Fluid Mechanics: Fundamentals
and Applications, 4th edition
Yunus A. Cengel, John M. Cimbala
Lecture slides by Mehmet Kanoglu
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Chapter 9
DIFFERENTIAL ANALYSIS OF FLUID
FLOW
©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education.
© Corbis RF
The fundamental differential equations of fluid motion are derived in
this chapter, and we show how to solve them analytically for some
simple flows. More complicated flows, such as the air flow induced
by a tornado shown here, cannot be solved exactly.
©McGraw-Hill Education.
Conservation of Mass Principle
The conservation of mass principle for a control volume: The net
mass transfer to or from a control volume during a time interval t is
equal to the net change (increase or decrease) in the total mass within
the control volume during t.
 Total mass entering   Total mass leaving  Net change of mass



 the CV during t   the CV during t   within the CV during t 
min  mout  mCV
m in  m out  dmCV /dt
mCV  mfinal  minitial
(kg)
(kg)
m in and m out the total rates of mass flow into
and out of the control volume
dmCV /dt
the rate of change of mass within the
control volume boundaries.
Mass balance is applicable to any control
volume undergoing any kind of process.
©McGraw-Hill Education.
Linear Momentum Equation (Steady Flow)
Steady linear momentum equation:




 F    m V    mV
out
in
The net force acting on the control volume during steady
flow is equal to the difference between the rates of
outgoing and incoming momentum flows.
The net force acting on the
control volume during steady
flow is equal to the difference
between the outgoing and the
incoming momentum fluxes.
©McGraw-Hill Education.
9–1 ■ INTRODUCTION
The control volume technique is useful when
we are interested in the overall features of a
flow, such as mass flow rate into and out of the
control volume or net forces applied to bodies.
Differential analysis, on the other hand,
involves application of differential equations of
fluid motion to any and every point in the flow
field over a region called the flow domain.
Boundary conditions for the variables must
be specified at all boundaries of the flow
domain, including inlets, outlets, and walls.
If the flow is unsteady, we must march our
solution along in time as the flow field changes.
(a) In control volume analysis, the interior of
the control volume is treated like a black box,
but (b) in differential analysis, all the details of
the flow are solved at every point
within the flow domain.
©McGraw-Hill Education.
9–2 ■ CONSERVATION OF MASS—
THE CONTINUITY EQUATION
Conservation of mass for a CV :
 

0
dV   V  n dA
CV t
CS

m  m
CV t dV  
in
out
The net rate of change of mass within the
control volume is equal to the rate at
which mass flows into the control volume
minus the rate at which mass flows out of
the control volume.
To derive a differential
conservation equation, we
imagine shrinking a control
volume to infinitesimal size.
©McGraw-Hill Education.
Derivation Using the Divergence Theorem
The quickest and most straightforward way to derive the differential form of
conservation of mass is to apply the divergence theorem (Gauss’s theorem).
Divergence theorem:
 
 
   GdV ∮G  ndA
V
A



0
dV     ( V ) dV
CV t
CV
 
  
CV  t    ( V )  dV  0
Continuity equation:

 
   ( V )  0
t
This equation is the compressible form of the continuity
equation since we have not assumed incompressible
flow. It is valid at any point in the flow domain.
©McGraw-Hill Education.
Special Cases of the Continuity Equation
Special Case 1: Steady Compressible Flow
Steady continuity equation:


  (V )  0
(9 - 13)
In Cartesian coordinates, Eq. 9–13 reduces to
(u ) ( ) ( w)


0
x
y
z
In cylindrical coordinates, Eq. 9–13 reduces to
1 (r ur ) 1 (u ) (u z )


0
r r
r 
z
©McGraw-Hill Education.
Incompressible continuity equation:
 
 V  0
Incompressible continuity equation in Cartesian coordinates:
u    w


0
x y z
Incompressible continuity equation in cylindrical coordinates:
1 (rur ) 1 (u ) (u z )


0
r r
r 
z
Special Case 2:
Incompressible
Flow
The disturbance from
an explosion is not
felt until the shock
wave reaches the
observer.
©McGraw-Hill Education.
9–5 ■ THE NAVIER–STOKES EQUATION
Introduction
  xx
 ij    yx
 
zx
Fluid at rest:
 xy  xz    P 0
0
 yy  yz    0  P 0 



 zy  zz   0
0  P
Moving fluids :
ij, called the
viscous stress
tensor or the
deviatoric stress
tensor
  xx
 ij    yx
  zx
 xy
 yy
 zy
 xz    P 0
0    xx  xy  xz 
 yz    0  P 0     yx  yy  yz 



 zz   0
0  P   zx  zy  zz 
Mechanical pressure:
1
Pm   ( xx   yy   zz )
3
Mechanical pressure is the
mean normal stress acting
inwardly on a fluid element.
For fluids at rest, the only
stress on a fluid element is
the hydrostatic pressure,
which always acts inward
and normal to any surface.
©McGraw-Hill Education.
Newtonian versus Non-Newtonian Fluids
Rheology: The study of the
deformation of flowing fluids.
Newtonian fluids: Fluids for which the
shear stress is linearly proportional to
the shear strain rate.
Newtonian fluids: Fluids for which the
shear stress is not linearly related to the
shear strain rate.
Viscoelastic: A fluid that returns (either
fully or partially) to its original shape
after the applied stress is released.
Rheological behavior of fluids—shear
stress as a function of shear strain rate.
In some fluids a finite stress called the
yield stress is required before the
fluid begins to flow at all; such fluids
are called Bingham plastic fluids.
©McGraw-Hill Education.
Some non-Newtonian fluids are called
shear thinning fluids or
pseudoplastic fluids, because the
more the fluid is sheared, the less
viscous it becomes.
Plastic fluids are those in which the
shear thinning effect is extreme.
Shear thickening fluids
or dilatant fluids: The
more the fluid is sheared,
the more viscous it
becomes.
When an engineer falls into quicksand (a
dilatant fluid), the faster he tries to move,
the more viscous the fluid becomes.
©McGraw-Hill Education.
Derivation of the Navier–Stokes Equation for
Incompressible, Isothermal Flow
The incompressible flow
approximation implies constant
density, and the isothermal
approximation implies constant
viscosity.
Viscous stress tensor for an incompressible Newtonian fluid with constant properties:
 ij  2 ij
(9 - 55)
  xx  xy

 ij    yx  yy
  
zx
zy
©McGraw-Hill Education.

 u  
u
 u w  
2
     

 z x 
x
 y x 


 xz 
   w  

     u 
 yz       
2
   
y
 z y  
   x y 

 zz  




w

u

w



w




2
  x  z    y  z 

z

 u   
u
 u w  
2




 




x
 y  x 
z x  

P 0 0  
    u 
   w  



 ij  0  P 0      
2
   

   x y 
y
 z y  
 0 0 P
  w u




w



w




2
  x  z    y  z 

z
Du
P
 2u
   u 
  w u 


  g x  2 2   
  
 
Dt
x
x
y  x y 
z  x z 

  2u  u    2u  w  2u 
Du
P


  gx    2 

 2
 2
Dt
x
x x x y y
x z z 
 x
   u  w   2u  2u  2u 
P

  gx     

 2  2  2 
x
y
z 
 x  x y z  x
The Laplacian operator, shown
here in both Cartesian and
cylindrical coordinates, appears
in the viscous term of the
incompressible Navier–Stokes
equation.
©McGraw-Hill Education.
  w 
  w 






z  x 
x  z 
Du
P


  g x   2u
Dt
x
D
P


  g y   2
Dt
y
Dw
P


  g z   2 w
Dt
z
Incompressible Navier –Stokes equation:



DV

2

  P   g   V
Dt
(9 - 60)
The Navier–Stokes equation is an
unsteady, nonlinear, second order, partial
differential equation.
Equation 9–60 has four unknowns (three
velocity components and pressure), yet it
represents only three equations (three
components since it is a vector equation).
The Navier–Stokes equation is the
cornerstone of fluid mechanics.
©McGraw-Hill Education.
Obviously, we need another equation to
make the problem solvable. The fourth
equation is the incompressible continuity
equation (Eq. 9–16).
Continuity and Navier–Stokes Equations
in Cartesian Coordinates
Incompressible continuity equation:
u   w


0
x  y  z
x-component of the incompressible Navier –Stokes equation:
  2 u  2 u  2u 
 u
u
u
u 
P
  u

w  
 gx    2  2  2 
 t
x
y
z 
x
y
z 
 x
(9 - 61a)
(9 - 61b)
y -component of the incompressible Navier –Stokes equation:
  2  2  2 
 


 
P
  u

w  
 g y    2  2  2 
 t
x
y
z 
y
y
z 
 x
(9 - 61c)
z -component of the incompressible Navier –Stokes equation:
 2 w 2 w 2 w 
 w
w
w
w 
P

u

w  
  g z    2  2  2  (9 - 61d)
 t
x
y
z 
z
y
z 
 x
©McGraw-Hill Education.
Continuity and Navier–Stokes Equations in
Cylindrical Coordinates
1 (rur ) 1 (u ) (u z )


0
r r
r 
z
r -component of the incompressible Navier –Stokes equation:
Incompressible continuity equation:
 u r
ur u ur u2
u r 

 ur

  uz
r
r 
r
z 
 t

 1   ur  ur 1  ur 2 u  ur 
P
 gr   
 2
 2 
 r
  2  2
2
r
r
r
r 
r 
z 
 r r
2
2
(9 - 62a)
(9 - 62b)
 -component of the incompressible Navier –Stokes equation:
u u u u u
u 
 u
 ur      r   u z  
 t
r
r 
r
z 


 1   u  u 1  u 2 ur  u 
1 P
  g   
 2
 2 
 r
  2  2
2
r 
r

r

r
r
r


r


z 

2
2
(9 - 62c)
z -component of the incompressible Navier –Stokes equation:
u u u
u 
 u z
 ur z   z  u z z 
 t
r
r 
z 


©McGraw-Hill Education.
 1   u z  1  u z  u z 
P
 gz   
 2 
 r
  2
2
z
r
r 
z 
 r r
2
2
(9 - 62d)
9–6 ■ DIFFERENTIAL ANALYSIS OF FLUID
FLOW PROBLEMS
There are two types of problems for which the differential equations
(continuity and Navier–Stokes) are useful:
1. Calculating the pressure field for a known velocity field
2. Calculating both the velocity and pressure fields for a flow of known
geometry and known boundary conditions
A general threedimensional but
incompressible flow field
with constant properties
requires four equations to
solve for four unknowns.
©McGraw-Hill Education.
Exact Solutions of the Continuity
and Navier–Stokes Equations
Boundary Conditions
No-slip boundary condition:


V fluid  V wall
Procedure for solving the
incompressible continuity and
Navier–Stokes equations.
©McGraw-Hill Education.
A piston moving at speed VP in a cylinder.
A thin film of oil is sheared between the
piston and the cylinder; a magnified view of
the oil film is shown. The no-slip boundary
condition requires that the velocity of fluid
adjacent to a wall equal that of the wall.