1. CEE 262A H YDRODYNAMICS Lecture 15 Unsteady solutions to the Navier-Stokes equation

2. Effects of unsteadiness Up to now we have assumed “ Steady State ” Poiseulle-Couette Flow H i.e. It takes a time >> for flow to reach a steady The “generic” estimate of the timescale for viscous effects to act over a lengthscale L is or we can say in time  the distance over which viscous forces are important is

3. Stokes' first problem 3 1 2 “Infinite” Moving Plate (i) Flow is invariant in x 1 direction (ii) No x 2 component

4. (iii) From (i) x 1 -Mom: x 3 -Mom: BC’s IC Need to solve

5. Can we convert PDE into an ODE by defining a new variable which combines x and t ? This is a self-similar solution Let and (a) Convert and to system U0U0

6. Substitution into the governing PDE Want ODE in f and Need to get rid of t or - (8) - (9) - (10) BC’s

7. Let Choose

8. From (9) From (10) By Definition (see, e.g. Abramowitz & Stegun)

9. By Definition: As we argued before is scaled by, the viscous length at time t Profile has same shape but height over which velocity varies grows like t 1/2

10. E.g. u 1 /U 0 =0.05 when This formula involving the error function looks like this:

11. What it looks like in terms of x 3 ( =1 m 2 /s) at different times (every 0.1 s) Note that if we had scaled the coordinate x 3 for each profile by (2 t) 1/2, we get the single “self-similar” profile seen on the previous slide

12. In terms of the vorticity: What we have calculated is the diffusion of a spike (sheet) of vorticity initially located at x 3 =0 into the fluid above, If we calculate the circulation around a contour that looks like this (but infinite in the x 3 direction): U0U0 l0l0 A C x3x3 Thus, the total vorticity is constant

13. Vertical Flux For x 3 =0 the flux = 0 except for t = 0 All the vorticity enters the fluid at t = 0 due to the impulsive onset of plate motion Summary: Setting the plate in motion imparts a fixed amount of vorticity to the fluid – This vorticity diffuses into the fluid, spreading vertically such that the layer containing a bulk of the vorticity is thick at time t What about the vertical flux of vorticity from the wall into the fluid?

14. Unsteady Couette flow The analysis of Stokes’ first problem showed a flow that was self- similar such that when viewed with a vertical scale that grew with time, the flow always looked the same. What happens if we confine the flow in a finite depth region? Qualitative: If t << D  / - erf solution near x 3 =D, zero elsewhere If t >> D 2 / - steady Couette flow as before

15. Exact method: Use of separation of variables/Fourier series This part is the steady solution, it satisfies the top and bottom BCs, but not the initial condition (at least by itself) This part is the deviation from the steady solution. It has two properties: (1)With steady solution it satisfies the initial conditions. (2) It satisfies the BCs that are left after we take away the steady solution, i.e Note: Both parts satisfy the p.d.e

16. Thus, we must solve subject to: Approaches: Laplace transforms, numerical methods, or Fourier series

17. We use a Fourier sine series with the amplitude of each Fourier component assumed to be a function of time: Why this form? (which works because of linearity) (1)The sine terms each satisfy the top and bottom boundary conditions on u’ (2) Any function of x 3 can be described as a sum of these sines. This property is known as “completeness”. (3) The sines are “orthogonal” (we’ll see why this is useful). This means that:

18. If we substitute our expansions into the governing PDE we find that How do we find U n (0)? At t = 0, we have the condition

19. Thus we need to find the U n (0) to make the sum = the negative of the steady state solution. This is done by multiplying both sides of this expression by sin(m  x 3 /D) and integrating from 0 to D, i.e. using the orthogonality condition:

20. Thus, when all is said and done we find that t =0 to 0.3 t t

21. Stokes' second problem A flow closely related to the flow of Stokes' first problem is Stokes' second problem! Now we consider the flow induced above a flat plate at x 3 =0 that oscillates according to Again we assume so that

22. First, what happens? If we move the plate to the right, we impart negative vorticity to the fluid – This diffuses upwards Now, when the plate reverses direction. A new front begins to diffuse having opposite sign vorticity. Thus as the plate oscillates for some time, we expect that the vorticity from successive fronts eventually cancels out one-another far enough away from the boundary. Near the boundary, we expect to see only the current forcing’s ‘front’. old front new front

23. To confirm this picture, let’s look at the forced or periodic soln. : We choose dimensionless variables using To create a parameter – free eqn. we set Expressing the fact that we expect the dominant length scale will be the distance vorticity can diffuse in one period Thus with Real part of complex #

24. We now set (and drop *) u is complex The P.D.E. gives We can eliminate one of these values of k by insisting that Eliminate since unbounded.

25. Here

26. These two components look like this (pic in page V39)