On Feb 21, 2011, at 1:40 PM, Joshua Cude wrote:
On Mon, Feb 21, 2011 at 11:52 AM, Horace Heffner
<[email protected]> wrote:
On Feb 21, 2011, at 5:50 AM, Joshua Cude wrote:
|One should also bear in mind that it takes only 2% steam by mass
to make up 97.5% of the expelled fluid by volume. And |since the
steam is created in the horizontal portion, it is forced up 50 cm
of pipe through liquid, which would |presumably turn the liquid
into a fine mist after a few minutes.
The above appears to to be a typo. It was probably meant to say:
"One should also bear in mind that it takes only 2% steam by
*volume* to make up 97.5% of the expelled fluid by *mass*.
| Well maybe a question of semantics, and some rounding errors.
| Try this: It takes only 2% of the H2O by mass, in the form of
steam, to make up 97% of the expelled water by volume.
Better. It is a matter of definitions. However, I think "2% steam
by mass" in your original statement means "2% wet steam" that is to
say 2% of the mass is water, 98% is steam, by mass. It wouldn't make
any sense vice versa, i.e. 2% by mass vapor, and 98% mass in liquid.
Such a "2% wet by mass steam" takes 98% of the vaporization energy to
create vs dry steam. What I provided were the numbers for 2% wet by
volume steam, that is to say 2% of the volume of the ejected fluid
being liquid. It is notable that the instrument used to measure
dryness actually measures the capacitance of the steam, and
capacitance is a function of the proportion by volume of liquid, not
by mass. This is why I produced the formula and table that convert
from portions by volume to portions by mass, back in January:
http://www.mail-archive.com/[email protected]/msg41703.html
That said, let's proceed on with your defined problem where 2% of the
water is vaporized, i.e. the ejecta is 98% liquid by mass, 98% wet by
mass.
|For an input flow rate of 300 cc/min = 300 mg/min,
The above should read g/min, i.e. grams per minute, not milligrams
per minute.
|2% of the water by mass means .02* 300 = 6 mg water per minute in
the form of steam.
Again, 6 grams per minute of steam vapor, not mg.
|The density of steam at 1 bar is .59 micrograms/cc, so that
amounts to 10,000 cc/minute steam.
I used density of steam at 100 C and 760 torr: 0.6 kg/m^3 = 0.0006 gm/
cm^3
http://hypertextbook.com/facts/2007/DmitriyGekhman.shtml
"Metz, Clyde R. Schaum's Outline of Physical Chemistry. McGraw-Hill,
1988."
""The density of steam at 100 °C and 760.0 torr is 0.5974 kg m-3."
This means the 6 gm/min represents (6 gm/min)/(0.0006 gm/cm^3) =
10,000 cc/minute, and we agree on that, so you used the rounded up
number as well, plus for you two wrongs in units (mg/cc vs ug/cc, and
mg vs g) made a right. The correct value for steam density is
5.974x10^-4 gm/cc, not 0.59 micrograms/cc. Maybe you mistook an
abbreviation for milligrams for micrograms in some reference?
The remaining liquid, 294 mg/ min = 294 cc/min, therefore makes up
2.8% of the volume.
The proportion of liquid in the total volume expelled, given your
definition of "2% of the H2O by mass" is 294/(10,000+294) = 2.856%.
The steam in this case is 2.856% wet by volume.
It is also neatly true, that if the total volume expelled is 2% wet
by volume, then the *vapor* by mass is 2.856%.
"If x is the liquid portion by volume, then x/((x+(1-x)*0.0006)) is
the portion by mass. If steam is 2% wet by volume, then x=0.02 and
the portion by mass is 0.97144, or 97.14%. It then takes only 2.856%
of the heat to produce the wet steam vs dry."
As a double check on concepts, if you plug x=0.02856 into x/((x+(1-x)
*0.0006)) then you get 0.98. That is to say, 98% of the mass of the
volume expelled is water, and 2% steam - your starting assumptions.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
