Yep, I couldn't find it either! You're number of 0.6kg was correct. When you wrote it as 0.6kg I just assumed it was in the g/m^3 as I had used... I didn't see the 'k'... my apologies! I did read your post completely, and I'll reread it more carefully... Are we in agreement with this statement: "If the conditions are such that we are at or less than saturation on the output side, then the mass of water vapor out will be equal to the mass of liquid water in, assuming as stated in the methodology, that there is no significant liquid water in the steam? Liquid Mass in = Mass out in vapor form (and no signif pressure increase inside). Josh wrote: "There was never a problem with steps 1 to 4. The problem came after step 6." Geez, ya could have just stated that single sentence up front and we could have gone on from there! :-) Thanks for indulging me, and I'll come back to your analysis on monday and read it more carefully... I've gotta get some work done and this is much more fun... -Mark
_____ From: Joshua Cude [mailto:[email protected]] Sent: Sunday, June 26, 2011 11:33 AM To: [email protected] Subject: Re: [Vo]: Proposed method for how Galantini measures steam quality... On Sun, Jun 26, 2011 at 1:11 PM, Mark Iverson <[email protected]> wrote: Josh: Your off by a factor of 1000 on the saturation mass of water vapor at 100.1 and 1 atm... So I'll assume that your calc was in kg/m^3, and you forgot to convert to grams... NIST has a really nice website for calculating physical properties here: http://webbook.nist.gov/chemistry/fluid/ >From that site, and plugging in 1.013 bar 100.1C we get a density of 0.597 KILOgrams/m^3: Temp(C) Pres(bar) Dens(kg/m3) Vol(m3/kg) Phase 100.10 1.0130 0.59729 1.6742 vapor I think your intuitive sense is on vacation :-)... When you calculated .6g/m^3 as the maximum mass of water in a cubic meter volume, your intuitive sense should have thrown up a red flag and said, "Geez, 0.6 grams is only a few drops at most, and knowing that air can hold more water as temperature goes up, and we're at 100C, I'd think that it could hold much more than a few drops of water." Where? I admit it's possible I might have said .6 g/m^3 somewhere by mistake, but I looked back at my posts and didn't spot it. What is clear though that in all the cases I found (and there were many), I said 0.6 kg/m^3. That's the same as 600 g/m^3. If you took the time to read the post in its entirety, you would know I had it right most of the time. (I still don't know where I made the error.) So, given the correct fact that the maximum mass of water in vapor form at 100.1 and 1013 mbar is 600g/m^3 -- this is WAY MORE than the flow rate I was using. But you clearly didn't understand the argument, if you think it depends on getting the number right. The point was, and still is, that it is always the same, and different from the number you gave. If it's always the same independent of the ratio of steam to liquid, then it can't tell you the ratio of steam to liquid. So please re-evaluate steps 1 to 4 of the methodology given the correct figures... There was never a problem with steps 1 to 4. The problem came after step 6. Try again.
