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
