Lecture 12 Principle of exchange of stabilities. As Ra is varied the unstable modes ( R > 0 ) are separated from the stable modes ( R > 0 ) by modes for which R = 0, I = 0 . N.B. This isn't usually the case in other physical problem
Consider the structure of a growing disturbance in a parallel shear flow. Streamlines are useful for visualization. Streamline pattern near critical level. u = [U + u, w] = + , - x z
Transform to coordinate system mov
Sturm Liouville Eigenvalue problem. Consider an equation y + q ( x ) - s ( x ) y = 0 with y ( x1 ) = y ( x2 ) = 0 where y is the eigenfunction. If everywhere in [ x1 , x2 ] we have p ( x ) > 0 and s ( x ) 0 , then there exists eigen
Inviscid instability of steady rotating flows Coordinate system cylindrical ( r , , z ) ( u, v, w)
or ( vr , v , vz )
Incompressible, Inviscid Governing Equations
1. 2. 3. 4. 5. 1 p Du v 2 - =- radial momentum r Dt r 1 1
University of Florida EOC 6934 Hydrodynamic Stability
Fall 2004 Civil and Coastal Engineering
Instructor: Don Slinn Office: 575 I Weil Hall, 392-9537 x 1431, firstname.lastname@example.org Office Hours: MWF 3:00 pm, TTh 9:30 am, and an open door policy Class Time:
Introduction A classic problem Reynolds Experiment (1883). Constant Head Tank
Introduced dye into inlet of pipe. Key parameter in interpretation of experiment was Reynolds number.
- density - viscosity
D - diameter
u - avg. velocity
uD = Re
Reynolds Supposition Assume u ( x ) steady solution to N. S. equations and B. C. Examine Behavior u ( x ) + u ( x, t ) Analytical Methods 1. Linear theory assume small u , linearize equations eigenvalue problem. Gives growth rate of disturbances
Lecture 4 Governing equation
1 u + u iu = - p - g t iu = 0
( or )
Note that in this inviscid formulation, a shear along the interface is allowable. 1. Assume flow is irrotational except at interface. Therefore ui , i = 1, 2
ui = i a velocity potential
Lecture 5 Solve Laplace's equation for velocity above and below the boundary condition. Assume Fourier series expansion
( x, y , t ) =
A ( t ) cos
,m =0 m ,m =0
x cos my x cos my
i ( x, y, z , t ) =
F ( z, t ) cos
Conservation of mass is guarantee
Modifying the Rayleigh-Taylor inviscid analysis to include surface tension at the interface will depress smaller scale motion. We expect that surface tension to suppress instability in the analysis it modifies the dynamic B.C. Before p1 - p2 = 0
The Kelvin Helmholtz Problem Instability of aVortex Sheet Examples:
- potential temperature - temperature a fixed parcel would have if moved adiabatically to a
given pressure level (1 bar).
P r T P
Instabilty: Kelvin-Helmholtz Continued Plug these into 2 Kinematic and 1 dynamic B.C.
t i( x + my )
) - kB = A ( e
t i( x + my )
) - 1 Ui A ( e 2
t i( x + my )
1 2. +kB2 = A + Ui A 2 1 1 3. 1 - B1 - gA + Ui B1 - 2 B2 - gA - Ui
Kelvin-Helmholtz Solution Continued For a given k, Re ( ) = R is greatest for k = , most unstable disturbances are 2-D. For a homogeneous fluid. 1 = 2 , T = 0, C0 = 0
1 U 2
Stability condition U 2 > 0
Any shear is unstable! If 2 > 1 , C02 ( k
Lecture 10 Convective (thermal) instability. Instability of a fluid layer heated from below. Basic Problem.
1. Assume initial instability is isentropic heat transfer by conduction is relatively slow (steady state). 2. Temperature an
Lecture 29 Consider the Reynolds stress of a perturbation.
1 2 u + u iu = -p + u Re t iu = 0 & B.C.'s. Assume 2-D, periodic in x. u ( x, t ) = u + u where u is u averaged over wavelength so u changes as u grows. Analagous to Reynold
Lecture 28 Rayleigh's Inflection Point Theorem Continued. Physical Mechanism: Suppose U 0 on z1, z2
d d U = ( - ) < 0 dz dz Say - background vorticity U = Vorticity increasing in z
Rayleigh's criterion says that U must have extre
*Do HW Problems 6.1 & 6.2 out of Drazin (pgs. 106-107) Rayleigh-Benard Convection: Continued.
- k 2 ) = -Rak 2
For neutral solutions with assumed solution becomes ( - n 2 2 - k 2 ) = - Rak 2 Particular values of Ra for different values
Experiments: 1. Benard (1900) Used a number of different liquids with high viscosity v . Thin layers, less than 1mm thick.
Water heated at the bottom Circulated hot water across solid surface to maintain constant temperature of wall. Observed r
Lecture 15 Rayleigh Benard Convection Continued. Ruby Krishnamurti (FSU). 1968, JFM, 33, 445. 1970, JFM, 42, 295. Looked at allowing the heat flux to slowly increase, causing the lower temperature to slowly increase as well. The tem
Lecture 16 HW Problem: For the Benard problem, prove that for neutral conditions
V fluctuating density and velocity
u dV - k z
horizontal average of fluctuating density.
That is, that the heat flux at the bottom is proportional
Lecture 17 Dispersion relation for double diffusion stability problem.
Pr 2 a 2 6 2 2 p 3 + ( Pr + + 1) k 2 p 2 + ( Pr + Pr + ) k 4 - p ( Ra - Rs ) p + Pr k + ( Rs - Ra ) Pr a = 0 2 k where k 2 = 2 ( a 2 + n 2 )
See figure 8.2, pg.
Lecture 18 New Topic: Chp.'s 2-3 of Drazin Determining the growth rates R = wR Example, Kelvin Helmholtz Instability for viscous fluid leading to numerical solution of the Orr-Sommerfeld Equation:
U ( z)
1. Start with Navier-Stokes
^ The Orr Sommerfeld equation replacing w with
IV - 2 2 + 4 = i Re (U - C ) ( - 2 ) - U
For no slip wall B.C.'s are = 0 and
= 0 at z = 0, L( z ) . z 2 = 0 at z = 0, L( z ) . 2 z
For free slip wall the B.C.'s are = 0 a
Lecture 20 Numerical solution of the Orr-Sommerfeld equation.
IV - 2 2 + 4 = i Re (U - C ) - 2 - U
Separate into A = CB The most obvious approach would be to use 2nd order accurate centered differences. But there is a complication
Lecture 21 www.coastal.ufl.edu/~slinn/Classes/Stability/OrrSommerfeld.f HW# 9: Plot stability curves for 2 velocity profiles, a free shear flow and a boundary layer type profile. Using the Orr-Sommerfeld solver for a relevant range
Lecture 22 Stability of density stratified shear flows.
Similar to K H Competition between shear resulting in instability, that does work against stable density stratification Assumptions Inviscid, non-diffusive, incompressible, Bou
Lecture 23 Stratified shear flows Continued. We derived w 2 w + U 2 w - 2U + U = -N 2 2 x x x x t t Plug in normal mode expansion. Note: Ci > 0 - Instability Ci 0 - Implies stability
(U - C )
D 2 - K 2 w - D 2U (U - C ) w + N 2
Lecture 24 Go back to 1. (W 2 F ) + ( N 2 - K 2W 2 ) F = 0 Now multiply by F and integrate over [ z1 , z2 ]
2 F ) F + ( N 2 - K 2W 2 ) F dz = 0
First term: by parts
2 (W F ) Fdz = F (W F )
- W 2 F d