2


The physical meaning of the terms in the Navier-Stokes equations can be interpreted as follows. Multiplying by and using continuity, the equations can be rewritten as

Term A ~ time rate of change of momentum

Term B ~ pressure force

Term C ~ net convective inflow rate of momentum ~ inertial force

Term D ~ viscous force ~ net diffusive inflow rate of momentum

Term E ~ gravitational force

ij

i

jji

ji

i gx

u

xuu

xx

p

t

u

A B C D E

3


We make the transformations (u1, u2, u3) = (u, v, w) and (g1, g2, g3) = (gx, gy, gz). Expanding out the equations we then obtain the following forms for the Navier-Stokes equations:

and the following form for continuity:

0z

w

y

v

x

u

z2

2

2

2

2

2

y2

2

2

2

2

2

x2

2

2

2

2

2

gz

w

y

w

x

w

z

p1

z

ww

y

wv

x

wu

t

w

gz

v

y

v

x

v

y

p1

z

vw

y

vv

x

vu

t

v

gz

u

y

u

x

u

x

p1

z

uw

y

uv

x

uu

t

u

4

LAMINAR PLANE COUETTE AND OPEN CHANNEL FLOWThe simplest flow we can consider is constant rectilinear flow. For example, consider a flow with constant velocity U in the x direction and vanishing velocity in the other directions, i.e. (u, v, w) = (U, 0, 0). This flow is an exact solution of the Navier-Stokes equations and continuity.

x

y

U0z

w

y

v

x

u

z2

2

2

2

2

2

y2

2

2

2

2

2

x2

2

2

2

2

2

gz

w

y

w

x

w

z

p1

z

ww

y

wv

x

wu

t

w

gz

v

y

v

x

v

y

p1

z

vw

y

vv

x

vu

t

v

gz

u

y

u

x

u

x

p1

z

uw

y

uv

x

uu

t

u

Thus for any constant rectilinear flow, all that needs to be satisfied is the hydrostatic pressure distribution (even though there is flow):

z

y

x

gz

p10

gy

p10

gx

p10

or ii

gx

p

5


For plane Couette flow we make the following assumptions:• the flow is steady (/t = 0) and directed in the x direction, so that the

only velocity component that is nonzero is u (v = w = 0);• the flow is uniform in the x direction and the z direction (out of the

page), so that /x = /z = 0;• the z direction is upward vertical;• the plate at y = 0 is fixed; and• the plate at y = H is moving with constant speed U

For such a flow the only component of the viscous stress tensor is

y

x

u

moving with velocity U

fixed

fluid

xy2112

H

6

LAMINAR PLANE COUETTE AND OPEN CHANNEL FLOWThat is, the components of the viscous stress tensor are

y

x

u


fixed

fluid

000

00dy

du

0dy

du0

z

w2

y

w

z

v

x

w

z

u

y

w

z

v

y

v2

x

v

y

u

x

w

z

u

x

v

y

u

x

u2

vij

Here we abbreviate dy

duxy

H

7

LAMINAR PLANE COUETTE AND OPEN CHANNEL FLOWThus u = u(y) only, and v = w = 0. This result automatically satisfies continuity:

0z

w

y

v

x

u

Momentum balance in the x, y and z directions (z is upward vertical)

z2

2

2

2

2

2

y2

2

2

2

2

2

x2

2

2

2

2

2

gz

w

y

w

x

w

z

p1

z

ww

y

wv

x

wu

t

w

gz

v

y

v

x

v

y

p1

z

vw

y

vv

x

vu

t

v

gz

u

y

u

x

u

x

p1

z

uw

y

uv

x

uu

t

u

g

8

LAMINAR PLANE COUETTE AND OPEN CHANNEL FLOWMomentum balance in the z direction (out of the page):

y

x

u


fixed

fluid

gdz

dpg

dz

dp10

That is, the pressure distribution is hydrostatic. Recall that the general relation for a pressure distribution ph obeying the hydrostatic relation is:

ii

gx

p

H

9

LAMINAR PLANE COUETTE AND OPEN CHANNEL FLOWMomentum balance in the x (streamwise) direction:

y

x

u


fixed

fluid

The no-slip boundary conditions of a viscous fluid apply:the tangential component of fluid velocity at a boundary = the velocity of the boundary (fluid sticks to boundary)

Uu,0uHy0y

dy

duwhere0

dy

dor

dy

ud0

2

2

H

10

LAMINAR PLANE COUETTE AND OPEN CHANNEL FLOWIntegrate once:

y

x

u


fixed

fluid

Thus the shear stress must be constant on the domain.

Uu,0uHy0y

dy

duwhereCorC

dy

du11

H

Integrate again:

Apply the boundary conditions to obtainC2 = 0, C1 = U/H and thus

21 CyCu

H

U

H

U,

H

yUu

11


For open-channel flow in a wide channel we make the following assumptions:• the channel has streamwise slope angle ;• x denotes a streamwise (not horizontal) coordinate, z denotes an

upward normal (not vertical) coordinate and y denotes a cross-stream horizontal coordinate;

• the flow is steady (/t = 0) and directed in the x direction, so that the only velocity component that is nonzero is u (v = w = 0);

• the flow is uniform in the x direction and the y direction (out of the page), so that /x = /y = 0;

• the bottom of the channel at z = 0 is fixed;• there is no applied stress at the free surface where z = H.

Hx

z

u

12


The channel width is denoted as B. It is assumed that the channel is sufficiently wide (B/H << 1) so that sidewall effects can be ignored.

Thus streamwise velocity u is a function of upward normal distance z alone, i.e. u = u(z).

H

BThe vector of gravitational acceleration is (gx, gy, gz) = (gsin, 0, -gcos)

Hx

z

u

g

gcos

gsin

13


Hx

z

u

Continuity is satisfied if u = u(z) and v = w = 0.

0z

w

y

v

x

u

The equations of conservation of streamwise and upward normal momentum reduce to:

cosgz

w

y

w

x

w

z

p1

z

ww

y

wv

x

wu

t

w

singz

u

y

u

x

u

x

p1

z

uw

y

uv

x

uu

t

u

2

2

2

2

2

2

2

2

2

2

2

2

14

Hx

z

u

The equations thus reduce to:


cosgdz

dp10

singdz

ud0

2

2

Since

dz

du

x

w

z

u3113

The first equation can thus be rewritten as

singdz

d0

where is an abbreviation for 13 = 31.

15

Hx

z

u

Assuming that a) pressure is given in gage pressure (i.e. relative to atmospheric pressure) and there is no wind blowing at the liquid surface, the boundary conditions on


cosgdz

dp,sing

dz

ud

dz

d2

2

are

0p

0

0u

Hz

Hz

0z

viscous fluid sticks to immobile bed

no applied shear stress as free surface

gage pressure at free surface = 0 (surface pressure = atmospheric)

16

Now the condition


cosgdz

dp

states that the hydrostatic relation prevails perpendicular to the streamlines (which are in the x direction). Integrating the relation with the aid of the boundary condition

0pHz

yields a pressure distribution that varys linearly in z:

H

z,)1(cosgHp

Hx

z

u

p

17


The equation

singdz

d

subject to

0Hz

similarly yields a linear distribution for shear stress in the z direction:

H

z,)1(singH

Hx

z

u

Note that the bed shear stress b at z = 0 is given as

singHb

18


Thus

H

z1singH

dz

du

subject to

0u0z

Integrates to give the following parabolic profile for u in z:

H

z,

2

1sinH

gu 22

Hx

z

u

19


The maximum velocity Us is reached at the free surface, where z = H and = 1);

Thus

H

z,

2

12

U

u 2

s

Hx

z

u

sinHg

2

1U 2

s

Depth-averaged flow velocity U is given as 1

0

H

0ududz

H

1U

Thus

2

3

U

U,sinH

g

3

1U s2

20


A dimensionless bed friction coefficient Cf can be defined as

Here Cf = f/8 where f denotes the D’arcy-Weisbach friction coefficient. Between the above relation and the relations below

Hx

z

u

2b

f UC

it can be shown thatHere Re denotes the dimensionless Reynolds No. of the flow, which scales the ratio of inertial forces to viscous forces.

,sinHg

3

1U 2

singHb

UH,

3Cf Re

Re

21


Now suppose that there is a wind blowing upstream at the free surface, exerting shear stress w in the – x direction. The governing equations of the free surface flow remain the same as in Slide 15, but one of the boundary conditions changes to

wHz

Hx

z

w

u

The corresponding solution to the problem is

,)r1(singH H

z,

2

1)r1(sinH

gu 22

where r is the dimensionless ratio of the wind shear stress pushing the flow upstream to the force of gravity per unit bed area pulling the flow downstream:

Hsingr w

22

LAMINAR PLANE COUETTE AND OPEN CHANNEL FLOWThe solution for velocity with the case of wind can be rewritten as

where und is a dimensionless velocity equal to 2u/(gsinH2).

,H

z,

2

1)r1(2u 2

nd

A plot is given below of und versus for the cases r = 0. 0.25, 0.5, 1 and 1.5.

Hsingr w

Velocity Distribution with Wind

0

0.2

0.4

0.6

0.8

1

-2 -1.5 -1 -0.5 0 0.5 1

und

r = 0(no wind)r = 0.25r = 0.5r = 1r = 1.5

laminar plane couette and open channel flow

Documents