raindrop model

8/10/2019 Raindrop Model

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Math 1270 HonorsODE I

Fall, 2008

Class notes # 2Revised, August 29

1 Applications of rst order ode's

A number of applications are given in section 1.1, which you should read over. HereI will describe a few more. The rst example in the text is that of an \object fallingin the atmosphere near sea level." The goal is to nd a dierential equation thatcan be used to predict the velocity of the object as a function of time, and then, ifpossible to solve this dierential equation with any appropriate initial conditions.

Finding the dierential equation from physical laws is called \modelling", and theresulting equation is a \mathematical model." (Anything referred to as a \computermodel" is really a mathematical model, with the required calculations done on acomputer.) The principle physical law used to model falling objects is Newton'ssecond law,

F=ma;

where the bold print indicates that F and a are vectors, while m is a scalar. Ofcourse, Fis force, m is the mass of the object, and a is acceleration. However, it isassumed that all vectors point straight up or down, and so the equation is writtenin scalar form

F =ma; (1)whereF andaare the magnitudes of the force and acceleration vectors.

We are assuming that there are two forces acting on this object, which add to-gether to produceF: The rst is the gravitational force, resulting from gravitationalacceleration,g . The statement that the object is falling \near sea level" is meant tosay that g remains constant. There is a contrast here with, say, a rocket sent intospace, which would cover a great altitude range and thus have a varying gravitationalacceleration to deal with.1

The second force on the object occurs because fall occurs \in the atmosphere",

meaning that there is air resistance. This tends to slow the object down, and so be1Another Newton's law is involved here, stating that the gravitational acceleration is inversely

proportional to the square of the distance from the center of the earth.

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in the opposite direction from the object. Unfortunately, it is at this stage that wecan begin to see the complications that usually arise in modeling. It is pretty clearthat the force depends on the velocity of the object. But, it clearly also depends on

the cross sectional area. If the object were at, like a penny, and stayed horizontal,it would encounter more resistance than if it were long and thin, like a nail, and wasfalling straight downward. On page 2, the text authors make the assumption thatthe air resistance force is proportional to the velocity, giving a term v, where isconstant. This results in a force termF =mg v. Substituting into Newton'slaw (1) gives us the ode

mdv

dt =mg v:

This is a rst order equation with constant coecients. We can put it into a familiarform by dividing bym, giving

dvdt =g mv:

We will assume the initial condition

v (0) = 0:

The resulting initial value problem is easily solved using the integrating factor method,giving the solution

v=gm

1 e

mt

(2)

Here are the assumptions of this model:

1. gravitational force is constant

2. mass is constant

3. force due to air resistance is proportional to velocity, with \" constant

4. there is no sideways wind

Maybe you can think of some more.But let's go back and look at our physical assumptions. In addition to assuming

that is constant, we also assumed that m is constant. But we can think of

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circumstances where this is not true. One is that of a falling raindrop, in whichwater evaporates as the object falls. What model do we get then?

I found this model discussed in another text.2. In that book, we read: \As araindrop falls, it evaporates while retaining its spherical shape. If we make thefurther assumptions that the rate at which the raindrop evaporates is proportionalto its surface area and that air resistance is negligible, then a model for the velocityv (t) of the raindrop is3

v0 +3 (k=)

r0+ k

tv= g:

Here, is the density of water andr0 is the initial radius of the raindrop. Noticethat this is a linear rst order nonhomogeneous equation with varying coecients.Again we can solve it by the integrating factor method. I will simply write down theanswer, which is pretty complicated. (I used Maple!) I assumed thatv (0) = 0;andgot

v (t) =1

4gt+

1

4

g

kr0

1

4

g

k4

r40

(kt+r0)3

:

So, what assumptions are being made in this model?

1. gravitational force is constant

2. no air resistance (unlike our text)

3. shape is always spherical

4. mass changes at a rate proportional to surface area

And maybe more. This is always a dicult thing in modeling. Maybe we aremaking some unconscious assumptions.

So I went on the web, and googled: \equations for a falling raindrop". The rstitem returned began:

2Dennis Zill, A First Course in Dierential Equations3The derivation of this is at the end of these notes

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\A standard undergraduate mechanics problem involves a raindrop which growsin size as it falls through a mist of suspended water droplets."4

Hmm. This is a whole new ball game. Rather than go o on this tangent, let'slook further. Changing the search to "shape of a raindrop", I found :

\.. in truth, raindrops are spherical in shape when they begin to fall. Then,unless they are very small, they take on shapes with attened bases and roundedtops, looking more like hamburger buns ...

\ ... In actual fact, even the hamburger-bun shape for raindrops , which is basedon the observation of single drops steadily falling through a nearly non-turbulent airstream, is idealized, particularly for heavy rain events. If we could isolate a singledrop during a rainstorm, and follow its formation until its nal splashdown, we wouldsee .... an ever-changing quasi-spherical shape throughout its lifetime."

No citation or evidence is presented for this claim. And it goes on to discusshow raindrops of dierent sizes fall at dierent speeds (this must be because of airresistance), and so are forever colliding, causing coalescing and breaking up of drops.It ends: \As drops change their size (their volume), they jiggle, joggle, and wobblethrough a variety of physical distortions.

Here is a graphic that accompanied this article:

Can we model this? Should we attempt to take into account all these com-plications? No way. The best we can do is make many simplifying assumptions,come up with a model we can analyze, and then test it against observations. I alsofound a website which tried to describe how to measure raindrop velocity with avideo camera and wind gauge, but it seemed to me this would require a much moreprofessional setup than was implied by this site.

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This was from a research paper written by people from the University of West Virginia, andpublished in 2001 in a reputable physics journal. In the abstract they claim that their solution

can be understood "by students of intermediate mechanics and nonlinear dynamics", so this paper

might be a good topic for your class project. Email me for the reference.

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I found another source discussing fall under gravity, this time in the contextof sky diving.5 The interesting thing about this model is that the author made acompletely dierent assumption about the eect of air resistance. The assumption

this time was that the force due to air resistance was proportional to thesquare

ofthe velocity. There was no discussion of where this assumption came from. (Nordid the text say where it got assumption 3 in our rst model.)

This new assumption results in the ode

dv

dt =g

k

mv2;

for some positive constantk:The equation is no longer linear. But we can still solveit, using separation of variables, and assuming thatv (0) = 0; we get

v (t) =r

gmk

tanhs

kgm

t;

where tanh is the so-called hyperbolic tangent function. This can be dened as

tanh x=et et

et +et:

Look at the graph of this function:

The interesting thing is that it levels out at the height y = 1: Hence, we see that

limt!1v (t) =rgm

k :

5Modelling with Ordinary Dierential Equations, by T. P. Dreyer.

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This is called the \terminal velocity" of the skydiver. Experiments are cited showingthat this is about 181 miles per hour.

This gives us a way to calculate k .

However, there is a simpler way to nd the terminal velocity. Return to the ode,

v0 =g k

mv2:

This time, we will plot the direction eld: (done in class).

We are only interested in the region v > 0: We can see that the arrows arepointing down ifv >

pgm

k ;and up ofv

raindrop model

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