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Modeling the velocity of a raindrop

Ben Lynch and Gavin Lommatsch

May 6, 2011

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Abstract

Mathematical modeling of a falling raindrop and the derivation of the motion of the raindrop as it is affected by air resistance. The nonspherical shape of the drop will be taken into consideration to producean accurate model

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1 Introduction

As a raindrop falls, it forms a shape different from popular belief; the result-ing shape from all of the forces causes the raindrop to form almost a perfectsphere at diameters under 1.25mm and at larger diameters the sphere at-tens out into an oblate spheroid which is somewhat similar in shape to thatof a hamburger bun. When a drop reaches diameters above 8mm it starts tobreak up into smaller pieces therefore giving us a range up to 8mm for thestarting diameter of a raindrop. As a rain drop falls, it attains a constantvelocity where it experiences no acceleration. This is called terminal velocity,when the forces of gravity are opposed equally by the resistive forces on theraindrop.

Figure 1: σdφ = − σsin (φ/x ) + 2 σ/R t + ∆ ρgz + ( pi )t − ( pa )t

1.1 Background

In this project, simplifying assumptions were made. One of these assump-tions was that the raindrop fell from a cumulonimbus cloud at low altitudeand therefore the changes in air pressure, air density and gravity are ignored.

The raindrop is also falling with other drops and therefore the surroundinghumidity is high enough that the evaporation of the raindrop as it falls willbe negligible and therefore ignored. The raindrop is also not affected by anyextraneous forces such as gusts of wind or gaining additional mass from co-hering with other raindrops.

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With these assumptions, the main force acting on the drop is air resistance.

It is caused by air molecules colliding with the water molecules, friction be-tween the air and water molecules as the air passes around the drop, andby the low pressure wake behind the object. The main factor in all of thesecases is the shape of the object. A raindrop, however, has a complicatedshape to determine because of the previously mentioned hamburger shape.The article Derivation of the Shape of Raindrops derived the shape of a dropfrom the Laplace-Young equation. The derivation is fairly lengthy and sincethe purpose of this project is to nd the velocity of the raindrop, we will notgo through the derivation. We will use much of the data generated in thisarticle.In uid mechanics, a number known as the Reynolds number is used to givea measure of the ratio of inertial forces to viscous forces and therefore sub-sequently quanties the relative importance of these two types of forces forgiven ow conditions. The Reynolds number is found from the equation

Re = ρv0 d

µawith ρ as the density of the uid, µa is the viscosity of the same

uid, and v which is velocity. As one may see, the Reynolds number is de-pendent on the velocity at which the object is traveling. For low Reynoldsnumbers ( Re << 1) associated with a sphere, the uid ows around thesphere without separating. Therefore, the drag coefficient, which is foundempirically, is inversely proportional to the Reynolds number. For moderateReynolds numbers (10 3 < Re < 2 105 ), a separation is formed between theuid and the sphere causing a helical wake. The drag coefficient continuesto decrease with the increasing Reynolds numbers until at 10 3 it becomes aconstant ( C d = .3) up to 2 105 . The Reynolds numbers for raindrops fallsinside of this range, so we will use C d = .3 as the drag coefficient.

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Figure 2: Air ow around a sphere at different Reynolds numbers; (a) Re <<1, (b) 103 < Re < 2 105 , (c) Re > 2 105

2 Equations

We will use Newton’s Second Law to develop our equation. It states thatthe net forces on a particle are equal to the time rate of change of its linearmomentum.

F n = ddt (mv)

Therefore,F n = mv̇

Now we can begin to sum the forces on the left side of the equation.

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The force due to gravity is dened as:

F g = mg

The other force acting on the drop is air resistance which is dened by thefollowing equation:

F d = 12

C dρa v2 A

C d = total drag coefficient.ρa =Air density (at 300 K , 1 bar)=1 .161kg/m 3

v = velocityA = cross sectional area

Here is where things get a bit tricky. The area that is used in the equationis the cross-sectional area, but due to the hamburger shape, the normalequation for the volume of a sphere can’t be used. Therefore, we had touse the area, volume, and Reynolds number calculated by the article for thediameters 1mm through 8mm.

Putting these equations together:

F n = F d − F g

mv̇ = 12

C dρa v2 A − mg

Dividing out m gives us our differential equation.

v̇ = C dρa v2 A

2m − g

2.1 Solving the equations

In order to solve the equation, we will make a few substitutions to make

things a little easier to manage. First of all, the terms, C d , ρa , and A are allconstants and therefore we will let r = 12 C dρa A

The equation then becomes

dvdt

= rm

v2 − g

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d(mm ) Re V (mm 3 ) A(mm 2 )

1.0 263 .510 .7912.0 863 4.238 3.3393.0 1593 14.956 8.0224.0 2267 36.822 15.2955.0 3012 73.805 25.7166.0 3625 129.603 39.1657.0 4230 204.228 56.1168.0 4834 315.783 80.270

Figure 3: Cross-sectional Area vs. Diameter

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Figure 4: Volume vs. Diameter

The resulting solution from the differential equation is an exponentialdecay equation which means that as time progresses on, the velocity willeventually become a constant. This constant velocity is the terminal velocityof the drop. This is expected to happen because of the increasing drag forcewith increasing velocity. We plugged in the appropriate constants to see if

our model is realistic and used the area and volume regressions to calculatethe area and mass for each diameter. We then graphed the solutions for thediameters from 1mm to 8mm on one graph and observed the velocities overtime. As expected, the solutions accelerated rapidly for about 2 to 4 seconds,then hit their terminal velocities. The terminal velocities increased withdiameter, but the increases were getting smaller. This suggests that theremay be a limit in the terminal velocities for large diameters, but accordingto our model, it is well outside a reasonable size for a drop.

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Figure 5: Velocity Vs. Time with diameters ranging from 1mm-8mm

3 Conclusions

With this model, we can predict the velocity of a raindrop with a givendiameter. The step would be to empirically test the velocity and compare it

with the models predictions. If the results didn’t match the model accurately,the model could be revised to take into account some of the assumptions thatwere ruled out. One way to reduce the error could be to take into account theevaporation of the drop as it falls. This would introduce a change in the massof the drop into the equation which would make it much more complicated,but it would be an interesting problem to solve.

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grid on

pauseendxlabel ( ’time (seconds)’ )ylabel ( ’velocity (m/s)’ )figure

plot (di,v_t, ’b’ )grid onxlabel ( ’diameter (m)’ )ylabel ( ’terminal velocity (m/s)’ )

3.1 References

Brian Lim. Derivation of the Shape of Raindrops. Cornell Uni-versity: School of Applied and Engineering Physics, 19 May2006. PDF.

E. Reyssat, F. Chevy, A.-L. Biance, L. Petitjean, and D. Quere.”Shape and Instability of Free-falling Liquid Globules.” A Let-ters Journal Exploring The Frontiers of Physics. EPL Journal,2007. Web.

Chean Chin Ngo and Kurt Gramoll. ”FLUID MECHANICS -THEORY.” ECourses. Web. ¡http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=¿.

John Polking, Albert Boggess, and David Arnold, DifferentialEquations with Boundary Value Problems, Second Ed., Pren-

tice Hall, New Jersey, 2006.

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