Transcript Chapter 5

MECH 221 FLUID MECHANICS
(Fall 06/07)
Chapter 5:
EQUATIONS OF MOTION
OF VISCOUS FLOWS
Instructor: Professor C. T. HSU
1
MECH 221 – Chapter 5
5.1 Flow Fields and Gradients

The continuum assumption allows the treatment of
fluid properties as fields, scalar vector or tensor, which
are function of space (r) and time (t)



 ( r ,t )
Scalar fields: density –
P( r ,t )
pressure –
temperature – T ( r ,t )
Vector fields: velocity – v( r ,t )
vorticity – ω( r ,t )
σ
τ
( r ,t )
Tensor fields: total stress –
viscous stress – ( r ,t )
S ( r ,t )
strain rate –
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MECH 221 – Chapter 5
5.1 Flow Fields and Gradients

Total change of a scalar field,  , due to change in
space only:
For the position vector given by r  xi  yj  zk ,
the change in  is only cause by the change in r
described by
dr  dxi  dyj  dzk
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MECH 221 – Chapter 5
5.1 Flow Fields and Gradients

The total derivative is then given by



d 
dx 
dy 
dz    dr
x
y
z
where



 
i
j
k
x
y
z
is the gradient of  . The gradient  is a vector
along the direction where the magnitude of 
has a maximum.
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MECH 221 – Chapter 5
5.1 Flow Fields and Gradients

Consider P & Q to be 2 points on a surface
where  ( x , y , z )  C . These points are chosen so
that Q is a small distance from P. Then dr is
tangential to the surface.

Now let’s move from P to Q. The change in
 ( x , y , z )  C is then given by,
d    dr  0
since the 2 points are on the surface with the
same C.
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MECH 221 – Chapter 5
5.1 Flow Fields and Gradients


Therefore,  is perpendicular to dr from P.
Since dr may be in any direction from P, as
long as it is tangential to the surface   C ,
we conclude that  has to be in normal to
the surface   C .
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MECH 221 – Chapter 5
5.1 Flow Fields and Gradients

Example:

For unsteady flows where  may change with
time, recall that the total derivative is
D 




u
v
w
Dt
t
x
y
z


 v  
t
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MECH 221 – Chapter 5
5.2 Conservation of Mass

We recall in Chapter 3 that the conservation of
mass can be understood more easily in the
Lagrangian frame. It states that the total mass m
in control volume V has to be conserved if the
control volume deformed with the flow to confine
the same fluid particles.
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MECH 221 – Chapter 5
5.2 Conservation of Mass

We now extend further to include the cases when
s
there is a mass source in the control volume. If m
represents the rate of mass source per unit
volume, the mass balance then read:
dm S
d 

 S dV


dV
m



V(t )
dt
t  V ( t )

V ( t ) t dV  S v  ds  V ( t ) m S dV
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MECH 221 – Chapter 5
5.2 Conservation of Mass

Above equation is the integral form of the mass
conservation in the Eulerian description. Note
that the control volume V can be either fixed or
varying.

The first term on the left hand side is the
contribution caused by the density change in V
and the second term caused by the mass flux
enter the surface that define the control volume.
The term on the right hand side then represents
the rate of mass created or annihilated in V
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MECH 221 – Chapter 5
5.2 Conservation of Mass

To obtain the differential form, we now
employ the divergence theorem to the
second term on the left of conservation of
mass equation to give:
 

V ( t )  t    ( v) - m S  dV  0
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MECH 221 – Chapter 5
5.2 Conservation of Mass

Again, as V(t)→0 the integrand is independent of
V and therefore,

   ( v )  m S
t
which is the differential form of equation for mass
conservation.

Note that in the above equation, all terms are in
rate of mass change per unit volume.
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MECH 221 – Chapter 5
5.2 Conservation of Mass

For flow fields without mass sources, the
integral and differential forms of conservation
of mass equation reduce to
and

V ( t ) t dV   S  v  ds  0

   ( v )  0,
t
respectively, which were given in Chapter 3
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MECH 221 – Chapter 5
5.2.1 Derivation of the Differential Equation
in Eulerian frame and Cartesian Coordinate

Consider the fixed control volume as shown below:


 
v   ( v )y xz
 y
 

uyz
y
x
 v x z


 
 u   ( u )x yz
 x
 

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MECH 221 – Chapter 5
5.2.1 Derivation of the Differential Equation
in Eulerian frame and Cartesian Coordinate

Net mass leaving the control volume/time



  u  ( u )x  u  yz
x





  v  ( v )y  v  xz
y





  w  ( w )z  w xy
z






  ( u )  ( v )  ( w ) xyz
y
z
 x

   ( v )xyz
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MECH 221 – Chapter 5
5.2.1 Derivation of the Differential Equation
in Eulerian frame and Cartesian Coordinate

Net mass increase in the control volume/time


 ( xyz) 
xyz
t
t

Conservation of mass states that the net
mass entering the control volume/unit time is
equal to the rate of increase of mass in the
differential control volume
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MECH 221 – Chapter 5
5.2.1 Derivation of the Differential Equation
in Eulerian frame and Cartesian Coordinate


   ( v)xyz  ( xyz) 
xyz
t
t

   ( v )  0
t
   ( v)    v  v  
D
   v  0
Dt
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MECH 221 – Chapter 5
5.2.2 Special Cases

Steady flow,   0    ( v)  0

Incompressible flow,   constant    v  0

t

Cartesian:

Polar:

Spherical:
u v w
 
0
x y z
1 ( ru r ) 1 u u z


0
r r
r 
z
1 ( r 2ur )
1 (sin u )
1 u


0
2
r
r
r sin 

r sin  

u v
0 

0
2-D flow,
z
x y
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

In Chapter 3, for inviscid flows, only pressure
forces act on the control volume V since the
viscous forces (stress) were neglected and the
resultant equations are the Euler’s equations. The
equations for conservation of momentum for
inviscid flows were derived based on Newton’s
second law in the Lagrangian form.
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

Here we should include the viscous stresses to
derive the momentum conservation equations.

With the viscous stress, the total stress on the
fluid is the sum of pressure stress( σ p   pI , here
the negative sign implies that tension is positive)
and viscous stress ( τ ), and is described by the
stress tensor given by:
2
σ σ p  τ   pI  S  (   )(  v ) I
3
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

Here, we generalize the body force (b) due to all
types of far field forces. They may include those
due to gravity ( g ) , electromagnetic force, etc.

As a result, the total force on the control volume
in a Lagrangian frame is given by
F  
σ  d s   bdV
S( t )
V(t)
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

The Newton’s second law then is stated as:
dM d 

F


v
dV

dt
dt  V ( t )

 
( v )dV   vv  ds
V ( t ) t
S

Hence, we have

S σ  ds  V ( t )bdV  V ( t ) t ( v )dV   S vv  ds
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

By the substitution of the total stress into the
above equation, we have

V ( t ) t ( v )dV   S vv  ds   S pds   S ( t )τ  ds  V ( t b) dV
which is integral form of the momentum
equation.
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

For the differential form, we now apply the divergence
theorem to the surface integrals to reach:

V ( t ) t ( v )dV  V ( t )  ( vv)dV  V ( t ) pdV  V ( t )  (τ)dV  V ( t b) dV

Hence, V→0, the integrands are independent of V.
Therefore,

( v )    ( vv )  p    (τ )  b
t
which are the momentum equations in differential form
for viscous flows. These equations are also named as the
Navier-Stokes equations.
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)


For the incompressible fluids where
 = constant.
If the variation in viscosity (  ) is negligible
(Newtonian fluids), the continuity equation
becomes   v  0 , then the shear stress tensor
reduces to τ  S.
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MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

The substitution of the viscous stress into the
momentum equations leads to:

 v     vv   p   2 v  b
t
2

    is the Laplacian operator which in a
where
Cartesian coordinate system reads
2
2
2



2      2  2  2
x y
z
26
MECH 221 – Chapter 5
5.3 Conservation of Momentum
(Navier-Stokes Equations)

For inviscid flow where   0 , the above
equation reduces to the Euler’s equation
given in Chapter 3 where the body force is
also taken the form due to gravity.
27