Chapter 2

Viscous flow (Navier-Stokes eq.)

2-1

(1) Continuity eq. (conservation of mass)

(2) Momentum eq (conservation of momentum)

v 0 v ,v,u wt

1 eq

2

4v

3

+ v v

+ v v

Dvg p

Dt

3 eqs’

Equation of motion of a viscous fluid

2-2

(3) Energy equation (conservation of energy) (Heat Transfer)

(4) equation of state

(5) material law

unknown : 9 unknowns

equation : 9 equations

1 eq

dissipation function

,P T

,

,

,p p

P T

k k P T

C C P T

,v, , , , , , , pu w T P k C

Equation of motion of a viscous fluid

2-3

'''p

DTC k T q

Dt

(1) Isothermal system (T= constant, no heat transfer)

(2) Constant viscosity , μ

(3) Constant ρ ( incompressible flow)

Continuity equation

(linear P.D.E.)

Momentum equation

0V

2 non-linear P.D.E.

4 unknows , v, ,

4 equations , , momention eq. & continuity eq.

DVP g V

Dt

u w p

x y z

First consider

2-4

(a) Initial condition

(b) Boundary conditions

, , , 0V x y z t

0 0 0 ( , , , )

(at all boundary points , t)

P x y z t

V

To solve must specify

I. pressure , 1st order

Π. velocity , 2nd order

2-5

x-component

y-component

z-component

2DVg P V

Dt

2

x

Du pg u

Dt x

2vvy

D pg

Dt y

2

z

Dw pg w

Dt z

2-6

2V V

Dg P

Dt

N-S equation with ρ, μ=constant

inertia gravity pressure force viscous force

Nervier-Stokes equetion

surface force T

2-7

Solid surface - no slip condition

- impermeable

Note: (1) porous surface-fluid injected or removed

through surface

(2) rarefied gases: can have slip at wall

tangent to surface 0V

normal to surface 0V

tangent 0V

Boundary Conditions

2-8

Here p’ is the difference between the pressure in a fluid in motion

and a fluid at rest.

'

hLet p = p p

h

2

: hydrostatic pressure

: deviation from hydrostatic due to the motion of the fluid

"motion pressure" or "modified pressur

p

p'

g ' '

DV'

e"

substit

VDt

ute

hp g p p p

p

2-9

Classes of exact solutions

(a) Parallel flow → nonlinear terms vanish

(b) a variable transformation allows the P.D.E.

to be written as O.D.E

Parallel flow ≡ all fluid particles moves in one directions

only one velocity component is non-zero

Exact solution of Navier-Stokes equetions

2-10

h

2

continuity equation

equation of

if 0 , v 0

0

0

, ,

in terms of thermodynamic pressure = '

u .

mo

t

tion

x

u w

u v w

x y z

u

x

u u y z t

p p p

p ux comp g

x y

2

2 2

z

. 0 =

. 0 = g

y

u

z

py comp g

y

pz comp

z

0 0

2-11

in turns of p’

∵

2 2

2 2

'

' 0

' 0

u dp u u

t dx y z

p

y

p

z

' ' onlyP P x

linear P.D.E. for u(y,z,t)

0hg P

2-12

0

steady 0t

2-D 0z

x

y

g

g g

U

h

x

y g

u= u (y) only

Steady Couette flow with pressure gradient

2-13

2

2

.

( )

( )

at most a function of x , independent of y

1. Constant

py comp g

y

P gy f x

P df x

x dx

d u px comp

dy x

at most a

function of y

at most a

function of x

Steady Couette flow with pressure gradient

2-14

(1) y=0 , u=0

(2) y=h , u=U

2

2

yIntegrate u(y)= 1

h 2

define p2

1 1

h p y yU

x h h

h p

U x

u y y yP

U h h

Boundary conditions

Steady Couette flow with pressure gradient

2-15

0, t>0U h U

U0

h

x

y g ,0 0 0p

u yx

0

0 =0 , =0

0 ,

t y u

t y h u U

x-component

Initial condition

2

2

u u

t y

Boundary condition

0 0 0t y h u

Developing Couette flow, no pressure gradient

2-16

1 2

1

1

1 2

2

2

2

0

1 1

, , ,

1 0 =0 , 0

2 0

B.C.

,

w

h

0

0

,0 ,0 ,0

= ,0

I.C.

ere

=

u y t u y t u y t

t y u

t y h u

u u

t y

t

u y u y u y

u y

yU

h

0

Assume non-homogeneous in y direction

Developing Couette flow, no pressure gradient

2-17

2

2

2

2 0

2 1 2

02 1

2 0

2

steady state solution

1 0 0

B.C.

soluti

2

where

on

B.C. 1 B.C.

0 ,

2

0

y u

y h u U

u

d u

dy

c y c

Uc c

h

yu y U

h

Developing Couette flow, no pressure gradient

2-18

2

1

1

22

2

2

2

1

2 2

2

2

Now to solve

let ,

1

take

need 0 at t

need a characteristic value problem in y direction

T

1

: 0

: 0

si

2

n

t

u

u y t T t Y y

dT d Y

T dt y dy

T t

dTT c e

dt

d YY y

dy

Y y c y

3 cosc y

﹖

Developing Couette flow no pressure gradient

2-19

2

1 3

1 2

1

0

B.C. 0 0 0

0 sin 0

0,1,2,3.......

0,1,2,3,......

, in

1

2

sn

n

t nn

n

y u c

y h u c h

h n n

nn

h

yu y t c e

Developing Couette flow no pressure gradient

2-20

1 0

0

0 0

00 0

0

I.C. : 0

sin sin

Determine C

sin sin sin

nn n

n n

n

h h

ny y

n

yt u U

h

yy n yU c c

h h

y m y n y m yU dy c dy

h h h h

0

2

m n

hm n

2 2

2

00

, 12sin

nnt

h

n

u y t y n ye

U h n h

Developing Couette flow no pressure gradient

2-21

t

increasest

0t

0U

Developing Couette flow, no pressure gradient

2-22

y

g 0 , 0

0

t u y t

v

00,U t U f t

0

2

2

0

D.E.

I.C. 0 u ,0 0

B.C. (1) , 0

2 0 0, constant - stokes 1st problem

u u

t y

t y

y u t

y u t U f t U

0 - stokes 2nd pcos t roblemU

: kinematic viscosity

Sudden accelerated flat plate

2-23

0

2 2 2

2 2 20

00 0

00

2 2

, , , 0

,

, ,

= , 0 ,

cos t

st

st

st st st

t t

u y t U y s u y t e dt s

u u d Ue dt y s

y y dy

u ue dt u y t e s u y t e dt

t t

u y t sU y s

UU

s

s

s

£

£

£

£

£

Solution by Laplace transform

NOTE：

(1)

(2)

(3)

(4)

(5)

2-24

2

2

2

2

1 2

transformed D.E.

, 0

0

,

s sy y

v v

d Uu y t sU

dy

d U sU

dy

U y s C e C e

Sudden accelerated flat plate

2-25

2

2

0 01

0

0

1 , 0 0

2 0 stokes 1st problem 0,

Apply transformed B.C.

transfor

,

, erfc 2

2 erfc =

med b

ack to & plane

where = 1 er

sy

v

y T

y U s C

U Uy U s C

s s

U eU y s

s

yu y t U

t

e d

2

0

2f 1 e d

Sudden accelerated flat plate

2-26

2

y

t

0

uU

Sudden accelerated flat plate

2-27

The flow near an oscillating flat plate NCKU Heat Exchanger LAB.

28

0

0

0

0

For stokes 2nd problem

0 0,

Solution

u(y,t)=U cos( )

cos

where , putting = ky = y /

t

22

u(y,t)=U cos( )

ky

ky

y u t U f t

e

k

U

t ky

e t

The flow near an oscillating flat plate NCKU Heat Exchanger LAB.

29

t 2y

vorticity curl y

v

0 irrotational flow

0 [ a = 0 ]

note: 1.

2.

i j k

Vx z

u w

V

Vorticity equation

2-30

Vorticity equation

N-S eqation

2

2

2

v

=2

2

left hand sid

e

DV Pg

Dt

DV VV V

Dt t

V VV V

t

DV VV

Dt t

V Vt

0

2-31

2

2

2

2

=

=

rig

=

ht hand

side

Combining

V V V

V Vt

D

Dt

2 V

Helmholtz’s eq

N-S eqation

Vorticity equation

0

0

2-32

2

2 2

2 2

v

v

2 D flow

vorticity tra

ns

port equ

ation

2-D

v

V ui j

uV k k

x y

D

Dt

ut x y x y

v 0

2 equations

2 unknows u, v

u

x y

Vorticity equation

2-33

, defined such that it exactly sa

2-D planar flowstream function-useful only for 2

tisfies conservation of

-D flow3-D axialy symmetric flow

i.e. , v incompressible flo

mas

c

w

s.

uy x

2 2v

h

eck continuity q.

0

e

u

x y y x x y

Stream function

2-34

By definition of the streamline

v

or v 0

0

constant (along a streamline)

dy

dx u

dx dyudy dx

u v

dy dx dy x

x

y

x

y f (x,y)

V

Physical meaning of the stream line

Stream function

2-35

flow

2 1

11

2

1

2 flow

2 1

1

1 2 2 1Q

volume flow rate

Stream function

2-36

2 2

2 2

2 2

2 2 24

4 4 44

4 2 2 4

vorticity

substitute into transport eq 2-D

=

irrotational flow 0

where 2

vorticity eq in term

v u

x y x y

t y y x y

x x y y

s of

Vorticity equation in terms of Ψ

2-37

1 unknown

1 eq.

NCKU Department of Mechanical Engineering