Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
On Sep 21, 2011, at 7:51 PM, Alan Fletcher wrote: 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. Most of the input water mass flow necessarily must have continued on out the exit port without being converted to steam. That presumes that the heat exchange takes place on the surface of the core. But the heat is (supposedly) produced by thermalization of gamma rays, which could be anywhere nearby. Rossi has said that it is partly in the copper tubing and partly in the lead shielding. The total available area is easily 10 times that of the core, so the delta T could be 780C, not 7800C. This is not likely, because no gammas were detected. 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) * 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. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
On Sep 21, 2011, at 7:40 PM, Jouni Valkonen wrote: [snip] I have snipped material, some of which I disagree with, but seems not worth debating at this late date. This is very good observation. If your calculations are correct, then it should be very good evidence for inconsistencies. Perhaps this is the reason why Rossi has said that his new E-Cat has more effective heat exchangers than stainless steal. However this issue has been discussed in Rossi's blog a lot and he is perfectly aware that stainless steal is poor heat conductor. Perhaps this issue is under control or perhaps not. Especially bad shadow this casts for Levi's observed 130 kW power surge during the 18 hour test. (I thought previously that it was measuring error, but now I am sure that it was measuring error) However it would have been good question to ask how thick walls steal reactor chamber had? Yes, but 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. === Very good analysis about the test anyway. Thanks for that. Thank you. Here is something else to analyze. REPORT ON HEAT PRODUCTION DURING PRELIMINARY TESTS ON THE ROSSI “NI- H” REACTOR. Dr. Giuseppe Levi http://22passi.blogspot.com/2011/01/report-ufficiale-esperimento- della.html?m=1 There does not seem to enough information there to analyze that does not have similar problems to the subsequent tests. Also, these tests have been discussed for a very long time and probably anything significant to be said about them has been said. –Jouni Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
Congrats! You are doing some good work. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Thursday, September 22, 2011 2:25 AM Subject: Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011 On Sep 21, 2011, at 7:51 PM, Alan Fletcher wrote: 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. Most of the input water mass flow necessarily must have continued on out the exit port without being converted to steam. That presumes that the heat exchange takes place on the surface of the core. But the heat is (supposedly) produced by thermalization of gamma rays, which could be anywhere nearby. Rossi has said that it is partly in the copper tubing and partly in the lead shielding. The total available area is easily 10 times that of the core, so the delta T could be 780C, not 7800C. This is not likely, because no gammas were detected. 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) * 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. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
On Sep 21, 2011, at 11:18 PM, Horace Heffner wrote: On Sep 21, 2011, at 7:40 PM, Jouni Valkonen wrote: [snip] I have snipped material, some of which I disagree with, but seems not worth debating at this late date. This is very good observation. If your calculations are correct, then it should be very good evidence for inconsistencies. Perhaps this is the reason why Rossi has said that his new E-Cat has more effective heat exchangers than stainless steal. However this issue has been discussed in Rossi's blog a lot and he is perfectly aware that stainless steal is poor heat conductor. Perhaps this issue is under control or perhaps not. Especially bad shadow this casts for Levi's observed 130 kW power surge during the 18 hour test. (I thought previously that it was measuring error, but now I am sure that it was measuring error) However it would have been good question to ask how thick walls steal reactor chamber had? Yes, but 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. There are other issues that complicate the situation, that could increase the heat transfer capabilities. We do not know if the insides of the E-cat are as presented. The temperature curve seems to indicate it is not. It is entirely possible there is more to it than a 50 cc stainless steel box. The big area might simply house a T fitting. There could be an internal heat pipe running throughout the device, even into the flue. Hydrogen can conduct a lot of heat by convection. Also, independent evaporative heat pipes of various kinds can operate at high temperatures. Rossi himself referred to the boiling occurring in the flue. The effective surface area could be very large. The effects on the thermometer itself could be large. We don't know. The problem is lack of information with regard to internal structure. Independent black box calorimetry is the only way to get convincing data. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
On Sep 22, 2011, at 5:16 AM, Joe Catania wrote: Congrats! You are doing some good work. Thanks Joe! - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Thursday, September 22, 2011 2:25 AM Subject: Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011 On Sep 21, 2011, at 7:51 PM, Alan Fletcher wrote: 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. Most of the input water mass flow necessarily must have continued on out the exit port without being converted to steam. That presumes that the heat exchange takes place on the surface of the core. But the heat is (supposedly) produced by thermalization of gamma rays, which could be anywhere nearby. Rossi has said that it is partly in the copper tubing and partly in the lead shielding. The total available area is easily 10 times that of the core, so the delta T could be 780C, not 7800C. This is not likely, because no gammas were detected. 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) * 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. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
You appear to have recovered. I am happy for you. -Original Message- From: Horace Heffner hheff...@mtaonline.net To: Vortex-L vortex-l@eskimo.com Sent: Wed, Sep 21, 2011 4:19 pm Subject: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011 INTRODUCTION This is a highly belated review of the Travel report by Hanno Essén and Sven Kullander, 3 April 2011, and created April 7, 2011, to be found at: http://www.lenr-canr.org/acrobat/EssenHexperiment.pdf FIG 1 2 NOTES It appears the thermometer wells, thermocouple holders, are only partially completed on the left 3 E-cats. Fig. 4 shows a completed thermocouple holder with probe inserted. Note that the exit for the thermocouple holder is located below the level of the steam/water exit port. If the brass fitting is not a pressure fit or o-ring sealing device, then if water leaks out of the exit port and down the hose, it should also leak around the probe. Steam should leak around the probe fitting as well. EK: ... according to Rossi, the reaction chamber is hidden inside in the central part and made of stainless steel. EK: Note that on the main heating resistor which is positioned around the copper tube and made of stainless steel (Figure 3) you can read the dimensions and nominal power (50mm diameter and 300W). EK: At the end of the horizontal section there is an auxiliary electric heater to initialize the burning and also to act as a safety if the heat evolution should get out of control. The auxiliary (preheater) element wire leads are clearly visible at the entrance to two of the three unused E-cats in Fig. 2. 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. FIG. 4 NOTES It is notable the 2 cm of lead is not evident. The insulation looks more like a polyester than fiberglass, and if so its melting temperature is around 250°C. FIG. 5 NOTES It looks like there may be two white power wires going into the E- cat. Hard to tell. It looks like some kind of straight white tubes of short length go through the insulation as well. No notes made of their function. FIG. 6 NOTES The slopes on leading side and trailing side of the inflection point (elbow or kink) at 10:37 are 6.7°C/min, 9.4 °C/min respectively. FIG. 7 NOTES The temperature reading might be interpreted to mean an equilibrium was possibly reached, but this not known since the water overflow was not measured and independent calorimetry was not performed on the output mass flow. The thermometer is subject to heat wicking through the well, other error producing effects which may or may not exist depending on the structure of the device inside which is not permitted to be observed. FIG. 10 NOTES There appears to be two power cords running from two receptacles on the rightmost (in the photo) back side of the blue box. This could indicate that both the main band heater and the auxiliary heater were in use, indicating more than 300 W was in use at some point. REVIEWING THE CALIBRATION CALCS EK: Calibrations. The flow of the inlet water was calibrated in the following way. The time for filling up 0.5 liters of water in a carafe was measured to be 278 seconds. Flow is thus (500 gm)/(278 s) = 1.80 gm/s. EK:Visual checks showed that the water flow was free from bubbles. Scaled to flow per hour resulted in a flow of 6.47 kg/hour (Density 1 kg/liter assumed). The water temperature was 18 °C. The specific heat of water, 4.18 joule/gram/ °C which is equal to 1.16 Wh/kg/ °C ... The heat capacity of water is 4.18 J/(gm K) near 45°C, but is above 4.2 J/(gm K) below 5°C and above 81°C. The average is probably closer to 4.19, but I have simply used 4.2 in my prior calcs, which gives a slight edge to a free energy conclusion. This I consider a non-issue, but simply of general
Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
On Sep 21, 2011, at 4:29 PM, fznidar...@aol.com wrote: You appear to have recovered. I am happy for you. Thanks! I won't know the full story for a week, but all is fine for now. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
Hello Horace, I appreciate the thorough and highly competent and experienced skill with which you have made exceedingly clear that so far the Rossi demos fail to prove any anomalies. Thanks, Rich
[Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
Just a few comments on your comments (parts I didn't comment on have been snipped away)... On 11-09-21 08:18 PM, Horace Heffner wrote: [ ... ] However, the graph makes no sense. There is no sign of things coming asymptotically to an equilibrium as would be expected. Yes, indeed. The graph is impossible. It was one of my Rossi moments when I realized that the (piecewise linear!) graph didn't correspond at all to the interpretation as showing that the reaction ignited when the temp reached sixty degrees. If there were no water flowing through the device -- or if the thermometer were isolated from the flowing water -- then the graph would be reasonable: Linear temperature rise is what we'd expect with constant heat applied (and constant thermal mass). The problem is in trying to reconcile the graph with *flowing* water, where what we might call the effective thermal mass of the system rises linearly with the effluent temperature, due to the linearly increasing cost of heating the water. Suppose for a moment it takes no energy to bring the temperature of the copper etc. up to equilibrium temperature of 60°C, that the copper is a perfect insulator (except for the heating chamber walls), and thus has no heat capacity, and that there is no heat loss through the insulation. If the device has no water in it initially, the outlet temperature would remain at room temperature until the water reached it, and then it would instantly jump to 60°C, because the heater *continually* provides enough heat to send the water out of the heating chamber at 60°C. Now, assume that the temperature rise is slow and constant because the device metal and possibly some residual water requires heating. There is a problem with this because it would be expected that as the copper comes up to equilibrium with the water temperature, the delta T between the water and copper at each point decreases, and the output temperature curve would asymptotically approach 60 degrees instead of heading there as a flat line. Sure, it would look a whole lot like a capacitor charge curve, where it's being charged through a resistor. Similarly, the elbow, the increase in temperature curve slope to a new constant value, appears to be an instantaneous increase in power output. If there were a sudden increase in power applied to the heating chamber, it would seem that the copper between the chamber and the temperature sensor would again have to be heated, as well as the water in the device. The temperature curve almost looks like what would be expected if a well stirred pot of water were being heated. It should not look like this. There is a water flow in and out. Right, there is supposed to be water flowing through the system ... but that's not what the graph says, is it? Very interesting. Suppose the device were initially full of a liter of water at 18°C, when the flow and power were turned on. This means, for the 300 W to bring the inflow of water to exactly 60°C in the 9 minutes, the existing pool of water could have been heated at all. It would all have to flow out the port at exactly 18°C until the 60°C water arrived at the thermometer. This does not happen, as shown in the temperature graph. The vicinity of the thermometer gradually warms up. It is questionable that the device should suddenly turn on hydrogen energy at 60°C. In the (later) water only test which was not public a mere 5°C sustained rise in water temperature was enough to maintain the supposed hydrogen energy production. THE KINK ELBOW OF HEAT RISE AT 60°C EK: Instead the temperature increases faster after 10:36, as can be seen as a kink occurring at 60 °C in the temperature-time relation. (Figure 6). A temperature of 97.5 °C is reached at 10:40. The time taken to bring the water from 60 to 97.5 °C is 4 minutes. The slope of the lower elbow section was (60°C)/(9 min) = 6.7°C/min. The slope of the upper elbow is (37.5°C)/(4 min) = 9.4 °C/min. If the first slope represents 300 W, then the second slope represents (9.4/6.7)*(300 W) = 421 W. A KEY STATEMENT EK: The 100 °C temperature is reached at 10:42 and at about 10:45 all the water is completely vaporized found by visual checks of the outlet tube and the valve letting out steam from the chimney. This means that from this point in time, 10:45, 4.69 kW power is delivered to the heating and vaporization, and 4.69 – 0.30 = 4.39 kW would have to come from the energy produced in the internal nickel-hydrogen container. This looks absurd. The power suddenly goes from 421 W to 4.69 kW when the water temperature reaches 100°C?? The most notable part of this EK statement is the phrase: visual checks of the outlet tube. This appears to mean periodic checks of the end of the rubber tube, in a manner similar to that done in the Krivit demonstration. Obviously only steam would come out of the top port, because the liquid water is separated, flowing
[Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
2011/9/22 Horace Heffner hheff...@mtaonline.net: EK: The system to measure the non-evaporated water was a certified Testo System, Testo 650, with a probe guaranteed to resist up to 550°C. The measurements showed that at 11:15 1.4% of the water was non-vaporized, at 11:30 1.3% and at 11:45 1.2% of the water was non-vaporized. This is nonsense because the Testo 650 is a relative humidity meter But it can also measure the enthalpy from steam by measuring the steam quality, if the mass flow of steam is known (We do not know it!). At least Galantini's DeltaOhm humidity measuring instrument can measure the enthalpy that steam caries. Jed Rothwell wrote: » In the first test, Galantini used a Delta Ohm monitor to measure the relative humidity of the steam. This is a model HD37AB1347 IAQ with a high temperature HP474AC SICRAM sensor. See: http://www.deltaohm.com/ver2010/uk/st_airQ.php?str=HD37AB1347 The brochure and the experts that Lewan and I have contacted say this instrument measures the enthalpy of steam. I expect they are right and the people who say otherwise here are wrong. I have no further comments on this issue. » DeltaOhm can measure the suspension of liquid water droplets and gaseous water medium, i.e. steam quality. I believe that Testo 650 can do the same trick. Problem is that measured steam quality (98.8%) does not tell us anything how much liquid water is overflowing, therefore we do not know the mass flow of steam and instrument is useless where they intended to use it. Your criticism is justified although you were incorrect whether Testo 650 can measure steam quality. === You are somewhat right that data from KE test is bad if not useless. Only way how we can estimate total enthalpy is that we assume, that we can use data from other E-Cat's where we have some better data. I.e. we assume that outlet hose is the same in all E-Cat's. This way we can estimate that there cannot be more heat generated than 2 kW, but lower limit is unknown. I estimated the lower limit to be 1.2 kW, I.e COP was 3-6. But I do not think that we can go for better accuracy than this. I think that your speculation about the auxiliary heater is not relevant, because if it would have provided total input heat that exceeds 300 watts, they would have included it into report. Also I think that your assumption that thermometer reading may be affected by heat that is conducted via copper, is unreasonable, because heat transfer between copper and stem/hot water is much much larger than what copper can conduct. This is that copper and water temperature are the same near thermometer. Also if thermometer would be affected by the heat, then the probe would be misplaced. And it would indicate that Rossi is lying, because he said that the probe measures the steam temperature. And we cannot expect that Rossi is lying, because if we do then we should assume that there is hidden conventional power source that provides alleged excess heat. This would end all discussion, because no-one has given scientific validation for E-Cat, because we cannot exclude hidden power sources. Further, this measurement ignores the water which can pour out of the exit port, or bubble or spurt out by a percolator effect, and which is not measured. This fact amazes me the most in this saga. How it is seriously possible, that such a grandiose mistake could have been made in assumptions. Not once, not twice but six times! I think Lewan got the message after discussion of his September demonstration. EK: Any chemical process for producing 25 kWh from any fuel in a 50 cm3 container can be ruled out. The only alternative explanation is that there is some kind of a nuclear process that gives rise to the measured energy production. This conclusion is without foundation because the 25 kWh number is without foundation. Due to inadequate instrumentation there is not even solid evidence the the power out is greater than the electrical power in. There are various inconsistencies in the report that can not be resolved without more detailed knowledge of the inside of the E-cat at the time, and better instrumentation for the test. That is correct conclusion. And further more. Fuel container does not need to be inside the core, but e.g. that wooden stand could be hollow. That could easily hide several kilograms of fuel, if needed. Only way we can make estimations, is to use other better demonstrations as reference and to _assume_ that this E-Cat behaved similarly as other E-Cats. 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
Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
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. Most of the input water mass flow necessarily must have continued on out the exit port without being converted to steam. That presumes that the heat exchange takes place on the surface of the core. But the heat is (supposedly) produced by thermalization of gamma rays, which could be anywhere nearby. Rossi has said that it is partly in the copper tubing and partly in the lead shielding. The total available area is easily 10 times that of the core, so the delta T could be 780C, not 7800C.
[Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
Am 22.09.2011 02:18, schrieb Horace Heffner: FIG. 4 NOTES It is notable the 2 cm of lead is not evident. The insulation looks more like a polyester than fiberglass, and if so its melting temperature is around 250°C. It could be silicone.
Re: [Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
On Sep 21, 2011, at 8:24 PM, Peter Heckert wrote: Am 22.09.2011 02:18, schrieb Horace Heffner: FIG. 4 NOTES It is notable the 2 cm of lead is not evident. The insulation looks more like a polyester than fiberglass, and if so its melting temperature is around 250°C. It could be silicone. Yes. Good thought. http://www.sz-wholesaler.com/p/860/863-1/insulation-ceramic-fiber- blanket-379821.html http://tinyurl.com/3ntc5xj Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
Am 22.09.2011 05:51, schrieb Alan Fletcher: That presumes that the heat exchange takes place on the surface of the core. But the heat is (supposedly) produced by thermalization of gamma rays, which could be anywhere nearby. Rossi has said that it is partly in the copper tubing and partly in the lead shielding. The total available area is easily 10 times that of the core, so the delta T could be 780C, not 7800C. Its early in the morning, not much time. Quick thought: If the heat is released inside the lead shield, then it must travel through the lead. I dont know the thermal conductivity of lead, but lead melts at 325 centigrade (if I remember correctly) and this also should limit the thermal flow. Peter
