57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
1
Chapter 6 Differential Analysis of Fluid Flow
Fluid Element Kinematics
Fluid element motion consists of translation, linear
deformation, rotation, and angular deformation.
Types of motion and deformation for a fluid element.
Linear Motion and Deformation:
Translation of a fluid element
Linear deformation of a fluid element
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
2
Change in  :
 u 
 x   y z   t

x


the rate at which the volume  is changing per unit
volume due to the gradient ∂u/∂x is
  u x   t  u
1 d 
 lim 

 t 0
 dt

t

 x
If velocity gradients ∂v/∂y and ∂w/∂z are also present, then
using a similar analysis it follows that, in the general case,
1 d   u v w



 V
 dt
x y z
This rate of change of the volume per unit volume is called
the volumetric dilatation rate.
  
Angular Motion and Deformation
For simplicity we will consider motion in the x–y plane,
but the results can be readily extended to the more general
case.
Angular motion and deformation of a fluid element
The angular velocity of line OA, ωOA, is
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
3

 t 0  t
OA  lim
For small angles
v x   x t v

tan    
 t
x
x
so that
  v x   t  v

 t 0
t

 x
Note that if ∂v/∂x is positive, ωOA will be counterclockwise.
OA  lim 
Similarly, the angular velocity of the line OB is
 u
OB  lim

 t 0  t
y
In this instance if ∂u/∂y is positive, ωOB will be clockwise.
The rotation, ωz, of the element about the z axis is defined
as the average of the angular velocities ωOA and ωOB of the
two mutually perpendicular lines OA and OB. Thus, if
counterclockwise rotation is considered to be positive, it
follows that
1  v u 
z    
2  x y 
Rotation of the field element about the other two coordinate
axes can be obtained in a similar manner:
1  w v 
x  
 
2  y z 
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
1  u
Chapter 6
4
w 
y   

2  z x 
The three components, ωx,ωy, and ωz can be combined to
give the rotation vector, ω, in the form:
1
1
ω   x i   y j   z k  curlV    V
2
2
since
i
j
k
1
1 
V 
2
2 x
u

y
v

z
w
1  w v  1  u w  1  v u 
 
 i  

 j    k
2  y z  2  z x  2  x y 
The vorticity, ζ, is defined as a vector that is twice the
rotation vector; that is,
  2ω    V
The use of the vorticity to describe the rotational
characteristics of the fluid simply eliminates the (1/2) factor
associated with the rotation vector. If  V  0 , the flow
is called irrotational.
In addition to the rotation associated with the derivatives
∂u/∂y and ∂v/∂x, these derivatives can cause the fluid
element to undergo an angular deformation, which results
in a change in shape of the element. The change in the
original right angle formed by the lines OA and OB is
termed the shearing strain, δγ,
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
5
    
The rate of change of δγ is called the rate of shearing strain
or the rate of angular deformation:
  v x   t   u y   t  v u

  lim  lim 
 
 t 0  t
 t 0

t

 x y
The rate of angular deformation is related to a
corresponding shearing stress which causes the fluid
element to change in shape.
The Continuity Equation in Differential Form
The governing equations can be expressed in both integral
and differential form. Integral form is useful for large-scale
control volume analysis, whereas the differential form is
useful for relatively small-scale point analysis.
Application of RTT to a fixed elemental control volume
yields the differential form of the governing equations. For
example for conservation of mass

dV
CV t
 V  A   
CS
net outflow of mass
across CS
=
rate of decrease
of mass within CV
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
6
Consider a cubical element oriented so that its sides are to
the (x,y,z) axes


u dxdydz
u 

inlet mass flux
udydz
x

outlet mass flux
Taylor series expansion
retaining only first order term
We assume that the element is infinitesimally small such
that we can assume that the flow is approximately one
dimensional through each face.
The mass flux terms occur on all six faces, three inlets, and
three outlets. Consider the mass flux on the x faces



x flux  ρu   ρu  dx  dydz outflux  ρudydz influx
x


=

(u )dxdydz
x
V
Similarly for the y and z faces

y flux  (v)dxdydz
y

z flux  (w )dxdydz
z
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
7
The total net mass outflux must balance the rate of decrease
of mass within the CV which is

 dxdydz
t
Combining the above expressions yields the desired result
  




(

u
)

(

v
)

(

w
)
 t x
 dxdydz  0

y

z


dV
 


 (u )  (v)  (w )  0
t x
y
z
per unit V
differential form of
continuity equations

   (V)  0
t
  V  V  
D
   V  0
Dt
D 
  V 
Dt t
Nonlinear 1st order PDE; ( unless  = constant, then linear)
Relates V to satisfy kinematic condition of mass
conservation
Simplifications:
1. Steady flow:   (V)  0
2.  = constant:   V  0
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
i.e.,
Chapter 6
8
u v w
 
0
x y z
3D
u v

0
x y
2D
The continuity equation in Cylindrical Polar Coordinates
The velocity at some arbitrary point P can be expressed as
V  vr er  v e  vz e z
The continuity equation:
 1   r  vr  1    v     vz 



0
t r
r
r 
z
For steady, compressible flow
1   r  vr  1    v     vz 


0
r
r
r 
z
For incompressible fluids (for steady or unsteady flow)
1   rvr  1 v vz


0
r r
r  z
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
9
The Stream Function
Steady, incompressible, plane, two-dimensional flow
represents one of the simplest types of flow of practical
importance. By plane, two-dimensional flow we mean that
there are only two velocity components, such as u and v,
when the flow is considered to be in the x–y plane. For this
flow the continuity equation reduces to
u v

0
x y
We still have two variables, u and v, to deal with, but they
must be related in a special way as indicated. This equation
suggests that if we define a function ψ(x, y), called the
stream function, which relates the velocities as


u
, v
y
x
then the continuity equation is identically satisfied:
         2  2

0

 

x  y  y  x  xy xy
Velocity and velocity components along a streamline
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
10
Another particular advantage of using the stream function
is related to the fact that lines along which ψ is constant are
streamlines.The change in the value of ψ as we move from
one point (x, y) to a nearby point (x + dx, y + dy) along a
line of constant ψ is given by the relationship:


d 
dx 
dy  vdx  udy  0
x
y
and, therefore, along a line of constant ψ
dy v

dx u
The flow between two streamlines
The actual numerical value associated with a particular
streamline is not of particular significance, but the change
in the value of ψ is related to the volume rate of flow. Let
dq represent the volume rate of flow (per unit width
perpendicular to the x–y plane) passing between the two
streamlines.


dq  udy  vdx 
dx 
dy  d
x
y
Thus, the volume rate of flow, q, between two streamlines
such as ψ1 and ψ2, can be determined by integrating to
yield:
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
11
2
q   d   2  1
1
In cylindrical coordinates the continuity equation for
incompressible, plane, two-dimensional flow reduces to
1   rvr  1 v

0
r r
r 
and the velocity components, vr and vθ, can be related to the
stream function, ψ(r, θ), through the equations
1 

vr 
, v  
r 
r
Navier-Stokes Equations
Differential form of momentum equation can be derived by
applying control volume form to elemental control volume
The differential equation of linear momentum: elemental
fluid volume approach
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
d
 F  dt  Vd    V V  dA
CV
CS

 =
Chapter 6
12
1-D flow approximation
 =   mi Vi out    mi Vi in
  AV  dydzu x-face
where m
mass flux
d
V dxdydz
dt




 =  u V   vV   w V  dxdydz
y
z
 x

x-face
y-face
z-face
combining and making use of the continuity equation yields
DV  V
DV

 V  V
dxdydz
F  
Dt
t
Dt
where  F   F body   Fsurface
Body forces are due to external fields such as gravity or
magnetics. Here we only consider a gravitational field; that
is,
 F body  d F grav  gdxdydz
and g  gk̂
for g z
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
13
i.e., f body  gk̂
Surface forces are due to the stresses that act on the sides of
the control surfaces
symmetric (ij = ji)
ij = - pij + ij
2nd order tensor
normal pressure
viscous stress
=
-p+xx
yx
zx
xy
-p+yy
zy
ij = 1
ij = 0
i=j
ij
xz
yz
-p+zz
As shown before for p alone it is not the stresses
themselves that cause a net force but their gradients.




dFx,surf =   xx    xy    xz  dxdydz
y
z
 x

 p 



=    xx    xy    xz  dxdydz
y
z
 x x

This can be put in a more compact form by defining
 x   xx î   xy ĵ   xz k̂
vector stress on x-face
and noting that
 p

dFx,surf =      x  dxdydz
 x

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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
fx,surf = 
p
   x
x
Chapter 6
14
per unit volume
similarly for y and z
p
fy,surf =      y
y
 y   yx î   yy ĵ   yz k̂
p
   z
z
 z   zx î   zy ĵ   zz k̂
fz,surf = 
finally if we define
ij   x î   y ĵ   z k̂
then
f surf  p    ij    ij
ij  pij  ij
Putting together the above results
DV
 f  f body  f surf  
Dt
f body  gk̂
f surface  p    ij
a
DV  V

 V  V
Dt
t
 a    gkˆ  p    ij
inertia
force
body
force surface
due to force due
gravity to p
surface force
due to viscous
shear and normal
stresses
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
15
For Newtonian fluid the shear stress is proportional to the
rate of strain, which for incompressible flow can be written
ij  ij
 = coefficient of viscosity
ij = rate of strain tensor
=
 v u 
  
 x y 
v
y
 v w 
 

 z y 
u
x
 u v 
  
 y x 
 u w 
 

 z x 

du
dy
 w u 
 


x
z 

 w v 
 


y
z 

w
z
1-D flow
rate of strain
a  gk̂  p    ij 


 ij    2 V
x i
a  gk̂  p   2 V
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
a  p  z    2 V
V  0
Chapter 6
16
Navier-Stokes Equation
Continuity Equation
Four equations in four unknowns: V and p
Difficult to solve since 2nd order nonlinear PDE
 2u  2u  2u 
 u
u
u
u 
p
x:   u  v  w       2  2  2 
x
y
z 
x
y
z 
 t
 x
2v 2v 2v
 v
v
v
v 
p
y:    u  v  w       2  2  2 
x
y
z 
y
y
z 
 t
 x
z:
 2w  2w  2w 
 w
w
w
w 
p

u
v
w
   z    2  2  2 

t

x

y

z
y
z 


 x
u v w
 
0
x y z
Navier-Stokes equations can also be written in other
coordinate systems such as cylindrical, spherical, etc.
There are about 80 exact solutions for simple geometries.
For practical geometries, the equations are reduced to
algebraic form using finite differences and solved using
computers.
Exact solution for laminar flow in a pipe
(neglect g for now)
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
17
use cylindrical coordinates: vx = u
vr = v
u = u(r) only
v = w = 0
Continuity:

rv   0  rv = constant = c
r
v = c/r
v(r = 0) = 0  c = 0
i.e., v = 0
Momentum:
  2 u 1  2 u 1 u  2 u 
Du
p

    2  2 2 
 2
Dt
x
r

r

x
r


r 

1 u  2 u 
u
u w u 
p
 u
  u  v 
 2
   
z
r r  
x
 t
 r r r 
1   u  1 p

r  
r r  r   x
r
u  2
 r A
r 2

u r   r 2  A ln r  B
4
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
u(r = 0)    A = 0
u(r = ro) = 0  u r  
i.e. u r  


 2 2
r  ro
4
1 p 2 2
r  ro
4 x

Chapter 6
18

parabolic velocity profile
Differential Analysis of Fluid Flow
We now discuss a couple of exact solutions to the NavierStokes equations. Although all known exact solutions
(about 80) are for highly simplified geometries and flow
conditions, they are very valuable as an aid to our
understanding of the character of the NS equations and
their solutions. Actually the examples to be discussed are
for internal flow (Chapter 8) and open channel flow
(Chapter 10), but they serve to underscore and display
viscous flow. Finally, the derivations to follow utilize
differential analysis. See the text for derivations using CV
analysis.
Couette Flow
boundary conditions
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
19
First, consider flow due to the relative motion of two
parallel plates
Continuity
u
0
x
u = u(y)
v=o
p p

0
x y
d2u
Momentum
0 2
dy
or by CV continuity and momentum equations:
u1y  u 2 y
u1 = u2
 Fx  uV  dA  Qu 2  u1   0

dp 
d 

 py   p  x y  x     dy x = 0
dx 

 dy 
d
0
dy
d  du 
i.e.
   0
dy  dy 
d 2u
 2 0
dy
from momentum equation
du
 C
dy
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
20
C
yD

u(0) = 0  D = 0
u
u(t) = U  C = 
U
t
U
y
t
du U
 
 constant
dy
t
Generalization for inclined flow with a constant pressure
gradient
u
Continutity
u
0
x
Momentum

d2u
0   p  z    2
x
dy
i.e.,
d 2u
dh
 2 
dx
dy
u = u(y)
v=o
p
0
y
h = p/ +z = constant
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
21
plates horizontal
plates vertical
dz
0
dx
dz
=-1
dx
which can be integrated twice to yield

du
dh
  yA
dy
dx
dh y 2
u  
 Ay  B
dx 2
now apply boundary conditions to determine A and B
u(y = 0) = 0  B = 0
u(y = t) = U
dh t 2
U
dh t
U  
 At  A 

dx 2
t
dx 2
 dh y 2 1  U
dh t 
u ( y) 
 

 dx 2   t
dx 2 
 dh
U
=
ty  y 2  y
2 dx
t


This equation can be put in non-dimensional form:
u
t 2 dh  y  y y

1   
U
2U dx  t  t t
define: P = non-dimensional pressure gradient
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
t 2 dh
=
2U dx
Y = y/t
Chapter 6
22
p
z

z 2  1 dp dz 


2U   dx dx 
h
u
 P  Y(1  Y)  Y
U
parabolic velocity profile

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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
23
u Py Py 2 y

 2 
U
t
t
t
t
q   udy
0
t
 U dy
q 0
u 
t
t
tu t P
P
y
   y  2 y 2   dy
U 0 t
t
t
Pt Pt t
=  
2 3 2
u P 1
t 2  dh  U
  u
   
U 6 2
12  dx  2
For laminar flow
ut
 1000

Recrit  1000
The maximum velocity occurs at the value of y for which:
d u
P 2P
1
du
0
 0  2 y
dy  U 
t t
t
dy
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
y
Chapter 6
24
t
P  1  t  t @ umax
2P
2 2P
 u max  u y max  
note: if U = 0:
u
u max

for U = 0, y = t/2
UP U U
 
4
2 4P
P P 2

6 4 3
The shape of the velocity profile u(y) depends on P:
dh
1. If P > 0, i.e.,  0 the pressure decreases in the
dx
direction of flow (favorable pressure gradient) and the
velocity is positive over the entire width

dh
d  p  dp
    z 
  sin 
dx
dx  
 dx
a)
dp
0
dx
b)
dp
  sin 
dx
dh
 0 the pressure increases in the
dx
direction of flow (adverse pressure gradient) and the
velocity over a portion of the width can become
negative (backflow) near the stationary wall. In this
1. If P < 0, i.e.,
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
25
case the dragging action of the faster layers exerted on
the fluid particles near the stationary wall is insufficient
to over come the influence of the adverse pressure
gradient
dp
  sin   0
dx
dp
  sin 
dx
2. If P = 0, i.e.,
u
a)
b)
or
 sin  
dp
dx
dh
 0 the velocity profile is linear
dx
U
y
t
dp
 0 and  = 0
dx
dp
  sin 
dx
For U = 0 the form
Note: we derived
this special case
u
 PY1  Y   Y is not appropriate
U
u = UPY(1-Y)+UY
t 2 dh
Y1  Y   UY
=
2 dx
t 2 dh
u
Y1  Y 
Now let U = 0:
2 dx
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
26
3. Shear stress distribution
Non-dimensional velocity distribution
u* 
u
 P  Y 1  Y   Y
U
u
is the non-dimensional velocity,
U
 t 2 dh
P
is the non-dimensional pressure
2U dx
where
u* 
Y
gradient
y
is the non-dimensional coordinate.
t
Shear stress
 
du
dy
In order to see the effect of pressure gradient on shear
stress using the non-dimensional velocity distribution, we
define the non-dimensional shear stress:
* 
Then
1
U 2
2
Ud  u U  2 du*
 


1
td
y
t
Ut dY
2


U
2
2

 2 PY  P  1
Ut
2

 2 PY  P  1
Ut
 A  2PY  P  1
*
where

A
2
0
Ut
1
is a positive constant.
So the shear stress always varies linearly with Y across any
section.
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
27
At the lower wall Y  0  :
 lw*  A 1  P 
At the upper wall Y  1 :
*
 uw
 A 1  P 
For favorable pressure gradient, the lower wall shear stress
is always positive:
1. For small favorable pressure gradient  0  P  1 :
*
 lw*  0 and  uw
0
2. For large favorable pressure gradient  P  1 :
*
 lw*  0 and  uw  0


 0  P  1
 P  1
For adverse pressure gradient, the upper wall shear stress is
always positive:
1. For small adverse pressure gradient  1  P  0 :
*
0
 lw*  0 and  uw
2. For large adverse pressure gradient  P  1 :
*
 lw*  0 and  uw
0
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
28


 1  P  0
 P  1
For U  0 , i.e., channel flow, the above non-dimensional
form of velocity profile is not appropriate. Let’s use
dimensional form:
 t 2 dh
 dh
u
Y 1  Y   
y t  y 
2 dx
2 dx
Thus the fluid always flows in the direction of decreasing
piezometric pressure or piezometric head because
dh

 0, y  0 and t  y  0 . So if
is negative,
dx
2
dh
positive; if dx is positive, u is negative.
u is
Shear stress:
 
Since
 1 
t  y  0 ,
 2 
du
 dh  1

t 
dy
2 dx  2

y

the sign of shear stress  is always
opposite to the sign of piezometric pressure gradient
dh
,
dx
and the magnitude of  is always maximum at both walls
and zero at centerline of the channel.
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
29
dh
 0,   0
dx
dh
0,  0
dx
For favorable pressure gradient,
For adverse pressure gradient,

dh
0
dx

dh
0
dx
Flow down an inclined plane
uniform flow  velocity and depth do not
change in x-direction
Continuity
du
0
dx
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
Chapter 6
30

d2u
x-momentum 0   p  z    2
x
dy

y-momentum 0   p  z   hydrostatic pressure variation
y
dp

0
dx
d 2u
 2    sin 
dy
du

  sin y  c
dy


y2
u   sin   Cy  D

2
du


 0   sin d  c  c   sin d
dy yd


u(0) = 0  D = 0

y2 
u   sin   sin  dy

2 
=

sin  y2d  y 
2
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
u(y) =
Chapter 6
31
g sin 
y2d  y 
2
d
 2 y3 

q   udy  sin dy  
2
3 0
0

d
=
V avg
discharge per
unit width
1 3
d sin 
3
q 1 2
gd 2
 
d sin  
sin 
d 3
3
in terms of the slope So = tan   sin 
gd 2So
V
3
Exp. show Recrit  500, i.e., for Re > 500 the flow will
become turbulent
p
   cos 
y
Re crit 
Vd
 500

p    cos  y  C
pd   p o   cos  d  C
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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2006
i.e.,
Chapter 6
32
p   cos  d  y  p o
* p(d) > po
* if  = 0
if  = /2
p = (d  y) + po
entire weight of fluid imposed
p = po
no pressure change through the fluid
32