# Evaporation of Water Mass Transfer

1. May 18, 2012

### Red_CCF

Hi

I had a question regarding an example about mass transfer. I already have the solutions to this but there is one thing I am confused about.

The problem just asks the rate of evaporation from the water-air surface to the atmosphere at a given temperature, pressure, and relative humidity. My question revolve around the fact that in the solutions, they say that the air and the water-air interface is saturated with water vapor and that the vapor pressure at the surface is the saturation pressure of water. I don't understand why this is valid?

Thanks

2. May 20, 2012

### rude man

On the surface the vapor pressure must equal saturation pressure, since otherwise there would not exist equilibrium conditions at the surface. But note that while saturation pressure is known if T is known, that is insufficient information to determine the rate of evaporation.

3. May 20, 2012

### Red_CCF

Hi

I'm still not getting this, so why would equilibrium not hold at the surface unless the relative humidity just above the water surface is 100% saturated with water vapor?

Thanks

4. May 21, 2012

### rude man

Because if the vapor pressure just above the surface is less than the saturated vapor pressure the water would continue to evaporate until the vapor is saturated. That's what the saturated vapor tables tell you.

OK, there is one caveat: if there is unsaturated water vapor air blowing at a good clip across the surface then equlibrium would indeed be such that surface vapor pressure would be kept at the vapor pressure of the wind.

There is a formula available to account for a finite wind velocity. It's strictly an empirical formula, far as I know. It's in some handbook or other.

Last edited: May 21, 2012
5. May 22, 2012

### Red_CCF

Hi

Thanks for the help

This may stem from my a poor understanding of thermodynamics, but I never understood why steady state cannot be reached such that the rate of evaporation from the surface water vapor into the atmosphere (assuming relative humidity is <100% far away) is equal to the rate at which liquid water at the surface turns into water vapor when the relative humidity right next to the surface is not 100%. In short, why must the relative humidity of water vapor near the surface be 100% before steady state can be reached? Why not at a lower relative humidity?

This doesn't have anything to do with the OP but, if a fan is blowing air over the water surface, would the total pressure of the air be 1atm (same as if there is no air blowing over the surface) at one or both the inlet and exit of the fan? If so I don't understand how this satisfies Bernoulli's principle if the air drawn from the atmosphere that is 1atm pressure

Thanks

6. May 22, 2012

### rude man

If the wind is sufficiently high (which isn't very high, actually) then the vapor ppressure just above the surface woud be that of the vapor pressure of the ambient air. Atmospheric pressure is not directly relevant. Atmospheric pressure is mostly due to nitrogen and oxygen content, not water vapor.

Bernoulli applies to total pressure, not just vapor pressure. Also, Bernoulli derives from mass and energy conservation, whereas a fan imparts a new source of energy to the air flow. I'm not sure how Bernoulli fits into this.

7. May 22, 2012

### Red_CCF

Hi

I was actually referring to total pressure not vapor pressure. If I have a fan and I pick three points, one point far away (such that I have V = 0, P = Patm), and the other two points at the inlet and exit of the fan, ignoring elevation changes, is either the inlet and/or exit pressures going to be equal to Patm? I'm considering the modified Bernoulli that includes head losses and work input.

I was also wondering if you can explain why steady state evaporation cannot occur if the RH of water vapor just above the water surface is not 100%?

Thanks very much

8. May 23, 2012

### rude man

At the moment I really don't feel competent to answer your question on Bernoulli. I'd have to review all that. However, the air RH is not affected by the fan. The fan just provides a steady-state vapor pressure equal to ambient at the surface. In the absence of the fan a thin layer of air just above the surface would have a vapor pressure approaching 100% RH.

So, to answer your second question directly - steady-state evaporation will occur if the air vapor pressure is below saturated, i.e. below 100% R.H. I hope I never indicated otherwise. Simple: evaporation occurs if the air is unsaturated, and won't if it's saturated.

Evaporation is essentially thermodynamically a state of disequilibrium, in the sense that whatever water you have gets evaporated slowly until it's gone. But if you look at a p-V diagram for water, the region between the saturation curves represents the equilibrium situation - where p and T are constant and the percentage of water vapor vs. liquid (the "quality") varies with the volume. A dish of water sitting on a shelf has the problem that volume is essentially infinite so equilibrium between liquid and vapor states is never reached.

BTW something to think about - if the R.H. is 100% and the air is at 100 F you'd die, unless you had access to water well below 100F! (Here in Phoenix, AZ we have recorded 122F = 50C. Fortunately our R.H. is seldom above 40% and we do have ice water, not to mention air conditioning!

9. May 23, 2012

### Red_CCF

Hi

As for my question regarding evaporation of RH = 100%, I was referring to the thin layer of air just above the water surface not the air far away. So is it possible for the thin layer of air above the water surface to have like RH = 90% or something and still be at steady state evaporation (assuming that the air far away has RH << 90%)? If so I don't get why my textbook made the assumption that this thin layer of air is saturated.

With regards to the second paragraph, when you refer to p-V diagram, does V represent volume or specific volume? When you say the region between saturation curves, do you mean the region where phase change occurs (the horizontal line between saturated liquid and saturated vapor)? Also, why is the volume of the dish of water infinite and why would this affect equilibrium?

My apologies for being so slow at getting this

Thanks very much for all your help

10. May 23, 2012

### rude man

You know, I'm not sure how to explain how the air immediately above the water can be at 100% humidity. There has to be < 100% for evaporation to take place. The layer must be very thin so that some water molecules can penetrate it & get evaporated just beyond the thin 100% layer. Maybe I don't see the complete picture here either.

I said V which usually denotes absolute volume. Obviously, any chart with numbers on the x and y axes are going to be specific volume and pressure respectively.

Yes, I mean the area on the p-v diagram where phase change occurs. Take any horizontal line; there T and p are constant from 0% quality to 100% (I defined quality before). Equilibrium can exist anywhere along that line with no evaporation taking place. v determines where you are along the horizontal line. Now, think of expanding v isothermally (you have to add heat to do this). Soon all the liquid is gone & we have 100% quality. If you further increase v isothermally, the vapor pressure starts to drop per your p-v diagram (just follow the isothermal). If you now were to add liquid water the liquid would evaporate.

Seems to me it's obvious the sky is essentially infinite volume? Like, what contains it if it isn't? They say the universe is infinite .... Anyway,mentally follow the isothermal out to large v and it's obvious you can't have any liquid water. So if you add some, like in a standing dish, it has to evaporate.

I forgot to add - you should feel good about asking lots of questions. Wanting to really know is what differentiates a poor student from a good one. Be curious and don't settle for anything less than the complete picture, or as complete a one as your level of studies allows.

11. May 25, 2012

### Red_CCF

Hi

I looked at some more examples in my textbook and discovered that the boundary condition of RH =100% at the air and water interface was used religiously for all problems. This includes steady state evaporation (no air flow), transient water vapor transfer (in an example of bubble rising from the bottom of a fish tank), and even problems with air blowing over a water surface. The last case seem to contradict what you posted earlier; my book basically used a heat transfer analogy with mass transfer and specified the densities of water vapor across the mass transfer boundary layer to be at RH=100% at the water surface to RH = some problem specific number at infinity.

However, intuitively I don't see any physical reasons why this would be true. They just state that RH=100% at the air-water interface as a fact as if it is obvious so right now I'm not sure if I missed something when I took thermodynamics or if the book is just conveniently assuming this boundary conditions for the ease of calculations.

Do you have any insights as to why this is? Is there a book specifically on evaporation (at the undergrad level) with solved problems you can recommend?

Is this setup similar to a closed piston cylinder that begins with just water at some T and P at 0% quality and allowing it to expand isobarically and isothermally to 100% quality, and then isothermally expand it further? Is the reason that liquid water input will evaporate because at this specific p and T the phase diagram says that water cannot exist as a liquid (provided the piston adjusts as the liquid evaporates to maintain this T and p)? Would the mechanism be evaporation or boiling?

When I was taught why a dish of water has to evaporate I learned it from a mass transfer point of view, whereby a density gradient (from water surface to infinity) drives the water vapor at the air-water interface to leave, and then liquid water will evaporate to replace it (if process is already at steady state). Is this correct as I don't see what drives liquid to just turn into vapor when some vapor above it diffuses to the sky and I don't really get the thermodynamic explanation you provided (why does following v to isothermally mean water has to evaporate and why is this valid as the phase diagram is only for closed systems)?

Thanks very much for all your help

12. May 25, 2012

### rude man

Looking at the layer microscopically, the demarcation between the water and vapor is itself vague. So I would accept the definition of the surface as being that layer at which the vapor is saturated. I'm afraid my knowledge more or less ends there.

That is exactly right. It's not boiling. Boiling is where the vapor pressure equals the total atmospheric pressure + the weight of the water column above where the bubbles form, so that vapor is liberated from within the liquid. Evaporation by definition is a surface process.

I don't know how to answer any better than I did. Yes, you description of mass transfer and density gradient sound right. Imagine your piston as having a very large or even infinite volume, then moving the piston to greater and greater v. It should be obvious that liquid water is an impossibility in the steady state.

Again - evaporation is not in thermodynamical equilibrium, not in the sense that your p-v diagram is. The p-v diagram allows no evaporation to take place at any point on it.

The p-v diagram obviously doesn't extend to v = ∞ but it should be obvious that, as you go to the extreme end of the v axis the impossibility of liquid water is evident. Sorry, I can't do better than that.