How to calculate evaporation rate of a liquid?

Question

Is there a way to calculate the evaporation rate of say, a droplet of water? From what I understand, there are several factors to be taken into account:
- temperature
- air humidity
- concentration of substances dissolved in the water
- the rate at which fresh air is brought in near the surface (Fick's law of diffusion?)
- effect of hydrogen bonds in the water
- surface area of the droplet
- Does the type of material the droplet lies on also have a significant effect on evaporation rate?

Answer 1

This is far too complicated.
As you correctly point out there is a lot things to take
into consideration. There are some some limiting cases,
however, where a nice looking solution is possible.
Let me give you a jump start with a couple of such cases:

A)##################
A droplet of water of radius R evaporates in air of pressure P
and temperature T and relative humidity β in absence of gravity.

The dominating limiting factor in this case is diffusion.
If relative humidity is α(x,u,z,t) = α(r,t), then diffusive current
is j(r,t) = -γ grad α(r,t), where γ(P,T) = const is diffusion
constant of water moleculaes in air.

Conservation of water molecules:
div j(r,t) = -∂α(r,t)/∂t
-γ div grad α(r,t) = -∂α(r,t)/∂t

Since density of liquid water is much grater than that of
vapor, the time deroivative ∂/∂t can be neglected,
and fianlly the equation is

**********************
Laplacian α(r) = 0
**********************

Thus we can see that density of vapor is the same as
potential of electrically charged sphere:
α(r) = (1-β) R/r + β

The speed of evaporation dm/dt of the droplet is eqaul to
the flux of j at the surface:

-dm/dt = 4πR² ρ(sat) j(r) = 4πR² ρ(sat) (1-β) R/R²
-dm/dt = 4π ρsat(T) (1-β) R
-4/3 π ρw dR³/dt = 4π ρsat(T) (1-β) R
-R ρw dR/dt = ρsat(T) (1-β)
-dR² = 2 ρsat(T)/ρw (1-β) dt
-dR² = 2 ρsat(T)/ρw (1-β) dt

**********
R² = Ro² - 2 ρsat(T)/ρw (1-β) t
**********




B)##################
A droplet of water of radius R evaporates in vacuum
starting from temperature T.

In this case the limiting factor is potential barrier of
latent heat = 500 cal/g = 2000J /g = 4500 Kelvin x k_B
of extra energy needed for water molecule to evaporate.

Lets assume that collisions of water molecules with
the surface are 100% inelastic, meaning that all
vapor molecules hitting the surface stuck to the
surface and never bounce. In this case the flux of
water molecules evaporating from the surface
would be equal to the pressure of saturated water
vapor in equilibrium:

-dm/dt = 4π ρsat(T) R² ,

where is the mean speed of water molucules.
can be calculated exactly, but I am already
too tired, and will simply assume that
= √(RT/μ), μ = 18 g/mole.

-4/3 π ρw dR³/dt = 4π ρsat(T) R² √(RT/μ)
-ρw dR/dt = ρsat(T) √(R/μ) √T

Now we need ρsat(T), which is given by
Clausius-Clapeyron relation:
ρsat(T) = ρsat(To) exp(-ΔH/RT)

-ρw dR/dt = ρsat(To) exp(-ΔH/RT) √(R/μ) √T

Last thing we need to get rid of either T or radius R
(not to be confuesed with gas constant R aleady lurking
around here):

Cv(water) Δ(TR³) + (latent heat) * Δ(R³) = 0
R = Ro ³√(To - ΔH/Cv)/³√(T-ΔH/Cv)

And fianlly:
-ρw Ro ³√(To - ΔH/Cv) √T exp(ΔH/RT) d[1/³√(T-ΔH/Cv)] = ρsat(To) √(R/μ) dt

Well, at least we managed to separate the variables....

Answer 2

Calculate Evaporation Rate

Answer 3

while water boils in a pan, warmth potential is being offered to the backside of the pan by utilising a burner, etc. while the vast majority of the water reaches one hundred*C any further warmth will reason steam bubbles to sort and upward push to the floor. while the action is stated it somewhat is named boiling. Evaporation is somewhat diverse and calls for a approaches much less potential. while warmth is extra to a water floor (by utilising the sunlight or by utilising heat air, etc.) individual molecules of water (somewhat of the finished bulk) get carry of sufficient potential to interrupt the bonds of floor rigidity and depart the liquid to flow away as vapor. If the room is at one hundred% relative humidity, approximately as many water molecules will return to the pan of water as depart it and it will seem as though no evaporation has befell. besides the shown fact that, if the air is dry (say 50% relative humidity) two times as many water molecules will depart the pan as return to it. Even chilly water will evaporate in dry air. Even ice will evaporate by utilising sublimation without melting. the sea water is continuously evaporating even with the reality that it somewhat is gentle sufficient so which you would be able to swim in.