This is exhausting. You're going to blindly believe any evidence supporting your conclusion, and if I were to give you 10 distinct reasons that the thermocouple placement is crap, you'll try to dismiss one, and assume it negates the rest. Rossi is using a herringbone liquid counterflow heat exchanger. It is meant for recovery of heat between two liquids. Even without phase change, it is difficult to produce point-specific analysis. It is worth mentioning that there are companies that produce proprietary software to analyze the liquid heat transfer in these units, and it's not something that Google calculator can do for free. So, I'm going to oversimplify this by design. Let's give you some numbers to show you how futile this is, and how Houke's method is insufficient to model the dynamic environment in which the thermocouples reside: 1) We don't know the flow rate of the primary, but "Rossi says" it's 15 l/h, and you've never known him to lie, so let's assume 15 l/h, or 4.17 g/s 2) We don't know the pressure is, while the steam is trying to force itself out of the E-Cat, through the criss-crossing walls of the exchanger, while there collects condensed water in front of it, being forced out of the exchanger, down the table, across the floor, under the doormat, pushing any slugs of water in the way, out into the parking lot, and down the drain, but you've said "it's about 1 ATM", so let's go with that. If the E-Cat is outputting 100% dry steam at 121.7C that condenses immediately, cooling to the output temperature at the secondary of 32.4C, it transmits: [((121.7C - 100C) x (.48cal/gram specific heat of steam)) + 540cal/gram latent heat from phase conversion + ((100C - 32.4C) x 1cal/gram specific heat of water) = 618 cal/gram x 4.17 g/sec = 2,577 kcal/sec Now, what if, I know this is a stretch, not all heat transfer occurs immediately? If steam is still present after the beginning of the manifold, the steam "rushing by" may only impart the energy it takes to cool to 100C: (121.7C - 100C) x .48cal/gram = 10.416 cal/gram x 4.17 g/sec = 43.43472 cal/sec That's a pretty big difference of heat energy imparted to the brass manifold. The manifold is one continuous metal block that BOTH hot and cold water flow through, albeit in their own dedicated channels. The two circumstances do not require any power output change in the E-Cat to occur. If any of the power available at the steam input is not immediately whisked away, it will necessarily heat up its environment (the manifold). Notice that it would take 650C water to impart the same amount of energy as 121.7C steam condensing to 32.4C.
Date: Thu, 8 Dec 2011 14:53:18 -0500 Subject: Re: [Vo]:Will tests surface mounted thermocouples on pipe From: [email protected] To: [email protected] Peter Heckert <[email protected]> wrote: If there is an air gap of 0.1mm between metal and thermoelement, then it is not nonsense. I doubt that. I would like to see you prove it. I do not think this would cause even a 0.1°C difference. Can you suggest a way to deliberately introduce such a small gap? Perhaps with a thin piece of paper instead of an air gap? Dont you see that Rossis arrangement was horrible and disqualifies him and Levi and Focardi to do such measurements? No, I do not. I have measured temperatures on pipes several times. As far as I know, this method works fine. Actually Rossi did a better job than most people do. Your other assertions about bubbles of air in the pipe are untrue. The metal of a steel or copper pipe averages out the temperature quite nicely. Miles and others showed this with a copper sheathed calorimeter with an air space at the top and thermal gradients inside. Probably braided pipe does not work as well. If you are so sure this was "horrible" I suggest you do a test and prove it. Even a rudimentary test such as the one I did shows it is not horrible. Rossi's methods were much better than mine. - Jed
