Hi David,
Thought experiments are good and I like the your analysis because it causes us to consider another possible effect. I believe I understand your calculation, but there are a few things that I think were different than in your thought experiment. First, the E-cat holds 30kg of water (AR was asked). I don’t think it was only half full at that time. The fins on the heat sink extend to within a few cm of the top and they are desirably kept submerged in water. I suspect that the E-cat was filled with more than 25kg of water at the time it was shut down. I think it is strange that there is not some kind of water level control in the E-cat – just open loop. Another thing that I think was different than in your experiment is the boundary matching. The instant that the water flow rate is increased, there will be no change in water temperature due to the heat capacity of the liquid. You are right that it will eventually change. Presuming that the output valve was open enough to allow steam to continuously escape, the rate of escape must have been about equal to the input flow rate or the pressure would have increased and caused the valve to open more. So the first effect that should have been seen would have been a slight pressure increase to allow the increased flux in the output. At that time, the steam output rate would have increased in proportion the new flow rate (about double) and since there was no instantaneous change in temperature, the heat exchanger will measure about double the previous heat output rate. This presumes that the slightly greater pressure does not significantly increase the leak rate (leak rate delta much less than the increase in input flow rate). Since the temperature data points taken at the heat exchanger are so sparse, it is hard to know exactly how the output tailed off from there and I think there is not enough sampling to see the backflow effect you describe. Another characteristic of the heat exchanger does not seem to be accounted for in your experiment. Imagine the heat exchanger as two long, highly coupled tubes (thermally coupled like coaxial) with nominal flow in opposite directions. The temperature at primary output is ideally the same as the secondary input and along the length of the tubes, temperature should monotonically change to match the primary input and the secondary output temperatures. So should any liquid water be sucked back toward the E-cat in the primary, it will be colder water than the temperature nominally at that distance from the primary inlet and should cool the water in the secondary at that distance. Since we are talking about small pressures, you also would need to consider the flowing mass suction effects of the primary water in the discharge. These may also compensate for any backflow effects. I am going to try to do some additional fitting to the heat exchanger data to remove some of the group delay from the current analysis. This will be hard because I suspect the measurement points are somewhat triggered – I.E. when the flow rate was changed, Mats recorded a temperature reading. Cheers, Bob From: David Roberson [mailto:[email protected]] Sent: Saturday, October 22, 2011 2:23 PM To: [email protected] Subject: Re: [Vo]:Possible mechanism-Excess Power Reading of ECAT Hi Bob, I appreciate your response to my post. It is important to me that I have a clear understanding of the relationship between the real output power delivered to the heat exchanger and the internal energy of the ECAT. My suspicion is that ultimately we will be able to correlate the temperature measured at the ECAT output thermocouple and the real power exiting the device. It seems to me that addition of extra input pump water would result in a lowering of total ECAT temperature. The water surrounding this newly introduced mass would immediately start to loose energy. This energy transfer is required in order for the newly inputted water mass to match the overall T2 temperature. I was thinking of a simple thought experiment as follows. The temperature T2 is at 116.6 (use 117) when deactivation occurs. Assume that the ECAT has 15 kilograms of water inside at that moment. The energy within the liquid would be 15000 grams x 491.08 joules/gram(NIST) = 7.3662 Megajoules. Saturation pressure at that temperature is 1.8052 bars. Lets pour a slug of water into the mix. Add 100 grams(Temp = 30 C) which is in the range of 30 seconds worth of extra water. That has an energy of 100 grams x 125.73 joules/gram = 12.573 kilojoules. First order approximation is that the total average temperature will remain near 117 after the addition so the new water will require 100 grams x 491.08 joules/gram = 49.108 kilojoules- 12573 kilojoules = net absorbed 36535 joules. Now we have a remaining energy of 7.3662 MJ - 36.535 KJ = 7.329665 MJ. When the water is thoroughly mixed and stabilized the final tally is 7.329665 Megajoules / 15100 grams = 485.4 joules/gram. This energy is associated with a temperature of slightly below 116 degrees centigrade(486.83 j/g). Saturation pressure would now be a bit less than 1.744 bars. This experiment suggests that the temperature will drop from 117 C to 116 C and the saturation pressure would drop from 1.8052 bars to ~ 1.744 bars or delta of .0612 bars(.887 psi). My theory suggests that this pressure difference will cause the valve to close slightly reducing the output vapor flow rate. As a result the water inside the exchanger will move closer to it input port. Consult my original post for details of how the thermocouple reading is distorted. I agree with you that the increase of pump input flow will result in an offset to my proposed effect to some degree, but the vapor density is many times less than that of the liquid and displacing 100 cc of vapor as per this experiment would not result in much condensed water delivery. Does this thought experiment appear logical? It is important to understand that I am assuming that the ECAT is not full of water and capable of overflow when this process occurs. Dave -----Original Message----- From: Higgins Bob-CBH003 <[email protected]> To: vortex-l <[email protected]> Sent: Sat, Oct 22, 2011 9:26 am Subject: RE: [Vo]:Possible mechanism-Excess Power Reading of ECAT Hi David, Yours was a very thoughtful post. It has taken some time to digest, and I can say I have not fully evaluated the implications across the whole experiment. However, I don’t think something so complicated need be invoked to explain the power spike immediately after shutdown. According to Mats’ data, as the hydrogen was released, the input flow rate at the peristaltic pump was increased – in fact, basically doubled. Since the reactor was boiling, the output at the time was pretty much steam and the reactor pressure was high enough to keep the valve open constantly discharging steam. The immediate effect of doubling the T3 input water rate is to double the VOLUME of effluent from the reactor output. Since the temperature at this time remained well above boiling, the output that was doubled was the volume of the steam. This simple explanation seems sufficient to explain the spike in measured temperature – double the steam volume at about the same temperature and you double the heat output measured at the heat exchanger. Most of this is heat already stored in the E-cat – this is not a burst in reactor output. Do you believe a more exotic explanation is necessary? Bob Higgins **** On 10/21/2011, David Roberson wrote: Another thorn is our paws has been the unusual behavior when the total power has been shut down and water flow maximized at the end of the test run. Look at the data from 19:22. About 14 minutes before this time the power was shut down, hydrogen eliminated and input water flow rapidly increased. A nice 2.1 degree drop is seen in the ECAT output temperature from the last reading. My thought is that the increased water input flow quickly reduces the rapid boiling within the ECAT and allows the vacuum effect to draw the exchanger hot water into the manifold. This water then leads to a large apparent power increase (Tout – Tin = 8.6 degrees) which is an illusion. Temperature just prior to this (Tout – Tin = 5.3 degrees) yields a lot less power. ________________________________ No virus found in this message. Checked by AVG - www.avg.com <http://www.avg.com> Version: 10.0.1411 / Virus Database: 1522/3968 - Release Date: 10/22/11
