cee 262a h ydrodynamics lecture 12 steady solutions to the navier-stokes equation
DESCRIPTION
Generally, the model flow illustrates 1 of 3 possible two-way force balances (2 terms involved): Force 1Force 2Examples InertiaFrictionEkman layer Stokes Layer (Boundary layers) InertiaPressure (buoyancy) Waves Geostrophy Bernoulli Stratified flows PressureFrictionPipe flow Porous media Creeping flows Three way force balances (3 terms) are much harder to deal withTRANSCRIPT
CEE 262A HYDRODYNAMICS
Lecture 12Steady solutions to the Navier-Stokes
equation
Exact solutions to the Navier Stokes Equations
Real flow problems are too complicated for us to be able to solve the NS equations subject to appropriate BCs – We must simplify matters considerably!
Our method:
1. Reduce real flow to a much simpler, ideal flow
2. Solve model problem exactly
3. Extract important dynamical features of solutions
4. Apply lessons learned to real flow
Generally, the model flow illustrates 1 of 3 possible two-way force balances (2 terms involved):
Force 1 Force 2 Examples
Inertia Friction Ekman layerStokes Layer(Boundary layers)
Inertia Pressure (buoyancy)
WavesGeostrophyBernoulliStratified flows
Pressure Friction Pipe flowPorous mediaCreeping flows
or
u u ut
or ' p g
Three way force balances (3 terms) are much harder to deal with
Poiseuille – Couette Flows
Consider the non-rotating, constant density flow between two infinite parallel plates, one of which moves the other of which is fixed:
U0
x3=0 Fixed plate:
Moving plate:x3= H
Constant pressure gradient1
px
Because the plates are infinite,
1 2
0
anything anything
x x
except pressure
1 0
2
3
00
u Uuu
1
2
3
000
uuu
everything
So, continuity tells us that
31 2
1 2 3
3
3
3
3
0
0 0 0
0
uu uu
x x xuxux
Thus, most generally, u3=fn(x1,x2). But since u3 = 0 on the top (or bottom plates), it must be 0 everywhere. Additionally since there is no pressure gradient or plate motion in the x2 direction, u2 = 0 . Thus, the only non-zero flow component is u1(x3)
2~0 0 ~~ ~ ~ ~
' 2
uu u p u u
t
Steady (by assumption)
~ 0
u
t~ ~
2 0 u
No rotation (by assumption)
Now we turn to the Navier Stokes equation to see what can be eliminated:
i
1 1 11 1 2 3
1 2 3
Since either or x 0 for all i:
0 0 0 0
iuu u uu u u u ux x x
' , 0,0 p
2 2 22 1 1 1
1 2 2 21 2 3
2 21 12 2
3 3
0 0
u u uux x x
u ux x
(a) u1=0 at x3=0 B=0
(b) u1=U0 at x3=H
2
0 2
HU HA
0
2
U HAH
3 01 3 3 3( )
2
x Uu x x H xH
So what is left is2
12
3
u
xSubject to the boundary conditions that
1 0
1 0 0
u H U
u
Thus if we integrate twice with respect to x3
And use the boundary conditions
We find that
BAxxu 3
23
1 2
2
3 31
0
1
x xu P PU H H
2
02
HP
Uwhere
If we divide both sides by U0 we can rewrite our velocity distribution in a convenient non-dimensional form:
23
12
1
0xu
xp
1xp
20
23
12
HU
xu
0
2
23
121
221
UH
xu
xp
P
Pressure-friction balance
The parameter P represents the relative importance of the imposed pressure gradient and the moving surface
-10-5
10 5 3
-31 -10
P=-10: dp/dx<0P=+10: dp/dx>0Flow Flow
We can calculate the flow rate
23 3
1 3 0 30 0
1 20 30
13 2
0 0 0
0
1
1
1 113 2 3 2 2 6
H H x xQ u dx U P P dxH H
U H P P d x H
PP PU H P P U H U H
Thus, Q = 0 when P = 3, whereas Q > 0 when P < 3, and Q < 0 when P > 3
P = 3 – wind driven flow in a lake
Wind
Two special cases
1. Plane Couette Flow: 0
0 31 3( ) U x
U xH
1 0 2 QU UH0 2Q U H
Stress on plane parallel to plates
x3=0
x3=2b 0U
~n
)1,0,0(~n
HUfi
xu
xu
xu
ennnnf
i
i
i
iiiiijiji
0131
3
3
3
33332211
1:
2
~f
Resultant force is inx1 direction and constant in x3.
2. Plane Poiseuille Flow: U0=0
31 3 3( )
2
xu x H x
3
12
HQwhence:
and2
1 12
HU
Tangential Stress =
Shear stress varies linearly with depth
x3
for 0
Wall stress opposes p.g.
1u
313
113 2
xHxp
xu
Darcy's law: velocity ~ pressure gradient
00,0: 13133131 fnnf
00,0: 13133131 fnnf