Fluid Dynamics: Equations and Streamlines for a Steady Two-Dimensional Flow

[Steph]

A steady two-dimensional flow (pure straining) is given by u = kx, v = -ky, for k constant.
(i) Find the equation for a general streamline of the flow, and sketch some of them.
(ii) At t = 0,the fluid on the curve x^2 = y^2 = a^2 is marked (by an electro-chemical technique). Find the equation for this material fluid cuve for t > 0.
For part (i), I used that the partial derivative of x with respect to s is kx, and that of y with respect to s is -ky. So integrating these, and using the initial condition x = xo, y = yo at s = i, I found that
ln x = ks + ln xo, and ln y = ln yo - ks.
But I don't understand what part (ii) of the question is asking. What equation does it mean for me to find? I know I could write x^2 + y^2 as (u/k)^2 + (v/k)^2, but I'm not sure that helps atall.
Thanks in advance for any help.

 

[HallsofIvy]

A steady two-dimensional flow (pure straining) is given by u = kx, v = -ky, for k constant.
(i) Find the equation for a general streamline of the flow, and sketch some of them.

Are we to assume that u and v are the components of velocity in the x and y directions respectively? If so then you have the two equations
dx/dt= kx and dy/dt= -ky which can be solved easily for x and y as functions of t. When you are drawing the streamlines, t is irrelevant- you want to eleminate t in the two equations, which is simple.

(ii) At t = 0,the fluid on the curve x^2 = y^2 = a^2 is marked (by an electro-chemical technique). Find the equation for this material fluid cuve for t > 0.

Did you mean x^2+ y^2= a^2, a circle?

For part (i), I used that the partial derivative of x with respect to s is kx, and that of y with respect to s is -ky. So integrating these, and using the initial condition x = xo, y = yo at s = i, I found that
ln x = ks + ln xo, and ln y = ln yo - ks.

What is s?? I would have thought x and y were functions of time, t. And since they are functions of a single variable, whether called s or t, you don't need partial derivatives. s= i?? You seem to mean s= 0.
Yes, your equations for ln x and ln y are correct. Now solve for x and y!

But I don't understand what part (ii) of the question is asking. What equation does it mean for me to find? I know I could write x^2 + y^2 as (u/k)^2 + (v/k)^2, but I'm not sure that helps atall.
Thanks in advance for any help.

The points are on a circle. You could think of each point as an initial value (the xo, yo you have above) and get the position of each point as a function of time. Once you have solved for x and y in (i) and eliminated t to get y as a function of x, remember that (yo)^2+ (x0)^2= a^2 and then see if you can write the function with a rather than x0 or y0.

 

[Steph]

Thanks for your reply to my question. I apologise for the typing errors.
- yes, u and v are the components of velocity in the x and y direction respectively
- I had intended to type x^2+ y^2= a^2, and also later that s = 0
s is a variable we used in lectures, although it was never defined, and so has confused me. I think I imagined it as the distance from the starting point, almost like curve length.
I'll work on the second part of the question with your suggestion later, and see what I can come up with!
Thanks again, Steph