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| Oct30-05, 04:49 AM | #1 |
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Fluid Dynamics
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. |
| Oct30-05, 08:40 AM | #2 |
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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. Yes, your equations for ln x and ln y are correct. Now solve for x and y! |
| Oct31-05, 03:55 AM | #3 |
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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 |
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