058:0160
Professor Fred Stern
Fall 2013
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 2013
Chapters 3 & 4
2
058:0160
Professor Fred Stern
Chapters 3 & 4
3
Fall 2013
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
As with Q and 𝑚̇, ∆𝐵̇ flux though A per unit time is:
𝑑𝑄 = 𝑉𝑅 . 𝑛 𝑑𝐴
𝑑𝑚̇ = 𝜌𝑉𝑅 . 𝑛 𝑑𝐴
𝑑∆𝐵̇ = 𝛽𝜌𝑉𝑅 . 𝑛 𝑑𝐴
058:0160
Professor Fred Stern
Chapters 3 & 4
4
Fall 2013
Therefore:
dBSYS d

  d  CS V R  n dA
dt
dt CV
General form RTT for moving deforming control volume
Special Cases:
1) Non-deforming CV
dBSYS

   d   V R  n dA
dt
t
CV
CS
2) Fixed CV
dBSYS

   d   V  n dA
dt
t
CV
CS
Greens Theorem:
  b d    b  n dA
CV
CS
dBSYS


        V  d
dt
t

CV 
Since CV fixed and arbitrary lim gives governing
differential equation.
d0
058:0160
Professor Fred Stern
Chapters 3 & 4
5
Fall 2013
3) Uniform flow across discrete CS (steady or unsteady)
  V
R
 n dA    V R  n dA
CS
CS
(- inlet, + outlet)
or for fixed CV, 𝑉𝑅 = 𝑉, 𝑉𝑆 = 0
4) Steady Flow:

0
t
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
058:0160
Professor Fred Stern
Chapters 3 & 4
6
Fall 2013
Note simplifications for 1) non-deforming and fixed CV
(  ≠  (t), 𝑉𝑆 = 0), 2) uniform flow across discrete CS
(∫=∑), 3) steady flow (

 0 ), and 4) incompressible fluid
t
d

(ρ = constant ⇒ dt  d   V R  n dA : “conservation of
CV
CS
volume”)
1) Non-deforming and fixed CV

CV t d  CS V  n dA  0
2) and uniform flow over discrete inlet/outlet

CV t d    V  nA  0
3) and steady flow
or
 V  nA  0
  VAin    VAout  0
Q  m    min    mout
4) and incompressible flow
 Qin   Qout  0
if non-uniform flow over discrete inlet/outlet
QCSi   V  n dA  Vav ACS
i
CS
Vav 
1
V  n dA

A CS
058:0160
Professor Fred Stern
Chapters 3 & 4
7
Fall 2013
Differential Form:
𝑑𝑀
𝜕𝜌
= 0 = ∫ [ + ∇. (𝜌𝑉)] 𝑑∀
𝑑𝑡
𝜕𝑡
𝐶𝑉

   V   0
t
𝛽=1

    V  V    0
t
D
   V  0
Dt
M  
dM   d   d  0
d d



1 D
1 D

 Dt
 Dt

1 D


V
0


Dt

rate of change  u  v  w 1 D  1 D
x y z 
Dt
 Dt
per unit 


rate of change 
per unit 
058:0160
Professor Fred Stern
Fall 2013
Chapters 3 & 4
8
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
Chapters 3 & 4
9
Fall 2013
P3.15 Water, assumed incompressible, flows steadily
through the round pipe in Fig. P3.15. The entrance
velocity is constant, u  U 0 , and the exit velocity
approximates turbulent flow, u  umax 1  r R  . Determine
17
the ratio U0/umax for this flow.
Steady flow, non-deforming, fixed CV, one inlet uniform
flow and one outlet non-uniform flow
̇
−𝑚𝑖𝑛
̇ + 𝑚𝑜𝑢𝑡
̇ = 0; 𝜌 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡;
− 𝑄𝑖𝑛 + 𝑄𝑜𝑢𝑡 = 0
0  U 0 R   umax 1  r R  2 rdr
2
R
0
0  U 0 R 2  umax
17
49 2
R
60
U0
49

umax 60
R


17


R
r
1
1
15 7
87
2 umax  1   rdr  2 umax 
1 r R 
1 r R 


0
1 
 R
 R 2  1  2 

R 2   1




7

7 
0
  7 7 
49
 2 umax R 2 0        umax R 2
60
  15 8  
058:0160
Professor Fred Stern
Chapters 3 & 4
10
Fall 2013
P3.12 The pipe flow in Fig. P3.12 fills a cylindrical tank
as shown. At time t=0, the water depth in the tank is
30cm. Estimate the time required to fill the remainder of
the tank.
h (0)=0.3m
Unsteady flow, deforming CV, one inlet one outlet
uniform flow
0
d
  d  Q1  Q2
dt CV
d
d2
d2
0
 d   V1
 V2

dt CV
4
4
 t   h t 
 D2
4
058:0160
Professor Fred Stern
Chapters 3 & 4
11
Fall 2013
0
 D 2 dh
4
dt

d2
4
V2  V1 
2
dh  d 
   V1  V2   0.0153
dt  D 
dh
0.7
dt 

 46s
0.0153 0.0153
Steady flow, fixed CV with one inlet and two exits with
uniform flow
Note: Q   V  n dA
A
L3
s
Volume of fluid per unit time through A=
∀
𝑄 = = 0  Q1  Q2  Q3
𝑑𝑡
𝜋𝑑 2
(𝑉1 − 𝑉2 )
𝑄3 = 𝑄1 − 𝑄2 =
4
𝜋𝐷 2
𝑑ℎ
∀
4
𝑑𝑡 = =
𝑄 𝜋𝑑 2
(𝑉1 − 𝑉2 )
4
2
D
dh  
d
  
V1  V2 
058:0160
Professor Fred Stern
Chapters 3 & 4
12
Fall 2013
P4.17 A reasonable approximation for the twodimensional incompressible laminar boundary layer on
 2y y 2 
the flat surface in Fig.P4.17 is u  U     2  for y   ,


where   Cx1 2 , C  const
(a) Assuming a no-slip condition at the wall, find an
expression for the velocity component v  x, y  for y   .
(b) Find the maximum value of v at the station x  1m , for
the particular case of flow, when U  3 m s and   1.1cm .
u v

0
x y
v
u


 U  2 y 2  2 y 2 3 
y
x
x
v  2U x 
y
0
 y2
y3 
(a) v  2U  x  2 2  3 3 


(b) Since v y  0 at y  
vmax  v  y    
 y
2
 y 2 3  dy
  Cx
12
x 
C 1 2 
x 
2
2x
2U   1 1  U  3  0.011

 0.0055 m s
  
2x  2 3  6x
6
058:0160
Professor Fred Stern
Fall 2013
Chapters 3 & 4
13
Momentum Equation:
B = MV = momentum, β = V
Integral Form:
d ( MV ) d

 V  d   CS V V R  n dA   F
dt
dt CV
3
1
2
F =
=
FB =
Fs =
vector sum of all forces acting on CV
FB + F s
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.
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.
058:0160
Professor Fred Stern
Chapters 3 & 4
14
Fall 2013
1 = rate of change of momentum in CV
2 = rate of outflux 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




CV  t
 d    F
Where  V   V    V
t
t
t
ˆ   wkˆ V is a tensor
and V V  VV   uiˆ V   vjV
  (V V )    ( VV ) 



(  uV )  (  vV )  (  wV )
x
y
z
 V  ( V )  V  V
  

 V

V




V



V

V




 d    F
CV   t

 t

= 0 , continuity
058:0160
Professor Fred Stern
Since
Chapters 3 & 4
15
Fall 2013
V
DV
 V V 
t
Dt

CV


DV
d   F
Dt
DV
f
Dt
a  f  f
b
f
f
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
ˆ
and g   gk
i.e. f body
  g dxdydz
z
for
   gkˆ
g
058:0160
Professor Fred Stern
Chapters 3 & 4
16
Fall 2013
Surface Forces are due to the stresses that act on the sides
of the control surfaces
   p  
ij
ij
Normal pressure
ij
Viscous stress
 p  
  
 
xx

xy
yx
 p 
zx

zy
yy





 p   
xz
yz
zz
Symmetric
 ij   ji
2nd order tensor
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
058:0160
Professor Fred Stern
Chapters 3 & 4
17
Fall 2013
^
^
^
   i  j  k
x
xx
xy
^
Resultant
stress
on each face
xz
^
^
   i   j  k
y
yx
yy
^
yz
^
^
   i   j  k
z
zx
zy
zz
 yx 



dy  dxdz
 yx
y


y
 xx dydz



dx  dydz
 
x


xx
x
z
 yx dxdz
xx
and similarly for z
𝜕
𝜕
𝜕
(
)
𝐹𝑠 = [
𝜎𝑥𝑥 +
(𝜎𝑦𝑥 ) + (𝜎𝑧𝑥 )] 𝑑𝑥𝑑𝑦𝑑𝑧 𝑖̂
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝜕
+ [ (𝜎𝑥𝑦 ) +
(𝜎 ) + (𝜎𝑧𝑦 )] 𝑑𝑥𝑑𝑦𝑑𝑧 𝑗̂
𝜕𝑥
𝜕𝑦 𝑦𝑦
𝜕𝑧
𝜕
𝜕
𝜕
(
)
+[
𝜎𝑥𝑧 +
(𝜎𝑦𝑧 ) + (𝜎𝑧𝑧 )] 𝑑𝑥𝑑𝑦𝑑𝑧 𝑘̂
𝜕𝑥
𝜕𝑦
𝜕𝑧




F s   ( x )  ( y )  ( z )  dxdydz
y
z
 x

058:0160
Professor Fred Stern
Chapters 3 & 4
18
Fall 2013
Divided by the volume:
fs 



( x )  ( y )  ( z )
x
y
z
f s     ij 

 ij
x j
Putting together the above results,
DV
a  
   gkˆ     ij
Dt
body force surface force = p + viscous terms
due to
(due to stress gradients)
gravity
Inertial force
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
V  dV  V  dr  V
 ux

dV  dr  V   vx
 wx
relative motion

1st order Taylor Series
uy
vy
wy
u z  dx 

vz  dy   eij dx j
wz   dz 
deformation rate
tensor =
e
ij
058:0160
Professor Fred Stern
eij 
Fall 2013
ui
1  u u j 
1  ui u j 
  i


  ij  ij




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
19
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 ) iˆ  (u z  wx ) ˆj  (vx  u y ) kˆ
2
2
2
= 2  angular velocity of fluid element
058:0160
Professor Fred Stern
Chapters 3 & 4
20
Fall 2013
 ij  rate of strain tensor

ux


1
  (v x  u y )
2
1
 ( wx  u z )
 2
1
(u y  vx )
2
vy
1
( wy  v z )
2
1

(u z  wx ) 
2

1
(v z  wy ) 

2


wz

u x  v y  wz    V = elongation (or volumetric dilatation)
of fluid element 
1 D
 Dt
1
(u  v ) = distortion wrt (x,y) plane
2
1
(u  w ) = distortion wrt (x,z) plane
2
1
(v  w ) = distortion wrt (y,z) plane
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 ω
volumetric dilatation described by  V
distortion in shape described by ij
ij
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
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Fall 2013
Chapters 3 & 4
21
observations and have been verified experimentally even
for extreme conditions. For a Newtonian fluid:
1) When the fluid is at rest the stress is hydrostatic and
the pressure is the thermodynamic pressure
2) Since there is no shearing action in rigid body
rotation, it causes no shear stress.
3) ij is linearly related to ij and only depends on ij.
4) There is no preferred direction in the fluid, so that
the fluid properties are point functions (condition of
isotropy).
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Chapters 3 & 4
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Fall 2013
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.
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.
kijmn   ij mn   im jn   in jm
( ,  ,  )  scalars
Lastly, the symmetry condition σij = σji requires:
kijmn = kjimn  γ = μ = viscosity
𝜎𝑖𝑗 = −𝑝𝛿𝑖𝑗 + 𝜇𝛿𝑖𝑚 𝛿𝑗𝑛 𝜀𝑖𝑗 + 𝜇𝛿𝑖𝑛 𝛿𝑗𝑚 𝜀𝑖𝑗 + 𝜆𝛿𝑖𝑗 𝛿𝑚𝑛 𝜀𝑖𝑗
 ij   p ij  2 ij    mm  ij
V
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
λ and μ can be further related if one considers mean
normal stress vs. thermodynamic p.
 ii  3 p  (2  3 ) V
p
1
  ii
3
2

       V
3

p mean
normal stress
2

p  p        V
3

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    Dt   Dt , e.g., within shock
waves.
Stokes Hypothesis also supported kinetic theory
monotonic gas.
   23 
p p
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Chapters 3 & 4
24
Fall 2013




2
 ij    p    V   ij  2 ij
3
Generalization   
 u
u 

   


 x x 
i
i j
j
ij
j
du
for 3D flow.
dy
relates shear stress to strain rate
i
 ui 
 1
ui 
   p  2     V 

xi 
 xi 
 3
normal viscous stress
2
3
 ii   p   V  2 
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.
3  x y z 
incompressible fluids, average = 0 i.e.
For
V  0 .
Non-Newtonian fluids:
   for small strain rates  , which works well for
air, water, etc. Newtonian fluids

ij
 ij 
ij

 ij
t
nonlinear history effect
 ijn

Non-Newtonian
Viscoeslastic materials
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Chapters 3 & 4
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Fall 2013
Non-Newtonian fluids include:
(1) Polymer molecules with large molecular
weights and form long chains coiled together
in spongy ball shapes that deform under shear.
(2) Emulsions and slurries containing suspended
particles such as blood and water/clay
Navier Stokes Equations:
a  

DV
   gkˆ     ij
Dt
DV

   gkˆ  p 
Dt
x j
2


2





V

ij
ij 

3

Recall μ = μ(T) μ increases with T for gases, decreases
with T for liquids, but if it is assumed that μ = constant:

DV

2 
   gkˆ  p  2
 ij  
 V
Dt
x j
3 x j


2
 ij 
x j
x j
 ui u j


 x j xi

 2ui
  2ui   2 V
 
 x j x j
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
𝐷𝑉    g kˆ  p    2 V  2    V 


𝜌
3

x
𝐷𝑡
j


For incompressible flow

DV

Dt
V  0
  gkˆ  p
  2V
pˆ where pˆ  p z
piezometric pressure
For μ = 0

DV
   g kˆ  p
Dt
Euler Equation
NS equations for ρ, μ constant

DV
 pˆ   2 V
Dt
 DV

 V V   pˆ   2 V
 Dt


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.
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Chapters 3 & 4
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Fall 2013
Application of CV Momentum Equation:
d
 F  dt  V  d    V V
CV
R
 n dA
CS
F  F  F (includes reaction forces)
B
S
Note:
1. Vector equation
2. n = outward unit normal: V R  n < 0 inlet, > 0 outlet
3. 1D Momentum flux
 V V  n dA    m V 
i
i out
   mi V i in
CS
Where V i , i are assumed uniform over discrete inlets
and outlets
mi  iVni Ai
F
net force
on CV

d
V  d

dt CV
time rate of change
of momentum in CV

m V 
i
i out

 m V 
i
i in
outlet momentum inlet momentum
flux
flux
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
4. Momentum flux correlation factors
 u V  n dA 
  u 2 dA
  AVav2
axial flow with
nonuniform
velocity profile
2
Where
1  u 
     dA
A CS  Vav 
1
Q
u
dA


A
A CS
Vav 
Laminar pipe flow:

r 
r

u  U 0 1  2   U 0 1  
 R
 R 
2
Vav  .53U 0
1
2
  43
Turbulent pipe flow:
r

u  U 1  
 R
m
0
Vav  U 0
2
1  m (2  m)
1 m 1
9
5
m 1
7
Vav =.82U0
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
1  m  2  m 
2

2
2(1  2m)(2  2m)
m=1/7 β = 1.02
5. Constant p causes no 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 with continuity and
energy equations. Careful practice is needed 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
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Chapters 3 & 4
30
Fall 2013
h. Nozzle
i. Propeller
j. Water-hammer
≈
First relate umax to U0 using continuity equation
R

0  U R   u 1  r
2
0
max
0
R

1
U0 
u
1 r
2  max
R
R 0

m
 2r dr
m
R
2 r dr  Vav
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Professor Fred Stern
Fall 2013
2
(1  m)(2  m)
Vav  umax
m = 1/2
m = 1/7
Chapters 3 & 4
31


U0 = .53umax
U0 = .82umax
m = 1/2
Vav = .53umax
m = 1/7
Vav = .82umax
umax = U0/.53
umax = U0/.82
Second, calculate F using momentum equation:
F = wall drag force =  2Rx (force fluid on wall)
-F = force wall on fluid
w
 F   p  p R  F   u ( u 2r dr )  U ( R U )
R
2
x
1
2
2
2
2
0
0
R
F   p1  p2   R  U  R    u22 2 r dr
2
2
0
2
0
 AV 2
av
F   p1  p2   R2  U02 R2  2  AVav2
= U02 from
continuity
2
1  u 
     dA
A  Vav 
momentum flux correction
factor , page 163 text
=
=
4/3 laminar flow
1.02 turbulent flow
0
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Professor Fred Stern
Chapters 3 & 4
32
Fall 2013
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
Complete analysis
using CFD!
2
0
Reconsider the problem for fully developed flow:
Continuity:
min  mout  0
or
m  min  mout
Q, V = constant
Momentum:
F
x
 ( p1  p2 ) R 2  F    u V  n  dA    u V  n  dA
in
out
2
2
   (  AVave
)in   (  AVave
)out
  QVave (  out   in )
0
 p1  p2   R 2   w 2 Rx  0
p
 
w
R  dp 
r  dp 



or
for
smaller
CV
r
<
R,


 
2  dx 
2  dx 
(valid for laminar or turbulent flow, but assume laminar)
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
du
du r  dp 
 
  
dy
dr 2  dx 
du
r  dp 
   
dr
2  dx 
 
y = R-r (wall coord.)
r  dp 
u      c
4  dx 
2
u ( r  R)  0

R  dp 
c
 
4  dx 
2
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  4 Vave
   

2  dx  2  R 2 
R
2
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Professor Fred Stern
f 
Chapters 3 & 4
34
Fall 2013
8 w
32
64
64



2
Vave
 RVave Vave D Re
Re 
Vave D

Exact solution of NS for laminar fully developed pipe
flow
Application of relative inertial coordinates for a
moving but non-deforming control volume (CV)
The CV moves at a constant velocity V CS with respect to
the absolute inertial coordinates. If V R represents the
velocity in the relative inertial coordinates that move
together with the CV, then:
VR  V  VCS
Reynolds transport theorem for an arbitrary moving deforming
CV:
dBSYS d

 d    VR  n dA

dt
dt CV
CS
For a non-deforming CV moving at constant velocity, RTT for
incompressible flow:
dBsyst


d      VR  ndA
dt
CV
t
CS
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
1. Conservation of mass
Bsyst  M , and   1 :
dM
   VR  ndA
dt
CS
For steady flow:
V
R
 ndA  0
CS
2. Conservation of momentum
Bsyst  M VR  V CS  and   dBsyst dM  VR  VCS


d [ M VR  V CS ]
dt
 F  


 VR  V CS
t
CV
 d  
 V
R

 V CS VR  ndA
CS
For steady flow with the use of continuity:
 F    V
R

 V CS VR  ndA
CS
   VR VR  ndA  V CS  VR  ndA
CS
F
0
CS
   VR VR  ndA
CS
Example (use relative inertial coordinates):
Ex) A jet strikes a vane which moves to the right at constant velocity 𝑉𝐶 on a
frictionless cart. Compute (a) the force 𝐹𝑥 required to restrain the cart and (b)
the power 𝑃 delivered to the cart. Also find the cart velocity for which (c) the
force 𝐹𝑥 is a maximum and (d) the power 𝑃 is a maximum.
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
Solution:
Assume relative inertial coordinates with non-deforming CV i.e. CV moves
at constant translational non-accelerating
𝑉𝐶𝑆 = 𝑢𝐶𝑆 𝑖̂ + 𝑣𝐶𝑆 𝑗̂ + 𝑤𝐶𝑆 𝑘̂ = 𝑉𝐶 𝑖̂
then V R  V  V CS . Also assume steady flow 𝑉 ≠ 𝑉(𝑡) with 𝜌 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 and
neglect gravity effect.
Continuity:
V
R
 ndA  0
CS
Bernoulli without gravity:
0
1
1
VR21  p2  VR22
2
2
VR1  VR 2
VR1 A1  VR 2 A2
A1  A2  A j
0
p1 
Since
Momentum:
∑𝐹 = 𝜌 ∫𝐶𝑆 𝑉𝑅 𝑉𝑅 ⋅ 𝑛𝑑𝐴
F
x
  Fx    VRx VR  ndA
CS
−𝐹𝑥 = 𝜌𝑉𝑅𝑥 1 (−𝑉𝑅1 𝐴1 ) + 𝜌𝑉𝑅𝑥 2 (𝑉𝑅2 𝐴2 )
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Professor Fred Stern
Chapters 3 & 4
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Fall 2013
0 = 𝜌 ∫𝐶𝑆 𝑉𝑅 ⋅ 𝑛𝑑𝐴
−𝜌𝑉𝑅1 𝐴1 + 𝜌𝑉𝑅2 𝐴2 = 0
𝑉𝑅1 𝐴1 = 𝑉𝑅2 𝐴2 = (𝑉
⏟𝑗 − 𝑉𝐶 )
𝐴𝑗
𝑉𝑅1 =𝑉𝑅𝑥 1 =𝑉𝑗 −𝑉𝐶
−𝐹𝑥 = 𝜌(𝑉𝑗 − 𝑉𝐶 )[−(𝑉𝑗 − 𝑉𝐶 )𝐴𝑗 ] + 𝜌(𝑉𝑗 − 𝑉𝐶 ) cos 𝜃 (𝑉𝑗 − 𝑉𝐶 )𝐴𝑗
2
𝐹𝑥 = 𝜌(𝑉𝑗 − 𝑉𝐶 ) 𝐴𝑗 [1 − cos 𝜃]
2
𝑃𝑜𝑤𝑒𝑟 = 𝑉𝐶 𝐹𝑥 = 𝑉𝐶 𝜌(𝑉𝑗 − 𝑉𝐶 ) 𝐴𝑗 (1 − cos 𝜃)
𝐹𝑥 𝑚𝑎𝑥 = 𝜌𝑉𝑗2 𝐴𝑗 (1 − cos 𝜃), 𝑉𝐶 = 0
𝑃𝑚𝑎𝑥 ⇒
𝑑𝑃
𝑑𝑉𝐶
=0
𝑃 = 𝑉𝐶 𝜌(𝑉𝑗2 − 2𝑉𝐶 𝑉𝑗 + 𝑉𝐶2 )𝐴𝑗 (1 − cos 𝜃)
= 𝜌(𝑉𝑗2 𝑉𝐶 − 2𝑉𝐶2 𝑉𝑗 + 𝑉𝐶3 )𝐴𝑗 (1 − cos 𝜃)
𝑑𝑃
= 𝜌(𝑉𝑗2 − 4𝑉𝐶 𝑉𝑗 + 3𝑉𝐶2 )𝐴𝑗 (1 − cos 𝜃) = 0
𝑑𝑉𝐶
3𝑉𝐶2 − 4𝑉𝑗 𝑉𝐶 + 𝑉𝑗2 = 0
+4𝑉𝑗 ± √16𝑉𝑗2 − 12𝑉𝑗2
𝑉𝐶 =
VC 
𝑃𝑚𝑎𝑥 =
6
=
4𝑉𝑗 ± 2𝑉𝑗
6
Vj
3
𝑉𝑗
3
4
𝜌(
2𝑉𝑗 2
3
) 𝐴𝑗 (1 − cos 𝜃)
= 27 𝑉𝑗3 𝜌𝐴𝑗 (1 − cos 𝜃)
Example (use absolute inertial and relative inertial
coordinates)
058:0160
Professor Fred Stern
Chapters 3 & 4
38
Fall 2013
Vout,R
Vin,R
ΩR
CV
Assume gravity force is negligible and the cross section
area of the jet does not change after striking the bucket.
Taking moving CV at speed Vs= ΩR î enclosing jet and
bucket:
Solution 1 (relative inertial coordinates)
Continuity: min, R  mout , R  0
mR  min, R  mout , R    V R  n dA
CS
Bernoulli without gravity:
0
0
1
1
2
Vin2, R  p2  Vout
,R
2
2
Vin , R  Vout , R
p1 
058:0160
Professor Fred Stern
Fall 2013
n  iˆ
Vin, R  V j  R
Inlet
Vout , R  (V j  R)
Outlet
Since
Chapters 3 & 4
39
^
ni
 Vin , R A1  Vout , R A2  0
A1  A2  A j
Momentum:
F
X
 Fbucket  mRVout , R  mRVin, R
Fbucket  mR  (V j  R)  (V j  R ) 
 2mR (V j  R)
 2  Aj (V j  R) 2
mR   Aj (V j  R)
P  RFbucket  2 Aj R(V j  R)2
dP
 0 for
d
R 
Vj
3
Pmax 
8
 AjV j3
27
If infinite number of buckets: mR   A jV j
all jet mass flow
result in work.
Fbucket  2  AjV j (V j  R)
P  2  AjV j R (V j  R )
dP
0
d
for R 
Vj
2
Pmax 
1
 A jV j3
2
058:0160
Professor Fred Stern
Chapters 3 & 4
40
Fall 2013
Solution 2 (absolute inertial coordinates)
Continuity: from solution 1
V in , R  V out , R  0
express in the absolute inertial coordinates
V
in , e
 R   (Vout ,e  R)  0
Vin,e  V j
Inlet
Vout ,e  V j  2R
Outlet
Note min,e  mout ,e
Momentum:
n  iˆ
F
X
^
ni
!!
 Fbucket   AjVout ,eVout ,R   AjVin,eVin,R
Fbucket    Aj (V j  R) (V j  2R)  V j 
 2  Aj (V j  R)2
Same as Solution 1.
058:0160
Professor Fred Stern
Chapters 3 & 4
41
Fall 2013
Application of CV continuity equation for steady
incompressible flow one inlet and outlet A = constant
  V  ndA    V  ndA  m  Q
in
out
Qin  Qout
Vave Ain  Vave Aout
Vave in  Vave out
For A = constant
 F    V V  n  dA    V V  n  dA
in
Pipe:
F
x
out
   u V  n  dA    u V  n  dA
in
out
2
     AVave
     AVave2 
in
out
 QVave  out  in 
Vane:
change in shape u
 F  m V V  V  V
 F  m u  u   m  2u 
out
x
out
in
out
in
in
in
change in direction u
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
058:0160
Professor Fred Stern
Chapters 3 & 4
42
Fall 2013
U
2. Laminar flow Recrit =   2000
Re > Recrit 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. 80 exact solutions for simple laminar flows are
mostly linear V V  0
a.
b.
c.
d.
e.
f.
g.
Couette flow = shear driven
Steady duct flow = Poiseuille flow
Unsteady duct flow
Unsteady moving walls
Asymptotic suction
Wind-driven flows
Similarity solutions. etc.
058:0160
Professor Fred Stern
Fall 2013
Chapters 3 & 4
43
6. Also many exact solutions for low Re Stokes and
high Re BL approximations
7. Can also use CFD for non simple laminar flows
8. 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, i.e., assume analytical solution = truth, called
analytical benchmark
LT(T)=LM(M)=LA(A)=0 similarly for IC&BC
δS = S – T = δSN + δSM
US2 = USN2 + USM2
058:0160
Professor Fred Stern
Chapters 3 & 4
44
Fall 2013
Energy Equation:
B = E = energy
β = e = dE/dm = energy per unit mass
Integral Form (fixed CV):
dE

dt

 t (e ) d    e V  n dA
CV
 Q W
CS
rate of change rate of outflux
E in CV
E across CS
Rate of
change E
Rate of heat
added CV
^
1
e  u  v  gz  internal + KE + PE
2
2
Q = conduction + convection + radiation
W
Wshaft
pump / turbine
dWp   p ndA V
Wp 
 p V  n  dA
CS

W p  W
pressure viscous
- pressure force  velocity
Rate work
done by CV
058:0160
Professor Fred Stern
Chapters 3 & 4
45
Fall 2013
dWv   dAV
- viscous force  velocity
Wv      V dA
CS
Q  Ws  W 

CV t (e ) d   CS  e  p /   V  n dA
For our purposes, we are interested in steady flow one
inlet and outlet. Also W 0 in most cases; since, V = 0 at
solid surface; on inlet and outlet only τn ~ 0; or for V  0
and τstreamline ~ 0 if outside BL.
Q  WS 
1 2


ˆ
u

V

gz

p
/


 V  n dA

2

inlet &outlet 
Assume parallel flow with
p /   gz
and û constant over
inlet and outlet.
= constant ie hydrostatic
pressure variation
Q  WS   uˆ  p /   gz 

inlet &outlet
V  n dA 

2 inlet &outlet
Q  WS   uˆ  p /   gz in (min ) 

V
2
in
3
V 2 (V  n) dA
dAin
in
  uˆ  p /   gz out (mout ) 

3
V
out
 dAout
2 out
058:0160
Professor Fred Stern
Chapters 3 & 4
46
Fall 2013
Define kinetic energy correction factor
3
1  V 
  
 dA 
A A  Vave 
Laminar flow:
0
β = 4/3
r

u  U 1  
 R
2



α=2
m
0
1  m   2  m 
3

2 
Vave
V (V  n) dA  
m
2 A
2
2
 r
u  U 1   
 R
Vave=0.5
Turbulent flow:

3
4(1  3m)(2  3m)
m=1/7
α=1.058
as with β, α~1 for
turbulent flow
2
2
Vave
Vave
Q Ws

 (uˆ  p /   gz  
)out  (uˆ  p /   gz  
)in
m m
2
2
Let in = 1, out = 2, V = Vave, and divide by g
058:0160
Professor Fred Stern
Fall 2013
Chapters 3 & 4
47
p1 1 2
p


V1  z1  hp  2  2 V22  z2  ht  hL
 g 2g
 g 2g
Ws Wt Wp


 ht  hp
gm gm gm
hL 
1
Q
(u2  u1 ) 
g
mg
hL = thermal energy (other terms represent mechanical energy
m   AV
1 1   A2V2
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.
058:0160
Professor Fred Stern
hL 
Chapters 3 & 4
48
Fall 2013
p1  p2
L  d

 ( z1  z2 ) 

(
p


z
)
L  dx


g
 g  dx

w
L  2 w 

f

1
 g  R 
2
Vave
8
2
L Vave
hL  f
D 2g
For laminar flow,
f 
hL 
Darcy-Weisbach Equation (valid for laminar or
Turbulent flow)
8 w
32 

2
Vave
 RVave
32 LVave
 D2
For turbulent flow,
f=f (Re, k/D)

Vave
Recrit ~ 2000, Retrans ~ 3000
Re = VaveD/ν, k = roughness
2
hL  Vave
In this case hL = hf = friction loss
more generally,
exact solution!
058:0160
Professor Fred Stern
Chapters 3 & 4
49
Fall 2013
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.
(a) First suppose 2D problem: D1 and D2 denotes width in
y instead of diameter and we take unit in z direction
F
x
  F  mV2  .79  989  0.02 1 V22  425 N

A2
V2  5.22 m / s, m  81.6 kg / s
Continuity equation between points 1 and 2
V1 A1  V2 A2  V1  V2
D2
 2.09m / s
D1
058:0160
Professor Fred Stern
Chapters 3 & 4
50
Fall 2013
Bernoulli neglect g, p2=pa
p1 
1
1
V12  p2  V22
2
2
1
p1  p2   V22  V12 
2

hL=0, z=constant
p1  101, 000 
.79  998
(5.222  2.092 )
2
p1  110,020 Pa

p2 
Note:
V22  p3 
2

2
V32  p4 

2
V42
p2  p3  p4  pa  V2  V3  V4
0   V  A  A2V2  A3V3  A4V4
CS
A A A
2
3
F
y
4
 0   VV  A  V3V3 A3   (V4 )V4 A4
CS
 V32 A3  V42 A4 
A A
3
4
(b) For the round jet given in the problem statement
F
x

  F  mV2  .79  989 .022 V22  425 N
4

A2

V2  41.4 m / s, m  10.3 kg / s
058:0160
Professor Fred Stern
Chapters 3 & 4
51
Fall 2013
Continuity equation between points 1 and 2
D 
V1 A1  V2 A2  V1  V2  2 
 D1 
2
V1  41.4  
5
2
2
V1  6.63 m / s
Bernoulli neglect g, p2=pa
p1 
1
1
V12  p2  V22
2
2
1
p1  p2   V22  V12 
2
p  760,000 Pa
1

hL=0, z=constant
p1  101, 000 
.79  998
(41.42  6.632 )
2
058:0160
Professor Fred Stern
(a)
Torricelli’s
expression
for speed of
efflux from
reservoir
Fall 2013
V22
z1 
 z2
2g
  1, h  0, z  11, z  0
2
V2  2 g ( z1  z2 )
Q2  A2V2 
Re 
(b)
Chapters 3 & 4
52
VD



4
L
 2 * 9.81*11
1
2
 14.7 m / s
(.01) 2 *14.7 *3600  4.16 m3 / h
14.7  0.01
 1.5 105
6
10
V22
z1   2
 z2  hL
2g
 2  2, hL 
32VL
,   106 m2 / s
2
D g
V22  3.2V2  107.8  0
V2 = 8.9 m/s
Q= 2.516 m3/s
Re  89,000  8.9 104
2000
058:0160
Professor Fred Stern
Chapters 3 & 4
53
Fall 2013
V22
L V22
z  2
 z2  f
(c) 1
2g
D 2g
α2=1
V22
z1  z2 
1  fL / D 
2g
V2   2 g ( z1  z2 ) /(1  fL / D)
V2   216 /(1  f *1000)
1
1
2
f  f (Re), Re 
2
guess f = 0.015
V2  3.7 m / s  Re  3.7 x104 ,
V2  2.94 m / s  Re  2.9 x104 ,
f  .024
f  .025
V2  2.88 m / s  Re  2.9 x104
VD
(d) Re    2000
2000
D
V
V22
32 LV2
( z 1  z2 )   2

20002 2
2g
g
V22
V22 32 LV23
( z 1  z2 )   2

2 g 20002 g
VD

058:0160
Professor Fred Stern
Chapters 3 & 4
54
Fall 2013
32LV23 V22

 11  0
2
2000  g g
V2  1.1 m / s
D  0.00182 m
Low U and small D to actually have laminar flow
Differential Form of Energy Equation:


dE


   (e )    (eV ) d   Q  W
dt CV t



e

De
 e

e
 e  (eV )  V e  
    V e 
t
t
Dt
 t

1
1
e  uˆ  V 2  gz  uˆ  V 2  g  r
2
2

De
DV
 Duˆ

 (Q  W ) /   q  w   
V
 g V 
Dt
Dt
 Dt

q    q    (k T )
Fourier’s Law
058:0160
Professor Fred Stern
Chapters 3 & 4
55
Fall 2013
w    (V   ij )  V      ij    ij
 DV

g 
 Dt


Where

ui
x j
momentum equation
 DV

V      ij     V
 g V 
 Dt

Duˆ
   ( k T ) 
Dt
 ij
ui
x j
 ij ui  pV
x j
continuity equation
p D
D
Dp
 
( p / ) 
 Dt
Dt
Dt

u
D
Dp
(uˆ  p /  )    (kT ) 
  ij i
Dt
Dt
x j
  0  dissipation function


Dh
Dp
   (kT ) 

Dt
Dt
Summary GDE for compressible non-constant property
fluid flow
Cont.

   ( V )  0
t
Mom.
DV

  g  p    ij '
Dt
 ij '   ij   V ij
g   gkˆ
058:0160
Professor Fred Stern
Energy 
Chapters 3 & 4
56
Fall 2013
Dh Dp

   (kT )  
Dt Dt
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)
6)
Continuum
Newtonian fluids
Thermodynamic equilibrium
f = -ρg k̂
heat conduction follows Fourier’s law
no internal heat sources
For incompressible constant property fluid flow
duˆ  cv dT
cv, μ, k , ρ ~ constant
c
v
DT
 k T  
Dt
For static fluid or V small
2
058:0160
Professor Fred Stern
c
Fall 2013
T
 k T
t
2
p
Chapters 3 & 4
57
heat conduction equation (also valid for solids)
Summary GDE for incompressible constant property fluid
flow (cv ~ cp)
V  0

DV
   gkˆ  p   2 V
Dt
c
DT
 k T  
Dt
2
p
“elliptic”
where   
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 .
058:0160
Professor Fred Stern
Chapters 3 & 4
58
Fall 2013
DV
1
  pˆ  2V
Dt

NS
  (NS )
p̂  p   z
(See derivation details on p.81)
ui u j
1 2
 D
2





V



p




x j xi
 Dt

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
Linear, “elliptic”
Most exact solutions NS; and for steady
flow superposition, elemental solutions and
separation of variables
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Chapters 3 & 4
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Fall 2013
V  0
V
1
  p  2V
t

  ( NS )   p  0
2
2) Boundary Layer Equations
For high Re >> 1 and attached boundary layers or
fully developed free shear flows (wakes, jets, mixing


layers), v  U ,  , p  0 , and for free shear flows
x
y
y
px = 0.
u v 0
x
y
ut  uu x  vu y   pˆ x   u yy
non-linear, “parabolic”
p y  0   pˆ x  U t  UU x
Many exact solutions; similarity methods
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Chapters 3 & 4
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Fall 2013
3) Inviscid Flow

    V   0
t
DV

  g  p
Euler Equation, nonlinear ," hyperbolic "
Dt
Dh Dp


   (k T ) p,V , T unknowns and  , h, k  f ( p, T )
Dt Dt
4) Inviscid, Incompressible, Irrotational
 V  0  V  
 V  0   2  0
linear ," elliptic "
 Euler Equation  Bernoulli Equation:
p

2
V 2   gz  const
Many elegant solutions: Laplace equation using
superposition elementary solutions, separation of
variables, complex variables for 2D, and Boundary
Element methods.
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Chapters 3 & 4
61
Fall 2013
Couette Shear Flows: 1-D shear flow between surfaces of
like geometry (parallel plates or rotating cylinders).
Steady Flow Between Parallel Plates: Combined Couette
and Poiseuille Flow.
 V  0
u x  v y  wz  0
ux  0

DV
 pˆ   2 V
Dt
u
 uu  vu  wu  0
t
x
y
z
0   pˆ x   u yy
DT
c
 k T  
Dt
0  kT   u
2
p
T
 uTx  vTy  wTz  0
t
2
yy
y
    2u x2  2v y2  2 wz2
 (vx  u y ) 2  ( wy  vz ) 2  (u z  wx ) 2 
  (u x  v y  wz )
  u y2
(note: inertia terms vanish identically and ρ is absent
from equations)
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Chapters 3 & 4
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Fall 2013
Non-dimensionalize equations, but drop *
T 
u  u /U
*
*
T T
T T
0
1
y*  y / h
0
u 0
(1)
h2
u yy 
pˆ x   B  cons.
U
(2)
x
T 
yy
U
2
(3)
2
k (T  T )



1
 u 
y
0
Pr Ec
B.C.
y=1
u=1
T=1
y = -1 u = 0
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.
 u
1
1
(1  y )  B(1  y 2 )
2
2
y=y/h
Note: linear
superposition since
Linear flow
due to U
Parabolic flow
due to px
V V  0
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Chapters 3 & 4
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Fall 2013
Solution depends on
h2
B
pˆ :
U x
pˆ x is opposite to U
B<0
B < -0.5
|B| >> 1
backflow occurs near lower wall
flow approaches parabolic profile
Pressure gradient effect
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
T rises due to
viscous dissipation
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
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Chapters 3 & 4
64
Fall 2013
Shear Stress
1) pˆ x  0
i.e. pure Couette Flow
Cf 


1
U 2
2

Uh


u y
1
U 2
2
1
Re
h
u *y*  1/ 2
(u / U )
 1/ 2
 ( y / h)
P0 = CfRe = 1: Better for non-accelerating flows
since ρ is not in equations and P0 = pure constant
058:0160
Professor Fred Stern
Fall 2013
2) U = 0
u* 
Chapters 3 & 4
65
i.e. pure Poiseuille Flow
1
B(1  y*2 )
2
u *y*   By *
B
Where
u 
 
1 h2 
u
pˆ x 1  y
h
2 

umax
u
Q 2
 u
2h 3
BU
upper
h
BU
 
lower
h
BU
2u
 

 3u
h
h
h
w
y y  h
U u
2
2u
h
pˆ x  max
U
U
Dimensional form
  u
y
BU
y
h
2

h
 Q   u dy  4 hu
max

h
3
max
 
max
w
C 
f


w
1
U
2
2
6
or
 uh
u
 u
lam.
2
turb.
P  C Re  6
0
f
h
Except for numerical constants same as for circular pipe.
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Professor Fred Stern
Chapters 3 & 4
66
Fall 2013
Rate of heat transfer at the walls:
q  kT
w
y yh
2
k
U
 (T  T )  
2h
4h
1
0
+ = upper, - = lower
Heat transfer coefficient:
  qw
Nu 
T1  T0 
2h
 1  Br
2
k
For Br >> 2, both upper & lower walls must be cooled to
maintain T1 and T0
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Chapters 3 & 4
67
Fall 2013
Conservation of Angular Momentum: moment form of
momentum equation (not new conservation law!)
B  H0 
 r V dm  angular momentum of system about
sys
inertial coordinate system 0.
  r V
dH0 d

 r V   d     r V V  n dA

dt
dt 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 )
out
mout   (r  V )in min
CS
M 
0
   dA  r


CS
surface force moment
   g d  r  M

 

CV
body force moment
M  moment of reaction forces
R
R
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Professor Fred Stern
Chapters 3 & 4
68
Fall 2013
Take inertial frame 0 as fixed to earth such that CS
moving at Vs= -Rω i
V 2  V0iˆ  Riˆ
r 2  Rjˆ
V 1  V0 kˆ
Retarding torque due to
bearing friction
M
Z
V0
T
 0 2
R  QR
V0  Q
Apipe
 T0 kˆ  (r 2 V 2 )mout  (r1 V 1 )min
mout  min  Q

r1  0 ˆj
To kˆ  R(V0  R)(kˆ) Q
interestingly, even for T0=0, ωmax=V0/R
(limited by ration such that large R small ; large V0 large )
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Professor Fred Stern
Chapters 3 & 4
69
Fall 2013
Differential Equation of Conservation of Angular
Momentum:
Apply CV form for fixed CV:
 z = angular acceleration
I = moment of inertia
I z  a dy
dx
dx
dy
dy
 b dy  c dx  d dx
2
2
2
2
I z   xy   yx  dxdy
Since
I


 dxdy 3  dydx3   dxdy  dx 2  dy 2 
12
12

 dx 2  dy 2  z   xy   yx
12
lim  xy   yx ,
dx 0, dy 0
i.e   
ij
ji
similarly
  ,  
xz
zx
yz
zy
stress tensor is symmetric (stresses
themselves cause no rotation)
058:0160
Professor Fred Stern
Fall 2013
Chapters 3 & 4
70
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
IC and BC for their solution, depending on physical
problem and appropriate approximations.
Types of Boundaries:
1. Solid Surface
2. Interface
3. Inlet/exit/outer
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Chapters 3 & 4
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Fall 2013
1. Solid Surface
a. Liquid
ℓ = mean free path << fluid motion; therefore,
maroscopic view is “no slip” condition, i.e. no
relative motion or temperature difference between
liquid and solid.
V liquid  V solid
T
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
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Professor Fred Stern
 
w
du
dy
Chapters 3 & 4
72
Fall 2013
l
w
3 
u 
2 a 

2 a
3
Ma  U
w
w
u / U  .75Ma C
w
a
low density limit
Cf 
w
1
U 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
Tgas  Tw
Low Re
 .87 MaC f
Tr  Tw  = driving ∆T
Ref. T
air
058:0160
Professor Fred Stern
Where
Chapters 3 & 4
73
Fall 2013
q
C  2C  2

 
 C U (T  T )


w
f
h
p
Reynolds Analogy
r
w
Ch = Stanton number, i.e. wall
heat transfer coefficient
2. Idealized gas/liquid interface (free surface problems
since interface is unknown and part of the solution, but
effect gas on liquid idealized).
Kinematic FSBC: free surface is stream surface
F   ( x, y )  z  surface function
1
n  F / F  ( x ,  y ,  1) /  x2   y2  1 2
DF
F
0
 V F
Dt
t
1 F
V  n  0
F t
Dynamic FSBC: stress continuous across free surface
(similarly for mass and heat flux)
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Professor Fred Stern
Chapters 3 & 4
74
Fall 2013
 ij n j   ij* n j  p  ij
Fluid 1 stress
Fluid 2 stress
Surface
tension pres.
(vector whose components are stress in direction of coordinate axes on surface with normal nj)
   p  Re (U  U )
1
ij
ij
i, j
j ,i
 ij*   pij  Re1 (Ui , j  U j ,i )  fluid 2
eg  p 
a
neglecting μair
ij
for
air
if
Atmospheric pressure
p  We1  KSN  KtN 
eˆS
K SN  nˆ 
s
eˆt
ˆ
K tN  n 
t
Curvature F for two mutually perp. directions.
Note: eˆS and eˆt normal to n = eˆn
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  Wz
incompressible flow
1+3+1=5 conditions for 5 unknowns = (V, p, ζ)
The first 4 conditions nonlinear
-Also need conditions for turbulence variables
058:0160
Professor Fred Stern
Chapters 3 & 4
75
Fall 2013
Many approximations, eg, inviscid approximation:
pa  p  0
small slope: ζx ~ ζy ~ 0
small normal velocity gradient: Wx ~ Wy ~ Wz = 0

(U ,V )  0
z
p=0
or
W  U  V
z
x
p̂   gz
y
or W  0
z
p̂ = 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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Professor Fred Stern
Chapters 3 & 4
76
Fall 2013
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).
V1  V 2
Vn1  Vn 2
required by KFSBC
V1 n V 2 n  
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”.
058:0160
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Fall 2013
Chapters 3 & 4
77
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Professor Fred Stern
Chapters 3 & 4
78
Fall 2013
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 are 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  (wy  vz )iˆ  (uz  wx ) ˆj  (vx  u y ) kˆ
i   ijk
 u u j
 k 
xk  x j xk
u j



123   321   231  1
 213   321  132  1
 ijk  0 otherwise
alternating tensor
= 2  the angular velocity of the fluid element
(i, j, k cyclic)
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Professor Fred Stern
Chapters 3 & 4
79
Fall 2013
A quantity intimately tied with vorticity is the circulation:
   V  dx
Stokes Theorem:
 a  dx    a  dA
A
   V  dx   V  dA    ndA
A
A
Which shows that if ω =0 (i.e., if the flow is
irrotational, then Γ = 0 also.
Vortex line = lines which are everywhere tangent to the
vorticity vector.
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Professor Fred Stern
Chapters 3 & 4
80
Fall 2013
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-free, 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 and using the
divergence theorem:
    d     n dA  0

A
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Professor Fred Stern
Chapters 3 & 4
81
Fall 2013
Application to a vortex tube results in the following
   n dA     n dA  0
Minus sign due to
outward normal
A1
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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Professor Fred Stern
Chapters 3 & 4
82
Fall 2013
D D

V  dx
Dt Dt 
DV
D

 dx   V 
dx
Dt
Dt
From the N-S equations we have
DV 1
p
 f
  2 V
Dt 

  F  p  2V


Also,
D

Dt

D
Dx
dx  d
 dV
Dt
Dt
   F  p /    d x 
  dF  

Define f  F for
the gravitational body
force F=ρgz.
2



  V   d x 
dp
 V  dV
1
d (V V )

2


dp 1
2

dF


dV
   2V  d x

 
 2

=0 since integration is around a closed
contour and F,p, & V are single valued!
D
    2 V  dx        dx
Dt
  V      V   2V

0
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Professor Fred Stern
Fall 2013
Chapters 3 & 4
83
Implication: The circulation around a material loop of
particles changes only if the net viscous force on those
particles gives a nonzero 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, incompressible, and irrotational) acted upon by
conservative forces (e.g., 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
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
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that is initially irrotational remains irrotational, which is
the justification for ideal-flow theory.
In a real viscous fluid, vorticity is generated by
viscous forces. Viscous forces are large 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:
 ij n j
where
 ij   ij
 11n1   12 n2   13n3   x
 21n1   22 n2   23n3   y
 u
v 
u
 12  12       
y
 y x 
 22   22  2
v
0
y
 w
v 
 32   32       0
 y z 
 31n1   32 n2   33n3   z
NOTE: the only component of
ω is ωz. Actually, this is a
general result in that it can be
shown that ωsurface is
perpendicular to the limiting
streamline.
Which shows that
u
 
dy
x
   0
y
z
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However from the definition vorticity we also see that
x  
u
  z
y
i.e., the wall vorticity is directly proportional to the wall
shear stress. This analysis can be easily extended for
general 3-d flow.
Rotation tensor
 ij n j   ij n j at a fixed solid wall
True since at a wall with coordinate x2,
from continuity



0
x1 x3
and
v
0
x2
Once vorticity is generated, its subsequent behavior is
governed by the vorticity equation.
N-S
V
 V V    p /    2V
t
Or
V
1

   V  V   V      p /     2 V
t
2

neglect f
The vorticity equation is obtained by taking the curl of
this equation. (Note     0 ).
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
  V      2 
t
Rate of change of ω
=
 V ( )  (V )  (V )  ( )V
Rate of deforming
vortex lines

 (V )    V  2 
Rate of viscous
Or t
diffusion of ω
D
Dt
Transport Eq. for ω

  

 


 
  u  v  w     x   y
 z  V  2 
t  x
y
z 
y
z 
 x
x



u
u
u
 u x  v x  w x  x
 y
 z
 2x
t
x
y
z
x
y
z
Stretching
 y
 y
 y
 y
turning
v
v
v
 y
 z 2 y
t
x
y
z
x
y
z
z



w
w
w
 u z  v z  w z  x
 y
 z
 2z
t
x
y
z
x
y
z
u
Note:
v
w
 x
(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
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In order to determine the pressure field in terms of the
vorticity, the divergence of the N-S equation is taken.
 V


 V V    p /     2 V 
 t

2 ( p /  )   V V 
Poisson Eq. for p
1
   2 V  V   V  2V    
2
does not depend explicitly on ν
Derivation of pressure Poisson equation:
Three vector identities to be used:
(1)
1
V V    V  V   V     V 
2
  a  b   b   a   a   b 
(2)
2
(3)   a   a    a 
Pressure Poisson equation in vector form:
 p
2       V V 

1

      V  V   V     V  
2

1
  2  V  V      V  ω 
2
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1
  2  V  V   ω    V   V    ω 
2
1
  2  V  V   ω  ω  V      V 
2
1
   2  V  V   ω  ω  V    2 V     V 


2
1
   2  V  V   V  2 V  ω  ω
2


Pressure Poisson equation in tensor form:
 p
1
 2      2  V  V   V  2 V  ω  ω
2

 2  uk ek 
1 2
 u j e j    uk ek    ui ei  

   V     V 

2 xi xi 
x j x j
 2 uk 
uk  
un 
1 2

u
u


u




e




 j k jk  i ik x x  ijk x i   lmn x el 
2 xi xi
j
j
j
m


 
2
 2ui 
u u 
1  u ju j 

 ui
   ijk  lmn k n   ei  el 
2 xi xi
x j x j 
x j xm 


 2ui
uk un 
1   

u
u

u



 j j   i x x  ijk lmn x x  il

2 xi  xi
j
j
j
m 


u j 
 2ui
uk un
1  

2
u

u








 j
 i
jm kn
jn km
2 xi 
xi 
x j x j
x j xm
 2ui
u u
u u
  u j 

  jm kn k n   jn km k n
uj
  ui
xi  xi 
x j x j
x j xm
x j xm
 u j u j
 2u j 
 2ui
u u u u j
 
 uj
 ui
 k k k

 x x
xi xi 
x j x j x j x j x j xk
i
 i

uk u j
x j xk
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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.

V  V V

   V 
Where
0   V

The irrotational part of
the velocity field can be
expressed as the gradient
of a scalar

 V  
If
Then
And


 V   V   V  0
 2  0

V   A
The GDE for  is the Laplace Eq.
Since     A  0
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
 V     A
Again, by vector identity
 2 A  (  A)
i.e
 A  
2
The solution of this equation is
A
Thus
1 
 d
4 R

V 
1
4

R 
R
3
d
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, high Re flow about slender bodies with
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Chapters 3 & 4
91
attached thin boundary layer and wake, 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).
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).
𝐷𝑉
𝑎𝑖 =
+ 𝑎𝑟𝑒𝑙
𝐷𝑡
𝐷𝑉
∑ 𝐹 = 𝑚𝑎𝑖 = 𝑚 (
+ 𝑎𝑟𝑒𝑙 )
𝐷𝑡
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∑ 𝐹 − 𝑚𝑎𝑟𝑒𝑙 = 𝑚
𝐷𝑉
𝐷𝑡 i.e Newton’s law
applies to noninertial frame with
addition of known
inertial force terms
𝑆𝑖 = 𝑅 + 𝑟
𝑑𝑅
𝑉𝑖 = 𝑉 +
+Ω×𝑟
𝑑𝑡
𝐷𝑉 𝑑 2 𝑅 𝑑Ω
𝑎𝑖 =
+
+
× 𝑟 + 2Ω × V + Ω × (Ω × 𝑟)
𝐷𝑡 𝑑𝑡 2 𝑑𝑡
𝐷𝑉
3rd term from fact that
=
+ 𝑎𝑟𝑒𝑙
𝐷𝑡
d2R
dt 2
(x,y,z) rotating at Ω(t).
= 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.
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CV form of Momentum equation for non-inertial
coordinates:
F 

a rel  d  
CV
d
 V  d   CS V V R  n dA
dt CV
where V R is the velocity of the CV relative to the noninertial coordinates (x,y,z).
Differential form of momentum equation for non-inertial
coordinates:
 V

 V V     arel    p   z    2V
 t
 body force

where
𝑎𝑟𝑒𝑙 = 𝑅̈ + 2Ω × V + Ω × (Ω × 𝑟) + Ω̇ × 𝑟
All terms in a seldom act in unison (e.g. geophysical
flows):
R ~0
earth not accelerating relative to distant stars
rel


 ~0
for earth
    r  ~ 0 g nearly constant with latitude
2V
𝑎𝑖 =
𝐷𝑉
𝐷𝑡
most important!
+
𝑅0−1 (2Ω
× V)
V02 L V0
R0  Rossby # 

V0 L
V
tV
V
,t  0
V0
L
if L is large, i.e., comparable to
the order of magnitude of the
earth radius, R0<1, then Coriolis
term is larger than the inertia
terms and is important.
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Example of Non-inertial Coordinates:
Geophysical fluids dynamics
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
  T  gkˆ term; and μ, k, Cp = constant) and thin layer on
W H
rotating surface assumption  ~  .
U
L
Differences between atmosphere and oceans: lateral
boundaries (continents) in oceans; currents in ocean (gulf
and Kuroshio stream) along western boundaries; clouds
and latent heat release in atmosphere due to moisture
condensation; Vocean = 0.1~1 or 2 m/s and Vatmosphere 10~20
m/s
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
v
y

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
f>0
f<0
f = 
f=0
= planetary
vorticity
= 2 * vertical
component Ω
northern hemisphere
southern hemisphere
at poles
at equator
Person translates with inertial
2
period Ti 
f
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Equations of Motion
V  0
Du
1 p
 fv  
  u
Dt
 x
2
0
Dv
1 p
 fu  
  v
Dt
 y
2
0
vertical component Ω negligible due
Dw
1 p g


  w to thin layer assumption, i.e.,
Dt
 z 
magnitude of 2 cos  u << other terms
   1   (T  T )
2
0
0
0
0
p, ρ = perturbation from hydrostatic condition
Geostrophic Flow: quasi-steady, large-scale motions in
atmosphere or ocean far from boundaries
 fv  
1 p
 x
fu  
0
U2 
DV
~ 0

Dt
 L 
f V ~ 0 ( fU )
0
U,L = horizontal scales
Rossby number =
Atmosphere:
Ocean:
1 p
 y
U
fL
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
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Therefore, neglect
DV
Dt
and since there are no boundaries,
2
neglect  V .
Z momentum 
p
  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.
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 iˆ  v ˆj  p  1 f   py iˆ  px ˆj    px iˆ  py ˆj 
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!
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Ekman Layer on Free Surface: effects of friction near
boundaries
Viscous layers:
u ( y , 0)  0
Sudden acceleration flat plate: ut   u yy
  3.64  t
Oscillating flat plate: u  u
t
Flat plate boundary layer:
0
u ( , t )  0
u v 0
x
y
uu  vu  u
x
  4.9  x / U
u ( , t )  0
u (0, t )  U cos t
yy
  6.5  / 
u (0, t )  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   uZZ
fu   vZZ
058:0160
Professor Fred Stern
Chapters 3 & 4
99
Fall 2013
u  
at z = 0
z
^
v 0
at z = 0
z
  i
  .002  (v
air
(u, v)  0
wind
 u (0))
at z =-∞
Multiply v-equation by i   1 and add to u-equation:
d 2V i f

V
dz 2

V  u iv
 complex velocity
V  Ae(1i ) z /   Be  (1i ) z / 

2

f
Ekman layer thickness
B = 0 for u(-∞), v(-∞) = 0

dV

dz
 A
at z = 0
 (1  i)
2 
i.e. u 
 /
 z 
e cos    and
f
  4
z /
z  xi y
058:0160
Professor Fred Stern
v
Fall 2013
Chapters 3 & 4
100
 /
 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
101
Fall 2013
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:
u0 (Vwind  6 m / s, T  200 C )  0.6
m/s after one min., 2.3 m/s
after one hour
turbulent t solution:(more realistic)
u0=0.2 m/s after 1 hr (3 % vwind)
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 νt, u0 = 2 cm/s and D =
100 m, which is more realistic.