Dear David, The evaporation rate depends on the exact conditions in the experiment. This is not a fluid property. The environment of the liquid should supply heat by conduction and/or radiation and/or dissipation (electric for example). Of course a layer of vapour will for around the surface of the liquid, which will partially bloc the evaporation.

In quite/slow evaporation: At the surface vapour saturation will be reached. The evaporation will depend on the rate at which the vapour can be evacuated by diffusion towards the environment. In stagnant surroundings this can be a very slow process. When however a "dry" gas is flowing over the liquid surface a much thinner layer of vapour will form, referred to as a diffusion boundary layer. The diffusion towards the surroundings also depends on the density and nature of the gas (diffusion coefficient of the vapour in the surrounding gas).

If the liquid is sufficiently overheated (temperature above "normal" boiling temperature), vapour bubbles form in the liquid first at the wall surfaces and at even larger overheating bubbles of vapour form in the bulk of the liquid around "dust" (nuclei). When boiling an bubble formation is rapid, the vapour escape the liquid through flow induced by gravity and if this is rapid, this will drastically change the surrounding (creating its own flow) so that diffusion in boundary layers is not controlling the evaporation any more. The flow induced by bubbles within the liquid will also affect the heat transfer from the energy sources to the liquid.

Finally when the liquid is very pure (clean) or the energy transfer is very efficient the overheating can become so large that the overheated liquid simply explodes into a vapour cloud. This vapour explosion can occur when you put a cup of water in the micro-wave oven. It is therefore extremely dangerous to heat clean water in the micro-wave oven.

I hope I convinced you that the question you had, has no single answer.

You should describe the experimental condition more carefully.

Yours sincerely

Mico

An interesting experiment is to drop some liquid on a very hot horizontal metal plate. The liquid will form droplets that are boiling but still "surf" above a vapour cloud formed by evaporation. This vapour cloud actually reduces the heat transfer from the plate to the droplets. Because of this reduction of heat transfer the droplets can "live" for some time "running" around the surface of the plate. This illustrates that bubble formation can decrease heat transfer from an energy source, while the flow induced by rising bubbles will in general increase the heat transfer from walls of a pan towards the boiling liquid.

Mr. David Wyatt

greeting ....

I believed in the scientific opinion of Mr. Avraham Hirschberg

The vaporization rates depending on many coefficient, one of them the physical properties of fluid other coefficient represented by the satiation of fluid stagnation in vessel or flow in pipe or shell, over pressure or under vacuum pressure, the fluid under heating condition or not .... these coefficients and more can be affected to the vaporization of fluid.

In evaporates the mass flow rate, working saturated temperature with corresponding saturation pressure can be affected to the rate of vaporization if the heat gain enough to evaporate the liquid under T and P.

also in the vessel (stagnation fluid) can be vaporization in hi rate if its work under vacuum pressure with heat gain.

you can fined more details in the references below

-(introduction in heat and mass transfer)

- THERMO-FLUID DYNAMICS OF TWO-PHASE FLOW (MAMORU ISHII)

- Convective Boiling and Condensation ( Collier )

Dr. Hirschberg,

Thank you for your response. I really appreciate your time and effort.

Let me restate the problem through an example:

You have water at 99.9 °C and 1 atm (101.325 kPaA). Under these conditions, there is no boiling and minimal mass conversion from liquid to vapor (for now, we’ll neglect diffusion, etc.). Once you drop the pressure to just below the water’s vapor pressure, say to 101.000 kPaA, the water will boil. What will the vaporization rate be (mol/s or g/s or some such unit) under those conditions (we’ll assume that nucleation sites are not in short supply)? Say we keep the temperature at 99.9 °C and lower the pressure to 50 kPaA, what is the vaporization rate under those conditions? I am assuming the liquid-to-vapor transition rate will be greater at 50 kPaA than it would be at 101 kPaA.

The next concern is a time element. The pressure of the 99.9 °C water is instantaneously dropped from 101.325 kPaA to 50 kPaA: How long will it take for nucleation (boiling) to begin? 1 ms? 100 ms? Is the “time-to-boiling” a function of how far below the vapor pressure the test fluid is? For instance, will boiling commence in 100 ms if the pressure is lowered to 100 kPaA, but start in 1 ms if the pressure is lowered to 50 kPaA (these values are arbitrary)? Or do we assume the mass conversion from liquid to vapor commences instantaneously, but at the rates outlined in the first paragraph, above?

Lastly, how do the answers for water relate to liquids with higher molecular weights, etc.? Will, for a given temperature, lowering the pressure 50 kPa below the vapor pressure (or some percentage of the vapor pressure, or the critical pressure) result in the same mol/s conversion rate for, say an alcohol, versus that of water?

By knowing the answers – or how to answer – these questions, then I believe the change in temperature should be a pretty straight-forward calculation.

I’ll say it again: I truly appreciate your insights on this.

Dear David, If there is a strong wind over your water, such that the diffusion boundary layer is very thin and the vapour is carried away by the flow (we call this "convection"), then the bottle neck for evaporation is the supply of energy for the evaporation. The efficiency of that process is dependent on the way you supply heat. In your kitchen you heat up the bottom of the pan (either with contact with a flame or an electrically heated plate or by inducing current in the electrically conducting bottom of the pan (induction heating)). This heat has to "travel" to the surface. If the water is perfectly stagnant, this occurs by conduction through the water and the metal side wall. That is a inefficient process. Due to temperature differences between the side walls and the center of the water, you will have density differences that will result in "free" convection currents. This will slightly enhance the process of heat transport to the top of the water (where evaporation occurs). When the pan is "sufficiently" large this thermally induced flow can even become turbulent, which also helps. This is possible without bubbles.

As I told you before the rising of bubbles will dramatically enhance this process of evaporation, both because bubbles of vapour reaching the water surface will directly contribute to the eva[poration and because the flow induced by rising bubbles enhances the heat transfer from the bottom heated plate to the water.

Of course when one uses micro-waves to heat directly the liquid in its bulk, then evaporation can be quite efficient and there is a very significant danger of vapour explosion, which can have dramatic consequences. Vapour condensing on you skins deliver the condensation heat results in terrible wounds! Do not try this experiment.

When you decrease the atmospheric pressure the same phenomena occurs at a lower temperature. Keep in mind that the liquid boiling line is an exponential function of the temperature. Hence the boiling point temperature decrease is not very large when you reduce the pressure. You do experience this when looking in the montains. At 5000 m height, the atmospheric pressure is half that at sea level. You will have to boil your pasta a few minutes longer than at sea level. Opposite situation occurs in a pressure cooker. With a pressure slightly higher than atmospheric the food is much faster cooked.

I hope this answer your questions.

Regards,

Mico