Momentum Balance
Steven A. Jones
BIEN 501/CMEN 513
Friday, March 17, 2006
Louisiana Tech University
Ruston, LA 71272
Momentum Balance
Learning Objectives:
1. State the motivation for curvilinear coordinates.
2. State the meanings of terms in the Transport Theorem
3. Differentiate between momentum as a property to be transported
and velocity as the transporting agent.
4. Show the relationship between the total time derivative in the
Transport Theorem and Newton’s second law.
5. Apply the Transport Theorem to a simple case (Poiseuille flow).
6. Identify the types of forces in fluid mechanics.
7. Explain the need for a shear stress model in fluid mechanics.
The Stress Tensor.
Appendix A.5
Show components of the stress tensor in Cartesian and cylindrical
coordinates. Vectors and Geometry
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Motivation for Curvilinear Coordinates
Fully developed pipe flow (Poiseuille)

r2 
u z  umax 1  2 
R 

Application: What is the
flow stress on an endothelial
cell?
Flow around a small particle (Stokes Flow)
Applications:
How fast does a blood cell settle?
What is the motion of a catalyzing particle?
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Cylindrical Coordinates: Examples
Fully developed pipe flow (Poiseuille)
Cylindrical coordinates are
simpler because of the
boundary conditions:

u  0 at r  R
r

In cartesian coordinates, there are three velocity components
to worry about.
In spherical coordinates, one of these components is zero (u).

 3R 1  R  3 
ur  v 1 
    cos 
 2r 2  r  
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3

3R 1  R  
u  v  1 
    sin 
4
r
4  r  

Stokes (Creeping Flow)
In cartesian coordinates, there are three velocity components to worry about.
To confirm the three components, consider the point (x, y, z) = (1, 1, 1).
Slice parallel to the equator (say the equator is in the xz plane):
Top View
z
x
y
z
x
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This velocity vector has an x and z
component (visible above) and a y
component (visible to the left).
Momentum Balance
Consider flow entering a
control volume:
The rate at which momentum is generated in a chunk of
fluid that is entering the control volume is governed by
the Reynolds Transport Theorem
d

dV  
dV   v  ndA

R
R
S m 
 m  t
dt  m 
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Momentum Balance
Consider flow entering a
control volume:
The property  in this case is momentum per unit
volume,  = v. Both  and v are bold (vectors).
d
dt



Rm  dV  Rm  t dV  Sm  v  ndA
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
Momentum Balance
It is useful to recall the meanings of the terms.
d
dt
 v
Rm   v dV  Rm  t dV  Sm   v  v  ndA
Rate at which the
momentum of the fluid
passing through the
sample volume
increases (production of
momentum).
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Rate at which
the momentum
increases inside
the sample
volume (partial
derivative)
Flux of
momentum
through the
surface of the
control volume.
Say What?
The momentum of the car passing through the
location of measurement is increasing.
The momentum at the location of measurement is
not increasing.
Rate at which the
momentum of the fluid
passing through the
sample volume
increases (production of
momentum).
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40 mph
25 mph
Location of
Measurement
Momentum Balance & Newton
The momentum balance is a statement of
Newton’s second law.
d
dt
 v
Rm   v dV  Rm  t dV  Sm   v  v  ndA
Production of Momentum
(Force per unit volume).
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Eulerian form of the time
derivative of momentum
(i.e. ma per unit volume).
Momentum Balance & Newton
Eulerian time derivative
d
dt
 v
Rm   v dV  Rm  t dV  Sm   v  v  ndA
Lagrangian time
derivative
d
dt

R m 
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Eulerian form of the time
derivative of momentum
(i.e. ma per unit volume).
 v dV   tdA  
S m 
R m 
fdV
Momentum Balance
It is also useful to note that this is three equations,
one for each velocity component.
d
dt
 v
Rm   v dV  Rm  t dV  Sm   v  v  ndA
For example, the v1 component of this equation is:
d
dt
 v1
Rm   v1 dV  Rm  t dV  Sm   v1  v  ndA
But note that the full vector v remains in the last integral.
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Momentum Surface Flux
The roles of the velocity components differ, depending
on which surface is under consideration.
  v  v  ndA
S m 
1
The momentum being
carried through the surface.
v1
v2
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The velocity vector that
carries momentum
through the surface.
In the figure to the left: The velocity
component perpendicular to the
plane (v2) carries momentum (v1)
through the plane.
Momentum Shell Balance
Fully developed pipe flow (Poiseuille)
dr
vr
Assumptions:
dz
1. Steady, incompressible flow (no changes with time)
2. Fully developed flow
3. Velocity is a function of r only (v=v(r))
4. No radial or circumferential velocity components.
5. Pressure changes linearly with z and is independent of r.
Note: 3, 4 and 5 follow from 1 and 2, but it takes a while to demonstrate the
connection.
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The Control Volume
The control volume is an annular region dz long
and dr thick. We will be concerned with 4
surfaces:
r 
rr
rz
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Outer Cylinder
The Control Volume
Inner Cylinder
r  
rr
rz
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The Control Volume
Left Annulus
z 
zr
zz
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The Control Volume
Right Annulus
z 
zr
zz
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Continuity
The mass entering the annular region = the mass exiting.
dr
dz
Thus:
2r dr vz (r, z)  2r dr vz r, z  dz 
This equation is automatically satisfied by assumption
3 (velocity does not depend on z).
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Momentum in Poiseuille Flow
The momentum entering the annular region - the
momentum leaving= momentum destruction. (Newton’s
2nd law – F=ma)
dr
dz
In fluid mechanics, we talk about momentum per unit volume
and force per unit volume.
D v 
F
Dt
For example, the force per unit volume caused by
gravity is g since F=mg. (Units are g cm/s2).
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Momentum
Rate of momentum flow into the annulus is:
uz z  uz z  r dr
Rate of momentum flow out is:
uz z  dz  uz z  dz  r dr
Again, because velocity does not change with z, these
two terms cancel one another.
dr
dz
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Shearing Force
Denote the shearing force at the cylindrical surface at r as (r).
The combined shearing force on the outer and inner cylinders
is:
2 r  dr  dz  rz r  dr   2r dz  rz r 
Note the signs of the two terms above.

r rz   0
r
dr
dz
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Pressure Force
The only force remaining is that cause by pressure on the two
surfaces at r and r+dr.
F  2r dr pz   2r dr pz  dz 
This force must balance the shearing force:
dr
dz
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Force Balance
2 r  dr  dz  rz r  dr   2r dz  rz r  
2r dr pz   2r dr pz  dz 
Divide by 2 dr dz:
r  dr   rz r  dr   r  rz r  
dr
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r p  z   r p  z  dz 
dz
Force Balance
From the previous slide:
r  dr   rz r  dr   r  rz r  
dr
r p z   r p z  dz 
dz
Take the limit as dr, dz  0.
 r rz 
p
r
r
z
Now we need a model that describes the relationship
between the shear rate and the stress.
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Shear Stress Model
The Newtonian model relating stress and strain rate is:
 u r u z 
 rz   


r 
 z
u r
0
In our case, u r  0 
z
u z
Thus,  rz  
r
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Differential Equation
So, with:
 r rz 
p
r
r
z
u z
and  rz  
r
  u z
The equation is:
 r
r 
r
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p

r
z

Differential Equation
If viscosity is constant,
  u z 
p
 r
r
r  r 
z
Since the pressure gradient is constant (assumption 4),
we can integrate once:
2
 u z  r p
 C1
r

 r  2 z
u z r 
 0 , so C1= 0.
By symmetry,
r r 0
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Differential Equation
If viscosity is constant,
  u z 
p
 r
r
r  r 
z
Since the pressure gradient is constant (assumption 4),
we can integrate once:
2
 u z  r p
 C1
r

 r  2 z
u z r 
 0 , so C1= 0.
By symmetry,
r r 0
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Differential Equation
With
2
u z
r p
 u z  r p
r






r
r 2 z

 2 z
Integrate again.
r 2 p
uz 
 C2
4 z
The no-slip condition at r=R is uz=0, so
R 2 p
R 2 p 
r2 
1  2 
C2  
 uz 
4 z
4 z 
R 
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Review, Poiseuille Flow
1. Use a shell balance to relate velocity to the
forces.
2. Use a model for stress to write it in terms of
velocity gradients.
3. Integrate
4. Use symmetry and no-slip conditions to
evaluate the constants of integration.
If you have had any course in fluid mechanics
before, you have almost certainly used this
procedure already.
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Moment of Momentum Balance
It is useful to recall the meanings of the terms.
d
dt

R m 
 p  v 
dV    p  v  v  ndA
R m 
S m 
t
 p  v  dV  
Rate at which the
moment of momentum of
the fluid passing
through the sample
volume increases
(production of
momentum).
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Rate at which
the moment of
momentum
increases inside
the sample
volume (partial
derivative)
Flux of moment
of momentum
through the
surface of the
control volume.
Types of Forces
1. External Forces
(gravity, electrostatic)
F    f e dV
2. Mutual forces (arise
from within the body)
F    f m dV
a. Intermolecular
b. electrostatic
3. Interfacial Forces (act
on surfaces)
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Rp
Rp
F   tdA
Sp
Types of Forces
1. Body Forces (Three Dimensional)
•
•
Gravity
Magnetism
2. Surface Forces (Two Dimensional)
a. Pressure x Area – normal to a surface
b. Shear stresses x Area – Tangential to the
surface
3. Interfacial Forces (One Dimensional)
e.g. surface tension x length)
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Types of Forces
4. Tension (Zero Dimensional)
The tension in a guitar string.
OK, really this is 2 dimensional, but it is
treated as zero-dimensional in the
equations for the vibrating string.
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The Stress Tensor
 12
 11  12  13 


τ   ijei e j   21  22  23 
i
j




 31 32 33 
The first subscript is the face on
which the stress is imposed.
The second subscript is the direction
in which the stress is imposed.
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 11
 13
The Stress Tensor
 12
  1  12  13 


τ   ij e i e j   21  2  23 
i
j




31
32
3


The diagonal terms (normal stresses)
are often denoted by i.
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1
 13
Exercise
For a general case, what is the momentum balance
in the -direction on the differential element shown
(in cylindrical coordinates)?
dz
d
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dr
Look at the r term
  r  r  dr   r  dr  d  dz    r  r   rd  dz
Divide by dr, d, dz

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  r r 
r
Contact Forces
Diagonal elements are often denoted as 
P
B-P
n
t(z,P)
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Stress Principle:
Regardless of how we
define P, we can find
t(z,n)
Contact Forces
Diagonal elements are often denoted as 
n
P
t(z,P)
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B-P
Stress Principle:
Regardless of how we
define P, we can find
t(z,n)
Cauchy’s Lemma
Stress exerted by B-P on P is equal and opposite to the
force exerted by P on B-P.
t(z, n)
P
n
BP
P
n
BP
t(z,n)
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Finding t(z,n)
If we know t(z,n) for some surface normal n, how
does it change as the orientation of the surface
changes?
P
n
t(z,n)
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BP
The Tetrahedron
We can find the dependence of t on n from a
momentum balance on the tetrahedron below.
Assume that we know the surface forces on the
sides parallel to the cartesian basis vectors. We
can then solve for the stress on the fourth
surface.
z2
A3
A1
n
z3
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z1
A2
The Stresses on Ai
Must distinguish between the normals to the
surfaces Ai and the directions of the stresses.
In this derivation, stresses on each surface
can point in arbitrary directions. ti is a vector,
not a component.
z
2
A3
A1
n
z1
t1
z3
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A2
Components of ti
Recall the stress tensor:
 11  12  13 


τ   21  22  23 




 31 32 33 
Surface (row)
z2
A3
A1
Direction of
Force (Column)
n
z1
t1
z3
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A2
Total Derivative
Let t1, t2, and t3 be the stresses on the three Ai
z2
A3
d
dt

R m 
 v dV   tdA  
S m 
R m 
fdV
A1
n
z1
z3
d m  v hA
d
 v dV  

R
dt  m 
dt 3
A2
Value is
constant if
region is small.
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Volume of the
tetrahedron.
Body Forces
Similarly,
z2
A3
d
dt

R m 
 v dV   tdA  
S m 
R m 
fdV
A1
n
z1
z3
d
dt
hA
Rm   f dV  f 3
A2
Value is
constant if
region is small.
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Volume of the
tetrahedron.
Surface Forces
z2
A3
d
dt

 v dV   tdA  
R m 
S m 
R m 
fdV
A1
n
z1

S m 
z3
t dA  At  A1t 2  A2 t 2  A3t 3
A2
Area of the
surface.
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t does not vary for
differential volume.
Find Ai
Each of the Ai is the projection of A on the coordinate
plane. Note that A projected on the z1z2 plane is just
An  e 3  An1 (i.e. dot n with the  coordinate).
z2
At  A1t 2  A2 t 2  A3 t 3 
A3
At  n1t1  n2 t 2  n3 t 3 
A1
n
z1
z3
A2
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Infinitessimal Momentum Balance
Thus:
 d m  v
 hA
 
 f 
 At  n1t1  n2 t 2  n3t 3 
dt

 3
z2
A3
Divide by A
A1
n
z1
z3
A2
 d m  v
h
 
 f   t  n1t1  n2 t 2  n3 t 3 
dt

3
Take the limit as the tetrahedron
becomes infinitessimally small (h 0)
t  n1t1  n2 t 2  n3t 3
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Meaning of This Relationship
If we examine the stress at a point, and we wish to
determine how it changes with the direction of the
chosen normal vector (i.e. with the orientation of the
surface of the body), we find that:
z2
t  n1t1  n2 t 2  n3t 3
A3
A1
n
z1
z3
A2
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Where t1 , t 2 , and t 3
are the stresses on the
surfaces perpendicular to
the coordinate directions.
Mohr’s Circle
Those students familiar with solid mechanics
will recall the Mohr’s Circle, which is a
statement of the previous relationship for
2-dimensions in solids.
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Symmetry
The stress tensor is symmetric. I.e. ij=ji
  1  12  13 


τ   21  2  23 




3
 31 32
 12
1
 13
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Kroneker Delta
The Kroneker delta is defined as:
0 if i  j
 ij  
1 if i  j
It can be thought of as a compact notation for the identity matrix:
 1 0 0


 ij  I   0 1 0 
 0 0 1


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Permutation Tensor
The permutation tensor is defined as:
 ijk
  1 if ijk is a positive rotation

  1 if ijk is a negative rotation
 0 if i  j, or j  k , or i  k

It is a sparse tensor, so the only components (of 27 possible)
that are not zero are:
1
1
 123   231   312  1
 321   213   132  1
3
Positive
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Ruston, LA 71272
2
2
3
Negative
Permutation Tensor and Delta
A well known result is:
 mjk njk  2 mn
(note sums over j and k)
This expression is a 2nd order tensor, each component of which
is the sum of 9 terms. For example, with m=1 and n=2.
 1 jk 2 jk   111 211   112 212   113 213
  121 221   122 222   123 223
  131 231   132 232   133 233
k’s are the same
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j’s are the same
Permutation Tensor and Delta
Consider the mn component of
 mjk njk
If m=n=1, then there are only 2 possibilities for j and k that do
not lead to zero values of the permutation tensor. They can
be 2 and 3, for if either is 1, then the value is zero.
 1 jk 1 jk   123 123   132 132  11   1 1  2
The same result occurs for m=n=2 and m=n=3. I.e. if m=n,
then the value us 2.
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Ruston, LA 71272
Permutation Tensor and Delta
If mn, then the first  is nonzero only if its j and k indices are
not m. But in that case, since n must be one of these other
two values and the second  must be zero. I.e., the
expression is zero when mn. The two results combine as
follows:
 1 jk 1 jk
2 if m  n

 2 mn
0 if m  n
This expression is valuable because it allows us to relate
something that looks complicated in terms of something that
is more readily understandable.
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Ruston, LA 71272
Permutation Tensor and Delta
Consider another expression:
 ijk mnk
This expression is frightening because it is a 4th order tensor.
It has 81 components, each of which is made of 3 terms.
Yet, all terms for which i=j or m=n or will be zero. Let i=1,
then j=2, k=3 or j=3, k=2 give nonzero results. If k=3, then
there are only two nonzero values of m and n. Overall, the
“important” values of the subscripts are:
i j k mnk
123 123, 132 132, 123 132, 132 123 for i  1
231 231, 213 213, 231 213, 213 231 for i  2
312 312, 321 321, 312 321, 321 312 for i  3
Gives +1
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Gives 1
Permutation Tensor and Delta
Consider another expression:
 ijk mnk
i j k mnk
123 123, 132 132, 123 132, 132 123 for i  1
231 231, 213 213, 231 213, 213 231 for i  2
312 312, 321 321, 312 321, 321 312 for i  3
Gives +1
Gives 1
Only 6 terms are non-zero (those for which ij and
either i=m, j=n or i=n and j=m. Two of these are:
 12k  12k   123 123  1
 k  21k   123 213  1
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Ruston, LA 71272
Permutation Tensor and Delta
Thus,
 ijk mnk
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Ruston, LA 71272
 0 for i  j

  1 for i  j, i  m, j  n
 1 for i  j, i  n, j  m

Permutation Tensor and Delta
It can be shown, through similar enumeration, that the delta
expression:
 im jn   im jn
Gives the same results and that therefore:
 ijk mnk   im jn   im jn
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Ruston, LA 71272