Calculation-Volume of Droplet

Very interesting question.

I have spent a short time looking and cannot find any text that is free. Obviously I am not sure if any of the ones you need to register for is about your subject as I cannot read them, and I am too stingy poor to pay for them

I would initially think there is more to this than water tension alone. The situation under which you have the droplet is also important i.e. sitting on surface, hanging of edge, falling short time or having fallen long time.

I know that rain can come in different shapes and forms and you can have a drizzle with little drops or those painfully large slow falling drops. I don't know how they can be different if water tension is the only parameter here.

I noticed, once on holiday in Spain, when i threw water out a glass down this little mountain, as my brother was walking down there so it had to be done, the droplets gathered speed up to a point and than suddenly they all splat out-wards and became all very many small drops as if the air pressure was holding them back and spreading them out. I don't think they would accelerate after that point but it meant that I could not throw a large amount of water over my brother as it would always spread out before it got to him down there.

Could you not use your burette to drop a measuring jug full to a point and count the drops?

While there are several methods to measure the number of droplets in a given volume and come up with an average droplet size, I would assume (maybe ass-u-me ?) that someone has developed an analytical calculation that relates the physical properties to the droplet size.

Like you, I am not in a position where I have access to the "pay for view" science papers and personally, I balk at the thought of paying to retrieve scientific literature from a journal. Whatever happened to free libraries? Over the years, I and many others have contributed to scientific literature and I (we) never received any remuneration from the publishers, but it was always open literature to the technical public. {But then my wife says I'm just a crabby old man.}

Hey Doc,

Just happens that I am working on the development of a performance calculation package for screw compressors and one of the problems in a machine that uses oil spray for cooling is to relate the mass flow of the oil as a fraction of the gas and its effect on compression.

So in one of my references I came across the concept of Sauter Mean Diameter. Not sure I fully understand it yet, but maybe it will help you too.

Nick Name--heavy but good looking answer. One additional point which significantly affects drop size is the effective diameter of the tip from which each drop is falling. A micro dropper gives perhaps 30 drops per ml while a larger one gives 15-20. Before the common usage of IV infusion pumps, nurses routinely had to use these values to calculate the flow rates of IV solutions based on drops per minute and the dropper type. Also, holding a dropper at an angle other than vertical changes its effective diameter, so accurate results always required holding the pipette or dropper vertically.

--JMM

I totally agree with you. I considered the most general case with an horizontal fluid separation surface before the droplet forming. I am also not very found of buying expensive books for such problems so that I try, respecting the laws of physics, to find an answer by myself even if not totally correct in general it is a good approximation.

In my previous computation I considered the "neck diameter" = 2*R. If we assume an other value in place of the factor "2" the relationship becomes:

R=(3*δ*σ/(4*ρ*g))^0.5 where δ=d/R.

It is also possible to introduce a value for "d" and use the equation→ R=(3*d*σ/(4*ρ*g))^(1/3).

As an example for a needle with d= 1.3 E-3 m the computation gives R= 1.93 E-3 m and the number of drops/ml n ≈ 104.

I have not the tools to make a true investigation but I tried with available tools to check the order of magnitude.

Shortly before leaving the needle the diameter of the droplet was ≈ 3.3 mm.

From the computation it should be 3.86 mm. I consider that the convergence is for the imprecision of my equipment quite ok.

I am impressed!

--JMM

Thank you for the kind words

This may be of interest ; http://www.islandnet.com/~see/weather/history/lenard.htm

Dear Guest,

it was very interesting to read the paper. I tried to check if the simplified model with which I computed the drop volume could bring coherent results with experimental ones mentioned in the text.

I considered some data available via Google as for instance the Cw of a sphere function of the Reynolds number and the "internal pressure" generated by the superficial tension.

For the fall speed I assumed as a limit the value for which the drag force and the weight are equal. It was necessary to proceed by successive approximations since Cw is function of Reynolds, Reynolds is function of speed and speed function of Cw.

The pint = σ/R for a sphere is generated by superficial tension and pstop = V²*ρ/2 is the static pressure where the air flow meets the drop surface and where the flow speed is=0.

The results are quite converging to the experimental values:

for small d≤ 2mm one sees that pint >> pstop so that the drop surface form is dictated by the internal pressure and for minimal energy it has to be a sphere.

If the diameter increases the difference between pint and pstop decreases so that the radius at the contact zone between droplet and air flow cannot be any more the same as in the opposite region where outer pressure is a lot less.

For a diameter between 3 and 4 mm both pressures are equal which suggests that the contact surface should be plane and for higher "d" pstop>pint. This means that for an equilibrium the surface has to change the sign of curvature i.e. it should become concave!

But such a drop form will for many reasons have a tendency to part which could explain why only one drop was found in the 5 mm range.

The different stages are sketched in above picture.

There is a last point to clarify: the maximal speed. In the graph the drop has been considered as spherical and at d=5mm a fall speed of ≈ 10m/s was obtained. If we consider as a rough approximation that the deformed drop is a full hemisphere the falling speed will be 6.85 m/s since the corresponding Cw is a lot higher than the one for the sphere. But the drop has a more rounded form so that its Cw will be less and it is coherent with the model that the fall speed will be found between 10 and 6.85 m/s value mentioned in the text is 8m/s which is well in the interval given by above limits.

Thank you for the link it was really interesting to make this check and pleasant to obtain a good convergence.

There are several factors that influence this, and I plan to do a bit of research to see if there's an "official" discussion available.