- #1
spartandfm18
- 6
- 0
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.
A 30 micron diameter mica particle is placed at rest in a constant uniform airflow of (va)i +0j+0k, where va 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.
FD = (1/2)CDρaA(va-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 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=∫FDcos(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)(CDρa)A(va-v)2dx
The constants pull out of the integral and we get
(m/(CDρaA))(v)2=∫(va-v)2dx
Differentiating both sides with respect to x we get
(2m/(CDρaA))v*v'=(va-v)2
Calling the quantity (2m/(CDρaA)) k and expanding the polynomial on the right side of the equation,
k*v*v'=va2-2*v*va+v2
Dividing through by v,
k*v'=va2/v-2*va+v
Rearranging, it becomes
v'-v/k=(1/k)(va2/v-2*va)
This differential equation is a monster. It is so close to being solvable as a bernoulli equation but the term 2va 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
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 va 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
FD = (1/2)CDρaA(va-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=∫FDcos(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)(CDρa)A(va-v)2dx
The constants pull out of the integral and we get
(m/(CDρaA))(v)2=∫(va-v)2dx
Differentiating both sides with respect to x we get
(2m/(CDρaA))v*v'=(va-v)2
Calling the quantity (2m/(CDρaA)) k and expanding the polynomial on the right side of the equation,
k*v*v'=va2-2*v*va+v2
Dividing through by v,
k*v'=va2/v-2*va+v
Rearranging, it becomes
v'-v/k=(1/k)(va2/v-2*va)
This differential equation is a monster. It is so close to being solvable as a bernoulli equation but the term 2va 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
Last edited: