Joshua wtote on Saturday, June 25, 2011 11:49 PM: "You've lost me here. Say the device is calibrated to measure this. How do you deduce that from 10 g/m^3 that all the water is converted to vapor? If the device measures 10 g/m^3, but the steam is coming out at 0.1 m^3/s, then clearly, the steam doesn't account for all the mass. So there must be some liquid coming out too. The only way that 10 g/m^3 tells you that all the water is being converted to vapor, is if you measure the output flow rate to be 1 m^3/s, but that is not measured. So you don't know." Okay, due to my randomly selecting an unrealisticly low flow-rate of 10g/sec, I can see where it could be confusing. Let me try to clear things up... We BOTH agree that the saturation mass of water vapor is 600g/m^3 for 100.1C steam at ambient pressure. My original example used 10g/sec which is no where near saturation, that is obvious, so I can see where that may have been confusing. I'll revise the example to be close to the saturated condition in the e-Cat tests... Here's a scenario which is just BELOW saturation... - 594g/sec liquid water in (this is 594/600 = 99% of saturation flowrate) - assume that the power level of e-Cat is such that 100% of the water is vaporized, but little or no excess power, so now the entire mass of water (594g/sec) is vaporized every second, but not superheated. - according to the methodology, the pressure inside the chimney is the same as ambient, thus, the flowrate of the dry steam out the chimney is equivalent to 594g/m^3. We do NOT need to measure the flow rate. The entire mass of the liquid water is now vapor, and that vapor HAS TO BE EXITING the chimney in order to maintain ambient pressure levels inside. If we were to slowly reduce the diameter of the outlet hole, we would eventually begin to restrict the vapor's exit and the pressure would begin to increase inside the chimney. If the pressure inside is ambient, then the exit hole MUST be large enough to not restrict the vapor's flowrate out of it. I also realize that if the pressure inside does increase significantly, then depending on the temperature, some vapor MIGHT begin to condense out which would REDUCE the volume of steam, but it would also INCREASE the steam's temperature. This is a multivariable situation that is modestly complex... I am very much aware of that.
For whatever reason, be it stability or fraud, Rossi has chosen to operate his reactor at this borderline of phase-transition which needlessly complicates the analysis and generates doubt. But it's what we have to deal with at this time, so let's just run with a more or less idealized example and see where it leads. To Be Continued.... On Monday. -Mark ________________________________ From: Joshua Cude [mailto:joshua.c...@gmail.com] Sent: Saturday, June 25, 2011 11:49 PM To: vortex-l@eskimo.com Subject: Re: [Vo]: Proposed method for how Galantini measures steam quality... On Sat, Jun 25, 2011 at 4:36 PM, Mark Iverson <zeropo...@charter.net> wrote: First, here is my conclusion based on the methodology and resoning below: "If certain conditions are present, one can reduce this to a mass-in, mass out problem, and you don't need to measure the volume of steam exiting in order to estimate dryness" I don't think anyone here was suggesting that the instrument used by Galantini could measure steam quality directly. What I am attempting to do is ascertain if there is a way to make an indirect measurement with what variables we DO have, and I think I may have the answer. Let's discuss whether this method will work, and under what conditions... 1) we know the flow-rate of water going in; can't remember but say its 10g/sec. 2) assume the entire mass of water IS vaporized; 3) we can easily calculate the volume of steam that would be generated each second; 4) *IF* the pressure inside the chimney is ambient, then the entire volume of steam is exiting the chimney each second, else pressure would build up inside; 5) assume that overall the process is relatively stable, with constant flow of steam out the chimney and only minor fluctuations in temp and pressure inside; 6) Given the above, the mass on the instrument's display of mass of water in the steam MUST equal the mass of water going in. If the instrument is reading 10g/m^3, then ALL the inlet water is being converted to vapor, and the steam is dry. You've lost me here. Say the device is calibrated to measure this. How do you deduce that from 10 g/m^3 that all the water is converted to vapor? If the device measures 10 g/m^3, but the steam is coming out at 0.1 m^3/s, then clearly, the steam doesn't account for all the mass. So there must be some liquid coming out too. The only way that 10 g/m^3 tells you that all the water is being converted to vapor, is if you measure the output flow rate to be 1 m^3/s, but that is not measured. So you don't know. It is very easy to show that this reasoning is completely wrong because it leads to an obvious contradiction: You are requiring the steam to be at 100C, and at atmospheric pressure, and contain 10g/m^3 mass. But steam at 100C and 1 atmosphere pressure has a density of 0.6 kg / m^3. It can't be 10 g/m^3. If you are producing pure steam at 100C, and the probe is reading the water vapor per unit volume correctly, then it will *always* read 0.6 kg/m^3, no matter what the input flow rate is. How can it be used to determine the fraction of input converted to steam, if it reads the same thing for different input flow rates? To put it another way, if you measure water content per unit volume of a gas at a single point, you cannot determine anything about the water content in the bulk of the gas, unless you know what the bulk is. The steam could be going past at any rate in any size pipe and your measurements (pressure, temperature, and humidity) will be the same no matter what. Same measurement, but obviously different mass flow rates, means the measurement cannot determine the mass flow rate. Volume is irrelevent *IF* the pressure inside the chimney is ambient. Wrong. If you only know density, you need volume to get the mass, regardless of pressure. The pressure can be ambient for any mixture of liquid and steam. This also assumes the instrument's humidity sensor isn't getting bombarded by liquid water, as we have discussed recently in our postings about how the polymer/capacitive sensor works. Again, if the capacitor isn't getting wet, and correctly measures the water vapor content of the dry steam component, then it always gives the same answer. So it can't be used to deduce anything. CONCLUSION: If certain conditions are present, one can reduce this to a mass-in, mass out problem, and you don't need to know the volume of steam exiting in order to estimate dryness... You have not shown anything that leads to this conclusion. In fact the contradiction your reasoning leads to shows that the conclusion is wrong.
