Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Sat, Jun 25, 2011 at 9:11 PM, Jeff Driscoll hcarb...@gmail.com wrote: On Sat, Jun 25, 2011 at 8:58 PM, mix...@bigpond.com wrote: In reply to Joshua Cude's message of Fri, 24 Jun 2011 16:20:48 -0500: Hi, [snip] I was talking about running it above boiling, but way below the level needed to boil it all. Different thing. And it's easy. The power can range within a factor of 7. In this case, anywhere between 600W and about 5 kW. BTW (the latent heat of steam) / (the heat energy required to bring water to the boil) is a factor of about 6.7 (depending on starting temperature of water), and curiously close to the COP Rossi claims to be aiming for. In short, if virtually none of the water were converted to steam, and he was assuming that it all was, then it would neatly explain the conversion factor he is claiming. You might be thinking of another scenario - but if I'm guessing what you are saying then the best anyone could do is about 1.86 to 1 ratio. But this assumes that any liquid hot water needed to cool water vapor in a heat exchanger is included in the calculation (otherwise the ratio would be worse, less than 1.86 to 1). I did this calculation, shown below, weeks ago. [...] There is no need for heat exchangers to arrive at the ratio of around 7. The argument goes, that if the water starts at 10C, then the amount of heat required to vaporize 1 g is 90 + 540 = 630 cal. The amount of heat required to bring it to the bp is 90 cal, and the ratio is 630/90 = 7. (Different starting temperatures give slightly different ratios) These two scenarios result in the same quantitative data reported in one of Rossi's steam producing demos, because he only reports temperature, and input flow rate. So the same data is consistent (in Krivit's run) with 600W and with 5 kW. Rossi does not provide quantitative evidence that it should be closer to, or at, the high end of that range. He only makes pronouncements based on things like visual inspections, or unreported RH measurements, which indicate nothing. Now, of course, the fact that there is some steam, means it is not at the bottom of the range either, but in the videos where he shows the steam, it is not impressive.
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Sun, Jun 26, 2011 at 3:23 AM, Joshua Cude joshua.c...@gmail.com wrote: On Sat, Jun 25, 2011 at 9:11 PM, Jeff Driscoll hcarb...@gmail.com wrote: On Sat, Jun 25, 2011 at 8:58 PM, mix...@bigpond.com wrote: In reply to Joshua Cude's message of Fri, 24 Jun 2011 16:20:48 -0500: Hi, [snip] I was talking about running it above boiling, but way below the level needed to boil it all. Different thing. And it's easy. The power can range within a factor of 7. In this case, anywhere between 600W and about 5 kW. BTW (the latent heat of steam) / (the heat energy required to bring water to the boil) is a factor of about 6.7 (depending on starting temperature of water), and curiously close to the COP Rossi claims to be aiming for. In short, if virtually none of the water were converted to steam, and he was assuming that it all was, then it would neatly explain the conversion factor he is claiming. You might be thinking of another scenario - but if I'm guessing what you are saying then the best anyone could do is about 1.86 to 1 ratio. But this assumes that any liquid hot water needed to cool water vapor in a heat exchanger is included in the calculation (otherwise the ratio would be worse, less than 1.86 to 1). I did this calculation, shown below, weeks ago. [...] There is no need for heat exchangers to arrive at the ratio of around 7. The argument goes, that if the water starts at 10C, then the amount of heat required to vaporize 1 g is 90 + 540 = 630 cal. The amount of heat required to bring it to the bp is 90 cal, and the ratio is 630/90 = 7. (Different starting temperatures give slightly different ratios) Why would you divide the energy to vaporize 1 g of water (starting at 10 C) by the energy to heat it from 10 C to 100 C (liquid)? Seems random to me. I wrote out a whole scenario with (I thought) clear steps. Give me your reasoning and steps. These two scenarios result in the same quantitative data reported in one of Rossi's steam producing demos, because he only reports temperature, and input flow rate. So the same data is consistent (in Krivit's run) with 600W and with 5 kW. Rossi does not provide quantitative evidence that it should be closer to, or at, the high end of that range. He only makes pronouncements based on things like visual inspections, or unreported RH measurements, which indicate nothing. Now, of course, the fact that there is some steam, means it is not at the bottom of the range either, but in the videos where he shows the steam, it is not impressive.
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Sun, Jun 26, 2011 at 10:02 AM, Jeff Driscoll hcarb...@gmail.com wrote: Why would you divide the energy to vaporize 1 g of water (starting at 10 C) by the energy to heat it from 10 C to 100 C (liquid)? Seems random to me. Because those are the two extremes of a situation that results in the *same* measured data that Rossi reports. I thought that was clear. I wrote out a whole scenario with (I thought) clear steps. Give me your reasoning and steps. These two scenarios result in the same quantitative data reported in one of Rossi's steam producing demos, because he only reports temperature, and input flow rate. So the same data is consistent (in Krivit's run) with 600W and with 5 kW. Rossi does not provide quantitative evidence that it should be closer to, or at, the high end of that range. He only makes pronouncements based on things like visual inspections, or unreported RH measurements, which indicate nothing. Now, of course, the fact that there is some steam, means it is not at the bottom of the range either, but in the videos where he shows the steam, it is not impressive.
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
In reply to Joshua Cude's message of Fri, 24 Jun 2011 16:20:48 -0500: Hi, [snip] I was talking about running it above boiling, but way below the level needed to boil it all. Different thing. And it's easy. The power can range within a factor of 7. In this case, anywhere between 600W and about 5 kW. BTW (the latent heat of steam) / (the heat energy required to bring water to the boil) is a factor of about 6.7 (depending on starting temperature of water), and curiously close to the COP Rossi claims to be aiming for. In short, if virtually none of the water were converted to steam, and he was assuming that it all was, then it would neatly explain the conversion factor he is claiming. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
Yep...
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Sat, Jun 25, 2011 at 8:58 PM, mix...@bigpond.com wrote: In reply to Joshua Cude's message of Fri, 24 Jun 2011 16:20:48 -0500: Hi, [snip] I was talking about running it above boiling, but way below the level needed to boil it all. Different thing. And it's easy. The power can range within a factor of 7. In this case, anywhere between 600W and about 5 kW. BTW (the latent heat of steam) / (the heat energy required to bring water to the boil) is a factor of about 6.7 (depending on starting temperature of water), and curiously close to the COP Rossi claims to be aiming for. In short, if virtually none of the water were converted to steam, and he was assuming that it all was, then it would neatly explain the conversion factor he is claiming. You might be thinking of another scenario - but if I'm guessing what you are saying then the best anyone could do is about 1.86 to 1 ratio. But this assumes that any liquid hot water needed to cool water vapor in a heat exchanger is included in the calculation (otherwise the ratio would be worse, less than 1.86 to 1). I did this calculation, shown below, weeks ago. Basically in this fraudulent set up, the Ecat would do the following: 1. Create 1 kg of 99.9 C water from 10 C water which requires (99.9 - 10) x 4.18 kJ/kg/C = 376 kJ 2. Using same water from step 1, make 1 kg of water *vapor* requiring 2257 kJ. Total input to Ecat required at this point is 376 + 2257 = 2633 kJ 3. Condense water vapor into micro droplets (i.e. fog) deep *inside* the Ecat using a heat exchanger and use this heat to heat 6.00 kg of cold liquid water from 10 C to 99.9 C.This is because 2257 kJ /376 kJ/kg = 6.00 kg (note that the units are correct). Also, note that at this point the total input energy is still 2633 kJ. The actual/real end result is 6.00 kg of 99.9 C water and 1 kg of micro liquid water *droplets* (fog or steam with 0% quality). A gullible observer would think that the Ecat just produced 6 kg of hot water and 1 kg of water *vapor* when it really made 6 kg of hot water and 1 kg of hot *liquid* water droplets. The gullible observer would think that the energy normally needed to create this is 4890 kJ because: (6 kg + 1 kg) x (99.9 - 10) x 4.18 kJ/kg + (1 kg) x 2257 kJ/kg = 4890 kJ While in *reality* it took the following amount of electrical energy: (6 kg + 1 kg) x (99.9 -10) x 4.18 kJ/kg = 2633 kJ So, the gullible observer would see 2633 kJ of electrical energy go into the Ecat and 4890 kJ of thermal energy leave the Ecat. This is a ratio of 4890/2633 = 1.86 I can't think of any way of increasing this ratio using any other similar method. Jeff Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
Nice... well, high time for a few of Rossi's trusted friends to ensure that he is alerted quickly...
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
Not convincing calculations. What is the density of water that can be sustained in droplets and what is the size of the droplets before they coalesce and rain back?
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Fri, Jun 24, 2011 at 8:19 PM, Jed Rothwell jedrothw...@gmail.com wrote: Joshua Cude joshua.c...@gmail.com wrote: Nope. When you put 800 W into something like this, a large fraction of it radiates from the cell into the surroundings. The cell is insulated. It is too hot to touch according to witnesses. Which witnesses are those? I have not seen any mention of the temperature of the insulation, but maybe I missed it. But it's definitely not too hot to touch. Check the Krivit video from 9:22 to 10:00, Rossi repeatedly touches and presses it. At 9:50 he lays two fingers across the top with considerable pressure, and leaves them there for nearly 10 seconds, without any indication of discomfort, and without any comment about its temperature. It's not too hot to touch. I think you made that up. In the next moments he touches the hose, and lets go quickly with a comment about how hot it is. The insulation means it takes longer to get hot on the outside; the difference between the inside and the outside is greater; and more heat transfers to the water. But there is still plenty being radiated out. A little maybe, but without a temperature measurement, we don't know if it's plenty or oodles or scads. Rossi does all his calculations without taking account of losses through the insulation, so presumably he doesn't think they're significant. That isn't much with a large object that is too hot to touch. But it is much for an object that appears to be at room temperature. And if you're claiming 50 - 75 % for any power, then at 5 kW, about 2.5 kW would have to radiate from the insulation. Are you claiming that? Dunno. Recovery rates change with temperature, flow rates and other conditions. Actually, they usually get worse. Probably this is producing ~4 kW and that makes the surface too hot to touch. If, as you believe, it is only producing 800 W then the insulation isn't very good, is it? Except it's not too hot to touch, so either the insulation is very good, or it's not producing 4 kW. No. It doesn't. Whatever the fluid is, and regardless of the shape, it's gonna flow through. It does it as a liquid, and it does it as a steam-liquid mixture. There's a pump forcing it through. If the water was overflowing out of the top and down the hose, cold water would be coming in to replace the boiling water and the temperature would drop below boiling, as I said. Probably down to around 95 deg C. That's ridiculous. Cold water is always coming in to replace the liquid water or the steam. It's pushed in at a steady rate by the pump. The cold water runs past the reactor and warms up. If the power is high enough, some of it changes phase, but always it gets pushed out by the inflowing cold water. It is very difficult to maintain a flow calorimeter outlet temperature of exactly 101 deg C unless the water is boiling, leaving as vapor, and only the vapor touches the temperature sensor. No. It is very easy to maintain the output of the ecat at exactly boiling temperature as long as the input power is above the value required to raise the water to boiling, and below the power required to boil it all. In the Krivit run, that range is between 600W and something over 4 kW. If the power is in that range, the output fluid must be a mixture of liquid and steam. What do you suggest would happen if the input power was 2 kW? That's more than enough to heat it all to boiling, and not enough to vaporize it all. In this case, the water would be overflowing, as you call it, and so you think the temperature would drop. But then the water is removing less than 600 W from the reactor, If there is 2 kW input, it has to heat up, and then of course the water would get hotter, so its temperature wouldn't drop. To repeat if the input is between 600W and 4 kW (or so) the output must be a mixture of liquid and gas. It is not difficult, but rather dead easy, to arrange power in that range. Rossi could tell it is overflowing by watching the temperature. When it falls below 100 deg C, he increases anomalous heat. Wait. How does he do this? You insisted earlier that he does not mess with the input power. If not enough water comes in and it dries up, the temperature would rise above 101 deg C, and he reduces it. When and how does he do this. The assumption in all the recent demos is that the input power is constant. That's why it's measured at the beginning, and then left alone (ostensibly). He makes no claims about adjusting the reaction, and there is no evidence that he does. He can control the strength of the anomalous heat. I do not know how he does that. Maybe telepathy. Apparently he has enough control to keep the bottom portion filled with boiling water but not overflowing. Have you looked at the photos of the ecats without the insulation. The water flows past the reactor horizontally, and then it reaches the chimney after it passes through the reactor. There is
[Vo]:Okay, suppose there is only 800 W input with no anomalous heat
Joshua Cude wrote: You only get a stable water/steam mixture in a closed vessel (a teapot). Why? If it takes say 1 kW to raise the temperature of the flowing water to 100C, and then you supply 1.5 kW (using only and electric heater), then only part of the flowing water will get converted to steam, and you will have to have a mixture of liquid and gas coming out. It is hard to arrange things so it transfers just enough heat to bring the temperature up to boiling, and boils some of the water in the time it takes the water to transit the hot surface. You can get it below that a little, or above it, but manually adjusting the flow rate or input power just right to hit that level is tough. (As I said, doing it with computer control is a piece of cake.) It usually ends up at ~95°C, as I said. That's what you see in data from people who run flow calorimetry close to boiling. Let's look at the facts here: 1. Rossi did not adjust the flow at all. Krivit would have said if he did. 2. Rossi did not adjust the input power. Krivit would seen this, too. 3. The video shows some steam coming out of the 3 m hose. 4. Input power was ~800 W. 5. The flow rate was ~7 L/h = ~1.9 g/s So the only way for Rossi to make it produce a little steam and a lot of hot water would be for him to adjust the anomalous heat output. It would be a miracle if Rossi has such good control over the anomalous heat that he can push the temperature up to 99°C and have mostly liquid water go through plus a little steam. If he can do that, he has truly mastered cold fusion! And if he can do that, why not just vaporize the whole stream of water? I realize you do not think there is any anomalous heat. You think the electric power input balances the heat output. That is barely possible with this test, assuming you can magically transfer all of the heat to the water without heating the vessel. But in previous tests the input power was lower and the water temperature would only be 60°C so there must have been anomalous heat. I realize you think there is some sort of trick or fraud at work, and the input power was really larger than shown. But let me suggest that for you assume for the sake of argument there were no tricks. Without tricks, there had to be anomalous heat in previous test runs, as I said. And with this run, ~800 W input and 1.9 ml/s flow rate, assuming not one joule of heat radiated from the cell and it remained miraculously at room temperature . . . 800 W = 190 calories per second. To bring 1.9 g from 25°C to 99°C takes 140 calories. That leaves 50 calories to vaporize some of the remaining water: 0.09 grams, to be exact. Not much! I don't think you could see 0.09 g/s of vapor. Do you? How much do you think would reach the end of the 3 m hose? In real life we know the cell got hot. It would have to get hot. There is no chance any of the water would vaporize with only ~800 W input. You would not any steam at all. Even with this high input power, any steam at all is proof there is anomalous heat. - Jed
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Fri, Jun 24, 2011 at 3:20 PM, Jed Rothwell jedrothw...@gmail.com wrote: ** Joshua Cude wrote: You only get a stable water/steam mixture in a closed vessel (a teapot). Why? If it takes say 1 kW to raise the temperature of the flowing water to 100C, and then you supply 1.5 kW (using only and electric heater), then only part of the flowing water will get converted to steam, and you will have to have a mixture of liquid and gas coming out. It is hard to arrange things so it transfers just enough heat to bring the temperature up to boiling, and boils some of the water in the time it takes the water to transit the hot surface. You can get it below that a little, or above it, but manually adjusting the flow rate or input power just right to hit that level is tough. What are you on about. We can calculate the heat needed to bring the water to the boiling point. I was suggesting we exceed it by 50%. There's nothing hard about that, and no computers are needed. It usually ends up at ~95°C, as I said. That's what you see in data from people who run flow calorimetry close to boiling. I was talking about running it above boiling, but way below the level needed to boil it all. Different thing. And it's easy. The power can range within a factor of 7. In this case, anywhere between 600W and about 5 kW. Let's look at the facts here: 1. Rossi did not adjust the flow at all. Krivit would have said if he did. We don't know that, but it's not relevant to this discussion. 2. Rossi did not adjust the input power. Krivit would seen this, too. Again, we don't know that, and again it's not relevant. 3. The video shows some steam coming out of the 3 m hose. 4. Input power was ~800 W. 5. The flow rate was ~7 L/h = ~1.9 g/s OK. So the only way for Rossi to make it produce a little steam and a lot of hot water would be for him to adjust the anomalous heat output. Wrong. As you showed, only 600 W is needed to bring the water to the boiling point.That leaves 200W to produce a little steam and a lot of hot water. It would be a miracle if Rossi has such good control over the anomalous heat that he can push the temperature up to 99°C and have mostly liquid water go through plus a little steam. I don't get your problem. The electrical power raises the water to 100C and produces a little steam on top of it. Simple so far. You can argue that the steam coming out represents more than 200W worth of steam, and therefore that the reactor must have contributed some heat. But there is no fine control needed for this. The more heat it produces, the more steam you would get. Nothing at all magic is needed here. And my guess is that the Ni-H produces a little chemical heat, but the evidence for even that is not convincing. I still think Rossi could easily have adjusted the power (and less likely the flow) without Krivit noticing. I also think his claim of the flow is wrong based on the esowatch evidence. I realize you do not think there is any anomalous heat. You think the electric power input balances the heat output. Then you haven't read my posts. I have frequently allowed the possibility of some heat production in the reactor; a few hundred watts seems to fit some of the data. That is barely possible with this test, assuming you can magically transfer all of the heat to the water without heating the vessel. It's not magic to transfer nearly all the heat to the water. The vessel gets hot sure, but it doesn't radiate much with all that insulation around it. Once it reaches equilibrium temperature, then the input heat goes to the water, or to radiation from the insulation. But in previous tests the input power was lower and the water temperature would only be 60°C so there must have been anomalous heat. Right. I've addressed those too. There are 3 obvious possibilities. The power is higher than claimed, the flow is lower than claimed, or the device produces a little chemical heat. It's really only needed in the EK run, and then only about 300 W. Without tricks, there had to be anomalous heat in previous test runs, as I said. In the EK run, without tricks or mistakes, it seems the reactor would have to produce a few hundred watts, yes. I've said this many times. But a few hundred watts does not convincingly exclude chemical heat. In the Lewan runs, less than 100 W were needed, if any at all. In the January run, I cannot exclude mistakes, because they are too obvious. The claimed flow rate exceeded the pump specs by a factor of 2. (Even Levi made such obvious mistakes in his written report as claiming the temperature was at 100C for 40 minutes, when it was only 18 minutes; it is hard to trust anything from that run.) But if you use the max flow rate of the pump, then no additional power from the reactor is needed to explain the data. And with this run, ~800 W input and 1.9 ml/s flow rate, assuming not one joule of heat radiated from the cell and it
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
While I am also a skeptical, even rough approximation gives a huge output gain. Above 100 degrees means gas, and pumping a mixture would require either another pump, by means of ventilation. Ventilation is noisy and would require a large opening. Even 1% of liquid is a thick fog, which is not the case, it doesn't change the end result as much as 1% of the volume were liquid, which would basically mean a foam as the output. The actual results are all more consistent with at least 2500KW than less than 1000W. You'd have to put a lot of things that are not there. It is much easier to suppose that Rossi is just draining energy from somewhere else.
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
Joshua Cude wrote: There is no chance any of the water would vaporize with only ~800 W input. You would not any steam at all. Even with this high input power, any steam at all is proof there is anomalous heat. What are you talking about. You just did the calculation yourself showing that it takes only 3/4 of that (600W) to bring the water to the boiling point. If you are putting 800W into the cell, and the only way you are taking it out is with water, some of the water would vaporize. Nope. When you put 800 W into something like this, a large fraction of it radiates from the cell into the surroundings. The recovery rate for the water flowing through will be maybe 50% to 75%. In other words, only 400 to 600 W reaches the water. Barely enough to boil it, and none of that steam would make it to the other end of the 3 m hose. Even if 700 W reached the water, I doubt you would see any steam emerge from the other side. I believe the most pessimistic estimate here is that the hose radiates ~100 W. What I am saying is that this is a poor design for a flow calorimeter. If you want to recover 80% or 90%, you need to run at a much lower temperature (below 50°C I think), and you need convolutions in the cooling water path. You can make a fantastic flow calorimeter, with 98% recovery, with the water below 30°C and lots of other techniques. That's what McKubre did. Rossi is claiming these things produce multi-kW, but only a few hundred watts are enough to explain all the quoted data. You have it backwards. Rossi is assuming the steam is dry, which it almost certainly is. Based on that assumption he estimates that it produces multiple kilowatts. He does not start off with that assumption and then work backwards. _You_ are doing that! You assume there must be only 800 W so there has to be some way to explain these temperatures and the appearance of the steam, and there must be hot water coming through. Rossi has spent a lot of time with teapot-shaped flow calorimeters, where the steam exit is placed well above the hot surface. That ensures dry steam, as long as you keep the flow rate reasonable. I have seen tons of calibration data from Fleischmann and Pons and the Italians who did boiling calorimetry to know that he is right. You're saying even those few hundred watts prove a nuclear effect, and maybe if they ran it long enough there would be something, if all the numbers were really nailed down with credible observers. But if it's really nuclear, why is this experiment, just like all CF experiments, in this pergatory, where it's even possible to quibble day after day? Why is there never enough power to make it obvious, and better, to power itself? There is enough power to make it obvious! The power is in kilowatts. Rossi's method of estimating power is correct. You imagine it is wrong, and you invent all kinds of improbable Just So Stories to explain how your fantasy might be true, but you are wrong. You know perfectly well why this one does not power itself. Because it is likely to explode. Ask Mizuno what that's like. There will be self-powered ones with electric power generation within a year or so. - Jed
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Fri, Jun 24, 2011 at 4:35 PM, Daniel Rocha danieldi...@gmail.com wrote: While I am also a skeptical, even rough approximation gives a huge output gain. Above 100 degrees means gas, A temperature reading within a degree or two of 100C is consistent with a mixture of gas and liquid. and pumping a mixture would require either another pump, by means of ventilation. Why? The pump is capable of pushing through pure liquid. It should have no trouble with a mixture of gas and liquid. Ventilation is noisy and would require a large opening. Why? Pure steam would require more ventilation than a mixture of steam and water. The volume is 1700 times higher. It would be louder and hotter. Even 1% of liquid is a thick fog, which is not the case, This depends on droplet size etc. There are papers on 2-phase flow that measure the size of the droplets; they're larger than in a fog. The actual results are all more consistent with at least 2500KW than less than 1000W. OK. We clearly disagree about this. That little puff of steam looks like a few hundred watts to me. I don't think we can resolve this by typing. In any case, it's not a quantitative measure. Using the quantitative measures (temperature and flow rate), only 600 W is needed. The rest is hand-waving arguments about steam dryness. Rossi could easily prove it's dry by heating it to 120C (by reducing the flow rate), or by measuring the flow rate of the steam. Wonder why he doesn't do it. It is much easier to suppose that Rossi is just draining energy from somewhere else. I think that's harder, actually.
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
This is going into an infinite loop. Trying to explain that with only 800 is just too hard for me. Thanks for trying.
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Fri, Jun 24, 2011 at 4:49 PM, Jed Rothwell jedrothw...@gmail.com wrote: ** Joshua Cude wrote: There is no chance any of the water would vaporize with only ~800 W input. You would not any steam at all. Even with this high input power, any steam at all is proof there is anomalous heat. What are you talking about. You just did the calculation yourself showing that it takes only 3/4 of that (600W) to bring the water to the boiling point. If you are putting 800W into the cell, and the only way you are taking it out is with water, some of the water would vaporize. Nope. When you put 800 W into something like this, a large fraction of it radiates from the cell into the surroundings. The cell is insulated. The recovery rate for the water flowing through will be maybe 50% to 75%. In other words, only 400 to 600 W reaches the water. I don't believe it. Then the insulation would be radiating 200W to 400W. Not plausible. But go ahead. Try to make it plausible. Estimate the area and the temperature necessary for this. And if you're claiming 50 - 75 % for any power, then at 5 kW, about 2.5 kW would have to radiate from the insulation. Are you claiming that? Rossi is claiming these things produce multi-kW, but only a few hundred watts are enough to explain all the quoted data. You have it backwards. Rossi is assuming the steam is dry, which it almost certainly is. Based on that assumption he estimates that it produces multiple kilowatts. He does not start off with that assumption and then work backwards. *You* are doing that! You assume there must be only 800 W so there has to be some way to explain these temperatures and the appearance of the steam, and there must be hot water coming through. No, I'm looking at the output, at the temperature curves and concluding that dry steam is laughably implausible, and therefore I do not accept the claim of multi-kW output. Rossi has spent a lot of time with teapot-shaped flow calorimeters, where the steam exit is placed well above the hot surface. That ensures dry steam, as long as you keep the flow rate reasonable. No. It doesn't. Whatever the fluid is, and regardless of the shape, it's gonna flow through. It does it as a liquid, and it does it as a steam-liquid mixture. There's a pump forcing it through. You're saying even those few hundred watts prove a nuclear effect, and maybe if they ran it long enough there would be something, if all the numbers were really nailed down with credible observers. But if it's really nuclear, why is this experiment, just like all CF experiments, in this pergatory, where it's even possible to quibble day after day? Why is there never enough power to make it obvious, and better, to power itself? There is enough power to make it obvious! Even several of the CF advocates here are skeptical, so it is clearly not obvious. There will be self-powered ones with electric power generation within a year or so. And will that be used to power the CF car you predicted would be built before the year 2000?
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
YOW -- WHAT YOU JUST SAID On 11-06-24 04:20 PM, Jed Rothwell wrote: So the only way for Rossi to make it produce a little steam and a lot of hot water would be for him to adjust the anomalous heat output. It would be a miracle if Rossi has such good control over the anomalous heat that he can push the temperature up to 99°C and have mostly liquid water go through plus a little steam. If he can do that, he has truly mastered cold fusion! Jed, man, think about that -- don't just jerk your knee at me in an automatic defense of Rossi, really think about it. Rossi has a factor of SEVEN in output level in the range he has to hit in order to produce SOME steam and SOME hot water, and you have just said it would be hard for him to control the anomalous heat well enough to do that. But Rossi's claiming to have produced exactly enough heat to EXACTLY vaporize all the input water, and NOT HEAT THE STEAM beyond boiling -- that target is orders of magnitude smaller than the target he'd need to hit to produce some steam and some hot water! If he overshoots his dry steam power level by even a little, the steam temperature will go up by a lot; the specific heat of steam is very small compared to the heat of vaporization of water. But the temperature never rises more than about a degree over boiling! Jed, the point you just made is the point that's been bugging me all along -- it would take a miracle of fine control to generate EXACTLY enough anomalous heat to EXACTLY vaporize all the input water, without superheating the steam, and without leaving wet steam or having the device spit water! There's no evidence of that degree of control, no evidence of a feedback loop which could be providing it, no reason except wishful thinking to believe such control exists ... so the conclusion is that he's actually got the power level set somewhere within the factor of 7 window, and he's producing very wet steam or a mix of steam and liquid water; he does *NOT* have it right on the edge, producing dry steam just over the boiling point. It's absurd to think he could exercise the level of precise control needed to produce exactly dry steam. (And that about uses up my Friday night send-some-useless-email time...)
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Fri, Jun 24, 2011 at 5:36 PM, Stephen A. Lawrence sa...@pobox.comwrote: YOW -- WHAT YOU JUST SAID On 11-06-24 04:20 PM, Jed Rothwell wrote: So the only way for Rossi to make it produce a little steam and a lot of hot water would be for him to adjust the anomalous heat output. It would be a miracle if Rossi has such good control over the anomalous heat that he can push the temperature up to 99°C and have mostly liquid water go through plus a little steam. If he can do that, he has truly mastered cold fusion! Jed, man, think about that -- don't just jerk your knee at me in an automatic defense of Rossi, really think about it. Rossi has a factor of SEVEN in output level in the range he has to hit in order to produce SOME steam and SOME hot water, and you have just said it would be hard for him to control the anomalous heat well enough to do that. But Rossi's claiming to have produced exactly enough heat to EXACTLY vaporize all the input water, and NOT HEAT THE STEAM beyond boiling -- that target is orders of magnitude smaller than the target he'd need to hit to produce some steam and some hot water! If he overshoots his dry steam power level by even a little, the steam temperature will go up by a lot; the specific heat of steam is very small compared to the heat of vaporization of water. But the temperature never rises more than about a degree over boiling! Jed, the point you just made is the point that's been bugging me all along -- it would take a miracle of fine control to generate EXACTLY enough anomalous heat to EXACTLY vaporize all the input water, without superheating the steam, and without leaving wet steam or having the device spit water! There's no evidence of that degree of control, no evidence of a feedback loop which could be providing it, no reason except wishful thinking to believe such control exists ... so the conclusion is that he's actually got the power level set somewhere within the factor of 7 window, and he's producing very wet steam or a mix of steam and liquid water; he does *NOT* have it right on the edge, producing dry steam just over the boiling point. It's absurd to think he could exercise the level of precise control needed to produce exactly dry steam. (And that about uses up my Friday night send-some-useless-email time...) Thanks. You put it better than I did.
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
On Jun 24, 2011, at 2:36 PM, Stephen A. Lawrence wrote: YOW -- WHAT YOU JUST SAID On 11-06-24 04:20 PM, Jed Rothwell wrote: So the only way for Rossi to make it produce a little steam and a lot of hot water would be for him to adjust the anomalous heat output. It would be a miracle if Rossi has such good control over the anomalous heat that he can push the temperature up to 99°C and have mostly liquid water go through plus a little steam. If he can do that, he has truly mastered cold fusion! Jed, man, think about that -- don't just jerk your knee at me in an automatic defense of Rossi, really think about it. Rossi has a factor of SEVEN in output level in the range he has to hit in order to produce SOME steam and SOME hot water, and you have just said it would be hard for him to control the anomalous heat well enough to do that. But Rossi's claiming to have produced exactly enough heat to EXACTLY vaporize all the input water, and NOT HEAT THE STEAM beyond boiling -- that target is orders of magnitude smaller than the target he'd need to hit to produce some steam and some hot water! If he overshoots his dry steam power level by even a little, the steam temperature will go up by a lot; the specific heat of steam is very small compared to the heat of vaporization of water. But the temperature never rises more than about a degree over boiling! Jed, the point you just made is the point that's been bugging me all along -- it would take a miracle of fine control to generate EXACTLY enough anomalous heat to EXACTLY vaporize all the input water, without superheating the steam, and without leaving wet steam or having the device spit water! There's no evidence of that degree of control, no evidence of a feedback loop which could be providing it, no reason except wishful thinking to believe such control exists ... so the conclusion is that he's actually got the power level set somewhere within the factor of 7 window, and he's producing very wet steam or a mix of steam and liquid water; he does *NOT* have it right on the edge, producing dry steam just over the boiling point. It's absurd to think he could exercise the level of precise control needed to produce exactly dry steam. (And that about uses up my Friday night send-some-useless-email time...) Hi Stephen, It is not difficult at all to achieve something that *looks* like perfectly regulated high thermal output by simply using percolator type effects to dump water into the output hose. I commented on this earlier in another thread, quoted below. On Jun 24, 2011, at 1:35 AM, Horace Heffner wrote: If you look at the E-cat design you can see that it has the potential to act similar to a coffee percolator. See: http://3.bp.blogspot.com/-VIn_mQi1H-M/TZ1ZIpKD4-I/LAE/ xo1T4ZRm41o/s1600/ECAT_explained.jpg http://www.wipo.int/patentscope/search/en/WO2009125444 It is a boiling chamber followed by a vertical tube and elevated ejection port. A relative humidity sensor will max out at 100%, and would not be capable of detecting a percolator style of operation. It is merely a polymer or metal oxide thin film protected by a porous metal electrode. It can not measure steam quality. There is no reason to expect that water on the surface of the protecting porous metal electrode will have a significant effect on an already 100% RH reading. A percolator can produce liquid mass flows far exceeding 1% by volume of gas. The amount of percolation obtained can be controlled by controlling the ratio of the flow of water to the amount of heat applied to the chamber. Active controllers exist in the Rossi device. Water has been seen coming out of the hose. Unless careful measurements are taken it is not known the quantity of water vs gas. On Jun 24, 2011, at 10:53 AM, Horace Heffner wrote: It is notable that the power input varies depending on the controller actions, that if the power input (plus any nuclear output heat if any) should become less than that required to convert all the input water to steam then the liquid excess will eventually simply overflow, i.e. be pumped out into the hose and down the drain. Note that the pump rate is small, on the order of a few cc per second, so it can take a while to fill up a hose held upright into the air, even if the device itself is full of water - which probably can not happen due to percolator type effects. Ironically, all that is required to get excess heat is to *reduce* the input power occasionally (or even permanently) so as to pump some excess water out instead of steam. Something that would obviously be helpful for demos would be the use of translucent tubing, such as polyamide (nylon) tubing, which is good up to 100 °C, instead of black rubber. See: http://www.graylineinc.com/tubing-materials/nylon.html A transparent U-trap just past the current steam
Re: [Vo]:Okay, suppose there is only 800 W input with no anomalous heat
Joshua Cude joshua.c...@gmail.com wrote: Nope. When you put 800 W into something like this, a large fraction of it radiates from the cell into the surroundings. The cell is insulated. It is too hot to touch according to witnesses. The insulation means it takes longer to get hot on the outside; the difference between the inside and the outside is greater; and more heat transfers to the water. But there is still plenty being radiated out. The recovery rate for the water flowing through will be maybe 50% to 75%. In other words, only 400 to 600 W reaches the water. I don't believe it. Then the insulation would be radiating 200W to 400W. Not plausible. That isn't much with a large object that is too hot to touch. And if you're claiming 50 - 75 % for any power, then at 5 kW, about 2.5 kW would have to radiate from the insulation. Are you claiming that? Dunno. Recovery rates change with temperature, flow rates and other conditions. Actually, they usually get worse. Probably this is producing ~4 kW and that makes the surface too hot to touch. If, as you believe, it is only producing 800 W then the insulation isn't very good, is it? All I know is that people have reported it is too hot to touch, as is the hose coming out of it. No. It doesn't. Whatever the fluid is, and regardless of the shape, it's gonna flow through. It does it as a liquid, and it does it as a steam-liquid mixture. There's a pump forcing it through. If the water was overflowing out of the top and down the hose, cold water would be coming in to replace the boiling water and the temperature would drop below boiling, as I said. Probably down to around 95 deg C. It is very difficult to maintain a flow calorimeter outlet temperature of exactly 101 deg C unless the water is boiling, leaving as vapor, and only the vapor touches the temperature sensor. Rossi could tell it is overflowing by watching the temperature. When it falls below 100 deg C, he increases anomalous heat. If not enough water comes in and it dries up, the temperature would rise above 101 deg C, and he reduces it. He can control the strength of the anomalous heat. I do not know how he does that. Apparently he has enough control to keep the bottom portion filled with boiling water but not overflowing. I thank Jouni Valkonen for that observation, by the way. There will be self-powered ones with electric power generation within a year or so. And will that be used to power the CF car you predicted would be built before the year 2000? The only reason we did not have cold fusion powered cars by 2000 was because of academic politics and irrational opposition by people like you. I believe Mallove predicted that, not me. I predicted that academic politics would destroy the field, and end all of the research. - Jed
