At 06:50 PM 6/22/2011, Mark Iverson wrote:
RE: Abd's comment about, "the instrument does not provide mass of
water as vapor, unfortunately,
that's not what it shows. It reads in g/m^3. To convert this to mass
we'd need to know the volume,
eh?"
NO, you do NOT need to know the volume to calculate the mixing ratio
if you have measurements of the
proper other variables.
Uh, g/m^3 *is* the "mixing ratio." If you have g/m^2, you have the
ratio already. But ratio of *what*?
It appears that what the instrument is presenting is a figure for
water vapor as weight of water per cubic meter, neglecting any liquid
water present, which it has no way of measuring directly.
In atmospheric science the mixing ratio is a very common and key variable.
It is always CALCULATED from other more common variables which do
NOT include flow rate or volume.
This is gas law and in gas law, concentrations do not matter... Its
all about mass or Mole fraction.
Again, fraction of what. Using CAPITAL LETTERS does not increase
cogency, the opposite.
The desired number here is how much water is exiting the E-Cat as
water, rather than as water vapor. That way, this can be deducted
from the water which is presumed to have evaporated, in order to
calculate the generated heat.
The instrument has no way of measuring this water. Some instrument
that determines the total grams per cubic meter of liquid water still
would not do it, unless we knew the total, or the "mixing ratio," and
the mixing ratio for water and steam is not a fixed value, and it is
not determinable, it seems, from relative humidity and temperature,
for wet steam has an RH of 100%, at 100 C. No matter how wet it is.
I will provide the explanation shortly, but I have to run an
errand... I may not get to it this eve,
but I will definitely respond.
I do suggest more thinking and less leaning on the caps lock.
