Rossi Lugano/early demo's revisited. (technical)

    For example, assuming that inflated temperatures where used on the Optris, then you can calculate that setting back the emissivity to 1 should have resulted in a temperature of 342 degree C, not the 366.6 degree C which is mentioned in the report.


    Also it was shown in post #548 that iterating to the end temperature on the Optris does not result in the shown iteration temperatures presented in the report.


    Ah - but you are not remembering the key (well hidden) sentence in the report which says that for the lower temperature emissivity values they corrected the book values based on empirical results from TCs. And we do not know how this was done, or what their inferred values were, hence we cannot tell anything about the method from the dummy temperature values.

    Ah - but you are not remembering the key (well hidden) sentence in the report which says that for the lower temperature emissivity values they corrected the book values based on empirical results from TCs. And we do not know how this was done, or what their inferred values were, hence we cannot tell anything about the method from the dummy temperature values.


    For sure I am rembering the text in the report.

    This is the text in the report :


    It was not possible to extract any sample of the material constituting the rods, as this is firmer than that of the reactor. The rods were made of pure alumina, crystallized however with a different degree of fineness due to the industrial origin of their manufacture.

    We therefore took the same emissivity trend found in the literature as reference; but, by applying emissivity reference dots along the rods, we were able to adapt that curve to this specific type of alumina, by directly measuring local emissivity in places close to the reference dots (Figure 7).


    The curve correction was as you can read done for the rods, not for the reactor body !



    Nevertheless, the calculation for setting back of the emissivity to 1 and the resulting temperature of 342 degree C only depend on the n factor and the formula the Optris uses.

    Broadband emissivities are not used in that calculation, also not when they would have been adjusted.


    And the iteration, if that was indeed done on the Optris, would still have shown intermediate values different from what was published.


    In addition we see from the published temperatures during the iteration and the emissivities belonging to those temperatures that those emissivity values where not adjusted for the ribbed area.

    The Lugano investigation


    I have currently no new idea's about research I can do to investigate the contents of the Lugano report any further. (but am willing to respond to suggestions if it is within my capabilities)

    If new idea's arise I will pick up this thread again, howver am planning to devote my time now to learn more about the theoretical aspects of LENR.

    (Also winter is nearing, so soon I can not paint the house outside anymore)

    When studying I learned quantum mechanics but can't remember anything about it anymore, no problem because the old quantum mechanics is depricated anyway.

    So I have to start from scratch again and would welcome any suggestions for a good introductory in atom theory. (really from the beginning)


    Thank you all for the discussions, comments.


    LDM

    Dear LDM

    if Lugano reactor's was made as i explained by a poster at ICCF with double compartment , what could be implication with heat released you measured ?


    Quote from Alan Smith

    LDM


    Thank you so much for all your painstaking investigative work on the Lugano experiment. Since it is scattered over a long time period, would you be able to write a couple of paragraphs describing your main conclusions. That would be useful to many members I'm sure.


    You have a PM in 'conversations' btw.


    I will take into consideration your suggestion to make a document summerizing the conlusions of the different posts if I think it can be done in a meaningfull way.

    So I want first to make an outline of such a document and then decide if it will suit its pupose.

    However I have currently many other activities (Indeed, also doing paintwork on the house) which prevent me to make a start soon.

    So if I decide to make the document, it for sure will take a long time (month's) to be completed.

    Quote from Cydonia

    Dear LDM

    if Lugano reactor's was made as i explained by a poster at ICCF with double compartment , what could be implication with heat released you measured ?

    Hello David,


    As far as Lugano is concerned I have a problem with the principle which you outlined on your poster.

    The problem is that when I calculate the total weight of the Lugano ECAT from the FEM model, I get alsmost exactly the weigth reported.

    This means that it is not possible to add a stainless steel tube because it would increase the weight too much.

    However if there where some cavities in the end caps then that might possibly compensate for the adition of a (thin) metal tube.

    This brings up another question : Does it need to be a tube or may it also be a very thin steel layer on the inside of the ceramic tube ?


    As far as the heat release is concerned, the outside temperatures are largely determined by the emissivity and convective heat transfer coefficient. This results in a dissipated power which must be equal to the generated power inside the reactor.

    So internal power with the emissivities and convective heat transfer coefficients determine the outside temperatures.


    Internal temperatures are determined by internal heat generated and the outside wall temperature.

    The internal heat need to pass the internal thermal resistance determined by the thermal conductivities of the internal materials used.

    The lower the thermal conductivities (Larger thermal resistance) the more drive needs to be applied to get the power out. This drive is due to the internal temperatures and lower thermal conductivities means higher internal temperatures to get the power out.


    So adding an internal metal tube does not change the external tempertures much.

    There is an exception, stainless steel has a relative high thermal conductivity at higher temperatures. This means that heating power can be tranfered by the tube to places with lower temperatures.

    This means that the temperature distribution both at the inside and outside of the reactor can change.

    As a result some places will get a higher temperature, other places a lower outside temperature. However the total dissipated power stays the same.


    If you want me to do a FEM simulation for a specific situation you have in mind then I need a cross section drawing with dimensions, materials used, applied heater power and possibly internal generated power not related to the heating element and the supposed location and distribution of that power source. (Also for non Lugano type reactors)

    I can then make a CAD drawing of it, create a FEM model and do a thermal simulation.

    It is quite some work so only if it really helps you with answering some questions you have you can ask me.

    Dear LDM thanks for your interest.

    I really appreciate your long term work so seriousness both in a difficult context.

    You are right, AR has used a SS tube as H+ membrane maker both with empty external compartments.

    Lugano's device remains only one more release from patent explanation i shared by an ICCF paper : US 2014/0326711 A1

    When you will have time available again, you can test your understanding by more tricks i will share with you but only by private msg..

    Indeed, none give money but all scrutinize.

    You will have to leave me your true identity too, you can easily understand why :)

    @LDM


    Success in getting LENR reactor to work like Lugano is to generate LENR fuel. I wonder if you are willing to try out Ohmasa gas as a LENR fuel. Ohmasa gas has been shown to carry lots if EVOs and it is a profific generator of transmutation.


    Ohmasa gas is produced by doing electrolysis of water while cavitation is ongoing. The hydrogen component of the gas is what you can load into the ceramic tube. Heat or RF can serve as a stimulant.


    If you are interested in this approach, we can discuss more details.

    Quote from Shane D.

    When Ahlfors posts, I pay attention. Not much for words, but he always has something important he is trying to get across. I found this interesting.


    The theory is interesting, the Lugano reference not so interesting.


    It is referenced as [1-3] where the others are Lipinski and Mills!!!!


    Anyone looking at theory positing LENR will quote as possible examples published claims of LENR. However it is a shame because the Lugano papers were never peer reviewed and the second one has been thoroughly demolished (wrong calculations). So somone doing a thorough review of the LENR experimental stuff would not quote either of them as evidence. But, to be fair, it is common for theoreticians to take any claims of LENR as "something to explain" without examining them much, and fair enough they do this since the theory is interesting of itself.


    I always pay attention to Ahlfors posted papers (and often read them) but they vary a lot. Sometimes I think he/she is just taking the mickey.

    Success in getting a reactor to work like Lugano is in conflating total emissivity with the IR camera in-band emissivity and using the wrong value for the IR setting. This has been demonstrated by myself and the MFMP quite thoroughly with physical demonstrations that have both matched and exceeded the ‘COP’ performance reported in Lugano.

    As I wrote earlier, if new idea's arise I would pick up this thread again.



    For several reasons I did not find any reason to do new Lugano based calculations nor find the time to continue studying atom theory as I intended.

    But in between I did a new investigative calculation on the second MFMP thermal dogbone test.

    The calculation is based on the DB_test3_TC test.

    Note that this calculation is indicative, not exact due to the unknown heat distribution in the MFMP dogbone for the used power of 300 Watt.


    see : https://drive.google.com/drive…xJkjesxe4kXzczVnlfajhjbDQ


    We start with a calculation on the heater coil of the MFMP dogbone.

    Specifications are :


    Diameter 11.1 mm

    Windingen under ribs : 76.5

    From the coil specificatione we derive a total wire length under the ribs of 265.05 cm

    Total wire length in the dogbone : 273.05 cm (two times 4 cm in the end caps added)


    This means that 265.05/273.05 = 0.971 of total coil power is dissipated under the ribs

    (Did not take into account the heater wire length outside the dogbone since it is not known)


    From the DB_test3 we use for our calculation the last value for the 300 watt applied power.

    (sample number 147)

    The important data :


    Optris temperature-----------------505.4 degree C

    Applied power------------------------300.24 Watt


    The power under the ribs is now calcualted as 0.971 x 300.24 = 291.5 Watt

    We have 10 measurement sections, thus the power per rib section = 29.15 watt


    For the temperature of 505.4 degree C with an ambient temperature of 17.5 degree C the calculated convective heat transfer coefficient is 13.49

    Due to the ribs being close which reduces the convective heat transfer we have to correct this with a factor 0.735 which brings the value to 0.735 x 13.49 = 9.92

    (The value of 0.735 was interpolated from two CFD simulations)


    Area of a section is 0.00263 m^2.

    Thus the disspated convective power is 9.92 x 0.00263 x (505.4 - 17.5) = 12.73 Watt


    Radiated power 29.15 -12.73 = 16.42 Watt (Discarding the limited lateral power dissipation)


    We now us the following formula to calculate the emissivity

    See for an explanation of the formula :


    Rossi Lugano/early demo's revisited. (technical)


    E = σ x Af x e x Fbg x { 1/(1 - (1-Fbg)(1-e))} x T^4


    with---E--------Radiated energy (11.84 Watt)

    ---------σ--------Stefan-Boltzmann constant (5.67·10-8 W/m2K4)

    ---------Af-------Fin area of a section (0.00263 m^2)

    ---------e--------Emissivity

    ---------Fbg-----View factor of ribs to background ( 0.572 )

    ---------T--------Temperature (Kelvin) ( 505.40 + 273.15 = 778.55 K )


    Before we apply the formula we have to correct the Optris temperature reading due to the increase in apparent emissivity caused by the reflection between the ribs.

    The apparent emissivity is

    e' = e /(1 - (1-Fbg)(1-e))


    For the used Optris in band emissivity setting of 0.95 the apparent emssivity becomes 0.97.

    The relationship between temperature and n value as used in the formula given by Optris is given in the following post :


    Rossi Lugano/early demo's revisited. (technical)


    Applying this to the formula given by Optris we find that the real temperature was 499 degree C (772.15 K) with a value of n = 2.311.


    Filling in the found values in the above equation for the radiated energy we find for the emissivity a value of 0.41.

    This calculated indicative value is much lower then the broad band emissivity value of Alumina for the temperature of 499 degree C, the value being 0.65.

    This becomes clear if we plot this value in the graph of the broad band emissivity as a function of temperature for Alumina.

    This is shown in the following figure :





    The much lower calculated emissivity then that of Alumina indicates that the MFMP dogbone was not casted of pure (> 98 %) Alumina, but instead must have contained other components or being based on an other type ceramic.

    The question is then if the MFMP as a part of their test did check if their Dogbone was casted conform their material specification.


    Note :


    As a verification I did a FEM simulation with broad band emissivities set at a value of 0.41 at the middle of the dogbone and got simulated center temperatures of 497 degree C, close to the 499 C calculated from the Optris. It confirms that to get a temperature of about 499 C the emissivity has to be much lower then that of Alumina.

    This post tries to come up with some theoretical background on the ECAT plasma reactors.

    It assumes that the plasmas in these reactors where thick plasmas.

    We have however no proof that this was the case.

    Nevertheless ...


    Plasma :


    There are basically two types of plasmas, thin Plasma and thick Plasma.


    In a thin Plasma the density of the Nuclei in the plasma is such that radiation inside the plasma can largely be radiated out of the plasma without being blocked by the nuclei.

    This results in that the spectral discharge lines are separately visible.


    In a thick plasma the density of the nuclei is such that most of the inner radiation is reabsorbed leading to an equilibrian of inner radiation and the radiation leaving the plasma is largely from the surface of the plasma.

    This leads to a special shape of the intensity as a function of the frequency profile.

    Such a profile is shown in the following figure.



    -------------------------------------------------------Figure 1


    The curve is largely flat.

    At the higher thermal cutoff frequncy the intensity drops of very fast.

    At the lower side the intensity drops off less fast.

    Note that spectral lines with frequencies higher then the upper cutt-off frequency will still be shown.


    Plasma temperature

    Plasma temperature is due to the mobility of the electrons in the plasma.

    Plasma temperature can be expressed in degree Kelvin (K) or in electron Volt (EV)

    relationship between both is 1 eV equals 11605 degree Kelvin.


    For the higher frequency cutt off as shown in the above figure the relationship between the cut off frequency and the plasma temperature is given by the following formula


    ν = 2.8E10 x T----------------------------1)


    This makes it possible to calculate the plasma temperature based on a measurement of the radiation as a function of the frequency using the upper cutt off frequency.


    If the radiation of a thick plasma is uniform from the surface then it can be theoretical shown that the total radiation can be calculated with the following formula.


    Q = A σ T4---------------------------------2)


    See "Plasma Physics and Engineering" by Alexander Fridman and Lawrence A.Kennedy

    Chapter 6.7.9


    Now wait a moment !

    Formula 2 is the same formula as for calculating the radiation of a black body !

    I wonder if nature wanted to play a trick on us because this can lead to some people believe that a thick plasma is a black body.

    However it isn't , as can be seen the shape of the spectrum of a thick plasma as shown in figure 1 is completely different from that of the radiated spectrum of a black body as given by Planck's law.

    Only the formula for calculating the radiated energy is the same.


    If you based on formula 2 wrongfully consider the thick plasma to be a black body then you are possibly also intended to calculate the peak wave length of the black body radiation using Wien's displacement law.

    However this calculated peak wavelength does not exist for the plasma, it is virtual and can not be measured or shown.

    It can only be used as a comparason value if you want to compare the thick plasma with a black body.

    And reporting such a virtual wavelength can lead to confusion when people do not see this peak frequency in the reported spectrum.

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