OK, gentlemen, now you have a steamless- Wasser uber alles experiment too. Peter
On Tue, Feb 22, 2011 at 9:50 AM, Horace Heffner <hheff...@mtaonline.net>wrote: > > 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><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/ > > > > > -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
