Hi Horace, I find your posts quite interesting and you seem to have a rational rather than emotional approach which makes for good reading.
I just read your reply to Dave and as it seemed to make the ECat (and my kettle) impossible I thought I'd double check some of your calculations and I think you've made a mistake on the heat flow from the reactor: R = (0.002 m)/(16 W/(m K)*(1.8x10^-2 m^2) = 1.78 °C/W By my calculations: R = 0.002/(16 * 0.018) = 0.002/.288 = 0.007 °C/W >From engineeringtoolbox.com *Fourier's Law* express conductive heat transfer as *q = k A dT / s (1)* *where* *A = heat transfer area (m2, ft2)* *k = thermal conductivity of the material<http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html>(W/m.K or W/m oC, Btu/(hr oF ft2/ft))* *dT = temperature difference across the material (K or oC, oF)* *s = material thickness (m, ft)* So A = 180 CM^2 = 0.018 M^2 K = 16 W/(m K) s = 0.002m Then q = 16*0.018*dT/0.002 = 144 * dT So for 2500W we'd have a temperature difference of 2500/144 = 17 C which is quite reasonable. This is all way out of my area of expertise so I could be messing up units somewhere. Best Regards, Colin On Wed, Oct 19, 2011 at 4:22 AM, Horace Heffner <hheff...@mtaonline.net>wrote: > > On 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 >> > > > Hi Dave, > > Welcome to vortex! > > I am happy to see your spreadsheet made it through the vortex filter. > Historically nothing made it through above 40KB without special processing > by Bill Beaty himself. Your post with spreadsheet was 55.4 KB. Either a new > limmit has been established or Bill Beaty is closely watching (the latter > seems to me unlikely.) > > The implications that gammas heat the lead and coolant do not make any > sense. If they had the energy to make it out of the stainless steel fuel > compartment used in prior tests, then they would have been readily detected > by Celani's counter. This was discussed here in relation to my "Review of > Travel report by Hanno Essén and Sven Kullander, 3 April 2011". > > http://www.mail-archive.com/**vortex-l@eskimo.com/msg51632.**html<http://www.mail-archive.com/vortex-l@eskimo.com/msg51632.html> > http://www.mail-archive.com/**vortex-l@eskimo.com/msg51644.**html<http://www.mail-archive.com/vortex-l@eskimo.com/msg51644.html> > http://www.mail-archive.com/**vortex-l@eskimo.com/msg51648.**html<http://www.mail-archive.com/vortex-l@eskimo.com/msg51648.html> > > Excerpted below are the most relevant notes I made regarding the April 3, > 2011 test: > > FIG. 3 NOTES > > It appears the heating chamber goes from the 34 cm to the 40 cm mark in > length, not 35 cm to 40 cm as marked. Maybe the band heater extends beyond > the end of the copper. It appears 5 cm is the length to be used for the > heating chamber. Using the 50 mm diameter above, and 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 volume 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 the > heating chamber appears to actually be 6.1 cm. The ID thus might be 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 volume > 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 radius 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. > > > HEAT FLOW THROUGH THE NICKEL CONTAINING STAINLESS STEEL COMPARTMENT > > If the stainless steel compartment has a surface area of approximately S = > 180 cm^2, as approximated above, and 4.39 kW heat flow through it occurred, > as specified in the report, then the heat flow 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 compartment > area is 180 cm^2 or 1.8x10^-2 m^2. If the wall thickness is 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 given > as: > > delta T = (1.78 °C/W) * (4390 W) = 7800 °C > > The melting point of Ni is 1453°C. Even if the internal temperature of the > chamber were 1000°C above water temperature then power out would 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 MeV, a > large portion of the gammas, on the order of 25%, will pass right through 2 > cm of lead. > > The lower the energy of the gammas, the more that make up a kW of gamma > flux. Consider the following: > > Energy Activity (in gammas per second) for 1 kW > -------- ---------- > 1.00 MeV 6.24x10^15 > 100 keV 6.24x10^16 > 10.0 keV 6.24x10^17 > > The absorption for low energy gammas is mostly photoelectic. The > photoelectric mass attenuation coefficient (expressed in cm^2/gm) increases > with decreasing gamma wavelength. Here are some approximations: > > Energy mu (cm^2/gm) > -------- ---------- > 1.00 MeV 0.02 > 100 keV 1.0 > 10.0 keV 80 > > We can approximate the gamma absorption qualities of the subject E-cat as > 2.3 cm of lead. > > Given a source gamma intensity I0, surrounded by 2.3 cm of lead we have 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 much > below that not so. However, it is also true that 0.2 cm of stainless will > absorb the majority of the low energy gamma energy, so we are back > essentially where we started, all the heat absorbed by the 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 100 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%. Down > near 10 keV all the gamma energy is captured in the stainless steel or in > the nickel itself. > > To support this hypothesis a p+Ni reaction set including all possibilities > for all the Ni isotopes in the catalyst would have to be found that emitted > gammas only in the approximately 50 kEV range or below, but well above 10 > keV, and yet emitted these at a kW level. This seems very unlikely. If such > were found, however, it would be a monumental discovery. And, it would be > easily detectible at close range by NaI detectors, easily demonstrated > scientifically. > > SUBSEQUENT COMMENTS > > ... can anything said about the inside of the E-cat be believed? There are > numerous self-inconsistencies in Rossi's statements, and behaviors. These > things may be justifiable in Rossi's mind to protect his secrets. Whether > justified or not, such things damage credibility. > > One thing is for sure: if the E-cat is operated at significant pressure > then 2 mm walls would be too thin at high temperatures. Also, there are > other limits to surface steam generation I have not discussed, that take > precedence at high power densities. One limiting factor is the ability of > the catalyst and hydrogen to transfer heat to the walls of the stainless > steel container, a process which would likely be mostly very small > convection cell driven. Again, we know too little about the internals. > Nothing much new about that. A heat transfer limit is reached if a stable > vapor film is formed between the walls of the catalyst container and the > water. The top of the catalyst container may be exposed to vapor, thereby > increasing the thermal resistance, the effective surface area. At high heat > transfer rates bubbles can limit transfer rates. It would be an interesting > and challenging, though now probably meaningless, experiment to put 4 kW > into a small stainless steel container under water and see what happens, see > if the element burns out, etc. > > > Best regards, > > Horace Heffner > http://www.mtaonline.net/~**hheffner/<http://www.mtaonline.net/%7Ehheffner/> > > > > >
