# Evaporation questions about saturation pressure

1. Feb 9, 2014

### Red_CCF

Hello

I have a few fundamental questions about evaporation

1. In Cengel's Heat and Mass Transfer, in all sample and practice problems the author assumed without explanation that for a system such as a pool of water exposed to air, the air immediately above the liquid surface is saturated (100% relative humidity) at the liquid's temperature and thereby gives one boundary condition for mass transfer/evaporation calculations. I was wondering what physically dictates that this be true?

2. For some specified temperature, why is it impossible physically for water vapour to exist at a partial pressure (in air) higher than its saturation pressure?

3. What are the physical mechanisms in which increasing liquid or water vapour temperature leads to higher evaporation rates? My suspicion is that hotter liquid means more energetic surface molecules (and quantitatively found as higher vapour pressure as Q1) and hotter gas means higher diffusion coefficient.

4. For a system like a closed piston cylinder with liquid water and air, keeping pressure constant, is it possible that evaporation rate reaches zero before the relative humidity reaches 100% (ie cold liquid and hot air)?

Thank you very much

2. Feb 10, 2014

### olivermsun

Saturation means the vapor is in equilibrium with the liquid phase at the (flat) interface. So if the saturation were less than 100%, then you'd expect more water to evaporate until it reached 100%.

You're right about the hotter liquid having more energetic molecules near the surface, hence some molecules are more likely to escape.

Hotter vapor molecules above colder liquid will heat the interface, leading to a hotter liquid. I suppose it's also true if the vapor is hotter that molecular diffusion will be higher, so that more heat will flow down to "re-heat" the cooler vapor particles near the surface. So okay, sure.

3. Feb 10, 2014

### dauto

The infinitesimal layer of air in contact with the water has an infinitesimal mass and requires an infinitesimal amount of water and heat in order to reach equilibrium with the water. That process will require an infinitesimal amount of time.

4. Feb 10, 2014

### Staff: Mentor

These are all great questions.
The velocity at the air at the interface with the liquid water is zero relative to the liquid water, by virtue of the non-slip boundary condition. So immediately at the interface, the air is essentially stagnant. This means that the water vapor at the interface will have have to come to virtually instantaneous equilibrium with the liquid water at the interface.
If it got higher than the saturation vapor pressure, it would condense out into droplets.

The driving force for evaporation is the difference between the equilibrium vapor pressure at the interface and the vapor pressure in the bulk of the air. If the temperature goes up, the vapor pressure at the interface goes up, and this results in a larger driving force (for diffusion). This is the main effect.
The evaporation rate goes to zero when the gradient of water vapor partial pressure at the interface goes to zero. That means that the water vapor pressure is not only the equilibrium value at the interface, but also at small distances away from the interface. Further from the interface, if the air temperature is higher than at the interface, the partial pressure can be higher than the equilibrium partial pressure at the interface (without condensation), because the equilibrium partial pressure in the higher temperature air is greater than at the colder interface temperature. So a positive temperature gradient away from the interface combined with a constant or increasing water vapor partial pressure with distance from the interface can stop the evaporation (or even reverse it).

5. Feb 10, 2014

### Red_CCF

Thank you very much for your responses

I'm a bit confused on why stagnant air leads to instantaneous saturation at the interface? Is it the same concept as what dauto described in the post above yours?

I'm wondering how the air "knows" that the saturation vapor pressure has been reached, and why is it that water only begins to condense once the saturation vapor pressure is reached and not at any other pressure.

In the practice problems in my heat and mass transfer book, the author uses the liquid temperature's P_sat as the vapor pressure at the interface and didn't consider the air temperature. Is this because the air and vapor at the interface is essentially equal to the liquid temperature regardless of air temperature to maintain a continuous temperature profile? Does this mean that liquid temperature is the main factor in generating a driving force/mass gradient?

If I wanted to think about this from a mass gradient perspective, would this scenario be equivalent to having a vapor mass fraction at the interface equal to that further away (evaporation = 0), but if the air further away is hotter than at the interface then for the same vapor mass fraction the corresponding relative humidity would be lower than at the interface?

Thanks very much for your help

6. Feb 10, 2014

### Staff: Mentor

Yes. He probably articulated it better than I did.

The air doesn't know from nothing. The air really isn't the key part of the story. The water vapor and the liquid water are the story. The vapor condenses when it is cooled to the point where its kinetic energy is not sufficient to overcome the attractive forces of the molecules.

If you are asking whether the temperature profile is continuous across the interface, then, yes, it is continuous. You can have temperature gradients in the liquid, as well as temperature gradients in the gas phase. But, throughout the liquid and throughout the gas, as well as across the interface, the temperature is continuous.
Yes.

7. Feb 11, 2014

### sophiecentaur

All the above will work with or without the presence of an atmosphere. You just have to include or leave out the partial pressures of the other gases, accordingly.

8. Feb 11, 2014

### Staff: Mentor

Except if you are calculating the evaporation rate, which depends on the rate of diffusion of the water vapor through the air in the vicinity of the interface.

Chet

9. Feb 12, 2014

### sophiecentaur

I didn't consider that. What is the difference, quantitatively? I suppose it could be very significant in a high pressure atmosphere - it would act like a porous brick.

The diffusion factor would apply to the movement of all gases, presumably.

10. Feb 12, 2014

### Staff: Mentor

Quantitatively its very important, even at ideal gas pressures (e.g., 1 atm). The diffusivities of gases through other gases is low, even at these pressures. By Fick's law, the molar flux of water vapor (per unit area) is equal to the diffusivity of water vapor in air times the concentration gradient of the water vapor in the air, where the molar concentration is equal to p/RT, with p being the partial pressure of the water vapor.
Here is an example that I got from google:

So, in practice, the evaporation rate is typically "mass transfer limited," as we ChEs call it.

11. Feb 12, 2014

### sophiecentaur

I looked at that link (too hard for me, actually) and a few others (noddy level). It seems that the diffusion factor is around 10e4 less for air/water vapour than for air dissolved on water. To me, that implies things would happen ten thousand times quicker in gases than in liquids. Sounds reasonable. Fast but not instantaneous.
This will account for the fact that your washing dries a lot quicker on a windy day, I suppose; the water takes a long time to diffuse away in stationary air. Also this must be what is at work with decompression sickness, when the bubbles that are formed after decompression may take weeks to go away, even when a dive has only lasted a few minutes.

12. Feb 13, 2014

### Red_CCF

So the presence of air in this case does not matter?

I'm wondering if I'm thinking about this correctly. If I had a closed container containing a (pure) liquid-water vapour mixture at some T, the pressure of the vapour phase would be P_sat since the system is at saturation. If an infinitesimal amount of liquid were to evaporate, the vapour pressure would exceed P_sat (by an infinitesimal amount) which would be thermodynamically unstable, leading to condensation so pressure goes back down to P_sat. For this same system but air is introduced (and the container size changed such that the water vapour partial pressure still at P_sat), the same would occur for the same reason regardless of the total pressure due to other gases be it 1atm or something else?

Would taking the water vapour properties at the liquid interface using the liquid's temperature (instead of the air) be more accurate then?

Thanks very much

13. Feb 14, 2014

### Staff: Mentor

Yes. As far as the equilibrium is concerned, this is correct.
Yes, as long as the pressure is not so high that you have to take into account non-ideal effects.
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The temperature is continuous across the interface, so the liquid's temperature is equal to the air temperature at the interface. However, you can't use the air temperature in the bulk of the air away from the interface. If the temperature is varying spatially either in the air, the liquid, or both, you need to include heat transfer effects in the analysis.

14. Feb 14, 2014

### Red_CCF

Sorry but I got confused by this concept again. I still can't understand the connection between a layer of stagnant air and instantaneous saturation right above the liquid interface.

What kind of effects would these include, and qualitatively speaking how would one account for these effects in evaporation calculations? What pressure range would be considered "high"?

Would heat transfer effects only matter when the system is not in steady state and the interface temperature profile are time varying? Is spatial variation referring to temperature variations in a plane parallel to the interface?

Thanks very much for all your help

Last edited: Feb 14, 2014
15. Feb 14, 2014

### Staff: Mentor

The effects would involve mostly the gas phase. The gas phase represents a solution of air and water vapor. At high total pressures, the total pressure influences the partial molar free energies (chemical potentials) of both the water and the air, and these are not simply determined by the total pressure times their mole fractions. They need to be corrected for non-ideal gas behavior and non-ideal solution behavior. At equilibrium, the chemical potential of water in the gas phase must match the chemical potential of water liquid. So the equilibrium of the water vapor with the liquid water is influenced by the total pressure, unlike at low total pressures. (There is also a small effect of the total pressure on the free energy of liquid water). Pressures at which the ideal gas law is not a good approximation for describing the behavior of water vapor would be considered "high." This would be more than about 10 atm.

As far as how to take this kind of thing into account, this would not be an appropriate venue for describing it, because too much material would be involved. Check out chapters 10 - 15 in Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.

Heat transfer effects matter when the system is not at equilibrium. Steady state means that things are not changing with time. It is possible for a system to be at steady state without being at equilibrium. So just because a system is at steady state does not mean that heat transfer effects do not matter. Spatial variation refers to any kind of variations of temperature with respect to location in 3D space. Such spatial variations guarantee that the system is not at equilibrium (and heat transfer is occurring).

16. Feb 15, 2014

### Red_CCF

Does the book you recommend also go through (at an undergrad level) the qualitative theory of real gases that you described above in more detail as I'm unfamiliar with the concept of chemical potentials, free energies etc.? Also, the effects mentioned above apply regardless of whether the gas phase contains air or purely water vapour or some other mixture? Do the steam tables/phase diagrams take into account these effects?

If the system is already in steady state but not thermal equilibrium, the temperature profiles along the interface would be the constant with time just not with space. In this case, how the fact that heat is being transferred change the calculations if the temperature at each point along the interface and of the system is constant w.r.t. time?

I'm also wondering for two identical systems (i.e. two glasses of water) with the same temperature exposed to the same atmosphere of air, if one glass was salt water and the other pure water how would that change the evaporation and interface water vapour partial pressure at steady state?

Thanks very much

17. Feb 15, 2014

### Staff: Mentor

The book I recommended is the one my daughter used as an undergraduate at the Univ. of Pennsylvania. For pure water vapor, of course, all you need to know (for vapor-liquid equilibrium) is the vapor pressure as a function of temperature (i.e.,for the purposes you were referring to). However, for air-water vapor mixtures, you need to understand the thermodynamics of multicomponent systems.

The steam tables and pure water phase diagrams are consistent with the non-ideal high pressure concepts I alluded to previously. But, to do mixtures at high pressures, you need to include the concepts related to multicomponent solution behavior. The psychometric charts for air-water mixtures are consistent with these concepts.
I don't understand this question. Sorry.

The vapor pressure of water vapor in equilibrium with salt water is lower than the equilibrium vapor pressure of pure water. So the interface partial pressure would be less. This would reduce the driving force for water vapor evaporation into the gas phase. So, all other things being equal, the rate of evaporation would be less.

18. Feb 16, 2014

### Red_CCF

My apologies for being a bit slow at grasping these concepts.

Sorry but I got confused by this again, what is the connection between stagnant air and near instantaneous saturation at the interface?

Is this because with knowledge that the system is a pure two phase mixture at saturation, P_sat becomes a function of T only, but specific volume is still needed to define the quality/state? Also, since psychrometric charts are generally at 1atm, would assuming ideal gas be a reasonable approximation to the chart's values?

I meant that if the system is at steady state, the temperature at each point along the interface is constant with time, so knowledge of how much heat is transferred wouldn't be needed since it is temperature that dictates the interface vapour pressure and evaporation which is constant?

For salt water, does this mean at the interface the vapour partial pressure is not equal to P_sat of pure water (at the salt water's temperature) and that there is <100% relative humidity at the interface unlike for a cup of pure water? Why does adding a solute cause this difference in interface partial pressure?

Thanks again

19. Feb 16, 2014

### Staff: Mentor

I think that Dauto explained this adequately in post #3.
Yes. Specific volume averaged over liquid and vapor.
yes. That's pretty much how they prepare these charts.
Yes, but to get the evaporation rate, you still need to know the partial pressure gradient. Also, the system has to be at thermal steady state so that the heat of vaporization is being supplied continuously at the interface. This dictates the relationship between the temperature gradients (normal to the interface) in the gas and liquid phases at the interface.
Yes.
If it was 100% salt, the water vapor partial pressure would be zero. Part of the answer is that there are fewer water molecules per unit area at the interface, and part of it is related to the energetic interaction between the water molecules and the ions of sodium and chlorine in the liquid phase.

20. Feb 18, 2014

### Red_CCF

Hi

One last question, does this mean that if I left a large cup of salt water in a closed box (everything isothermal), the partial vapour pressure in the bulk air will never reach saturation pressure corresponding to the temperature of the system unlike if the cup was pure water?

Thanks very much for all your help