1
2000, W. E. Haisler
Chapter 14 - Fluid Mechanics
Fluid mechanics problems, like solid mechanics problems,
must satisfy all the conservation principles. The major
differences are that in fluid mechanics problems:
1. The continuum (the fluid) generally has a velocity.
2. The constitutive relations relate stress to velocity
gradients [instead of stress to displacement gradients (strain)
as in a solid].
3. The final solution of the problem (after combining all
applicable principles) is in terms of velocities [instead of
displacements as in a solid].
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2000, W. E. Haisler
Conservation of Mass


  
(v )
t
Conservation of Linear Momentum

  



[v  v v ] g  T
t
  
S  T  PI
where

P  1 (Txx  Tyy  Tzz) 1/ 3tr(T )
3
  
  

[v  v v ] g  S P
t
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2000, W. E. Haisler
Incompressible Fluids (=constant in time and space). Conservation
of mass (continuity equation) becomes
 
 v  0 Continuity equation for =constant
Constitutive equations for Newtonian Fluids

 
The strain tensor has been defined as   1 [(u )  (u )T ] . We
2
can define the rate of deformation tensor by taking the time derivative
of the strain tensor to obtain

 
D  1 [(v )  (v )T ] =
2

v

x



x


u
 1 v
y
x
 (


 2 y
x


v
u
1
x
 z
 (

 2 z
x
)
1 v y v x
(

)
2 x
y
v
y
y
)
1 v y uz
(

)
2 z
y

1 v z v x 
(

) 
2 x
z 

1 v z v y 
(

)

2 y
z 

v

z

z


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2000, W. E. Haisler
Fluids are generally isotropic and we can observe
experimentally the following linear relation between stress
and velocity gradients <-- defined to be a Newtonian
fluid.

 

S  tr(D)I  2D
<-- defined to be a Newtonian fluid
where  and  are 2 experimentally observed material
constants (similar to E and  in isotropic solids).
The above can also be written as
  

 

T  S  PI   PI  tr(D)I  2D
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2000, W. E. Haisler
Meaning of  and 
Consider the case where all velocity gradients are zero
except that in the x-y plane (as might occur with steady flow
between two parallel plates with vx  vx (y) only).
y
fluid flow
x
dv x
1
where vx  vx ( y)
Dxy 
2 dy
Dxx  Dyy  Dzz  Dxz  Dyz  0 (flow only in x direction)
Substituting above into the constitutive equation gives
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2000, W. E. Haisler
Txx  T yy  Tzz   P
Txz  T yz  0
,
, Txy  
dv x
dy
The material constant  relates the shearing stress to the
velocity gradient in the y direction.  is called the
coefficient of viscosity.
Txy

units of 
MPa sec
psi sec
dvx/dy
2000, W. E. Haisler
7
Stokes proposed a relation between  and  based on the
following argument (and physical observation). First take
the sum of the diagonal terms of the stress tensor [S]

 

 

tr(S )  tr(tr(D)I  2D)  tr(D)tr( I )  2tr(D)
note 



tr(S )  Sxx  Syy  Szz ; tr( I )  3 ; tr(D)  Dxx  Dyy  Dzz
Thus
Sxx  Syy  Szz  3(Dxx  Dyy  Dzz )  2(Dxx  Dyy  Dzz )
or
(1/ 3)(Sxx  Syy  Szz )  0 (  2 / 3)(Dxx  Dyy  Dzz )

tr(S )  Sxx  Syy  Szz  0 and thus (  2 / 3)  0. (   /  )
is called the coefficient of bulk viscosity. Thus   2 / 3 .
2000, W. E. Haisler
Substitute Stokes’ assumption back into the constitutive
relation (stress-velocity gradient relation) to obtain


 
S  2[D  (1/ 3)tr(D)I ]
 

But tr(D)  vx / x v y / y vz / z  v  0
from


 
S  2D  [(v )  (v )T ]
conservation of mass. Thus
 
Note, we need S in linear momentum equation:


  
    T 

S  (2 )D  2 ( 1 [(v )  (v ) ])
2









2
2
 [ v ( v )]   v
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2000, W. E. Haisler
9
 
 
Note: ( v )  0 from incompressibility. Substitute for S
into Linear Momentum equation to obtain the
Navier-Stokes Equation (constant density and viscosity):
  
  

[v  v v ]  g  2v P
t
For body forces which are conservative, it convenient to rewrite the body force term in terms of a potential function
(means the force is the gradient of a potential function). The

body force due to gravity can be written as g  gh and
h is the distance above a reference in the direction of the
gravitational vector. If we assume g is a constant,
then



g  gh (gh). Thus the two terms g P can be
2000, W. E. Haisler

combined and written (gh)P  (P  gh) 
where   P  gh is called the modified pressure potential
function. With this the Navier-Stokes equation becomes
  
  

[v  v v ]  g  2v P
t
or
  
  
[v  v v ]  2v 
t
10
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2000, W. E. Haisler
Summary - Navier-Stokes Equations for Newtonian Fluid
Constant Mass (), Constant Viscosity ()
 
Continuity: ( v )  0

 
Deformation Gradient: D  1 [(v )  (v )T ]
2


 
Constitutive (Stress-Velocity): S  2D  [(v )  (v )T ]
Linear Momentum:
  



[v  v v ]  g  2v P
t
  


or [v  v v ]  2v  where   P  gh
t
12
2000, W. E. Haisler
Cylindrical (r, , z) coordinates.
y
v

v
r
v
z
r

x
Recall
 
 1  

 er  e r  ez
r  
z
2
2
2
 
2     r1   1   
2 r 2 2
2
r
r  z
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2000, W. E. Haisler
1. Steady-State, Laminar Flow in a Tube of radius R.
axial flow
z
r
 
Continuity: ( v )  0
Assume no radial flow and that flow is axisymmetric
(velocity is independent of ), ie, vr  v  0 . Continuity

vz
 0 . Thus vz  vz (r ) C .
reduces to
z
Linear Momentum:
  



[v  v v ]  g  2v P
t
2000, W. E. Haisler



v
 0. Neglect body forces: g  0.
For steady-state:
t

In cylindrical coordinates, components of v [vr ,v ,vz].

14
From continuity, vr  v  0. Components of gradient


 



operator are  er   e r1   ez  . Thus [v v ] becomes
r  
z
in matrix notation





v


  
z


0


0

 r 





r









1

 vr v vz  r vr v vz    0 0 vz 0 0 0   0 0 0








  




0
0
0


  






 z 


The two remaining terms on the right side of L.M. become:
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2000, W. E. Haisler


































P 
0
r 


0   2v P  
0
 r1 P 
 
2
P 
d v
dv
z1 z
z 
2
dr
r dr











Thus we have 3 equations:
P  0
r
1P  0
r 











2
--> Implies that P=P(z)











d v z 1 dv z dP

r

dr
dz
2
dr
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2000, W. E. Haisler
To solve the last equation for vz(r), we assume that dP/dz is
a given value (and not a function of r). Integrating the above
differential equation twice gives
dv z 1 dP r C1

r

dr
dz 2
2
and
vz  1 dP r  C lnr  C
 dz 4 1
2
C1 must be zero since at r=0, the velocity gradient must be
finite (not infinity). Assume BC that axial flow velocity is
2
zero at the outer wall: vz(r=R) = 0. Thus C   1 dP R .
2  dz 4




2
2
 r
1
dP
R
1 .
The axial velocity solution is thus vz(r) 

dz 4  R 2 
The velocity distribution is parabolic and the maximum
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2000, W. E. Haisler
2
  dP R . The shear stress
dz 4
max
distribution (from constitutive equation) is given by
dv z
Srz  
  (dP r )  r dP . The shear stress varies
dr
dz 2 2 dz
linearly with the radius, is zero at r=0, and is a maximum at
the wall where r=R. The volumetric flow rate is given by
value occurs as r=0: vz


R
R


2
2
4
 r

dP
R

R
dP
Q   v dA   v 2rdr  

1
2

rdr



z
z
dz 8
dz 4  R2 
A
0
0
Note: Both flow rate Q and velocity vz are proportional to
axial pressure gradient, dP/dz.
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2000, W. E. Haisler
2. Plane Couette Flow Between Two Parallel Plates
y
d
z
 
Continuity: ( v )  0
Assume fluid flow is only in the z direction, ie, vx  vy 0 .
vz
 0 . Thus vz  vz ( y) C .
Continuity reduces to
z
Linear Momentum:
  



[v  v v ]  g  2v P
t
2000, W. E. Haisler
19
vx
vx
vx
vx
 2vx  2vx  2vx P
(
 vx
 vy
 vz
)  gx  (


)
2
2
2
x
t
x
y
z
x
y
z
v y
v y
v y
v y
 2v y  2v y  2v y P
(
 vx
 vy
 vz
)  g y  (


)
t
x
y
z
x 2 y 2 z 2 y
vz
vz
vz
vz
 2vz  2vz  2vz P
(
 vx
 vy
 vz
)  gz  (


)
t
x
y
z
x 2 y 2 z 2 z



v
 0. Neglect body forces: g  0.
For steady-state:
t
Linear Momentum reduces to
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2000, W. E. Haisler
0   P
x
0   P
y
d 2vz dP
0 

dy 2 dz
The first two equations imply that P=P(z). Solving the third equation
for vz(y) and taking dP/dz as a given (and not a function of y), then
2
dP ) d
vz  ( 1 dP ) y  C y  C
B.C.:
v
z(0)= vz(d)=0 --> C  ( 1
 dz 2 1
 dz 2
2
1

2  2





y
y
dP
1
d





Thus vz ( y)   ( )       which is parabolic in y. The
dz 2  d   d  


2
max velocity occurs at the center: vz
 vz (d / 2)   d ( dP ).
8 dz
max
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2000, W. E. Haisler
3
The flow rate (vol/time) is Q   vz dA  0dvzWdy  Wd (dP )
12 dz
A
where W is the width in the x direction of the plates. The
dv z dP
 ( )( y  d / 2) .
shear stress is Syz  
dy
dz
It should be kept in mind that the definition of P is given
by P = -(1/3) tr[T] = -(1/3) (Txx+Tyy+Tzz) = - average
hydrostatic stress. Thus, a negative dP/dz causes flow in the
positive z direction.
Some units:
force: 1 N = 1 kg m/s2
pressure or stress: 1 Pa = 1 N/m2 = 1 kg/(m s2),
1 psi = 6,894.76 Pa, 1 Mpa = 145.04 psi
viscosity, : 1 Pa s = 1 N s/ m2 = 1 kg/(m s)
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2000, W. E. Haisler
3. Annular Flow between two pipes of radius R2 and R1.
flow
R
1
z
r
solid (no flow)
R
2
flow
The general solution for this problem is identical to the first
example (laminar flow in a tube) except for geometry and
boundary conditions.
2
vz  1 dP r  C lnr  C .
 dz 4 1
2
B.C.: vz(R1)= vz(R2)=0
2000, W. E. Haisler
R2
vz (R )  0 1 dP 1 C ln R C
 dz 4
1
1 1 2
R 2
vz (R )  0 1 dP 2 C ln R C
 dz 4
2
1 2 2
Solving for C1 and C2 gives
R 2 R 2
1 )
1
C   1 dP ( 2
1  dz
4
ln(R / R )
2 1
R2
R 2 R 2
ln R
1 )
1
C   1 dP ( 1 )  1 dP ( 2
 dz
2  dz 4
4
ln(R / R )
2 1
23
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2000, W. E. Haisler
Substituting the constants of integration back into vz gives











2












2


R2


vz   dP ( 1 ) 1  r   ( 1)ln(r / R ) where   R / R
2 1
R 
dz 4
ln
1



1
This is the same as equation (13-80) except that (13-80) is
written in terms of the pressure potential  where
 
dP  L
0 = change in potential  (pressure) from z=0
L
dz
to z=L. Note: ln()=-ln(1/). Note also that as 1 (thin
annulus), the ln term goes to zero and the above solution is
similar to plane couette flow.