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 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? > 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. Or maybe I'm just tired. I should go to bed.
