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