Algebra of Vectors and Tensors

06-703

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, 2015 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. 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, 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"♦ ♠ 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.

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.

Copyright© 2015 by Dennis C. Prieve

♦ The “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 second vector (v); the thumb then points in the direction with the correct sense.

PDF generated December 4, 2015

06-703

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.♥ 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. 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: 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) 9. sab = (sa)b = a(sb) ♥ 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.


2

Fall, 2015

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: v = v1e1 + v2e2 + v3e3 where the vi are called the scalar components of v. Usually, for convenience, we choose orthonormal vectors as the basis:

1 if i = j ei.ej = dij =  0 if i ≠ j although this is not necessary. Kronecker delta.

dij 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 decomposed as a linear combination of the 9 unit dyads: ♣ “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.


06-703

3

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

Fall, 2015

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. A unit vector normal to the lower plate is n = ey. Problem: Calculate the force per unit area exerted by the fluid on the lower plate. Solution: 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 52), 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

µU e x − pe y L

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 ♣ 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 (n.T )upper = L plate


06-703

4

Fall, 2015

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.

“Magnitude” of a Vector or Tensor

Next, we present an example which illustrates the formulas for calculating the result of some multipli– cation operation using scalar components.

While other definitions are sometimes used, the most common definition for the magnitude (or norm) of a vector is

EXAMPLE: problem: show that the dot product of an arbitrary vector v and an arbitrary tensor T can be calculated from their scalar components according to the formula:

v.τ =

3

3

∑ ∑ vi τij e j

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

   

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 =

dij

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 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:

= v.τ

3

3

∑ ∑ v j τ jk ek

=j 1 = k 1

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


3

3

∑ ∑ vi τij e j

=i 1 =j 1

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

=i 1 =j 1

= v.τ

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 26). 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 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


06-703

5

I1 =

3

∑ Tkk

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

Fall, 2015 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 )

I2 =

I3 = T =λ1λ 2 λ3

and Scalar Fields

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: ∂f ∂f ∂f df = 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 Lecture #2 begins here

(2)

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 ∇f = (∇f)xi + (∇f)yj + (∇f)zk

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

(3)

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

(4)

According to the definition of the gradient, (3) and (4) are identical. Equating them and collecting terms: [(∇f)x–(∂f/∂x)]dx + [(∇f)y–(∂f/∂y)]dy + [(∇f)z–(∂f/∂z)]dz = 0

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 Copyright© 2015 by Dennis C. Prieve

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

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


06-703

6

way this can be true is if each individual term separately vanishes:**

Fall, 2015

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

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

df = dr.∇f (5)

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

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

2. magnitude: |∇f | is the rate of change of f with distance along this direction

Substituting this into the definition of gradient:

What do we mean by an "f =const surface"? Consider an example.

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.

EXAMPLE: Suppose the steady state temperature profile in some heat conduction problem is given by: T(x,y,z) = x2 + y2 + z2 Perhaps we are interested in ∇T at the point (3,3,3) where T=27. ∇T is normal to the T=const surface: x2 + y2 + z2 = 27 which is a sphere of radius

df = 0 = dr.∇f = dr∇fcosθ

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? Applications of Gradient •

determine a normal to some surface (needed to apply b.c.’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

27 .

∂f n.∇f = ∂s

** 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.


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


06-703

•

7

Fall, 2015

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

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. PDE becomes ODE). 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 in cylindrical coordinates, we write

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: r = (x2+y2)1/2 θ = tan–1(y/x) z=z

x = rcosθ y = rsinθ z=z

v = vrer + vθeθ + vzez 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):

The position vector in cylindrical coordinates is given by♣ p = rer + zez Above is another attempt at rendering the 3-D view defining the cylindrical coordinates.


(6)

♣ 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).


06-703

8

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.

Fall, 2015

EXAMPLE: taken from an old exam.

We also have a different set of nine unit dyads for decomposing tensors: erer, ereθ, erez, eθer, etc. 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)

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 +

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. 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

(7)

eθ = ( − sin θ ) e x + ( cos θ ) e y

(8)

ez = ez

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): choose:

f = z ( r, θ) = δ +

Evaluating this result at r=R ∇f = er

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

∂f ∂f 1 ∂f er + eθ + ez r ∂θ ∂r ∂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. Copyright© 2015 by Dennis C. Prieve

r2 2R

r ∇f = e r R

( cos θ ) er + ( − sin θ ) eθ

e y = ( sin θ ) er + ( cos θ ) eθ

(9)

The third coordinate, θ (not shown in figure) is an angle measured in the z=0 plane between an arbitrary rz plane and the xz plane (θ=0 is the plane of the page to the right of the z-axis).

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

r2 2R

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 (9) for the constant d:

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


06-703

9

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

r= + x2 + y2 + z2

(

y r sin θ sin φ =

= θ tan −1

z r cos θ =

= φ tan −1 ( y x )

x2 + y2 z

Fall, 2015

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.

)

z θ

(r,θ,φ) r y

x

φ

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.

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


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. 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)


06-703

10

Fall, 2015

ez

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.

θ

z

unit circle on f = const surface eθ

er

θ

eθ y x

φ

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 = 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

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

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

( cos θ ) er + ( − sin θ ) eθ

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 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

This is the same result given by (10). Differentiation of Vectors w.r.t. Scalars Suppose we have a vector v which depends on the scalar parameter t: v = v(t)

or their inverse relationships:

ey =

er

For example, the velocity time. What do we mean vector with respect to Fundamental Theorem of derivative as:

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

lim  v (t +∆t )-v (t )  dv =   ∆t →0  dt ∆t 

(10)

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

Note that dv/dt is also a vector.

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.

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:

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



06-703

11

Fall, 2015

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).

Find the velocity of the object.

Using the parallelogram law, we obtain the difference vector. Its magnitude is:

Solution: According to (6), the position vector in cylindrical coordinates is

∆er = 2sin ∆θ 2

p(r,θ,z) = rer(θ) + zez

r = r(t), θ = θ(t), and z = z(t)

Its direction is parallel to eθ(θ+∆θ/2), so:

(

= ∆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: der = eθ dθ

This result can also be obtained by differentiating (7) and comparing the result with (8). Similarly,

deθ = −er dθ

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

eθ

Lecture #3 begins here

= 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   (( vr

vθ

vz

Vector Fields 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)

dp v= dt


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)♠

♠ 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.


06-703

12

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:

Fall, 2015

direction of n. In this case, the mathematical surface having area ∆a is stationary.

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 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.♣

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 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:

Suppose the trajectory of a material point is given by: or

r = r(t) Then the fluid velocity at any time is

v= r t

()

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

(11)

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 

dq n.v = da

(13)

dq = n.v da

(14)

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

(15)

A

(12)

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

♦ 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. ♣ 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 Chapter 4. For now, we just state that by “location of a material point” we mean the location of the center of mass of the molecules. The “point” needs to contain a statistically large number of molecules so that r(t) converges to a smooth continuous function.


where one choice of the surface A is shown in the figure above. Time Out: Notation Involving “d” The integral in (15) might need some explanation and further elaboration. In (15) 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 (15) 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 (15) is eventually evaluated in some coordinate system [see (18) below for an example] da


06-703

13

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

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

Fall, 2015

Df  df  = = Dt  dt dr = vdt

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 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 gives the total differential (in invariant form) as df ≡ f ( r + dr ) − f ( r ) =dr.∇f

When t is a 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 ) =

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.

∂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 =

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.

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

∂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

We will define ∇v in the next section.

material derivative (a.k.a. substantial derivative) rate of change within a particular material point (whose spatial coordinates vary with time):



06-703

14

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. Notation ∇.v ∇×v ∇v

Divergence Curl Gradient

Result scalar vector tensor

Fall, 2015

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.)

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:

To emphasize the general shape of the surface, we have drawn it above to resemble a potatoe. What is meant by this surface integral? Imagine A to be the skin of the potato. To compute the integral:

Gradient of a Scalar (Explicit)

1) Carve the skin into many small elements. Each element must be sufficiently small so that

Recall the previous definition for gradient: df = dr.∇f

f = f (r):

Such an implicit definition is like defining f ′ (x) as that function associated with f(x) which yields: 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 definition of ∇f ∇f ≡

 lim  1  ∫ nfda  V → 0 V   A

• 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 This same type of definition can be used for each of the three spatial derivatives of a vector field: Divergence, Curl, and Gradient Divergence

Curl

Gradient

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

(16)

(17)

 lim  1  nvda    ∫ V → 0 V   A 

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 A = a set of points which constitutes any closed♦ surface enclosing the point r at which ∇f is to be evaluated ♦ The circle over the integral sign in the definition denotes integration over a closed surface.


• n.v da = dq = volumetric flowrate out through da (cm3/s). This quantity is positive for outflow and negative for inflow. • ∫A n.v da = net volumetric flowrate out of enclosed volume (cm3/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)


06-703

15

 > 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.♣ We will now use the surface-integral definition (16) to evaluate the divergence in Cartesian coordinates.

Fall, 2015

x=const surface, the appropriate unit vector is i which points in the direction of increasing x. This is the opposite direction from the outward pointing normal n, so we assign: 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 mean yo ≤ y ≤ yo+∆y and zo ≤ z ≤ zo+∆z. The following figure shows the area we will be integrating over

Given: v = vx(x,y,z)i + vy(x,y,z)j + vz(x,y,z)k Evaluate ∇.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 (might be curved). One corner of A is the point at which the divergence is to be evaluated.

Our differential area patch (shaded) is a rectangle of height dz and width dy. It’s area 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 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 (18)

For R.C.C.S., we choose A to be surface of rectangular parallelepiped of dimensions ∆x,∆y,∆z with one corner at xo,yo,zo (see figure above). So we partition A into the six faces of the parallelepiped: A1, A2 … A6. The integral will be computed separately over each face:

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

A1

A2

A6

We choose A1 to 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 ♣ R.C.C.S. stands for rectangular Cartesian coordinate system.



06-703

16

Using the Mean Value Theorem,♣ we can replace the integral 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

(19)

A1

The Mean Value Theorem guarentees 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

(20)

although it doesn’t tell us which point takes on the mean value. Now let’s move to another face of the parallelepiped. Surface A2 is the x=xo+∆x face. On the opposite side of the parallelepiped from the shaded face, the unit normal points in the opposite direction: in the direction of increasing x so n = +i n.v = i.v = vx(xo+∆x,y,z)

= ∫ n.v da

zo +∆z

yo +∆y

zo

yo

∫

A2

∫

vx ( xo + ∆x, y, z ) dy dz

Using the Mean Value Theorem:

∫ n.v da = vx(xo+∆x,y″,z″)∆y∆z

(21)

A2

where

yo ≤ y″ ≤ yo+∆y

and

zo ≤ z″ ≤ zo+∆z

♣ Mean Value Theorem: If f (x) is continuous over the interval a < x < b, then there exists some x in this interval such that

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 (19) and (21) is:

∫

+= ∫

A1

A2

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

Dividing by V = ∆x∆y∆z: 1 V

∫

A1 + A2

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

Letting ∆y and ∆z tend to zero. Notice that as ∆y→0, the interval (20) 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  n.vda  =  ∫ ∆y , ∆z → 0  V   A1 + A2  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 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

Similarly, from the two y=const surfaces, we obtain: lim  1  V → 0 V 

 ∂v y .vda  n = ∫ y ∂  xo , yo , zo A3 + A4 

and from the two z=const surfaces:

b

(b a ) f (x) ∫ f ( x ) dx =−

a

f (x) is called the “mean value” of f (x) over this interval. Similarly, if f (x,y) is continuous over a bounded area A, then there exists some point (x,η) in A such that

da ∫ f ( x, y ) =

A f ( x, η)

A

where

Fall, 2015

A = ∫ da

lim  1  V → 0 V 

  ∂v n.vda  = z ∂z x , y , z  A5 + A6 o o o 

∫

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

(22)

A



06-703

17

This formula for calculating divergence in Cartesian coordinates is sometimes cited as the definition of divergence instead of (16). Like its counterpart (5) for the gradient, formula (22) can be used to calculate divergence in other coordinate systems. Yet it is hard to infer physical significance using (22) whereas we were able (see Physical Interpretation of Divergence on page 14) to infer physical significance using (16).

Fall, 2015

x = x(u,v,w)

In the same way, we could use the definition to determine expressions for the curl and the gradient.

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

j ∂ ∂y vy

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

(23)

 j 

But remember that the determinant is just a mnemonic device, not the definition of curl. Finally, the gradient of the vector v is 3

3

∂v j

i =1

j =1

∂xi

Σ Σ

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

= ∇f

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

divergence of vector:

 ∂v y ∂v x  + − k ∂y   ∂x

∇v =

h1 ( u, v, w ) =

gradient of scalar

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

z = z(u,v,w)

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

Lecture #4 begins here Evaluation of ∇×v and ∇v in R.C.C.S.

y = y(u,v,w)

ei e j

∇.v =

1 × h1h2 h3

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

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

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

where x1 = x, x2 = y, and x3 = z, v1 = vx, etc. Evaluation of ∇.v, ∇×v and ∇v in Curvilinear Coordinates Ref: Greenberg, p175 These surface-integral definitions can be applied to any coordinate system. On HWK #2, we obtain ∇.v in cylindrical coordinates.

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.

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



06-703

18

Fall, 2015

Ω

eθ θ + ∆θ e θ θ + ∆θ 2

Ω

r

r

z

side view

∆θ r eθ(θ(t)) v(t)

er(θ(t))

r v(0)

∆s = r ∆θ eθ θ

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

eθ(θ(0)) er(θ(0))

θ( t )

∆r

r top view

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. 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. 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.

Dp r d θ eθ dθ lim r= = = eθ r Ωeθ dt dt Dt

Dt →0

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 (24)

 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 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

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 11: dr ( t ) d θ (t ) dz ( t ) r Ωe θ v= er + r eθ + ez = dt dt dt    (( 0

rΩ

0


At some radius r from the axis of rotation the translational velocity is r ωeθ(+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 PDF generated December 4, 2015

06-703

19

radians per second and a direction which is normal to all the translational velocities r ωeθ(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.

Fall, 2015

∇× v ≡

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

(17)

For our closed surface A, let’s choose a cylinder of radius r and height h. We can partition A into three parts: the curved wall plus the top (at z = h) and bottom (at z = 0) of the cylinder. 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. The integral over A becomes:

∫ = A

∫

+

curved

∫

∫

+

top

(27)

bottom

Time In: Physical Meaning of Curl (Cont’d) Returning to our fluids problem of solid-body rotation in cylindrical coordinates, the angular velocity is  Ω = Ωez (25) and the position vector is = p re r + ze z

Taking the cross product of these two vectors [keeping the order the same as in (24)]: v ( p ) =r Ω e z × er + z Ω e z × e z =r Ωeθ   eθ

On the curved surface, the outward unit normal and translational velocity are given by n = er(θ) and v = rΩeθ from (26). The integrand of the curl in (17) is n × v =r Ω ( er × eθ ) =r Ωe z

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

∫

n × vda =Ω r ez

da =π 2 r 2 hΩe z (28)

curved   

curved

(26)

∫

2 πrh

0

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).

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

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 (17):


n × v z =0 =r Ω ( −e z × eθ ) =+ r Ωer ( θ )

We now see that the integrand depends on both r and θ, so we no longer are integrating a constant. 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:

∫ top

+

∫ bottom

=

−r Ωer + r Ωer ) da = ∫ (((((

0 (29)

0

Substituting (28) and (29) into (27):

∫ n × vda =2πr

2

hΩe z

A


06-703

20

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

2 πr 2 h Ωe z n × v da = = 2Ωe z ∫ πr 2 h A

 lim  1  ∫ n × vda  = 2Ωe z V → 0 V  A 

Comparing with (25), we conclude that  ∇ × v = 2Ω

(31)

(32)

Thus the curl is 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 (23) given for RCCS. In cylindrical coordinates, the formula for the curl is again obtained from the definition of curl (17). We will not derive this formula in the notes or in the homework, but the results are given in Appendix III: Vector Operations in Cylindrical Coordinates (r, θ, z):

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

  eθ 

Substituting vr = 0 we obtain

vθ = rΩ

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

(30)

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 (17): ∇× v ≡

Fall, 2015

vz = 0

∇ × v = 2Ωe z

which is identical to (31). 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 (30) was independent of the dimensions of the cylinder — even before taking the limit.

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 52), 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 (23) for calculating curl in RCCS, we obtain: ∇× v = −

U ez L

According to our general interpretation of curl represented by (32), 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 18 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).

Example #2: Linear Shear Flow 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.



06-703

21

Vector Field Theory There are three very powerful theorems which constitute “vector field theory:” • Divergence Theorem • Stokes Theorem • Irrotational ⇔ Conservative ⇔ Derivable from potential

Fall, 2015

  .v da = .v dV = .v )  dV  = n ∇ ∇ ( ∇.v ) V ( ∫ ∫ ∫  A V V 

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

1 ∇.v = ∫ n.vda V A

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= ∫ ∇.v dV A

V

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 

(33)

where V is the region enclosed by A and n is the outward pointing unit normal to the differential surface element having area da. Although we will not attempt to prove this theorem, we can 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

♣ 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. *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.


Let's now apply this principle to an arbitrary system whose boundaries are fixed spatial points (see figure above). 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 these mathematical boundaries. Subdividing V into many small volume elements: dm = ρdV M=

∫ dm= ∫ ρ dV V


06-703

22

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

(34)

where we have switched the order of differentiation and integration. This last equality is only valid if the boundaries are independent of t.♥ Now mass enters through the surface A. Subdividing A into small area elements: n = outward unit normal n.v da = vol. flowrate out through da (cm3/s) ρ(n.v)da = mass flowrate out through da (g/s)  rate of  r ( n.v ) da = n.( rv ) da  =  ∫ ∫ mass leaving   A A

values of the integrand. Since V is arbitrary, and since this integral must vanish for all V, the integrand must vanish at every point:* ∂ρ + ∇.( ρv ) = 0 ∂t

(35)

where in the second equality, we have taken advantage of changing the order of scalar multiplication (which doesn’t matter).

(37)

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. 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.:

We need to evaluate the rate entering rather than leaving, but this is just the same quantity with opposite sign:  rate of  − ∫ n.( rv ) da  = mass entering   A

Fall, 2015

∇.v = 0

(38)

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:

We now substitute (34) and (35) into our word statement of the mass balance (33):

∂ρ

− ∫ n.( ρv ) da ∫ ∂t dV =

V

or

A

∂ρ 0 ∫ ∂t dV + ∫ n.( ρv ) da = V A (( . ∫ ∇ ( ρv )dV

(36)

Then the accumulation term in the balance will involve time derivatives of volume integrals whose

V

Too bad the limits of integration are different for these integrals: if they 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 the Thm and combining the two integrals, we have

 ∂ρ



∫  ∂t + ∇.( ρv ) dV = 0

V

For one particular choice of V, this integral might be zero because of cancellation of positive and negative

*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: 2π

∫0

cos= θ dθ

2π

∫0

sin= θ dθ 0

2π

♥ To see what happens when the boundaries are moving, see the following section entitled “Reynolds Transport Theorem”.


∫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.


06-703

23

Fall, 2015

{

limits change with time. Similar to Leibnitz rule* for differentiating an integral whose limits depend on the differentiation variable, it turns out that:♣

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

∫

or (39)

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. If w = 0 at all points on the boundary, the boundary is stationary and this equation 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.♦ EXAMPLE: rederive the continuity equation using a control volume whose boundaries move with the velocity of the fluid. 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:

}

d mass of =0 dt system

dM d = dt dt

∫

ρ dV = 0

(40)

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 (39) with w=v (i.e. applying the Reynolds Transport Theorem):   d  ∂ρ  ρ ( ρ , t ) = dV dV + ∫ ρ (n.= v )da 0 ∫ ∫  ((   dt  ∂t V ( t ) V ( t ) A ( t ) dq  

which must vanish by (40) at all times t. Notice that the surface integral above is just the net rate of mass leaving the system though the boundaries (if the boundaries of A were stationary). The fact that the boundaries of A are moving with time does not matter as long as the second equality above must hold for all V — which is the same as this equality holding at all times. In short, the second equality above is equivalent to (36). The rest of the derivation is the same as before, which leads again to (37). Stokes Theorem

*Consider an integral which depends on some parameter a:

= I (a)

b( a )

∫

a( a )

f ( x, a ) dx

Leibnitz’ rule for differentiating this integral is

dI = da

b( a )

∫

a( a )

∂f ( x, a ) ∂a

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

(41)

C

dx +

db da + f b ( a ) , a  − f  a ( a ) , a  da da ♣ For a proof, see Greenberg, page 163-4. ♦ 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. * 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.


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”. 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


06-703

24

the crepe is the contour C. (41) 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 segment. Velocity Circulation: Physical Meaning The contour integral appearing in Stokes’ Theorem is an important 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: Stokes' Theorem

dr ∫ v.=

C Mean Value Theorem

=

∫ n.( ∇ × v ) da A

n.( ∇ × v ) A= 2 Ω n A

Finally, we note from the meaning of curl 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.

Fall, 2015

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 18): v = r Ωeθ ∇ × 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

According to L&L Vol I♣ page 25, the angular momentum L of a mass m undergoing motion at velocity v is the lever arm r times 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: L=

∫ ( ρ × v ) ρ dV = V

ρ∆z ∫ ( ρ × v ) da 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 R

L = r∆z ∫ ( r 2 Ωe z ) da = r∆z Ωe z ∫ r 2 2πr dr 0

A

π 4 R r∆z Ωe z = 2

Dividing this by the velocity circulation integral:

♥ 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).


♣ Landau & Lifshitz, Mechanics and Electrodynamics (Course of Theoretical Physics: Vol. 1), Pergamon, 1959. ♦ 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.


06-703

25

π 4 R ρ∆z Ω Lz z 1 π R 2 ∆z ρ = 2 = 1 R 2 ρ∆= 2 .dρ 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

∫ v.dr =

C

L 4π z M

(42)

So the velocity-circulation integral is just proportional to the angular momentum per unit mass. 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:

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

If f (r) exists such that: v(r) = ∇f 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.

Fall, 2015

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

∫ v.dr = 0 C

for any closed contour 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

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) 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 C is open, v=∇φ is still quite useful:

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

Property III: let Co be an open contour connecting points A and B.

Fgrav = –Mgk = –∇φ

If v=∇φ then ∫Cov.dr = φ(rB)–φ(rA)

and the associated potential energy is: φ(z) = Mgz Note that

∇φ = Mgk

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.”


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

∫ Co

v ⋅ dr =

∫ d φ = φ ( rB ) − φ ( rA ) 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.


06-703

26

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 it here. We can distill these properties and their converse into a single statement:

 f = f ( r ) exists   ∇× v =0       for all r  ⇔ such that v =∇f  in Region   in Region       v.dr = 0 for every  ∫ ⇔ C  closed C in Region 

    

Fall, 2015

of as the derivative of the position vector with respect to itself. In R.C.C.S., recall that: r = xi + yj + zk ∇r =∑ ∑

and

i

ei e j

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

then

∂xi

= δij

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

The transpose of a tensor τ is denoted τt and is defined so that:

and

∂xi

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”. Transpose of a Tensor, Identity Tensor

j

∂r j

e j ∑ e i ei * ∑ ∑ δij ei= i

j

i

v.τ = τt.v

so

τ.v = v.τt

As a partial proof that I has the desired properties which make it the identity tensor, consider dotting it with an arbitrary vector v:

for all vectors v. For example: if

τ = ab

then

τt

= ba

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

Symmetric Tensor:

=τ

An example of a symmetric tensor is the dyad aa. Lecture #5 begins here Identity Tensor: Also known as the Idem Factor. Denoted as I and defined so that:

I = ii + jj + kk

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. Divergence of a Tensor This quantity is defined just like divergence of a vector. lim  1 .  ∇.τ ≡ n τda   V → 0 V ∫  A 

v.I = v = I.v 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: ∂r I =∇r = ∂r

where r is the position vector expressed in terms of the unit vectors in that coordinate system. Recalling from p6 that gradient can be thought of as the partial derivative with respect to position, I can be thought Copyright© 2015 by Dennis C. Prieve

* This expression for the identity tensor is valid for any set of orthonormal unit vectors (not just the Cartesian ones for which we have derived it here).


06-703

27

Fall, 2015

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

Corollaries of the Divergence Theorem Note that this definition uses a pre-dot not a post-dot. In R.C.C.S.

τ = ΣΣτijeiej

∫ nf da= ∫ ∇f dV

On the x=xo face:

A

n = –e1 n.τ = ∑

∑

i

On page 21 we wrote the Divergence Theorem for vectors (tensors of rank 1). It can also be applied to tensors of other rank: V

∫ n.τ da = ∫ ∇.τ dV A

τij e1.ei e j = − ∑ τ1 j

j

j

V

ej xo

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

∑

n. = τ

τ1 j

j

ej xo +∆x

After integrating over the area, we obtain:

∫

n.τ da=

A1 + A2

∑ {τ1 j |xo +∆x −τ1 j |xo }e j ∆y ∆z j

Dividing by V:

1 V

 τ1 j |x +∆x −τ1 j |xo  n.τda = e j ∑  o ∆x    j A1 + A2

∫

Taking the limit as V→0:   ∂τ1 j lim  1  n.τda  = ej  ∑ ∫ V → 0 V ∂x  j  A1 + A2 

Adding on similar contributions from the y=const and z=const faces: ∇.τ =∑ ∑ i

j

∂τij ∂xi

ej



06-703

28

Fall, 2015

Introduction to Continuum Mechanics* 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.

r

ρ(ρ ) =

 1  V → 0 V lim



∑ mi  i



Fluids are composed of molecules. In principle, if you tell me the initial location of every molecule in the fluid and 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:

where mi is the mass of molecule i, the sum is over all the molecules inside surface A, and V is the volume.

1 cm3 of water → 3.3x1022 molecules → 1 million years

In fluid mechanics, as in heat and mass transfer, we make an assumption known as the "continuum hypothesis."

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 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 region: * Reference: G.E. Mase, "Continuum Mechanics," Schaum's Outline Series, McGraw-Hill, 1970. ♠ During the 19th century, philosophers argued that Newton’s laws of motions implied that the future was ordained (i.e. pre-determined). We now know that this would require prescribing the initial conditions with infinite precision. 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 one (of position or velocity) is prescribed, the less certain is the other.


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 subsets of the fluid (called fluid elements). This average velocity manifests itself macroscopically as the velocity profile v(r).

molecular granularity

Basically this assumption is that the above limit will converge long before the dimensions of V shrink to molecular size. Similarly, we don't need to know the translational, rotational, vibrational and electronic energy of each molecule. We usually need only to know the internal energy per unit volume as a function of position, which in turn, manifests itself macroscopically as temperature. A more precise statement of the continuum hypothesis is: 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 2. is large enough to contain a statistically large number of molecules.


06-703

29

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.

Fall, 2015

where g is the acceleration of gravity (a vector pointing toward the center of the earth). At the other end of the spectrum are forces which have very short range. If the range is of molecular dimensions, then only fluid elements at the surface experience a nonzero interaction with the universe outside the system.

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.” The basic problem in continuum mechanics is to describe the response of material to stress. A quantitative statement of that response is known as: Constitutive Equation - model which describes how a material will respond to stress. Familiar examples of constitutive equations include: • Hooke's law of elasticity (solids) • Newton's law of viscosity (fluids) Many 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.

Although interior fluid elements might interact with one another through this short-range force, this interaction is not considered in a force balance, because the “action” and “reaction” forces cancel, leaving no net contribution to the force on the system. When only surface elements feel a particular force from outside, that force is called a “surface force.”

inside system

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 list the forces which can 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 molecular size) forces are computed as volume integrals (called “body forces”) and short-range forces are computed as surface integrals (called “surface forces”).

outside system

At the molecular scale, pressure arises from the momentum transferred during collisions between molecules outside and molecules inside the system. Since only surface molecules will be struck from outside, pressure is a surface force.

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, this is a long-range force. Thus gravity is a “body force.” body forces: those which act on every fluid element in body (e.g. gravity): dFg = (dm)g = ρg dV


surface forces: those which act only on surfaces (including mathematical boundaries)


06-703

30

Fall, 2015

One example is hydrostatic pressure: dFp = –pn da where dFp is the force exerted on the system (through the surface element da) by the fluid outside, and n is a unit outward (to system) normal. For a proof that this is the correct form for hydrostatic pressure, see Batchelor, Section 1.3.



06-703

31

Fall, 2015

Hydrostatic Equilibrium Newton’s♣ 2nd 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 Fg = ∫V ρgdV Fp = –∫A npda = – ∫V ∇pdV To obtain this last result, we applied one of the corollaries of the Divergence Theorem. Substituting back into the force balance and combining the two volume integrals leads to: Fg + Fp = ∫V [ρg–∇p]dV = 0 Since V is arbitrary, we conclude that the integrand vanishes at every point in the fluid, or ∇p = ρg

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.

♣ 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.



06-703

32

Flow of Ideal Fluids Now let's consider fluids in motion. The simplest analysis is for:

Fall, 2015

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=

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 an ideal fluid 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

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 ρ Dt

(43)

which is called Euler's Equation (1755).* Another relationship among the unknowns is the continuity equation (see page 22), which comes from the mass balance. Let r(t) denote the trajectory of one particular fluid element inside the system. Then the velocity of the fluid element is: v=

For an incompressible fluid, ρ = const. w.r.t. both time and position. Then the continuity equation reduces to:

Dr Dt

while the acceleration is:

∇.v = 0,

Dv a= 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 )

∂ρ + ∇.( ρv ) = 0 ∂t

Dv Dt


(44)

To see that we now have as many equations as unknowns, note that the unknowns in (43) and (44) are

* 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.


06-703

33

v and p which represents 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 (43) and (44):

Fall, 2015

the tank; so Dv/Dt is just a. Solving for the gradient, we have

1 ∇p = g − a ρ

Euler + continuity but Euler’s equation (43) 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 (43) and (44) “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 the geometric meaning of gradient (see page 6) of a scalar 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 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

♣ 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.


We were lucky in the previous example, because we knew the left-hand side of (43), so instead of 4 scalar unknowns, we only had one: p. The solution was relatively easy. In the more general problem, the left-hand side of (43) is an unknown nonlinear partial differential equation:

∂v . 1 + v ∇v = g − ∇p ∂t ρ

(45)

In this form, we have expanded Dv Dt using the relationship between material derivative and partial derivative (see page 13). Now we have four scalar unknowns: the three scalar components of v and pressure: vx, vy, vz and p. Coupled with the continuity equation (for an incompressible fluid) ∇.v = 0 Euler’s equation also gives us 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 physical 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 Velocity Circulation: Physical Meaning on page 24 we defined a contour integral called the velocity circulation:


06-703

34

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

(46)

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

Fall, 2015

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: ρ=const p 1 ∇p = ∇ ρ ρ

This is only possible for an incompressible fluid. Turning now to the first term of the integrand in (47): the gravitational acceleration g is just a constant vector, which can be written as the gradient of a scalar: g = –∇φg

(49)

φg = gz ,

where

C(t2)

(48)

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. trajectory of a material point

Substituting (48) and (49) into (47), then combining the two gradients into one leads to

a material point C(t1)

Partial proof:♥ Since the contour C is composed of material points, the time derivative of this contour integral can be expected to be like the contour integral of the material derivative: ? dΓ d Dv . = = v.dr ∫ 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 (43) to obtain dΓ = dt

  1 ∫  g − ρ ∇p .dρ C

(47)

♦ 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.


  dΓ p p =  −∇φ g − ∇ .dρρ =− ∫ ∇  φ g + .d 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 (46). Integrating (46) 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 ideal flow of incompressible fluid. Since C is composed of material points (which in general move with different velocities), the contour may change shape or move. Given the meaning of this contour integral (Γ = angular momentum), this result implies that (in the absence of friction; i.e. for an ideal fluid) angular momentum of 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


06-703

35

rest and an object undergoing rotation continues to rotate with the same angular velocity unless acted upon by an unbalanced torque. Comparing this analogous statement with Kelvin’s theorem (which basically says that the angular velocity remains a constant), we conclude that there is no way to apply a torque to an ideal fluid. To understand why, consider the following thought experiment: I wish to rotate a dinner plate. The usual way to apply a torque (with our hands to 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, 2015

C which causes v to be nonzero. Kelvin’s Theorem still requires

∫ v.dr = 0

for all t

Despite this,

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). Applying Theorem III: ∇×v = 0 for all r,t

(50)

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

(51)

where φ is called the velocity potential. Significance: Knowing that the solution has the form given by (51) allows us to decouple the four scalar equations represented by (43) and (44): 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, there can be no torque. Lecture #6 begins here Irrotational Flow of an Incompressible Fluid As an example, consider towing a submerged object through an “ideal fluid” which is otherwise stagnant.

Substituting (51) into (44) yields Laplace’s equation in the velocity potential: ∇.∇φ ≡ ∇2φ = 0 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 (43). Although Euler was the first to suggest this approach, this is called Laplace's Equation* after the French mathematician who solved this 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 Outside the region disturbed by the moving object, the fluid is completely stagnant: 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. Now suppose at some later time the submerged object moves into the vicinity of


(52)

We will now integrate this vector equation to obtain a single scalar equation for the pressure profile. Each term in (52) can be expressed as a gradient of something. For example, we’ve already seen in (49) that: g = –∇φg Similary, from (48): * 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.


06-703

36

Fall, 2015

ρ=const  p  1 ∇p = ∇   ρ ρ

stagnant fluid ( v = 0)

For potential flow, the unsteady term becomes

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

(53)

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 

(54)

0

where

where v = v. In our particular case, the second term vanishes because ∇×v = 0 for potential flow. Substituting (49), (48), (53) and (54) into (52):

 ∂φ v 2 p ∇ + + φg +  =0 ρ  ∂t 2 which is called Bernoulli’s Equation.1♣ It implies that p ∂φ v 2 + + φ g + = const w.r.t. r ∂t 2 r

Solution: If the fluid behaves ideally, it undergoes potential flow and the velocity profile must satisfy Laplace’s equation: P.D.E.:

(55)

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 to obtain the velocity profile and then we can evaluate the pressure profile using Bernoulli's equation. 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.


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 surface element of area da is given by: dq =n.( v − U ) da =0

For an impenetrable sphere, the flowrate must vanish 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 avoid the unsteadiness, we adopt a new reference frame in which the origin moves along with the center of the sphere. Now let the reference frame move with the object, so that the object appears stationary but the fluid moves around it:

It turns out that the PDE does not change upon this shift in reference frame for velocity: the new velocity potential must also satisfy Laplace's equation: P.D.E.:

♣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

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

v.v = v2

U

∇2 φ = 0

We have dropped the prime (') on the potential to emphasize that this is not the same scalar potential which appeared in the previous PDE, although both potentials satisfy Laplace’s equation. In particular, φ turns out to be independent of time t whereas φ' is not. While the PDE is the same, the boundary conditions are changed. In this moving coordinate


06-703

37

system, the sphere appears to be stationary and the fluid at infinity is undergoing uniform flow b.c. #1:

v → –U ≡ Uk as r→ ∞

(56)

where U is the velocity of the sphere in the original stationary reference frame and k 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:

Fall, 2015

Referring to the figure above (see page 8), we note that er, eθ, and ez = k 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 bottom figure above, from trigonometry of the right triangle it is apparent that k=

( cos θ ) er − ( sin θ ) eθ

Thus the b.c. can be written as as r→ ∞:

∇φ → U cos θ er − U sin θ eθ

The components on of the vector on the right-hand side are of course the 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) For a stationary sphere, no fluid entering the sphere means dq = n.v da = 0, or b.c. #2:

n.v = vr = 0 at r=R

Now the potential has a steady state solution. Next we rewrite the b.c.’s in terms of the velocity potential in spherical coordinates. In terms of velocity potential, (56) becomes: ∇φ = Uk

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

Equating corresponding components: ∂φ = U cos θ ∂r

and

1 ∂φ = U sin θ r ∂θ

Integrating either PDE leads to b.c. #1:

φ → Urcosθ + const

as r→ ∞

where we have arbitrarily selected a value of zero for this “const”.* We then translate b.c. #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 vr er + vθ eθ + vΦ eΦ = er + eθ r ∂θ ∂r 1 ∂φ + eΦ r sin θ ∂Φ

Dotting both sides by the unit vector n = er, then using b.c. #2: b.c. #2:

n.v = vr = (∇φ)r = ∂φ/∂r = 0

at r=R

Since neither the azimuthal angle Φ nor the velocity component vφ appears in these boundary conditions, we will try to find a solution which is independent of Φ: φ = φ(r,θ)

* Here Φ is the spherical coordinate, while φ is the velocity potential. For more on spherical coordinates, see page 7.


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


06-703

38

Fall, 2015

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.

different values at different r's. So we try a third guess which is slightly more general than the second:

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.

Substituting (60) into (57)-(59), we find that cosθ cancels out, leaving:

φ(r,θ) = f (r)cosθ

(60)

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

f = Ur as r→ ∞ Thus we have reduced the problem of solving the P.D.E. to one of solving an O.D.E. We recognize this O.D.E. to be 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 When the potential function is independent of the angle Φ, each trajectory is confined to one of these planes. 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 

(57)

∂φ/∂r = 0 at r=R

(58)

φ = Urcosθ as r→ ∞

(59)

Let’s now check on our assumption that the potential φ is independent of the angle Φ. Since neither the PDE nor the b.c.’s depend on Φ, the particular solution is indeed independent of Φ. Although there are systematic procedures, like “separation of variables,” for solving P.D.E.'s 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? satisfies the P.D.E. and the b.c. at r=R. Unfortunately, it does not satisfy the b.c. at r→ ∞, so we will have to try another guess. Since the failure occurred with the b.c. at r→ ∞, we look there for the form of our second guess: φ = Arcosθ ? This too satisfies the P.D.E. and the b.c. at r→ ∞ (provided A=U) but to satisfy the b.c. at r=R, A must be chosen as 0. The tells us that A should have


Using the b.c.'s, we can evaluate A and B. particular solution to this problem is:

The

 R3  f= ( r ) U  r + 12 2  r   Substituting this back into (60) leaves:

 R3  f ( r ,= θ ) f ( r ) cos = θ U  r + 12  cos θ r2     R 3  ∂φ vr = U 1 −    cos θ = r ∂r    

 1 ∂φ vθ = = −U 1 + 12 r ∂θ 

R3 r3

  sin θ 

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

aN x N

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

At least one of the N linearly independent solutions has the form y(x) = Axa. Substituting this form and dividing out the common factor Axa, we obtain an Nth-order polynomial for a:

a0 + a1a + a2 a ( a − 1) +  + a N a ( a − 1)( a − 2 ) ( a − N + 1) = 0 Each distinct root of the polynomial leads to a separate solution. In this example, the roots are a = -2 and +1.


06-703

39

= vΦ

∂φ 1 = 0 r sin θ ∂Φ

Below are several attempts to display this solution graphically: vr = 0

Above is a plot of the trajectories of materials points around the sphere. These are called streamlines. In the next chapter, we will see how to calculate them. The actual solution will be delayed until Prob. 5 in Hwk #5. Having solved for the velocity profile, we can determine the pressure profile from Bernoulli's equation (55). Assuming steady state:

vθ (r , θ = π ) 2

vr (r , π) vθ = 0

Fall, 2015

v2 p + + φ g = const 2 ρ

vr (r ,0) vθ = 0

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♥:

vr = 0 vθ (r , θ = 3π ) 2

at r→∞, φg = 0:

Recall that θ is defined as follows: U θ

v2 p U 2 p∞ + + φ= + g 2 ρ 2 ρ

Thus

p ( r,= θ ) p∞ − rφ g + 12 r (U 2 − v 2 )

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

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

Substituting the known velocity profile, we obtain the pressure profile. Let's focus on the pressure profile over the surface of the sphere: for r=R: p ( R, θ= ,Φ)

1.5

p = p∞ and v2 = U2

p − rφ g ∞ (( 

hydrostatic head ph ( x ) 1

+ 18 rU 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 37). 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.

1

0.5

On the surface of the sphere, the dynamic pressure depends solely on θ.

0

1

2

3

4

r/R

vr/Ucos(theta) vtheta/Usin(theta) 1.5

1

0.5

0 0

0

0.5

1

1.5

2

2.5

3

3.5


4

4.5

5

♥ This 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.


06-703

40

pd 1 ρU 2 2

Fall, 2015

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

(61)

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]. 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. Proof: first consider the contribution from dynamic pressure:

pd ( R, θ ) ≡ 18 ρU 2 ( 9cos 2 θ − 5 ) The plot above indicates the local pressure using color (red = high, green = moderate, blue = low). The contours are isobars (pressure = const). 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.

The n in (61) is a unit vector normal to the surface, pointing outward. In spherical coordinates, with their origin at the center of the sphere, this vector is the unit vector in the r-direction: n = er(θ,φ) which 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 (61) by k: Fp = .Fp = −

∫ A

.n p da   cos θ

For the contribution from dynamic pressure, pd(r=R,θ) depends solely on θ, so we choose the strip of width R dθ and length 2πRsinθ as our differential area da. On this strip θ is virtually constant.



06-703

41

dθ

as an exercise for the reader (HWK #4, Prob. 2). The net force due to pressure is

n

Fp = −ρgV

θ

r

The net force on the sphere is the sum of its weight and the net pressure force:

k

Fp + Fg = −ρgV + ρs gV = ( ρs − ρ ) gV ≠ 0

rsinθ

p

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

0

da

p

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

0

x cos θ =

=

−1

2pR 2 × 18 ρU 2

∫ x ( 9 x 2 − 5) dx=

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.

This can be achieved by selecting the red rectangle above as our da. The dimensions of this rectangle are Rdθ (the same as the width of the yellow annulus) and (Rsinθ)dΦ (the length of the arc subtended by the angle dΦ). The surface integral now becomes a definite double integral:

∫ f ( θ, Φ )

da 

=

( R sin θd Φ )( Rd θ )

π  2π  = R 2 ∫ sin θ  ∫ f ( θ, Φ ) d Φ d θ 0  0 

The nonzero contribution from (61) comes from the hydrostatic head. We will leave this calculation Copyright© 2015 by Dennis C. Prieve

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 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:

0

1 (( (( 0

A

Fall, 2015

Lecture #7 begins here 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 >

or

∆p = − 12 ρ ∆v 2

(83)

ρ ∆v 2 2 c2

The largest change in density corresponds to the largest change in v2, which is vmax2 – 0:

 ∆ρ  1v =  max    ρ max 2  c

  

l c

where τ = time over which significant changes in v occur l = distance over which changes in v occur l/c = time for sound to propagate a distance l For steady flow τ = ∞ and Criteria 2 is always satisfied. Any fluid can be considered incompressible if both criteria are met. Thus in our solution for potential flow over an airfoil, the fluid can be treated as incompressible provided the velocity of the aircraft remains well below the speed of sound. However, in the analysis of the propagation of sound, we cannot treat the fluid as incompressible. To see this, note that the characteristic time τ is the period of oscillations which is related to the frequency ω of sound being produced by the speaker [recall (69)]: τ=

2π ω

The distance between adjacent wavefronts (surfaces on which the velocity vx is constant) is the wavelength of the sound: l=

∆ρ ≈ −

(83) into (82):

Fall, 2015

2πc ω

In this case, the characteristic time τ = l/c and we cannot assume an incompressible fluid. And we did not made this assumption: we used (65) as the approximation for continuity whereas ∇.v = 0 is the approximate continuity equation for an incompressible fluid.

2

If the fraction change in density is small enough, then it can be neglected: vmax 0.

x-component:

τyx(x,y=d) = 0

y-component:

p(x,y=d) = patm

The simplest form of solution which is consistent with these constraints is:

With this result, we can evaluate the integration constant in (108):

v = vx(y)ex

p(x,y=d) = –ρgdsinβ + c(x) = patm

Note that this form automatically satisfies the requirement of continuity for an incompressible fluid:

Equating separate components: (110)

c(x) = patm + ρgdsinβ

Thus (108) becomes:

∇.v = ∂vx/∂x = 0

p(x,y) = patm + ρg(d–y)sinβ

Substituting this velocity field into the Navier-Stokes equations:♦ x:

0 = –∂p/∂x + µd2vx/dy2 + ρgx

(106)

y:

0 = –∂p/∂y + ρgy

(107)

z:

0 = –∂p/∂z

Thus ∂p/∂x = 0 and (106) becomes: µd2vx/dy2 = –ρgcosβ

No-slip at the wall requires: at y=0:

vx = 0

where

gx = g.ex = gcosβ

whereas if the stress in (110) is evaluated using Newton's law of viscosity, we also require:

and

gy = g.ey = –gsinβ

at y=d:

♦ Unlike the last problem, we do not decompose the total pressure p into ph + P. With a free surface, one of the boundary conditions involves the total pressure.


τyx = µ dvx/dy = 0

* For a Newtonian fluid, the nine scalar components of the stress tensor are expressed in terms of the derivatives of the velocity field: see NLV_RCCS.pdf under Course Documents /Tables /Newton’s Law of Viscosity.


06-703

65

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

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

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

d

∫ vx (y) dy

Fall, 2015

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. 2

To answer this question, let’s consider a vertical film of water in contact with a vertical film of air, as shown in the sketch at right. Let’s re-solve the problem and see how large da must be for a given dw before we can neglect the effect of the air. For fully developed flow, the velocity and pressure profiles should have the form:

0

vx = vx ( y ) v= v= 0 y z p ( x, y, z ) = 1 atm

NSEx becomes: 0 = µw

for water: Q=

ρgW δ3 cos β 3µ

where W is the width of the plane. 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 =

4ρQ µW

d 2 vxw dy 2

0 = µa

for air:

+ ρw g

d 2 vxa 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=dw:

vxw ( δ w ) = vxa ( δ w ) ≡ U vxa ( δ w + δa ) = 0

at y= dw+ da:

The particular solutions to the two ODE’s are 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

vxw (= y)

ρw g y ( δw − y ) y + U 2µ w δw

δ + δa − y vxa ( y ) = U w δa

(111)

(112)

* Perry’s Handbook, 4th edition, page 5-55.

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



06-703

66

Fall, 2015

τwyx ( δ w ) = τayx ( δ w )

δ= 0.074 δ w a

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

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.

at y=dw:

µw

dvxw dy

dv a = µa x dy

y= dw y= dw

Using the velocity profiles of (111) and (112), this stress matching yields: U U − 12 ρw g δ w + µ w = −µ a δδ w a

Solving for U:

Time In! Problems with Non-Zero Inertia L&L list only three flow problems in which both including viscous and inertial terms are important and in which exact solutions are known:

ρ gδ U = 12 w w µ w µa + δδ w a

1. rotating disk

The flowrate of the water is dw

Q = W

∫

vxw ( y )= dy

0

ρ w g d3w 1 + 2 U dw 12µ w

µ µ 4 w + a d w da = 12µ w µ w µ a + d w da

(113)

ρ w g d3w

If the air film is very thick, the flowrate becomes ρw g δ3w Q∞ ≡ lim {Q} = W 3m w δa →∞

(114)

which is the same expression we had in the Notes when the viscosity of air was completely ignored. Dividing (113) by (114):

2. converging (or diverging) nonparallel planes

flow

between

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

Q = 0.99 , for Q∞

3. the submerged jet

which the air film thickness must be



06-703

67

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

Fall, 2015

reason, convective heat and mass transfer problems becomes one-dimensional. 1.0 0.8

G

-H

0.6 0.4 F

0.2 0

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

z =z

0

1

2

3

4

z

This is perhaps the only problem in which there is flow normal to a wall where the mass-transfer coefficient can be determined analytically. For this reason, the rotating disk is a favorite tool of researchers in mass transfer.

ω ν

ν = µ/ρ [=] cm2/s

is called the kinematic viscosity. Then:

vr(r,z) = rωF(z) vθ(r,z) = rωG(z)

vz ( z )=

νω H ( z )

p(z) = µωP(z) Continuity and N-S become: 2F + H' = 0 r:

F2+F'H–G2–F" = 0

θ:

2FG+HG'–G" = 0

z:

P'+HH'–H" = 0

where the prime (') denotes differentiation with respect to z. Boundary conditions take the form: z=0:

F=H=P=0, G=1

z=∞:

F=G=0

An important property of this solution is that the zcomponent of velocity depends only on z. For this * See Schlichting, page 93 (6th Ed); also Landau & Lifshitz, page 79.



06-703

68

Non-Newtonian Behavior 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 rotating rod for which polymer solutions illustrate a classic feature of elastic fluids called “rod climbing.”

Fall, 2015 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 circular cylinder, there is no reason for anything to vary with θ, so p = p(r,z) The Navier-Stokes equations become

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

NSE r : NSE θ : NSE z :

−r

vθ2

= −

∂p ∂r

r ∂ 1 ∂ 0= µ ( rvθ )  ∂r  r ∂r  ∂p 0= − + rg ∂z

Integrating NSEθ subject to the boundary conditions, we obtain

vθ ( r ) =

R2 ω r

Find the steady state velocity and pressure profiles. In cylindrical coordinates, the boundary conditions are at r = R:

vr = vz = 0

as r→∞:

vθ = Rω

and

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

and

The first b.c. is no-slip on the surface of the rotating cylinder. The solid material of the rotating cylinder is undergoing solid-body rotation (see page 18): 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 arises from the belief that, far from the disturbance caused by the rotating cylinder, the fluid remains quiescent. Clearly, vθ must be nonzero and depend on r. We will guess the other components can be neglected: vθ = vθ(r)

and

vr = vz = 0

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

0

∂p R 4 ω2 = r ∂r r3 Integrating once:

p ( r= , z) c (z) − r

(115)

Notice that this form for the guess automatically satisfies continuity:

= ∇.v

Substituting this result into NSEr:

2r 2

(116)

where c(z) is an integration constant. Substituting this result into NSEz: dp dc = = ρg dz dz

0

Integrating:


R 4 ω2

c(z) = ρgz + const


06-703

69

Fall, 2015

Substituting this into (116):

p ( r , z ) = rgz − r

R 4 ω2 2r 2

+ const

(117)

This integration constant is evaluated by imposing continuity of surface forces across the air water interface. Treating the air as inviscid, we obtain (109) at the free surface: n.Tliq ≈ –npgas = –npatm For a flat horizontal interface at z = 0 we have n = -k and (109) becomes n.Tliq = –k.Tliq = τzrer + τzθeθ + (τzz – p)ez Using the assumed form (115) for the velocity profile, Newton’s law of viscosity yields (formulas taken from Appendix V): ∂v   ∂v 0 τ zr = µ z + r  = ∂z   ∂r  ∂v 1 ∂vz  0 τ zθ = µ θ + =  ∂z r ∂θ  τ zz

∂v =2µ z =0 ∂z

The shape of this deformation 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

for r ≥ R

2 gr 2

r z

The only nonzero component of the surface force 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 (117) at z=0 and r→∞, we are able to evaluate the only remaining integration constant: p(∞,0) = const = patm leaving (117) as

p ( r= , z ) patm + rgz − r

R 4 ω2 2r 2

Near the rotating cylinder but at z=0, the pressure is predicted to subatmospheric — provided we are still in the liquid phase. In experiments, we see the free surface is deformed near the rotating cylinder:*

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

* photos taken from page 63 of Bird, Armstrong & Hassager, Dynamics of Polymeric Liquids. Vol I, 1987.



06-703

70

Fall, 2015

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: yI ≡ τ11 – τ22 = τθθ – τrr < 0

[to be continued]

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. As 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

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

∂vr =0 ∂r  1 ∂vθ vr  τθθ =2µ  +  =0 r   r ∂θ ∂vz τ zz =2µ =0 ∂z τrr =2µ



06-703

71

Fall, 2015


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, and an inverted cone which is lowered into the liquid until its apex just touches the plate. The cone is then rotated at some angular velocity, Ω, and the torque required to turn the cone or to keep the plate stationary is measured.

while the angular velocity is (see figure above and page 10): Ω = Ωez = Ω[(cosθ)er – (sinθ)eθ]

(118)

Substituting into the expression for solid body rotation: θ d

• d 0 there.

f = f ' = 0 at η=0 f ' → 1 as η→ ∞


♠ Paul Richard Heinrich Blasius (1883–1970), German physicist. One of Prandtl’s first PhD students.


06-703

110

Boundary-Layer Thickness The definition of boundary layer thickness is somewhat arbitrary. Although we are tempted to say that d 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 d is ∞. There are several ways to assign a more meaningful finite value to d(x).

Fall, 2015

• 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, d, is that value of y at which the velocity is 99% of the asymptotic value. • Def'n 1:

vx(x,d1) = 0.99Us(x)

 v y ( x, y )   ν  = lim  =  lim 0.8604  0 Ux  y →∞   U  x→∞  x →∞

This means that the streamline moves away from the plate. A third definition for d 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 d 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: d3 = b–a

Graphically, this d 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

y =0

c dc = −D = −D ∞ d dy y =0

If we define d 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: y =

Q = aU= W

b( x )

∫

v x dy

0

where W is the width of the plate (assumed to be arbitrarily wide). Adding d3U to both sides: b

( a + d= 3 )U ∫ v x dy + d3U

(233)

0



06-703

111

The left-hand side of this expression can be rewritten as: = ( a + d3 ) U

b

∫ U dy

bU =

(234)

Fall, 2015

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

0

δ≈

νx U

Re ≡

Ux ν

b

b

(233) becomes: 0

0 b

which is consistent with (235) and with Prandtl's more general result.

0

Drag on Plate

U dy ∫ v x dy + d3U ∫= d3U =

∫ (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 d3 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:

d3U =

∫ (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) ∝

(235)

We showed (see page 102) for any 2-D flow (which this is) that:

δ≈

R Re

=

νR U

(236)

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

1.72 δ3 ( x ) =

and

x  ∂v  dx Fx = 2µ  x  ∂y y =0   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:

Fx 1.328 2 CD ≡ Wx = 1 ρU 2 Re 2 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.


06-703

112

Fall, 2015

vx

∂v x ∂v ∂2 vx dU s + vy x − ν =U s 2 ∂x ∂y dx ∂y ∂v x ∂v y + = 0 ∂x ∂y

Substituting (237):


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 vr = U 1 −     r 

∂v x ∂v ∂2 vx U2 x x + vy x − n =4 sin cos 2 ∂x ∂y R R R ∂y

vx

Solution for a Symmetric Cylinder

2

 cos θ 

∂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 → ∞ 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.

  R 2  −U 1 +    sin θ vθ =   r  

From this potential-flow solution, we can calculate the Us appearing in Prandtl's equation:

Us ( x) = − vθ ( R , θ ) = 2U

sin θ= 2U sin

x x  sin π−  =sin R R  

x R

(237)

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.

dU s 1 dp = − ≠0 dx ρ dx

• Step 3: re-write the boundary-layer equations in terms of the streamfunction

U2 x x sin cos y y y xy − y x y yy − ny yyy = 4 R R R   Us

at y=0:

yx = yy = 0

as y→ ∞:

y → 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:

The two are related by x= R ( π − θ )

Us

or

θ= π−

x R

• Step 2: apply Prandtl’s boundary-layer equations (see page 103)

Us(x) = u1x + u3x3 + u5x5 + ... Then Blasius obtained a solution with the form (see Schlichting, pages154-158): y(x,y) = (ν/u1)1/2[u1xf1(η) + 4u3x3f3(η) +

+ 6u5x5f5(η;u1,u3,u5) + ...] where


η = y(u1/ν)1/2


06-703

113

Fall, 2015

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 a=X or θ). In particular notice that the initial slope ( ∂v x ∂y ) is positive for a=0 and becomes y =0

zero for a=108.8°. At even larger a’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



06-703

115

Fall, 2015

Drag Coefficient and Behavior in the Wake of the Cylinder 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.

y(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 ≈ 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.

Re = 1.1

1 < Re < 6. Streamlines are no longer symmetric about θ=π/2. Streamlines are somewhat farther from the cylinder on the downstream side. Drag coefficients are significantly smaller than predicted by Lamb.



06-703

116

Fall, 2015

Measurements of the drag coefficient for flow normal to a circular cylinder are summarized in the large figure above.



06-703

117

Fall, 2015

the inclination, h is no longer independent of x, so our guess leads to:


The Lubrication Approximation

Q = 12= Uh ( x ) W

Sliding between Parallel Plates Consider a plate sliding on a lubricating film past a second stationary surface:

f ( x)

(238)

which violates continuity. It turns out that continuity is preserved by a nonzero pressure gradient, dp/dx, which causes pressure-driven flow. Thus even the primary component of the velocity profile is affected by this slight inclination. More significantly, it turns out that this inclination will produce a different pressure in the film from the fluid outside the slider block which, tends to push the two surfaces together or apart. Fy ≠ 0 Let’s try to estimate this force.

If the distance h separating the two plates is small compared to the dimensions of the plate (L and W), we can assume fully developed flow applies throughout most of the film. Then: vx ( y ) = U

for h>1:

Q = −πa 2

εz 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. Example: A typical value of the z potential for aqueous solutions is about 50 mV. Plugging in typical values (for water) of the other parameters leads to εζmm/s = 3.5 m V/cm

Electrophoresis of Large Spheres Consider a rigid sphere of radius R, constructed of some insulating 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.

♠ Marian Smoluchowski (1872-1917), ethnic Polish physicist in Austro-Hungarian Empire.



06-703

169

Fall, 2015

2 

1   2   r  r 2 r  r 

1

     sin  0 2     r sin 

Fig. 12: 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 (393) below] that the electrophoretic velocity of a large sphere is given by R>>1:

U

 E 

which appears to be a straightforward application of Smoluchowski’s slip velocity (382). 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. 12) would be predicted if the particle is negatively charged (i.e.  < 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 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 (348) or Poisson’s equation (349). If the charge layer is very thin, the fluid is electrically neutral nearly everywhere: thus the charge density e vanishes and Poisson’s equation (349) reduces to Laplace’s equation:

at r=R:

 0 r

as r:

 –E rcos

The b.c. at r=R assumes that the sphere is an insulator and that no electrical current can be conducted into it. The b.c. 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 36). The particular solution to (383) and its boundary conditions is:

   r ,     E  r  12 

R3   cos  r2 

In particular, we will need the tangential component of the electric field at the surface of the sphere:

3  1     E sin  at r = R: E      2 r   rR

(384)

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. 12). Smoluchowski’s equation (382) 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. 13.

Fig. 13: Same situation as in Fig. 12 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 v and replacing Ex by E, Smoluchowski’s equation (382) yields: at r = R+-1:


(383)

v  

 E 

(385)


06-703

170

for the fluid’s velocity just outside the diffuse cloud in a reference frame in which the sphere is stationary (see Fig. 14). (385) and (382) on which it is based is sometimes called the “slip” boundary condition (as opposed to no-slip). (382) is called that because -1 is so thin compared to R that the fluid appears to change discontinuously with r or “slip” over the surface of the sphere.

Fall, 2015 counterion cloud, we will have a tangential component of the fluid velocity as specified by (385); however there is still no normal component of fluid velocity: at r = R+-1:

vr = 0 v  

(389)

 E 

(390)

If there is no pressure gradient imposed on the distant fluid (we are neglecting any effect of gravity), 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+-1:

v = 

and

p = p

Recall from Theorem III, that the existence of a potential implies v = 0. Then the left-hand side of (386) 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 (386) and (387) becomes: 2  = 0 Fig. 14: Magnified view of very thin diffuse cloud next to particle’s surface in Fig. 13.

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 an electric field. Both the speed U and particle radius R are quite small in such experiments (m/s and m, respectively) so the Reynolds number is always very small and Stokes equation (129) then applies: for r>R+-1:

2v = p

(386)

.v = 0

(387)

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, then the appropriate boundary conditions (in the moving reference frame of the sphere) are: as r : at r = R:

v = U

and

p = p

(388)

v=0

which requires no slip of the fluid immediately in contact with the sphere. But just outside of the thin Copyright© 2015 by Dennis C. Prieve

(391)

b.c. (388) translates to: as r:

 = Uz = Urcos

and b.c. (389) becomes: at r=R

 0 r

The particular solution to (391) subject to these boundary conditions also turns out also to be axisymmetric:

   r ,    U   r  12 

R3   cos  r2 

We still need to satisfy the Smoluchowski’s slip condition (390): at r = R:

3  1     U  sin  v    2  r   r  R

(392)

Using (384) and (392) in (390), we conclude that  3   3  U  sin      E sin     2 2 


06-703

or

171

U  

Comparing (393) and (395), we see that both electrophoretic velocities are proportional to electric field. The proportionality constant is electrophoretic mobility:

 E 

In vector form, we can write:

U  

 E 

electrophor. mobility 

In the stationary reference frame of the laboratory, U = –U (recall Fig. 13) and this result becomes R>>1:

U

 E 

(393)

The absence of the minus sign can be rationalized as follows: a positively charged particle (i.e. >0) is predicted to move in the same direction as the electric field. This result for the electrophoresis of large spheres was first derived by Smoluchowski in 1914. Except for sign, this electrophoretic velocity of a charged sphere through a stagnant fluid is identical to the osmotic velocity of fluid through a stationary solid membrane. Although we have derived this result for a sphere, this 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 Now let's consider the opposite extreme of a very small particle compared to the thickness of the counterion cloud. In other words, let R  0. In this limit, we can treat the particle as a point charge. The force acting on the particle by the applied electric field is given by the definition of electric field: Fel = QE The particle will begin to move creating a drag force which can be calculated from Stokes' law: Fdrag = –6RU

(394)

The particle continues to accelerate until the drag force exactly balances the electrostatic force and the net force is zero: Fel + Fdrag = 0 This terminal velocity represents the electrophoretic velocity in the induced electric field:

U

Fall, 2015

Q E 6R

(395)

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 (395) different from (393)? To compare them, we must first relate Q to . One relationship between charge and zeta potential is given by the Gouy-Chapman equation (369). 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. R>>1). However in the current problem, the sphere is very small compared to the Debye length (i.e. R 7 lie below both of these asymptotes. Analytical Approximation for Large Particles When the zeta potentials are large, κR has to be very large indeed in order for Smoluchowski’s formula (399) for be accurate. The curve corresponding to κR =100 in Fig. 15 deviates from the straight line corresponding to (399) for ζ greater than about 3. For ζ greater than about 6, U E

actually goes through a maximum. So how large does κR have to be to use (399)? O’Brien has suggested the following criterion:  zez  kR 2 cosh    2kT 

(400)

which means that the minimum value of κR needed to use (399) grows exponentially with ζ , regardless of sign (cosh is an even function).

Fig. 16: Solution to Example on this page.

O’Brien & Hunter♦ developed an improved approximation for electrophoretic mobility of large particles in z-z electrolytes:

for ζ >0:

where

 z ln 2 (1 − e− zz ) 6 − 3 2 z   U E = z −  R κ 2 2+ e − zz 2 3m 1+ 2 z

m ≡

(401)

2εRgasT z 12.86 = (402) 0 3m Λ Λ0

is the dimensionless drag coefficient for the counterions (assumed to be anions), Λ0 is the ionic conductance of counterions at infinite dilution (conductivity divided by molar concentration) and Rgas is the universal gas constant (k times Avogadro’s number). The second equation above applies for aqueous solutions at 25ºC provided that the units of Λ0 are Scm2/mol. ♦ R.W. O’Brien & R.J. Hunter, Canadian J. Chem. 59, 1878 (1981).


Example: compare predictions of (401) for U E versus ζ in solutions of KCl having κR = 150 with corresponding results of Fig. 15. Solution: Since (401) is only valid for positive zetas, the counterion is the anion of the salt. The ionic conductance of Cl– at 25ºC is♠ Λ0 = 76.31

S-cm 2 mol

(KCl was probably chosen for Fig. 15 because its two ions have nearly identical values of the conductance.) (402) gives m = 0.169. (401) gives the blue curve in Fig. 16. Notice that the blue curve has a maximum mobility of 7.15 which occurs at ζ = 6.8 which is

♠ from Table 05-09-86 (“Ionic Conductivity and Diffusion at Infinite Dilution”) in the CRC Handbook available (free when using CMU acct) online at www.hbcpnetbase.com.


06-703

175

Fall, 2015

very close to the maximum of the curve in Fig. 15 for κR = 150. The mobilities calculated from approximation (401) for this and other κR values are compared in Fig. 17 with the exact numerical results from Fig. 15. While the comparison for κR = 150 is quite close, the comparisons for κR = 50 and 30 show greater deviation. Streaming Potential So far in our discussion of electrokinetic phenomena, we have considered the motion of either fluid or solid that results from the application of an electric field: Electrophoresis, Electro-osmosis: E≠0 → U≠0 Now we would like to consider the inverse situation in which relative motion across a charged interface generates an electric field: Streaming Potential, Sedimentation Potential: U≠0 → E≠0

Consider the pressure-driven flow through a large circular capillary. κ–1 > exp    2kT 

♣ Overbeek, 1952.



06-703

178

Surface Tension Driven Flows The video Surface Tension in Fluid Mechanics 530.427 S961 (which we viewed) contained many demonstrations of bubble, drop and particle motion caused by surface tension. In this chapter, we will try to analyze one such motion. But first we need to understand the nature of surface tension. Surface Tension In thermodynamics, the Gibbs free energy is an extensive property of material whose value is proportional to the mass or volume of material. When the system consists of a mixture of two or more phases, the free energy also contains a contribution which is proportional to the surface area between the phases. This contribution is called surface tension (units are energy per unit area or force per unit length). Such a contribution arises from intermolecular forces. All molecules experience van der Waals attraction for one another. Consider a single water molecule buried deep in a pool of water. Such a molecule is represented by the blue dot in the lower right of the figure below. This interior water molecule is surrounded on all sides by other water molecules and is pulled equally in all directions (indicated by the four arrows). As a result it feels no air water

Fall, 2015

phase will rearrange themselves to minimize the interfacial area. Thus small droplets tend to be spherical because that geometry has the smallest area for a given volume. Expanding Area of Soap Films From a continuum point of view, the tendency to minimize interfacial area is achieved by assigning a contribution to the free energy of the air-water mixture which is proportional to the interfacial area. That contribution is called surface tension which we will denote by γ.

To understand the consequences of surface tension, consider a soap film (colored area in sketch above) supported by a rectangular wire frame, in which the right side of the wire frame is a slide wire which can be moved either right or left. Suppose we apply a force F and drag the slide wire to the right a distance dx. The amount of work done is F dx. This work goes into increasing the surface area of the soap film by an equal to 2l dx which increases the free energy by 2 γ l dx.♣ The factor of two arises because the soap film has two air-water interfaces: one each on the top and bottom. Assuming there are no irreversible losses of energy: dW = F dx = 2 γ l dx

where γ is the surface tension. Dividing out the dx we have F = 2γ l

(409)

net force. Next, consider a second water molecule located next to the air/water interface. Such a molecule is represented by the blue dot in the upper left of the figure above. Owing to the very low density of air compared to water, there are virtually no molecules located above our interfacial water molecule. Consequently, the intermolecular forces acting on it are no longer balanced: it feels a net force pulling it normal to the interface into the pool. All liquid molecules near the interface experience a net force normal to the interface pulling them back into the liquid. Each interfacial molecule then moves away from the interface, but conservation of mass causes another molecule to take its place. In the absence of other forces, the molecules of the liquid Copyright© 2015 by Dennis C. Prieve

This force must be continuously applied to overcome the thermodynamic tendency to reduce the interfacial area and keep the film from shinking in area. Values of the surface tension for various liquids are shown at several temperatures in Table 1. Strictly speaking, the term “surface tension” refers to the interfacial free energy between a liquid and its vapor. However, when the density of the gas phase is much less than the liquid, replacing the vapor by a mixture of vapor and air usually doesn’t change the value of the interfacial free energy.

♣ The volume of liquid held in the film is a constant, so the film also thins a little as the area is increased.


06-703

179

Fall, 2015

Table 1 Surface Tension for Common Liquids*

*Taken from CRC Handbook of Chemistry & Physics, 87th edition, 2006-7.

Table 2 Surface vs Interfacial Tensions*

*taken from Hunter, Vol. 1, page 233. Replacing the gas phase by second immiscible liquid does however change the interfacial free energy. The term interfacial tension is the generic term used for the interfacial free energy per unit area between two arbitrary (immiscible) phases. Table 2 shows the effect of replacing the gas phase in contact with various organic liquids by liquid water. At least in these examples, the resulting interfacial tension between liquids lies between the surface tensions for the two liquids.


Laplace Pressure This derivation of the Young-Laplace equation is adapted from Hunter, Vol 1.


06-703

180

Fall, 2015

must exert a force to keep the area of the spherical cap from shinking. This force is exerted tangent to the surface as shown in the figure and its magnitude for a differential arc length ds (the analog of l) is dF = γ ds

The component of this force normal to the surface (i.e. in the direction of the line XY) is dFvertical = γ ds sin θ

Consider a drop of liquid (see figure above) having surface tension γ in air. A spheroidal cap is bounded by a contour (the thick pink line) which is everywhere the same distance d from some arbitrary point X on the surface. The three distances denoted d might not look equal in the diagram owing to perspective.

but when d→0, this angle is that subtended by an arc of length d and radius r1: lim θ =

d →0

and

d r1

lim sin θ = θ =

d →0

d r1

dFvertical d = γ ds r1

Thus

This is the contribution to the vertical component of the force from point A. Averaging similar contributions from points B, C and D yields dFvertical γd  1 1  γd  1 1  = + = + ds 2  r1 r2  2  R1 R2 

In the sketch above, the edge of the spheroidal cap has been re-drawn and re-scaled as the pink line labelled ACBDA. Contours AB and CD (which lie completely on the surface of the drop) both contain the point X and form a right angle at point X. In the limit that d→0, both AB and CD become circular arcs From the of radius r1 and r2, respectively. differential geometry of surfaces (Weatherburn, 1930), the directions of AB and CD (which are perpendicular) can be chosen such that r1 is a maximum (R1) and r2 is a minimum (R2). Furthermore the sum of the reciprocals is independent of the orientation of the two perpendicular axes AB and CD: 1 1 1 1 + = + = H r1 r2 R1 R2

This sum of the reciprocals of the radii is called the mean curvature of the surface at point X, r1 and r2 are called the radii of curvature and R1 and R2 are called the principal radii of curvature. In the same way that a force F was needed to overcome the thermodynamic tendency of the soap film to shink in (409), the remainder of the surface Copyright© 2015 by Dennis C. Prieve

Since the sum of the reciprocals of the radii is independent of orientation, we can integrate over the entire edge of the spherical cap. In the limit d→0, the edge is just a circle of radius d.

F= vertical

gd 2

1   1 R +R  2   1

∫

ds

edge of cap

   2 pd

or

1   1 + Fvertical = γπd 2   R R 2   1

This vertical component of force on the cap is acting downward. But the drop is not is motion; so another force of equal magnitude but opposite direction must be acting on the cap. This force is hydrostatic pressure: the pressure inside the drop is higher than the pressure outside by ∆p. The upward force due to this higher pressure is ∆p multiplied by the area of the disk (πd2): 1   1 ∆p pd 2 = γpd 2  +   R1 R2 


06-703

or

181

1   1 ∆p =γ  +   R1 R2 

which is called the Young-Laplace equation. The vector generalization of this equation is given by the stress boundary condition found in (98).

Fall, 2015

Capillary Rise

Marangoni Motion of Bubbles

Contact Angle



06-703

182

Fall, 2015

Appendix I: Useful Identities in Vector Notation In the equations below, s and t are arbitrary scalar fields; u, v and w are arbitrary vector fields; and T is an arbitrary tensor field.



06-703

183

Fall, 2015

Appendix II: Vector Operations in Rectangular Coordinates (x, y, z)



06-703


184

Fall, 2015


06-703

185

Fall, 2015

Appendix III: Vector Operations in Cylindrical Coordinates (r, θ, z)



06-703


186

Fall, 2015


06-703

187

Fall, 2015

Appendix IV: Vector Operations in Spherical Coordinates (r, θ, φ)



06-703


188

Fall, 2015


06-703

189

Fall, 2015

Appendix V: Newton’s Law of Viscosity

(

)

t t = µ ∇v + ( ∇v )  + κ − 23 µ ( ∇.v ) I  

In Cartesian Coordinates

In Cylindrical Coordinates



06-703

190

Fall, 2015

In Spherical Coordinates



06-703

191

Fall, 2015

Appendix VI: Navier-Stokes Equations ρ

∂v + ρv.∇v = ρg − ∇p + µ∇ 2 v ∂t

In Cartesian Coordinates



06-703

192

Fall, 2015

In Cylindrical Coordinates



06-703

193

Fall, 2015

In Spherical Coordinates



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, 2015.

Copyright © 2015 by Dennis C. Prieve

06-703

ii

Fall, 2015

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 DIVERGENCE THEOREM The Continuity Equation Reynolds Transport Theorem 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 INTRODUCTION TO CONTINUUM MECHANICS*

1 1 1 2 2 4 5 5 6 6 7 7 9 10 11 12 12 13 14 14 14 14 15 17 17 17 21 21 21 22 23 24 25 26 26 26 27 28

CONTINUUM HYPOTHESIS CLASSIFICATION OF FORCES

28 29

HYDROSTATIC EQUILIBRIUM

31

FLOW OF IDEAL FLUIDS

32



06-703

iii

EULER'S EQUATION KELVIN'S THEOREM IRROTATIONAL FLOW OF AN INCOMPRESSIBLE FLUID Potential Flow Around a Sphere d'Alembert's Paradox Propagation of Weak Sound Waves STREAM FUNCTION TWO-D FLOWS AXISYMMETRIC FLOW (CYLINDRICAL) AXISYMMETRIC FLOW (SPHERICAL) ORTHOGONALITY OF y=CONST AND φ=CONST STREAMLINES, PATHLINES AND STREAKLINES PHYSICAL MEANING OF STREAMFUNCTION 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 CREEPING FLOW APPROXIMATION CONE-AND-PLATE VISCOMETER Velocity Profile for Shallow Cones. Torque Shear Stress and Torque for Shallow Cones. FLOW AROUND A SPHERE CREEPING FLOW AROUND A SPHERE (RE→0) Velocity Profile Displacement of Distant Streamlines Pressure Profile CORRECTING FOR INERTIAL TERMS Perturbation Expansion WALL EFFECTS ON MOTION OF SPHERE Copyright© 2015 by Dennis C. Prieve

Fall, 2015 32 33 35 36 40 42 45 45 46 46 46 47 47 48 50 50 53 53 53 54 55 56 57 58 58 59 59 61 61 61 64 66 67 68 68 71 71 72 73 74 75 76 77 79 79 81 82 84 PDF generated December 4, 2015

06-703

iv

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

Fall, 2015 85 85 85 86 86 86

SOUND PROPAGATION IN VISCOUS FLUIDS

89

BOUNDARY-LAYER APPROXIMATION

92

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 GENERAL ANALYSIS OF SLIDING FOR 2-D FLOW Special Case: Two Parallel Plates Special Case #2: Slightly Nonparallel 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 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


92 93 95 95 96 97 98 99 99 101 102 104 106 108 108 110 111 112 113 115 117 117 117 119 120 121 122 125 129 129 130 132 133 134 136 136 136 138 138 139 139


06-703

v

PRANDTL’S MIXING LENGTH THEORY PRANDTL’S “UNIVERSAL” VELOCITY PROFILE LAMINAR SUBLAYER PRANDTL’S UNIVERSAL LAW OF FRICTION 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 ELECTROKINETIC PHENOMENA HYDROSTATIC EQUILIBRIUM SMOLUCHOWSKI'S ANALYSIS (CA. 1918) ELECTRO-OSMOSIS IN CYLINDRICAL PORES ELECTROPHORESIS OF LARGE SPHERES ELECTROPHORESIS OF SMALL PARTICLES ELECTROPHORESIS OF A SPHERE OF ARBITRARY SIZE STREAMING POTENTIAL

Fall, 2015 140 142 144 144 148 158 158 158 159 159 165 164 165 168 168 171 172 173

SURFACE TENSION DRIVEN FLOWS

178

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

178 178 179 181 181 181

APPENDIX I: USEFUL IDENTITIES IN VECTOR NOTATION

182

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

183

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

185

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

187

APPENDIX V: NEWTON’S LAW OF VISCOSITY

189

IN CARTESIAN COORDINATES IN CYLINDRICAL COORDINATES IN SPHERICAL COORDINATES APPENDIX VI: NAVIER-STOKES EQUATIONS IN CARTESIAN COORDINATES IN CYLINDRICAL COORDINATES IN SPHERICAL COORDINATES

189 189 190 191 191 192 193

TABLE OF CONTENTS

I

TABLE OF LECTURES

I

TABLE OF CROSS REFERENCES TO PAGE NUMBERS

II



Table of Lectures Lecture #2 begins here

5


11


17

06-703

ii

Fall, 2015

Table of Cross References to Page Numbers Page

Item

2 26 56 57 55 60

double-dot product Identity Tensor Poisson’s ratio Modulus for pure shear Uniaxial stress figure stress boundary condition

