps. On page 14 the authors stated that Joule heating for Table 7 for the active run was calculated in the same manner as for the dummy run; “We may calculate the dissipated heat of the limited extent of the dummy reactor: the results relevant to the E-Cat will be given in Table 7, due to the fact the average current values changed from day to day”.
            Dr. mike

          • Thomas Clarke

            Ged,

            The Joule heating is calculated as shown in the dummy calculation by multiplying the square of the RMS current by the effective wire resistance, which is constant for the dummy and active runs.

            The current is measured by the PCE-830, which also multiples currents by voltages and displays total power.

            I think you are incorrect in imagining the testers to be unable to perform a simple multiplication on the real test that they have correctly performed on the dummy test.

            However, were this the case, I do not understand why you trust their much more complex calculations to determine output power on the active test.

        • fact police

          Obvious> A doubling of the current input should result in a 4 times increase in Joule heating, due to I^2.

          Agreed.

          Obvious> Since the highest input is nearly double the dummy run,

          That’s the highest input *power* is nearly double. That means the highest input current (according to this) is root(2) times the dummy run.

          Obvious> then normally I would expect to see something like ~28 W, rather than something in the 40 W range for the Joule Heating values.

          No. If the current increases by the root(2), then the joule heating should increase by a factor of 2, to 14 W or so.

          In other words, the joule heating and the input power should increase by the same factor.

          Obvious> The possibility remains that a very skinny pulse of AC is used for the Joule heating of the dummy (not at full power). This skinny pulse would not conduct power through all the C2 leads, since the full 120° is not used, and the normal Kirchhoff law applies, as though it were DC.

          Then (possibly) when higher power levels are delivered, then the AC square root of three rule applies, when the width of the pulse is wide enough to use all three phases through the C2 wires, increasing the current load, and therefore Joule heating. There is no indication of this calculation methodology in the report, however.

          And they don’t even know what the special waveform is. Their calculation of the joule heating could only be based on the current measurement in the live run, just as it was based on the current measurement in the dummy run (19.7 A), and using the same resistance. So, to get 6 times higher power, means the current is root(6) times higher. And that should produce 6-fold increase in the coil power.

        • Ged

          Well, looking back at the paper, we see they have a controller range of 40-50 amps. On page 3: “The regulator is driven by a potentiometer used to set the operating point (i.e. the current through the resistor coils, normally 40-50 Amps)”.

          That means we have an I^2 range of 56% variance (1600 vs 2500 for I^2 at the two ends of that range). Let’s add just 50% to the joule heating calculated to the dummy, thus the 7 becomes 10.5 W. Now let’s take 4 times 10.5 W = 42W.

          Hey look, everything is now explained if we simply assume the dummy run was near the 40 amp end of their range and the experimental run was up at the 50 amp end of their range. Meanwhile, volts can vary and input power stay the same 900 W while joule heating can change substantially.

          Nothing strange at all guys, just us forgetting variance, that they themselves report, is a thing.

          • fact police

            Ged> Well, looking back at the paper, we see they have a controller range of 40-50 amps.

            For the live run. In the dummy run they used 19.7 A.

            Ged> On page 3: “The regulator is driven by a potentiometer used to set the operating point (i.e. the current through the resistor coils, normally 40-50 Amps)”.

            That means we have an I^2 range of 56% variance

            You’re misreading. That does not represent variance, which they said elsewhere was much smaller. That represents the normal range of currents used.

            Ged> Let’s add just 50% to the joule heating calculated to the dummy, thus the 7 becomes 10.5 W. Now let’s take 4 times 10.5 W = 42W.

            Even if this were reasonable (and it’s not), if the joule heating is 4 times higher, then the reactor heating should also be 4 times high as well. But the reactor heating is only 2 times higher.

            Ged> Hey look, everything is now explained if we simply assume the dummy run was near the 40 amp end of their range

            They said the current in the dummy run was *measured* as 19.7 A. So, you’re way off here.

            Ged> Meanwhile, volts can vary and input power stay the same 900 W while joule heating can change substantially.

            No it can’t, unless the resistors change in value. A change in voltage necessarily produces a change in power the resistance doesn’t change.

            Ged> Nothing strange at all guys, just us forgetting variance, that they themselves report, is a thing.

            No. The discrepancy is still there. A 6-fold increase in joule heating, but a 2-fold increase in input power. If that can be attributed to variance, then the COP of 3 can be attributed to variance.

        • Thomas Clarke

          That cannot be true if the report power measurements are correct. the input powers are stated as 500W/800W/900W and that is at most a 50% change in duty cycle. we also have the evidence from the photograph and the spectral analysis that the pulses are as expected.

          The ratio of Joule heating to total power depends only on the ratio of the resistances. This is a very strong result.

      • Dr. Mike

        Fact police,
        Well said!
        Dr. Mike

    • Dr. Mike

      Obvious,
      I would agree with your second sentence if you would add: those latter calculations are grossly wrong OR THE POWER MEASUREMENT IS INCORRECT! The Cu wire Joule heating measurements are based on a calculation of resistance and a measurement of current. The report states that the current measurements were verified with independent clamp-on ammeters so I have to assume that if a difference were ever seen between the clamp-on ammeters and the PCE-830, it would have been investigated immediately. If you make calculations using only the ratios of the Joule heating numbers, that ratio represents the square of the current ratio, a parameter that was measured by two independent meters. So do you believe a parameter that was measured with two independent meters, or a number that was just read off a single meter.
      If the experimental set-up had really been wired as shown in Figure 4, it would have been hard to imagine how the PCE-830 could have had an erroneous power reading. However, the authors say on page 1 that “specific electromagnetic pulses” were added to the coils, and the picture in Figure 3, page 4 shows another brown box hooked up to the reactor (presumably a pulse generator). My hypotheses is that the connection of this pulse generator resulted in an erroneous power readings on the PCE-830 for the active runs.
      Please continue to look over the data and post your future thoughts.
      Dr. Mike

      • Obvious

        Sorry if my post came across with attitude. I have been dragging my brain through all the calculations and inserting various possibilities and tracking the changes through the math.
        I think the specific pulses are after the control box, so they should be included in the mean power consumption. I don’t think the power consumption measurements are wrong. They are the easiest to do and verify.
        The pulses are not likely to be high powered, but the pulse box could use a highly variable amount of power. There is no way to determine the characteristics of the pulse box.
        The calculated values are prone to various assumptions, many of which are neither falsifiable nor verifiable, unless the math jives.

        • Dr. Mike

          Obvious,
          You are entitled to post with attitude! If the repetition rate of the pulses is low (no data on this) the pulses won’t be adding much power to the reactor. It would have been good for the authors to state that the “specific electromagnetic pulses” contributed very little power to the reactor if this really is true.
          The power consumption measurements should have been easy, but the reported values for the active runs do not “jive” with the other data in the report. What is needed is further explanation from the authors on why the Joule heating in the Cu wires does not correlate with the Joule heating in the Inconel wires for the active runs.
          Dr.Mike

          • Obvious

            Thanks for sticking around to defend your theory. We will all learn something from this discussion, eventually.

      • Ged

        Or their power measurement is correct and we are missing something about their table 7 -calculation-. I find it far easier to assume a math error or lack of our knowledge of all their math parameters, than an instrumentation error; especially since our example is the dummy run and we are given only a vague “may actually be higher than 40” for amperage and nothing on voltage.

        We’re making an awful lot of assumptions to scream that their reported input power is wrong when just a slight increase in amps and drop in volts could explain everything. Also recall their “joule heating” is not simply the Inconel heater wires, but the power line loss of the copper wires feeding the system.

        That’s the point I’ve been trying to make — especially since they are measuring from the main line, as well as after the controller, so this isn’t subject to calculation error like calculating joule heating is, and 2 or 3 KW is a little high for their set up to handle (based on watt densities I’m finding for similar wires and inconel), but 900 W is totally reasonable.

        • fact police

          Ged> Or their power measurement is correct and we are missing something about their table 7 -calculation-.

          I agree. Those are the options.

          Ged> I find it far easier to assume a math error or lack of our knowledge of all their math parameters, than an instrumentation error; especially since our example is the dummy run and we are given only a vague “may actually be higher than 40” for amperage and nothing on voltage.

          The current can be determined from the power, so it’s only a question if we trust their calculation of the power. And even that only depends on them using the *same* value for the wire resistance in both calculations, and a measure of the current. Using the same resistance seems reliable enough, so this also comes down to an instrumental question of their measurement of the current.

          And others have shown that a very simple reversal of a current clamp can produce a factor of 3 error in the power, so that is not that difficult to imagine.

          Ged> We’re making an awful lot of assumptions to scream that their reported input power is wrong when just a slight increase in amps and drop in volts could explain everything.

          I disagree. The only assumption is that they measure the current correctly, and use the same value for the wire resistance in both cases. Trusting the authors on that requires no more confidence than trusting they didn’t miss an inverted current clamp. And the latter involves a very rare scientific revolution.

  • LuFong

    FYI from the report, “The other IR camera was primarily used to frame the hollow rods containing the power cables, and its position was changed often in the course of the test. When experimental conditions were seen to be constant, it would be pointed towards various parts of the reactor as well as of the rods, in order to verify the symmetry of heat emission and thus yield a more comprehensive picture of the thermal behavior of the system.” [Emphasis mine]

    Good post. Will give me something to read and think about over the weekend!

    • Dr. Mike

      LuFong,
      Thanks for pointing this out. Their statement seems to say that they saw no obvious non-uniformity of the active runs as compared to the dummy run (or they probably would have reported it). Looking forward to seeing your comments on the rest of the post.
      Dr. Mike

      • LuFong

        Yes, I think we can reasonably infer that they looked for temperature gradients and hotspots over the reactor core and that if they found them they would have pointed these out since it would affect the heat calculations.

    • Obvious

      Using the formula and currents described in the dummy calculations, and inserting a current value 2.5 times higher (C1 current at 49.25 A [2.5x 19.7), and C2 current at 24.63 A [2.5 x 9.85]), then the cable Joule loss result is (31.84 + 10.23) = 42.07 W.

      • fact police

        And if the current is 2.5 times higher, then the power input (and the power in the coils) will be 2.5^2 = 6.25 times higher than in the dummy run, which is about 2800W, not 900W.

        • Andreas Moraitis

          You assume that the resistances of the coils are constant, or nearly constant. But that is not what one might expect in case that strong magnetic fields and/or superconductivity are involved (see the posts below).

          • fact police

            Yes. I’ve repeated the assumption of constant resistance over and over, but in this case I didn’t. But a variable resistance is not consistent with what happens on day 10 when they increase the power from 785.79 W to 923.71 W, and the joule heating increases by almost exactly the same proportion, indicating in this range the resistance is constant. And it would take a reduction of the resistance by a factor of 3 to account for the discrepancy.

          • Andreas Moraitis

            Maybe the supposed magnetic (or whatever) effect remains stable during the operation. At least, it could explain the difference between the dummy run and the active runs. In any case, we need more information from the authors. I hope that they have recorded both currents and voltages for the whole period.

          • Obvious

            What if we use pure nickel wire for the reactor resistor, rather than Inconel? All standard alloys of Inconel melt below 1425° C anyways.

          • AlbertNN

            The currents measured are to low to give high magnetic fields where anything exotic might happen. And the resistance wires used according to the paper are not superconducting at high temperatures.

          • Andreas Moraitis

            The magnetic fields which have been measured by DGT are far too strong that you could explain them as a result of the supplied currents. They must have been caused by the reaction. High-temp superconductivity in the context of LENR has been hypothesized by various authors. However, I don’t know if there are reliable data on this effect.

        • Obvious

          Except that we have a measured power consumption that does not agree with that. If we cannot trust the measured quantities, then the entire Joule heating discussion is a waste of time, since those measurements are part of the Joule calculation.

      • Dr. Mike

        fact police,
        Again thank you for your careful explanation to Ged.
        Dr. Mike

    • Andreas Moraitis

      Yes, but if the reactor itself generates electric energy that is transferred back to the power supply unit, this would contribute also to the joule heating of the wires. One could not deduce the supplied energy from the measured currents.

  • Bob Greenyer

    Thanks Dr. Mike,

    We have been planning to work off Ikegami’s proposal of the operation of the low temp ECat in the powder cell.

    A hybrid interim test in the sparky cell.

    We have already agreed that in the short term we add extra coils further from the core to do RF – Bob H says at a later time he is experienced in how to add these onto the power waveform.

    • Dr. Mike

      Bob,
      That’s great! Good luck with your work!
      Dr. Mike

      • Bob Greenyer

        And keep up yours!

        I have a question for you.

        You must first assume that one of the heating elements or supply phases has failed in some way. This would result in an “open delta” whose two remaining heater elements can only generate 33% of the required heat output. Now assume that the controller compensates by changing the phase angle on the controller to address the loss. The net result would be a move away from very short pulses to longer ones. The input power to the CompactFusion would be the same.

        So the question is, would this affect the dissipation in the supply wires?

        You will need this

        http://www.chromalox.com/catalog/resources/technical-information/Electrical-Wiring-Theory-Three-Phase-Equations-Wiring-Diagrams.pdf

        • Dr. Mike

          Bob,
          I assume the scientists were regularly measuring the current in each phase of the output from the TRIAC with the supplemental clamp-on ammeters and would have discovered an open heater coil. I checked your link above. This article states that if one coil in the delta wiring opens up the output power drops by 33% (midway down in the first column) so the coils would continue to generate 67% of their power. There is not enough information in the report to determine how the thermocouple output was used to make fine adjustments to the output power, but I assume that if the temperature of the reactor started to drop due to a loss of a coil, the feedback from the TC would cause the power regulator to increase the current to the remaining 2 coils to try to maintain temperature. I think that an event such as this would have been easily detected and would have been included in the report if it had happened.
          Dr. Mike

          • Bob Greenyer

            You are absolutely right about it being a 33% drop – 50% if a loss of a Y. Reminds me that I can’t always expect to perform best in the middle of the night!

            I love Live Open Science – no mistake lives long!

  • Bob Greenyer

    Thanks Dr. Mike,

    We have been planning to work off Ikegami’s proposal of the operation of the low temp ECat in the powder cell.

    A hybrid interim test in the sparky cell.

    We have already agreed that in the short term we add extra coils further from the core to do RF – Bob H says at a later time he is experienced in how to add these onto the power waveform.

    • Dr. Mike

      Bob,
      That’s great! Good luck with your work!
      Dr. Mike

      • Bob Greenyer

        And keep up yours!

        I have a question for you.

        You must first assume that one of the heating elements or supply phases has failed in some way. This would result in an “open delta” whose two remaining heater elements can only generate 33% of the required heat output. Now assume that the controller compensates by changing the phase angle on the controller to address the loss. The net result would be a move away from very short pulses to longer ones. The input power to the CompactFusion would be the same.

        So the question is, would this affect the dissipation in the supply wires?

        You will need this

        http://www.chromalox.com/catalog/resources/technical-information/Electrical-Wiring-Theory-Three-Phase-Equations-Wiring-Diagrams.pdf

        • Thomas Clarke

          The Joule heating figures were derived from line currents. Unless the testers were electrically incompetent, in which case we cannot trust any of the report results, they would use the sum of the three RMS currents, multiplied by the resistance.

          This figure is proportional to the total input power, independently of the number of lines active. So lines open circuit with more power on other lines would not change the ratio of Joule heating to input power.

          There are only two possibles for this:

          (1) The input power was wrongly measured (reverse clamp configuration, or perhaps something else) in the active test but not the dummy test

          (2) The heater wire is not Inconel as stated, but some unusual semiconductor that has resistance exactly 3X smaller at 1250C than at 500C. And has resistance nearly identical between 1250C and 1400C. Silicon Carbide is quite close to this but the resistance reduction happens at a much lower temperature.

          The testers have the data from the PCE-830 needed to resolve this and make the test much more solid. They have not yet revealed what that data tells us. I find this surprising – any scientist, when such cogent questions are raised about a possible error in their published work, would normally feel it important to answer the criticism as soon as possible if they are able.

          • Dr. Mike

            Thomas,
            Well said! I think that even with the data from the PCE-830 there still may be a problem. If the Insertion of the pulse generator affected the power calculation by the PCE-830, the power data from the PCE-830 may not be any good. I believe that they reported the data they read off the meter (column 2, Table 7, page 22 of the report). i also would hope they are carefully going over all of their data and their electrical set-up so that they can eventually explain a possible error in their work, or better yet, why there is no error.

          • Obvious

            A much simpler version would be to have a few mechanical thermo switches (bimetallic or something along those lines) in the reactor that could series/parallel the resistances differently, depending on temperature. I doubt that something like that would survive long in that heat….

        • Dr. Mike

          Bob,
          I assume the scientists were regularly measuring the current in each phase of the output from the TRIAC with the supplemental clamp-on ammeters and would have discovered an open heater coil. I checked your link above. This article states that if one coil in the delta wiring opens up the output power drops by 33% (midway down in the first column) so the coils would continue to generate 67% of their power. There is not enough information in the report to determine how the thermocouple output was used to make fine adjustments to the output power, but I assume that if the temperature of the reactor started to drop due to a loss of a coil, the feedback from the TC would cause the power regulator to increase the current to the remaining 2 coils to try to maintain temperature. I think that an event such as this would have been easily detected and would have been included in the report if it had happened.
          Dr. Mike

          • Bob Greenyer

            You are absolutely right about it being a 33% drop – 50% if a loss of a Y. Reminds me that I can’t always expect to perform best in the middle of the night!

            I love Live Open Science – no mistake lives long!

  • Obvious

    A doubling of the current input should result in a 4 times increase in Joule heating, due to I^2. Since the highest input is nearly double the dummy run, then normally I would expect to see something like ~28 W, rather than something in the 40 W range for the Joule Heating values.

    • Ged

      I think they just miscalculated the joule heating for table 7, or we’re missing that the joule heating is a huge summation of all the wires that are conducting electricity anywhere along the path, and something is being added in that wasn’t in the dummy example. But I think they just misreported since the input power is an actual measured quantity but the joule heating is not. And we do not know the data used to calculate the joule heating during the live run, only their example on the dummy (i.e. we get no amperage or voltage data for the live run, at least that I can find).

      • Obvious

        The Compact Fusion Triac uses 1.3W per amp load per phase. Maybe that got added in to the Joule heating for the fueled run? (wild guess)

      • Dr. Mike

        Ged,
        The Cu wire Joule heating in table 7 actually agrees with the statement on page 3 that the normal operating current was 40-50 Amps. On page 14 the authors stated that Joule heating for Table 7 for the active run was calculated in the same manner as for the dummy run; “We may calculate the dissipated heat of the limited extent of the dummy reactor: the results relevant to the E-Cat will be given in Table 7, due to the fact the average current values changed from day to day”.
        Dr. mike

    • Ged

      Well, looking back at the paper, we see they have a controller range of 40-50 amps. On page 3: “The regulator is driven by a potentiometer used to set the operating point (i.e. the current through the resistor coils, normally 40-50 Amps)”.

      That means we have an I^2 range of 56% variance (1600 vs 2500 for I^2 at the two ends of that range). Let’s add just 50% to the joule heating calculated to the dummy, thus the 7 becomes 10.5 W. Now let’s take 4 times 10.5 W = 42W.

      Hey look, everything is now explained if we simply assume the dummy run was near the 40 amp end of their range and the experimental run was up at the 50 amp end of their range. Meanwhile, volts can vary and input power stay the same 900 W while joule heating can change substantially.

      Nothing strange at all guys, just us forgetting variance, that they themselves report, is a thing.

  • Dr. Mike

    kasom,
    I would guess IH is concentrating on the 1MW plant because getting this plant running is a contractual obligation. I believe that everyone involved would rather be working on the development of the Hot-Cat.
    Dr. Mike

    • Omega Z

      Focus on the 1Mw Lt-Cat is just the logical next step.
      Both from a business and R&D point of view.

      Focusing on the Ht-Cat is just an observers wish point of view…
      It is a lot of R&D away from useful work.

  • Observer

    You assume the product of the reaction is heat. How can the input be heat and the output be heat and not have positive feedback? Periodic depletion of resources? Maybe in the pulsed mode, but not in the steady state mode.

    For stable operation, the reaction temperature must only be a product of the heater temperature.

    The question is, if the energy output of the e-cat is not heat, then what is it, and what and where is it converted to heat.

    • Ophelia Rump

      The input heat is a driver not a limiter. As you indicated feedback would prohibit it acting as a limiter.

      There is another driver which acts independently of the heat which it produces and can be used as both a driver and a choke, so there is no logical requirement for any other form of output than heat. Which is not to say that there is not any other form of output energy.

      • Observer

        Find the negative feedback mechanism, and you will understand Rossi’s reactor (if not Rossi’s reaction).

  • Ophelia Rump

    The input heat is a driver not a limiter. As you indicated feedback would prohibit it acting as a limiter.

    There is another driver which acts independently of the heat which it produces and can be used as both a driver and a choke, so there is no logical requirement for any other form of output than heat. Which is not to say that there is not any other form of output energy.

    • Ged

      Resistance need not change, the controller need only drop the voltage and voila. We already have an amperage range of 40-50 going on here, so is resistance constantly changing to drive these amp changes?

      I have tried to find where they use current to calculate input power, but I haven’t located it. I see they measured it with the PCE directly, not calculating it just from current (which wouldn’t work without voltage anyways).

      Also, we see the control system varying the amperage to hold input power steady in the paper itself, invalidating your claims that that can only change by resistance changes. To wit (page 7): “After this initial period, we noticed that the feedback system had gradually cut back the input current, which was yielding about 790 W. We therefore decided to increase the power, and set it slightly above 900 W”

      Even further, we see on page 3, that the power supply itself would not be able to handle the kilowatts you guys are trying to claim, such as : “its maximum nominal power consumption is 360 W” (I am assuming that’s for each of the three channels independently, so 1080 W total).

      Your claims that voltage and amperage to the system can’t fluctuate while holding input power the same without changing resistance is demonstrably mistaken according to the paper itself when they saw current being cut back and had to up the power to counter it. I guess that is your Waterloo?

      • Ged

        They don’t report only a 2-fold increase in power dissipated by the coils. I do not see where you get that from, so if I’m mistaken there, please show me the reference.

        What they do show, according to page 7 for instance, is the input power directly detected going to the reactor — this is not the dissipation by the coils, but the electric draw of the system. Therefore, you can have a 2-fold increase in input power and a 6 fold (50% more than 4 fold, and if you look at page 3, you’ll notice their I^2 is varying by at least 56%) in joule heating, simply by having their feedback controller drop voltage to keep current pumped. We see them having to modulate this during the actual run.

    • Dr. Mike

      Thomas,
      I believe you are correct that Rossi’s patent application is not going to provide protection for a Hot-Cat design device. I don’t think Rossi had a good enough theory when the patent was applied for to write up a good all-encompassing patent. Also, I don’t see how he thought he could get a patent without disclosing the entire invention, including the catalyst. I agree that there is probably enough disclosed in the Lugano repiort on the Hot-Cat design and fuel to make it non-patentable. Your question is a good one!
      Dr. Mike

  • Andreas Moraitis

    I agree with Ged’s statement that you cannot draw conclusions about the input energy on the basis of the values for joule heating. You would have to know either all the resistances or the applied voltages. One hypothetical explanation for the results is that the reaction produces a magnetic field which has an effect on the wires. It is known that LENR can generate very strong fields; for example, DGT have measured 1.6 Tesla at 20 cm distance (!) from their reactor. However, it is unclear if that field was static, pulsating or alternating. A non-static field should cause the reactor to act as an electric generator. Therefore, the effective resistance of the coils would be altered significantly, so that any calculation on the basis of the behaviour of inconel wires under normal circumstances would be invalid.

    • Dr. Mike

      Andreas,
      Good point about the magnetic fields. It would be interesting if anyone reading these comments knows if Inconel wire has a large magneto-resistance effect. It would require the resistance to change by a factor of 3 going from the dummy run to the active run, but then have the resistance in the Inconel wires increase by about 1% when the active power was increased after 10 days into the experiment.
      Dr. Mike

  • Andreas Moraitis

    I agree with Ged’s statement that you cannot draw conclusions about the input energy on the basis of the values for joule heating. You would have to know either all the resistances or the applied voltages. One hypothetical explanation for the results is that the reaction produces a magnetic field which has an effect on the wires. It is known that LENR can generate very strong fields; for example, DGT have measured 1.6 Tesla at 20 cm distance (!) from their reactor. However, it is unclear if that field was static, pulsating or alternating. A non-static field should cause the reactor to act as an electric generator. Therefore, the effective resistance of the coils would be altered significantly, so that any calculation on the basis of the behaviour of inconel wires under normal circumstances would be invalid.

    • Dr. Mike

      Andreas,
      Good point about the magnetic fields. It would be interesting if anyone reading these comments knows if Inconel wire has a large magneto-resistance effect. It would require the resistance to change by a factor of 3 going from the dummy run to the active run, but then have the resistance in the Inconel wires increase by about 1% when the active power was increased after 10 days into the experiment.
      Dr. Mike

    • fact police

      Moraitis> DGT have measured 1.6 Tesla at 20 cm distance (!) from their reactor.

      But DGT have been discredited.

      A non-static field should cause the reactor to act as an electric generator. Therefore, the effective resistance of the coils would be altered significantly,

      Any change in current through the coils because of an emf generated by a magnetic field will also cause a change in the current through the wires. So the ratio of powers should not be affected.

      • Andreas Moraitis

        Yes, but if the reactor itself generates electric energy that is transferred back to the power supply unit, this would contribute also to the joule heating of the wires. One could not deduce the supplied energy from the measured currents.

  • jousterusa

    What is so exciting about E-Catworld these days is reading of the global coverage of the TIP report, ideas about the E-Cat’s implementation, Rossi’s reaction to it, and the general freshet of news bytes from all over the globe welcoming a device that will forever change the world. I am extremely interested in learning how the Chinese developers of the E-Cat industrial park and fabrication plant are doing, thinking and building – news that is hard to come by!

    • Ophelia Rump

      Yes.

      • jousterusa

        Are you a single woman, Ophelia, by any chance? I’m a single guy…

        “Nymph, in thy orisons, be all my sins remembered?”

  • jousterusa

    What is so exciting about E-Catworld these days is reading of the global coverage of the TIP report, ideas about the E-Cat’s implementation, Rossi’s reaction to it, and the general freshet of news bytes from all over the globe welcoming a device that will forever change the world. I am extremely interested in learning how the Chinese developers of the E-Cat industrial park and fabrication plant are doing, thinking and building – news that is hard to come by!

    • Ophelia Rump

      Yes.

      • jousterusa

        Are you a single woman, Ophelia, by any chance? I’m a single guy…

        “Nymph, in thy orisons, be all my sins remembered?”

  • georgehants

    Dr. Mike I understand what you are saying, do you agree that is a terrible situation for science to be in?
    Do you agree that it is up to scientists to put that faulty situation right?
    Why are they not on strike or protesting at the White House or Parliament?
    From these pages it appears very few of them have any sort of care or concern for the terrible life harming crimes being committed by science.

    • Dr. Mike

      georgepants,
      I think that science has always been in this terrible situation. New ideas always are hard to accept. It certainly is up to scientists to spread the word on LENR, but I think it will continue to progress slowly until results get published in journals that have a large audience. Perhaps a successful installation of Rossi’s 1MW plant will be the catalyst to finally get LENR moving forward.
      Dr. Mike

  • Giuliano Bettini

    fact police> “However, compared to the dummy run, the joule heating increases
    6-fold, whereas the claimed heater power increases only 2-fold. For a constant
    resistance the heater power and the joule heating should be proportional. So
    something clearly changes from the dummy run to the real run. That’s the
    conundrum.”

    Maybe the resistance changes from “dummy” to “run” because there are two different heater helices? Images apparently show two different helices.

    https://www.facebook.com/photo.php?fbid=754649834605839&set=gm.1001924113156826&type=1

    • Fortyniner

      Lots of very informed speculation there – probably one or two guesses are very close to the truth. More info is required to home in on the actual design, but as that was the last published test, we are not now likely to learn much more.

    • Ophelia Rump

      Is there a rule which says the nickel in the resistors cannot also react?

      • Dr. Mike

        Ophelia,
        The coils are on the outside of the alumina so they would not be exposed to any hydrogen.
        Dr. Mike

        • Ophelia Rump

          Hydrogen penetrates. I also believe there was some question of the role of the hydrogen if the nickel is actually the fuel.

          If you want to explain anomalous heat in a nickel resistor in an inexplicable reaction which has nickel for a fuel, it might be unwise to rationalize away the obvious.

          • Dr. Mike

            Ophelia,
            The conversion of the isotopes of Ni all to Ni62 contributes some output heat. However, the reactor was working well when it was shut down so the conversion of the Ni all to Ni62 isn’t the fuel of the reactor unless the reactor just happened to be stopped as the last of the Ni was converted. It is more likely that the Ni reactions are actually competing with the primary reactions, and possibly the reactor works better when all of the Ni has been converted to Ni62. (Just a guess.) It is more likely that hydrogen is the real fuel (and maybe also the lithium). I would also guess that if the reactor had even a small leak, the hydrogen would leak out and the reaction would stop.
            Dr. Mike

      • not clear what is the structure of the e-cat…
        given the apparent change in resistance (beware it can be just bad assumption as people used Rapp=P/Ieff^2…with switched triphase in V mode not delta) one hypothesis is a negative temperature coefficient, and in an E-cat the evidence hypothesis is that it is a LENR hydride, the “mouse”…

        speculation, but at least it is not a conspiracy like to inverted current clamps on only the active test, with strange controlbox that in that crazy context, consume energy in both blank (right wiring) and active (wrong wiring)…
        all without any of the testers having any suspicion…

        at least an international conspiracy involving Exxon, Putin, with Total boss assassination, Ukraine MH17 and ISIS/CIA collusion, is more probable.

    • Dr. Mike

      Giuliano,
      There are actually 3 helices wound on the alumina, one for each phase of the 3 phase power supply. The idea behind the dummy run was to see how the reactor heated with just electrical power supplied so that its effect could be subtracted out of the LENR effect. An ideal control for the experiment would have been to heat the reactor up to the temperature during the active run just using the electrical heating.
      Dr. Mike

  • Giuliano Bettini

    fact police> “However, compared to the dummy run, the joule heating increases
    6-fold, whereas the claimed heater power increases only 2-fold. For a constant
    resistance the heater power and the joule heating should be proportional. So
    something clearly changes from the dummy run to the real run. That’s the
    conundrum.”

    Maybe the resistance changes from “dummy” to “run” because there are two different heater helices? Images apparently show two different helices.

    https://www.facebook.com/photo.php?fbid=754649834605839&set=gm.1001924113156826&type=1

    • Lots of very informed speculation over there – probably one or two guesses are very close to the truth, but everyone is just spinning their wheels until more information is available. As that was the last published test, we are probably out of luck, unless MFMP manage to pull a rabbit out of the hat.

    • Ophelia Rump

      Is there a rule which says the nickel in the resistors cannot also react?

      • Dr. Mike

        Ophelia,
        The coils are on the outside of the alumina so they would not be exposed to any hydrogen.
        Dr. Mike

        • Ophelia Rump

          Hydrogen penetrates. I also believe there was some question of the role of the hydrogen if the nickel is actually the fuel.

          If you want to explain anomalous heat in a nickle resistor in an inexplicable reaction which has nickel for a fuel, it might be unwise to rationalize away the obvious. For all we know, once the reaction begins there may be field effects in play.

          • Dr. Mike

            Ophelia,
            The conversion of the isotopes of Ni all to Ni62 contributes some output heat. However, the reactor was working well when it was shut down so the conversion of the Ni all to Ni62 isn’t the fuel of the reactor unless the reactor just happened to be stopped as the last of the Ni was converted. It is more likely that the Ni reactions are actually competing with the primary reactions, and possibly the reactor works better when all of the Ni has been converted to Ni62. (Just a guess.) It is more likely that hydrogen is the real fuel (and maybe also the lithium). I would also guess that if the reactor had even a small leak, the hydrogen would leak out and the reaction would stop.
            Dr. Mike

      • not clear what is the structure of the e-cat…
        given the apparent change in resistance (beware it can be just bad assumption as people used Rapp=P/Ieff^2…with switched triphase in V mode not delta) one hypothesis is a negative temperature coefficient, and in an E-cat the evidence hypothesis is that it is a LENR hydride, the “mouse”…

        speculation, but at least it is not a conspiracy like to inverted current clamps on only the active test, with strange controlbox that in that crazy context, consume energy in both blank (right wiring) and active (wrong wiring)…
        all without any of the testers having any suspicion…

        at least an international conspiracy involving Exxon, Putin, with Total boss assassination, Ukraine MH17 and ISIS/CIA collusion, is more probable.

    • Dr. Mike

      Giuliano,
      There are actually 3 helices wound on the alumina, one for each phase of the 3 phase power supply. The idea behind the dummy run was to see how the reactor heated with just electrical power supplied so that its effect could be subtracted out of the LENR effect. An ideal control for the experiment would have been to heat the reactor up to the temperature during the active run just using the electrical heating.
      Dr. Mike

  • DocSiders

    In less time than it has taken “us” here to calculate power in’s/out’s, the TIP group or IH could have done (or still could do) a dummy run at the same power inputs as the “charged runs”. Providing data that is needed. Required actually.
    It is unforgivable that the control run wasn’t run like any normal control run.

  • DocSiders

    In less time than it has taken “us” here to calculate power in’s/out’s, the TIP group or IH could have done (or still could do) a dummy run at the same power inputs as the “charged runs”. Providing data that is needed. Required actually.
    It is unforgivable that the control run wasn’t run like any normal control run.

    • Nigel Appleton

      Absolutely right. Would have removed a large target for the skeptics. Experiments should not be run without controls. Controls and tests should be run under exactly the same conditions. Standard scientific practice.

  • Thomas Clarke

    Correct. If you want to split hairs:

    A nonlinear component – e.g. a diode, or a varistor, could in principle change the ratio at different input powers. But a X3 change is asking a lot and there is no conceivable nonlinear component that could operate as the heater does.

    The beauty of the total power/joule heating power ratio is that it is insensitive to almost everything! And the data from which it is derived is securely stored by the testers direct from the PCE-830. So any possible mistake here can be resolved.

  • Thomas Clarke

    Several have done so. You might like to check out andrea.s for detailed knowledge of the PCE-830 on Mats Lewan’s comment thread [url=https://matslew.wordpress.com/2014/10/09/interview-on-radio-show-free-energy-quest-tonight/]here[/url] where the whole matter is debated at great length.

    The power calculations needed here require somone with basic knowledge of ohms law, KCL, KVL and phasors. That many have. On this thread you will find that fact police has been entirely correct.

  • georgehants

    Thomas, I am unable to understand what your comment has to do with what I have written.
    Please answer, who is meant to Research and investigate new possible phenomenon in this World if not main-line academic science etc.
    Or should they just sit on their arses and wait for that phenomenon to be researched and proven by small people like Mr. Rossi et al.
    If you think that, then what the devil are these wasters being paid for?

  • georgehants

    AlbertNN, you appear to be saying these academic clowns only look at something after it has been fully proven by other people and never do their own Research etc.
    As I ask above, then what are these wasters there for, just to spend taxes on there entertainments and refreshment.
    Maybe you are saying on unproven rubbish like GW or hot fusion.

  • kasom

    One could have done a simple check for plausability at first.

    The output over 32 days was aprox. 1.5MWh

    Ask http://www.officineghidoni.ch/en/ on readings of their grid’s powermeter in the office during these days. If you find excess heat, stop persnicketiness on the input.

    Go check the output measurements again. If correct, there is indeed excess heat.

    • roseland67

      Or,

      They could have used a simple single phase heater for the required energy input, and
      they could have simply modified the controls to run on 12vdc, and
      they could also have used simple marine battery(s) for the power source.

      Simple, no “hidden wires” energy source, the battery has a fixed energy and power density to compare reactor output to.
      Simple pid controller, no 3 phase phase angles, triac switching and
      joule heating effect to interpret.

      “No tricks, no weapons, skill against skill alone”
      (couldn’t resist).

      Still waiting in Chicago.

  • Dr. Mike

    Thomas,
    Well said! I think that even with the data from the PCE-830 there still may be a problem. If the Insertion of the pulse generator affected the power calculation by the PCE-830, the power data from the PCE-830 may not be any good. I believe that they reported the data they read off the meter (column 2, Table 7, page 22 of the report). i also would hope they are carefully going over all of their data and their electrical set-up so that they can eventually explain a possible error in their work, or better yet, why there is no error.

  • Ged

    Way too many replies so have to do one here.

    The whole powering of the e-cat is controlled by a microcontroller, a little mini-computer, which is likely a smaller version of this one http://ccipower.com/system/files/brochures/Compact%20FUSION%20Datasheet_0.pdf based on visual inspection. Notice that with this controller: “Output is controlled linearly with respect to command
    signal and can be set to the average or RMS value of the voltage or current, as well as true signal and can be set to the average or RMS value of the voltage or current, as well as true instantaneous power or external feedback”.

    – People are conflating joule heating by the intrinsic resistance of the conductive wires (this is -not- the voltage of the entire circuit), and the voltage of the entire circuit controlled by the microcontroller. It can change the voltage and amperage however it wants to hold whichever parameter, or total power, steady according to what is set. It has potentiometers and other sensors as you can see that allows it to monitor amperage, voltage, and power (that’s exactly what the brochure says, which sounds a little redundant on the surface), and vary them all.

    When they talk about input power, they are talking about the microcontroller. On page 7 we see them set the microcontroller to hold the power at slightly above 900 W after the current -dropped- due to the microcontroller’s feedback system deciding it needed to cut back on the amperage when the power was set near 790 W, something the authors didn’t want and which prompted them to let the microcontroller to 900 W. They then measured with PCE’s and saw that indeed the controller, a computer with its own internal systems to monitor and modify voltage/amps/power, was drawing and outputting roughly 900 W just as it was set to do.

    That’s all well and good. We see the controller in action, and we see that it can -dynamically change voltage and amperage- being pushed to the e-cat’s circuitry based on feedback.

    – Joule heating is -not- the input power the controller is modifying. This is being horribly conflated. Joule heating is due to the intrinsic resistance of the wire conductors per length by cross sectional area, this has nothing to do with the circuit’s resistors and over all voltage driving power on the circuit, and is a loss of power due to transmission. If one pumps the input power by pumping the voltage being applied by the controller across the circuits, then joule heating doesn’t change, as it’s a function of amperage. This is why US houses are 20 A and 115 V circuits mostly, and why high tension lines are high voltage — this is why neither of these two things commonly burst into flames despite the high amounts of wattage flowing through them.

    The controller is entirely capable of upping the input power by driving up voltage, that is part of its specification and dynamic control system. But only the amperage being applied to the circuit counts for the power lost as heating, not of the resistors, but of the conducting copper.

    – A super conductor has a joule heating of 0. This is because the voltage drop intrinsic to the superconducting wires is 0. The voltage applied to the circuit is -non 0- and input power is -non 0- when you are charging a super conductive magnet to 14.6 T. But there is a 0 joule heating due to a 0 voltage drop by the wire itself. Completely different voltage under discussion between the circuit and power supply voltage versus the intrinsic voltage drop of a length of conductive wire. Again, superconductors have 0 joule heating, no matter how many amps or volts you apply to a circuit containing them. That’s why they are the holy grail of transmission technology.

    – The joule heating we see in the paper in the dummy reactor is very clearly explained: 7 W at ~ 20 A and when you jump it up to the amps in the live run the error range easily includes the 41W reported. The steadiness of the input power is due to the controller being dynamic and attempting to hold its set point however it feels like, nothing mysterious here. The voltage applied to the circuit by the controller -is not the same voltage drop intrinsic to the wires per length by cross section that is used to calculate joule heating-: that is an intrinsic material property of the wires and if they had used super conductors the joule heating would have been 0.

    There is no mystery here, nothing funky, and nothing amiss as far as I can see.

    • Obvious

      Just a quick note. The amperage clearly exceeds 40 A total in the live reactor run, since each C1 lead in the dummy run is carrying 19.7 A. Therefore a total of 59.1 A is being delivered to the dummy.

      • LuFong

        The total of all currents in 0 for 3 phase, that is at any one time one current is the opposite of the sum of the other two? I pose this as a question because I’m trying to reconcile what you just said with my understanding of a 3 phase power in a delta configuration.

        • Obvious

          The total of all currents cannot be zero is any work is being done.

      • Dr. Mike

        Obvious,
        The current to each of the 3 Inconel coils in the dummy run is 19.7/2 = 9.85A (See the joule heating discussion on pages 13-14 of the report.). This current is set at a value about 2.35 times greater for the first 10 days of the active run and then increased to about 2.5 times greater than the dummy run for the rest of the active run.
        Dr. Mike

        • Obvious

          I should have said the total for amps consumed by the entire device wiring assembly (including the reactor), before subtracting the Joule heating. The total amps, using the three phase power equations should then be 34.12 A, for the entire device wiring assembly of the dummy run (rather than adding the sums of the currents in the three leads).

  • Ged

    Way too many replies so have to do one here.

    The whole powering of the e-cat is controlled by a microcontroller, a little mini-computer, which is likely a smaller version of this one http://ccipower.com/system/files/brochures/Compact%20FUSION%20Datasheet_0.pdf based on visual inspection. Notice that with this controller: “Output is controlled linearly with respect to command
    signal and can be set to the average or RMS value of the voltage or current, as well as true signal and can be set to the average or RMS value of the voltage or current, as well as true instantaneous power or external feedback”.

    – People are conflating joule heating by the intrinsic resistance of the conductive wires (this is -not- the voltage of the entire circuit), and the voltage of the entire circuit controlled by the microcontroller. It can change the voltage and amperage however it wants to hold whichever parameter, or total power, steady according to what is set. It has potentiometers and other sensors as you can see that allows it to monitor amperage, voltage, and power (that’s exactly what the brochure says, which sounds a little redundant on the surface), and vary them all.

    When they talk about input power, they are talking about the microcontroller. On page 7 we see them set the microcontroller to hold the power at slightly above 900 W after the current -dropped- due to the microcontroller’s feedback system deciding it needed to cut back on the amperage when the power was set near 790 W, something the authors didn’t want and which prompted them to let the microcontroller to 900 W. They then measured with PCE’s and saw that indeed the controller, a computer with its own internal systems to monitor and modify voltage/amps/power, was drawing and outputting roughly 900 W just as it was set to do.

    That’s all well and good. We see the controller in action, and we see that it can -dynamically change voltage and amperage- being pushed to the e-cat’s circuitry based on feedback.

    – Joule heating is -not- the input power the controller is modifying. This is being horribly conflated. Joule heating is due to the intrinsic resistance of the wire conductors per length by cross sectional area, this has nothing to do with the circuit’s resistors and over all voltage driving power on the circuit, and is a loss of power due to transmission. If one pumps the input power by pumping the voltage being applied by the controller across the circuits, then joule heating doesn’t change, as it’s a function of amperage. This is why US houses are 20 A and 115 V circuits mostly, and why high tension lines are high voltage — this is why neither of these two things commonly burst into flames despite the high amounts of wattage flowing through them.

    The controller is entirely capable of upping the input power by driving up voltage, that is part of its specification and dynamic control system. But only the amperage being applied to the circuit counts for the power lost as heating, not of the resistors, but of the conducting copper.

    – A super conductor has a joule heating of 0. This is because the voltage drop intrinsic to the superconducting wires is 0. The voltage applied to the circuit is -non 0- and input power is -non 0- when you are charging a super conductive magnet to 14.6 T. But there is a 0 joule heating due to a 0 voltage drop by the wire itself. Completely different voltage under discussion between the circuit and power supply voltage versus the intrinsic voltage drop of a length of conductive wire. Again, superconductors have 0 joule heating, no matter how many amps or volts you apply to a circuit containing them. That’s why they are the holy grail of transmission technology.

    – The joule heating we see in the paper in the dummy reactor is very clearly explained: 7 W at ~ 20 A and when you jump it up to the amps in the live run the error range easily includes the 41W reported. The steadiness of the input power is due to the controller being dynamic and attempting to hold its set point however it feels like, nothing mysterious here. The voltage applied to the circuit by the controller -is not the same voltage drop intrinsic to the wires per length by cross section that is used to calculate joule heating-: that is an intrinsic material property of the wires and if they had used super conductors the joule heating would have been 0.

    There is no mystery here, nothing funky, and nothing amiss as far as I can see.

    • Obvious

      Just a quick note. The amperage clearly exceeds 40 A total in the live reactor run, since each C1 lead in the dummy run is carrying (edit: delivering is a better word) 19.7 A. Therefore a total of 59.1 A is being delivered to the dummy.

      • AlbertNN

        You can’t add currents in that way in a three phase system without a common neutral wire.

        • Obvious

          Hmmm.

          • Obvious

            Then the total current into the dummy (edit: including cables) must be 34.12 A.

          • AlbertNN

            The sum of the instantanous currents going to the E-cat at any one time is always zero. It is not possible to add RMS-currents in this way, as the RMS value only give you the magnitude, and not the phase. It is possible to add the three time-variant currents in phasor representation, and then the sum would still be zero.

          • Obvious

            19.7 amps average current is flowing towards the reactor through each of the C1 wires. Presumably this current returns down one or two of the other two C1 wires, depending on phasing. What is the total average forward current flowing into the wiring (heater and all) by which heat is made? (IE: what amount of current is the electrical company going to charge you for? Certainly not the instantaneous zero current (that would be nice).)

            The sum of the power in each phase is equal to the total power of the system.
            And therefore the sum of the currents is appropriate, when the voltage is constant between phases (balanced resistive load).

          • AlbertNN

            The sum of the currents in the three wires going to the E-cat must be zero. This is a consequence of the law of charge conservation, and expressed by Kirchoff’s laws.

            The electrical company is not charging for current, they are charging for energy, J or Ws. Also commonly expressed as Wh.

          • Obvious

            The 3 phase electrical line has constant voltage.

            The electrical company cannot charge for energy if they cannot measure the total current moving through 3 phases and add them together and get a number bigger than zero.

            The three phases can be treated as a single phase.

          • AlbertNN

            Please study the theory of electrical engineering and three phase systems. None of your statements this time are correct.

          • Obvious

            I’m obviously not describing my point or question clearly, so I’m not getting any help here. Clearly I do not deal with 3 phase power often.
            Is the total current consumption of the entire dummy assembly, including cables:
            A) 19.7 A (the average of three phases, which are balanced)
            B) a factor of 19.7 A due to three phases (times 1.73)
            C) a sum of 19.7 A due to three phases
            D) a factored sum of some sort

          • AlbertNN

            Current is never consumed. It is talked about like that sometimes, but what you actually are talking about, and what you should add, is power, as in Watts.

          • Obvious

            Summing the power in each phase will give the total power for the system.
            But that requires knowing either U or R. If I’m trying to solve for R or U (V), I would like to know what the total I is. We have a P value.
            Is the 486 W for the dummy run the instantaneous Power, or the total power?

      • Dr. Mike

        Obvious,
        The current to each of the 3 Inconel coils in the dummy run is 19.7/2 = 9.85A (See the joule heating discussion on pages 13-14 of the report.). This current is set at a value about 2.35 times greater for the first 10 days of the active run and then increased to about 2.5 times greater than the dummy run for the rest of the active run.
        Dr. Mike

        • Obvious

          I should have said the total for amps consumed by the entire device wiring assembly (including the reactor), before subtracting the Joule heating.

          The total amps, using the three phase power equations should then be 34.12 A, for the entire device wiring assembly of the dummy run (rather than adding the sums of the currents in the three leads).(?)

    • AlbertNN

      Short comment: Most of your conclusions is based on that the heating elements of the reactor is not behaving as a classic resistor. Bus based on what is stated in the report they are exactly that.

    • Thomas Clarke

      Jed. The issue is not the amount of Joule heating, which depends on the wire resistance. It is the ratio between Joule heating and total power. which is equal to the ratio of wire resistance to total resistance.

      In electrical circuits you cannot vary power and current independently. They are related by power = current ^2 X resistance. When you take power ratios. because the current isknown to be the same, you get resistance ratios.

      Voltage drop is not intrinsic to a material. Resistance is. Voltage drop is then current X resistance.

      Therefore what you say is wrong. I should point out that I teach university analysis of circuits – at a good university – and know what I’m taking about. As do many others here and elsewhere.

      There is either a most unusual heater element with variable resistance, or something amiss.

    • fact police

      Ged> – People are conflating joule heating by the intrinsic resistance of the conductive wires (this is -not- the voltage of the entire circuit), and the voltage of the entire circuit controlled by the microcontroller.

      The wires and the heating coils form a passive resistance network, so the ratio of the currents through the wires and the coils remains constant, as long as the resistance stays the same. Any change to the input voltage controlled by the micro controller changes the current through the wires and and the coils by the same factor.

      What you call “conflation” is justified by Ohm’s law.

      Ged> It can change the voltage and amperage however it wants to hold whichever parameter, or total power, steady according to what is set.

      You can’t make this true just by saying it. The current in the resistors depends on the voltage according to Ohm’s law, and the power is determined by the current. The controller, as you said yourself can control the voltage *or* current *or* power. It can’t control them independently.

      Ged> It has potentiometers and other sensors as you can see that allows it to monitor amperage, voltage, and power (that’s exactly what the brochure says, which sounds a little redundant on the surface), and vary them all.

      It monitors current and voltage, and calculates power. But it can only control the voltage, which produces a change in the current and power. It can “control” current or power by adjusting voltage to produce a desired current or power. But the circuit responds to V = IR, so for a constant R, V fixes I, and the power by P=I^2R.

      Ged> When they talk about input power, they are talking about the microcontroller.

      I don’t know what you mean by that sentence.

      On page 7 we see them set the microcontroller to hold the power at slightly above 900 W after the current -dropped- due to the microcontroller’s feedback system deciding it needed to cut back on the amperage when the power was set near 790 W, something the authors didn’t want and which prompted them to let the microcontroller to 900 W.

      We can all read the report. Your paraphrase adds nothing, and I fail to see what point you’re trying to make.

      Ged> They then measured with PCE’s and saw that indeed the controller, a computer with its own internal systems to monitor and modify voltage/amps/power, was drawing and outputting roughly 900 W just as it was set to do.

      That’s a paraphrase of this: “We therefore decided to increase the power, and set it slightly above 900 W.” The actual sentence does not say how the 900 W was set. Probably, they simply turned the dial until the PCE read 900W.

      Ged> That’s all well and good. We see the controller in action, and we see that it can -dynamically change voltage and amperage- being pushed to the e-cat’s circuitry based on feedback.

      We don’t see any of that. We see that they can increase the power by adjusting the controller, and the PCE reads a higher power. Anything beyond that is speculation.

      And I still don’t know what point you’re trying to make that is relevant to the central inconsistency at issue.

      Ged> Joule heating is -not- the input power the controller is modifying. This is being horribly conflated. Joule heating is due to the intrinsic resistance of the wire conductors per length by cross sectional area, this has nothing to do with the circuit’s resistors and over all voltage driving power on the circuit, and is a loss of power due to transmission.

      You are right that joule heating is not equal to the input power. And you are right that what they call “joule heating” is the power lost due to the wire resistance.

      But you are wrong that it has nothing to do with the input voltage. The input voltage is proportional to the input current and the input current determines the joule heating. So the joule heating is proportional to the input power. It is a result of elementary electrical principles.

      Ged> If one pumps the input power by pumping the voltage being applied by the controller across the circuits, then joule heating doesn’t change, as it’s a function of amperage.

      This is incorrect. If the input power is increased, the input current increases according to Ohm’s law. If the current didn’t change, then the power dissipated by the coils could not increase, because it is also determined by the current according to I^2R.

      Ged> This is why US houses are 20 A and 115 V circuits mostly, and why high tension lines are high voltage — this is why neither of these two things commonly burst into flames despite the high amounts of wattage flowing through them.

      But the high tension lines are not connected to US houses by passive resistor networks. There are transformers involved. When a voltage source is connected to a resistor network, Ohm’s law fixes the relationship between the voltage and current. I don’t know if you really have difficulty understanding this elementary concept, or are simply writing a lot of muddled arguments to avoid admitting that the paper contains a glaring contradiction

      Ged> The controller is entirely capable of upping the input power by driving up voltage, that is part of its specification and dynamic control system.

      Right. But this will result in a proportional increase in the joule heating.

      Ged> But only the amperage being applied to the circuit counts for the power lost as heating, not of the resistors, but of the conducting copper.

      I don’t follow. The current through the wires and the resistor coils remains proportional. And the power dissipated by each is I^2R. So if the power dissipated in the coils increases, the joule heating increases by the same amount, and vise versa, as long as the resistances are fixed.

      Ged> Completely different voltage under discussion between the circuit and power supply voltage versus the intrinsic voltage drop of a length of conductive wire.

      They are different, yes, but proportional, according to elementary circuit analysis.

      Ged> Again, superconductors have 0 joule heating,

      I fail to see what superconductors add to this discussion. The joule heating is still proportional to the input power. Zero times anything is still zero.

      Ged> The joule heating we see in the paper in the dummy reactor is very clearly explained: 7 W at ~< 20 A. When they jumped up the input power to 900 W (this is the power set on the CONTROLLER and measured coming from it), that implies they simply doubled the amperage, and thus joule heating would have gone up by the square of this, to ~28 W.

      No, your analysis is wrong here. When they doubled the input power, that implies the current was increased by root(2). So the joule heating would have increased by the square of that, which of course is just the amount they increased the power by: a factor of 2.

      So the joule heating should have gone to ~14 W, not 28 W. Check your math.

      Ged> Or so it would be if they simply doubled the amps. But they didn’t simply double them.

      No. Doubling the current would have produced a factor of 4 increase in the input power, but the input power only increased by a factor of 2.

      Ged> They clearly report on page 3 that during the experimental run the controller is “normally” holding the amperage within a range from 40 – 50 A,

      That’s not what they say. They say it is normally set somewhere between 40-50 amps. It doesn’t fluctuate within that range, because if it did, the power would not be constant to within a few per cent. And they say explicitly elsewhere that the power is determined to 5 %.

      Ged> But the authors also report the amperage went above 40 A, in vague terms. That’s all well and good, since the controller isn’t going to compensate unless we hit that 50 A or above. So what happens if the amperage, which was slightly lower than 20 A in the dummy run, drifted more towards the 50 A?

      The joule heating change from 40 A to 50 A is an increase by 56% (from 1600 to 2500). Let’s increase the joule heating calculated by simply increasing from 20 to 40 A by just 50%, well within our envelope.

      -We get a joule heating of 42 W.

      But if 20 A in the dummy run corresponds to 450W input, and you are saying the current in the live run was 50 A, then the power in the live run should be (50/20)^2 * 450W = 2800 W. You see? That’s the discrepancy, because they’re saying it’s only 900 W.

      Ged> 1) the controller crept the amperage above 40 though they didn’t say by how much, and 2) the dummy run’s amperage was slightly lower than 20.

      If you agree with this, then you have agreed with the lions share of the discrepancy.

      If you admit the current doubles from the dummy run, then the power should be 4 times higher in the live run, but they claim it only doubles. See the problem?

      And if it is actually closer to 50A (to explain the 41 W joule heating), then the input power should be 2800W.

      Ged> Meaning our error in calculating a straight joule heating from the dummy run is well within the error propagation that includes 42 W, which is more than the 41 W they report.

      Whatever error you want to invoke to explain 42 W joule heating would also require 2700W total power input, and not 900W as they report.

      That’s the problem.

      Ged> There is no problem with their values. They make complete sense.

      No. There is a factor of 3 discrepancy. If the joule heating increases 6-fold, then the coil heating must increase 6-fold as well, unless the resistance changes.

      Ged> And we see on page 13 that the amperage could very well go above 40 A, since the controller only cares to keep the amps it’s pushing to the circuit between 40 and 50 A when at 900 W.

      If the power is 450 W at 20 A, then at 50 A it’s 2800W, not 900W, because of the I^2R law. So, again, a factor of 3 discrepancy.

      Ged> Summary: They set he controller to 900 W, they measure 900 W, the controller has free reign to vary amperage between 40-50 A (a 56% joule heating error range) at 900 W and it varies it above 40 A by some amount, the dummy is ~> 20 A and when you jump it up to the amps in the live run the error range easily includes the 41W reported.

      If the current in the dummy run of 20 A produces a power of 450W, then the current you agree is between 40 and 50 A, produces a power of 1800W to 2800W, not 900W. And if you choose a current close to 50 A to explain 42 W joule heating, then the power has to be 2800 W.

      The power in the coils and the joule heating have to remain in constant proportion, unless the resistance values change.

      It’s clear you didn’t read any of the responses to your previous comments, because you are repeating the same error.

      • Dr. Mike

        fact police,
        Again thank you for your careful explanation to Ged.
        Dr. Mike

    • Dr. Mike

      Ged,
      I hope that “fact police” and Thomas Clarke have clarified the issue with the discrepancy in the Lugano data with their comments listed below. Let me add to their comments with a simple analogy to the data and measurements from the Lugano report. Suppose we have a simple circuit consisting of a variable power supply and connected across that supply are two series connected wire-wound resistors. The only measurement instrument we have is an ammeter that is connected in the circuit to read the current. Also suppose we don’t know the value of either resistor or know the voltage being supplied. However, we know what kind of wire was used in the first wire-wound resistor, its diameter, and its length. Therefore, we were able to calculate the value of resistor #1 – let’s assume we calculated its resistance as about 1 Ohm. Assume the power supply is turned on and the control knob turned up until the current reads 2 amps in the ammeter. What can be calculated at this point? The only thing we can calculate is the joule heating in the first resistor- I*I*R = 4W, but this value is uncertain to to accuracy of the calculated resistor value. Now suppose the power supply knob is turned up to where the ammeter read 5 amps. What knowledge would we have? 1)The power in the first resistor would now be about 25W. 2)The power in the first resistor went up by a factor of exactly (5/2)^2 = 6.25 (assuming the currents are measured exactly). 3)The power in the second resistor went up by the same factor of 6.25.
      What would happen if we didn’t get to see the ammeter readings, but were just told that when the power supply was turned up, the power in resistor #1 increased by a factor of 6.25. We would know that 1) the current increased by a factor of exactly 2.5, independent of the value of resistor #1’s resistance and 2) the power also increased by a factor of 6.25 in resistor #2, independent of its resistance. This is exactly the information we have in the Lugano report. We are told the power in resistor #1 (Cu wire joule heating) went up from 6.7W in the dummy run to 42W in the second part of the active run, a factor of about 6.25 (6.27 actual). This means we also know the power in resistor #2, the Inconel coils, also went up by a factor of 6.25. If we know the power to the coils was 479W in the dummy run, the power to the coils in the second part or the active run was 479*6.25 = 2994W. Add in the 42 W of Cu wire Joule heating and the power supplied was 3036W. However, the supplied power was measured as only about 920W in the second part of the active run.
      The question that needs answered by the Lugano authors is how did the current increase by a factor of 2.5 form the dummy run to the active run (second part), but the measured supplied power only increase from 479W to about 920W? My argument is that since the current is measured by two independent meters, it is most likely that there is an error in the measured power.
      Dr. Mike

  • Obvious

    A much simpler version would be to have a few mechanical thermo switches (bimetallic or something along those lines) in the reactor that could series/parallel the resistances differently, depending on temperature. I doubt that something like that would survive long in that heat….

  • Obvious

    Using the formula and currents described in the dummy calculations, and inserting a current value 2.5 times higher (C1 current at 49.25 A [2.5x 19.7), and C2 current at 24.63 A [2.5 x 9.85]), then the cable Joule loss result is (31.84 + 10.23) = 42.07 W.

    • fact police

      And if the current is 2.5 times higher, then the power input (and the power in the coils) will be 2.5^2 = 6.25 times higher than in the dummy run, which is about 2800W, not 900W.

      • Andreas Moraitis

        You assume that the resistances of the coils are constant, or nearly constant. But that is not what one might expect in case that strong magnetic fields and/or superconductivity are involved (see the posts below).

        • fact police

          Yes. I’ve repeated the assumption of constant resistance over and over, but in this case I didn’t. But a variable resistance is not consistent with what happens on day 10 when they increase the power from 785.79 W to 923.71 W, and the joule heating increases by almost exactly the same proportion, indicating in this range the resistance is constant. And it would take a reduction of the resistance by a factor of 3 to account for the discrepancy.

          • Andreas Moraitis

            Maybe the supposed magnetic (or whatever) effect remains stable during the operation. At least, it could explain the difference between the dummy run and the active runs. In any case, we need more information from the authors. I hope that they have recorded both currents and voltages for the whole period.

          • Obvious

            What if we use pure nickel wire for the reactor resistor, rather than Inconel? All standard alloys of Inconel melt below 1425° C anyways. (suppose that Inconel is only the attachment lead portion). Just tossing this out there…

        • AlbertNN

          The currents measured are to low to give high magnetic fields where anything exotic might happen. And the resistance wires used according to the paper are not superconducting at high temperatures.

          • Andreas Moraitis

            The magnetic fields which have been measured by DGT are far too strong that you could explain them as a result of the supplied currents. They must have been caused by the reaction. High-temp superconductivity in the context of LENR has been hypothesized by various authors. However, I don’t know if there are reliable data on this effect.

          • AlbertNN

            I am not aware that any high magnetic fields was measured at this experiment. And what materials has been demonstrated to go superconducting at almost 1700K?

          • Andreas Moraitis

            In this experiment magnetic fields have either not been measured, or they did not report about them – for whatever reason. But there has been a report of a scientist from Defkalion Green Technologies, which have developed a similar reactor some time ago.
            I think Celani has, among others, supported the superconductivity hypothesis. You might also want to have a look at http://www.google.com/patents/DE102008047334B4?cl=en

          • AlbertNN

            Then we do not know anything about any large magnetic fields in this case. And I am still very interested in information on materials that have demonstrated superconductivity at 1700K.

          • Andreas Moraitis

            We cannot know what we will know in some decades, years, months, or even tomorrow. This keeps life exciting – just enjoy it.

          • hunfgerh

            No one speaks from superconductivity at 1700 K. In the context here we speak from superconductivity at max. 400 K. The superconductor in a hot-cat must therfore be isolated from the rest of the arrangement be a thermally insulating ceramic layer. I myself developed during my industrial activity such layers, wich allowed in mm-thickness temperature Deltas of more than 1000 K. A further tip, such layers can be baked from a mix of Al(OH)3 and a special polyphosphat at 850 K. Maybe in a hot-cat the superconductor and the ceramic layer are baked parallel.

          • AlbertNN

            In what part of the arrangement would this superconductor be used? And what material is superconducting at 400K?

          • hunfgerh

            The superconductor is in the center of the hot-cat. It is formed by heating (600 oC) of a mixture of nickel powder and lithiumhydrid. Result NiLiH.
            The same result is obtained by heating Palladium powder with lithiumhydrid. Result PdLiH. Or by deposition of Lithium on a Pd surface by elecrolysis of LiH in a hydrogen atmospere Google patent mentioned by Andreas Moraitis. End of discussion.

          • AlbertNN

            600 degree C equals 873 K. I am not aware of any material that has demonstrated superconductivity at that temperature.

            A statement in a patent application is not valid as a scientific publication. Patent publications have very low value in a scientific discussions, as they do not even have to been demonstrated as actually working.

            And even if the center of the E-cat would be cooler than the surface, and the fuel would get superconducting during the experiment, this does not explain the discrepancy of the losses in the heating coils of the reactor. Which was the topic of the discussion regarding superconductivity and the E-cat above.

      • Obvious

        Except that we have a measured power consumption that does not agree with that. If we cannot trust the measured quantities, then the entire Joule heating discussion is a waste of time, since those measurements are part of the Joule calculation.

  • Nicholas Payne

    Just remember with a triac power cpntroller the peak to average ratio of the current will change with phase angle. Whenever you are trying to do a heat calculation make sure you are using rms current and the meter has sufficient bandwidth to accurately measure it

    • Thomas Clarke

      The currents are measured by the PCE-830, which samples and performs calculations. It is very capable for the job, with a sampling rate more than high enough to deal with the harmonics from the triac switching. It will display RMS current.

  • Thomas Clarke

    Dear moderator,

    I made a comment below NP’s post which it seems vanished saying that the PCE-830 sampling rate was easily good enough to catch the triac harmonics (as the authors show) and that it could not therefore misread rms in the way that a simple meter might do. In fact measuring RMS current exactly in this situation – triac switched 3 phase power – is precisely what it is designed to do. I can justify these comments by boring you with sections from the PCE-830 manual if that would make you happier, please say. Otherwise in the interests of fair comments don’t delete this courteous post.

    The manual is here;

    http://www.industrial-needs.com/technical-data/power-anlayser-PCE-830.htm
    The crest current would have to be above 100A rms (140A) for the meter to saturate on peaks. As the authors say, that is not possible because it is around 40A.

    Thank you.

    • Mark Szl

      Finally, someone who knows about curcuit analysis. Thanks Thomas.

      Moderator, please get rid of this article and posts. It is a confused and confusing/misleading mess!!!

    • Dr. Mike

      Thomas,
      Thanks for the link. I wanted to see what the manual had to say about how the meter needed to be hooked up.
      Dr. Mike

    • Dr. Mike

      Thomas,
      The manual is clear that you need to hook up the current sensors correctly. Are you familiar enough with the PCE-830 to know what happens to the power reading if one of the sensors was hooked up backwards? Would the measured power to a 3-phase delta connected resistive load only be about 33% of the real value if one of the current sensors was backwards? (or maybe somewhere in the range of 30-36% if the load resistances varied by up to 10%?)
      Dr. Mike

  • Omega Z

    Rossi doesn’t need to raise money. That is Industrial Heats department now that they are the owners of the technology.

    As to how convincing the test was, I don’t’ think it is Rossi’s intent to provide indisputable results to the gallery. Not in best interests of a business development.

    Most of the complaints & questions about the test would be valid if this were strictly a scientific endeavor, But it is not. We need to keep this in mind in the discussion. There is a ton of data that we Don’t, and Wont have access to. It will be provided only to a few with a need to know.

    I have no doubt that If Rossi/IH made every detail available, Dozens of large entities would setup a small R&D lab to replicate it, But not a single one of them would build any products for the market.

    They would focus on developing an alternate technique that could be patented to protect their Billions it will cost to bring such a product to market. This would delay LENR introduction to market by many years.

  • ivanc

    “If you want to push more amps across each individual resistor, while keeping power steady, you have to drop the voltage.” Fact police.

    -Joule heating is directly, geometrically, dependent on amperage.-

    Input
    power is dependent linearly on amperage And voltage. You can up the
    input power by upping the voltage without significantly affecting joule
    heating. Again, this is exactly why we use high voltage lines to carry
    MW of electricity, not high amperage.” Ged.

    Both statements are wrong!!!!

    We dealing with a resistive static load, so the inductance will be very small,

    For a single phase. load.
    total power = current^2 * drop voltage cu + current^2 * drop voltage resistor.
    total voltage=drop voltage cu + drop voltage resistor.
    Current is same for cu and resistor.

    V=RI, or I=V/R, or R=V/I

    Then if you drop voltage you drop current.
    if you drop resistance you increase current.
    you can not increase/drop current at will , you need to change voltage or resistance.

    In both cases power = VI, but v will be different for the cu wire that the resistance, but will follow the relationship that Vt=Vcu+Vr

    In case of three phase the law still fulfills but the three phase is modeled as single phase and a factor of sqr root of 3 is used.

    But in this case thing complicate even more as the 240v 60 Hz are cut using TRIAC.
    a rms meter shoul still read voltage and current as any wave is made of armonics of sin waves.

    bat average current may not be used, the instruments have to be certified for this kind of application.

    Is better for this kind of lab test to use pure sin wave using an autotransformer to control the voltage, that will cause change in current.

    • fact police

      ivanc>”If you want to push more amps across each individual resistor, while keeping power steady, you have to drop the voltage.” Fact police.

      -Joule heating is directly, geometrically, dependent on amperage.-

      Input power is dependent linearly on amperage And voltage. You can up the input power by upping the voltage without significantly affecting joule heating. Again, this is exactly why we use high voltage lines to carry MW of electricity, not high amperage.” Ged.

      Both statements are wrong!!!!

      I agree, but I didn’t make either of them. They are both Ged’s. I quoted Ged, but the attribution was clear.

  • JC

    We don’t really know what the input current was for the active run. It’s not clear to me that it was actually running 40-50 amps during the active run, but that the statement relating to 40-50 amps might have had something to do with the device specs. We need the authors to clarify the input current, which I suspect may be something closer to 25 amps. Otherwise, their input power calculations don’t make sense as noted by Dr. Mike. I really don’t see how they would not have been able to multiply the current and voltage values to get the correct power level, and I would guess the most likely explanation is that the actual current level was not 40-50 amps.

  • Thomas Clarke

    (1) fact police was right. However more current less voltage means less resistance. You don’t control the resistance, so more current means more voltage and more power.

    (2) The current for the active run was measured and used to calculate power. We therefore can calculate it exactly. The ratio of currents is sqrt the ratio of powers, which is of order 6 => roughly 2.5 X current. The dummy current was 19.7A, so the active current 50A.

    (3) It is true the powers and currents don’t match unless you suppose a X3 change in heater resistance. That has not been ruled out. Why assume the testers measured current wrong? And why think that measuring current wrong is more likley than measuring power wrong?

    There is a proposed mechanism which explains the wrong measured powers precisely – that when the clamps were reattached before the active run obe was the wrong way around. i know of nothing to explain a X3 current error.given the dummy and active calculations are indentical.

    (4)
    ivanc – the currents and powwrs are all measured by a PCE-830 which can happily deal correctly with a triac waveform.

    • Dr. Mike

      Thomas,
      How do you think the pulse generator was hooked up to the circuit? If I were designing the Hot-Cat, i would have put the pulse generator on a separate winding. Do you think it was just connected across 2 phases of the 3 phase TRIAC output? Could the way the pulse generator was hooked up effect the TRIAC output power measurement?
      Dr. Mike

  • Thomas Clarke

    (1) fact police was right. However more current less voltage means less resistance. You don’t control the resistance, so more current means more voltage and more power.

    (2) The current for the active run was measured and used to calculate power. We therefore can calculate it exactly. The ratio of currents is sqrt the ratio of powers, which is of order 6 => roughly 2.5 X current. The dummy current was 19.7A, so the active current was around 50A.

    (3) It is true the powers and currents don’t match unless you suppose a X3 change in heater resistance. That has not been ruled out, though I think it unlikely.

    Why assume the testers measured current wrong? And why do you think that measuring current wrong is more likely than measuring power wrong?

    There is a proposed mechanism which explains the wrong measured powers precisely – that when the clamps were reattached before the active run one was the wrong way around. i know of nothing to explain a X3 current error.given the dummy and active calculations are identical.

    (4)
    ivanc – the currents and powers are all measured by a PCE-830 which can happily deal correctly with a triac waveform.

    • Dr. Mike

      Thomas,
      How do you think the pulse generator was hooked up to the circuit? If I were designing the Hot-Cat, i would have put the pulse generator on a separate winding. Do you think it was just connected across 2 phases of the 3 phase TRIAC output? Could the way the pulse generator was hooked up effect the TRIAC output power measurement?
      Dr. Mike

    • Mark Szl

      Thomas here is a two part analysis of the report. Part of your comment in Part 2 is below. Are you now convinced that there was a serious/significant error in the test set up?

      Part 1:

      http://blog.stepchange-innovat

      Part 2:

      http://blog.stepchange-innovat

      Comments on Part 2 include:

      Thomas Clarke:

      “A careful analysis of the input power and current shows the strong
      likelihood of a X3 error, mediated by a misplaced current clamp probably
      the responsibility of Rossi, that explains the apparent high COP. (See
      my comment on Part 1). The testers have the data needed to confirm or
      deny this, so absence of further comment from them must be taken as
      confirmation (perhaps we will soon get some comment).

      This test overall provides very strongly convincing evidence. But it points in a
      direction far different from what Rossi would like everyone to believe.”

  • bkrharold

    Reading this post and all the comments is very perplexing and unsettling. Out of all these posts Ophelia Rumps one about not second guessing test results makes the most sense to me.
    Here is what we know :-

    Brouillon Energy have signed a $multimillion contract with a South Korean company to deliver hot water boilers based on LENR.

    Every year for the past three years Drs Peter Hagelstein and Mitchell Swartz have taught a course “Cold Fusion 101” at MIT in which they build a working cold fusion device which is on display to the public for several months.

    Dr Michael Mc Kubre for the Stanford Research Institute; said he has tested between 12 and 15 low energy nuclear reaction (LENR) technologies for the United States government (DARPA). McKubre said his team was able to replicate at least five of those technologies.

    Mitsubishi Industries have been a cold fusion LENR patent for nuclide transmutation.

    Industrial Heat have invested a large sum in the intellectual property rights to the ecat, and they have a commercial customer for one of the devices.

    It is clear that LENR is real. The only question is the reality of LENR+ and the “hot cat”.

    I notice one of the admitted skeptopaths on this thread coined a term “credopath”, I had never thought of myself in those terms, however I am open to new information and much as I would like to believe LENR+ is true, apparently there is not yet conclusive evidence for this.

    Once the scientific community invest in the research to understand LENR, it might be possible to use that knowledge to produce LENR+ from solid scientific principles, rather than random experimentation. I still believe LENR+ is possible, even at the risk of being called a credopath.

    • Dr. Mike

      bkrharold,
      I’m sorry that you find this post perplexing. The electrical engineers among our audience all realize that there is a discrepancy in the Lugano data between the measured input power on the PCE-830 meter and power generated in the heater coils based on the current flowing through them. I don’t remember seeing a single comment where anyone disputed the analysis of the fuel and ash showing that nuclear reactions had taken place.
      My other major point of this post is that doesn’t seem possible to generate 3000W of power in 0.55 grams of Ni without melting the Ni. Since my background in thermal engineering is limited to one course in thermodynamics that i took many years ago, I was hoping someone in the audience would have the background to calculate or estimate a maximum power generation per gram of Ni using assumptions based on the environment of the Hot-Cat. Do you think 3000W could be generated in .55 grams of Ni without melting the Ni? For reference, a Ni wire that’s 1mm in diameter and 8cm in length would be about .55 grams of Ni.
      Dr. Mike

      • Mike Henderson

        Dr. Mike,

        Let’s keep this in perspecitive. I believe the total thermal output was 2.6 MW over 32 days. That is (roughly) the same as two typical 1500 W space heaters I can buy at Target or Home depot for less than $50 each. This is impressive, but it isn’t a massive heat output.

        Yes, I could melt a nickel wire with that amount of heat, but a dispersed bed of nanoparticles? Not so much. Nobody seems to know whether the heat-generating reaction is a surface effect, in the gas phase, or deep within the crystal structure. Since the nanostructure seems important to the activity, perhaps it occurs at a surface or between grains. If so, the thermal dissipation across the bulk material is less important than heat transfer from near the surface by radiation or convection.

        We also don’t know if NI-AL alloys, which have somewhat higher melting points, are formed in the process.

        Pure nickel melting could provide a self-modulating feedback loop. As the crystal lattice loses its integrity, it would also lose its capacity for supporting Bloch waves, deuterium loading, phononic resonance or other lattice effects that may be involved. As the material cools and recrystallizes, the effect returns.

        I have a concern to add to your list. I don’t think the photos of the exterior surface of the reactor support the calculation used for radiated energy calculation. The spiral grooves aren’t that big, I suspect the calculated energy is overstated. These demonstrations clearly need to be performed in a calorimeter to eliminate these sorts of questions.

        • Dr. Mike

          Mike,
          I certainly agree with you that a dispersed bed of nanoparticles or microparticles are a lot different than any solid piece of Ni. However, Rossi has told us that once the Ni particles melt the reaction stops. (Also I have melted a lot of metal in an evaporator and the first thing the metal does in clump into a ball from surface tension.) I like your statement “Pure nickel melting could provide a self-modulating feedback loop. As
          the crystal lattice loses its integrity, it would also lose its capacity
          for supporting Bloch waves, deuterium loading, phononic resonance or
          other lattice effects that may be involved. As the material cools and
          recrystallizes, the effect returns.” If you look at the ash particle #1 in Figure 2, page 43 of the report, and compare it to particle #1 in the fuel in Figure 1 on the same page, it appears that there has been a slight amount of melting on the surface of the Ni. (It would have been nice if Figure 1 had the same magnification as Figure 2.)
          I didn’t study the radiant heat calculation since it is out of my field of knowledge so I really don’t have a comment on the radiant energy calculation. However, since you do have some knowledge in this field maybe you could answer a couple of questions. First, if the Ni particles were spread in a line at the bottom of the alumina cylinder and were generating 1660W of power, what would you expect the difference in temperature between the top and bottom of the alumina cylinder? Second, if the Ni particles were uniformly distributed inside of the alumina cylinder, what temperature would the Ni particles have to be at to see a 1400C temperature on the outside of the alumina cylinder? it would be helpful if you could even make an estimate for the answers to these two questions!
          Dr. Mike

      • bkrharold

        My consternation stems from the possibility that the Independent test might not be the conclusive proof we were all hoping for. I am really not qualified to judge from a technical standpoint, but I was under the impression the test generated 1.5Mwh of excess energy in the form of heat to over a period of 32 days. The temperatures recorded were very close to, but still less than the melting point of Nickel. The report mentions 1g of powder, but this was probably a mixture of powdered nickel with other substances like lithium hydride. Based on the analysis by theoretical physicist Carl-Oscar Gullstrom the amount of heat generated was compatible with a process called Bound Neutron Tunneling. The scientists who conducted the third party test, which lasted almost a year, are all highly qualified. Hopefully they double checked their testing protocols and heat measurements carefully, since their reputations are on the line and they have nothing to gain from any deception.

        • Thomas Clarke

          Whatever their qualifications, and competence in some of the needed areas, they have been provably deficient in two areas relating to electrical power:
          (1) they incorrectly stated the current ration between C1 and C2 circuits as 1/2 when it could be no more than 1/sqrt(3). This does not affect the results of the test, because it cancels out – though there is a very small second order change. Nevertheless it is a mistake.

          (2) they did not notice the glaring anomaly between input powrs measured and currents measured. At the very least they should have note this and revised their statement about the heating wire being inconel, which does not have an unusual negative temperature coefficient.

          Competent testers would also have been suspicious of such an unusual discrepancy and cross-checked whether the resistance really changed in such a way with temperature (from the active test warmup data) and whether the total input power was measured correctly by checking the powers for each of the lines to ensure no clamp was reversed.

        • AlbertNN

          The analysis by Gullstrom is not compatible with the amount of fuel, 1g, that was used in the latest test.

          • bkrharold

            Gullstrom mistakenly wrote there were 10g instead of one. However he calculated that only 5% of that amount would have been needed to supply the amount of excess heat observed.

          • AlbertNN

            Gullstrom calculates that he needs to burn 2g of fuel to get the amount of energy that was reported. Which is clearly not compatible with the latest test, as they reportedly only loaded the E-cat with 1g of fuel.

        • Dr. Mike

          bkrharold,
          I don’t believe that any errors in the Lugano test are due to deception. However, it doesn’t appear that they had anyone on the team that was an expert in electrical engineering although several probably had some knowledge in this field. I don’t really think that the reputations of anyone on the team will be damaged if there was an error in something that no one on the team had expertise in. Remember the group of physicists that claimed to have measured neutrinos traveling at faster than the speed of light. This group needed outside help to determine what was wrong with their measurements.
          Double checking testing protocols does not help if no one realizes that there is a fundamental problem with the some of data. The important outcome of all this discussion is that the Lugano team will go back over the data and set up, determine what is causing the discrepancy in the data, and issue a revised report that addresses that discrepancy. They also have the opportunity to add to the report many of the suggestions that have been made for improvement.
          Dr. Mike

          • bkrharold

            I would like to see them clear up these discrepancies, regardless of the final result. I only want to know the truth.

      • LCD

        From any model where the heat is generated as a black body from the nickel internally it’s not even close, if guess 3000 C minimum.

        • LCD

          Just doesn’t make sense

  • Dr. Mike

    Thomas,
    Thanks for the link. I wanted to see what the manual had to say about how the meter needed to be hooked up.
    Dr. Mike

  • Dr. Mike

    bkrharold,
    I’m sorry that you find this post perplexing. The electrical engineers among our audience all realize that there is a discrepancy in the Lugano data between the measured input power on the PCE-830 meter and power generated in the heater coils based on the current flowing through them. I don’t remember seeing a single comment where anyone disputed the analysis of the fuel and ash showing that nuclear reactions had taken place.
    My other major point of this post is that doesn’t seem possible to generate 3000W of power in 0.55 grams of Ni without melting the Ni. Since my background in thermal engineering is limited to one course in thermodynamics that i took many years ago, I was hoping someone in the audience would have the background to calculate or estimate a maximum power generation per gram of Ni using assumptions based on the environment of the Hot-Cat. Do you think 3000W could be generated in .55 grams of Ni without melting the Ni? For reference, a Ni wire that’s 1mm in diameter and 8cm in length would be about .55 grams of Ni.
    Dr. Mike

    • LCD

      From any model where the heat is generated as a black body from the nickel internally it’s not even close, if guess 3000 C minimum.

      • LCD

        Just doesn’t make sense

  • Obvious

    Mike, not that this affects the discussion regarding it much, but where did the .55 grams come from?

    • Dr. Mike

      Obvious,
      On page 29 in about the middle of the 4th paragraph, the authors found that the 1 gram of “fuel” contained .55 gram of Ni.
      Dr. Mike

      • Obvious

        Thanks.

  • Obvious

    Mike, not that this affects the discussion regarding it much, but where did the .55 gram number come from?

    • Dr. Mike

      Obvious,
      On page 29 in about the middle of the 4th paragraph, the authors found that the 1 gram of “fuel” contained .55 gram of Ni.
      Dr. Mike

      • Obvious

        Thanks.

  • Dr. Mike

    Thomas,
    The manual is clear that you need to hook up the current sensors correctly. Are you familiar enough with the PCE-830 to know what happens to the power reading if one of the sensors was hooked up backwards? Would the measured power to a 3-phase delta connected resistive load only be about 33% of the real value if one of the current sensors was backwards? (or maybe somewhere in the range of 30-36% if the load resistances varied by up to 10%?)
    Dr. Mike

  • Thomas Clarke

    Whatever their qualifications, and competence in some of the needed areas, they have been provably deficient in two areas relating to electrical power:
    (1) they incorrectly stated the current ration between C1 and C2 circuits as 1/2 when it could be no more than 1/sqrt(3). This does not affect the results of the test, because it cancels out – though there is a very small second order change. Nevertheless it is a mistake.

    (2) they did not notice the glaring anomaly between input powrs measured and currents measured. At the very least they should have note this and revised their statement about the heating wire being inconel, which does not have an unusual negative temperature coefficient.

    Competent testers would also have been suspicious of such an unusual discrepancy and cross-checked whether the resistance really changed in such a way with temperature (from the active test warmup data) and whether the total input power was measured correctly by checking the powers for each of the lines to ensure no clamp was reversed.

  • DickeFix

    Michael Lammerts question how such large heat generation can occur in such a small volume nickel without it melting is very relevant. In conclusion, we need now explanations to several “mysteries” regarding the E-Cat to make the stated heat generation credible:

    1. An explanation how nuclear reactions can occur at low temperature
    2. An explanation how nuclear reactions can occur without any radiation
    3. An explanation how all Ni isotopes and most other metals in the fuel can be converted to almost pure Ni62 in the ash
    4. An explanation why the generated power didn´t decrease with time despite the fuel was almost burnt out at the end
    5. An explanation why the fuel consisted of natural Ni while Rossi repeatadly stated that enriched Ni-62 was an essential ingredient in the fuel
    6. An explanation for the completely different ash compositions in ITP test 1 and 2
    7. An explanation why the electrical Joule heating increased 6 times when the stated input power only increased 2 times (Giancarlo)
    8. An explanation of the unexpected measured current shapes which indicate a reversed current clamp (Andrea S.)
    9. An explanation why the Ni didn´t melt if the power was generated by a reaction in the fuel (Dr. Mike)
    10. An explanation why the E-Cat gave COP=1 when SP Technical Research Institute (the swedish measurement calibration authority) tested it.

    All mysteries 1-10 are solved or circumvented if something happened during the loading of the E-cat so that the fuel inserted had the same composition as the ash and one of the three current clamps of both power meters were reversed. That would also explain the results of first independent test. The COP would then be about unity in all three tests (including SP) and all points 1-10 are solved.

    Does anyone here have a simpler and more likely explanation to ALL the ten mysteries without assuming more than one natural miracle; that LENR works?

    I am hence NOT a skeptopath! I don´t rule out that LENR is possible; it would be a fantastic gift to mankind. When I initially saw the results of the recent E-cat test I was excited and believed that maybe LENR was possible. That is why I have followed the E-Cat story and this excellent blog regularly. However, now after more careful analysis, the recent test has created more new questions (points 3-9) than it answered. It is no longer sufficient that LENR works to explain all mysteries regarding the E-cat. And I can not accept more natural miracles, apart from LENR, as an explanation. Explanations with heating coils made of semiconductors, mysterious magnetic fields or fantastic superconduction at temperatures far above the present record (138K) are just not realistic.Then I much rather accept the disappointing conclusion that my dream was just an illusion…

    • Hank Mills

      The clamps were not reversed and the fuel was not tampered with. The hot cat really does work. In fact, it is capable of infinite COP in self sustain mode. I do not think it involves any miracles (in the negative, hostile sense) but only that we pathetic humans have huge gaps in our understanding of the universe and physics. When several years from now thousands of successful E-Cat replications have taken place – showing self sustain mode so the skeptopaths will no longer be able to come up with bogus arguments to keep the current physics paradigm intact – these chasms will have bridges over them.

      • DickeFix

        “When several years from now thousands of successful E-Cat replications have taken place – showing self sustain mode”

        Even if I personally find the chances for that very low, you are of course free to believe in it if it makes you happy. You don´t need to lose your faith in Rossi and his fantastic E-Cat only because I have lost mine. Lets hope you are correct and I am wrong because it would be wonderful for all of us.

      • Dr. Mike

        Hank,
        I also don’t believe there was any tampering with the fuel or the ash- there is no reason for a tampering that results in something not readily explainable. Thomas Clarke claims that if the clamp was reversed, the evidence would still exist for this in the PCE-830 data. Do you know of the authors have reviewed the PCE-830 data to confirm that the data shows no clamp reversal?
        Do you understand the issue with the data in the Cu wire “Joule heating” that indicates heater coil currents are much higher than what is indicated by the power readings. If so, what are your thoughts on why the power and Joule heating data do not correlate? My theory on this was that the way the pulse generator is hooked up is causing some problems in the data. Do you know how the pulse generator is hooked up and why it was not included in the wiring diagram?
        I’m not sure what your technical background is, but do you have any background in thermal calculations? If you have some knowledge in this area, would you estimate the temperature at which the .55 grams of Ni would have to be to produce a temperature of 1400C on the outside surface of the main reactor body.
        Let me use this opportunity to thank you for all of the support you have given to promoting LENR.
        Dr. Mike

        • Hank Mills

          I am doing a write up about an idea of mine that might explain the current and joule heating issue. However, even if my idea is not correct, there is no reason to jump to the idea of a clamp reversal (unless you really think the testers were stupid enough to connect the clamps correctly for the dummy run and the control box test and then reverse them to screw up the results) when there are lots of unknowns about what was happening inside the reactor.

          • Dr. Mike

            Hank,
            Looking forward to your thoughts on the joule heating issue.
            Dr. Mike

    • fact police

      DickeFix wrote> “In conclusion, we need now explanations to several “mysteries” regarding the E-Cat to make the stated heat generation credible:”

      I would add another mystery to the list:

      11) Where does all the energy go?

      The authors make a very crude estimate of the energy from nickel transmutation, based on 58Ni + p -> 59Cu + gamma (3.4 MeV), and assume the other reactions have similar energy, but they neglect the 3.8 MeV from positron emission (of the 59 Cu) and the 1 MeV from positron annihilation, for 8.2 MeV.

      In fact, it’s a simple matter to calculate the energy released producing 62Ni from Ni58 and protons, from the mass change. It works out to about 35 MeV per atom. For one gram of nickel, then, that amounts to 57 GJ, compared to the 5.5 GJ they claim to have measured. That means that if the fuel is largely consumed (not even counting the lithium), 90% of the energy must escape the ecat, and yet none is measured in the form or radiation.

      Now, if the nucleons to make 62Ni come from the lithium, then the energy per atom of nickel is much reduced to less than half, but there is still a great excess. And the problem is that at 1% abundance, there’s not enough lithium to provide the necessary nucleons, so the majority will presumably have to come from hydrogen.

  • Dr. Mike

    Mike,
    I certainly agree with you that a dispersed bed of nanoparticles or microparticles are a lot different than any solid piece of Ni. However, Rossi has told us that once the Ni particles melt the reaction stops. (Also I have melted a lot of metal in an evaporator and the first thing the metal does in clump into a ball from surface tension.) I like your statement “Pure nickel melting could provide a self-modulating feedback loop. As
    the crystal lattice loses its integrity, it would also lose its capacity
    for supporting Bloch waves, deuterium loading, phononic resonance or
    other lattice effects that may be involved. As the material cools and
    recrystallizes, the effect returns.” If you look at the ash particle #1 in Figure 2, page 43 of the report, and compare it to particle #1 in the fuel in Figure 1 on the same page, it appears that there has been a slight amount of melting on the surface of the Ni. (It would have been nice if Figure 1 had the same magnification as Figure 2.)
    I didn’t study the radiant heat calculation since it is out of my field of knowledge so I really don’t have a comment on the radiant energy calculation. However, since you do have some knowledge in this field maybe you could answer a couple of questions. First, if the Ni particles were spread in a line at the bottom of the alumina cylinder and were generating 1660W of power, what would you expect the difference in temperature between the top and bottom of the alumina cylinder? Second, if the Ni particles were uniformly distributed inside of the alumina cylinder, what temperature would the Ni particles have to be at to see a 1400C temperature on the outside of the alumina cylinder? it would be helpful if you could even make an estimate for the answers to these two questions!
    Dr. Mike

  • Dr. Mike

    bkrharold,
    I don’t believe that any errors in the Lugano test are due to deception. However, it doesn’t appear that they had anyone on the team that was an expert in electrical engineering although several probably had some knowledge in this field. I don’t really think that the reputations of anyone on the team will be damaged if there was an error in something that no one on the team had expertise in. Remember the group of physicists that claimed to have measured neutrinos traveling at faster than the speed of light. This group needed outside help to determine what was wrong with their measurements.
    Double checking testing protocols does not help if no one realizes that there is a fundamental problem with the some of data. The important outcome of all this discussion is that the Lugano team will go back over the data and set up, determine what is causing the discrepancy in the data, and issue a revised report that addresses that discrepancy. They also have the opportunity to add to the report many of the suggestions that have been made for improvement.
    Dr. Mike

  • Hank Mills

    Two facts:

    1 – The dummy test found perfect unity.

    2 – The correctly measured the power consumption of the control box that matched the documentation.

    If the clamps had been inverted the above would have been impossible.

    This totally rules out the possibility that the clamps were inverted.

    • DickeFix

      You are correct. For the clamp theory to be true, the clamp(s) need to have been reversed during the loading of the fuel. According to the paper, the electrical data was recorded every 2 second so the research team can easily debunk or confirm this theory by comparing the recorded data of effective load resistance R=P/I^2 before and after the current was turned off for fuel change. Unfortunately, the research group has remained silent for weeks despite that Giancarlo had this question sent to Prof. Hanno Essén soon after the ITP report was realeased and Prof. Essén forwarded the question to Prof. Bo Höistad.

      Despite this silence and regardless of the outcome, I respect the scientists involved and the funding agencies that supported them. Due to the controversial subject of LENR, the researchers were brave to do this experiment since there are research colleagues that regards LENR with the same skeptic attitude as they regard telephaty and telekinesi. I personally think it is a scientific duty to investigate phenomena that seem to contradict established theory until one understands the results fully, especially where the outcome has such a potential importance for mankind. This is equally true regardless if the error is found in the theory or in the measurement procedure.

      The latter seem to have been the case when SP Technical Research Institute during summer 2012 tested the old version of E-Cat using a true RMS instrument to measure power and found the input power two to three times higher than the input power Rossi had measured. SP concluded then that the COP was close to unity.

      http://www.e-catworld.com/2012/09/10/nyteknik-reports-on-halted-swedish-investment-in-hydrofusion-following-tests/

      Hence, even if the post analysis of the current data would change the conclusion of the paper and instead again indicate a COP close to unity, it is an important scientific result. Hence, if that turns out to be the case, I sincerely hope the scientists involved are not ridiculed. If they followed good scientific conduct and didn´t consciously change the measurement setup between the calibration and active test, it is naturally that they didn´t recheck the clamp orientations again. Still, in retrospect, it is strange that neither the scientists involved, nor external reviewers who have read the paper, discovered the discrepancy regarding the Joule heating that Giancarlo found.

      • good point.
        with logged data, if Ieff is logged, we should see current phase and apparent impedance and observe that if follow a process incompatible with a student error.

        I am afraid however that IH will ask for some trade secret to be protected

  • Hank Mills

    Two facts:

    1 – The dummy test found perfect unity.

    2 – The correctly measured the power consumption of the control box that matched the documentation.

    If the clamps had been inverted the above would have been impossible.

    This totally rules out the possibility that the clamps were inverted.

    • DickeFix

      You are correct. For the clamp theory to be true, the clamp(s) need to have been reversed during the loading of the fuel. According to the paper, the electrical data was recorded every 2 second so the research team can easily debunk or confirm this theory by comparing the recorded data of effective load resistance R=P/I^2 before and after the current was turned off for fuel change. Unfortunately, the research group has remained silent for weeks despite that Giancarlo had this question sent to Prof. Hanno Essén soon after the ITP report was realeased and Prof. Essén forwarded the question to Prof. Bo Höistad.

      Despite this silence and regardless of the outcome, I respect the scientists involved and the funding agencies that supported them. Due to the controversial subject of LENR, the researchers were brave to do this experiment since there are research colleagues that regards LENR with the same skeptic attitude as they regard telephaty and telekinesi. I personally think it is a scientific duty to investigate phenomena that seem to contradict established theory until one understands the results fully, especially where the outcome has such a potential importance for mankind. This is equally true regardless if the error is found in the theory or in the measurement procedure.

      The latter seem to have been the case when SP Technical Research Institute during summer 2012 tested the old version of E-Cat using a true RMS instrument to measure power and found the input power two to three times higher than the input power Rossi had measured. SP concluded then that the COP was close to unity.

      http://www.e-catworld.com/2012/09/10/nyteknik-reports-on-halted-swedish-investment-in-hydrofusion-following-tests/

      Hence, even if the post analysis of the current data would change the conclusion of the paper and instead again indicate a COP close to unity, it is an important scientific result. Hence, if that turns out to be the case, I sincerely hope the scientists involved are not ridiculed. If they followed good scientific conduct and didn´t consciously change the measurement setup between the calibration and active test, it is naturally that they didn´t recheck the clamp orientations again. Still, in retrospect, it is strange that neither the scientists involved, nor external reviewers who have read the paper, discovered the discrepancy regarding the Joule heating that Giancarlo found.

      • good point.
        with logged data, if Ieff is logged, we should see current phase and apparent impedance and observe that if follow a process incompatible with a student error.

        I am afraid however that IH will ask for some trade secret to be protected

    • Thomas Clarke

      Not if they checked the control box power used during the dummy run and not the active run.

  • Oystein Lande

    Looking at the picture the control box looks like something from Control Concepts (CCIPOWER.com), which delivers SCR controllers. The testers possibly used a CCI Fusion 3 Phase power SCR (from the picture may be a Compact Fusion SCR power controller), and phase angle control, either with in-line or hybrid configuration.

    In that case: In addition to the PCE’s they could also get data acquisition from the control box. That one would definitely tell them delivered power. They would be able to do diagnostics, charting, do logging with the control box hooked up to a PC

  • Oystein Lande

    Looking at the picture the control box looks like something from Control Concepts (CCIPOWER.com), which delivers SCR controllers. The testers possibly used a CCI Fusion 3 Phase power SCR (from the picture may be a Compact Fusion SCR power controller), and phase angle control, either with in-line or hybrid configuration.

    In that case: In addition to the PCE’s they could also get data acquisition from the control box. That one would definitely tell them delivered power. They would be able to do diagnostics, charting, do logging with the control box hooked up to a PC

  • Oystein Lande

    DickeFix:

    Possible answers to your questions:

    “1.An explanation how nuclear reactions can occur at low temperature”

    This is what has haunted the Cold Fusion / LENR science since it all started in 1989. Many theories from many nuclear scientists have been proposed. Any Rossi suggested theory, will be just that – a theory until other have confirmed the theory by experiments and measurements.

    “2. An explanation how nuclear reactions can occur without any radiation”

    Yet another issue that haunted the Cold Fusion / LENR science since it all started in 1989. Many theories from many nuclear scientists have been proposed. Any Rossi suggested theory, will be just that – a theory until other have confirmed the theory by experiments and measurements.

    “3. An explanation how all Ni isotopes and most other metals in the fuel can be converted to almost pure Ni62 in the ash”

    The ASH analysis is based on a sample representing 0,2% of total ASH weight. In my opinion we cannot base any types of conclusions on a 0,2% sample.
    Other than that “this 0,2% sample show high concentration of 62Ni.” Most likely this is NOT representative of the total. We may speculate that there has been some separation happening at 1400 degC.

    “4.An explanation why the generated power didn´t decrease with time despite the fuel was almost burnt out at the end”

    Concluding that the whole ASH sample had turned to Ni62 based on a 0,2% weight sample, is way too much of a assumption.

    “5. An explanation why the fuel consisted of natural Ni while Rossi repeatedly stated that enriched Ni-62 was an essential ingredient in the fuel”

    Did he? Why not ask Rossi on JONP

    “6. An explanation for the completely different ash compositions in ITP test 1 and 2”

    Comparing analysis based on 0,2% weight sample is not very scientific. But in this case also answers to question 1 & 2 apply.

    “7. An explanation why the electrical Joule heating increased 6 times when the stated input power only increased 2 times (Giancarlo)”

    This is strange. It seems there is a calculation error. Both in dummy and In real test. Could one have a larger calculation error? ANYHOW: In addition to the PCE’s they could also get data acquisition from the control box used from Control Concepts. That one would definitely tell them delivered power. They would be able to do diagnostics, charting, do logging with the control box hooked up to a PC. Hope they did…

    “8. An explanation of the unexpected measured current shapes which indicate a reversed current clamp (Andrea S.)”

    The picture shows OL all over. Probably not hooked up when picture was taken. But having two positive spikes followed by two negative spikes is possible in a three phase with angle control.

    “9. An explanation why the Ni didn´t melt if the power was generated by a reaction in the fuel (Dr. Mike)”

    This is one of the best questions I have seen. The 1 gram powder must have been distributed very evenly on the internal surface to achieve good heat exchange and not melt….

    “10. An explanation why the E-Cat gave COP=1 when SP Technical Research Institute
    (the swedish measurement calibration authority) tested it.”

    Not sure what this is related to.

    My own thoughts on the Ni-H LENR matter:

    My interest in Rossi is based upon what was done by Professor Sergio Focardi
    in the early 1990’s at the University of Bologna.

    He did nickel-hydrogen reactor experiments,and got Heat out larger than what could be explained by any chemical reactions.

    Focardi further published a few papers in the 1990’s on the subject in a scientific Journal (peer-reviewed ;-)… )

    Focardi S, Habel R, Piantelli F (January 1994): “Anomalous Heat
    Production in Ni-H Systems”. Il Nuovo Cimento A, Volume 107 A, Number 1, 163–167

    Focardi S, Gabbani V, Montalbano V, Piantelli F, Veronesi S (November 1998). “Large excess heat production in Ni-H systems”. Il Nuovo Cimento A, 111(11): 1233–1242. OCLC 204819206.

    Neutron emission in Ni-H systems. Il Nuovo Cimento A (1971-1996), Volume 112, Number 9,
    921–931. Authors: Battaglia, Daddi, Focardi, Gabbani, Montalbano, Piantelli, Sona, Veronesi. Retrieved on SpringerLink.

    Later Rossi contacted Focardi with some creative ideas to possible increase power
    levels….

    So I’m still hopeful for LENR, even if the HOT cat turns out to be junk 😉

    Anyone knows if the 1 MW plant IH have delivered is based on HOT cats, or on the first version of lower temp cats..?
    regards
    Lande

    • about 4)
      there is an obvious answer… because the fuel is not nickel.
      note also that only the surface seems fully transmuted.
      the difference between the surface and bulk isotopic measurement le clear evidence of a real and complex phenomenon.

      note for the spikes timing question, that it seems clear that the E-cat is single phase, like for TPR1… there are two coils, probably for some stabilisation reason…

      anyway there are many question, but since we dn’t know the process but we have to stand on solid facts :

      1- inverting clamps polarity is a student error you can fix in second. it seems that even the PCE830 beep if you invert clamps. you can detect it because of abnormal power, negative or reactive, incoherent with logic… in that case the power of the controller would not look logic as it did. So PEC830 measured electric power correctly.

      2- Industrial Heat would never have risked their credibility in providing reactors they know did not work, to scientists, expecting they do incredibly improbable error, plus bad measurement.

      I admit however; following McKubre sad report, that the calibration is very insufficient.
      there is no rational possibility given what we know that COP=1 (emissivity change from 450 to 900W should be of 6x from nearly 1 to nearly 0) , but there is enough uncertainty to allow deniers to continue fooling the innocent readers with FUD.

      Unless the testers make a correction report, we will have to wait for Rossi’s factory delivery and LENR-Cities industrial partners names outing.

    • Dr. Mike

      Oystein Lande and DickeFix,
      My brother just sent me a link to another theory on LENR- thought you might be interested in reading it:
      http://vixra.org/pdf/1401.0169v1.pdf
      I have looked at it it, but haven’t had time to digest it.

      Dr. Mike

    • fact police

      Lande wrote> “Possible answers to your questions:
      “3. An explanation how all Ni isotopes and most other metals in the fuel can be converted to almost pure Ni62 in the ash”
      The ASH analysis is based on a sample representing 0,2% of total ASH weight. In my opinion we cannot base any types of conclusions on a 0,2% sample.”

      But the point of the analysis is that we *should* base conclusions on a 0.2% sample. The exercise was meant to show that the nuclear reactions are happening. If the sample is not representative, then we can’t know there isn’t simple isotopic fractionation going on, and all the Ni-58 is in a different part of the sample. Such an effect without a theoretical mechanism to explain it seems crazy doesn’t it? But radiation-free complete transmutation of Ni-58 to Ni-62 (even if only in isolated places) is a far less plausible without a theoretical mechanism to explain it. At least fractionation doesn’t require the concentration of MeV energies into single atomic sites to explain it.

      Lande> “Other than that “this 0,2% sample show high concentration of 62Ni.” Most likely this is NOT representative of the total. We may speculate that there has been some separation happening at 1400 degC.”

      Yes, and if that’s possible, then you have neutralized evidence that nuclear reactions are happening.

      Lande> “”9. An explanation why the Ni didn´t melt if the power was generated by a reaction in the fuel (Dr. Mike)”

      This is one of the best questions I have seen.”

      This question applies and was raised for Levi2013 as well, and was raised in the context of Levi2014 a few days after the report was released:

      (see ecatnews.com/?p=2669&cpage=4#comments, and search for the post titled “Cats Have Nine Lives”)

      Lande> “The 1 gram powder must have been distributed very evenly on the internal surface to achieve good heat exchange and not melt….”

      That’s not good enough. The rate of heat flow is proportional to a coefficient of heat transfer and the temperature difference, and there is no coefficient large enough to account for such a power with a few tens of degrees temperature difference. For comparison, consider a 1 kW water kettle. In that case the source of heat is some 1000 degrees above the material it is transferring its heat to, and the contact area is if anything much greater. You could not transfer power at twice that rate from 1 gram of material that is only 55 degrees higher in temperature. It’s nonsense.

      Lande> “My interest in Rossi is based upon what was done by Professor Sergio Focardi in the early 1990’s at the University of Bologna. He did nickel-hydrogen reactor experiments,and got Heat out larger than what could be explained by any chemical reactions. Focardi further published a few papers in the 1990’s on the subject in a scientific Journal (peer-reviewed ;-)… )”

      The 1994 Focardi paper did not use any sort of reliable calorimetry. Instead, they just measured the temperature at one or two places of a device cooled unpredictably by the ambient air.

      A group from CERN reproduced their observations and explained them without excess heat by thermal properties that depend on the uptake of hydrogen (Cerron-Zeballos et al, Il Nuovo Cimento 109A (1996) 1645).

      This error was confirmed by their own work in the 1998 Focardi paper, but instead of correcting the error in calorimetry, they continued to make claims based on the same method, even if the temperature was measured at more points. To this day, that experiment has not been reproduced with reliable calorimetry.

  • Oystein Lande

    DickeFix:

    Possible answers to your questions:

    “1.An explanation how nuclear reactions can occur at low temperature”

    This is what has haunted the Cold Fusion / LENR science since it all started in 1989. Many theories from many nuclear scientists have been proposed. Any Rossi suggested theory, will be just that – a theory until other have confirmed the theory by experiments and measurements.

    “2. An explanation how nuclear reactions can occur without any radiation”

    Yet another issue that haunted the Cold Fusion / LENR science since it all started in 1989. Many theories from many nuclear scientists have been proposed. Any Rossi suggested theory, will be just that – a theory until other have confirmed the theory by experiments and measurements.

    “3. An explanation how all Ni isotopes and most other metals in the fuel can be converted to almost pure Ni62 in the ash”

    The ASH analysis is based on a sample representing 0,2% of total ASH weight. In my opinion we cannot base any types of conclusions on a 0,2% sample.
    Other than that “this 0,2% sample show high concentration of 62Ni.” Most likely this is NOT representative of the total. We may speculate that there has been some separation happening at 1400 degC.

    “4.An explanation why the generated power didn´t decrease with time despite the fuel was almost burnt out at the end”

    Concluding that the whole ASH sample had turned to Ni62 based on a 0,2% weight sample, is way too much of a assumption.

    “5. An explanation why the fuel consisted of natural Ni while Rossi repeatedly stated that enriched Ni-62 was an essential ingredient in the fuel”

    Did he? Why not ask Rossi on JONP

    “6. An explanation for the completely different ash compositions in ITP test 1 and 2”

    Comparing analysis based on 0,2% weight sample is not very scientific. But in this case also answers to question 1 & 2 apply.

    “7. An explanation why the electrical Joule heating increased 6 times when the stated input power only increased 2 times (Giancarlo)”

    This is strange. It seems there is a calculation error. Both in dummy and In real test. Could one have a larger calculation error? ANYHOW: In addition to the PCE’s they could also get data acquisition from the control box used from Control Concepts. That one would definitely tell them delivered power. They would be able to do diagnostics, charting, do logging with the control box hooked up to a PC. Hope they did…

    “8. An explanation of the unexpected measured current shapes which indicate a reversed current clamp (Andrea S.)”

    The picture shows OL all over. Probably not hooked up when picture was taken. But having two positive spikes followed by two negative spikes is possible in a three phase with angle control.

    “9. An explanation why the Ni didn´t melt if the power was generated by a reaction in the fuel (Dr. Mike)”

    This is one of the best questions I have seen. The 1 gram powder must have been distributed very evenly on the internal surface to achieve good heat exchange and not melt….

    “10. An explanation why the E-Cat gave COP=1 when SP Technical Research Institute
    (the swedish measurement calibration authority) tested it.”

    Not sure what this is related to.

    My own thoughts on the Ni-H LENR matter:

    My interest in Rossi is based upon what was done by Professor Sergio Focardi
    in the early 1990’s at the University of Bologna.

    He did nickel-hydrogen reactor experiments,and got Heat out larger than what could be explained by any chemical reactions.

    Focardi further published a few papers in the 1990’s on the subject in a scientific Journal (peer-reviewed ;-)… )

    Focardi S, Habel R, Piantelli F (January 1994): “Anomalous Heat
    Production in Ni-H Systems”. Il Nuovo Cimento A, Volume 107 A, Number 1, 163–167

    Focardi S, Gabbani V, Montalbano V, Piantelli F, Veronesi S (November 1998). “Large excess heat production in Ni-H systems”. Il Nuovo Cimento A, 111(11): 1233–1242. OCLC 204819206.

    Neutron emission in Ni-H systems. Il Nuovo Cimento A (1971-1996), Volume 112, Number 9,
    921–931. Authors: Battaglia, Daddi, Focardi, Gabbani, Montalbano, Piantelli, Sona, Veronesi. Retrieved on SpringerLink.

    Later Rossi contacted Focardi with some creative ideas to possible increase power
    levels….

    So I’m still hopeful for LENR, even if the HOT cat turns out to be junk 😉

    Anyone knows if the 1 MW plant IH have delivered is based on HOT cats, or on the first version of lower temp cats..?
    regards
    Lande

    • DickeFix

      Oystein Lande,

      The possible explanations you have to 1-4 and 6 are the same that I have considered. However, even if the fuel and ash samples were small, it is strange that the ash contained natural Ni with some natural Cu in test 1 while it was almost pure Ni62 in test 2.

      Regarding 5, Rossi has claimed repeatedly, both in his patent applications and on his blog that Ni62 is an important part of the _fuel_ and that it is transformed to Cu63.

      Dec. 6 2011 Rossi wrote:
      “As I have explained many times, we use Ni enriched of 62 and 64 Ni, which are the
      sole to react, and 63 and 65 Cu are stable. Our process has been developed upon a theory that became stronger in time, based on the results of the thousands of our tests we made with our apparatuses. At this point we have a solid theory which is leading our R&D, making progress by the day. The problem is that the theory leads directly to the industrial confidential IP and since we have not a granted patent we deem opportune not to disclose the theory.”

      I have, as you, difficulties to find explanations to 7-10. Question 8 is not only based on the picture in the present report but also the pictures in previous report. Maybe one can find some natural explanation to each one of these question, it is just when you take it together I think it is easier to explain it with that the fuel contained enriched Ni62 (as Rossi previously stated) and that the input power was wrongly measured so the real COP is 1 as in the SP test.

      Since I want to be objective I admit that the “reversed clamp theory” does also not explain all experimental evidence:

      1. Sanjeev pointed out from Fig 12, page 25 in the report it appears that most of the heat is coming from the inner reactor and not from the heating coils. The heating coils seem to be so cold compared to the inner reactor that they cast a shadow. I first thought that it could be due to that the alumina is translucent and that the Inconel coil had lower emissivity than the fuel and reactor. However, the difference in emissivity between polished metals and ceramics in visible wavelength range is not so large so I am hesitant if this can be an explanation

      2. The reversed clamp can also not explain why the measured output power increased so much almost 700W when the input power increased only around 100W. Maybe, the relation between convection and radiation changes so the output power measurement is wrong but it is still strange that the differential error is so large.

      I agree that the work on LENR in Italy before Rossi gave credibility to the E-cat. In the nice overview of LENR research in Italy from 2008 entitled “COLD FUSION-The history of research in Italy” by ENEA there is a a review article by Focardi and Piantelli where they describe the excess heat seen in Ni-H system and which has a comprehensive litterature list. I have in vain tried to find the most recent publication in that list by Campari et al. from The 7 International Workshop on Anomalies in Hydrogen/
      Deuterium loaded Metals, Asti 23-27 September 2006

      I didn’t find the article but found the opening address in that conference by Bill Collis

      http://www.infinite-energy.com/iemagazine/issue70/summary7thinternational.html

      Collis writes “Of course, during the last 17 years some researchers have managed to
      throw the switch. By patient observation and refinement of their protocols, they have noticed tantalizing signs that metals loaded with isotopic hydrogen can become self-sustaining. I don’t mean to imply a perpetual motion machine, but merely that such systems can remain hot without any external energy. Fleischmann called this “Heat after Death.” This phenomenon has been observed by many others. In fact, I was phoned by one Italian researcher in August whose system remained hot, producing 34 watts for some two days without any apparent energy source—no input energy. Obviously more work needs to be done to verify,control, and replicate these results. But if verified, who needs calorimetry? Can we now expect non-specialists to accept the reality of anomalous heat production? Can we expect that the door will open to new funding and commercial interest? Time will tell. In the meantime we should feel encouraged.”

      Viewgraph 31 of that presentation had title “Self sustaining heat in Ni / H System” and said:

      August 2006 64 years after Fermi
      34 Watts 70 x
      2 days 100 x
      ~4 kg 10-5 x
      ~$27,000 1%
      No energy input!

      Was this a precursor to the E-Cat? After 2006 not much was heard until beginning of 2011 when Rossi filed the first patent application and arranged the public demonstration.

      I have also a hard time to see that all this research by Focardi, Piantelli and others were just illusions based on poor measurements. Hence, I still do not rule out that LENR is possible. However, the many questions and inconcistencies with the recent ITP test have made me more of a skeptic than a believer in the E-Cat.

    • about 4)
      there is an obvious answer… because the fuel is not nickel.
      note also that only the surface seems fully transmuted.
      the difference between the surface and bulk isotopic measurement is clear evidence of a real and complex phenomenon.

      note for the spikes timing question, that it seems clear that the E-cat is single phase, like for TPR1… there are two coils, probably for some stabilization reason…

      anyway there are many question, but since we don’t know the process but we have to stand on solid facts :

      1- inverting clamps polarity is a student error you can fix in second. it seems that even the PCE830 beep if you invert clamps. you can detect it because of abnormal power, negative or reactive, incoherent with logic… in that case the power of the controller would not look logic as it did. So PCE830 measured electric power correctly.

      2- Industrial Heat would never have risked their credibility in providing reactors they know did not work, to scientists, expecting they do incredibly improbable error, plus bad measurement.

      I admit however; following McKubre sad report, that the calibration is very insufficient.
      there is no rational possibility given what we know that COP=1 (emissivity change from 450 to 900W should be of 6x from nearly 1 to nearly 0) , but there is enough uncertainty to allow deniers to continue fooling the innocent readers with FUD.

      Unless the testers make a correction report, we will have to wait for Rossi’s factory delivery and LENR-Cities industrial partners names outing.

      • Thomas Clarke

        Alain – the beep is an optional function. The standard screens do not obviously show when a clamp is reversed, though if you look closely you will notive that overall power is roughly eual to voltage times current on one line.

        The testers might have noticed. Or not.

        It is unwise to make assumptions about what Rossi would do. He is CTO at IH and if he has always measured his devices reversed clamp I guess that IH will not know anything about it. You think they have another 3 phase electrical engineer who can challenge Rossi?

      • DickeFix

        Giancarlo has posted a test of PCE 830 on Mats Lewans blog to show what happens if you reverse one clamp:

        https://dl.dropboxusercontent.com/u/66642475/PCE%20Clamp%20test.docx

        No, beep, no warning, just a factor of three too low power. The only way to see that the clamp is reversed is either to look at the orientation of the physical clamp or by checking the phases of the currents in the polar diagram. The reversed clamp may hence be easily overlooked by the research group, especially if they had checked it during the dummy run and didn´t suspect that it may be reversed during active run.

    • Dr. Mike

      Oystein Lande and DickeFix,
      My brother just sent me a link to another theory on LENR- thought you might be interested in reading it:
      http://vixra.org/pdf/1401.0169v1.pdf
      I have looked at it it, but haven’t had time to digest it.

      Dr. Mike

      • DickeFix

        Thank you Dr. Mike,

        It is indeed a well written paper. However, the paper does not explain how the gamma radiation becomes heat (which is maybe the greatest mystery). It also assumes enriched Ni62 and Ni64 in the fuel and Cu63, Cu64, Zn63 and Zn66 in the ash. This contradicts the present test assuming the fuel and ash samples analyzed were representative.

        • Dr. Mike

          DickeFix,
          I got a chance to read it more carefully. The theory seems to be applicable to the original E-Cat, but I didn’t see anything in the theory that could explain the Lugano “ash” results with no Cu being found. The one good thing that I saw in the theory is that it gives a means of overcoming the Coulomb barrier. Perhaps the understanding of the Coulomb barrier has been advanced.
          Dr. Mike

    • fact police

      Lande wrote> “Possible answers to your questions:
      “3. An explanation how all Ni isotopes and most other metals in the fuel can be converted to almost pure Ni62 in the ash”
      The ASH analysis is based on a sample representing 0,2% of total ASH weight. In my opinion we cannot base any types of conclusions on a 0,2% sample.”

      But the point of the analysis is that we *should* base conclusions on a 0.2% sample. The exercise was meant to show that the nuclear reactions are happening. If the sample is not representative, then we can’t know there isn’t simple isotopic fractionation going on, and all the Ni-58 is in a different part of the sample. Such an effect without a theoretical mechanism to explain it seems crazy doesn’t it? But radiation-free complete transmutation of Ni-58 to Ni-62 (even if only in isolated places) is a far less plausible without a theoretical mechanism to explain it. At least fractionation doesn’t require the concentration of MeV energies into single atomic sites to explain it.

      Lande> “Other than that “this 0,2% sample show high concentration of 62Ni.” Most likely this is NOT representative of the total. We may speculate that there has been some separation happening at 1400 degC.”

      Yes, and if that’s possible, then you have neutralized evidence that nuclear reactions are happening.

      Lande> “”9. An explanation why the Ni didn´t melt if the power was generated by a reaction in the fuel (Dr. Mike)”

      This is one of the best questions I have seen.”

      This question applies and was raised for Levi2013 as well, and was raised in the context of Levi2014 a few days after the report was released:

      (see ecatnews.com/?p=2669&cpage=4#comments, and search for the post titled “Cats Have Nine Lives”)

      Lande> “The 1 gram powder must have been distributed very evenly on the internal surface to achieve good heat exchange and not melt….”

      That’s not good enough. The rate of heat flow is proportional to a coefficient of heat transfer and the temperature difference, and there is no coefficient large enough to account for such a power with a few tens of degrees temperature difference. For comparison, consider a 1 kW water kettle. In that case the source of heat is some 1000 degrees above the material it is transferring its heat to, and the contact area is if anything much greater. You could not transfer power at twice that rate from 1 gram of material that is only 55 degrees higher in temperature. It’s nonsense.

      Lande> “My interest in Rossi is based upon what was done by Professor Sergio Focardi in the early 1990’s at the University of Bologna. He did nickel-hydrogen reactor experiments,and got Heat out larger than what could be explained by any chemical reactions. Focardi further published a few papers in the 1990’s on the subject in a scientific Journal (peer-reviewed ;-)… )”

      The 1994 Focardi paper did not use any sort of reliable calorimetry. Instead, they just measured the temperature at one or two places of a device cooled unpredictably by the ambient air.

      A group from CERN reproduced their observations and explained them without excess heat by thermal properties that depend on the uptake of hydrogen (Cerron-Zeballos et al, Il Nuovo Cimento 109A (1996) 1645).

      This error was confirmed by their own work in the 1998 Focardi paper, but instead of correcting the error in calorimetry, they continued to make claims based on the same method, even if the temperature was measured at more points. To this day, that experiment has not been reproduced with reliable calorimetry.

  • Thomas Clarke

    DickieFix –

    in answer to the two additional matters you raise:

    (1) I’m not sure anyone can distinguish between flat tape with gaps, and heating wire. The idea that heating wire would shade the internal heat source is also I think wrong, because the power density from the wire would raise the local Al2O3 temperature to more than the surrounding body temperature. While the centre would be very hot, the light you would see (because – it is claimed – the Al2O3 is opaque) would come from the hot body, not the hotter central heat source. think of a lamp with a diffuser.

    (2) the difference in power out does I agree need explanation. The reason, I believe, is because the assumptions made about Al2O3 are not very accurate. Specifically I believe from references that Al2O3 is transparent at higher frequencies (which there will be more power in at higher temperatures). The effect of Al2O3 being transparent, is complex. For example, if the emissivity of the flat tape that is heating the body is low, as is true of metals, then the radiation at higher frequencies will be less than at IR frequency. Thus the total power out will be less than the apparent power out calculated from the IR power by the camera. This effect will then be more important at higher temperatures where radiation is a larger proportion of total output and also higher frequency light a higher proportion of radiation.

    One issue here is that the depth of Al2O3 outside the heating tape is unknown and could be very small. The thinner this is the lower the frequency at which the Al2O3 is effectively transparent.

    I’m not saying this is necessarily a correct argument. Just that the real situation, and how it changes with different power in, is very complex and could easily result in this +20% reading as calculated from temperature estimated from IR emissions and an assumed emissivity of 0.4.

    This is a structural flaw in the experiment, since if they had measured the dummy run at the same temperature as the active run they would have clear information about how accurate is their theory at much higher temperatures.

    • DickeFix

      Thomas, thank you for those possible explanations. I fully agree that it is a pity that they didn´t do measurements at same temperature or input power.

      They could have started with a first dummy test 1 at the certain power input, then run the active test at same power input and finally run a dummy test 2 at an input power that gave the same temperature as in the active test. Then they could have divided the power input in dummy test 2 with the power input in dummy test 1 and got the COP without any complex assumptions and calculations. As an extra check they could have compared the RMS current^2 in all three tests.

  • Thomas Clarke

    I would add one more thing that requires explanation.

    11. Given that the IH patent does not in any way either cover or protect the hot e-cat tested, because it has a different construction as well as not fully revealing how the device works, why does Rossi allow his powder – both fuel and ash – to be analysed by third parties and published? That surely would reveal IP and also prevent the composition from being patented in the future.

    my reference for the patent comment is here:
    http://blog.stepchange-innovations.com/2014/10/excess-heat-isotope-changes-e-cat-lenr-reactor-part2/#.VFeWVfmsWik

    “David French, a retired patent attorney who frequently contributed to a cold fusion website, correctly commented on Rossi´s US patent examination[13] “these independent claims stipulate for the presence of a metal tube. In the absence of such a component, a competing construction would not infringe these claims. For example, if a ceramic tube were employed, it would not fall under the language of the claim”. This is the case in the Lugano-hotcat.”

    Indeed it is not clear why, if IH are seriously expecting to exploit a revolutionary and extremely valuable new technology, it is being demonstrated in this manner at all. After all, IH can convince customers or backers that the technology works with internal tests under NDA without revealing any secrets? And without an external test competitors will most likely not believe the technology is real and therefore give IH first mover advantage.

    Some here will perhaps think that the whole external test setup is designed by a cunning Rossi specifically to appear to fail in such a way as to prevent any competitor from looking seriously at the technology. The other unexplained matters could then be explained as Rossi wanting it to look as though he has nothing and therefore deliberately lying to make himself look unbelievable.

    • Dr. Mike

      Thomas,
      I believe you are correct that Rossi’s patent application is not going to provide protection for a Hot-Cat design device. I don’t think Rossi had a good enough theory when the patent was applied for to write up a good all-encompassing patent. Also, I don’t see how he thought he could get a patent without disclosing the entire invention, including the catalyst. I agree that there is probably enough disclosed in the Lugano repiort on the Hot-Cat design and fuel to make it non-patentable. Your question is a good one!
      Dr. Mike

  • GreenWin

    While there is a ton of speculation and demands to know more about details of the Lugano test, I would remind all that Industrial Heat has invested in a commercial venture. Its value is dependant on commercial performance, NOT meeting scientific method. Expecting either the authors or IH to release proprietary information prior to achieving their commercial goals is wishful thinking. Scientists have had their opportunities since P&F — and they’ve blown it.

    • Thomas Clarke

      If the report is correct then proprietary information – the fuel composition – has been released? In fact it is difficult to see what has not been released about the hot-cat design.

      I think what people expected (because that was the purpose) was a black box test that would validate the technology. No details of the technology were required.

      The details asked for now relate to the testing, not the technology

    • Donk970

      I couldn’t agree more. At the end of the day, no matter how much people want LENR to be true (or not true), all that will really matter is wether or not a commercial product works and ships to the customer. The only people who need convincing is the board of directors and investors in IH and so far they appear to be convinced.

  • Donk970

    We have two distinct groups of people talking about the Lugano test. One group believes the observations are real and wants to understand why. The other group fundamentally believes that the observations cannot be real and wants to know how the incorrect observations were produced. The problem for the second group is that they require ever more convoluted explanations that involve both truly gross incompetence on the part of the testers and truly brilliant and sustained deception on Rossi’s part. The idea that Rossi could maintain this level of deception while working closely with engineers and scientists at IH for several years is absurd. This means that either everyone at IH is in on the deception or there is no deception. A conspiracy that involves more than one person is bound to be discovered in short order so the only conclusion I can come to is that there is no deception. I also find it highly unlikely that the testers were so incompetent that they would miss something obvious like an inverted current clamp. What I see here is a group of people who have gone down a rabbit hole of speculation about errors in testing that could lead to virtually any conclusion because none of the people doing the speculating was actually in the lab when the tests were done. My feeling is that all those who are convinced that the test was so badly done that it reported a COP of 3+ instead of 1 need to set up a test rig to show how a dummy reactor could be made to show a COP of 3+ without the testers being aware of the error.

    • I agree, but the most crazy is that meanwhile serious LENr scientists agree that the calorimetry is inconcusive because of inssuficient calibration, and too much dependence on theory and assumption.

      sure something is working, but we cannot shut up deniers with such an imprcise test.

      I hope the testers can gather data in their test to clear the question of COP=1 or not…

      for COP>3 a new test is needed I fear.

      • Obvious

        If the COP is shown to be 1, then the isotope change process consumes no power.

        • Dr. Mike

          Obvious,
          The isotope change process could easily be contributing some power, but perhaps that power level is in the noise of the total power. When the COP is claimed to be 1, it is really 1 +/- the measurement error.
          Dr. Mike

          • Obvious

            OK.
            The isotope change process is so energy neutral, that it is within the range of error of the heat measurement of a large resistor, should the COP of the Lugano test unit be shown to be 1, within the range of error of power measurement, and making certain assumptions about the measurement of power, which may or may not be correct.
            LOL

          • Dr. Mike

            Obvious,
            Let’s wait for the revision of the report to see if the authors need to revise their COP calculation. Sometime in the future when we have both the corrected report and a solid LENR theory we will have (at least we better have) a good explanation for the Lugano results.
            Dr. Mike

          • Obvious

            I check for a newer, appended or modified version every day.

    • Dr. Mike

      Donk970,
      I certainly believe you are correct that we have two distinct groups of people talking about the Lugano test, however, I believe you are mistaken about how you categorize those two groups. I believe there is one group that is examining the data from the Lugano test carefully, seeing an inconsistency, and asking why. The second group wants to believe the final results of the Lugano test, but really don’t feel they need to ask why or understand how the results were achieved (or maybe feel they don’t have the background to understand the results). (There are others that really don’t fit into either of these groups.)
      I believe that most of the electrical engineers that read the Lugano report carefully, looked at the data in Table 7 on page 22 and asked themselves why wasn’t the Joule heating in the Cu wires about 11W for the first 5 files and about 12.5W for Files #6-16 (that is, directly proportional to the reported supplied powers)? The Lugano authors showed us they know how to calculate Joule heating with the calculations for the dummy run shown on pages 13-14. Either the reported power “consumption” is incorrect or the second time the Joule heating was calculated, it was done incorrectly in the Table 7 data. (This data would have also fit if the heater wire had a large negative temperature coefficient of resistance, however, not only does Inconel wire not have this property, the data from the report shows the resistance not to change when the active run temperature was increased from 1260C to 1400C.)
      So the issue is not whether some people believe or don’t believe in observations made by the Lugano authors, the issue is why the supplied power data is not consistent with the Joule heating data in Table 7. Why are several electrical engineers suggesting that one of the current clamps may have been reversed? Because if one clamp had been reversed, then you would expect to see the numbers presented in Table 7 for for the power “consumption” and the “Joule heating”. Their theory is a good fit to the data. There easily can be another explanation for the data presented in Table 7. We just need to get that explanation from the authors and verify that their explanation also fits the data presented, or maybe the authors will determine there is an error in Table 7, and they will correct that error.
      I believe that most of those people that are pointing out the error in the Table 7 data are actually good supporters of Rossi and his efforts to advance LENR.
      Dr. Mike

  • Donk970

    We have two distinct groups of people talking about the Lugano test. One group believes the observations are real and wants to understand why. The other group fundamentally believes that the observations cannot be real and wants to know how the incorrect observations were produced. The problem for the second group is that they require ever more convoluted explanations that involve both truly gross incompetence on the part of the testers and truly brilliant and sustained deception on Rossi’s part. The idea that Rossi could maintain this level of deception while working closely with engineers and scientists at IH for several years is absurd. This means that either everyone at IH is in on the deception or there is no deception. A conspiracy that involves more than one person is bound to be discovered in short order so the only conclusion I can come to is that there is no deception. I also find it highly unlikely that the testers were so incompetent that they would miss something obvious like an inverted current clamp. What I see here is a group of people who have gone down a rabbit hole of speculation about errors in testing that could lead to virtually any conclusion because none of the people doing the speculating was actually in the lab when the tests were done. My feeling is that all those who are convinced that the test was so badly done that it reported a COP of 3+ instead of 1 need to set up a test rig to show how a dummy reactor could be made to show a COP of 3+ without the testers being aware of the error.

    • I agree, but the most crazy is that meanwhile serious LENr scientists agree that the calorimetry is inconcusive because of inssuficient calibration, and too much dependence on theory and assumption.

      sure something is working, but we cannot shut up deniers with such an imprcise test.

      I hope the testers can gather data in their test to clear the question of COP=1 or not…

      for COP>3 a new test is needed I fear.

      • Obvious

        If the COP is shown to be 1, then the isotope change process consumes no power.

        • Dr. Mike

          Obvious,
          The isotope change process could easily be contributing some power, but perhaps that power level is in the noise of the total power. When the COP is claimed to be 1, it is really 1 +/- the measurement error.
          Dr. Mike

          • Obvious

            OK.
            The isotope change process is so energy neutral, that it is within the range of error of the heat measurement of a large resistor, should the COP of the Lugano test unit be shown to be 1, within the range of error of power measurement, and making certain assumptions about the measurement of power, which may or may not be correct.
            LOL

          • Dr. Mike

            Obvious,
            Let’s wait for the revision of the report to see if the authors need to revise their COP calculation. Sometime in the future when we have both the corrected report and a solid LENR theory we will have (at least we better have) a good explanation for the Lugano results.
            Dr. Mike

          • Obvious

            I check for a newer, appended or modified version every day.

          • fact police


            Mike wrote: “The isotope change process could easily be contributing some power, but perhaps that power level is in the noise of the total power.”

            This is only possible if the sample is grossly unrepresentative of the ash, in which case, it can’t be used as evidence of nuclear reactions either.

            If 99% of the nickel is converted into Ni-62, using nucleons from hydrogen or lithium, then the total energy released, based on the mass difference, is many times the output heat they claimed even if the COP is 3. It would not be in the noise.

          • Dr. Mike

            fact police,
            Thanks for the correction- these calculations are outside my field of knowledge. What mechanisms did you assume for the conversion of the Ni isotopes to Ni62 Would we need a solid theory to really know the mechanisms? Finally, using your mechanisms, how much total energy should the reactor have produced form converting all 0.55 grams of Ni to Ni62? Note: even though the ash sample size was really small, the conversion of the Ni to Ni62 was confirmed by two measurement techniques.
            Dr. Mike

          • fact police

            Mike> “What mechanisms did you assume for the conversion of the Ni isotopes to Ni62 “

            The mechanism doesn’t matter. Only the masses of the reactants and the reaction products matter. The simplest is to use a Q-value calculator such as the one at http://www.nndc.bnl.gov/qcalc/qcalcr.jsp

            Enter 58Ni as the target, and n as the projectile, and hit submit, and in the resulting table you see the reaction product 59Ni + gamma has a Q-value of 9 MeV.

            If you do it 3 more times to get to 62Ni, and subtract the .8 MeV times 4 to make neutrons from protons, you get 35 MeV per atom of Ni to go from 58Ni to 62Ni using free protons. It works out to about 30 GJ for a half a gram of nickel. They claimed 5 GJ output.

            If you enter 58Ni as the target, 7Li as the projectile, and 6Li as the ejectile, you get the neutron transfer that student talked about with a Q-value of 1.75 MeV. If you do that 3 more times, you get

            9.8 MeV per nickel atom, which is still close to 10 GJ. But there aren’t enough Li for this to be the only path, and I don’t see any other path consistent with the claimed isotope ratios.

            Mike> “Finally, using your mechanisms, how much total energy should the reactor have produced form converting all 0.55 grams of Ni to Ni62? “

            See above. I only did the calculation from 58Ni to 62Ni, but that’s the main one. going from 64 to 63 is endothermic, but there aren’t that many of them.

    • Dr. Mike

      Donk970,
      I certainly believe you are correct that we have two distinct groups of people talking about the Lugano test, however, I believe you are mistaken about how you categorize those two groups. I believe there is one group that is examining the data from the Lugano test carefully, seeing an inconsistency, and asking why. The second group wants to believe the final results of the Lugano test, but really don’t feel they need to ask why or understand how the results were achieved (or maybe feel they don’t have the background to understand the results). (There are others that really don’t fit into either of these groups.)
      I believe that most of the electrical engineers that read the Lugano report carefully, looked at the data in Table 7 on page 22 and asked themselves why wasn’t the Joule heating in the Cu wires about 11W for the first 5 files and about 12.5W for Files #6-16 (that is, directly proportional to the reported supplied powers)? The Lugano authors showed us they know how to calculate Joule heating with the calculations for the dummy run shown on pages 13-14. Either the reported power “consumption” is incorrect or the second time the Joule heating was calculated, it was done incorrectly in the Table 7 data. (This data would have also fit if the heater wire had a large negative temperature coefficient of resistance, however, not only does Inconel wire not have this property, the data from the report shows the resistance not to change when the active run temperature was increased from 1260C to 1400C.)
      So the issue is not whether some people believe or don’t believe in observations made by the Lugano authors, the issue is why the supplied power data is not consistent with the Joule heating data in Table 7. Why are several electrical engineers suggesting that one of the current clamps may have been reversed? Because if one clamp had been reversed, then you would expect to see the numbers presented in Table 7 for for the power “consumption” and the “Joule heating”. Their theory is a good fit to the data. There easily can be another explanation for the data presented in Table 7. We just need to get that explanation from the authors and verify that their explanation also fits the data presented, or maybe the authors will determine there is an error in Table 7, and they will correct that error.
      I believe that most of those people that are pointing out the error in the Table 7 data are actually good supporters of Rossi and his efforts to advance LENR.
      Dr. Mike

  • Thomas Clarke

    Donk970

    Your position means that you don’t need an independent test. You already know it works. That is not the normal case. It puts you in the same position of disinterest in the test as somone who is certain that Rossi must be a fraud.

    I agree uncommitted observers will vary in how likely they think it is a priori. Those who are convinced LENR is proven will have a higher prior probability than those who do not think there is as yet any credible evidence for LENR, so to see that would be extraordinary.

    Both groups, if rational, will be swayed by strong evidence. They will already have been swayed by past statements and actions of Rossi, as well as the past tests. A good third party test would be very strong evidence and would sway both groups towards the “Rossi has something” conclusion. And vice versa.

    Saying that the test does not matter because Rossi must have it is as bad as saying the test does not matter because you know Rossi does not have it.

    On Mats thread Giancarlo who works with PCE-830s has shown screenshots of them with one clamp reversed and normal. It is not obvious whether the clamp is reversed. If the testers had checked various things then it would have been obvious. We do not at the moment know whether they checked these things, but if they did not then indeed the active test would seem to have COP 3 when it only had COP = 1.

    You are forgetting the issue of the reported Joule powers, and Inconel wire. Something must be wrong. It is then a matter of balancing probabilities of different errors.

    In a third party test meant to be credible that is not tenable – no-one would take it seriously. Which is why I would expect the testers to resolve the matter. They say they have the data to do this, it does not disclose any secrets to do so.

    The only reasons I can think for them not doing this are sinister. But it might take them a while – having egg on your face once means you want to be very sure to get things right second time round.

    • Dr. Mike

      Thomas,
      Very balanced and very relevant!
      Dr. Mike

  • Thomas Clarke

    it seems my balanced and relevant reply to donk970 has been moderated away? Or is this some strange bug of the site that posts vanish?

    • Frank Acland

      Thomas, I just found your post in the moderation queue and approved it. I hope it stays that way — sometimes for reasons unknown to me, posts get unapproved after approval.

  • Dr. Mike

    Thomas,
    Very balanced and very relevant!
    Dr. Mike

  • Mark Szl

    Here is another theory of what happened. The Internal Conflict Theory.

    Rossi did not want to let information about his IP to leak out. He placed a sample of diluted or different powder which still did work, not nearly as well as what his IP was based on but did work well enough for the test. This would also temporarily mislead other until IH was ready to go to market.

    The scientists testing found out! Ayayayaya!

    As payback they would teach IH/Rossi a lesson by connecting the clamp in reverse and generating more COP than expected by IH/Rossi.

    The full report was then released by “accident.” Ooopsy!!

    That caused all kinds of concerns because of undisclosed information and anomalous results beyond what people expected in both camps.

    This now would force IH/Rossi to hand over the real goods and let scientists do the test over like originally planned.

    And everyone lived happily ever after.

    • Dr. Mike

      Mark,
      I hope you are joking! Let me ask you and others this question. Why does the original E-Cat need to be shielded, but the Hot Cat used in Lugano not require shielding? One could assume that Rossi has determined this experimentally., but this would mean he knows the reactions in the Hot-Cat are different from the original E-Cat. Why was he surprised the the “ash” analysis?
      Dr. Mike

    • and rossi is a good prestidigitator able to swap sample in front of observers…
      He’d make money at las vegas.

      I call that extraordinary claim.

  • Mark Szl

    Here is another theory of what happened. The Internal Conflict Theory.

    Rossi did not want to let information about his IP to leak out. He placed a sample of diluted or different powder which still did work, not nearly as well as what his IP was based on but did work well enough for the test. This would also temporarily mislead other until IH was ready to go to market.

    The scientists testing found out! Ayayayaya!

    As payback they would teach IH/Rossi a lesson by connecting the clamp in reverse and generating more COP than expected by IH/Rossi.

    The full report was then released by “accident.” Ooopsy!!

    That caused all kinds of concerns because of undisclosed information and anomalous results beyond what people expected in both camps.

    This now would force IH/Rossi to hand over the real goods and let scientists do the test over like originally planned.

    And everyone lived happily ever after.

    • Dr. Mike

      Mark,
      I hope you are joking! Let me ask you and others this question. Why does the original E-Cat need to be shielded, but the Hot Cat used in Lugano not require shielding? One could assume that Rossi has determined this experimentally., but this would mean he knows the reactions in the Hot-Cat are different from the original E-Cat. Why was he surprised the the “ash” analysis?
      Dr. Mike

      • Mark Szl

        Yes I am joking.

    • and rossi is a good prestidigitator able to swap sample in front of observers…
      He’d make money at las vegas.

      I call that extraordinary claim.

  • Dr. Mike

    DickeFix,
    I got a chance to read it more carefully. The theory seems to be applicable to the original E-Cat, but I didn’t see anything in the theory that could explain the Lugano “ash” results with no Cu being found. The one good thing that I saw in the theory is that it gives a means of overcoming the Coulomb barrier. Perhaps the understanding of the Coulomb barrier has been advanced.
    Dr. Mike

  • It is time to add up all the crazy conspiracy claims about lugano test, to defend the COP=1

    First the dummy was tested normally, and all was right.
    the COP was about1 and thus emissivity estimated at 0.7 was right
    the two powermeter were well wired and the two clamps were not inverted.

    then there is the electric conpiracy theory.

    for the second run, with powder, one clamp on the front power meter, and one in the middle powermeter were inverted.
    this led to an apparent COP of 3, because 2.7kWwas looking as 900W
    the problem is that there is still 33% of error, 33% more heat than measured
    note also that with time COP increase.

    note also that from 800W to 900W the apparent power increase much more tha linearily…

    this mean that IR cam have wrong emissivity… but since it was 0.7, correctly measured on the dummy at 450C, and since only twice the power is produced if COP=1 , to show a similar effect as 1400C the emissivity have to be very low.
    The lack of calibration at high temperature let doubt on such an effect but not to a point that COP=1 is possible…

    strangely the box behave as if it was consuming energy as written on the box…

    strangely the reactor don’t behave like a triphase load but as a single phase load driven by a triphase dimmer.

    note also that none of the testers detect the problem on any of the 2 powermeter, and that Industrial Heat have provided a reactor that is broken, hoping for such a mistake…

    this is why I call that a conspiracy theory.

    anyway the test is incomplete, not on the electric side, but on the calorimetry side by lack of high temp calibration, ruling out very low emissivity at top temperature.
    depending on litterature to estimate alumina emissivity is not enough.

    • AlbertNN

      We do not know if they disconnected the measurement equipment between the dummy run and the real one. Wo do not either know if the dummy run was done with two or three active phases. We do not know if they compared the power readings of the two power meters during the active run. We do know that one of the meters at one point gave an OL reading, and thus did not give any measurements to compare with at all.

  • It is time to add up all the crazy conspiracy claims about lugano test, to defend the COP=1

    First the dummy was tested normally, and all was right.
    the COP was about1 and thus emissivity estimated at 0.7 was right
    the two powermeter were well wired and the two clamps were not inverted.

    then there is the electric conpiracy theory.

    for the second run, with powder, one clamp on the front power meter, and one in the middle powermeter were inverted.
    this led to an apparent COP of 3, because 2.7kWwas looking as 900W
    the problem is that there is still 33% of error, 33% more heat than measured
    note also that with time COP increase.

    note also that from 800W to 900W the apparent power increase much more tha linearily…

    this mean that IR cam have wrong emissivity… but since it was 0.7, correctly measured on the dummy at 450C, and since only twice the power is produced if COP=1 , to show a similar effect as 1400C the emissivity have to be very low.
    The lack of calibration at high temperature let doubt on such an effect but not to a point that COP=1 is possible…

    strangely the box behave as if it was consuming energy as written on the box…

    strangely the reactor don’t behave like a triphase load but as a single phase load driven by a triphase dimmer.

    note also that none of the testers detect the problem on any of the 2 powermeter, and that Industrial Heat have provided a reactor that is broken, hoping for such a mistake…

    this is why I call that a conspiracy theory.

    anyway the test is incomplete, not on the electric side, but on the calorimetry side by lack of high temp calibration, ruling out very low emissivity at top temperature.
    depending on litterature to estimate alumina emissivity is not enough.

    • AlbertNN

      We do not know if they disconnected the measurement equipment between the dummy run and the real one. Wo do not either know if the dummy run was done with two or three active phases. We do not know if they compared the power readings of the two power meters during the active run. We do know that one of the meters at one point gave an OL reading, and thus did not give any measurements to compare with at all.

  • DickeFix

    I read the report again and don´t really understand the measured values of the currents.

    1. In the dummy run they measure an “average” (I assume they mean RMS) current of 19.7A per phase and a measured power of 486W. During active run they claim 40-50A current and 800-900W total input power. Since the current is 2-2.5 times as large and P=R*I^2 one would expect 4-6 times input power in active run, i.e, around 2000-3000W which would yield COP around 1.

    2. I also wonder why the RMS values of the current are so large if they use line voltage. In Giancarlos PCE 830 test the current was 5.7A at an input power of 2.477 kW, almost an order of magnitude less. How could the difference be so large? Maybe with “average current” they mean the average of the peak current in the three phases? In that case the Joule heating calculation is completely wrong.

    Someone who can explain this?

    • Dr. Mike

      DickeFix,
      For your #1 the authors will have to claim that the heating coil resistance dropped by a factor of 3 for the first part of the active run. This is a good answer, but they need to tell us how this happened. They also need to tell us why the resistance of the coils did not change further when when the power was increased a the 10 day point in the active test. Maybe they will have a good theory for the resistance drop, and maybe not. (Of course, they may find that they made an error in the active power measurements.)
      For #2- If the current was 5.7A and the power 2,477W, the resistance was R= P/I^2 = 76.2 ohms. The resistance of each heating coil used in Lugano is only about 1.24 ohms. The TRIAC power controller reduces the RMS line voltage to the level required to supply the heater coils with the current required to produce a particular power.
      Dr. Mike

      • DickeFix

        Thank you for the reply. I understand that the regulator will regulate the voltage and current by changing phase and duty cycle. However, the high current and low impedance imply that the controller works close to its regulating limits.

        The maximum power would be at full duty cycle Pmax=3*Urms^2/R=3*400^2/1.25 =384kW. Of course it is just a theoretical value since the system is not designed for these kind of powers. However, this implies that the regulators working point in the E-cat (around 1kW input power) is less than 1% of its maximum. It makes me question if the low load is really within the specification of the regulator and the power meter.

        Moreover if RMS current is 40A and duty cycle small, the peak current must be several times higher. At these exteme currents I would also expect quite a large voltage drop in the power outlet.

  • Thomas Clarke

    Let us hope they mean RMS – but it is likely – this is what the PCE-830 will give.

    The 40-50A is not a clearly measured figure but you are right – it is consistent with the current they use for the Joule wire heating figures, which is mesured precisely, and shows COP 1.

    The waveforms they use are triac switched with a low duty cycle. That means the RMS voltage is less then the mains voltage by a factor of very roughly the duty cycle. That accounts for the difference.

  • Thomas Clarke

    Let us hope they mean RMS – but it is likely – this is what the PCE-830 will give.

    The 40-50A is not a clearly measured figure but you are right – it is consistent with the current they use for the Joule wire heating figures, which is mesured precisely, and shows COP 1.

    The waveforms they use are triac switched with a low duty cycle. That means the RMS voltage is less then the mains voltage by a factor of very roughly the duty cycle. That accounts for the difference.

  • Thomas Clarke

    @alain,

    I’ll highlight where we disagree as a set of comments on your arguments

    >First the dummy was tested normally, and all was right.
    >the COP was about1 and thus emissivity estimated at 0.7 was right
    >the two powermeter were well wired and the two clamps were not inverted.

    Correct.

    > for the second run, with powder, one clamp on the front power meter, and one in
    > the middle powermeter were inverted.[/quote]

    We have no evidence that the testers used two power meters for the active test. If they did, and cross-checked data, then yes, two clamps were reversed.

    > this led to an apparent COP of 3, because 2.7kWwas looking as 900W
    > the problem is that there is still 33% of error, 33% more heat than measured
    > note also that with time COP increase.

    That is not true. In the 1250C test – the first part of the active data – the COP – once the Joule heating is corrected – comes out at almost exactly 3. For the second part (1400C) the COP comes out as 20% higher than 3. The discrepancy is well within experiment berrors, mainly the assumption about Al2O3 emissivity. That is highly variable with temperature, frequency, and Al2O3 microstructure. The researchers use a single value from a reference and ignore transparency. A 20% error in this calculation is about what you would expect.

    > note also that from 800W to 900W the apparent power increase much more tha linearily…

    The error in the optical measurement relates to frequencies at which the Al2O3 is transparent. These increase more than linearly with temperature. So this is expected.

    > this mean that IR cam have wrong emissivity… but since it was 0.7, correctly
    > measured on the dummy at 450C, and since only twice the power is produced
    > if COP=1 , to show a similar effect as 1400C the emissivity have to be very low.

    I don’t understand this argment. We have a 20% error in power which would surely correspond to a 20% change in effective luminosity? So the real luminosity would need to be 0.4*1.2=0.48.

    > The lack of calibration at high temperature let doubt on such an effect
    > but not to a point that COP=1 is possible…

    The numbers above don’t seem to bear this out

    > strangely the box behave as if it was consuming energy as written on the box…

    I’m not sure in what way this is relevant?

    > strangely the reactor don’t behave like a triphase load but as a
    > single phase load driven by a triphase dimmer.

    I must disagree with that. The only difference between the two cases is the delta configuration of resistors. those are clearly present from the wiring, and measured by the testers. But in any case I do not see the relevance.

    > note also that none of the testers detect the problem on any of
    > the 2 powermeter, and that Industrial Heat have provided a reactor
    > that is broken, hoping for such a mistake…

    I note the lack of cross-checking from the testers. Clearly they did not do this or they would have noticed the Joule heating/input power discrepancy.

    Presumably IH tests this reactor in the same manner and observes the same apparent COP?

    > this is why I call that a conspiracy theory.

    You need Rossi involved and no-one else, if they believe him. He is CTO IH. Who would contradict his reading of power meters – it is not obvious?

    > anyway the test is incomplete, not on the electric side, but on the
    > calorimetry side by lack of high temp calibration, ruling out very low
    > emissivity at top temperature. depending on litterature to estimate
    > alumina emissivity is not enough.

    That is true, and introduces an unknown extra error. The X3 issue is different, it is a known error.

    • Dr. Mike

      fact police,
      Thanks for the correction- these calculations are outside my field of knowledge. What mechanisms did you assume for the conversion of the Ni isotopes to Ni62 Would we need a solid theory to really know the mechanisms? Finally, using your mechanisms, how much total energy should the reactor have produced form converting all 0.55 grams of Ni to Ni62? Note: even though the ash sample size was really small, the conversion of the Ni to Ni62 was confirmed by two measurement techniques.
      Dr. Mike

    • on the transparency the affair is closed since the alumina is opaque at the considered wavelength

      there was 2 powermeter

      anyway you assume that industrial heat sent a broken reactor hoping such a stupid error would be done, while most of the time they were not present.

      forget it.

      at worst there is a minor error in the report about some losses estimations, and sure an error in not calibrating at high temperature.
      Anyway the most probable is that critics are based on error and false assumptions.

  • Thomas Clarke

    @alain,

    I’ll highlight where we disagree as a set of comments on your arguments

    >First the dummy was tested normally, and all was right.
    >the COP was about1 and thus emissivity estimated at 0.7 was right
    >the two powermeter were well wired and the two clamps were not inverted.

    Correct.

    > for the second run, with powder, one clamp on the front power meter, and one in
    > the middle powermeter were inverted.[/quote]

    We have no evidence that the testers used two power meters for the active test. If they did, and cross-checked data, then yes, two clamps were reversed.

    > this led to an apparent COP of 3, because 2.7kWwas looking as 900W
    > the problem is that there is still 33% of error, 33% more heat than measured
    > note also that with time COP increase.

    That is not true. In the 1250C test – the first part of the active data – the COP – once the Joule heating is corrected – comes out at almost exactly 3. For the second part (1400C) the COP comes out as 20% higher than 3. The discrepancy is well within experiment errors, mainly the assumption about Al2O3 emissivity. That is highly variable with temperature, frequency, and Al2O3 microstructure. The researchers use a single value from a reference and ignore transparency. A 20% error in this calculation is about what you would expect.

    > note also that from 800W to 900W the apparent power increase much more tha linearily…

    The main error in the optical measurement relates to frequencies at which the Al2O3 is transparent. Power at these increases more than linearly with temperature. So this is expected.

    > this mean that IR cam have wrong emissivity… but since it was 0.7, correctly
    > measured on the dummy at 450C, and since only twice the power is produced
    > if COP=1 , to show a similar effect as 1400C the emissivity have to be very low.

    I don’t understand this argument. We have a 20% error in power which would surely correspond to a 20% change in effective luminosity? So the real luminosity would need to be 0.4*1.2=0.48.

    > The lack of calibration at high temperature let doubt on such an effect
    > but not to a point that COP=1 is possible…

    The numbers above don’t seem to bear this out

    > strangely the box behave as if it was consuming energy as written on the box…

    I’m not sure in what way this is relevant?

    > strangely the reactor don’t behave like a triphase load but as a
    > single phase load driven by a triphase dimmer.

    I must disagree with that. The only difference between the two cases is the delta configuration of resistors. Those are clearly present from the wiring, and measured by the testers. But in any case I do not see the relevance.

    > note also that none of the testers detect the problem on any of
    > the 2 powermeter, and that Industrial Heat have provided a reactor
    > that is broken, hoping for such a mistake…

    I note the lack of cross-checking from the testers. Clearly they did not do this or they would have noticed the Joule heating/input power discrepancy.

    Presumably IH tests this reactor in the same manner and observes the same apparent COP?

    > this is why I call that a conspiracy theory.

    You need Rossi involved and no-one else, if they believe him. He is CTO IH. Who would contradict his reading of power meters – it is not obvious? I should add that some people here seem pretty convinced Rossi is telling the truth, so why should you expect others to be less convinced?

    > anyway the test is incomplete, not on the electric side, but on the
    > calorimetry side by lack of high temp calibration, ruling out very low
    > emissivity at top temperature. depending on litterature to estimate
    > alumina emissivity is not enough.

    That is true, and introduces an unknown extra error. The X3 issue is different, it is a known error.

    • on the transparency the affair is closed since the alumina is opaque at the considered wavelength

      there was 2 powermeter

      anyway you assume that industrial heat sent a broken reactor hoping such a stupid error would be done, while most of the time they were not present.

      forget it.

      at worst there is a minor error in the report about some losses estimations, and sure an error in not calibrating at high temperature.
      Anyway the most probable is that critics are based on error and false assumptions.

  • Dr. Mike

    DickeFix,
    For your #1 the authors will have to claim that the heating coil resistance dropped by a factor of 3 for the first part of the active run. This is a good answer, but they need to tell us how this happened. They also need to tell us why the resistance of the coils did not change further when when the power was increased a the 10 day point in the active test. Maybe they will have a good theory for the resistance drop, and maybe not. (Of course, they may find that they made an error in the active power measurements.)
    For #2- If the current was 5.7A and the power 2,477W, the resistance was R= P/I^2 = 76.2 ohms. The resistance of each heating coil used in Lugano is only about 1.24 ohms. The TRIAC power controller reduces the RMS line voltage to the level required to supply the heater coils with the current required to produce a particular power.
    Dr. Mike