Re: [Vo]:Steam Sparge
On Sep 26, 2011, at 11:04 AM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: It is nice to see our views so closely aligned. They are indeed. I think running the steam and water through a condensing heat exchanger works very well, provided *all* the flow and temperature variables are recorded very frequently - more frequently than a bucket test would allow . . . For practical purposes, I do not think it is a good idea to generate steam and water inside the machine. This erodes the pipes and pumps. Defkalion's method is much better. They use fluid that boils at a high temperatures and they leave it in liquid state. You can generate steam in the secondary loop. Yes, much better and more stable. When evaluating the device, I do not see any reason to measure the temperatures in the primary loop. This depends highly on the feedback mechanism for the hot water, if there is one, and whether the ability to control input water temperature is needed. Rossi stated the hot water would be fed back to E-cat. The most stable way to do that would be to create an insulated reservoir, at room air pressure, and pump water from the reservoir as is done now. The reservoir would present a good probability of unmeasured heat loss. The varying temperature of the input water creates a varying load on the condenser. The most stable configuration would be to pump cold water into the E-cat and dump the primary loop water, measuring its heat content first though. A system of this kind, with feedback, would then have two inputs to measure: (M1) cold water into the E-cat and (M2) cold water into the heat exchanger secondary. The system would have two heat outputs: (M3) hot water out of the heat exchanger secondary, and (M4) hot water out of the heat exchanger primary, which could and probably would involve a substantial amount of power. The measuring stations Mi have to measure flow and temperature. Summing the two outputs would be essential if a clean curve of (nearly instantaneous) power out vs power in were desirable, or timely energy in vs energy out curves. Measuring all four heat flows provides a means to decouple the primary circuit output temperature (and pressure) from the E-cat input temperature (and pressure). Water storage in the E-cat itself need not be taken into account until the end of the run when cold water is used to run out the numbers for final total energy calculations. It is feasible to build a calorimeter using only three measuring stations, and I think this approach might be especially good for the 1 MW E-cat. This is accomplished by eliminating the heat exchanger and simply merging the primary output and secondary input flows into a continuous sparging condenser, with a single output into one measuring station M3.The nice thing about this approach for the 1 MW E-cat is all that is needed for cooling is a 5 gal/min pump, some big hose, and a lake or river. Water to the input measuring station M1 could be pumped from any desired source. The output water could be dumped, or air cooled and recycled to the secondary input, with some recycled to the E-cat input reservoir. If a special coolant is used, or high pressure primary operation is desired, the four measuring station approach seems to me necessary to maintain control of the E-cat, and to avoid heat loss errors for the system as a whole. As I said before, in a test to prove the thing is producing excess energy, I see no reason to generate steam at all. Why not just use hot water even if it is inefficient? Harry Veeder said that Rossi is devoting all of this time to steam tests. Perhaps he is but he can certainly spare a day to have someone do water tests. Since people will be in the lab taking up space and interfering with his work anyway, they might as well do a flowing water test. Since the cooling solution is isolated from the catalyst it would be possible to use car coolant solution (antifreeze). There are pressure and temperature instability problems with recycling the fluid, but not nearly as much as with boiling water. The principle expense I would expect is in accurate digitally interfaced flow meters. It is always good to have an independent method to confirm results and to provide confidence in control run calibrations. Yes, this is essential. [snip] I think the temperature in such a bucket falls, or at least can fall, significantly, considering a delta T measurement is being made. The more accurate the delta T the longer the test and the bigger the delta T, but then the more error due to heat loss unless the bucket is insulated. Also, there is not just one calorimetry constant at higher temperatures. There is a calorimetry function by temperature (vs constant) due to nonlinear losses due to evaporation and radiation
Re: [Vo]:Steam Sparge
Some corrected thoughts. For the secondary circuit only one flow meter is needed, and two thermometers. For the primary circuit, input and output flow meters should be used, and two thermometers. It is important not to assume the pump outputs at a constant rate against all pressures into the E- cat. Knowing both primary circuit flow rates, which can be very different, due to the water storage capacity of the E-cat, provides some information about what is happening inside the E-cat, and decouples the output flow and pressure from the input. If only one primary side flow meter is available it should be on the output of the condenser, where heat measurement is most critical. A comparatively inexpensive manually read accumulating water meter could be used on the input side, for verification purposes and to manually diagnose problems like low input flow. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Steam Sparge
I wrote: The nice thing about this approach for the 1 MW E-cat is all that is needed for cooling is a 5 gal/min pump, some big hose, and a lake or river. That should have said: The nice thing about this approach for the 1 MW E-cat is all that is needed for cooling is a 200 gal/min pump, some big hose, and a lake or river. I don't now recall where I posted the calculation for that. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Inexpensive steam/water calorimeter
, and secondary heat exchanger coil, a stirrer, and two precision volume pumps, one primary, the other secondary. Temperatures would be monitored frequently for the secondary in and out flows, perhaps less frequently for room temperature, barrel temperature, and test device input temperature. As a second level, ordinary integrating water meters could be added, for flow confirmation, on both the primary and secondary circuits. Ideally, precision digital flow meters should be used for both the primary and secondary circuit input flows. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Inexpensive steam/water calorimeter
On Sep 27, 2011, at 10:49 AM, Peter Gluck wrote: Dear Jouni, I have described this method long ago, for individual e-Cats A key part of this idea is the reliability obtained by the averaging performed by the large thermal mass of the water container. I am suggesting a hybrid design, a hybrid flow and partial isoperibolic method. It would of course be feasible to employ a mixer and extra thermometer just prior to the water container which does the averaging, but that would also require an extra pump, and flow meter. I should also note this idea was initially largely for my own use. I have a potential use for calorimetry in the multiple kW range. I optimize the cheap variable when designing for my own purposes, with some constraints regarding reliability and accuracy. This is because I am so tight with money the little birdies say cheep cheep when they fly over me. 8^) I have a 5 thermometer system that should work OK with this approach even with manual recording and spreadsheet analysis. In winter I have the advantage of practically unlimited cooling capacity here in Alaska. Unfortunately my two peristaltic pumps are too small for this power range. I can readily afford the barrel, blue board insulation, copper pipe and hose, fittings etc. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Inexpensive steam/water calorimeter
On Sep 27, 2011, at 9:35 AM, Jouni Valkonen wrote: 2011/9/27 Peter Gluck peter.gl...@gmail.com: The simplest solution is to use a Steam Water mixing valve,in which the heated mixture coming out from the demo is mixed with a constant flow of cold water, you can know the enthalpy performance in any moment. Indeed, continuous experiments easiest way is to use enthalpy sensors, that gives as total enthalpy for any given moment. Even more simple is to measure the steam pressure inside E-Cat, because it gives directly the total enthalpy, but of course we need to first calibrate this kind of enthalpy sensors. –Jouni You have again not specified the precise method you would use. It would appear you have a case of missing variables. The principle missing variable is mass flow, m dot, which is best to isolate and measure directly. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Inexpensive steam/water calorimeter
On Sep 27, 2011, at 9:27 PM, Peter Gluck wrote: Dear Horace, The missing variable is cooling water flow- to be established by Rossi- water that carries the excess heat generated by the 52 (?) Fat Cats and is partially transformed in steam- F1. To achieve accuracy in delta T measuring the condensing water flow rate should be adjusted to the flow rate of the steam. If the flow is too high the delta T is small and even very small errors in measuring T translate into very large errors in delta T. If the device enthalpy varies rapidly then it is much easier to adjust the cooling water flow to a longer term moving average than to instantaneous measurements. The flow of mixing water- condensing the steam is say, 5-10 times greater than F1 see please the formula given in my paper. Like most people I don't generally go looking for a URL if it is not provided in a reference. What matters is not the mixer cooling water flow rate but its combined temperature and flow rate. The flow has to be matched to the steam thermal power, mass flow, and cooling water temperature in order to achieve a significant delta T. This problem does not exist when the steam is condensed into a very large thermal mass of water - provided the large mass is kept in a useful temperature range, and the thermal power from the secondary cooling circuit is matched to the device thermal power. If the thermal mass is large enough such matching can take place gradually and even manually, provided it is properly recorded. No peristaltic but other types of positive displacement pumps to be used, I said, Unfortunately my two peristaltic pumps are too small for this power range. This does not imply that I would even consider trying to buy large peristaltic pumps. Perhaps we have a language barrier. Also, the flow rate for the cooling water should ideally be adjustable to the thermal power output of the device if that is variable and unpredictable. An adjustable flow rate pump, or a selection of pumps, would thus be useful for driving the secondary cooling circuit. e,g. gear pumps- for which the flow is not influenced by counterpressure. The flow rate of gear pumps is influenced by a pumping into a large pressure head, both due to rpm loss (slip) for AC induction motors under load, and due to rotor seal leakage under high pressure. In the case of the new Rossi device, it looked like perhaps the water flow was entirely blocked towards the end of the test. This would create as large a pump pressure head as required to terminate flow. The evidence for flow blockage was the high pressure the device was under at the end. This system measures the enthalpy in any moment, Including the start up period and possibly the heat after death. The mass flow measurement depends on measuring the mixer exit mass flow. This flow likely contains bubbles, is not well thermally mixed, and has fast dynamics requiring fast sampling times. Some degree of smoothing increases reliability of the numbers and reduces the required sampling rate. A large degree of smoothing provides a first principle check on the flow calorimetry numbers. Of course, in the case of Rossi's device any even low precision mass flow calorimetry is an improvement. In the case of my own work I would like some degree of consistency checking. A hybrid method provides this consistency check. The formula for efficiency is actually O/3I because electrical energy is at least 3 times more valuable or expensive than thermal energy That is not a formula for efficiency but relative value. Peter On Wed, Sep 28, 2011 at 7:38 AM, Horace Heffner hheff...@mtaonline.net wrote: On Sep 27, 2011, at 9:35 AM, Jouni Valkonen wrote: 2011/9/27 Peter Gluck peter.gl...@gmail.com: The simplest solution is to use a Steam Water mixing valve,in which the heated mixture coming out from the demo is mixed with a constant flow of cold water, you can know the enthalpy performance in any moment. Indeed, continuous experiments easiest way is to use enthalpy sensors, that gives as total enthalpy for any given moment. Even more simple is to measure the steam pressure inside E-Cat, because it gives directly the total enthalpy, but of course we need to first calibrate this kind of enthalpy sensors. –Jouni You have again not specified the precise method you would use. It would appear you have a case of missing variables. The principle missing variable is mass flow, m dot, which is best to isolate and measure directly. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Inexpensive steam/water calorimeter
When I say precise method I mean the inclusion of the specific data to be obtained, where it is obtained, and the formulas applied. You wrote: Indeed, continuous experiments easiest way is to use enthalpy sensors, that gives as total enthalpy for any given moment. Even more simple is to measure the steam pressure inside E-Cat, because it gives directly the total enthalpy, but of course we need to first calibrate this kind of enthalpy sensors. There is no such thing as an actual enthalpy sensor. Only specific enthalpy is sensed. Only incremental enthalpies (delta H) of a system can be measured. To obtain energy of a mass of steam, relative to that mass at some temperature, you need to know the mass of the steam. The mass of an army tank differs from the mass of a small car. Measuring only pressure, or specific enthalpy, provides an insufficient amount of information. To obtain thermal power you need to know the mass flow. The water overflow is a significant part of the flow by volume, more than 2% in some cases by volume. This means the specific enthalpy of the steam is almost insignificant in those cases. If x is the liquid portion by volume, then x/((x+(1-x)*0.0006)) is the portion by mass. This gives the following table which I posted here last January: Liquid LiquidGas PortionPortion Portion by Volume by Mass by Mass - --- --- 0.000 0. 100.00 0.001 0.6252 0.3747 0.002 0.7695 0.2304 0.003 0.8337 0.1662 0.004 0.8700 0.1299 0.005 0.8933 0.1066 0.006 0.9095 0.0904 0.007 0.9215 0.0784 0.008 0.9307 0.0692 0.009 0.9380 0.0619 0.010 0.9439 0.0560 0.011 0.9488 0.0511 0.012 0.9529 0.0470 0.013 0.9564 0.0435 0.014 0.9594 0.0405 I consider the big deal about the definition of steam quality to be a red herring, a diversion from the important issues of measurement of the thermal power carried by the mass flow of a water steam mixture. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ On Sep 27, 2011, at 9:16 PM, Jouni Valkonen wrote: First I would add to my previous message, that I think that Peter's method is more accurate than measuring pressure. That is because in order to find out correlation between pressure and enthalpy we need to do very careful calibration. In short run high accuracy may be difficult to archieve, but if experiment lasts for example 10 years continuously, then of course calibrating pressure sensor for enthalpy calculations will give great pay off. Horace wrote: « You have again not specified the precise method you would use. It would appear you have a case of missing variables. The principle missing variable is mass flow, m dot, which is best to isolate and measure directly. » Actually I have defined but it is so simple that you have probably missed it. First of course, we need to know that system is at equilibrium, i.e. water massflow in and massflow out are both matching. If water inflow rate varies a lot then calculations and calibrations are difficult, if system is overflowing. That means that for sure massflow must be known and it must be measured in calibration. But if system is a kettle boiler that does not overflow, then calibration is very easy. In industrial water boilers this is the most reasonable situation because this ensures high steam quality because we can easily superheat steam to remove that 1-2% natural wettness of steam. This reduces the corrosion. Superheating can also be considered in calculations so this does not reduce the accuracy of method. Pressure can be measured either directly with pressure sensor (easiest and most reliable and it is always available in pressure boilers.) or in kettle boilers boiling water temperature can be measured or last method is to measure steam temperature (this works only if steam is not superheated and is thus wet. I.e. steam quality must be measured, therefore this method is not universal). —Jouni On Sep 28, 2011 7:41 AM, Horace Heffner hheff...@mtaonline.net wrote: On Sep 27, 2011, at 9:35 AM, Jouni Valkonen wrote: 2011/9/27 Peter Gluck peter.gl...@gmail.com: The simplest solution is to use a Steam Water mixing valve,in which the heated mixture coming out from the demo is mixed with a constant flow of cold water, you can know the enthalpy performance in any moment. Indeed, continuous experiments easiest way is to use enthalpy sensors, that gives as total enthalpy for any given moment. Even more simple is to measure the steam pressure inside E-Cat, because it gives directly the total enthalpy, but of course we need to first calibrate this kind of enthalpy sensors. –Jouni You have again not specified the precise method you would use. It would appear you have a case
Re: [Vo]:Re: Regarding Rossi and NASA (+ some Piantelli news)
On Sep 28, 2011, at 12:20 PM, OrionWorks - Steven V Johnson wrote: Rizzi sez: ... I think that the end of the hoax is approaching. I doubt we are witnessing a hoax, though it's possible I am in error. Another thought came to mind in regards to the megawatt reactor design: Why for their first generation of products are they building a 1 MW module? Many have stated many times that a smaller less complicated configuration that generates a more modest amount of heat of say 10 - 50 kilowatts of energy would be more than sufficient to prove their point. One theory as to why the 1 MW reactors is being designed for prime time is to prove to prospective investors that the technology can be scaled up immediately. That may be true, but perhaps a more subtle point might be that by assembling a bunch of eCat cores under one hood the engineers increase their chances that at least a decent number of the individual reactors will work. Maybe there are far more individual eCat cores than what ought to be necessary in order to generate 1 MW of heat under the hood. Maye the engineers have discovered the fact that statistically speaking only about 50% - 75% of the individually assembled reactor cores work. I wonder if they have installed enough additional reactor cores to more-or-less guarantee that the entire module will, statistically speaking, generate at a minimum 1 Megawatts of heat. Just a thought... and I suspect it has already been raised by others here. Regards, Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks Looking at the other side of the coin, the probability of catastrophic failure, suppose there is a 0.1% chance per hour one of the E-cats can blow up spreading steam throughout the container. There is thus a 0.999 probability of success, i.e. no explosion for one E-cat, operating for one hour.The probability that all 52 E- cats perform successfully for a 24 hour test period is then 0.999^ (52*24) = .287. That means there is a 71.3% chance of an explosion during a 24 hour test. The fact it is more difficult to manually monitor 53 E-cats than a single E-cat also means the probability a single E-cat of the 53 blows up in a given hour would be higher than it would be for that E- cat operated singly. It is not even clear facilities to monitor individual critical E-cat conditions, like internal pressure or flow, are present in the 1 MW E-cat. If no individual monitoring is feasible then the probability of individual failure in a given hour should be much larger than when independently operated. Then there is the feasibility of the 1 MW unit producing over a MW just from the huge thermal mass it has, even if all nuclear reactions are shut down. A significant back pressure due to the seam vent pipe being too small could reduce input water flow resulting in suddenly increased boil off of the water in the E-cats resulting in a catastrophic feedback loop and multiple E-cat explosion. The individual probabilities of failure can be made larger in a combined configuration due to additional shared parts. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: Regarding Rossi and NASA (+ some Piantelli news)
On Sep 29, 2011, at 4:02 AM, Man on Bridges wrote: Hi, On 29-9-2011 8:27, Horace Heffner wrote: Looking at the other side of the coin, the probability of catastrophic failure, suppose there is a 0.1% chance per hour one of the E-cats can blow up spreading steam throughout the container. There is thus a 0.999 probability of success, i.e. no explosion for one E-cat, operating for one hour.The probability that all 52 E-cats perform successfully for a 24 hour test period is then 0.999^(52*24) = .287. That means there is a 71.3% chance of an explosion during a 24 hour test. Me thinks you are wrong. Your statistical probability calculation is based upon the fact that the chance of a single Ecat exploding is influenced by it's behaviour earlier, This is false. The probability in each time increment is assumed to be independent. For there to be success there must be no failures for any time increment. If there are T time increments, and the probability of failure in any time increment is p, the probability of success q=1-p in each time increment is independent of the other time increments, and the probability of success in all time increments is q^T (only possible if what happens in each time increment is independent event), and the probability of any failure having occurred is thus 1-(q^T). which of course is not true. Statistically each Ecat has it's own independent chance of explosion at any given moment which does not change over time. The instantaneous probability of failure is zero. Zero time results in zero probability because lim t-0 q^t = 1 for for all 0=q=1 and positive t. Therefore lim t-0 1-(q^t) = 0. Note that I provided an assumption of 0.001 percent probability of failure *per hour*. With your probability of 0,1% chance per hour this would result for the whole of 52 Ecats then in a chance of explosion at any given moment of 1 - (0.999^52) = .05 or 5%. No. The probability of at least one E-cat failure in the 52 E-cat system, based on the assumption of 0.001 probability of failure of an individual E-cat in an hour is 1-(0.999)^52 = 0.506958 = 5%. Your number 5% is right, but your interpretation of it representing an instantaneous moment is wrong. Looking even a bit more closer again this would mean that if the chance of explosion is 0.1% per hour then the chance of explosion is 2,77e-7 per second at any given moment for a single Ecat, which would result for 52 Ecats into 1-((2,77e-7)^52) = 0,134 or 0,00144% at any time. The phrase at any time makes the above statement nonsensical. An hour represents 3600 seconds, which are 3600 independent events of 1 second duration. Let a be the probability of failure in 1 second, and b=(1-a) be the probability of success in 1 second. We have the given probability p of failure for 3600 seconds being 0.001, and the probability of success of one E-cat for one hour being q = 0.999. The probability of success (no failures) for the 3600 1 second independent time increments is q = 0.999 = b^3600 b = q^(1/3600) = 0.999^(1/3600) a = 1 - 0.999^(1/3600) = 2.779x10^-7 Note that a is the probability of failure in one second, not at any time. This is totally consistent with the probability of failure in one E-cat in one hour being 5%. In other words, going backwards: p = 1-(1-a)^3600 = 1-(1-2.779x10^-7)^3600 = 1-0.999 = 0.001 My calculations are therefore self consistent. The time intervals are all treated as independent events. Your interpretation of moment is perhaps a conceptual problem. Kind regards, MoB Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: Regarding Rossi and NASA (+ some Piantelli news)
The sentence below: This is totally consistent with the probability of failure in one E-cat in one hour being 5%. should read: This is totally consistent with the probability of failure of at least one E- cat (of 52) in one hour being 5%. On Sep 29, 2011, at 11:34 AM, Horace Heffner wrote: On Sep 29, 2011, at 4:02 AM, Man on Bridges wrote: Hi, On 29-9-2011 8:27, Horace Heffner wrote: Looking at the other side of the coin, the probability of catastrophic failure, suppose there is a 0.1% chance per hour one of the E-cats can blow up spreading steam throughout the container. There is thus a 0.999 probability of success, i.e. no explosion for one E-cat, operating for one hour.The probability that all 52 E-cats perform successfully for a 24 hour test period is then 0.999^(52*24) = .287. That means there is a 71.3% chance of an explosion during a 24 hour test. Me thinks you are wrong. Your statistical probability calculation is based upon the fact that the chance of a single Ecat exploding is influenced by it's behaviour earlier, This is false. The probability in each time increment is assumed to be independent. For there to be success there must be no failures for any time increment. If there are T time increments, and the probability of failure in any time increment is p, the probability of success q=1-p in each time increment is independent of the other time increments, and the probability of success in all time increments is q^T (only possible if what happens in each time increment is independent event), and the probability of any failure having occurred is thus 1-(q^T). which of course is not true. Statistically each Ecat has it's own independent chance of explosion at any given moment which does not change over time. The instantaneous probability of failure is zero. Zero time results in zero probability because lim t-0 q^t = 1 for for all 0=q=1 and positive t. Therefore lim t-0 1-(q^t) = 0. Note that I provided an assumption of 0.001 percent probability of failure *per hour*. With your probability of 0,1% chance per hour this would result for the whole of 52 Ecats then in a chance of explosion at any given moment of 1 - (0.999^52) = .05 or 5%. No. The probability of at least one E-cat failure in the 52 E-cat system, based on the assumption of 0.001 probability of failure of an individual E-cat in an hour is 1-(0.999)^52 = 0.506958 = 5%. Your number 5% is right, but your interpretation of it representing an instantaneous moment is wrong. Looking even a bit more closer again this would mean that if the chance of explosion is 0.1% per hour then the chance of explosion is 2,77e-7 per second at any given moment for a single Ecat, which would result for 52 Ecats into 1-((2,77e-7)^52) = 0,134 or 0,00144% at any time. The phrase at any time makes the above statement nonsensical. An hour represents 3600 seconds, which are 3600 independent events of 1 second duration. Let a be the probability of failure in 1 second, and b=(1-a) be the probability of success in 1 second. We have the given probability p of failure for 3600 seconds being 0.001, and the probability of success of one E-cat for one hour being q = 0.999. The probability of success (no failures) for the 3600 1 second independent time increments is q = 0.999 = b^3600 b = q^(1/3600) = 0.999^(1/3600) a = 1 - 0.999^(1/3600) = 2.779x10^-7 Note that a is the probability of failure in one second, not at any time. This is totally consistent with the probability of failure in one E-cat in one hour being 5%. In other words, going backwards: p = 1-(1-a)^3600 = 1-(1-2.779x10^-7)^3600 = 1-0.999 = 0.001 My calculations are therefore self consistent. The time intervals are all treated as independent events. Your interpretation of moment is perhaps a conceptual problem. Kind regards, MoB Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: Regarding Rossi and NASA (+ some Piantelli news)
If you look at my text you will see I wrote catastrophic failure not just failure. This means an E-cat blows up spreading steam throughout the container, injuring anyone present, and preventing access to the container, causing the test to fail. I think I was clear on this point. I did not refer to anything bout an E-cat performance dropping. The other side of the coin to increased probability of some E-cat working when multiple devices run together is the increased probability of catastrophic failure. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ On Sep 29, 2011, at 12:26 PM, Axil Axil wrote: The failure of one module of the Rossi 1 MW reactor will not cause the entire 1 MW reactor to fail. Its performance will only degrade gracefully. When the core of the module overheats or melts, the surface of the nickel nanopowder will fail before the nanopowder enclosure will fail since the enclosure will be cooled by low temperature steam or water which would remove heat, effectively cool the enclosure, and support its structural strength. The failure of the nanopowder will cause the individual module to cool and be ineffective at generating thermal power. It would be analogous to a failure of one pixel of your computer screen; if one such pixel grows dark, your screen will not fail but its performance would degrade. You would still be able to use the screen, just the picture would not be as sharp. So too with the Rossi reactor; it would still generate heat, but not so much as before. Its capacity would be reduced until its performance would eventually degrade below a certain predefined lower threshold. When this low bound threshold is reached, the entire reactor is considered to have failed. On Thu, Sep 29, 2011 at 3:34 PM, Horace Heffner hheff...@mtaonline.net wrote: On Sep 29, 2011, at 4:02 AM, Man on Bridges wrote: Hi, On 29-9-2011 8:27, Horace Heffner wrote: Looking at the other side of the coin, the probability of catastrophic failure, suppose there is a 0.1% chance per hour one of the E-cats can blow up spreading steam throughout the container. There is thus a 0.999 probability of success, i.e. no explosion for one E-cat, operating for one hour.The probability that all 52 E-cats perform successfully for a 24 hour test period is then 0.999^(52*24) = .287. That means there is a 71.3% chance of an explosion during a 24 hour test. Me thinks you are wrong. Your statistical probability calculation is based upon the fact that the chance of a single Ecat exploding is influenced by it's behaviour earlier, This is false. The probability in each time increment is assumed to be independent. For there to be success there must be no failures for any time increment. If there are T time increments, and the probability of failure in any time increment is p, the probability of success q=1-p in each time increment is independent of the other time increments, and the probability of success in all time increments is q^T (only possible if what happens in each time increment is independent event), and the probability of any failure having occurred is thus 1-(q^T). which of course is not true. Statistically each Ecat has it's own independent chance of explosion at any given moment which does not change over time. The instantaneous probability of failure is zero. Zero time results in zero probability because lim t-0 q^t = 1 for for all 0=q=1 and positive t. Therefore lim t-0 1-(q^t) = 0. Note that I provided an assumption of 0.001 percent probability of failure *per hour*. With your probability of 0,1% chance per hour this would result for the whole of 52 Ecats then in a chance of explosion at any given moment of 1 - (0.999^52) = .05 or 5%. No. The probability of at least one E-cat failure in the 52 E-cat system, based on the assumption of 0.001 probability of failure of an individual E-cat in an hour is 1-(0.999)^52 = 0.506958 = 5%. Your number 5% is right, but your interpretation of it representing an instantaneous moment is wrong. Looking even a bit more closer again this would mean that if the chance of explosion is 0.1% per hour then the chance of explosion is 2,77e-7 per second at any given moment for a single Ecat, which would result for 52 Ecats into 1-((2,77e-7)^52) = 0,134 or 0,00144% at any time. The phrase at any time makes the above statement nonsensical. An hour represents 3600 seconds, which are 3600 independent events of 1 second duration. Let a be the probability of failure in 1 second, and b=(1-a) be the probability of success in 1 second. We have the given probability p of failure for 3600 seconds being 0.001, and the probability of success of one E-cat for one hour being q = 0.999. The probability of success (no failures) for the 3600 1 second independent time increments is q = 0.999 = b
Re: Aw: [Vo]:H2 and O2 bubbles .15 micrometer burn, damaging electrodes in AC electrolysis -- could complicate cold fusion devices: Rich Murray 2011.09.28
On Sep 28, 2011, at 11:03 PM, peter.heck...@arcor.de wrote: - Original Nachricht Von: Rich Murray rmfor...@gmail.com An: vortex-L@eskimo.com Datum: 29.09.2011 03:04 Betreff: [Vo]:H2 and O2 bubbles .15 micrometer burn, damaging electrodes in AC electrolysis -- could complicate cold fusion devices: Rich Murray 2011.09.28 H2 and O2 bubbles .15 micrometer burn, damaging electrodes in AC electrolysis -- could complicate cold fusion devices: Rich Murray 2011.09.28 It would be interesting to know the frequencies and current densities used. I am still looking for a simple experiment that I could do myself at home to prove LENR effects ;-) Now I had this idea: Use a NiMH battery. The positive electrode consists out of Nickel +Nickeloxide nanoparticles, so far I know. The electrolyte is KOH. The negative electrode is an unkown alloy that is optimized to form metalhydrides, it has high hydrogen adsorption capacity. Charge a NiMH battery reverse, of course with very low current, otherwise it would explode. For the current use AC + a DC bias. Then bubbles should form at the positive Nickel electrode, that contain HH + O, but if the charging AC has a negative bias, the bubbles should contain more hydrogen than necessary to burn. This should happen: A microbubble forms inside the Nickel Nanomaterial. H2+O combustion ignites. The Bubble expands and because the combustion product is water, the bubble should then collapse rapidly. Because we have a surplus of Hydrogen, the Hydrogen + the Nickel Nanomaterial should now be under high pressure inside the bubble. Because electrolysis forms atomar hydrogen, I hope that Nickel-Hydrogen LENR reacions happen inside the NiMH battery. ;-) Peter NiMH batteries have been tested for excess heat both in forward current and reverse current mode, with null results. To my knowledge no testing for transmutation or occasional high energy radiation has been made. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: Regarding Rossi and NASA (+ some Piantelli news)
On Sep 29, 2011, at 4:37 AM, OrionWorks - Steven Vincent Johnson wrote: From MoB: ... Looking even a bit more closer again this would mean that if the chance of explosion is 0.1% per hour then the chance of explosion is 2,77e-7 per second at any given moment for a single Ecat, which would result for 52 Ecats into 1-((2,77e-7)^52) = 0,134 or 0,00144% at any time. Ah! Understanding the mathematics of Probability can occasionally be a useful talent to possess! ;-) Yes. It would be nice if MoB had the above correct though, i.e. understood what the numbers mean. Thanks MoB Regards, Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Some personal thoughts on NET Krivit
On Sep 29, 2011, at 12:53 PM, Jouni Valkonen wrote: And Krivit started vicious ad hominem attacking against Rossi and Levi. By for what reason? Here the definition of ad hominem seems distorted. Criticizing a paper or posting or experimental approach is not ad hominem. Calling someone a derogatory name, like fool, snake or clown is. Where is an example of ad hominem attack by Krivit? It may well exist, but I don't recall seeing such. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Defkalion GT forum appears to be open again
On Sep 29, 2011, at 7:56 PM, Rich Murray wrote: Lots of situations on V-L, I don't have or else don't share an opinion: 1 not worth my time 2 too much effort to gather and comprehend enough fractured details 3 can't bother to write up an adequate discussion 4 the issue is polarized, so each side listens to their own chorus 5 maybe a newbie would benefit 6 maybe can encourage other skeptics 7 maybe can get someone else to provide better analyses 8 like to show paying some cursory attention 9 welcome being wrong, especially when there's a big benefit for all of us -- albeit an hugely impossible to assess possibly catastrophic weapons hazard 10 prefer to wave flag clearly, if wave at all, while being brief 11 would hate to see Rossi harm himself 12 do enough appraisals to become convinced -- yes, Rossi is confidently caught up in delusion 13 as is Piantelli 14 seems like Krivit is saying NASA also saw the darkness at the end of the stairs 15 I really enjoy Horace Heffner these weeks, and Joshua Cude in recent months Thanks. I thought maybe it was all for nothing. 16 believe that Krivit, like me, basically wants to find for himself what the case actually is 17 many big players have goofed -- Defkalion, the USA firm 18 it's valuable that a fairly tolerant discussion has evolved on V-L and H-Ni_Fusion and Krivit's site 19 basic focus is seeing how the entire shebang, universes on all levels, arises unaccountably within awareness-being, without any actual casuality, space, time, or actuality -- always changing, never satisfying, never establishing a self within any event cluster neither Rich nor not-Rich... Schrödinger Rich? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Some personal thoughts on NET Krivit
there are no general rules in argumentation. . No general rules??? How about specific rules? Gee, what are all those latin phrases about?? . But it is always depended on the context where arguments are presented. . It? What are you talking about. . –Jouni Ps. Still waiting an apology. . . . P.S. Still waiting for (a) specific details on what you find so offensive about my remarks and (b) specific occurrences of ad hominem attack by Krivit ... 2011/9/30 Horace Heffner hheff...@mtaonline.net: These remarks provide an excellent pedagogical example! Your argument below is an ad hominem attack. The statements:you that you are not able for normal social interaction and you do not have ability to understand sarcasm or hostile intentions, if they are hidden behind formally correct language. are both attacks on the person, and not the person's statement. I am certainly capable of making such attacks, but I usually avoid them. At least I know when I am using ad hominem and when I am not. Ad hominem is a fallacious argument based an irrelevant attack on an opponents ability to make an argument vs the opponents argument itself. On Sep 29, 2011, at 4:48 PM, Jouni Valkonen wrote: It is understandable, for you that you are not able for normal social interaction, therefore you do not have ability to understand sarcasm or hostile intentions, if they are hidden behind formally correct language. See, no derogatory names used, but the content was EXTREMELY insulting. At least it was meant as such. Insulting comments and ad hominem are two different things. Therefore your theory about ad hominem is flawed. This is a fallacious argument, based on the false assumption that ad hominem and insulting comments are necessarily the same thing. You really does not use derogatory names, such as senile, blind, idiot in order to insult persons. I did not imply using such names were *necessary*, only that the use of such names attacking the person instead of his arguments is *sufficient* for an ad hominem. Such an attack is not a logical argument. Try to understand that. It is completely irrelevant what words you are using, but only thing that matters is how people perceive your writings. You do not seem to understand the meaning of ad hominem. Do you find this remark insulting? It is not an attack on you in order to discredit your argument. If it said you are too stupid or too ignorant to discuss logic then that would be ad hominem. Saying your remarks indicate a lack of understanding of a definition is not an ad hominem attack. It is a relevant statement based on the content of your argument. If you did not meant to be insulting, then it is your fault if someone feels your writing as offending. Some people take any disagreement as insulting. When the degree of disagreement is extreme the insult is then extreme. Most often the most insulting thing is that the one who is writing is just ignorant and too stupid to admit his ignorance. –Jouni Insult, like beauty, is in the mind of the beholder. Ad hominem is another thing altogether. It is a fallacious form of argument. I am still waiting for an example from you of Krivit using an ad hominem attack against Rossi and Levi. If you have none then your statement, ... Krivit started vicious ad hominem attacking against Rossi and Levi, is in error. 2011/9/30 Horace Heffner hheff...@mtaonline.net: On Sep 29, 2011, at 12:53 PM, Jouni Valkonen wrote: And Krivit started vicious ad hominem attacking against Rossi and Levi. By for what reason? Here the definition of ad hominem seems distorted. Criticizing a paper or posting or experimental approach is not ad hominem. Calling someone a derogatory name, like fool, snake or clown is. Where is an example of ad hominem attack by Krivit? It may well exist, but I don't recall seeing such. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Some personal thoughts on NET Krivit
I wrote: My intentions are not obviously not malicious. To what end would that serve? That was a typo. It should have read: My intentions are obviously not malicious. To what end would that serve? Corollary to Murphy's law: The probability of a typo is proportional to its importance. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Some personal thoughts on NET Krivit
These remarks provide an excellent pedagogical example! Your argument below is an ad hominem attack. The statements:you that you are not able for normal social interaction and you do not have ability to understand sarcasm or hostile intentions, if they are hidden behind formally correct language. are both attacks on the person, and not the person's statement. I am certainly capable of making such attacks, but I usually avoid them. At least I know when I am using ad hominem and when I am not. Ad hominem is a fallacious argument based an irrelevant attack on an opponents ability to make an argument vs the opponents argument itself. On Sep 29, 2011, at 4:48 PM, Jouni Valkonen wrote: It is understandable, for you that you are not able for normal social interaction, therefore you do not have ability to understand sarcasm or hostile intentions, if they are hidden behind formally correct language. See, no derogatory names used, but the content was EXTREMELY insulting. At least it was meant as such. Insulting comments and ad hominem are two different things. Therefore your theory about ad hominem is flawed. This is a fallacious argument, based on the false assumption that ad hominem and insulting comments are necessarily the same thing. You really does not use derogatory names, such as senile, blind, idiot in order to insult persons. I did not imply using such names were *necessary*, only that the use of such names attacking the person instead of his arguments is *sufficient* for an ad hominem. Such an attack is not a logical argument. Try to understand that. It is completely irrelevant what words you are using, but only thing that matters is how people perceive your writings. You do not seem to understand the meaning of ad hominem. Do you find this remark insulting? It is not an attack on you in order to discredit your argument. If it said you are too stupid or too ignorant to discuss logic then that would be ad hominem. Saying your remarks indicate a lack of understanding of a definition is not an ad hominem attack. It is a relevant statement based on the content of your argument. If you did not meant to be insulting, then it is your fault if someone feels your writing as offending. Some people take any disagreement as insulting. When the degree of disagreement is extreme the insult is then extreme. Most often the most insulting thing is that the one who is writing is just ignorant and too stupid to admit his ignorance. –Jouni Insult, like beauty, is in the mind of the beholder. Ad hominem is another thing altogether. It is a fallacious form of argument. I am still waiting for an example from you of Krivit using an ad hominem attack against Rossi and Levi. If you have none then your statement, ... Krivit started vicious ad hominem attacking against Rossi and Levi, is in error. 2011/9/30 Horace Heffner hheff...@mtaonline.net: On Sep 29, 2011, at 12:53 PM, Jouni Valkonen wrote: And Krivit started vicious ad hominem attacking against Rossi and Levi. By for what reason? Here the definition of ad hominem seems distorted. Criticizing a paper or posting or experimental approach is not ad hominem. Calling someone a derogatory name, like fool, snake or clown is. Where is an example of ad hominem attack by Krivit? It may well exist, but I don't recall seeing such. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Upcoming October 6th test location revealed
On Sep 30, 2011, at 1:23 AM, Akira Shirakawa wrote: Hello group, It definitely looks like that the next test will be performed in Bologna. Have a look at this scan here, from Rossi's EPO patent application page: http://goo.gl/3MhxO This is the usual location we've seen many times in images and videos over the last months. It's Rossi's lab in Bologna. I found this information on Passerini's blog here, who confirms that the test will last 24 hours. This time he won't report in real time what will happen during the test, but he will post his story about it at a later time and together with NyTeknik and Focus (an Italian science and technology news magazine): http://22passi.blogspot.com/2011/09/fatti-non-parole.html Cheers, S.A. Interesting! Thanks for posting that. So the 1 MW E-cat will be tested in Bologna Oct. 6 with Teknik, and Focus information agency, using a heat exchanger. http://www.focus-fen.net/ That certainly saves shipping costs. That must be a very large heat exchanger! It is excellent for credibility of the calorimetry that a heat exchanger will be used. Hopefully the flows will be measured in a credible fashion. Apparently the primary coolant will be fed back to the E-cat somewhat hot. There is a notable lack of comment on the presence of a big american company. I see no attribution to the published quote. I hope this goes off very successfully. If so this would be a great blessing to the LENR field. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Upcoming October 6th test location revealed
On Sep 30, 2011, at 9:47 AM, Peter Heckert wrote: Am 30.09.2011 19:33, schrieb Horace Heffner: On Sep 30, 2011, at 1:23 AM, Akira Shirakawa wrote: Hello group, It definitely looks like that the next test will be performed in Bologna. Have a look at this scan here, from Rossi's EPO patent application page: http://goo.gl/3MhxO This is the usual location we've seen many times in images and videos over the last months. It's Rossi's lab in Bologna. I found this information on Passerini's blog here, who confirms that the test will last 24 hours. This time he won't report in real time what will happen during the test, but he will post his story about it at a later time and together with NyTeknik and Focus (an Italian science and technology news magazine): http://22passi.blogspot.com/2011/09/fatti-non-parole.html Cheers, S.A. Interesting! Thanks for posting that. So the 1 MW E-cat will be tested in Bologna Oct. 6 with Teknik, and Focus information agency, using a heat exchanger. No they will test one module, taken from the 1MW plant. This is what I understand. Oh, I see the letter to the patent office does indeed say on a module. The google translation of the passi.blogspot article says, Data-based EPO (European Patent Office) has been published ( here ) an invitation to attend the examiner Andrea Rossi, on October 6, to test a 1 MW module plant (!) in the presence several scientists around the world: the test will take place in Bologna and will last 24 hours. I mistakenly assumed a 1 MW module plant (!) to mean a one megawatt module located in a container. The exclamation point is what convinced me. I did not read the letter. Perhaps the explanation point was Passini's. Hmm no, it's the Italian science and technology magazine Focus: http://www.focus.it/ I wonder if other Italian mainstream media that covered this matter during the past months (RAI, Radio24/Il Sole 24 Ore, etc.) will be there too. Thanks for the correction. I was under the mistaken impression the italian Focus was a subsidiary of the Focus information agency (FIA). Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:The faster than light neutrino speed should be determined in a non rotating frame
On Sep 30, 2011, at 11:16 AM, David Jonsson wrote: I made a calculation in an inertial system and found that the CERN- OPERA neutrino speed was by some percent due to the rotation of the Earth around its own axis. Do you agree that the calculation should be made in a non rotating system? By the time CERN sends and OPERA receives the Earth rotation makes OPERA to come a bit closer. How many of you agree or disagree with this? Silvertooth, Bryan G. Wallace, GPS and laser gyroscopes also supports this view. It is not suitable to apply the principle of relativity in a non inertial rotating frame. David David Jonsson, Sweden, phone callto:+46703000370 This hypothesis appears to me to be false. I calculate the motion at latitude of Gran Sasso to be 0.833 m in the 2.435x10^-3 S it takes light to travel the 730 km distance. I estimate the angle to lattitude to CERN to be about 33.6 degrees. This means the path length between CERN and Gran Sasso is elongated by cos(33.5°)*0.833 m = 0.615 m. The anomaly is 18.1 m early arrival. Assuming this was not taken into account by the team (unlikely) then it could only account for a 0.615/18.1 = 3.4% error. I can provide detailed calcs later. I have to go for an MRI and other tests today which may take the rest of the day. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:The faster than light neutrino speed should be determined in a non rotating frame
On Sep 30, 2011, at 11:16 AM, David Jonsson wrote:I made a calculation in an inertial system and found that the CERN-OPERA neutrino speed was bysome percentdue to the rotation of the Earth around its own axis. Do you agree that the calculation should be made in a non rotating system? By the time CERN sends and OPERA receives the Earth rotation makes OPERA to come a bit closer. How many of you agree or disagree with this? Silvertooth, Bryan G. Wallace, GPS and laser gyroscopes also supports this view. It is not suitable to apply the principle of relativity in a non inertial rotating frame.David David Jonsson, Sweden, phone callto:+46703000370 The OPERA experiment neutrino beam is directed from CERN,46°14'N6° 3'E, to Gran Sasso LNGS lab, 42°25'N13°31'E. The geometry of this is shown in Fig.1, in OPERA.jpg, attached.Point C is CERN, the neutrino origin. Point S is San Sasso at the time of neutrino departure. Since San Sasso is east of CERN, the earth rotates away, eastward, from CERN during the time of flight of the neutrino. This makes the distance longer than would be estimated by distance between geodetic coordinates. The neutrino arrives at the new San Sasso location S', which is eastward from S by distance d. Only the neutrinos initially aimed at point S' arrive there. Assume the distance C to S is 130 km stated in the Adam et al. OPERA article. Assume point B to be 130 km from point C on the line from C to S'. The neutrino thus has to travel the additional distance x from B to S' due to the eastward motion of the earth during its time of flight.Let point A be the point due south of CERN and due wet of San Sasso, i.e. at 42°25'N, 6°3' E. The distance C to A s then about 923 km, and A to S 1385 km. The angle of the direction of CERN from due wast as seen from San Sasso is thus roughly ATAN(923/1385) = 42.4°. The earth's radius if 6371 km. San Sasso is located at latitude 42.42°N. Its radius of rotations is thus cos(42.4)*(6371 km) = 4720 km. Its speed of rotation is thus 2*Pi*(4720 km)/(24 hr) = 343 m/s.The speed of CERN due to earth's rotation is2*Pi*cos(46.2°)* (6371 km)/(24 hr) = 321 m/s. The 22 m/s speed difference between CERN and San Sasso is not enough to relativistically affect the measurements, especially given the extreme effort put into clock synchronization and geodetic coordinate location. The relative motion however, is enough. A non-rotating linear motion approximation is sufficient to approximate the expected effect.Light travels 730 km in (730 km)/(3x10^8 m/s) =2.435x10^-3 s. In that time San Sasso moves d = (2.435x10^-3 s) * (343 m/s) = 0.835 m eastward. The distance x added to the travel can thus be approximated as x = cos(42.4°) * d = 0.7386 * (0.853 m) = 0.63 m. The travel time of the neutrinos should be increased by (0.63 m)/(3x10^8 m/s) = 2.1x10^-9 s = 2.1 ns. The neutrinos were observed arriving 60.7 ns early. This extra 0.63 m,2.1 ns,had it not been taken into account, would have made the neutrino arrival time60.7 ns +2.1 ns = 62.8 ns early vs speed of light. Failure to account for earth's rotation thus provides approximately a 2.1/60.7 = 3.46 % error. However, this error is in a direction which makes the anomaly even greater. Best regards,Horace Heffnerhttp://www.mtaonline.net/~hheffner/
Re: [Vo]:The faster than light neutrino speed should be determined in a non rotating frame
On Sep 30, 2011, at 11:16 AM, David Jonsson wrote:I made a calculation in an inertial system and found that the CERN-OPERA neutrino speed was bysome percentdue to the rotation of the Earth around its own axis. Do you agree that the calculation should be made in a non rotating system? By the time CERN sends and OPERA receives the Earth rotation makes OPERA to come a bit closer. How many of you agree or disagree with this? Silvertooth, Bryan G. Wallace, GPS and laser gyroscopes also supports this view. It is not suitable to apply the principle of relativity in a non inertial rotating frame.David David Jonsson, Sweden, phone callto:+46703000370 The OPERA experiment neutrino beam is directed from CERN,46°14'N6° 3'E, to Gran Sasso LNGS lab, 42°25'N13°31'E. The geometry of this is shown in Fig.1, in OPERA.jpg, attached.Point C is CERN, the neutrino origin. Point S is San Sasso at the time of neutrino departure. Since San Sasso is east of CERN, the earth rotates away, eastward, from CERN during the time of flight of the neutrino. This makes the distance longer than would be estimated by distance between geodetic coordinates. The neutrino arrives at the new San Sasso location S', which is eastward from S by distance d. Only the neutrinos initially aimed at point S' arrive there. Assume the distance C to S is 730 km stated in the Adam et al. OPERA article. Assume point B to be 730 km from point C on the line from C to S'. The neutrino thus has to travel the additional distance x from B to S' due to the eastward motion of the earth during its time of flight.Let point A be the point due south of CERN and due wet of San Sasso, i.e. at 42°25'N, 6°3' E. The distance C to A s then about 404 km, and A to S 608 km. The angle of the direction of CERN from due wast as seen from San Sasso is thus roughly ATAN(404/608) = 33.6°. The earth's radius if 6371 km. San Sasso is located at latitude 42.42°N. Its radius of rotations is thus cos(42.4)*(6371 km) = 4720 km. Its speed of rotation is thus 2*Pi*(4720 km)/(24 hr) = 343 m/s.The speed of CERN due to earth's rotation is2*Pi*cos(46.2°)* (6371 km)/(24 hr) = 321 m/s. The 22 m/s speed difference between CERN and San Sasso is not enough to relativistically affect the measurements, especially given the extreme effort put into clock synchronization and geodetic coordinate location. The relative motion however, is enough. A non-rotating linear motion approximation is sufficient to approximate the expected effect.Light travels 730 km in (730 km)/(3x10^8 m/s) =2.435x10^-3 s. In that time San Sasso moves d = (2.435x10^-3 s) * (343 m/s) = 0.835 m eastward. The distance x added to the travel can thus be approximated as x = cos(33.6°) * d = 0.833 * (0.853 m) = 0.71 m. The travel time of the neutrinos should be increased by (0.71 m)/(3x10^8 m/s) = 2.36x10^-9 s = 2.36 ns. The neutrinos were observed arriving 60.7 ns early. This extra 0.71 m,2.36 ns,had it not been taken into account, would have made the neutrino arrival time60.7 ns +2.4 ns = 63.1 ns early vs speed of light. Failure to account for earth's rotation thus provides approximately a 2.4/60.7 = 4 % error. However, this error is in a direction which makes the anomaly even greater. Best regards,Horace Heffnerhttp://www.mtaonline.net/~hheffner/
Re: [Vo]:The faster than light neutrino speed should be determined in a non rotating frame
Hopefully this one is correct. Sorry for the multiple posts on this. I am surprised and happy to see the archives now save and show jpgs.On Sep 30, 2011, at 11:16 AM, David Jonsson wrote:I made a calculation in an inertial system and found that the CERN-OPERA neutrino speed was bysome percentdue to the rotation of the Earth around its own axis. Do you agree that the calculation should be made in a non rotating system? By the time CERN sends and OPERA receives the Earth rotation makes OPERA to come a bit closer. How many of you agree or disagree with this? Silvertooth, Bryan G. Wallace, GPS and laser gyroscopes also supports this view. It is not suitable to apply the principle of relativity in a non inertial rotating frame.David David Jonsson, Sweden, phone callto:+46703000370 The OPERA experiment neutrino beam is directed from CERN,46°14'N6° 3'E, to Gran Sasso LNGS lab, 42°25'N13°31'E. The geometry of this is shown in Fig.1, in OPERA.jpg, attached.Point C is CERN, the neutrino origin. Point S is San Sasso at the time of neutrino departure. Since San Sasso is east of CERN, the earth rotates away, eastward, from CERN during the time of flight of the neutrino. This makes the distance longer than would be estimated by distance between geodetic coordinates. The neutrino arrives at the new San Sasso location S', which is eastward from S by distance d. Only the neutrinos initially aimed at point S' arrive there. Assume the distance C to S is 730 km stated in the Adam et al. OPERA article. Assume point B to be 730 km from point C on the line from C to S'. The neutrino thus has to travel the additional distance x from B to S' due to the eastward motion of the earth during its time of flight.Let point A be the point due south of CERN and due wet of San Sasso, i.e. at 42°25'N, 6°3' E. The distance C to A s then about 404 km, and A to S 608 km. The angle of the direction of CERN from due wast as seen from San Sasso is thus roughly ATAN(404/608) = 33.6°. The earth's radius if 6371 km. San Sasso is located at latitude 42.42°N. Its radius of rotations is thus cos(42.4)*(6371 km) = 4720 km. Its speed of rotation is thus 2*Pi*(4720 km)/(24 hr) = 343 m/s.The speed of CERN due to earth's rotation is2*Pi*cos(46.2°)* (6371 km)/(24 hr) = 321 m/s. The 22 m/s speed difference between CERN and San Sasso is not enough to relativistically affect the measurements, especially given the extreme effort put into clock synchronization and geodetic coordinate location. The relative motion however, is enough. A non-rotating linear motion approximation is sufficient to approximate the expected effect.Light travels 730 km in (730 km)/(3x10^8 m/s) =2.435x10^-3 s. In that time San Sasso moves d = (2.435x10^-3 s) * (343 m/s) = 0.835 m eastward. The distance x added to the travel can thus be approximated as x = cos(33.6°) * d = 0.833 * (0.853 m) = 0.71 m. The travel time of the neutrinos should be increased by (0.71 m)/(3x10^8 m/s) = 2.36x10^-9 s = 2.36 ns. The neutrinos were observed arriving 60.7 ns early. This extra 0.71 m,2.36 ns,had it not been taken into account, would have made the neutrino arrival time60.7 ns +2.4 ns = 63.1 ns early vs speed of light. Failure to account for earth's rotation thus provides approximately a 2.4/60.7 = 4 % error. However, this error is in a direction which makes the anomaly even greater. Best regards,Horace Heffnerhttp://www.mtaonline.net/~hheffner/
[Vo]:Russian Roulette, Murphy, and Independent Events
When playing dice the probability p of rolling a 1 in a single roll is 1/6, the probability q of not rolling a 1 is q = 1-p = 5/6. The probability of rolling two 1's in two rolls is p^2 = 1/36, because the two rolls are independent events. More interesting is the probability of not rolling any 1's at all. Call this success. Similarly, for two rolls, this probability is q^2, or 5/36, because the rolling events are independent. It does not matter the order in which order the die are rolled or whether they are rolled simultaneously. For n casts of the die, the probability of success is q^n. The probability of failure is thus 1-q^n = 1-(1-p)^n. Solo Russian Roulette can involve spinning a cylinder of a pistol with a single round in it, the player aiming the gun at his own head, and pulling the trigger. Call this an event. In the case of a six shooter, the probability p of the player killing himself, failure, in one event is 1/6. The probability of success in a one event game is then q = 1-p = 5/6. If a cylinder spin occurs in each event, then the probabilities of success or failure in any two events are independent of each other. The probability of success when playing a two event game, given both events are independent, is as in dice, q^2 = 25/36. The probability of success in an n event game is q^n. The probability of failure is thus 1-q^n. If p is the probability of failure in a single event, then 1-(1-p)^n is the probability of failure in an n event game. The only difference between this form of Russian roulette and dice is the events of a game must stop upon the first failure event. The aggregate events can be looked upon as dependent in that sense. The events always occur one at a time and the latter events do not occur once a failure is encountered. However, the same can be true of the dice game. If the dice are rolled one at a time, then once a 1 is encountered it is not necessary to continue rolling the dice because the outcome is then already determined. The formula 1-q^n provides some not common understanding of games of chance, risk taking, and product liability. This understanding comes from the fact that in the limit, as n approaches infinity, for *any* positive q less than 1, q^n rapidly approaches zero. This is expressed as: lim n-inf q^n = 0 for all q=01 In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of no catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. Similarly, if a condom brand has a 1% chance of failure, then 100 uses results in a probability of failure at some time in those uses of 1-.99^100 = 0.634 = 63%. If every critical component of a rocket has a very small chance of failure, say 1/1, but there are 1000 such components, then the probability of system failure is 1-(0.)^1000 = 0.095, or about 10%. If a product, like a vehicle, is used N=50 minutes a day by M=10,000,000 people, and the probability of failure in any given minute is p, then the probability of some failure P in a year is given by: P = 1-(1-p)^N*M*365 = 1-q^1825 Is is easy to see then, that for a product used by many people that, as time goes on, the number of opportunities for failure, n=(years) *N*M*365, becomes very large. No matter how close q is to 1, q^n then approaches zero. If anything can go wrong it will. This is Murphy's law. Similarly, each of the bets of a gambler is an independent event. However, all gamblers have a limited purse, even when credit is available. When a purse runs out then the gambler is done. If a gambler plays against less than favorable odds, no matter how small the margin, he eventually loses his purse, all he has. This makes the independent events dependent in that context, even though a single failure is not a total failure. This happens much more quickly than often understood, even with typical house odds, and in a psychologically unfortunate manner. This is described in detail here: http://mtaonline.net/~hheffner/Gambling.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:PESN description of Rossi October 6 test is accurate
eggs. In high school in college you learn that all technical papers should include the make and model. But they did not do this, so I thought I should tell them. To give another example, they should have reported the readings from the flow meter in the 18-hour test. It appears to be an analog, non-electronic meter. In that case, they should have reported the instantaneous readings every 10 minutes, and the final cumulative reading. Of course it is better to use an electronic meter and record the data along with input power and temperatures every minute. The instruments typically measure these values thousands of times a second, before recording an average value periodically, one to five times per minute. There is no need to record more frequently than this in a test that lasts an hour or longer. - Jed Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:PESN description of Rossi October 6 test is accurate
quirks of a calorimeter arrangement prior to a demonstration test, especially a test requiring many people to travel long distance, is only common sense. A test of this importance should not be rushed and hopefully will not be underfunded. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rubber
Vortex-l has been a great teacher for me in the art of technical writing. I still have long way to go. Sometimes I run into things that make me want to be a much better writer. Below is one of them. This has to be one of my favorite movie reviews. It is a review of the film Rubber. I love the clarity and passion. It made me laugh quite a while. Wow! This movie works on so many levels! Not only is it totally boring, but it's really stupid and completely pointless and quite poorly executed to boot! Imagine a movie that is really, really terrible and it's also horribly acted and directed. Then imagine that really bad movie has a terrible script and doesn't actually go anywhere. Then imagine the director is a complete poseur who spectacularly fails at making some sort of half-realized pathetic statement. Then, my friends, you would have Rubber. This movie is the single worst piece of self-indulgent garbage I've ever seen. A bunch of high school kids with a video camera and some cough medicine would make a better movie than this. Heck, a bunch of hamsters with a video camera and some cough medicine would make a better movie than this. It's awful. It's really, really, awful. And it's not awful in the, oh, my God, this is hilarious way. It's really, really awful in the, it certainly is a shame that those kids had to get cancer way. Don't waste a download. Don't waste a DVD. For the love of all that's clean and decent in this universe, do not watch this movie. It is a pox on the behind of creation. It, and I am being completely serious here, is not worth you to even take the time to read this review. Do yourself a favor and watch some paint dry instead. I guarantee you'll have a better time than if you watch Rubber. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:[OT] Rubber
On Oct 3, 2011, at 4:04 AM, Terry Blanton wrote: I know Horace posted this just for the delightful prose; but, we must also be fair to the movie: http://www.imdb.com/title/tt1612774/ So, here is a review from someone who enjoyed it: [snip unexciting though probably more accurate review] end review A pox on you Horace as I have now added this movie to my Netflix queue and will probably watch it. But, then again, I did watch Bubba Ho-tep. T I actually watched Rubber after reading the review. Who could resist? I didn't dislike it so I rated it three stars. I watch a lot of independent films. This was not one of the best I've seen, but I certainly watched it to the end. I thought reading the funny review was much more enjoyable. I have often thought I would like to write some fiction, but I just don't have any talent for it. I would never have been able to write something so creative and funny as that review. It seems like it would be a lot easier to write without a calculator in hand, but it is not easier for me. I've been working over 15 years here now trying to improve my expository prose, with limited success. I admire people like Jed who are so good at clearly, accurately, and fully saying what they have to say, and usually without a tone that puts a person to sleep. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 8:38 AM, Man on Bridges wrote: Hi, On 2-10-2011 14:25, Horace Heffner wrote: When playing dice the probability p of rolling a 1 in a single roll is 1/6, the probability q of not rolling a 1 is q = 1-p = 5/6. The probability of rolling two 1's in two rolls is p^2 = 1/36, because the two rolls are independent events. More interesting is the probability of not rolling any 1's at all. Call this success. Similarly, for two rolls, this probability is q^2, or 5/36, because the rolling events are independent. It does not matter the order in which order the die are rolled or whether they are rolled simultaneously. For n casts of the die, the probability of success is q^n. The probability of failure is thus 1-q^n = 1-(1-p)^n. This should of course be read as: Similarly, for two rolls, this probability is q^2, or 25/36, Yes. Thanks. In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of no catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. This should of course be read as: then the probability of a catastrophe in 100 such events is 1-. 999^100 = 0.095, or about 1%. As earlier said: The probability of failure is 1-q^n = 1-(1-p)^n. Yes. Also it should say a catastrophe in 100 such events is 1-. 999^100 = 0.095, or about 10% BTW, use of color and other special features not in plain text is very useful for correcting typos like this, but some people do not realize only plain text ends up in the archives. See: http://www.mail-archive.com/vortex-l%40eskimo.com/msg52043.html Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. No, this suggests that the pistol has a memory, which is of course false. This does no more suggest the pistol has a memory than the dice do. Rolling the dice one at a time in the first example is completely analogous to one event in the suggested form of Russian roulette. As noted, The only difference between this form of Russian roulette and dice is the events of a game must stop upon the first failure event. The aggregate events can be looked upon as dependent in that sense. The events always occur one at a time and the latter events do not occur once a failure is encountered. However, the same can be true of the dice game. If the dice are rolled one at a time, then once a 1 is encountered it is not necessary to continue rolling the dice because the outcome is then already determined. The Russian roulette effect is a mental process that happens between the ears of the person (with his a memory!) spinning the cylinder. The cumulative effect of repeated dangerous but independent events on the probability of catastrophe is not psychological at all. The failure probability 1-q^2 is memory-less. Admittedly, the effect of catastrophe in the case of Russian roulette is likely primarily between the ears. 8^) This brings back memories during my student time when we had a similar discussion among students about throwing a dice so we decided to perform an experiment and it turned out that statistically after throwing a thousand times the dice this resulted in an almost equal number of times that all the numbers of the dice were thrown. Yes. However this is of course very different from the no occurrence of conditions we are discussing. That's also the reason why well performed surveys always require a representative amount of people to be interviewed. Similarly, if a condom brand has a 1% chance of failure, then 100 uses results in a probability of failure at some time in those uses of 1-.99^100 = 0.634 = 63%. Correct, but as you know condoms are for obvious reasons meant for single use only ;-) The probability given is for a brand, and is for single use, followed by disposal. The probability of failure for a given use would clearly increase with re-use of a single condom. Looking again at my wording, I can see your how the funny interpretation can be made. This is a great example of how difficult it is sometimes to write brief yet precisely communicating prose, how dependent communicating is on the mutual assumption set of the writer and reader. The writer has to guess or expect to some extent what the reader's lifetime accumulated assumption set is in order to communicate in a brief manner. I might have said 100 uses and disposals of a condom. The problem there is the use of a. I might have said 100 one time uses of a condom, but that too is ambiguous. The only way I see at the moment to avoid the ambiguity problem is to define use, as applied
[Vo]:Re: Russian Roulette, Murphy, and Independent Events
When playing dice the probability p of rolling a 1 in a single roll is 1/6, the probability q of not rolling a 1 is q = 1-p = 5/6. The probability of rolling two 1's in two rolls is p^2 = 1/36, because the two rolls are independent events. More interesting is the probability of not rolling any 1's at all. Call this success. Similarly, for two rolls, this probability is q^2, or 25/36, because the rolling events are independent. It does not matter the order in which order the die are rolled or whether they are rolled simultaneously. For n casts of the die, the probability of success is q^n. The probability of failure is thus 1-q^n = 1-(1-p)^n. Solo Russian Roulette can involve spinning a cylinder of a pistol with a single round in it, the player aiming the gun at his own head, and pulling the trigger. Call this an event. In the case of a six shooter, the probability p of the player killing himself, failure, in one event is 1/6. The probability of success in a one event game is then q = 1-p = 5/6. If a cylinder spin occurs in each event, then the probabilities of success or failure in any two events are independent of each other. The probability of success when playing a two event game, given both events are independent, is as in dice, q^2 = 25/36. The probability of success in an n event game is q^n. The probability of failure is thus 1-q^n. If p is the probability of failure in a single event, then 1-(1-p)^n is the probability of failure in an n event game. The only difference between this form of Russian roulette and dice is the events of a game must stop upon the first failure event. The aggregate events can be looked upon as dependent in that sense. The events always occur one at a time and the latter events do not occur once a failure is encountered. However, the same can be true of the dice game. If the dice are rolled one at a time, then once a 1 is encountered it is not necessary to continue rolling the dice because the outcome is then already determined. The formula 1-q^n provides some not common understanding of games of chance, risk taking, and product liability. This understanding comes from the fact that in the limit, as n approaches infinity, for *any* positive q less than 1, q^n rapidly approaches zero. This is expressed as: lim n-inf q^n = 0 for all q=01 In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of no catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. Similarly, if a condom brand has a 1% chance of failure, then 100 uses results in a probability of failure at some time in those uses of 1-.99^100 = 0.634 = 63%. If every critical component of a rocket has a very small chance of failure, say 1/1, but there are 1000 such components, then the probability of system failure is 1-(0.)^1000 = 0.095, or about 10%. If a product, like a vehicle, is used N=50 minutes a day by M=10,000,000 people, and the probability of failure in any given minute is p, then the probability of some failure P in a year is given by: P = 1-(1-p)^N*M*365 = 1-q^1825 Is is easy to see then, that for a product used by many people that, as time goes on, the number of opportunities for failure, n=(years) *N*M*365, becomes very large. No matter how close q is to 1, q^n then approaches zero. If anything can go wrong it will. This is Murphy's law. Similarly, each of the bets of a gambler is an independent event. However, all gamblers have a limited purse, even when credit is available. When a purse runs out then the gambler is done. If a gambler plays against less than favorable odds, no matter how small the margin, he eventually loses his purse, all he has. This makes the independent events dependent in that context, even though a single failure is not a total failure. This happens much more quickly than often understood, even with typical house odds, and in a psychologically unfortunate manner. This is described in detail here: http://mtaonline.net/~hheffner/Gambling.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 9:28 AM, Stephen A. Lawrence wrote: Both wrong. Where'd you guys get those 1% values, anyway? 1 - 0.001 = 0.999, sure enough. But 0.999 ^ 100 = 0.904792147, which means there's about 90% chance of no catastrophe, or about 10% chance of a mess. Strangely, neither of you got the 10% number, even though it's not just correct, it's also what you'd guess if you didn't know anything (1 chance in 1000, repeated 100 times, give about 1 chance in 10 of hitting). Have you never heard of a typo or clerical error? I make them all the time. It is nice to see someone actually read that post. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Re: Russian Roulette, Murphy, and Independent Events
When playing dice the probability p of rolling a 1 in a single roll is 1/6, the probability q of not rolling a 1 is q = 1-p = 5/6. The probability of rolling two 1's in two rolls is p^2 = 1/36, because the two rolls are independent events. More interesting is the probability of not rolling any 1's at all. Call this success. Similarly, for two rolls, this probability is q^2, or 25/36, because the rolling events are independent. It does not matter the order in which order the die are rolled or whether they are rolled simultaneously. For n casts of the die, the probability of success is q^n. The probability of failure is thus 1-q^n = 1-(1-p)^n. Solo Russian Roulette can involve spinning a cylinder of a pistol with a single round in it, the player aiming the gun at his own head, and pulling the trigger. Call this an event. In the case of a six shooter, the probability p of the player killing himself, failure, in one event is 1/6. The probability of success in a one event game is then q = 1-p = 5/6. If a cylinder spin occurs in each event, then the probabilities of success or failure in any two events are independent of each other. The probability of success when playing a two event game, given both events are independent, is as in dice, q^2 = 25/36. The probability of success in an n event game is q^n. The probability of failure is thus 1-q^n. If p is the probability of failure in a single event, then 1-(1-p)^n is the probability of failure in an n event game. The only difference between this form of Russian roulette and dice is the events of a game must stop upon the first failure event. The aggregate events can be looked upon as dependent in that sense. The events always occur one at a time and the latter events do not occur once a failure is encountered. However, the same can be true of the dice game. If the dice are rolled one at a time, then once a 1 is encountered it is not necessary to continue rolling the dice because the outcome is then already determined. The formula 1-q^n provides some not common understanding of games of chance, risk taking, and product liability. This understanding comes from the fact that in the limit, as n approaches infinity, for *any* positive q less than 1, q^n rapidly approaches zero. This is expressed as: lim n-inf q^n = 0 for all q=01 In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of a catastrophe in 600 such events is 1-.999^600 = 0.45, or about 45%. Of course the probability of a catastrophic event is much larger if the drinking is very heavy, perhaps 0.1. The probability of a catastrophe in 20 such events is 1-0.9^20 = 0.878, or about 88%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. Similarly, if a condom brand has a 1% chance of failure, then 100 uses results in a probability of failure at some time in those uses of 1-.99^100 = 0.634 = 63%. If every critical component of a rocket has a very small chance of failure, say 1/1, but there are 1000 such components, then the probability of system failure is 1-(0.)^1000 = 0.095, or about 10%. If a product, like a vehicle, is used N=50 minutes a day by M=10,000,000 people, and the probability of failure in any given minute is p, then the probability of some failure P in a year is given by: P = 1-(1-p)^N*M*365 = 1-q^1825 Is is easy to see then, that for a product used by many people that, as time goes on, the number of opportunities for failure, n=(years) *N*M*365, becomes very large. No matter how close q is to 1, q^n then approaches zero. If anything can go wrong it will. This is Murphy's law. Similarly, each of the bets of a gambler is an independent event. However, all gamblers have a limited purse, even when credit is available. When a purse runs out then the gambler is done. If a gambler plays against less than favorable odds, no matter how small the margin, he eventually loses his purse, all he has. This makes the independent events dependent in that context, even though a single failure is not a total failure. This happens much more quickly than often understood, even with typical house odds, and in a psychologically unfortunate manner. This is described in detail here: http://mtaonline.net/~hheffner/Gambling.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 9:28 AM, Stephen A. Lawrence wrote: In general, it's only after things dip well below 90% that the naive formula starts going seriously wrong. Good point, though I suppose I should have more fully expressed what I meant. I changed the referenced text to read: For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of a catastrophe in 600 such events is 1-.999^600 = 0.45, or about 45%. Of course the probability of a catastrophic event is much larger if the drinking is very heavy, perhaps 0.1. The probability of a catastrophe in 20 such events is 1-0.9^20 = 0.878, or about 88%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. I think part of the problem the typical person, who has little or no knowledge of probability at all, faces is a lack of appreciation for the cumulative effect of repeated risk. This was more my jist than the naive formula error. Perhaps this stems from an intuitive Bayesian model of reality. If a person does not or can not update his own mental Bayesian model from the experiences of others then he tends to expect no possibility of catastrophe for himself, until it happens. This is a gross over simplification but it is my general impression. It is a different situation for an experienced gambler, but there are of course a number of other logical fallacies and biases a gambler has to overcome, as I pointed out in my A Perspective on Gambling article: http://mtaonline.net/~hheffner/Gambling.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:LENR-CANR.org total downloads exceed 2 million
On Oct 3, 2011, at 11:53 AM, Jed Rothwell wrote: See: http://lenr-canr.org/News.htm#Downloads Congratulations! All your hard work is paying off, if not for you directly, for everyone. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 1:39 PM, Man on Bridges wrote: Hi, On 3-10-2011 23:00, Horace Heffner wrote: I changed the referenced text to read: For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of a catastrophe in 600 such events is 1-.999^600 = 0.45, or about 45%. Of course the probability of a catastrophic event is much larger if the drinking is very heavy, perhaps 0.1. The probability of a catastrophe in 20 such events is 1-0.9^20 = 0.878, or about 88%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. Which very clearly demonstrates that drinking and driving don't go together. Though many people believe that they are not negatively influenced by any alcohol or drugs and are still able to drive a car, truck, train, ship or airplane. Therefore in most countries a limit of 0.5 promille is used for alcohol levels while driving a car. Kind regards, MoB This is excellent. In my case I would not dare to drink at all and drive due to the compounding effect of being a doddering old man. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:NyTeknik report on October 6th test
not have been shut down there, but re- energized. To be shown to have any commercial value the device should be shown producing net energy for an extended period, like the 24 hours originally touted for the test. The claim is it can run for 6 months without refueling. This test was apparently not meant to be a demonstration of commercial value. More to come on the numbers. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
introduce water into output stream, as before. If the thermocouple within the E-cat is subject to thermal wicking, the water temperature may actually be 100°C, as before. This sudden flow of 100°C water could then account for increased temperature from the Tout thermocouple, which is located close to the hot water/steam input. In any case, it is nonsensical that when power is cut that output power quickly momentarily rises. This kind of mystery can be, should be, unravelled using a dummy or inactive E-cat during calorimeter calibration sessions. If the heat exchanger were 70% efficient as estimated by some individuals, then the condensed steam water temperature should have been above Tin. 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 if the coupling of the two circuits were imperfect. The insulated condenser itself and the insulated flow lines do not appear to be a significant source of loss of energy, and thus low measurement efficiency. Further, the low temperature of the condensed steam water upon output from the primary circuit 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. 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. Even if it is real, a COP of 3 is marginal for commercial application. It is much more difficult to achieve self powering with a cop of 3 vs 6. Unfortunately the temperature data is unreliable, and the COP does not look to be anywhere near the advertised 6 or even 3. Further, the temperature tailed off after less than 4 hours of no power input. The device should not have been shut down there, but re-energized. To be shown to have any commercial value the device should be shown producing net energy for an extended period, like the 24 hours originally touted for the test. The claim was the E-cat can run for 6 months without refueling. This test was not useful as demonstration of commercial value. As in the numerous prior demonstrations of the E-cats, we are left tantalized by the indication of possible excess energy, and disappointed that with a little extra effort the evidence might have finally been at hand. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rossi statement regarding 7 Oct 2011 results
From: http://nextbigfuture.com/2011/10/nyteknik-information-on-rossi- energy.html?utm_source=feedburnerutm_medium=feedutm_campaign=Feed%3A +blogspot%2Fadvancednano+(nextbigfuture)#comment-329052535 http://goo.gl/5QrM1 THANK YOU VERY MUCH, AND, SINCE I HAVE ABSOLUTELY NOT TIME TO ANSWER (I MADE AN EXCEPTION FOR YOU) PLEASE EXPLAIN THAT BEFORE THE SELF SUSTAINING MODE THE REACTOR WAS ALREADY PRODUCING ENERGY MORE THAN IT CONSUMED, SO THAT THE ENERGY CONSUMED IS NOT LOST, BUT TURNED INTO ENERGY ITSELF, THEREFORE IS NOT PASSIVE. ANOTHER IMPORTANT INFORMATION: IF YOU LOOK CAREFULLY AT THE REPORT, YOU WILL SEE THAT THE SPOTS OF DRIVE WITH THE RESISTANCE HAVE A DURATION OF ABOUT 10 MINUTES, WHILE THE DURATION OF THE SELF SUSTAINING MODES IS PROGRESSIVELY LONGER, UNTIL IT ARRIVES TO BE UP TO HOURS. BESIDES, WE PRODUCED AT LEAST 4.3 kWh/h FOR ABOUT 6 HOURS AND CONSUMED AN AVERAGE OF 1.3 kWh/h FOR ABOUT 3 HOURS, SO THAT WE MADE IN TOTAL DURING THE TEST 25.8 kWh AND CONSUMED IN TOTAL DURING THE TEST 3.9 kWh. iN THE WORST POSSIBLE SCENARIO, WHICH MEANS NOT CONSIDERING THAT THE CONSUME IS MAINLY MADE DURING THE HEATING OF THE REACTOR DURING THE FIRST 2 HOURS, WE CAN CONSIDER THAT THE WORST POSSIBLE RATIO IS 25.8 : 3.9 AND THIS IS THE COP 6 WHICH WE ALWAYS SAID. OF COURSE, THE COP IS BETTER, BECAUSE, OBVIOUSLY, THE REACTOR, ONCE IN TEMPERATURE, NEEDS NOT TO BE HEATED AGAIN FROM ROOM TEMPERATURE TO OPERATIONAL TEMPERATURE. WARMEST REGARDS TO ALL, ANDREA ROSSI I show a net energy balance of zero at 15:56, 284 minutes into run. Unless I have a mistake in my spreadsheet at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf the COP never rises above 3.3, or 2.7 if no correction to thermocouple reading delta T is made. The energy to heat the reactor is recovered when the power is turned off. If the power off and cool down period were extended well beyond 19:58 the COP might have been much larger. Some energy is lost to the environment. This amount could have been determined if a calibration run had been made. With a little patience, a little more data recording, and moved thermocouples, this test might have been a stunning success. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Hustedt graph proves there is energy generation
On Oct 7, 2011, at 1:33 PM, Jed Rothwell wrote: [snip] In this discussion, it took Hustedt a while to figure out that the condensed water from the primary loop is being flushed down the drain rather than recycled back into the cell. The original plan called for it to be recycled back into the cell. In his latest comment he notes correctly that heat lost with the warm condensate going down the drain from the primary loop would only add to the performance of the eCat. . . . Excess heat wasted out of the condensate side will be additional heat output from the e cat not included above, ie it will only make the ecat look better when this is included. [snip] - Jed 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. There is a serious problem with the output temperature recorded for manual measurement of the condensed steam. It was repeated. Perhaps that was a repeated recording 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. This is not possible. However, if even close to true, the efficiency of the heat exchanger is very high. Since it is a commercially built model that can be expected. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 7, 2011, at 5:03 PM, Alan J Fletcher wrote: I preliminarily agree with your Preliminary Data Analysis. What I DON'T understand from Hustedt's graph http://www.facebook.com/photo.php?fbid=10150844451570375set=o. 135474503149001type=1theater (and your spreadsheet) is why there was NO heat transfer to the secondary circuit until 13:22. Maybe they didn't turn on the eCat's input pump until then. 19:22: Measured outflow of primary circuit in heat exchanger, supposedly condensed steam, to be 345 g in 180 seconds, giving a flow of 1.92 g/s. Temperature 23.2 °C. This indicates pump flow is probably 1.82 ml/s. The heat showed up in the exchanger at about 130 min, or 7800 seconds into the run. See graph attached, or spreadsheet at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf This means the flow filled a void of (7800 s)*(1.82 ml) = 14200 ml = 14.2 liters before hot water began to either overflow or percolate out of the device, and thus make it to the heat exchanger. If you look at the graph you clearly see the Pout data points are all over the place when Pin ~= 0. As I noted in my Preliminary Data Analysis: http://www.mail-archive.com/vortex-l%40eskimo.com/msg52405.html it is notable that when the power is turned off, for example at time 14:20, and 14:51, and 15:56, the power Pout actually rises. This may be a confirmation that the Tout thermocouple is under the influence of the temperature of the incoming water/steam in the primary circuit. Water carries a larger specific heat. Cutting the power may introduce water into output stream, as before. If the thermocouple within the E-cat is subject to thermal wicking, the water temperature may actually be 100°C, as before. This sudden flow of 100°C water could then account for increased temperature from the Tout thermocouple, which is located close to the hot water/steam input. Further, the fact the data is highly variable is an indication the hot water arrives at the heat exchanger in slugs. That's my take on it. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ inline: RossiGraph.jpg
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 7, 2011, at 5:03 PM, Alan J Fletcher wrote: I preliminarily agree with your Preliminary Data Analysis. What I DON'T understand from Hustedt's graph http://www.facebook.com/photo.php?fbid=10150844451570375set=o. 135474503149001type=1theater (and your spreadsheet) is why there was NO heat transfer to the secondary circuit until 13:22. Maybe they didn't turn on the eCat's input pump until then. 19:22: Measured outflow of primary circuit in heat exchanger, supposedly condensed steam, to be 345 g in 180 seconds, giving a flow of 1.92 g/s. Temperature 23.2 °C. This indicates pump flow is probably 1.82 ml/s. The heat showed up in the exchanger at about 130 min, or 7800 seconds into the run. See graph sent with separate email, or see spreadsheet at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf This means the flow filled a void of (7800 s)*(1.82 ml) = 14200 ml = 14.2 liters before hot water began to either overflow or percolate out of the device, and thus make it to the heat exchanger. If you look at the graph you clearly see the Pout data points are all over the place when Pin ~= 0. As I noted in my Preliminary Data Analysis: http://www.mail-archive.com/vortex-l%40eskimo.com/msg52405.html it is notable that when the power is turned off, for example at time 14:20, and 14:51, and 15:56, the power Pout actually rises. This may be a confirmation that the Tout thermocouple is under the influence of the temperature of the incoming water/steam in the primary circuit. Water carries a larger specific heat. Cutting the power may introduce water into output stream, as before. If the thermocouple within the E-cat is subject to thermal wicking, the water temperature may actually be 100°C, as before. This sudden flow of 100°C water could then account for increased temperature from the Tout thermocouple, which is located close to the hot water/steam input. Further, the fact the data is highly variable is an indication the hot water arrives at the heat exchanger in slugs. That's my take on it. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
inline: RossiGraph.jpg
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 7, 2011, at 4:33 PM, Jouni Valkonen wrote: horace, you have two flaws in reasoning. T3 is inlet water temperature. Not the temperature of output of primary circuit. You are correct, it should be the value what you thought it to be, but this is the main flaw in the test. This also means that we do not have any means to know what was the efficiency of heat exchanger, because we do not know how much heat went down the sink from open primary circuit. Primary circuit should have been closed. I did not reference T3 in this regards, as far as I know. If you think I did in some relevant way please provide a quote of the material to which you refer. Here again are the quotes I think are important with regards to *measuring* the outflow of the primary circuit: 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. 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. The water coming out of the primary circuit should not be cooler than the cooling water going into the heat exchanger, but the difference may be just thermometer error. My point here is there is no wasted heat going down the drain if this is correct. Second flaw in your reasoning is that it pointless to calculate COP from the beginning of the temporarily limited test. That is because initial heating took 18 MJ energy before anything was happening inside the core. Therefore COP bears absolutely no relevance for anything because after reactor was stabilized, it used only 500 mA electricity while outputting plenty. And self-sustaining did not show unstability. Even when they reduced the hydrogen pressure, E- Cat continued running for some 40 minutes. This is not a flaw in reasoning. I have done many similar calculations and I typically like like Ein Eout and COP as final columns. COP is very meaningful, and helpful to quick interpretation, but you have to wring out the latent heat in the system at the end of the test. I have posted a test of mine where the COP ended at 1, and another where it ended significantly above 1. You are making the unwarrented assumption above that the thermometry can be relied upon. I don't think it can. The thermometers were improperly located and no manual checks were provided, no calibration run. Of course you can calculate the COP, and it has it's own interesting value, but it has zero relevance for commercial solutions, because E-Cat is mostly self-sustaining. There is no evidence provided of that at this point. Real long running COP should be something between 30 and 100, but we do not have no way of knowing how long frequency generator can sustain E-Cat. My guess is that it far longer than 4 hours, perhaps indefinitely. Again, there is no evidence provided of that at this point. But your calculations were absolutely brilliant. Thanks, but they are just standard operating procedure for this kind of thing I think. It was something that I wanted. It also confirmed my estimation of 100-150 MJ for total output, including 30 MJ of electricity. Although I did consider also something for the innefficiency of heat exchanger. for Mats Lewan, I would like to ask did anyone measure the temperature of primary circuit after the heat exchanger? This would be very important bit of information. I provided quote of a couple of such measurements above. —Jouni lauantai, 8. lokakuuta 2011 Horace Heffner hheff...@mtaonline.net kirjoitti: The following is in regard to the Rossi 7 Oct E-cat experiment as reported by NyTeknic here: http://www.nyteknik.se/nyheter/energi_miljo/energi/ article3284823.ece http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of +E-cat+October+6+%28pdf%29 A spread sheet of the NyTecnik data is provided here: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf Note that an extra 0.8°C was added to the delta T value so as to avoid negative output powers at the beginning of the run. This compensates to some degree for bad thermometer calibration and location, buy results in a net energy of 22.56 kWh vs 16.62 kWh for the test, and a COP of 3.229 vs 2.643. The 22.56 kWh excess energy amounts to 81.2 MJ excess above the 36.4 MJ input. If real this is extraordinary scientifically speaking. However, the lack of calibration and placement of the thermocouples makes the data unreliable. The experiment was closer than ever before to being credible. Just a few things might have made all the difference. First, a pre-experiment run could have been made to iron out calorimetry problems. A lower flow rate and thus larger delta T would have improved reliability of the power out values. Second, the lack of hand
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 7, 2011, at 10:04 PM, Mark Iverson-ZeroPoint wrote: The Tout thermocouple being within an inch or two of the hot steam flow into the heat exchanger does not sit well w/me... From watching Lewan's video again, the external heat exchanger (XHX) was operated in counter-current flow, where the steam from the primary circuit flowed opposite to the water flow in the secondary circuit. Yeah, yeah, we don't really know how that XHX is constructed, but let's just look at the inlet/outlet physical locations on both sides of it. The steam entered the same side of the XHX as did the out-flowing heated water from the secondary side. So we are assuming that the metal fitting to which the thermocouple was attached, was at the temperature of the water flowing inside and was not influenced by the 120+C steam that was entering only an inch or two away from the thermocouple??? Just doesn't sit well w/me... ...now I can go to bed. -m You may be an undigested bit of beef, a blot of mustard, a crumb of cheese, a fragment of underdone potato. There's more of gravy than of grave about you, whatever you are! External heat and cold had little influence on Scrooge. No warmth could warm, no wintry weather chill him. No wind that blew was bitterer than he, no falling snow was more intent upon its purpose, no pelting rain less open to entreaty. A Christmas Carol, Charles Dickens Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 7, 2011, at 10:10 PM, Rich Murray wrote: Hey, Horace, I don't see anyone calling YOU a pathological skeptic -- thanks muchly for doing my homework for me... Gratefully, Rich Murray Well, I am admittedly a member of the free energy lunatic fringe. What would be the point? 8^) I still am on the fence on this one. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 7, 2011, at 4:33 PM, Jouni Valkonen wrote: Second flaw in your reasoning is that it pointless to calculate COP from the beginning of the temporarily limited test. That is because initial heating took 18 MJ energy before anything was happening inside the core. Therefore COP bears absolutely no relevance for anything because after reactor was stabilized, it used only 500 mA electricity while outputting plenty. And self-sustaining did not show unstability. Even when they reduced the hydrogen pressure, E- Cat continued running for some 40 minutes. The format I used I think is very useful for calibration runs, where it is known there is no excess heat. If the protocol is good and sufficiently long, and the measurements good, then at the end of the run the COP ends up at 1. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:frequency generator
On Oct 7, 2011, at 4:57 PM, Jouni Valkonen wrote: frequency generator was shutdown 19:00, but E-Cat continued runing still some 40 minutes before reactions stopped because of increased water inflow rate. Curiously hydrogen pressure seems not to be that important for E-Cat. It does seem that frequency generator is not necessary, but it certainly boosts the output power. When it was initialized when the main input power was reduced to zero, output power jumped from 3 kW to 6 kW although electric power was reduced by 2.7 kW. That could be simply due to a slug of hot water, pumped out due to reduced boiling rate, heating up the Tout thermometer via the metal to which it is held. —Jouni lauantai, 8. lokakuuta 2011 Axil Axil janap...@gmail.com kirjoitti: Does anybody know if the frequency generator(I am assuming a 50 watt microwave source) was powered and functioning all throughout the self-sustaining phase of the Rossi demo. This seems to be something new in the Rossi design and may be how the self-sustaining mode was engineered. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
An extended review of the Rossi 6 Oct 2011 test, with a better format graph, is located at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rossi 6 Oct 2011 Spreadsheet without 0.8°C bias
I can't afford to go to sleep any more. Too many messages on vortex for me to keep up. I am way behind. I had to add 0.8°C to the Tout in order to compensate for the bad thermometer calibration. Here is the spreadsheet with the bias removed: http://www.mtaonline.net/~hheffner/Rossi6Oct2011noBias.pdf This demonstrates what a large effect a small error in delta T can make. The computed data is probably wrong both with the bias and without it. Correct numbers, assuming no systematic error on Tout measurement, are probably somewhere in between. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 8, 2011, at 7:14 AM, Akira Shirakawa wrote: On 2011-10-08 01:28, Horace Heffner wrote: The following is in regard to the Rossi 7 Oct E-cat experiment as reported by NyTeknic here: A knowledgeable user on italian discussion board Energeticambiente.it made a few impressive charts regarding the 7 Oct experiment. Everybody, have a look at the following link: http://goo.gl/gm0D0 Cheers, S.A. I don't see any charts. What am I doing wrong? Is there a link there I am missing? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 7, 2011, at 11:57 PM, Jouni Valkonen wrote: Horace, you were correct. I did error with the temperature (one example how easy it is to jump into conclusions when you thought to be certain, but actually reasoning was flawed). Temperature after the heat exchanger was indeed measured in primary circuit. But we have just two datapoints which had mass flow rate of 3.3 kg/h and 6.9 kg/h. This is rather variable. However, I do not think that this variation could explain temperature fluctuations in secondary loop, because most of the enthalpy was caried out by steam and and that should not have no other fluctuations than what are caused by power fluctuations. 95°C water without steam did not cause notable temperature change in secondary loop. Therefore we can just assume high efficiency for the heat exchanger. Something like 90% or above. Or we can just ignore it. Jouni wrote: Of course you can calculate the COP, and it has it's own interesting value, but it has zero relevance for commercial solutions, because E-Cat is mostly self-sustaining. Horace wrote: There is no evidence provided of that at this point. We do not have any evidence against it either. All evidence that we have is pointing into this direction that E-Cat is mostly self- sustaining after initial heating. Jouni wrote: Real long running COP should be something between 30 and 100, but we do not have no way of knowing how long frequency generator can sustain E-Cat. My guess is that it far longer than 4 hours, perhaps indefinitely. Horace wrote: Again, there is no evidence provided of that at this point. There is no evidence against either, because test was scheduled to be short (8 hours). Here you are making the point I made in my report. The evidence presented is insufficient to determine one way or another if there is excess heat. This is poor experiment design. It wouldn't be so horrific if many people had not suggested in advance ways to get good evidence, like combined use of isoperibolic calorimetry methods. lauantai, 8. lokakuuta 2011 Horace Heffner hheff...@mtaonline.net kirjoitti: On Oct 7, 2011, at 4:33 PM, Jouni Valkonen wrote: Second flaw in your reasoning is that it pointless to calculate COP from the beginning of the temporarily limited test. That is because initial heating took 18 MJ energy before anything was happening inside the core. Therefore COP bears absolutely no relevance for anything because after reactor was stabilized, it used only 500 mA electricity while outputting plenty. And self- sustaining did not show unstability. Even when they reduced the hydrogen pressure, E-Cat continued running for some 40 minutes. The format I used I think is very useful for calibration runs, where it is known there is no excess heat. If the protocol is good and sufficiently long, and the measurements good, then at the end of the run the COP ends up at 1. For this this is useful, but it is not meaningful to extrapolate long term COP, what you were trying to do, when you thought that COP was rather low for industrial applications: »Even if it is real, a COP of 3 is marginal for commercial application. It is much more difficult to achieve self powering with a cop of 3 vs 6.» This is just utterly false reasoning, because initial heating of E- Cat consumed most of the input and it does not need to be done more than once. But perhaps your mistake was with this misunderstanding: »Further, the temperature tailed off after less than 4 hours of no power input. The device should not have been shut down there, but re- energized.» Temperature tailed off when the hydrogen pressure was reduced and frequency generator was shutdown in 19:00. after that it took some 40 mins to stop heat production at kilowatt scale. that is, reactor was shutdown in 19:00 as was scheduled. The test was advertised to be 24 hours. Then it was advertised to be at least 12 hours. It would be nice to know when the 19:00 shutdown time was scheduled. Therefore E-Cat test was phenomenal success that surpassed even our wildest dreams. I find this viewpoint unimaginable. I guess I am short on imagination. 8^) I think we need David Copperfield to explain the illusion, because no less skilled illusionist can not do such a convincing demonstration, if it was the gratest hoax in history of cold fusion. We have positive evidence against hidden power sources Hidden power sources are not needed to explain the results. A misplaced Tout thermometer provides all the explanation that is necessary. and positive evidence for huge amounts of excess heat with only 50-80 watts input for frequency generator. —Jouni Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 8, 2011, at 12:23 PM, Taylor J. Smith wrote: Hi Horace, 10-8-11 I don't understand the two attached captions for your graph. Would you please put them in plain text (ascii) for me? Also, I would appreciate any explanation of the graph you can give me. Thanks, Jack Smithrossi106.jpgr2os106.jpg I have updated my review with a DISCUSSION OF GRAPH section, and other corrections. Thanks for the question! Here it is: DISCUSSION OF GRAPH The legend tags are: red circle - Pin (kW) [power in] blue diamond - Pout (kW) [power out] yellow square - Ein (kWh) [energy in] brown triangle - Eout (kWh) [energy in] The x axis shows elapsed time in minutes. The Y axis shows kw for Pin and Pout, kWh for Ein and Eout. It is important to show these values all on the same graph because it clearly shows that once hot water is flowing, i.e. power is turned off, quickly eliminating much steam volume, the excess heat values show up immediately. Eout only crosses Ein, i.e. COP1 occurs, only once the electric power is mostly shut down. During the first 130 minutes there is no hot water flow into the heat exchanger because the E-cat is still filling up, and still heating up, thus the blue line remains flat near zero. Once the flow begins the over unity power begins. It is quickly elevated when the power is turned off. Notice the steep decline trend of the blue curve from 350 minutes to 550 minutes. This corresponds to the nearly linear drop in T2 (not shown), which likely corresponds to a drop in the internal temperature of the huge thermal mass of hot metal inside. It is most notable the experiment was terminated when that temperature approached 100°C. Due to bad calorimetry, there is an excess energy explanation for all the Rossi tests if one thinks in terms of how the output thermometer can be affected by thermal wicking - an old problem discussed many years ago with regards to metal thermometer wells in CF cells. The thermometer attached to the heat exchanger is right next to the water/steam input to the heat exchanger. There is an insulated thick metal heat conduit from the steam inlet to the Tout thermometer. When steam goes into the heat exchanger it does not have enough specific heat to provide a large false reading for Tout, which is maintained at a lower temperature by the competing cold water flow. However, when power is cut back, and pure nearly 100°C water is pumped to the heat exchanger from the E-cat, that water has the thermal power to drive up a large false temperature reading for Tout. This explains why there is an upward temperature movement almost immediately every time the electric power is cut back. The steam quickly abates, leaving only a water flow due to the pump. The Tout thermocouple is placed directly on the metal and under insulation, not placed in the water, so this is a perfect situation in which to obtain false temperature readings. This placement was described by Rossi in NyTechnik video shown in the URL referenced above. There is still enough energy stored in the metal thermal mass to produce a bit of steam for 3.5 hours, on the order of 100 W or so. This is enough to generate a percolator effect which makes the blue line erratic as shown, due to slugs of water moving through the line. It is notable that if a calibration run were made then this kind of measuring error, if it exists, would show up as soon as the test device were full and up to temperature and then the power cut back. In the case of the thermometer hidden inside the Rossi device, and previous devices, they are likely subject to direct wicking from a large insulated metal thermal mass which heats up well beyond 100°C. Also, steam present above the water line in the device, especially in the chimney of the earlier devices, when the flow is reduced, is subject to superheating to some degree. The 120°C temperature recorded may just be a thermometry problem - easily solved by measuring outlet temperature a small distance down the hose away from the device itself, where the thermometer is not subject to direct metal to metal thermal wicking. It is notable that in this test the primary flow circuit is open. Pressure should not build up inside the E-cat, unless valves are present inside which close or partially close automatically near 100° C. However, the water condensed steam flow through the heat exchanger was manually verified, indicating a significant flow was present, indicating the pressure should not be high inside the E- cat. Yet a higher than 100°C reading was present for the thermometer inside the E-cat. That indicates a good possibility that this high reading is merely a systematic false reading. This is a hypothetical explanation of the graph. Others, involving genuine excess energy, have been made. Best regards, Horace Heffner http://www.mtaonline.net
Re: [Vo]:Thermometer used to measure cooling water
On Oct 8, 2011, at 1:51 PM, Jed Rothwell wrote: This was shown in the video on the table. Lewan says this was a Termometro a 4 canali TM-947 SD. 4 canali evidently means you can attach up to 4 thermocouples. http://www.bergamomisure.it/parametri-ambientali/termometri/ termometro-datalogger-4-canali-tm-947-con-scheda-sd.html - Jed Not very accurate for this purpose. Risoluzione: 1°C , 1°F / 0.1°C,0.1°F I take it this means 1°C absolute, and 0.1°C relative. This means for a 5°C delta t there could be 2°C error, or 40%. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Thermometer used to measure cooling water
On Oct 8, 2011, at 3:10 PM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: On Oct 8, 2011, at 1:51 PM, Jed Rothwell wrote: This was shown in the video on the table. Lewan says this was a Termometro a 4 canali TM-947 SD. 4 canali evidently means you can attach up to 4 thermocouples. http://www.bergamomisure.it/parametri-ambientali/termometri/ termometro-datalogger-4-canali-tm-947-con-scheda-sd.html - Jed Not very accurate for this purpose. Risoluzione: 1°C , 1°F / 0.1°C,0.1°F I take it this means 1°C absolute, and 0.1°C relative. This means for a 5°C delta t there could be 2°C error, or 40%. No it does not mean that. I have used many handheld meters and K- type thermocouples in other configurations. At a normal temperature range for a 5°C Delta T, the error is less than 0.1°C. That is to say, every single meter or thermocouple you use will report the same temperature difference to the limits of the display. They may start off at different absolute temperatures, but they will all show the same change. Even the cheapest red liquid thermometer Omega offers, the GT 736590, does not have a 2°C error in precision. I have 3 of them, and they are all accurate to within a degree, and precise to 10 or 20 degrees. (That is, when 1 goes up 18 deg C, so do the other 2, and so do the thermocouples.) A thermocouple or thermometer with a 2 deg C error in the normal range of use, 0 to 100 deg C, would be absurd. It would be useless. This particular one goes much higher, and you might see a 2 deg C error at higher temperatures, but not around room temperature. Trust me, a 250 Euro instrument would never do that. I have seen and used dozens of them. - Jed Two thermocouples were used to measure the delta T. If one is off 1 degree hot absolute and the other is off 1 degree cold absolute, then delta T is off by 2°C, systematically. It looked to me the actual error was on the order of 0.8°C, but who knows without calibration. It is possible, even easy, to make a thermocouple (pair) specifically for measuring delta T. One means to greatly simplify and improve the delta T measurement is to use paired type T thermocouples at the Tin and Tout measuring points. (See The Temperature Handbook by Omega Corp. p z-31.) Copper leads are used from the data aquisition device to and from the Tin and Tout measuring points, but a constantin lead connects the Tin and Tout points. Therefore, there is no junction compensating voltage involved in interfacing with the thermocouple, provided only copper leads are used. Delta T = alpha*(Tout - Tin) is then purely a product of a thermocouple constant alpha times the thermocouple voltage. There are no zeroing problems or non-linear functions. I don't know why this method was not applied for CF experiments where delta T was small. Of course this will not help the situation at all if one of the measuring points is under the influence of the primary circuit hot water supply, i.e. connected to that nearby heat source by a thick high thermal conductivity material like brass or copper. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 6 Oct Experiment Data - Preliminary Data Analysis
On Oct 8, 2011, at 5:15 PM, Mark Iverson-ZeroPoint wrote: Horace wrote: Yet a higher than 100°C reading was present for the thermometer inside the E-cat. That indicates a good possibility that this high reading is merely a systematic false reading. Horace, The T2 thermometer (inside the E-Cat) started out at nearly the same temp as the peristaltic pump water (T3), T3: 25.6C @ 11:22 T2: 29.9C @ 11:22 Ok, so worst case is that T2 is reading ~4C higher than T3, but then, T3 has water flow over it, whereas T2 is INSIDE the E-Cat and supposedly above any liquid water (in order to measure steam temperatures). So as the reactor is heating up, the air inside the reactor is also heating up and we see a steady rise in T2. So far the behavior of T2 is not anomalous. Therefore, I don't see any justification for your saying that the 120C readings for almost 2 hours were merely systematic false readings. T3, which is the water temp going into the Reactor core, remained quite stable (+-0.7 C) for nearly the whole test. T2 on the other hand, spent a lot of time above 120C, and was also reasonably stable... obviously, measurements significantly above boiling temp would indicate superheated steam if the pressure inside was not much above ambient. Lewan also put his hand on the E-Cat during Self- sustain mode several times and could feel the rumblings indicating significant boiling. Why do you think that all of the time that T2 is 100C, that it must be false readings??? Are you saying that the T2 thermometer was reading 10's of degrees off??? -Mark I m not saying the T2 thermometer is reading its local temperature wrong. I am saying that its local temperature could be under the influence of the huge thermal mass of lead and steel, which is located within the insulation jacket. In the case of the earlier E- cats this appeared to be likely. In the case of the E-cat in this test we simply do not know where the T2 thermocouple is located. The thermal mass is on the order of 3 J/K, as I computed in the STORED HEAT section of my review. At a delta T of 200 °C this is about 6 MJ of thermal storage. If there is some thermal resistance R1 to the T2 thermocouple, and a thermal resistance R2 to the 100°C water, then the thermocouple will be at a temperature of 100°C + (R2/ (R1+R2))*200°C. To get a 30°C difference all is needed is for r=(R2/ (R1+R2)) to satisfy: r * 200°C = 30°C r = 0.15 The interpretation made regarding the earlier E-cats was the steam/ water had to be under pressure to permit a 120°C temperature near the exit port. My point was this does not necessarily follow. The high temperature could merely be a systematic artifact. In the case of the current test it does not matter if it is due to superheated steam in the locality or due to direct high thermal conductivity to the thermocouple from the large metal thermal mass which is directly in contact with the heater. Because it is above 100°C I take it as an indication heat is stored which can provide a stream of hot water to influence the close by Tout location. At the heat exchanger side of things, a similar formula applies, but the water does not even have to be 100°C, merely hot enough to obtain a small delta T to the Tout temperature. If we designate Thot to be the temperature of the water arriving at the steam/hot water entry port, then there is some composite thermal resistance R1 from the Tout water to the Tout thermocouple, and a similar thermal resistance R2 to the Thot water/steam, then the thermocouple will be at a temperature of 24°C + (R2/(R1+R2)*100°C. To get an 8°C difference all is needed is for r=(R2/(R1+R2)) to satisfy: r * (100°C-24°C) = 8°C r = 8/76 = 0.1 We see the Tout temperature decline with the E-cat temperature at the end. This could be an indication the water temperature was actually less than 100°C. It could also be a partial indication of flow reduction. It is notable that the T2 thermometer inside the E-cat could possibly be under the influence of the low power frequency device at the end of the run, and thus maintain an artificially high temperature. It would be useful to have some form of thermometer at the Thot location. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rossi heat exchanger fitting
Attached is a jpg of the fitting for the hot end of the Rossi heat exchanger. The finger points to where the Tout themocouple was located. The other side of this big brass fitting was the entry point for the steam/water from the E-cat. You can see white streak marks on the tape both sides of the fitting. I wonder if those are footprints of the thermocouples used. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ inline: Tout.jpg
Re: [Vo]:Rossi heat exchanger fitting
On Oct 8, 2011, at 10:39 PM, Mark Iverson-ZeroPoint wrote: Alan: Thx for doing the calcs... I too saw the TC lead wires going under the black tape which is on the fitting where they push on the flexible hose. However, if you look closely, the lead wires continue for at least another 2 inches after the black tape, so I think the actual TC was mounted closer to the center of the heat exchanger manifold. Jed, can you contact Mats, and include the pic being referred to, and see if he can locate exactly where the Tout TC was mounted??? -m Mark, In the video Rossi points to the spot. Attached is a clip showing where he pointed. Not very definitive, but pretty close to the top of nut I would say, right where the wire length puts it. inline: Tout.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Missing posts
I sent a number of posts last night which have not shown up. I'll resend and see what happens. Sorry if they end up being duplicates. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi heat exchanger fitting
that if you actually did a 2D or 3D FEM calculation would come out a LOT smaller. I don't know, because, due to the nut protruding on top and the narrowing of the cross section, I wonder just how tight the silicon wool was in that vicinity. If only there had been good calibration sessions. - Original Message - Attached is a jpg of the fitting for the hot end of the Rossi heat exchanger. The finger points to where the Tout themocouple was located. The other side of this big brass fitting was the entry point for the steam/water from the E-cat. You can see white streak marks on the tape both sides of the fitting. I wonder if those are footprints of the thermocouples used. Best regards, Horace Heffner [image/jpeg:Tout.jpg] Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi heat exchanger fitting
On Oct 8, 2011, at 10:39 PM, Mark Iverson-ZeroPoint wrote: Alan: Thx for doing the calcs... I too saw the TC lead wires going under the black tape which is on the fitting where they push on the flexible hose. However, if you look closely, the lead wires continue for at least another 2 inches after the black tape, so I think the actual TC was mounted closer to the center of the heat exchanger manifold. Jed, can you contact Mats, and include the pic being referred to, and see if he can locate exactly where the Tout TC was mounted??? -m Mark, In the video Rossi points to the spot. Attached is a clip showing where he pointed. Not very definitive, but pretty close to the top of nut I would say, right where the wire length puts it. inline: Tout.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rossi T2 and Pout Charts
Here are some charts of possible interest. RossiT2Pout.jpg shows a scaled plot of T2 overlaid on a plot of Pout and Pin. In addition, an exponential moving average (EMA) of Pout is shown in yellow. The RF (frequency device) on and off times are denoted on the graph as well, using magenta lines. See: http://www.mtaonline.net/~hheffner/RossiT2Pout.jpg RossiT2_RF.png shows a scaled plot of T2 (i.e. T2/1000) overlaid on a plot of Pin for the period in which the RF source was on. See: http://www.mtaonline.net/~hheffner/RossiT2_RF.png It appears the RF power was ramped up at 16:38 (326 min)and down at 18:53 (461 min). The T2 curve mysteriously responds, despite the input RF power being nominal. The thermal mass of the metal and water is huge. This response of T2 to RF Pin should not be possible unless the T2 thermocouple reading is directly affected by the RF. Some obvious questions arise. Did Rossi manually adjust the RF power at 16:38? Did the RF device have controls? Is there a photo or video of it? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Re: Rossi T2 and Pout Charts
My graphs are now present at these URLs: http://www.mtaonline.net/~hheffner/RossiGraph.png http://www.mtaonline.net/~hheffner/RossiT2Pout.png http://www.mtaonline.net/~hheffner/RossiT2_RF.png The last one was updated to provide a better Y axis. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi T2 and Pout Charts
On Oct 9, 2011, at 6:25 PM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: Here are some charts of possible interest. Thanks! Put the first one up there too, in jpg or pgn format. I don't understand why your renditions come out so small but the images are sharp so that's good. - Jed They are not small on my computer. It must be the way you are displaying them. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi heat exchanger fitting
On Oct 9, 2011, at 5:52 PM, Mark Iverson-ZeroPoint wrote: When you zoom in on the end of the sensor lead wire, where the frayed insulation is, you clearly see the bare metal thermocouple wires. And from the length of that section of lead wire (~1.5 to 2 inches), the most likely location for the actual TC was on one of the flat surfaces on the shiny steel nut. They probably laid it on one of the flats, and wrapped black tape around the circumference of that shiny nut, more or less covering the entire shiny surface. Horace, I doubt if they would have just assumed the insulation would hold the TC against the nut; I vaguely remember reading that ...the TCs were held tightly against the outer metal surface by tape. But then, that would be one less thing for us to get frustrated about! Can't have that, now can we... -Mark Well we can always figure out more to worry about! 8^) Putting a metal thermocouple up against a metal surface sounds like a prescription for variable but systematic error, depending on vibrations, touching the wire, humidity, etc. The steel nut can short out at least some some of the potential. This means requiring a high bias. However, if the short is removed or reduced, then the bias is too high. When playing with the bias in my spreadsheet I settled on 0.8°C. However, it looked as if only one bias was not sufficient to fit the numbers. In any case, it seems to me to be just bad technique. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Oct 9, 2011, at 7:05 PM, Robert Leguillon wrote: Alright, if it's conclusive without the thermocouples Does anyone have a decent water capacity for the E-Cat? I see that H.H. calculated 14.2 liters, but has there been any confirmed number out of the Rossi camp? I only ask, because multiple references have been made to tons of cooling water to quench the reaction during H.A.D. In reality, the water flowing through the E-Cat (as the heat exchanger primary-side output) was measured twice: The first time, it was .91 grams/sec and the second time it was just shy of 2 g/s. If the E-Cat were indeed 14.2liters (14.2 kg), the entire contents of the E-Cat would take 2-4 hours to be completely replaced. All the while, a device that generates frequencies is still running. When it is turned off, the E-Cat temp begins declining. S many questions. Somewhere I think I saw a statement that the new E-cat has 30 liters water volume. I don't see how that is possible if the dimensions provided in the NyTeknik report are correct. I have made changes to my data review in this area. It is located at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Following are the related sections. VOLUME CALCULATIONS The Lewan report says: The E-cat model used in this test was enclosed in a casing measuring about 50 x 60 x 35 centimeters. After cooling down the E-cat, the insulation was eliminated and the casing was opened. Inside the casing metal flanges of a heat exchanger could be seen, an object measuring about 30 x 30 x 30 centimeters. The rest of the volume was empty space where water could be heated, entering through a valve at the bottom, and with a valve at the top where steam could come out. This gives an external volume of (50 x 60 x 35) cm^3 = 105000 cm^3 = 105 liters. The heat exchanger etc. is (30 x 30 x 30) cm^3 = 27 liters. This should give an internal volume of 105 liters - 27 liters = 78 liters. The prior similar E-cat weighed in at 85 kg. Looking at the open E-cat photo it looks like about (1/9)*30 cm = 3.3 cm is cooling fins. About 50% of the 3.3 cm x 30 cm x 30 = 2.97 liters should be water, giving a total water volume of 78 liters + 3 liters = 81 liters. NO HEAT TRANSFER TO HEAT EXCHANGER UNTIL 13:22 19:22: Measured outflow of primary circuit in heat exchanger, supposedly condensed steam, to be 345 g in 180 seconds, giving a flow of 1.92 g/s. Temperature 23.2 °C. This indicates the pump primary circuit flow is probably about 1.92 ml/s, as it was in the Krivit demo. The heat showed up in the exchanger at about 130 minutes, or 7800 seconds into the run. See appended graph, or see spreadsheet at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf This means the flow filled a void of (7800 s)*(1.92 ml/s) = 15 liters before hot water began to either overflow or percolate out of the device, and thus make it to the heat exchanger. If overflow started after 15 liters then it would appear 81 - 15 = 66 liters were already present. The device weighed in at 98 kg before the test and 99 kg after, when the water was drained, making this impossible. If the E-cat cold water input is 24°C and 15 liters were input, it takes (4.2 J/(gm K)) *(15,000 gm))*(76K) = 4.79 MJ to heat the water to boiling. Looking at the spread sheet this input energy Ein was indeed reached at about 13:22. This means steam probably reached the heat exchanger at this time, about 130 minutes into the test. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi T2 and Pout Charts
On Oct 10, 2011, at 6:07 AM, Stephen A. Lawrence wrote: On 11-10-09 09:39 PM, Horace Heffner wrote: Here are some charts of possible interest. ... It appears the RF power was ramped up at 16:38 (326 min)and down at 18:53 (461 min). The T2 curve mysteriously responds, despite the input RF power being nominal. The thermal mass of the metal and water is huge. This response of T2 to RF Pin should not be possible unless the T2 thermocouple reading is directly affected by the RF. Mysterious RF oscillators with undocumented connections and functions add so much interest to the question of How It Works Has Rossi become the New Ron Stiffler? Uhh.. I think that is Dr. Stiffler to us. Rossi is a mere engineer. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi T2 and Pout Charts
On Oct 10, 2011, at 6:07 AM, Stephen A. Lawrence wrote: On 11-10-09 09:39 PM, Horace Heffner wrote: Here are some charts of possible interest. ... It appears the RF power was ramped up at 16:38 (326 min)and down at 18:53 (461 min). The T2 curve mysteriously responds, despite the input RF power being nominal. The thermal mass of the metal and water is huge. This response of T2 to RF Pin should not be possible unless the T2 thermocouple reading is directly affected by the RF. Mysterious RF oscillators with undocumented connections and functions add so much interest to the question of How It Works Say, maybe the oscillations are from a digital recording of boiling sounds, sent to an underwater transducer. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi T2 and Pout Charts
I continue to update the review at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf I found out I need to make the graphs small to not lose font readability in the report pdf. I made the separate graphs much larger now: http://www.mtaonline.net/~hheffner/RossiGraph.png http://www.mtaonline.net/~hheffner/RossiT2Pout.png http://www.mtaonline.net/~hheffner/RossiT2_RF.png I hope they are not too large. I noticed I had to reduce size to print them. The letters come out small. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:No Control
May people have made this comment. Some, like Jed, directly to Rossi. Use of experimental controls is such a basic science concept it is taught in grade school science. Still, Rossi rejects the approach. I've made similar statements about controls myself: http://www.mail-archive.com/vortex-l@eskimo.com/msg50706.html Meaningful data can be obtained through the performance of well calibrated, and preferably dual method, calorimetry on the device, as a black box, that establishes a complete energy balance for each run. Use of control runs is also a standard method, and useful for calibrating the calorimetry. A thermal pulse method is also a useful check on calorimetry functions during run times. Anything less than this kind of professional calorimetry can not be relied upon. Anyone who has actually done calorimetry is keenly aware of the difficulty of getting it right. The format of the data spread sheet I provided is useful to evaluate control runs: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf You run the experiment protocol, and fix problems, until the control run COP is 1. Then when you run live you know a COP not 1 is a sign of excess energy. Without a control run, the data is meaningless. Calorimetry is subject to many kinds of artifacts - about as many as there are specific calorimeters. On Oct 10, 2011, at 10:36 AM, Joe Catania wrote: I made nearly the same post about a week ago. - Original Message - From: OrionWorks - Steven V Johnson svj.orionwo...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Monday, October 10, 2011 1:14 PM Subject: [Vo]:No Control I'm reminded of something recently stated over at the PESN web site, author, Hank Mills: See: http://pesn.com/2011/10/08/9501929_E- Cat_Test_Validates_Cold_Fusion_Despite_Challenges/ http://tinyurl.com/6a7zcw2 Specifically: No Control One of the most useful tools in the scientific method is a control. A control is an object or thing that you do not try to change during the experiment. For example, if you were giving an experimental drug to a hundred people, you might want to have a number of additional people who do not receive the drug. You would compare how the drug effects the people who consumed it, to those who did not receive the drug at all. By comparing the two sets of people, those who consumed the drug and those who did not, you could more easily see the effectiveness of the drug -- or if it was doing harm. In Rossi's test, a control system would have been an E-Cat module that was setup in the exact same way, except it would have not been filled with hydrogen gas. It would have had the same flow of water going through it, the same electrical input, and it would have operated for the same length of time as the E-Cat unit with hydrogen. By comparing the two, you could easily see the difference between the control E-Cat (that was not having nuclear reactions take place), and the real E-Cat (that was producing excess heat). If a control had been used in the experiment, the excess heat would be even more obvious. It would have been so obvious, that it could have made the test go from a major success (with some flaws), to the most spectacular scientific test in the last hundred years. Couldn't agree more. Hope someone suggests this to Rossi. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Oct 10, 2011, at 5:01 PM, Jed Rothwell wrote: Ed Storms wrote: A careful examination of the attached graph reveals an interesting conclusion. This refers to Heffner's graph 1: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf - Jed BTW, I finally figured out how to make the charts look better in the pdf. I simply had to make them a bit more narrow. I did so and moved the legend to the bottom. I'm still not happy with it. I guess I need to update or change my software. The review is still a work in progress. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
-cat temperature T2. The red line is the power applied to the blue box and RF generator. Assuming the power to the blue box is constant at this point, it is the change in power that is of interest. A change in input power of a mere 25 W has a large effect on the T2 decline. T2 is located inside the E-cat, in the midst of a very large thermal mass. Yet it appears to respond immediately to the mere 25 W increase in Pin. Between times 450 and 470 a response to a mere 3 W change can be seen. This is a reactor under the finest imaginable control! However, when we look a Graph 2, we see the heat exchanger view of this is very different. The blue Pout line varies wildly. After about time 340 on the graph the trend of the blue Pout line (represented roughly by the yellow Pout exponential moving average line) begins to mimic to some degree the T2 line. It appears the variability of the blue line in Graph 2 is not due to reaction rate changes, but to calorimeter transients. However, if the major Pout increase upon cut off of electric power is not all due to calorimetry error due to thermocouple placement, but possibly is due to the application of the RF signal, the yellow line has much significance with regard to actual reaction power generation, and Ed's conclusion 1 is valid. The tail off of the yellow curve along with the tail-off of the T2 temperature do indicate a limited time of reaction, as note in Ed's conclusion 2. It appears to me the reaction, if real, is not nearly as difficult to control as the Pout lines would indicate. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
On Oct 10, 2011, at 11:10 PM, Axil Axil wrote: The hyperlink to graph 3 is mistakenly pointing to graph 2 I think. Right you are. Thanks! Should have been: http://www.mtaonline.net/~hheffner/RossiT2_RF.png On Tue, Oct 11, 2011 at 2:44 AM, Horace Heffner hheff...@mtaonline.net wrote: On Oct 10, 2011, at 4:57 PM, OrionWorks - Steven Vincent Johnson wrote: Ed Storms said it was ok for me to post the following analysis he made: * * * * * * A careful examination of the attached graph reveals an interesting conclusion. The Pout (power out) and the Eout (Energy out) appear to describe the net excess, not the total as everyone seems to assume. Power is applied to the internal heater, showed by the red dots, until extra power starts to increase starting at 140 min. The power to the heater is turned off for a short time at 160 min because the excess power starts to rise. This interruption of applied power and the resulting reduced temperature of the Ni caused the excess to decrease and excess power production is again brought under control. Applied power is interrupted several more times to test the stability of the power-producing reaction. Finally, applied power was turned off at 280 min whereupon the extra power increased and reached a relatively stable value. The variations in excess power production after 280 min are expected as the nuclear reaction responds to variations in local temperature in the Ni. The nuclear reaction slowly decayed away and the test was terminated before it stopped all together. I make two conclusions from this behavior. 1. The amount of energy produced was far in excess of any possible chemical source. 2. The energy-producing reaction is unstable and difficult to control. It also slowly becomes less productive unless the temperature is increased by an external source of power that can increase the temperature of the Ni, thereby causing a greater output of energy. This means the energy-producing reaction has a limited life-time, which is what Rossi has indicated. If the Pout and E out are interpreted as net excess, the graph makes perfect sense and is consistent with how such a device must behave. Ed I provided two spreadsheets from which the graphs were produced: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf http://www.mtaonline.net/~hheffner/Rossi6Oct2011noBias.pdf The latter one uses the raw data, the former has an 0.8°C bias applied to Delta T to compensate for probable thermocouple error, as noted in the DISCUSSION OF GRAPH 4 section of the review: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf The graphs were taken from the spread sheet with the bias. The above seems to refer to Graph 1, which is in the review, but also in higher resolution here: http://www.mtaonline.net/~hheffner/RossiGraph.png Graph 2 in high resolution is here: http://www.mtaonline.net/~hheffner/RossiT2Pout.png Graph 3 in high resolution is here: http://www.mtaonline.net/~hheffner/RossiT2Pout.png I do not see how Pout and Eout can be interpreted as net excess. I am possibly missing the intended meaning of this phrase. Delta Eout is the thermal energy detected by the heat exchanger for the time period of a given row. Pout and Eout are created from this number. Pout is determined by a ratio of Delta Eout to the time period. Eout is just a sum of all the Delta Eout values to the end of the individual time periods each row represents. These numbers represent the thermal output. The net output, i.e. output energy - input energy, is not in the graph. It is in the spread sheet column Net E. One way to interpret Ed's phrase net excess is to consider the thermal energy still stored in the E-cat as part of the total thermal energy generated. That which has escaped and been measured by the heat exchanger is the net of total thermal energy generated minus the still stored energy. However, this interpretation does not seem to add anything to understanding discussion. When cold water is run through the E-cat sufficiently long that it cools, and if there is no nuclear energy generated, and the calorimetry works well, then Net E should be zero at the end of the run, and COP should be 1. No energy is then left stored in the E-cat at the end. This is how a control run should be evaluated, and a live test done. The power Pin applied to the heater in Graph 1 is indeed the red line. In Graph 2 it is the brown line. I think Graphs 2 and 3 have much to say about how well controlled the reaction is, if there indeed is one. In Graph 2 we can see the E-cat temperature is very well controlled. In the time 220 - 280 the red line T2 is fairly flat. There is no sign of any runaway reaction - even though the power was applied for a long period. T2 even looks fairly flat for the period 200-280. The output power Pout detected at the heat exchanger, however
Re: [Vo]:Re: Rossi Steam Quality Updates
I am way behind on reading. I hope this is not redundant. It appears the same pump is being used 6 Oct 2011 as in all the prior tests. See: http://www.nyteknik.se/nyheter/energi_miljo/energi/article3284823.ece I wonder what the 4 pumps are for? Eventually pumping water into the heat exchanger? That would be excellent, if the flow rate could be reduced, and highly controlled. It appears there is tubing on the pumps. Could the pumps have been used in an earlier test and discarded? At 2:37 the old yellow pump can be seen. It appears it was used to pump the E-cat input water as usual. At 0:21 in video at this point we have been going for several hours. One hour or so ago we went into self sustained mode. At this point in the Lewan video I counted 41 strokes per minute of the pump. Based on Matiia Rizzi's comments below, that is a maximum flow rate of (41 str/min)*(2 ml/Str)*(1 min)/(60 sec) = 1.37 ml/sec, or 4.9 liters per hour. I wonder how a pump with a maximum flow rate of 12 liters/hr could pump 15 liters/hr? On Oct 11, 2011, at 8:19 AM, Peter Heckert wrote: Rossi wrote: 15kg/h here: http://www.journal-of-nuclear-physics.com/? p=510cpage=20#comment-94236 Peter Heckert October 10th, 2011 at 1:17 AM Mr. Rossi, Could you tell the primary flow rate of the peristaltic pump? Unfortunately this was not documented. From this we could get an optimistic upper limit for the energy generated, if we assume all water was vaporized. Andrea Rossi October 10th, 2011 at 4:48 AM Dear Peter Heckert: Good question. The primary circuit flow rate has been 15 kg/h of water. Warm Regards, A.R. On Aug 23, 2011, at 12:46 PM, Mattia Rizzi wrote: And the water flow can’t be 7 liter/h since the pump is pumping every 2.5-3 seconds, so the true water flow is lower than 3 liter/h LMI P18 pump has a maximum flow of 12 l/h at 100 strikes/minutes. With 25 strikes/minute is (maximum) 3 l/h. It can be lower than 3 liter/h. On Aug 24, 2011, at 6:59 AM, Mattia Rizzi wrote: Again, if you write “7 l/h flow” you are talking about the test done in june, with Krivit. In june, there wan’t a weight scale, only a “Rossi said” that he controlled the flow by weighting it. But is a “Rossi said”. What we really know is the pump used, an LMI P18, and we know that the maximum flow is 12 l/h at 100 strokes/min. Since you can hear in Krivit video that the pump is stroking every 2.5-3 seconds, according to the manual the maximu flow rate achievable with 25 strokes/min is 3 liter/h. That’s a fact, not a “Rossi said”. On Aug 25, 2011, at 9:32 AM, Mattia Rizzi wrote: It’s a dosimetric pump. In every stroke it can inject a maximum volume of 2ml of water (volume is regulable) It’s regulable from 20 to 100 strokes/minute. So with a 100 strokes/min and a volume of 2ml, the pump is running witha flow of 12 liter/h. With 25 strokes/min, the pump is running up to 3liter/h (but it can be lower since volume is adjustable). kind regards, Peter Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Horace Heffner report
On Oct 11, 2011, at 10:27 AM, Alan J Fletcher wrote: At 01:37 AM 10/11/2011, Horace Heffner wrote: .. In the section : NO HEAT TRANSFER TO HEAT EXCHANGER UNTIL 13:22 19:22: Measured outflow of primary circuit in heat exchanger, supposedly condensed steam, to be 345 g in 180 seconds, giving a flow of 1.92 g/s. Temperature 23.2 °C. you're using the flow after it was increased to cool down the system. You should use Lewan's 0.9 g/s .. (or the pump's 2ml * 40 strokes/ minute = 1.33 g/s) 0.91 * 7800 seconds = 7.1 litres. This system supposedly has one eCat, while Lewan's Sept version had two or three, and needed 25 litres to start overflowing, so this data is more consistent than most that we've seen from October. I have significantly changed my review at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf The following sections I think are relevant. Only very rough estimates are provided because there is not enough data to go on for accurate calculations. However, hopefully the basic concepts are OK. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO HEAT TRANSFER TO HEAT EXCHANGER UNTIL 13:22 The heat showed up in the exchanger at about 146 minutes, or 8760 seconds into the run. See appended graph, or see spreadsheet at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf See appended graphs, or see spreadsheet at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf In the ecat.com video at: http://www.youtube.com/watch?v=EhvD4KuAEmo at time 0:29, there were 30 strokes in 40 seconds, or about 45 strokes per minute. That is a maximum flow rate of (30 str/(40 s))*(2 ml/str) = 1.5 ml/sec, or 5.4 liters per hour, if the pump stroke were set at 2 ml. The earlier noted flow measurement of 0.9 g/s, by Lewan, was at the output of the water/steam from the condenser heat exchanger. It might have had nothing to do with with the actual pump rate. It only had to do with the volume of steam being output, which is independent of the volume of water being pumped in - unless overflow is occurring, which seems unlikely at the early stage. A flow of 1.5 ml/sec means the flow filled a void of (8760 s)*(1.5 ml/s) = 13.1 liters, or about 13 liters before hot water began to either overflow or percolate out of the device, and thus make it to the heat exchanger. If overflow started after 13 liters then it would appear 81 - 12 = 68 liters were already present. The device weighed in at 98 kg before the test and 99 kg after, when the water was drained, making this impossible. If the E-cat cold water input is 24°C and 12 liters were input, it takes (4.2 J/(gm K)) *(13,000 gm))*(76K) = 4.15 MJ = 1.15 kWh to heat the water to boiling. CHARACTERISTICS OF THE CENTRAL MASS Looking at the spread sheet, by time 146 the input energy Ein reached was 4.446 kWh. This implies about 4.446 kWh - 1.15 kWh = 3.3 kWh = 11.88 MJ was required to heat up the thermal mass of metal in the center of the E-cat, and immediately surrounding area. Suppose there is a mass of iron between the cooling fins and heater. There might also be a layer of higher thermal resistance between the iron and the cooling fins. Use 50 kg as a rough guess at the mass of the iron. The specific heat capacity of iron is 0.46 J/(gm °C). The heat capacity of 50 kg of iron is thus (0.46 J/(gm °C)) * ( 50,000 gm) = 2.3x10^4 J/°C. Storing the 11.88 MJ requires a mean storage Delta T of (1.188x10^7 J)/(2.3x10^4 J/°C) = 516°C. Assuming the metal started out at 27°C that means an iron temperature of 543°C. This sets a limit on the period of heat after death boiling that can occur. If the central metal is heated to 543°C, then energy stored for boiling is 443°C * (2.3x10^4 J/°C) = 10.2 MJ. To last through the heat after death period from 284 min. to 476 min. = 192 min., the water boiling power output is limited to an average of 10.2 MJ/(192 min.) = 885 W. Limiting the mean thermal output of the stored thermal mass to a mean output of 885 W requires a significant degree of thermal resistance between the thermal mass and the water heat exchanger above the thermal mass. At a midpoint of heat after death, thus a thermal mass delta T of 443° C/2 = 222°C, i.e. delta T of 22°C to the boiling water, the thermal resistance required between the thermal mass and the water is (222°C)/ (885 W) = 0.025 °C/W. Registering a multi-kilowatt heat output at the heat exchanger then requires that the Tout thermocouple be under the influence of the steam/water mix, and that a mean output of 885 W provides a steam/ water mix that can drive the Tout reading up about 8°C. - - - - - - - - - - - - - - - - - - - - - Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Water meters
Does anyone know what units are displayed on the odometer type display and sweep hands on the water meters used? It says m^3 on the face, and the dials have x0.1, x 0.01, x0.001 and one with a value I can't read but assume is x0.0001. No instrument specs were provided. At the top it looks like A-B Pn16 If the smallest unit is 1 m^3 I can see why the data was not recoded, especially for the E-cat input water. The units are inappropriate. However, the following Sensu PN 16 meter data sheet looks interesting: http://delvin.co.nz/datasheets/WMU-data.htm It states: The black digits on the roller counter indicate whole cubic meters. Parts of a cubic meter are indicated by the red roller counter digits or by the red sweep hands. This means one of the little sweep hands, the x0.001 hand, should read in liters, which would be very good. A x0.0001 hand would be even better! The possibility such great data was not taken is extremely disappointing. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi: fat-cat architecure
On Oct 11, 2011, at 2:44 PM, Alan J Fletcher wrote: At 01:50 PM 10/11/2011, Jed Rothwell wrote: Anyway, there are multiple cells and only one cell was in use during this test. I assume each of the 4 cells has its own reservoir, so 30 L / 4 = 7.4 L. That's not my interpretation. The fat-cat is a big tub of overall volume 110 litres. The wafer is 20 x 20 x 4 cm = 1.6 litres Lewan's report says: Inside the casing metal flanges of a heat exchanger could be seen, an object measuring about 30 x 30 x 30 centimeters. Was that a typo? It's not stated whether each core has its own wafer, or if multiple cores are in the same wafer. The rest of the space is taken up by steel wings -- presumably we can see one of them -- the corrugated object at the top of Lewan's picture. What's left over is 30 liters, which doesn't belong to any core. (Again, 2 hours to fill at 15 litres/hour doesn't match Lewan's 0.9 g/s) I've seen that 30 liter number somewhere before too. I don't recall where. It does not jive with the apparent dimensions though. I looked for a hose between the core and the outlet of the ecat -- but couldn't see it -- because there isn't one. The outlet is just a bent tube coming out of the top of the eCat -- with a hole at the top serving as the instrument port. Again, I couldn't see any trailing wires in the lid, so the thermocouple must be right in the outlet tube. The thermocouple used has a handle on it and a long probe. It may well reach the radiator fins. One thing that concerns me is the seal holding the probe in place would have to be very good to keep the probe from blowing out of the hole, or the hole from leaking, if the device is under pressure. I'll have to think what this means for Lewan's September case, with 120C at the outlet, and 50% water !!! The only rational explanation is 2 bar 50% dry, with excess water just bubbling up through the outlet tube. The primary circuit is open. How can the E-cat maintain a 2 bar internal pressure? If there is a pressure restriction it would have to be in the horizontal part of the thermocouple mounting T, on the leg away from the thermocouple. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:pictures, data, analysis of July 7 test of fatCat
On Oct 11, 2011, at 6:45 PM, Harry Veeder wrote: On Oct.6th Passerini posted info and pics on his blog of the July 7th test with Stremmenos. http://22passi.blogspot.com/2011/10/test-e-cat-7-luglio-2011.html You can see two fat eCats wrapped in black insulation. Evidently the fat eCat has been around since atleast the beginning of July. This experiment also uses a flow of 15kg/hr. Harry Say, I wonder, what is the little white box and the black box behind it, in the third picture down. RF generator in that picture? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:More calcs.
My review of the Rossi 7 Oct 2011 experiment has been updated. http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Also, the following sections were added: VOLUME CALCULATIONS The Lewan report says: The E-cat model used in this test was enclosed in a casing measuring about 50 x 60 x 35 centimeters. These appear to be external measurements with wrapping, etc. After cooling down the E-cat, the insulation was eliminated and the casing was opened. Inside the casing metal flanges of a heat exchanger could be seen, an object measuring about 30 x 30 x 30 centimeters. The rest of the volume was empty space where water could be heated, entering through a valve at the bottom, and with a valve at the top where steam could come out. This gives an external volume of (50 x 60 x 35) cm^3 = 105000 cm^3 = 105 liters. The heat exchanger etc. is (30 x 30 x 30) cm^3 = 27 liters. This should give an internal volume of 105 liters - 27 liters = 78 liters. The disagrees with Rossi’s prior statements. Rossi states: “The volume free for the water is about 30 liters, so that to fill up it are necessary about 2 hours ( the pump of the primary circuit pumps about 15 liters per hour), but, as a matter of fact, the water begins to evaporate before the box is full of water, so usually the “Effect” of the reactor starts before 2 hours.” Using the photo in the NyTeknik report, an estimate of internal dimensions can be made. The width of the finned structure is 134 pixels, giving in that line 134 px/(30 cm) = 4.467 pix/cm. The box width on that line is 209 pix, giving a true dimension of (209 px)/ (4.467 px/m) = 46.8 cm. The length of the finned structure is 253 px, giving in that line (253 px)/(30 cm) = 8.43 px/cm. The inside length of the box is 376 px, giving a true length of 44.6 cm. The lip appears to be 35 px/(4.467 px/cm) = 7.8 cm wide. Judging from the lip width, the top of the finned structure appears to be about 4 cm below the lid. The gross inner volume of the box is (44.6 cm x 46.8 cm x 34 cm) = 71 liters. The gross volume of the finned structure is (30 cm x 30 cm x 30 cm) = 27 liters. It looks like about (1/9)*30 cm = 3.3 cm is cooling fins. About 50% of the 3.3 cm x 30 cm x 30 cm = 3 liters should be water, giving a total finned structure volume of 27 liters - 3 liters = 24 liters. The net water occupiable volume of the box is thus 71 liters - 27 liters = 44 liters. The prior similar E-cat weighed in at 85 kg. The current E-cat weighed 95 kg before water was added. ESTIMATING THE PRIMARY CIRCUIT WATER FLOW RATE The extreme instability of Pout begins at about 169 minutes into the run. If we assume this means percolator effects begin then the device should be almost full. It should contain close to 44 liters of water. The flow rate to accomplish this is (44 liters)/(169 minutes) = 4.34 ml/s or 15 liters per hour. This is a familiar number as a pump limit, but not as the primary circuit flow rate. Percolator effects could happen at a lesser volume if ripples are made in the water level . If the stated water volume of 30 liters is correct then the flow rate to accomplish percolator effects is (30 liters)/(169 minutes) = 3 ml/ s or 10.7 liters per hour. This is not consistent with the flow rate 1.5 ml/sec, or 5.4 liters per hour estimated earlier. Note that if this flow rate is correct then the stored energy calulated in prior sections is reduced. It is also true that more iron could be used to increase the thermal capacity, and space for such is available. The numbers provided here are only for concept checking. A sophisitcated model and knowledge of actual measurements is needed for an accurate consistency check. Unfortuantely measurement of flow rate into the E- cat was not made, even though a water meter was in the circuit. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:T2 thermocouple anomalies
The various anomalies associated with the T2 thermocouple, like its disconnect from the Pout, may be explained by the close distance, a few cm, of the internal fins to the lid. The T2 thermocouple rod, which protrudes down through a fitting in the lid, may be long enough to reach the fins. Dennis Cravens noted that RF affects thermocouples. They rectify the signal. This has been observed in CF experiments. This would mean the RF signal could bias T2 upwards during the heat after death period. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Water meters
On Oct 11, 2011, at 5:06 PM, Horace Heffner wrote: Does anyone know what units are displayed on the odometer type display and sweep hands on the water meters used? It says m^3 on the face, and the dials have x0.1, x 0.01, x0.001 and one with a value I can't read but assume is x0.0001. MoB has kindly confirmed the values on the dials, based on his water meter. This means the x0.0001 dial registers 0.1 liter increments. This I confirmed by looking at time 0:57 to 0:59 in the video: http://www.youtube.com/watch?v=2-5cFOsisAo You can see the 0.1 liter sweep hand going very fast. In the two seconds it moved from 0 to 0.5 liters, giving a flow rate of very approximately 0.25 liters/sec = 15 liters/min. A bit fast based on the given average of 10.7 liters/min! This means the flow meters are excellent for this purpose. Maybe they weren't read because it is difficult? Time stamped digital photos taken periodically might be a good idea. Reading the sweep hands can then be done later. Interesting. The secondary circuit flow meter can be read at the end of the test here: http://www.redmatica.com/media/Thermo1.jpg I read the meter as 13.1403 m^3, or 1314.3 liters. Given the test lasted 526 minutes that is 1314.3 liter/(536 min.) = 2.45 liters/min = 0.0409 liters per second = 40.9 ml/s. Strange. The secondary flow rate was given as 178 ml/s, or 10.7 liters/min. In 526 minutes that would be 5628 liters, or 5.62 m^3. It appears the meter began the test at 7.25 m^3. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Primary loop inflow stable, outflow varies! The two readings agree.
On Oct 12, 2011, at 5:59 AM, Jed Rothwell wrote: The answer is been staring at us the whole time. I have been thinking of the Rossi reactor as something like a US water heater where inflow must always equal outflow, because a reservoir is always full. I have been thinking that if the inflow is a steady 0.9 mL/s, the outflow has to be the same. But there is a reservoir that can hold different amounts, unlike a water heater. This is what I was talking about when I wrote: The earlier noted flow measurement of 0.9 g/s, by Lewan, was at the output of the water/ steam from the condenser heat exchanger. It might have had nothing to do with with the actual pump rate. It only had to do with the volume of steam being output, which is independent of the volume of water being pumped in - unless overflow is occurring, which seems unlikely at the early stage. Lewan measured the outflow at 18:57. It was 0.91 g/s. That indicates output power of around 2 kW. Looking at the power in the vs power out graph, at 18:57 indicated power was 2.5 kW. Close enough! http://a2.sphotos.ak.fbcdn.net/hphotos-ak- ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg Earlier, at 16:51, indicated power was 8 kW. I show a Pout of 8.073 kW in the spread sheet without correcting bias: http://www.mtaonline.net/%7Ehheffner/Rossi6Oct2011.pdf but 8.673 with correcting delta T bias. Here is my graph with bias. http://www.mtaonline.net/%7Ehheffner/RossiT2Pout.png Perhaps I should just do away with the bias correction. If Lewan had measured the outflow at that time he would have found it much faster, 3.5 g/s. We have no idea what average inflow was during this test. It was probably steady the whole time. Probably, at 16:51 when the power was high, the water level in the reservoir was falling, and at 18:54 the water level was rising. - Jed It may be that at 16:51 a slug of hot water was affecting the Pout thermocouple, and it 18:54 not so much. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:More calcs.
On Oct 12, 2011, at 6:17 AM, Jed Rothwell wrote: Susan Gipp wrote: I'm sure that for Rossi numbers are pretty meaningless. He often use them just as nice words to emphasize his speeches. We don't have to take them to make calculations. Let's talk instruments (when they work properly) No, let's talk human reflexes. At around 18:00, someone touched the hose going to the heat exchanger. That person jumped back because the connection with hot. that was four hours after the power was turned out. No, that was 2 hours 7 minutes after the power was turned off, at 15:53. No matter how you analyze it, there is no way any part of the system could have been even warm at that time, unless there was kilowatt levels of heat being generated in the system. This is simply wrong. people also held there hands over the reactor and determined that it was very hot. Again there is no way this could be true unless heat was being generated inside the reactor. The reactor was well insulated at this point. Let us talk human hearing. People heard boiling inside the reactor. four hours after the power was turned off. Yes. They should hear boiling, as I showed there should be a few hundred watts steam generation at that point. It would have been great to have the hose off momentarily at that point to see what was actually coming out of the E-cat. You do not need to believe Rossi and you do not need to believe any of the instruments to be sure the thing was producing anomalous heat. You have first principle irrefutable proof right there, in what the witnesses felt and heard. That is merely proof a thermal storage mechanism is available. It would be nice if we had more reliable instrument readings, but we do not. However, that is no reason for us to ignore witness accounts, or to imagine that a person who is burned and feels pain has not touched something hot. Do not let your anger at Rossi cloud your judgment and make you ignore first principle proof. You should look at the evidence you have, not evidence you do not have, or that you wish you had instead. As it happens the Rossi numbers are not meaningless. As I just showed you can reconcile the condensate flow rate with the inlet and outlet temperature readings. It is likely that the outlet temperature was affected by the steam pipe, but that the effect was small and the numbers are basically correct. Innumerate arm waving. Instruments such as the Termometro meter are extremely reliable and there is no way Rossi could open up to meter and change it so that it produces fake numbers. Instruments such as this have microelectronics, like digital watches, and the only thing you can do is break them. - Jed It may be useful to read my review, though it is still in draft form: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Water meters
On Oct 12, 2011, at 6:36 AM, Andrea Selva wrote: So why, as suggested by many, he didn't use another kind of flowmeter so we could have now instantaneous values to plot along the temperature ones ? Couldn't he afford it ? He sold home ! Having such data we could have explained the apparent instability and crazyness of the output power curve. Probably not, if the two existing meters were replaced. The input flow was likely steady. However, it is true, an extra water meter located at the primary circuit exit of the heat exchanger would have been very informative. Flow meters were used but apparently no one thought to record 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. Because they monitor continuously, they catch brief excursions that might otherwise be missed. Unfortunately the flow meters were ignored. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Primary loop inflow stable, outflow varies! The two readings agree.
On Oct 12, 2011, at 6:38 AM, Jed Rothwell wrote: Jouni Valkonen jounivalko...@gmail.com wrote: Water inflow rate was calibrated and it was 13 kg/h. For this test? Where does it say that? Anyway, that comes to 3.6 g/ s which is enough to sustain the highest power without having the reservoir run dry. It could be the reservoir was overflowing during most of the test. That would make no difference to the readings in the heat exchanger. The flow could be steam, or a mixture of steam and hot water; the heat exchanger will read it the same way. I think that Hefner said that there may be intermittent bursts of hot water entering the heat exchanger and this might explain some of the outlet thermocouple variations. I do not think so: 1. We have very few data points from the thermocouples. You would have to have readings taken every few seconds to see a momentary heat burst. Groups of slugs can last a while - until the water level drops. 3. Hot water coming through the pipe would lower the temperature not raise it. Wrong! Even if the steam were hotter, the high specific heat of water means the heat transfer will be greater at that point. However, because T2 readings can not be trusted, we don't even know if the steam temperature is 120°C. 3. Putting the outlet thermocouple on the pipe is a good way to blur out momentary variations and heat bursts. It is a recommended technique. Its baloney. The thermocouples should have been located in wells in the water flow a few cm down the rubber tubing. The NRL went to a lot of trouble to make sure their pipes are good heat sinks in their test bed system. However, I think installation instructions recommend you put the thermocouple about 2 feet from the boiler or heat exchanger on a straight segment of pipe. I read somewhere that you are supposed put at strap on pipe thermocouple about 2 feet away. This is different. Rubber tubing is used. I cannot find that document. I think that is what it said. If that is true, the readings may be a little bit high because of the steam pipe. As I said before, the thermal mass of the cooling water is so much larger than that of the steam that even if the thermocouple is picking up the average temperature right between them -- which is highly unlikely given its position -- the temperature would still be pretty close to the correct value. There are many strap on thermometers available that are intended to be used on the outside of pipes. This one is a general-purpose thermocouple. However, this technique will work fine as long as the thermocouple is wrapped in insulation. - Jed More arm waving. Your voice recognition system needs to be adapted to sense arm waves and footnote the text with reader warnings. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Primary loop inflow stable, outflow varies! The two readings agree.
On Oct 12, 2011, at 6:46 AM, Jed Rothwell wrote: Horace Heffner wrote: This is what I was talking about when I wrote: The earlier noted flow measurement of 0.9 g/s, by Lewan, was at the output of the water/steam from the condenser heat exchanger. It might have had nothing to do with with the actual pump rate. . . . So you did say that! You are way ahead of me. It only had to do with the volume of steam being output, which is independent of the volume of water being pumped in - unless overflow is occurring, which seems unlikely at the early stage. I don't get what you have in mind about overflowing, and the slug of hot water idea. Total enthalpy would be the same whether it overflows or not, wouldn't it? I don't see how it would affect the outlet thermocouple temperature. As I said, putting the thermocouple on the pipe which is a large heat sink will blur out any fluctuations. Perhaps I should just do away with the bias correction. How much is your correction? You probably indicate it but I don't see the number. Is at 0.5°C? - Jed The bias adjustment is 0.8 °C. The relevant graph is here: http://www.mtaonline.net/~hheffner/dTbias.png The discussion in my paper is quoted here: DISCUSSION OF GRAPH 4 Graph 4 shows Pout for the intial period before any steam came from the E-cat. The red line in the graph shows about a negative 0.5 kW Pout for no heat input. The blue line shows Pout after a 0.8°C adjustment to Delta T. No negative power is produced. However, some nonexistent positive power is produced. The net effect Ebias on total energy out of the 0.8C bias over the 526 minutes of the test is Ebias = (0.8K)*(178gm/s)*(4.2 J/(gm K))*(526 min)*(60 s/min) = 19 MJ = 5.3 kWh Without the bias the COP for the test drops from 3.2 to 2.6. http://www.mtaonline.net/~hheffner/dTbias.png Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Primary loop inflow stable, outflow varies! The two readings agree.
On Oct 12, 2011, at 6:15 AM, Jouni Valkonen wrote: Water inflow rate was calibrated and it was 13 kg/h. However, as in September, when water starts boiling and pressure is generated, it will reduce the flow rate, like it did in September. Therefore we can assume that water inflow rate was something like 10 kg/h. As was mentioned several times before, indeed steam mass flow when measured (0.9 g/s and 1.9 g/s) corresponds to current output of E- Cat. Therefore when measurements were made, E-Cat was not overflowing. Because E-Cat was not overflowing, this method could have been used for checking the calibration of heat exchanger, but this opportunity was missed by the observers. —Jouni There is no way to know if it was overflowing or not. No one took the hose off to look. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Need a break
I need to take a break from this for a while. Snow line is coming down the mountains. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Water meters
On Oct 12, 2011, at 7:20 AM, Man on Bridges wrote: Hi, On 12-10-2011 16:10, Horace Heffner wrote: Interesting. The secondary circuit flow meter can be read at the end of the test here: http://www.redmatica.com/media/Thermo1.jpg I read the meter as 13.1403 m^3, or 1314.3 liters. Given the test lasted 526 minutes that is 1314.3 liter/(536 min.) = 2.45 liters/ min = 0.0409 liters per second = 40.9 ml/s. Strange. The secondary flow rate was given as 178 ml/s, or 10.7 liters/min. In 526 minutes that would be 5628 liters, or 5.62 m^3. It appears the meter began the test at 7.25 m^3. Hmmm, I read that as 13.1403 m^3 is equal to 13,140.3 liter. Yes. Thanks! I didn't get much sleep. Gives you 24.5 liters/min = 0.409 liters per second or 509 ml/s. I get 13,140.3 liter/(526 min) = 24.98 liter/min = 416 ml/s. Clearly the meter had a large value on it to start. Also, it was apparently not recorded before the experiment began. Kind regards, MoB On Oct 12, 2011, at 6:26 AM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: I read the meter as 13.1403 m^3, or 1314.3 liters. Given the test lasted 526 minutes that is 1314.3 liter/(536 min.) = 2.45 liters/ min = 0.0409 liters per second = 40.9 ml/s. Lewan says the meter accumulated a total of 4,554 L from 11:57 to 19:03 (7 hours, 6 minutes; 426 minutes). With these meters, it is easier to read the accumulated amount than the instantaneous flow. Yes. There is no instantaneous flow value on the meter. It was nonsensical to not simply record the total flows and times. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rossi and Palin?
See current cartoon: http://freeenergytruth.blogspot.com/2011/04/e-cat-to-be-official- launch-name.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Possible systematic thermometry errors
There were questions about this but I can't find the post. I'm lost in a sea of vortex posts. The POSSIBLE SYTEMATIC THERMOMETRY ERRORS section of: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf has been updated with new URLs pointing to pictures of the T2 type probe. Here is the section as currently written: POSSIBLE SYTEMATIC THERMOMETRY ERRORS Regarding the T2 probe, examine the two photos to the right of this article: http://www.nyteknik.se/nyheter/energi_miljo/energi/article3284823.ece The top one shows the E-cat with the T2 thermocouple probe inserted down through the T fitting located on top. The second photo shows the E-cat without insulation and the cover removed. The T fitting can clearly be seen. The top of the cooling fins almost reach the bottom of the lid when it is on. The long probe may be resting on the cooling fins when it is in the fitting. The length of the probe can be seen in the photos here: http://www.energydigital.com/green_technology/e-cat-device-commercial- cold-fusion-finally-reality and here: http://newenergytimes.com/v2/sr/RossiECat/ AndreaRossiEnergyCatalyzerPhotoGallery-June.shtml more specifically here: http://newenergytimes.com/v2/sr/RossiECat/img/June2011/DSC_0025- BlueBox.JPG Regarding the Tout thermocouple, examine these photos of the hot end of the heat exchanger: http://www.redmatica.com/media/Thermo1.jpg http://www.redmatica.com/media/Thermo2.jpg The central brass fitting is very thick. Given the hose ID is about 1.5 cm, perhaps over a cm thick. It appears from the wire length the thermocouple was placed not far from it. The intermediate section looks to be at least 0.75 cm thick From the location of the tape, and the protruding thermocouple, in: http://www.redmatica.com/media/Thermo2.jpg it appears the thermocouple may have been taped to the large steel nut, possibly extending into the air beyond it. Note: the steam/water inters the heat exchanger at the same end where the Tout thermocouple is located. If we designate Thot to be the temperature of the water/steam arriving at the steam/hot water entry port, then there is some composite thermal resistance R1 from the Tout water to the Tout thermocouple, and a similar thermal resistance R2 to the Thot water/ steam, then the thermocouple will be at a temperature of 24°C + (R2/ (R1+R2)*100°C. To get an 8°C difference all is needed is for r=(R2/(R1 +R2)) to satisfy: r * (100°C-24°C) = 8°C r = 8/76 = 0.1 Here are photos that show the thermocouple before removing the tape: http://lenr.qumbu.com/111010_pics/111010_1_crop.jpg http://lenr.qumbu.com/111010_pics/111010_2_crop.jpg http://lenr.qumbu.com/111010_pics/111010_3_crop.jpg http://lenr.qumbu.com/111010_pics/111010_4_crop.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Plot / Spreadsheet request
On Oct 12, 2011, at 3:43 PM, Alan J Fletcher wrote: At 04:19 PM 10/12/2011, Alan J Fletcher wrote: Does anyone have a link to a plot (or a spreadsheet) showing T2 (output of eCat), Input power and Output power superimposed. Coincidentally, a few of us just got a BIG spreadsheet analysis via Mats Lewan. I haven't even looked at it, but the summary says that the possible Primary-in/Secondary-out thermocouple placement isn't a problem. See this graph: http://www.mtaonline.net/~hheffner/RossiT2Pout.png Copies of this and other graphs: http://www.mtaonline.net/~hheffner/RossiGraph.png http://www.mtaonline.net/~hheffner/RossiT2_RF.png http://www.mtaonline.net/~hheffner/dTbias.png are in my review with descriptions: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Spread sheet with 0.8° bias correction: http://www.mtaonline.net/~hheffner/Rossi6Oct2011.pdf Original data spread sheet: http://www.mtaonline.net/~hheffner/Rossi6Oct2011noBias.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Possible systematic thermometry errors
On Oct 12, 2011, at 3:39 PM, Alan J Fletcher wrote: At 04:22 PM 10/12/2011, Horace Heffner wrote: POSSIBLE SYTEMATIC THERMOMETRY ERRORS Regarding the T2 probe, examine the two photos to the right of this article: http://www.nyteknik.se/nyheter/energi_miljo/energi/article3284823.ece The top one shows the E-cat with the T2 thermocouple probe inserted down through the T fitting located on top. The second photo shows the E-cat without insulation and the cover removed. The T fitting can clearly be seen. The top of the cooling fins almost reach the bottom of the lid when it is on. The long probe may be resting on the cooling fins when it is in the fitting. The length of the probe can be seen in the photos here: more specifically here: http://newenergytimes.com/v2/sr/RossiECat/img/June2011/DSC_0025- BlueBox.JPG No, the long thermocouple described there is the probe which Galantini (was it) used, both on the original and mini eCat's. We've never seen the probe/thermocouple on the fat-cat. I think it just has a thermocouple wired in to the top of the T-fitting. The lid was never opened wide enough to see its underside. I enhanced the dark areas of Lewan's open-lid photo, but I can't see anything like a probe. Did you look at the NyTeknik photos? It is the same kind of probe. Attached is a small clip from he NyTeknik photo referenced. The barrel of that probe is used to seal into the T fitting. If it is not sealed then any possibility of above atmospheric pressure operation is absurd. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ inline: T2probe.png
Re: [Vo]:Analysis by Bob Higgins
A belated welcome to Bob Higgins and all the other newcomers brought here by the Rossi extravaganza. On Oct 13, 2011, at 3:35 PM, Mark Iverson-ZeroPoint wrote: Hi Bob, Having some basic dimensioning (height, width, length, etc.) would have been helpful for many of the analyses done to date, however, at this late stage it probably isn't much needed. I am still trying to get Horace to read my articles about the dimensions of the spreader, since Lewan's 30x30 must be an error, ... [snip] Mark, I am working on getting better or confirming estimates. I am also working on multiple other things at the moment so please be patient. The 30 x 30 x 30 cm numbers are indeed just rough estimates provided to Mats Lewan, not measurements. I have not found anything yet that permits accurate scale determination for the photos I have, in any photos or video video frames. I'll post an analysis of multiple photos soon. Here I provide evidence the T2 probe comes down right on the side of the fins. The probe may actually rest on the horizontal extension where the reactor housing is bolted to the bottom of the E-cat housing. Here is a Mats Lewan photo with some pixel length measurements superimposed: http://www.mtaonline.net/~hheffner/lewan_bw.jpg It is not clear which way the top bolts onto the box. It turns out it does not matter. The T is located a relative distance of 59 px / 296 px = 0.1993 from the edge of the top (see yellow lines with black numbers.) Lengthwise (see blue lines) the relative distance of the fin edges is 49 px/ 246 px = 0.1992. In the width direction (see brown lines) the relative distance is 81 px / 405 px = 0.020. It appears from various analyses the fin tops are located between 3 and 4 cm below the bottom of the top cover. The probe itself is very long (see below). It does not appear there are any fins on the bottom of the reactor housing. For convenience, my earlier comments regarding the T2 probe follow: POSSIBLE SYTEMATIC THERMOMETRY ERRORS Regarding the T2 probe, examine the two photos to the right of this article: http://www.nyteknik.se/nyheter/energi_miljo/energi/article3284823.ece The top one shows the E-cat with the T2 thermocouple probe inserted down through the T fitting located on top. The second photo shows the E-cat without insulation and the cover removed. The T fitting can clearly be seen. The top of the cooling fins almost reach the bottom of the lid when it is on. The long probe may be resting on the cooling fins when it is in the fitting. A careful analysis of the photo shows the center of the T fitting located right at the edge of the fin location. The probe should thus touch the fins or even the base of the reactor structure where it is bolted to the bottom of the E-cat. The length of the probe can be seen in Steve Krivit’s New Energy Times photos here: http://newenergytimes.com/v2/sr/RossiECat/ AndreaRossiEnergyCatalyzerPhotoGallery-June.shtml more specifically here: http://newenergytimes.com/v2/sr/RossiECat/img/June2011/DSC_0025- BlueBox.JPG Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Analysis by Bob Higgins
On Oct 13, 2011, at 4:11 PM, Higgins Bob-CBH003 wrote: I think the resistor network and finite element approaches discussed below are a great track for understanding the possible magnitude of the Tout error. The big uncertainty is the pipe thread. It may take experiments to estimate the thermal resistance across the pipe thread - particularly if it is NPT instead of NPTF because NPT will require Teflon tape to seal which would provide greater thermal isolation of the outlet pipe. Can anyone discern the thread type or whether Teflon tape has been used? Bob Higgins Hi Bob, The use of teflon tape may be important, but there may be issues with larger effects involved. There may have been electrician's tape between the thermocouple and the steel nut. Note the tape still present after the thermocouple is removed: http://lenr.qumbu.com/111010_pics/111010_2_crop.jpg I do not know to whom the credit goes for the photos referenced here. Possibly Mats Lewan. It appeared in one video that Rossi pulled on that wire quickly and firmly when unwrapping the heat exchanger. Sorry I don't know which video right now, but it might be the one in Italian. It was very fast and at the end of a clip, so difficult to determine exactly what happened. In a later photo it appeared the wire was bent, not taught, between the end tape and the tape on the nut. See: http://lenr.qumbu.com/111010_pics/111010_2_crop.jpg It also appeared the wire segment after the end tape was long enough to locate at least part of the Tout thermocouple out in the air between the nut and the brass manifold edge. See: http://www.redmatica.com/media/Thermo2.jpg It appears to me there is enough room for the sensor to extend out over the top of the big steel nut. You might have to blow up the section next to the red arrow to see the sensor tip. It is unfortunate there are not photos of the Tout location prior to wrapping with insulation. If the tip extended out into the air pocket under the insulation, then it was exposed to the temperature on the edge of the big brass manifold, or possibly even touched it. All this uncertainty obviously could have been avoided if the thermocouple had been located a few cm down the hose away from the heat exchanger, preferably in a protective well exposed to the water. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Analysis by Bob Higgins
On Oct 13, 2011, at 5:14 PM, Alan J Fletcher wrote: At 05:58 PM 10/13/2011, Horace Heffner wrote: It does not appear there are any fins on the bottom of the reactor housing. My confidential observer said that there ARE fins on the bottom. I pleaded with him to get photos of everything WITH a ruler but here we are again pixel-peeping! Anyway, I don't think the eCat body is a problem. I don't understand what Anyway, I don't think the eCat body is a problem. means. For sure it is necessary to determine the sizes of the reactor and the interior compartment to put good limits on the interior water volume. Here are some photos by Mats Lewan of NyTeknik: http://www.mtaonline.net/~hheffner/LewanEcatFront.jpg http://www.mtaonline.net/~hheffner/LewanEcatTop.jpg From the top view pipe entry points it seems clear the reactor compartment is, if not bolted to the floor of the outer container, very close to the bottom. Considering the location of the entry pipes on the exterior front, there does not appear to be room for fins. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Analysis by Bob Higgins
On Oct 13, 2011, at 8:15 PM, Mark Iverson-ZeroPoint wrote: Hi Horace, Sorry! I didn't mean to be a pest, You are not a pest! but I didn't even get an ACK that indicated you had seen my post Here you mean the following I assume? Look at all the pictures of the heat-spreader inside the E-Cat, and tell me that the shape of the finned structure is a square and not a rectangle… Now, do you think that Lewan’s dimensions for that structure 30 x 30 x 30 are right? -mark (or personal email) Sorry, I did not realize I got a personal email from you on this. There vas a [Vo] in the subject line, which routes it to my vortex-l in box. I am way behind on both reading and posting. There are lots of posts I flagged to respond to but did not get to because of time. If a subject is one which I have already dealt with and not posted or am dealing with in my survey I tend to just delay any response, because I can simply quote or refer to the section which deals with the issue. In this specific case I have been looking at photos and film to try to determine the dimensions. This is a lengthy process. BTW, it appears to me the rectangular look of the finned structure is just a matter of perspective. It looks more square from the top: http://www.mtaonline.net/~hheffner/LewanEcatTop2.jpg in this modified photo by Mat Lewans of NyTeknik. The thing most bothersome to me is the lack of any reference item with which to pin down *actual dimensions*. about this issue... next time, if possible, just a quick note to indicate that you saw the post or email and are working on it... for a trivial issue I wouldn't care, but this definitely affects some of your calculations in your report, so I wanted to be sure you at least saw that there might be an error in Mat's dimensions. I was keenly aware of the problem and working on it. I saw no sense in disagreeing with your post, which I anticipated, until I had better data put together. No Hurry... let us know when you've updated your report. -Mark Here is a preview based on this modified photo by Mat Lewans of NyTeknik: http://www.mtaonline.net/~hheffner/LewanEcatTop2.jpg All lengths below in pixels unless otherwise given. The ratio of the magneta lines is 244/195 = 1.319. The ratio of the red lines is 179/154 = 1.162. This is due to perspective, assuming all the angles are right angles. The mid-line width of the inside of the container box (magenta lines) is (195+244)/2 =219.5. The mid-line width of the reactor box (red lines) is (179+154)/2 = 166.6 The ratio of the width of the box to the width of the reactor is 219.5/166.6 = 1.3145. If the reactor is 30 cm wide then the box interior is 1.3145*(30 cm) = 39.4 cm wide. This gives a mean sideways gap width of (39.4 cm - 30 cm)/2 = 4.7 cm. The average length of the reactor (blue lines) is (155+154)/2 =154.5. The average length of the inside of the container box (orange lines) is (229+237)/2 = 233. Adjusting the orange line lengths for perspective, we have a length of (1.162/1.319)*233 = 205. The ratio of length of the interior of the container to the reactor box is 205/154.5 = 1.6181. If the reactor length is 30 cm then the length of the box is 48.5 cm. This gives a mean lengthwise gap width of (48.5 cm - 30 cm)/2 = 9.25 cm. Using a gap between the top of the reactor and the bottom of the lid of 3.5 cm, determined elsewhere, we have a container interior dimensions of 34.9 cm x 48.5 cm x 33.5 cm, for a volume of 56703 cm^3 = 56.7 liters. The volume of the reactor box is (30 cm^3) = 27000 cm^3 = 27 liters. T this we need to subtract the water spaces between the fins. It looks like about (1/9)*30 cm = 3.3 cm is cooling fins. About 50% of the 3.3 cm x 30 cm x 30 cm = 3 liters should be water, giving a total finned structure volume of 27 liters - 3 liters = 24 liters. The net water occupiable volume of the box is thus 56.7 liters - 27 liters = 29.7 liters. Rossi stated in his blog that this value is 30 liters. The measurements estimated for the device based on a 30 cm^3 reactor appear to be inconsistent with Rossie’s statement. It is of course important to obtain accurate measurements of these values to make consistent sense of the data. -Original Message- From: Horace Heffner [mailto:hheff...@mtaonline.net] Sent: Thursday, October 13, 2011 5:59 PM Subject: Re: [Vo]:Analysis by Bob Higgins Mark, I am working on getting better or confirming estimates. I am also working on multiple other things at the moment so please be patient. [snip] Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Analysis by Bob Higgins
SLight correction made at end On Oct 13, 2011, at 8:15 PM, Mark Iverson-ZeroPoint wrote: Hi Horace, Sorry! I didn't mean to be a pest, You are not a pest! but I didn't even get an ACK that indicated you had seen my post Here you mean the following I assume? Look at all the pictures of the heat-spreader inside the E-Cat, and tell me that the shape of the finned structure is a square and not a rectangle… Now, do you think that Lewan’s dimensions for that structure 30 x 30 x 30 are right? -mark (or personal email) Sorry, I did not realize I got a personal email from you on this. There vas a [Vo] in the subject line, which routes it to my vortex-l in box. I am way behind on both reading and posting. There are lots of posts I flagged to respond to but did not get to because of time. If a subject is one which I have already dealt with and not posted or am dealing with in my survey I tend to just delay any response, because I can simply quote or refer to the section which deals with the issue. In this specific case I have been looking at photos and film to try to determine the dimensions. This is a lengthy process. BTW, it appears to me the rectangular look of the finned structure is just a matter of perspective. It looks more square from the top: http://www.mtaonline.net/~hheffner/LewanEcatTop2.jpg in this modified photo by Mat Lewans of NyTeknik. The thing most bothersome to me is the lack of any reference item with which to pin down *actual dimensions*. about this issue... next time, if possible, just a quick note to indicate that you saw the post or email and are working on it... for a trivial issue I wouldn't care, but this definitely affects some of your calculations in your report, so I wanted to be sure you at least saw that there might be an error in Mat's dimensions. I was keenly aware of the problem and working on it. I saw no sense in disagreeing with your post, which I anticipated, until I had better data put together. No Hurry... let us know when you've updated your report. -Mark Here is a preview based on this modified photo by Mat Lewans of NyTeknik: http://www.mtaonline.net/~hheffner/LewanEcatTop2.jpg All lengths below in pixels unless otherwise given. The ratio of the magneta lines is 244/195 = 1.319. The ratio of the red lines is 179/154 = 1.162. This is due to perspective, assuming all the angles are right angles. The mid-line width of the inside of the container box (magenta lines) is (195+244)/2 =219.5. The mid-line width of the reactor box (red lines) is (179+154)/2 = 166.6 The ratio of the width of the box to the width of the reactor is 219.5/166.6 = 1.3145. If the reactor is 30 cm wide then the box interior is 1.3145*(30 cm) = 39.4 cm wide. This gives a mean sideways gap width of (39.4 cm - 30 cm)/2 = 4.7 cm. The average length of the reactor (blue lines) is (155+154)/2 =154.5. The average length of the inside of the container box (orange lines) is (229+237)/2 = 233. Adjusting the orange line lengths for perspective, we have a length of (1.162/1.319)*233 = 205. The ratio of length of the interior of the container to the reactor box is 205/154.5 = 1.6181. If the reactor length is 30 cm then the length of the box is 48.5 cm. This gives a mean lengthwise gap width of (48.5 cm - 30 cm)/2 = 9.25 cm. Using a gap between the top of the reactor and the bottom of the lid of 3.5 cm, determined elsewhere, we have a container interior dimensions of 34.9 cm x 48.5 cm x 33.5 cm, for a volume of 56703 cm^3 = 56.7 liters. The volume of the reactor box is (30 cm^3) = 27000 cm^3 = 27 liters. T this we need to subtract the water spaces between the fins. It looks like about (1/9)*30 cm = 3.3 cm is cooling fins. About 50% of the 3.3 cm x 30 cm x 30 cm = 3 liters should be water, giving a total finned structure volume of 27 liters - 3 liters = 24 liters. The net water occupiable volume of the box is thus 56.7 liters - 24 liters = 32.7 liters. This volume should be reduced some for the many bolt heads that bolt the reactor case to the box, or bolt the reactor case together. Rossi stated in his blog that this internal volume is 30 liters. The measurements estimated for the device based on a 30 cm^3 reactor appear to be roughly inconsistent with Rossie’s statement. It is of course important to obtain accurate measurements of these values to make consistent sense of the data. -Original Message- From: Horace Heffner [mailto:hheff...@mtaonline.net] Sent: Thursday, October 13, 2011 5:59 PM Subject: Re: [Vo]:Analysis by Bob Higgins Mark, I am working on getting better or confirming estimates. I am also working on multiple other things at the moment so please be patient. [snip] Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Thermocouple extends beyond steel nut?
This photo by Mats Lewan of Nyteknik of the 6 Oct Rossi Tout thermocouple leaves little doubt in my mind that the Tout thermocouple can and probably did extend beyond the steel nut, toward the brass manifold. It was thus subject to the air temperature near that manifold. http://www.mtaonline.net/~hheffner/LewanTcoupleClose.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
