louisiana tech university ruston, la 71272 momentum balance steven a. jones bien 501/cmen 513...

Louisiana Tech UniversityRuston, LA 71272

Momentum Balance

Steven A. Jones

BIEN 501/CMEN 513

Friday, March 17, 2006


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


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?


Cylindrical coordinates are simpler because of the boundary conditions:

Cylindrical Coordinates: Examples

Rru at0

cos2

1

2

31

3

r

R

r

Rvur


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


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


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


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


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.


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).


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).


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


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.


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.


Momentum Shell Balance


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


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


The Control Volume

Inner Cylinderr rr

rz


The Control Volume

z zr

zz

Left Annulus


z zr

zz

The Control Volume

Right Annulus


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).


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).


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


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


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


Force Balance

rdzrdrrdzdrr rzrz 22

dzzpdrrzpdrr 22

Divide by 2 dr dz:

dr

rrdrrdrr rzrz r p z r p z dz

dz


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:


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


Differential Equation

r

uzrz

So, with:

z

pr

r

ur

rz

The equation is:

and

z

pr

r

r rz



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.



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


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.


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.


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


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)


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.


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.


The Stress Tensor

33231

23221

13121

ji j

iij eeτ 1

13

12

The diagonal terms (normal stresses) are often denoted by i.


Exercise

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

drd

dz


Look at the r term

r rr dr r dr d dz r rd dz

Divide by dr, d, dz

r r

r


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


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


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


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


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


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


Components of ti

Recall the stress tensor:

A3

A2

A1

nz1

z3

z2

t1

333231

232221

131211

τSurface (row)

Direction of Force (Column)


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.


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.


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.


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).


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)


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.


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.


Symmetry

33231

23221

13121

τ

1

13

12

The stress tensor is symmetric. I.e. ij=ji


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:


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


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



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.



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.



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



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:



mjniji

njmiji

ji

mnkijk

,,1

,,1

0

for

for

for

Thus,



jnimjnim

It can be shown, through similar enumeration, that the delta expression:

Gives the same results and that therefore:

jnimjnimmnkijk

louisiana tech university ruston, la 71272 momentum balance steven a. jones bien 501/cmen 513...

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