Talk:Evaporation/Archive 1

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RE: Evaporation of water
The Bowen´s equation was modified by Sartori (1987) who included a parameter that allows the calculation of the three cases of mass flow that can occur when a free water surface is exposed to the air, which situations cannot be calculated with the Bowen´s equation alone. Then the Bowen-Sartori equation becomes:


 * $${Q_{c,pa} \over Q_e} = {0.46(T_p -T_a) \over P_{wp} - (rh X P_a)}{p \over 760}$$

where rh is the relative humidity.

References:

Sartori, E. "A mathematical model for predicting heat and mass transfer from a free water surface". Proc. ISES Solar World Congress, Germany (1987).

Sartori, E. “Solar still versus solar evaporator: a comparative study between their thermal behaviors”. Solar Energy, 56/2 (1996). Sartori, E. "A critical review on equations employed for the calculation of the evaporation rate from a free water surface". Solar Energy, 68/1 (2000).

Sartori, E. "Letter to the Editor", Solar Energy Journal, 73/6, 2003.

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==Does the size of a container effect the rare of which a substance evaporates? ==

I've never understood why a liquid evaporates below the boiling point. If soeone could explain this, it would be a boon to the article. --Spikey 02:27, 27 Jan 2004 (UTC)

Liquids evaporate below the boiling point because temperature (see article for the technical stuff) is basically a measure of the average kinetic energy of the particles. Some particles will be moving much faster, and if certain other conditions are met, will enter the gaseous state. Brianjd 06:52, 2004 Nov 14 (UTC)

Factors influencing rate of evaporation
Shouldn't surface area be included? Brianjd 06:52, 2004 Nov 14 (UTC)

Yes, the area also influences the amount of evaporation, but normally the evaporation rate is given per unit of area. The air temperature is another parameter that influences the evaporation rate. The evaporation decreases as the air temperature increases. (Ernani Sartori, Aug 07, 2005).

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I don't know how "concentration of other substances in the air" would affect the evaporation rate - is there a reference for this, or is this correct? If this is true, it would be nice to have a simple explanation of why. (Anonymous).

Dear Anonymous,

I don´t know what you mean by "concentration of other substances in the air", but if this refers to solid particles then it might occur the following, as I wrote in http://noparadoxes.tripod.com:

"In a fictional limit case, where the sky would be completely covered and opaque as well as “solid” due to the polluting solid particles in suspension, no energy would be reflected and transmitted by this cover, the solar energy would be only absorbed and part of it subsequently transferred by radiation emission and free convection to the interior of the atmospheric greenhouse". And the evaporation would be reduced, although the temperatures would increase. (Ernani Sartori, Apr 16, 2007).

---

I also wonder whether the rate of water evaporation is in any way dependent on the chemical composition of the gas into which the water is being evaporated. For example, given the same temperature and atmospheric pressure, and assuming 0% humidity (i.e., no water vapor already in a volume of gas), will the rate of water evaporation into the gas (or the amount of water vapor the gas can hold before condensation occurs) always be the same if the gas is 100% oxygen as if it is 100% nitrogen or 100% carbon dioxide? Embram (talk) 20:47, 3 December 2012 (UTC)

Articles in Wikipedia helping with evaporation concepts
Vapor pressure | Water vapor ... and other links on those pages will help as well --Hard Raspy Sci 03:20, 13 Feb 2005 (UTC)

The sentence "water boils at 100C at STP" can't be true because if it was 100C, it wouldn't be STP. Could someone fix that. I do not know the right answer for it so i can't really fix it


 * Good point. I have changed it.--GregRM 17:49, 4 November 2005 (UTC)

When we say a phenomenon occurs at stp it means the surroundings external to that of the system under consideration where the phenomenon takes place satisfies stp conditions. So when u say water boils at 100 C at stp it is correct because what we actually mean is Only the temperature outside the vessel of water is at stp and the temperature of the water in the vessel is at 100 C. If the temperature/pressure of the atmosphere around the vessel was to change then the boiling point of water in the vessel would also change. So the statement water boils at 100 C at stp is correct.


 * The anonymous explanation is correct, but limited. Thermodynamics determines evaporation, so each molecule that has evaporated, or leaves the liquids surface, has reached the boiling point.  While at STP, also not necessarily that the average-liquid-mass-temperature is anywhere near boiling.  Evaporation is only a surface thing!  Boiling, typically, refers to the average temperature of a liquid. - HRS IAM 01:07, 12 August 2007 (UTC)

Freezing (ha-ha!) the evaporation article?
For reasons that I personally can't fathom, the evaporation article attracts the attention of an undue number of vandals. Should we try to somehow limit editing of it? I know that's been done for some controversial topics. The article is already in quite decent shape. Alison Chaiken 17:55, 26 March 2006 (UTC)


 * Good point. Not sure it is worth doing yet.  I'm interested in why people come here to vandalise.  Do they think "Hmm, I feeling some lame wiki vandalism, I know! Evaporation is the article to do."  What goes through these children's minds? njh 22:39, 26 March 2006 (UTC)

RE: Factors influencing rate of evaporation
The third bullet item read: "Concentration of other substances in the liquid. If the liquid contains other substances (such as salts), it will have a lower capacity for evaporation. This is due to osmotic pressure."

The linked article on "osmotic pressure" correctly states that "Osmotic pressure or turgor (also called turgor pressure) is the pressure produced by a solution in a space that is enclosed by a differentially permeable membrane." Since there is no differentially permeable membrane present in simple evaporation, the connection is in error.

The writer may have confounded osmotic pressure with Raoult's Law which describes the dependence of the vapor pressure of a solution upon its composition.

Definitions of "evaporation" and "vaporization"
A couple of of months ago, an addition to the introduction (which remains in the current version of the article) said that evaporation was one of two forms of vaporization. I am assuming that the author intended that the other form of vaporization be sublimation. However, the next sentence seems to imply that sublimation is one type of evaporation. Does anyone know the technical definitions for the terms "evaporation" and "vaporization" and whether they include sublimation? (Note also that vaporization is a redirect to the evaporation article.)--GregRM 23:12, 11 June 2006 (UTC)
 * Also, I am questioning the relationship of "boiling" and "evaporation"; the article presents them as being separate phenomenon, but Britannica Online (link to beginning of "vaporization" article) presents boiling as a specific type of evaporation. Perhaps someone could clarify this discrepancy?(Aside: On the issue of "evaporation" vs. "vaporization" that I was wondering about in the above post, the Britannica article seems to suggest that they are equivalent.)--GregRM 02:15, 6 February 2007 (UTC)

Why Doesn't Oil Evaporate? / Influence of Surface Tension on Evaporation
I may have missed it, but it looks like the article does not go into why some liquids evaporate, and some do not. It certainly needs an explanation of this. I believe it's a factor of the cohesive force (and therefore surface tension) of the liquid, with some liquids being so cohesive that their individual molecules never accidentally reach the right state of energy and escape, but I'm not certain that's the only reason. --Kaz 17:35, 8 July 2006 (UTC)


 * As far as I know, vapor pressure is the main factor. I'm not sure about surface tension. However, I just took a look at the article and it seems to suggest that surface tension plays a role, with low surface tension being correlated with high vapor pressure. I'm not sure if this is correct, however. I tend to doubt the validity of a surface tension-vapor pressure correlation. (For instance, I think oil has both lower vapor pressure and lower surface tension than water.) Perhaps someone else could comment on these issues.--GregRM 17:56, 8 July 2006 (UTC)
 * Minor update: I forgot that surface tension can affect vapor pressure (see Surface tension), but as far as I know, this is only significant for surfaces with small radii of curvature like small droplets.--GregRM 19:10, 8 July 2006 (UTC)
 * Also, in response to your original question, I think oil does vaporize, but not very much, due to its low vapor pressure.--GregRM 18:07, 8 July 2006 (UTC)
 * I just found this site, which may be helpful (although the two replies to the question seem to contradict one another).--GregRM 19:42, 8 July 2006 (UTC)
 * As for the issue in the article of whether surface tension affects evaporation, I wonder if it would be more accurate to reword references to surface tension to instead refer to intermolecular forces. This Encarta article links evaporation to intermolecular forces, while this page links intermolecular forces to vapor pressure.--GregRM 20:17, 8 July 2006 (UTC)
 * Surface tension is nothing but the effect of the molecular cohesion of the liquid, at the point where that liquid ends. In other words, the molecules pull at each other, which is why stay a "liquid" instead of flying off to become a gas. Where the liquid ends, like the surface of water or oil meeting air, the top layer of the liquid's molecules are pulled at from behind, and from each side, by each other. But they are not pulled at from the "air" side. This creates a stronger layer of apparent adhesion toward the liquid, ergo the surface tension.


 * That's all it is. But it also must, inherently, have relevance to evaporation, because it's that adherence of the liquid's molecules which both creates the surface tension, and yet must be overcome in order for a molecule to escape and join the gas past the surface tension itself. --Kaz 05:44, 9 July 2006 (UTC)


 * I'm not so sure that surface tension plays a significant role in many cases. For instance, surface tension is related to the energy needed to create new surface area, and surface area does not necessarily change when a liquid is evaporating.  I will try to look into it some more to see what I can find. (I recently found this, but I haven't looked at it closely yet.)--GregRM 14:53, 9 July 2006 (UTC)
 * I haven't been able to find much in the way of references, but I still think the effect of surface tension is questionable. The way I look at it, intermolecular forces affect both surface tension and vapor pressure, but it is vapor pressure that is the relevant quantity for evaporation; surface tension can effect vapor pressure in cases such as small droplets (as I noted in a previous reply), but in these cases, surface tension promotes evaporation rather than inhibiting it. Minor note: I found this article which seems to suggest that surfactants reduce evaporation by reducing surface tension. However, I do not really see the mechanism by which this would occur (I would think the monolayer would instead act as a physical barrier).  Note however, that I have not really given much effort into locating/reading the references that this article cites for this. I will look into adding an expert tag or finding some other way to have others weigh in on this discussion.--GregRM 02:20, 11 July 2006 (UTC)

In the interrum, what can we say about why oil, mercury, et cetera don't evaporate? Does it not, for example, have a boiling point? I'm guessing it does (I know mercury does, and at some point of course something would happen to the oil), so then why aren't individual molecules happening to bump into each other in such a way that one of them temporarily is given enough energy to break free, as with water? --Kaz 17:37, 11 July 2006 (UTC)
 * Yes, I think you are right about the boiling point (although there are some materials that may chemically degrade before reaching a high enough temperature to boil). Higher boiling points are generally (but not always, because the slopes of the ln(P) vs 1/T curves can also be different) associated with lower vapor pressures at a given temperature, leading to reduced tendency to evaporate. As I recall, larger/higher molecular weight molecules tend to have higher boiling points because of greater intermolecular forces; this probably explains the high boiling point of oil (which would be triglycerides in the case of vegetable oil or probably appropriately-sized alkanes in the case of petroleum-type oil). There are other trends, too...for example, I think more branched hydrocarbons tend to have lower intermolecular forces than "straighter" hydrocarbons of the same molecular weight, leading to lower boiling point.  Also, I believe that one of the factors that determines if a fat/oil is solid or liquid at room temperature is the presence or absence of "kinks" in the molecules caused by double bonds (unsaturation). The molecules with the kinks are not able to develop intermolecular forces as high as the flexible, saturated molecules and tend to be liquid, rather than solid at room temperature. I am not as familiar with metals, but I am thinking that the strong bonds ("interatom forces", analogous to the intermolecular forces above) that make most metals a solid at room temperature, persist to a more limited extent after the solid passes its melting point. Such forces result in high boiling point/low vapor pressure. Although water is a small/light molecule, the polar nature of water (and its associated tendency to form hydrogen bonds) keeps it a liquid at room temperature.  Molecules with comparable weight that are nonpolar, like methane, have such a low boiling point that they are gas at room temperature. Keep in mind that this is my interpretation of the mechanisms at work, and I could have remembered some facts incorrectly and I could be wrong in these descriptions for oils, metals, water, etc.--GregRM 00:48, 13 July 2006 (UTC)
 * OK, at the very least, I do think we can safely say that it's a high boiling point which makes evaporation appear negligable at room temperature, right? Does this mean that if you were to heat mercury up to near its boiling point, it would begin to evaporate? I suppose the relative temperature of the air is a factor; if the air is still room temp, then any almost-escaping mercury molecules would probably be cooled before they'd succeeded, but if the air were near its boiling point, they might succeed in escaping. Oil probably does change chemically before it reaches its own boiling point. --Kaz 20:01, 13 July 2006 (UTC)
 * The way I look at it, vapor pressure is still the relevant quantity for evaporation at a given temperature, but boiling point provides a general rule of thumb for tendency to evaporate. I think significantly different molar heats of vaporization (or other complicating factors) can reverse this trend (vapor pressure can be approximated by the Clausius-Clapeyron equation...see this, for example). The Clausius-Clapeyron equation predicts a that a plot of ln P (where P is the vapor pressure) versus 1/T (with T being absolute temperaute) will be a straight line, with slope determined by the molar heat of vaporization. (Therefore, for example, if two pure compounds have a normal boiling point at the same temperature, the Clausius-Clapeyron equation predicts that their vapor pressures will differ at other temperatures if their heats of vaporization are different.) Mercury will evaporate to a greater degree near its boiling point, but I think it's vapor pressure can have important consequences even at room temperature; for instance, see this report of an apparent mercury vapor poisoning and this warning. The Mercury (element) article has a table of vapor pressures at various temperatures. (Note that a pure compound's normal boiling point occurs when the vapor pressure reaches normal atmospheric pressure.)--GregRM 23:35, 13 July 2006 (UTC)

May somebody be interested in the fact I made several oil evaporators? Intended to remove oil from solids or higher boiling liquids. Just a question of boiling point - call it vapour pressure, I don't see that much of a difference. The main difference is that oils, being usually not polar, do not dissolve much, an therefore the typical treatment is a liquid-liquid separation like the classical API separator. Other products, like monoethylenglycol, not an oil for chemists, but viscous and sticky, are typycally evaporated for example in natural gas treatment plants.--UbUb 14:27, 27 August 2006 (UTC)

I decided to go ahead and remove references to surface tension, which, as I mentioned in some of my above posts, seemed to be inaccurate. If I am mistaken, please correct. (Assuming my edit is accurate, if anyone knows of a better reference, please feel free to replace the one I used.) Thanks.--GregRM 02:03, 6 February 2007 (UTC)

But why??????
{helpme}
 * Why does evaporation occur?
 * Shouldn't this be added ??
 * I do not know why that is why I am asking.
 * --The one and only Cloud

entropy and enthalpy and evaporation
The following statement is incorrect

"Gas has less order than liquid or solid matter, and thus the entropy of the system is increased, which always requires energy input."

It is true that gas has less order than liquid or solid matter and evaporation would lead to an increase in entropy. However, an increase in entropy does not require energy input, a decrease in entropy requires energy input. The change in Gibbs free energy is equal to the change in the enthalpy minus the temperature times the change in entropy. So an increase in entropy will lead to a decrease in the Gibbs free energy while a decrease in the entropy will lead to an increase in the Gibbs free energy.

I believe that evaporation is endothermic because of the enthalpy change on evaporation- the enthalpy is equal to the internal energy plus pressure times volume. I think the internal energy change when you move from the liquid state to the gas state should be the thing that requires so much input heat as you are breaking many intermolecular interactions like hydrogen bonds in water or van der Waals forces in nonpolar liquids.

check out

http://en.wikipedia.org/wiki/Heat_of_vaporization

In any event, please fix the part about increasing entropy requiring energy input. Decreasing entropy requires energy input.

Forced Evaporation
I believe forced evaporation is drying, not distillation. Distillation is a flash process and does not involve running hot air over an entrained liquid to separate it. It's similar, but I think a better example of forced evaporation is industrial drying, such as in the food science and technology industry (to expediate sun drying, reduce the capital costs involved in storage of fresh foods) and the pharmaceutical industry (the final process step of patch production). —The preceding unsigned comment was added by 68.32.27.194 (talk) 19:33, 15 December 2006 (UTC).

Drying is really evaporation from a solid, and distillation is evaporation from a liquid mixture. Most air-drying processes do indeed use forced convection, and are thus good examples of forced evapoation, but there are a lot of dryers where the heat comes from a hot surface and there is no forced airflow. Hyperman 42 01:05, 20 December 2006 (UTC) ewwww


 * Uh-hem, a dessicant is a dryer, but it does not use heat...or require the exchange of heat... HRS IAM 01:16, 12 August 2007 (UTC)

Evaporation of water
I'm dumping this text which is repetitive and seems out of place for the moment. Evaporation is when a liquid turns into a gas because of heat energy allowing the atoms to escape. However, when levels of heat energy decrease, the gas cools down, thus condensing or turning into a liquid. For example as the tea cools, the water vapor condenses. An example of both condensation and evaporation is a nuclear reactor which smashes the molecules, produces heat energy to heat the water and turns it into water vapor by evaporation to move turbines, and when the water vapor moves out of the turbines it condenses. Another example of both is when you have a shower or a bath. When the water in the bathtub heats, it starts to evaporate but when it comes in contact with the cooler surface of the mirror, it condenses.

The ratio of the heat loss from a pond by evaporation to the heat loss due to convection, independent of wind speed, is given by:


 * $${Q_{c,pa} \over Q_e} = {0.46(T_p -T_a) \over P_{wp} - P_a}{p \over 760}$$

where $$Q_{c,pa}$$ is the heat loss from the pond by convection, in W/(m2·K), $$Q_e$$ is the heat loss from the pond by evaporation, in W/(m2·K), $$T_p$$ and $$T_a$$ are the Kelvin (or Celsius) temperatures of the water and air, and $$P_{wp}$$ and $$P_a$$ are the vapor pressures of the pond surface and air, and $$p$$ is the barometric pressure, with all pressures in mm Hg. (Bowen, 1926) --Rifleman 82 20:43, 4 February 2007 (UTC)

I removed references to solids evaporating and edited the initial description to mention sublimation as the equivalent process in solids. Evaporation is, unless I'm sorely mistaken, not a catch all term for both processes, it is essentially the same as sublimation but still only used in reference to liquids.