Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Oct 10, 2011, at 4:57 PM, OrionWorks - Steven Vincent Johnson wrote: Ed Storms said it was ok for me to post the following analysis he made: * * * * * * A careful examination of the attached graph reveals an interesting conclusion. The Pout (power out) and the Eout (Energy out) appear to describe the net excess, not the total as everyone seems to assume. Power is applied to the internal heater, showed by the red dots, until extra power starts to increase starting at 140 min. The power to the heater is turned off for a short time at 160 min because the excess power starts to rise. This interruption of applied power and the resulting reduced temperature of the Ni caused the excess to decrease and excess power production is again brought under control. Applied power is interrupted several more times to test the stability of the power-producing reaction. Finally, applied power was turned off at 280 min whereupon the extra power increased and reached a relatively stable value. The variations in excess power production after 280 min are expected as the nuclear reaction responds to variations in local temperature in the Ni. The nuclear reaction slowly decayed away and the test was terminated before it stopped all together. I make two conclusions from this behavior. 1. The amount of energy produced was far in excess of any possible chemical source. 2. The energy-producing reaction is unstable and difficult to control. It also slowly becomes less productive unless the temperature is increased by an external source of power that can increase the temperature of the Ni, thereby causing a greater output of energy. This means the energy-producing reaction has a limited life-time, which is what Rossi has indicated. If the Pout and E out are interpreted as net excess, the graph makes perfect sense and is consistent with how such a device must behave. Ed I provided two spreadsheets from which the graphs were produced: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf http://www.mtaonline.net/~hheffner/Rossi6Oct2011noBias.pdf The latter one uses the raw data, the former has an 0.8°C bias applied to Delta T to compensate for probable thermocouple error, as noted in the DISCUSSION OF GRAPH 4 section of the review: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf The graphs were taken from the spread sheet with the bias. The above seems to refer to Graph 1, which is in the review, but also in higher resolution here: http://www.mtaonline.net/~hheffner/RossiGraph.png Graph 2 in high resolution is here: http://www.mtaonline.net/~hheffner/RossiT2Pout.png Graph 3 in high resolution is here: http://www.mtaonline.net/~hheffner/RossiT2Pout.png I do not see how Pout and Eout can be interpreted as net excess. I am possibly missing the intended meaning of this phrase. Delta Eout is the thermal energy detected by the heat exchanger for the time period of a given row. Pout and Eout are created from this number. Pout is determined by a ratio of Delta Eout to the time period. Eout is just a sum of all the Delta Eout values to the end of the individual time periods each row represents. These numbers represent the thermal output. The net output, i.e. output energy - input energy, is not in the graph. It is in the spread sheet column Net E. One way to interpret Ed's phrase net excess is to consider the thermal energy still stored in the E-cat as part of the total thermal energy generated. That which has escaped and been measured by the heat exchanger is the net of total thermal energy generated minus the still stored energy. However, this interpretation does not seem to add anything to understanding discussion. When cold water is run through the E-cat sufficiently long that it cools, and if there is no nuclear energy generated, and the calorimetry works well, then Net E should be zero at the end of the run, and COP should be 1. No energy is then left stored in the E-cat at the end. This is how a control run should be evaluated, and a live test done. The power Pin applied to the heater in Graph 1 is indeed the red line. In Graph 2 it is the brown line. I think Graphs 2 and 3 have much to say about how well controlled the reaction is, if there indeed is one. In Graph 2 we can see the E-cat temperature is very well controlled. In the time 220 - 280 the red line T2 is fairly flat. There is no sign of any runaway reaction - even though the power was applied for a long period. T2 even looks fairly flat for the period 200-280. The output power Pout detected at the heat exchanger, however, is anything but flat. This variation looks to me to be likely due to periods of water slugs moving through the exchanger, not variations from the nuclear reaction output. The most interesting relationship, however, I think is shown in Graph 3. The blue line shows a scaled version of the
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
The hyperlink to graph 3 is mistakenly pointing to graph 2 I think. On Tue, Oct 11, 2011 at 2:44 AM, Horace Heffner hheff...@mtaonline.netwrote: On Oct 10, 2011, at 4:57 PM, OrionWorks - Steven Vincent Johnson wrote: Ed Storms said it was ok for me to post the following analysis he made: * * * * * * A careful examination of the attached graph reveals an interesting conclusion. The Pout (power out) and the Eout (Energy out) appear to describe the net excess, not the total as everyone seems to assume. Power is applied to the internal heater, showed by the red dots, until extra power starts to increase starting at 140 min. The power to the heater is turned off for a short time at 160 min because the excess power starts to rise. This interruption of applied power and the resulting reduced temperature of the Ni caused the excess to decrease and excess power production is again brought under control. Applied power is interrupted several more times to test the stability of the power-producing reaction. Finally, applied power was turned off at 280 min whereupon the extra power increased and reached a relatively stable value. The variations in excess power production after 280 min are expected as the nuclear reaction responds to variations in local temperature in the Ni. The nuclear reaction slowly decayed away and the test was terminated before it stopped all together. I make two conclusions from this behavior. 1. The amount of energy produced was far in excess of any possible chemical source. 2. The energy-producing reaction is unstable and difficult to control. It also slowly becomes less productive unless the temperature is increased by an external source of power that can increase the temperature of the Ni, thereby causing a greater output of energy. This means the energy-producing reaction has a limited life-time, which is what Rossi has indicated. If the Pout and E out are interpreted as net excess, the graph makes perfect sense and is consistent with how such a device must behave. Ed I provided two spreadsheets from which the graphs were produced: http://www.mtaonline.net/~**hheffner/Rossi6Oct2011.pdfhttp://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf http://www.mtaonline.net/~**hheffner/Rossi6Oct2011noBias.**pdfhttp://www.mtaonline.net/~hheffner/Rossi6Oct2011noBias.pdf The latter one uses the raw data, the former has an 0.8°C bias applied to Delta T to compensate for probable thermocouple error, as noted in the DISCUSSION OF GRAPH 4 section of the review: http://www.mtaonline.net/~**hheffner/Rossi6Oct2011Review.**pdfhttp://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf The graphs were taken from the spread sheet with the bias. The above seems to refer to Graph 1, which is in the review, but also in higher resolution here: http://www.mtaonline.net/~**hheffner/RossiGraph.pnghttp://www.mtaonline.net/~hheffner/RossiGraph.png Graph 2 in high resolution is here: http://www.mtaonline.net/~**hheffner/RossiT2Pout.pnghttp://www.mtaonline.net/~hheffner/RossiT2Pout.png Graph 3 in high resolution is here: http://www.mtaonline.net/~**hheffner/RossiT2Pout.pnghttp://www.mtaonline.net/~hheffner/RossiT2Pout.png I do not see how Pout and Eout can be interpreted as net excess. I am possibly missing the intended meaning of this phrase. Delta Eout is the thermal energy detected by the heat exchanger for the time period of a given row. Pout and Eout are created from this number. Pout is determined by a ratio of Delta Eout to the time period. Eout is just a sum of all the Delta Eout values to the end of the individual time periods each row represents. These numbers represent the thermal output. The net output, i.e. output energy - input energy, is not in the graph. It is in the spread sheet column Net E. One way to interpret Ed's phrase net excess is to consider the thermal energy still stored in the E-cat as part of the total thermal energy generated. That which has escaped and been measured by the heat exchanger is the net of total thermal energy generated minus the still stored energy. However, this interpretation does not seem to add anything to understanding discussion. When cold water is run through the E-cat sufficiently long that it cools, and if there is no nuclear energy generated, and the calorimetry works well, then Net E should be zero at the end of the run, and COP should be 1. No energy is then left stored in the E-cat at the end. This is how a control run should be evaluated, and a live test done. The power Pin applied to the heater in Graph 1 is indeed the red line. In Graph 2 it is the brown line. I think Graphs 2 and 3 have much to say about how well controlled the reaction is, if there indeed is one. In Graph 2 we can see the E-cat temperature is very well controlled. In the time 220 - 280 the red line T2 is fairly flat. There is no sign of any runaway reaction - even though the power was applied
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Oct 10, 2011, at 11:10 PM, Axil Axil wrote: The hyperlink to graph 3 is mistakenly pointing to graph 2 I think. Right you are. Thanks! Should have been: http://www.mtaonline.net/~hheffner/RossiT2_RF.png On Tue, Oct 11, 2011 at 2:44 AM, Horace Heffner hheff...@mtaonline.net wrote: On Oct 10, 2011, at 4:57 PM, OrionWorks - Steven Vincent Johnson wrote: Ed Storms said it was ok for me to post the following analysis he made: * * * * * * A careful examination of the attached graph reveals an interesting conclusion. The Pout (power out) and the Eout (Energy out) appear to describe the net excess, not the total as everyone seems to assume. Power is applied to the internal heater, showed by the red dots, until extra power starts to increase starting at 140 min. The power to the heater is turned off for a short time at 160 min because the excess power starts to rise. This interruption of applied power and the resulting reduced temperature of the Ni caused the excess to decrease and excess power production is again brought under control. Applied power is interrupted several more times to test the stability of the power-producing reaction. Finally, applied power was turned off at 280 min whereupon the extra power increased and reached a relatively stable value. The variations in excess power production after 280 min are expected as the nuclear reaction responds to variations in local temperature in the Ni. The nuclear reaction slowly decayed away and the test was terminated before it stopped all together. I make two conclusions from this behavior. 1. The amount of energy produced was far in excess of any possible chemical source. 2. The energy-producing reaction is unstable and difficult to control. It also slowly becomes less productive unless the temperature is increased by an external source of power that can increase the temperature of the Ni, thereby causing a greater output of energy. This means the energy-producing reaction has a limited life-time, which is what Rossi has indicated. If the Pout and E out are interpreted as net excess, the graph makes perfect sense and is consistent with how such a device must behave. Ed I provided two spreadsheets from which the graphs were produced: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf http://www.mtaonline.net/~hheffner/Rossi6Oct2011noBias.pdf The latter one uses the raw data, the former has an 0.8°C bias applied to Delta T to compensate for probable thermocouple error, as noted in the DISCUSSION OF GRAPH 4 section of the review: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf The graphs were taken from the spread sheet with the bias. The above seems to refer to Graph 1, which is in the review, but also in higher resolution here: http://www.mtaonline.net/~hheffner/RossiGraph.png Graph 2 in high resolution is here: http://www.mtaonline.net/~hheffner/RossiT2Pout.png Graph 3 in high resolution is here: http://www.mtaonline.net/~hheffner/RossiT2Pout.png I do not see how Pout and Eout can be interpreted as net excess. I am possibly missing the intended meaning of this phrase. Delta Eout is the thermal energy detected by the heat exchanger for the time period of a given row. Pout and Eout are created from this number. Pout is determined by a ratio of Delta Eout to the time period. Eout is just a sum of all the Delta Eout values to the end of the individual time periods each row represents. These numbers represent the thermal output. The net output, i.e. output energy - input energy, is not in the graph. It is in the spread sheet column Net E. One way to interpret Ed's phrase net excess is to consider the thermal energy still stored in the E-cat as part of the total thermal energy generated. That which has escaped and been measured by the heat exchanger is the net of total thermal energy generated minus the still stored energy. However, this interpretation does not seem to add anything to understanding discussion. When cold water is run through the E-cat sufficiently long that it cools, and if there is no nuclear energy generated, and the calorimetry works well, then Net E should be zero at the end of the run, and COP should be 1. No energy is then left stored in the E-cat at the end. This is how a control run should be evaluated, and a live test done. The power Pin applied to the heater in Graph 1 is indeed the red line. In Graph 2 it is the brown line. I think Graphs 2 and 3 have much to say about how well controlled the reaction is, if there indeed is one. In Graph 2 we can see the E-cat temperature is very well controlled. In the time 220 - 280 the red line T2 is fairly flat. There is no sign of any runaway reaction - even though the power was applied for a long period. T2 even looks fairly flat for the period 200-280. The output power Pout detected at the heat exchanger,
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
“As already speculated by a few here, Rossi continues to give me the impression that he operates very much on intuition. Recording scientific data is almost incidental to him, a characteristic I suspect probably drives a few of his colleagues to distraction. “ After watching Rossi for some months now, it looks to me like he is the quintessential edistonian trial and error development type of guy. He learns by doing. It might be that he is learning how to do demos by trial and error. After about a dozen more demos he may get his act together and come up with a demo that is generally acceptable to the majority. He doesn’t take the time required to plan things because he is working mostly in the dark of a largely unknown technology. He operates in a “mouse in a maze” methodology; what the software people case a rapid prototyping mode. It must drive the thoughtful planers amongst his colleagues and many of us here amongst our number right up a wall. On Sun, Oct 9, 2011 at 7:12 PM, OrionWorks - Steven Vincent Johnson orionwo...@charter.net wrote: Thanks for the analysis, Jed. Will be interesting to read what others have to say. ** ** BTW, what did Rossi have to say? ** ** * * * * * ** ** When I look at the graph I continue to be drawn to the curious fact that the input power is cycled on and off a total of three or four times starting from around 13:59 to finally ending at 15:50 when it is permanently turned off. Looks to me as if Rossi's team may have been trying to get their eCat airborne way before the time stamp of 15:50. ** ** My apologies if the following has already been discussed or speculated since there has been so much discussion in the past three days – I can't keep track of it all. The characteristics of the input data gives me the impression that Rossi's team is trying to capitalize on what I would describe as the Sweet Spot, where Rossi feels that the core reaction is finally beginning to take off without further need for an input power source to sustain the output reaction. It's analogous to the Wright Brothers hand cranking the propeller of their first air craft where the first couple of spins don't necessarily catch on with the engine. ** ** As already speculated by a few here, Rossi continues to give me the impression that he operates very much on intuition. Recording scientific data is almost incidental to him, a characteristic I suspect probably drives a few of his colleagues to distraction. Rossi has probably acquired a reasonable amount of instinctual horse sense as to when he thinks his mysterious eCats are likely to take off in self-sustain mode. ** ** The following is what I speculate is happening between 13:59 to 15:50: ** ** Rossi initially tries at 13:59... *It's catching It's catching... Ah, shoot! It petered out. Ok guys! Crank her up again.* Input Power turned back on 14:11. ** ** Rossi tries again at around 14:24... *Well, Shoot. Still didn't catch! Maybe I need to prime the pistons. Where's my canister of Ether. Ok guys. Crank her up again.* Input power turned back on at 14:36 ** ** Looks like Rossi tries for the third time at around 14:48, but I suspect the there are data anomalies here (human error?) and Rossi actually turns off the input power at around 15:00. It's turning... It's turning... *Come on! Come on You can do it Shoot it's going down again. We're close guys! I can feel it in my ancient Italian bones! Ok, let's crank'er up again.* Input Power turned back on at 15:25. ** ** For the fourth time, Rossi turns off the input power around 15:50. Meanwhile the output signal has been strong and rapidly rising starting at around 15:40 or so. Rossi’s Italian bones sense that this is probably the Big One. …*TURN THE INPUT POWER OFF Got it! Hand me my goggles, guys! It's Steam Punk Rock'N'Role time!* ** ** Here are some final personal interpretations: ** ** It looks to me as if in every case input power is turned off several minutes after the Rossi senses that the output power is on a steady rise (in self-sustain mode) towards the 3000 mark and above. ** ** I wonder if Rossi may have initially been trying to hit that sweet spot early in the data recordings starting at 13:59. Perhaps he initially turned input power off back then in order to help minimize the potential of introducing skeptical arguments such as those presented over at Krivit's blog having to do with the total accumulation of input power from the start of the experiment and how the entire collection of data appears to be greater than the total accumulation of recorded output power. In any case, it looks to me as if Rossi had three false starts before he finally hit pay dirt on the fourth crank. ** ** Comments? ** ** Regards, Steven Vincent Johnson www.OrionWorks.com http://www.orionworks.com/
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On 11-10-10 04:35 PM, Robert Leguillon wrote: If someone Couldn't care less, it means that they care so little that it's impossible for them to care any less than they do right now. If someone Could care less, it means that they care enough that it's possible to care less. _*Irregardless*_, people will continue to use the phrase to the four corners of the earth. Supposably, it's commonplaced. Oops. In fact, regardless is the word you want there. Irregardless is not an official word but if it were it would mean the opposite of what you intended. (Some modern dictionaries have surrendered to the ungrammatical hordes and now define it as a synonym for regardless, but all us nit-pickers out here know they're wrong.) Date: Mon, 10 Oct 2011 15:24:41 -0500 Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof From: svj.orionwo...@gmail.com To: vortex-l@eskimo.com Terry sez: ... I'm sure you could care less. whisper: . . . not care less g, d r Really? I wuz never good at grammar. Grammatically speaking I always thought it is better form to avoid cluttering up one's literary intent with the use of double negatives. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
That was intentional - just keeping you guys on your toes. Irregardless: should be regardless Four corners of the Earth: the earth does not have four corners Supposably: should be SupposeDLy Commonplaced: should be commonplace (no d) Stephen A. Lawrence sa...@pobox.com wrote: On 11-10-10 04:35 PM, Robert Leguillon wrote: If someone Couldn't care less, it means that they care so little that it's impossible for them to care any less than they do right now. If someone Could care less, it means that they care enough that it's possible to care less. _*Irregardless*_, people will continue to use the phrase to the four corners of the earth. Supposably, it's commonplaced. Oops. In fact, regardless is the word you want there. Irregardless is not an official word but if it were it would mean the opposite of what you intended. (Some modern dictionaries have surrendered to the ungrammatical hordes and now define it as a synonym for regardless, but all us nit-pickers out here know they're wrong.) Date: Mon, 10 Oct 2011 15:24:41 -0500 Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof From: svj.orionwo...@gmail.com To: vortex-l@eskimo.com Terry sez: ... I'm sure you could care less. whisper: . . . not care less g, d r Really? I wuz never good at grammar. Grammatically speaking I always thought it is better form to avoid cluttering up one's literary intent with the use of double negatives. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
That appears to be a graph of power noy yemperature. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 9:24 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: No the band heater is at 900C but that metal block talk was only for illustrative purposes. Newtons LAw is irrelevant. Newton's law governs passive heat loss, which is what this has to be if there is not energy input and the flow rate does change. An insulated metal block that loses heat at a rate of 1W loses heat at the rate of 1W. You mention lack of monotonicity but what's the example (be specific, post link). Right here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg The temperature rises several times after the power is turned off. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
If its passive cooling? Excuse me but are we discussing something here? - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 9:41 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Excuse me I meant to say that the cooling rate must obey Newton's law if there is NO energy generation and the flow rate does NOT change. In other words, if it passive cooling in unchanging conditions. Lewan's observations and report show that the flow rate and other essential parameters did not change. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Oct 9, 2011, at 7:05 PM, Robert Leguillon wrote: Alright, if it's conclusive without the thermocouples Does anyone have a decent water capacity for the E-Cat? I see that H.H. calculated 14.2 liters, but has there been any confirmed number out of the Rossi camp? I only ask, because multiple references have been made to tons of cooling water to quench the reaction during H.A.D. In reality, the water flowing through the E-Cat (as the heat exchanger primary-side output) was measured twice: The first time, it was .91 grams/sec and the second time it was just shy of 2 g/s. If the E-Cat were indeed 14.2liters (14.2 kg), the entire contents of the E-Cat would take 2-4 hours to be completely replaced. All the while, a device that generates frequencies is still running. When it is turned off, the E-Cat temp begins declining. S many questions. Somewhere I think I saw a statement that the new E-cat has 30 liters water volume. I don't see how that is possible if the dimensions provided in the NyTeknik report are correct. I have made changes to my data review in this area. It is located at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Following are the related sections. VOLUME CALCULATIONS The Lewan report says: The E-cat model used in this test was enclosed in a casing measuring about 50 x 60 x 35 centimeters. After cooling down the E-cat, the insulation was eliminated and the casing was opened. Inside the casing metal flanges of a heat exchanger could be seen, an object measuring about 30 x 30 x 30 centimeters. The rest of the volume was empty space where water could be heated, entering through a valve at the bottom, and with a valve at the top where steam could come out. This gives an external volume of (50 x 60 x 35) cm^3 = 105000 cm^3 = 105 liters. The heat exchanger etc. is (30 x 30 x 30) cm^3 = 27 liters. This should give an internal volume of 105 liters - 27 liters = 78 liters. The prior similar E-cat weighed in at 85 kg. Looking at the open E-cat photo it looks like about (1/9)*30 cm = 3.3 cm is cooling fins. About 50% of the 3.3 cm x 30 cm x 30 = 2.97 liters should be water, giving a total water volume of 78 liters + 3 liters = 81 liters. NO HEAT TRANSFER TO HEAT EXCHANGER UNTIL 13:22 19:22: Measured outflow of primary circuit in heat exchanger, supposedly condensed steam, to be 345 g in 180 seconds, giving a flow of 1.92 g/s. Temperature 23.2 °C. This indicates the pump primary circuit flow is probably about 1.92 ml/s, as it was in the Krivit demo. The heat showed up in the exchanger at about 130 minutes, or 7800 seconds into the run. See appended graph, or see spreadsheet at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf This means the flow filled a void of (7800 s)*(1.92 ml/s) = 15 liters before hot water began to either overflow or percolate out of the device, and thus make it to the heat exchanger. If overflow started after 15 liters then it would appear 81 - 15 = 66 liters were already present. The device weighed in at 98 kg before the test and 99 kg after, when the water was drained, making this impossible. If the E-cat cold water input is 24°C and 15 liters were input, it takes (4.2 J/(gm K)) *(15,000 gm))*(76K) = 4.79 MJ to heat the water to boiling. Looking at the spread sheet this input energy Ein was indeed reached at about 13:22. This means steam probably reached the heat exchanger at this time, about 130 minutes into the test. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Joe Catania zrosumg...@aol.com wrote: ** That appears to be a graph of power noy yemperature. It is derived from Lewan's temperature readings. The flow rate was unchanged so correspondence to the temperature is unchanged for the entire dataset. In other words, you could replace the vertical axis power numbers with the corresponding temperatures and it would look exactly the same. ** If its passive cooling? Excuse me but are we discussing something here? You claim the heat comes from heat storage with no input power. That would mean it is passive cooling, by definition. It has to follow Newton's law of cooling. That is how heat storage and release works. You keep talking about thermal inertia. I suggest you learn what that is, how it works, and what laws of physics govern it. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Robert Leguillon robert.leguil...@hotmail.com wrote: The rapid overfilling was at .91 grams/second (It turns out the 1.92 g/s was for quenching) The rapid overfill I refer to is the quenching, at 1.92 g/s. I believe 0.91 was the rate during the test when Lewan checked it. 1.92 isn't very rapid, is it? Apparently it worked though. With Pd-D I have heard of researchers taking the cathode out and plunged into a cold bath. It can be difficult to quench the reaction. An additional 2,056 watts is required for the phase-change, but, of course, we have no idea how much is boiling away. Greater than 2,437 watts would completely vaporize the input water. Since the temperature is 120°C I believe it has to be completely vaporized. I do not think there can be any hot water at that temperature in the system. Of course, this means that the water in the E-Cat would be running dry and getting super-heated if there were prolonged excursions over 2.5 kW. But, of course, this didn't happen, did it? No, it means that Rossi has to keep an eye on the water level. He has to adjust it to keep the reactor full but not overflowing. This is no more difficult than it is for me to keep an eye on the level of pinot noir when I simmer a pot roast for 5 hours. (The trick to a good pot roast is keep the water level low so that it forms at thick brown sauce, but not so low that it burns.) Of course I can open the pot and look, but I can also tell from the sound and smell. With my miniature steam engine I can tell the boiler level from the sound as well. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Newton's Law is irrelevant. Your the type of buffoon who believes that since there's an Ohms LAw every conductor obeys it. The temperature law the e-cat obeys is ostensibly written in the temperature data if we can consider that valid. Whether that confirms its Newton's Law or notr is not relevant to the dubunking of the CF myth. Cf is not being assumed and since it hasn't been shown we are correct in not assuming it. You still aren't able to show me the temperature data you say exists and is increasing. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 10:28 AM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: That appears to be a graph of power noy yemperature. It is derived from Lewan's temperature readings. The flow rate was unchanged so correspondence to the temperature is unchanged for the entire dataset. In other words, you could replace the vertical axis power numbers with the corresponding temperatures and it would look exactly the same. If its passive cooling? Excuse me but are we discussing something here? You claim the heat comes from heat storage with no input power. That would mean it is passive cooling, by definition. It has to follow Newton's law of cooling. That is how heat storage and release works. You keep talking about thermal inertia. I suggest you learn what that is, how it works, and what laws of physics govern it. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Newton's Law is irrelevant. Your the type of buffoon who believes that since there's an Ohms LAw every conductor obeys it. The temperature law the e-cat obeys is ostensibly written in the temperature data if we can consider that valid. Whether that confirms its Newton's Law or notr is not relevant to the dubunking of the CF myth. Cf is not being assumed and since it hasn't been shown we are correct in not assuming it. You still aren't able to show me the temperature data you say exists and is increasing. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 10:28 AM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: That appears to be a graph of power noy yemperature. It is derived from Lewan's temperature readings. The flow rate was unchanged so correspondence to the temperature is unchanged for the entire dataset. In other words, you could replace the vertical axis power numbers with the corresponding temperatures and it would look exactly the same. If its passive cooling? Excuse me but are we discussing something here? You claim the heat comes from heat storage with no input power. That would mean it is passive cooling, by definition. It has to follow Newton's law of cooling. That is how heat storage and release works. You keep talking about thermal inertia. I suggest you learn what that is, how it works, and what laws of physics govern it. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
From Joe Catania: On Mon, Oct 10, 2011 at 10:04 AM, Joe Catania wrote: Newton's Law is irrelevant. Your the type of buffoon who believes that since there's an Ohms LAw every conductor obeys it. The temperature law the e-cat obeys is ostensibly written in the temperature data if we can consider that valid. Whether that confirms its Newton's Law or notr is not relevant to the dubunking of the CF myth. Cf is not being assumed and since it hasn't been shown we are correct in not assuming it. You still aren't able to show me the temperature data you say exists and is increasing. Ok, Mr. Catania, TIME OUT Starting to call other people derogatory names is only going to come back and bite you in the parte posteriore. Keep this up and your rhetoric will eventually come to the attention of the benevolent vortex-l dictator, Mr. Beaty, and he will most likely deal with you in any what he sees fit. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Jed, I said: An additional 2,056 watts is required for the phase-change, but, of course, we have no idea how much is boiling away. Greater than 2,437 watts would completely vaporize the input water. You said: Since the temperature is 120°C I believe it has to be completely vaporized. I do not think there can be any hot water at that temperature in the system. The 120 degrees C must be boiling with back pressure or an incorrect reading from thermal wicking of the metal. If it were truly superheated steam, you would see large variances, especially when it allegedly surpassed 6 or 8 kW. It is at boiling. I said: Of course, this means that the water in the E-Cat would be running dry and getting super-heated if there were prolonged excursions over 2.5 kW. But, of course, this didn't happen, did it? You said: No, it means that Rossi has to keep an eye on the water level. He has to adjust it to keep the reactor full but not overflowing. This is no more difficult than it is for me to keep an eye on the level of pinot noir when I simmer a pot roast for 5 hours. Anything over 2.5 kW would have been boiling more water than was being introduced into the E-Cat. But, you're saying that Rossi was constantly adjusting flow from the paristaltic pump to ensure that all of the input water was being evaporated, but no more? When the E-Cat power allegedly teiples, he also triples the input water, so as to maintain exact 120 degree temperature of the above-boiling-temperature steam? I didn't see this anywhere in the reports. This wasn't worth mentioning? I'm just saying that the calorimetry does not jive. I find the comments from Defkalion to be refreshing, though. I am optimistic towards Ni-H research, but I have not seen convincing data from any of Rossi's public tests. We can argue the points of each until the cows come home, but I assure you that I want nothing more than to be convinced. His actions would lead us to believe that he has seen it work, and other reports indicate that he is not alone. But, in my own opinion, this was certainly not a conclusive test. I think that extraordinary claims require extraordinary evidence; it's just not here. This could have very easily been a conclusive test, but it went just as predicted. Date: Mon, 10 Oct 2011 10:39:58 -0400 Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof From: jedrothw...@gmail.com To: vortex-l@eskimo.com Robert Leguillon robert.leguil...@hotmail.com wrote: The rapid overfilling was at .91 grams/second (It turns out the 1.92 g/s was for quenching) The rapid overfill I refer to is the quenching, at 1.92 g/s. I believe 0.91 was the rate during the test when Lewan checked it. 1.92 isn't very rapid, is it? Apparently it worked though. With Pd-D I have heard of researchers taking the cathode out and plunged into a cold bath. It can be difficult to quench the reaction. An additional 2,056 watts is required for the phase-change, but, of course, we have no idea how much is boiling away. Greater than 2,437 watts would completely vaporize the input water. Since the temperature is 120°C I believe it has to be completely vaporized. I do not think there can be any hot water at that temperature in the system. Of course, this means that the water in the E-Cat would be running dry and getting super-heated if there were prolonged excursions over 2.5 kW. But, of course, this didn't happen, did it? No, it means that Rossi has to keep an eye on the water level. He has to adjust it to keep the reactor full but not overflowing. This is no more difficult than it is for me to keep an eye on the level of pinot noir when I simmer a pot roast for 5 hours. (The trick to a good pot roast is keep the water level low so that it forms at thick brown sauce, but not so low that it burns.) Of course I can open the pot and look, but I can also tell from the sound and smell. With my miniature steam engine I can tell the boiler level from the sound as well. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On 11-10-10 11:04 AM, Joe Catania wrote: Newton's Law is irrelevant. Your the type of buffoon who ... And you, /Mister/ Catania, are apparently the type of poster who resorts to ad hominems when he's having trouble expressing himself clearly enough to get his point across. Jed's may be a lot of things, possibly even including wrong, but he's no buffoon. And you, /Mister/ Catania, are plonked. I don't need to see this kind of stuff on Vortex. (You are also apparently the type of poster who can't be bothered to proof read his posts for obvious typos before sending them, which also contributes needlessly to the annoyance level of this list.)
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On 11-10-10 12:33 PM, Stephen A. Lawrence wrote: On 11-10-10 11:04 AM, Joe Catania wrote: Newton's Law is irrelevant. Your the type of buffoon who ... And you, /Mister/ Catania, are apparently the type of poster who resorts to ad hominems when he's having trouble expressing himself clearly enough to get his point across. Jed's may be a lot of things, possibly even including wrong, but he's no buffoon. Jed's --- Jed [oops] And you, /Mister/ Catania, are plonked. I don't need to see this kind of stuff on Vortex. (You are also apparently the type of poster who can't be bothered to proof read his posts for obvious typos before sending them, which also contributes needlessly to the annoyance level of this list.)
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Jed Rothwell is a serious, intelligent, dedicated, honorable, careful, scientific layman with the highest motives to benefit our world -- he always acknowledges his bias clearly and openly. I think it would be much to his credit to agree that the term pathological skeptic is as unworthy in public discourse as buffoon. within infinite patience, Rich Murray
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Quit picking on Catania who does not know the difference between 'your' and 'you're'. He passed away some time ago as is evidenced by this piccy of him surrounded by flowers. RIP JOE! http://www.theeestory.com/posts/199540 T
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
LOL. That's hypocritical. - Original Message - From: Rich Murray rmfor...@gmail.com To: vortex-l@eskimo.com; Rich Murray rmfor...@gmail.com; Rich Murray rmfor...@comcast.net Sent: Monday, October 10, 2011 12:49 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Jed Rothwell is a serious, intelligent, dedicated, honorable, careful, scientific layman with the highest motives to benefit our world -- he always acknowledges his bias clearly and openly. I think it would be much to his credit to agree that the term pathological skeptic is as unworthy in public discourse as buffoon. within infinite patience, Rich Murray
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
LOL. That's hypocritical. - Original Message - From: Rich Murray rmfor...@gmail.com To: vortex-l@eskimo.com; Rich Murray rmfor...@gmail.com; Rich Murray rmfor...@comcast.net Sent: Monday, October 10, 2011 12:49 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Jed Rothwell is a serious, intelligent, dedicated, honorable, careful, scientific layman with the highest motives to benefit our world -- he always acknowledges his bias clearly and openly. I think it would be much to his credit to agree that the term pathological skeptic is as unworthy in public discourse as buffoon. within infinite patience, Rich Murray
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Funny, you don't seem annoyed. All Jed is capable with regard to this matter is condescension. - Original Message - From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 12:33 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof On 11-10-10 11:04 AM, Joe Catania wrote: Newton's Law is irrelevant. Your the type of buffoon who ... And you, /Mister/ Catania, are apparently the type of poster who resorts to ad hominems when he's having trouble expressing himself clearly enough to get his point across. Jed's may be a lot of things, possibly even including wrong, but he's no buffoon. And you, /Mister/ Catania, are plonked. I don't need to see this kind of stuff on Vortex. (You are also apparently the type of poster who can't be bothered to proof read his posts for obvious typos before sending them, which also contributes needlessly to the annoyance level of this list.)
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
No that was part of the decor in a restaurant in Taormina. Its nice to know that the only thing that counts here is spelling (and self-affected narcissists). - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 1:41 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Quit picking on Catania who does not know the difference between 'your' and 'you're'. He passed away some time ago as is evidenced by this piccy of him surrounded by flowers. RIP JOE! http://www.theeestory.com/posts/199540 T
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Congratulations, Mr. Catania. Further posts from you will be routed to my block list. I'm sure you could care less. I guess the feeling is mutual. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
What do my posts matter anyway? Yes please block me. - Original Message - From: OrionWorks - Steven V Johnson svj.orionwo...@gmail.com To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 2:50 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Congratulations, Mr. Catania. Further posts from you will be routed to my block list. I'm sure you could care less. I guess the feeling is mutual. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Mon, Oct 10, 2011 at 2:50 PM, OrionWorks - Steven V Johnson svj.orionwo...@gmail.com wrote: Congratulations, Mr. Catania. Further posts from you will be routed to my block list. I'm sure you could care less. I guess the feeling is mutual. whisper: . . . not care less g, d r -the narcissist
RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
From one narcissist to another... Seems ol Joe thinks he's converted the lot of us... http://www.theeestory.com/users/1681/posts# 80kgs of metal can easily store over 40MJ. It's not on the level of a discussion. My arguments have been extremely convincing as I think you can tell by the recent conversion of vortex members and Krivit. Joe Catania states, The band heater temp is ~900C. In September test my calculations show that boiling could be produced for many hours. There is certainly a massive amount of metal in the e-cat. Joe: So your reasoning is based on the band heater being 900C, and therefore the majority of the massive amount of metal in the E-Cat is at or near that same temperature. You sincerely think that everything underneath the insulation is anywhere near that temp? The melting point of lead is 327C, so we certainly know that the lead is no more than one-third 900C, or else we'd have a mass of molten lead on the table. In addition, with the irregularity of the shape of the plumbing, at least with the old, tubular design, it is unlikely that there is much physical contact between the lead shielding and the plumbing (water jacket), ergo, poor heat conduction between the plumbing and the lead, ergo, not much heat storage in the lead. Finally, the only thing that could be anywhere near 900C is the (stainless steel) core container that is the transfer medium between the reaction material (Ni-powder-hydrogen-catalyst) and the water outside the core container. Conclusion: Being that the only mass that could possibly be anywhere near 900C is the reactor core container, which might be a few kilograms, would you care to revise your ... not on the level of a discussion heat storage estimate??? -Mark
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
If that were the approach you would use graphite inductively heated to 3500 deg C in a graphite foil/foam insulated vacuum flask, add hydrogen to start convective heat transfer. Stores about 1.3kWh/kg and about 2.7kWh/liter, so would need about 10 liters for 80MJ of latest demo. Note I am sure this wasn't done, but would work better than iron On 10 October 2011 20:44, Mark Iverson-ZeroPoint zeropo...@charter.netwrote: From one narcissist to another... Seems ol Joe thinks he's converted the lot of us... http://www.theeestory.com/users/1681/posts# 80kgs of metal can easily store over 40MJ. It's not on the level of a discussion. My arguments have been extremely convincing as I think you can tell by the recent conversion of vortex members and Krivit. Joe Catania states, The band heater temp is ~900C. In September test my calculations show that boiling could be produced for many hours. There is certainly a massive amount of metal in the e-cat. Joe: So your reasoning is based on the band heater being 900C, and therefore the majority of the massive amount of metal in the E-Cat is at or near that same temperature. You sincerely think that everything underneath the insulation is anywhere near that temp? The melting point of lead is 327C, so we certainly know that the lead is no more than one-third 900C, or else we'd have a mass of molten lead on the table. In addition, with the irregularity of the shape of the plumbing, at least with the old, tubular design, it is unlikely that there is much physical contact between the lead shielding and the plumbing (water jacket), ergo, poor heat conduction between the plumbing and the lead, ergo, not much heat storage in the lead. Finally, the only thing that could be anywhere near 900C is the (stainless steel) core container that is the transfer medium between the reaction material (Ni-powder-hydrogen-catalyst) and the water outside the core container. Conclusion: Being that the only mass that could possibly be anywhere near 900C is the reactor core container, which might be a few kilograms, would you care to revise your ... not on the level of a discussion heat storage estimate??? -Mark
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Terry sez: ... I'm sure you could care less. whisper: . . . not care less g, d r Really? I wuz never good at grammar. Grammatically speaking I always thought it is better form to avoid cluttering up one's literary intent with the use of double negatives. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
If someone Couldn't care less, it means that they care so little that it's impossible for them to care any less than they do right now. If someone Could care less, it means that they care enough that it's possible to care less. Irregardless, people will continue to use the phrase to the four corners of the earth. Supposably, it's commonplaced. Date: Mon, 10 Oct 2011 15:24:41 -0500 Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof From: svj.orionwo...@gmail.com To: vortex-l@eskimo.com Terry sez: ... I'm sure you could care less. whisper: . . . not care less g, d r Really? I wuz never good at grammar. Grammatically speaking I always thought it is better form to avoid cluttering up one's literary intent with the use of double negatives. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Since you know nothing of the e-cat your remarks have been dismissed. Yes it was prooveable in the September e-cat that the effects were purely based on thermal inertia. I suspect the same here. Rothwwell has not been able to substantiate his position which seems to be a blind acceptance of CF before aanyone heard of Rossi. I never made the claims you say I made. Yes there has been conversion and elaborate journalism on this point. You seem to confuse your total ignorance with lack of merit. You will regret that. - Original Message - From: Mark Iverson-ZeroPoint zeropo...@charter.net To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 3:44 PM Subject: RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof From one narcissist to another... Seems ol Joe thinks he's converted the lot of us... http://www.theeestory.com/users/1681/posts# 80kgs of metal can easily store over 40MJ. It's not on the level of a discussion. My arguments have been extremely convincing as I think you can tell by the recent conversion of vortex members and Krivit. Joe Catania states, The band heater temp is ~900C. In September test my calculations show that boiling could be produced for many hours. There is certainly a massive amount of metal in the e-cat. Joe: So your reasoning is based on the band heater being 900C, and therefore the majority of the massive amount of metal in the E-Cat is at or near that same temperature. You sincerely think that everything underneath the insulation is anywhere near that temp? The melting point of lead is 327C, so we certainly know that the lead is no more than one-third 900C, or else we'd have a mass of molten lead on the table. In addition, with the irregularity of the shape of the plumbing, at least with the old, tubular design, it is unlikely that there is much physical contact between the lead shielding and the plumbing (water jacket), ergo, poor heat conduction between the plumbing and the lead, ergo, not much heat storage in the lead. Finally, the only thing that could be anywhere near 900C is the (stainless steel) core container that is the transfer medium between the reaction material (Ni-powder-hydrogen-catalyst) and the water outside the core container. Conclusion: Being that the only mass that could possibly be anywhere near 900C is the reactor core container, which might be a few kilograms, would you care to revise your ... not on the level of a discussion heat storage estimate??? -Mark
RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Joe: Is that the way to rebut someone who has only questioned some of your reasoning regarding the heat storage capacity of the E-Cat? Your rebuttal is to claim they know nothing about the E-Cat and dismiss their points with no facts or explanation! Then you go on continuing to claim that all your conclusions are right... hmmm, that sounds a bit, dare I say, narsissistic. Welcome to the club! :-) Rothwell has not been able to substantiate his position which seems to be a blind acceptance of CF before anyone heard of Rossi. What the F* does Jed's strong opinions on CF have to do with my questions to you??? 1) My reasoning has nothing to do with Rothwell or anything other than your insistence that the E-Cat's performance can be entirely explained by heat storage by the massive amounts of metal in the E-Cat and the band heater being 900C. 2) You say, I never made the claims you say I made. There are only two quotes which I attribute to you: 80kgs of metal can easily store over 40MJ. It's not on the level of a discussion. My arguments have been extremely convincing as I think you can tell by the recent conversion of vortex members and Krivit. And, The band heater temp is ~900C. In September test my calculations show that boiling could be produced for many hours. There is certainly a massive amount of metal in the e-cat. Well, I just saved the webpage at TheEEStory.com where I COPIED these quotes from... I would be happy to email it to you. I'd attach the JPEG of the screen capture I made but it's too big and the vortex-l server will not allow the posting. I suppose that someone could have hacked into the EEStory.com website and changed the wording on your forum postings... but who would bother, if it's even possible. Finally, you resort to attacking and threatening me... You seem to confuse your total ignorance with lack of merit. You will regret that. Bring it on Joe... Having been in several small to medium sized startups, and on the Board of Directors for two of them, I've been threatened numerous times with jail and lawsuits and other nasty and unpleasant things. What would really be nice is for you to simply answer my question from the original posting, and I'll repeat it here: So your reasoning is based on the band heater being 900C, and therefore the majority of the massive amount of metal in the E-Cat is at or near that same temperature. You sincerely think that everything underneath the insulation is anywhere near that temp? Being that the only mass that could possibly be anywhere near 900C is the reactor core container, which might be a few kilograms, would you care to revise your ... not on the level of a discussion heat storage estimate??? Well, I guess there are two questions in that... Clearly, these are NOT statements attributed to you, but legitimate, reasonable QUESTIONS to you. All I expect is for you to clear up your reasoning regarding HOW MUCH of all that massive metal is at the very high temperatures that you constantly use in your examples to prove that metal can store megajoules of heat! -Mark -Original Message- From: Joe Catania [mailto:zrosumg...@aol.com] Sent: Monday, October 10, 2011 12:56 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Since you know nothing of the e-cat your remarks have been dismissed. Yes it was provable in the September e-cat that the effects were purely based on thermal inertia. I suspect the same here. Rothwell has not been able to substantiate his position which seems to be a blind acceptance of CF before anyone heard of Rossi. I never made the claims you say I made. Yes there has been conversion and elaborate journalism on this point. You seem to confuse your total ignorance with lack of merit. You will regret that. - Original Message - From: Mark Iverson-ZeroPoint zeropo...@charter.net To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 3:44 PM Subject: RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof From one narcissist to another... Seems ol Joe thinks he's converted the lot of us... http://www.theeestory.com/users/1681/posts# 80kgs of metal can easily store over 40MJ. It's not on the level of a discussion. My arguments have been extremely convincing as I think you can tell by the recent conversion of vortex members and Krivit. Joe Catania states, The band heater temp is ~900C. In September test my calculations show that boiling could be produced for many hours. There is certainly a massive amount of metal in the e-cat. Joe: So your reasoning is based on the band heater being 900C, and therefore the majority of the massive amount of metal in the E-Cat is at or near that same temperature. You sincerely think that everything underneath the insulation is anywhere near that temp? The melting point of lead is 327C, so we certainly know that the lead is no more than
RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Ed Storms said it was ok for me to post the following analysis he made: * * * * * * A careful examination of the attached graph reveals an interesting conclusion. The Pout (power out) and the Eout (Energy out) appear to describe the net excess, not the total as everyone seems to assume. Power is applied to the internal heater, showed by the red dots, until extra power starts to increase starting at 140 min. The power to the heater is turned off for a short time at 160 min because the excess power starts to rise. This interruption of applied power and the resulting reduced temperature of the Ni caused the excess to decrease and excess power production is again brought under control. Applied power is interrupted several more times to test the stability of the power-producing reaction. Finally, applied power was turned off at 280 min whereupon the extra power increased and reached a relatively stable value. The variations in excess power production after 280 min are expected as the nuclear reaction responds to variations in local temperature in the Ni. The nuclear reaction slowly decayed away and the test was terminated before it stopped all together. I make two conclusions from this behavior. 1. The amount of energy produced was far in excess of any possible chemical source. 2. The energy-producing reaction is unstable and difficult to control. It also slowly becomes less productive unless the temperature is increased by an external source of power that can increase the temperature of the Ni, thereby causing a greater output of energy. This means the energy-producing reaction has a limited life-time, which is what Rossi has indicated. If the Pout and E out are interpreted as net excess, the graph makes perfect sense and is consistent with how such a device must behave. Ed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Ed Storms wrote: A careful examination of the attached graph reveals an interesting conclusion. This refers to Heffner's graph 1: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Mon, Oct 10, 2011 at 8:57 PM, OrionWorks - Steven Vincent Johnson orionwo...@charter.net wrote: Ed Storms said it was ok for me to post the following analysis he made: Isn't PoutE a bit funny? T
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Oct 10, 2011, at 5:01 PM, Jed Rothwell wrote: Ed Storms wrote: A careful examination of the attached graph reveals an interesting conclusion. This refers to Heffner's graph 1: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf - Jed BTW, I finally figured out how to make the charts look better in the pdf. I simply had to make them a bit more narrow. I did so and moved the legend to the bottom. I'm still not happy with it. I guess I need to update or change my software. The review is still a work in progress. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Or if it is refutable, let's see someone make a serious effort to refute it. Stop quibbling about details. Get the heart of the matter, and tell us how a box of this size with no input power can boil water for 3 hours and remain at the same high temperature while you cool it with 1.8 tons of water. I wrote to some friends complaining about the test. My conclusion: Despite these problems . . . I think this test produced irrefutable proof of anomalous heat. Here is why I think so -- Look at the graph here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg Nothing happens until 13:22 when the steam begins to flow through the heat exchanger. At 15:13 output is a little higher than input, even though there is a great deal of heat unaccounted for, especially the water from the condensed steam, which they poured down the drain. At 15:50 the power is cut off. If there had been no source of anomalous heat, the power would have fallen off rapidly and monotonically, at the same rate it did after 19:55. It would have approached the zero line by 17:25. Actually, it would have approached zero before that, based on Newton's law of cooling. In other words, it would have been stone cold after 3 hours. During that time, 1.8 tons of water went through the cooling loop. It is inconceivable that an object of this size with no power input could have remained at the *same high temperature* the whole time. Yet Lewan reports that the surface of the reactor was still hot, and boiling could still be heard inside it. As you see, the temperature did not fall. It went up at 16:26. The cooling water flow rate was unchanged, so only a source of heat could have caused this. You can ignore the thermocouple data, and look only at the fact that it continued to boil for more than 3 hours after the power was turned off, and the reactor surface remained hot. That alone is rock solid proof. It is possible that the placement of the outlet thermocouple was flawed, and it recorded a value midway between the outlet cooling water temperature and the steam in the pipe next to that. I do not think much heat can cross from the steam pipe to the water pipe next to it. Alan Fletcher did a rigorous analysis to demonstrate this. The thermal mass of the cooling water was much larger than the steam, so the average temperature was closer to the water than the steam. However, for sake of argument let us assume the temperature was too high. In that case, we can ignore the actual temperature and look only at the temperature trends. We can look at relative temperatures. Whatever the temperature was, it goes *up* after the power turns off. It stays up. It stays at a higher level than it was when the power was on! Even if the actual temperature was half this value, it still should have fallen monotonically, as I said. This behavior is simply impossible without some source of heat, at some power level. I think that very little wicking from the hot water pipe occurred, so I expect the peak anomalous power was ~8 kW as shown in this graph. (I also ran this analysis and my complaints past Rossi himself.) - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
With 40MJ of heat in the system it would be impossible for the temperature to drop suddenly. I heat a block of steel to 900C, then I stop heating it, and drop a gram of water on it. What's the temperature? 900C. Notice there was no precipitous drop. Nor would there be after many grams of water. In fact 40MJ is stored in the metal. This is enough to boil ~20kg of water. Where are you getting 1.8 tons? - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 4:59 PM Subject: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Or if it is refutable, let's see someone make a serious effort to refute it. Stop quibbling about details. Get the heart of the matter, and tell us how a box of this size with no input power can boil water for 3 hours and remain at the same high temperature while you cool it with 1.8 tons of water. I wrote to some friends complaining about the test. My conclusion: Despite these problems . . . I think this test produced irrefutable proof of anomalous heat. Here is why I think so -- Look at the graph here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg Nothing happens until 13:22 when the steam begins to flow through the heat exchanger. At 15:13 output is a little higher than input, even though there is a great deal of heat unaccounted for, especially the water from the condensed steam, which they poured down the drain. At 15:50 the power is cut off. If there had been no source of anomalous heat, the power would have fallen off rapidly and monotonically, at the same rate it did after 19:55. It would have approached the zero line by 17:25. Actually, it would have approached zero before that, based on Newton's law of cooling. In other words, it would have been stone cold after 3 hours. During that time, 1.8 tons of water went through the cooling loop. It is inconceivable that an object of this size with no power input could have remained at the same high temperature the whole time. Yet Lewan reports that the surface of the reactor was still hot, and boiling could still be heard inside it. As you see, the temperature did not fall. It went up at 16:26. The cooling water flow rate was unchanged, so only a source of heat could have caused this. You can ignore the thermocouple data, and look only at the fact that it continued to boil for more than 3 hours after the power was turned off, and the reactor surface remained hot. That alone is rock solid proof. It is possible that the placement of the outlet thermocouple was flawed, and it recorded a value midway between the outlet cooling water temperature and the steam in the pipe next to that. I do not think much heat can cross from the steam pipe to the water pipe next to it. Alan Fletcher did a rigorous analysis to demonstrate this. The thermal mass of the cooling water was much larger than the steam, so the average temperature was closer to the water than the steam. However, for sake of argument let us assume the temperature was too high. In that case, we can ignore the actual temperature and look only at the temperature trends. We can look at relative temperatures. Whatever the temperature was, it goes up after the power turns off. It stays up. It stays at a higher level than it was when the power was on! Even if the actual temperature was half this value, it still should have fallen monotonically, as I said. This behavior is simply impossible without some source of heat, at some power level. I think that very little wicking from the hot water pipe occurred, so I expect the peak anomalous power was ~8 kW as shown in this graph. (I also ran this analysis and my complaints past Rossi himself.) - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On 2011-10-09 22:59, Jed Rothwell wrote: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg This is another graphical analysis which shows an overall energy gain: http://imgur.com/a/oix51 (conveniently grouped in a single image gallery with swapped colors for clarity by me. Original source with downloadable data: http://www.scribd.com/doc/68116335/Temp-Data-Ecat-6-10-11-Edited-by-MAP-v2 ) Cheers, S.A.
RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Thanks for the analysis, Jed. Will be interesting to read what others have to say. BTW, what did Rossi have to say? * * * * * When I look at the graph I continue to be drawn to the curious fact that the input power is cycled on and off a total of three or four times starting from around 13:59 to finally ending at 15:50 when it is permanently turned off. Looks to me as if Rossi's team may have been trying to get their eCat airborne way before the time stamp of 15:50. My apologies if the following has already been discussed or speculated since there has been so much discussion in the past three days - I can't keep track of it all. The characteristics of the input data gives me the impression that Rossi's team is trying to capitalize on what I would describe as the Sweet Spot, where Rossi feels that the core reaction is finally beginning to take off without further need for an input power source to sustain the output reaction. It's analogous to the Wright Brothers hand cranking the propeller of their first air craft where the first couple of spins don't necessarily catch on with the engine. As already speculated by a few here, Rossi continues to give me the impression that he operates very much on intuition. Recording scientific data is almost incidental to him, a characteristic I suspect probably drives a few of his colleagues to distraction. Rossi has probably acquired a reasonable amount of instinctual horse sense as to when he thinks his mysterious eCats are likely to take off in self-sustain mode. The following is what I speculate is happening between 13:59 to 15:50: Rossi initially tries at 13:59... It's catching It's catching... Ah, shoot! It petered out. Ok guys! Crank her up again. Input Power turned back on 14:11. Rossi tries again at around 14:24... Well, Shoot. Still didn't catch! Maybe I need to prime the pistons. Where's my canister of Ether. Ok guys. Crank her up again. Input power turned back on at 14:36 Looks like Rossi tries for the third time at around 14:48, but I suspect the there are data anomalies here (human error?) and Rossi actually turns off the input power at around 15:00. It's turning... It's turning... Come on! Come on You can do it Shoot it's going down again. We're close guys! I can feel it in my ancient Italian bones! Ok, let's crank'er up again. Input Power turned back on at 15:25. For the fourth time, Rossi turns off the input power around 15:50. Meanwhile the output signal has been strong and rapidly rising starting at around 15:40 or so. Rossi's Italian bones sense that this is probably the Big One. .TURN THE INPUT POWER OFF Got it! Hand me my goggles, guys! It's Steam Punk Rock'N'Role time! Here are some final personal interpretations: It looks to me as if in every case input power is turned off several minutes after the Rossi senses that the output power is on a steady rise (in self-sustain mode) towards the 3000 mark and above. I wonder if Rossi may have initially been trying to hit that sweet spot early in the data recordings starting at 13:59. Perhaps he initially turned input power off back then in order to help minimize the potential of introducing skeptical arguments such as those presented over at Krivit's blog having to do with the total accumulation of input power from the start of the experiment and how the entire collection of data appears to be greater than the total accumulation of recorded output power. In any case, it looks to me as if Rossi had three false starts before he finally hit pay dirt on the fourth crank. Comments? Regards, Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On 2011-10-10 01:12, OrionWorks - Steven Vincent Johnson wrote: In any case, it looks to me as if Rossi had three false starts before he finally hit pay dirt on the fourth crank. I haven't thought of this before, but after pondering a bit about it I believe it really might have been the case. I wonder if Rossi himself could confirm this. It would make the experiment outcome even more positive if he did, in my opinion. Cheers, S.A.
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
I don't know if Rossi would consider them false starts. From what he has said in the past it seems that cycling the input on and off is now standard operating procedure to run the E-Cat in a stable mode. He has said that in commercial models this cycling will be automated. On Sun, Oct 9, 2011 at 6:28 PM, Akira Shirakawa shirakawa.ak...@gmail.comwrote: On 2011-10-10 01:12, OrionWorks - Steven Vincent Johnson wrote: In any case, it looks to me as if Rossi had three false starts before he finally hit pay dirt on the fourth crank. I haven't thought of this before, but after pondering a bit about it I believe it really might have been the case. I wonder if Rossi himself could confirm this. It would make the experiment outcome even more positive if he did, in my opinion. Cheers, S.A. -- Frank Acland Publisher, E-Cat World http://www.e-catworld.com Author, The Secret Power Beneath https://www.smashwords.com/books/view/
RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
From Akira: This is another graphical analysis which shows an overall energy gain: http://imgur.com/a/oix51 The I/O energy values listed at Imgur certainly bear little resemblance the values reported over in Mr. Krivit's blog: http://blog.newenergytimes.com/ Of particular interest to me, Krivit's blog claims that according to Lewan the total recorded output energy in self sustain mode (3.5 hours) was around 31.5 MJ of energy, whereas over at the flashy imgur site a total of 101.3 MJ had been recorded. That's a HUGE discrepancy in recorded values. I can't make heads or tails of this. Does this discrepancy have something to do with who may and who may not have been taking into consideration the energy speculated to have been produced from BOTH the primary and secondary heat exchangers? I hope someone with more math skills than I can clarify why there appears to be such a discrepancy. Who is taking WHAT into account. Likewise, who is NOT taking WHAT into account? Regards, Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Joe Catania zrosumg...@aol.com wrote: ** With 40MJ of heat in the system it would be impossible for the temperature to drop suddenly. I heat a block of steel to 900C, then I stop heating it, and drop a gram of water on it. What's the temperature? 900C. Notice there was no precipitous drop. Please see Newton's law of cooling: https://www.math.duke.edu/education/ccp/materials/diffcalc/ozone/ozone1.html The other point you are overlooking is the drop is monotonic, that is Varying in such a way that it either never decreases or never increases. When heat is released from a system the way you describe, the temperature can only drop. It NEVER NEVER RISES. That is a fundamental physical law. Note also that this device was at 80 deg C, not 900 deg C. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
No the band heater is at 900C but that metal block talk was only for illustrative purposes. Newtons LAw is irrelevant. An insulated metal block that loses heat at a rate of 1W loses heat at the rate of 1W. You mention lack of monotonicity but what's the example (be specific, post link). - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 8:14 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: With 40MJ of heat in the system it would be impossible for the temperature to drop suddenly. I heat a block of steel to 900C, then I stop heating it, and drop a gram of water on it. What's the temperature? 900C. Notice there was no precipitous drop. Please see Newton's law of cooling: https://www.math.duke.edu/education/ccp/materials/diffcalc/ozone/ozone1.html The other point you are overlooking is the drop is monotonic, that is Varying in such a way that it either never decreases or never increases. When heat is released from a system the way you describe, the temperature can only drop. It NEVER NEVER RISES. That is a fundamental physical law. Note also that this device was at 80 deg C, not 900 deg C. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Joe Catania zrosumg...@aol.com wrote: ** No the band heater is at 900C but that metal block talk was only for illustrative purposes. Newtons LAw is irrelevant. Newton's law governs passive heat loss, which is what this has to be if there is not energy input and the flow rate does change. An insulated metal block that loses heat at a rate of 1W loses heat at the rate of 1W. You mention lack of monotonicity but what's the example (be specific, post link). Right here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg The temperature rises several times after the power is turned off. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Sun, Oct 9, 2011 at 9:24 PM, Jed Rothwell jedrothw...@gmail.com wrote: Joe Catania zrosumg...@aol.com wrote: No the band heater is at 900C but that metal block talk was only for illustrative purposes. Newtons LAw is irrelevant. Newton's law governs passive heat loss, which is what this has to be if there is not energy input and the flow rate does change. An insulated metal block that loses heat at a rate of 1W loses heat at the rate of 1W. You mention lack of monotonicity but what's the example (be specific, post link). Right here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg The temperature rises several times after the power is turned off. - Jed Good point. Harry
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Excuse me I meant to say that the cooling rate must obey Newton's law if there is NO energy generation and the flow rate does NOT change. In other words, if it passive cooling in unchanging conditions. Lewan's observations and report show that the flow rate and other essential parameters did not change. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Jed, I hate to ask, really. You seem to be impressed by that graph. If you look closely at the Ny Teknik results, the output at the heat exchanger doesn't seem to track the logged E-Cat temperatures in any meaningful way. A quick example is between 19:03 and 19:22: In that time frame, E-Cat temp is steadily decreasing, hydrogen is purged, the frequency generator is turned off, and water flow increased (in the primary). But in the following 20 minutes, the output supposedly increases from 3.9 kW to 6.1 kW. This doesn't seem to be an issue for you, so I was wondering if you could explain it to the rest of us. Jed Rothwell jedrothw...@gmail.com wrote: Joe Catania zrosumg...@aol.com wrote: ** No the band heater is at 900C but that metal block talk was only for illustrative purposes. Newtons LAw is irrelevant. Newton's law governs passive heat loss, which is what this has to be if there is not energy input and the flow rate does change. An insulated metal block that loses heat at a rate of 1W loses heat at the rate of 1W. You mention lack of monotonicity but what's the example (be specific, post link). Right here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg The temperature rises several times after the power is turned off. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Robert Leguillon robert.leguil...@hotmail.com wrote: You seem to be impressed by that graph. If you look closely at the Ny Teknik results, the output at the heat exchanger doesn't seem to track the logged E-Cat temperatures in any meaningful way. It cannot track them. The eCat is boiling water at a given pressure, somewhat above 1 atm. The temperature cannot rise. If power increases, it will boil more water but the temperature will not rise. If you capture the steam from the pot on your stove in a heat exchanger, and you turn the gas light up, you will see no change in the boiling water temperature but the heat exchanger will capture more heat. There are minor fluctuations in the eCat steam temperature. I do not know what causes them. Perhaps hot water, or just instrument noise. Note also that the cooling water outlet thermocouple of attached to the outside of the pipe. A pipe is a large heat sink, and a way to average out or blur the heat signal. This has been talked to death here, but people have not noted that this is actually a recommended technique. It prevents rapid fluctuations and local hot spots in the water from affecting the thermocouple. In this case, it may be picking up heat from the steam pipe as well, so it may be a little too high, but it is still an excellent way to smooth out the signal and be sure that the heat is homogeneous and real. If it turns out to be a little high that has no impact on the overall conclusions. Note that it can only be a little too high. Not a lot. Compare the thermal mass of 10 kg/min of cooling water to 55 g/min of steam. Try it! Sparge 55 g of steam at 120 deg C in 10 kg of tap water and you will see that the final temperature is a lot closer to the tap water than the steam. Or just do it in your head. It takes roughly 34,000 calories to raise water from 25 deg C to steam at 120 deg C. Divide that into 10,000 g of water, and the water goes up about 3.4 deg C. For most of the test, the temperature rose 5 deg C. That's in the same ballpark. Maybe the actual temperature rise was only 3.4. So what? An hour after the power was cut it would have been 0.000 deg C, in the absence of anomalous heat. There may have been more than 55 g of steam per minute at times. No one kept track of the input water to the reactor. There was no need to. That is not relevant to the calorimetry, in this case. A quick example is between 19:03 and 19:22: In that time frame, E-Cat temp is steadily decreasing, hydrogen is purged, the frequency generator is turned off, and water flow increased (in the primary). But in the following 20 minutes, the output supposedly increases from 3.9 kW to 6.1 kW. That is a different issue. That is when the eCat is being degassed and the flow through the eCat is turned up, according to Lewan's log. Conditions are no longer stable and the calorimetry no longer works. Calorimetry requires steady state conditions, in which only the heat flux varies. When you open valves or change flow rates, conditions are not in steady state. It is difficult to model the system. Also, there may have been a burst of heat then. It is hard to judge. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Alright, if it's conclusive without the thermocouples Does anyone have a decent water capacity for the E-Cat? I see that H.H. calculated 14.2 liters, but has there been any confirmed number out of the Rossi camp? I only ask, because multiple references have been made to tons of cooling water to quench the reaction during H.A.D. In reality, the water flowing through the E-Cat (as the heat exchanger primary-side output) was measured twice: The first time, it was .91 grams/sec and the second time it was just shy of 2 g/s. If the E-Cat were indeed 14.2liters (14.2 kg), the entire contents of the E-Cat would take 2-4 hours to be completely replaced. All the while, a device that generates frequencies is still running. When it is turned off, the E-Cat temp begins declining. S many questions. Jed Rothwell jedrothw...@gmail.com wrote: Robert Leguillon robert.leguil...@hotmail.com wrote: You seem to be impressed by that graph. If you look closely at the Ny Teknik results, the output at the heat exchanger doesn't seem to track the logged E-Cat temperatures in any meaningful way. It cannot track them. The eCat is boiling water at a given pressure, somewhat above 1 atm. The temperature cannot rise. If power increases, it will boil more water but the temperature will not rise. If you capture the steam from the pot on your stove in a heat exchanger, and you turn the gas light up, you will see no change in the boiling water temperature but the heat exchanger will capture more heat. There are minor fluctuations in the eCat steam temperature. I do not know what causes them. Perhaps hot water, or just instrument noise. Note also that the cooling water outlet thermocouple of attached to the outside of the pipe. A pipe is a large heat sink, and a way to average out or blur the heat signal. This has been talked to death here, but people have not noted that this is actually a recommended technique. It prevents rapid fluctuations and local hot spots in the water from affecting the thermocouple. In this case, it may be picking up heat from the steam pipe as well, so it may be a little too high, but it is still an excellent way to smooth out the signal and be sure that the heat is homogeneous and real. If it turns out to be a little high that has no impact on the overall conclusions. Note that it can only be a little too high. Not a lot. Compare the thermal mass of 10 kg/min of cooling water to 55 g/min of steam. Try it! Sparge 55 g of steam at 120 deg C in 10 kg of tap water and you will see that the final temperature is a lot closer to the tap water than the steam. Or just do it in your head. It takes roughly 34,000 calories to raise water from 25 deg C to steam at 120 deg C. Divide that into 10,000 g of water, and the water goes up about 3.4 deg C. For most of the test, the temperature rose 5 deg C. That's in the same ballpark. Maybe the actual temperature rise was only 3.4. So what? An hour after the power was cut it would have been 0.000 deg C, in the absence of anomalous heat. There may have been more than 55 g of steam per minute at times. No one kept track of the input water to the reactor. There was no need to. That is not relevant to the calorimetry, in this case. A quick example is between 19:03 and 19:22: In that time frame, E-Cat temp is steadily decreasing, hydrogen is purged, the frequency generator is turned off, and water flow increased (in the primary). But in the following 20 minutes, the output supposedly increases from 3.9 kW to 6.1 kW. That is a different issue. That is when the eCat is being degassed and the flow through the eCat is turned up, according to Lewan's log. Conditions are no longer stable and the calorimetry no longer works. Calorimetry requires steady state conditions, in which only the heat flux varies. When you open valves or change flow rates, conditions are not in steady state. It is difficult to model the system. Also, there may have been a burst of heat then. It is hard to judge. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Robert Leguillon robert.leguil...@hotmail.com wrote: Does anyone have a decent water capacity for the E-Cat? I see that H.H. calculated 14.2 liters, but has there been any confirmed number out of the Rossi camp? I only ask, because multiple references have been made to tons of cooling water to quench the reaction during H.A.D. I do not know of any any such references. H.A.D. is quenched by degassing the cell and then subjecting it to a cold thermal shock. That's how you do with Pd-D anyway. I assume it is the same with Ni-H. Lewan indicates that's what they did. It does not take much water to give it a shock. The water capacity of the eCat reservoir does not matter. Just fill it up rapidly and keep overfilling it to cool down the cell in a hurry. S many questions. But many are not germane. Many are distractions. Many are the result of people asking so many questions about trees they fail to see the forest. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
The rapid overfilling was at .91 grams/second (It turns out the 1.92 g/s was for quenching) I've wanted to look at these numbers, and back-of-the-envelope, 381 watts would raise the water entering the E-Cat by 100 degrees (from 24 to 124 degrees C). An additional 2,056 watts is required for the phase-change, but, of course, we have no idea how much is boiling away. Greater than 2,437 watts would completely vaporize the input water. Of course, this means that the water in the E-Cat would be running dry and getting super-heated if there were prolonged excursions over 2.5 kW. But, of course, this didn't happen, did it? Hmmm. If my numbers are off, I apologize. I didn't recheck. Jed Rothwell jedrothw...@gmail.com wrote: Robert Leguillon robert.leguil...@hotmail.com wrote: Does anyone have a decent water capacity for the E-Cat? I see that H.H. calculated 14.2 liters, but has there been any confirmed number out of the Rossi camp? I only ask, because multiple references have been made to tons of cooling water to quench the reaction during H.A.D. I do not know of any any such references. H.A.D. is quenched by degassing the cell and then subjecting it to a cold thermal shock. That's how you do with Pd-D anyway. I assume it is the same with Ni-H. Lewan indicates that's what they did. It does not take much water to give it a shock. The water capacity of the eCat reservoir does not matter. Just fill it up rapidly and keep overfilling it to cool down the cell in a hurry. S many questions. But many are not germane. Many are distractions. Many are the result of people asking so many questions about trees they fail to see the forest. - Jed
