Fluid dynamics - find distribution of a conserved variable

[chilge]

I accidentally posted this to the "Calculus & Beyond" forum when I meant to post it to the physics forum. If someone can tell me how to move this post, I will get rid of it here!

Homework Statement

Consider a property, for example temperature θ, that is conserved during advection (i.e. Dθ/Dt = 0 where D/Dt represents the total derivative). Suppose the velocity of the field advecting the fluid is:

u = ax, v = -ay, w = 0

Find the form of the temperature for t > 0 when at t=0, θ(t=0) = θ₀y/L.

Homework Equations

Dθ/Dt = ∂θ/∂t + u*(∂θ/∂x) + v*(∂θ/∂y)

The Attempt at a Solution

So far, I've calculated the trajectories:
x(t) = C₁e^(at)
y(t) = C₂e^(-at)

I also know that since θ is conserved, θ is a function of where in space the parcel originated. In other words, θ=θ(Y) where Y is the Lagrangian variable.

I've expanded the total derivative and set it equal to zero since θ is conserved:
(1) Dθ/Dt = ∂θ/∂t + u*(∂θ/∂x) + v*(∂θ/∂y) = 0

Then, I thought to say that θ(t)=(θ₀y(t)/L)*f(t), where f(t) is some function of time that we want to find out so we can see what the function θ(t) looks like.

I took this θ(t)=(θ₀y(t)/L)*f(t) and plugged it back into equation (1), but don't think I'm getting anywhere. Can anyone tell me if I'm on the right track or if I'm going about this completely the wrong way?

 

[Chestermiller]

You started off OK, but then you started going in a fruitless direction. Let's focus on time zero. Let x(0) and y(0) be the coordinates of a material particle at time zero. So, at time zero, the "temperature" of the material particle is θ(x(0),y(0))=θ₀y(0)/L. What do the lines of constant θ look like at time t = 0? When the fluid stretches the way it is described, does θ at later times vary with x? What do the lines of constant θ look like at later times t? If a material particle is at x(0) and y(0) at time zero, what are the coordinates of this same material point x(t) and y(t) at time t? What is the value of θ for the material point x(t) and y(t) at time t?

Chet

 

[chilge]

Hi Chet, thank you very much for your reply - it helped get me out of my rut and thinking in a new way. Here's what I've come up with:

We've already found that
x(t)=C₂e^at
y(t)=C₃e^-at

Let's say that the particle is at (x₀,y₀) at t=0. Then
x(t)=x₀e^at
y(t)=y₀e^-at

Therefore we can find out where the particle started by inverting the above equations:
x₀=x(t)e^-at
y₀=y(t)e^at

Now, let's think about a particle that has traveled to the point (x(t),y(t)). Since θ is conserved, in order to get the particle's temperature θ(t) all we need to know is its original y-coordinate (and we are lucky here because the equations for x(t) and y(t) are not coupled). In other words,
θ(x,y,t)=θ₀*y(t=0)/L=θ₀*y₀/L

Thus the final answer is
θ(x,y,t)=θ₀*y(t)*exp(at)/L

Do you think this looks better now? I think before I was getting caught up in trying to plug things into the total derivative instead of thinking about the picture of an individual parcel.

 

[Chestermiller]

Excellent job. Very nice.

Chet

 

[paranormal]

Hi there! It looks like you are on the right track. However, instead of plugging in your proposed solution for θ(t), try using the given velocity field to express the total derivative in terms of y and t. Then, set it equal to zero and solve for θ(y). This will give you the distribution of temperature as a function of y and time. You can then use your initial condition, θ(t=0) = θ0y/L, to determine the specific form of the function f(t). Let me know if you need any further assistance. Good luck!