1 MAE 5130: VISCOUS FLOWS Stokes’ 1 st and 2 nd Problems Comments from Section 3-5 October 21, 2010 Mechanical and Aerospace Engineering Department Florida Institute of Technology D. R. Kirk

2 COMMENTS ON SECTION 3-5: STOKES’ 1 st PROBLEM Stokes’ first problem or the Rayleigh problem: Impulsively started flat plate –Analogous to a conducting solid whose bottom plane is suddenly changed to a different temperature At first particles near wall are accelerated by an imbalance of shear forces As time proceeds, this effect is felt farther and farther from plate, inducing more and more fluid to move along with plate Examining y-direction momentum equation shows that pressure is not a function of x Density, , and absolute viscosity, , do not enter problem independently, but only in combination =  / , which is kinematic viscosity. We will continue to see through this course that is a much more important parameter in fluid mechanics than absolute viscosity (except in low-Reynolds number flows). Heat or diffusion equation is parabolic. Proper conditions to prescribe parabolic equations: –An initial condition for all space –Boundary conditions at two positions in space for all time Solution is an example of a similarity solution. A similarity solution is one where the number of independent variables in a PDE is reduced by one –If choice of a trail similarity variable does not produce an ODE, trail is unsuccessful –Even if similarity variable is found for a PDE, must also make boundary conditions on original problem transform so that new problem makes physical sense

3 ADDITIONAL COMMENTS ON SECTION 3-5: STOKES’ 1 st PROBLEM Velocity profile shows that influence of plate extends to infinity immediately after plate starts moving At large distances error function vanishes ‘exponentially’ (actually, erf(  ~  -1 exp(-  2 ) as  → ∞), but there is still a minute viscous influence through out flow. This can be rationalized influence at infinity by considering molecular model of gas viscosity. Molecules that collide with plate absorb some extra momentum before returning to fluid. Although for most part, molecules collide with other molecules several times before getting very far from plate, in principle, there is some possibility (probability) of molecules traveling to infinity without a collision. Diffusion: –Slows as time goes on –Depends on kinematic (not the absolute) viscosity –Independent of plate velocity Rayleigh problem also applies to flow above a stationary plate when fluid is started impulsively with a uniform velocity. Solutions are related by a Galilean transformation History: –Mathematical solution to this problem was first given by Stokes (1851), but now more often called Rayleigh problem because Rayleigh (1911) used result in a creative way to derive a skin friction law

4 STOKES’ 1 st PROBLEM, 3-5: IMPULSIVELY STARTED FLAT PLATE Plate’s effect diffuses into fluid at a rate proportional to square root of kinematic viscosity Define shear layer thickness, , as point where wall effect on fluid has dropped to 1% –u/V w = 0.01,  ~ 3.64√( t) Example: –Diffusion length after 1 min for water  ~ 2.8 cm ( ~ cm 2 /s) –Diffusion length after 1 min for air  ~ 10.8 cm ( ~ cm 2 /s) –In terms of viscous diffusion, air is more viscous than water by a factor of about 10

5 STOKES’ 1 st PROBLEM, 3-5: IMPULSIVELY STARTED FLAT PLATE In general diffusion is a slow process Compare: Boeing 747 flying at 500 MPH, particle travels from nose to tail in ⅓ s

6 STOKES’ 2 nd PROBLEM, 3-5.1: FLUID OSCILLATION ABOVE AN INFINITE PLATE

7 Plots show period T = , 2 , and 6  Can also develop expression for transient for a Stokes oscillating plate –Requires detailed understanding of complex variables

8 SOLUTION AND PLOT COMMENTS The first exponential term provides a damping in the y-direction –Note that the depth at which the viscosity makes itself felt is proportional to √, just as it was in Stokes’ 1 st problem The second term exhibits a wave-like behavior, with an apparent wave velocity of √2  –This interpretation is tied up in the oscillating boundary condition –The physical process that is occurring is viscous diffusion, and Stokes’ 2 nd problem does not have a physical wave velocity, although we could trace the depth of penetration of the diffusion effect as a function of time –Note how the effect of the wall motion is delayed –When wall reverses its motion and generates a net shear in the opposite direction, say at T=  /2, a net acceleration force from viscosity still exists deeper in fluid, say at Y=2. –Only after some time delay does the net shear force within the fluid change sign and begin to decelerate the fluid.