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.
