Promoter Andrea Rossi’s most recent test of his “energy catalyzer” failed to demonstrate the production of excess heat.
On Oct. 7, “the ‘E-cat’ invented by Andrea Rossi ran in a completely stable self-sustained mode for over three hours,” journalist Mats Lewan of Ny Teknik wrote.
According to Lewan, Rossi’s device released an average of 2.5 kilowatts of heat in 3.5 hours. This amounts to 31.5 megaJoules of energy.
However, Rossi heated the device with 2.7 kilowatts of electricity for four hours in advance. This amounts to 38.88 megaJoules of energy. He also heated the device during the phase which Lewan called “self-sustaining.” The input was 115 Watts for 3.5 hours. That’s 1.44 megaJoules of energy.
Let’s do the math of the total energy input versus energy output: (38.88 + 1.44) – 31.5 = 8.82.
That’s a total energy loss of 8.82 MJ during a 7.5-hour period*. In other words, Rossi has demonstrated a water heater that is 78 percent efficient. This is inconsistent with his claims of having a device that produces substantial amounts of excess heat in comparison to input energy.
On Oct. 7, New Energy Times asked Lewan how he justified his and Rossi’s implication that the Oct. 6 experiment showed a net energy gain, as well as their claim that the device was self-sustaining.
“I don’t believe I claimed anything at all,” Lewan wrote. “And I don’t think I reported a net energy gain. I only reported the energy developed during self- sustained mode.”
These are glaring inconsistencies.
Furthermore, in his technical report, Lewan ignores the input energy from the first four-hour warm-up period.
“The E-cat was considered to be completely operating only after reaching self-sustained mode,” he wrote.
Lewan’s and Rossi’s choice to ignore the input energy from the four-hour warm-up period is like saying a sailplane can fly without power, so long as it is first carried aloft by a tow plane.
* Minor radiative losses are not included in this rough calculation.
Received via e-mail:
Rossi was in total control of the test yesterday. He said it would run for 12 hours or more, but he shut it down after three and one half hours running on almost no external power input. In early September he tried a similar trick by running with almost no power for two hours, When he was caught, he reacted badly.
This time he could have concealed enough lithium ion batteries to provide power for 3.5 hour operation.
Nobody weighed the ‘secret box’ and nobody looked inside the ‘SECRET BOX even though he said everything would be available to them to discover.
All the attendees should consider themselves UNINDICTED CO-CONSPIRATORS in Rossi’s latest fraudulent endeavor. This makes 11 inconclusive tests in succession for Rossi.
He could have removed all doubt by running long enough to remove all doubt about batteries, but that was clearly not his goal. Anybody who believes this test was sufficient should examine their motivations.
Brian Ahern
While I am a skeptic about Rossis claims as well, I think that a battery is out of question. If Rossi had a batter of this size and weight capable of storing that much energy, it would be a sensation in itself. The casing of the “reactor” was opened too and there was really not much inside. The actual e- cats are too small to contain that many KWh of energy.
ok se credi [Moderator Edit: Posting Guidelines #13]
“Unindicted co-conspirator” seems a bit harsh to apply to those who may well have attended for no reason beyond simple curiosity.
But now we’ve added a “device producing frequencies.” You could say a microwave element would fall within that broad description, for whatever purpose it might accomplish with ~100 watts of power.
It is true that the battery issue cannot be ruled out, but the argument for 78% efficiency in the original post seems odd. If indeed the 4-hour ‘self sustained run’ heating were generated by cooling of a previously-heated heat store, then one would expect the deltaT to fall gradually (exponentially) and, after four hours, the device to be almost at room temperature, unless it were also seriously pre-heated. However, the observers weighing the device didn’t observe any heat from it, so chances are that that’s not it. Meanwhile, deltaT fluctuated wildly, but roughly around the same mean.
The intensity of the spotlight is becoming such that the truth will out soon, one way or the other, but this particular test doesn’t seem damning; in fact, it seems to leave open only two possibilities: a successful device and an elaborate, long-running fraud (probably involving many different ‘artificial turk’ designs to fool individual tests of varying durations and types). A measurement error can certainly be discarded as a possibility by now.
One thing is for sure: Rossi’s secretiveness and grumpiness has done him no favors if he is a real inventor.
“Any chemical process should be ruled out for producing 25 kWh from whatever is in a 50 cubic centimeter container. The only alternative explanation is that there is some kind of a nuclear process that gives rise to the measured energy production.” Professor Emeritus at Uppsala University Sven Kullander, also chairman of the Royal Swedish Academy of Sciences’ Energy Committee
[Moderator Edit: Posting Guidelines #2, #4]
Krivit wrote:
“However, Rossi heated the device with 2.7 kilowatts of electricity for four hours in advance. This amounts to 38.88 megaJoules of energy.”
The implication here appears to be that during 4 hours in advance, the 33.88 MJ of input were somehow stored in the device. If that had been the case, the device would have remained at room temperature. There would a heat deficit. There was no such deficit, and it is physically impossible for there to be one. Nearly the entire 33.88 MJ that went in during this period came right out again. There was a balance. Actually it was slightly exothermic. In any case, even if 38 gigawatts had been input before the event, that would make no difference if all of that heat came out as soon as it went in.
The heat after death event can only have been caused by heat generated internally during the event.
“According to Lewan, Rossi’s device released an average of 2.5 kilowatts of heat in 3.5 hours. This amounts to 31.5 megaJoules of energy.”
That must be a typo. The PDF report states “2 to 3 kilowatts” at the end. But, further up the calculations use 3.5 kW. The report sets a lower bound of 3.5 kW by taking the minimum delta-T observed. This gives us 44 MJ. Please double check the report.
“However, Rossi heated the device with 2.7 kilowatts of electricity for four hours in advance.”
Not exactly. Again, the PDF report describes exactly when the resistor and the “frequency generator” were turned on/off. There were several on/off cycles so I cannot go into details here. But, when I integrate the power I get 33 MJ of input energy including the 0.55A load at 230V during the self-sustain phase. Please double-check my calculation.
“Furthermore, in his technical report, Lewan ignores the input energy from the first four-hour warm-up period.”
Yup. He also ignored the output energy during that time as well which was substantial.
When I do an energy balance calculation from the unadulterated raw data in the PDF report I get 33 MJ in and 101 MJ out. The report ends at 19:58 with T3 = 103.1C and T2 = 27.5 for a delta-T = ~75C. That is 75C worth of energy for 100kg of metal and 30kg of water for a total of 13 MJ stored using specific heat constants of 0.5 and 4.18 respectively. This brings our grand total energy balance to +81 MJ net. Again, I encourage everyone to double check this figure.
Disclaimer: This post is not intended to be an endorsement of the demonstration or the data contained within the PDF report. It is intended only as a presentation of the energy balance calculations using the data in the PDF report.
“That must be a typo. The PDF report states “2 to 3 kilowatts” at the end. But, further up the calculations use 3.5 kW. The report sets a lower bound of 3.5 kW by taking the minimum delta-T observed. This gives us 44 MJ. Please double check the report.”
Err…lowest delta-T was 4.7 which is 3500w. But, he then shaved 0.5 off that for 4.2 to be conservative. At 178g/s (lowest rate observed) we get 178 * 4.2 * 4.18 = 3124w. At 3.5 hours that is 39 MJ.
It is interesting to note that either the inlet or outlet probe had a 0.5 to 0.8 degree bias with the outlet probe recording lower than the inlet before the demonstration began.
Again, please double check my math against the PDF report.
[Moderator Edit: Posting Guidelines #2]
It would seem that Brian has not read any of the reports posted by first hand observers, nor watch the videos. From what I have read and seen from those who were there:
1) After the cool-down period, the device was opened up and it was clear that there was no room for concealing enough batteries of any kind to provide the energy required for the self-sustain mode.
2) The unit was weighed before and after the test, with only a 1kg increase (99kg vs 98kg?) after the test, which may have been some residual water.
Have all possible forms of fraud been eliminated at this time? No. But I suggest everyone take some time to visit different websites and do some reading and watching of videos before making any attempt to conclude anything…
-Mark
The e-cat is not working stable, it is necessary 4 hours preparing it and it worked for only a 3 and a half hour. The start up time is greater than the working time.
From Brian Ahern’s email:
“This time he could have concealed enough lithium ion batteries to provide power for 3.5 hour operation.”
Maybe. According to wikipedia typical lithium ion chemistry has a specific energy of about 0.9 MJ/kg. You would need a 35 kg battery to cover the “31.5 megaJoules of energy”.
I did my own energy balance calculation from the PDF report. I came up with +81 MJ net. That would indicate an 90 kg battery. Note, the report indicates the device weighed 98 kg.
According to wikipedia lithium-air batteries are said to have a “practical” specific energy of 3.6 MJ/kg. Unfortunately, they require an oxidizer not unlike many chemical fuels (gasoline, ethanol, etc.) which would have to be hidden as well.
By the way, according to wikipedia lithium ion chemistry has an energy density of 2.2 MJ/L. The report said the volume of the device was 105 L. We also know that the device holds at least 25 L of water. This gives us 80 L to play with. So we could cram 80 / 2.2 = 36.4 MJ worth of li-ion energy if nothing else were in the device. Unfortunately, the visual inspection showed that a large internal heat exchanger was occupying a lot of that space.
Please double check my numbers.
Disclaimer: This post is not intended to be an endorsement of the demonstration or the data contained within the PDF report. Likewise, it is not intended to refute Mr. Ahern’s email. There may, in fact, be batteries of an even more exotic nature with higher specific energies and energy densities than I am ware of (perhaps nanowire batteries?).
[Moderator Edit: Posting Guidelines #2]
The cat used about 5kwh and produced 26 kwh total. Thats 21 kwh net generated.
It is blatantly obvious now that Rossi [Moderator Edit: Posting Guidelines #4.]
There’s no decay curve in the self sustained mode so no. There are two water circuits and the primary was being dumped not recirculated.
Brian may be right, he could be hiding some fuel or chemical heat source and that is indeed the question and the problem with the test among others (such as deltaT credibility) but your energy calculations are just silly.
Steven,
[Moderator Edit: Posting Guidelines #2, #4] Consider the following: How do you take into account the energy that remained stored within the device when the test concluded? I also calculated the total energy inputed into the ECAT during the entire test and agree that it was in the range of 30 Megajoules. The energy consumed by the control system should not be included in the total if we are to measure the LENR effect. Did you consider the heat that was carried out by the main loop into the drain while the device was heating up? I thought not. I did not make that calculation either as it would have taken a lot of extra time and information to perform.
I guess that you were able to tell how much energy was being extracted from the ECAT after the input was removed by the slope in temperature. I was not able to do that due to the modest decline visible. And, how do you know that the device would cease to generate excess heat if it were allowed to operate for another hour or more in this mode? You need to stare into a mirror and ask yourself why you have such a strong bias against the ECAT.
As usual, I know you will not allow a negative comment such as this to be posted. [Moderator Edit: Posting Guidelines #2, #4] Forgive me for being angry, but I think you need to hear the truth.
>When he was caught, he reacted badly.
If this was the 5-6 Sept demo, somebody reported that he left the building slamming very loudly the door.
The guy that reported this rumor is one of the “quite well informed ones” from the italian energeticambiente forum.
I disagree with this, I think the e cat works works here is why. You are only measuring the energy out put for 7.5 hours which shows a 78% energy produced from the energy input. This is just a test, and in real life use the e cat will be used for longer the 7.5 hours. If you were to increasing the time to a longer times. Lets say 14 hours then the calculations produce a more then 100% energy output compared to input. The reason why it looks like energy loss is because the ecat is not run long enough to produce more energy as output then the energy used to start the system including the preparation energy. The point is that when its running it produces more energy then what you put in by a large amount, so the longer you run the system the more energy is produces, and eventually even more then the energy input including the preparation energy. Think about it.
Good job Krivit… The skepticism you are displaying is actually very helpful in both cases:
1) if the e-cat doesn’t work as you say you would be right
2) if the e-cat would work, you would contribute till the official and incontroverible moment to keep big stakeholders of fossiles away from this thing, making them thing it’s rubbish.
where is, in your data, the heat produced during the pre-self sustaining phase??
Hi Steve,
You are completely right here: the moment output temperature of the cell reaches around 100 C, the heater is shut down. But all that energy (put into it in the first 4 hours) is used later on to produce the energy required to heat up the water. When that energy is depleted (seen by a sudden drop in temperature of the steam) the experiment is quickly ended.
It’s just silly people would fall for that. Because it’s just all talk and make believe.
The only remaining question is: how do they do store the energy, and release it later?
Regards,
Jeffrey
[Moderator Edit: Posting Guidelines #4]
Put times/amperes/voltages from Lewan’s report on an Excel spreadsheet and you will see that the energy consumption of the electrical resistance throughout the day was an overall 8.78 kWh (or 31.6 MJ), between hrs 11.52 and 15.53. This alone should suggest you that there is something terribly wrong in your calcs. Basically, the device stored heat for hours and later managed to spit all of it (31.5 MJ, you state) out with no losses! this alone would be big step for manking.
But let’s go on. Mats Lewan calculated 38 MJ output, instead of your 31.5 MJ figure. Where does your number come from? But who cares, let’s stick to your discounted figure. Let’s check something else.
You importantly choose to ONLY consider the output of the device during the self-sustaining mode (between 15:53 and 19:22) and to “FORGET” to calculate it for the hours before. Let’s for simplicity take into account the data available between around 13:48 and 15:53, say 2 hours, just those two and nothing else. During such time there was a consistent temperature difference of at least 3°C between inlet and outlet of the secondary circuit, while the average was over 4°C. So as a minimum, over those 2 hours prior to the selfsustaining mode, there was an additional energy output – released to the secondary circuit – of around 4.5 kWh (or 16.1 MJ) to be added to your original calculation. If we were to consider the average temp difference over those two hours (over 4°C, instead of just 3°C), we would instead have around 6.6 kWh (or 23.7 MJ) extra.
So, as a minimum, 16.1 + 31.5 = 47.6 MJ out (between 13:48 and 19:22, no waste heat during the test and no output figures prior to 13:48!), to be compared to 31.6 MJ consumed by the electrical resistance throughout its use. There is an extra 16 MJ, which is also extremely conservative and based on available data. If we consider Lewan’s 38 MJ figure and add the 23.7 MJ average I gave above, we already are around 52 MJ output vs 31.6 MJ input…
[Moderator Edit: Posting Guidelines #4. If you fail to be civil next time, your post will be completely deleted.]
Regards
Carlo Ombello
Nobody can replies to math. I agree with you, just a question to clear all doubts: was there no measure of output energy while the input energy was provided? Why have you consider just the 31.5 MJoule of Energy during the self-sustaining phase? The right balance of energy is different from what your equation describes.
[Moderator Edit: Posting Guidelines #2,3,4]
Rossi’s Oct 6 test is not conclusive, but Krivit’s summary of of Mats Lewan’s report does not tell the whole story. Please link this or show this chart: [Moderator Edit: Posting Guidelines #11] which is a reasonable interpretation of the whole time line of the test.
If you are going sum all the energy input then you must also sum all the energy output and the quick calculation by Krevit above has not done that.
One should also point out that one does not need any concealed lithium ion battery to commit a fraudulent test. You could simply put a variable resistor in series with the thermistors and create an appearance of a temperature rise of a few degrees. Some initial spurious reading in the heat exchanger temperature rise measurements may give a clue as to whether something like that took place.
@Brian Ahern (Krivit)
[Moderator Edit: Posting Guidelines #12]
Obviously, 38+24=62 MJ, not 52 as I wrote… that is twice as much as the input from the electrical resistance.
Mr. Krivit, this time your numbers are totally wrong!
If you take Lewan’s numbers to the spreadsheet and do some (easy) number crunching, you’ll see it yourself. I did it yesterday and got the following result:
Total heater energy in ~33MJ, total energy out ~108MJ, net energy out ~75MJ.
And I have not accounted for residual heat left in the system when reactor was shut down (too early, it’s a pity, agreed) and data logging was finished, so the net energy is actually considerably larger.
Average output power while in “self sustain” mode was ~5.9kW and what is interesting, the internal temperature (T2) did not decline at all between 16:38 and 19:08 when eCat was shut down, it got even slightly higher (~1 degree).
If you want to check my simple calculations, I can send you my spreadsheet.
The calculation that finds an 8.82 MJ loss assumes the E-Cat produced zero energy during the 4 hr warm-up period. Is this really the case? It seems unlikely. I was unable to find graphs or test data online but I would think that the e-cat output was ramping up during the 4 hr period. The output may have been less than the input during the 4 hr period but I can’t imagine it was zero. As long as it was greater than 8.82 MJ, you would have a net positive energy for the 7.5 hr period.
Another mistake made by some is they assume the primary (steam) loop is closed. That is not the case. The condensed steam simply exits the heat exchanger and, via a hose, flows into a sink drain. That means whatever heat was still present in the condensed steam coming out of the heat exchanger was NOT accounted for, and that would have to be added to the total heat generated by the reactor. Since no attempt was made to monitor the temperature and flow rate of this primary circuit discharge, we cannot estimate how much energy was going down the drain, unaccounted for.
One of my main concerns so far is that the Tout thermocouple on the output side of the secondary circuit of the heat exchanger was too close to the steam input on the primary side… perhaps two to three inches at most. If some heat from the steam was affecting that thermocouple, then it would give a higher reading than actual.
You forgot the output power during the heating step…
If Mr. Rossi has been caught engaging in fraudulent behavior visa vie use of batteries or any other mechanism, this information needs to be made public knowledge IMMEDIATELY. To continue to remain silent about this is akin to a crime in and of itself. Regardless of any legal implications, to allow Rossi to continue to engage in fraudulent behavior is a great disservice to legitimate LENR researchers everywhere and he needs to be exposed before more damage is done.
there is something wrong
(38.88 + 1.44) – 31.5 = 8.82
eletric heating makes energy too?
Dear Steven and Ahern, i just wanna reply some statements
“Rossi was in total control of the test yesterday”
I suppose that if Rossi didn’t do anything, nothing could happen at all.
“This time he could have concealed enough lithium ion batteries to provide power for 3.5 hour operation. Nobody weighed the ‘secret box’ and nobody looked inside the ‘SECRET BOX even though he said everything would be available to them to discover.”
The measureament have been performed for 3.5 hours, but the sistem continuosly work for over 5, the system was weighted and opened to let people check inside.
“All the attendees should consider themselves UNINDICTED CO-CONSPIRATORS in Rossi’s latest fraudulent endeavor”
[Moderator Edit: Posting Guidelines #2]
Enrico Billi
Dear Steven,
You are absolutely right the total Energy OUT / IN data should be accounted, to be more specific here is some data analysis along all test time line = [ "heating phase" AND "self-sustained phase" ]
[Moderator Edit: Posting Guidelines #11]
“Energia” graph shows the total Energy OUT / IN ratio ~= 3
Thanks
Stefan
[Moderator Edit: Posting Guidelines #13]
[Moderator Edit: Posting Guidelines #2]? If I understand correctly, [Rossi] merely heats up a volume of water under pressure until it is well above its boiling point, then mostly turns off the heating process, then slowly releases the steam and hot water stored up and claims it is a self-sustaining nuclear reaction. When I use my 22 quart pressure cooker on top of the stove, I get the same effect. It takes an amazingly long time to cool down that volume of water heated to about 115 degrees C, and if the outlet valve on top is slightly opened, it will blow steam vigorously for a very long time, maybe hours if it were wrapped with insulation. What gives? It should be so easy to demonstrate a real effect of the sort claimed. The really interesting questions here would seem to be whether anybody involved really believes they are getting nuclear power this way.
Mr. Krivit, your calculation seems based on the assumption that the Ecat’s electric resistance was switched on at full power and then powered without interruption for four hours.
Indeed, the report published by Lewan show that the total “power on” time is 202 minutes, not 240 as you suggest, because of on-off cycling between 14.01 and 14.59.
Furthermore, the resistance was switched on at 61% of full power and gradually set to full power between 11.25 and 12.32.
So, by using the actual V-A values measured through resistance (reported as zero in self-sustaining mode), the energy input in the system amounts to 30.89 MJ. The RF device did not feed current through the resistance.
If indeed the Ecat produced 31.5 MJ of energy, its efficiency is 102%.
While Rossi failed to show the high efficiency he has been claiming since the first test, the last test did not demonstrate the energy loss stated in your article.
Best wishes
Gianni Bedini
Hello!
Allow me to start by pointing out a typo.When you say “Rossi’s device released an average of 2.5 kilowatts of heat in 3.5 hours” it should be …kilowatts of heat during 3.5hours,as power[kw] is inherently stretched over time.
And claiming this test shows Rossi’s device produces heat losses is at best deceiving.
Taking such a long time to heat up “shows” that the device has already reached a stable condition,so it wouldnt matter if it had been 4 ,8 or 100 hours.However ,in your calculations it matters very much.
This last paragraph depends totally on water flowing through the ecat during that time.
If it werent,your calculations would be fine.
If you want to take into account how much energy the heated box (ecat) gives back to the flowing water,as to show no real energy is produced,I guess you could take the Ecats weight,suppose certain material percentages ,look their heat densities up (Cp is the common symbol) and put that value instead of 4hours*x kw in your formula.
[Moderator Edit: Posting Guidelines #2]
How about talking to the famous scientists who al least did support Rossi to design a test for him and put it through,in order to clean his image?
Thank you for so many months of dedication
Hi Steve,
Here is how I think they might have done it:
- if you look at the image of the box ( http://www.nyteknik.se/incoming/article3284879.ece/BINARY/original/DSC_0089_600.jpg ) you see flinges on a slab of some material heavely bolted to the sides of the approx 50 x 40 x 30 cm metal box.
- Lets say this slab is 15 x 30 x 30 cm made out of iron (roughly 100kg)
- Lets say this slab is suspended 10 cm above the bottom of the box (and the water inlet is at the bottom)
- it takes several hours of pumping water (at around 0.91 g/s) for the water to reach the iron slab
- during this time, the iron slab is heated to 900 degrees Celsius, using electrical heaters
- The moment the water hits the slab, it vaporizes instantly (and when this is detected the electrical input current is stopped)
- Since the slab has 900 * 100000 * 0.45 = 40.5 MJ of energy, slowly pumping water into the box will keep vaporizing water for several hours.
- When the slab reaches a temperature around 100 degrees Celsius the vaporization stops, and the experiment is (quickly) stopped.
End of magic trick.
Regards,
Jeffrey
the measures wasn’t so correct, however also when the device was in heating mode it was releasing power to the exchanger. so dubbious measures + dubious interpretations = caos
who won?
Before you can decide if there was a energy gain don’t you have to know if there was any output in the warm up period. I don’t think that there’s enough information to conclude that this demonstration proves anything one way or the other. My conclusion is I have none. It is perfectly possible this works and Rossi is just eccentric and doesn’t want to prove beyond doubt it works.
Dan,
You are correct, this is an assumption that is ambiguous. Thank you for pointing this out.
Lewan wrote: “the start-up was effected by heating the E-cat with an electrical resistor at about 2.7 kilowatts, this time for about four hours, in order to achieve, according to Rossi, sufficient stability.”
Lewan mentioned nothing about measuring energy output (not just temperature, you need to know the flow, of course) during the heating phase.
I have just asked Lewan the following questions. Hopefully he can shed some light.
*********
Can you please tell me the kj of heat released during the heating phase?
And also how you derive it?
I don’t see any evidence that you measured rate of outflow of primary circuit during the heating phase. Have I missed this information?
********
SBK
[Oct. 8. 2:44pm Pacific. Lewan has provided responses. I have sent him back questions to clarify his responses. Hope to publish answers on Sunday. SBK]