• GreenWin

    HAIL to Chief Slad!

  • Axil Axil

    Heat is light, infrared light. LENR needs light as a stimulant. Heat is converted to magnetic energy. It is magnetic energy that produces muons and mesons. I suspect that muon will be found coming out of all LENR reactions, even very low powered reactions.

    The assumptions behind this article and what Ed Storms is says are wrong. Its light that is important in LENR.

    • Slad

      Axil. Where’s this infra-red coming from? Storms gets a straight line when he plots power out vs. temperature. Wouldn’t the infra-red radiation increase according to Temperature^4? Hence, why does excess heat not increase in a similar fashion in Storms’ experiment?

      • Axil Axil

        Storms in not producing nano particles(Hydrogen Rydberg matter). Hydrogen Rydberg matter is produced by using a lithium or potassium dispenser. For example, Rossi uses large amounts of lithium,

        In the Lugano test at one point when the temperature was increased to the max level, the COP on that temperature increase was 7.

        • Slad

          According to Holmlid (of muon fame), his experiments are more akin to hot fusion than cold?

    • Obvious

      Phonons in and photons out , the reaction is supraluminous, and requires constant heat in or it will freeze. Well, micro particles that just reacted freeze anyways, and steal heat from the rest of the particles, causing the device to drop below its operating range, killing further reactions. A device with molten metals is not functioning, and is most certainly dead.
      IMO

      • Axil Axil

        When the reaction is subcritical, the SPPs require constant pumping. But when the positive feedback loop that transfers energy into the SPP produces more heat than is required for pumping, the reaction become supercritical and the reactor melts down. The ideal case is when the energy required for pumping plus the energy needed to boil water is equal to the energy produced by the positive feedback loop.

        • Obvious

          Insufficient particle size distribution causes meltdown. Since the output reaction is a collective effect, only an identical particle can directly resorb the entire energy “packet” emitted by another reacted particle. When an identical stationary, zero Kelvin particle accepts a full packet, it causes an explosion of the particle (rare), initiates a fusion event, or immediate re-emission. When a hot identical particle accepts a full packet it melts immediately, and melts its neighbours subsequently. Too many identical particles thereby cause a resonant sequence which is fatal to the device. This is why nano particles are bad for the reactor. The odds of having a large number of functionally identical particles is greatly increased with decreasing particle size.
          IMO

          • Axil Axil

            You are correct. The reaction will destroy nano particles…hydrogen particles. That is why they must be constantly recreated dynamically to replace the ones that are destroyed. Fixed nano particles won’t work because of the destruction issue you raise.

  • Axil Axil

    Heat is light, infrared light. LENR needs light as a stimulant. Heat is converted to magnetic energy. It is magnetic energy that produces muons and mesons. I suspect that muon will be found coming out of all LENR reactions, even very low powered reactions.

    The assumptions behind this article and what Ed Storms is says are wrong. Its light that is important in LENR.

    • Slad

      Axil. Where’s this infra-red coming from? Storms gets a straight line when he plots power out vs. temperature. Wouldn’t the infra-red radiation increase according to Temperature^4? Hence, why does excess heat not increase in a similar fashion in Storms’ experiment?

      • Axil Axil

        Storms in not producing nano particles(Hydrogen Rydberg matter). Hydrogen Rydberg matter is produced by using a lithium or potassium dispenser. For example, Rossi uses large amounts of lithium,

        In the Lugano test at one point when the temperature was increased to the max level, the COP on that temperature increase was 7.

        • Slad

          According to Holmlid (of muon fame), his experiments are more akin to hot fusion than cold?

    • Obvious

      Phonons in and photons out: the reaction is supraradiant and requires constant heat in or it will freeze. Well, micro particles that just reacted freeze anyways ( 0 K ), and steal heat from the rest of the particles, causing the device to drop below its operating range, killing further reactions. A device with molten metals is not functioning, and is most certainly dead.
      IMO

      • Axil Axil

        When the reaction is subcritical, the SPPs require constant pumping. But when the positive feedback loop that transfers energy into the SPP produces more heat than is required for pumping, the reaction become supercritical and the reactor melts down. The ideal case is when the energy required for pumping plus the energy needed to boil water is equal to the energy produced by the positive feedback loop.

        • Obvious

          Insufficient particle size distribution causes meltdown. Since the output reaction is a collective effect, only an identical particle can directly resorb the entire energy “packet” emitted by another reacted particle. When an identical stationary, zero Kelvin particle accepts a full packet, it causes an explosion of the particle (rare), initiates a fusion event, or immediate re-emission. When a hot identical particle accepts a full packet it melts immediately, and melts its neighbours subsequently. Too many identical particles thereby cause a resonant sequence which is fatal to the device. This is why nano particles are bad for the reactor. The odds of having a large number of functionally identical particles is greatly increased with decreasing particle size.
          IMO

          • Axil Axil

            Yes, wide size distribution is critical for magnetic power amplification reasons. It is not the nanoparticle that absorbs the energy, it is the SPP that the nanoparticle creates that absorbs the energy. The particle that is destroyed is the one hit by the SPP magnetic beam. You are correct. The reaction will destroy nano particles…hydrogen particles. That is why they must be constantly recreated dynamically to replace the ones that are destroyed. Fixed nano particles won’t work because of the destruction issue you raise.

  • bachcole

    What did he say, in one or two sentences?

    • GreenWin

      He says Yugo and Josh Cude are deluded whackos.

      • bachcole

        Nice. Only one sentence. Both links resulted for me in blank screens. And the embedded version was very difficult to read. Thanks.

      • Slad

        …A fine summation.

        • builditnow

          The MaryYugo team are not deluded whackos, they are deliberately and strategically spreading doubts. The MaryYugo team knows the reality of LENR is real and LENR+ is very very likely. Just google all the blog entries and judge for yourself if you really think they believe their “delusions”.

          Why is the MaryYugo team so persistent. Well, there is something in it for them. One could speculate.
          One such speculation is that they have invested in Rossi and now want to give Rossi the best possible lead in the LENR race.
          No matter what the MaryYugo team’s actual purpose is, no doubt they are now helping Rossi, big time.
          Rossi is going to be much richer due to the MaryYugo team’s help in keeping serious competitive investments in LENR lower.
          My advice to the MaryYugo team …. double down and Rossi might even acknowledge your work by naming one of his reactors the MY-E-Cat in your honor.

  • Jarea

    I find that info new about the Optris.

    In the document you can find:

    DON’T YOU KNOW THAT THE CoP WAS = 1.07?
    ‘here are two types of Optris PI-160 camera. The standard version, and a manufacturer modified, high temperature version. The Lugano team used the latter, however the authour of the CoP = 1.07 paper based his arguments on the datasheet for the former.

    • Slad

      I prefer the Bob Higgins COP figure based on his emissivity recalculation (2.36), over the Lugano figures, however it’s certainly greater than 1.07

      • Obvious

        Although several people looked into it, I am not aware of anyone getting to the bottom of the high temperature calibration Optris 160 and the low temperature calibration Optris 160 difference. Due to the need for Optris to have the camera in their factory for some extended period ( I forget how long, more than a week), there was a sense that the microbolometer array needed to be replaced. What this change does ( whatever it is ) to the sensitivity, response, emissivity setting, etc. is an unknown I would like to un-unknown a bit more.

  • Jarea

    I find that there is new info about the Lugano measurement reported problems.

    In the document you can find:

    DON’T YOU KNOW THAT THE CoP WAS = 1.07?
    Here there are two types of Optris PI-160 camera. The standard version, and a manufacturer modified, high temperature version. The Lugano team used the latter, however the author of the CoP = 1.07 paper based his arguments on the datasheet for the former.

    So the COP was correctly calculated and it was >3

    • Slad

      I prefer the Bob Higgins COP figure based on his emissivity recalculation (2.36), over the Lugano figures, however it’s certainly greater than 1.07

      • Obvious

        Although several people looked into it, I am not aware of anyone getting to the bottom of the high temperature calibration Optris 160 and the low temperature calibration Optris 160 difference. Due to the need for Optris to have the camera in their factory for some extended period ( I forget how long, more than a week), there was a sense that the microbolometer array needed to be replaced. What this change does ( whatever it is ) to the sensitivity, response, emissivity setting, etc. is an unknown I would like to un-unknown a bit more.

  • fact police

    Slad wrote:

    Ye gods! Where to start with this one… The E-Cat fuel is not “heated” in the sense you claim above – it is loosing more heat than is being provided by the electric heater. It makes no difference what the temperatures of the two transfer surfaces are, if its cooler, heat will flow to it, whether it’s at 1300ºC or 200ºC.

    It’s obviously true that if the ecat works as claimed, heat flows out of it. That is admitted in the 2nd sentence of your quotation: “That is to say the E-Cat fuel has to transfer its heat to the wall…”

    The relevant difference is that in the ecat, *additional* heat is deliberately generated within the reactor (slowing heat loss), whereas within a fission reactor, heat is deliberately removed (accelerating heat loss). Again the next sentence makes it clear: “That is to say the E-Cat fuel has to transfer its heat to the wall at a temperature even you put at 1300C, whereas the rim of the fuel pellet in a fission reactor is typically 200C.”

    The direction of heat flow is indeed determined by the sign of the temperature gradient, as you say, but the issue is clearly the *rate* of heat flow, and that *is* determined by the magnitude of the gradient, and it is 5 to 10 times larger in the fission reactor.

    What you seem to be suggesting is that at any boundary, for a given ∆T temperature (1000ºC), the same amount of heat will flow. Never heard of a heat transfer coefficient?

    Actually, the suggestion is spelled out in the following paragraphs, which was that for these particular boundaries similar temperature differences will give similar heat flows.

    It is absolutely wrong to say the have similar effective wall thickness. The two shapes have entirely different thermal characteristics, and need to be treated as such. …. What is the inner radius of a MOX fuel pellet? Zero.

    They’re both ceramic, they’re both cylindrical, and a typical outer diameter of fuel pellets is 2 cm, the same as the ecat.

    For that equation to apply with a zero inner radius, all the fuel would have to be contained on the axis in a cylinder of radius zero. That’s not the case for a fuel pellet.

    Rather, the fuel is distributed uniformly throughout, so the *effective* radius would be some value in between. And since the circumference increases with radius, there’s more fuel nearer the walls than nearer the axis, so the effective radius would be more than 5 mm, compared to the 4 mm you assumed for the ecat. So, that exasperates the implausibility of the ecat temperature difference being only 100C or so compared to 1000C in a fuel pellet.

    As a matter of fact the temperature distribution inside a pellet decreases only slowly from the axis to about 4 mm, where it’s still about 1200C, and then it starts to drop faster.

    So, the effective radius is rather similar, and any difference makes the ecat even more implausible.

    • Slad

      Nope, one is cylinder, and one is a hollow cylinder. Look at the ‘heat flow through a hollow cylinder’ equation, and see how Q alters when the tube becomes a solid cylinder (ie. ‘r inner’ starts to get smaller, eventually becoming zero)

      And the rate of heat flow through the tube is inconsequential… The real issue is the how the power density of the nickel can be 100x that of the MOX without melting.

      And 2cm dia. is pretty big for a fuel pellet. Got a reference for that?

      The whole point is that similar temperature differences do not give similar heat flows, it’s much more complex than that, and simple statements based on ‘just thinking about it’ don’t capture the well understood characteristics of thermal flow.

      And you are mixing up inner and outer radius when forming your argument that the effective radius of the fuel pellet would be zero.

      You’re never going to make detective at this rate, officer.

      • fact police

        Slad wrote:

        Nope, one is cylinder, and one is a hollow cylinder. Look at the ‘heat flow through a hollow cylinder’ equation, and see how Q alters when the tube becomes a solid cylinder (ie. ‘r inner’ starts to get smaller, eventually becoming zero)

        Again, that equation applies to a tube in which all the heat is generated within the tube. But the heat in a fuel pellet is not generated on the axis in a cylinder with a zero radius. It is generated throughout the pellet, with more of near the rim than the axis, so there is an effective radius probably larger than 5 mm. That’s certainly consistent with the temperature profile within pellets.

        And 2cm dia. is pretty big for a fuel pellet. Got a reference for that?

        In the “nuclear fuel cycle” article in Wikipedia, they give a temperature profile for pellets with a 1 cm radius (2 cm diameter).

        The whole point is that similar temperature differences do not give similar heat flows,

        No, but similar temperature differences with similar thermal properties do. And these are similar, and if anything the conduction out of a pellet is *better* than for the ecat because much of the fuel is nearer to the rim.

        And you are mixing up inner and outer radius when trying to suggest the effective radius of the fuel pellet would be zero.

        No, I think you are. Your zero radius does not apply, because most of the heat is not generated at zero radius.

        • fact police

          Nope, one is cylinder, and one is a hollow cylinder.

          Just to elaborate on this. A solid cylinder is a hollow cylinder with a smaller solid cylinder inside. If the heat is generated entirely in the inner solid cylinder, then the heat from that is lost through the hollow cylinder enclosing it.

          For a fuel pellet, the fuel is distributed throughout, so the heat from any selected cylinder is lost through the remaining hollow cylinder. This would lose heat like a tube with an some effective radius representing a weighted average of all the solid cylinders, which again is larger than half the radius, because more of the fuel is near the rim than near the axis.

          • Slad

            Nearly right… It is a like weighted average, but because your theoretical outer hollow cylinder of the fuel pellet is also generating heat, the temperature profile going towards the centre climbs much faster than you are allowing for.

          • john

            Slad i am very impressed .I was going to keep quiet and respond but you explaines all the key points.. Be alite less heavy on fact police ..hes only learning and its a good things…

          • Slad

            I find it annoying having to repeat myself.

            Feel free to take over my friend.

          • Slad

            And only a troll calls themself ‘fact police’

          • bachcole

            Since “troll” seems to have no clear definition, I cannot agree. But “skeptopath” has been well defined here and elsewhere, and only a skeptopath would call themselves “fact police”.

            In the fringe, there are no facts, only views and perspectives and ideas that we share with each other cordially. Pushy, bossy know-it-alls who try to debunk everything that seems to be positive are not well received here. “fact police” indeed!

          • Slad

            John: ‘fact police’ is actually one of the fine fellows who inspired the article in the first place…. Hopefully Mr Hody and Mr Clarke might soon join our discussion here too?

          • doug marker

            FWICT – Fact police is actually George Hody. (Mary Yugo, Al Potenza etc: etc: etc: etc: ) – I may be wrong but I doubt it. Joshua Schroeder (IMHO) has a far more condescending and arrogant style. Hody, when it suits, can be polite but among ‘friends’ (the pack), he is not shy about what he really thinks and the insults/innuendo/assasinations flow freely.

            Doug

          • Slad

            I also doubted it was ‘Tan nee Schroeder’ at first, as he generally comes across as being more intelligent, but he eventually admitted it above*. I think his style has to be toned down here to pass moderation.

            Hody is a different kettle of fish, who’s style is also very obvious. In fact from what I have read here, I think he also posts on here, however that’s only a hunch of mine, and I’m not entirely convinced.

            *Well, to be fair, he disagrees.

          • fact police

            Slad wrote:

            because your theoretical outer hollow cylinder of the fuel pellet is also generating heat, the temperature profile going towards the centre climbs much faster than you are allowing for.

            Ah, this is progress. You are admitting that not all the heat is generated at zero radius. The point is the heat generated by the fuel has to be transferred through the ceramic. The distance it has to be transferred has a distribution from zero to the outside radius. So the effective radius would be somewhere in between, and given that more fuel is near the rim than the axis, it would be more than half the outside radius. And that makes the Lugano ecat claims totally implausible.

          • Slad

            I’m not “admitting” anything. Your whole argument is ridiculous:

            The point is that ‘effective wall thickness’ is dependent on shape, not on thickness and/or radius, which become dimensionless. The fuel pellet has a much larger effective radius, which drastically limits it’s heat transfer.

            If that’s too hard to understand, just google ‘fuel rod temperature profile’, rather than trying to reinvent the wheel.

            But please, try not to let any more weird graphs you find on Wikipedia fool you.

          • fact police

            Slad wrote:

            The point is that ‘effective wall thickness’ is dependent on shape, not on thickness and/or radius, which become dimensionless. The fuel pellet hasa much larger effective radius, which drastically limits it’s heat transfer.

            Slad, you’re not making any sense. Of course it depends on thickness and radius. You’re still hung up on your equation with zero inner radius, which says nothing about this situation, because all the fuel is supposed to within that radius, and it’s *not*.

            The pellet does not have a much larger effective radius. The outer radius is simply the outer radius. The effective inner radius for your formula is *increased* (beyond say the 0.4 ro in Lugano) because much more of the fuel is outside 0.4 ro, and therefore closer to the coolant, with less thermal insulation in between.

            Look, heat from the reaction has to flow through ceramic to the coolant. In a fuel pellet, there is *less* ceramic it has to flow through, and yet the temperature difference is 10 times higher. The claim that you can dissipate the same power through *more* ceramic with a *smaller* temperature difference is just plain nonsense.

          • Slad

            You are only trying to ignore and misrepresent my arguments in order to salvage some dignity. Read the report again: My points are very clear.

            You are trying to argue against a mathematical equation. This is insanity. No wonder it is very easy to write articles that mock your statements.

            Your misconception that “there is more fuel near the rim so the effective radius is less than the actual radius” is an entirely ridiculous statement, as demonstrated by the equation mentioned above, and as explained in the report.

            It is clear that you have not yet understood the only important factor is the ratio between ‘inner’ and outer radius… Not the actual size of either. Hence, “dimensionless”.

            My calculations are commensurate with real-life MOX fuel rod temperature data, which can be found online.

          • fact police

            Slad wrote:

            You are only trying to ignore and misrepresent my arguments in order to salvage some dignity. Read the report again: My points are very clear.

            No, I’m trying to show you that your arguments are illogical, and do not apply.

            You are trying to argue against a mathematical equation.

            No. I’m arguing against your use of the equation where it does not apply (which you admitted). The equation applies to a hollow cylinder in which all the energy is generated inside the hollow portion. That doesn’t apply to a fuel pellet, and as the inner radius approaches zero, less and less of the energy is generated inside the hollow portion, and the relevance of the equation also approaches zero.

            My approximate reasoning suggested that a solid cylinder with fuel distributed evenly would be like a hollow cylinder with a radius something larger than half the outer radius with the fuel inside the hollow portion. And this can be justified analytically.

            Here’s a book chapter (looks more like lecture notes) that solves the relevant differential equations to develop equations of heat transfer for different geometries:

            http://wwwme.nchu.edu.tw/Enter/html/lab/lab516/Heat%20Transfer/chapter_2.pdf

            The equation for a solid cylinder with power generated uniformly throughout is 2-71. The max temperature difference from the axis to the surface is given by

            deltaT = e-dot * ro^2 / (4k)

            where e-dot is the power generated per unit volume, and ro is the outside radius.

            The relevant comparison is between cylinders with the same power per unit length, Qout/L using your labels. Now, since

            Qout = e-dot*volume = e-dot*(pi*r^2*L), the above equation can be rewritten as

            Qout/L = pi* deltaT * 4k

            Setting that equal to the Qout/L in your equation leads to ln (ro/ri) = 1/2, or ri = .61 ro. So a hollow cylinder containing fuel and generating the same power per unit length as a solid cylinder with uniform fuel, with the same temperature difference would have an inner radius 0.6 times that of the outer radius. So, the Lugano reactor by your assumption with ri = 0.4 ro would be less effective at conducting heat to the rim. This is consistent with my previous qualitative arguments.

            No wonder it is very easy to write articles that mock your statements.

            It’s obviously not that easy or you would have the courage to post it in the forum from which you extracted the quotes.

            Your misconception that “there is more fuel near the rim so the effective radius is less than the actual radius” is an entirely ridiculous statement, as demonstrated by the equation mentioned above, and as explained in the report.

            First, I didn’t say the effective radius is less. I said the effective inner radius is *more* than in the alleged Lugano geometry (as a ratio of the outer radius), and that conception turns out to be spot on.

            Second, that equation is not valid for a solid cylinder with uniform fuel.

            It is clear that you have not yet understood the only important factor is the ratio between ‘inner’ and outer radius… Not the actual size of either.

            For a hollow cylinder with the fuel contained within which is independent of the inner radius, the temperature difference depends on the ratio. But that condition is not satisfied in a solid cylinder with uniformly distributed fuel. In that case, another equation applies, and as demonstrated, corresponds thermodynamically to a hollow cylinder with a larger inner to outer ratio.

            My calculations are commensurate with real-life MOX fuel rod temperature data, which can be found online.

            What calculations? You just said that, by your formula, the temperature should tend toward infinity.

          • Slad

            The calculations where I compared the heat transfer through the alumina tube, to the heat transfer through a MOX pellet.

            As I have said, please stop wasting my time, and answer the three simple questions I have asked you, rather than trying to re-invent the wheel by introducing other pointless formulae.

            As my report says:

            “We can draw an analogy by looking at the Heat Flow Through a Hollow Cylinder equation from the original report

            “Whilst it is true to say this equation should not be used to analyse the heat emanating from a working nuclear fuel rod, this is merely an argument based on the shapes under consideration

          • fact police

            Slad wrote:

            The calculations where I compared the heat transfer through the alumina tube, to the heat transfer through a MOX pellet.

            Right, where you predicted the temperature tending toward infinity (confirmed in this post before editing), contrary to the finite and rather flat temperature of a pellet near the axis.

            As I have said, please stop wasting my time, and answer the three simple questions

            There was never disagreement about the first two, and I answered the third, but it’s not geometry. It’s that there is more heat per unit length generated in a fuel pellet, and the thermal conductance is lower. I demonstrated explicitly that the fuel pellet and lugano ecat have rather close thermodynamic properties, but the ecat presents somewhat greater thermal resistance from geometry alone.

            I have asked you, rather than trying to re-invent the wheel by introducing other pointless formulae.

            Surely, that’s not what they taught you in engineering. If your formula doesn’t apply, as you agree yours doesn’t, then it makes good sense to introduce a formula that *does* apply.

            “We can draw an analogy by looking at the Heat Flow Through a Hollow Cylinder equation from the original report

            “Whilst it is true to say this equation should not be used to analyse the heat emanating from a working nuclear fuel rod, this is merely an argument based on the shapes under consideration”

            But the argument is not valid, as shown in detail in another post.

          • Slad

            But the argument is not valid, as shown in detail in another post.

            So you claim It’s just that, to me, you have very little credibility.

          • fact police

            The argument does not depend on my credibility. Just verifiable information and simple logic.

          • Slad

            Maybe, but how much attention I pay to your arguments does depend on it.

          • fact police

            I don’t mind if you don’t pay attention to my arguments. But simple courtesy would suggest that you pay attention to them before you respond to them.

          • bachcole

            For me:

            1. Anyone with a handle “fact police”

            2. Anyone has denies LENR

            has not credibility.

          • fact police

            Now, since I’m such a nice guy, I’m gonna help you out, and show you what you should have argued.

            Your arguments about inner radius and so on for a solid cylinder are clearly all wet, but the formula should work for a hollow cylinder containing fuel, so at least assuming uniform contact to the inside surface, the temperature gradient for an alleged power should be calculable, even if it doesn’t give the fuel temperature itself.

            You may remember my puzzlement back in July, when last we discussed this, as to why a comparison to fission reactor fuel gave a rather different answer than the straight calculation gave for the inner surface temperature. So, I dug a little, and believe I have resolved it.

            The problem was that it’s not easy to find a simple statement of power density. I used total weight of fuel and total reactor output to get an estimate something on the order of 10 W/g, or about 200 times less than claimed for the ecat by the Lugano authors.

            But now I found better data using this reference:

            https://www.oecd-nea.org/science/docs/2007/nsc-doc2007-6.pdf

            You can deduce from this paper that the power density is more like 60 W/g for pellet temperatures of 1200C (on axis), but the really relevant metric to compare with Lugano is the power per unit length, which for this temperature is about 35 kW/m. That’s about 7 times higher than the 5 kW/m in the Lugano ecat assuming 1 kW output (and a 100 C temperature gradient) or 3.5 times the 10 kW/m corresponding to the 2 kW claimed by the authors (and a 200C gradient).

            So, even though the alleged power density would be much higher in the ecat, the fuel is more spread out along the axis, giving a *lower* power output per unit length along the reactor. So, that, combined with a factor 2 lower conductivity of MOX pretty well explains the different temperature gradients across the ceramic, given very similar thermodynamics otherwise, as shown separately.

            This by no means suggest that the Lugano claims just became plausible. As I’ve argued before, all this does is determine the inner temperature of the alumina, assuming it is uniform. And according to the report, that already puts the alumina *surface* above the melting point. With your revised numbers, it’s in spitting distance.

            But the temperature is clearly not gonna be uniform. Even in your description the powder covers only about 1/5 of the alumina cylinder, and since it’s a powder, the actual contact area is even less. This means the temperature of the inner surface of the alumina where it makes contact will have to be much higher to produce a delta T several times higher, and this would cause the fuel to melt.

            And this doesn’t take account of the temperature difference needed to transfer a large amount of heat from a very small amount of fuel. In an electric kettle, the heat source contact area is far larger, and *still* the element is hundreds of degrees above the water temperature to which it transfers about the same 1 kW of heat. So it is still highly implausible that a gram of nickel could transfer 1 kW power to a surface without being many hundreds of degrees hotter.

          • Slad

            First, I want to see a reference for that nonsense about a kitchen kettle (Ever heard of Leidenfrost?)…

            May I suggest you fill your kettle up, turn it on, and see how long you can hold your hand against the element for. No, seriously, you will be surprised.

            Or, just divide power rating of your kettle, by the surface area of it’s element, and read superheat temperature off a pool boiling curve.
            (* Hint: Nucleate boiling regime)
            You will then understand, why I can tell you are talking nonsense.

            Second, a nice guy, who seeks to waste my time with his silly solipsisms? Well in that case, I’m going to really help you out now…

            All you need to do is answer these three very simple questions.… The answer to the first two questions is a number, not a sentence. The third can be answered however you please.

            1) What temperature difference (across the wall) is required to transmit 2150W through a 200mm long alumina tube that has external diameter 20mm and wall thickness of 6mm?

            *Hint: If you are struggling, the working is in my first report.

            2) What is the temperature difference across a working 5mm radius MOX fuel rod, from centre to rim?

            *Hint: Google is your friend here.

            3) Now, why is the temperature difference seen in the fuel rod 2 to 3 times higher that seen in the alumina tube, despite the alumina tube having a 20% longer apparent conduction length and similar conductivity?

            *Hint: The answer should include the word “shape” and “effective radius”.

            Seriously. Stop wasting my time, and please just answer these questions.

          • fact police

            Slad wrote:

            First, I want to see a reference for “the element is hundreds of degrees above the water” in a kitchen kettle

            Maybe later. Do you really think a nichrome heating element in a kettle is *not* several hundred degrees C?

            Now, why is the temperature difference seen in the fuel rod 2 to 3 times higher that seen in the alumina tube, despite the alumina tube having a 20% longer apparent conduction length?

            This was already answered in detail. First the temperature difference is more than 2 to 3 times, but that’s a quibble. The reason is that the power per unit length of the fuel rod is about 5 times higher, and the fuel has a lower thermal conductivity than alumina by about a factor of 2.

            I showed explicitly that shape is *not* the reason. The solid fuel rod is thermodynamically similar (as I argued from the outset) to a hollow cylinder containing the same fuel with an inner radius 0.6 times that of the outer radius.

          • Slad

            I specifically mentioned the wire inside, and also explained to you why it got so hot.

            This turned out be be a waste of time, because it was just a bad analogy, tossed in at the end of apparently unrelated argument, seemingly only there to muddy the water. Or maybe not, but it’s just too ambiguous to parse.

            No more kettles please, unless you can clearly describe the point you are trying to make.

          • fact police

            Slad wrote:

            May I suggest you fill your kettle up, turn it on, and see how long you can hold your hand against the element for. No, seriously, you will be surprised.

            The element is not accessible in my kettle, so I did a simpler experiment using a pot of water on a stove top.

            With the element at maximum heat, it took 6 minutes to boil one liter of water. So that’s pretty close to 1 kW thermal power.

            The temperature of the element can be estimated from the color. I turned the element off, then removed the pot, and the element that had made contact was still glowing faint red, so about 500C. (My pocket IR thermometer maxes out at 300C, and it overloaded.)

            The bottom of the pot immediately after removal was about 200C, measured with an IR thermometer.

            So, there’s a 300 C temperature difference between the element and the pot, for a 1 kW (at least) thermal transfer.

            The area of the element under the pot was 135 cm * .8cm = 107 cm^2, about twice the maximum possible area in the Lugano ecat, and 10 times the area of contact you estimated.

            Now the actual interface temperature during contact would be different, but a steep gradient is necessary for the heat to flow out of the element at the necessary rate, and the measured temperature a half second after contact would be a pretty good estimate of the element temperature just inside the interface.

            So, to sum up, 10 times the area with similar power transfer requires a temperature difference (from the source) of about 300C. A proportionally higher temperature difference seems likely for the smaller area, and so it is totally implausible that the nickel would not have melted in either of your scenarios.

          • Slad

            but a steep gradient is necessary for the heat to flow out of the element at the necessary rate

            I disagree. Google ‘water pool boiling curve’, and look at the wall superheat temperature at the critical heat flux. It’s thirty or so degrees above boiling point.

            There is a very steep gradient inside the element (in the old fashioned ones, where you can touch it)… The inner wire is very hot but the surrounding material is an insulator, it’s designed to ‘step down’ the temperature, to better fit water’s boiling curve.

            You proabably managed to put your boiling pan into a different regime (ie. not nucleate boiling). Yes the heat still flows, but not in a controlled manner like a kettle, which aims for an efficient transfer.

            You used an IR thermometer, did you calibrate the emmisivity for for the pans surface? (joke… well, sort of)

          • fact police

            Slad wrote:

            Come on, there’s a huge air gap between your pan and the element.

            No, the surfaces I’m considering were in contact. And both are pretty flat, so the percentage contact would be pretty high, and with some force — at least as high as the percentage between a powder loosely sitting on alumina.

            Also did you measure the bottom of the pan to get the 200C?

            Yes. That’s the surface that’s in contact with the element.

            For a realistic way of comparing it to a kettle elements temperature, you should have measured inside part of the pan touching the water.

            Water is beside the point. The kettle was just an example of heat transfer from a heating element to another metal before it heats the water.

            I’m looking at the temperature difference between the stove element and the pot surface. And this was about 300 C to transfer 1 kW across an area larger than that available in the ecat.

            And you used an IR thermometer, did you calibrate it for the pans surface?

            It’s not sensitive enough to the surface to affect the argument.

            but a steep gradient is necessary for the heat to flow out of the element at the necessary rate

            I disagree. Google ‘water pool boilng curve’, and look at the saturation temperature at the critical heat flux.

            I’m not talking about heating water here (except as a measure of the overall power transfer). I’m talking about heating the pot with the element.

          • Slad

            I’m not talking about heating water here (except as a measure of the overall power transfer). I’m talking about heating the pot with the element.

            No, the surfaces I’m considering were in contact. And both are pretty flat, so the percentage contact would be pretty high,

            Flatness is a very relative thing in heat transfer. Unless you’ve polished your surfaces, a slight roughness can matter a lot.
            Like ‘factor of 100’ a lot, on a heat transfer coefficient.

            That’s the point about the fine nickel particles, they can contact a rough surface as if two highly polished (~5um) surfaces are in contact.

            And your hob was likely dry. Liquid metals at the interface in the E-Cat.

            Regarding what you originally said:

            In an electric kettle, the heat source contact area is far larger, and *still* the element is hundreds of degrees above the water temperature to which it transfers about the same 1 kW of heat

            You must have meant “In a modern kettle, with a hidden element not immersed in water, the air gap to the visible heat transfer plate causes the element to get very hot”

            Which I agree with. But it is a really bad analogy, not to your pan, but to the E-Cat.
            It is also an example the tangential reasoning, extraneous to your main point, that unfortunately takes time to reply to. I should have skimmed over it, but now, I suggest we avoid the topic.

          • fact police

            Slad wrote:

            I’m not talking about heating water here (except as a measure of the overall power transfer). I’m talking about heating the pot with the element.

            No, the surfaces I’m considering were in contact. And both are pretty flat, so the percentage contact would be pretty high,

            Flatness is a very relative thing in heat transfer. Unless you’ve polished your surfaces to <5um, the roughness matters a lot. Like 'factor of 100' a lot.

            That's the point about the nickle particles, they can contact a rough surface as if two highly polished surfaces are in contact.

            I agree a powder could make more intimate contact, if pressed into place with a large force. But I’m skeptical that simply pouring it in would make the contact even as well as two unpolished solids. Moreover, the powder itself presents more internal thermal resistance even when pressed to the equivalent density, but when loose represents more boundaries and more air gaps for the heat to be conducted through. So, for such a high power transfer form so little material, it seems highly implausible that a large temperature gradient within the nickel and between the nickel and ceramic would not be needed.

          • Slad

            I think a goodportion of large force would also be replicated by the effect.

            And for sure, there’s a porosity, which I included, but the pores are filled with hydrogen (I modeled air) so I’m happy I used a somewhat conservative figure.

          • fact police

            I think a goodportion of large force would also be replicated by the effect.

            by which I assume you mean you would like this to be the case.

            And for sure, there’s a porosity, which I included, but the pores are filled with hydrogen (I modeled air) so I’m happy I used a somewhat conservative figure.

            Air or hydrogen will make it less thermally conductive than solid.

          • fact police

            In an electric kettle, the heat source contact area is far larger, and *still* the element is hundreds of degrees above the water temperature to which it transfers about the same 1 kW of heat

            You must have meant “In a modern kettle, with a hidden element not immersed in water, the air gap to the visible heat transfer plate causes the element to get very hot”

            Which I agree with. But it is a really bad analogy, not to your pan, but to the E-Cat.

            The specific design is hidden, and that’s why I went to a simpler analogy, which was not meant to be compared to a kettle, but directly to an ecat. I considered heat conduction between two solids. And it served the purpose well. Ten times the contact area, probably more than 10 times the mass, and it still has a temperature difference of 300C to transfer 1 kW of power.

        • Slad

          No… The heat struggles the most to conduct away of the centre of the pellet. This is the hottest area, and the area that would melt first if the power density was increased.

          You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot. (In fact, the same very point I am having to repeatedly make to you)

          The wikipedia graph is only there to illustrate another point about the poor heat conductance of uranium oxide compared to other materials. I don’t believe a fuel pellet that size exists in real life: the person who drew that chart just created it as a theoretical example to make their point. The red line is set to 1.0cm for comparison purposes only. 3000C is meltdown temperature, why would anyone want a pellet that big?

          “No, but similar temperature differences with similar thermal properties do. And these are similar, and if anything the conduction out of a pellet is *better* than for the ecat because much of the fuel is nearer to the rim.”

          ** a) The thermal properties are different, due to the shape, which I have repeatedly explained to you… The equation I quoted explains this.

          ** b) As I explained above, it’s what is happening at the centre of the MOX pellet that is the problem.

          “I think you are”

          It doesn’t matter what you ‘think’, because you are arguing against the science of thermodynamics and heat transfer… Not against me.
          Go read a textbook, not the ravings of Mr Tan.

          • fact police

            Slad wrote:

            No… The heat struggles the most to conduct away of the centre of the pellet. This is the hottest area, and the area that would melt first if the power density was increased.

            Actually the temperature is pretty well flat at zero radius, and decreases only slightly by 4 mm. And it’s not about where it most struggles to dissipate from, it’s about the overall dissipation that determines the necessary temperature gradient. The contribution from the zero radius fuel is infinitesimal.

            You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot. (In fact, the same very point I am having to repeatedly make to you)

            But the point is only valid if all the fuel is contained in the cylinder with the shrinking radius. Since it’s not, the point you are trying to illustrate is wholly invalid.

            The wikipedia graph is only there to illustrate another point about the poor heat conductance of uranium oxide compared to other materials. I don’t believe a fuel pellet that size exists in real life: the person who drew that chart just created it as a theoretical example to make their point. The red line is set to 1.0cm for comparison purposes only. 3000C is meltdown temperature, why would anyone want a pellet that big?

            True, 10 to 14 mm seems to be more typical, but as you said yourself, smaller pellets with a 10 times larger delta T makes the ecat geometry *less* plausible, not more.

            “No, but similar temperature differences with similar thermal properties do. And these are similar, and if anything the conduction out of a pellet is *better* than for the ecat because much of the fuel is nearer to the rim.”

            ** a) The thermal properties are different, due to the shape, which I have repeatedly explained to you… The equation I quoted explains this.

            No, it doesn’t because it doesn’t apply.

            ** b) As I explained above, it’s what is happening at the centre of the MOX pellet that is the problem.

            No, you merely asserted that. You did not explain it, nor is it true. Heat is generated throughout the pellet, and it all has to be conducted out. Considering only the infinitesimal fuel at zero radius is simply wrong.

            It doesn’t matter what you ‘think’, because you are arguing against the science of thermodynamics and heat transfer… Not against me.

            No. It’s definitely you. You are applying the science incorrectly. You are using an equation for a tube containing fuel and applying it to a cylinder with uniformly distributed fuel as though all the fuel were on the axis. Just wrong. Not even a good approximation.

            If you have trouble thinking of an effective inner radius, or have difficulty considering a continuum, I understand. But trying to put it all at zero radius is not the answer. The average or effective distance of the fuel from the rim is smaller than half the radius, and so it would lose heat faster than if all the fuel were in a cylinder with the inside radius half of the outside radius, and that makes the ecat claims completely implausible.

            Go read a textbook, not the ravings of Mr Tan.

            I’ve read text books, and I have no idea who Mr Tan is.

          • Slad

            I don’t believe that for a second. He goes by the name of Popeye, you’d like him, you share a lot of traits. However, if I were to list them, I would definitely fall foul of the commenting policy here.

            And, as I have explained to you three times now:

            “You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot.”

          • fact police

            Slad wrote:

            I don’t believe that for a second. He goes by the name of Popeye,

            You can believe what you want, but I’ll tell you what I know. I write as “fact police” here, popeye on ECN, and Joshua Cude on vortex and some other places. And I have no idea who Mr Tan is.

            And, as I have explained to you three times now:

            You need to learn the difference between an assertion and an explanation.

            “You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot.”

            As I have *explained* to you, letting r tend toward zero in that equation assumes all the heat is generated at a radius less than r, which is completely wrong. And while shape matters, if you assume an irrelevant shape, it doesn’t inform the question of the actual fuel pellet. A much better approximation, if you must use that equation, is one in which all the fuel is contained inside a cylinder with a radius somewhere between zero and the outside radius. Given that 3/4 of the fuel is at more than half the radius from the axis, an effective radius of something larger than half the outside radius is conservative. Of course, you don’t like that because it makes the ecat look totally implausible.

          • Slad

            Assertion/explanation/condescension whatever.

            Yes, the fuel pellets shape gives it a larger effective radius. Hence, the heat from the centre of the pellet is insulated more. Hence… A nuclear reactors power is limited (with regard to its potential power), to avoid melting the uranium oxide. This is one of the reasons the E-Cat can tolerate a much higher power density.

            And yes, your style is fairly obvious. Also, don’t forget your other pseudonyms of ScienceApologist and Joshua Cude.

            As you insist… This is Mr ‘Popeye’ Tan nee Schroeder:

            https://groups.yahoo.com/neo/groups/newvortex/conversations/topics/648

            http://web.archive.org/web/20120901213441/http://scienceapologist.wikinet.org/wiki/Joshua_P._Schroeder_%28ScienceApologist%29

            https://twitter.com/astrophysically

            Nice bow tie!

          • fact police

            Slad wrote:

            Yes, the fuel pellets shape gives it a larger effective radius. Hence, the heat from the centre of the pellet is insulated more.

            No, the distribution of the fuel in the pellet gives the inner diameter of an equivalent tube containing the fuel a larger radius. Hence the fuel is insulated *less*. Look, it’s not complicated. Fully 75% of the fuel is situated at more than half the outer radius from the axis, making it *closer* to the rim, and therefore *less* insulated than the fuel in the ecat.

            This is the reason the fuel in the ecat would be *less* able to tolerate the same power density with the same temperature difference. Now the 100 times higher power density of the ecat is more or less offset by less fuel, but you still have an order of magnitude lower temperature difference in the case of the ecat, which makes it implausible by comparison.

            And yes, your style is fairly obvious. Also, don’t forget your other pseudonyms of ScienceApologist and Joshua Cude.

            I listed Joshua Cude, but I’m afraid ScienceApologist is someone else.

            As you insist… This is Mr Tan:

            …but it’s not me.

          • Slad

            You’ve left an indelible trail all over Wikipedia! (currently as ‘jps’)
            Same style. Same arguments. Same attitiude. Same topics of interest. Unbroken timeline between the various pseudonyms. Stand tall and proud man!

          • fact police

            Slad wrote:

            You’ve left an indelible trail all over Wikipedia! (currently as ‘jps’)

            Same style. Same arguments. Same attitiude. Same topics of interest. Unbroken timeline between the various pseudonyms.

            Given you (and people like Lomax) can reach such certainty without good evidence about something I happen to know you are wrong about, it’s not surprising you are so certain about other things with a similar lack of good evidence.

            I know about ScienceApologist from Lomax’s writing, but I have never participated in wikipedia editing, nor have I read any of SA’s writings (unless they were quoted in cold fusion forums, of which I don’t remember any specific instances, or of course what ends up in wikipedia articles). If you have some examples of passages or arguments of his that are so similar to mine, I’d be very interested to see them.

            It’s a trivial exercise to identify verbatim passages from fact police, popeye, and cude, since I rely heavily on the cut and paste function.

            The errors (and laziness) in this sort of thing are obvious from above where doug marker and you admit uncertainty about my identity, when what I posted here (to begin with) was extracted almost verbatim from what I posted in ECN as popeye.

          • Slad

            I’m starting to understand why you have been banned from Wikipedia so many times.

  • fact police

    The above arguments are based on the fallacious reasoning that the rate of an LENR reaction must have the same exponential dependence on temperature as chemical reactions (The Arrhenius Equation).

    I don’t recall anyone claiming an exponential dependence, nor does the argument depend on one. Thermal runaway can occur with a linear dependence if the slope is steeper than the heat loss slope. Even in your graph, beyond the intersection, thermal runaway would happen, and adding heat would not stop it. So it’s only a question of whether it could reach that point by some error or accident. And that intersection could happen at a lower temperature, depending on the reaction slope which depends (in your argument) on the number of “active” environments, and on the cooling slope.

    The simple argument was that if one unit of thermal power from outside the reactor is enough to initiate the reaction, then 3 units from inside the cell will be enough to keep it going, and in any case, can be made to keep it going with controlled heat loss. This argument does not depend on the Arrhenius equation.

    • Slad

      The exponential dependence is implicit in the argument that there could be thermal runaway in the system as designed. It’s the unstated basis behind their whole reasoning.

      Regarding your final paragraph, what happens if you design the heat transfer surface to loose four ‘units’?

      Answer – a stable and easily controllable reactor.

      • fact police

        The exponential dependence is implicit in the argument that there could be thermal runaway in the system as designed. It’s the unstated basis behind their whole reasoning.

        No. Thermal runaway does not require an exponential dependence. It only requires that the heat generated exceed the heat loss, and the rate of increase of heat generation with temperature is the same or higher than that of heat loss, whether linear or not.

        In your graph that condition is met everywhere (above the alleged turn-on) if the heater generates 13 W/m^2 or more, and is met above the intersection for no input heat at all. And at the operating point, if the power increased slightly, it would go into thermal runaway as well, if it was not immediately turned down. It’s always on knife-edge.

        Regarding your final paragraph, what happens if you design the heat transfer surface to loose four ‘units’?

        If it loses 4 units, then it will never reach the operating temperature. On your graph, the 5 W/m^2 from the heater at the operating point would not have been enough to reach the alleged turn-on point where the heat loss is already 13 W/m^2.

        And if you used 13 W/m^2 to reach that point, then it would have to be turned down by more than a factor of 2 to prevent thermal runaway at the operating temperature. The ecat input power for the Lugano experiment does not follow this description.

        Anyway, the more important point is that in the interest of eliminating the need for external input, it *could* be made to self-sustain, and if it could, they would do it.

        Answer – a stable and easily controllable reactor.

        No. It would actually be rather unstable, because a slight increase in heater power could send it into runaway, and that slope is unlikely to be stable, depending on the particular NAE density. If it changed suddenly, even turning off the heater would not prevent runaway.

        (There’s more detailed discussion of this on ECN, in the “Rossi Sells eCat” thread.)

        • Slad

          “Thermal runaway does not require an exponential dependence. It only requires that the heat generated exceed the heat loss”

          ** Well obviously this is true, but in real life, you would have to be pretty dumb to design a system that suffers from this issue. Unless your job description is ‘bomb designer’…

          “5 W/m^2 from the heater at the operating point would not have been enough to reach the alleged turn-on point where the heat loss is already 13 W/m^2… …And if you used 13 W/m^2 to reach that point, then it would have to be turned down by more than a factor of 2 to prevent thermal runaway at the operating temperature.”

          ** Which is exactly how a normal heater control would work!
          Come on man!

          As for the “discussion” on ECN, what you see there is a butt-hurt young astrophysicist trying to argue about thermodynamics and control systems with a doctor of mechanical engineering. It’s fairly amusing, however, it’s also unlikely to get a response, as the time required to fully unpick his lengthy mistakes (and purposeful misunderstandings) just takes too long. Plus… your man Mr Tan prefers to argue, rather than listen, which is a shame, as in this case, he would learn something.

          Do I attempt to lecture him on Cosmology?

          • fact police

            Well obviously this is true, but in real life, you would have to be pretty dumb to design a system that suffers from this issue.

            Thermal runaway is possible in your design, according to your graph. In fact, it would be tricky to avoid, and impossible to stop after a certain tipping point. Controlled cooling would be much safer.

            ” 5 W/m^2 from the heater at the operating point would not have been enough to reach the alleged turn-on point where the heat loss is already 13 W/m^2… …And if you used 13 W/m^2 to reach that point, then it would have to be turned down by more than a factor of 2 to prevent thermal runaway at the operating temperature.”

            ** Which is exactly how a normal heat control would work!

            But that’s *not* how the ecat works. The input power in all the reports is simply increased monotonically to the operating point. In your design, it would have to be decreased by more than a factor of 2.

            As for the “discussion” on ECN, what you see there is a butt-hurt young astrophysicist trying to argue about thermodynamics and control systems with a doctor of mechanical engineering.

            I’m not impressed by credentials. I’m impressed by logic. And no, popeye is not an astrophysicist, and the arguments are logical.

          • Slad

            Listen. It’s my graph… I could easily have drawn it so it wouldn’t ever have run away, I was going to make a further point based on that, but took it out to keep things *relatively* concise.

            We’ve moved beyond the report to the patent, however, even the Lugano E-Cat had an internal thermocouple running in a feedback loop.

            And… I’m not going to keep on endlessly doxxing the poor chap. So lets just say he is absolutely and definitely an astrophysicist (of some kind). You’ll have to take my word for it. Why do you think he isn’t?

            I suppose you think your arguments are logical too? The problem, (and unfortunately this does come down to credentials) is that neither of you has the theoretical knowledge or experience to realise the simple mistakes you both keep making.

          • fact police

            Slad wrote:

            Listen. It’s my graph…

            Yes, but I think you misinterpreted it.

            I could easily have drawn it so it wouldn’t ever have run away,

            And yet you didn’t, and you went to some effort to justify the way you did draw it. It doesn’t support your case is the problem.

          • Slad

            Of course it supports my case.

          • fact police

            Slad wrote:

            Of course it supports my case.

            It is very vulnerable to runaway, which can be uncorrectable by turning off the heater. And it depends on a reaction slope that is unpredictable. And it requires the heater to be increased to start the reaction, and then reduced by a factor of more than 2 for stable operation, a sequence not used in the Lugano ecat.

          • Slad

            Only if operated it near the runaway point of 1366C.

          • fact police

            Wrong. With external input heat of 13 W/m^2, it would runaway at any temperature above your alleged turn-on point.

          • Slad

            Oh dear. Are you really pretending not to understand the concept of a proportional heat controller?

          • fact police

            Slad wrote:

            Oh dear. Are you really pretending not to understand the concept of a proportional heat controller?

            No. But you talked a lot about an entirely passive system. Now you’ve changed your paper to require a robust and responsive control system, when elsewhere you argue that the need for a control system for self-sustained mode is too dangerous.

            I didn’t say it *would* runaway below the intersection. I said it was vulnerable to it. If the control system is not sufficiently robust or responsive, from your operating point, a mere 5 W/m^2 decreasing too slowly would cause thermal runaway. Or if the output spiked because of diffusion of the hydrogen to an anomalously high density of NAEs, simply turning the heater off would not help.

            The use of a proportional controller for a simple heater is quite different. Regardless of excursions above the set temperature, reducing power *at any rate* will always bring it back. Not so, with an ecat following your graph. And it doesn’t have to exceed 1366C either. As the temperature begins to increase, the lines converge and less heat is needed to exceed the rate of dissipation. If this convergence is faster than the decrease in the input, you get runaway.

            This kind of knife-edge control is bad design, and it’s not necessary. What you want is a system in which the rate of cooling increases faster from the operating point than the reaction rate, to provide a stable point. That’s what stabilizes the sun, and many other exothermic reactions. And in the ecat theory, presumably, the location of this stable point is adjustable by changing the input heater power.

            The problem is you can’t get it with linear cooling and linear reaction rates. What’s more, with linear dependencies, as you’ve shown them, the operation would be very different from the way the Lugano ecat was in fact operated. It would have to be heated much more to get it started than to keep it going, whereas with the Lugano ecat, the heat was simply ramped up. So your speculated dependencies are not in evidence from the Lugano data.

            A year ago, after the release of Lugano, Paul Stout and I hashed over this in some detail. Paul actually contrived a cooling dependence and a reaction rate dependence that fit the Lugano operating procedure, and that would give stable operation determined by the input power, and could never runaway. But the dependencies were highly non-linear, and also highly unrealistic.

            But more important than whether such a system could be contrived, is the question of why anyone would do it. The important point is that if one unit of energy (from the outside) can initiate the reaction, then 2 or 3 units from within can be made to sustain it. That means that with suitable insulation and controlled cooling, it could run without input power, which aside from being far more practical, and useful for making electricity, would make for one hell of a demonstration.

          • Slad

            Oh, so you are pretending not to understand the concept… OK

            I don’t ascribe much credibility to your personal heat flow analyses, if you hadn’t guessed. Do they always involve misinterpreted Wikipedia graphs?

            And you seem to be working off the assumption that things are non-linear. Read the report, my views are clear.

          • fact police

            Slad wrote:

            And you seem to be working off the assumption that things are non-linear.

            No, I’m not making any assumptions. I’m just interpreting yours, since you don’t seem to understand your own graph. The point is your graph does not fit the behavior reported for the Lugano ecat.

            Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat, and it would quench without input, and never run away. Then the risk would simply involve the unpredictability of the reaction slope. A higher than expected slope would still produce runaway.

            But, as it happens, in the Lugano ecat, most of the heat loss from the device is by radiation, and that *is* highly non-linear, so at least to that extent, your graph is not realistic.

            And again, the more important question is why bother trying to control an exothermic reaction with heat. The important point is that if one unit of energy (from the outside) can initiate the reaction, then 2 or 3 units from within can be made to sustain it. That means that with suitable insulation and controlled cooling, it could run without input power, which aside from being far more practical, and useful for making electricity, would make for one hell of a demonstration. It’s inconceivable that they wouldn’t do that if it were possible. And if the claims have validity, it would be possible.

            And again, the

            Read the report, my views are clear.

            Unfortunately, it’s the logic that is muddled.

          • Slad

            Just to remind you what you actually said:

            “But the dependencies were highly non-linear”

            But now you say, you didn’t make that assumption.

            Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat

            Oh dear. It seems you are mixing the axes of my graph up.*
            Do you see why I doubt your personal heat transfer analyses?

            Just answer the three questions I asked you below, and stop wasting our time.

            *I feel another report update coming on…

          • fact poice

            Slad wrote:

            Just to remind you what you actually said:

            “But the dependencies were highly non-linear”

            Now you say, you didn’t make that assumption.

            Right, those were the dependencies used by Paul Stout. The reason for the “But” was because the dependencies *he* used were different from yours.

            Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat

            Oh dear. It seems you are mixing the axes of my graph up.*

            No, I have the axes straight. Your mocking would be more effective if you revealed the logic that led to the idea that I might be mixing them up. But I’m not.

            Try this: Redraw your line representing the heat generated by the fuel from the same starting point, but with a shallower slope than the cooling line. Then the lines will of course diverge.

            Now, add a constant value (the input thermal power) to the heat line so that the starting point is above the cooling line, say 20 or 30 W/cm^2. The slope is not changed by adding a constant, so now it will eventually intersect the cooling line at some higher temperature.

            But *this* intersection is one of natural stability. At temperatures above, cooling exceeds total heating so the temperature decreases. At temperatures below, total heating exceeds cooling and the temperature increases.

            Furthermore, this intersection is selectable by adjusting the input heat, there is no runaway condition, and with the removal of the heat, it cools to ambient.

            So, as I said, the only danger of runaway is from the uncertainty in the reaction slope. A higher slope (larger density of NAEs) would produce runaway that could not be stopped by removing the input heat.

            Of course, it would be difficult to justify that cooling curve with heat loss by radiation, and as I said, if this scenario were possible, self-sustained mode would be too, and if it were, it would be daft not to demonstrate it.

          • fact police

            We’ve moved beyond the report to the patent,

            The patent doesn’t even claim LENR. It’s hardly logical to use the patent configuration to defend the implausible thermodynamics of the Lugano configuration.

            even the Lugano E-Cat had an internal thermocouple running in a feedback loop.

            Although the TC was operating well outside its rated temperature range, and there is no sign of feedback in the power record. Anyway, you made a point of the constraints *water* placed on your assumptions, so that is completely irrelevant to the Lugano implausibilities.

            So lets just say he is absolutely and definitely an astrophysicist (of some kind). You’ll have to take my word for it. Why do you think he isn’t?

            You’re asking me why I think I’m not an astrophysicist? A better question is why you think I am. There is no evidence for it, although I do realize that that is no an impediment to what you believe.

            The problem, (and unfortunately this does come down to credentials) is that neither of you has the theoretical knowledge or experience to realise the simple mistakes you both keep making.

            First, you have no way to know that, but more importantly, I’ve never been convinced by someone who can’t defend their arguments except by saying: “I know better.” Especially when their paper is full of simple logical errors.

          • Slad

            My assumptions about your level of theoretical knowledge regarding thermodynamics are based on our discussions over the past nine months or so, including my favourite exchange where you explained Fouriers Law to me. This lead on to a discussion about transient heat modelling, of which I think we/you eventually came to an understanding of.

            And I ‘believe’ you are an astrophysicist because your twitter account is ‘astrophysically’ and you studied astrophysics. The reasons I beliveve this can be found a few posts below.

          • fact police

            Slad wrote:

            I ‘believe’ you are an astrophysicist because your twitter account is ‘astrophysically’ and you studied astrophysics.

            Again, the logic escapes me. That’s not my twitter account. Do you think all people who use the same first name are the same people? I’m not Josh Groan either.

            The reasons I beliveve this can be found a few posts below.

            I couldn’t find reasons anywhere. It’s amazing how you can be so certain about things without evidence.

          • bachcole

            fact police, I suggest that you leave and come back with a different handle. The arrogance of “fact police” and similarity to the attitudes of skeptopaths is such that you have next to no credibility here. I am trying to do you a favor. I never read your comments, really.

          • fact police

            bachcole wrote:

            I never read your comments, really.

            Your loss.

          • fact police

            Slad wrote:

            My assumptions about your level of theoretical knowledge regarding thermodynamics are based on our discussions over the past nine months or so, including my favourite exchange where you explained Fouriers Law to me. This lead on to a discussion about transient heat modelling, of which I think we/you eventually came to an understanding of.

            Yes, I remember it well. It’s difficult to imagine someone who couldn’t even use terms like transient and gradient correctly could have a degree in engineering, let alone hold out credentials as a reason to trust them. You said incomprehensible things like “The transient temperature gradient in the alumina would be minimised, if the (increasing) rate of change of the fuel’s power output is minimised.” and “It must be transient, heat isn’t conducted instantaneously from the inside to the outside.” Seemed like rookie misunderstandings to me.

            By the way, changing your text (or adding to it) after I’ve replied to it, makes the chronology a little weird. If you have something to add, a new post would be more sensible.

          • Slad

            I prefer to be concise.

            So you think heat is conducted instantaneously?!!

            Indeed, the transient temperature gradient – the temperature distribution seen inside a material whilst it is warming up. (maybe profile would have been a better word than gradient). Incomprehensible to some perhaps, but then, it’s a fairly obscure area of heat transfer. Maybe you would like to try to explain the differential relationship in approximately twenty words here? …Please.

          • fact police

            Slad wrote:

            So you think heat is conducted instantaneously?!!

            No, and the question indicates you’re still confused. Heat is transferred at a finite rate (not instantaneously) with a steady state (not transient) temperature gradient. So, your suggestion that a transient was necessary, or heat would have to be conducted instantaneously is all wet.

            Indeed, the transient temperature gradient – the temperature distribution seen inside a material whilst it is warming up.

            Sorry, but I can’t decipher a complete thought from that sentence fragment. (nor from what follows…)

            The problem was that the thermodynamic implausibility that I identified was during the steady-state operation of the ecat as claimed by the Lugano authors. That would correspond to a steady temperature gradient and a constant power output from the fuel. It had nothing to do with transient temperature gradients or with the rate of change of the power output, and so that’s why that other sentence made no sense in the context.

          • Slad

            If you choose to dodge the question, that’s OK, most people probably would too. Just don’t pretend you don’t understand the concept, seeing as how you are such an expert in heat transfer and all. Like I said above, it is hard to formulate a coherent argument, when one is out of one’s depth.

            Transient heat flow, Google it.

            Another thing I quoted is that “if you argue with an idiot, there are two idiots”. Now I don’t believe you are an idiot, but you are incredibly arrogant.

            It’s a waste of my time talking to someone who only wants to argue, when it is very apparent (to me at least) that you could learn something, if only you would choose to.

          • fact police

            Slad wrote:

            Just don’t pretend you don’t understand the concept, seeing as how you are such an expert in heat transfer and all. …

            Transient heat flow, Google it.

            I didn’t say I don’t understand the concepts. I said your use of the terms made no sense. It makes no sense to say that there must be a transient temperature gradient or the heat conduction would have to be instantaneous. The other sentence was simply irrelevant.

            Like I said above, it is hard to formulate a coherent argument, when one is out of one’s depth.

            You have provided ample proof of that.

            Now I don’t believe you are an idiot, but you are incredibly arrogant.

            It’s a waste of my time talking to someone who only wants to argue, when it is very apparent (to me at least) that you could learn something, if only you would choose to.

            That sounds kind of arrogant itself, but in your case, it’s not justified.

          • Slad

            “It makes no sense to say that there must be a transient temperature gradient or the heat conduction would have to be instantaneous. The other sentence was simply irrelevant.”

            * * No, I was simply pointing out two daft things that you said.

          • fact polive

            Slad wrote:

            Now I don’t believe you are an idiot, but you are incredibly arrogant.

            This from someone who seems to think he is Galileo…

          • Slad

            Zing!

          • bachcole

            fact police, there is no way that you are not arrogant with the handle “fact police”. Wikipedia thinks that it is the fact police, and they are about to be shown how arrogant they are. I hope that you are not so arrogant that you cannot be told that you are arrogant and to accept my suggestion to tone down your arrogance.

          • fact police

            Do I attempt to lecture him on Cosmology?

            So, you’re suggesting we accept your arguments on your authority instead of your logic?

            I’m not in the habit of doing that. And it that’s what you want, why bother writing a paper? Since you did write a paper, it seems reasonable that objections to the logic be raised, and that you try to defend them.

          • Slad

            Well, given your other habit, of misrepresenting yourself as a professor, I am not suprised by this.

            I’m suggesting you go read a Heat Transfer textbook instead of all those Cosmology ones. The logic isn’t in question (if the assumptions are correct)

            I suppose you also argue with your lawyer about the law, and with your doctor about medicine?

          • fact police

            Slad wrote:

            I’m suggesting you go read a Heat Transfer textbook instead of all those Cosmology ones.

            I am not a cosmologist. And I have read thermodynamics texts, enough to know that you completely misuse the equation for transfer through a cylinder in applying to a fuel pellet as though all the fuel were concentrated at zero radius.

            The logic isn’t in question (if the assumptions are correct)

            Definitely the logic. It’s an illogical use of an equation in a situation it was not developed for.

            I suppose you also argue with your lawyer about the law, and with your doctor about medicine.

            If they make blatant errors in logic that I can understand, of course. You don’t?

            Again, you seem to telling us to trust you because you are a mechanical engineer. But I don’t because you make obvious errors that an undergraduate can spot.

          • Slad

            “you completely misuse the equation for transfer through a cylinder in applying to a fuel pellet as though all the fuel were concentrated at zero radius.”

            As I have explained to you already:

            “You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot.”

          • fact police

            Slad wrote:

            You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot.

            But as I have explained to you, letting r tend toward zero in that equation assumes all the heat is generated at a radius less than r, which is completely wrong. And while shape matters, if you assume an irrelevant shape, it doesn’t inform the question of the actual fuel pellet. A much better approximation, of you must use that equation, is one in which all the fuel is contained inside a cylinder with a radius somewhere between zero and the outside radius. Given that 3/4 of the fuel is at more than half the radius from the axis, an effective radius of something larger than half the outside radius is conservative. Of course, you don’t like that because it makes the ecat look totally implausible.

  • john does

    Read through this and went through calculation and if the assumptions are correct this writeup is faultless. Terse analysis

  • Slad

    The exponential dependence is implicit in the argument that there could be thermal runaway in the system as designed. It’s the unstated basis behind their whole reasoning.

    • fact police

      The exponential dependence is implicit in the argument that there could be thermal runaway in the system as designed. It’s the unstated basis behind their whole reasoning.

      No. Thermal runaway does not require an exponential dependence. It only requires that the heat generated exceed the heat loss, and the rate of increase of heat generation with temperature is the same or higher than that of heat loss, whether linear or not.

      In your graph that condition is met everywhere (above the alleged turn-on) if the heater generates 13 W/m^2 or more, and is met above the intersection for no input heat at all. And at the operating point, if the power increased slightly, it would go into thermal runaway as well, if it was not immediately turned down. It’s always on knife-edge.

      Regarding your final paragraph, what happens if you design the heat transfer surface to loose four ‘units’?

      If it loses 4 units, then it will never reach the operating temperature. On your graph, the 5 W/m^2 from the heater at the operating point would not have been enough to reach the alleged turn-on point where the heat loss is already 13 W/m^2.

      And if you used 13 W/m^2 to reach that point, then it would have to be turned down by more than a factor of 2 to prevent thermal runaway at the operating temperature. The ecat input power for the Lugano experiment does not follow this description.

      Anyway, the more important point is that in the interest of eliminating the need for external input, it *could* be made to self-sustain, and if it could, they would do it.

      Answer – a stable and easily controllable reactor.

      No. It would actually be rather unstable, because a slight increase in heater power could send it into runaway, and that slope is unlikely to be stable, depending on the particular NAE density. If it changed suddenly, even turning off the heater would not prevent runaway.

      (There’s more detailed discussion of this on ECN, in the “Rossi Sells eCat” thread.)

      • Slad

        “Thermal runaway does not require an exponential dependence. It only requires that the heat generated exceed the heat loss”

        ** Well obviously this is true, but in real life, you would have to be pretty dumb to design a system that suffers from this issue.

        ” 5 W/m^2 from the heater at the operating point would not have been enough to reach the alleged turn-on point where the heat loss is already 13 W/m^2… …And if you used 13 W/m^2 to reach that point, then it would have to be turned down by more than a factor of 2 to prevent thermal runaway at the operating temperature.”

        ** Which is exactly how a normal heat control would work!

        As for the “discussion” on ECN, what you see there is a butt-hurt young astrophysicist trying to argue about thermodynamics and control systems with a doctor of mechanical engineering. It’s fairly amusing, however, it’s also unlikely to get a response, as the time required to fully explain his mistakes just takes too long. Plus… your man Mr Tan prefers to argue, rather than listen, which is a shame, as in this case, he would learn something.

        • fact police

          Do I attempt to lecture him on Cosmology?

          So, you’re suggesting we accept your arguments on your authority instead of your logic?

          I’m not in the habit of doing that. And it that’s what you want, why bother writing a paper? Since you did write a paper, it seems reasonable that objections to the logic be raised, and that you try to defend them.

          • Slad

            I’m suggesting you go read a Heat Transfer textbook instead of all those Cosmology ones.

          • fact police

            Slad wrote:

            I’m suggesting you go read a Heat Transfer textbook instead of all those Cosmology ones.

            I am not a cosmologist. And I have read thermodynamics texts, enough to know that you completely misuse the equation for transfer through a cylinder in applying to a fuel pellet as though all the fuel were concentrated at zero radius.

            The logic isn’t in question (if the assumptions are correct)

            Definitely the logic. It’s an illogical use of an equation in a situation it was not developed for.

            I suppose you also argue with your lawyer about the law, and with your doctor about medicine.

            If they make blatant errors in logic that I can understand, of course. You don’t?

            Again, you seem to telling us to trust you because you are a mechanical engineer. But I don’t because you make obvious errors that an undergraduate can spot.

          • Slad

            “you completely misuse the equation for transfer through a cylinder in applying to a fuel pellet as though all the fuel were concentrated at zero radius.”

            As I have explained to you already:

            “You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot. (In fact, the same very point I am having to repeatedly make to you)”

          • Slad

            An argument also cannot be made coherently if one is out of one’s depth

          • psi2u2

            Slad, somehow I find it difficult to take seriously someone called fact police. Police rarely work with facts, and pretty often their “facts” are deemed wrong or irrelevant by scientists, judges, and lawyers. I admit this is a prejudice. Sorry “fact police.” You may be correct, but you would be more effective under a less presumptively silly name.

          • Slad

            Police Chief, perhaps?

  • Slad

    Hello Josh?

    Nope, one is cylindrical, and one is a tube.

    And you are mixing up inner and outer radius when trying to suggest the effective radius of the fuel pellet would be zero.

    • fact police

      Slad wrote:

      Nope, one is cylinder, and one is a hollow cylinder. Look at the ‘heat flow through a hollow cylinder’ equation, and see how Q alters when the tube becomes a solid cylinder (ie. ‘r inner’ starts to get smaller, eventually becoming zero)

      Again, that equation applies to a tube in which all the heat is generated within the tube. But the heat in a fuel pellet is not generated on the axis in a cylinder with a zero radius. It is generated throughout the pellet, with more of near the rim than the axis, so there is an effective radius probably larger than 5 mm. That’s certainly consistent with the temperature profile within pellets.

      And 2cm dia. is pretty big for a fuel pellet. Got a reference for that?

      In the “nuclear fuel cycle” article in Wikipedia, they give a temperature profile for pellets with a 1 cm radius (2 cm diameter).

      The whole point is that similar temperature differences do not give similar heat flows,

      No, but similar temperature differences with similar thermal properties do. And these are similar, and if anything the conduction out of a pellet is *better* than for the ecat because much of the fuel is nearer to the rim.

      And you are mixing up inner and outer radius when trying to suggest the effective radius of the fuel pellet would be zero.

      No, I think you are. Your zero radius does not apply, because most of the heat is not generated at zero radius.

      • Slad

        No… The heat struggles the most to conduct away of the centre of the pellet. This is the hottest area, and the area that would melt first if the power density was increased.

        I wish I had seen the reference for 2cm, as it would have helped make an even stronger argument.

        “No, but similar temperature differences with similar thermal properties do. And these are similar, and if anything the conduction out of a pellet is *better* than for the ecat because much of the fuel is nearer to the rim.”

        ** a) The thermal properties are different, due to the shape, which I have repeatedly explained to you… The equation I quoted explains this.

        ** b) As I explained above, it’s what is happening at the centre of the MOX pellet that is the problem.

        “I think you are”

        It doesn’t matter what you ‘think’, because you are arguing against the science of thermodynamics and heat transfer… Not against me.

  • fact police

    Slad wrote:

    Assuming that the cooling system is not highly pressurized, this temperature is likely around 120-150ºC, as this allows the most efficient transfer of heat into boiling water, with a small safety margin. (120ºC is 70% of the ‘critical temperature’ on water’s pool-boiling curve, which corresponds to a heat transfer of 70 W/cm^2 at 1 atm.

    I thought this was about the Lugano experiment. There was no water cooling in that experiment.

    • Slad

      Please. It’s about the E-Cat, a point which everyone else seems to get. See *purposeful misunderstandings* below

      • fact police

        Slad wrote:

        Please. It’s about the E-Cat, a point which everyone else seems to get.

        It is about the ecat. But the objections related to thermodynamic implausibility arose from the Lugano configuration, so it doesn’t make sense to answer the objections by saying they don’t apply to a configuration completely different from Lugano.

        See *purposeful misunderstandings* below

        The only purposeful misunderstanding so far is your misunderstanding that I suggested heat flows into the ecat fuel.

        • Slad

          * Apart from the one you made above*

          If you can’t formulate a clear and unambiguous argument, then It shouldn’t surprise you if people misinterpret it.

          Your exact words are readable in the document above.

          The temperature difference across the alumina tube is easily calculated with the equation quoted in the report. The temperature profile of a mox pellet is easily searchable.

          • fact police

            Slad wrote:

            If you can’t formulate a clear and unambiguous argument, then It shouldn’t surprise you if people misinterpret it.

            No argument can be made so clear that someone won’t misinterpret it. But as I said, this was a purposeful misunderstanding, because it’s the only way you could dream up a reply of any kind.

          • Slad

            An argument also cannot be made clearly, if one is out of one’s depth

          • psi2u2

            Slad, somehow I find it difficult to take seriously someone called fact police. Police rarely work with facts, and pretty often their “facts” are deemed wrong or irrelevant by scientists, judges, and lawyers. I admit this is a prejudice. Sorry “fact police.” You may be correct, but you would be more effective under a less presumptively silly name.

          • Slad

            Police Chief, perhaps?

          • fact police

            It’s strange that no one seems bothered by the far more pompous and arrogant tone of the subject article, where Slad seems to think he is channeling Galileo. Given the many errors in the piece, an arrogant response seems entirely appropriate.

          • Slad

            Slad seems to think he is channeling Galileo

            If you practice hard enough, at some point, you may develop a sense of humour too…

            The only errors in the report are yours, as quoted verbatim.

          • fact police

            If you practice hard enough, at some point, you may develop a sense of humour too…

            You mean the kind people here exhibit?

            The only errors in the report are yours, as quoted verbatim.

            You were wrong about the need for exponential dependence for thermal runaway, or that anyone had assumed it.

            You were wrong that letting the inner radius go to zero provided a better or even realistic approximation to a cylinder with fuel distributed uniformly. Indeed, since more fuel is nearer the rim than in the Lugano configuration, letting the inner radius increase is a better approximation.

            You were wrong that your graph suggests thermal runaway won’t happen, and that it is a realistic representation of the ecat; particularly the Lugano ecat.

            You were wrong that the presence of water cooling in your description has any relevance to the thermodynamic objections to the Lugano configuration.

            You were wrong about the fuel in ecats being 30 cm square, particularly in the configuration relevant to the thermodynamic objections you were trying to refute.

            You were also wrong about Fukushima and forced cooling as described in more detail on ECN.

            In short, there was little, if anything, you were right about.

          • Slad

            This is a collection of your misinforming opinions, stated as fact. The idea being that if you throw enough muck, some must stick. By tirelessly arguing and expounding on these made up points, you are seeking to win the argument only by wearing the responder out.

            Indeed, it’s your usual M.O. and the reason I included the Robert Kiyosaki quote at the top of the report.

            We have already covered all these points either in the report, or below.

            And please keep the comments coming at the Circular Temple of Onan that is ECN:

            If I am bored one evening, I might actually read them, then select a few more prime cuts of your’s and Hody’s weird ranting, then write another report, for whomever’s amusement…

            “Principia Skeptopathia” maybe?

          • fact police

            Slad wrote:

            This is a collection of your misinforming opinions, stated as fact.

            No. Those are not opinions. They are interpretations, and far more justified than yours.

            You agreed that zero radius in your formula did not describe the fuel pellet, yet you keep on referring to it. It is self-evident that the fuel in a fuel pellet is much less insulated from the surrounding coolant, being on average much closer, *and* the temperature of the surrounding coolant is much lower. Therefore, the Lugano ecat is thermodynamically implausible by comparison.

            You agreed that linear reaction dependence can produce thermal runaway. Indeed, it’s the rate of change that’s important (the slope), and how it compares to the cooling slope, more than the nature of the function that produces conditions for thermal runaway. What you said about the exponential dependence was just wrong in the first place, and the idea that anyone made that assumption manifestly so.

            And it is also clear that your introduction of water cooling and the geometry of the fuel described in the patent and the forced cooling at Fukushima are totally irrelevant to the objections you were trying to refute, and therefore amount to some sort of smoke screed. How you got water cooling in your analysis in reference to Lugano is anyone’s guess.

            Your paper does not succeed at either of its main goals, as I understand them. It does not justify the use of heat to control an exothermic reaction, or explain why self-sustained operation would not be (1) possible, (2) safe, or (3) highly desirable, if the claims reported were real. Nor does it make the 100 times higher power density of the Lugano fuel compared to fission reactor fuel plausible (without melting) by direct comparison of the fuel geometries.

          • Slad

            I think it’s time for your milk and cookies

          • psi2u2

            I will leave you and Slad to debate the science. Its your handle I was commenting on. I did not personally find the tone pompous, but I’m just a lit. prof. Sorry, but your chosen name evokes a prejudice in me. Police logic is the reasoning of stage 4 in the Kohlberg typology. Sometimes police have valid facts and tell the truth. In my experience, many do not, most of the time. Consider that the primary reason we are still arresting people for cannabis in this country is that police departments are funding vacations and buying Humvees with the stolen proceeds. Yikes.

          • fact police

            psi2u2 wrote:

            Its your handle I was commenting on. I did not personally find the tone pompous, but I’m just a lit. prof. Sorry, but your chosen name evokes a prejudice in me. Police logic is the reasoning of stage 4 in the Kohlberg typology. Sometimes police have valid facts and tell the truth. In my experience, many do not, most of the time. …

            A literature professor so constrained by prejudice that he can’t appreciate a generic, informal, and light use of a word, is not part of my target audience. The American Heritage dictionary recognizes this informal usage as a group that admonishes, cautions, or reminds. Presumably Joan Rivers also evoked prejudice from you instead of the intended levity with her fashion police.

            Maybe your department has a refresher course you could enroll in. If you do, pay particular attention to the idea of context, and maybe you can slay that prejudice and learn to keep an open mind.

        • GreenWin

          Wait. Is “fact police” in fact… MY or JoshCude???

          • ‘fact police’ is the self-admitted (http://ecatnews.com/?p=2655#comments ) ECW incarnation of ECN’s resident and incredibly verbose adherent of scientism, ‘popeye’, aka elsewhere as ‘ScienceApologist’ and ‘joshua cude’.

            Abd ul-Rahman Lomax identifies ‘cude’ as Joshua P. Schroeder (http://archive.is/jSp9B ). This and other interesting insights about the ECN trolls here: https://groups.yahoo.com/neo/groups/newvortex/conversations/topics/361

          • bachcole

            A change of a label/handle is not a change of attitude. Until I see a change of attitude, I will continue to not even read their junk.

          • fact police

            Fact police is the self-admitted popeye and joshua cude. I have not admitted to being, nor am I, ScienceApologist or Joshua Schroeder. Lomax is mistaken about that.

  • Slad

    Please. It’s about the E-Cat, a point which everyone else seems to get. See *purposeful misunderstandings* below

  • Slad

    Listen. It’s my graph… I could easily have drawn it so it wouldn’t ever have run away, I was going to make a further point based on that, but took it out to keep things *relatively* concise

  • Slad

    Nearly right… It is a like weighted average, but because your theoretical outer hollow cylinder of the fuel pellet is also generating heat, the temperature profile going towards the centre climbs much faster than you are allowing for.

    • john

      Slad i am very impressed .I was going to keep quiet and respond but you explaines all the key points.. Be alite less heavy on fact police ..hes only learning and its a good things…

      • Slad

        I find it annoying having to repeat myself.

        Feel free to take over my friend.

        • Slad

          And only a troll calls themself ‘fact police’

      • Slad

        As I had suspected: ‘fact police’ is actually one of the fine fellows who inspired the article in the first place…. Hopefully Mr Hody and Mr Clarke might soon join our discussion here too?

        • doug marker

          FWICT – Fact police is actually George Hody. (Mary Yugo, Al Potenza etc: etc: etc: etc: ) – I may be wrong but I doubt it. Joshua Schroeder (IMHO) has a far more condescending and arrogant style. Hody, when it suits, can be polite but among ‘friends’ (the pack), he is not shy about what he really thinks and the insults/innuendo/assasinations flow freely.

          Doug

          • Slad

            I also doubted it was ‘Tan nee Schroeder’ at first, as he generally comes across as being more intelligent, but he eventually admitted it above. I think his style has to be toned down here to pass moderation, but the arrogance still comes through IMO.

  • Slad

    If you can’t formulate a clear and unambiguous argument, then It shouldn’t surprise you if people misinterpret it.

  • Slad

    Of course it supports my case.

  • GreenWin

    Wait. Is “fact police” in fact… MY or JoshCude???

    • Agaricus

      ‘fact police’ is the self-admitted ECW incarnation of ECN’s incredibly verbose bigot, ‘popeye’, aka ‘joshua cude’: http://ecatnews.com/?p=2655#comments.

      Abd ul-Rahman Lomax thinks ‘cude’ is actually Joshua P. Schroeder. This and other interesting insights about the ECN trolls here: https://groups.yahoo.com/neo/groups/newvortex/conversations/topics/361

    • Slad

      Report updated to clarify some continued misunderstandings…

      • Slad

        And I will probably do so again, after this beauty:

        Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat (fact police)

        Oh dear. It seems you are mixing the axes of my graph up.

        • fact police

          Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat… (‘fact’ police)

          Oh dear. It seems you are mixing the axes of my graph up.

          No, I have the axes straight. Your mocking would be more effective if you revealed the logic that led to the idea that I might be mixing them up. But I’m not. See my other reply for details.

          Please pay more attention to the graphs you quote;

          You might consider this advice yourself. Your graph was intended to show immunity from runaway, but failed. If you modified according to my advice, you would have what you were looking for.

          this is even worse than that “20mm fuel rod” graph from the other day.

          …which did not affect the argument at all, as shown explicitly now using the appropriate formula. And you still haven’t explained why you think I’ve confused your axes.

          Did you manage to answer my three questions regarding the “more complex bits” yet?

          These questions were answered before you asked them. All you have to do is read. And it’s not because of the geometry.

          • Slad

            But anyway, let’s avoid the excess verbosity…
            Did you manage to answer my three questions regarding the “more complex bits” yet?

          • Slad

            In fact, you are right, you didn’t mix the axes up. My bad.

            It wasn’t such as an egregious error compared to the ’20mm fuel rod’ stuff after all.

            So yes, you could increase the slope of the cooling graph, to increase it’s stability at high temperature, but then, I like my conjecture about self-sustain mode, which is why the graph is drawn how it is.

            Regarding your “appropriate formula”… Have you adapted it to a hollow tube yet, for the sake of a comparison?

          • Slad

            And I agree with you that:

            These questions were answered before you asked them. All you have to do is read.

            The answer to the question, about the temperature across the alumina, is “~200C”. It is very simple to calculate.

            So why do you also say:

            the temperature difference needed should be half of the value in the pellet, or about 600C across the ceramic wall

            Which is a quote from yourself, that was included in the original version of the report. All this discussion, of shape, and effective radius, revolves around that single statement. Do you now agree that 600C is wrong?

          • Slad

            What you calculated in the paper was the temperature difference to the coil (assumed to be at 7 mm radius), which was 92C. But these are quibbles.

            You just didn’t read down far enough, that’s just an intermediate stage in the calc.

            OK I see where the 600C comes from… I assume full contact with the inner surface. Short and sweet.

            So lets say: The LiAlH bubbles up at 700C or so, and pushes the nickle powder out towards the the inner face of the tube, which it then clings to, due to surface tension forces. Full contact restored.

            Or does that also sound completely implausible to you?

          • Slad

            I’m not talking about heating water here (except as a measure of the overall power transfer). I’m talking about heating the pot with the element.

            No, the surfaces I’m considering were in contact. And both are pretty flat, so the percentage contact would be pretty high,

            Flatness is a very relative thing in heat transfer. Unless you’ve polished your surfaces, a slight roughness can matter a lot.
            Like ‘factor of 100’ a lot.

            That’s the point about the fine nickel particles, they can contact a rough surface as if two highly polished (~5um) surfaces are in contact.

            And your hob was likely dry. Liquid metals at the interface in the E-Cat.

          • fact police

            Slad wrote:

            What you are saying above, about power per unit length, is the same point I made in my original report.

            I don’t recall any mention of it in your report.

            (I also calculate the thermal conductivity of the nickel as being higher). If we restore the full surface area, my point still stands.

            Yes, I remember that, but it was the transfer of heat through the ceramic that was the issue. The fuel conductivity is lower by a little more than a factor of 2 compared to alumina.

            If we restore the full surface area, then the temperature of the *surface* is not implausible based on a comparison to fission fuel. But that point stands not because of shape differences, but because of linear power density differences. The question of the fuel temperature is still implausible to me.

            If you don’t mind me saying, it is not easy to pick out the bones of your arguments, due to the lengthy surrounding material, and bombardment of several ideas that they often contain (a counter example being my short summation above) You risk a reader skimming over them. Yes, the radius arguments were a blind alley, but I thought that was the only reason you could arrive at the apparently nonsensical idea of a 600C gradient in the alumina tube.

            If you don’t mind my saying so, it seems you don’t actually make the necessary effort to consider what I’ve written. You admitted that yourself, when you said you don’t pay attention to what I write. And that was abundantly clear when you repeated questions I had answered several times already.

            And the ultimate point, that shape matters (to some degree), is one that I don’t believe you can completely disagree with,

            Obviously shape matters to a degree, but the point was the difference in the fuel shape *reduced* the thermal resistance for the fuel pellet compared to the ecat geometry — it did not increase it. And this was reasonably plausible from simple considerations, and verified by quantitative analysis.

    • Slad

      I think it’s time for your milk and cookie

  • Slad

    He goes by the name of Popeye, you’d like him, you share a lot of traits.

    And, as I have explained to you three times now:

    “You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot.”

  • Slad
  • Slad

    I ‘believe’ you are an astrophysicist because your twitter account is ‘astrophysically’ and you studied astrophysics. The reasons I beliveve this can be found a few posts below.

  • Slad

    You left an indelible slug trail all over wikipedia! Same style. Same arguments. Same atitiude. Same topics of interest. Unbroken timeline between the various pseudonyms. Stand tall and proud man!

  • Slad

    So you think heat is conducted instantaneously?

    • Slad

      OK thanks, I meant 1240W. The 200C remains unchanged.

      So, you are saying that you prefer to guesstimate the temperature difference of the alumina as 600C, by comparing it to a fuel pellet, despite the fact that a simple standard equation can calculate this value much more accurately?

      Yes or no?

  • Slad

    If you choose to dodge the question, that’s OK, most people probably would too. Just don’t pretend you don’t understand the concept, seeing as how you are such an expert in heat transfer and all. Like I said above, it is hard to formulate a coherent argument, when one is out of one’s depth.

    Transient heat flow, Google it.

    Another thing I quoted is that “if you argue with an idiot, there are two idiots”. Now I don’t believe you are an idiot, but you are incredibly arrogant.

    It’s a waste of my time talking to someone who only wants to argue, when it is very apparent (to me at least) that you could learn something, if only you would choose to.

    • fact police

      Slad wrote:

      Just don’t pretend you don’t understand the concept, seeing as how you are such an expert in heat transfer and all. …

      Transient heat flow, Google it.

      I didn’t say I don’t understand the concepts. I said your use of the terms made no sense. It makes no sense to say that there must be a transient temperature gradient or the heat conduction would have to be instantaneous. The other sentence was simply irrelevant.

      Like I said above, it is hard to formulate a coherent argument, when one is out of one’s depth.

      You have provided ample proof of that.

      Now I don’t believe you are an idiot, but you are incredibly arrogant.

      It’s a waste of my time talking to someone who only wants to argue, when it is very apparent (to me at least) that you could learn something, if only you would choose to.

      That sounds kind of arrogant itself, but in your case, it’s not justified.

      • Slad

        “It makes no sense to say that there must be a transient temperature gradient or the heat conduction would have to be instantaneous. The other sentence was simply irrelevant.”

        * * No, I was pointing out two daft things that you said.

  • Slad

    I’m starting to understand why you have been banned from Wikipedia so many times.

  • Slad

    Zing!

  • GreenWin

    WARNING: 2fp/1my = 1/2potenza + 1/2hoady + spurious radiation. Danger Will Robinson!

    • Slad

      Hot air flow calorimeter suggests major excess heat.

  • GreenWin

    WARNING: 2fp/1my = 1/2potenza + 1/2hoady + spurious radiation. Danger Will Robinson!

    • Slad

      Hot air flow calorimeter shows excess heat!

  • Slad

    If you practice hard enough, at some point, you may develop a sense of humour too…

    The only errors in the report are yours, as quoted verbatim.

    • fact police

      If you practice hard enough, at some point, you may develop a sense of humour too…

      You mean the kind people here exhibit?

      The only errors in the report are yours, as quoted verbatim.

      You were wrong about the need for exponential dependence for thermal runaway, or that anyone had assumed it.

      You were wrong that letting the inner radius go to zero provided a better or even realistic approximation to a cylinder with fuel distributed uniformly. Indeed, since more fuel is nearer the rim than in the Lugano configuration, letting the inner radius increase is a better approximation.

      You were wrong that your graph suggests thermal runaway won’t happen, and that it is a realistic representation of the ecat; particularly the Lugano ecat.

      You were wrong that the presence of water cooling in your description has any relevance to the thermodynamic objections to the Lugano configuration.

      You were wrong about the fuel in ecats being 30 cm square, particularly in the configuration relevant to the thermodynamic objections you were trying to refute.

      You were also wrong about Fukushima and forced cooling as described in more detail on ECN.

      In short, there was little, if anything, you were right about.

      • Slad

        This is a collection of your opinions, stated as fact. The idea being that if you throw enough muck, some must stick. By tirelessly arguing and expounding these made up points, you are seeking to win the argument only by wearing the responder out.

        Indeed, it’s your usual M.O. and the reason I included the Robert Kiyosaki quote at the top of the report.

        We have already covered all these points either in the report, or these comments.

  • Slad

    Only if operated it near the runaway point of 1366C.

    • fact police

      Wrong. With external input heat of 13 W/m^2, it would runaway at any temperature above your alleged turn-on point.

      • Slad

        Oh dear. Are you really pretending not to understand the concept of a proportional heat controller?

        • fact police

          Slad wrote:

          Oh dear. Are you really pretending not to understand the concept of a proportional heat controller?

          No. But you talked a lot about an entirely passive system. Now you’ve changed your paper to require a robust and responsive control system, when elsewhere you argue that the need for a control system for self-sustained mode is too dangerous.

          I didn’t say it *would* runaway below the intersection. I said it was vulnerable to it. If the control system is not sufficiently robust or responsive, from your operating point, a mere 5 W/m^2 decreasing too slowly would cause thermal runaway. Or if the output spiked because of diffusion of the hydrogen to an anomalously high density of NAEs, simply turning the heater off would not help.

          The use of a proportional controller for a simple heater is quite different. Regardless of excursions above the set temperature, reducing power *at any rate* will always bring it back. Not so, with an ecat following your graph. And it doesn’t have to exceed 1366C either. As the temperature begins to increase, the lines converge and less heat is needed to exceed the rate of dissipation. If this convergence is faster than the decrease in the input, you get runaway.

          This kind of knife-edge control is bad design, and it’s not necessary. What you want is a system in which the rate of cooling increases faster from the operating point than the reaction rate, to provide a stable point. That’s what stabilizes the sun, and many other exothermic reactions. And in the ecat theory, presumably, the location of this stable point is adjustable by changing the input heater power.

          The problem is you can’t get it with linear cooling and linear reaction rates. What’s more, with linear dependencies, as you’ve shown them, the operation would be very different from the way the Lugano ecat was in fact operated. It would have to be heated much more to get it started than to keep it going, whereas with the Lugano ecat, the heat was simply ramped up. So your speculated dependencies are not in evidence from the Lugano data.

          A year ago, after the release of Lugano, Paul Stout and I hashed over this in some detail. Paul actually contrived a cooling dependence and a reaction rate dependence that fit the Lugano operating procedure, and that would give stable operation determined by the input power, and could never runaway. But the dependencies were highly non-linear, and also highly unrealistic.

          But more important than whether such a system could be contrived, is the question of why anyone would do it. The important point is that if one unit of energy (from the outside) can initiate the reaction, then 2 or 3 units from within can be made to sustain it. That means that with suitable insulation and controlled cooling, it could run without input power, which aside from being far more practical, and useful for making electricity, would make for one hell of a demonstration.

          • Slad

            Oh, you are… OK

          • fact police

            Slad wrote:

            And you seem to be working off the assumption that things are non-linear.

            No, I’m not making any assumptions. I’m just interpreting yours, since you don’t seem to understand your own graph. The point is your graph does not fit the behavior reported for the Lugano ecat.

            Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat, and it would quench without input, and never run away. Then the risk would simply involve the unpredictability of the reaction slope. A higher than expected slope would still produce runaway.

            But, as it happens, in the Lugano ecat, most of the heat loss from the device is by radiation, and that *is* highly non-linear, so at least to that extent, your graph is not realistic.

            And again, the more important question is why bother trying to control an exothermic reaction with heat. The important point is that if one unit of energy (from the outside) can initiate the reaction, then 2 or 3 units from within can be made to sustain it. That means that with suitable insulation and controlled cooling, it could run without input power, which aside from being far more practical, and useful for making electricity, would make for one hell of a demonstration. It’s inconceivable that they wouldn’t do that if it were possible. And if the claims have validity, it would be possible.

            And again, the

            Read the report, my views are clear.

            Unfortunately, it’s the logic that is muddled.

          • Slad

            I’ll just paste what you actually said above: “But the dependencies were highly non-linear”

            Now you say, you didn’t make that assumption. What?

      • Slad

        Please don’t pretend you didn’t mix the axes up.

      • Slad

        But the argument is not valid, as shown in detail in another post.

        So you claim It’s just that, to me, you have very little credibility.

        • fact police

          The argument does not depend on my credibility. Just verifiable information and simple logic.

          • Slad

            Maybe, but how much attention I pay to your arguments does depend on it.

          • fact police

            I don’t mind if you don’t pay attention to my arguments. But simple courtesy would suggest that you pay attention to them before you respond to them.

          • Slad

            You are the last person who should ever be lecturing anyone regarding courtesy or manners.

        • bachcole

          For me:

          1. Anyone with a handle “fact police”

          2. Anyone has denies LENR

          has not credibility.

  • Slad

    I’m not “admitting” anything. As I have said, for the fourth time now:

    “You are right to say It is incorrect to use that equation to analyse a fuel pellet… The “r tending toward zero” section is merely there to help illustrate a point that shape matters a lot.”

    Just google fuel rod temperature profile, rather than trying to reinvent the wheel.

    …But please try not to let any more odd graphs on Wikipedia fool you.

    • fact police

      Slad wrote:

      The point is that ‘effective wall thickness’ is dependent on shape, not on thickness and/or radius, which become dimensionless. The fuel pellet hasa much larger effective radius, which drastically limits it’s heat transfer.

      Slad, you’re not making any sense. Of course it depends on thickness and radius. You’re still hung up on your equation with zero inner radius, which says nothing about this situation, because all the fuel is supposed to within that radius, and it’s *not*.

      The pellet does not have a much larger effective radius. The outer radius is simply the outer radius. The effective inner radius for your formula is *increased* (beyond say the 0.4 ro in Lugano) because much more of the fuel is outside 0.4 ro, and therefore closer to the coolant, with less thermal insulation in between.

      Look, heat from the reaction has to flow through ceramic to the coolant. In a fuel pellet, there is *less* ceramic it has to flow through, and yet the temperature difference is 10 times higher. The claim that you can dissipate the same power through *more* ceramic with a *smaller* temperature difference is just plain nonsense.

      • Slad

        You are trying to argue against a mathematical equation. This is insanity. No wonder it is very easy to write reports that mock you.

  • Slad

    Hey up! The Inquisition has arrived!

    Report Updated …To clarify some continued misunderstandings.

    Double Updated! …My own misunderstandings apparently. Just like the olden days.

    • Slad

      And, I will probably have to do so again, after receiving this latest beauty today:

      Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat… (‘fact’ police)

      Oh dear. It seems you are mixing the axes of my graph up.

      Did you base half your arguments over the last five days on this reasoning?
      No wonder we are still struggling with the more complex bits.

      And please pay more attention to the graphs you quote: This is even worse than the ’20mm fuel rod’ foolery you subjected me to, a couple of days ago.

      But anyway, let’s avoid the excess verbosity
      Did you manage to answer my three questions regarding the “more complex bits” yet?

      • fact police

        Your graph would actually serve your purpose better if the slope of the reaction were *less* steep than the cooling curve. That would produce a stable operating point dependent on the input heat… (‘fact’ police)

        Oh dear. It seems you are mixing the axes of my graph up.

        No, I have the axes straight. Your mocking would be more effective if you revealed the logic that led to the idea that I might be mixing them up. But I’m not. See my other reply for details.

        Please pay more attention to the graphs you quote;

        You might consider this advice yourself. Your graph was intended to show immunity from runaway, but failed. If you modified according to my advice, you would have what you were looking for.

        this is even worse than that “20mm fuel rod” graph from the other day.

        …which did not affect the argument at all, as shown explicitly now using the appropriate formula. And you still haven’t explained why you think I’ve confused your axes.

        Did you manage to answer my three questions regarding the “more complex bits” yet?

        These questions were answered before you asked them. All you have to do is read. And it’s not because of the geometry.

        • Slad

          In fact, you are right, you (probably) didn’t mix the axes up. My bad. No more updates for now.

          It wasn’t such as an egregious error compared to the ’20mm fuel rod’ stuff after all. So, forgive me if I’m still a little dismissive of you.

          So yes, you could increase the slope of the cooling line, to increase it’s stability at high temperature, but then, I like my conjecture about self-sustain mode, which is why the graph is drawn how it is.

          • Slad

            And I agree with you that:

            These questions were answered before you asked them. All you have to do is read.

            In fact, the answer to the question, about the temperature across the alumina wall, is “~200C”. It is very simple to calculate.

            So why do you also say:

            the temperature difference needed should be half of the value in the pellet, or about 600C across the ceramic wall

            Which is a quote from yourself, that was included in the original version of the report. All this discussion, of shape, and effective radius, revolves around that single statement (and the weird sentence preceding it).
            Do you now agree that 600C is wrong?

          • fact police

            Slad wrote:

            The answer to the question, about the temperature across the alumina, is “~200C”. It is very simple to calculate.

            That’s for ~ 2 kW generated inside the tube, which is what the authors claimed based on 1400C on the outside, and that would cause the nickel to melt.

            For your revised scenario, the fuel only produces about 1 kW, so the delta T would be more like ~100 C, based on the formula.

            So why do you also say:

            the temperature difference needed should be half of the value in the pellet, or about 600C across the ceramic wall

            That’s all explained in the essay that starts out, “Now, since I’m such a nice guy”. You should read it.

            The 600C was based on an assumption of similar power per unit length in the fuel pellet, and the similarity in the thermodynamic properties. The thermodynamic properties *are* similar, based on geometry, but the power per unit length in the fuel pellet is much higher than than in the ecat.

            All this discussion, of shape, and effective radius, revolves around that single statement. Do you now agree that 600C is wrong?

            No. the discussion of shape and effective radius was right as shown explicitly. Like you yourself said, you’re not arguing against me — your arguing against math. It was the power in the fuel pellet that was underestimated. The power density is more like 60 W/m^2.

            It seems I have to argue both sides of this, because you’re so hung up on the geometry, and you’ve quite simply got that wrong.

          • Slad

            Ignore the revised scenario, it has no place here. This is based solely on your response to the first report, as merely quoted in the second report.

            The heat from the fuel in my original model was 2140W. I also add the ~900W heat from the coils, at half wall thickness. This equates to a ~200C difference between the faces of the tube. This is found by using a simple equation, itself found in every heat transfer textbook.

            Why do you disagree, by saying this temperature should be 600C?

            Please be concise, it’s a simple question. We can come back to the effective radius bit later.

          • fact police

            Slad wrote:

            The heat from the fuel in my original model was 2140W. I also add the ~900W heat from the coils, at half wall thickness. This equates to a ~200C difference between the inside and outside of the tube. This is found by using a simple equation, itself found in every heat transfer textbook.

            I actually quote this equation itself in the first report.

            Not that it makes much difference, but let’s get it straight.

            In the first report, which I have, you quote Qfuel = 1240 W and Qout = 2150W, using the model based on the GSVIT/MFMP re-calculations. That’s what gave rise to July discussions you quoted from.

            If you use 2140 W from the Lugano report itself, then that corresponds to 1400C on the outside of the alumina cylinder. When you add the 200C across the alumina, you’re already above the melting point of nickel, even without considering the flow from the fuel to the alumina. That’s why you said in your first report that based on data reported in Lugano, “the nickel particles would in my opinion definitely have melted”. That’s the end of discussion, because that’s the thermodynamic inconsistency I raised from the beginning.

            Why do you disagree, by saying this temperature should be 600C?

            I answered this in the previous reply. All you have to do is read it. The 600C was based on an assumption of similar power per unit length in the fuel pellet, and the similarity in the thermodynamic properties. The thermodynamic properties *are* similar, based on geometry, but the power per unit length in the fuel pellet turns out to be much higher than than in the ecat.

          • Slad

            OK thanks, I meant 1240W. The 200C remains unchanged.

            So, you are saying that you prefer to guesstimate the temperature difference of the alumina as 600C, by comparing it to a fuel pellet?

            Despite the fact that a simple standard equation can calculate this value much more accurately?

            Yes or no? Please be concise. I don’t want to have to find and pick apart some excessively long post, containing much extraneous information, many screens below.

          • fact police

            Slad wrote:

            OK thanks, I meant 1240W. The 200C remains unchanged.

            Well, for 1240 through the alumina ignoring the heater requires 150C. What you calculated in the paper was the temperature difference to the coil (assumed to be at 7 mm radius), which was 92C. But these are quibbles.

            So, you are saying that you prefer to guesstimate the temperature difference as 600C, by comparing it to a fuel pellet,

            No. You really should read what I wrote. I was serious when I said I was being a nice guy, and giving you a better argument. I like things to give the same answer regardless of the approach. It gives confidence in the result. That’s why I was puzzled back in July when the comparison to fuel pellets gave a different result than the heat loss equation for transfer through a hollow cylinder, which should be straightforward.

            And if you read what I wrote, you’d see that I found the discrepancy. It was not your equation or the result and had nothing to do with shape. It was an incorrect assumption of power density in the fuel pellet, based on a global average power density in a reactor, instead of the local density for a given temperature gradient (which I found in the reference provided). When the correct power per unit length is used, which is 5 times higher than alleged in the ecat, the comparison (using similar thermal properties) gives the same answer.

            You’ve been so focused on proving your crazy ideas about shape and zero radius that you missed the real reason the gradient is higher in a fuel pellet. The power per unit length is higher. Simple as that. The thermal properties are in fact quite similar.

            So, in case you’re still too daft to understand what I’m saying, comparison to fission fuel pellets does not make the inner surface of the revised lugano ecat (GSVIT) higher than the melting point of nickel. It just puts it in spitting distance. But that doesn’t change the implausibility of requiring the fuel itself to be much above the melting point to transfer its heat *to* the surface at a power higher than 1 kW from one gram of nickel.

          • Slad

            What you calculated in the paper was the temperature difference to the coil (assumed to be at 7 mm radius), which was 92C. But these are quibbles.

            You just didn’t read down far enough, that’s just an intermediate stage in the calc.

            OK, I see where the 600C comes from: I assume contact between the nickel powder and the whole inner surface. There, short and sweet…

            I don’t agree that your way of calculating the 600C is a good idea, but short of some proper FEA modelling, it is reasonable.

            So lets say: The LiAlH bubbles up at 700C or so, and pushes the nickel powder out towards the the inner face of the tube, which it then clings to due to surface tension forces. Full contact restored.

            Or does that also sound completely implausible to you?

            What you are saying above, about power per unit length, is the same point I made in my original report. (I also calculate the thermal conductivity of the nickel as being higher). If we restore the full surface area, my point still stands.

            If you don’t mind me saying, it is not easy to pick out the bones of your arguments, due to the lengthy surrounding material, and bombardment of several ideas that they often contain (a counter example being my short summation above) You risk a reader skimming over them. Yes, the radius arguments were a blind alley, but I thought that was the only reason you could arrive at the apparently nonsensical idea of a 600C gradient in the alumina tube.

            And the ultimate point, that shape matters (to some degree), is one that I don’t believe you can completely disagree with, however, I imagine this section will be stripped from the report in due course.

          • fact police

            Slad wrote:

            OK I see where the 600C comes from… I assume full contact with the inner surface. Short and sweet.

            No, I don’t think you do. The 600C in that quote came from comparison with fuel pellets with an underestimated power density, and it was assuming full contact with the inner surface. With the higher power density, the estimated temperature by comparison is in line with the calculation from the formula, and is therefore not a source of implausibility.

            The contact area, however, *is* reason for the implausibility of the inner wall temperature, and the necessary of heat from the fuel to the wall is another.

            So lets say: The LiAlH bubbles up at 700C or so, and pushes the nickle powder out towards the the inner face of the tube, which it then clings to, due to surface tension forces. Full contact restored.

            Or does that also sound completely implausible to you?

            You guessed it. Implausible for reasons already argued. According to one of the appendices of the Lugano report, and quoted in the main body of the report, the mass fraction of the Li is only 1.2% at the beginning and only 0.03% at the end. The rest of it presumably separated from the fuel for some reason — possibly leaked out. Even with a density lower by a factor 16, the volume would only be one fifth of the nickel at the start (one 200th at the end). And at 700C, there may be melting, but not boiling. That means that there is simply not enough to coat the inner surface.

            Anyway, this is all in the category of over-interpretation for an experiment that is not even regarded seriously by most of Rossi’s advocates anymore. Until he can produce accessible evidence that the ecat even works, this is all just mental gymnastics.

          • Slad

            The 100x power density is a (paraphrased) quote from yourself, that’s why the original report uses it. Underestimated or not, it doesn’t matter to me, I’m only aiming for ballpark figures, in order to assess plausibility.

            the mass fraction of the Li is only 1.2% at the beginning and only 0.03% at the end.

            From a tiny sample. Maybe it’s all stuck to the tube wall? My report assumed 10% Li, and similar amounts are mentioned in rossi’s patent.

            Maybe you are right about over interpretation, but the two main arguments about the credibility of Lugano are based on Thomas Clarkes notions*, and what essentially amounts to slurs against Rossi or the researchers.

            *Which I covered, in a somewhat provocative fashion, and I’m interested to hear a response to.

          • fact police

            Slad wrote:

            The 100x power density is a (paraphrased) quote from yourself, that’s why the original report uses it.

            Yes, I know. The power density I first estimated came from the total fuel in a reactor and the total power output. But the actual local power density within an active pellet at 1200C is much higher than this, by a factor of 6 or so. And the power per unit length along a pellet (or rod made up of pellets end to end) is around 5 times higher than the alleged power density in the Lugano ecat, because the fuel is more concentrated.

            And that additional power explains why the temperature difference in a pellet is larger than the calculated difference for the ecat using the alleged ecat power. So, based on a comparison to fuel pellets, the temperature of the inside surface of the alumina tube (about 1300C) is not implausible with the alleged power from the fuel in your calculations, *if* you assume uniform temperature over the entire surface. The temperature required for partial contact and the temperature needed for the fuel itself to deliver that power level however is implausible without melting the nickel.

            the mass fraction of the Li is only 1.2% at the beginning and only 0.03% at the end.

            From a tiny sample. Maybe it’s all stuck to the tubes walls? My report assumed 10% Li, and similar amounts are mentioned in rossi’s patent.

            I would think a measurement is better than an assumption, and the fuel before insertion should be homogenous. (The patent is for a completely different geometry, and doesn’t even claim LENR, so I regard it as irrelevant.) Moreover, if the picture you described in the first report were accurate, the Li would be infused with the Ni, so a sample of used fuel should show the proper amount of Li, especially with a bulk technique like ICP MS.

            Maybe you are right about over interpretation, but the two main arguments about the credibility of Lugano are based on Thomas Clarkes notions*, and what essentially amounts to slurs against Rossi or the researchers.

            When there are openly voiced suspicions about a claim, the purpose of a validation is to dispel them. That is to exclude the possibility that deception may be involved, whether such suspicions represent slurs or not. The paper fails to dispel such suspicions.

            The credibility of Lugano was seriously in doubt among many advocates (and all skeptics) even before Clark’s paper. And for me, just the fact that it can *have* different interpretations that vary by as much as a factor of 3 in the COP is one reason among many that the report is not useful as a validation of the ecat. Others include (1) that the evidence is not accessible, and must be accepted on faith, and (2) (as mentioned above) that even if you accept the observations, deceptions on the input are not properly excluded, and (3) that the paper contains clearcut inconsistencies by a factor of 3, which are in fact explainable by one possible deception on the input, (4) the thermodynamic implausibility even you agree with using the original data, but I argue is present even with the revised data, and (5) the use of heat to control an exothermic reaction, when if the claims had validity, self-sustain mode would be easy to implement and far far more persuasive.

          • Slad

            TLDR. You are entitled to your own opinions, and these obviously differ from mine. Your replies have been noted, and in the cases where you have been able to make a cohesive argument that I was wrong, I have now edited the report to reflect this.

            As for the rest, I disagree with nearly everything you have said, but I will not be drawn into any further never-ending discussions, as frankly, it takes up too much time, and the lack clarity in your writing just allows you more wriggle-room in your arguments:

            The kettle analogy was a prime example of this. I honestly cannot tell if you changed your position halfway through the discussion, as your original statement was so unclear.

            Maybe you should write your own report, in order to help define your ideas concisely, and in an unambiguous way.

            Robert Kiyosaki was right.

          • fact police

            Slad wrote:

            TLDR.

            Again, I don’t mind if you don’t read what I write, but none of my posts are as long as your paper, and they are much more logical. I would just appreciate the courtesy of your not responding to what you don’t read.

            As for the rest, I disagree with nearly everything you have said,

            You mean the parts you bothered reading, and took the time to understand, I presume.

            I similarly disagree with the essential points from your paper, and you have failed to defend them here. In case you’re curious, there are additional disagreements we haven’t covered over at the other place.

            The kettle analogy was a prime example of this. I honestly cannot tell if you changed your position halfway through the discussion, as your original statement was so unclear.

            There was no change in position. I just simplified the analogy to make it more relevant. Transferring more than 1 kW from a gram of fuel to a surface already at 1300C is almost certain to put the fuel above its melting point at 1455C.

            Maybe you should write your own report, in order to help define your ideas concisely, and in an unambiguous way.

            Like I said, this is all gross over-interpretation, and as such, little more than mental gymnastics. I like gymnastics, but really dislike marathons. But if you write another paper, I’ll be delighted to tell you what’s wrong with it.

          • Slad

            Transferring more than 1 kW from a gram of fuel to a surface already at 1300C is almost certain to put the fuel above its melting point at 1455C.

            I think the opposite is entirely plausible, especially if the whole inner surface of the tube is considered i.e. if the foaming LiAlH moves the powder. However, I acknowledge your disagreement, and I would be interested to see some numbers.

            I’m sorry to labour the point, but direct and concise communication is a good thing. Especially if I am at work… Mainly because it avoids misunderstandings, but also so your main points are not obfuscated, which can only be a good thing.

            And any final report will be likely be written after this sorry saga is over, and will be unlikely to feature yourself heavily, as frankly, it is far more enjoyable to needle another person, who is less wordy, and more irrational.

          • fact police

            Slad wrote:

            Transferring more than 1 kW from a gram of fuel to a surface already at 1300C is almost certain to put the fuel above its melting point at 1455C.

            I think the opposite is entirely plausible, especially if the whole inner surface of the tube is considered i.e. if the foaming LiAlH moves the powder. However, I acknowledge your disagreement, and I would be interested to see some numbers.

            I should think you would need numbers before you consider that plausible. Boiling doesn’t begin until 1342C in atmosphere, and so probably higher than that under pressure.

            I’m sorry to labour the point, but direct and concise communication is a good thing.

            I write to communicate my thoughts, and do so in as effective a manner as I can. If you don’t want to read it, that’s ok. But it seems kind of odd that you still want to respond to it.

            And any final report will be likely be written after this sorry saga is over,

            If this saga ends, I suspect it will not be for a long time yet. It looks to me like it’s following the Randell Mills BLP game plan. And if it ever does end, it is nearly certain that you won’t be writing a report, because it would almost certainly be about why you were wrong.

            and will be unlikely to feature yourself heavily, as frankly, it is far more enjoyable to needle another person, who is less wordy, and more irrational.

            You mean someone who won’t show where you screwed up.

          • fact police

            Slad wrote:

            Maybe you should write your own report, in order to help define your ideas concisely, and in an unambiguous way.

            Actually, shortly after Lugano, I wrote a 5000 word critique with 11 specific criticisms, and posted it at the other place. It stands up pretty well, though I would make a few changes a year later. In particular, the discovery of the joule heating factor of 3 discrepancy, and the two revised analyses of the thermographic data post-date it.

          • fact police

            Slad wrote:

            Maybe you are right about over interpretation, but the two main arguments about the credibility of Lugano are based on Thomas Clarkes notions*, and what essentially amounts to slurs against Rossi or the researchers.

            *Which I covered, in a somewhat provocative fashion, and I’m interested to hear a response to.

            He has responded briefly.

          • fact police

            Slad wrote:

            What you are saying above, about power per unit length, is the same point I made in my original report.

            I don’t recall any mention of it in your report.

            (I also calculate the thermal conductivity of the nickel as being higher). If we restore the full surface area, my point still stands.

            Yes, I remember that, but it was the transfer of heat through the ceramic that was the issue. The fuel conductivity is lower by a little more than a factor of 2 compared to alumina.

            If we restore the full surface area, then the temperature of the *surface* is not implausible based on a comparison to fission fuel. But that point stands not because of shape differences, but because of linear power density differences. The question of the fuel temperature is still implausible to me.

            If you don’t mind me saying, it is not easy to pick out the bones of your arguments, due to the lengthy surrounding material, and bombardment of several ideas that they often contain (a counter example being my short summation above) You risk a reader skimming over them. Yes, the radius arguments were a blind alley, but I thought that was the only reason you could arrive at the apparently nonsensical idea of a 600C gradient in the alumina tube.

            If you don’t mind my saying so, it seems you don’t actually make the necessary effort to consider what I’ve written. You admitted that yourself, when you said you don’t pay attention to what I write. And that was abundantly clear when you repeated questions I had answered several times already.

            And the ultimate point, that shape matters (to some degree), is one that I don’t believe you can completely disagree with,

            Obviously shape matters to a degree, but the point was the difference in the fuel shape *reduced* the thermal resistance for the fuel pellet compared to the ecat geometry — it did not increase it. And this was reasonably plausible from simple considerations, and verified by quantitative analysis.

  • psi2u2

    I will leave you and Slad to debate the science. Its your handle I was commenting on. I did not personally find the tone pompous, but I’m just a lit. prof. Sorry, but your chosen name evokes a prejudice in me. Police logic is the reasoning of stage 4 in the Kohlberg typology. Sometimes police have valid facts and tell the truth. In my experience, many do not, most of the time. Consider that the primary reason we are still arresting people for cannabis in this country is that police departments are funding vacations and buying Humvees with the stolen proceeds. Yikes.

  • Slad

    Of course it tends towards infinite. That’s why the report clearly says:

    “We can draw an analogy by looking at the Heat Flow Through a Hollow Cylinder# equation from the original report”

    And….

    “Whilst it is true to say this equation should not be used to analyse the heat emanating from a working nuclear fuel rod, this is merely an argument based on the shapes under consideration

    The only point of this section in the text is to help you to understand why…

    “it is wrong to say they both have similar effective wall thickness”

    Understand now?

    • Slad

      The calculations where I compared the heat transfer through the alumina tube, to the heat transfer through a MOX pellet.

      As I have said, please stop wasting my time, and answer the three simple questions I have asked you, rather than trying to re-invent the wheel by introducing other pointless formulae.

      As my report says:

      “We can draw an analogy by looking at the Heat Flow Through a Hollow Cylinder equation from the original report

      “Whilst it is true to say this equation should not be used to analyse the heat emanating from a working nuclear fuel rod, this is merely an argument based on the shapes under consideration

    • Slad

      The 100x power density is a (paraphrased) quote from yourself, that’s why the original report uses it. Underestimated or not, it doesn’t matter to me, I’m only aiming for ballpark figures, in order to assess plausibility or not.

      the mass fraction of the Li is only 1.2% at the beginning and only 0.03% at the end.

      From a tiny sample. Maybe it’s all stuck to the tubes walls? My report assumed 10% Li, and similar amounts are mentioned in rossi’s patent.

      Maybe you are right about over interpretation, but the two main arguments about the credibility of Lugano are based on Thomas Clarkes notions*, and what essentially amounts to slurs against Rossi or the researchers.

      *Which I covered, in a somewhat provocative fashion, and I’m interested to hear a response to.

  • Slad

    Care to sum up this this latest brainwave of yours in a few sentences, without throwing in extraneous references to confuse the matter, causing me to loose the will to type?

    And a figure number in that 61 page document would be most useful.

  • Slad

    First, I want to see a reference for “the element is hundreds of degrees above the water” in a kitchen kettle.

    Second, a nice guy, who seeks to waste my time with his solipsisms?

    I’m going to really help you out now…

    All you need to do is answer these three very simple questions.… The answer to the first two questions is a number, not a sentence. The third can be answered however you please.

    1) What temperature difference (across the wall) is required to transmit 2150W through a 200mm long alumina tube that has external diameter 20mm and wall thickness of 6mm?

    *Hint: If you are struggling, the working is in the first report.

    2) What is the temperature profile of a working 5mm radius MOX fuel rod?

    *Hint: Google is your friend here.

    3) Now, why is the temperature difference seen in the fuel rod 2 to 3 times higher that seen in the alumina tube, despite the alumina tube having a 20% longer apparent conduction length?

    *Hint: The answer should include the word “shape”.

  • Slad

    The heat from the fuel in my original model was 2140W. This equates to a ~200C difference between the inside and outside of the tube. This is found by using a simple equation, itself found in every heat transfer textbook. I actually quote this equation in the report.

    Why do you disagree, by saying this temperature should be 600C?

    Please be concise.

  • Slad

    Come on, there’s a huge air gap between your pan and the element. This will make the element get hotter than it needs to.

    Also did you measure the bottom of the pan to get the 200C? For a realistic way of comparing it to a kettle elements temperature, you should have measured inside part of the pan touching the water.

    And you used an IR thermometer, did you calibrate it for the pans surface?

    but a steep gradient is necessary for the heat to flow out of the element at the necessary rate

    I disagree. Google ‘water pool boilng curve’, and look at the saturation temperature at the critical heat flux.

    There is a much steeper gradient inside the element… The inner wire is very hot, but surrounding material is an insulator, it’s designed to ‘step down’ the temperature, to better fit water’s boiling curve.

  • Slad

    I specifically mentioned the wire inside, and also explained to you why it got so hot.

    Discussion over.

  • Slad

    I think a goodportion of large force would also be replicated by the effect.

    And for sure, there’s a porosity, which I included, but the pores are filled with hydrogen (I modeled air) so I’m happy I used a somewhat conservative figure.