slow viscous flows

7/29/2019 Slow Viscous Flows

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Slow Viscous Flows

John Lister(Zachary Ulissi)

Michaelmas Term, 2009-2010


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Contents

1 Summary of Fluid Mechanics 31.1 Mass Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Stress Tensor Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 The Momentum Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.5 Newtonian Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.6 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6.1 Kinematic boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . 61.6.2 Rigid boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6.3 Dynamic boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.7 The Reynolds Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.7.1 Notes about the Reynolds Number . . . . . . . . . . . . . . . . . . . . . . 7

2 Stokes Equations 82.1 Simple Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.1 Instantaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.2 Linear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.3 Reversibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.4 Forces and Torque Balance . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.5 Work Balances Dissipation (since there is no KE) . . . . . . . . . . . . . . 9

2.2 Three theorems based on dissipation integrals . . . . . . . . . . . . . . . . . . . . 92.2.1 Example of the dissipation theorem . . . . . . . . . . . . . . . . . . . . . 102.2.2 Example with rigid particles . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.3 Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Representation by potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3.1 Complex Variables in 2D . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3.2 Papkovich-Neuber Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3.3 Notes on the Papkovich-Neuber Solutions . . . . . . . . . . . . . . . . . . 12

2.4 Solutions for points, spheres, and cylinders . . . . . . . . . . . . . . . . . . . . . 132.4.1 Pseudo-tensors vs true tensors . . . . . . . . . . . . . . . . . . . . . . . . 132.4.2 Solution due to a point force . . . . . . . . . . . . . . . . . . . . . . . . . 142.4.3 Fundamental Solutions of Stokes Flow . . . . . . . . . . . . . . . . . . . . 142.4.4 Point Source Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4.5 Force Dipoles, Stresslets, Rotlets, Etc. . . . . . . . . . . . . . . . . . . . . 152.4.6 Example: Rigid Sphere in Translation . . . . . . . . . . . . . . . . . . . . 162.4.7 Simple Derivation of Force/Stress on a Translating Sphere . . . . . . . . . 172.4.8 Gravitational Settling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.4.9 2D singularities and Flow Past a Cylinder . . . . . . . . . . . . . . . . . . 172.4.10 Uniform Flow Past a Stationary Cylinder . . . . . . . . . . . . . . . . . . 18

2.5 The Motion of Rigid Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.5.1 The Resistance Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5.2 The Faxen Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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3 Applications 213.1 Marangoni Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.1 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.1.2 Thermophoresis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.1.3 Surfactant Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Deformation of Drops (Rallison 1984, Stow 1994, Ann. Rev. Fluid. Mech.) . . . 25

3.2.1 Small deformations (Ca 1) in uniform streams (U =Ex . . . . . . . . 253.2.2 Larger Deformations in Pure Strain . . . . . . . . . . . . . . . . . . . . . 263.2.3 Larger Deformations for 1 (Taylor 1964) . . . . . . . . . . . . . . . . 26

3.3 The Rayleigh Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.3.1 Neglect of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4 Long Thin Flows and Immersed Bodies 294.1 Long thin flows I: Lubrication theory . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Lubrication Theory for Pigs and Slugs . . . . . . . . . . . . . . . . . . . . . . . . 354.3 Long Thin Flows II: Extensional Flows . . . . . . . . . . . . . . . . . . . . . . . . 394.4 Long Thin Flows III: The Two Fluid Case . . . . . . . . . . . . . . . . . . . . . . 404.5 Slender Body Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.6 Elastohydrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.6.1 The Beam equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.6.2 Example: Thin Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.6.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.6.4 Example: Compression of a rod of length l, radiusr . . . . . . . . . . . . . 474.6.5 Winkler Elastic Foundation Model . . . . . . . . . . . . . . . . . . . . . . 494.6.6 Example: Falling Cylinder Onto a Winkler Substrate (Skotheim+Mahadevan

PRL, 2004) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5 Handouts 525.1 Summary of Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.2 Integral Representations of Stokes Flow . . . . . . . . . . . . . . . . . . . . . . . 605.3 2D Singularities and Flow Past a Cylinder . . . . . . . . . . . . . . . . . . . . . . 62

5.4 The Greens Function for Stokes Flow . . . . . . . . . . . . . . . . . . . . . . . . 645.5 Two Solutions in Lubrication Theory . . . . . . . . . . . . . . . . . . . . . . . . . 655.6 Extensional Flow Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665.7 Small Deformations of a Viscous Drop by Uniform Pure Strain E . . . . . . . . . 695.8 Example: Swimming Flagella . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

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Chapter 1

Summary of Fluid Mechanics

The continuum assumption deals with infinitesimally small fluid particles that have a welldefined density (x, t), pressure p(x, t), and velocity u(x, t), where x(t) is the position of theparticle. From the Eulerian perspective (static frame of reference), a time derivative will be /t,but in a frame of reference following a fluid parcel the derivative will be D/Dt = /t + u

.

1.1 Mass Conservation

In general, Eulers equation is:

t+ (u) = 0 D

Dt+ u = 0

For incompressible fluids (everything in this course):

D

Dt= 0 u = 0

1.2 Stress TensorStress is a force per unit area across a surface. For example, a surface force balance on aninfinitesimal tetrahedron shows is linearly related to surface normal n by:

= nWhere is the stress tensor and is the stress by the outside acting on the inside of a surfacewith outward normal n. An angular momentum balance, for most materials (except, for example,magnetic ones), yields:

ij = ji

1.2.1 Stress Tensor SheetIn the continuum approximation, the forces on a small of fluid at (x, t) can be divided into thoseacting on and proportional to the volume, like gravity, and those acting on and proportional tothe surface area, like pressure or friction due to relative motion. (In reality, surface forces areexerted over a thickness of a few intermolecular distances so, for example, molecules in relativemotion travel through a mean-free path before depositing their momentum in a collision). Sincethe volume forces on the small blob and its inertia are both proportional to its volume V, whereasthe surface forces on the blob are of order V2/3, the surface forces must balance by themselvesin the limit V 0. This observation can be used as follows to show that the stress (x, t; n)(force per unit area) acting on a surface with normal n is linearly related to n by a second-ranktensor .

Consider a small tetrahedron of fluid aligned with local, rectangular coordinate axes e(1), e(2), e(3).

Denote the forces exerted by the exterior of the tetrahedron on the surfaces of the tetrahedron byF(k) acting on the three faces with outward normals in the three negative coordinate directionse(k) and F acting on the sloping face, which has outward normal n and area A.

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Since the surface forces must balance by themselves in the limit as V 0, we obtain:

F = k=1,3

F(k)

=k=1,3

F(k) (By Newtons third law)

A =k=1,3

A(k)(k)

where A(k) is the area of the kth surface of the tetrahedron. From geometry, A(k) = An e(k).Thus the stress can be written as:

=

k=1,3

e(k)(k))

n

= n

where

(x, t) =k=1,3

e(k)(k))

is the stress tensor, which is independent of the direction of n. The components of the stresstensor are given by:

ij =

k=1,3e(

k)j

(k)i

But e(k)j = jk , and hence ij =

(j)i is the ith component of the force per unit area exerted

on the fluid by a surface with normal in the jth coordinate direction.The most important statement about the stress tensor is that the force per unit area (stress)

exerted on the fluid by a surface with unit normal n pointing out of the fluid (equal, of course,to minus the stress exerted by the fluid on the surface) is given by:

= n

1.3 The Momentum Equation

Consider an arbitrary fixed control volume used for mass conservation. The rate of change of

the total momentum within the control volume is due to the outflow of momentum through the

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boundary, and to the forces acting on the fluid, which comprise body forces (per unit volume)and surface forces (per unit area). Thus:

d

dt

V

udV = V

(u)u ndS Momentum flux

+ V

F dV Body forces

+

V

ndS Surface forces

Use of the divergence theorem givesV

t(ui)dV = V

xj(uiuj)dV +

V

FidV +

V

xj(ij)dV

Also, since this holds for arbitrary control volumes, the integrands must equate to give

Du

Dt+ u

t+ (u)

= F +

(actually (T

),but well soon show that = T

). The second term is zero by conservationof mass, so

Du

Dt= F +

1.4 Energy

Taking the integral of the momentum dotted with velocity of the fluid, the rate of internal viscousdissipation is obtained:

D =

V

udV = V

eijijdV =

V

e : dV

where e is the strain rate tensor

eij =1

2

uixi

+ujxi

Notice that e is symmetric, that is eij = eji and eii = 0. The rate of work by surface forces onthe fluid is:

dV

u ndS

1.5 Newtonian Fluids

Stresses are produced by fluid deformation. If relationship between stress/deformation rate

ui/xj is local, linear, instantaneous, and isotropic, the fluid is Newtonian. If it is also incom-pressible

= pI + 2e = pI + dWhere is the dynamic viscosity and d is teh deviation stress, which depends only on e =

1/2[(u + (u)T] and not on the vorticity = u. For incompressible Newtonian fluids withconstant viscosity:

Du

Dt= F p + 2u u = 0

which together are known as the Navier-Stokes equations. The rate of viscous dissipation is then:

D = 2V

e : edV

Often, F = and F can be incorporated into a modified pressure p + = P.

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1.6 Boundary Conditions

1.6.1 Kinematic boundary conditions

At fluid/fluid interfaces, using mass conservation and no mixing:

[u

n] = 0, Mass conservation

[u n] = 0, No stress

The brackets indicate terms across the interface (i.e. matching, but not necessarily 0, velocities).

1.6.2 Rigid boundary

A hard impermeable boundary yields instead:

u n = 0, No fluxu n = 0, No slip

1.6.3 Dynamic boundary conditions

In the absence of surface tension [ n] = 0. With surface tension:

[ n] = n s

where = s n (specified since interfaces may be curved) and is the coefficient of surfacetension.

1.7 The Reynolds Number

Once again, the Navier-Stokes equations are:

DuDt = p + f + 2u u = 0

Supposed L, U, and L/U are representative length, velocity and time scales of a given body:

Examining the scaling of each term

Du

Dt U

2

L2u U

L2

The ratio of these two terms is defined as the Reynolds number Re:

Re Inertial ForcesViscous Stresses

=U L

=

U L

where = /. If Re 1, we have the Stokes equations, which neglect the small inertia term

2u = p F u = 0

The Stokes equations are very useful for:

1. Large - Magma, glass

2. Small L - microorganisms, microfluidics

3. Thin long films - lubrication theory

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1.7.1 Notes about the Reynolds Number

1. There is an assumption that Re scaled with L, but u and 2u may not have the samelength scale:

2. L1 will be the lengthscale for the viscous term (2u), while L2 will be the length scale forthe boundary term (u )

3. The length scales may very throughout the flow, such as in the case of a flow past a body(the reynolds number in the far field will be different)

4. The time scale may not be L/U if there is an externally imposed time scale (e.g. anoscillating body, T

1)

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Chapter 2

Stokes Equations

The Stokes equations are:

= 2u p = F u = 0 (2.1)

2.1 Simple Properties

2.1.1 Instantaneous

Since there is no time derivative term (no /t), there is no inertia or memory. The flow onlyknows about the current boundary conditions and applied forces, and responds instantaneously.If the boundary conditions or forces are changing w/ time, the flow is called quasi-steady.

2.1.2 Linear

The response of the flow is linearly proportional to the forcing (double the force, double thevelocity field). Additionally, it is possible to superimpose solutions onto the same geometry.

2.1.3 Reversibility

If the sign of all forces is changed, the sign of the velocity field will change ( F u). If the sign of the forces is changed and history of application is reversed, the flow retraces its ownhistory exactly.

This can be used with symmetry arguments to rule something out. For example:

Migration is impossible due to the contradiction. Three more phenomena to be described:

1. There will be no rotation when falling.

Reflection across both axes, then a reversal of force/velocity will yield a contradiction.

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2. There will be no migration in the boundary layer.

Reversing the flow, then reflecting along the y axis yields a contradiction

3. Swimming is not possible.

2.1.4 Forces and Torque Balance

Becaues there is no inertia, all forces must balance

= F V

ndS+V

F dV = 0

This allows for a consistency check on any stress boundary conditions. Similarly,

u = 0 V

u ndS = 0

is a consistency check on velocity BC. Likewise, moments/torques balance.

2.1.5 Work Balances Dissipation (since there is no KE)

The dissipation (as defined previously):

D = 2

V

eijeijdV

=

(ij + ij)eijdV

=

ij

uixj

+peii

dV

=

xj(ijui) ui

xjij

dV

= V

u

ndS+ u F dV

work balances dissipation

2.2 Three theorems based on dissipation integrals

Lemma 2.2.1. If uI is incompressible flow, uI = 0 and uS is a Stokes flow w/ body forceFS( S = FS) then:

2

V

eI : eSdV =

uI S ndS+V

uI FSdV

Proof. fill this in later, using the the same method as for the dissipation of work.

Theorem 2.2.1. Solutions of Stokes equation are unique

Proof. Suppose u1 and u2 are Stokes flows with the same boundary conditions and body force(i.e. F1 = F2 in V and either u1 = u2 or 1 n = 2 n on V), let:

u = u1 u2u is also a Stokes flow with F = 0 and either u = 0 or n = 0 on V. From the previouslemma:

2

e : edV = 0 e = 0 in V

Therefore, u is a solid body motion (no strain) with (u + x), and usually u = 0. Therefor:u1 = u2 Solution is unique

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Theorem 2.2.2. Reciprocal Theorem: If u1,u2 are Stokes flows (in a certain domain), then:V

u1 2 ndS+V

u1 F dV =

u2 1 ndS+V

u2 F1dV

Proof. Apply previous lemma to show that both sides are:

2

e1

: e2

dV

This is essentially Greens 2nd theorem.

Theorem 2.2.3. Maximum Dissipation Theorem: Among al l flows in a certain domain, thatsatisfy given velocity boundary conditions and u = 0, the dissipation is minimized by stokesflow w/e F = 0.

Proof. First,

0 2

(e eS) : (e eS)dV = 2

(e : e eS : eS)dV + 4

eS : (eS e)dV

The second term is zero by the previous lemma, with:

uI = uS u

where uI = 0 on the boundary and FS = 0. Therefore, the remaining term is:

0 2

(e : e eS : eS)dV = D DS

2.2.1 Example of the dissipation theorem

The dissipation theorem allows for bounds to placed on the drag of irregularly shaped objects.For example, in the case of an arbitrary volume using inscribed and circumscribed circles:

If body A is inside body B then an inequality can be derived by letting uS be a Stokes flowpast A and uI be the Stokes flow past B plus the solid body motion in the gap between A and B.This inequality can be derived for the inner and outer circles, allowing the appropriate boundsto be calculated:

(6a1U)U F u (6a2U)U

2.2.2 Example with rigid particles

Adding rigid particles to a Stokes flow with given external velocity boundary conditions increasesthe dissipation and, if the particles are force free and couple free, increases the apparent viscosity.This does not say anything about more or less particles, just that the addition of some particles

will increase the viscosity. For example, does not say anything about shear thickening suspensionsof silica.

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2.2.3 Inertia

Inertia will increase drag (for example, consider that Du/Dt is part of F).

2.3 Representation by potentials

Assuming no body force (F = 0), a conservative flow with a modified pressure:

2u = p (1) u = 0 (2)

Taking the divergance of (1):

(1) 2p = 0 (1) 2 = 02(1) 4u = 0

In 2D, if we use a velocity field:

u = (0, 0, ) z = 2

4

= 0

Using axisymmetric sphericals (r,,) with:

u =

0, 0,

r sin

This results in:

0 = E2

r sin E4 = 0 where E2 =

2

r2+

sin

r2

1

sin

Many explicit solutions can be found in coordinate systems in which the operators 2, 4, E2,and E4 are separable (see Happel and Brenner).

2.3.1 Complex Variables in 2D

Writing z = x + iy and z = x iy allows:

2 = 4 2

zz

If f(x, y) is analytic, meaning:

f = f(z)f

z

then Ref and Imf are harmonic. Similarly,4 = 0 implies can be written as:

= Im(z(z) + (z))

where and are analytic. Some very clever solutions can be found by conformal mappingtechniques. An example of a very tricky solution is the combination of two bubbles:

Of course, the problem is that the complex methods only work in 2D.

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2.3.2 Papkovich-Neuber Solution

This will be the main way to solve for Stokes flows. Let:

p = 2 can always be solved with an integral representation

(x) =14

p(x)

|x x|dV

Using an equivalence with the normal definition of p:

2(u ) = 0u = where 2 = 0

But:

u = 0 2 = The idea is then to think of a function for that satisfies this relationship:

=1

2 x + 2 = 0

This comes essentially from completing the squares with the :2(A) = (2A) B + A 2B + 2A B

Hence any Stokes flow (where F = 0) can be written in terms of a harmonic vector and aharmonic scalar :

2u = ( + x) 2 p = (2.2)

which is called the Papkovich-Neuber representation of Stokes flow.

2.3.3 Notes on the Papkovich-Neuber Solutions

1. Any potential flow of the form:

2u = is also a Stokes flow but p = 0 and ij = u/2 + uT/2 = ij differ from thecorresponding inviscid flow.

2. Sometimes it is possible to find another harmonic scalar with:

= x 2

If so, one can dispense of by:

= + For example, if has a spherical harmonic expansion then we can do all the terms exceptfor the uniform strain as

=1

2x x

where is the far field strain. In this case we can write

2u = x + (x ) 2p =

2e = + () xwhere 2 = 0 and is a constant, symmetric, traceless matrix.

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3. Conversely, if = , then u can be calculated using: = x 2

which is generally easier than (2)

4. Lambs Solution is frequently referred to in literature and uses three harmonic scalars

expanded in spherical harmonics, which is essentially the same as the method just above.

2.4 Solutions for points, spheres, and cylinders

Points and spheres have no intrinsic orientation or direction, so spherical harmonics areused. Let:

r = |x| 2 1r

= 0 (when r = 0)

All other functions where 0 as r 0 can be obtained from:1

r

,

1

r

,

1

r

,

1

r

, . . .

Harmonic functions which are bounded at r = 0 can be obtained from:

r1

r, r31

r= x, r51

r, . . . r2n+1()n 1

r

This is a generalization of other schemes for generating harmonic functions, such as using Leg-endre polynomials:

rn

rn1

Pmn ()

cos n

sin n

Recall that:

x = Ior xi

xj=

jxi

= ij

r = xr

from 2rr = r2 = (xjxj)[same as /xi(xjxj)] = 2x

f(r) = x

rf(r)

For example:

1r

= xr3

1r

=Ir3

+3xx

r5

i

j

k =

3(ijxk + jkxi + kixj)r5

15xixjxk

r7

These functions depend only on x and r and have no preferred orientation, which is useful forproblems with spherical symmetry. It is possible to form Papkovich-Neuber potentials , bymultiplying by constant tensors and taking an appropriate number of dot products. For example:

A1

r, B1

r, C 1

r, (D )1

r, 1

rare all vectors

2.4.1 Pseudo-tensors vs true tensors

Tensors have no right-handedness defined. We can distinguish between true and pseudo tensorsby checking whether they keep or change signs under reflection. Examples of true tensors andpseudo-tensors are:

True: U , F , x , Psuedo: , U x, = u

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All of the pseudo-tensors have a right-handedness obtained form some time of cross product.The product of two true tensors is pseudo, and vice versa. An important distinguishing propertyis behavior under transformations. Under rotation, both psuedo and true tensors stay the same,but under reflection they change:

Tijk = RijRjmRkn, . . . T lmnwhere the sign depends on whether it is a pseudo or true tensor and R is the reflection matrix.

We need u to be true, so , should also be true.

2.4.2 Solution due to a point force

Consider:

= F(x) u = 0, u() 0Because of the linearity of Stokes flow, the answer must be linear in F. In addition, there is noorientation of the flow, due to rotational symmetry. So, try:

=F

r

Since F(1/r) is pseudo, using (F)(1/r) would yield = F(1/r) which is too singular(blows up at r = 0). The velocity field is then:

2u =

F xr

2F

r

=

F I

r (F x)x

r3 2F

r

=

F

r (F x)x

r3

The stress can then be calculated

= 3(F x)xx

r5

On any sphere r = R,n = x/R then:

u n = 22

F nR

n = 3(F n)nR2

Checking the mass flux: r=R

u ndS = R

F

r=R

ndS = 0

= 0

As expected, there is no net flow from the point force. Since we want to specify a force:

F =r=R

ndS = 3F

ninjdS

R2=

4ij3

Thus, is chosen to be:

=14

The solution to this problem was achieved using only some thinking, and the rest came from thePapkovich-Neuber solution. All that really needed to be done was identify the coefficient.

2.4.3 Fundamental Solutions of Stokes Flow

The fundamental solutions of Stokes flow, upon which other solutions are linearly dependent,are as follows:

u = F J(x) = F K(x) p = F x4r3

J =1

8I

r +xx

r3

K = 3

4

xxx

r5

Where J is known as the Oseen tensor.

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Example: Flow due to a point force

From u = 0 and = F (x): J = 0 K = I(x)

Thus, the velocity profile will fall off proportionally to 1 /r

2.4.4 Point Source Flow

From mass conservation and radial symmetry, the velocity field must be:

u =Qx

4r3

This could also be derived from:

Q

r or Q1

r

2.4.5 Force Dipoles, Stresslets, Rotlets, Etc.

Further solutions can be obtained by taking gradients of the stokeslet and source solutions. Forexample, consider:

The solution for small d, due to the linearity of the Stokes equations, will be:

u = F J(x d) F J(x)Taking a Taylor expansion:

uj = Fi (di )Jij + . . .As the displacement goes to zero, the force will go to infinity while maintaining a constant F d.The term Fidj can be split into three parts:

1. An isotropic part with no flow (

J = 0)

13

Fkdkij

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2. A symmetric traceless part, representing a stresslet of strength S:

Sij = 12

(Fidj + Fjdi) +1

3Fkdkij

3. An antisymmetric part, representing a Rotlet of point couple G:

12

ijkGk where G = d F

2.4.6 Example: Rigid Sphere in Translation

Consider:

2u = p u = 0 u = U(on r = a) u(r = ) = 0

We need harmonic functions of U and x that are linear in U, decay at r = , and are truevectors (not pseudo). First, some possibilities for U and :

1

2 =

U

r, U 1

r = U 1

r

The second possibility for yields the same behavior as , so it must be the first possibility.

Higher order terms are not possible because any additional gradients will require more dotproducts with U and x, which would break linearity. Thus:

u = A

u

r+

(u x)xr3

+ B

U

r3+

3(U x)xr5

u =

u

r+

(u x)xr3

+ U

1r

With the boundary condition u = U on r = a. Solving for the coefficients:

a

a3

= 1, a

+3

a3= 0 = a3/4, = 3a/4

Thus, the velocity field is:

u =3

4U

a

r+

a3

3r3

+

3

4(U x)x

a

r3 a

3

r5

=

1

r

Stokeslet +

1

r3

quadropole

In order to calculate the pressure and vorticity, either the Stokes relations can be used, orthe form can be inferred. They must be of the form (due to dimensional and scaling/vectorarguments):

p a U xr3

= 3

2a

U xr3

U xr3

= a3

2

U xr3

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Force and Stress on a Translating Sphere

The next step is calculating the stress on the sphere from the Stokes equations and availableinformation:

= pI + [u + (u)T] x = I r = xr

rn = nrn1 xr

This leads to:

= 32

U xr3

I+3

4

ar3

a3

r5

(xU + U x)

+

U x + xU + 2(x U)I

a

r3 a

3

r5

+ (U x)2xx3a

r5+

5a3

r7

Evaluating at the surface of the sphere:

n|r=a = x

r r=a= 3

2

a(u n)n + 3

4a[2((uU n)n + U) + O + 4(U n)n] = 3U

2a

This remarkably leads to a constant across the sphere. Hence: ndS = 3U

2a4a2 = 6aU

2.4.7 Simple Derivation of Force/Stress on a Translating Sphere

The far-field is dominated by the 1/r stokeslet term with coefficient F = 6aU. Since the forceexerted at r = is dominated by this term and the force must balance the total exerted acrossr = a, we can deduce without any further work that the drag on a sphere is:

Drag = 6aU

2.4.8 Gravitational SettlingA force balance for small particles using Stokesian drag and gravity is:

4

3(s )a3g 6aU U = 2

9

(s )ga2

This relation to the square of the particle radius allows small particles to stay suspended insolution while larger ones may fall out.

2.4.9 2D singularities and Flow Past a Cylinder

The harmonic functions in 2D are:

ln r, (ln r), . . .and when bounded at the origin:

1, r2 ln r = x, = r2nn(ln r)Using a potential = F /2 yields:

u = F J2D p = F x2r2

= F K2D

where

J2D =1

4

ln rI + xx

r2

K2D =

1

xxx

r4

This is the 2D Stokeslet solution, which corresponds to a line force F per unit length. Note thatu does not converge to 0 at infinite distance for a line force, although it will converge to 0 forline dipoles, line quadropoles, etc.

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2.4.10 Uniform Flow Past a Stationary Cylinder

Motivated by the previous solutions for uniform flow past a sphere, we try:

2= Uln r + U ln r + U

to solve the Stokes equations with

u(r = a) = 0 u(r ) U

(The U is partly motivated by the change in reference frame.) However we can find that, whilewe can satisfy the boundary conditions on the cylinder with:

u =

U

ln

r

a+

1

2

1 a

2

r2

+

(U x)xa2

a4

r4 a

2

r2

p7 = 2U x

r2

Drag = 4U

There is no choice of that allows u U as r . The problem is that inertia cannot be p7?neglected far from the cylinder since

(u )u2u

U2/r

U/r2=

U r

1 as r

This scaling suggests that inertia needs to be brought back into the equations when r =O(aRe1).

After scaling lengths with a and velocities with U, the steady Navier-Stokes equations give:

Re(u )u = p + 2u u = 0, r > 1 u = 0, r = 1 u z, r

Where Re = Ua/

1. A somewhat ad hoc way of obtaining an approximate solution is to

recognize that the (u )u term only has a leading-order effect where u z. This motivateesthe Oseen equations:

Rezu = p + 2u u = 0

which are linear (with solutions by fourier transform), no less accurate than the Stokes equationsfor small r, and correct at leading order for large r.

A rather better way of proceeding is to solve the original equations by matching asymptoticexpansions for Re 1 between r = O(1) and r = O(Re1). The zeroth-order solution forr = O(1) is the above Stokes solution with determined by matching to be:

=1

ln 1/Re+ O

1

[ln1/Rey]2 F = 4U

ln 1/Ren

cn[ln1/Re]n

This depends only weakly on a and corrections are quite large. Intertia cannot be neglectedfar from a sphere either, or from any other 3D body, but the solution for r = O(Re1) does notaffect the zeroth-order Stokes solution for r = O(1), and comes in at first order instead.

2.5 The Motion of Rigid Particles

We can get some information for particles of various shapes using the dissipation theorems andbounding the results by inscribing and circumscribing spheres. An alternate, more detailedformulation is:

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2.5.1 The Resistance Matrix

A rigid particle moving with velocity U+ x through a fluid, otherwise at rest, exerts a forceF and couple G on the fluid. By linearity, both F and G must be linear to U , . The grandresistance matrix is:

FG

=A B

C D U

where the tensors A , . . . , D depend on the size, shape, orientation, etc of the body. The dissipa-tion is then:

u ndS =

(U + x) ndS U F + G

Hence, from the reciprocal theorem (with F = 0):

[U1, 1]

A BC D

U22

= [U2, 2]

A BC D

U11

For all U1, 1, U2, 2. Thus A = AT, D = DT means the matrix is diagonalisable and also

that BT = C. Thus, forces due to pure rotation balance forces associated with translation. Sincethe dissipation must be positive, A , B , C , D must all be positive, definite, and invertible.

A few examples:

1. If a particle has 3 independent planes of reflectional symmetry, then B = C = 0. (i.e. ifthree independent vectors dotted with B,C are zero, B,C must be zero.

2. A cube falls with the same speed and no rotation in all orientations because A I (fromsymmetry).

3. A and D are known for ellipsoids (w/ B, C = 0) and hence also known for discs and rods(which are just limiting cases).

The resistance matrix (with R1 being the mobility matrix) is sometimes extended to:FG

S

=U U

E

and is called the grand resistance matrix.It is possible to describe the force, couple, and stresslet exerted by a particle placed in a

background linear flow with:

U = U + x + E xA linear flow L gives leading order effects for small particles of characteristic length a placed insome background flow U(x) of larger lengthscale L:

U(x) = U(0) + (x )U(0) + O

a2L2

2.5.2 The Faxen Relations

These relations concern a particle placed in an arbitrary unbounded flow U(x). The goal is tocalculate the particle velocity uP

uP = U + xLet u be the perturbation flow u(x) U(x), which satisfies u() 0 and uP = U +

x u(x) on the particle. let u be a test flow due to translation with arbitrary velocty V. Bylinearity, (x) = V for some . The reciprocal theorem for u, u gives

V Part.

ndS = V

(U + x U(x)) ndS

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But V is arbitrary and the contribution from U+ x is given by the resistance matrix. Hence

F = A U + B

U ndS= Unknown + Unknown + Background Flow

which is Faxens first relation. Similar formulas can be derived for the couple and stresslet usingother test flows.

2.5.3 Example

For a sphere

n = 32a

I

The matrices A, B are

A = 6aI B = 0 (Spherical Symmetry)

which, used with Faxens first relation, gives

U =F

6a+

1

4a2

r=a

U(x)dS

where the second term is just the average velocity on the surface. The expansion of the far fieldis:

U(x) = U(0) + (x )U(0) + 12

xx : U(0) + . . .

which leads to

1

4a2

UdS = U(0) + 0 +

1

8a2

xxdS

: (U(0)) + . . .The odd terms in this series (e.g.

xxxdS) are zero by symmetry. The even terms are isotropic,

and give (2)nU(0). But 4U(0) through the harmonic requirement of the Stokes equation(same for 6,8,etc). Thus, the Taylor series naturally terminates after just a few terms.

u =F

6a+ u(0) +

a2

62U(0)

One application of this is two falling bodies. Its possible to calculate the fall velocity toO(a3/R3) with a 1/r4 dependence on each other.

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Chapter 3

Applications

3.1 Marangoni Flow

Marangoni flow is driven by gradients in surface tension () gradients between immiscible fluids.

A few examples:

1. Temperature effects: As the temperature increases, the surface tension decreases. For awater/air interface, (d/dt)/ = 1/50K. A classic example is

2. Surfactant gradients:

surfactant

diagram

3. Alcohol/water wineglass

3.1.1 Boundary Conditions

Surface tension can be represented as a surface stress (units N/m) derived from a surface energyJ/m2, which acts across lines in an interface. The surface stress tensor is (I nn) s, whichessentially just kicks out the relevant tangential directions.

A force balance on a small bit of interface gives

[ n]+ + s s = 0

where

s (I nn)

is the gradient operator in the surface. Hence, the jump in the fluid stresses is

[ n]+ = S ((I nn)= (I nn) s+ (s n)n+ (n s)n= s+ n

which is the surface tension crossed with the curvature.

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3.1.2 Thermophoresis

An immiscible drop in a temperature field migrates from a cold region to a hot region. Weassume a background temperature gradient H = (T). From a scaling argument, T Haimplies Ha, and = d/dT < 0. The driving force from a surface tension difference isproportional to F Ha2. The viscous stress U/a acts proportional to the area. Thus,the drag is proportional to Ua F and U Ha/.This leads to a second thermal problem within the bubble and the field. The heat equationis

T

t+ u T = 2T

The temperature gradient approaches the gradient H at infinite distance. Since the temperaturemust be continuous at the surface of the particle

[T] = [kn T] = 0 On |x| = a

where k is the thermal conductivity (q = kT) and is the thermal diffusivity = k/Cp.We assume that the Peclet number is small

Pe =U a

=

Advection

Diffusion 1

so that the heat equation becomes

2T = 0

Thus, for an insulating drop (ki = 0)

T = T0 + H x(1 + a3/2r3)

which is a harmonic scalar linear in H. On the surface

T = T0 + 32 H x

We can expand around the temperature difference, and take the first term of the temperaturedifference is small

(T) (T0) + (T T0)(T0)

The next step is to connect the surface tension gradients to the fluid problem. The surface stressis driven by an interfacial force density

Fs = (1 n 2 n) = [ n]+

= (I nn) (3H/2) (sn)n

The last term is also dependent on H through the surface tension. The divergence of the normalfor a sphere is just 2/a (as expected). The force acting on the droplet is then

F =3

2H (I 3nn) 0 2

an

For the simplest case of a drop with equal viscosity to the surrounding fluid ( = 1), the velocityprofile is

u(y) =

r=a

J(y x) Fs(x)dS

=3

2

1

8H (I 3nn)

I

R+

RR

R3 dS=

3a

16H G(y)

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where R = y x and G is a dimensionless function of y. By the symmetry of the integral

G(y) = r

a

I + r

a

yy

where y = y/r. Continuinc

u y = 3a16

( + )H y

Thus, a sphere will remain spherical and translate with a velocity

U =3a

16( + )H = 1

5

aH

Its possible to find the coefficients , from

Gii = 3 +

yiGij yj = +

3.1.3 Surfactant EffectsThe concentration C of surfactant in a small material area of surface changes due to variousphysical processes

1. Surface diffusion with

Flux = DssCWhere the diffusion coefficient is on the order of 109 m2/s

2. Adsorption from bulk with rate

r = k(C C0)

Where C0 is the equilibrium concentration and the rate constant is on the order of 1000/s.

3. Changing surface area

(AC) = 0 C = AA

C

to calculate the rate of change of surface area its necessary to consider a material volume

The volume of the fluid element is then

V = L An dVdt

= ( u)V ddt

L = (L )u

where L is an arbitrary length vector (notice in the next step it cancels on both sides).

d

dt(An) = [( u)I u] An

We now take n

d/dt(nA) and use n

n = 0 since n

n = 1.

d

dt(A) = A( u nn : u] = (s u)A

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Expanding this last term

d

dt(A) = As (us + (u n)n)

= A(s us + (s n)(u n) + (n s)(u n))= Dilation of Flow in Surface + Dilation of Surface + 0

(since s is in the tangential direction only)Combining all of these effects into a total time derivative on the concentration

DC

Dt= C[s us + (u n)s n] + DS2sC k(C C0)

Example: Rigidification of InterfacesConsider a rising bubble with surfactants present. Any flow leads to a concentration gradient

in surfactant that creates a gradient in surface tension thus driving a reverse flow. A steadysolution in the frame of the bubble has a steady concentration profile of surfactant concentrationand no mass flux into or out of the surface (u n = 0). Consider a linearized calculation withC = C0 + C(x) with C C0 (to be checked later). Thus

Ds2sC kC = C0s usAfter neglecting the nonlinear terms u C and Cs us. Since the solution should be linearin u and have spherical symmetry,

us = A(I nn) u

for some A (O(1)). Then

s n = (I nn) (x/a) =I nn

a

which is the same as the divergence of the normal vector, (x/r), and the nn (x/a) termvanishes. Thus

s n =2

a

Which isnt surprising as this is the curvature of a sphere. Plugging this in

s us = As (I nn) U = 2

aAU n

Simplifying the surface divergence

s = (I nn) s (nn) = (s n)n + (n s)n = (s n)n

Guessing that the surface concentration is proportion to the normal velocity

C (u n)

Since

2s(u n) = s

I nna

U

= 2a2

U n

The solution for C is then

C =2C0A(u n)/a

k + 2Ds/a2

The assumption of small C is thus fine as long as surface diffusion is fast (Ua/Ds 1) orsurface absorption is fast (U/ka 1), or if the flow is very slow. The problem is completed

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by the Papkovich-Neuber representation of flow inside and outside the bubble. The new set ofboundary conditions for this example are

u(r = ) = U u(r = 0) bounded

and on the surface

u = us [n n] = n Cd

dC

Solving the problem, the drag is

F = 4aU

3( + )/2 + 1

( + ) + 1

= aC0

ddC

3(a2k + 2Ds)

So the main effect is to increase the apparent viscosity of the bubble. As the effect becomesstronger, the solution approaches that of a rigid sphere.

3.2 Deformation of Drops (Rallison 1984, Stow 1994, Ann.

Rev. Fluid. Mech.)

To leading order, a small drop in an external flow with velocity U is advected with velocityU(0) and is deformed by gradients of the flow x U(0) against the restoring effect of surfacetension.

The strength of this behavior can be represented in a new dimensionless quantity

Viscous Stress

Surface Tension Pressure=

E

/a= Ca =

U

Which is just the capillary number or perhaps U/. Low capillary numbers lead strong strongsurface tensions.

3.2.1 Small deformations (Ca 1) in uniform streams (U

= E xThere is deformation due to a symmetric and traceless flow E (ellipsoidal symmetry). Thus, anew shape might be expected

r = a

1 +

x D xr2

+ O(D2)

where D is also a symmetric, traceless (to conserve volume) tensor and D = O(Ca). In a generaltime dependent problem, D need not be proportional to E. Only in the equilibrium case is thetime evolution of D determined by E.

n = (r a())

= xr

2a

D xr2

(x D x)xr4

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We note that |n| = 1 + O(D2) as required. Since |n| = 1,

n (n )n = 0

s n = n =2

r+

6x D xr4

a

=2

a +4(x

D

x)

a3 + O(D2)

Solution using Papkovich-Neuber gives

D =5

2 + 3E 6

a

40( + 1)

(2 + 3(19 + 16)D

3.2.2 Larger Deformations in Pure Strain

However, beyond a critical capillary number the bubble breaks

However, bubbles with small values of can handle a large amount of deformation.

3.2.3 Larger Deformations for 1 (Taylor 1964)

With parameters

L R0 LR0

Ca3 Cacrit 1/6

A preliminary argument can be constructed from a scaling argument. Since 1, the shearstress exerted by the bubble on the flow can be neglected. Thus

uext Ez EL

We first start by neglecting the pressure gradients within the bubble. Moving along the bubblewith a slice of fluid, we see a hole of radius R(z), collapsing with a velocity of EzdR/dz, witha strain rate proportional to vel/R, due to the normal stress difference /R between the insideand the outside. Thus the energy is related to

E

R0 a

R0 Ca

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The volume of the bubble is approximately

R20L a3 L

R

a

R0

3 C3a

The flow inside the bubble is a Poiseuille flow, but with wall velocity Ez and no net flux.

The pressure drop along the length can be estimated from pipe flow

p

L ()EL

R20

The pressure drop must be p /R0. The viscosity ratio must then be

ER0

R0L

2 Ca6

3.3 The Rayleigh Instability

Consider a small asymmetric perturbation to a cylinder of one fluid immersed in another

If the radius is only changed slightly (|| a, |z| 1), the normal is

n = (r ()) (1, 0, z)

which is basically normalized since z is very small. The divergence of the normal is then

n = 1r

zz 1a

a2

zz

The shape is stabilized by the axial curvature zz but destabilized by the azimuthal curvature/a2. Consider disturbances of the form

= cos(kz)est n

n = 1a

1 a2k2

a2

The flow will be unstable if 1 > a2k2 (that is, for sufficiently long wave length disturbance). Thegrowth rate s depends on the dynamics. In the general case, with inertia, and 0, i, 0, i is inTomotika (PRS1935). We will consider the simple case with no inertia and no viscosity in theexternal flow. Linearized kinematic boundary conditions are used u|r=a = t = s. The dynamicboundary conditions

[ n] = ( n)n = 0 (outside)

The tangential stress is then

u

z+

w

r= 0

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The normal stress is

p0 p + 2 wr

= a

+

1 a2k2

a2

u

s

To solve for the Stokes flow inside the cylinder, harmonics functions are needed of the form cos(kz) (see Morse/Fesbach or derive by separation of variables). The Papkovich-Neuber rep-

resentation is then

= AI0(kr)cos(kz)

= (BI1(kr)cos(kz), 0, 0)

where I0, I1 are the modified Bessel function (which are well behaved at r = 0. Using I0(x) =

I1(x), (xI1) = xI0) and thus find that

u = (AkI1 + BkrI0 2BI1)cos(kz)w = (AkI0 BkrI1)sin(kz)p = 2BkI0 cos(kz)

Substituting these equations into the boundary conditions,

s = 2a

(1 k2a2) I1(ka)2k2a2[I0(ka)]2 (1 + k2a2)[I1(ka)]2

The first part of the term could have been guessed from intuition, but the second requires thefull model (which Lord Reyleigh was able to do in 1892). Plotting this function as a function ofka

Interestingly, the most unstable distrubance has infinite wavelength, because this minimizes theinternal stretching w/z and external drag is ignored (while a small effect, at long enough lengthscales it becomes important). A full model would take into account changes in the external flow.

A cylindrical bubble in a viscous fluid is most unstable at infinite wavelength because thisminimizes du/dz and ignores internal Poiseuille flow. For a more realistic model with 0 < =

B fl2

12

f12

= Bl2f2

23/2L 1

0 L 1This ties in with reversibility, as L 1 the rod behaves like a straight filament and the motionis not reversible, while when L 1 there is a nonreversible traveling wave.

Unfortunately, this does not give the flagellum shape that is observed in experiments (a

continuous wave), so Machin concluded that the flagellum must be active, rather than passiveelastic filaments (i.e. not driven by the attachment to the cell). Therefore, the flagella mustgenerate moments along its length (and this has been confirmed experimentally.

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4.6.5 Winkler Elastic Foundation Model

Consider a thin, compressible layer of a linear elastic material bonded to a rigid substrate.

This is applicable to a thin layer of cartilage attached to a bone in a joint. Imagine applyingsome pressure distribution p(x) to the surface of the elastic layer (z = 0), e.g. some king ofviscous flow over the surface layer.

We assume that p(x) has a typical value p and acts over some horizontal lengthscale lp, sowe want to know the deformation within the layer in response to the pressure. We assume that

1. hl lp (the elastic layer is thin)2. w hl where w is a typical vertical displacement (small deformations).We recall that Naviers equation for the displacement field u = (u, w), in a linearly elastic

isotropic solid.

= 0 ij = ( u)ij + 2 12

uixj

+ujxi

We take the vertical component of Naviers equation with the assumptions h?l lp and w hl

0 = ( + 2)2w

z2+ ( + )

2u

xz+

2w

z2= O

w

h2l

+ O

u

hllp

+ O

w

l2p

Usiung these order of magnitude estimates, the thinness assumption and compressibility we seethat the leading order term in the small parameter hl/lp is given by the term containing 2w

/z2

so

2w

z2= 0 w = A(x)z + B(x)

Since the base is rigid (w(z = hl) = 0)

w(x, z) = A(x)[z + hl]

So, by continuity of the vertical traction at the surface (z = 0)

p(x) = n ( n)z=0For small deformations the normal is approximately vertical so

p(x) = n ( n)z=0 = ( + 2) w

z

Solving for w

w(x, z) = p(x) hl + 2

z

hl+ 1

The real quantity of interest is the displacement of the free surface, so setting z = 0 we find that

w(x, 0) =hl

+ 2p(x) = wp(x)

which is basically an elastic spring (and w isthe Winkler stiffness compliance). Since w(x, 0) p(x), the layer behaves like a matress of linearly elastic springs.

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4.6.6 Example: Falling Cylinder Onto a Winkler Substrate (Skotheim+MahadevanPRL, 2004)

Consider a rigid cylinder translating parallel and close to a Winkler substrate in a viscous flow

We still assume a rigid body below the elastic layer and no slip conditions on the surface. Whenthere is no elastic layer, we expect no vertical/normal force from the reversibility argument forStokes flow. However, the elasticity breaks reversibility in this case, as well see. In steady state

dQ

dx= 0

Lubrication theory in a fluid (Reynolds equation) tells us

0 =

d

dx h3

12

dp

dx+

1

2Uh

Moving to dimensionless variables

H hh0

X x(2h0R)1/2

P pU(2R)1/2/h

3/20

Reynolds equation becomes

d

dx

H3

P

dx+ 6H

= 0

The gap thickness is given by

H = 1 + x2 + P U(2R)1/2

h5/20

w

We can now eliminate H and the dimensionless Reynolds equation becomes an equation just forP. We need two BCs for P, names p() = 0. Were interested in the limit of a small amountof elasticity (small deformations) so that 1 and w is small. A power expansion in is thenappropriate for P.

P(X) = P0(X) + P1(X) + . . .

At O(0

dP0dX

=A

(1 + X2)3 6

(1 + X2)2

P0 = A

3

8arctan X+

X

8

(5 + 3X2

(1 + X2)2

6

1

2arctan X+

X

2(1 + X)2

+ A

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Through the BC, A = 0 and A = 8, so the arctan terms cancel. Finally

P0 =2X

(1 + X2)2

As expected, it integrates to zero so there are no net normal forces (consistent with viscoelasticity)

and is antisymmetric. At the first order in ,

dP1dX

=B

(1 + X2)3+

24X(X2 1)(1 + X2)6

P1(X) =3

5

(3 5X2)(1 + X2)5

Since B = B = 0 from the BC. To calculate the magnitude of the normal force, we integrateover the pressure.

F

(P0 + P1 + . . . )dx =3

8 + O(2)

Where the P0 term cancels from antisymmetry. Thus, the presence of the elastic layer gives riseto a normal force, which would not exist without the elastic layer present. Using numerics wecan calculate P in terms of without the assumption of small . The result is

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Chapter 5

Handouts

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5.1 Summary of Fluid Mechanics

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5.2 Integral Representations of Stokes Flow

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5.3 2D Singularities and Flow Past a Cylinder

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5.4 The Greens Function for Stokes Flow

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5.5 Two Solutions in Lubrication Theory

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5.6 Extensional Flow Example

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5.7 Small Deformations of a Viscous Drop by UniformPure Strain E

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5.8 Example: Swimming Flagella

slow viscous flows

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