Hi Horace, Thank you for the kind welcome into vortex. I suspect that my oversized attachment tunneled through the barrier; maybe using the same path as Rossi's device.
You have written an interesting description of the old ECAT version and I plan to review it thoroughly as time allows. I guess I may have needed a hit on the head to believe everything(or anything) that I read on line about ECAT operation. As you know, I was just parroting what I saw in the journal. My understanding of nuclear physics is lacking as my field is electronics design. Allow me a lot of slack when I suggest something totally whacko as sometimes I have good ideas and approach problems from different perspectives. Now, let me ask you a few questions that I suspect you can answer easily. You have presented some interesting calculations concerning the penetration of gamma rays and x rays through lead. The new ECAT design has 5 cm of lead according to reports. That is more than twice the original thickness of 2 cm in the earlier version. Could you help me to reverse engineer the ECAT shielding and figure out what energy of gamma rays would be just barely shielded enough to be undetected? And, if this thickness is not adequate, is any amount of shielding able to stop gammas to that degree? I would suspect that if you answer no amount of lead is within reason, then you must think that the ECAT is a scam since the shielding is arbitrary. Rossi has stated that all of the energy released by the LENR process is in the form of photons. Do you think that this is possible? Do you know of any process that releases gammas or high energy x rays but not heat directly? You answers to these simple questions would be most appreciated. Dave -----Original Message----- From: Horace Heffner <[email protected]> Sent: Tue, Oct 18, 2011 4:25 pm n Oct 18, 2011, at 10:36 AM, David Roberson wrote: > Rossi has stated that the energy released by the LENR reaction is in the form of moderate energy gamma rays(X-Rays?) These rays are converted into heat within the lead shielding and coolant. If this is true, heat to activate the core could be made to exit into the coolant to slow down the reaction. The actual temperature within the core section is perhaps 600(?) C degrees or more. You can find his statement within his journal if it is important to you. The 60 degree figure probably refers to the temperature of the water bath when the core reaches its starting value. Dave i Dave, Welcome to vortex! I am happy to see your spreadsheet made it through the vortex ilter. Historically nothing made it through above 40KB without pecial processing by Bill Beaty himself. Your post with spreadsheet as 55.4 KB. Either a new limmit has been established or Bill Beaty s closely watching (the latter seems to me unlikely.) The implications that gammas heat the lead and coolant do not make ny sense. If they had the energy to make it out of the stainless teel fuel compartment used in prior tests, then they would have been eadily detected by Celani's counter. This was discussed here in elation to my "Review of Travel report by Hanno Essén and Sven ullander, 3 April 2011". http://www.mail-archive.com/[email protected]/msg51632.html ttp://www.mail-archive.com/[email protected]/msg51644.html ttp://www.mail-archive.com/[email protected]/msg51648.html Excerpted below are the most relevant notes I made regarding the pril 3, 2011 test: FIG. 3 NOTES It appears the heating chamber goes from the 34 cm to the 40 cm mark n length, not 35 cm to 40 cm as marked. Maybe the band heater xtends beyond the end of the copper. It appears 5 cm is the length o be used for the heating chamber. Using the 50 mm diameter above, nd 5 cm length we have heating chamber volume V: V = pi*(2.4 cm)^2*(5 cm) = 90 cm^3 If we use 46 mm for the internal diameter we obtain an internal olume of: V = pi*(2.4 cm)^2*(5 cm) = 83 cm^3 Judging from the scale of picture, determined by the ruler, the OD of he heating chamber appears to actually be 6.1 cm. The ID thus might e 5.7 cm. This gives: V = pi*(2.85 cm)^2*(5 cm) = 128 cm^3 The nickel container is stated to be about 50 cm^3, leaving 78 cm^3 olume in the heating chamber through which the water is heated. If the Ni containing chamber is 50 cm^3, and 4.5 cm long, then its adius r is: r = sqrt(V/(Pi L) = sqrt((50 cm^3)/(Pi*(4.5 cm)) = 3.5 cm total surface are S is: S = 2*Pi*r^2 + 2*Pi*r*L = 2*Pi*(r^2+r*L) = 2*Pi*((3.5 cm)^2 + 3.5 cm)*(4.5 cm)) S = 180 cm^2 The surface material is stainless steel. EAT FLOW THROUGH THE NICKEL CONTAINING STAINLESS STEEL COMPARTMENT If the stainless steel compartment has a surface area of pproximately S = 180 cm^2, as approximated above, and 4.39 kW heat low through it occurred, as specified in the report, then the heat low was (4390 W)/(180 cm^2) = 24.3 W/cm^2 = 2.4x10^5 W/m^2. The thermal conductivity of stainless steel is 16 W/(m K). The ompartment area is 180 cm^2 or 1.8x10^-2 m^2. If the wall thickness s 2 mm = 0.002 m, then the thermal resistance R of the compartment is: R = (0.002 m)/(16 W/(m K)*(1.8x10^-2 m^2) = 1.78 °C/W Producing a heat flow of 4.39 kW, or 4390 W then requires a delta T iven as: delta T = (1.78 °C/W) * (4390 W) = 7800 °C The melting point of Ni is 1453°C. Even if the internal temperature f the chamber were 1000°C above water temperature then power out ould be at best (1000°C)/(1.78 °C/W) = 561 W. COMMENT ON GAMMAS As I have shown, if the gamma energies are large, on the order of an eV, a large portion of the gammas, on the order of 25%, will pass ight through 2 cm of lead. The lower the energy of the gammas, the more that make up a kW of amma flux. Consider the following: Energy Activity (in gammas per second) for 1 kW ------- ---------- .00 MeV 6.24x10^15 00 keV 6.24x10^16 0.0 keV 6.24x10^17 The absorption for low energy gammas is mostly photoelectic. The hotoelectric mass attenuation coefficient (expressed in cm^2/gm) ncreases with decreasing gamma wavelength. Here are some pproximations: Energy mu (cm^2/gm) ------- ---------- .00 MeV 0.02 00 keV 1.0 0.0 keV 80 We can approximate the gamma absorption qualities of the subject E- at as 2.3 cm of lead. Given a source gamma intensity I0, surrounded by 2.3 cm of lead we ave an activity: I = I0 * exp(-mu * rho * L) where rho is the mass density, and L is the thickness. For lead rho 11.34 gm/cm^3. For 1 kW of MeV gammas we have: I = (6.24x10^15 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) * 2.3 cm)) I = 3.7x10^15 s^-1 For 1 kW of 100 keV gammas we have: I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) * 2.3 cm)) I = 2.9x10^5 s^-1 For 1 kW of 10 keV gammas we have: I = (6.24x10^17 s^-1) * exp(-(0.80 cm^2/gm) * (11.34 gm/cm^3) * 2.3 cm)) I = ~0 s^-1 So, we can see that gammas at 100 keV will be readily detectible, but uch below that not so. However, it is also true that 0.2 cm of tainless will absorb the majority of the low energy gamma energy, so e are back essentially where we started, all the heat absorbed by he stainless, and even the catalyst itself, in the low energy range. If the 2 mm of stainless is equivalent to 1 mm of lead, for 1 kW of 00 keV gammas we have: I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) * 0.1 cm)) I = 2x10^16 s^-1 and an attenuation factor of (2x10^16 s^-1)/(6.24x10^16 s^-1) = 32%. own near 10 keV all the gamma energy is captured in the stainless teel or in the nickel itself. To support this hypothesis a p+Ni reaction set including all ossibilities for all the Ni isotopes in the catalyst would have to e found that emitted gammas only in the approximately 50 kEV range r below, but well above 10 keV, and yet emitted these at a kW level. his seems very unlikely. If such were found, however, it would be a onumental discovery. And, it would be easily detectible at close ange by NaI detectors, easily demonstrated scientifically. SUBSEQUENT COMMENTS ... can anything said about the inside of the E-cat be believed? here are numerous self-inconsistencies in Rossi's statements, and ehaviors. These things may be justifiable in Rossi's mind to rotect his secrets. Whether justified or not, such things damage redibility. One thing is for sure: if the E-cat is operated at significant ressure then 2 mm walls would be too thin at high temperatures. lso, there are other limits to surface steam generation I have not iscussed, that take precedence at high power densities. One imiting factor is the ability of the catalyst and hydrogen to ransfer heat to the walls of the stainless steel container, a rocess which would likely be mostly very small convection cell riven. Again, we know too little about the internals. Nothing much ew about that. A heat transfer limit is reached if a stable vapor ilm is formed between the walls of the catalyst container and the ater. The top of the catalyst container may be exposed to vapor, hereby increasing the thermal resistance, the effective surface rea. At high heat transfer rates bubbles can limit transfer rates. t would be an interesting and challenging, though now probably eaningless, experiment to put 4 kW into a small stainless steel ontainer under water and see what happens, see if the element burns ut, etc. est regards, Horace Heffner ttp://www.mtaonline.net/~hheffner/
