On Feb 21, 2011, at 6:50 PM, Stephen A. Lawrence wrote:
On 02/21/2011 09:48 PM, Horace Heffner wrote:
On Feb 21, 2011, at 1:40 PM, Joshua Cude wrote:
On Mon, Feb 21, 2011 at 11:52 AM, Horace Heffner
<hheff...@mtaonline.net> 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.
But that last formulation makes perfect sense, I think,
.. and I said I thought it was better.
and is surely what Joshua wrote.
In Joshua's scenario, each gram of effluent consists of 980
milligrams of liquid water, in the form of tiny droplets taking up
just under a milliliter of total volume, and just 20 milligrams of
vapor, in the form of gas. None the less, the 20 milligrams of
vapor, being enormously less dense, constitute nearly all the
volume of the effluent -- thus, it's 97.5% vapor, by volume,
because the vapor is taking up about 39 milliliters of space, to
the single ml being consumed by the liquid.
By volume, this stuff Joshua is describing would be 2.5% liquid
water, or, one might say, 97.5% dry steam.
Only 2.5% of the mass of the water has been vaporized in this
scenario, so the heat of vaporization required will be about 40
times smaller than that required to fully vaporize the water.
What doesn't make sense? Is it that the expansion factor for
liquid->vapor Joshua used is too large?
No, it is a matter of definitions, as I said.
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.
I think you and Joshua were talking about the same thing, really.
Yes, I merely pointed out what appeared to be a typo - of the kind I
make often, exchanging terms.
Or maybe I'm just tired. I should go to bed.
I think Joshua and I both have a grasp on the basic principles
involved, and both of us know it. I provided both forwards and
backwards calculations of the values in question (but which were
unfortunately cut above), so that should be good enough to
demonstrate that I understand the principles I think. Below are the
values discussed regarding this experiment in tabular form.
Liquid Liquid Gas
Portion Portion Portion
by Volume by Mass by Mass
--------- ------- -----------
0.010 0.9439 0.0560
0.020 0.97144 0.02856
0.02856 0.98 0.02
The problem, to me, centered on the meaning of "2% steam". When this
phrase is used it typically (AFAIK) means 2% wet steam, i.e 2% of the
steam is water. That can be by 2% water of total mass or 2% water of
total volume, but I think is usually expressed in terms of water by
mass. Therefore, when I saw "2% steam by mass", it appeared Joshua
was talking about 2% water by mass, and 98% vapor by mass. I doubt
that anyone normally talks abut 98% steam, especially when talking
about "dry steam", because that quickly will be pure water, i.e. it
is 98% water by mass, and probably unmistakable to the eye as dry
steam. In the case of Rossi's experiment there was some doubt and
discussion about how accurate the measurement could be, because the
value was determined by steam capacitance, and thus might be by
volume. All talk of relative humidity (RH), which the instrument
actually measured in a limited range which did not include 99-100%,
seemed nonsensical when applied to "dry steam". A 1% error by volume
could mean a 94.4% error in heat, and the instrument was rated as
only 2.7% accurate in its valid range.
In any case Joshua's statement did not make sense to me as written,
but made total sense as corrected, given a very small error in the
third place. Note in the table that 2% steam by volume is
coincidentally 97.144 % steam by mass (but not 98% or 97.5%). That
is to say, if 2% of the *volume* of the H2O is liquid, then 97.144%
of the H2O is liquid by mass. This matches up very well with what
Joshua was saying provided *volume* and *mass* are exchanged in his
sentence: "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*." This
is why I thought it was a typo and posted.
Then Joshua posted his calculation, which I followed totally, line by
line, and noted the units mistakes which self corrected by
cancellation. He nicely changed his defining statement and it all
totally made sense to me what his meanings were, which I demonstrated
by my calculations. He was clearly talking about a mixture that was
2% vapor by mass, i.e. 98% wet steam. I showed that this was 2.856%
wet steam (2.856% liquid) by volume, and thus 97.144% gas by
volume. I further showed my formula was correct and other
calculations correct by back calculating the 98% wet steam number,
as shown in the last line of the table above.
As far as I know everyone is in synch with regard to definitions and
calculations. When 2% of the total mass ejected is vapor, then
97.144% of the total volume ejected is vapor. Also, when 2% of the
total volume ejected is water, 97.144% of the total mass ejected is
water, and 2.856% of the total mass ejected is vapor. Steam which
is 2% wet by volume requires only 2.856% of the heat of vaporization
to form compared to dry steam. Steam which is 2% wet by mass
requires 98% of the heat of vaporization to form compared to dry
steam of the same volume. Steam which is 2.856% wet by volume
requires only 2% of the energy required to vaporize compared to dry
steam of the same volume. Steam in which the vapor comprises 2% of
the total mass requires only 2% of the energy to form compared to dry
steam of the same mass. I'm essentially now rephrasing what I wrote
earlier:
On Feb 21, 2011, at 5:48 PM, Horace Heffner wrote:
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/
