Louisiana Tech University

1
Momentum Balance

• Steven A. Jones
• BIEN 501/CMEN 513
• Friday, March 17, 2006

2
Momentum Balance

• Learning Objectives
• State the motivation for curvilinear coordinates.
• State the meanings of terms in the Transport
  Theorem
• Differentiate between momentum as a property to
  be transported and velocity as the transporting
  agent.
• Show the relationship between the total time
  derivative in the Transport Theorem and Newtons
  second law.
• Apply the Transport Theorem to a simple case
  (Poiseuille flow).
• Identify the types of forces in fluid mechanics.
• 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

3
Motivation for Curvilinear Coordinates
Fully developed pipe flow (Poiseuille)
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?
4
Cylindrical Coordinates Examples
Fully developed pipe flow (Poiseuille)
Cylindrical coordinates are simpler because of
the boundary conditions
In cartesian coordinates, there are three
velocity components to worry about. In spherical
coordinates, one of these components is zero (u?).
r
?
?
5
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
z
x
x
y
This velocity vector has an x and z component
(visible above) and a y component (visible to the
left).
z
x
x
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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
7
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).
8
Momentum Balance

• It is useful to recall the meanings of the terms.

Rate at which the momentum increases inside the
sample volume (partial derivative)
Flux of momentum through the surface of the
control volume.
Rate at which the momentum of the fluid passing
through the sample volume increases (production
of momentum).
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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).
40 mph
Location of Measurement
25 mph
10
Momentum Balance Newton

• The momentum balance is a statement of Newtons
  second law.

Production of Momentum (Force per unit volume).
Eulerian form of the time derivative of momentum
(i.e. ma per unit volume).
11
Momentum Balance Newton
Eulerian time derivative
Eulerian form of the time derivative of momentum
(i.e. ma per unit volume).
Lagrangian time derivative
12
Momentum Balance

• It is also useful to note that this is three
  equations, one for each velocity component.

For example, the v1 component of this equation is
But note that the full vector v remains in the
last integral.
13
Momentum Surface Flux

• The roles of the velocity components differ,
  depending on which surface is under consideration.

The velocity vector that carries momentum through
the surface.
The momentum being carried through the surface.
In the figure to the left The velocity
component perpendicular to the plane (v2) carries
momentum (?v1) through the plane.
v2
?v1
14
Momentum Shell Balance
Fully developed pipe flow (Poiseuille)
dr
vr
dz

• Assumptions
• Steady, incompressible flow (no changes with
  time)
• Fully developed flow
• Velocity is a function of r only (vv(r))
• No radial or circumferential velocity components.
• 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.

15
The Control Volume
The control volume is an annular region dz long
and dr thick. We will be concerned with 4
surfaces
?r?
?rr
Outer Cylinder
?rz
16
The Control Volume
Inner Cylinder
17
The Control Volume
Left Annulus
18
The Control Volume
Right Annulus
19
Continuity
The mass entering the annular region the mass
exiting.
dr
dz
Thus
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.
(Newtons 2nd law Fma)
dr
dz
In fluid mechanics, we talk about momentum per
unit volume and force per unit volume.
For example, the force per unit volume caused by
gravity is ?g since Fmg. (Units are g cm/s2).
21
Momentum
Rate of momentum flow into the annulus is
Rate of momentum flow out is
Again, because velocity does not change with z,
these two terms cancel one another.
dr
dz
22
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
Note the signs of the two terms above.
dr
dz
23
Pressure Force
The only force remaining is that cause by
pressure on the two surfaces at r and rdr.
This force must balance the shearing force
dr
dz
24
Force Balance
Divide by 2? dr dz
25
Force Balance
From the previous slide
Take the limit as
Now we need a model that describes the
relationship between the shear rate and the
stress.
26
Shear Stress Model
The Newtonian model relating stress and strain
rate is
In our case,
Thus,
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Differential Equation
So, with
and
The equation is
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Differential Equation
If viscosity is constant,
Since the pressure gradient is constant
(assumption 4), we can integrate once
, so C1 0.
By symmetry,
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Differential Equation
If viscosity is constant,
Since the pressure gradient is constant
(assumption 4), we can integrate once
, so C1 0.
By symmetry,
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Differential Equation
With
Integrate again.
The no-slip condition at rR is uz0, so
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Review, Poiseuille Flow

• Use a shell balance to relate velocity to the
  forces.
• Use a model for stress to write it in terms of
  velocity gradients.
• Integrate
• 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.

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.
Rate at which the moment of momentum of the fluid
passing through the sample volume increases
(production of momentum).
33
Types of Forces

• External Forces (gravity, electrostatic)
• 2. Mutual forces (arise from within the body)
• Intermolecular
• electrostatic
• 3. Interfacial Forces (act on surfaces)

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Types of Forces

• 1. Body Forces (Three Dimensional)
• Gravity
• Magnetism
• 2. Surface Forces (Two Dimensional)
• Pressure x Area normal to a surface
• Shear stresses x Area Tangential to the surface
• 3. Interfacial Forces (One Dimensional)
• e.g. surface tension x length)

35
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.

36
The Stress Tensor
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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The Stress Tensor
The diagonal terms (normal stresses) are often
denoted by ?i.
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Exercise

• For a general case, what is the momentum balance
  in the q-direction on the differential element
  shown (in cylindrical coordinates)?

dz
dr
dq
39
Look at the qr term
Divide by dr, dq, dz
40
Contact Forces
Diagonal elements are often denoted as ?
Stress Principle Regardless of how we define P,
we can find t(z,n)
P
B-P
n
t(z,P)
41
Contact Forces
Diagonal elements are often denoted as ?
Stress Principle Regardless of how we define P,
we can find t(z,n)
n
P
B-P
t(z,P)
42
Cauchys 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
B?P
t(z,n)
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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
z1
A2
z3
45
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.

z2
A3
A1
n
z1
t1
A2
z3
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Components of ti

• Recall the stress tensor

Surface (row)
z2
A3
A1
Direction of Force (Column)
n
z1
t1
A2
z3
47
Total Derivative

• Let t1, t2, and t3 be the stresses on the three Ai

z2
A3
A1
n
z1
A2
z3
Value is constant if region is small.
Volume of the tetrahedron.
48
Body Forces

• Similarly,

z2
A3
A1
n
z1
A2
z3
Value is constant if region is small.
Volume of the tetrahedron.
49
Surface Forces
z2
A3
A1
n
z1
A2
z3
Area of the surface.
t does not vary for differential volume.
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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

(i.e. dot n with the ? coordinate).
z2
A3
A1
n
z1
A2
z3
51
Infinitessimal Momentum Balance
Thus
z2
A3
Divide by A
A1
n
z1
Take the limit as the tetrahedron becomes
infinitessimally small (h ?0)

A2
z3
52
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
A3
A1
Where
n
are the stresses on the surfaces perpendicular to
the coordinate directions.
z1
A2
z3
53
Mohrs Circle

• Those students familiar with solid mechanics will
  recall the Mohrs 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
55
Kroneker Delta
The Kroneker delta is defined as
It can be thought of as a compact notation for
the identity matrix
56
Permutation Tensor
The permutation tensor is defined as
It is a sparse tensor, so the only components (of
27 possible) that are not zero are
1
1
3
2
2
3
Negative
Positive
57
Permutation Tensor and Delta
A well known result is
(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 m1 and n2.
js are the same
ks are the same
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Permutation Tensor and Delta
Consider the mn component of
If mn1, 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.
The same result occurs for mn2 and mn3. I.e.
if mn, then the value us 2.
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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
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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Permutation Tensor and Delta
Consider another expression
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 ij or mn or will be zero. Let i1, then
j2, k3 or j3, k2 give nonzero results. If
k3, then there are only two nonzero values of m
and n. Overall, the important values of the
subscripts are
Gives ?1
Gives 1
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Permutation Tensor and Delta
Consider another expression
Gives ?1
Gives 1
Only 6 terms are non-zero (those for which i?j
and either im, jn or in and jm. Two of these
are
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Permutation Tensor and Delta
Thus,
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Permutation Tensor and Delta
It can be shown, through similar enumeration,
that the delta expression
Gives the same results and that therefore