Re: [Vo]:NyTeknik report on October 6th test
Castro? Castro! Just as I suspected. Rossi is part of a commie plot to undermine our way of life. Harry On Fri, Oct 7, 2011 at 9:20 AM, Jed Rothwell jedrothw...@gmail.com wrote: Craig Haynie wrote: I would like to point out that if it were a battery, then it would have been hidden and pre-charged before anyone came into the room. There would be no need to charge it up in front of everyone then. If there was a battery than when they opened the device they would have seen it. Someone else suggested that there might be a Castro gas hidden in the table leg. This is ruled out. Videos of previous tests show the observers picking the device off the table and put on weight scale as they did this time. Videos also show them sliding the device across the table. A hose connecting the device to the hidden source of gas would be revealed when they do this. - Jed
Re: [Vo]:NyTeknik report on October 6th test
Some preliminary notes about the test. The weight of E-Cat before test: 98kg and after the test 99 kg. I think that this may be explained with inaccuracy of the scale and remaining water residuals. Therefore no chemical combustion inside E-Cat! Of course metal-oxide production is still possible, if there is hidden oxygen bottle inside E-Cat (very unlikely). Could lithium be volatile enough to enable controlled burning? I think that most of the other metals, such as aluminium or beryllium require such a high temperature that they are not practical for sustained burning. I hope that someone measured the outlet water temperature after steam/hot water went through heat exchanger. This would be important bit of information to exclude the possibility that heat exchanger does not absorb all the out flowing energy. It looks like 3 kW is the lower limit for heat output. But total heating power was somewhere between 3-6 kW. And if we assume 60% efficiency for the heat exchanger, heating power was something like 5-8 kW. For single core I think that this is reasonable performance. With full power it would thus produce 15-24 kW that is reasonable power production. One important detail to notice that steam temperature was around 120°C. This is significantly less than in September test, where temperature was above 130°C. This would imply that September E-Cat was producing around 6-10 kW energy, when self-sustaining. I think that this is surprisingly close to my steam pressure calculations. Also, Rossi's business continues as usual, because he cancelled the contract with one of the greatest entities in Americas... The company that was rumored to be behind the contract was here http://www.e-cat.com/ But I think that there is not much substance for this rumor. –Jouni 2011/10/7 Jouni Valkonen jounivalko...@gmail.com: TV: New test of the E-cat enhances proof of heat http://www.nyteknik.se/nyheter/energi_miljo/energi/article3284823.ece Test of Energy Catalyzer Bologna October 6, 2011 http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29
Re: [Vo]:NyTeknik report on October 6th test
On 2011-10-07 13:37, Jouni Valkonen wrote: Test of Energy Catalyzer Bologna October 6, 2011 http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 This must be the secret sauce: 15:53 Power to the resistance was set to zero. A device “producing frequencies” was switched on. Overall current 432 mA. Voltage 230 V. Current through resistance was zero, voltage also zero. From this moment the E-cat ran in self sustained mode Cheers, S.A.
Re: [Vo]:NyTeknik report on October 6th test
Maybe the secret source was charging a battery for around 4 hours with an energy above 2KW coupled with some other kind of auxiliary battery... 2011/10/7 Akira Shirakawa shirakawa.ak...@gmail.com On 2011-10-07 13:37, Jouni Valkonen wrote: Test of Energy Catalyzer Bologna October 6, 2011 http://www.nyteknik.se/**incoming/article3284962.ece/** BINARY/Test+of+E-cat+October+**6+%28pdf%29http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 This must be the secret sauce: 15:53 Power to the resistance was set to zero. A device “producing frequencies” was switched on. Overall current 432 mA. Voltage 230 V. Current through resistance was zero, voltage also zero. From this moment the E-cat ran in self sustained mode Cheers, S.A.
Re: [Vo]:NyTeknik report on October 6th test
Maybe the secret source was charging a battery for around 4 hours with an energy above 2KW coupled with some other kind of auxiliary battery... This test was almost as ludicrous as the Steorn waterways test. There, they kept things running by periodically swapping out the devices, presumably to replace the batteries; and they absurdly claimed that they were demonstrating OU. Here, there was just one battery, charged up for 4 hours, and then depleted by heating water for 3.5 hours.
Re: [Vo]:NyTeknik report on October 6th test
This must be the secret sauce: 15:53 Power to the resistance was set to zero. A device “producing frequencies” was switched on. Overall current 432 mA. Voltage 230 V. Current through resistance was zero, voltage also zero. From this moment the E-cat ran in self sustained mode Interesting... Frank, can you predict the frequency? Craig
Re: [Vo]:NyTeknik report on October 6th test
On Fri, 2011-10-07 at 08:59 -0400, vorl bek wrote: Maybe the secret source was charging a battery for around 4 hours with an energy above 2KW coupled with some other kind of auxiliary battery... This test was almost as ludicrous as the Steorn waterways test. There, they kept things running by periodically swapping out the devices, presumably to replace the batteries; and they absurdly claimed that they were demonstrating OU. Here, there was just one battery, charged up for 4 hours, and then depleted by heating water for 3.5 hours. I would like to point out that if it were a battery, then it would have been hidden and pre-charged before anyone came into the room. There would be no need to charge it up in front of everyone then. Craig
Re: [Vo]:NyTeknik report on October 6th test
But that was what happened... 2011/10/7 Craig Haynie cchayniepub...@gmail.com On Fri, 2011-10-07 at 08:59 -0400, vorl bek wrote: Maybe the secret source was charging a battery for around 4 hours with an energy above 2KW coupled with some other kind of auxiliary battery... This test was almost as ludicrous as the Steorn waterways test. There, they kept things running by periodically swapping out the devices, presumably to replace the batteries; and they absurdly claimed that they were demonstrating OU. Here, there was just one battery, charged up for 4 hours, and then depleted by heating water for 3.5 hours. I would like to point out that if it were a battery, then it would have been hidden and pre-charged before anyone came into the room. There would be no need to charge it up in front of everyone then. Craig
Re: [Vo]:NyTeknik report on October 6th test
Craig Haynie wrote: I would like to point out that if it were a battery, then it would have been hidden and pre-charged before anyone came into the room. There would be no need to charge it up in front of everyone then. If there was a battery than when they opened the device they would have seen it. Someone else suggested that there might be a Castro gas hidden in the table leg. This is ruled out. Videos of previous tests show the observers picking the device off the table and put on weight scale as they did this time. Videos also show them sliding the device across the table. A hose connecting the device to the hidden source of gas would be revealed when they do this. - Jed
Re: [Vo]:NyTeknik report on October 6th test
I would like to point out that if it were a battery, then it would have been hidden and pre-charged before anyone came into the room. There would be no need to charge it up in front of everyone then. I guess I should have referred to it as a 'battery'. That cylinder of nickel powder could have been 'charged' before the demo for all we know; maybe without the 'pre-charge', it would only have lasted for one hour instead of 3.5. Here is an inventor and entrepreneur, who intends to have a 1MW system running within 3 weeks, and who gives a demo of one of the modules, and does it in a way that can only engender skepticism and ridicule. There is no reason I can think of why Rossi would not do his best for this demo, but what he came up with was almost a joke, presumably because he could not come up with anything better.
Re: [Vo]:NyTeknik report on October 6th test
I wrote: Someone else suggested that there might be a Castro gas hidden in the table leg. A canister of gas, for crying out loud. There is no gas, no wires and no batteries. Get that through your heads. That is nonsense. - Jed
Re: [Vo]:NyTeknik report on October 6th test
I made some initial calculations for the COP. They are just rough estimations. Electricity provided to the E-Cat was approximately 30 MJ (average input power when electricity was on, was 2 kW). It was little tricky to calculate, because input power was variable. Here we can see that most of the energy was consumed initial heating of the device. 100 kg Metal + 25 kg water alone takes 18 MJ to heat up by 80 °C and in addition that there was ca. 13 kg/h water inflow during the initial heating. Therefore, It is safe to say that almost all the electricity was absorbed by heating E-Cat to 100 °C. However, as E-Cat was producing ca. 5-8 kW power (60% efficiency for heat exchanger is assumed) around 6-7 hours, we can calculate that total energy output was around 100-150 MJ. That is directly more than 10 times more output than input, if we ignore the initial heating. And when E-Cat is running on all three cores, I think that COP is more than 30, in sustained mode. We can say that this test was not only the great success, but it was phenomenal success that surpassed even our wildest dreams! So where I can reclaim the price for guessing correctly the COP?^^ –Jouni
RE: [Vo]:NyTeknik report on October 6th test
My Two Cents: Whiskey. Tango. Foxtrot. Most of the previous experimental problems were solved in this setup. We could've seen measurable, stable, power gains completely unaffected by phase-change or water overflow. We should have been presented with an operating E-Cat producing 6 or more times input power. Instead, we were asked to evaluate a temperature decay of an E-Cat, whose power output was at or near parity with the input, while a new device produces frequencies. The only explanation that I can come up with is this: Rossi was originally claiming a MINIMUM of 6x power gains. Skeptics said, Then why do you need two heaters? Even with an 80% loss in thermoelectric power generation, it should be able to run self-sustained. Of course skeptics MEANT that you could generate electricity with the output heat, and use it to power the heaters. This would close the loop, and allow it to run ad infinitum; much crow would be eaten. Rossi couldn't be bothered with thermoelectric generators, and tried to come up with a way for it to run without input. I'd predicted that the self-sustained or heat after death mode of operation would be a bone of contention. Let's just look hard at Heat-Before-Death. It's obviously not his promised 6x power gain, but there may be something to this, yet. Donating to the World - Two Cents at a Time, R.L. There is no reason I can think of why Rossi would not do his best for this demo, but what he came up with was almost a joke, presumably because he could not come up with anything better.
Re: [Vo]:NyTeknik report on October 6th test
Now that Jed has told me my utility pension is at risk and I have vested interests. I will have to agree there is probably something wrong with the tests. Perhaps a laser was heating it from the ceiling? Frank Z
Re: Re: [Vo]:NyTeknik report on October 6th test
Von: Jouni Valkonen jounivalko...@gmail.com However, as E-Cat was producing ca. 5-8 kW power (60% efficiency for heat exchanger is assumed) If the heat exchanger has only 60% efficieny, then the energy loss is 5kW * 0.4 = 2kW. Where does the enrgy go? Energy cannot vanish magically, it must go into the ambient. I think even if the heat exchanger at this size (as visible in the video) has no insulation, it cannot lose 2kW. It is well isolated and the loss must be much lower. The output temperature delta of course is lower then te input temperature delta, but at the secondary circuit the flow can higher. A heat exchanger has always a loss of temperature, but this does not necessary mean a loss of energy. If the isolation against ambient is 100% perfect, then it will reduce the temperature but not loose energy, because the secondary flow must be higher than the primary flow. Peter
Re: [Vo]:NyTeknik report on October 6th test
Frank sez: Now that Jed has told me my utility pension is at risk and I have vested interests. I will have to agree there is probably something wrong with the tests. Perhaps a laser was heating it from the ceiling? ...will have to agree I can't tell if Frank is being serious or not. If Frank is not being serious, I'd say Frank has a wicked sense of humor. Well played, Frank! OTOH, if Frank was being dead serious... Well, let's just say that disliking the ramifications someone else draws should not in itself become the primary reason to decide it must therefore be wrong. Most of us try to come to terms with those kinds of hurdles during the terrible twos of our lives. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:NyTeknik report on October 6th test
peter.heck...@arcor.de wrote: If the heat exchanger has only 60% efficieny, then the energy loss is 5kW * 0.4 = 2kW. Where does the enrgy go? Energy cannot vanish magically, it must go into the ambient. Correct. It radiates into the surroundings, from the reactor and the heat exchanger. Lewan reported the reactor surface was between 60°C and 85°C. I presume he means at different times. I do not know how he measured that. It has a lot of surface area so it is radiating a lot of heat. Someone better physics and I can estimate how much. With a really good calorimeter having a high recovery rate, nearly all the heat ends up captured by the calorimeter. With the flow calorimeter it ends up heating the water. With a Seebeck calorimeter it may radiate out into the room, or if there is a water bath on the outside shell of the chamber to ensure a stable background, it will be captured by the water bath. This reactor is too large for most Seebeck calorimeters. I think even if the heat exchanger at this size (as visible in the video) has no insulation, it cannot lose 2kW. It is well isolated and the loss must be much lower. I believe the heat exchanger plus the reactor itself can radiate 2 kW. They look crude to me. Such things are inefficient. See photo of the two of them (in one box): http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 - Jed
RE: [Vo]:NyTeknik report on October 6th test
On Fri, 2011-10-07 at 09:01 -0500, Robert Leguillon wrote: My Two Cents: Whiskey. Tango. Foxtrot. Most of the previous experimental problems were solved in this setup. We could've seen measurable, stable, power gains completely unaffected by phase-change or water overflow. We should have been presented with an operating E-Cat producing 6 or more times input power. Instead, we were asked to evaluate a temperature decay of an E-Cat, whose power output was at or near parity with the input, while a new device produces frequencies. I disagree with this. During the 'power phase', you can measure the power coming out of the system as heat. The conclusion is far away from a 4 hour 'charging phase' followed by a 3 1/2 hour 'discharging phase' of near equal parity. Craig
Re: [Vo]:NyTeknik report on October 6th test
The lastest version of Steorn's 'orbo' technology also produces steam and uses nickel. I think Rossi and Steorn are both exploiting the same underlying phenomena, or they are both mistaken or ... Harry On Fri, Oct 7, 2011 at 8:59 AM, vorl bek vorl@antichef.com wrote: Maybe the secret source was charging a battery for around 4 hours with an energy above 2KW coupled with some other kind of auxiliary battery... This test was almost as ludicrous as the Steorn waterways test. There, they kept things running by periodically swapping out the devices, presumably to replace the batteries; and they absurdly claimed that they were demonstrating OU. Here, there was just one battery, charged up for 4 hours, and then depleted by heating water for 3.5 hours.
RE: [Vo]:NyTeknik report on October 6th test
I have not yet had time to compile the four hours of warm up. Obviously, we don't have all of the data required to even remotely show a balanced energy equation. The at or near parity statement was referring to E-Cat performance before it was turned off. One would expect an operating E-Cat that is consuming 2 kW input power, to be displaying 12 kW output power during operation. This does not appear to be what was demonstrated. If the E-Cat was running at a high enough core temperature to produce 3.5 kW output, while 2.5 kW was being introduced to the heater (230V x 11A), then why did the output not immediately drop to 1 kW when the power was removed? Why did it not slowly decline and stabilize at a new baseline that represented the E-Cat's output power? How does it maintain the same output power, when you've removed 2 kW of input? Is he claiming that the E-Cat isn't producing its own heat for the first 4 hours, and now it only operates when you REMOVE power from the heaters? These questions would never have to be asked if we were only evaluating 8 hours of operating gains, and that's point in its entirety. Subject: RE: [Vo]:NyTeknik report on October 6th test From: cchayniepub...@gmail.com To: vortex-l@eskimo.com Date: Fri, 7 Oct 2011 11:21:18 -0400 On Fri, 2011-10-07 at 09:01 -0500, Robert Leguillon wrote: My Two Cents: Whiskey. Tango. Foxtrot. Most of the previous experimental problems were solved in this setup. We could've seen measurable, stable, power gains completely unaffected by phase-change or water overflow. We should have been presented with an operating E-Cat producing 6 or more times input power. Instead, we were asked to evaluate a temperature decay of an E-Cat, whose power output was at or near parity with the input, while a new device produces frequencies. I disagree with this. During the 'power phase', you can measure the power coming out of the system as heat. The conclusion is far away from a 4 hour 'charging phase' followed by a 3 1/2 hour 'discharging phase' of near equal parity. Craig
Re: [Vo]:NyTeknik report on October 6th test
Lewan's report states that hydrogen pressure was lowered during shut-down. This is the angle they should have exploited. With constant heating and water flow conditions they should vary the hydrogen pressure and record the results. They should also try an inert gas like helium. - Original Message - From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com Sent: Friday, October 07, 2011 10:59 AM Subject: Re: [Vo]:NyTeknik report on October 6th test peter.heck...@arcor.de wrote: If the heat exchanger has only 60% efficieny, then the energy loss is 5kW * 0.4 = 2kW. Where does the enrgy go? Energy cannot vanish magically, it must go into the ambient. Correct. It radiates into the surroundings, from the reactor and the heat exchanger. Lewan reported the reactor surface was between 60°C and 85°C. I presume he means at different times. I do not know how he measured that. It has a lot of surface area so it is radiating a lot of heat. Someone better physics and I can estimate how much. With a really good calorimeter having a high recovery rate, nearly all the heat ends up captured by the calorimeter. With the flow calorimeter it ends up heating the water. With a Seebeck calorimeter it may radiate out into the room, or if there is a water bath on the outside shell of the chamber to ensure a stable background, it will be captured by the water bath. This reactor is too large for most Seebeck calorimeters. I think even if the heat exchanger at this size (as visible in the video) has no insulation, it cannot lose 2kW. It is well isolated and the loss must be much lower. I believe the heat exchanger plus the reactor itself can radiate 2 kW. They look crude to me. Such things are inefficient. See photo of the two of them (in one box): http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 - Jed
Re: [Vo]:NyTeknik report on October 6th test
2011/10/7 Robert Leguillon robert.leguil...@hotmail.com: I have not yet had time to compile the four hours of warm up. Obviously, we don't have all of the data required to even remotely show a balanced energy equation. I disagree. Calculating energy input is straight forward and it is ca. 30 MJ. Calculating energy output is more tricky, but we have sufficient data to do it in reasonable accuracy. If we examine the secondary circulation data carefully. I calculating for the output something between 100-150 MJ. The at or near parity statement was referring to E-Cat performance before it was turned off. One would expect an operating E-Cat that is consuming 2 kW input power, to be displaying 12 kW output power during operation. This does not appear to be what was demonstrated. Where did you get that 12 kW? On average input power was 1 kW during the demonstration. As we see that Average output power was close to 6 kW, then during this test COP was 6. But if consider that most of the electric input was for initial heating of E-Cat to 95°C when cold fusion reactions were kicked in. Then we get more than 10 for COP. If the E-Cat was running at a high enough core temperature to produce 3.5 kW output, while 2.5 kW was being introduced to the heater (230V x 11A), then why did the output not immediately drop to 1 kW when the power was removed? Why did it not slowly decline and stabilize at a new baseline that represented the E-Cat's output power? How does it maintain the same output power, when you've removed 2 kW of input? Is he claiming that the E-Cat isn't producing its own heat for the first 4 hours, and now it only operates when you REMOVE power from the heaters? This is good observation. And it is good to read healthy skepticism, because this is not obvious Here is the temperature graph. http://www.facebook.com/photo.php?fbid=231409333581939set=o.135474503149001type=1theater and direct link to the picture. http://a3.sphotos.ak.fbcdn.net/hphotos-ak-ash4/296467_231409333581939_11386229231_643956_806537009_n.jpg It is good to see, that after the input power was cut from E-Cat, Temperature of secondary circuit was increased to maximum power output! This explains very clearly why there was not a drop in the output when power was cut. Also 3,5 kW is too low figure for power. Because it does not include inefficiency of heat exchanger. Therefore reasonable figure is 5 kW typically and in peak after power was cut it was more close to 6-8 kW. –Jouni These questions would never have to be asked if we were only evaluating 8 hours of operating gains, and that's point in its entirety. Subject: RE: [Vo]:NyTeknik report on October 6th test From: cchayniepub...@gmail.com To: vortex-l@eskimo.com Date: Fri, 7 Oct 2011 11:21:18 -0400 On Fri, 2011-10-07 at 09:01 -0500, Robert Leguillon wrote: My Two Cents: Whiskey. Tango. Foxtrot. Most of the previous experimental problems were solved in this setup. We could've seen measurable, stable, power gains completely unaffected by phase-change or water overflow. We should have been presented with an operating E-Cat producing 6 or more times input power. Instead, we were asked to evaluate a temperature decay of an E-Cat, whose power output was at or near parity with the input, while a new device produces frequencies. I disagree with this. During the 'power phase', you can measure the power coming out of the system as heat. The conclusion is far away from a 4 hour 'charging phase' followed by a 3 1/2 hour 'discharging phase' of near equal parity. Craig
Re: [Vo]:NyTeknik report on October 6th test
2011/10/7 Joe Catania zrosumg...@aol.com: Lewan's report states that hydrogen pressure was lowered during shut-down. This is the angle they should have exploited. With constant heating and water flow conditions they should vary the hydrogen pressure and record the results. They should also try an inert gas like helium. Of course, but unfortunately there was not time to do such thing (doing such correlative analysis would take several days) . And also, reaction speed did not react too much for the reducing the hydrogen pressure. But test excluded all possible hidden power sources (E-Cat was weighted before and after the test). Therefore what would be the point of injecting helium? –Jouni
Re: [Vo]:NyTeknik report on October 6th test
Am 07.10.2011 16:59, schrieb Jed Rothwell: peter.heck...@arcor.de wrote: If the heat exchanger has only 60% efficieny, then the energy loss is 5kW * 0.4 = 2kW. Where does the enrgy go? Energy cannot vanish magically, it must go into the ambient. I think even if the heat exchanger at this size (as visible in the video) has no insulation, it cannot lose 2kW. It is well isolated and the loss must be much lower. I believe the heat exchanger plus the reactor itself can radiate 2 kW. They look crude to me. Such things are inefficient. See photo of the two of them (in one box): http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 I cannot calculate this, I can only estimate it by comparison with known devices: I live in rooms directly under the roof. I have a gas boiler 10 kW. This heats water, that is pumped through copper pipes and these are connected to 5 radiators. Because I live under the roof, the pipes are partially on the outside of the walls. They are still under the roof but exposed to the cold winter air. They are isolated by glass wool and alu foils, just as the e-cat. The isolated copper pipes are in a length of about 5m exposed to the winter air. The water temperature is max. 80 centigrade. Now imagine it is outside under the roof -10 centigrade. Then I must loose several kilowatts of heating energy in winter! Possibly I should check this. Peter
Re: [Vo]:NyTeknik report on October 6th test : disappointed again
Inaccurate calorimetry? Thermocouples INSIDE the box, provided by Ross? Do I understand that the thermocouples were attached to the OUTSIDE of the heat exchanger in well-established positions -- and not IN the water flow? Where they could be affected by the ambient heat from the eCat ? And not recorded continuously? Just the lid taken off ... can't even see nekkid eCats inside? Digital bathroom scale used for weighing the E-cat. It was calibrated by two persons knowing their weight. (Before or after lunch?) This isn't even science-fair quality science ... it's more like bar-bet science. My bet : The test will be conclusive. *NO* My expectation : and positive. *YES* I don't even know what volume to use for the Ecat I guess I'll have to redo my calculation with energy/mass.
Re: [Vo]:NyTeknik report on October 6th test
BTW, if the heat exchanger is inside the housing of the e-cat, then its energy loss is zero, if we compare the steam measurement in the september test to the water measurement in october. The output temperature will of course be lower, but the thermal mass flow in the secondary circuit must be higher.
Re: [Vo]:NyTeknik report on October 6th test
On Oct 7, 2011, at 3:37 AM, Jouni Valkonen wrote: TV: New test of the E-cat enhances proof of heat http://www.nyteknik.se/nyheter/energi_miljo/energi/article3284823.ece Test of Energy Catalyzer Bologna October 6, 2011 http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E- cat+October+6+%28pdf%29 Once again no kWh meter was used to measure the total input energy. It is far better to record E(t) frequently and then drive power P(t) by P(t) = d E(t)/dt than to occasionally and sporadically take power measurements and integrate to obtain E(t). Flow meters were used but apparently no one thought to record the time stamped volume data! It is much more accurate, depending on flow variations, to calculate flow f(t) from volume v(t) as: f(t) = d V(t)/dt than to integrate: V(t) = integral f(t) dt (or a similar integration to obtain energy) using occasional sporadic short interval flow measurements. This is the value of using volume meters. This appears to actually be a small point in this case, however, because fortunately overall flow volume was measured, and total volume vs sum of periodic flows does not appear to be an issue, at least compared to the other issues. The flow rate chosen was (once again) too large, resulting in a max delta T of about 8°C and thus very unreliable accuracy in the heat measurements. The measurements might have been more reliable if the thermocouples had not been placed on insulated metal parts, i.e. connected directly, metal to metal, to the heat exchanger itself. They should have been separated from the heat exchanger by low conductivity material, such as a short length of rubber hose, to avoid thermal wicking problems through the metal. The same applies to the output temperature measurement for the E-cat. This is the same old problem as before, but compounded. This makes the temperature data highly unreliable. From Report: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Room temperature was between 28.7 °C and 30.3 °C. 18:53 Tin = 24.3 °C Tout = 29.0 °C T3 = 24.8 °C T2 = 116.4 °C 18:57 Measured outflow of primary circuit in heat exchanger, supposedly condensed steam, to be 328 g in 360 seconds, giving a flow of 0.91 g/s. Temperature 23.8 °C. 19:22 Tin = 24.2 °C Tout = 32.4 °C T3 = 25.8 °C T2 = 114.5 °C 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 a significant problem with temperature measurement. The most serious problem, however, is the output temperature recorded for the condensed steam. Perhaps that was a repeated recoding error. The condensed steam is measured leaving the heat exchanger at a temperature lower than room temperature by at least 5°C, and lower than the Tin of the exchanger by 1°C. If the heat exchanger were 70% efficient as estimated by some individuals, then the condensed steam water temperature should have been closer to Tout. Given a delta T of the cooling water of 32.4°C - 24.2°C = 8.2°C, we might expect a condensed steam temperature more like 34.8°C, not 23.2°C. The condenser itself and the highly insulated flow lines do *not* appear to be a source of loss of energy, and thus low measurement efficiency. Further, the low temperature of the condensed steam water indicates no loss of energy in the heat exchange process due to dumped heat in the form of condensed steam going down the drain. Based on all the above, the temperature measurements lack the degree of credibility required to make any reliable assessment of commercial value. A rough estimate of energy in: 11:52 to 12:02 8.07 A * 225 V * 10/60 hr = 0.302 kWh 12:02 to 12:12 9.22 A * 226 V * 10/60 hr = 0.347 kWh 12:22 to 12:32 11.24 A * 224 V * 10/60 hr = 0.420 kWh 12:32 to 14:00 12.05 A * 224 V * 88/60 hr = 3.959 kWh 14:10 to 15:53 11.90 A * 221 V * (10+13+54)/60 hr = 3.419 kWh 15:53 to 19:08 0.50 A * 230 V * 195/60 hr = 0.168 kWh Total energy in: 8.615 kWh = 31 MJ Noted in report: 15:53 Power to the resistance was set to zero. A device “producing frequencies” was switched on. Overall current 432 mA. Voltage 230 V. The power measurement during this period may be highly flawed, depending on the circuits involved and where the measurement was taken. Filtering between the power measurement and E-cat is essential, unless a fast response meter, like the Clarke-Hess is used. The heat after death was estimated in the report to be between 38 MJ (based on secondary circuit water flow) and 21.7 MJ (based on steam mass). This indicates some possible energy gain, but the temperature data is highly unreliable, and the COP does not look to be anywhere near 6. Further, the temperature tailed off after less than 4 hours. The device should not
Re: [Vo]:NyTeknik report on October 6th test : disappointed again
Fake paper updated : http://lenr.qumbu.com/rossi_ecat_proof_frames_v401.php I used Lewan's size of the box as t 50 x 60 x 35 centimeters = 105 liters From his (only) photo I estimated that about 60 litres is still hidden. Power : 3.125 kW Time : 4 hours Based on this, even Lithium-ion batteries would have run for 19 hours ... with no weight change (before OR after lunch). Boron/Compressed oxygen also has no weight-change : 86 hours. (OK,OK .. not feasible) Obvious fake : a small bleed in the heat exchanger from the steam to the water circuit could account for the temperature change. OUTPUT water volume was NOT measured.
Re: [Vo]:NyTeknik report on October 6th test
On 11-10-07 09:30 AM, Jed Rothwell wrote: I wrote: Someone else suggested that there might be a Castro gas hidden in the table leg. A canister of gas, for crying out loud. A... Thanks for the correction. I was thinking this must be yet another odd thingy which I'd never heard of before: Castro gas. Sort of like Brown's gas, I suppose, but with higher energy density. There is no gas, no wires and no batteries. Get that through your heads. That is nonsense. - Jed
Re: [Vo]:NyTeknik report on October 6th test
I have to disagree that the change in hydrogen pressure wouldn't be almost immediately obvious. IYou should get an immediate rise in delta T across the reactor which would immediately boost heat flow. Helium should confirm a null result- ie no CF and would be used as a control. You should be able to subtract out the helium data to account for thermal inertia and warm up and cool down w/ the heater.--- Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Friday, October 07, 2011 12:14 PM Subject: Re: [Vo]:NyTeknik report on October 6th test 2011/10/7 Joe Catania zrosumg...@aol.com: Lewan's report states that hydrogen pressure was lowered during shut-down. This is the angle they should have exploited. With constant heating and water flow conditions they should vary the hydrogen pressure and record the results. They should also try an inert gas like helium. Of course, but unfortunately there was not time to do such thing (doing such correlative analysis would take several days) . And also, reaction speed did not react too much for the reducing the hydrogen pressure. But test excluded all possible hidden power sources (E-Cat was weighted before and after the test). Therefore what would be the point of injecting helium? –Jouni
Re: [Vo]:NyTeknik report on October 6th test
Peter Heckert wrote: BTW, if the heat exchanger is inside the housing of the e-cat, then its energy loss is zero, That can't be. That would violate CoE. All heat exchangers lose heat. If the heat exchanger is inside the housing, that means the housing is hotter and radiates more heat than it would if there were no heat exchanger inside it. It does not matter where you put the thing must produce heat. - Jed
Re: [Vo]:NyTeknik report on October 6th test
At 11:23 AM 10/7/2011, Jed Rothwell wrote: Peter Heckert wrote: BTW, if the heat exchanger is inside the housing of the e-cat, then its energy loss is zero, That can't be. That would violate CoE. All heat exchangers lose heat. If the heat exchanger is inside the housing, that means the housing is hotter and radiates more heat than it would if there were no heat exchanger inside it. It does not matter where you put the thing must produce heat. The radio24 pics show the heat exchanger outside. The corrugated section inside the eCat is part of its internal core-to-steam heat exchanger.
Re: [Vo]:NyTeknik report on October 6th test
Golly... I finally looked, very briefly, at the Nyteknik report. (I've been, and am, tied up with other stuff these days.) For some reason I had assumed it was friendly to Rossi. The report is eight pages long, and uses the word supposedly seven times. I'm not used to seeing that word used /at all/ in papers. That doesn't seem very friendly, after all. Interesting... On 11-10-07 10:59 AM, Jed Rothwell wrote: [ ... ] I believe the heat exchanger plus the reactor itself can radiate 2 kW. They look crude to me. Such things are inefficient. See photo of the two of them (in one box): http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 - Jed
Re: [Vo]:NyTeknik report on October 6th test
Alan J Fletcher wrote: The radio24 pics show the heat exchanger outside. The corrugated section inside the eCat is part of its internal core-to-steam heat exchanger. I don't get it. Please explain. Are there two heat exchangers? One to condense the steam maybe?? I thought that's what the secondary loop exchanger does. - Jed
Re: [Vo]:NyTeknik report on October 6th test
Am 07.10.2011 20:23, schrieb Jed Rothwell: Peter Heckert wrote: BTW, if the heat exchanger is inside the housing of the e-cat, then its energy loss is zero, That can't be. That would violate CoE. All heat exchangers lose heat. If the heat exchanger is inside the housing, that means the housing is hotter and radiates more heat than it would if there were no heat exchanger inside it. It does not matter where you put the thing must produce heat. I politely ask to disagree. What I mean is, if the heat exchanger is inside the housing, then the outer surface is unchanged. So the thermal resistance to ambient air and the thermal infrared radiation to ambient is unchanged. So in this case the heatexchanger does not cause /additional/ energy loss. I know, you will not believe me. Ask an expert for thermal machines. I am not an expert, but I know how to calculate cooling dissipators for power transitors and so on, based on the data that is given by the manufacturers. Have done this many times successfully and measured temperatures afterwards. I think, I have basic understanding. Best, Peter
Re: [Vo]:NyTeknik report on October 6th test
At 11:44 AM 10/7/2011, Jed Rothwell wrote: Alan J Fletcher wrote: The radio24 pics show the heat exchanger outside. The corrugated section inside the eCat is part of its internal core-to-steam heat exchanger. I don't get it. Please explain. Are there two heat exchangers? One to condense the steam maybe?? I thought that's what the secondary loop exchanger does. This is the pic Lewan posted : http://www.radio24.ilsole24ore.com/Foto/articoli/ecat071011-3.jpg corrugated I presumed (wrongly) that THAT was the heat exchanger between the primary (steam) and secondary (water) circuit This one shows the actual steam-water exchanger : http://www.radio24.ilsole24ore.com/Foto/articoli/ecat071011-1.jpg The radio24 video won't show on my system.
Re: [Vo]:NyTeknik report on October 6th test
Eric Hustedt made new graph that shows power output without considering the efficiency of heat exchanger, what is probably 60-80% http://www.facebook.com/photo.php?fbid=10150844451570375set=o.135474503149001type=1theater http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg This is very informative, that I significantly underestimated the total output of E-Cat in my previous estimates. If we correct the heat loss, then max output was ca. 10 kW. Although as was pointed out by Horace and others, we really do not know how trustworthy actually this calorimetry is. There are too many unknowns like was the water flow constant and were the temperature probes correctly placed and what was the primary circuit water temperature after it exited from heat exchanger? –Jouni Ps. here is the temperature graphs: http://www.facebook.com/photo.php?fbid=231409333581939set=o.135474503149001type=1theater and for the heat exchanger: http://www.facebook.com/photo.php?fbid=202076193195962set=o.135474503149001type=1theater
Re: [Vo]:NyTeknik report on October 6th test
Jouni Valkonen wrote: Eric Hustedt made new graph that shows power output without considering the efficiency of heat exchanger, what is probably 60-80% http://www.facebook.com/photo.php?fbid=10150844451570375set=o.135474503149001type=1theater http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg This is very helpful. Thanks for pointing it out, and thanks to Eric as well. - Jed
Re: [Vo]:NyTeknik report on October 6th test
Am 07.10.2011 13:37, schrieb Jouni Valkonen: TV: New test of the E-cat enhances proof of heat http://www.nyteknik.se/nyheter/energi_miljo/energi/article3284823.ece Test of Energy Catalyzer Bologna October 6, 2011 http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 I think, the delta_th was too low (4.5 centigrade). A small error in temperature measurement gives a big error in energy. It is impossible to feel the difference. It is very difficult to verify. This is the same problem as Levi had in his private undocumented 18 hour test. I have now measured the heat radiator in my main living room. Water input temperature is 60 centigrade. (I can read this from the boilers thermometer) I cant touch the input for a long time. Output temperature is 30-40 centigrade. (I cant measure it, but I estimate, it is comfortable bathing temperature) I can touch it for unlimited time. Rossi said some time ago, the setup was proposed by the scientists. Why doesnt he ask his plumber? These guys know how to build a heating system, and calculate the temperatures, water flow and energy needs. Or, if they dont know, they have tables and software that give the needed data. He should let his plumber design the system, then the only thing he must do, is measure and demonstrate this. ;-)
