Dave--
Where did the pressure of 15.75 psi abs come from? I thought the pressure of the 102C dry steam (assumed) was 1 atmos.--not 15.75 abs. I think your assumed conditions above 1 atmos. were never measured. Bob Cook ________________________________ From: David Roberson <dlrober...@aol.com> Sent: Monday, August 22, 2016 3:49 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Interesting Steam Calculation I followed the calculation below with an additional one to further my research. For the second calculation I used the flow rate information supplied by Engineer48 for the 24 pumps that were manually set from the front panel. With this data I determined that the power delivered to the customer would be 30.1 kW under the following assumptions: Twenty two of the pumps were delivering full flow of 18 kg per hour while two were operating at 1/2 full rate of 9 kg per hour. The total was therefore 414 kg per hour which translates to .115 kg/second. The temperature of the water inside all of the ECAT sections was controlled at 130 C, which is in line with what was seen during several of Rossi's single unit demonstrations. All of this water then escaped through a restrictive, pressure dropping orifice such that some of the liquid flashed into steam according to the below analysis. The resulting water filled vapor flow was sent to the customer with a pressure reading of approximately atmospheric and a temperature of 102 C as below. In this case the gauges would read correctly. Water finally returned from the customer at 68 C, in liquid form, back to the Rossi system. A further calculation of the power delivered to the customer if it is assumed that all of the water is in the form of vapor with zero water at 102 C and atmospheric pressure would be 275 kW. Within this scenario the water returns at 68 C as before. The purpose of these calculations is to seek a possible hypothesis as to how the power being sent to the customer could be dramatically less than one might calculate if he depended upon the gauge readings and did not have a method to verify that the mass supplied to the customer was dry steam. If it can be shown that a steam quality measuring device was located between the Rossi system and the customer that indicated dry steam then the power delivered would be much closer to the 275 kW level. If not, then 30.1 kW could well be possible. Detailed calculation are available upon request. Dave 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
