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