Momentum Balance
description
Transcript of Momentum Balance
Louisiana Tech UniversityRuston, LA 71272
Momentum Balance
Steven A. Jones
BIEN 501/CMEN 513
Friday, March 17, 2006
Louisiana Tech UniversityRuston, LA 71272
Momentum Balance
Learning Objectives:1. State the motivation for curvilinear coordinates.2. State the meanings of terms in the Transport Theorem3. 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.5Show components of the stress tensor in Cartesian and cylindrical
coordinates. Vectors and Geometry
Louisiana Tech UniversityRuston, LA 71272
Motivation for Curvilinear Coordinates
Fully developed pipe flow (Poiseuille)
2
2
max 1R
ruuz
Flow around a small particle (Stokes Flow)
Applications:
How fast does a blood cell settle?
What is the motion of a catalyzing particle?
Application: What is the flow stress on an endothelial cell?
Louisiana Tech UniversityRuston, LA 71272
Cylindrical coordinates are simpler because of the boundary conditions:
Cylindrical Coordinates: Examples
Rru at0
cos2
1
2
31
3
r
R
r
Rvur
Fully developed pipe flow (Poiseuille)
In cartesian coordinates, there are three velocity components to worry about.
In spherical coordinates, one of these components is zero (u).
sin4
1
4
31
3
r
R
r
Rvu
r
Louisiana Tech UniversityRuston, LA 71272
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):
This velocity vector has an x and z component (visible above) and a y component (visible to the left).
Top View
x
z
x
z
x
y
x
z
Louisiana Tech UniversityRuston, LA 71272
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
mmm SRR
dAdVt
dVdt
dnv
Louisiana Tech UniversityRuston, LA 71272
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).
mmm SRR
dAdVt
dVdt
dnv
Louisiana Tech UniversityRuston, LA 71272
Momentum Balance
It is useful to recall the meanings of the terms.
mmm SRR
dAdVt
dVdt
dnvv
vv
Rate at which the momentum of the fluid passing through the sample volume increases (production of momentum).
Rate at which the momentum increases inside the sample volume (partial derivative)
Flux of momentum through the surface of the control volume.
Louisiana Tech UniversityRuston, LA 71272
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.
Location of Measurement
25 mph
40 mph
Rate at which the momentum of the fluid passing through the sample volume increases (production of momentum).
Louisiana Tech UniversityRuston, LA 71272
Momentum Balance & Newton
The momentum balance is a statement of Newton’s second law.
mmm SRR
dAdVt
dVdt
dnvv
vv
Production of Momentum (Force per unit volume).
Eulerian form of the time derivative of momentum (i.e. ma per unit volume).
Louisiana Tech UniversityRuston, LA 71272
Momentum Balance & Newton
mmm SRR
dAdVt
dVdt
dnvv
vv
Lagrangian time derivative
Eulerian form of the time derivative of momentum (i.e. ma per unit volume).
Eulerian time derivative
mmm RSRdVdAdV
dt
dftv
Louisiana Tech UniversityRuston, LA 71272
Momentum Balance
It is also useful to note that this is three equations, one for each velocity component.
mmm SRR
dAdVt
dVdt
dnvv
vv
mmm SRR
dAvdVt
vdVv
dt
dnv1
11
For example, the v1 component of this equation is:
But note that the full vector v remains in the last integral.
Louisiana Tech UniversityRuston, LA 71272
Momentum Surface Flux
The roles of the velocity components differ, depending on which surface is under consideration.
mSdAv nv1
The momentum being carried through the surface.
The velocity vector that carries momentum through the surface.
v1v2
In the figure to the left: The velocity component perpendicular to the plane (v2) carries momentum (v1) through the plane.
Louisiana Tech UniversityRuston, LA 71272
Momentum Shell Balance
Fully developed pipe flow (Poiseuille)
dr
dzAssumptions:
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.
vr
Louisiana Tech UniversityRuston, LA 71272
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
Outer Cylinder
Louisiana Tech UniversityRuston, LA 71272
The Control Volume
Inner Cylinderr rr
rz
Louisiana Tech UniversityRuston, LA 71272
The Control Volume
z zr
zz
Left Annulus
Louisiana Tech UniversityRuston, LA 71272
z zr
zz
The Control Volume
Right Annulus
Louisiana Tech UniversityRuston, LA 71272
Continuity
The mass entering the annular region = the mass exiting.
dr
dz
Thus:
dzzrvdrrzrvdrr zz ,2),(2
This equation is automatically satisfied by assumption 3 (velocity does not depend on z).
Louisiana Tech UniversityRuston, LA 71272
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.
FDt
vD
For example, the force per unit volume caused by gravity is g since F=mg. (Units are g cm/s2).
Louisiana Tech UniversityRuston, LA 71272
Momentum
drrzuzu zz
Rate of momentum flow into the annulus is:
Again, because velocity does not change with z, these two terms cancel one another.
drrdzzudzzu zz
Rate of momentum flow out is:
dr
dz
Louisiana Tech UniversityRuston, LA 71272
Shearing Force
0
rzrr
Denote the shearing force at the cylindrical surface at r as (r). The combined shearing force on the outer and inner cylinders is:
dr
dz
Note the signs of the two terms above.
rdzrdrrdzdrr rzrz 22
Louisiana Tech UniversityRuston, LA 71272
Pressure ForceThe only force remaining is that cause by pressure on the two surfaces at r and r+dr.
dr
dz
This force must balance the shearing force:
dzzpdrrzpdrrF 22
Louisiana Tech UniversityRuston, LA 71272
Force Balance
rdzrdrrdzdrr rzrz 22
dzzpdrrzpdrr 22
Divide by 2 dr dz:
dr
rrdrrdrr rzrz r p z r p z dz
dz
Louisiana Tech UniversityRuston, LA 71272
Force Balance
.0, dzdrTake the limit as
dr
rrdrrdrr rzrz dz
dzzprzpr
z
pr
r
r rz
Now we need a model that describes the relationship between the shear rate and the stress.
From the previous slide:
Louisiana Tech UniversityRuston, LA 71272
Shear Stress Model
00
z
uu r
r
The Newtonian model relating stress and strain rate is:
r
u
z
u zrrz
Thus,
In our case,
r
uzrz
Louisiana Tech UniversityRuston, LA 71272
Differential Equation
r
uzrz
So, with:
z
pr
r
ur
rz
The equation is:
and
z
pr
r
r rz
Louisiana Tech UniversityRuston, LA 71272
Differential Equation
0
0
r
z
r
ru
If viscosity is constant,
z
pr
r
ur
rz
By symmetry,
Since the pressure gradient is constant (assumption 4), we can integrate once:
1
2
2C
z
pr
r
ur z
, so C1= 0.
Louisiana Tech UniversityRuston, LA 71272
Differential Equation
0
0
r
z
r
ru
If viscosity is constant,
z
pr
r
ur
rz
By symmetry,
Since the pressure gradient is constant (assumption 4), we can integrate once:
1
2
2C
z
pr
r
ur z
, so C1= 0.
Louisiana Tech UniversityRuston, LA 71272
Differential Equation
2
2
4C
z
pruz
z
pr
r
u
z
pr
r
ur zz
22
2
Integrate again.
The no-slip condition at r=R is uz=0, so
With
2
222
2 144 R
r
z
pRu
z
pRC z
Louisiana Tech UniversityRuston, LA 71272
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. Integrate4. 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.
Louisiana Tech UniversityRuston, LA 71272
Moment of Momentum Balance
It is useful to recall the meanings of the terms.
mmm SRR
dAdVt
dVdt
dnvvp
vpvp
Rate at which the moment of momentum of the fluid passing through the sample volume increases (production of momentum).
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.
Louisiana Tech UniversityRuston, LA 71272
Types of Forces
1. External Forces (gravity, electrostatic)
2. Mutual forces (arise from within the body)
a. Intermolecular
b. electrostatic
3. Interfacial Forces (act on surfaces)
dVFpR m f
dAFpS t
dVFpR e f
Louisiana Tech UniversityRuston, LA 71272
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)
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
The Stress Tensor
333231
232221
131211
ji j
iij eeτ 11
13
12
The first subscript is the face on which the stress is imposed.The second subscript is the direction in which the stress is imposed.
Louisiana Tech UniversityRuston, LA 71272
The Stress Tensor
33231
23221
13121
ji j
iij eeτ 1
13
12
The diagonal terms (normal stresses) are often denoted by i.
Louisiana Tech UniversityRuston, LA 71272
Exercise
For a general case, what is the momentum balance in the -direction on the differential element shown (in cylindrical coordinates)?
drd
dz
Louisiana Tech UniversityRuston, LA 71272
Look at the r term
r rr dr r dr d dz r rd dz
Divide by dr, d, dz
r r
r
Louisiana Tech UniversityRuston, LA 71272
Contact Forces
Diagonal elements are often denoted as
B-PP
t(z,P)
Stress Principle: Regardless of how we define P, we can find t(z,n)
n
Louisiana Tech UniversityRuston, LA 71272
Contact Forces
Diagonal elements are often denoted as
B-P
t(z,P)
Stress Principle: Regardless of how we define P, we can find t(z,n)
nP
Louisiana Tech UniversityRuston, LA 71272
Cauchy’s Lemma
Stress exerted by B-P on P is equal and opposite to the force exerted by P on B-P.
BP
t(z,n)
nP
BP
t(z, n)
Pn
Louisiana Tech UniversityRuston, LA 71272
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?
BP
t(z,n)
nP
Louisiana Tech UniversityRuston, LA 71272
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.
A3
A2
A1
nz1
z3
z2
Louisiana Tech UniversityRuston, LA 71272
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.
A3
A2
A1
nz1
z3
z2
t1
Louisiana Tech UniversityRuston, LA 71272
Components of ti
Recall the stress tensor:
A3
A2
A1
nz1
z3
z2
t1
333231
232221
131211
τSurface (row)
Direction of Force (Column)
Louisiana Tech UniversityRuston, LA 71272
Total Derivative
Let t1, t2, and t3 be the stresses on the three Ai
mmm RSRdVdAdV
dt
dftv
A3
A2
A1
nz1
z3
z2
3
hA
dt
ddV
dt
d m
R m
vv
Value is constant if region is small.
Volume of the tetrahedron.
Louisiana Tech UniversityRuston, LA 71272
Body Forces
Similarly,
mmm RSRdVdAdV
dt
dftv
A3
A2
A1
nz1
z3
z2
3
hAdV
dt
dmR
ff
Value is constant if region is small.
Volume of the tetrahedron.
Louisiana Tech UniversityRuston, LA 71272
Surface Forces
mmm RSRdVdAdV
dt
dftv
A3
A2
A1
nz1
z3
z2
332221 ttttt AAAAdA
mS
Area of the surface.
t does not vary for differential volume.
Louisiana Tech UniversityRuston, LA 71272
Find Ai
332211
332221
ttt
tttt
nnntA
AAAA
A3
A2
A1
nz1
z3
z2
Each of the Ai is the projection of A on the coordinate plane. Note that A projected on the z1z2 plane is just 13 AnA en (i.e. dot n with the coordinate).
Louisiana Tech UniversityRuston, LA 71272
Infinitessimal Momentum Balance
332211 tttt nnn
3322113ttttf
vnnnA
hA
dt
d m
A3
A2
A1
nz1
z3
z2
Thus:
Divide by A
3322113ttttf
vnnn
h
dt
d m
Take the limit as the tetrahedron becomes infinitessimally small (h 0)
Louisiana Tech UniversityRuston, LA 71272
Meaning of This Relationship
332211 tttt nnn A3
A2
A1
nz1
z3
z2
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:
Where
321 ,, ttt andare the stresses on the surfaces perpendicular to the coordinate directions.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Symmetry
33231
23221
13121
τ
1
13
12
The stress tensor is symmetric. I.e. ij=ji
Louisiana Tech UniversityRuston, LA 71272
Kroneker Delta
100
010
001
Iij
ji
jiij if
if
1
0
The Kroneker delta is defined as:
It can be thought of as a compact notation for the identity matrix:
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor
1
1
132213321
312231123
kikjji
ijk
ijk
ijk
ororif
rotationnegativeaisif
rotationpositiveaisif
,,0
1
1
The permutation tensor is defined as:
It is a sparse tensor, so the only components (of 27 possible) that are not zero are:
1
23
1
32
Positive Negative
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor and Delta
233133232132231131
223123222122221121
21311321211221111121
jkjk
mnnjkmjk 2A well known result is:
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.
(note sums over j and k)
j’s are the same
k’s are the same
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor and Delta
2111113213212312311 jkjk
njkmjkConsider the mn component of
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.
The same result occurs for m=n=2 and m=n=3. I.e. if m=n, then the value us 2.
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor and Delta
mnjkjk nm
nm 2
0
211
if
if
If mn, 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 mn. 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.
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor and Delta
3312321,321312,321321,312312
2231213,213231,213213,231231
1123132,132123,132132,123123
i
i
i
for
for
for
mnkijkConsider 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 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:
Gives +1 Gives 1
mnkkji
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor and Delta
3312321,321312,321321,312312
2231213,213231,213213,231231
1123132,132123,132132,123123
i
i
i
for
for
formnkkji
mnkijkConsider another expression:
Gives +1 Gives 1
1
1
21312321
1231231212
kk
kk
Only 6 terms are non-zero (those for which ij and either i=m, j=n or i=n and j=m. Two of these are:
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor and Delta
mjniji
njmiji
ji
mnkijk
,,1
,,1
0
for
for
for
Thus,
Louisiana Tech UniversityRuston, LA 71272
Permutation Tensor and Delta
jnimjnim
It can be shown, through similar enumeration, that the delta expression:
Gives the same results and that therefore:
jnimjnimmnkijk