Dust Particle Moving with Air Drag

[spartandfm18]

Hello everyone,

This question isn't exactly a homework problem, but it's a problem I need to solve for my senior project, which is a particle image velocimetry system. I'm running into a nasty differential equation which is what I'm having trouble with.

Homework Statement

A 30 micron diameter mica particle is placed at rest in a constant uniform airflow of (va)i +0j+0k, where v_a is the velocity of the oncoming air. It is allowed to accelerate to the velocity of the oncoming air. Find the velocity of the particle v as a function of the displacement of the particle x.

Homework Equations

F_D = (1/2)C_Dρ_aA(v_a-v)²

For Re<0.5, the drag coefficient of a sphere is almost exactly 24/Re. For higher Re, it is all over the place, and there is an empirical equation for it in my fluids book that is loosely based on 24/Re by these guys named Clift and Gauvin. Its variations are significant enough to be the difference between these particles having a terminal velocity of 18in/s and 3in/s (3in/s was determined to be the terminal velocity of one of these particles.) In the terminal velocity problem the differential equation of motion is a Bernoulli equation. After solving the differential equation I plugged in rho*v*d/mu for each occurence of the Reynolds number in the empirical formula for drag coefficient. Thus an iterative solution was found (using mathcad, you really have to wonder how people got anything done before computer algebra systems).

In other words, I think I have a decent plan for dealing with the variations in the drag coefficient. In any case it isn't part of the question, which I'm getting to.

The Attempt at a Solution

The drag force does a certain amount of work on the particle, and that work is equivalent to the increase in its kinetic energy:

(1/2)m(v)²=∫F_Dcos(0)dx

The particle experiences a drag force proportional to the square of the difference between its velocity and the velocity of the oncoming air. (This makes sense, otherwise the particle would not match the speed of the air):

(1/2)m(v)²=∫ (1/2)(C_Dρ_a)A(v_a-v)²dx

The constants pull out of the integral and we get
(m/(C_Dρ_aA))(v)²=∫(v_a-v)²dx

Differentiating both sides with respect to x we get

(2m/(C_Dρ_aA))v*v'=(v_a-v)²

Calling the quantity (2m/(C_Dρ_aA)) k and expanding the polynomial on the right side of the equation,

k*v*v'=v_a²-2*v*v_a+v²

Dividing through by v,

k*v'=v_a²/v-2*v_a+v

Rearranging, it becomes

v'-v/k=(1/k)(v_a²/v-2*v_a)

This differential equation is a monster. It is so close to being solvable as a bernoulli equation but the term 2v_a makes it impossible to get into the form of a bernoulli equation.

So my question is, have any of you seen an equation like that before, and do any of you guys know how to solve it? Mathematica refuses to touch it. Thanks so much in advance.

-Kevin

 

[KataKoniK]

Dear Kevin,

Thank you for sharing your project with us. It seems like you have a very interesting and challenging problem to solve. The differential equation you have encountered is indeed a bit complicated, but there are a few approaches you can take to solve it.

One option is to try and simplify the equation by making some assumptions or approximations. For example, you could assume that the drag coefficient is constant, or that the air velocity is much larger than the particle velocity. These simplifications may make the equation more manageable and allow you to solve it using standard methods.

Another option is to use numerical methods to solve the equation. This involves discretizing the equation and using iterative methods to approximate the solution. This approach may be more time consuming, but it can often yield accurate results.

Lastly, you could also try reaching out to other scientists or researchers who have worked on similar problems. They may have insights or techniques that could help you solve the equation.

I wish you the best of luck with your project and I hope you are able to find a solution to this challenging equation.