Could you show me a reference to level gauges in each of the devices? I do not recall seeing one so far.
Dave -----Original Message----- From: a.ashfield <[email protected]> To: vortex-l <[email protected]> Sent: Sat, Aug 20, 2016 3:00 pm Subject: Re: [Vo]:Interesting Steam Calculation That would mean the Tiger E-Cats would have to be completely flooded. But the level gauges don't show that. Why not suggest pixie dust? On 8/20/2016 1:51 PM, David Roberson wrote: Today I made an interesting calculation that some may find relevant to the ongoing discussions. According to steam tables, the following could be possible, assuming that I did not make a mistake in my calculations. Assume you have 1kg of water inside a solid container at 130 C and 39.2 psi absolute. Then you place a restriction device that allows all of the liquid to eventually escape. Some of the liquid will immediatly flash into vapor while most of the 1 kg remains in the liquid form as it exits the restriction. If you assume that the resulting mixture ends up at 102 C and 15.75 psi absolute then it is possible to calculate the amount of vapor and liquid that is present at that location. The internal energy of the initial liquid at 130 C is 546.388 kj/kg which in this case yields 546.388 thousand joules. I am assuming that this same amount of energy remains within the liquid and vapor combintation of the lower temperature and pressure stream. When I solved the equation relating the quality of the mixture to the various heat contents I determined that there would be .053 kg or vapor and .947 kg of liquid water at the output. On first glance, this result suggests that it should be easy to separate the water from the steam, but actually calculating the two volumes makes that not so evident. The volume of the vapor would be .053 kg * 1.565 cubic meters per kg = .0826 cubic meters. The volume of the liquid water would be .947 kg* .001045 cubic meters per kg = .000989 cubic meters. Using the above numbers it appears that you would have 83.488 times as much vapor by volume as liquid. This is quite a large ratio which suggests that it might well be possible to mistake a stream of mass with this consistency as consisting of only vapor. Especially if a visual technique were used. I am not saying that this calculation reveals the source of the Rossi test confusion, but that perhaps it might open discussions that have not been considered so far. I do recall that on earlier demonstrations that the temperature within the ECATs was reported to be in the range of 130 C. Perhaps some of our mathematically inclined vortex residents can take a few moments to verify that my assumptions and calculations make sense. Dave
