Hi Ehsan...

i'm sorry i don't have any information about your question. but maybe Mohammad Asrardel can solve this problem.

Hello!

You have a very interesting question. 

What condition is your experiment?

I think this law is still valid in a two dimensional evaporating drople, because, when the image processing, you'll be in a two-dimensional case. Then you go from two- to three-dimensional go when you get results.

Dear Max, Thanks for your response. I'm interested in solving a diffusion case where the droplet starts to evaporate by the diffusion mechanism.

Thanks for your help. that's very kind of you. I got your file.

Dear Vitalii Pertsevyi, The article is an experimental study which implies that the problem has a 3D nature. do you think that the results are analogous with a 2D case?

Dear Remi Cornwall, you have done an interesting analogy but as you see, this doesn't lead to the correct result. because squared value of diameter should be decreasing linearly with respect to time and not the diameter itself. the reason is that according to the Maxwell theory, the rate of volume loss (hence mass) is proportional to the diameter (or radius) and not the area.

Eslam Gomaa ِAl-Sakkari, Thanks for your comment.

The classic Maxwell theory reads:

dm/dt=-4*pi*D*r*(c0-ci) where D is the diffusion coefficient and r is the radius of a spherical droplet going under evaporation in an infinite medium. if you continue by this formula, you will see that r^2=const1-const2*t and V^(2/3)=const1-const2*t .

This law comes from a round drop, where variations are only allowed in the radial direction., and there are no convective or gravitational effects. All properties are assumed constant, cp, diffusion constant, heat of vaporization, and heating value for burning drops. Classic combustion text give the derivation of this, for example "Introduction to Combustion" by S. R. Turns, in my opinion one of the better combustion books around.  It applies to both burning and only evaporating drops, with different values for the constant in the D-squared law. 

In practice, experiments have been done with suspended drops burning in air.  The attached file shows the results for different fuels, and in the case where bouyancy is active, it still looks to apply, even though this is not a strictly one dimensional case.  The reference for the figure is a book by N. Chigier, it has been "borrowed" so I don't have the title. This basic model, modified for convection, is also the starting point for the modelling of diesel sprays for CFD calculations in diesel engines.

So in terms of a practical model, it looks to be useful.  To answer the theoretical 2- dimensional case, one would have to solve the partial differential equations, in contrast to the ordinary differential equations that give the above model.  At the present state, given all other inaccuracies in combustion modeling, the D-squared model look to be a good starting point, that provides realistic trends.

The difference between the droplet and the puddle is the geometry, in the absence of convection, both are one dimensional, the differential equations are different, do the math for the diffusion equations in linear coordinates and see what happens. I don't off hand know of references for this case, it's probably in the literature somewhere.  The basic model says nothing about the movement of the liquid, it is assumed to be uniform in properties throughout the evaporation.  I don't know what you mean about "intuitively obvious", there's nothing intuitively obvious to me about it.

The original  question was about the D-squared law for a droplet in 2-dimensions. The origin of the "law" is the one-dimensional case as described above.  The empirical data show that it is useful in at least one case which is not strictly one-dimensional, which is relevant to the original question. This simplified model has also been proven to be a useful starting point for modelling droplet evaporation in several practical combustion systems involving evaporation droplets (not one-dimensional).  Possibly a  bit surprising considering that convection and buoyancy were not included, but  it has been proven very useful nevertheless. I hope that is useful to the original poster.

>>What is c0, ci? How can the rate of mass loss be just proportional to r and not the area? Is it even dimensionally correct? Show the full derivation and not just the end result, please.

Ci and C0 are vapor concentrations at the interface and far from the interface respectively. Of course it is dimensionally correct:

[C]=kg/m3

[D]=m2/s

[r]=m

and multiplying these three parameters results in kg/s which is the dimension of dM/dt.

I don't know the derivation, but you can find the expression in many papers.

Dear S.C. Sorenson, Thanks for your comments. I think that you misunderstood my question. I don't think that solving partial differential equations is necessary because the evaporation is still one-dimensional.

As you said, however the droplet has a spherical (3D) geometry, but it is assumed that the evaporation is taking place only in radial direction and hence the problem is one-dimensional in spherical coordinates and ordinary differential equations would have to be solved.

But my question is about the case where the droplet has a 2D geometry (suppose a cylinder with a unit depth instead of a sphere) with one-dimensional evaporation (just radial evaporation). So what I meant by two-dimensional was the geometry (cylinder instead of a sphere) and not the evaporation. 

Ok, when one says droplets, the assumption is normally spherical.  When you say unit depth, it sounds to me like you mean an infinitely long cylinder and the evaporation per unit length. In that case, it would still be one-dimensional mathematically, as everything would be uniform in the circumferential and axial directions, and the variations would be in the radial direction. That would give the diffusion equations in cylindrical coordinates instead of the usual spherical case.  I don't recall seeing that solution either, but look at the book by Turns, or another basic text, and all the parameters and solution technique will be explained.  Make the assumptions that they use for the droplet and solve the cylindrical equations, it should be solvable. Let us know what you get!. We used Turns' book in our combustion course for several years, I find it one of the best combustion texts around.  

If the cylinder is short (non-infinite length) , then it is really two dimensional (radial and axial directions), you need to solve the PDE's.  There would also be another dimension involved, so a first guess would be that the D squared law may not hold as such.

I really appreciate your help. Actually your mentioned book was really helpful at first sight. I will consider your comments and let you know if I get anything.

I would like to start with a simple "example" very realistic.

Think to a droplet of liquid close to its boiling temperature. To have evaporation you need to provide energy, that is arriving by convection,

the energy provided is proportional to surface A=r^2, a convection coefficient h, r^2, deltaT which is constant. So Q' = h r^2 (pi deltaT and other constants omitted), hence

dQ = h r^2 dt

This energy heats-up and turns to vapor a layer of liquid with

volume = surface * dr (dr = thickness )

that requires energy dE= r^2 *dr (density, evaporation entalpy and other constants neglected)

finally r^2 * dr = h * r^2 * dt

(dr is negative because the radius is decreasing)

IF h=constant, than dr/dt = constant. Experimentally it is not the case, so it means that h is not constant. If you have a look to any convection correlation for a sphere, h is function of radius. You may use existing correlation to predict your behavior, but you may also use experimental results to obtain h as function of r. It is in fact what was done in the past, with different experiment. You may finally get that h= r^-1, so similar to 64/Re that I'm not surprised.

Let's go to 2D

Note that the evaporating droplet is a monodimensional problem (the only variable is the radius), in a 3D word.

I do not understand what kind of 2D we are talking about.

- The problem may be like evaporation from a very long cilinder. In that case i would simply see what are the correlation for convection around a long cilynder, and go on with it. I would expect h = D^something , depending also on the regime, pure molecular diffusion or turbulent diffusion.

- The problem may be like a "flat droplet", that is evaporation from a pool. i would see what are the correlation for convection on a circular plate, and go on with it (natural convection at low Gr = pure diffusivity)