MFMP Analyzes Fuel Mix, Causes of Reactor Failure

A document written by Alan Goldwater of the Martin Fleischmann Memorial Project provides some analysis of the performance of the reactor that was recently tested which ended with an explosion.

The document can be accessed here:

The document reports that in the fuel mix used in the experiment the ratio of nickel powder to lithium aluminum hydride was 2.5:1 (which is higher than the Parkhomov mix, which was around 10:1), and that there was no free space inside the reactor.

Alan calculates that normally the gas pressure at 1000 C should is 19861 psi., but that the tensile strength of alumina is 35244 psi, which suggest that the pressure inside the reactor was unusually high during this test, as the alumina wall of the reactor ruptured.

The document continues:

Further corrections for absorption of hydrogen into the Nickel should be considered. But even if my calculation is high by a factor of ten, the key take-away is to use much less LiAlH4 and/or allow for more free volume in the cell. Great caution should be used if the remaining cells loaded with the same mixture are heated, and inclusion of a pressure sensor in future experiments should be considered mandatory to avoid such dangerous failures.

Finally, the decomposition is apparently non-reversible, and at ambient temperature the internal pressure would still be 600 bar. This will be a problem when trying to open a cell that has been heated and cooled without breaking, like the mini-*GlowStick* that saw 500 C in the initial leakage test.

This statement indicates that used reactors need to be treated with care!

When asked on their Facebook page about how Alexander Parkhomov avoided having similar explosions due to high pressure inside his reactor, the MFMP posted:

Our Alumina tube wall thickness in that reactor core was 1.18mm compared to Parkhomov’s 2.5mm wall thickness, also he might have left a good deal more free volume. Bob Higgins is preparing to run some experiments with 2.5mm wall thickness cheap tubes purchased from China, we will message more on that tomorrow.

  • Jarea1

    Hi good report!! 🙂
    I am looking forward the new test with 2.5mm wall thickness and less LiAIH4.
    Do you know when MFMP will schedule the next test?

    • Bob Greenyer

      Team member Bob Higgins is preparing such a test but hopefully with an apparatus that will allow gas analysis, pressure profiling and purging. We will message on that later, perhaps today.

  • Alan DeAngelis

    Maybe they could use a pressure relief valve (perhaps not this one but something like it).

  • Bob Greenyer

    UPDATE: Mark Jurich calculates the pressure to be exactly half (H2 not H) so his estimate is that the pressure was ‘only’ 9,930.5 psi.[email protected]…/msg101570.html

    This is now over 3.5X less than the break strength, could there have been a sudden energy release, or could it be as we and others have suggested, there is a temperature gradient and a localised pressure on the swagelok that might contribute to non-uniform stresses.

    • Anon2012_2014


      I agree closer with Alan’s calculation.

      0.134 gms of LiAlH4 at 37.95 gm/mole = 0.003740962 moles of LiAlH4
      =0.00706 moles of H2 when released by heating, i.e. 2 H2 per LiAlH4

      H2 molar mass is 2.015 gm/mole
      H2 mass is 0.1423 grams.

      Reaction chamber volume = 0.586 cc as per Alan’s calculation
      rho=H2 Density = 24.28 kg/m^3
      T=1000K = 1273K
      R_specific(H2) = 4120 J/kg/K

      P = rho * R_specific * T
      127 MPa = 1270 bar.

      I am not sure about other elements of Alan’s calculation — I just took his reaction chamber volume as a given, but we are within 6% — maybe something with the space taken by the reactants in the chamber.

      • Nigel Appleton

        If your figures for moles of H2 released is correct at 0.00706 moles, then I make the mass of H2 released 0.0142259 g

        • Anon2012_2014

          Agreed and corrected in my comment above. Thank you. (Misplaced decimal by 10x when I retyped from the spreadsheet.)

      • Bob Greenyer

        Thanks Anon, it is great that this risk assessment debate is finally happening before anyone got badly hurt.

        • Anon2012_2014

          Looking at the risk profile (and ignoring the poison gas aspect of LiAlH4):

          1) The amount of hydrogen in the glowstick is about 2000 joules when combusted in air, i.e. about the same as a an AK-47 or M-16 round cartridge. Put another way, if burnt in .1 second it is going to poof in 20 kW. That is a decent amount of energy.

          2) I think Alan is right that the glowstick, I get when cooled, probably is at around 300 bar. Alan may get more (600 bar) because he figures more volume is taken by the Nickel and the LiAl byproduct. 300 bar of room temperature gas.

          E = nRT ln(P/P0) [when from isothermal expansion]

          If P0 = 1 bar, and P = 300 bar, n = 0.007 moles of H2, R=8.314 J/K/mole, and T=300K then

          E = 100 joules, i.e. there is not a lot of energy in the cooled glowstick from the pressure.

          3) When the glowstick is hot, it has energy due to compression and energy due to heat in the gas. I got tired of thinking about the thermodynamics (someone else here has to be better) but I would guess that

          E = nRT ln(P/P0) + the heat energy in the H2.

          I get something like 1300 bar and T=1300 at the bursting point = 542 joules plus the extra energy because it is hot. Not certain here and have to come back to this later.

          4) Velocity of fragments: There can either be a gun barrel effect (a projectile/piston accelerating in a cylinder) — or what I would assume – a break that sends a “blowout” in one direction but where the hot gas is not contained. I am still looking for some equations.

          The problem is that the pressure falls as soon as the device is open, so that the acceleration falls. Obviously the rate of pressure fall depends on the blowout geometry, and it has to be integrated to get the final velocity imparted to the blowout fragment.

          I’m in favor of the lexan screen. How much more you need because the fragments are very light, I don’t know. If the new swaglock bolt comes off it would carry more momentum, but wouldn’t accelerate as much because it is heavy.

  • Leonard Weinstein

    The tube strength is given for room temperature, not at elevated temperature. The strength drops considerably as temperature increases, so a much stronger structure is needed for a good system. Depending on type of Nickel powder and percent of LiALH4, free space may be needed. If the spikey powder is used, it’s powder specific gravity is about 1/4 that of solid grain powder, so a space of 75% is available even for a full chamber. The type of Nickel power to use is T123 from Vale Canada. If non-spikey powder is used, the surface area is also far lower for a given mass, and I don’t know if it even would work as well.

  • Anon2012_2014

    (Repost from prior thread)

    Alan, thanks for the NIST link. Here’s the whole table:

    1) tensile strength falls with temperature increase:

    1000 C = 243 MPa

    1200 C = 140 MPa

    1400 C = 22 MPa (!!!)

    1500 C = 13 MPa

    The strength essentially falls by half at 1200 and then by another 6.5x 1400C, and then by about half again at 1500C.

    2) I needed to relate bursting pressure to tensile strength. I found this formula — a better materials/mechanical engineer can find a better formula


    P = (2*t*s)/(D – .8*t)
    P=bursting pressure
    t = thickness
    s=tensile strength

    Seems to be independent of the units (cm or inches) because they cancel out.


    P= 2t/(D-.8t) s
    Parkhomov has a 10mm outer diameter and 5 mm inner diameter, thus t=5, D=10

    P=1.66 s
    Using Parkhomov’s 1190C experiment, the bursting pressure is 232 MPa.
    As I have already calculated, Parkhomov’s internal pressure is around 46 MPa at that temperature, i.e. 5x margin.

    I am looking for the OD and ID of the boom test tube to calculate bursting pressure of the material, i.e. to see if it models.

    Just guessing for the glowstick that blew up

    D=5 mm
    t= 2 mm
    P=1.18 s.

    The unit failed when the Williamson IR pyrometer measured 1027 C. I think we can assume that the pressure vessel was hotter since it appears it was hidden from the Williamson sensor by the outer reactor wall. Let’s say it was at 1100 C.

    Then by interpolation, s = 191.5 MPa, bursting pressure = 226 MPa at 1100C, and using ideal gas law we get a reactor pressure of 137 MPa.

    At 1200C, s=140 MPa, bursting pressure = 165 MPa, and reactor pressure = 147 MPa.

    Again, by interpolation, at 1250C, s= 110 MPa, bursting pressure = 130 MPa, and reactor pressure = 152 MPa.

    Thus, by interpolation we would expect the reactor to fail at 1225 C.

    We don’t know the reactor wall temperature, but it could have been near that.

    • Anon2012_2014


      Thickness is 1.18 mm. What is outer (or inner) diameter? Also, am I correct that the pressure vessel/reaction chamber wall temperature is higher than where you have the Williamson aimed due to insulation of the dogbone outside, particularly if the pressure vessel is reacting on the inside between the H2 and the Ni (chemically or LENR), and further even if it was not reacting, the pressure vessel is inside the SiC heater element and thus has no way to radiate to the cooler lab around it??

    • Anon2012_2014


      I think Barlow’s formula is better to use which can be written

      P = s * t/r


      P=bursting pressure
      s=tensile strength
      r=outer radius

      As the highest t/r is 1 (i.e. a nearly solid rod with an infinitesimal reaction chamber drilled in it) we can see that bursting strength can only be reduced by using thinner walls relative to the overall oouter radius of the cylinder.

      For Parkhomov we have t= 5mm and R_outer = 7.5 mm or a 66% lower bursting strength relative to tensile strength.

      For Glowtube (the banger) I am guessing t/r = .7 ish.

      I will apply the failure pressure data accordingly.

      • nickec

        AGP tube is 5mm ID, 10mm OD.

        • Anon2012_2014

          AGP (Parkhomov) tube: you’re right, 10 mm OD, 5 mm ID = 2.5 mm inner radius, 5mm outer radius, 2.5 mm thickness. Barlow would have P = s * 2.5/5. My mistake.

          I need MFMP tube dimensions.

  • Alain Samoun


    To start the CF in these experiments we have two related factors: Pressure and Temperature. It seems to me that you want a higher temperature so you need to have less LAH. I suggest maybe to go the other way: More LAH and lower temperature. Referring to the thermal decomposition of LAH that is complete at about 500C you may try to keep the reactor at 600 C and use 200mg of LAH?

    • Bob Greenyer

      Now that we have a reliable, fast, cheap, sealing method and there is a good debate about safety and a move to have proper calculations of the risks – everything is wide open.

      • Alain Samoun

        Yep! the new reactor makes it possible,lowering the temperature would have many advantages for improving the COP and would simplify the development if it works.

      • Pekka Janhunen

        My advice would be to stick to faithful Lugano/Parkhomov replication until it succeeds or proves impossible, which means that temperature should be at least 1000 C.

        Having said that, it’s open to speculation whether or not Rossi’s low-temp reactors use some different fuel mix which might work at lower core temperature. In some of his video interviews, Sergio Focardi said that the reaction starts at 60 C. Potassium is the unique element whose melting point is close to 60 C.

        • Bob Greenyer

          We have tried to do as faithful a Lugano experiment as possible – but there are many unknowns, because we chose to stick to that, we held off doing a Ni+LiAlH4 experiment (we had bought 100g of LiAlH4 in the days after the report for just such a test). Then Parkhomov decides to “just do it” and due credit to him for that. As soon as he reported we started contact. After he supplied his approach for sealing we tried over the month to make it work without success. Only after reporting our failures at repeated attempts did he add extra detail – all of this we have published in the hope that someone else can be successful. It might be like our experience with Celani, in that he does not know what is important to share, because he just does something automatically.

          As I have said before, on December the 14th 2012 in Rome, I was told by an anonymous person to add Alkali metals. Currently we are focussed on those suggested by atomic ratio calculations in Lugano report and tested by Parkhomov.

          • Andreas Moraitis

            Despite of the idea that alkali metals like lithium could participate in the reaction, one should keep in mind that they are strong reducing agents, or electron donors. Some theories postulate that LENR require a surplus of electrons. Anyway, the best electron donor is electric current – it may even reduce alkali metals. So one could guess that supplying a current to the fuel (as a cathode) might work even better than relying on the lithium. This method has been applied in the ‘classical’ electrolysis-based LENR experiments, but it could work as well in dry cells. (There are already corresponding attempts, some of them have been described in patent applications.) In case that a high electron density is the key, strong currents at low voltage would outperform low currents at high voltage.

            In addition, some organic substances called “super electron donors” (you can Google them) might be more effective than alkali metals. Maybe they are suitable for low temperature LENR devices. (I guess at high temperatures they would decompose.)

          • Andreas Moraitis

            „Electron density“ seems to be the wrong term here. It should rather be something like “the number of electrons that pass the cross-section of the fuel per time unit.”

          • uDevil

            That is electron flux. Usually called current density.

          • Andreas Moraitis

            Thanks. Did not find it in my dictionary.

        • Omega Z


          On JONP, Rossi said that the fuel charge was different for the Low temp E-cat.
          I don’t know if that is still available. Sometimes if Rossi slips, he will delete certain posts or at least edit them.

          In a more recent post he gave his normal response. He could not comment on such matters.
          He has also said that the controls are also different. Mostly the software control I believe.
          And, they are both of the same principle.

          • Pekka Janhunen

            OmegaZ: yes I remember that as well.

        • LCD

          It would help if we knew what the activation parameter was, pressure, temp, or H absorption level, etc. we are shooting in the dark. Changing anything now is statistically unwarranted if parkhamov is right but darn it would really help to see the reaction at lower temps.

  • Anon2012_2014

    Agreed and thank you. I think Roark’s formula which includes the compressive stress as well as the hoop strain is better than Barlow’s. I believe that they will come out the same within 30%, but I need to prove it.

  • Omega Z

    Maybe of interest of MFMP
    Let me sum up. 1, 2, 3 ALL related to the big bang. #3 especially interesting.

    1) In the previous thread-How to Make a MFMP ‘Glowstick’ the pressure is being discussed. I mentioned this before a few days ago. I thought I read where MFMP or Parkhomov was using about 10%( LiAlH4 ).
    I was thinking that in Rossi’s Lugano charge, it was a much smaller proportion of ( LiAlH4 ) verses Nickel. 1% or 0.1 Something to that effect.

    That’s a substantial difference given the Lugano test ran for 30 days on a much smaller quantity. Given the quantity, I would speculate that it released much more hydrogen which would create much more pressure then the calculations would indicate in that short period of time before the Big Bang. However, I’ll admit this is a little over my head.
    2) Another post expanded on.

    Have you taken into account that the swagelok ferrule itself creates a pressure point to achieve it’s seal.

    This would produce a bolt cutter effect, but instead of increasing pressure from the cutter, you’ve increased the pressure from within.

    Swagelok’s, Flares joints, They create stress points. Even hair line fractures. You psi rating would no longer be valid.
    You’ve also paired a ceramic with metal which have different expansion contraction rates
    3) The Lugano Test

    Rossi sent additional backup parts. In case something broke.
    We learned latter that it was actually additional complete reactors.
    We learned that Rossi was present part of the time.
    At Start up & Shut down times.
    The Lugano dummy test wasn’t allowed to exceed 500’C to avoid damaging the reactor.

    Enough teasing if you haven’t figured it out. MFMP & Parkhomov are not alone in reactor breakage. Rossi’s reactors can break/explode also. Enough so that extra'(s) were sent.

    This explains why Rossi’s presence was needed at the start up & shut down. It is a delicate process that the testers wouldn’t have experience at.. Rossi also has stated that the Hot-cat is still in R&D. And, he has indicated on JONP that it has undergone changes since the Lugano test. Breakage may be problematic enough to try & find different materials to eliminate this problem.

  • Albert D. Kallal

    That pressure sounds somewhat high??

    At room temperature, we assume about 25C or (298K). At 1000C (1274K) means we increased absolute temperature by 1273/298 = 4.2 times.

    So we ONLY increased pressure from static room to about 4 times.

    Based “just” on temperature increase and an “ideal” gas, then pressure goes up by about 4-5 times – not really a lot at all.

    So 20,000 lbs. sounds REALLY rather high.

    Is this a pressure a reading, or some estimate based on “gas” boiling or being released from the fuel inside?

    And as for the pressure ratings of that tube? Again they are at room temperatures. At 1000C that metal and fittings become RATHER soft and fragile – such tubes and pressure rated fittings can fail at rather low pressures.

    While the pipe made a nice bang such a pipe blowing up at room temperature is FAR MORE loud and the pipe rips apart and twists many different ways – trust me, I blown up metal tubes that size at room temperature.

    The pipe in the video more “popped” like a loud balloon then what cold steal pipe at room temperature does when they explode!

    Does anyone know how the 19,000 psi number was arrived at? While a simple calculation of going from 25C to 1000C is a factor of 40, the Celsius scale does not start from absolute zero so that is NOT a linear increase of 40 times.

    So the actual “linear” or absolute increase in temperature is from 298K to 1274K is only a factor of 4 times increase in the temperature (and hence pressure) from room static, not 40 times.

    That was more of a balloon pop then a real pipe exploding…

    Albert D. Kallal
    Edmonton, Alberta Canada

    • Anon2012_2014


      The pressure goes up more than the temperature because above 500C the H2 has been released from the LiAlH4. Kind of like an alka-seltzer in a bottle. The molar calculations for the amount of H2 released are in this thread.

      The tube is really really small on the inside — long and thin and only of the order of 1 CC and 3 mm in diameter. That is why the pop is smaller than your “pipe”. It also likely explosively combusted the hot H2 when it hit the O2 in the air as the autoignition point is 536 C.

      • Albert D. Kallal

        Thanks for the follow up. Much agree the “lack” of a larger bang is likely due to such a small fuel chamber (a good thing).

        And as I noted in my post about pressure:

        or some estimate based on “gas” boiling or being released from the fuel inside?


        So certainly a combination of gas released from the fuel + temperature will give the resulting pressure. However I still maintain that at higher temperatures, strength of the pipe will diminish.

        Albert D. Kallal
        Edmonton, Alberta Canada

        • Anon2012_2014


          Property [unit] 20 °C 500 °C 1000 °C 1200 °C 1400 °C 1500 °C
          ———————————— ———- ———- ———- ———- ———- ———-

          Tensile Strength [MPa]………….. 267(30) 267 243 140 22 13

          Compressive Strength [GPa]………. 3.0(5) 1.6 0.7 0.4 0.3 0.28

          Flexural Strength [MPa]…………. 380(50) 375 345 300 210 130


          This suggests to me that above around 1300 C, alumina cylinders are weaken considerable. The mechanical engineers among us use something called “Roark’s Formula for Stress and Strain” as “the” handbook for calculating these items.

          I believe (I didn’t buy Roark — I am physics guy, not a ME) the formula is the same as this _thick_ cylinder from page 396/397 of this:

          Finally the simple Barlow’s formula (shown in Wikipedia) is on page 397 for thin walled cylinder. I believe if you play with the thick walled solution, you will see it reduces to the Barlow solution if t much less than r_interior.

          I see a possible sign error in the pdf formula when I check against this:

          I am tired of doing the algebra — I would say check your work to see which is right (engineeringtoolbox or “”).

          Maybe the sign is by convention, but doesn’t make sense to me for circumferential stress.

          • Anon2012_2014

            The guy at switched the equations on p397 for thick walled radial stress and tangential stress. Engineeringtoobox equations are right.

            See Page 7 of this paper from RPI on the external pressure=0 case )which is close enough, as the ratio of internal to external pressure is more than 400 times at the failure area).


            Note that there is also lengthwise (longitudinal) stress because of the end caps.

            I believe a complete engineering would figure out the displacement, the shear stress, and the failure mode (by cracking).

  • uDevil

    Thanks! Using the Redlich-Kwong equation, I get a similar number, 11,500 psi. The deviation from ideal may not be worth worrying about here.

  • Anon2012_2014

    Hint to Frank: Start a blowout photo’s commentary thread. The new photos are up on FB and

    It appears to me that the initial crack formed in one place, and that the H2 which was past the ignition point of 600C immediately combusted upon meeting the outside air, causing the burned spot and the melt on the SiC heater. I do not see that as evidence in itself of a hotspot — just the initial fracture point.

    Perhaps at that point, the SiC shorted out where for a brief period it was running 2 kW — this might explain the heat-up/cool-down on the optical camera observed in the Bang! video.

    I find it interesting that subsequently the inner pressure vessel shattered like a dropped ceramic cup into 100 pieces, and that it took the remainder of the SiC heater element with it. Was this from the hot H2 overpressure, or from the hot H20 combustion product overpressure as it tried to escape the outer tube.

    I also find it interesting how the fuel powder was sintered into multiple broken cylinders.

    Next time — only 1/4th material and try again. All was very interesting. Looking for some calorimetry as well.

    • AlanG

      Thanks to all who offered very useful comments and suggestions. Here are some further details of the fuel loading process and my revised estimate of the pressure (available at

      The powders were already inside the glove box when the scale (also
      inside the box) failed, so volume measurements were the only data I had
      available. As a result, precise mass measurement was not possible, nor
      was determination of exact densities by measurement. The relative
      density of the powders was taken from the bulk densities as given in the
      respective Wikipedia entries. Unknowns include the packing ratio of each
      of the powders. They are both finely divided but not nano scale, so
      assuming a similar packing seems reasonable in the absence of other data.

      The volumes were calculated from dimensions of the actual components
      used, measured with a digital caliper. The space between the filler rod
      and the ID of the tube is significant and was included in my
      calculation. The possible vacant volume within the powder mass was not
      included, nor was the possible absorption of H2 into the nickel, which
      we think was minimal given the time scale of the experiment.

      Regarding the calculation itself, the mass of the fuel was determined
      accurately by weighing the loaded cell after sealing and removal from
      the glove box. This was divided by the volume mix ratio, then by the
      estimated relative density ratio of the two powders to get the mass of
      the LiAlH4 in the cell. The amount of H was then found simply by the
      ratio of standard atomic weights. As you correctly pointed out earlier,
      the equivalent molar amount must be based on the H2 molecules in the
      gas, and that was the final figure used to calculate the pressure.

      If I missed something important in my analysis, I’d be happy to know,
      and make further corrections.