Advanced Fluid Mechanics with Vector Field Theory

Advanced Fluid Mechanics with Vector Field Theory by Dennis C. Prieve Department of Chemical Engineering Carnegie Mellon University Pittsburgh, PA 15213

An electronic version of this book in Adobe PDF® format was made available to students of 06-703, Department of Chemical Engineering, Carnegie Mellon University, Fall, 2016.

Copyright © 2016 by Dennis C. Prieve

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Table of Contents

Contents ALGEBRA OF VECTORS AND TENSORS VECTOR ADDITION VECTOR MULTIPLICATION Definition of Dyadic Product DECOMPOSITION INTO SCALAR COMPONENTS “MAGNITUDE” OF A VECTOR OR TENSOR SCALAR FIELDS GRADIENT OF A SCALAR Geometric Meaning of the Gradient Applications of Gradient CURVILINEAR COORDINATES Cylindrical Coordinates Spherical Coordinates DIFFERENTIATION OF VECTORS W.R.T. SCALARS VECTOR FIELDS Fluid Velocity as a Vector Field Time Out: Notation Involving “d” PARTIAL & MATERIAL DERIVATIVES CALCULUS OF VECTOR FIELDS GRADIENT OF A SCALAR (EXPLICIT) DIVERGENCE, CURL, AND GRADIENT Physical Interpretation of Divergence Calculation of ∇.v in R.C.C.S. Evaluation of ∇×v and ∇v in R.C.C.S. Evaluation of ∇.v, ∇×v and ∇v in Curvilinear Coordinates Physical Interpretation of Curl VECTOR FIELD THEORY

I 1 1 2 2 4 5 5 6 7 7 7 9 11 12 12 13 13 15 15 15 15 16 18 18 19 23

DIVERGENCE THEOREM The Continuity Equation Reynolds Transport Theorem Time Out: Leibnitz’ Rule Time In STOKES THEOREM Velocity Circulation: Physical Meaning DERIVABLE FROM A SCALAR POTENTIAL THEOREM III TRANSPOSE OF A TENSOR, IDENTITY TENSOR DIVERGENCE OF A TENSOR COROLLARIES OF THE DIVERGENCE THEOREM

23 23 24 24 25 25 26 27 28 28 29 29

INTRODUCTION TO CONTINUUM MECHANICS

30

CONTINUUM HYPOTHESIS CLASSIFICATION OF FORCES HYDROSTATIC EQUILIBRIUM

Copyright© 2016 by Dennis C. Prieve

30 32 33

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FLOW OF IDEAL FLUIDS EULER'S EQUATION KELVIN'S THEOREM IRROTATIONAL FLOW OF AN INCOMPRESSIBLE FLUID Potential Flow Around a Sphere d'Alembert's Paradox STREAM FUNCTION Two-D Flows Axisymmetric Flow (Cylindrical) Axisymmetric Flow (Spherical) Orthogonality of ψ=const and φ=const Streamlines, Pathlines and Streaklines Physical Meaning of Stream function PROPAGATION OF WEAK SOUND WAVES INCOMPRESSIBLE FLUIDS VISCOUS FLUIDS TENSORIAL NATURE OF SURFACE FORCES GENERALIZATION OF EULER'S EQUATION Sign Convention for Stress Resolving Total Stress into Pressure and Deviatoric Stress MOMENTUM FLUX RESPONSE OF ELASTIC SOLIDS TO NORMAL (UNIAXIAL) STRESS RESPONSE OF ELASTIC SOLIDS TO SHEAR STRESS GENERALIZED HOOKE'S LAW RESPONSE OF A VISCOUS FLUID TO PURE SHEAR GENERALIZED NEWTON'S LAW OF VISCOSITY NAVIER-STOKES EQUATION BOUNDARY CONDITIONS EXACT SOLUTIONS OF N-S EQUATIONS PROBLEMS WITH ZERO INERTIA Flow in Long Straight Conduit of Uniform Cross Section Flow of Thin Film Down Inclined Plane PROBLEMS WITH NON-ZERO INERTIA Rotating Disk* NON-NEWTONIAN BEHAVIOR ROTATING ROD IN NEWTONIAN FLUID ROTATING ROD IN NON-NEWTONIAN FLUID DIE SWELL RECOIL OF A SUDDENLY “CUT” JET TUBELESS SIPHON CREEPING FLOW APPROXIMATION CONE-AND-PLATE VISCOMETER Velocity Profile for Shallow Cones. Calculation of Torque Shear Stress and Torque for Shallow Cones. No Exact Solution for Pressure Profile FLOW AROUND A SPHERE CREEPING FLOW AROUND A SPHERE (RE→0)


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Velocity Profile Displacement of Distant Streamlines Pressure Profile CORRECTING FOR INERTIAL TERMS Perturbation Expansion WALL EFFECTS ON MOTION OF SPHERE Motion Normal to Wall Motion Tangent to Wall SPHERE ENTRAINED IN LINEAR SHEAR FLOW Faxen’s Law Effect of a Nearby Wall FLOW AROUND CYLINDER AS RE→0

86 88 89 90 91 94 94 95 95 96 96 96

SOUND PROPAGATION IN VISCOUS FLUIDS

99

BOUNDARY-LAYER APPROXIMATION FLOW AROUND CYLINDER AS RE → ∞ MATHEMATICAL NATURE OF BOUNDARY LAYERS MATCHED-ASYMPTOTIC EXPANSIONS The outer problem. The inner problem. Matching MAE’S APPLIED TO 2-D FLOW AROUND CYLINDER Outer Expansion Inner Expansion Boundary Layer Thickness PRANDTL’S B.L. EQUATIONS FOR 2-D FLOWS ALTERNATE METHOD: PRANDTL’S SCALING THEORY SOLUTION FOR A FLAT PLATE Time Out: Flow Next to Suddenly Accelerated Plate Time In: Boundary Layer on Flat Plate Boundary-Layer Thickness Drag on Plate SOLUTION FOR A SYMMETRIC CYLINDER Boundary-Layer Separation Drag Coefficient and Behavior in the Wake of the Cylinder THE LUBRICATION APPROXIMATION SLIDING BETWEEN PARALLEL PLATES SLIDING BETWEEN ALMOST-PARALLEL PLATES GENERAL ANALYSIS OF SLIDING FOR 2-D FLOW Special Case #1: Slightly Nonparallel Plates Special Case #2: Two Parallel Plates PERFORMANCE OF A JOURNAL BEARING SQUEEZING FLOW REYNOLDS EQUATION Example #1: 2-D Sliding Motion Example #2: Squeezing Flow between Disk and Plate Electrokinetic Lift Example #3: Sliding between Cylinder and Plate Cavitation Example #4: Sliding between Sphere and Plate TURBULENCE Copyright© 2016 by Dennis C. Prieve

102 102 102 105 106 106 108 108 109 109 112 113 115 117 118 119 121 122 123 124 126 128 128 129 129 131 132 133 134 137 141 141 142 144 145 146 148 PDF generated December 6, 2016

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GENERAL NATURE OF TURBULENCE TURBULENT FLOW IN PIPES TIME-SMOOTHING TIME-SMOOTHING OF CONTINUITY EQUATION TIME-SMOOTHING OF THE NAVIER-STOKES EQUATION ANALYSIS OF TURBULENT FLOW IN PIPES PRANDTL’S MIXING LENGTH THEORY PRANDTL’S “UNIVERSAL” VELOCITY PROFILE LAMINAR SUBLAYER PRANDTL’S UNIVERSAL LAW OF FRICTION ELECTROSTATICS OF CONTINUA COULOMB’S LAW GENERALIZATION OF COULOMB’S LAW TO MANY POINT CHARGES GAUSS'S LAW: INTEGRAL FORM CONTINUOUS DISTRIBUTION OF CHARGE POLARIZATION OF DIELECTRICS Electronic Polarization Orientational Polarization Polarization of Continua BOUNDARY CONDITIONS AT A CHARGED INTERFACE ELECTROHYDRODYNAMICS ORIGIN OF CHARGE ELECTROSTATIC FORCES IN WATER VS. COULOMB’S LAW GOUY-CHAPMAN MODEL OF DOUBLE LAYER Boltzmann’s Equation: Thermodynamic Derivation Boltzmann’s Equation:Transport Derivation Special Case #1: Small Potentials. Special Case #2: Symmetric Binary Electrolyte HYDROSTATIC EQUILIBRIUM ELECTROKINETIC PHENOMENA SMOLUCHOWSKI'S ANALYSIS ELECTRO-OSMOSIS IN CYLINDRICAL PORES ELECTROPHORESIS OF LARGE SPHERES ELECTROPHORESIS OF SMALL PARTICLES ELECTROPHORESIS OF A SPHERE OF ARBITRARY SIZE ANALYTICAL APPROXIMATION FOR LARGE PARTICLES STREAMING POTENTIAL

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SURFACE TENSION DRIVEN FLOWS

191

SURFACE TENSION EXPANDING AREA OF SOAP FILMS LAPLACE PRESSURE CONTACT ANGLE CAPILLARY RISE MARANGONI MOTION OF BUBBLES

191 191 192 194 194 194

APPENDIX I: USEFUL IDENTITIES IN VECTOR NOTATION

195

APPENDIX II: VECTOR OPERATIONS IN RECTANGULAR COORDINATES (X, Y, Z)

196

APPENDIX III: VECTOR OPERATIONS IN CYLINDRICAL COORDINATES (R, θ, Z)

198

APPENDIX IV: VECTOR OPERATIONS IN SPHERICAL COORDINATES (R, θ, φ)

200



06-703 APPENDIX V: NEWTON’S LAW OF VISCOSITY IN CARTESIAN COORDINATES IN CYLINDRICAL COORDINATES IN SPHERICAL COORDINATES APPENDIX VI: NAVIER-STOKES EQUATIONS IN CARTESIAN COORDINATES IN CYLINDRICAL COORDINATES IN SPHERICAL COORDINATES APPENDIX VII: MOVING REFERENCE FRAME TABLE OF CONTENTS


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Algebra of Vectors and Tensors Whereas heat and mass are scalars, fluid mechanics concerns transport of momentum, which is a vector. Heat and mass fluxes are vectors, momentum flux is a tensor. Consequently, the mathematical description of fluid flow tends to be more abstract and subtle than heat and mass transfer. In an effort to make the student more comfortable with the mathematics, we will start with a review of the algebra of vectors (which should be familiar) and an introduction to tensors and dyads (which is probably new).

1

Fall, 2016 coincides with head of a (see above).♠ A brief review of vector addition and multiplication can be found in Greenberg,♣ pages 132-139. Vector Multiplication Having learned how vectors add, let’s talk about multiplication of two vectors.

Scalar - a quantity having magnitude but no direction (e.g. temperature, density) Vector - (a.k.a. 1st rank tensor) a quantity having magnitude and direction (e.g. velocity, force, momentum)♥ (2nd rank) Tensor - a quantity having magnitude and two directions (e.g. momentum flux, stress) In case you are wondering how any physical quantity could have two directions associated with it, let me give you a sneak preview of what we will eventually develop in these Notes: the two directions associated with momentum flux are 1) the direction of the momentum being transported (momentum is a vector) and 2) the direction in which that momentum is being transported. For example, you could have xmomentum being transported in the y-direction; hence momentum flux involves two independent directions.

Given two arbitrary vectors a and b, three types of vector products are defined: Dot Product Cross Product Dyadic Product

Notation a.b a×b ab

Result scalar vector tensor

Definition ab cosθ absinθn —

where θ is an interior angle (0 ≤ θ ≤ π) and n is a unit vector which is normal (perpendicular) to both a and b. In the figure above, this means n is normal to plane of the page, which contains both vectors. Of course, there are two unit vectors which are normal to this plane: one pointing into the page and another pointing out of the page. The sense of n is determined from the “right-hand-rule:”

Vector Addition So let’s now review some of the algebra of vectors, starting with addition of two vectors.

Given two arbitrary vectors a and b, by a+b we mean the vector formed by connecting the tail of a to the head of b when b is moved such that its tail

♥Mathematicians define “vector” as an element in a vector space, which is a set satisfying about 10 rules. While this broader definition includes what we listed above, it also includes functions like the Legendre polynomials defined on an infinite domain, which are also “vectors” according to this broader definition.


Right-hand rule: with the fingers of the right hand initially pointing in the direction of the first vector (u), rotate the fingers to point in the direction of the

♠ The origin of a vector (i.e. the location of its tail) is not part of the definition of the vector. Thus vectors can always be moved around the page without changing the vector (provided orientation and length are preserved). ♣ Greenberg, M.D., Foundations Of Applied Mathematics, Prentice-Hall, 1978.


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second vector (v); the thumb then points in the direction with the correct sense. In the above definitions, we denote the magnitude (or length) of vector a by the scalar a. Boldface will be used to denote vectors and italics will be used to denote scalars. Second-rank tensors will be denoted with double-underlined boldface; e.g. tensor T. Definition of Dyadic Product Reference: Appendix B from Happel & Brenner.♥

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9. sab = (sa)b = a(sb) Decomposition into Scalar Components Three vectors (say e1, e2, and e3) are said to be linearly independent (LI) if none can be expressed as a linear combination of the other two (e.g. i, j, and k which are the three unit vectors in Cartesian coordinates). Given such a set of three LI vectors, any vector (belonging to E3)♣ can be expressed as a linear combination of this basis:

This is the first concept which is new for nearly everyone. The word “dyad” comes from Greek: “dy” means two while “ad” means adjacent. Thus the name dyad refers to the way in which this product is written: the two vectors are written adjacent to one another with no space or other operator in between: for example ab.

where the vi are called the scalar components of v. Usually, for convenience, we choose orthonormal vectors as the basis:

There is no geometrical picture that I can draw which will explain what a dyadic product is. It's best to think of the dyadic product as a purely mathematical abstraction having some very useful properties:

although this is not necessary. Kronecker delta.

Dyadic Product ab - that mathematical entity which satisfies the following properties (where a, b, v, and w are any four vectors): 1. ab.v = a(b.v) [which has the direction of a; note that ba.v = b(a.v) which has the direction of b. Thus ab ≠ ba since they don’t produce the same result on post-dotting with v.] 2. v.ab = (v.a)b [thus v.ab ≠ ab.v] 3. ab×v = a(b×v) which is another dyad 4. v×ab = (v×a)b 5. ab:vw = (a.w)(b.v) which is sometimes known as the inner-outer product or the double-dot product.* 6. a(v+w) = av+aw (distributive for addition) 7. (v+w)a = va+wa 8. (s+t)ab = sab+tab (distributive for scalar multiplication--also distributive for dot and cross product) ♥ Happel, J., & H. Brenner, Low Reynolds Number Hydrodynamics, Noordhoff, 1973. * Brenner defines this as (a.v)(b.w). Although the two definitions are not equivalent, either can be used -- as long as you are consistent. In these notes, we will adopt the definition in the body of the text above and ignore Brenner's definition.


v = v1e1 + v2e2 + v3e3

1 if i = j ei.ej = δij =  0 if i ≠ j δij is called the

Just as the familiar dot and cross products can written in terms of the scalar components, so can the dyadic product: vw = (v1e1+v2e2+v3e3)(w1e1+w2e2+w3e3) Distributing the product of the sums: vw = (v1e1)(w1e1)+(v1e1)(w2e2)+(v1e1)(w3e3)+ (v2e2)(w1e1)+(v2e2)(w2e2)+(v2e2)(w3e3)+ (v3e3)(w1e1)+(v3e3)(w2e2)+(v3e3)(w3e3) Changing the order of scalar multiplication while preserving the order of vector multiplication: vw = v1w1e1e1+v1w2e1e2+v1w3e1e3+ v2w1e2e1+v2w2e2e2+v2w3e2e3+ v3w1e3e1+v3w2e3e2+v3w3e3e3 where the eiej are nine distinct unit dyads. We have applied the definition of dyadic product to perform these two steps: in particular items 6, 7 and 9 in the numbered list above. More generally any nth rank tensor (in E3) can be expressed as a linear combination of the 3n unit nads. For example, if n=2, 3n=9 and an n-ad is a dyad. Thus a general second-rank tensor can be

♣ “E3” is shorthand notation for three-dimensional Euclidean space (the particular set of vectors used in these notes). “Euclidean” implies that the magnitude of the vector is calculated in a particular way.


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decomposed as a linear combination of the 9 unit dyads:

= T T11e1e1 + T12 e1e2 += 

3

3

∑ ∑ Tij ei e j

=i 1 =j 1

To facilitate the notation, we have used numerical subscripts to denote the basis vectors and scalar components. In Cartesian coordinates the basis vectors are sometimes denoted by i, j and k. Usually the following relationships are implied: e1 = i,

e2 = j

and

e3 = k

Although a dyad (e.g. vw) is an example of a secondrank tensor, not all 2nd rank tensors T can be expressed as a dyadic product of two vectors. Thus the set of all dyads is a subset of all tensors:

For linear shear flow between two infinite parallel plates (see figure above), the stress tensor is given by = T

µU ex e y + e y ex − p ex ex + e y e y + ez ez L

) (

(

)

where µ is the viscosity of the fluid and p is the local pressure.♦ Problem: Calculate the force per unit area exerted by the fluid on the lower plate. Solution: A unit vector normal to the lower plate is n = ey. Applying our formula:

 µU = n.T e y . ex e y + e y ex  L

(

(

)

)

− p ex ex + e y e y + ez ez   To see why, note that a general second-rank tensor has nine scalar components which need not be related to one another in any way. By contrast, the 9 scalar components of dyadic product above involve only six distinct scalars (the 3 components of v plus the 3 components of w). After a while you get tired of writing the summation signs and limits. So an abbreviation was adopted whereby repeated appearance of an index implies summation over the three allowable values of that index: ♣ T = Tijeiej This is sometimes called the Cartesian (implied) summation convention. EXAMPLE: Much later in the course (see page 56), we will show that the local force per unit area exerted by the fluid on a solid surface is given by n.T, where T is the total stress tensor and n is the unit vector normal to the surface pointing out of the solid.

Distributing the dot product over each of the dyads, preserving the order of vector multiplication:

= n.T

(

µU e y .e x e y + e y .e y e x L

(

− p e y .e x e x + e y .e y e y + e y .e z e z


)

Applying the definition of dyadic product by first performing the dot operation on the two vectors surrounding the dot

µU  e y .e x e y + e y .e y e x    L   1  0  − p  e y .e x e x + e y .e y e y + e y .e z e z       1 0  0 

= n.T

Evaluating the dot products using the orthonormal property of the basis vectors, we finally obtain

(n.T= )lower plate

♣ A vertical bar in the left margin denotes material which (in the interest of time) will be omitted from the lecture.

)

µU e x − pe y L

♦ This expression for the viscous part of T is obtained by substituting the velocity profile (vx = Uy/L and vy = vz = 0) into Newton’s law of viscosity (see p8 of Appendices.pdf). The derivation of NLV and its use in examples like this will be covered later.


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If we repeat this calculation for the upper plate, the only thing which changes is the outward normal, which now points downward: n = –ey. The result is µU − e x + pe y ( n.T )upper = L

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remaining vector. Recognizing that the dot product is zero for all terms in which i ≠ j, we replace i by j wherever it appears and drop the sum over i:

plate

The point of the above example is to show how Rules 1-9 associated with the definition of dyad can be used to calculate the direction as well as the magnitude of the force exerted by the fluid on a solid. Next, we present an example which illustrates the formulas for calculating the result of one multiplication operation using scalar components. EXAMPLE: problem: show that the dot product of an arbitrary vector v and an arbitrary tensor τ can be calculated from their scalar components according to the formula: 3

3

Solution: First we substitute the component representations of the vector and tensor, making sure each index is given a distinct name:    3 3 v.τ  ∑ vi ei . ∑ ∑ τ jk e j ek =    = i 1  =j 1 =k 1 3

   

The next step is to distribute the multiplication over the sums, so instead of the sum of 3 terms dotted with the sum of 9 terms, we have the sum of 27 dot products: 3

3

3

∑ ∑ ∑ vi τ jk ei .e j ek

k 1 =i 1 =j 1 = def'n of dyad 3

≡

3

3

.e ) e (ei ∑ ∑ ∑ vi τ jk  j k 

k 1 =i 1 =j 1 =

δij

Although in the first equation above we have changed the order of scalar multiplication (which doesn’t matter), we have preserved the original order of the vectors in the dot product. In the second equation above, we employ one of the properties of the dyad: namely, that when you dot a vector with a dyad, we first perform the dot-product on the two vectors on either side of the dot, then multiply by the Copyright© 2016 by Dennis C. Prieve

The names of the indices are not important and can be changed, as long as all appearances of a given name are changed consistently:

= v.τ

3

3

∑ ∑ vi τij e j

=i 1 =j 1

“Magnitude” of a Vector or Tensor While other definitions are sometimes used, the most common definition for the magnitude (or norm) of a vector is v=

v.v =

vx2 + v 2y + vz2

which is the usual Euclidean length. For a secondrank tensor T, the most common definition for magnitude is tr ( T ) ≡ T:I = Txx + Tyy + Tzz

∑ ∑ vi τij e j

=i 1 =j 1

= v.τ

3

∑ ∑ v j τ jk ek

=j 1 = k 1

Note that the shear stress exerts a force parallel to the direction of main flow (x), dragging the lower plate in the direction of flow and opposing the motion of the upper plate; the pressure acts to push the plates apart.

= v.τ

3

= v.τ

which is called the trace of the tensor. In the above expression I is the identity tensor or unit tensor which — in Cartesian coordinates — is given by I = ii + jj + kk

We offer a more general definition and computational formula for I below (see page 28). Of course, the trace of a tensor involves only the diagonal components Tii of the matrix of coefficients of the unit dyads; it does not depend on the offdiagonal components Tij (where i≠j). The trace is one of three scalar invariants of any second-rank tensor quantity. These invariants arise in the determination of the three eigenvalues associated with any 3×3 matrix of coefficients. You might recall that the eigenvalue λ is defined as that value which gives zero value for the following determinant: T − λI = 0

(1)

where T and I are matrices (not tensors) of the coefficients of unit dyads in the tensors T and I. To distinguish tensors from matrices, you should realize that the tensor T represents a physical quantity which does not depend on the coordinate system used to describe it, whereas the matrix of coefficients does depend on those coordinates. Denoting the three PDF generated December 6, 2016

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invariants as I1, I2 and I3, the determinant in (1) for a 3×3 matrix can written as

λ3 − I1λ 2 + I 2 λ − I3 = 0 which is a cubic equation in λ. In general, there will be three roots to any cubic (although some of the roots might be repeated). Denote these roots as λ1,

λ2 and λ3. The three invariants are given by 3

∑ Tkk

I1 =

k =1

= λ1 + λ 2 + λ3

2   3 3  1  3  ∑ Tkk  − ∑ ∑ Tij T ji    2  =k 1  =i 1 =j 1   = λ1λ 2 + λ 2 λ3 + λ3 λ1

= I2

I3 = T =λ1λ 2 λ3

and Scalar Fields

The scalar field can be expressed as a function of a vector argument, representing position, instead of a set of three scalars: f = f (r) Consider an arbitrary displacement away from the point r, which we denote as dr to emphasize that the magnitude dr of this displacement is sufficiently small that f (r) can be linearized as a function of position around r. Then the total differential can be written as df = f (r + dr ) − f (r )

Suppose I have some scalar function of position (x,y,z) which is continuously differentiable, that is f = f (x,y,z) and ∂f /∂x, ∂f /∂y, and ∂f /∂z exist and are continuous throughout some 3-D region in space. This function is called a scalar field. Now consider f at a second point which is differentially close to the first. The difference in f between these two points is called the total differential of f : f (x+dx,y+dy,z+dz) - f (x,y,z) ≡ df For any continuous function f (x,y,z), df is linearly related to the position displacements, dx, dy and dz. That linear relation is given by the Chain Rule of differentiation: df =

∂f ∂f ∂f dx + dy + dz ∂x ∂y ∂z

Instead of defining position using a particular coordinate system, we could also define position using a position vector r: r = xi + yj + zk

Gradient of a Scalar We are now is a position to define an important vector associated with this scalar field. The gradient (denoted as ∇f ) is defined such that the dot product of it and a differential displacement vector gives the total differential: df ≡ dr.∇f

(2)

Lecture #2 begins here EXAMPLE: Obtain an explicit formula for calculating the gradient in Cartesian* coordinates. Solution:

r = xi + yj + zk

r+dr = (x+dx)i + (y+dy)j + (z+dz)k subtracting:

dr = (dx)i + (dy)j + (dz)k

*Named after French philosopher and mathematician René Descartes (1596-1650), pronounced "day-cart", who first suggested plotting f (x) on rectangular coordinates



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6 ∇f = (∇f)xi + (∇f)yj + (∇f)zk

EXAMPLE: Suppose the steady state temperature profile in some heat conduction problem is given by:

dr.∇f = [(dx)i + ...].[(∇f)xi + ...] df = (∇f)xdx + (∇f)ydy + (∇f)zdz

T(x,y,z) = x2 + y2 + z2 (3)

Perhaps we are interested in ∇T at the point (3,3,3) where T=27. ∇T is normal to the T=const surface:

(4)

x2 + y2 + z2 = 27

Using the Chain rule: df = (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz

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According to the definition of the gradient, (3) and (4) are identical. Equating them and collecting terms:

which is a sphere of radius

27 .

[(∇f)x–(∂f/∂x)]dx + [(∇f)y–(∂f/∂y)]dy + [(∇f)z–(∂f/∂z)]dz = 0 Think of dx, dy, and dz as three independent variables which can assume an infinite number of values, even though they must remain small. The equality above must hold for all values of dx, dy, and dz. The only way this can be true is if each individual term separately vanishes:** So (∇f)x = ∂f/∂x, (∇f)y = ∂f/∂y, and (∇f)z = ∂f/∂z, leaving

∂f ∂f ∂f ∇f= i+ j+ k ∂x ∂y ∂z

Proof of 1). Let's use the definition (2) to show that these geometric meanings are correct. df = dr.∇f

(5)

Other ways to denote the gradient include: ∇f = gradf = ∂f /∂r

(6)

Geometric Meaning of the Gradient Formula (5) is sometimes given as the definition for gradient ∇f instead of (2). Indeed one can use (5) to derive the formula for ∇f in other coordinate systems. However it is hard to infer the geometric significance of ∇f from (5). In the following paragraphs, we will infer the geometric significance of ∇f from (2). 1. direction: ∇f (r) is normal to the f =const surface passing through the point r in the direction of increasing f. ∇f also points in the direction of steepest ascent of f. 2. magnitude: |∇f | is the rate of change of f with distance along this direction What do we mean by an "f =const surface"? Consider an example.

** For any particular choice of dx, dy, and dz, we might obtain zero by cancellation of positive and negative terms. However a small change in one of the three without changing the other two would cause the sum to be nonzero. To ensure a zero-sum for all choices, we must make each term vanish independently.


Consider an arbitrary f. A portion of the f =const surface containing the point r is shown in the figure at right. Choose a dr which lies entirely on f =const. In other words, the surface contains both r and r+dr, so f (r) = f (r+dr) and

df = f (r+dr)–f (r) = 0

Substituting this into the definition of gradient: df = 0 = dr.∇f = dr∇fcosθ Since dr and ∇f are in general not zero, we are forced to the conclusion that cosθ=0 or θ=90°. This means that ∇f is normal to dr which lies in the surface. 2) can be proved in a similar manner: choose dr to be parallel to ∇f. Does ∇f point toward higher or lower values of f?


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Applications of Gradient •

determine a normal to some surface (needed to apply boundary condition’s like n.v = 0 for a boundary which is impermeable)

•

determine the rate of change along some arbitrary direction: if n is a unit vector pointing along some path, then

Fall, 2016

r = (x2+y2)1/2 θ = tan–1(y/x) z=z

x = rcosθ y = rsinθ z=z

∂f n.∇f = ∂s

is the rate of change of f with distance (s) along this path given by n. ∂f ∂s is called the directed derivative of f. Note that the largest value of the derivative is obtained by choosing n to be in the same direction as ∇f ; thus we can also •

find a vector pointing in the direction of steepest ascent of some scalar field

Above is another attempt at rendering the 3-D view defining the cylindrical coordinates.

Curvilinear coordinates In principle, all problems in fluid mechanics and transport could be solved using Cartesian coordinates. Often, however, we can take advantage of symmetry in a problem by using another coordinate system. This advantage takes the form of a reduction in the number of independent variables (e.g. partial differential equation becomes ordinary differential equation). A familiar example of a non– Cartesian coordinate system is cylindrical coordinates: When choosing which coordinate system to use in a given problem, we are interested in the shape of coordinate = constant surfaces. For example, r = constant surfaces are cylinders (see figure above), while θ = constant surfaces are planes passing through the z-axis. z = constant surfaces are also planes just as in Cartesian coordinates. Vectors are decomposed differently. Instead of writing a vector in R.C.C.S.♠ as v = vxi + vyj + vzk Cylindrical Coordinates Position is denoted by the three scalars (r, θ, z) instead of (x, y, z). The three new coordinates are defined by the figure above and are related to the three old coordinates by:

in cylindrical coordinates, we write v = vrer + vθeθ + vzez

♠ R.C.C.S. stands for rectangular Cartesian coordinate system.



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where er, eθ, and ez are new unit vectors pointing in the r, θ and z directions as shown in the figure below (eθ is not shown but is orthogonal to both er and ez and points in the direction of increasing θ):

Fall, 2016

The unit vectors in cylindrical coordinates (er, eθ and ez) are related to the unit vectors in Cartesian coordinates by

er = ( cos θ ) e x + ( sin θ ) e y

(8)

eθ = ( − sin θ ) e x + ( cos θ ) e y

(9)

ez = ez Or we can turn these equations inside-out to obtain e= x

( cos θ ) er + ( − sin θ ) eθ

e y = ( sin θ ) er + ( cos θ ) eθ ez = ez The position vector in cylindrical coordinates is given by♣ p = rer(θ) + zez

(7)

Notice the absence of any θ-component to this vector. Nonetheless you can confirm from the figure above that this vector sum points to any arbitrary point whose Cartesian coordinates are x,y,z. The θdependence is embedded in the unit vector er who orientation depends on θ. We also have a different set of nine unit dyads for decomposing tensors: erer, ereθ, erez, eθer, etc.

In Hwk #2, we will show that the gradient in cylindrical coordinates is given by f ∇ =

∂f ∂f 1 ∂f er + eθ + ez ∂r r ∂θ ∂z

This serves as the analog to formula (5), but in cylindrical coordinates. A common boundary condition in fluid flow problems arises because solid objects are impenetrable to the fluid. At the boundary, the fluid velocity v must satisfy n.v = 0 where n is a unit vector normal to the boundary at any point. The next example shows now to obtain an expression for n in a particular coordinate system. EXAMPLE: taken from an old exam.

Like the Cartesian unit vectors, the unit vectors in cylindrical coordinates also form an orthonormal set of basis vectors for E3. Unlike Cartesian unit vectors, the orientation of er and eθ depend on position. In other words: er = er(θ)

and

eθ = eθ(θ)

When any of the unit vectors is position dependent, we say the coordinates are: curvilinear - at least one of the basis vectors is position dependent This will have some profound consequences which we will get to shortly.

♣ Recalling that p = xi + yj + zk in Cartesian coordinates, you might be tempted to write p = rer + θeθ + zez in cylindrical coordinates. Of course, this temptation gives the wrong result (in particular, the units of length in the second term are missing).


A paraboloid (a curved surface) is formed by rotating a parabola around the z-axis as shown above. In cylindrical coordinates, the surface of the paraboloid is given by

z ( r, θ) = δ +

r2 2R

(10)

The third coordinate, θ (not shown in figure) is an angle measured in the z=0 plane between an arbitrary


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Fall, 2016

rz plane and the xz plane (θ=0 is the plane of the page to the right of the z-axis). Problem: find any vector which is normal to the paraboloid above the point (r,θ) = (R,0). Solution: we construct a function f which is constant on the surface and compute its gradient. Incorrect solution (submitted by about a third of the class):

f = z ( r, θ) = δ +

choose:

r2 2R

r ∇f = e r R

Evaluating this result at r=R ∇f = er

Inspection of the drawing (where a unit outwardpointing normal is denoted by n) clearly shows that this result is not in the direction of the normal. The problem with this solution is that the surface of the paraboloid is not described by z(r) = const. Correct solution: solve (10) for the constant δ:

f ( r , θ, z ) = z−

r2 = δ 2R

One correct answer turns out to be the gradient of this function: ∇f =−

r=R r er + e z =− er + e z R

or any scalar multiple of this vector, like er − e z which is the n shown in the figure. Spherical Coordinates Spherical coordinates are related to Cartesian coordinates by x= r sin θ cos φ

+ x2 + y2 + z2 r=

(

= y r sin θ sin φ

= θ tan −1

= z r cos θ

= φ tan −1 ( y x )

x2 + y2 z

)

Above is another attempt to render the 3D image. The green surface is the xy-plane, the red surface is the xz-plane, while the blue-green surface is the yzplane. These three planes intersect at the origin (0,0,0), which lies deeper into the page than (1,1,0). The blue quarter-disk is a φ = constant surface. The straight red line, drawn from the origin to the point (r,θ,φ)♣ has length r, The angle θ is the angle the red line makes with the z-axis (the red circular arc labelled θ has radius r and is subtended by the angle θ). The angle φ (measured in the xy-plane) is the angle the second blue plane (actually it’s one quadrant of a disk) makes with the xy-plane (red). This plane which is a quadrant of a disk is a φ=const surface: all points on this plane have the same φ coordinate. The second red (circular) arc labelled φ is also subtended by the angle φ.

♣ This particular figure was drawn using r = 1, θ = π/4 and φ = π/3.



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These unit vectors are shown in the figure above. Notice that the surface φ=const is a plane containing the point r itself, the projection of the point onto the xy-plane and the origin. The unit vectors er and eθ lie in this plane as well as the Cartesian unit vector k (sometimes denoted ez). The unit vectors in spherical coordinates are related to the unit vectors in Cartesian coordinates by

er = Of course, r = constant surfaces are spheres. One such sphere (r=1) is shown in the figure above. φ=const surfaces are planes which intersect along the z-axis, which also passes through the center of the sphere.

e= θ

( sin θ cos φ ) e x + ( sin θ sin φ ) e y + ( cos θ ) e z ( cos θ cos φ ) e x + ( cos θ sin φ ) e y + ( − sin θ ) e z eφ = ( − sin φ ) e x + ( cos φ ) e y

or their inverse relationships:

e x = ( sin θ cos φ ) er + ( cos θ cos φ ) eθ + ( − sin φ ) eφ

ey =

( sin θ sin φ ) er + ( cos θ sin φ ) eθ + ( cos φ ) eφ e= z

( cos θ ) er + ( − sin θ ) eθ

(11)

EXAMPLE: decompose the unit vector k (also known as ez) into its spherical components.

Finally, θ = constant surfaces are cones. Shown above is the θ = π/8 surface (innermost), θ = π/4, θ = 3π/8 and finally θ = π/2 which is the green disk which coincides with the xy plane.

Solution: Note (from the figure below) that k, er and eθ all lie in the same plane, which is a φ=const surface having eφ as its normal.

The position vector in spherical coordinates is given by p = xi+yj+zk = rer(θ,φ) In this case all three unit vectors depend on position: er = er(θ,φ), eθ = eθ(θ,φ), and eφ = eφ(φ) where er is the unit vector pointing the direction of increasing r, holding θ and φ fixed; eθ is the unit vector pointing the direction of increasing θ, holding r and φ fixed; and eφ is the unit vector pointing the direction of increasing φ, holding r and θ fixed. Copyright© 2016 by Dennis C. Prieve

If we tilt this φ=const plane into the plane of the page (as in the sketch above), we can more easily see the relationship between these three unit vectors. We project ez onto er by drawing the line perpendicular to er. This forms a right triangle with k as the PDF generated December 6, 2016

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hypotenuse, whose length is unity. The lengths of the other two sides are sinθ and cosθ. We then can produce the vector k by adding a vector of length cosθ having direction er to a second vector of length sinθ having direction –eθ; thus k=

( cos θ ) er − ( sin θ ) eθ

Fall, 2016

(

= ∆er 2sin ∆θ e θ + ∆θ 2 θ 2

Recalling that sinx tends to x as x→0, we have lim {∆er } = ∆θ eθ ( θ )

∆θ→0

Dividing this by ∆θ, we obtain the derivative:

This is the same result given by (11).

de r = eθ dθ

Differentiation of Vectors w.r.t. Scalars Suppose we have a vector v which depends on the scalar parameter t: v = v(t) For example, the velocity time. What do we mean vector with respect to Fundamental Theorem of derivative as:

)

This result can also be obtained by differentiating (8) and comparing the result with (9). Similarly,

of a satellite depends on by the “derivative” of a a scalar. As in the Calculus, we define the

deθ = −er dθ

Lecture #3 begins here

lim  v(t +∆t )-v(t )  dv =   ∆t →0  dt ∆t  Note that dv/dt is also a vector. In curvilinear coordinates, the orientation of the unit vectors varies from point to point. This has some profound consequences, which will be demonstrated in the next two examples. EXAMPLE: coordinates.

Compute

der/dθ

in

cylindrical

Solution: From the definition of the derivative:

der dθ

lim  er (θ + ∆θ) − er (θ)  lim  ∆er  =     ∆θ→0  ∆θ  ∆θ→0  ∆θ 

One important application of “differentiation with respect to a scalar” is the calculation of velocity, given position as a function of time. In general, if the position vector is known, then the velocity can be calculated as the rate of change in position: p = p(t)

v=

dp dt

Similarly, the acceleration vector a can be calculated as the derivative of the velocity vector v:

a=

dv dt

EXAMPLE: Given the trajectory of an object in cylindrical coordinates Since the location of the tail of a vector is not part of the definition of a vector, let's move both vectors to the origin (keeping the orientation fixed). Using the parallelogram law, we obtain the difference vector. Its magnitude is:

∆er = 2sin ∆θ 2

r = r(t), θ = θ(t), and z = z(t) Find the velocity of the object. Solution: According to (7), the position vector in cylindrical coordinates is p(r,θ,z) = rer(θ) + zez

Its direction is parallel to eθ(θ+∆θ/2), so: Copyright© 2016 by Dennis C. Prieve


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Fall, 2016 would obtain the velocity.♦ A mathematical “tracer particle” is called a “material point:”♥ Material point - an ideal tracer particle or fluid element - a given set of fluid molecules whose location may change with time.♣ Suppose the trajectory of a material point is given by: r = r(t) Then the fluid velocity at any time is v= r t

()

Now we can apply the Chain Rule:

 ∂p   ∂p   ∂p  = dp   dr +   d θ +   dz  ∂r θ, z  ∂θ r , z  ∂z r ,θ          er

r

der ( θ )  dθ 

ez

= er dr + reθ d θ + e z dz Dividing by dt, we obtain the velocity: d θ (t ) dz ( t ) dp dr ( t ) v= =er + r eθ + ez dt  dt dt dt    vθ

vz

(12)

A second way to measure fluid velocity is similar to the “bucket-and-stopwatch method.” We measure the volume of fluid crossing a surface per unit time:

 ∆q  n.v = lim   ∆a →0  ∆a 

eθ

vr

dr ( t ) r ( t + dt ) − r ( t ) = dt dt

(13)

where ∆a is the area of a surface element having a unit normal n (see figure below) and ∆q is the volumetric flowrate of fluid crossing ∆a in the direction of n. In this case, the mathematical surface having area ∆a is stationary.

Vector Fields A vector field is defined just like a scalar field, except that it's a vector. Namely, a vector field is a position-dependent vector: v = v(r)♠ Common examples of vector fields include force fields, like the gravitational force or an electrostatic force field. Of course, in this course, the vector field of greatest interest is the fluid velocity: Fluid Velocity as a Vector Field Consider steady flow around a submerged object. What do we mean by “fluid velocity?” There are two ways to measure fluid velocity. First, we could add tracer particles to the flow and track the position of the tracer particles as a function of time; differentiating position with respect to time, we

♠ We will use r (instead of p) to denote the position vector in situations in which there is no chance of confusing its magnitude r with the radial coordinate in either cylindrical or spherical coordinates.


When ∆a is small enough so that this quotient has converged in a mathematical sense and ∆a is small enough so that the surface is locally planar so we can ♦ Actually, this only works for steady flows. In unsteady flows, pathlines, streaklines and streamlines differ (see Streamlines, Pathlines and Streaklines on page 45). ♥ By contrast, a real tracer particle might not move with the same velocity as the fluid. For example, when a rigid sphere is within a few of its radii of the wall, it will generally lag behind the fluid (see Sphere Entrained in Linear Shear Flow on page 88). ♣ In a molecular-level description of gases or liquids, even nearby molecules have widely different velocities which fluctuate with time as the molecules undergo collisions. We will reconcile the molecular-level description with the more common continuum description in Introduction to Continuum Mechanics on page 29. For now, we just state that “location of a material point” means the location of the center of mass of a set of nearby molecules. The “point” needs to contain a statistically large number of molecules so that r(t) converges to a smooth continuous function.


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Fall, 2016

denote its orientation by a unique unit normal n, we say that ∆a is differential and denote its value as da; Similarly, the corresponding ∆q is denoted dq and rewrite this definition as:

what do we mean by the derivative with respect to time t? In general, we might define the time derivative this way:

dq n.v = da

(14)

dq = n.v da

 f ( r + ∆r, t + ∆t ) − f ( r, t )  df = lim   dt ∆t →0  ∆t 

(15)

or

This is particularly convenient to compute the volumetric flowrate across an arbitrary curved surface, given the velocity profile. We just have to sum up the contribution from each surface element: Q = ∫ n.v da

(16)

A

generic derivative:

where df is the “total differential” arising from changes in both position and time (see page 5). The trajectory r = r(t) is called the “path” along which the derivative is evaluated and ∆r ≡ r(t+∆t) – r(t)

Once again, if ∆t is small enough so this quotient converges, then we denote ∆t as dt and the total change in f as df ; then rewrite the definition as

f ( r + dr, t + dt ) − f ( r, t ) df = dt dt where

where one choice of the surface A is shown as the dashed lined in the figure above. Time Out: Notation Involving “d” The integral in (16) might need some explanation and further elaboration. In (16) we are NOT integrating a function of a “with respect to a”, which is the type of integral that is most familiar to students. Instead, the integral in (16) represents a surface integral. “da” represents a differential element of surface area. The “A” written under the integral denotes the set of points constituting the macroscopic surface (possibly curved) which serves as the domain of integration. When (16) is eventually evaluated in some coordinate system [see (22) below for an example], da will be expressed in terms of two differential changes in coordinates: for example, da = dx dy in RCCS or da = r dθ dz in cylindrical coordinates. The surface integral itself becomes a double integral. The domain A translates into the limits of integration over x and y or over θ and z. Partial & Material Derivatives Let

f = f (r,t)

represent some unsteady scalar field (e.g. the unsteady temperature profile inside a moving fluid). If the position can also depend on time, i.e. r = r(t), Copyright© 2016 by Dennis C. Prieve

dr = r(t + dt) – r(t)

There are two special paths for which the time derivative is given a special name. In the example in which f represents temperature, these two time derivatives correspond to the rate of change measured with a thermometer which either is held stationary in the moving fluid or drifts along with the local fluid. partial derivative — rate of change at a fixed spatial point: ∂f  df  = = ∂t  dt dr =0

f ( r, t + dt ) − f ( r, t ) dt

In other words, there is no displacement in position during the time interval. As time proceeds, different material points occupy the spatial point r. Another name for the partial derivative is the Eulerian derivative. material derivative — rate of change within a particular material point (whose spatial coordinates vary with time): Df  df  = = Dt  dt dr = vdt

f ( r + vdt , t + dt ) − f ( r, t ) dt

where v denotes the local fluid velocity. As time proceeds, the moving material occupies different spatial points, so r is not fixed. In other words, we are following along with the fluid as we measure the rate of change of f. Other names for the material derivative include 1) total derivative, 2) substantial derivative or 3) Lagrangian derivative


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A relation between these two derivatives can be derived using a generalized vectorial form of the Chain Rule. First recall that for steady (independent of t) scalar fields, the Chain Rule (2) gives the total differential (in invariant form) as df ≡ f ( r + dr ) − f ( r ) =dr.∇f

When time t is another independent variable, we just add another contribution to the total differential which arises from changes in t, namely dt. The Chain Rule becomes df =

f ( r + dr, t + dt ) − f ( r, t ) =

∂f dt + dr∇f ∂t

The first term has the usual Chain-Rule form for changes due to a scalar variable; the second term gives changes due to a displacement in vectorial position r. In our definition for material derivative, the displacement in position is proportional to dt: dr = v dt so df =

∂f dt + vdt ∇f ∂t

Dividing by dt yields the material derivative:

Df  df  ∂f ≡  = + v.∇f Dt  dt dr = vdt ∂t This relationship holds for a tensor of any rank. For example, the material derivative of the velocity vector is the acceleration a of the fluid, and it can be calculated from the velocity profile according to a=

Dv ∂v . = + v ∇v Dt ∂t

(17)

We will define ∇v in the next section.



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Calculus of Vector Fields Just like there were three kinds of vector multiplication which can be defined, there are three kinds of differentiation with respect to position. Divergence Curl Gradient

Notation ∇.v ∇×v ∇v

Result scalar vector tensor

Shortly, we will provide explicit definitions of these quantities in terms of surface integrals. Let me introduce this type of definition using a more familiar quantity: Gradient of a Scalar (Explicit) Recall the previous definition for gradient: f = f (r):

df = dr.∇f

Such an implicit definition is like defining the derivative f ′ (x) as that function associated with f(x) which when multiplied by dx gives the corresponding change in f denoted as df: f = f (x):

df = (dx) f ′ (x)

An equivalent, but explicit, definition of derivative is provided by the Fundamental Theorem of the Calculus:

f ′( x ) ≡

lim  f ( x + ∆x ) − f ( x )  df  = ∆x → 0  ∆x  dx

We can provide an analogous explicit definition of ∇f: lim  1 ∇f ≡  V → 0 V 

  f da n  ∫  A 

Fall, 2016 A = a set of points which constitutes any closed♦ surface containing the point r at which ∇f is to be evaluated V = volume of region enclosed by A da = area of a differential element (subset) of A n = unit normal to da, pointing out of region enclosed by A lim (V→0) = limit as all dimensions of A shrink to zero (in other words, A collapses about the point at which ∇f is to be defined.) To emphasize the general shape of the surface, we have drawn it above to resemble a potato. What is meant by this surface integral? Imagine A to be the skin of the potato. To compute the integral: 1) Carve the skin into many small elements. Each element must be sufficiently small so that • element can be considered planar (i.e. n is practically constant over the element) • f is practically constant over the element 2) For each element of skin, compute nf da 3) Sum of nf da over all elements yields integral Divergence, Curl, and Gradient This same type of definition can be used for each of the three spatial derivatives of a vector field: Divergence

Curl

Gradient

lim  1 ∇.v ≡  V → 0 V 



∫ n.vda  A



lim  1   ∫ n × vda  V → 0 V  A   lim  1  ∇v ≡ nvda    ∫ V → 0 V   A 

∇× v ≡

(18)

(19)

(20)

Physical Interpretation of Divergence Let the vector field v = v(r) represent the steadystate velocity profile in some 3-D region of space. What is the physical meaning of ∇.v? where f = any scalar field

• n.v da = dq = volumetric flowrate out through da (m3/s). This quantity is positive for outflow and negative for inflow. ♦ The circle over the integral sign in the definition denotes integration over a closed surface.



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• ∫A n.v da = net volumetric flowrate out of enclosed volume (m3/s). This is also positive for a net outflow and negative for a net inflow. • (1/V) ∫A n.v da = flowrate out per unit volume (s-1)  > 0 for an expanding gas  (perhaps T ↑ or p ↓)  • ∇ v =  = 0 for an incompressible fluid  (no room for accumulation)  < 0 for a gas being compressed 

• ∇.v = volumetric rate of expansion of a differential element of fluid per unit volume of that element (s–1) Calculation of ∇.v in R.C.C.S.

Fall, 2016

v da ∫ n.v da + ∫ n.v da +  + ∫ n.v da ∫ n.= A

A1

A2

(21)

A6

Let A1 be the x=xo face, which is the shaded surface in the figure above. The unit vector n always points out of the region. Since this is an x=const surface, the associated unit vector is i which points in the direction of increasing x. The normal n shown in the figure points in the direction of decreasing x, so we choose: n = –i n.v = –i.v = –vx(xo,y,z)

and

Now this result must be integrated over the range of y and z values corresponding to shaded area in the figure above: in particular this means that yo ≤ y ≤ yo+∆y and zo ≤ z ≤ zo+∆z. The following figure shows the area we will be integrating over

We will now use the surface-integral definition (18) to evaluate the divergence in Cartesian coordinates. Given: Evaluate:

v = vx(x,y,z)i + vy(x,y,z)j + vz(x,y,z)k ∇.v at (xo, yo, zo).

Solution: the first step is to choose the shape of the surface A. In general we choose a six-sided shape in which each side (or face) is a coordinate-equalconstant surface (some are curved). One corner of A is the point at which the divergence is to be evaluated.

Our differential area (shaded region in the figure above) is a rectangle of height dz and width dy. Its area is given by da = dy dz We multiply this area by the local value of the integrand –vx and repeat this for each tiny rectangle in the row of total width ∆y (the integral with respect to y); this is then repeated for each row (the integral with respect to z):

= ∫ n.vda A1

zo +∆z yo +∆y

∫

∫

zo

yo

−vx ( xo , y, z ) dy dz (22)

Time Out: Mean Value Theorem For R.C.C.S., we choose A to be surface of rectangular parallelepiped of dimensions ∆x, ∆y and ∆z with one corner at the point (xo, yo, zo) (see figure above).

If f(x) is continuous over the interval a < x < b, then there exists some ξ also bounded by this interval such that

So we partition A into the six faces of the parallelepiped: A1, A2 … A6. The integral will be computed separately over each face:

(b − a ) f (ξ) ∫ f ( x ) dx =


b

(23)

a

f (ξ) is called the “mean value” of f(x) over this interval


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b

b−a = ∫ dx

where

= ∫ n.v da

a

A2

In other words, by replacing the function f(x) in the integral by a constant, we are able to evaluate the integral. Of course, without evaluating the original integral of f(x), we don’t know the value of f(ξ). The advantage provided by the MVT is that we now have bounds on the values. In particular, since ξ is bounded by a≤ξ≤b then f (ξ) is similarly bounded by the maximum and minimum values of f(x) in this interval. The is a 2-D analog of (23). If f (x,y) is continuous over a bounded area A, then there exists some point (ξ,η) in A such that da ∫ f ( x, y ) =

A f ( ξ, η)

A

A

is the total area of the 2-D region. Just a reminder, the integrals appearing ust above are surface integrals — not the definite integral with respect to a. Time In: Using the Mean Value Theorem, we can replace the integral (22) over the surface area by some mean value of the integrand multiplied by the area of the surface:

∫ n.vda =

–vx(xo, y′,z′) ∆y∆z

yo +∆y

zo

yo

∫

∫

vx ( xo + ∆x, y, z ) dy dz

Using the Mean Value Theorem:

∫ n.v da = vx(xo+∆x,y″,z″)∆y∆z

(26)

A2

where

yo ≤ y″ ≤ yo+∆y

and

zo ≤ z″ ≤ zo+∆z

We denoted the point where the integrand takes on the mean by (y", z") [instead of (y', z')] since there is no reason to suppose the two mean values occur at the same point. The sum of these two integrals (24) and (26) is:

∫

+= ∫

A1

A2

[v x ( xo + ∆x, y ′′, z ′′) − v x ( xo , y ′, z ′)] ∆y ∆z

Dividing by V = ∆x∆y∆z:

A = ∫ da

where

zo +∆z

(24)

1 V

∫

A1 + A2

v ( x + ∆x, y ′′, z ′′) − v x ( xo , y ′, z ′) n.v da = x o ∆x

Now let ∆y and ∆z tend to zero. Notice that as ∆y→0, the interval (25) of possible values of y' collapses to a single value: y' = yo. Similarly for the intervals of z', y" and z". Thus the ratio of the surface integral to the volume becomes   lim 1 .vda  n  = ∆y , ∆ z → 0  V ∫  + A A  1 2 

A1

The Mean Value Theorem guarantees that this “mean value” –vx(xo,y′,z′) is assumed by the integrand at some point within the domain of integration: where

yo ≤ y′ ≤ yo+∆y

and

zo ≤ z′ ≤ zo+∆z

(25)

although it doesn’t tell us which point takes on the mean value. Now let’s move to another face of the parallelepiped. Let surface A2 be the x = xo+∆x face. On the opposite side of the parallelepiped from the shaded face, the unit normal points in the opposite direction: n = +i

v ( x + ∆x, yo , zo ) − v x ( xo , yo , zo ) = x o ∆x

Finally, we take the limit as ∆x tends to zero. The limit of the above quotient (on the right-hand side) becomes the partial derivative according to the fundamental theorem of the calculus: lim  1  V → 0 V 

∫

A1 + A2

  ∂v n.vda  = x ∂x  

xo , yo , zo

Now that ∆x→0 as well as ∆y, ∆z→0, the limit on the left-hand side above corresponds to V→0 and we have written it this way above. In the same way we obtained the above expression for the contribution to (21) from the two

n.v = i.v = vx(xo+∆x,y,z) Copyright© 2016 by Dennis C. Prieve


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x=const faces, we can also obtain the contribution from the two y=const surfaces: lim  1  V → 0 V 

∫

A3 + A4

 ∂v y  n.vda  = y ∂  

Fall, 2016

In particular, the above formula does not directly apply to cylindrical or spherical coordinates (see formulas in next section). Finally, the formula for calculating the gradient of the vector v in RCCS is

xo , yo , zo

∇v =

and from the two z=const surfaces: lim  1  V → 0 V 

∫

A5 + A6

  ∂v n.vda  = z ∂z  

3

3

∂v j

i =1

j =1

∂xi

Σ Σ

ei e j

where x1 = x, x2 = y, and x3 = z, v1 = vx, etc. xo , yo , zo

Summing these three contributions into (21) yields the formula for calculating the divergence in Cartesian coordinates: ∂v x ∂v y ∂v z ∇.v= + + ∂x ∂y ∂z

(27)

This formula for calculating divergence in Cartesian coordinates is sometimes cited as the definition of divergence instead of (18). Like its counterpart (5) for the gradient of a scalar, formula (27) can be used to calculate divergence in other coordinate systems. Yet it is hard to infer the physical significance using (27). By contrast we were able to infer the physical significance using (18) (see Physical Interpretation of Divergence on page 15). Evaluation of ∇×v and ∇v in R.C.C.S. In the same way, we could use the surfaceintegral definitions provided by (19) and (20) to determine formulas to calculate the curl and the gradient in R.C.C.S.:

∂v y   ∂v x ∂v z   ∂v ∇= ×v  z − − j i + ∂z   ∂z ∂x   ∂y  ∂v y ∂v x  + − k ∂y   ∂x

Lecture #4 begins here Evaluation of ∇.v, ∇×v and ∇v in Curvilinear Coordinates Ref: Greenberg, p175 These surface-integral definitions can be applied to any coordinate system. In Homework Set #2, we derive the formula for calculating ∇.v in cylindrical coordinates. Alternatively, we can express divergence, curl and gradient in terms of the metric coefficients for the coordinate systems. If u,v,w are the three scalar coordinate variables for the curvilinear coordinate system, and functions needed to transform into RCCS x = x(u,v,w)

z = z(u,v,w)

can be determined, then the three metric coefficients — h1, h2 and h3 — are given by

(28)

The formula for curl in R.C.C.S. turns out to be expressible as a determinant of a matrix:

y = y(u,v,w)

h1 ( u, v, w ) =

xu2 + yu2 + zu2

h2 ( u, v, w ) =

xv2 + yv2 + zv2

h3 ( u, v, w ) =

xw2 + yw2 + zw2

where letter subscripts denote partial derivatives while numerical subscripts denote components. The general expressions for evaluating divergence, curl and gradient are given by gradient of scalar

i ∂ ∂x vx

j ∂ ∂y vy

k ∂v y   ∂v x ∂v z   ∂v ∂ j = z − − i + ∂z  ∂y ∂z   ∂z ∂x  vz  ∂v y ∂v x  + − k ∂y   ∂x

But remember that the determinant is just a mnemonic device — it is not the definition of curl.


= ∇f

1 ∂f 1 ∂f 1 ∂f e1 + e2 + e3 h1 ∂u h2 ∂v h3 ∂w

divergence of vector:

∇.v =

1 × h1h2 h3

∂ ∂ ∂ ×  ( h2 h3v1 ) + ( h1h3v2 ) + ( h1h2 v3 ) ∂v ∂w  ∂u 


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curl of vector:

h1e1 h2 e 2 1 ∂ ∂ ∇×v = ∂v h1h2 h3 ∂u h1v1 h2 v2

Fall, 2016

Then the orbit lies in a z=const plane. The radius of the orbit is the radial coordinate r which is also constant. Only the θ-coordinate changes with time and it increases linearly so that d θ dt = const = Ω.

h3e 3 ∂

∂w h3v3

Formulae for divergence, curl and gradient have been evaluated for a number of common coordinate systems. The results are tabulated in Appendices II (for RCCS), III (for cylindrical) and IV (for spherical coordinates). Physical Interpretation of Curl To obtain a physical interpretation of ∇×v, let’s consider a particularly simple flow field which is called solid-body rotation. Solid-body rotation is simply the velocity field a solid would experience if it was rotating about some axis. This is also the velocity field eventually assumed by viscous fluids undergoing steady rotation.

In parametric form in cylindrical coordinates, the trajectory of a material point is given by r(t) = const, z(t) = const, θ(t) = Ωt The velocity can be computed using the formulas developed in the example on page 12: dr ( t ) d θ (t ) dz ( t ) r Ωe θ v= er + r eθ + ez = dt dt dt     0

rΩ

0

Ω

Ω

Alternatively, we could deduce v from the definition of derivative of a vector with respect to a scalar:

r

Dp ∆p r d θ eθ dθ r= = = = v= lim eθ r Ωeθ Dt ∆t →0 ∆t dt dt

z

side view r eθ(θ(t)) v(t)

More generally, in invariant form (i.e. in any coordinate system) the velocity profile corresponding to solid-body rotation is given by  v (p ) = Ω × p (29)

er(θ(t))

r v(0)

eθ(θ(0)) er(θ(0))

θ( t )

r top view

 where Ω is called the angular velocity vector and p is the position vector whose origin lies somewhere  along the axis of rotation. The magnitude of Ω is the rotation speed in radians per unit time. It’s direction is the axis of rotation and the sense is given by the “right-hand rule.” Time Out: Vector Nature of Angular Velocity

Imagine that we take a container of fluid (like a can of soda pop) and we rotate the can about its axis. After a transient period whose duration depends on the dimensions of the container, the steady-state velocity profile becomes solid-body rotation.

In classical mechanics, we have (translational) velocity and angular velocity. For a rotating body like the bicycle wheel in the illustration below, we have both types of vectors drawn as the green arrows

A material point imbedded in a solid would move in a circular orbit at a constant angular speed equal to Ω radians per second. The corresponding velocity is most easily described using cylindrical coordinates with the z-axis oriented perpendicular to the plane of the orbit and passing through the center of the orbit. Copyright© 2016 by Dennis C. Prieve


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Fall, 2016

To obtain this result we have used the fact that the cross product of any two parallel vectors vanishes (because the sine of the angle between them is zero — recall definition of cross product on p1).

At some radius r from the axis of rotation the translational velocity is rωeθ(+π/2) at the top of the wheel and rωeθ(–π/2), where the argument of eθ is the theta coordinate the point in cylindrical coordinates and where ω denotes the angular rotation speed in this illustration.  The angular velocity ω is defined as a vector having magnitude equal to the speed of rotation in radians per second and a direction which is normal to all the translational velocities rωeθ(θ). This direction also coincided with the axis of rotation. The sense of  ω is given again by the “right-hand rule:” curling the fingers of your right hand in the direction of the translational velocities, your thumb points in the  direction of ω . Just like we have (translational) momentum mv associated with translation of a mass m at velocity v,  we also have angular momentum L = I ω associated with rotation of an object having a moment of inertia  I with an angular velocity ω . Thus the moment of inertia I is the angular analog of the inertial mass m. The vector L is also shown in the figure above.

The cross product of two distinct unit vectors in any right-handed coordinate system yields a vector parallel to the third unit vector with a sense that can be remembered using the figure above. If the cross product of the two unit vectors corresponds to a “clockwise” direction around this circle, the sense is positive; in a “counter-clockwise” direction, the sense is negative.♦ In this case, we are crossing ez with er which is clockwise; hence their cross product is +eθ. Now that we have the velocity field, let’s compute the curl using the definition given by (19): ∇× v ≡



∫ n × vda  

A

(19)

For our closed surface A, let’s choose a cylinder of radius r and height h. Of course, we orient this cylinder with its axis coincident with the axis for our cylindrical coordinates which is also coincident with the axis of rotation. We partition A into three parts: the curved wall plus the top (at z = h) and bottom (at z = 0) of the cylinder. The integral over A becomes:

∫ = A

Time In: Physical Meaning of Curl (Cont’d)

lim  1  V → 0 V 

∫ curved

+

∫ top

+

∫

(32)

bottom

Returning to our fluids problem of solid-body rotation in cylindrical coordinates, the angular velocity is  Ω = Ωez (30)

On the curved surface, the outward unit normal and translational velocity are given by n = er(θ) and v = rΩeθ from (31). The integrand of the curl in (19) is

and the position vector is

Since r = const over the entire curved wall of the cylinder, this integrand is a constant. So the integral over the curved wall is just the product of the constant integrand and the area

= p re r + z e z

Taking the cross product of these two vectors [keeping the order the same as in (29)]: v ( p ) =r Ω e z × er + z Ω e z × e z =r Ωeθ   eθ

0


(31)

n × v =r Ω ( er × eθ ) =r Ωe z

♦ This convention assumes a “right-handed” coordinate system.


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∫

∫

n × vda =Ω r ez

da =π 2 r 2 hΩe z (33)

curved   

curved

2 πrh

Similarly the integral over the top and bottom surfaces can be evaluated, keeping in mind that n always points out of the region: n = +ez on top and n = –ez on the bottom: n × v z = h = r Ω ( e z × eθ ) =−r Ωer ( θ )

and

n × v z =0

Fortunately, we can avoid the integration entirely by realizing that, when the integrals over the top and bottom are combined, the contribution to the integral from corresponding surface elements da on the top and bottom cancel out:

∫

+

bottom

=

−r Ωer + r Ωer ) da = ∫ (

0 (34)

0

2

hΩe z

A

Now we divide by the volume of the cylinder which is V = πr2z: 1 V

2 πr 2 h Ωe z

∫ n × vda =

πr 2 h

A

= 2Ωe z

lim  1  V → 0 V 



∫ n × vda  = A



Comparing with (30), we conclude that  ∇ × v = 2Ω

2Ωe z

(36)

(37)

Thus the curl is vectorially equal to twice the local angular velocity of the fluid elements. While we have only shown this for one particular flow field, this result turns out to be quite general. As an alternative to using the definition, we could have evaluated the curl using a formula like (28) given for RCCS. In cylindrical coordinates, the formula for the curl is again obtained from the definition of curl (19). We will not derive this Copyright© 2016 by Dennis C. Prieve

we obtain

vθ = rΩ

vz = 0

∇ × v = 2Ωe z

which is identical to (36). From this formula we see that ∇×v takes on the same value at each point (r,θ,z) in this particular flow field. This explains why the quotient in (35) was independent of the dimensions of the cylinder — even before taking the limit. Example #2: Linear Shear Flow

Let’s look at a second example: linear shear flow which we introduced on page 3:

(35)

Both dimensions of the cylinder (r and z) cancelled out so this quotient is already independent of the dimensions. Taking the limit V→0 changes nothing but yields the curl according to the definition given by (19): ∇× v ≡

  eθ 

In solid-body rotation, the entire fluid and its container (e.g. soda can) is undergoing rotation, so it isn’t surprising that the fluid elements themselves are rotating. However rotation of fluid elements is quite common.

Substituting (33) and (34) into (32):

∫ n × vda =2πr

 1 ∂vz ∂vθ   ∂vr ∂vz = ∇× v  − −  er +  r z ∂θ ∂ ∂r  ∂z    1 ∂ ( rvθ ) 1 ∂vr  + −  ez  r ∂r r ∂θ  

vr = 0

We now see that the integrand depends on both r and

∫

formula in the notes or in the homework, but the results are given in Appendix III: Vector Operations in Cylindrical Coordinates (r, θ, z):

Substituting

=r Ω ( −e z × eθ ) =+ r Ωer ( θ )

θ, so we no longer are integrating a constant.

top

Fall, 2016

Unlike solid-body rotation where the fluid elements undergo motion along curved streamlines, the streamlines in linear shear flow are straight. Are the fluid elements rotating as well as translating? When we eventually analyze linear shear flow (see page 56), we will find that the velocity profile is given by v=

Uy ex L

In other words, vx = Uy L while vy = vz = 0. Using the formula (28) for calculating curl in RCCS, we obtain: ∇× v = −

U ez L


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Fall, 2016

According to our general interpretation of curl represented by (37), the fluid elements are rotating with an angular velocity given by  U Ω=− ez 2L

In the Cartesian coordinate system used in the figure above, the z-axis (not shown) points out of the page (need to use right-hand rule). Thus according to equation above, the angular velocity of fluid elements points into the page. Referring to the bicycle schematic on 20 used to relate the direction of angular velocity to the translational motion, we must orient the thumb of our right hand into the page. This means that the rotation of fluid elements in the figure of linear shear flow is clockwise (as indicated by the red arrow).



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Vector Field Theory Reference: Chapter 9 of Greenberg, M.D., Foundations of Applied Mathematics, Prentice-Hall, 1978. Reprinted in 2013 by Dover and is now available as an inexpensive Kindle edition. There are three very powerful theorems which constitute “vector field theory:” • (Gauss) Divergence Theorem • Stokes Theorem • Irrotational ⇔ Conservative ⇔ Derivable from potential Divergence Theorem This is also known as Gauss’♣ Divergence Theorem or Green’s Formula (by Landau & Lifshitz♠). Let v be any (continuously differentiable) vector field and choose A to be any (piecewise smooth, orientable) closed surface; then

∫ n.v da= A

∫ ∇.v dV

(38)

V

where V is the region enclosed by A (see Fig. 1) and n is the outward pointing unit normal to the differential surface element having area da.

Fall, 2016

Although we will not attempt to prove this theorem (proof provided in Sect. 9.4 of Greenberg), we will offer the following rationalization. Consider the limit in which all dimensions of the region are very small, i.e. V→0. When the region is sufficiently small, the integrand (which is assumed to vary continuously with position)* is just a constant over the region: ∇.v = const. inside V 

 dV  = ( ∇.v ) V  V 

( ∇.v )  ∫ ∫ n.v da = ∫ ∇.v dV = A

V

Solving for the divergence, we get the definition back (recalling that this was derived for V→0):

1 ∇.v = ∫ n.vda V A

Thus the divergence theorem is at least consistent with the definition of divergence. One application of the divergence theorem is to simplify the evaluation of surface or volume integrals. However, we will use GDT mainly to derive invariant forms of the equations of motion: Invariant: independent of coordinate system. To illustrate this important application, let’s use GDT to derive the continuity equation in invariant form. The Continuity Equation Let ρ(r,t) and v(r,t) be the density and fluid velocity. What relationship between them is imposed by conservation of mass? For any system, conservation of mass means:

 rate of acc.   net rate of   =  of total mass  mass entering  Fig. 1 Schematic showing elements appearing in the Divergence Theorem.

♣ Carl Friedrich Gauss (1777-1855), German mathematician, physicist, and astronomer. Considered the greatest mathematician of his time and the equal of Archimedes and Isaac Newton, Gauss made many discoveries before age twenty. Geodetic survey work done for the governments of Hanover and Denmark from 1821 led him to an interest in space curves and surfaces. ♠ Landau, L.D., & E.M. Lifshitz, Fluid Mechanics (Vol. 6 of the 10-volume book series Course of Theoretical Physics), Pergamon, 1959. For some history and impact of this book series see https://en.wikipedia.org/wiki/Course_of_Theoretical_Physi cs


(39)

Let's now apply this principle to an arbitrary system whose boundaries are fixed spatial points (see Fig. 1). Note that this system, denoted by V, can be macroscopic (it doesn’t have to be differential). The boundaries of the system are the set of fixed spatial points denoted as A. Of course, fluid may readily cross this mathematical boundary. Subdividing V into many small volume elements: dm = ρdV

*This is a consequence of v being “continuously differentiable”, which means that all the partial derivatives of all the scalar components of v exist and are continuous.


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M=

∫ dm= ∫ ρ dV

values of the integrand. Since V is arbitrary, and since this integral must vanish for all V, the integrand must vanish at every point:*

V

 ∂ρ dM d   ∫ ρ dV  = ∫ = dV  dt dt  ∂t V  V

n = outward unit normal n.v da = vol. flowrate out through da (m3/s) ρ(n.v)da = mass flowrate out through da (kg/s) (41)

(42)

(44)

which is called the equation of continuity. Comment: Note that we were able to derive this result in its most general vectorial form, without recourse to any coordinate system and using a finite (not differential) control volume. This illustrates the power of GDT. In the special case in which ρ is a constant (i.e. depends on neither time nor position), the continuity equation reduces to: for ρ=const.:

where in the second equality, we have taken advantage of changing the order of scalar multiplication (which doesn’t matter). Now we need to evaluate the rate entering rather than leaving, but this is just the same quantity with opposite sign:  rate of  − ∫ n.( ρv ) da  = mass entering   A

∂ρ + ∇.( ρv ) = 0 ∂t

(40)

where we have switched the order of differentiation and integration in the last step. This is only valid if the boundaries are independent of t.♥ Now mass enters through the surface A. Subdividing A into small area elements:

 rate of  n.( ρv ) da ρ ( n.v ) da =  =   ∫ ∫ mass leaving   A A

Fall, 2016

∇.v = 0

(45)

Recall that ∇.v represents the rate of expansion of fluid elements. “∇.v = 0” means that any flow into a fluid element is matched by an equal flow out of the fluid element: accumulation of fluid inside any volume is negligible small. Reynolds Transport Theorem In the derivation above, the boundaries of the system were fixed spatial points. Sometimes it is convenient to choose a system whose boundaries move:

Next we substitute (40) and (42) into our word statement of the mass balance (39):

∂ρ

− ∫ n.( ρv ) da ∫ ∂t dV =

V

or

A

∂ρ

0 ∫ ∂t dV + ∫ n.( ρv ) da =

V

(43)

A  ∇.( ρv ) dV

∫

V

Then the accumulation term in the balance [i.e. the first equation in (40)] will involve the time derivative of a volume integral whose limits change with time. Time Out: Leibnitz’ Rule

If these integrals had the same limits, we could combine them into one integral. We can make the limits the same by applying the Divergence Thm to the surface integral. After applying GDT and combining the two integrals, we have

A simple example of this situation is the definite integral of an ordinary function of a single variable x and a parameter α:

 ∂ρ  ∫  ∂t + ∇.( ρv ) dV = 0 V

*If the domain V were not arbitrary, we would not be able to say the integrand vanishes for every point in the domain. For example:

For one particular choice of V, this integral might be zero because of cancellation of positive and negative ♥To see what happens when the boundaries move, see the following section: Reynolds Transport Theorem.


2π

∫0

cos= θ dθ

2π

∫0

sin= θ dθ 0

2π

∫0 ( cos θ − sin θ ) d θ = 0 does not imply that cosθ = sinθ since the integral vanishes over certain domains, but not all domains.


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= I (α)

b( α )

∫

a( α )

EXAMPLE: rederive the continuity equation using a control volume whose boundaries move with the velocity of the fluid.

f ( x, α ) dx

Notice that the limits of the integral also depend on the parameter α. Clearly the integral depends on α both as a result of the integrand’s dependence on α as well the dependence of the limits themselves. The derivative can be calculated using Leibnitz’ Rule (see §1.7 in Greenberg): dI d = dα dα b( α )

=

∫

a( α )

b( α )

Solution: If the boundaries of the system move with the same velocity as local fluid elements, then fluid elements near the boundary can never cross it since the boundary moves with them. Since fluid is not crossing the boundary, the system is closed.* For a closed system, conservation of mass requires:

{

a( α )

∂f ( x, α ) ∂α

or dx +

db da + f b ( α ) , α  − f  a ( α ) , α  dα dα

The first term in the second equation is the simple change in order of the derivative and integral operations; the second and third terms represent additional contributions from the limits’ dependence on α. Time In The analog for a differentiating a volume integral whose boundaries are moving with time t is (see p163 of Greenberg):

  ∂S d  = S t dV dV r ( , ) ∫ ∫   dt ∂t V ( t ) V ( t )  

∫

(46)

S (r, t )(n.w )da

A(t )

where w is the local velocity of the boundary and S(t) is a tensor of any rank. Again, the first term on the right-hand side represents the simple change in order of the differential and integral operations; the second term represents an additional contribution arising from the time-dependence in the domain V(t). If w = 0 at all points on the boundary, the boundary is stationary and (46) reduces to that employed in our derivation of the continuity equation. In the special case in which w equals the local fluid velocity v, this relation is called the Reynolds Transport Theorem.♦ ♦ Osborne Reynolds (1842-1912), Engineer, born in Belfast, Northern Ireland, UK. Best known for his work in hydrodynamics and hydraulics, he greatly improved centrifugal pumps. The Reynolds number takes its name from him.


}

d mass of =0 dt system

f ( x, α ) dx

∫

+

Fall, 2016

dM d = dt dt

∫

ρ dV = 0

(47)

V (t )

Notice that we now have to differentiate a volume integral whose limits of integration depend on the variable with respect to which we are differentiating. Applying (46) with w=v (i.e. applying the Reynolds Transport Theorem):   ∂ρ d   r v )da 0 ρ t = dV dV + ∫ ρ (n.= ( , ) ∫ ∫      ∂t dt A(t ) dq  V (t )  V (t )

The total of these two integrals must vanish at all times t according to (47). Notice that the surface integral is just the net rate of mass leaving the system though the boundary A calculated as if the boundary were stationary [the local mass flowrate across boundary moving with velocity w equals n.(v-w) da]. The fact that the boundaries of A are moving with time does not affect the value of this surface integral, except to make its value dependent on t. The rest of the derivation [see second equation in (41)] is the same as before, which leads again to (44). Stokes Theorem Let v be any (continuously differentiable) vector field and choose A to be any (piecewise smooth, orientable) open surface. Then

∫ n.( ∇ × v ) da = ∫ v.dr

A

(48)

C

* When we say “closed,” we mean no net mass enters or leaves the system; individual molecules might cross the boundary as a result of Brownian motion. However, in the absence of concentration gradients, as many molecules enter the system by Brownian motion as leave it by Brownian motion. v is the mass-averaged velocity.


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Fall, 2016

Stokes' Theorem

dr ∫ v.=

C Mean Value Theorem

A

n.( ∇ × v ) A= 2 Ω n A

=

where C is the closed curve forming the edge of A (has direction) and n is the unit normal to A whose sense is related to the direction of C by the “righthand rule”.

∫ n.( ∇ × v ) da

Finally, we recall from (37) that ∇×v is twice the angular velocity of fluid elements, so that n.∇×v is the normal component of the angular velocity (i.e. normal to the surface A). Thus velocity circulation is twice the average angular speed of fluid elements times the area of the surface whose edge is the closed contour C.

My rendition above of an arbitrary curved open surface might require some elaboration: imagine a differentially thin crepe (pancake) lying on a hillside; the surface of the curved crepe is A and the edge of the crepe is the contour C. (48) is called Stokes Theorem.♥ Evaluation of contour integrals is similar to evaluating surface and volume integrals: we divide the contour C into a large number of small segments. One such segment is shown in the figure above as the differential displacement vector dr. Each segment is small enough so that its direction is unique (given by the direction of dr) and the integrand is virtually constant over this differential segment. Velocity Circulation: Physical Meaning When the velocity v represents fluid velocity, the contour integral appearing in Stokes’ Theorem

∫ v.dr

Example: Compare “velocity circulation” and “angular momentum” for a thin circular disk of fluid undergoing solid-body rotation about its axis. Solution: Choose cylindrical coordinates with the zaxis aligned with axis of rotation. Solid-body rotation corresponds to the following velocity profile (see page 19):

(49)

v = r Ωeθ

C

is a quantity called velocity circulation. We will encounter this quantity in a few lectures when we discuss Kelvin’s Theorem. For now, I’d like to use Stokes Theorem to provide some physical meaning to velocity circulation. Using Stokes Theorem and the Mean Value Theorem, we can write the following:

∇ × v = 2Ωe z

and

Finally the unit normal to the disk surface is n = ez. Then the velocity-circulation integral becomes dr ∫ n.( ∇ × v ) da ∫ v.=

C

A

=

∫ e z .( 2Ωe z ) da =

2ΩπR 2

A

♥ Sir George Gabriel Stokes (1819-1903): British (Irish born) mathematician and physicist, known for his study of hydrodynamics. Lucasian professor of mathematics at Cambridge University 1849-1903 (longest-serving Lucasian professor); president of Royal Society (18851890).


The angular momentum L of a mass m undergoing motion at velocity v is the lever arm r crossed with the linear momentum (p = mv): i.e. L = r×p.♣ Summing this over differential fluid mass in our disk with dm = ρ dV, the net angular momentum of the disk is: ♣ see p25 of Landau & Lifshitz, Mechanics and Electrodynamics (Course of Theoretical Physics: Vol. 1), Pergamon, 1959.


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L=

∫ ( r × v ) ρ dV =

ρ∆z ∫ ( r × v ) da

V

A

Since the disk is of uniform thickness ∆z and density ρ, we can write the second equation above in which the volume dV of a cylinder of length ∆z and crosssectional area da is dV = da ∆z. If the disk is sufficiently thin that we can neglect the z contribution to the position vector, then we can approximate r = rer in cylindrical coordinates.♦ Substituting into the second integral above

=

An example of a vector field which is derivable from a potential is the gravitational force near sea level:

0

Fgrav = –Mgk = –∇φ

π 4 R ρ∆z Ωe z 2

φ(z) = Mgz

π 4 R ρ∆z Ω Lz z 1 π R 2 ∆z ρ = 2 = 1 R 2 ρ∆= 2 .dr 4 4π  2 R Ωπ v V ∫ C

=M 4π

where M is the mass of fluid in the disk. This could be rewritten as

C

L 4π z M

(50)

So the velocity-circulation integral is just proportional to the angular momentum Lz per unit mass M. Lecture #5 begins here Derivable from a Scalar Potential A very special class of vector fields consists of those vectors for which a scalar field exists such that the vector can be represented as the gradient of the scalar: Suppose:

(51)

and the associated potential energy is:

Dividing this by the velocity circulation integral:

∫ v.dr =

for all r in some domain, then f (r) is called the scalar potential of v and v is said to be derivable from a potential in that domain.

R

L = ρ∆z ∫ ( r 2 Ωe z ) da = ρ∆z Ωe z ∫ r 2 2πr dr A

Fall, 2016

v = v(r) and f = f (r)

If f (r) exists such that: v(r) = ∇f

∇φ = Mgk

Note that

is identical to the force, except for the sign (introduced by convention). This example also suggests why φ is called the “potential” of v (because φ represents a potential energy associated with the force). In other examples where v is not a force, φ will not have units of energy but φ is still called the “potential.” Not every vector field has a potential. Which do? To answer this, let's look at some special properties of such vector fields. if v=∇φ then ∇×v=0

Property I:

Proof: Recall that ∇×(∇φ) = 0 (see HWK #2, Prob. 7e). A vector whose curl vanishes throughout some region is said to be irrotational. This name is an allusion to ∇×v representing the rotation rate if v is the fluid velocity. ∇×v=0 means the fluid elements are not rotating. Property II: if v=∇φ then

C

for any closed contour C in the region. Proof: Using Property I, we know that ∇×v=0. Then we can deduce the value of this closed-contour integral from Stokes’ Theorem: dr ∫ n.( ∇ × v )= da ∫ v.= 

C

♦ Actually this assumption isn’t necessary since any zcomponent of r will produce an r-component in the crossproduct and this r-component will integrate to zero as long as V is a body of rotation about the same axis.


∫ v.dr = 0

A

0

0

A vector field which has this property is said to be conservative. This name is an allusion to the special case in which v represents a force, like gravity. Then v.dr (force times displacement) PDF generated December 6, 2016

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represents the work required to move the object through the force field. Saying that the contour integral vanishes means that the work required to lift a weight can be recovered when the weight falls. In other words, energy is conserved. If a contour C is open, v=∇φ is still quite useful:

Fall, 2016

of the tensor by the same vector (or vice versa). For example: if

τ = ab

then

τt = ba

In terms of scalar components of τ, we can write the relationship between a tensor and its transpose as: τtij = τji τt = τ

Symmetric Tensor: Property III: let Co be an open contour connecting points A and B. If v=∇φ then ∫Cov.dr = φ(rB)–φ(rA)

An example of a symmetric tensor is the dyad aa. Identity Tensor: Also known as the Idem Factor. Denoted as I and defined so that: v.I = v = I.v

for any contour connecting A and B. Proof: Note that ∇φ.dr = dφ [from our definition of gradient (2)]. Then

∫

v ⋅ dr =

Co

∫ d φ = φ ( rB ) − φ ( rA )

∂r I =∇r = ∂r

Co

We call this property path independence since any path connecting A and B gives the same result. Of course, Property II is just a special case of this for which A=B so that φ(rB) – φ(rA) = 0. Theorem III We showed that Properties I, II and III are implied by v = ∇φ; it turns out that the converse of these three statements is also true (although I’m not going to prove that here). We can distill these properties and their converse into a single statement:  φ = φ ( r ) exists   ∇× v =0       for all r  ⇔ such that v =∇φ  in Region   in Region       v.dr = 0 for every  ∫ ⇔ C  closed C in Region 

    

and

∇r =∑ ∑ i

then

j

∂r j ∂xi

ei e j

∂r j ∂xi

= δij

which is 0 if i≠j or 1 if i=j. This leaves: τt

and is

v.τ = τt.v τ.v = v.τt

for all vectors v. In other words, pre-dotting a tensor by a vector is the same as post-dotting the transpose Copyright© 2016 by Dennis C. Prieve

r = xi + yj + zk

r1 = x1 = x, r2 = x2 = y, and r3 = x3 = z:

Transpose of a Tensor, Identity Tensor The transpose of a tensor τ is denoted defined so that:

where r is the position vector expressed in terms of the unit vectors in that coordinate system. Recall (6) from page 6 that gradient can be thought of as the partial derivative with respect to position, I can be thought of as the derivative of the position vector with respect to itself. In R.C.C.S., recall that:

where rj is the jth component of the position vector r and xi is the ith coordinate. In Cartesian coordinates, the position vector components are related to the coordinates according to:

which we will refer to as “Theorem III”.

and

for any vector v. Clearly I is symmetric, but in addition, dotting it with another vector gives that vector back (like multiplying by one).♣ In any coordinate system, I can be calculated from:

♣ According to the Merriam-Webster’s 11th Collegiate Dictionary, “idem” means “the same as something previously mentioned” — used chiefly in bibliographies. Somewhat consistent with this definition, by multiplying something by I we obtain the same thing.


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r ∇ =

i

j

Fall, 2016

After integrating over the area, we obtain:

i

This expression for the identity tensor turns out to be valid for any set of orthonormal unit vectors (not just the Cartesian ones for which we have derived it here). For the special case of RCCS: I = ii + jj + kk As a partial proof that I has the desired properties which make it the identity tensor, consider dotting it with an arbitrary vector v: v.(ii+jj+kk) = v.ii + v.jj + v.kk = (v.i)i + (v.j)j + (v.k)k = vxi + vyj + vzk = v Thus we have shown that v.I = v, as advertised.

∫

j

Dividing by V:

1 V

∫

A1 + A2

 τ1 j − τ1 j  xo +∆x xo   n.τda = e ∑   j ∆ x j

Taking the limit as V→0:  ∂τ1 j lim  1  . n ej τ da  = ∑ ∫ V → 0 V ∂x  j + A A  1 2 

Adding on similar contributions from the y=const and z=const faces:

∇.τ =∑ ∑

Divergence of a Tensor This quantity is defined just like divergence of a vector. ∇.τ ≡

∑ {τ1 j xo +∆x − τ1 j xo }e j ∆y∆z

n.τ da =

A1 + A2

 lim  1  ∫ n.τda  V → 0 V   A

i

j

∂τij ∂xi

ej

∂τ yx ∂τ zx   ∂τ = ∇.τ  xx + + i ∂y ∂z   ∂x  ∂τ xy ∂τ yy ∂τ zy  +  + + j ∂y ∂z   ∂x ∂τ yz ∂τ zz   ∂τ +  xz + + k ∂y ∂z   ∂x

Corollaries of the Divergence Theorem (38) gives the Divergence Theorem for vectors (tensors of rank 1). It can also be applied to tensors of other rank: including for scalars

∫ nf da= ∫ ∇f dV A

Note that this definition uses a pre-dot not a post-dot. In R.C.C.S.

or for tensors

∫ n.τ da = ∫ ∇.τ dV A

τ = ΣΣτijeiej

(52)

V

V

On the x=xo face: n = –e1

n.τ = ∑∑ τij e1.ei e j = −∑ τ1 j i

j

j

xo

ej

Similarly, on the x=xo+∆x face, we obtain: n = +e1

n.τ=

∑ τ1 j xo +∆x e j j



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Introduction to Continuum Mechanics Reference: G.E. Mase, Continuum Mechanics, Schaum's Outline Series, McGraw-Hill, 1970. In this course, we will treat fluids like continua; in other words, we are going to ignore the molecular granularity of matter. This is an assumption which we, as engineers, often make in describing transport of heat, mass, or momentum although we don’t always state this assumption explicitly. To make the nature of this assumption clearer, it might help to discuss the alternative. Fluids are composed of molecules. In principle, if you tell me the initial location of every molecule in the fluid along with its initial velocity, I can compute the position and velocity at some later time by applying Newton’s laws of motion (i.e. F = ma) to each molecule during each collision with another molecule.♠ The difficulty with this approach is that the number of molecules in any volume of fluid of interest to us is so large as to make such a detailed calculation impractical. For example: 1 cm3 of water → 3.3x1022 molecules → 1 million years Even with a computer capable of performing 109 multiplications per second, it would take 1 million years to do just one multiplication for each molecule. Molecules of a liquid collide on the average of once every 10–12 seconds. To describe one second of real behavior, I would need 1012 million years. Clearly, this is an absurd length of time. Although computers get faster every year, this will remain an absurdly long time for the foreseeable future. The alternative is: Continuum Hypothesis A detailed description at the molecular level is not required in order to predict macroscopic behavior of any material. For example, it is not necessary to ♠ During the 19th century, some philosophers argued that Newton’s laws of motions implied that the future was predetermined. Of course, this would have profound philosophical consequences (even if the future is not known). We now know that we can only predict the future for a relatively short “event horizon” whose duration increases with the precision with which we know the initial location and velocity. However Heisenberg’s uncertainity principle (circa 1927) states that you cannot know both the position and velocity of even one particle with infinite precision. The more certain either position or velocity is prescribed, the less certain is the other. So the distant (beyond a couple of pico-seconds) future is not calculable nor pre-determined.


Fall, 2016

know the precise location of every molecule of fluid; it turns out that all that is needed for most applications is the distribution of mass described by the density profile ρ(r) of molecules in some 3-D region V:

ρ(r ) =

 1  V → 0 V lim



∑ mi  i



(53)

where mi is the mass of molecule i, the sum is over all the molecules inside the region having volume V. Similarly, we don’t need to know the velocity of every single molecule; it is sufficient to know the average velocity of molecules in small fluid elements (also known as material points).

v (r ) =

 1  V → 0  N lim



∑ vi  i



where vi is the velocity of molecule i and N is the number of molecules inside V. This average velocity manifests itself macroscopically as the velocity profile v(r). Finally, we don't need to know the translational, rotational, vibrational and electronic energy of each molecule. We usually only need to know the internal energy per unit volume as a function of position, which in turn, manifests itself macroscopically as temperature of the fluid. In fluid mechanics, as in heat and mass transfer, we make an assumption known as the “continuum hypothesis.” To understand it, consider calculating the numerator (mass) and denominator (volume) in (53) for a sequence of regions V of different size with all regions in the sequence having at lease one point r in common. Suppose the initial region in the sequence is macroscopically nonuniform in density. As the volume of the region shrinks, both numerator and denominator are reduced but not at the same rate. The resulting quotient depends on V as shown in the right side of the graph below (for V between 10–5 to 10+5 cm3):


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molecular granularity

When the volume becomes small enough (say V between 10–12 and 10–8 cm3), the quotient has converged to a constant value over several decades of volumes; this constant value of the quotient is defined to be the local value of the density ρ(r). However, if we continue to shrink the volume to even smaller dimensions (as the limit seems to suggest), the quotient eventually begins to vary erratically as the volume decreases further. This erratic behavior always occurs when the linear dimensions of the volume approach the average distance between molecules.

Fall, 2016

2. is large enough to contain a statistically large number of molecules. In other words, we are assuming that dV exists such that the two conditions above are both satisfied. Materials which obey this “hypothesis” as said to behave as a continuum. Generally, the continuum hypothesis works well provided all the dimensions of the system are large compared to the mean free path between molecular collisions. An example of a situation in which the continuum hypothesis does not work is the flow of dilute gases in small pores, where the mean free path (for the collision of molecules) is comparable to the dimensions of the pore. This situation is called Knudsen diffusion. Example: compare the mean-free path λ and Lennard-Jones collision diameter d with the average intermolecular spacing for air at the standard temperature (0ºC) and pressure (1 atm) (STP). Solution: the mean-free path can be calculated from [see (1.4-3) on page 24 of BSL♥] v = 70 nm 2 πd 2

= λ

Suppose the black dots in the figure above each represent one molecule [one of the is in (53)] whose location is “frozen” in time. Consider the quotient in (53) as we shrink the volume from the boundary denoted in black to the boundary denoted in blue. The denominator (i.e. the volume decreases very little compared to the change in the numerator (from 10 molecular masses to 9). This is the source of the erratic behavior in the quotient.

where d = 0.3617 nm is the Lennard-Jones collision diameter of air molecules [see Table E.1 on page 864f of BSL]♦ and v is the average volume of air occupied by one molecule. According to the ideal gas law, the latter is given by

Basically the continuum hypothesis is that the above limit converges before the dimensions of V become so small as to reveal the molecular granularity of matter. In other words, by the mathematical limit V→0 in (53), we don’t really mean zero volume at the molecular scale. We simply mean to let V get just small enough so that the quotient has converged (i.e. becomes independent of V ) in the usual mathematical sense.

where k is Boltzmann’s constant (the ratio of the universal gas constant R to Avogadro’s number). Finally, the average distance between gas molecules under these conditions can be estimated as

A more precise statement of the continuum hypothesis is:

The basic problem in continuum mechanics is to describe the response of material to stress. A quantitative statement of that response is known as:

Continuum Hypothesis - the region to be described can be subdivided into a set of (infinitesimal) volume elements, each of which simultaneously: 1. is small enough to be considered uniform (i.e. any spatial variations in properties -- such as ρ, v, T, p -- inside the volume element are negligible); and


J   1.3807 × 10−23  ( 273 K ) kT  K v = = = 40.6 nm3 p 1.013 × 105 Pa

3

v = 3.4 nm

which lies between the collision diameter d = 0.3617 nm and the mean-free path λ = 70 nm.

♥ R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, 2nd Ed., Wiley, 2002. ♦ This is an average of the values of d for oxygen (0.3433 nm) and nitrogen (0.3667 nm).


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Fall, 2016

Constitutive Equation - model which describes how a material will respond to stress.

Other examples of body forces are electric and magnetic forces. In Lecture #23 begins here

Some familiar examples of constitutive equations include:

Electrohydrodynamics on page 170, we will consider electrical body forces.

• Hooke's law of elasticity (solids): stress is proportional to strain • Newton's law of viscosity (fluids): stress is proportional to the rate of strain

At the other end of the spectrum are forces which have very short range, such as those exerted through intermolecular collisions. The most common examples are hydrostatic pressure and viscous stress.

Some materials, like toothpaste and polymer melts, have characteristics of both solids and fluids and do not obey either of these simple “laws.” Such fluids are called viscoelastic. For an example of the strange behavior which such materials can display, see Lecture #12 begins here Non-Newtonian Behavior starting on page 75 below. Classification of Forces Having derived an equation (the Continuity equation) to describe the relationship among the variables which is imposed by conservation of mass, the remaining fundamental principle of physics is Newton’s second law (ΣiFi = ma) which, as it turns out, is equivalent to conservation of momentum. To apply this principle, we will need to calculate the forces which act on fluid systems. Forces tend to fall into one of two different categories, depending on the range over which they act: long-range (compared to intermolecular distances) forces are computed as volume integrals (called “body forces”) and shortrange forces are computed as surface integrals (called “surface forces”). Let’s start with the long-range forces, the most common of which is gravity. All fluid elements (not just those at the system boundary) feel a gravitational force of interaction with the rest of the universe outside the system boundaries; in particular, with the earth itself. Because the center of the earth is far from us, gravity is a long-range force. body forces: those which act on every fluid element in body (e.g. gravity): dFg = (dm)g = ρg dV where g is the acceleration of gravity (a vector pointing toward the center of the earth), dm and dV are the mass and volume of a fluid element. The total gravitation force on the system is just the sum of the force on each fluid element composing the system:

Fg =

∫ dFg = ∫ ρg dV V

Fig. 2

To understand how molecular collisions give rise to hydrostatic pressure, consider the elastic collision between two gas molecules shown in Fig. 2. Molecule #1 is outside the system (system boundary denoted by vertical black line with gray shading denoting the system-side of the boundary) and Molecule #2 lies just inside the system. As a result of the collision, Molecule #2 feels an impulse of force that causes a sudden change in its momentum (and velocity) shown as the black arrow. Since Molecule #2 is inside the system, this collision caused a change in momentum of the system as a whole. We have drawn the black arrow in Fig. 2 to be exactly normal to the boundary. This is only true on average. During a single collision, the direction of the force exerted on molecule #2 would coincide with the direction of a line drawn between the molecules’ centers at the moment of impact. In general, this direction will have components that are tangent to the surface as well as normal to the surface. However, averaging over many such collisions, the tangential components vanish whereas the normal component does not. Similarly, we have treated molecule #2 as if it is stationary when in fact both molecules are moving. Again if we average over many such collisions, the average velocity of molecule #2 is zero.

(54)

V



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Fall, 2016

the surface area da: the larger the area the more collision occur per unit time. To obtain the net pressure force acting on the system, we simply add up the contribution from each surface element da: Fp = − ∫ pn da

(55)

A

Similar collisions occur between molecules which are both inside the system. These interior collisions do not result in any change in the momentum of the system as a whole because the “action” and “reaction” forces cancel out (recall Newton’s third law of motion: for every action, there is an equal but opposite reaction). Only collisions between interior and exterior molecules result in a change of momentum of the system: While a single collision imparts only a sudden impulse of force, such collisions occur continuously at very high frequency. When integrated over a short time interval, these collisions collectively exert a continuous force on the surface which we call “pressure.” In other words, pressure is the continuum manifestation of intermolecular collisions.

where Fp is the net force exerted on the system enclosed by the closed surface A. Hydrostatic Equilibrium Newton’s♣ Second Law generally requires

∑ Fi = Ma , i

For a fluid at rest (both velocity and acceleration vanish) this reduces to

∑ Fi = 0 i

which is also equilibrium.”

referred

to

as

“mechanical

Gravity and hydrostatic pressure forces are usually the only forces acting on a fluid at rest, so Fg + Fp = 0 Substituting (54) and (55):

0 ∫ ρg dV − ∫ pn da =

V

surface forces: those which act only on surfaces (including mathematical boundaries) The hydrostatic pressure force exerted on a system by pressure p through an element of surface having area da and an outward pointing normal n is given by

We can combine these two integrals into one by making the domain of integration the same. This can be accomplished using one of the corollaries of the Divergence Theorem. Applying (52) with f = –p, the surface integral becomes

∫ ρg dV − ∫ ∇p da =0 V

dFp = –pn da where dFp is the force exerted on the system. For a proof that this is the correct form for hydrostatic pressure, see §1.3 in Batchelor.♠ We can rationalize the form as follows. The direction of the force is always normal to the surface and acting inward: this is consistent with the direction of the black arrow in Fig. 2. The magnitude of the force is proportional to

♠ G.K. Batchelor, An Introduction To Fluid Dynamics, Cambridge Univ. Press, 1970.


A

or

V

∫ [ρg − ∇p ] dV = 0 V

Since V is arbitrary, we conclude that the integrand vanishes at every point in the fluid, or ∇p = ρg

(56)

♣ Sir Isaac Newton (1642–1727), English mathematician and natural philosopher (physicist); considered by many the greatest scientist of all time; invented differential calculus and formulated the theories of universal gravitation, terrestrial mechanics and color.


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This says that the pressure increases in the direction of the acceleration of gravity (downward), which correctly describes (for example) how the pressure increases with depth in an ocean. Similarly, gravity causes the gas pressure to decrease as we move up in elevation to a mountain top or in an airplane. We have just introduced the topic of continuum mechanics by providing a bridge between molecular collisions and their continuum manifestation as hydrostatic pressure. When the fluid undergoes motion, a similar bridge can be provided between molecular collisions and viscous stress for Newtonian fluids. The interested student is referred to §1.1 in BSL.♥ A similar bridge can also be provided between molecular collisions and the diffusivity appearing in Fick’s law (see §17.1 in BSL) or the thermal conductivity appearing in Fourier’s law of heat conduction (see §9.1 in BSL).

♥ R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, 2nd Ed., Wiley, 2002.



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Flow of Ideal Fluids

Fall, 2016

To get the net inertia of the entire system, we must repeat this calculation for each of the fluid elements composing the system and add them up:

Ma =

Now let's consider fluids in motion. The simplest analysis is for an ideal fluid: ideal fluid - deformation of fluid elements is an isentropic process (i.e. adiabatic and reversible): k = 0 and µ = 0 where µ is the viscosity and k is the thermal conductivity. Generally this means that any viscous forces are negligible (since viscous forces represent friction arising between fluid elements and friction gives rise to irreversibility). Furthermore, to keep the process adiabatic, the thermal conductivity must also be negligible. Euler's Equation Suppose these conditions on the fluid are met. Thus consider the isentropic deformation of a fluid element for an arbitrary macroscopic system. In addition to pressure and gravity, we must also consider inertia when the system accelerates. Newton's law requires: Ma = ΣiFi

Dv

∫ ρ Dt dV

V

Newton's second law requires us to equate this with the net force acting on the system:

Dv

∫ ρ Dt dV =Fp + Fg =∫ ρg dV − ∫ np da

V

A   ∫ ∇p dV

V

V

where Fp and Fg are given by (54) and (55). Using the divergence theorem to convert the surface integral into a volume integral, we have three volume integrals over the same domain. Combining these three volume integrals leaves:

 Dv



∫ ρ Dt − ρg + ∇p  dV = 0

V

Since this must hold for any choice of V, the integrand must vanish at each point in the domain. After dividing by ρ:

Dv 1 = g − ∇p Dt ρ

(57)

which is called Euler's Equation (1755).* Thus any deviation from the mechanical equilibrium described by (56) causes acceleration in the fluid.

Let r(t) denote the trajectory of one particular fluid element inside the system. Then the velocity of the fluid element is: v=

Dr Dt

Another relationship among the unknowns is the continuity equation (44), which expresses conservation of mass. For an incompressible fluid, the continuity equation reduces to (45): ∇.v = 0,

(58)

To see that we now have as many equations as unknowns, note that the unknowns in (57) and (58) are

while the acceleration is:

v and p a=

Dv Dt

We use the material derivative here, since we are following a particular material point. Multiplying the acceleration by the mass of the fluid element gives the inertia:

( dm ) a = ρ ( dV )

Dv Dt


* Euler, Leonhard, 1707-83, Swiss mathematician. The most prolific mathematician who ever lived, he worked at the St. Petersburg Academy of Sciences, Russia (1727-41, 1766-83), and at the Berlin Academy (1741-66). He contributed to areas of both pure and applied mathematics, including calculus, analysis, number theory, topology, algebra, geometry, trigonometry, analytical mechanics, hydrodynamics, and the theory of the moon's motion.


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which represent the 3 scalar components of v plus p, for a total of 4 scalar unknowns (ρ and g are considered to be known). To evaluate these unknowns, we have equations (57) and (58):

1 ∇p = g − a ρ

Euler + continuity but Euler’s equation (57) is a vector equation, which can be expanded into 3 independent scalar equations — one for each component. When added to continuity (a scalar equation), we obtain a total of four independent scalar equations, the same number as of scalar unknowns. Thus we are now in position to begin solving problems involving fluid flow. We will call (57) and (58) “Euler’s equations of motion for incompressible fluids.” Using vector addition in the drawing above, we can see that the angle of inclination of the free surface (relative to the horizon) is just

a θ = tan −1   g

EXAMPLE. Water in a partially filled tank undergoes uniform♣ acceleration a in the horizontal plane. Find the angle θ of inclination of the water’s surface with respect to the horizontal plane. Solution. The key to solving this problem is to recognize that, regardless of the angle of inclination, the pressure is equal to 1 atm everywhere on the free surface, separating water from air. Recall from Geometric Meaning of the Gradient on page 6 that the gradient of any scalar field (e.g. p) is a vector ∇p which is normal to a p=const surface. Since the air/water interface is the p=1 atm surface, the pressure gradient ∇p must be normal to this plane. If we can find the orientation of ∇p we will also have the orientation of the free surface. Recall Euler's equation of motion for an ideal fluid: Dv 1 = g − ∇p Dt ρ  a

Dv/Dt is just the acceleration of the fluid in a stationary reference frame. At steady state, all of the fluid will undergo the same uniform acceleration as the tank; so Dv/Dt is just a. Solving for the gradient, we have

We were lucky in the previous example, because we knew the left-hand side of (57), so instead of 4 scalar unknowns, we only had one: p. The solution was relatively easy. In the more general problem, (57) is an unknown nonlinear partial differential equation:

∂v . 1 + v ∇v = g − ∇p ∂t ρ

(59)

In this form, we have expanded Dv Dt using (17). Now we have four scalar unknowns and four scalar equations. One important class of solutions has the form v = ∇φ, which is called “potential flow.” In the next section, we discuss how this form comes about and identify which fluid flow problems have this form. Kelvin's Theorem An important precursor to the theory of potential flow is the principle of conservation of circulation. Recall in (49) we defined a contour integral called the velocity circulation: Γ≡ ∫ v.dr C

where C is any closed contour. We showed that this contour integral is associated with the average angular momentum of fluid elements located on the

♣ By “uniform acceleration” I mean the same acceleration is experienced at each point and at each time. For a fluid, uniformity at each position occurs only in the steady state after a initial transient which is nonuniform.



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surface whose edge is C. Kelvin♦ showed that this velocity circulation is conserved: dΓ =0 dt

(60)

for any set of material points forming a closed contour in an ideal, incompressible fluid. This result is called Kelvin's Theorem. contours of material points at different times

Turning now to the first term of the integrand in (61): the gravitational acceleration g is just a constant vector, which can be written as the gradient of a scalar [also see (51)]:

g is the magnitude of g and z is the elevation of the mass. The quantity φg represents the gravitational potential energy per unit mass. Substituting (62) and (63) into (61), then combining the two gradients into one leads to

a material point Fig. 3

Partial proof:♥ Since the contour C is composed of material points, the time derivative of this contour integral is like the contour integral of the material derivative: ? dΓ d Dv . .dr v = = dr   ∫ ∫ dt dt ( ) Dt C t C(t )

Of course, we have not rigorously shown the second equality to be valid, thus we only claim the proof is “partial.” Next, we substitute Euler’s equation (57) to obtain 

1



(61)

C

Each term in the integrand can now be written as the gradient of a scalar field. The second term is already in the form of a gradient, provided we can treat ρ as a constant and bring it inside the gradient operator: ♦ Lord Kelvin (William Thomson), 1st Baron, 1824-1907, British mathematician and physicist; b. Ireland. He was professor (1846-99) of natural philosophy at the Univ. of Glasgow. His work in thermodynamics coordinating the various existing theories of heat established the law of the conservation of energy as proposed by James Joule. He discovered what is now called the “Thomson effect” in thermoelectricity and introduced the Kelvin scale, or absolute scale, of temperature. His work on the transmission of messages by undersea cables made him a leading authority in this field. ♥ For a more rigorous proof, see Batchelor p269. For a more intuitive proof, see Landau & Lifshitz, page 15.


(63)

φg = gz ,

where

C(t1)

∫  g − ρ ∇p .dr

(62)

This is only possible for an incompressible fluid.

g = –∇φg C(t2)

trajectory of a material point

dΓ = dt

ρ=const p 1 ∇p = ∇ ρ ρ

  dΓ p p =  −∇φ g − ∇ .dr =− ∫ ∇  φ g + .dr ρ ρ dt ∫   C C 

The integrand now has the form ∇s.dr which equals the total differential ds according to the implicit definition of gradient (2). Integrating any total differential around a closed contour yields zero (Theorem III). Thus the above equation reduces to (60). Integrating (60) yields Γ = const. or

∫ v.dr = const. w.r.t. time

C

Keep in mind that this applies only if C is composed of material points and only for an incompressible and ideal fluid. Since C is composed of material points (which in general move with different velocities), the contour may change shape or move as suggested by the schematic in Fig. 3. In the analysis leading to (50), we showed that the velocity circulation integral Γ is proportional to the angular momentum per unit mass. Thus Kelvin’s theorem implies that the angular momentum of ideal, incompressible fluid is a constant. Compare Kelvin’s theorem with Newton’s first law of motion: An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force The analogous statement of Newton’s first law for angular motion would be: An object at rest stays at rest and an object undergoing rotation continues to rotate with the same angular velocity unless acted upon by an unbalanced torque. PDF generated December 6, 2016

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Comparing this analogous statement of Newton’s first law with Kelvin’s theorem, we conclude that there is no way to apply a torque to an ideal fluid. To understand why this might be true, consider the following thought experiment: I wish to rotate a dinner plate. The usual way to apply a torque (with our hands to the plate, say) is to hold the plate between our hands and then move our hands in opposite directions, as shown in the sketch below.

Fall, 2016

Despite the distortion of the shape of the contour and the nonzero velocity of the material points forming it, Kelvin’s Theorem still requires

∫ v.dr = 0

for all t

although v ≠ 0

C

This is also true for every closed contour in the region (since every contour initially had zero value for this integral, when that contour lie far away from the moving sphere). Applying Theorem III: This results in a force-couple or torque being applied to the plate. In this thought experiment, we rely on friction between our hands and the plate to exert the torque. If the plate were greased and our hands slipped over its surface, we would not be able to apply the torque. This is the essence of what Kelvin’s theorem is saying: in the absence of friction (recall we are dealing with an ideal fluid), there can be no torque. Irrotational Flow of an Incompressible Fluid As an example, consider towing a submerged object through an “ideal fluid” which is otherwise stagnant.

∇×v = 0 for all r,t

which is sometimes called the persistence of irrotationality. Also from Theorem III, we know φ(r,t) exists such that: v = ∇φ

v(r,t=0) = 0 Any integral over a closed contour integral outside the disturbed region vanishes since v vanishes:

∫ v.dr = 0

for t=0

since v = 0

C

for every such contour C in the stagnant region. Now suppose at some later time the submerged object moves into the vicinity of C which causes v to be nonzero:

(65)

where φ is called the velocity potential. Significance: Knowing that the solution has the form given by (65) allows us to decouple the four scalar equations represented by (57) and (58) as we will now show. Substituting (65) into (58) yields Laplace’s equation in the velocity potential: ∇.∇φ ≡ ∇2φ = 0

Consider some region far ahead of the moving object. Outside the region disturbed by the moving object, the fluid is completely stagnant:

(64)

(66)

where ∇2 ≡ ∇.∇ is called the Laplacian (operator). Instead of 4 equations in 4 unknowns, we now have a single equation which (together with suitable boundary conditiions) can be solved for φ and v = ∇φ, without any coupling to Euler’s equation (57). Although Euler was the first to suggest this approach, (66) is called Laplace's Equation* after the French mathematician who solved this partial differential equation in so many cases. Knowing the velocity profile v, we can now determine the pressure profile p from Euler's equation:

∂v 1 + v.∇v = g − ∇p ∂t ρ

(67)

* Pierre Simon de Laplace (1749-1827), French mathematician and astronomer, noted for his theory of a nebular origin of the solar system and his investigations into gravity and the stability of planetary motion.



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We will now integrate this vector equation to obtain a single scalar equation for the pressure profile. Each term in (67) can be expressed as a gradient of something. For example, we’ve already seen in (62) that:

Fall, 2016

the pressure profile using Bernoulli's equation (70). Let's illustrate the procedure using an example: EXAMPLE: Find the velocity and pressure profiles for potential flow caused by a sphere of radius R moving through a stagnant fluid with velocity U.

ρ=const  p  1 ∇p = ∇   ρ ρ

and in (63) that:

g = –∇φg

For potential flow, the unsteady term becomes

∂v ∂  ∂φ  = ( ∇φ ) = ∇   ∂t ∂t  ∂t 

(68)

P.D.E.:

To evaluate v.∇v, we use one of the vector identities tabulated in Appendix I: Useful Identities in Vector Notation. In particular, identity A.3 gives:  v2  v ⋅∇v = 12 ∇ ( v ⋅ v ) − v × ( ∇ × v ) = ∇     2 

(69)

0

where

v.v = v2

where v = v. In our particular case, the second term vanishes because ∇×v = 0 for potential flow. Substituting (63), (62), (68) and (69) into (67):

Boundary conditions can be formulated by recognizing that fluid far from the sphere is unperturbed: b.c. #1:

v’ = 0 far from sphere

while fluid near the sphere cannot penetrate the sphere. To express this mathematically, recall our “bucket-and-stopwatch” method for defining fluid velocity (see page 12). Modifying it slightly to account for the movement of the surface element at velocity U, the flowrate across a moving surface element of area da is given by:

For an impenetrable sphere, the flowrate must vanish

which is called Bernoulli’s Equation.♣ It implies that (70)

is spatially uniform, but it might depend on time. Once the velocity profile is obtained (by solving Laplace’s equation), both φ and v are known, leaving p as the only unknown. Potential Flow Around a Sphere To solve a typical problem involving potential flow, we would first solve Laplace's equation (66) to obtain the velocity profile and then we can evaluate ♣ Daniel Bernoulli (1700-1782, Swiss) has often been called the first mathematical physicist; the teacher of Leonhard Euler. His greatest work was his Hydrodynamica (1738), which included the principle now known as Bernoulli's principle and anticipated the law of conservation of energy and the kinetic-molecular theory of gases developed a century later. He also made important contributions to probability theory, astronomy, and the theory of differential equations.


∇2φ' = 0

dq = n.( v ′ − U ) da = 0

 ∂φ v 2 p ∇ + + φg +  =0 2 ∂ t ρ  

p ∂φ v 2 + + φ g + = const w.r.t. r ∂t 2 ρ

Solution: If the fluid behaves ideally, it undergoes potential flow and the velocity profile must satisfy Laplace’s equation (66):

b.c. #2:

n.( v ′ − U ) = 0 on sphere

Owing to the moving boundary, the solution to this problem is inherently unsteady; i.e. φ' = φ' (r,t). To see that, recall that fluid elements initially far upstream have no velocity; as the moving object gets near these same fluid elements, they acquire a velocity; then as the object moves far past them, their velocity decays again to zero. This the velocity profile is inherently unsteady. To avoid the unsteadiness, we adopt a new reference frame in which the origin moves along with the center of the sphere. We denote the new velocity as v (without the “prime”), which is related to the old velocity by

v= v ′ − U

(71)

Now the object appears stationary but the fluid moves around it:


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where U is the velocity of the sphere in the original stationary reference frame and ez is the unit vector along the z-axis which we choose to align with the direction of the uniform flow in the moving coordinate system:

Let’s also align the z-axis with the direction of uniform flow (to the right); then the original sphere velocity (directed to the left) can be written as

U = −Ue z Regardless of the reference frame used to calculate velocity, the flow corresponds to potential flow and both velocities can be written as the gradient of some scalar potential. Then (71) can be written as

∇φ = ∇φ′ + Ue z

(72)

Regardless of the reference frame, the fluid velocity profile has to satisfy continuity (58). Taking the divergence of both sides of the above equation leads to ∇ φ =∇ φ′ =0 2

2

For a stationary sphere, no fluid enters the sphere which means dq = n.v da = 0, or b.c. #2:

n.v = vr = 0 at r=R

(74)

Next we rewrite the boundary conditions in terms of the velocity potential in spherical coordinates. In terms of velocity potential, (73) becomes: as r→ ∞:

∇φ = Uez

(75)

Now we need to translate ez into the unit vectors in spherical coordinates.*

since the divergence of a constant vector is zero:

∇.(Ue z ) = U ∇.e z =0 So it turns out that the differential equation does not change upon this shift in reference frame for velocity: the new velocity potential must also satisfy Laplace's equation: P.D.E.:

∇2 φ = 0

In particular, φ turns out to be independent of time t whereas φ' is not. If desired, we could also integrate (72) to obtain♥

φ (r ) = φ′ ( r,t ) + Uz As a check, you can take the gradient of both sides and recover (72).

Referring to the figure above (see page 9), we note that er, eθ, and ez all lie in the same Φ=const plane (shaded region of top-most figure above). If we shift all three unit vectors to the origin (recall that the origin is not part of the definition of any vector) and re-orient the Φ=const plane to coincide with the plane of the page, then we get the next figure:

While the differential equation is the same, the boundary conditions have changed. In this moving coordinate system, the sphere appears to be stationary and the fluid at infinity is undergoing uniform flow b.c. #1:

v → –U ≡ Uez as r→ ∞

(73)

♥ In writing this equation, we have dropped an integration “constant” which might be a function of t. For more detail see Appendix VII: Moving Reference Frame.


* Here Φ is the spherical coordinate, while φ is the velocity potential. For more on spherical coordinates, see page 9.


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we will try to find a solution which is independent of Φ: φ = φ(r,θ)

This implies that the fluid does not have any Φcomponent of velocity. In other words, the trajectory of any fluid element remains entirely on a single Φ=const surface. From trigonometry of the right triangle it is apparent that ez =

Here is a sketch (looking along the z-axis which passes through the center of the sphere into the page) showing some edge views of Φ=const planes.

( cos θ ) er − ( sin θ ) eθ

Thus (75) can be written as as r→ ∞:

∇φ → U cos θ er − U sin θ eθ

The components of the vector on the left-hand side are of course the components of the gradient in spherical coordinates, whose general formulas can be found in Appendix IV: Vector Operations in Spherical Coordinates (r, θ, φ) Equating corresponding components: ∂φ = U cos θ ∂r

and

1 ∂φ = U sin θ r ∂θ

When the potential function is independent of the angle Φ, each trajectory is confined to one of these planes.

Integrating either differential equation leads to b.c. #1:

φ → Urcosθ + const

as r→ ∞

where we have arbitrarily selected a value of zero for this “const”.* We next translate boundary condition #2 into spherical coordinates using Appendix IV. In particular, (D) through (F) of this table can be used to express the gradient in spherical coordinates: v = ∇φ ∂φ 1 ∂φ er + eθ vr er + vθ eθ + vΦ eΦ = r ∂θ ∂r 1 ∂φ eΦ + r sin θ ∂Φ

Dotting both sides by the unit vector n = er, then using (74):

b.c. #2:

∂φ n.v = vr = ( ∇φ )r = = 0 ∂r

at r=R

Since neither the azimuthal angle Φ nor the velocity component vΦ appears in these boundary conditions,

In Appendix IV, we find the formula for ∇2φ in spherical coordinates:

∂φ  1 ∂  2 ∂φ  1 ∂  r + 2  sinθ  =0 2 ∂θ  r ∂r  ∂r  r sinθ ∂θ 

(76)

∂φ = 0 at r=R ∂r

(77)

φ = Urcosθ as r→ ∞

(78)

Let’s now check on our assumption that the potential φ is independent of the angle Φ. Since neither the differential equation nor the boundary conditions depend on Φ, the particular solution is indeed independent of Φ. Although there are systematic procedures, like “separation of variables,” for solving partial differential equations which work in many cases, one should first check to see if there is a simple solution. The problem appears formidible, but notice that the trivial solution: φ=0?

* Like most energies in thermodynamics, the reference state for potential is arbitrary and can be chosen solely for convenience.


satisfies the partial differential equation and the boundary condition at r=R. Unfortunately, it does not satisfy the boundary condition at r→ ∞, so we will have to try another guess. Since the failure PDF generated December 6, 2016

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occurred with the boundary condition at r→ ∞, we look there for the form of our second guess: φ = Arcosθ ? This too satisfies the differential equation and the boundary condition at r→ ∞ (provided A=U) but to satisfy the boundary condition at r=R, A must be chosen as 0. This tells us that A should have different values at different r's. So we try a third guess which is slightly more general than the second: φ(r,θ) = f (r)cosθ

(79)

Fall, 2016

  R 3  ∂φ = vr = U 1 −    cos θ ∂r   r  

 1 ∂φ vθ = = −U 1 + 12 r ∂θ  vΦ =

∂φ 1 = 0 r sin θ ∂Φ

Below are several attempts to display this solution graphically:

Substituting (79) into (76)-(78), we find that cosθ cancels out, leaving:

vr = 0 vθ (r , θ = π )

r 2 f ′′ + 2rf ′ − 2 f = 0 f ′ = 0 at r=R

2

vr (r , π) vθ = 0

f = Ur as r→ ∞ Thus we have reduced the problem of solving the partial differential equation to one of solving an ordinary differential equation. We recognize this differential equation as a Cauchy-Euler equation,♣ which always has at least one solution of the form f = rn. The general solution turns out to be: f (r) = Ar–2 + Br Lecture #7 begins here

R3   sin θ r3 

vr (r ,0) vθ = 0 vr = 0 vθ (r , θ = 3π ) 2

Recall that θ is polar angle measured from the north pole, where the north pole is aligned with the z-axis and pointing the direction of flow distant to the sphere:

Using the boundary conditions, we can evaluate A and B. The particular solution to this problem is:

 R3  f= ( r ) U  r + 12 2  r   Substituting this back into (79) leaves:

 φ ( r ,= θ ) f ( r ) cos = θ U  r + 12 

R3   cos θ (80) r2 

♣ The general form of an Nth-order Cauchy-Euler equation is

aN x N

Notice that for θ=π and θ=0, vθ=0 but vr decreases (in absolute magnitude) from ±U at r=∞ to 0 at r=R. On the equatorial plane (θ=π/2) vr=0 for all r but vθ decreases from (3/2)U at r=R to U as r increases. This increase is necessary to make up for the decrease in flow caused by the sphere blocking part of the flow path.

dN y d N −1 y dy 0. + a N −1 x N −1 +  + a1 x + a0 y = 1 N N − dx dx dx

At least one of the N linearly independent solutions has the form y(x) = Axα. Substituting this form and dividing out

the common factor Axα, we obtain an Nth-order polynomial for α:

a0 + a1α + a2 α ( α − 1) +  + a N α ( α − 1)( α − 2 ) ( α − N + 1) = 0 Each distinct root of the polynomial leads to a separate solution. In this example, the roots are α = -2 and +1.



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Fall, 2016 p ( r,= θ ) p∞ − ρφ g + 12 ρ (U 2 − v 2 )

or

1.5

2 2 where v 2 = v.v = vr ( r , θ )  + vθ ( r , θ ) 

1 1

Substituting the known velocity profile, we obtain the pressure profile. Let's focus on the pressure profile over the surface of the sphere:

0.5

for r=R: p ( R, θ= ,Φ)

0

1

2

3

4

r/R

vr/Ucos(theta) vtheta/Usin(theta)

Below is a plot of the trajectories of materials points around the sphere.

p − ρφ g ∞  

hydrostatic head ph ( x )

+ 18 ρU 2 ( 9 cos 2 θ − 5 )  dynamic head pd ( R ,θ )

The hydrostatic pressure varies linearly with elevation (the x-axis is oriented vertically in our figure on page 40). The force arising from this contribution to pressure will be the subject of a problem on Hwk 4. We will here focus on the contribution from remainder, which is called the dynamic pressure (or the dynamic head). On the surface of the sphere, the dynamic pressure depends solely on θ.

These are called streamlines. In the next chapter, we will see how to calculate them. The actual solution for the stream lines will be delayed until Prob. 5 in Hwk #5. Having solved for the velocity profile, we can now determine the pressure profile from Bernoulli's equation (70). Assuming steady state: v2 p + + φ g = const 2 ρ

In the sketch above, we plot the dynamic pressure (after dropping the contribution from hydrostatic equilibrium).

independent of position. The "const" can be evaluated using any point where we know both the velocity and the pressure. Suppose the pressure of the undisturbed fluid is known in the reference plane for the gravitational potential♥: at r→∞, φg = 0: Thus

p = p∞ and v2 = U2

v2 p U 2 p∞ + + φ= + g ρ 2 ρ 2

♥ The reference plane is any horizontal plane in which we choose to define φg = 0. It might pass through the center of the sphere or it might be the air-water interface at some distance above the sphere.


The plot above indicates the local pressure using color (red = high, green = moderate, blue = low). The contours are isobars (pressure = const). PDF generated December 6, 2016

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Note the location of regions having high and low pressure. The sphere is being pushed in at the poles and allowed to expand at the equator. This is why a large bubble rising through stagnant water tends to become distorted from spherical shape (see below). Rising bubbles tend to become extended in the horizontal plane and compressed in the vertical direction by the higher pressure.

Fall, 2016

origin at the center of the sphere, this vector is the unit vector in the r-direction: n = er(θ,φ) whose direction depends on location on the surface. Although its length is a constant, its direction varies with position of the sphere; thus n cannot be treated as a constant. Anticipating that any net force will be parallel to the direction of fluid flow, we dot both sides of (81) by ez: Fp = e z .Fp = −

∫ e.n p da A

z cos θ

Since ez is a constant with respect to both θ and Φ, we can bring the dot product inside the surface integral as shown in the second equation above. Both ez and n are unit vectors, so their dot product is just the cosine of the angle between them, which can be seen to be θ in the figure below.

d'Alembert's Paradox What is the net force on a rigid sphere owing the pressure profile developed by potential flow around it? The answer turns out to be:

Fp = − ∫ n p da = − ∫ n ( ph + pd ) da A

A

= − ∫ n ph da − ∫ n pd da = −ρgV

(81) Fig. 4

A   A    −ρgV

0

where V is the volume of the sphere. The integral of the dynamic pressure leads to zero force while the integral of the hydrostatic pressure leads to a net upward force equal to the weight of the fluid displaced by the sphere (see Prob. 3 of Hwk #4); this last result is known as Achimedes Law (212 B.C.). So the total force is just the weight of the displaced fluid, regardless of the speed U of the motion. In particular, there is no drag, which might be expected to act to retard motion.

On the surface r=R the dynamic pressure pd(r=R,θ) depends solely on θ, so we choose the yellow strip (see Fig. 4) as our differential area da. This yellow strip has width R dθ and circumference 2πRsinθ so its area is

da =

( R d θ )( 2πR sin θ )

On this narrow (i.e. small dθ) strip θ and the dynamic pressure pd are constants. To obtain the net force, we just need to sum the contributions from each yellow strip for θ between 0 and π:

Proof: first consider the contribution from dynamic pressure:

pd ( R, θ ) ≡ 18 ρU 2 ( 9cos 2 θ − 5 ) The n in (81) is a unit vector normal to the surface, pointing outward. In spherical coordinates, with the Copyright© 2016 by Dennis C. Prieve


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Fdp = − ∫ cos θ pd ( R, θ ) ( 2πR sin θ )( R d θ )      f ( cos θ )

0

da

π

= 2πR 2 ∫ cos θ f ( cos θ ) ( − sin θ d θ )  d cos θ

0

x cos θ =

=

−1

2πR 2 × 18 ρU 2

∫

x ( 9 x 2 − 5) dx= 0

1   0

In the second equation above, we brought the constant 2πR2 outside the integral and also brought the minus sign inside. In the third equation, we made a change of dummy integral variable: x = cosθ. This results in the integrand being a simple polynomial in x which can be easily integrated. Since there was no Φ-dependence in the integrand, we choose the entire yellow strip as da in the evaluation of the above surface integral. If instead, our integrand was f (θ,Φ), we would need both θ and Φ to be nearly constant over da.

Fall, 2016

The net force on the sphere is the sum of its weight (i.e. gravity) and the net pressure force:

Fp + Fg = −ρgV + ρs gV = ( ρs − ρ ) gV ≠ 0 which is the difference in weight of the sphere and the weight of the fluid displaced by it. Because the pressure force is independent of the speed of motion of the sphere through the fluid, the particle will continue to accelerate forever, without ever reaching a “terminal velocity.” Of course, experiments show that falling particles do indeed reach a terminal velocity which implies that the gravitational force is balanced by some other force. The other force is fluid drag (friction), which is not predicted by potential flow. This serious discrepancy between the predictions of potential flow theory and experiment is known as: d'Alembert's♣ paradox - potential flow predicts no drag but experiments indicate drag. Despite this, potential flow is still useful:

Uses of potential flow – predicts lift (but not drag) on streamline objects moving through stagnant fluid at high Reynolds numbers (but still sub-sonic, i.e. v

l c

where τ = time over which significant changes in v occur



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Viscous Fluids Up to this point, we have considered only the flow of ideal fluids which deform reversibly with no conversion of mechanical energy into heat. Our analysis of flow of an ideal fluid around a symmetric body like the sphere (see Potential Flow Around a Sphere) showed that the force exerted by the fluid on the sphere was the same as for hydrostatic equilibrium (see page 45). In other words, motion of an ideal fluid is predicted to exert no drag force although drag is evident in experiments. This contradiction is known as d’Alembert’s paradox. To resolve d’Alembert's paradox, we need to introduce another force into our force balance. This force can be thought of as friction between the fluid elements. Friction gives rise to viscous heating which represents an irreversible conversion of mechanical energy into heat. Indeed, friction is the main difference between an ideal fluid and a real fluid. friction → irreversible deformation → nonideal flow

Fall, 2016

Besides giving rise to tangential forces on surfaces, friction is also the cause of irreversibility. For purely elastic collisions, both energy and momentum are conserved; but intermolecular forces can cause the collisions to be inelastic. In other words, kinetic energy is not conserved. Some of the mechanical energy of the collision is converted into heat. This degrading of the mechanical energy into heat is the source of irreversibility. Tensorial Nature of Surface Forces So friction is a surface force like hydrostatic pressure, but unlike pressure, friction is not isotropic:♣ isotropic - independent of orientation (direction) We say that pressure is isotropic because the magnitude of the pressure force is independent of the orientation of the surface on which it acts. Recall: dFp = –npda |dFp| = pda

which is independent of n (orientation).

When fluid molecules strike a z=const surface, they always impart a z-component of momentum, which we have previously identified as the molecular origin of “pressure” (see page 32). These same collisions can also impart x- and y-components of momentum to the z=const surface.

Viscous friction does not have this property: |dFf| depends on n

When the fluid is stagnant (see above), the average of many such collisions results in no net x- and ycomponents of momentum transferred.

But if there is a net flow in the x-direction (see above), then collisions with a z=const surface will results in a nonzero x-component of momentum transferred, as well as a nonzero z-component. Thus the shaded region feels a net force in the x-direction arising from x-motion of the fluid in the unshaded region. This is molecular origin of “friction”.


But it turns out that it's magnitude is proportional to da: dFf ∝ da

Thus it makes sense to talk about the force per unit area, which we usually call pressure. In the more general case in which the magnitude of the force per unit area might depend on the orientation of the surface, we use the name stress. In general, all surface forces can be lumped together and written as dFsurf ≡ t(n) da where t(n) denotes the surface force per unit area acting on the body through the surface element

♣ don’t confuse “isotropic” with “isentropic” which means no entropy is generated.


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whose orientation is given by the unit outward normal n. We call this quantity the stress vector.* In the analysis which follows, I will try to convince you that the effect of orientation of the surface can be expressed as t(n) = n.T where T is called the stress tensor. While T depends on position and possibly time, it does not depend on the orientation of the differential surface element da, which is given by n. T is sometimes called the “state of stress” of the fluid. First, let’s try to understand better what we mean when we say a material experiences some “stress”. Consider a block of some material which bears some externally applied equilibrium load (see figure below).

Fall, 2016

t(n1)A = +F

(104)

while block “1” exerted an equal but opposite force on block “2” t(n2)A = –F

(105)

When expressed per unit area, this internal force between the two blocks (when they are physically connected) is what we mean by the “stress” experienced by the material under load. Eliminating F and A between (104) and (105) yields t(n2) = –t(n1) Next we substitute n2 = –n1: −t ( n1 ) t ( −n1 ) =

More generally, for any mathematically surface having orientation n, we can write t ( −n ) =−t ( +n )

By “equilibrium” I mean there is no net force and no net torque applied to the body: thus it is at mechanical equilibrium and has no tendency to accelerate. The material might be a fluid or a solid, but what we are about to say is easier to imagine if we think of material as a solid block. Consider some mathematical surface inside the material (indicated by the dotted line in the figure above). What are the forces exerted across this mathematical surface? To answer this question, suppose we actually separated the block into two pieces by physically cutting along this surface without changing the loading.

(106)

In essence, this is just a statement of Newton’s Third Law (for every action, there is an equal but opposite reaction). Now let’s generalize the load force to three dimensions. Suppose we know the distribution of stress (i.e. the “loading”) on all six faces of a block of material (see sketch below):

t(-j)Ay t(k)Az t(-i)Ax

t(i)Ax t(-k)Az t(j)Ay

Block “1” now experiences an unbalanced load –F while block “2” experiences an unbalanced load +F. Once separated, both blocks would tend to accelerate in opposite directions. Why don’t the two blocks accelerate when connected? Apparently, block “2” must have exerted a force on block “1” which we denote as t(n1)A, where A is the area of the cut face, and which equals *see Whitaker, Introduction to Fluid Mechanics, Chapt. 4.


Furthermore, suppose this loading is applied uniformly over each face. In other words, the x=0 face has a stress of t(–i) distributed uniformly over the face having an area Ax; thus the force on that face is t(–i)Ax. Furthermore, suppose this loading is an equilibrium loading (no net force and no net torque on the block). In other words, the force t(–i)Ax applied to the x=0 face is exactly balanced by an equal but opposite force t(i)Ax applied to the other x=const face so that no net force is applied along the x-axis, and similarly for the forces on the other four faces. Let’s try to calculate the internal stress on the diagonal plane indicated as the shaded surface in the


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sketch above. The orientation of this plane, denoted by the unit outward pointing normal n (see sketch below), can be made arbitrary by varying the lengths (OA, OB and OC) of the three mutually perpendicular edges of the parallelopiped.

n A

z

x

t(i), t(j), and t(k)

Find:

∫ t ( n ) da = 0 A

(107)

Next we apply (106). Of course, we can replace n in (106) with i or j or k. Using (106) three times in (107): t(n) = (n.i)t(i) + (n.j)t(j) + (n.k)t(k)

t(n) = n.it(i) + n.jt(j) + n.kt(k) Factoring out the common dot product from each term: t(n) = n.[it(i) + jt(j) + kt(k)] = n.T

We evaluate this surface integral by subdividing the surface A into the four faces of the tetrahedron: planes ABC, BCO, AOC, and ABO. For each surface, we need to evaluate the outward normal n, the stress vector t and the surface area A. The results are tabulated below: Outward Normal n –i –j –k

t(n) = –(n.i)t(–i) – (n.j)t(–j) – (n.k)t(–k)

Recalling the definition of dyadic product, we could re-write this expression as

t(n)

Solution: We choose the tetrahedron ABCO as our “system.” For the surface forces to be balanced*

Plane ABC BCO AOC ABO

+ ( n.k ) t ( −k )  A

Solving for t(n):

Problem: given the surface stress on mutually perpendicular planes [i.e. given t(i) for x=const plane, t(j) for y=const plane, and t(k) for z=const plane], calculate the surface stress on a plane of arbitrary orientation specified by the unit normal n. Given:

A

t(n) + (n.i)t(–i) + (n.j)t(–j) + (n.k)t(–k) = 0

O

C

∫ t da =0 = t ( n ) + ( n.i ) t ( −i ) + ( n.j) t ( − j) Dividing by A cancels the A out of our expression for the integral. Taking the limit as all the dimensions of the block vanish (i.e. as A→0) allows us to replace the unknown averages with their limit, which is the value of the vector at the point that the tetrahedron collapses about.

y B

Fall, 2016

Area A . n iA n.jA n.kA

Stress Vector t(n) t(–i) t(–j) t(–k)

Applying the Mean Value Theorem to our surface integral:

The sum (inside square brackets) of these three dyadic products is some second-rank tensor, which is independent of n. We denote this tensor by T. Significance: to calculate the surface force on some differential surface area da having orientation n, we just multiply this expression for the stress vector by the area: dFsurf = n.T da

(108)

where T is called the stress tensor and dFsurf represents the net surface force (all contributions) acting on the body whose outward normal is n. While T does not depend on n, it might depend on position inside a solid which is nonuniformly stressed; thus T = T(r)

*More generally, in our force balance, we should include body forces and inertia as well as surfaces forces. If we let V→0 then the volume integrals for body forces and inertia vanish more rapidly than surface forces. Thus for the surface A enclosing a tiny volume V, we must still require

∫ t da = 0

for the surface of every differential volume

represents the state of stress of the fluid. In terms of the components of the tensor, we say that Tij = jth component of the force/area acting on a ri = const surface

A

element.



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To make this more understandable, it might help to express T in terms of other variables in a familiar problem, say an ideal fluid.

For a real fluid, there is an additional contribution to T from friction:

For an ideal fluid:

where τ is called the viscous stress or the deviatoric stress (since it represents the deviation from ideal).

T = –pI

(109)

To show that this is correct, we will compute the surface force and show that it reduces to the familiar force due to pressure: dFsurf = n.Tda = n.(–pI)da = –p(n.I)da = –pnda = dF p

Lecture #10 begins here Generalization of Euler's Equation Recall that Euler’s equation was derived by applying Newton's Second Law (F=ma) to any fluid element. To generalize this result to include friction, we replace the “pressure force” by the surface force.

Ma= Fg + Fp

Instead of:

Dv

∫ ρ Dt dV = ∫ ρg dV − ∫ np da

or

V

V

Ma= Fg + Fsurf

we have: or

A

Dv

∫ ρ Dt dV = ∫ ρg dV + ∫ n.T da

V

V

Applying the Divergence Theorem to the surface integral and combining the resulting three volume integrals:

 Dv  ∫  ρ Dt − ρg − ∇.T  dV = 0 V

(110)

which is a more generalized version of Euler's Equation.

and

T = –pI

( )

∇ ⋅ T = −∇ ⋅ p I

identity C.11

=

− ∇p

For a proof of identity C.11, see Hwk #3, Prob. 5. Substituting the above result into (110) reduces to Euler’s equation for an ideal fluid (57). Copyright© 2016 by Dennis C. Prieve

Sign Convention for Stress By convention in these notes, Tij and τij are positive for tensile stresses (which arise for example inside a solid rod being stretched). This convention is common in material science which often involves experiments exerting tensile stresses in material. This is also the convention used by Whitaker.♥ Other books (e.g. BS&L♠) choose the convention that compressive stresses are positive. In contrast to Tij and τij, the isotropic pressure p (as we have defined it) is positive for compressive stress. This makes sense for fluids because the thermodynamic pressure of a gas or liquid can never become negative: no matter how much you expand a gas (in a piston for example), the gas pressure will just tend to zero. This opposite-sign convention for p and Tij is evident from the minus sign appearing in (109). This opposite-sign convention for p and Tij is also evident in the formulas for calculating the net surface force: compare

to the net pressure force

Fp = − ∫ np da A

Since this must result for all choices of V in some region, the integrand must vanish at every point in that region, or:

For an ideal fluid:

(111)

Fsurf = ∫ n.T da

A

Dv 1 = g + ∇.T Dt ρ

T = –pI + τ

exerted by the fluid on the body enclosed by A, where n is the unit vector to the differential surface element which points out of the body. Notice the opposite signs in the two equations above. Resolving Total Stress into Pressure and Deviatoric Stress In our derivation of (108), T represents the total stress from all surface forces. In solving problems, we customarily separate this total stress into an isotropic pressure p and a deviatoric stress τ as in (111). ♥ S. Whitaker, Introduction to Fluid Mechanics, PrenticeHall, 1968. ♠ R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, Wiley, 1960.


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When written in terms of scalar components, (111) represents 9 equations in 10 unknowns (the nine components of τ plus p). Clearly p and τ cannot yet be uniquely determined from (111) alone, given the state of stress T. One additional relationship is required. For a real fluid, we somewhat arbitrarily define p as the average of the three diagonal components of T:

(

p ≡ − 13 T: I

)

(112)

Fall, 2016

Momentum Flux Up to this point we have spoken of T as the stress on the fluid. T can also be thought of as momentum flux. To convince yourself that momentum is being transported (like heat and mass), consider the problem of unsteady simple shear flow. At time t=0, an initially quiescent fluid confined between infinite parallel plates is disturbed by imposing motion on the upper plate. The velocity profile gradually develops into linear shear flow.

where

T: I =  ∑ ∑ Tij ei e j : ∑ ek ek    i j   k    = ei .ek ) ( e j .ek ) ∑ Tkk ∑ ∑ ∑ Tij (     k i j k δ δ jk

ik

or, in Cartesian coords.: T:I = Txx+Tyy+Tzz In any coordinate system, T:I is called the trace of T. Then p can be thought of as the isotropic contribution to stress (that part of the normal stress which acts equally in all directions) while τ represents the remainder, or the nonisotropic part. (111) might be regarded as decomposing τ into an isotropic part and a nonisotropic part, which is a common thing to do with tensors. Although we will prove it, the choice of p made in (112) also represents the thermodynamic pressure; i.e. the p appearing in the thermodynamic equation of state:

Note that, like all other fluid elements, the fluid element at y=h/2 undergoes acceleration:

∂vx >0 ∂t at y=h/2

ρ = ρ(p,T)

In other words, the fluid element is gaining momentum. How can it acquire momentum? Answer: momentum is transported to y=h/2 from the moving plate through friction between fluid elements.

To summarize, we decompose the state of stress into two contributions: an isotropic pressure:

There are two directions associated with transport of momentum:

(

p = − 13 T: I

)

1) direction of the momentum being transported (in this example, x-momentum is transported) 2) direction in which transported (y)

and a deviatoric stress: τ = T – (1/3)(T:I)I

∇.T = –∇p + ∇.τ

is

–T = diffusive flux of momentum*

and substituting it into Euler's Eq. (110): Dv = ρg − ∇p + Dt

momentum

For this reason momentum flux must be a 2nd rank tensor. It turns out that:

Taking the divergence of (111):

ρ

the

∇.τ 

new for real fluids


and

ρvv = convective flux of momentum

(113) * "Diffusive" means it results from random collisions between molecules.


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In multicomponent systems, the flux of any species i due to convection is written as the product of concentration and velocity: civ [=] mol-cm–2-s–1 [=] rate/area ci [=] mol/cm3

where

More generally, by “flux,” I mean: flux —that tensor (of whatever rank), which when pre-dotted by nda, gives the rate of transport through the surface element having area da in the direction of its unit normal n. For example, the flux of fluid mass by convection is ρv [=] g-cm–2-s–1 [=] rate/area Proof: pre-dotting by nda we obtain: (nda).(ρv) = ρn.vda = ρ(dq) [=] g/s which represents the mass flowrate across da. Indeed the convective flux of total fluid mass also has the form of concentration times velocity since ρ = concentration of mass [=] mass/vol. It might come as less of a surprize that ρvv is the convective flux of momentum if you realize that ρv = conc. of momentum [=] momentum/vol. After all, momentum is mass times velocity. So (ρv)v = ρvv = conv. flux of momentum

Fall, 2016

∂ ( ρv ) + ∂ t   

∂ρ ∂v v +ρ ∂t ∂t

∇.( ρvv )   

− ∇.T − ρg = 0

∇.( ρv )  v +ρv.∇v

Expanding the partial time-derivative of the product and expanding the divergence of the dyadic product using Identity C.8. Now we have two scalars which are each multiplied by v; collecting them together: ∂v  ∂ρ  .  ∂t + ∇ ( ρv )  v + ρ ∂t + ρv.∇v − ∇.T − ρg = 0    0

Finally, we recognize the factor inside square brackets must vanish according to the general continuity equation. After this term is dropped, the remainder is Euler's equation (110): ρ

∂v + ρv.∇v = ∇.T + ρg ∂t

Comment: In writing the statement of conservation of momentum, we have a term representing transport of momentum into the system by the action of external forces. Clearly, an object falling from rest in a gravitational field acquires momentum through the action of gravity. Does this acquisition represent transport to the body from outside, or does it represent a "generation" term? If it is generation, then momentum is not conserved.

Solution: We choose a system that has fixed boundaries in the laboratory reference frame: in other words, V ≠ V(t).

With a little reflection, we can convince ourselves that the action of gravity is “transport” and not “generation.” Some of the earth's momentum is being transported to the falling object. When the object eventually collides with the earth and comes to rest again, that momentum is transported back to the earth. The total momentum of the universe has not changed: momentum is conserved.

Referring to the discussion above in which ρv is the concentration of momentum, ρvv is the momentum flux by convection, and –T is the momentum flux by diffusion, conservation of momentum requires

Actually, the term we call “diffusion of momentum” also arises from the action of an external force: this time it's the action of “surface” forces, rather than “body” forces.

Example: Apply conservation of linear momentum to an arbitrary fluid system. Thus prove that T and ρvv are momentum fluxes, as claimed.

( )

d ρv dV =−  ∫ n.ρvv da − ∫ n. −T da + ∫ ρg dV dt ∫ V A A  V        

accumulation

in by convection

in by diffusion

in by external forces

Applying the divergence theorem, combining the resulting volume integrals, and invoking the result for arbitrary V, we would obtain


Response of Elastic Solids to Normal (Uniaxial) Stress By generalizing, the force balance plus the mass balance now contain more unknowns than equations: τ, v, and p → 9+3+1 = 13 unknowns Euler + Continuity → 3+1 = 4 equations Missing is the constitutive equation which is an empirical description of how the fluid or solid material responds to stress. Obtaining this


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relationship is an important objective of that field of science known as continuum mechanics. Let's start by considering solids, whose response is more familiar. Suppose I try to stretch a rectangular bar by applying a tensile force, F.

We will attempt to describe the response of the bar under conditions of mechanical equilibrium. To have the bar stretch instead of accelerating as a result of the force, I must apply an equal but opposite force to the other end.

Experiments show that the strain increases with the applied stress as shown in the figure above. At low stress levels, the strain is directly proportional to the applied stress: Hooke’s law for purely elastic solids is: Txx = Eεxx E [=] force/area

Let the x-axis be aligned with the direction of the applied force:

where E is called Young’s Modulus. It turns out that deformation also occurs in the y and z directions:

Fx = |F|

εyy = εzz = –νεxx

At equilibrium, the length of the bar will increase by an amount δx. Hooke’s* experiments revealed that:

where ν is called Poisson’s ratio and these new components of the strain tensor are

F L δx ∝ x x L y Lz

but

Fx = Txx L y Lz

is the applied stress. The two subscripts on stress denote the two directions associated with it: the first subscript denotes the orientation of the surface the force is applied to (x=const) while the second denotes the direction of the applied force (x). Since the deformation is proportional to the original length of the bar, it makes sense to define the deformation per unit length, which is called: strain:

δy ε yy = Ly

and

δ ε zz =z Lz

For most materials, Poisson’s ratio lies in the interval 0.28 < ν < 0.33 Response of Elastic Solids to Shear Stress Instead of a normal force, suppose I apply a tangential force on the upper face. To keep the object from translating, I must apply an equal but opposite force on the lower face.

δ ε xx =x Lx

Later we will see that strain is a second-rank tensor. For now, we will interpret the order of the subscripts as follows: the first subscript gives surface (in the above case, an x=const surface), while the second subscript gives the direction of the deformation (in the above case, the x-direction).

This generates a force couple or torque, which will cause the body to spin. To prevent a steady increase in rotation speed, I must apply an equal but opposite torque. Recalling that torque is the force times the lever arm: FxLy = FyLx Dividing by LxLyLz:

* Robert Hooke (1635-1703). English experimental physicist. Hooke’s law was first stated in 1660.


Fy Fx = Lx Lz L y Lz


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Tyx = Txy

or

where Tyx = x-component of force/area acting on y=const surface. Note that this implies that the stress tensor is symmetric. This symmetry of the stress tensor turns out to be true for virtually all equilibrium loadings (no net force or torque).* After the load is applied, both material points might move as shown by the dashed lines in the figure above. After the load the position of the second material point relative to the first becomes x+δ, where δ(x) = relative displacement vector This loading produces a shear deformation in which δx ∝ Ly

of two material points initially located at x. resulting strain can be generalized as: t = ε 12 ∇δ + ( ∇δ )   

δy ∝ Lx

and

 ∂δ j ∂δi = εij 12  +  ∂xi ∂x j 

The corresponding definition of shear strain is:



1  δx 2L  y

+

or

δy  ε xy = ε yx = Lx 

Since we applied a symmetric stress, by applying forces in both the x and y directions, we average the strains in the x and y directions to obtain a symmetric strain. Moreover, this definition yields a strain which is invariant to rotation of the xy axes. As with normal stress, shear stress produces a shear strain in direct proportion to the stress: Txy = 2Gεxy

The

(pure shear)

Note that this general definition of strain reduces to earlier expression for strain in the cases of pure normal or pure shear stress. For unixial normal stress: x  δx 1  ∂δ x + ∂δ= ε= xx  2  ∂x ∂x  Lx 

Txx = E εxx

and For pure shear:

 ∂δ y ∂δ x  1  δ y δ x 1 ε= + = +   xy 2  ∂x ∂y  2  Lx L y 

where G is call the modulus of elasticity for pure shear. Although the units are the same, the value of 2G is different from that of E. Generalized Hooke's Law Now let's try to generalize to some arbitrary loading which might involve both shear and normal stresses. Consider two arbitrary material points in the material. Let x denote the position of the second material point relative to the first, before the load is applied (see figure below).

   

   

Txy = 2G εxy

and

If the strain components are all small, we might reasonably suppose that Hooke’s law generalizes into a linear relationship between any component of strain and the nine components of stress (or vice versa):

= Tij

3

3

∑ ∑

k 1 =l 1 =

Cijkl ε kl

(114)

There are then nine coefficients for each component of stress, making a total of 81 possible coefficients. But just by requiring: •

symmetry of T (Tij=Tji) and

* For a proof based on the assumption of local equilibrium, see Whitaker, Section 4.3



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isotropy of material (e.g. same Young’s modulus applies to uniaxial stress in xdirection as for y and z)

it can be shown that the number of independent constants is reduced from 81 to only 2. One way to generalize Hooke’s law with two constants is to partition the strain tensor ε into two parts: one is the isotropic part [(ε:I)I] and the second part is the remainder [ε – (ε:I)I]. Denote the two independent constants as k1 and k2 and use them as weighting factors in the sum:

( )

symmetric stress tensor. From our experience with fluids in the lab, we know that eventually the upper plate will slide past the lower plate at some steady By performing many such relative speed Ux. experiments, varying the magnitude of the applied force and each of dimension, Newton concluded that Ux ∝

isotropic

remainder

This can be rearranged to

( )

T= k1 ε + ( k2 − k1 ) ε: I I Instead of k1 and k2 as the two coefficients, most textbooks use the following notation

( )

T= 2ηε + λ ε: I I This equation represents the generalization of Hooke’s law of elasticity. Thus of G, E and ν, only two are linearly independent properties of the material; using the above relationship we can show (Hwk Set #5, Prob. 1) that E −1 2G

Response of a Viscous Fluid to Pure Shear Suppose I have a thin layer of fluid held between parallel plates and I apply a force Fx to the upper plate. To have an equilibrium loading, I must apply an equal but opposite force to the lower plate –Fx.

Fx L y Lx Lz

(115)

This speed represents a rate of deformation: = Ux

( )

= T k1 ε − ε: I I  + k2 ε: I I     

= ν

Fall, 2016

d δx ∝ Ly dt

Noting that the speed is proportional to the thickness Ly, the speed per unit thickness represents a rate of strain:

Ux = Ly

d δx  δ x  d ε yx dt d = =   Ly dt  L y  dt

(116)

Finally, the applied force divided by the area over which it's applied represents a shear stress: Fy F τ xy = Txy = = x = Tyx = τ yx L y Lz Lx Lz

Rewriting (115) as an equality using (116) and the equation immediately above, we have: τ yx = µ

d ε yx dt

where µ is a material property call the viscosity. The above equation is called Newton's Law of Viscosity (1686). Noting the linear velocity profile in the figure above, the slope of this linear profile is an alternate measure of U x dvx = Ly dy

which is called the shear rate. Newton’s law of viscosity is sometimes also written as τ yx = µ

dvx dy

Generalized Newton's Law of Viscosity As before, this produces a force couple or torque. To keep the system from rotating, I must apply an opposing torque using the force-couple of ±Fy. Once again this equilibrium state corresponds to a Copyright© 2016 by Dennis C. Prieve

This generalization of Newton's law of viscosity to arbitrary loading parallels that of Hooke’s Law with the strain tensor replaced by the rate of strain.


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The main difference is that now the deformation is a function of time:

Fall, 2016

Writing in terms of the velocity v in RCCS:

 ∂v ∂v j = dij 12  i +  ∂x j ∂xi 

   

 ∂v ∂v  ∂v1 d11 = 12  1 + 1  =  ∂x1 ∂x1  ∂x1 so δ(x,t) = v(x) t = displacement of material pt.

= deformation The trajectory of a material point initially located at x(0) is given by:

So the trace of the rate of deformation tensor is just the divergence of the velocity. Newton's law of viscosity becomes:

(

Keeping the material point fixed (i.e. x(0) is constant) while we take the time derivative is the same as taking the material derivative of position. Thus the rate of deformation of fluid elements is: d δ Dx = = v dt Dt

which is given by the fluid velocity. Just like we define the gradient of the deformation to be the strain: = ε

Dε Dt

∇.v = 0

leaving

(117)

Newton’s law of viscosity is a linear relationship between the stress and the rate of strain. Generalizing that linear relationship again leads to a relationship like (114), except that the strain tensor εkl is replaced by the rate of strain tensor dkl. Once again, in order for the stress tensor T or τ to be symmetric and for the material to be isotropic, only two of the 81 coefficients Cijkl are independent. We use the two independent coefficients to multiply the isotropic part of d and its remainder:

τ = 2µd +

(

κ − 23 µ

t τ = 2µd = µ ∇v + ( ∇v )   

(119)

where the second equation above is obtained by substituting (117). Navier-Stokes Equation Once again, let’s return to Euler's equation, generalized to account for the tensorial nature of viscous friction, which is given by (113):

∇δ + ( ∇δ )t  ,  

t = 12 ∇v + ( ∇v )   

(118)

For an incompressible fluid:

the gradient of the rate of deformation must be the rate of strain: d=

)

τ = 2µd + κ − 23 µ ( ∇.v ) I

x[x(0),t] = x(0) + δ[x(0),t]

1 2

∂v1 ∂v2 ∂v3 d :I = + + = ∇.v ∂x1 ∂x2 ∂x3

) (d:I ) I

ρ

Dv = ρg − ∇p + ∇.τ Dt

(113)

For an incompressible fluid, ∇.v = 0 and Newton’s law of viscosity (118) simplifies to (119). Taking the divergence of both sides of (119): ∇.τ = µ[∇.∇v + ∇.(∇v)t]

but using identity C.10: ∇.(∇v)t = ∇(∇.v) = 0

leaving

∇.τ = µ∇.∇v = µ∇2v

Euler's equation becomes: ρ

Dv = ρg − ∇p + µ∇ 2 v Dt

(120)

where µ is the usual viscosity and κ is called the second coefficient of viscosity. d:I has a special significance, which we will now point out. As we showed on page 58, d:I is the trace of d: d:I = d11 + d22 + d33 Copyright© 2016 by Dennis C. Prieve


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64

which is known as the Navier-Stokes Equation (1822).♣ Now we have as many equations as unknowns: v, p → 4 scaler unknowns NSE, Continuity → 4 scaler eqns Boundary Conditions But to successfully model a flow problem, we need more than a sufficient number of differential equations. We also need boundary conditions.

A typical boundary is the interface between two immiscible phases -- either two fluids or a fluid and a solid. One such boundary condition which can often be applied is the no slip condition: vI = vII

Fall, 2016

vθ = 0

at r=R:

Note that our potential flow solution did not satisfy this second equation: for pot’l flow:

vr(R,θ) = 0

vθ(R,θ) = –(3/2)Usinθ ≠ 0 In addition to d'Alembert's paradox, potential flow fails to satisfy the no-slip condition.* For a fluid-solid interface, in which the velocity of the solid phase is known, the no-slip condition is sufficient. But in the case of fluid-fluid interface, the velocity of the second fluid is usually also unknown. Then no-slip just relates one unknown to another unknown.

A second boundary condition can be obtained by considering the stresses acting on the material on either side of the interface. Suppose we were to apply a loading as on page 55 to a two-phase region straddling the interface, as suggested in the figure at right. Note that the loading is balanced: that is there is no net force on the system.

For example, in the problem of uniform flow around a stationary solid sphere, this requires: at r=R:

vf = vs = 0

Of course, this also implies that there is no flow of fluid across the boundary: at r=R:

vr = 0

We also assumed this boundary condition when we solved the potential flow problem. But “no slip” also means:

♣ Named for Claude Louis Navier (1785-1836): French engineer and physicist who specialized in mechanics. He formulated the general theory of elasticity in 1821. and Sir George Gabriel Stokes (1819-1903): British (Irish born) mathematician and physicist, known for his study of hydrodynamics. Lucasian professor of mathematics at Cambridge University 1849-1903 (longest-serving Lucasian professor); president of Royal Society (18851890).


If we were to split the system into two parts along the interface, each of the two halves would tend to accelerate. This suggests that, when the two phases are in contact, each exerts an “internal” force on the other, as shown in the figure at right. These forces are equal but opposite: F1on2 = – F2on1 = F which might be thought of as “Newton’s Third Law”: for every action, there is an equal but opposite reaction. Using (108) (on page 56) to express the forces in terms of the stress tensor:

* Although "no slip" is usually applicable, there are at least two situations where no slip might not be applicable: 1) when the mean-free path of gas molecules is comparable to the geometric dimension, and 2) when a liquid does not wet the solid.


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n 2 .T A = −n1 .T A = F 1

Fall, 2016

completely immersed in an infinite liquid at hydrostatic equilibrium (v = 0).

2

where A is the area of the interface and where ni is the normal to the interface pointing out of phase i. However, from the geometry, we can deduce that

Solution: Selecting phase “1” as the exterior and phase “2” as the interior of the sphere, the two unit vectors are shown below:

−n 2 ≡ n n1 =

Newton’s Third Law becomes:

n.T = n.T 1

2

(121)

In terms of the stress vector, this condition is identical with (106): The general boundary condition is given by (122):

t(n) = – t(–n) When the interface is highly curved (e.g. a small oil droplet in water), then surface tension can produce a discontinuity in the normal components of the above force, which has not been included in the above analysis [see L&L, Chapt. 7 or Hunter, Vol I., p237f]. The more general form of this boundary conditon is n2.(T2 – T1) = γ(∇s.n2)n1 – ∇sγ

(122)

where γ is another property of the fluid called the surface tension and ∇s ≡ (I – nn).∇

is the surface component of the ∇ operator and ni is the unit normal pointing out of phase “i ”. To calculate the surface divergence, we make use of property C.12: ∇s.n = I:∇sn

n2.(T2 – T1) = γ(∇s.n1)n2

where we have dropped the ∇sγ term because surface tension is uniform over the surface of the sphere in this problem.♦ The surface gradient is given by ∇ s n 2 =( I − n 2n 2 ).∇n 2

In spherical coordinates, the normal to the sphere is n2 = er

or

vr = 1, vθ = vφ = 0

Substituting the corresponding components of the gradient read from Appendix IV: ∇n 2=

( eθ eθ + eφ eφ ) r

1

To get the surface gradient, we dot the full gradient with I–nn: ∇ s n 2 =( I − n 2n 2 ).∇n 2

When n is the unit normal to a curved surface, its divergence represents the curvature of the surface ∇s.n = curvature of surface [=] m–1

(123)

(

= e r e r + eθ eθ + eφ eφ − e r e r =

).( eθeθ + eφeφ ) 1r

( eθ eθ + eφ eφ ) r

1

In this case, the surface gradient is the same as the full gradient. To obtain the surface divergence, we apply Identity C.12 from Appendix I:

For a flat surface, n is independent of position along the surface so that ∇s.n = 0 and (122) reduces to (121). Example: Calculate the pressure drop across the interface of a spherical gas bubble of radius R


♦ The ∇ γ term becomes important in problems involving s temperature or solute concentration gradients in the liquid. These cause the ∇sγ term to be non-zero and can drive motion of the bubble. This is called the “Marangoni effect.”


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∇ s.n 2 = I :∇ s n 2 = ( e r e r + eθ eθ + eφ eφ ) : ( eθ eθ + eφ eφ ) = ( 0 + 0 + 1 + 0 + 0 + 1)

1 r

1 2 = r r

For the hydrostatic case in this example, the stress tensor is given simply by

T = −I p

and

n 2.Ti = −n 2 pi

Then (123) yields

2 −n 2 ( p2 − p1 ) = γ n1 R 

−n 2

or

2γ pin − pout = R

Significance: The surface tension of water at room temperature is 0.072 N/m. For a 1 mm (2R) bubble of water vapor, this yields a pressure difference of only 0.003 atm (pressure is higher inside the bubble), which is usually not very important. But when the bubbles are microscopic in size, this effect can be very significant. For example, ∆p = 1 atm when 2R = 3 µm. For homogeneous nucleation of bubbles to occur upon boiling, we would first have to form bubbles that are less than 1 nm in diameter. For such small bubbles, ∆p is enormous: 3000 atm or more. Of course such pressures are far about the vapor pressure of water under normal boiling conditions (1 atm); thus homogeneous boiling virtually never occurs (except for temperatures near the critical value, where γ vanishes). Instead, boiling occurs heterogeneously on the surface of the container by pinching off of air pockets trapped in pits in the wall. We will have more to say about surface tension effects near the end of the course: see Surface Tension Driven Flows on page 191. For now, we will neglect surface tension effects. Lecture #11 begins here

Exact Solutions of N-S Equations The exact solution of the Navier-Stokes equations presents a formidible mathematical problem. By “exact” I mean:

Fall, 2016

One common difficulty is the non-linear inertial term. Most of the powerful mathematical techniques (such as eigenfunction expansions, used in “separation of variables”) only work when the partial differential equation is linear. Of course, if we can neglect the higher-order viscous terms, then we can cope with the nonlinearity using a velocity potential, as we did earlier in solving problems in potential flow. However, viscous terms are seldom completely negligible and leaving them in the equation makes the problem much more difficult by increasing the order of the partial differential equation Nevertheless, it is possible to find exact solutions in certain cases, usually when the inertial terms vanish in some natural way. We will now examine a few of these problems having exact solutions. Problems with Zero Inertia First, let’s consider problems in which the fluids elements travel along straight streamlines at constant velocity. Then their acceleration vanishes identically. Flow in Long Straight Conduit of Uniform Cross Section

Suppose we have pressure-driven flow in a long straight conduit whose cross section does not vary along the flow. In mathematical terms, the conduit is a cylinder of arbitrary cross-section. Define Cartesian coordinates such that the axis of the cylinder corresponds to the z-axis. Then the no-slip boundary condition can be written as at wall:

vx = vy = vz = 0

In a very long pipe, we expect that vz must be nonzero away from the walls (otherwise there could be no flow) and vz might also depend on z (as well as x and y) near the entrance and exit of the pipe.

exact — neither viscous nor inertial terms are neglected (i.e. approximated by zero, as opposed to being identically zero)



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∂p ∂x ∂p 0= − ∂y 0= −

x: y: z: In particular, near the entrance, we say the velocity profile is “developing”; i.e. evolving with z. Outside the entrance and exit regions vz will be independent of z. This situation is called fully developed flow. For fully developed, steady flow: vz = vz(x,y)

(124)

 ∂2 v ∂ 2 vz z 0 = µ +  ∂x 2 ∂y 2 

 ∂p − + ρg z  ∂z 

Thus pressure does not depend on x or y: p = p(z) but more generally for a horizontal or inclined pipe, the gravitational acceleration g might have components in all three directions. Then:

vx = vy = 0

p = p(x,y,z)

Note that this automatically satisfies the Continuity Equation:

where the dependence on x and y arises from the contribution to pressure from the hydrostatic head (i.e. from g).

∇.v = ∂vz/∂z = 0

Next, recall that the acceleration of fluid elements equals the material derivative of velocity which is related to the partial derivative by (17):

a=

Dv = Dt

∂v ∂t 

0 for steady flow

+

v.∇v 

For this and some other problems, it’s helpful to decompose the total pressure into contributions from gravity (i.e. hydrostatic pressure, ph) and from flow (called the dynamic pressure, P) = p ph + P

=0

∇p = ∇ph + ∇P 

0 for straight streamlines

ρg

Since we have both steady flow and (in the fully developed region) have straight streamlines, the inertial terms vanish identically. This can also be concluded by substituting the assumed form of the velocity profile (124) into the general component expansion of the Navier-Stokes equation, given in Appendix VI.

From our earlier analysis of hydrostatic equilibrium [see (56)], we know that ∇ph = ρg . Note that ∇p − ρg = ∇ph + ∇P − ρg = ∇P  ρg

Next, we substitute this into the Navier-Stokes Equation (125).

Strictly speaking, Reynolds number is not always equal to the ratio of inertial to viscous forces in this problem, since inertia is identically zero for fully developed laminar flow although the Reynolds number is not zero.

Expanding them in component form:

For steady, fully-developed flow in a uniform conduit, inertial terms are zero and the Navier-Stokes equations (120) become:

y:

0 = µ∇2v – ∇p + ρg

(125)

Note that for a vertical pipe, the gravitational acceleration g has only a z-component. Expanding the Navier-Stokes equation for (124):♣

♣ Each component of these equations can be found in Appendix VI.


(126)

0 = µ∇2v – ∇P

x:

z:

(127)

∂P  ∂x   P = P (z) ∂P   0= − ∂y  

0= −

0 = µ∇ 2 vz −

dP dz

where we have written the last term as the total derivative instead of the partial derivative because the first two equations require that P be a function of z alone. Notice that dP/dz is a function of z only while vz and the Laplacian of vz are a function of x and y only.


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The the z-component of Navier-Stokes (127) becomes  ∂ 2 vz ∂ 2 vz  dP const. µ + = = dz ∂x 2 ∂y 2      

or

1 dP 2 r + c1 ln r + c2 4µ dz

Requiring the solution to be bounded at the center of the tube (as r→0, lnr→-∞) forces us to choose c1=0 while the remaining constant can be chosen to satisfy the no-slip condition (128). The particular solution is the familiar parabolic velocity profile:

f ( z)

g ( x, y )

vz ( = r)

g(x,y) = f (z) = const. w.r.t. x,y,z

which can only hold for all x,y,z if both functions equal the same constant. Integrating the second equation above implies that P(z) must be a linear function.

= vz ( r )

(

1 dP 2 r − R2 4µ dz

)

To proceed further without resorting to a numerical solution, let’s restrict our attention to a circular cylinder. For this shape of conduit, it makes sense to switch to cylindrical coordinates r,θ,z instead of RCCS. Then (124) becomes vz = vz(r,θ) vr = vθ = 0 and the z-component of the Navier-Stokes equation (127) becomes♣

 1 ∂  ∂v  1 ∂ 2 v  dP z z µ = const. = r +  r ∂r  ∂r  r 2 ∂θ2  dz

The volumetric flowrate through the conduit is computed from: = Q

R

n.v da ∫ ∫= 0

vz ( r ) ( 2πr ) dr

A

where the surface A is chosen as a circular disk of For steady flow in a circular conduit of radius R, the “no-slip” boundary condition requires vz = 0

at r = R:

(128)

radius R having the unit normal n = ez. Then n.v = vz(r) is independent of θ. Then it makes sense to choose da to be a circular annulus of radius r and thickness dr (see “end view” of figure above) whose area is (2πr)(dr). Evaluating the resulting definite integral shown above leads to Q =

Since the boundary condition (128) contains no dependence on θ, we expect the solution to be axisymmetric about the z-axis so the θ-dependence of vz can be dropped, leaving:

This derivation was been repeated for a number of different cross sections. In general, one can show that = Q

which satisfies

The general solution of this equation is:

(129)

which is called Poisueille's Formula.

vz = vz(r)

1 d  dvz  1 dP = const r = r dr  dr  µ dz

πR 4  dP  −  8µ  dz 

where

kA2 µ

 dP  −   dz 

A = cross-sectional area of duct and

where k is some constant which depends on the shape of the duct; for example circle:

k = 1/8π = 0.0398

square:

k = 0.0351

♣ see Appendix VI



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ε

ellipse:

4π (1 + ε 2 )

where ε = b/a ≤ 1 is the ratio of the minor to major axis. For a given cross-sectional area A and a given pressure drop dP/dz, the shape which gives the greatest flowrate is the circle. Pressure drop versus flowrate is often represented in dimensionless form as the friction factor versus Reynolds number. Friction factor is a dimensionless pressure drop: λ =( 2 R )

− dP dz 1 ρU 2 2

To make the pressure drop dimensionless, we have choosen the diameter 2R as the characteristic length and (1/2)ρU2 as the characteristic pressure. This latter expression represents the characteristic kinetic energy of the fluid per unit volume (which also has units of force/area. One motivation for this is Bernoulli’s equation (70) which tends to apply at high Reynolds number where friction is small.

Fig. 5. Friction factor versus Reynolds number for pressure-driven flow through long circular pipes. Curve (1) is (130).

Flow of Thin Film Down Inclined Plane

The Reynolds number is

= N Re where

ρUD ρU ( 2 R ) = µ µ

U=

Q Q = A πR 2

is the cross-sectional average velocity. When expressed in terms of these dimensionless parameters, (129) becomes λ=

64 N Re

(130)

Fig. 5 (below) is a plot of friction factor versus Reynolds number for pressure-driven flow through long circular pipes. Curve (1) is (130), which is seen to accurately represent the experimental data up to a Reynolds number of 2100 where the flow becomes turbulent.

Suppose we have fluid overflowing some reservoir and then flowing down an inclined plane surface. Although there might be some entrance or exit effects (at the upstream and downstream ends of the plane), if the plane is sufficiently long compared to the film thickness, then we again obtain fully developed flow in a film of uniform thickness. Let's try to analyze this central region in which the film thickness is uniform and the flow is 2-D. Let the x-axis be oriented parallel to the inclined plane in the direction of flow and let the y-axis be perpendicular to the plane. One of the boundary conditions will be “no-slip” at the solid surface: at y=0:

vx = vy = vz = 0

At the very least, there must be flow in the xdirection: otherwise there could be no viscous force to balance gravity. In addition, the x-component must vary with y since vx must vanish at y=0 (no slip) and be nonzero for y>0. The simplest form of solution which is consistent with these constraints is: v = vx(y)ex


(131)


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Since the only nonzero component of velocity is vx which does not depend on x, this assumed velocity profile automatically satisfies continuity for an incompressible fluid:

∇.v=

∂vx = 0 ∂x

Next we will substitute this assumed form for the velocity profile into the Navier-Stokes equations (120). As in the previous section, if the flow is steady and fully developed, (120) simplifies to (125). Further substituting (131) and decomposing into scalar components leads to x:

d 2 vx ∂p 0= − +µ + ρg x ∂x dy 2

(132)

y:

∂p + ρg y 0= − ∂y

(133)

z:

∂p 0= − ∂z

(134)

Fall, 2016

n.T

liq

at y=δ:

(

gas

= n. − p gas + τ gas

)

In the second equation above, we have substituted (111). Now both the viscosity and density of air are about 0.001 times those of water. Then it is reasonable to treat the air as an inviscid fluid (i.e. neglect the viscouse stress τgas) and further it is reasonable to treat pgas as a constant: pgas = 1 atm. This leaves at y=δ:

n.Tliq ≈ –npgas = –npatm

(136)

where patm = 1 atm. Now n = ey is the unit normal to the interface in this problem ey.Tliq = ey.(–pI + τ)

at y=δ:

The gravitational acceleration vector g can be decomposed into its separate components using the trigonometry of a right-triangle (shown in teal below) having unit length for the hypotenuse (aligned with g):

= n.T

= –pey + τyxex + τyyey + τyzez

(137)

where p and τ are now the pressure and viscous stress in the liquid. The viscous stress τ is related to the velocity profile by Newton’s law of viscosity (118); the nine scalar components of this equation in RCCS can be found in Appendix V. When vx = vx(y) and the fluid is incompressible, the only nonzero components of the viscous stress are τxy=τyx. Dropping the other terms, (136) and (137) combine to yield: at y=δ:

–pey + τyxex = –eypatm

Equating corresponding components:

p(x,y=δ) = patm

c(x) = patm + ρgδsinβ

Thus

gz = 0

(135) becomes:

Substituting gy into (133) and integrating with respect to y: (135)

where the integration constant might depend on x, but cannot depend on y. This constant also cannot depend on z in order to satisfy (134). Now let's turn to the boundary conditions. Continuity of stress (121) across the interface yields:


y:

(138)

p(x,y=δ) = –ρgδsinβ + c(x) = patm

gy = g.ey = –gsinβ

p(x,y) = –ρgysinβ + c(x)

τyx(x,y=δ) = 0

With this result, we can evaluate the integration constant in (135):

gx = g.ex = gcosβ

and

x:

p(x,y) = patm + ρg(δ–y)sinβ Comment: In the previous problem, separating pressure into hydrostatic and dynamic contributions simplified the analysis. This separation is not advantageous in this problem. In this problem, gravity is what is driving the flow. Now that the pressure profile is determined, we can proceed with finding vx(y). Substituting ∂p/∂x = 0 into (132):


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µ

d 2 vx dy 2

Q=

= −ρg x = −ρg cos β

No-slip at the wall requires: vx = 0

at y=0:

Fall, 2016

whereas if the stress in (138) is evaluated using Newton's law of viscosity, we also require:

This relationship between flowrate and film thickness is generally found to be accurate for Reynolds numbers up to about 1000 and for lowviscosity fluids (µ < 5 cp).* Here Reynolds number is defined as

N Re =

dv 0 τ yx = µ x = dy

at y=δ:

Using these two boundary conditions, the velocity profile can be uniquely determined: = vx ( y )

2 ρg δ2 cos β   y   y   2   −    2µ   δ   δ  

which is sketched below:

ρgW δ3 cos β 3µ

4ρQ µW

For more viscous fluids, surface waves form (at a Froude number of about unity). At Reynolds numbers above 2000, turbulent flow is usually observed. Time Out: The main novelty of this problem is the treatment of the free surface. We treated the air as if it was inviscid, although it has some viscosity. How important is the drag imposed by the air? This is the subject of Hwk #6, Prob. 1

By integrating over a plane perpendicular to the flow of width W >> δ (so da = W dy), we can evaluate the flowrate: Q = ∫ n.v da = W A

δ

∫ vx (y) dy

0

where W is the width of the plane, as measured along the z-axis. Since nothing depends on z in this 2-D flow problem, it makes sense to choose as da a narrow strip of height dy and width W along the zaxis:

To answer this question, let’s consider a vertical film of water in contact with a vertical film of air (β = 0), as shown in the sketch at right. Let’s re-solve the problem and see how large δa must be for a given δw before we can neglect the effect of the air. For fully developed flow, the velocity and pressure profiles should have the form: vx = vx ( y ) v= y v= z 0 p ( x, y, z ) = 1 atm

NSEx becomes: for water:

This strip is an x=const surface having unit normal ex. Evaluating the resulting definite integral leads to:


0 = µw

d 2 vxw dy 2

+ ρw g

* see page 5-55 of R.H. Perry, C.H. Chilton and S.D. Kirkpatrick (eds), Chemical Engineers’ Handbook, 4th Ed., McGraw-Hill, 1963. Or see page 6-43 of the 7th Ed, 1997.


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ρ w g δ3w Q∞ ≡ lim {Q} = W 3µ w δa →∞

d 2 vxa

0 = µa

for air:

dy 2

where we have taken ρa = 0. Applying “no slip” at each of the three interfaces at y=0:

vxw ( 0 ) = 0

at y=δw:

vxw ( δ w ) = vxa ( δ w ) ≡ U

The particular solutions to the two ordinary differential equation’s are ρw g y vxw (= y) ( δw − y ) y + U δw 2µ w

(139)

δ + δa − y vxa ( y ) = U w δa

(140)

which is the same expression we had in the Notes when the viscosity of air was completely ignored. Dividing (141) by (142):

When the viscosity of water is 1000 times larger than air (i.e. µw = 1000 µa ), this gives δ 4000 + w δa Q 1 = δ Q∞ 4 1000 + w δa

To reduce the flowrate by 1% means

The interfacial speed U is unknown, but is choosen to match the shear stress at the interface:

δ= a 0.074 δ w

For Newtonian fluids, the stresses can be related to velocity profile: µw

dvxw dy

dv a = µa x dy

y= δw y= δw

Using the velocity profiles of (139) and (140), this stress matching yields: − 12 ρ w g δ w + µ w

Solving for U:

Q = 0.99 , Q∞

for which the air film thickness must be

τwyx ( δ w ) = τayx ( δ w )

at y=δw:

(142)

µ µ µ δ 4 w+ a 4 w+ w δ δ µ δa Q 1 1 w a a = = 4 µ w δw Q∞ 4 µ w µ a + + δ w δa µ a δa

vxa ( δ w + δa ) = 0

at y= δw+ δa:

Fall, 2016

So even if the air is stilled by a nearby boundary, the drag of the air on the free surface of the water will be negligible (provided the boundary is not too close). In the absence of a rigid boundary in the air, negligible error is made by treating the air as inviscid. Time In!

U U = −µ a δw δa

ρ gδ U = 12 w w µ w µa + δ w δa

The flowrate of the water is δw

Q = W

dy ∫ vxw ( y )=

0

ρ w g δ3w 1 + 2 U δw 12µ w

µ µ 4 w + a δ w δa = µ 12µ w w µa + δ w δa

(141)

ρ w g δ3w

If the air film is very thick, the flowrate becomes



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Problems with Non-Zero Inertia Landau & Lifshitz list only three flow problems in which both viscous and inertial terms are important and in which exact solutions are known: 1. rotating disk

An infinite plane disk immersed in a viscous fluid rotates uniformly about its axis. Determine the motion of the fluid caused by this motion of the disk. This problem was first solved by von Karmen (1921) using cylindrical coordinates with the z-axis coinciding with the axis of rotation. Define: where 2. converging (or diverging) nonparallel planes

flow

between

ζ =z

ω ν

ν = µ/ρ [=] cm2/s

is called the kinematic viscosity. Then:

vr(r,z) = rωF(ζ) vθ(r,z) = rωG(ζ)

vz ( z )=

νω H ( ζ )

(143)

p(z) = µωP(ζ) Continuity and Navier-Stokes become: 2F + H' = 0 r:

F2+F'H–G2–F" = 0

θ:

2FG+HG'–G" = 0

z:

P'+HH'–H" = 0

3. the submerged jet

where the prime (') denotes differentiation with respect to ζ. Boundary conditions take the form:

Rotating Disk*

z=0:

F=H=P=0, G=1

We will now examine the solution of the first problem — the rotating disk — because it is used as a model system for transport experiments.

z=∞:

F=G=0

The solution to the above simultaneous set of nonlinear, ordinary differential equations is given in Schlichting’s book and is plotted below.

* See Schlichting, page 93 (6th Ed); also Landau & Lifshitz, page 79.



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An important property of this solution is that the zcomponent of velocity depends only on z: see (143). For this reason, convective heat and mass transfer problems becomes one-dimensional. The rotating disk is perhaps the only physical situation in which there is flow normal to a wall and where the mass-transfer coefficient can be determined analytically. For this reason, the rotating disk is a favorite tool of researchers in mass transfer.



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= ∇.v

Non-Newtonian Behavior

∂v 1 ∂ 1 ∂v + z 0 ( rvr ) + θ = r ∂ r  ∂θ  ∂z  r 0

0

0

Now that we have seen a few solutions of the equations of motion for simple Newtonian fluids, I’d like to illustrate some profoundly non-Newtonian behavior. In the next example, we will solve a simple flow problem of a long, rotating, vertical rod partially immersed in a fluid. After first obtaining the solution for a purely Newtonian fluid, we will describe a profoundly different solution — call “rod climbing” — exhibited by non-Newtonian fluids.

So what can we say about the pressure p? Any θdependence will have to be periodic since if we go all the way around the cylinder from θ=0 to θ=2π, we are back where we started: any increase in pressure for some θ’s would have to be accompanied by a decrease at other θ’s. For a symmetric circular cylinder, there is no reason for anything to vary with θ, so we will assume

Rotating Rod in Newtonian Fluid

The Navier-Stokes equations become (see Appendix VI)

Consider a very long vertical rod of radius R rotating at ω radians per second, partially immersed in a large container of liquid:

p = p(r,z)

NSE r : NSE θ : NSE z :

v2 ∂p −ρ θ = − ∂r r ∂ 1 ∂ 0= µ ( rvθ ) ∂r  r ∂r  ∂p + ρg 0= − ∂z

Integrating NSEθ subject to the boundary conditions, we obtain

vθ ( r ) = Find the steady state velocity and pressure profiles. In cylindrical coordinates, the boundary conditions are at r = R: as r→∞: and

vr = vz = 0

and

R2 ω r

which is sketched below:

vθ = Rω

vr = vθ = vz = 0 p → patm as r→∞ for z=0

The first boundary condition is no-slip on the surface of the rotating cylinder. The material of the rotating cylinder is undergoing solid-body rotation (see page 19): v = r ωeθ for r ≤ R, so on the surface of the cylinder, we have vθ = Rω and vr = vz = 0. The second boundary condition assumes that, far from the disturbance caused by the rotating cylinder, the fluid remains quiescent. To satisfy these boundary conditions, vθ must be nonzero and it depend on r.

Substituting this result into NSEr:

∂p R 4 ω2 = ρ ∂r r3

We will guess the other components can be neglected: vθ = vθ(r)

and

vr = vz = 0

(144)

Notice that this form for the guess automatically satisfies continuity:


Integrating once:

p ( r= , z) c (z) − ρ

R 4 ω2 2r 2

(145)


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where c(z) is an integration constant (with respect to r) which might depend on z. Substituting this result into NSEz:

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in the liquid phase. In experiments, we see the free surface is deformed near the rotating cylinder:*

dp dc = = ρg dz dz

c(z) = ρgz + const

Integrating:

Substituting this into (145):

p ( r , z ) = ρgz − ρ

R 4 ω2 2r 2

+ const

(146)

The remaining “const” is evaluated by imposing continuity of normal stress across the air-water interface. Treating the air as inviscid, we obtain (136) at the free surface: n.Tliq ≈ –npgas = –npatm For a flat horizontal interface at z = 0 we have n = -k and (136) becomes n.Tliq = –k.Tliq = τzrer + τzθeθ + (τzz – p)ez

The shape of this deformed surface can be inferred from our expression for the pressure profile by recognizing that on this surface the pressure is a constant: p = patm:

z=

R 4 ω2

Using the assumed form (144) for the velocity profile, Newton’s law of viscosity yields (formulas taken from Appendix V): ∂v   ∂v 0 τ zr = µ z + r  = r ∂ ∂z    ∂v 1 ∂vz  0 τ zθ = µ θ + =  ∂z r ∂θ  ∂v τ zz =2µ z =0 ∂z

The only nonzero component of the normal stress in either fluid is the z-component, leaving at free surface:

p = patm

Far from the rotating cylinder, the free surface is located at z=0. Applying (146) at z=0 and r→∞, we are able to evaluate the “const”: p(∞,0) = const = patm For arbitrary r, (146) becomes

, z ) patm + ρgz − ρ p ( r=

R 4 ω2 2r 2

Near the rotating cylinder but at z=0, the pressure is predicted to subatmospheric — provided we are still

for r ≥ R

2 gr 2

r z

The circular streamlines of the fluid cause a centrifugal force which throws the fluid outward near the rotating cylinder. Our predicted shape for the free surface closely resembles that observed above for simple fluids. Rotating Rod in Non-Newtonian Fluid The photograph above — showing the free surface lower near the rod than far away — was taken in an experiment in which the fluid was glycerin (usually obtained by boiling animal fat). Although much more viscous than water (µ up to 10 poise), depending on how much water is mixed with it), glycerin is well known to behave in a Newtonian fashion. The profoundly different photograph below was obtained in the same apparatus under about the same * photos taken from page 63 of Bird, Armstrong & Hassager, Dynamics of Polymeric Liquids. Vol I, 1987.



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∂vr =0 ∂r  1 ∂vθ vr τθθ =2µ  + r  r ∂θ

experimental conditions: the main difference was that some high-molecular-weight polyacrylamide was dissolved in the glycerin:

τrr =2µ

τ zz =2µ

∂vz =0 ∂z

      =0      

(147)

For non-Newtonian elastic fluids, the primary normal stress (in the direction of main flow: the θ-direction here) is usually less positive than the secondary normal stress (in the direction in which the primary component of fluid velocity varies: the r-direction here), leading to a negative value for the first normal stress difference: ψI ≡ τ11 – τ22 = τθθ – τrr < 0

Instead of being depressed near the rotating rod, the fluid actually appears to climb up the rod. This “rod climbing” behavior is sometimes referred to as the Weissenberg effect. It is one of several classic examples of profoundly non-Newtonian behavior exhibited by “elastic fluids” such as this polymer solution or molten polymers. For a simple qualitative molecular explanation of the Weissenberg effect, suppose the beaker contained rubber bands rather than a liquid. Rotating a rod in a bed of rubber bands would cause the rubber strands to wind around the rod and climb, just as this liquid does. In a stagnant fluid, even linear polymer molecules often assume an equilibrium shape which is globular:

stagnant fluid

Clearly no normal-stress difference appears in (147) for a Newtonian fluid. To obtain a nonzero normal-stress difference, a different constitutive equation must be used. The elastic energy storage suggested in the above explanation implies that the constitutive equation should be a blend of that for a Newtonian viscous fluid and a Hookean-elastic solid — a so called visco-elastic fluid. Die Swell Another example of non-Newtonian behavior is die-swell, which results when a non-Newtonian fluid flows out of the end of a rigid tube called the die. In the photograph below, the tube is oriented vertically and the flow is downward and steady.

sheared fluid

Even simple shear flow perturbs this equilibrium shape, thereby storing elastic energy like a stretched rubber band. When the flow stops, the deformed polymer molecules relax to their equilibrium globular state, giving up their stored potential energy. In terms of quantitative constitutive equations, such elastic behavior is usually associated with unequal normal stresses arising from the flow. The Newtonian normal stresses for this rotating rod experiment are all zero


In the left tube [labelled (N)], we have a viscous Newtonian fluid (e.g. glycerin). The jet of fluid formed upon exiting the tube has a diameter equal to that of the “die”. In the right tube [labelled (P)], we have a visco-elastic fluid. Upon exiting the die, the jet swells to a diameter perhaps twice as large as the diameter of the die. This is what is meant by “die swell.”


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Recoil of a Suddenly “Cut” Jet Below is a sequence of photos of a viscoelastic fluid being slowly poured out of a jar into a beaker immediately below. In the left-most photo (a), the falling jet appears to be normal (same as would be seen for a highly viscous fluid). In the middle photo (b), a pair of scissors is used to cut the jet into two halves. In the right-most photo (c), the lower half continues to fall into the beaker as would a viscous jet. But the upper half of the jet unexpectly snaps back upward into the jar in apparent violation of gravity.

This behavior can be rationalized by thinking of the falling jet as a long rubberband being stretched downward by gravity. When you cut the rubber band, the lower half would fall but the upper half would recoil upward to the end of the rubberband which is pinned in the jar. Tubeless Siphon A siphon is a open tube used to transfer liquid by gravity from one open tank to a second tank. The second tank needs to be lower in elevation than the first tank, as shown below.

Usually a vacuum needs to be applied to the downstream end of the tube to suck fluid into the tube and completely fill it. Then if the downstream end of the tube is lower in elevation than the liquid level in the upstream tank, the fluid will flow


78

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spontaneously by gravity between the two tanks, as shown above. The flow is driven by a hydrostatic pressure difference arising between the tube ends. The downstream end of siphon is at 1 atm while the upstream end is at a higher pressure owing to its submergence in the liquid of the upper tank. The resulting pressure-difference pumps fluid through the tube. If the fluid is Newtonian and the upstream end of the tube is raised above the liquid level in the upper tank, air will be sucked into the tube and the flow will stop [see schematic (a) below].

If instead the fluid is elastic enough, it will continue to flow, as shown in schematic (b) above. This is called a tubeless siphon because the upstream end of the fluid path is not enclosed by any tube, yet the fluid flows upward. A somewhat extreme example is shown below in which no tube is used. Initially the upper beaker is tilted downward to start the flow.

Once a continuous jet of fluid is flowing from the upper beaker to the lower beaker, the upper beaker is tilted upward (as shown above). The fluid continues


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to flow into the lower beaker although the fluid in the upper beaker is actually flowing uphill.



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Creeping Flow Approximation Cone-and-Plate Viscometer The cone-and-plate viscometer consists of a flat plate, on which is placed a pool of liquid whose viscosity is to be measured. An inverted cone is lowered into the liquid until its apex just touches the plate with the axis of the cone normal to the plate. The cone is then rotated at some angular velocity Ω around its axis and the torque required to turn the cone (or to keep the plate stationary) is measured.

while the angular velocity is (see figure above): Ω = Ωez = Ω[(cosθ)er – (sinθ)eθ]

(150)

The second equation substituting (11) for ez. Combining (148) - (150), we obtain θ0) εn

where x0 is the location of the boundary at which the boundary layer arises.* In this transformation the distance from the boundary (x–x0) is magnified (“stretched”) by an amount depending on the small parameter. In our problem, the boundary layer arises at x0=0, so the transformation becomes X =

x εn

Step 1: Substitute the “stretch” transformation into the original ordinary differential equation. Rewriting (211) in terms of the new variables X and Y(X) ≡ y(x): y′ =

Similarly where

dy dY dX = = ε − nY dx dX dx

y ′′ = ε −2nY Y denotes dY/dX

Substituting into (211):

ε1− 2nY + ε − nY + Y = 0

(218)

* The stretched variable should be positive: X≥0. If the above transformation does not yield a positive X, x −x . then use X = 0 εn PDF generated December 6, 2016

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The purpose of the stretch transformation is to make Y , Y , Y all O(ε0): as ε→0 (X=const): Y , Y , Y = O(ε0)

whereas

y,y',y" = O(ε0),O(ε–n),O(ε–2n)

Step 2: Select a value for the parameter n in the stretch transform. Recall from the exact solution [see (210)] that the second-derivative term is not negligible inside the boundary layer. This is generally the most important consideration in choosing n: we select the value of the parameter n such that the term containing the highest-order derivative is not zero (after all, dropping the highest order derivative is what lead to the outer expansion, which we have shown fails in the inner region). To keep the highest order derivative, •

This term must be lowest order♣ in ε (otherwise it will be swamped by a lower order term)

•

This term must not be the only term which has this order (if this is the only term of that order, ODE requires it to vanish identically)

Fall, 2016

Try 1-2n -n 0 Criterion #1 Criterion #2 –n=0 1 0 0 No No 1-2n=0 0 ½ 0 No Yes or n=1/2 1-2n=-n -1 -1 0 Yes Yes or n=1 Step 3: Substitute this value of the parameter into the stretch-transformed ordinary differential equation. Using n=1, (218) becomes (after multiplying by ε):

Y + Y + εY = 0

Step 4: Substitute a regular perturbation expansion (using the new stretch-transformed independent variable X) into the transformed inner ordinary differential equation. Instead of (212), we seek a solution inside the boundary layer which has the following form: for x≤δ:

which is impossible for any single value of n. Next, we might try balancing first and third: 1–2n = 0 or n=1/2 Orders (i.e. the exponent of ε) of the three terms then are: 0, –1/2, 0 This is no good, because highest order derivative (i.e. the first term) is not lowest order in ε. Finally we try to balance first and second term: 1–2n = –n, or n=1

yi(x,ε) ≡ Y(X,ε) = Y0(X) + Y1(X)ε + Y2(X)ε2 + ...

(Y0 + εY1 + ) + (Y0 + εY1 + ) + ε (Y0 + ) =0 Step 5: Collect terms of like order in ε.

(Y0 + Y0 ) + (Y1 + Y1 + Y0 ) ε +  = 0 Step 6: Set coefficients equal to zero. ε0 :

Y0 + Y0 = 0

whose general solution is: Y0(X) = A + Bexp(–X) Step 7: Substitute the inner expansion into the inner boundary condition; expand any remaining function of ε into a Taylor series in ε. In this case, the inner boundary condition is: Y(0) = 0 Expanding this in a Taylor series about ε=0: Y0(0) + Y1(0)ε + Y2(0)ε2 + ... = (0)ε0 + (0)ε1 + (0)ε2 + ...

Now the orders of each term are: –1, –1, 0 which satisfies both of the criteria. These results are summarized in the table below:

(220)

(220) into (219):

As a first attempt, we might try to make all terms of the same order: 1–2n = –n = 0

(219)

Collecting terms of like power of ε and setting their coefficients to zero: Y0(0) = 0

♣The “order” of a term with respect to a small parameter ε (as opposed to the order of the derivative) refers to the exponent (power) to which ε is raised. For example, we say that a term which tends to vanish like εn as ε→0 is “of order n.”


Y1(0) = 0 and so on. Step 8: Find the particular solutions to each of the ordinary differential equation’s.


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Y0(X) = A[1–exp(–X)]

(221)

Similarly, we could determine Y1(X), Y2(X) and so on. Step 9: Construct the inner solution. If we stop after the leading term, we have for our inner solution: yi = A[1–exp(–X)] + O(ε)

Fall, 2016

to obtain a single smooth function over the entire domain: yc = yi + yo – (yi)o which is called the composite solution. For the present example, this is obtained by adding (216) and (221), expressing them in the same independent variable, and subtracting e: c e1− x + e (1 − e − x / ε ) − = y= e e ( e− x − e− x / ε )

Matching This remaining integration constant A must be chosen so as to match the inner and outer solutions. One possible choice is to take the outer limit of the inner solution [denoted (yi)o] and equate it with the inner limit of the outer solution [denoted (yo)i]: lim  y i= ( X , ε)  lim  y o ( x, ε)  X →∞   x →0  

( y i )o

( y o )i

which is called the Primitive Matching Principle and was originally used by Prandtl. Taking the limit of (217) as x→0 and the limit of (221) as X→∞: A=e The inner expansion becomes: yi = e[1–exp(–X)] + O(ε) which means that for x≤δ: y ≈ e[1–exp(–x/ε)] whereas for x≥δ:

y ≈ exp(1–x)

The plot above compares yc (the complementary solution), and the exact y. Note that yc is a reasonably good approximation to y. The agreement gets much better as ε → 0. Better agreement could also be obtained for any ε by including the next order term in the expansions. Greenberg (p508f) obtains the second term in both the inner and outer expansions; the corresponding composite solution for ε = 0.05 is virtually indistinguishable from the exact solution (see Hwk #7). MAE’s Applied to 2-D Flow Around Cylinder

A convenient choice of δ is where yi and yo intersect. Of course, this intersection point depends on ε: δ = δ(ε)

which is O(ε) in this problem. From the figure at right, neither yi nor yo is a good approximation to y in the vicinity of δ. However, in most transport problems, the quantity of greatest interest is dy/dx at x=0 and dyi/dx does seem to match dy/dx at x=0 quite well. If required, yi and yo can be blended together Copyright© 2016 by Dennis C. Prieve

Let’s now try to apply MAE’s to solve the problem of flow around a cylinder at high Reynolds number. First, let’s write the equation in dimensionless form: we will denote the dimensionless variables using a “hat” (^):


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v U p − p∞ pˆ ≡ ρU 2

r ˆ ≡ R∇ ∇ R µ ε≡ =Re−1 ρUR

vˆ ≡

whereas other terms are O(ε0). The outer expansion is required to satisfy the outer boundary condition:

rˆ ≡

(222)

This nondimensionalizing differs from our previous attempt which was for the opposite limit of small Reynolds number (see p91 of Notes). First, we have used ρU2 to nondimensionalize the pressure instead of µU/R. This is because the disturbance to pressure caused by flow is proportional to ρU2 in the potential flow solution (see HWK #5, Prob. 2). The second difference is that ε is defined as the reciprocal of the Reynolds number, rather than the Reynolds number itself. This choice makes ε a small parameter in the limit Re→∞. The Navier-Stokes equations for steady flow become:

vˆ .∇ˆ vˆ = ε∇ˆ 2 vˆ − ∇ˆ pˆ ˆ .vˆ = ∇ 0

and

Fall, 2016

as r→∞:

v0 → ex and

vri = O(ε1/2) and vθi = O(ε0) inside the boundary layer. This means that any lower-order terms are zero throughout the boundary layer, including the O(ε0) part of vr. Matching the inner limit of the outer solution with this zero will impose the following condition on the outer solution:♥ vr0 = 0

vˆ → e x

at rˆ → 1 :

r=1

(230)

v 0 = ∇φ

where φ(r,θ) is chosen to satisfy (228): ∇2 φ = 0

and

pˆ → 0

vˆ → 0

(225)

Outer Expansion In the outer region, the regular perturbation expansion in powers of ε uses rˆ as the position variable:

vˆ ( rˆ, θ= ; ε ) v 0 ( rˆ, θ ) + εv1 ( rˆ, θ ) + ε 2 v 2 ( rˆ, θ ) +  Similarly for the pressure profile:

pˆ ( rˆ, θ= ; ε ) p0 ( rˆ, θ ) + εp1 ( rˆ, θ ) + ε 2 p2 ( rˆ, θ ) + Substituting this perturbation expansion into (223) through (225), collecting terms of like order in ε, and setting their coefficients to zero, produces a series of well-posed mathematical problems for the coefficient functions. The first problem (the only one we will worry about in this analysis) is (dropping the ^’s): v0.∇v0 = –∇p0

(227)

∇.v0 = 0

(228)

Note that the viscous term does not appear in this result because the lowest order viscous term is O(ε1), Copyright© 2016 by Dennis C. Prieve

For potential flow, ∇×v0=0 and (227) becomes:

(

)

∇ p0 + 12 vo2 = 0

(226)

Expecting a boundary layer to arise at rˆ = 1 as Re→∞ (or as ε→0), we will now use the technique of MAE to solve the above problem:

ε0:

at

It turns out that (227) through (230) is the same problem we previously solved by potential flow:

Boundary conditions are as rˆ → ∞ :

(229)

When we eventually analyze the inner solution in the next section, we will find that

(223) (224)

p0 → 0

which is just Bernoulli's equation for the pressure profile. After imposing the outer boundary conditions in (229) (and vr=0 at r=1) we get the potential flow solution (see HWK #5, Prob. 2). Eventually, we will need the inner limit of the outer solution for matching: vro ( r = 1, θ ) = 0

(231)

vθo ( r =1, θ ) =−2sin θ

(232)

p o ( r = 1, θ ) =

1 − 4sin 2 θ 2

(233)

Lecture #17 begins here Inner Expansion For the inner expansion, we use a “stretched” radial coordinate. The radial distance from the boundary (located at r=R) is ♥ (230) is also one of the inner boundary conditions. Generally, we do not impose the inner boundary condition on the outer solution. But in this case, (230) represents a matching condition.


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y = r–R which will be stretched by a multiplicative factor of ε–n and made dimensionless by dividing by R: Y=

r−R εn R

r = R(1 + εnY)

or

(234)

Although we could continue to work with the remaining cylindrical coordinate θ (the angle measured from the rear stagnation point), later we will see advantages in also converting this coordinate (to the distance measured from the forward stagnation point):

Fall, 2016 versus r and θ) are being held fixed as we take the limit of ε→0. We have three unknown functions (vr, vθ and p) inside the boundary layer and three partial differential equations: continuity, NSEr and NSEθ. We seek a solution for these three unknowns in the form of the perturbation expansions in (235) - (237). The order in which we apply the differential equations is important: the usual order is 1) continuity, 2) the primary component of NSE and finally 3) the secondary component of NSE. 1) Continuity Equation. So we start with the continuity equation, which for 2-D flow in cylindrical coordinates (in the original variables with dimensions) is

1 ∂vθ 1 ∂ =0 ( rvr ) + r ∂θ r ∂r x = (π−θ)R

and

We need to re-express this in terms of the inner variables. From (234) we have

X = x/R

where x is the arc length measured from the forward stagnation point. θ is the angle measured from the rear stagnation point (see figure above); X is the angle measured from the forward stagnation point. To transform the NSE in cylindrical coordinates we will also need to use dX = – dθ and

or

r = R(1 + εnY) dr = R εn dY

so that

Substituting this, plus (235) through (237) into (238) yields (after multiplying through by r/U)

(

∂ U R ε ∂Y n

vθ = –vx

since increasing values of θ moves in the opposite direction to increasing values of x.

+

(235)

vr v y = = v0 ( X , Y ) + ε n v1 ( X , Y ) + U U +ε 2n v2 ( X , Y ) +

(236)

)

1 −∂vx 1 ∂vx ∂u0 n ∂u1 = = +ε + U −∂X U ∂X ∂X ∂X

With these changes, our perturbation expansion of the inner solution takes the following form: vθ vx − = = u0 ( X , Y ) + ε n u1 ( X , Y ) + U U +ε 2n u2 ( X , Y ) +

) (

 R 1 + ε nY U v + ε n v +   + 0 1        v r r 1 ∂vθ (239) 0 = U ∂θ 

1

dθ = –dX

p − p∞ = p0 ( X , Y ) + ε n p1 ( X , Y ) ρU 2

(238)

Expanding the product in the first term, collecting terms of like order in ε:

(1 + ε Y ) ( v n

0

)

+ ε n v1 + = v0 + ε n (Yv0 + v1 ) + 

Differentiating with respect to Y:

(237)

+ε 2n p2 ( X , Y ) +

(

)(

)

∂  1 + ε nY v0 + ε n v1 +   =  ∂Y  ∂v0 ∂v ∂v   = + ε n  v0 + Y 0 + 1  + O ( ε 2n ) ∂Y ∂Y ∂Y   After multiplying through by εn the continuity equation (239) becomes

We are scaling the dependent variables (vr, vθ and p) the same as we did in the outer solution. The main difference is which set of position variables (X and Y



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∂v0 ∂v ∂v   + ε n  v0 + Y 0 + 1  +   + ∂Y ∂Y ∂Y  

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vr

O( ε2 n )

NSEθ:

∂u0 +ε +   =0 ∂X n

O( ε2 n )

∂vθ vθ ∂vθ vr vθ 1 ∂p + + = − + ∂r r ∂θ r r ∂θ

 ∂  1 ∂ ( rvθ )  1 ∂ 2 vθ 2 ∂vr  ε   + 2 2 +  ∂r  r ∂r  r ∂ θ r ∂θ 

Transforming each term using (234) through (237):

Collecting terms of like order in ε and setting their coefficients to zero leads to ∂v0 =0 ∂Y

ε0 :

at Y=0:

= 0 u= 0 u= 1 

and

v0= v= = 0 1 

(242)

The particular solution to (240) can be found immediately to be for all Y

(243)

including the outer limit Y→∞. This is the basis for (230), which sets to zero the inner limit of the outer variable (vr0) corresponding to the same component of fluid velocity.

O ( ε0 )

∂u0 ∂v1 0 + = ∂X ∂Y

Besides this approximation to the continuity equation (238), the above analysis reveals the scaling of the normal and tangential components of fluid velocity. After substituting (243) (235) leads to

vθ = O(ε0)

inside b.l.

(236) leads to

vr = O(εn)

"

2) Primary component of NSE. The “primary” component of the velocity vector is that component having the largest characteristic value (i.e. lowest order in ε) in the limit ε→0. Based on the scaling above, the θ-component is primary as ε→0. For steady 2-D flow, the θ-component of (223) is (see Appendix VI):

O( ε1− 2 n )

n = 12 ♦

1–2n = 0:

(246)

3) Secondary component of NSE. For steady 2-D flow, the r-component of (223) is (see Appendix VI): vr

(244)

(245)

The primary component of the NSE is used to choose the stretch parameter n. As a general rule, we don't want to lose the highest-order derivative inside the boundary layer, so we want this term to be lowest-order in ε, but not the only term with this order. The highest-order derivative is ∂2u0/∂Y2 which is O(ε1-2n). Assuming 0 < n ≤ 1, the largest of the remaining terms are O(ε0), so we choose n such that 1–2n = 0:

Substituting v0 = 0 into (241) leads to εn:

O ( ε0 )

∂p ∂ 2u0 =− 0+   + ε1− 2n +   2 ∂X ∂Y    n ( ) ( O ε O ε1− n )

(241)

To satisfy “no slip” at the inner boundary, we must require:

v0 ( X , Y ) = 0

O ( ε0 )

(240)

∂v0 ∂v1  ∂u0  0 + =  v0 + Y + ∂Y ∂Y  ∂X 

εn :

∂u ∂u v1 0 + u0 0 +   = Y  ∂ ∂X  O( εn )

NSEr:

∂vr vθ ∂vr vθ2 1 ∂p + − = − + ∂r r ∂θ r r ∂r

 ∂  1 ∂ ( rvr )  1 ∂ 2 vr 2 ∂vθ  ε  −  +  ∂r  r ∂r  r 2 ∂ 2 θ r ∂θ 

As with NSEθ, we transform each term using (234) through (237) plus now we include (246). After collecting like-power terms, NSEr becomes: ∂p ∂p NSEr: u02 +  = − ε − 1 2 0 − 1 +     ∂ ∂Y Y    O( ε1 2 ) O( ε0 ) O( ε1 2 ) 0 O( ε−1 2 )

O( ε

)

The lowest-order term in this equation is ∂p0/∂Y. Since there is no other term with this order, we must require that: ε–1/2 of NSEr:

∂p0 =0 ∂Y

♦ Had we performed the matching instead using NSE , we r would have obtained a different value for n, but eventually this alternative choice would lead to an absurdity.



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Fall, 2016

which integrates to yield: p0 = c(X) where the integration “constant” c(X) can be evaluated by matching the outer limit (Y→∞) of this inner solution with the inner limit (r→R) of the outer solution (233):

(X ) = p0 c=

1 − 4sin 2 X 2

(247)

We still have two unknowns (u0 and v1) and two remaining equations. First we have the lowest-order term [O(ε0)] of (245). After substituting (247), this becomes

u0

∂u0 ∂u ∂ 2u0 + v1 0 − = 4sinX cos X (248)  ∂X ∂Y ∂Y 2  − dp0 dX

The second remaining equation is (244) ∂u0 ∂v1 0 + = ∂X ∂Y

(249)

These equations are second-order (derivative) in u0 and first-order in v1 so I need two boundary conditions on u0 and one on v1. One boundary condition on each component is provided by requiring the inner solution to satisfy the inner boundary condition: no slip at the surface of the cylinder (226) becomes at Y = 0:

u0 = v1 = 0

(250)

The second boundary condition on u0 is obtained from matching. The inner limit of the outer solution for the tangential component is given by (232); matching this with the outer limit of u0: as Y→∞:

u0 → –2sinX

(251)

Note: we have already matched the leading term of the normal component velocity when we imposed (243) to justify (230).

A typical profile for the tangential component of velocity inside the boundary layer is sketched above. On the scale of this sketch, the velocity appears to

( )

approach vθo

which is the inner limit of the outer

solution. Actually, the nearly horizontal dotted line (denoting the outer solution) isn’t quite horizontal. For example, see the curve corresponding to Re = 1000 in Fig. 10 on page 102. After going through a maximum, the exact solution becomes tangent to the potential-flow (outer) solution which has a small negative slope, but its slope is so small compared to the slope inside the boundary layer, that the outer solution appears to be almost flat on the expanded scale of the drawing above. We might define the thickness of the boundary layer as the distance we have to go away from the surface to reach the apparent plateau. From the geometry of this sketch, this distance (which we denote as δ) is approximated using

( vθo )

i

δ

≈

∂vθi ∂y

(252) y =0

where all of the quantities in this equation have units. Let’s try to “scale” (252) and solve for δ. Recall that the outer solution corresponds to potential flow. From HWK #5, Prob. 2: 2  R  vθo ( r , θ= ) U 1 +    sin θ  r 

Boundary Layer Thickness Recall the definition for “boundary layer” from page 102:

so that

boundary layer: a very thin region near to a boundary in which the solution has a gradient which is orders of magnitude larger than its characteristic value outside the region.

or

So how thick is the boundary layer for 2-D flows like the one we just analyzed?

i

o ( v= θ) i

lim vθo ( r= , θ ) 2U sin θ

r →R

( vθo )

i

≈U

To scale the derivative, we need the inner solution, which is given by (235). Substituting n = ½ into (235), we have u (= X , Y ; ε ) u0 ( X , Y ) + ε1 2u1 ( X , Y ) +



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The relationship between the tangential component of inner and outer solutions is

vθ ( r , θ ) =−Uu ( X , Y ) where Y is related to r or y = r-R by (234) ∂vθi ∂r

=

∂vθi ∂y

= −U

∂u ∂Y

∂Y ∂y  R −1ε

−1

and X is the dimensionless version of x (see figure above):

2

∂  1 u0 ( X , Y ) + ε 2 u1 ( X , Y ) +  =− 1 2   ε R ∂Y U  ∂u0  = − +   1 2 ε R  ∂Y   O ( ε0 )    U

X = x/R

Similarly v1 is given by (236) as (recall v0 = 0): 12 vy  ρUR  v y = v1 =   ε1 2U  µ  U

Dropping any multiplicative constants, we obtain

∂vθi ∂y

≈ ε−

1

2

U R

and Y is given by (234) and (246) as 12

 ρUR  y Y = ε−1 2 =  R  µ 

Substituting this result into (252): U 1 U ≈ ε− 2 δ R

and solving for the boundary-layer thickness δ:

δ ≈ εR =

R

∂u0 R ∂v x = ∂X U ∂x

so

where

y=r–R

∂v1 So = ∂Y

Re

y R

( ρUR µ )1 2 R ∂v y = ( ρUR µ )1 2 U ∂y

R ∂v y U ∂y

Thus δ is proportional to R and inversely proportional to the square-root of Reynolds number. The boundary layer gets ever thinner as the Reynolds numbers increases. This is true for all boundary layers in 2-D flows.

Thus the leading-order approximation (249) to the continuity equation becomes (after dividing through by R/U):


which we recognize as the usual continuity equation for 2-D flow (which this is: vz=0 and ∂/∂z=0). Similarly transforming the terms in (248) — which is the leading-order approximation to NSEθ — we obtain

Prandtl’s B.L. Equations for 2-D Flows Let’s now summarize the mathematical problem which must be solved to obtain the velocity profile inside the boundary layer. The mathematical problem is represented by equations (248)-(251). For clarity, let’s rewrite these equations using variables having dimensions [recall (222), (235)-(237) and (246)]. For example, u0 is given by (235) as

v u0 = x U

∂v x ∂v y 0 + = ∂x ∂y

∂v  ∂ 2 v x dps  ∂v ρ  vx x + v y x  = µ − dx ∂x ∂y  ∂y 2  where

(253)

(254)

ps ( x ) ≡ lim  p PF ( r , θ )  r →R

The subscript “s” was added to remind us that this is the surface-limit of the potential-flow solution. (254) looks like NSEx instead of NSEθ (compare equations in Appendix VI), except one of the viscous terms in NSEx is missing: ∂ 2 v x ∂x 2 . This resemblance to the equations of motion in Cartesian



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coordinates becomes even more surprising when we compare the relationship between (x,y) and (r,θ) in this problem:

• x = arc length measured from the forward stagnation point along the surface in the direction of flow

x = (π − θ)R y=r–R

• y = distance from the surface measured along a normal to the surface

with the usual relationship between Cartesian and cylindrical coordinates (see page 7): x = rcosθ y = rsinθ

To summarize: the general boundary-layer equations for any 2-D flow are given by (253) and (254): ∂v  ∂ 2 v x dps   ∂v ρ  vx x + v y x  = µ −  dx  ∂x ∂y  ∂y 2   (256) ∂v x ∂v y  0 + =  ∂x ∂y 

Clearly the (x,y) in our problem are not equivalent to Cartesian coordinates. Instead (253) and (254) represent approximations to continuity and NSEθ which apply inside the boundary layer. Fluid elements inside the boundary layer are flowing in the +x direction along the curved surface of the cylinder (y=0). Since the boundary layer is very thin compared to the radius of the cylinder, the fluid elements do not “see” the curvature of the cylinder (just like we don’t see the curvature of the earth). They behave as if they are moving along a straight-line trajectory instead of a curved path. Finally, we list the boundary conditions which the solutions to (253) and (254) must also satisfy: (250) becomes vx = vy = 0

at

y=0

at

y→∞

v= x ( x, 0 ) v= y ( x, 0 ) 0

at y=0:

Remaining integration constants are evaluated by matching the outer limit of the inner solution with the inner limit of the outer solution. A generalization of (255) is v x ( x, y ) → U s ( x )

as y→∞:

where Us (x) is the inner limit of the outer solution for the tangential component of velocity: U s ( x ) ≡ lim  vθPF ( r, θ )  r→R

while (251) becomes vx = 2U sinX

For boundary conditions, we impose “no slip” at the wall:

(255)

Although the above equations for the velocity profile inside the boundary layer were derived for the specific geometry of a circular cylinder, it turns out that very similar equations are obtained for any 2-D flow, provided we express them in terms of a local Cartesian coordinate system (x,y).

is the inner limit of the potential flow solution. For a circular cylinder, Us (x) = 2U sinθ. If the cylinder is not circular in cross-section, a different potential-flow solution will arise: From Bernoulli’s equation, the inner limit of the pressure profile from potential flow ps(x) is related to the inner limit of the tangential component of fluid velocity Us(x) by ps ( x ) U s2 ( x ) const + = 2 ρ

Differentiating with respect to x and rearranging: −

dU s 1 dps 1 dU s2 = 2 = Us ρ dx dx dx

Substituting into (256): In this local Cartesian coordinate system, x and y are defined as



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vx

∂vx ∂v ∂ 2 vx dU s + vy x − ν = Us 2 ∂x ∂y ∂y dx   

known f ( x )

(257)

∂vx ∂v y + = 0 ∂x ∂y

Fall, 2016 • outside the b.l., viscous δ

  R 2  −U 1 +    sin θ vθPF =   r  

• δ0:

vx = 0

at y=0 for all t>0:

vx = U

as y→∞:

vx → 0

vx ( δ, t ) = 0.01U

vx ( δ, t )  δ  = = erfc   0.01 U  4ν t 

erf

1 − erfc ( 0.99 ( )= )= δ 4 νt

δ 4 νt

Looking up the appropriate value in a table of error functions, we obtain:

 y  vx ( y, t ) = U erfc    4ν t 

δ 4νt or

= 1.825

= δ 3.65 νt

(271)

s

1 as s → 0 e− s′2 ds ′ →  erfc s ≡ 1 − ∫ π0 0 as s → ∞    2

3

At any fixed t, the velocity decays monotonically from U at y=0 to zero as y→∞ (see Fig. 14). As t gets larger, more and more fluid begins to move; we say “the motion penetrates deeper into the fluid.” Suppose we wanted to know how far from the wall we have to go before the velocity drops to 1% of the wall value. Let’s define this distance to be δ(t). Then:

∂vx ∂ 2 vx = ν ∂t ∂y 2

where

2 x

Based on these boundary conditions, the solution can be expected to have the form: vx = vx(y,t), vy = vz = 0. The NSE becomes:

where ν ≡ µ/ρ. The solution to this problem is well known:*

1

erf s

is the complementary error function; the integral itself is the error function. These functions are plotted in the figure below:

This gives us some idea how far the motion “penetrates” into the stagnant fluid. Notice that the penetration depth increases with the square-root of time. Lecture #19 begins here Time In: Boundary Layer on Flat Plate Now let’s apply this result to our original problem: the boundary-layer next to a semi-infinite flat plate. Fluid farther downstream from the leading edge has been in contact with the moving plate longer; and estimate of the “contact time” is: t = x/U

(272)

(272) into (271): * see p115f in Bird, Stewart & Lightfoot, Transport Phenomena, 2nd ed., Wiley, 2002.


δ ∝ νt ∝ νx

U

(273)


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Now we have an ordinary two-point boundary-value problem. While this problem is no simple analytical solution, it can be solved numerically. This numerical solution was first obtained by Blasius (1908),♠ who was a Ph.D. student of Prandtl.

• Step 4: solve b.l. equation using “similarity transform” Judging from the boundary conditions alone, we would expect vx to vary from 0 at y=0 to U at y=δ with sort of a parabolic shape, with δ getting larger as we move downstream. Notice that the basic shape of this profile is not really changing with x, only the range of y-values over which the solution departs from potential flow is increasing as we move downstream. This suggests a solution of the form: vx  y = f ′  U δ

(274)

Let's define a new independent variable which is scaled to the boundary-layer thickness:

y y ≈ ≡ s ( x, y ) δ( x) νx U

Let

This is called a similarity transform because it leads to a solution in which the profiles in y at each different value of x are geometrically similar (i.e. have the same shape).

Notice that the velocity component normal to the plate (i.e. vy) does not vanish far from the plate:

To transform this guess of the solution from velocity to stream function, recall: vx =

∂ψ ∂y

x, y

ψ ( x, y ) = x, y ) dy x ( ∫ v  thus

x, y =0

Uf ′

δ ds

s

 v y ( x, y )  ν lim = ≠0  0.8604 U Ux y →∞   

= U δ ∫ f ′ ( s ′ ) ds ′ = U δ f (s) 0

substituting δ from (273):

ψ ( x, y ) = νUx f ( s )

(275)

In terms of f (s), the boundary-layer equations (270) become: f f " + 2f "' = 0 subject to:

f = f ' = 0 at s=0 f ' → 1 as s→ ∞


Consider a fluid balance around the shaded rectangle in the figure at right. Owing to the need to meet the “no slip” condition on the surface of the plane, the flowrate out the right side of the system is less than the flow in the left side. The excess has to go out the top, causing vy > 0 there. ♠ Paul Richard Heinrich Blasius (1883–1970), German physicist. One of Prandtl’s first PhD students.


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Boundary-Layer Thickness The definition of boundary layer thickness is somewhat arbitrary. Although we are tempted to say that δ is that value of y at which the boundary-layer solution for vx(y) equals U (potential flow), vx approaches U only asymptotically as y→∞; therefore, this "definition" yields the unhelpful result that δ is ∞. There are several ways to assign a more meaningful finite value to δ(x).

Fall, 2016

• Def'n 2:

∂vx ∆vx U s ( x ) − 0 = = ∂y y =0 ∆y δ2 − 0

For flow tangent to a flat plate, this definition yields: δ2 ( x ) = 3.0

νx U

Recall that the normal component of the velocity profile was nonzero far from the plate:

 v y ( x, y )  ν ≠0 lim =  0.8604 U Ux y →∞    except far downstream from the leading edge: One way to define the boundary-layer thickness, δ, is that value of y at which the velocity is 99% of the asymptotic value. • Def'n 1:

vx(x,δ1) = 0.99Us(x)

 v y ( x, y )  = lim   y →∞   U  x →∞

 ν  = lim 0.8604  0 Ux  x →∞  

This means that the streamline moves away from the plate. A third definition for δ is the distance the streamline is displaced away from the plate.

For flow over a flat plate, this convention yields: δ1 ( x ) = 5.0

νx U

Another way to define δ was suggested by Nernst who was concerned with boundary layers which arose in mass transfer problems. Nernst chose the diffusion boundary-layer thickness as the thickness of a hypothetical stagnant film which has the same diffusion resistance.

Suppose that, far upstream, a streamline is given by x→–∞:

y=a

Thus the streamline is initially a distance a from the x-axis. If downstream from the leading edge of the plate, the streamline is a distance b from the x-axis, then the displacement of the streamline is given by: δ3 = b–a

Graphically, this δ can be determined by drawing a tangent to the concentration profile at the surface (y=0). By analogy, concentration of mass is like concentration of momentum, which is just the fluid velocity.

Ny

c dc = −D = −D ∞ y =0 dy y =0 δ

If we define δ for momentum transfer in the same way, replacing concentration by vx, then:

To evaluate this displacement, recall (from the definition of streamline) that the flowrate in the xdirection between y=0 and the streamline is the same all along the streamline. Thus: ψ =

Q = aU= W

b( x )

∫

v x dy

0

where W is the width of the plate (assumed to be arbitrarily wide). Adding δ3U to both sides: b

( a + δ= 3 )U ∫ v x dy + δ3U

(276)

0



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The left-hand side of this expression can be rewritten as: = ( a + δ3 ) U

b

∫ U dy

bU =

(277)

0

Fall, 2016

plate does not significantly effect the boundary layer thickness for a plate of length L (i.e. “exit effects” don’t propagate upstream). For an semi-infinite plate, the same result is obtained by choose x as the characteristic length; then δ≈

νx U

Re ≡

Ux ν

b

b

(276) becomes: 0

0 b

which is consistent with (278) and with Prandtl's more general result.

0

Drag on Plate

U dy ∫ v x dy + δ3U ∫= δ3U =

∫ (U − v x ) dy

Of course, the upper limit b will depend on the particular streamline we have chosen. In other words, different streamlines are displaced differently depending on how close to the plate they were initially. Thus the displacement distance δ3 is not unique. However, distant streamlines are all displaced by the same distance (since the integral converges as b→∞). So let's define the boundarylayer thickness as the displacement of external streamlines: ∞

• Def'n 3:

δ3U =

∫ (U − v x ) dy

For flow tangent to a flat plate, this yields: νx U

Notice that all three of these definitions yield a boundary-layer thickness which is proportional to νx U although the proportionality constant varies considerably. νx U

δ( x) ∝

(278)

We showed (see page 113) for any 2-D flow (which this is) that:

δ≈

R Re

=

νR U

(279)

where R is the radius of the cylinder. For noncircular cylinders, R is some characteristic dimension of the cross section (e.g. the major or minor axis of an ellipse). A semi-infinite flat plate is somewhat unusual in that it has no characteristic dimension. However, if the plate were finite with length L along the direction of flow, it would seem natural to choose L as the characteristic length. If one can reasonably assume that what happens downstream with a longer


The net force exerted by the fluid on the plate is calculated from: F = ∫An.Tda = –∫Apnda + ∫An.τda

Choosing n = +j on the upper surface and n = –j on the lower surface, the integral involving the pressure vanishes owing to cancellation of the contributions from the upper and lower surfaces (the pressure is the same, but n has opposite direction on the two surfaces). This leaves

0

δ3 ( x ) = 1.72

and

x ∂v Fx = 2µ  x  ∂y 0

∫

  dx  y =0 

The “2” comes from addition of the two contributions from the upper and lower surface. Evaluating the integral and expressing the result in dimensionless form:

CD ≡

(

Fx 1 ρU 2 2

)

1.328 = Re ( 2Wx )

where Re ≡ Ux/ν. In defining drag coefficient this way, we have departed somewhat from the convention which uses the projected area along the direction of flow (i.e., the area of the shadow cast by the object if the light source were located very far upstream). In the case of a plate this projected area is zero, so we have used the area of the plate instead.


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vx

dU s ∂v x ∂v ∂2 vx + vy x − ν =U s 2 dx ∂x ∂y ∂y ∂v x ∂v y 0 + = ∂x ∂y

Substituting (280):

Solution for a Symmetric Cylinder Let’s now return to the problem of flow around a cylinder at large Reynolds number. We follow the same general steps as we did in solving the problem of flow tangent to a flat plate: •

Step 1: find the potential flow solution and Us(x)

We accomplished this in HWK #5, Prob. 2. The velocity profile in the potential flow solution is   R 2  vr = U 1 −    cos θ   r  

∂v x ∂v ∂2 vx U2 x x + vy x − ν =4 sin cos 2 ∂x ∂y R R R ∂y

vx

∂v x ∂v y + = 0 ∂x ∂y

Appropriate boundary conditions are no slip at the wall: vx = vy = 0 at y=0 and, just outside the boundary layer, we obtain potential flow: vx → Us(x) as y → ∞

  R 2  −U 1 +    sin θ vθ =   r  

From this potential-flow solution, we can calculate the Us appearing in Prandtl's equation:

The main difference between a circular cylinder and the flat plate is the left-hand side of the first equation, which represents a pressure gradient along the direction of flow inside the boundary layer.

Us ( x) = −vθ ( R, θ ) = 2U

Us

sin θ= 2U sin

x x  sin π− =sin R  R

x R

dU s 1 dp = − ≠0 dx ρ dx

(280)

where x is the arc length, measured from the forward stagnation point, and θ is the polar coordinate, which is measured from the rear stagnation point.

• Step 3: re-write the boundary-layer equations in terms of the stream function

U2 x x ψ y ψ xy − ψ x ψ yy − νψ yyy = 4 sin cos R  R R   Us

The two are related by x= R ( π − θ )

or

θ= π−

x R

• Step 2: apply Prandtl’s boundary-layer equations (see page 115)

at y=0:

ψx = ψy = 0

as y→ ∞:

ψ → Us(x) y

dU s dx

Blasius (1908) obtained the solution to this problem, which in fact works for cylinders of much more general shape — not just circular cylinders. What is required is that the potential flow solution must be an odd function of x: Us(x) = u1x + u3x3 + u5x5 + ... Then Blasius obtained a solution with the form (see Schlichting, pages154-158): ψ(x,y) = (ν/u1)1/2[u1xf1(η) + 4u3x3f3(η) +

+ 6u5x5f5(η;u1,u3,u5) + ...] where


η = y(u1/ν)1/2


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The velocity profile obtained this way for a circular cylinder is sketched below:

This result is quite different from what was obtained with the flat plate. With the flat plate, the velocity profile vx as a function of y had the same shape for different x’s. Indeed the shape at different x’s were mathematically similar and we were able to use a “similarity transform.” Here, for flow perpendicular to a circular cylinder, the basic shape changes with x (or α=X or θ). In particular notice that the initial is positive for α=0 and becomes slope ( ∂v x ∂y ) y =0

zero for α=108.8°. At even larger α’s, the initial slope becomes negative.

This turning around of the velocity profile coincides with a profound event called boundary-layer separation. Separation does not occur on the flat plate. The main difference in Prandtl’s boundarylayer equations which causes separation is the form of the pressure gradient. The x-component of flow is toward the separation point on either side of S. Thus at y=0 dvx/dy>0 for x0 2. irreversible losses of energy



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Drag Coefficient and Behavior in the Wake of the Cylinder

Fall, 2016

coefficients are significantly smaller than predicted by Lamb.

The shape of the velocity profile undergoes several profound changes as the Reynolds number is increased from very small values. Let’s first describe these changes using flow visualization studies.

Re = 19

6 < Re < 40. Two stationary vortexes are formed in the wake of the cylinder -- a consequence of boundary-layer separation.

Re = 0.038

Re 40. Periodic vortex shedding. Large vortices, the size of the cylinder, are created on a periodic basis. These vortices detach and move downstream. Vortex shedding alternates between the top and bottom of the cylinder. Vortexes shed from the top have a vorticity (∇×v) with a sign opposite from vortexes shed from the bottom.

ψ(r,θ) = f (r)sinθ

Although inertial terms are never negligible, the measured drag coefficient agrees well with the predictions of Lamb. As we increase the Reynolds number, inertia becomes more important. Generally large inertia has a tendency to make streamlines straight. Re=140

These vortices persist many cylinder diameters downstream from the cylinder. Most of the irreversible losses of energy occur in forming these vortices, whose ultimate fate is to dissipate their kinetic energy as heat.

Re = 1.1

Re ≈ 4×105. Onset of turbulence in the boundary layer. Point of separation moves further back toward rear stagnation point. Drag is significantly reduced -almost a discontinuous drop.

1 < Re < 6. Streamlines are no longer symmetric about θ=π/2. Streamlines are somewhat farther from the cylinder on the downstream side. Drag



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Measurements of the drag coefficient for flow normal to a circular cylinder are summarized in the large figure above.



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The Lubrication Approximation Sliding between Parallel Plates To introduce this topic, let’s consider a very simple example: Example: Find the force acting between two parallel plates undergoing steady 2-D sliding motion. Let the lower plate be moving at speed U relative to the upper plate which remains stationary. We are interested in the limit in which the separation distance h is small compared to the dimensions of the plate (L and W).

Fall, 2016

Integrating the velocity profile across the vertical gives the flowrate per unit width (into the page) of the plate: h

Q = W

v x ( y ) dy ∫=

1 Uh 2

0

With the velocity profile known, we can calculate the shear stress from Newton’s law of viscosity (see Appendix V): τ xy = τ yx = µ

dvx µU = − dy h

which is constant throughout the fluid (for 0>1:

κ −1

εζ Ez µ

Q = −πa 2

εζ Ez µ

So a measurement of the flowrate for a known applied electric field and a known radius allows us to determine the zeta potential. Ex

In the analysis above, we took the solid wall to be stationary [i.e. we took vx(0) = 0]. We can easily generalize our result to a moving wall [i.e. let vx(0) = vx,f ]. Then (432) can be rewritten as


Example: A typical value of the ζ potential for aqueous solutions is about 50 mV. Plugging in typical values (for water) of the other parameters leads to

♠ Marian Smoluchowski (1872-1917), ethnic Polish physicist in Austro-Hungarian Empire.


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εζ µm/s = 3.5 µ V/cm Electrophoresis of Large Spheres Consider a rigid sphere of radius R, constructed of some insulating (dielectric) material, immersed in a quiescent electrolyte solution in which we have applied a uniform electric field E∞. As a result of any charge on the surface of the sphere, the sphere will be propelled at some terminal velocity U which will turn out to be parallel to E∞. We now seek that velocity for the case in which the counterion cloud is very thin compared to sphere’s radius; i.e. κR >> 1.

Fall, 2016

component. So the problem is not as straightforward as the simple result above suggests. We will first try to solve for the disturbance to the electric field. Changes in the electric field are determined by Gauss’s equation (397) or Poisson’s equation (398). If the charge layer is very thin, the fluid is electrically neutral everywhere outside of this thin film next to the surface of the sphere: thus the charge density ρe vanishes and Poisson’s equation (398) reduces to Laplace’s equation:

= ∇2ψ

1 ∂  2 ∂ψ  r + r 2 ∂r  ∂r 

∂  ∂ψ  0  sin θ = 2 ∂θ  r sin θ ∂θ  1

Fig. 24: Electrophoresis of a negatively charged dielectric sphere through an otherwise stagnant fluid, induced by an externally applied electric field E∞.

To a first approximation, the main difference between electrophoresis and electroosmosis is the reference frame for velocity. In electroosmosis, the solid is stationary while the fluid moves. In electrophoresis, the fluid is stationary but the solid particle moves through it. In the analysis below, we will eventually show [see (444) below] that the electrophoretic velocity of a large sphere is given by κR>>1:

U=

εζ E∞ µ

which appears to be a straightforward application of Smoluchowski’s slip velocity (434). Taking vx,f = 0 and vx,s = U, we obtain the scalar form of the above equation. The opposing directions of U and E∞ (shown in Fig. 24) would be predicted if the particle is negatively charged (i.e. ζ < 0). While the externally applied electric field E∞ is uniform very far from the sphere, the presence of the dielectric sphere perturbs that uniformity within a few radii of the sphere. The scalar component of E∞ which is tangent to the surface of the sphere at r = R varies in magnitude from 0 at the poles to 1.5E∞ at the equator; and the direction of the tangent is only parallel to E∞ along the equator of the sphere. Elsewhere the tangent has a significant vertical


at r=R:

∂ψ =0 ∂r

as r→∞:

ψ → –E∞ rcosθ

(435)

The boundary condition at r=R assumes that the sphere is an insulator and that no electrical current can be conducted into it. The boundary condition as r→∞ requires that the electric field E tend to the uniform value E∞ externally applied. This problem is closely related to potential flow around a sphere (see example starting on page 39). The particular solution to (435) and its boundary conditions is:

 R3  ψ ( r , θ ) =− E∞  r + 12  cos θ r2   In particular, we will need the tangential component of the electric field at the surface of the sphere:

3  1 ∂ψ  − = − E∞ sin θ at r = R: Eθ =  2  r ∂θ  r = R

(436)

Now let’s turn to the hydrodynamics. If our spherical particle is negatively charged, then the sphere is expected to move (through a stagnant fluid) in a direction parallel to the electric field but in the opposite direction (see Fig. 24). Smoluchowski’s equation (434) applies inside the counterion cloud next to a charged sphere. We will solve the hydrodynamic problem in a moving reference frame (so the sphere appears stationary); see Fig. 25.


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Fall, 2016

an electric field. Both the speed U and particle radius R are quite small in such experiments (µm/s and µm, respectively) so the Reynolds number is always very small and Stokes equation (164) then applies: for r>R+κ-1:

Fig. 25: Same situation as in Fig. 24 except the reference frame is now moving with the particle. In this reference frame, the particle appears stationary but the distant fluid moves.

Taking vx,s = 0 and replacing vx,f by vθ and replacing δEx by Eθ, Smoluchowski’s equation (434) yields: at r = R+κ-1:

vθ = −

εζ Eθ µ

(437)

for the fluid’s velocity just outside the diffuse cloud in a reference frame in which the sphere is stationary (see Fig. 26). (437) and (434) on which it is based is sometimes called the “slip” boundary condition (as opposed to no-slip). (434) is called that because κ-1 is so thin compared to R that the fluid appears to change discontinuously with r or “slip” over the surface of the sphere.

µ∇2v = ∇p

(438)

∇.v = 0

(439)

Let's choose a reference frame which moves with the center of the particle. If we define U as the electrophoretic velocity of the sphere in the laboratory reference frame (see Fig. 24), then the appropriate boundary conditions in the moving reference frame of the sphere (see Fig. 25) are: as r→ ∞:

v = –U ≡ U∞

at r = R:

and

p = p∞

(440)

v=0

which requires no slip of the fluid immediately in contact with the sphere. But just outside of the thin counterion cloud, we will have a tangential component of the fluid velocity as specified by (437); however there is still no normal component of fluid velocity: at r = R+κ-1:

vr = 0

vθ = −

(441)

εζ Eθ µ

(442)

If there is no pressure gradient imposed on the distant fluid (i.e. if ∇p∞ = 0), then the velocity profile outside of the counterion cloud turns out to be potential flow (this is not obvious, but we will now show that it is true): for r≥R+κ-1:

v = ∇φ

and

p = p∞

Recall from Theorem III, that the existence of a potential implies ∇×v = 0. Then the left-hand side of (438) vanishes identically: using Identity F.1 (see Appendix I)

∇ 2 v = ∇ ( ∇.v ) − ∇ × ( ∇ × v) = 0  0

0

If the pressure is constant everywhere, then potential flow automatically satisfies (438) and (439) becomes: Fig. 26: Magnified view of very thin diffuse cloud next to particle’s surface in Fig. 25.

Outside of the counterion cloud, the fluid is electrically neutral and there are no electrostatic body forces acting on the fluid, even though there might be


∇2 φ = 0

(443)

Boundary condition (440) translates to: as r→∞:

φ = U∞z = U∞rcosθ

and boundary condition (441) becomes:


06-703

at r=R

184

∂φ =0 ∂r

The particular solution to (443) subject to these boundary conditions also turns out also to be axisymmetric:

 3 εζ  3 − U ∞ sin θ = −  − E∞ sin θ  2 µ  2 

U∞ = −

εζ E∞ µ

U=

εζ E∞ µ

(445)

The absence of the minus sign can be rationalized as follows: a positively charged particle (i.e. ζ>0) is predicted to move in the same direction as the electric field. This is exactly the same direction we would expect a point charge to be pulled by the electric field. (445) for the electrophoresis of large spheres was first derived by Smoluchowski in 1914. Although we have derived this result for a sphere, this same result applies for any shape of particle provided the counterion cloud is thin compared to any radius of curvature of the particle.* Electrophoresis of Small Particles

(447)

*HWK: Derive Smoluchowski's result for a particle of arbitrary shape. This was accomplished by F.A. Morrison, Jr., J. Colloid Interf. Sci. 34:210-214 (1970).

U E

which is positive if the particle moves in the same direction as the electric field and negative if it moves in the opposite direction. Is the mobility of (447) different from (445)? To compare them, we must first relate Q to ζ. One relationship between charge and zeta potential is given by the Gouy-Chapman equation (418). This was derived for a uniformly-charged, flat plate. It also can be applied near the surface of a sphere which is very large compared to the Debye length (i.e. κR>>1). However in the current problem, the sphere is very small compared to the Debye length (i.e. κR 0 is the scalar magnitude of the vector U. Substituting (463) into (462): r = r' + Ut k

(464)

Expanding this equation into its scalar components in Cartesian coordinates x = x′ y = y′

(468)

According to (80), the steady solution is

 φ ( r , θ= ) U  r + 12 

R3   cos θ r 2 

(469)

Substituting (469) into (468) and solving for φ':

′, θ′, t ) φ′ ( r=

UR3 2r 2

cos θ − c ( t )

Bernoulli’s equation (70) contains the partial time derivative: ∂φ/∂t. In the moving reference frame, this partial derivative is zero because the flow is steady:  ∂φ   ∂t  = 0  r

(470)

z= z ′ + Ut

However, this is not true in the stationary reference frame:

Similar to (462) and (464), the velocities in the two reference frames are related by

 ∂φ′   ∂t  ≠ 0  r ′

v = v' – U = v' + U k

(465)

(466)

Expressing each velocity as the gradient of a potential: Copyright© 2016 by Dennis C. Prieve

But what is this nonzero value? The total differential can be written as


06-703

208

 ∂φ′  d φ′  =  dt + dr ′ ⋅ ∇φ′  ∂t r′

Fall, 2016

Bernoulli’s equation in the moving reference frame in which the flow is steady and ( ∂φ′ ∂t )r′ = 0 is

Divide both sides by dt and evaluating the resulting derivatives along a path on which r (instead of r') is held fixed:

p v ′2 p′ U2 + φ g + = const. = + φg + h ρ ρ 2 2

 ∂φ′   ∂φ′   dr ′  + ⋅ ∇φ′ =    ∂t r  ∂t r′  dt r        v′

Once again, far from the disturbance, the pressure tends toward hydrostatic equilibrium:

(471)

as r:

Solving for p': p ′ = ph +

Differentiating (468) with respect to t holding r fixed: dc  ∂φ   ∂φ′   ∂Ur cos θ    =   +  + t r  ∂t r  ∂t ∂ r dt     0

so

From (462):

p'ph and v'2

U

dc − dt

U2 

(

ρ 2 U − v ′2 2

)

(477)

To compare (476) and (477) we use (466): v ′2 = v ′ ⋅ v ′ = ( v − U ) ⋅ ( v − U ) = v ⋅ v − 2 U ⋅ v + U ⋅ U = v 2 − 2 U ⋅ v + U 2

(478)

0

dc  ∂φ′    = − dt  ∂t r

(472)

 dr ′   dt  = U  r

(473)

Substituting (478) into (477): p ′ = ph +

(

)

ρ ρ 2U ⋅ v − v 2 = ph + ρU ⋅ v − ( v ⋅ v ) 2 2

Thus (476) and (477) are identical.

Substituting (472), (473) and ∇φ' = v' into (471), we have dc  ∂φ′   ∂t  =−U ⋅ v ′ − dt  r′

(474)

Bernoulli’s equation in the stationary reference frame is

ph v ′2 p′  ∂φ′   ∂t  + 2 + φ g + ρ =const. =φ g + ρ  r ′

(475)

We evaluate the constant by requiring that far from the disturbance, the pressure tends toward hydrostatic equilibrium, the velocity vanishes and the partial derivative of potential from (474) when v' = 0): as r'→∞: p'→ph, v'2 →0 and ( ∂φ′ ∂t )r′ → –dc/dt Substituting these values into evaluate the integration constant (results are already shown as the right-hand side of (475)), then solving (475) for p:

= p ph − ρ

∂φ ρ 2 − v ∂t 2

Substituting (474): = p ph + ρU ⋅ v −

ρ ( v ⋅ v) 2


(476)


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