058:0160
Professor Fred Stern
Fall 2005
Chapters 3 & 4
1
Chapters 3 & 4: Integral Relations for a Control Volume
and Differential Relations for Fluid Flow
058:0160
Professor Fred Stern
Fall 2005
Chapters 3 & 4
2
058:0160
Professor Fred Stern
Chapters 3 & 4
3
Fall 2005
Reynolds Transport Theorem (RTT)
Need relationship between
Bcv    dm    d .
CV
 
d
Bsys and changes in
dt
CV
dBcv d
1 = time rate of change of B in CV =
   d
dt
dt
CV
2 = net outflux of B from CV across CS =
 V
R
 n dA
CS
dBSYS d

  d  CS V R  n dA
dt
dt CV
General form RTT for moving deforming control volume
058:0160
Professor Fred Stern
Chapters 3 & 4
4
Fall 2005
Special Cases:
1) Non-deforming CV moving at constant velocity
dBSYS

   d   V R  n dA
dt
t
CV
CS
2) Fixed CV
dBSYS

   d   V  n dA
dt
t
CV
CS
  b d    b  n dA
Greens Theorem:
CV
CS
dBSYS


        V  d
dt
t

CV 
Since CV fixed and arbitrary lim gives differential eq.
d0
3) Steady Flow:

0
t
4) Uniform flow across discrete CS (steady or unsteady)
 V  n dA   V  n dA
CS
CS
(- inlet, + outlet)
058:0160
Professor Fred Stern
Chapters 3 & 4
5
Fall 2005
Continuity Equation:
B = M = mass of system
β=1
dM
 0 by definition, system = fixed amount of mass
dt
Integral Form:
dM
d
0
  d  CS  V R  n dA
dt
dt CV

d
 d    V R  n dA

dt CV
CS
Rate of decrease of mass in CV = net rate of mass outflow across CS
Note simplifications for non-deforming CV, fixed CV,
steady flow, and uniform flow across discrete CS
Incompressible Fluid: ρ = constant

d
 d  CSV R  n dA
dt CV
“conservation of volume”
058:0160
Professor Fred Stern
Chapters 3 & 4
6
Fall 2005
Differential Form:

   V   0
t
M  
dM   d   d  0

    V  V    0
t
d d



1 D
1 D

 Dt
 Dt

D
   V  0
Dt
1 D

Dt


rate of change 
per unit 


V

0
u v w
 

x
y z

rate of change 
per unit 
Called the continuity equation since the implication is that
ρ and V are continuous functions of x.
Incompressible Fluid: ρ = constant
 V  0
u v w


0
x y z
058:0160
Professor Fred Stern
Fall 2005
Chapters 3 & 4
7
Momentum Equation:
B = MV = momentum, β = V
Integral Form:
d ( MVv) d

v vRR  n dA   F
 Vv d   CS VρV
dt
dt CV
3
1
2
F =
=
vector sum of all forces acting on CV
FB + F s
FB =
Body forces, which act on entire CV of fluid due
to external force field such as gravity or
electrostatic or magnetic forces. Force per unit
volume.
Fs =
Surface forces, which act on entire CS due to
normal (pressure and viscous stress) and
tangential (viscous stresses) stresses. Force per
unit area.
When CS cuts through solids Fs may also include FR =
reaction forces, e.g., reaction force required to hold nozzle
or bend when CS cuts through bolts holding nozzle/bend
in place.
1 = rate of change of momentum in CV
058:0160
Professor Fred Stern
+
Chapters 3 & 4
8
Fall 2005
2 = rate of change of momentum across CS
3 = vector sum of all body forces acting on entire CV
and surface forces acting on entire CS.
Many interesting applications of CV form of momentum
equation: vanes, nozzles, bends, rockets, forces on bodies,
water hammer, etc.
Differential Form:










V    
 V V d   F
CV  




t 

tensor 
 

  V
V

 t

t
^
^
^
V V  u i V   j V  w k V
  ( V V ) 



( u V )  (  V )  ( wV )
x
y
z
 V  ( V )  V  V
= 0 , continuity
  

 V



V




V



V


V



d   F

CV   t

t



058:0160
Professor Fred Stern


CV
Fall 2005
DV
d   F
Dt
DV
f
Dt

a  f  f
b
f
f
Chapters 3 & 4
9
b
s
=
=
per elemental fluid volume
s
body force per unit volume
surface force per unit volume
Body forces are due to external fields such as gravity or
magnetic fields. Here we only consider a gravitational
field; that is,
F
body
 dF
grav
  g dxdydz
z
^
and g   g k for
i.e. f
g
^
body
  g k
Surface Forces are due to the stresses that act on the sides
of the control surfaces
Symmetric   
ij
   p  
ij
Normal pressure
ij
ij
2nd order tensor
Viscous stress
ji
058:0160
Professor Fred Stern
Chapters 3 & 4
10
Fall 2005
 p  
  
 

xx
xy
yx
 p 
zx

zy
yy





 p   
xz
yz
zz
As shown before, for p alone it is not the stresses
themselves that cause a net force but their gradients.
Symmetry condition from requirement that for elemental
fluid volume, stresses themselves cause no rotation.
f  f  f
s
p
Recall f  p based on 1st order TS. f is more
complex since  is a 2nd order tensor, but similarly as for
p, the force is due to stress gradients and are derived
based on 1st order TS.

p
ij
^
^
^
   i  j  k
x
xx
xy
^
xz
^
^
   i   j  k
y
yx
yy
^
yz
^
^
   i   j  k
z
zx
zy
zz
Resultant
stress
on each face
058:0160
Professor Fred Stern
Chapters 3 & 4
11
Fall 2005




F s   ( x )  ( y )  ( z )  dxdydz
y
z
 x

fs 



( x )  ( y )  ( z )
x
y
z
 yx 

dy  dxdz
  yx 

y


y
 xx dydz





dx  dydz

x


xx
x
z
xx
 yx dxdz

^


f s   ( xx )  ( yx )  ( zx )  i
y
z
 x


^


  ( xy )  ( yy )  ( zy )  j
y
z
 x


^


  ( xz )  ( yz )  ( zz )  k
y
z
 x

f s     ij 

 ij
x j
058:0160
Professor Fred Stern
Chapters 3 & 4
12
Fall 2005



( xx )  ( xy )  ( xz )
x
y
z



f sy  ( yx )  ( yy )  ( yz )
x
y
z



f sz  ( zx )  ( zy )  ( zz )
x
y
z
f sx 
Putting together the above results,
a  
motion
^
DV
  g k     ij
Dt
body force
due to
gravity
surface force = p + viscous terms
(due to stress gradients)
Next, we need to relate the stresses σij to the fluid motion,
i.e. the velocity field. To this end, we examine the
relative motion between two neighboring fluid particles.
B
A
@ B:
dr
(u,v,w) = v
1st order Taylor Series
V  dV  V  dr  V
 ux

dV  dr  V   vx
 wx

relative motion
u z  dx 

vz  dy   eij dx j
wz   dz 
uy
vy
wy
deformation rate
tensor =
e
ij
058:0160
Professor Fred Stern
eij 
Fall 2005
ui
1  u u 
1  ui u j 
  i j





x j
2  x j xi 
2  x j xi 
symmetric part anit symmetric part
 ij  ji
 
ij ji




0


1
ij   (vx  u y )
2



1
 ( wx  u z )
2

where
Chapters 3 & 4
13
1
(u y  vx )
2
0
1
( wy  v z )
2





1

(u z  wx ) 
2

1
(vz  wy )   rigid body rotation
 of fluid element
2



0



= rotation about x axis
 = rotation about y axis
ς= rotation about z axis
Note that the components of ij are related to the vorticity
vector define by:
^
^
^
    v  ( wy  vz ) i  (u z  wx ) i  (vx  u y ) k
2
2
2
= 2  angular velocity of fluid element
058:0160
Professor Fred Stern
Chapters 3 & 4
14
Fall 2005
1
 ij  rate of strain tensor
2
1

u
(u y  vx )
x

2

1
  (v x  u y )
vy
2
1
 ( wx  u z ) 1 ( wy  vz )
2
 2
1

(u z  wx ) 
2

1
(v z  wy ) 

2


wz

u  v  w    v = elongation (or volumetric dilatation)
x
y
z
of fluid element
1
(u  v ) = distortion wrt (x,y) axis
2
1
(u  w ) = distortion wrt (x,z) axis
2
1
(v  w ) = distortion wrt (y,z) axis
2
y
x
z
x
z
y
Thus, general motion consists of:
1)
2)
3)
4)
pure translation described by v
rigid-body rotation described by ½ ω
volumetric dilatation desribed by   v
distortion in shape desribed by ij
ij
058:0160
Professor Fred Stern
Fall 2005
Chapters 3 & 4
15
It is now necessary to make certain postulates concerning
the relationship between the fluid stress tensor (σij) and
rate-of-deformation tensor (eij). These postulates are
based on physical reasoning and experimental
observations and have been verified experimentally even
for extreme conditions.
1) When the fluid is at rest the stress is hydrostatic and
the pressure is the thermodynamic pressure
2) σij is linearly related to eij and only depends on eij.
3) Since there is no shear strain in rigid body rotation,
it causes no shear stress.
4) There is no preferred direction in the fluid, so that
the fluid properties are point functions (condition of
isotropy).
Using statements 1-3
   p  k 
ij
ij
ijmn
ij
kijmn = 4th order tensor with 81 components such that each
stress is linearly related to all nine components of εij.
058:0160
Professor Fred Stern
Chapters 3 & 4
16
Fall 2005
However, statement (4) requires that the fluid has no
directional preference, i.e. σij is independent of rotation of
coordinate system, which means kijmn is an isotropic
tensor = even order tensor made up of products of δij.
k
ijmn
        
( ,  ,  )  scalars
ij
nm
im
jn
in
jm
Lastly, the symmetry condition σij = σji requires:
kijmn = kjimn  γ = μ = viscosity
   p  2   
ij
ij
ij
mm
ij
v
λ and μ can be further related if one considers mean
normal stress vs. thermodynamic p.
  3 p  (2  3 )  v
ii
1
2

p           v
3
3


ii
p  mean
normal stress
2

p  p        v
3

058:0160
Professor Fred Stern
Chapters 3 & 4
17
Fall 2005
Incompressible flow: p  p and absolute pressure is
indeterminant since there is no equation of state for p.
Equations of motion determine p .
Compressible flow: p  p and λ = bulk viscosity must be
determined; however, it is a very difficult measurement
1 D 1 D
requiring large   v  
, e.g., within shock

 Dt  Dt
waves.
Stokes Hypothesis also supported kinetic theory
monotonic gas.
   23 
p p


2


 ij    p    v   ij  2 ij
3
Generalization   
du
for 3D flow.
dy
 u u 

   

x 
 x
i j
i
j
j
i
ij
strain rate
relates shear stress to
058:0160
Professor Fred Stern
Chapters 3 & 4
18
Fall 2005
 u
 x
2
3
   p    v  2 
ii
i
i

 1
u 
   p  2    v 

3

x





i
i
normal viscous stress
Where the normal viscous stress is the difference between
the extension rate in the xi direction and average
expansion at a point. Only differences from the average =
1  u v w 
  
 generate normal viscous stresses. For
3  x y z 
incompressible fluids, average = 0 i.e.   v  0 .
Non-Newtonian fluids:

   for small strain rates  , which works well for
ij
ij
air, water, etc. Newtonian fluids
 
ij

N

ij
nonlinear



t

ij
Non-Newtonian
history effect
Viscoeslastic materials
Non-Newtonian fluids include:
(1) Polymers molecules with large molecular
weights and form long chains coiled together
in spongy ball shapes that deform under shear.
058:0160
Professor Fred Stern
Chapters 3 & 4
19
Fall 2005
(2) Emulsions and slurries containing suspended
particles such as blood and water/clay
Navier Stokes Equations:
^
Dv
a  
  g k    
Dt

^
Dv

  g k  p 
Dt
x
i
ij
2


2





v



3
ij
ij
Recall μ = μ(T) μ increases with gases, decreases with
liquids, but if its assumed that μ = constant:

^
Dv

2 
  g k  p  2      v
Dt
x
3 x
ij
i
i


2 
  g k  p    v 
  v
3 x


^
2
i


2
 ij 
xi
x j
 ui u j 
 2ui

  2ui   2 v

 
 x j xi  x j x j
058:0160
Professor Fred Stern
Chapters 3 & 4
20
Fall 2005
For incompressible flow   v  0
^
Dv

  g k  p   v

Dt
2
^
^
 p where p pz
piezometric pressure
For μ = 0
^
Dv

  g k  p
Dt
Euler Equation
NS equations for ρ, μ constant

Dv
 p   v
Dt
2
 
Dv

 v  v   p   v
 Dt

2
Non-linear 2nd order PDE, as is the case for ρ, μ not constant
Combine with   v for 4 equations for 4 unknowns v, p
and can be, albeit difficult, solved subject to initial and
boundary conditions for v, p at t = t0 and on all boundaries
i.e. “well posed” IBVP.
Application of Momentum Equation:
058:0160
Professor Fred Stern
F 
Chapters 3 & 4
21
Fall 2005
d
 v d   v v  n dA
dt
R
CV
CS
F F F
B
S
Note:
1. Vector equation
2. n = outwards unit normal: v  n < 0 inlet, > 0 outlet
R
3. 1D Momentum flux
 

 v v  n dA   m v
i
CS
 

i out
 m v
i
i in
Where vi, pi are assumed uniform over discrete inlets
and outlets

m v A
i
F 


net force
on CV
d
 v d
dt

CV
time rate of change
of momentum in CV
i
ni
i
 

 

 mv
 mv





i
i out
outlet momentum
flux
4. Momentum flux correlation factors
i
i in
inlet momentum
flux
058:0160
Professor Fred Stern
Chapters 3 & 4
22
Fall 2005
  u dA  Av
 u v  n dA  

2
2
av
axial flow with
nonuniform
velocity profile
Where

1 u

A  v
CS
av

 dA

2
1
Q
 u dA  A
A
v 
av
CS
Laminar pipe flow:
r 
r


u  U  1    U 1  
 R 
 R
2
0
0
2
v  .53U
av
1
2
  43
0
Turbulent pipe flow:
r

u  U 1  
 R
m
0
v U
av
0
2
1  m(2  m)
1 m 1
9
5
m 1
7
vav =.82U0
058:0160
Professor Fred Stern
Chapters 3 & 4
23
Fall 2005
1  m  2  m 
2

2
2(1  2m)(2  2m)
m=1/7 β = 1.02
5. Constant p causes net force; Therefore,
Use pgage = patm-pabsolute
F p    pn dA    p d   0
CS
for p = constant
CV
6. At jet exit to atmosphere pgage = 0.
7. Choose CV carefully with CS normal to flow and
indicating coordinate system and  F on CV similar
as free body diagram used in dynamics.
8. Many applications, usually careful practice is
needed with continuity and energy equations for
mastery.
a. Steady and unsteady developing and fully
developed pipe flow
b. Emptying or filling tanks
c. Forces on transitions
d. Forces on fixed and moving vanes
e. Hydraulic jump
f. Boundary Layer and bluff body drag
g. Rocket or jet propulsion
058:0160
Professor Fred Stern
Chapters 3 & 4
24
Fall 2005
h. Nozzle
i. Propeller
j. Water-hammer
First relate umax to U0 using continuity equation
R


m
0  U R   u 1  r
2r dr
R
2
0
max
0


1
U 
 u 1  r R 2r dr  v
R
R
0
2
max
0
n
av
058:0160
Professor Fred Stern
Fall 2005
2
(1  m)(2  m)
vav  umax
m = 1/2
m = 1/7
Chapters 3 & 4
25


U0 = .53umax
U0 = .82umax
m = 1/2
vav = .53umax
m = 1/7
vav = .82umax
umax = U0/.53
umax = U0/.82
pg. 150 text
Second, calculate F using momentum equation:
 F   p  p R  F   u ( u 2r dr )  U ( R U )
R
2
x
1
2
2
2
2
0
0
0
F   p  p R  U R   u 2r dr

R
2
1
2
2
2
2
0
2
0
Av2
av
F   p  p R  U R  Av
2
1
2
2
2
0
2
= U02 from
continuity
av
2
1  u 
     dA
A  vav 
momentum flux correction
factor , page 163 text
=
=
1.02 turbulent flow
1
F  ( p  p )R  U R
3
2
2
lam
1
2
2
0
F  ( p  p )R  .02 U R
2
2
turb
1
2
4/3 laminar flow
0
2
Complete analysis
using CFD!
Reconsider the problem for fully developed flow:
058:0160
Professor Fred Stern
Chapters 3 & 4
26
Fall 2005
Continuity:


m m



mm m
in
out
in
or
Q, V = constant
out
Momentum:
 F  ( p  p )R   u v  A
 u (u A)  u (u A)
2
x
1
2
1
1
2
2
 pQ(u  u )
2
1
0
F = wall drag force =  2Rx
-F = force wall on fluid
w
( p  p )R   2Rx  0
2
1
 
w
2
w
R  dp 
r  dp 
   or for smaller CV r < R,     
2  dx 
2  dx 
(valid for laminar or turbulent flow, but assume laminar)
 
du
du r  dp 
 
  
dy
dr 2  dx 
y = R-r (wall coord.)
058:0160
Professor Fred Stern
Chapters 3 & 4
27
Fall 2005
du
r  dp 
   
dr
2  dx 
r  dp 
u      c
4  dx 
2
R  dp 
c
 
4  dx 
2

u ( r  R)  0
R  r  dp 
u (r ) 
 
4  dx 
2
2
r 

u ( r )  u 1  
 R 
R  dp 

 
4  dx 
2
2
u
max
max
R
Q   u (r )2r dr 
8
R
0
vave
4
 dp 
 
 dx 
Q R 2  dp  umax
 
  
2
A 8  dx 
w 
R  dp  R  8 vave 
   

2  dx  2  R 2 
4 vave

R
2
058:0160
Professor Fred Stern
f 
Chapters 3 & 4
28
Fall 2005
8 w
32
64


2
 vave
 Rvave  vave D
Re 
vave D

 64
Re
Exact solution NS for laminar fully developed pipe
flow
058:0160
Professor Fred Stern
Chapters 3 & 4
29
Fall 2005
Taking moving CV at speed Vs= ΩR î enclosing jet and
bucket:


Continuity:  m  m  0
in
out
 

m  m  m    v  n dA
in
out
R
CS
Inlet
Outlet
Momentum:
v  v  R
R
j
v  (v  R)
R
j
^
ni
^
n  i
058:0160
Professor Fred Stern
Chapters 3 & 4
30
Fall 2005
 F  F
X
F
bucket
bucket


 mu  mu
out
in

  m (v  R )  (v  R )
j
j

 2 m(v  R )
j
 2 A (v  R )
j
2
j

m  A (v  R)
j
j
P  RF
bucket
dP
 0 for
d
 2 A R(v  R)
R 
j
2
j
v
3
P 
j
max
8
A v
27
j

If infinite number of buckets: m  A v
j
F
bucket
j
 2 A v (v  R )
j
j
j
P  2 A v R (v  R )
j
dP
0
d
j
j
for R 
v
2
j
1
P  A v
2
max
j
3
j
3
j
058:0160
Professor Fred Stern
Chapters 3 & 4
31
Fall 2005
Application of differential momentum equation:
1. NS valid both laminar and turbulent flow; however,
many order of magnitude difference in temporal and
spatial resolution, i.e. turbulent flow requires very
small time and spatial scales
2. Laminar flow Recrit =
U
Re > Recrit

 1000
instability
3. Turbulent flow Retransition > 10 or 20 Recrit
Random motion superimposed on mean coherent
structures.
Cascade: energy from large scale dissipates at
smallest scales due to viscosity.
Kolmogorov hypothesis for smallest scales
4. No exact solutions for turbulent flow: RANS, DES,
LES, DNS (all CFD)
5. So exact solutions for simple laminar flows are
mostly linear v  v  0
a. Couette flow = shear driven
b. Steady duct flow = Poiseuille flow
058:0160
Professor Fred Stern
c.
d.
e.
f.
g.
Chapters 3 & 4
32
Fall 2005
Unsteady duct flow
Unsteady moving walls
Asymptotic suction
Wind-driven flows
Similarity solutions. etc.
6. Can also use CFD for non simple laminar flows
7. AFD or CFD requires well posed IBVP; therefore,
exact solutions are useful for setup of IBVP,
physics, and verification CFD since modeling errors
yield USM = 0 and only errors are numerical errors
USN.
δS = S – T = δSN + δSM
US2 = USN2 + USM2
Energy Equation:
B = E = energy
β = e = dE/dm = energy per unit mass
Integral Form (fixed CV):
058:0160
Professor Fred Stern
Chapters 3 & 4
33
Fall 2005
dE

  (e ) d   e v  n dA  Q  W
dt 
t  

CV

CS
rate of outflux
E across CS
rate of change
E in CV
Rate of
change E
Rate of heat
added CV
Rate work
done by CV
^ 1
e  u  v  gz  internal + KE + PE
2
2

Q = conduction + convection + radiation


W


 W

 W
W

shaft
p
pump / turbine
pressure
viscous
viscous
d W   p dA  v  n 

pressure force  (vnormal)in
p
W   pv  n  dA

p
CS

dW v   dA v
viscous force  velocity

W      v dA
v
CS



Q W  W  
s
v
CV

(e ) d   e  p /   v  n dA
t
CS
For our purposes, we are interested in steady flow one
inlet and outlet. Also Wv ~ 0 in most cases; since, v = 0 at
058:0160
Professor Fred Stern
Chapters 3 & 4
34
Fall 2005
the solid surface; on inlet and outlet any τn ~ 0; or for v
 0 and τstreamline ~ 0 if outside BL.

^ 1

  u  v  gz  p /    v  n dA
 2


Q W 
2
S
inlet&outlet
^
Assume parallel flow with 
p /   gz d u constant over inlet


= constant ie hydrostatic
pressure variation
and outlet.
^


Q  W   u  p /   gz    v  n dA 
 v (v  n) dA
2




2
S
inlet&outlet
inletoutlet

^


Q  W S   u  p /   gz  ( min )   vin 3 dAin
2 in

 m



^


  u  p /   gz  (mout )   vout 3 dAout
2 out

out
Define kinetic energy correction factor
1  v
   
A v
3
A
ave
Laminar flow:

 dA 


2
v
v
(
v

n
)
dA


m

2
2
2
A
 r
u  U 1   
 R
0
2



ave

058:0160
Professor Fred Stern
Chapters 3 & 4
35
Fall 2005
β = 4/3
vave=0.5
r

u  U 1  
 R
Turbulent flow:
1  3 2  m 
3
4(1  3m)(2  3m)
α=1.058
m=1/7

m
0
3

α=2
as with β, α~1 for
turbulent flow

2
2
^
^
Q W
v
v

 (u  p /   gz   )  (u  p /   gz   )
2
2
m m
s

ave

ave
out
Let in = 1, out = 2, v = vave, and divide by g
p1 1 2
p


v1  z1  hp  2  2 v22  z2  ht  hL
 g 2g
 g 2g

W

s

gm

W

t

gm

W
p

gm

1
Q
hL  (u2  u1 )  
g
mg
h h
t
p
in
058:0160
Professor Fred Stern
Chapters 3 & 4
36
Fall 2005
hL = thermal energy (other terms represent mechanical energy

m  A v  A v
1
1
2
2
Assuming no heat transfer mechanical energy converted
to thermal energy through viscosity and can not be
recovered; therefore, it is referred to as head loss > 0,
which can be shown from 2nd law of thermodynamics.
1D energy equation can be considered as modified
Bernoulli equation for hp,ht, and hL.
Application of 1D Energy equation fully developed pipe
flow without hp or ht.
hL 
p1  p2
L  d

 ( z1  z2 ) 

(
p


z
)
L  dx

g
 g  dx
w
L  2 w 

f

1 2
 g  R 
 vave
8
2
L vave
hL  f
D 2g
Darcy-Weisbach Equation (valid for laminar or
turbulent)
For laminar flow,
8 w
32 
f  2 
 vave  Rvave
058:0160
Professor Fred Stern
Fall 2005
32Lv
h 
D
L
Chapters 3 & 4
37
2
ave
2
For turbulent flow,
F=f (Re, k/D)

exact solution!
vave
Recrit ~ 2000, Retrans ~ 3000
Re = vaveD/υ, k = roughness
hL α vave2
In this case hL = hf = friction loss
more generally,
hL = hf + Σhm
where
v2
hm  K
2g
K  loss coefficient
hm = “so called” minor losses, e.g. entrance/exit,
expansion/contraction, bends, elbows, tees, other
fitting, and valves.
Differential Form of Energy Equation:
058:0160
Professor Fred Stern
Chapters 3 & 4
38
Fall 2005




dE


   (e )    (e v) d   Q W
dt CV t



e 
De
 e

e  e  (ev)  ev  e  
    v  e 
t t
Dt
 t

^
^ 1
1
e  u  v  gz  u  v  g  r
2
2
2
2
^


D
Du
Dv


 (Q  W ) /   q  w  
v
 g  v
 Dt

Dt
Dt







q    q    (kT )

w    (v   )  v 
ij
Fourier’s Law

  

ij
 Dv


g 
Dt

 Dv

   v  g v 
 Dt


ij
u
x
i
j
058:0160
Professor Fred Stern
Chapters 3 & 4
39
Fall 2005
^
Du

   ( k T ) 
Dt
 ij
ui
x j
 ij ui  pv
x j
p D
D
Dp
 
( p / ) 
 Dt
Dt
Dt

D ^
D
(u  p /  )    (kT ) 

Dt
Dt
u
x


i
ij
j
  0  dissipatio n function

Dh
Dp
   (kT ) 

Dt
Dt
Summary GDE for compressible non-constant property
fluid flow
Cont.

   (  v)  0
t
Mom.
Dv

  g  p     '
Dt
Energy 
ij
Dh Dp

   (kT )  
Dt Dt
 '      v
ij
ij
^
g  g k
ij
058:0160
Professor Fred Stern
Chapters 3 & 4
40
Fall 2005
Primary variables:
p, v, T
Auxiliary relations:
ρ = ρ (p,T)
h = h (p,T)
μ = μ (p,T)
k = k (p,T)
Restrictive Assumptions:
1)
2)
3)
4)
5)
Continuum
Thermodynamic equilibrium
f = -ρg k̂
heat conduction follows Fourier’s law
no internal heat sources
For incompressible constant property fluid flow
^
d u  c dT
cv, μ, k , ρ ~ constant
v
c
v
DT
 k T  
Dt
2
For static fluid or v small
c
T
 k T
t
2
p
heat conduction equation (valid for solids too)
058:0160
Professor Fred Stern
Chapters 3 & 4
41
Fall 2005
Summary GDE for incompressible constant property fluid
flow (cv ~ cp)
v  0
^
Dv

  g k  p   v
Dt
“elliptic”
2
c
DT
 k T  
Dt
where   
2
p
ij
u
x
i
j
Continuity and momentum uncoupled from energy;
therefore, solve separately and use solution post facto to
get T.
For compressible flow, ρ solved from continuity equation,
T from energy equation, and p = (ρ,T) from equation of
state (eg, ideal gas law). For incompressible flow, ρ =
constant and T uncoupled from continuity and momentum
equations, the latter of which contains p such that
reference p is arbitrary and specified post facto (i.e. for
incompressible flow, there is no connection between p
and ρ). The connection is between p and   v  0 , i.e. a
solution for p requires   v  0 .
NS
Dv
1 ^
   p   v
Dt

2
^
p  p  z
058:0160
Professor Fred Stern
Chapters 3 & 4
42
Fall 2005
  (NS )
D
 

 Dt
2
1
u u



v



p



x x

2
i
j
j
i
For   v  0 :
 p  
2
u u
x x
i
j
j
i
Poisson equation determines pressure up to additive
constant.
Approximate Models:
1) Stokes Flow
For low Re 
UL

 1, v  v ~ 0
v  0
v
1
  p   v
t

2
  ( NS )   p  0
2
Linear, “elliptic”
Most exact solutions NS; and for steady
flow supposition, elemental solutions and
separation of variables
058:0160
Professor Fred Stern
Chapters 3 & 4
43
Fall 2005
2) Boundary Layer Equations
For high Re >> 1 and attached boundary layers or
fully developed free shear flows (wakes, jets, mixing


 , p  0 , and for free shear flows
x
y
layers),   U ,
y
px = 0.
u v 0
x
y
^
u  uu  vu   p  u
t
x
x
y
yy
non-linear, “parabolic”
^
p  0   p  U  UU
y
x
t
x
Many exact solutions; similarity methods
3) Inviscid Flow

    v   0
t
Dv

  g  p
Euler Equation, nonlinear, " hyperbolic"
Dt
Dh Dp


 (kT ) p, v, T and  , h, k  f ( p, T )
Dt Dt
058:0160
Professor Fred Stern
Chapters 3 & 4
44
Fall 2005
4) Inviscid, Incompressible, Irrotational
  v  0  v  
v  0     0
2
linear, " elliptic"
 Euler Equation  Bernoulli Equation :
p

2
v  gz  constant
2
Many elegant solutions: Laplace equation using
superposition elementary solutions, separation of
variables, complex variables for 2D, and Boundary
Element methods.
3.2 Couette Shear Flows: 1-D shear flow between
surfaces of like geometry (parallel plates or rotating
cylinders).
3.21 Steady Flow Between Parallel Plates: Combined
Couette and Poiseuille Flow.
058:0160
Professor Fred Stern
Chapters 3 & 4
45
Fall 2005
1st non-dimensionalize the equations:
T 
u  u /U
*
*
T T
T T
0
1
y  y/h
0
(note: inertia terms vanish and ρ is absent)
u 0
(1)
x
2
h ^
u 
p   B  cons.
U
T 
yy
(2)
x
yy
U
2
(3)
2
k (T  T )



1
 u 
y
0
Pr Ec
B.C.
y=1
y = -1
u=1
u=0
T=1
T=0
(1) is consistent with 1-D flow assumption. Simple
form of (2) and (3) allow for solution to be
obtained by double integration.
v  0
u v w 0
x
y
u 0
x
z
058:0160
Professor Fred Stern
Chapters 3 & 4
46
Fall 2005
u
 uu  vu  wu  0
t
^
Dv

  p   v
Dt
2
x
y
z
0   pˆ x   u yy
T
 uTx  vTy  wTz  0
t
DT
 k T  
Dt
0  kT   u
c
2
p
2
yy
y
    2u x2  2v y2  2wz2  (vx  u y ) 2  ( wy  vz ) 2  (u z  wx ) 2 
  (u x  v y  wz )
  u y2
1
1
 u  (1  y )  B(1  y )
2
2
2
Linear flow
due to U
Parabolic flow
due to px
Solution depends on
B<0
B < -0.5
|B| >> 1
y=y/h
Note: linear
superposition since
v  v  0
h2 ^
B
p
U x :
^
p is opposite to U
x
backflow occurs near lower wall
flow approaches parabolic profile
058:0160
Professor Fred Stern
Chapters 3 & 4
47
Fall 2005
Pr Ec
Pr Ec B
Pr Ec B 2
1
2
3
T  (1  y ) 
(1  y ) 
(y  y ) 
(1  y 4 )
2
8
6
12
Pure
conduction
n
Time due to
viscous dissipation
Pressure
gradient effect
Dominant term
for B ∞
Note: usually PrEc is quite small
Substance
Air
PrEc
0.001
Water
0.02
Crude oil
20
dissipation
very small
Br  Pr E
c
 Brinkman #
large
058:0160
Professor Fred Stern
Chapters 3 & 4
48
Fall 2005
Shear Stress
^
1) p  0
i.e. pure Couette Flow
x
C 
f

1
u
2

u
u  1/ 2
y
1
u
2

1


Uh Re
2
y
2
(u / U )
 1/ 2
 ( y / h)
h
P0 = CfRe = 1: Better for non-accelerating flows
since ρ is not in equations and P0 = pure constant
2) U = 0
i.e. pure Poiseuille Flow
1
u  B (1  y )
2
u  B
2
Where
y
B
u 
y
y
BU
y
h
2
U u
h ^
2u
p 
U
U
max
x
1u ^ y  
u
p 1  
 
2   h 



u
2
2
4
Q   u dy  hu
3
h
h
max
u
Q 2
 u
2h 3
max
max
058:0160
Professor Fred Stern
Chapters 3 & 4
49
Fall 2005
BU
upper
h
BU
 
lower
h
BU
2u
 

 3u
h
h
h
  u
w
y y  h
 
 u lam.
 u
2
turb.
max
w
C 
f


w
1
U
2
2
6
or
 uh
P  C Re  6
0
f
h
Except for numerical constants some are for circular pipe.
Rate of heat transfer at the walls:
q  kT
w
y yh
2
k
U
 (T  T )  
2h
4h
1
+ = upper, - = lower
0
Heat transfer coefficient:
  q T T
w
1
Nu 
0
2h
 1  Br
2
k
For Br >> 2, both upper & lower walls must be coated to
maintain T1 and T0
058:0160
Professor Fred Stern
Chapters 3 & 4
50
Fall 2005
Conservation of Angular Momentum: moment form of
momentum equation (not new conservation law!)
B  H   r  v dm  angular momentum of system about
0
sys
inertial coordinate system 0.
  rv
dH
d
  r  v  d   r  v  v  n dA
dt
dt
0
CV
CS
  M  vector sum all external moments applied CV
due to both FB and FS.
0
For uniform flow across discrete inlet/outlet:


 (r  v)  v  n dA   (r  v) m   (r  v) m
out
out
in
CS
   dA  r


M 
0
CS
surface force moment
   g d  r  M

 

CV
body force moment
M  moment of reaction forces
R
R
in
058:0160
Professor Fred Stern
Chapters 3 & 4
51
Fall 2005
The inertial frame 0 is fixed to earth such that CS moving
at vs= -Rω i
^
^
^
v  v i  R i
2
r Rj
2
0
^
^
v v k
1
r 0j
1
0
v Q
Retarding torque due to
bearing friction
0
A
pipe
^


 M  T k  (r  v ) m  (r  v ) m
Z


m  m  Q
out

in
v0
T
 0 2
R  QR
0
2
out
2
1
^
1
int
^
 T k  R(v  R )( k ) Q
o
0
interesting, even for T0=0, ωmax=v0/R
058:0160
Professor Fred Stern
Chapters 3 & 4
52
Fall 2005


79
9890
 F   F   m v  .


 4 .02 v  425 N




A
x
2
2
2
2

v  41.4 m / s, m  10.3 kg / s
2
D
v A  v A  v  v 
D
2
1
1
2
2
1
2
1
1
1
p  v  p  v
2
2
2
1
1
2
2
2
v  41.4 
5
2
2
2
1
2
v  6.63 m / s
1
1
hL=0, z=constant
2
2
1
p  p   v  v
2
1



2

p  101,000 
1
.79 * 998
(41.4  6.63 )
2
2
2
p  760,000 Pa
1
 F  0   vv  A  v v A   (v )v A
y
3
CS
3
3
4
 v A  v A
2
3
2
3
4
4

4
4
A A
3
4
058:0160
Professor Fred Stern
Note:
Fall 2005
p 
2
Chapters 3 & 4
53

v p 
2
2
2
3

2
v p 
2
3
4

2
v
2
4
p p  v v v
a
2
3
4
0   v  A  A v  A v  A v
2
CS
A A A
2
3
4
2
3
3
4
4
058:0160
Professor Fred Stern
Chapters 3 & 4
54
Fall 2005
2
v
(a) z 
z
2g
Torricelli’s
v  2g(z  z )
  1, h  0, z  11, z  0
2
expression
for speed of
efflux from
reservoir
1
2
2
1
Q  Av 
2
(b)
2
z1   2
2
2
4
1
 2 * 9.81*11
2

L
2
 14.7 m / s
(.01) *14.7 * 3600  4.16 m / h
2
v22
 z2  hL
2g
3
w/   2, h 
2
L
32vL
,   10 m / s
D g
6
2
v  3.2v  107.8  0
2
2
2
v2 = 8.9 m/s
Q= 2.516 m3/h
2
2
2
2
v
L v
(c) z  
z h  f
2g
D 2g
1
3
2
2
L
v
1  fL / D 
z z 
2g
2
1
2
α3=1
2
058:0160
Professor Fred Stern
Chapters 3 & 4
55
Fall 2005
v  2 g ( z  z ) /(1  fL / D)
2
v  216 /(1  f *1000)
f  f (Re), Re 
1
2
1
2
1
2
2
guess f = 0.015
v  3.7 m / s  Re  3.7 x10 ,
f  .024
4
2
v  2.94 m / s  Re  2.9 x10 ,
4
2
v  2.88 m / s  Re  2.9 x10
f  .025
4
2
(d) Re 
vD

 2000
D
v
32Lv
(z z )  

2 g g 2000 
v
2000
v
2
2
1
2
2
2
2
2
2
2
v
32Lv
(z z )  

2 g 2000 g
1
2
2
2
3
2
2
2
32 Lv23
v22
  11  0
20002 g g
v  1.1 m / s
2
D  0.00182 m
vD

058:0160
Professor Fred Stern
Chapters 3 & 4
56
Fall 2005
Differential Equation of Conservation of Angular
Momentum:
Apply CV form for fixed CV:

I w  a dy
z
dx
dx
dy
dy
 b dy  c dx  d dx
2
2
2
2
I w    

z
xy


Iw 
z
12


Iw 
z
lim
12
dx 0, dy 0
yx
 dxdy
dxdy
dx
2
3
 dydx  

3
12
 dy     
dxdydx  dy
2
2

2
xy
yx
   , similarly    ,   
xy
yx
i.e   
ij
xz
ji
zx
yz
zy
stress tensor is symmetric (stresses
themselves cause no rotation)
058:0160
Professor Fred Stern
Fall 2005
Chapters 3 & 4
57
1.4 Boundary Conditions for Viscous-Flow Problem
The GDE to be discussed next constitute an IBVP for
a system of 2nd order nonlinear PDE, which require
BC for their solution, depending on physical problem
and appropriate approximations.
Types of Boundaries:
1. Solid Surface
2. Interface
3. Inlet/exit/outer
058:0160
Professor Fred Stern
Chapters 3 & 4
58
Fall 2005
1. Solid Surface
a. Liquid
ℓ = mean free path << fluid motion; therefore,
microscopic view is “no slip” condition, i.e. no
relative motion or temperature liquid and solid.
v
liquid
v
T
solid
liquid
T
solid
Exception is for contact line for which analysis is
similar to that for gas.
b. Gas
Specular reflection
Conservation of
tangential momentum
uw=0=fluid velocity at
wall
Smooth wall
Diffuse reflection.
Lack of reflected
tangential momentum
balanced by uw
Rough wall
u l
w
du
dy
w
058:0160
Professor Fred Stern
 
w
u 
w
du
dy
Chapters 3 & 4
59
Fall 2005
l
w
3 
2 a 

2 a
3
Ma  U
w
u / U  .75Ma C
w
a
low density limit
C 
f

w
1
v
2
2
f
High Re:
Cf ~ 0.005
Say Ma ~ 20
Low Re:
Cf ~ .6Rex-1/2
u
 0.01
U
w
Rex=Ux/υ
u .4 Ma

U
Re
w
1
2
x
Significant slip possible at low Re, high Ma:
“Hypersonic LE Problem”
Similar for T:
High Re:
Tgas = Tw
Low Re
T T
 .87 MaC
T  T = driving ∆T
gas
r
Ref. T
w
w
f
air
058:0160
Professor Fred Stern
Where
Chapters 3 & 4
60
Fall 2005
C  2C  2

 

f
h
Reynolds Analogy
q
C U (T  T )


w
p
r
w
Ch = Stanton number, i.e. wall
heat transfer coefficient
Deformable fluid/solid interface: fluid structure
interaction
2. Gas/liquid interface (free surface problems since
interface is unknown and part of the solution).
Kinematic FSBC: free surface is stream surface
F   ( x, y )  z  surface function
n  F / F  ( ,  ,  1) / D
x
y
D      1
2
2
x
y
1
2
DF
F
0
 v  F
Dt
t
1 F
 v  n  0 (1)
F t
058:0160
Professor Fred Stern
Chapters 3 & 4
61
Fall 2005
Dynamic FSBC: stress continuous across free surface
(similarly for mass and heat flux)
 n  *n  p 
ij
j
ij
Fluid 1 stress

j
ij
Surface
tension pres.
Fluid 2 stress
(vector whose components are stress in direction of coordinate axes on surface with normal nj)
   p  Re (U  U )
1
ij
ij
i, j
*  

ij
fluid 2
j ,i
eg  p 
a
for air if neglecting μair
ij
Atmospheric pressure
p  We1  KSN  KtN 
^
e
K  n
s
^
s
SN
^
^
K  n
St
e
t
Curvature F for two mutually perp. Directions
t
^
Note:
e
^
s
and
e
^
t
tangent to n =
e
We  U L /   Weber Number
2
Surface tension
(2)  x   11n1   12 n2   13n3  ( pa  p )n1
(3)  y   21n1   22 n2   23n3  ( pa  p )n2
(4)  z   31n1   32 n2   33n3  ( pa  p )n3
(5)  v  0  U x  Vy  Wx
1+3+1=5 conditions for 5
unknowns = (v, p, ζ)
-Also need conditions for
turbulence variables
incompressible flow
n
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Many approximations, eg, inviscid approximation:
pa = p δ = 0
small slope: ζx ~ ζy ~ 0
small normal velocity gradient: Wx ~ Wy ~ Wz = 0

(U ,V )  0
z
p=0
W  U  V
z
or
x
^
y
p  gz
or W  0
z
^
p = piezometric pres.
3) Inlet/exit/outer
a) inlet: v, p, T specified
b) outer: v, p, T specified
eg. constant Temp.,
uniform stream:
v = U î , p = 0 , T = Ti,o
c) exit: depends on the problem, but often use U  0 ,
(i.e. zero stream wise diffusion for external
flow and periodic for fully developed
internal flow).
XX
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Interface Velocity Condition
Just as with solid surface, there can be no relative
velocity across interface (i.e. exact condition for
liquid/liquid and gas/gas or gas/liquid non-mixing fluids).
v v
1
v v
n1
n2
2
required by KFSBC
v n  v n  
1
2
1 F
F t
Tangential should also match, but usually due to
different approximations used in fluid 1 or 2, (eg fluid 1
liquid and fluid 2 gas do not). Often, in fact, motions in
gas are neglected and therefore v is not continuous.
Also liquid/liquid interfaces are not stable for large
Re and one must consider “turbulent interface”.
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C. Vorticity Theorems
The incompressible flow momentum equations focus
attention on v and p and explain the flow pattern in terms
of inertia, pressure, gravity, and viscous forces.
Alternatively, one can focus attention on ω and explain
the flow pattern in terms of the rate of change, deforming,
and diffusion of ω by way of the vorticity equation. As
will be shown, the existence of ω generally indicates the
viscous effects and is important since fluid particles can
only be set into rotation by viscous forces. Thus, the
importance of this topic is to demonstrate that under most
circumstances, an inviscid flow can also be considered
irrotational.
1. Vorticity Kinematics
^
^
^
    v  ( w  v ) i  (u  w ) j  (v  u ) k
y
i   ijk
z
 u u j
 k 
xk  x j xk
u j
z



= 2 * the angular velocity of
the fluid element
x
x
y
123   321   231  1
 213   321  132  1
 ijk  0 otherwise
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A quantity intimately tied with vorticity is the circulation:
   v  dx
Stokes Theorem:
 a  dx     a  dA
A
   v  dx    v  dA     n  dA
A
A
Which shows that if ω (i.e., if the flow is irrotational,
then Γ also.
Vortex line = lines which are everywhere tangent to the
vorticity vector.
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Next, we shall see that vorticity and vortex lines must
obey certain properties known as the Helmholtz vorticity
theorems, which have great physical significance.
The first is the result of its very definition:
  v
      (  v)  0
Vector identity
i.e. the vorticity is divergence-less, which means that
there can be no sources or sinks of vorticity within the
fluid itself.
Helmholtz Theorem #1: a vortex line cannot end in the
fluid. It must form a closed path (smoke ring), end at a
boundary, solid or free surface, or go to infinity.
Propeller vortex is
known to drift up
towards the free surface
The second follows from the first:
    d     n dA  0
A
A
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Application to a vortex tube results in the following
   n ds     n ds  0


 




A1
Minus sign due to
inward normal
A2
1
2
Or Γ1= Γ2
Helmholtz Theorem #2:
The circulation around a given vortex line (i.e., the
strength of the vortex tube) is constant along its length.
This result can be put in the form of a simple onedimensional incompressible continuity equation. Define
ω1 and ω2 as the average vorticity across A1 and A2,
respectively
ω1A1 = ω2A2
which relates the vorticity strength to the cross sectional
area changes of the tube.
2. Vortex dynamics
Consider the substantial derivative of the circulation
assuming incompressible flow and conservative body
forces
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D D

 v  dx
Dt Dt
Dv
D

 dx   v  dv
Dt
Dt
From the N-S equations we have
Dv 1
p
 f
  v
Dt 

Define f  F for
the gravitational body
force F=ρgz.
2
  F  p    v


2
Also,
D
Dx
dx  d
 dv
Dt
Dt
D

Dt
   F  p /    d x    v   d x   v  d v
2
  dF  
dp
1
d (v  v )
2

dp 1 

   dF 
 dv      v  d x
 2 
 


=0 since integration is around a closed
contour and F,p, & v are single valued!
2
2
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D
    2 v  dx        dx
Dt
    v     v    v
 


0

2
Implication: The circulation around a material loop of
particles changes only if the net viscous force on those
particles gives a homogenous integral.
If   0 or   0 (i.e., inviscid or irrotational flow,
respectively) then
D
0
Dt
The circulation of a
material loop never
changes
Kelvins Circulation Theorem: for an ideal fluid (i.e.
inviscid and incompressible) acted upon by conservative
forces (eg, gravity) the circulation is constant about any
closed material contour moving with the fluid.
Which leads to:
Helmholtz Theorem #3: No fluid particle can have
rotation if it did not originally rotate. Or, equivalently, in
the absence of rotational forces, a fluid that is initially
irrotational remains irrotational. In general, we can
conclude that vortices are preserved as time passes. Only
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through the action of viscosity can they decay or
disappear.
Kelvins Circulation Theorem and Helmholtz
Theorem #3 are very important in the study of inviscid
flow. The important conclusion is reached that a fluid
that is initially irrotational remains irrotational, which his
the justification for ideal-flow theory.
In a real viscous fluid, vorticity is generated by
viscous forces. Viscous forces are larger near solid
surfaces as a result of the no-slip condition. On the
surface there is a direct relationship between the viscous
shear stress and the vorticity.
Consider a 1-D flow near a wall:
The viscous stresses are given by:
 n where   
'
ij
'
j
ij
ij
 11' n1   12' n2   13' n3   x
 21' n1   22' n2   23' n3   y
 31' n1   32' n2   33' n3   z
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 u
v 
u
 12 '  12       
y
 y x 
 22 '   22  2
v
0
y
 w
v 
 32 '   32       0
 y z 
NOTE: the only component of
w is wz. Actually, this is a
general result in that it can be
shown that wsurface is
perpendicular to the limiting
streamline.
Which shows that
u
 
dy
x
   0
y
z
However from the deformation vorticity we also see that
 
x
u
  w
y
z
i.e., the wall vorticity is directly proportional to the wall
shear stress. This analysis can be easily extended for
general 3-d flow.


True since at a wall with coordinate x2, x  x  0 and
1
from continuity
v
0
x2
Rotation tensor
 ij ' n j   ij n j
at a fixed solid wall
3
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Once vorticity is generated, its subsequent behavior is
governed by the vorticity equation.
N-S
v
 v  v   p /     v
t
Or
v
1

  v  v   v     p /     v
t
2

2
neglect f
2
The vorticity equation is obtained by taking the curl of
this equation. (Note    )  0.

   v  w  2 
t
 v( )  ( v)  (v )  ( )v
Rate of change of ω
Or
=
Rate of
deforming vortex
lines

 (v  )     v   
t
Rate of viscous
2
Transport Eq. for ω
diffuision of ω
Note:
(1) Equation does not involve p explicitly
(2) for 2-D flow (  )v  0 since ω is perp. to v
and there can be no deformation of ω, ie
D
  
2
Dt
In order to determine the pressure field in terms of the
vorticity, the divergence of the N-S equation is taken.
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 v
    v  v   p /    
 t
 ( p /  )    v  v 
2

v

Poisson Eq. for p
1
   v  v  v   v   
2
2
2
2
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3. Kinematic Decomposition of flow fields
Previously, we discussed the decomposition of fluid
motion into translation, rotation, and deformation. This
was done locally for a fluid element. Now we shall see
that a global decomposition is possible.
Helmholtz’s Decomposition: any continuous and finite
vector field can be expressed as the sum of the gradient of
a scalar function  plus the curl of a zero-divergence
vector A. The vector A vanishes identically if the original
vector field is irrotational.
The irrotational part of
the velocity field can be
expressed as the gradient
of a scalar
v v v


  v
Where
0  v



 v  
If
v  v v  0
Then
 0
The G.d.e. for  is the Laplace Eq.
And
v   A
Since     A  0


2

v     A

  A  (  A)
2
i.e
 A  
2
Again, by vector identity
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The solution of this equation is
A
1 
 d
4 R
v 

Thus
1 R 
d

4 R
3
Which is known as the Biot-Savart law.
The Biot-Savart law can be used to compute the velocity
field induced by a known vorticity field. It has many
useful applications, including in ideal flow theory (e.g.,
when applied to line vortices and vortex sheets it forms
the basis of computing the velocity field in vortex-lattice
and vortex-sheet lifting-surface methods).
The important conclusion from the Helmholtz
decomposition is that any incompressible flow can be
thought of as the vector sum of rotational and irrotational
components. Thus, a solution for irrotational part v
represents at least part of an exact solution. Under certain
conditions, to be discussed next, v is small over much of
the flow field such that v is a good approximation to v .
This is probably the strongest justification for ideal-flow
theory. (incompressible, inviscid, and irrotational flow).



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Non-inertial Reference Frame
Thus far we have assumed use of an inertial reference
frame (i.e. fixed with respect to the distant stars in
deriving the CV and differential form of the momentum
equation). However, in many cases non-inertial reference
frames are useful (e.g. rotational machinery, vehicle
dynamics, geophysical applications, etc).
a 
i
 F  ma
i
dv
a
dt
 dv

 m
 a rel 
 dt

 F ma
S Rr
i
rel
rel
m
dv
dt
i.e Newton’s law
applies to noninertial frame with
addition of known
inertial force terms
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dR
3rd from fact that (x,y,z)
  r
rotating at Ω(t).
dt
dv d R d
a 


 r  2  v    (  r )
dt dt
dt
dv

a
dt
v v
i
2
i
2
rel
2
d R
dt
2
= acceleration (x,y,z)
d
r
dt
= angular acceleration (x,y,z)
2  v
= Coriolis acceleration
  (  r )
= centripetal acceleration (=-Ω2L, where L = normal
distance from r to axis of rotation Ω).
Since R and Ω assumed known, although more
complicated, we are simply adding known
inhomogeneities to the momentum equation.
CV form of Momentum equation for non-inertial
coordinates:
F   a
CV
rel
 d 
d
v d    v v R  n dA

dt CV
CS
Differential form of Momentum equation for noninertial
coordinates:
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^

 Dv

   a    g k  p 
x
 Dt

rel
2


2




v



3
ji
j
ij
a  R   r      r   2  v


rel
All terms in a seldom act in unison (e.g. geophysical
flows):
rel

R ~0
earth not accelerating relative to distant stars

 ~0
for earth
    r  ~ 0 g nearly constant with latitude
2  v
a 
i
most important!
dv
 R (2  v)
dt
0
v
R  Rossby # 
L
0
0
v
1
v
tv
,t 
v
L
0
0
i.e. for L << earth radius, the
Coriolis term is small and vice
verse for L large,
small
Example of Non-inertial Coordinates:
Geophysical fluids dynamics
dv
term is
dt
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Atmosphere and oceans are naturally studied using noninertial coordinate system rotating with the earth. Two
primary forces are Coriolis force and buoyancy force due
to density stratification ρ = ρ(T). Both are studied using
Boussinesq approximations (ρ = constant, except ρ(T) g k
term; and μ, k, Cp = constant) and thin layer on rotating
W H
surface assumption  ~  .
U
L
Differences: lateral boundaries (continents) in oceans;
currents in ocean (gulf stream) along western boundaries;
clouds and latent heat release in atmosphere due to
moisture condensation.
H << L = 0 (radius of earth = 6371 km)
Therefore, one can neglect curvature of earth and replace
spherical coordinates by local Cartesian tangent plane
coordinates.
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Coriolis force = 2  v
^
^
^
i
j
k
=
x

u
y
v

z
w
0 since w << v
= 2 i w cos  v sin    j u sin   k u cos 


^
^
^
^
^
^
f  2 sin 
=  fv i  fu j  2 cos u k
Person
spins at
Ω
= planetary
vorticity
= 2 * vertical
component Ω
f>0
f<0
f = 
f=0
northern hemisphere
southern hemisphere
at poles
at equator
Person translates with inertial
2
period Ti 
f
Equations of Motion
v  0
Du
1 p
 fv  
  u
Dt
 x
2
0
Dv
1 p
 fu  
  v
Dt
 y
2
0
Dw
1 p g


  w
Dt
 z 
2
0
0
Only vertical component Ω
due to thin layer assumption
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   1   (T  T )
0
p, ρ = perturbation from hydrostatic
condition
0
Geostrophic Flow: quasi-steady, large-scale motions in
atmosphere or ocean far from boundaries
 fv  
1 p
 x
fu  
0
Dv
U 
~0 
Dt
 L
1 p
 y
0
2
f v ~ 0 ( fv)
U,L = horizontal scales
Rossby number =
Atmosphere:
Ocean:
U ~ 10 m/s; f = 10-4 Hz; L ~ 1000 km;
and R0 = 0.1
U ~ 0.1 m/s; f = 10-4 Hz; L ~ 1000 km;
and R0 = 0.01
Therefore, neglect
neglect  v .
U
fL
Dv
and since there are no boundaries,
Dt
2
Z momentum 

  g
z
baroclinic (i.e. p = p(T))
and can be used to eliminate p in above equations
whereby (u,v) = f(T(z)), which is called thermal wind but
not considered here.
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If we neglect ρ=ρ(T) effects, (u,v) = f(p) and can be
determined from measured p(x,y). Not valid near the
equator (+ 3o) where f is small.
 u ^i  v ^j   p  1   p ^i  p ^j     p ^i  p ^j 


 f  y x   x y 


0
=0
i.e v is perpendicular to p  horizontal velocity is along
(and not across) lines of constant horizontal pressure,
which is reason isobars and stream lines coincide on a
weather map!
Ekman Layer on Free Surface: effects of freedom near
boundaries
Viscous layers:
Sudden acceleration flat plate: u  u
  3.64 t
t
Oscillating flat plate: u  u
  6.5  / 
t
yy
yy
u ( y , 0)  0
u (0, t )  U
u ( , t )  0
u (0, t )  U cos t
0
u ( , t )  0
058:0160
Professor Fred Stern
Chapters 3 & 4
84
Fall 2005
Flat plate boundary layer:
u v 0
x
y
uu  vu  u
x
  4.9  x / U
y
yy
u ( x ,0)  0
u ( x,  )  U
For Ekman layer viscous effects due to wind shear τ(x).
Assume horizontal uniformity (i.e px = py = 0), which is
justified for L ~ 100 km and H ~ 50 m. However, can be
included easily if assume p = p(z) such that geostropic
solution is additive and combined solution recovers
former for large depths z    .
 fv  u
fu  u
ZZ
u  
z
ZZ
at z = 0
^
v 0
z
at z = 0
  i
  .002  (v
air
(u, v)  0
wind
 u (0))
at z =-∞
Multiply v-equation by i   1 and add to u-equation:
2
dv if
 v
dt

2
v  u  iv
 complex velocity
058:0160
Professor Fred Stern
v  Ae

Fall 2005
( 1 i ) z / 
 Be
Chapters 3 & 4
85
 ( 1 i ) z / 
2
 Ekman layer thickness
f
B = 0 for u(-∞), v(-∞) = 0
p
dv

dt
 A
at z = 0
 (1  i)
2 
i.e. u 
v
 /
 z 
e cos    and
f
  4
z /
 /
 z 
e sin    
f
  4
z /
F. Nansen (1902) observed drifting arctic ice drifted 20400 to the right of the wind, which he attributed to
Coriolis acceleration. His student Ekman (1905) derived
the solution.
Recall f < 0 in southern hemisphere, so the drift is to the
left of τ.
058:0160
Professor Fred Stern
Chapters 3 & 4
86
Fall 2005
Similar solution for impulsive wind:
u  u , u  
t
zz
z
z  0, u  0 z  , u( z,0)  0
u 
0
2 t


laminar solution:
u (v
0
wind
 6 m / s,T  20 C )  0.6 m/s after one min., 2.3 m/s
0
after one hour
turbulent solution:(more realistic)
u0=0.2 m/s after 1 hr (3 % vwind)
058:0160
Professor Fred Stern
Fall 2005
Chapters 3 & 4
87
For Ekman layer similar conditions θ = 400 N ,
Laminar solution u0 = 2.7 m/s at D = 45 cm, which are too
high/low; however, using turbulent υt, u0 = 2 cm/s and D =
100 m, which is more realistic.