This report extends the insights described in Report #5 and shows that several common conclusions about LENR are wrong. These errors have handicapped efforts to achieve reproducibility and have lead several theories in the wrong direction.

PROGRESS-REPORT-6 Additional behavior of pure PdD (1.3Mb)

Temperature plays a significant role in affecting the amount of power produced by LENR. The activation energy for power production is very similar to the activation energy for diffusion of D in PdD. This behavior is consistent with my theory in which temperature is described as helping D reach the NAE by diffusion through the surrounding lattice.

Comments are welcome.

Figure 9 from Progress Report #6 showing surface of the Pd cathode after the study.
Figure 9 from Progress Report #6 showing surface of the Pd cathode after the study.
Figure 10 from Progress Report #6 shows surface of Pd before the study.
Figure 10 from Progress Report #6 shows surface of Pd before the study.

Read more from PROGRESS-REPORT-6 (1.3Mb)

See also:

Progress Report #6

Progress Report #5

Progress Report #4

Progress Report #3

Progress Report #2

Progress Report #1

22 thoughts on “Progress Report #6

  1. Dave, for some reason, we are not understanding each other and describing Nature in the same way. You seem to have difficulty understanding what I describe and I have the same problem with your descriptions.

    I find that temperature is a variable that can change the rate of heat production. The rate is charged because temperature allows the D to reach the site of the nuclear reaction more rapidly. The temperature is determined only by how fast heat is added to the system compared to how fast it can leave. In my case, I make no effort to slow the rate energy leaves. Nevertheless, the amount of energy needed to achieve the temperature is arbitrary and depends only on cell design.

    I hope this explanation is clear.

    1. Ed, I think I realize why we seem to be on different pages. You are mainly interested in characterization of the metal containing the deuterium whereas I have been mainly considering the overall system.

      It is important to understand the behavior of the materials as you are so well determining. That information will need to be incorporated into a complete product some day, which is where I have been concentrating my efforts.

      Both of these paths are vitally important and necessary. Please keep me informed about your progress. I hope that you will find time to answer any future question which may arise from my research pertaining to your work. Thanks!

  2. Dave, you are attempting to use the data to create an isoperibolic calorimeter. This can be done and I have data making such an approach is possible. However, this method is filled with error because temperature gradients exist in the system. Therefore, such an approach is a waste of time unless I can create much more excess power- enough to overwhelm the errors. The Seebeck method does not rely on the temperature in the cell or in the calorimeter. Therefore, it is accurate and is the preferred method.

    1. Ed, the graph that I am proposing assumes that adequate control of the temperature throughout the boundary is maintained. But, the temperature is acting as a proxy for power output from the system. If an accurate measurement of total output power exists then that information substitutes equally well.

      I was viewing your figure 6 and it appears to demonstrate an accurate conversion of your defined temperature variable to your definition of excess power. Your curve does not seem to be particularly noisy and temperature has been the variable that I have used to model devices such as Rossi’s ECAT. I have no problem changing variables if that makes your task easier.

      Is it possible that we are considering different system boundaries? I include everything that draws power, radiates, conducts, or convects energy from the outside world as part of the LENR device. Does that match your boundary? Are you including the power required to maintain the device at 63 C as part of the input power when you calculate excess power? I would normally include that component.

      Of course, life after death would not begin until the source of heat keeping the cell at 63 C were cut off according to my boundary choice. Are you aware of any standard that defines the proper system boundary to use when testing LENR devices?

  3. Yes, a heater in the cell is used to hold the temperature at 63°.

    Yes, the temperature and excess power dropped when the heater was turned off.

    The cut off time of 1000 minutes is arbitrary. I expect the cell would have behaved the same way regardless of when the temperature was caused to drop.

    The amount of D was determined by weighing the sample after the study to determine the weight of D contained therein. A better measurement after the Report was written places the D/Pd near 0.65.

    1. Thanks Ed, now I have a much clearer picture of your system. I have developed an interesting graph that I would like very much to use with your inputs. The graph is fairly simple to produce yet reveals a great deal about how an LENR device behaves.

      I could attempt to take your data and generate one of these but it would be much better for you to ensure that the proper data points are graphed. Is it possible for us to work together on this issue? You will find it quite simple to construct since you appear to have all of the information in one form or the other.

      My graph is constructed using temperature as the independent value lying along the X-axis. For the Y-axis I plot the total input power. By total input power I refer to all forms of input including DC power used in the electrolysis as well as any heating power. This could also be determined by taking the total power as measured by your calorimeter and subtracting away the excess power due to LENR activity at each temperature point.

      Your figure 6 appears to have most of the data required to derive the Y-axis information provided the input powers are associated with each point on it.

      When you compare two curves of my graphs, you can immediately see how the device will perform. The first curve needs to display a calibration plot for the cell where LENR does not take place. The second curve will then fall below the first one and its shape will indicate whether or not a negative resistance region exists for the LENR device.

      Even in the absence of a negative resistance region, the presence of LENR is apparent with this type of visual indication.

      I believe that your cell is what I refer to as a type 1 system that does not have sufficient positive feedback to generate a negative resistance band. That means it will always be stable and that you can set the total input power to a single level to reach any desired output power level within its operational range. This graph will clearly show that status if I am correct.

      Can you take the time to either generate these graphs with your data or just send me a set of data points in (X,Y) fashion so that I can produce it? I am confident that this information will further prove that your system is generating LENR as we expect.

  4. I am having difficulty understanding a few issues regarding your figure 7. After carefully reading the text several times I am left with the understanding that you are using an input heating source, which is independent of the electrolysis circuit, to keep the cell at a nearly constant temperature(63 C). Is this true? Has every source of input heating energy been removed during the slow, long thermal decay process?

    Then, do I understand correctly that the very rapid drop in temperature and internally generated thermal power occur when that extra heating device is turned off after around 1000 minutes?

    If so, would you expect the same rapid falloff of temperature to be present if, for instance, all of the external heating device were to be cut off at 500 minutes? Where in time would the rapid drop be observed?

    My thermal model would have predicted a rapid falloff at the time all types of heating is removed in the absence of some unknown time constant which I was referring to within my last posting.

    Could you clarify how you determine the amount of deuterium remaining within the sample at the conclusion of the experiment from figure 7? Is this measured or calculated?


  5. Dave, the excess power continues as long as the temperature remains high. Once the temperature is allowed to drop, the amount of power decreases. The effect of temperature is shown in Fig. 6. The amount of fuel has not been exhausted at the end of Fig.7. The amount of power was simply responding the the falling temperature.

  6. I was just reviewing an impressive chart showing the continuation of heat production after the input power is removed. Figure 7 additionally shows that you are observing excess output power that is greater than the amount of input required to produce it.

    This is very important from my perspective, especially when you demonstrate that the output continues to proceed for over 1000 minutes with zero input. Have you had the opportunity to calculate the amount of energy that would have been produced if most of the trapped hydrogen reacted?

    I suspect that only a fraction of the potential energy is actually produced before the reaction ceases. Your NAE most likely contain a tiny proportion of the total absorbed hydrogen.

    My general thermal model predicts that the temperature will begin to fall at the moment when the input power is eliminated as a result of positive feedback activity. It is also possible under certain other conditions for the temperature to begin to rise provided sufficient positive feedback is present. Fortunately, your system is restrained to cool down.

    The temperature slope appears to be influenced by the thermal mass of the overall system, or perhaps exhibits a time constant that is related to the diffusion of deuterium out of the NAE structures. Have you reached any final conclusions regarding these shapes in time?

    I also find the abrupt falloff in excess power near the end of figure 7 quite interesting. If you determine that the fuel is exhausted at that point in time due to fusion within the NAE having run it course then the answer is fairly straight forward. Can you verify that your measurement system is capable of responding rapidly enough to follow that falloff rate and display a reasonably accurate indication of its shape?

    I want to thank you again for offering us access to your excellent work.

  7. Thanks for the excellent report Dr. Storms. Perhaps my work with thermal models will help guide system engineering one day.

    In your 5th report you say that the internal pressure due to the absorbed hydrogen is not comparable to the extreme value calculated from your external data source. I have always assumed that the RMS velocity of the gas atoms more or less directly relates to the pressure of that gas due to collisions with the container. As more and more gas atoms are packed into the same volume, it makes sense that the pressure must go up at a fixed temperature.

    Could the hydrogen trapped within your relatively large NAE actually exhibit the equivalent enormous pressure since they exist at the temperature of the metal? This assumes that several atoms are trapped within one volume, which is required if they are to react together.

    1. The issue is how pressure is defined. Pressure in this case only applies to a gas. If gas is trapped in a void, then it will exhibit a pressure described by PV=nRT. If the atom is instead dissolved in the Pd, its condition is described as activity, which is given a value of pressure only for convenience. MY NAE does not contain gas. The atoms of D or H are present as a covalent linear compound. Therefore, the concept of pressure does not apply in my case. Nevertheless, a high activity is required to force the D into the PdD structure to achieve a high D/Pd ratio.

    2. What is the cost barrier to gargae LENR experiments? I did some wikipedia-level research ,and it looks like the components are all available on an open market. A dewar and palladium? Palladium is $700 an ounce. Even with special active site treatment, that can’t put it outside the price range of a dedicated gargae scientist. Can’t hydrogen be chemically treated to produce deuterium? So, why can’t anybody just conjure up their own LENR experiment in their backyard until it works? That’s one way to get around a scientific community who refuses to peer review research papers on the basis of one media snafu 40 years ago.

  8. “A well insulated cell can be made self-sustaining but the system would be very unstable and prone to run away”

    Possibly not: If the palladium sample is thin enough, there could/should be a natural negative feedback:

    As the D2O begins to boil, the different bubble formation regimes increase the heat flux away from the palladium surface, which would have the effect of cooling it.

    Exceeding the ‘critical heat flux’ would be a bad thing though, but that would be limited by the samples thickness…

    1. * Assuming excess heat isn’t only a surface based effect.

      My hunch is that the grain boundaries play a role.

  9. The only issue from an engineering view point is how much NAE can be created in the material. Once the NAE is created, energy can be extracted simply by putting a little D in the lattice using gas loading or electrolysis and then heating the sample under the required equilibrium pressure. When enough insulation is used, the sample will self-heat.

    The same requirements have been found to apply to Ni+H2. In other words, gas loading and electrolysis will produce identical behavior and this same behavior applies to the Ni-H2 system.

    Once temperature is accepted as being the important variable, rather than the electrolytic current, all of the behavior shown by Pd-D, regardless of the method used, and Ni-H start to make sense. Both systems are able to make energy with a high out/in ratio.

  10. Dr. Storms,

    During the excess heat production phase, you reduce the current to 0.1A, in order to keep the D inside the Pd.

    I am interested to know what voltage is necessary to create this amount of current.

    Best Regards.

      1. Huh! So would it be correct to say that (during the excess heat phase) you are introducing a maximum of 0.35W into the cell, and seeing *excess* heat of 0.55W (at 70C)?

        This is based on 3.5V times 0.1A, and figure 6 in report 6.

        If so, that is some Pd worth holding on to!

        Also, in report 5 you say ” Instead, small changes in temperature determine how much power can be produced by the nuclear reaction, which is an amazing realization. Now theory must explain how this effect is possible.”

        To my mind*, the linear dependence on temperature you have seen means diffusion must be a key part of this… The Einstein diffusion relation and the Teorell formula both highlight this linear relationship of the diffusion coefficient to temperature.

        *Which frequently gets things wrong…

        1. The amount of electrolytic power is irrelevant. Only enough current needs to be applied to keep the sample from deloading. Use of 0.1 A in my case is completely arbitrary. On the other hand, the temperature controls the amount of excess power. The amount of power needed to achieve a useful temperature depends on how well insulated the cell is. A well insulated cell can be made self-sustaining but the system would be very unstable and prone to run away, as F-P discovered when the cube they were studying melted through the table.

          I agree, temperature affect the rate of diffusion, which controls the rate of power production.

          1. Yes, I agree with what you say. I was just surprised by the ratio of excess heat to the input. It’s a superbly clear signal.

            Maybe I look at this from slightly different angle. I understand the aims of your research, but as an engineer, I can’t help but drill down into the energy flow numbers, in order to imagine the feasibility, or the potential, of the system outside a laboratory.

            1. That is to say ‘input’, if you assume any other internal heaters in you apparatus were replaced by insulation..

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