Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
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 materialhttp://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.netwrote: 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.**htmlhttp://www.mail-archive.com/vortex-l@eskimo.com/msg51632.html http://www.mail-archive.com/**vortex-l@eskimo.com/msg51644.**htmlhttp://www.mail-archive.com/vortex-l@eskimo.com/msg51644.html http://www.mail-archive.com/**vortex-l@eskimo.com/msg51648.**htmlhttp://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) *
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
On Oct 18, 2011, at 4:25 PM, David Roberson wrote: 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. It must have a long de Broglie wavelength. My guess is the mass of paper involved is very low. 8^) 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. No problem there! Most discussion here is based on second hand information. 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. New ideas are most welcome here. Slack is sometimes hard to come by! 8^) 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? Here is a spread of results, but for 5 cm thick lead: again using for I0: EnergyActivity (in gammas per second) for 1 kW -- 1.00 MeV 6.24x10^15 100 keV 6.24x10^16 10.0 keV 6.24x10^17 and using: I = I0 * exp(-mu * rho * L) where mu is given by: Energymu (cm^2/gm) -- 1.00 MeV 0.02 100 keV 1.0 10.0 keV 80 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) *(5 cm)) I = 2.07x10^15 s^-1 ( was 3.7x10^15) 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) *(5 cm)) I = 1.4x10^-8 s^-1 = ~0 s^-1 (I posted 2.9x10^5 s^-1, but this was my calc. error - correct value was 3.04x10^4 s^-1, still readily observable with a counter) For 1 kW of 10 keV gammas we have: I = (6.24x10^16 s^-1) * exp(-(80 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = ~0 s^-1 (was ~0 s^-1) So, the answer is the 5 cm of lead vs 2.3 cm of lead essentially eliminates gammas in about the 100 keV range that were readily countable with 2.3 cm of lead. The extra lead does nothing, however, for converting more gamma energy to heat. It still does not prevent MeV order gammas from being readily detected, so they are still ruled out as a heat source.The 3.04x10^4 counts/sec of 100 keV gammas eliminated represent the unobservable 4.8x10^-10 J/s ~= one trillionth of a watt, so no added heat is obtained there, but even close up counting is prevented. And, if this thickness is not adequate, is any amount of shielding able to stop gammas to that degree? The extra lead does nothing for 1 MeV gammas, but a lot for 100 keV gammas, but also nothing for producing more heat. The lead can however, add to the heat-after-death time significantly depending on configuration. Frankly I suspect the added shielding is iron, not lead. It can sustain heat-after-death much better and provide stability at high temperatures that lead can not. The goal of this test was heat-after-death. The mu for lead at 100 keV is 0.372. For 1 kW of 100 keV gammas through 2.7 cm of iron I get: I = (6.24x10^16 s^-1) * exp(-(0.372 cm^2/gm) * (7.87 gm/cm^3) *(3 cm)) I = 4.66 s^-1 The counts still effectively disappear with iron on the inside and lead on the outside or vice versa. If it is a hoax then there is of course no need for the lead at all. 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. I see no reason to go from 2.3 cm to 5 cm, since the prior counts were already nominal with regard to safety. 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? Anything is possible. Viable prospective nuclear reactions have not been identified. Anything that produces energy primarily from positron emission is not viable due to the large annihilation energy. Also, positron annihilation energies were looked for but not found. Do you know of any process that releases gammas or high energy x rays but not heat directly? That is somewhat of an inconsistent condition. If the energy output is in the form of high energy photons then it produces
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
On Oct 18, 2011, at 10:50 PM, Colin Hercus wrote: 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 Yes you are right! Another one of my clerical mistakes. The above should be written: R = (0.002 m)/((16 W/(m K)*(1.8x10^-2 m^2)) = 6.94x10^-3 °C/W Producing a heat flow of 4.39 kW, or 4390 W then requires a delta T given as: delta T = (6.94x10^-3 °C/W) * (4390 W) = 30.46 °C Using Fourier's law to check, I get q = (16 W/(m K))*(1.8x10^-2 m^2)*(30.47 K)/(0.002 m) = 4387 W which is well within tolerance. I did not put the review up on my site. I should correct it and put it there. Thanks for the correction! I wish my calculations were checked more often. 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 (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 Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
Focus Italia has now english translation in their video, so it is rewarding to watch it again. http://www.focus.it/scienza/e-cat-l-energy-catalizer-di-andrea-rossi-il-video-del-test-del-6-ottobre-2011-975_C7.aspx The interviews are interesting. Lewan thinks the reaction is triggered, when the temperature reaches 60 degree. Also Kullander and Essen observed this. What I still dont understand: If the outer surface reaches 66° then everything inside must be hotter than 60 degrees. Why is electric heating still needed? Ok, there is cold water flowing in. They could run this through an internal heat exchanger and avoid electric preheating. Am 18.10.2011 18:56, schrieb Peter Heckert: Rossi claims here, the FAT CAT has a surface of 5000 square cm and the surface heated up to 66-80 degrees in the october demonstration: http://www.journal-of-nuclear-physics.com/?p=516cpage=1#comment-93550 There was only 1 reactor active (of three) and the full 20 kW where not reached. Now consider the 1 MW container with all these 52 tightly packed FAT-CAT's in full operation. I think, if he closes the door, it should reach 100° inside. So must he let the door open? .. It would be nice, if persons that have seen the 1MW plant could clarify this, possibly by photos showing a cooling system. - Peter
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
Rossi is always saying that he needs the electric power to stabilize the reaction. Maybe it stars at 60 degrees, but needs to be hotter, or needs to be heated for a long time, to stabilize. Whatever he means by stabilize. 2011/10/18 Peter Heckert peter.heck...@arcor.de What I still dont understand: If the outer surface reaches 66° then everything inside must be hotter than 60 degrees. Why is electric heating still needed? Ok, there is cold water flowing in. They could run this through an internal heat exchanger and avoid electric preheating.
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
They could have two cold water inputs: One feeds into an internal heat exchanger and then feeds the preheated water into the reactor, and the other feeds the cold water directly into the reactor. Then the electric energy needs should be strongly reduced but cooling down the reactor is still possible. With a COP of 6 this plant will be soon obsolete, when the prices for electric energy rise as expected. Why didnt they implement gas heating? Then the cooling problem for the 1 MW container is still unanswered. 52 boxes at 66 degree or more surface temperature in this container, this needs cooling. There are so much obvious and unanswered questions . Its impossible to believe for me. Am 18.10.2011 20:04, schrieb Bruno Santos: Rossi is always saying that he needs the electric power to stabilize the reaction. Maybe it stars at 60 degrees, but needs to be hotter, or needs to be heated for a long time, to stabilize. Whatever he means by stabilize. 2011/10/18 Peter Heckert peter.heck...@arcor.de mailto:peter.heck...@arcor.de What I still dont understand: If the outer surface reaches 66° then everything inside must be hotter than 60 degrees. Why is electric heating still needed? Ok, there is cold water flowing in. They could run this through an internal heat exchanger and avoid electric preheating.
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
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 From: Bruno Santos besantos1...@gmail.com Sent: Tue, Oct 18, 2011 2:04 pm Rossi is always saying that he needs the electric power to stabilize the reaction. Maybe it stars at 60 degrees, but needs to be hotter, or needs to be heated for a long time, to stabilize. Whatever he means by stabilize. 2011/10/18 Peter Heckert peter.heck...@arcor.de What I still dont understand: If the outer surface reaches 66° then everything inside must be hotter than 60 degrees. Why is electric heating still needed? Ok, there is cold water flowing in. They could run this through an internal heat exchanger and avoid electric preheating.
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
Yes, this might be true. Of course, if they use an internal heat exchanger this would not change the energy bilance. But, in any case the water is pre-heated when they preheat the core. If they would use and internal heat exchanger, then cheap E-Cat energy would be used to preheat the water and only this energy to heat the core above 60° must be made electrically. This would not change the energy bilance, but the cost bilance. Am 18.10.2011 20:36, schrieb David Roberson: 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 From: Bruno Santos besantos1...@gmail.com Sent: Tue, Oct 18, 2011 2:04 pm Rossi is always saying that he needs the electric power to stabilize the reaction. Maybe it stars at 60 degrees, but needs to be hotter, or needs to be heated for a long time, to stabilize. Whatever he means by stabilize. 2011/10/18 Peter Heckert peter.heck...@arcor.de mailto:peter.heck...@arcor.de What I still dont understand: If the outer surface reaches 66° then everything inside must be hotter than 60 degrees. Why is electric heating still needed? Ok, there is cold water flowing in. They could run this through an internal heat exchanger and avoid electric preheating.
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
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/msg51644.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: EnergyActivity (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: Energymu (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)
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
Hatchet Heffner hacks hence... On Tue, Oct 18, 2011 at 1:22 PM, 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/msg51644.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 =
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
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 hheff...@mtaonline.net 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/vortex-l@eskimo.com/msg51632.html ttp://www.mail-archive.com/vortex-l@eskimo.com/msg51644.html ttp://www.mail-archive.com/vortex-l@eskimo.com/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
