Creeping flows
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Creeping Flows. Steven A. Jones BIEN 501 Wednesday, March 21, 2007 Start on Slide 53. Creeping Flows. Major Learning Objectives: Compare viscous flows to nonviscous flows. Derive the complete solution for creeping flow around a sphere (Stokes’ flow).

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Creeping flows

Creeping Flows

Steven A. Jones

BIEN 501

Wednesday, March 21, 2007

Start on Slide 53

Louisiana Tech University

Ruston, LA 71272


Creeping flows1

Creeping Flows

Major Learning Objectives:

  • Compare viscous flows to nonviscous flows.

  • Derive the complete solution for creeping flow around a sphere (Stokes’ flow).

  • Relate the solution to the force on the sphere.

Louisiana Tech University

Ruston, LA 71272


Creeping flows2

Creeping Flows

Minor Learning Objectives:

  • Examine qualitative inertial and viscous effects.

  • Show how symmetry simplifies the equations.

  • Show how creeping and nonviscous flows simplify the momentum equations.

  • Give the origin of the Reynolds number.

  • Use the Reynolds number to distinguish creeping and nonviscous flows.

  • Apply non-slip boundary conditions at a wall and incident flow boundary conditions at .

Louisiana Tech University

Ruston, LA 71272


Creeping flows3

Creeping Flows

Minor Learning Objectives (continued):

  • Use the equations for conservation of mass and conservation of momentum in spherical coordinates.

  • Use the stream function to satisfy continuity.

  • Eliminate the pressure term from the momentum equations by (a) taking the curl and (b) using a sort of Gaussian elimination.

  • Rewrite the momentum equations in terms of the stream function.

  • Rewrite the boundary conditions in terms of the stream function.

  • Deduce information about the form of the solution from the boundary conditions.

Louisiana Tech University

Ruston, LA 71272


Important concepts

Important Concepts

  • Flow Rate

  • Cross-sectional average velocity

  • Shear Stress (wall shear stress)

  • Force caused by shear stress (drag)

  • Pressure loss

Louisiana Tech University

Ruston, LA 71272


Creeping flows4

Creeping Flows

Minor Learning Objectives (continued):

  • Discuss the relationship between boundary conditions and the assumption of separability.

  • Reduce the partial differential equation to an ordinary differential equation, based on the assumed shape of the solution.

  • Recognize and solve the equidimensional equation.

  • Translate the solution for the stream function into the solution for the velocity components..

  • Obtain the pressure from the velocity components.

  • Obtain the drag on the sphere from the stress components (viscous and pressure).

Louisiana Tech University

Ruston, LA 71272


Creeping vs nonviscous flows

Creeping vs. Nonviscous Flows

Nonviscous Flows

Viscosity goes to zero (High Reynolds Number)

Left hand side of the momentum equation is important.

Right hand side of the momentum equation includes pressure only.

Inertia is more important than friction.

Creeping Flows

Viscosity goes to  (Low Reynolds Number)

Left hand side of the momentum equation is not important.

Left hand side of the momentum equation is zero.

Friction is more important than inertia.

Louisiana Tech University

Ruston, LA 71272


Creeping vs nonviscous flows1

Creeping vs. Nonviscous Flows

Nonviscous Flow Solutions

Use flow potential, complex numbers.

Use “no normal velocity.”

Use velocity potential for conservation of mass.

Creeping Flow Solutions

Use the partial differential equations. Apply transform, similarity, or separation of variables solution.

Use no-slip condition.

Use stream functions for conservation of mass.

In both cases, we will assume incompressible flow.

Louisiana Tech University

Ruston, LA 71272


Flow around a sphere

Flow around a Sphere

Nonviscous Flow

Creeping Flow

Velocity

Larger velocity near the sphere is an inertial effect.

Louisiana Tech University

Ruston, LA 71272


Flow around a sphere1

Flow around a Sphere

A more general case

Incident velocity is approached far from the sphere.

Increased velocity as a result of inertia terms.

Shear region near the sphere caused by viscosity and no-slip.

Louisiana Tech University

Ruston, LA 71272


Stokes flow the geometry

Stokes Flow: The Geometry

Use Standard Spherical Coordinates, (r, , and )

Far from the sphere (large r) the velocity is uniform in the rightward direction. e3 is the Cartesian (rectangular) unit vector. It does not correspond to the spherical unit vectors.

Louisiana Tech University

Ruston, LA 71272


The objective

The Objective

  • Obtain the velocity field around the sphere.

  • Use this velocity field to determine pressure and drag at the sphere surface.

  • From the pressure and drag, determine the force on the sphere as a function of the sphere’s velocity, or equivalently the sphere’s velocity as a function of the applied force (e.g. gravity, centrifuge, electric field).

Louisiana Tech University

Ruston, LA 71272


Some applications

Some Applications

  • What electric field is required to move a charged particle in electrophoresis?

  • What g force is required to centrifuge cells in a given amount of time.

  • What is the effect of gravity on the movement of a monocyte in blood?

  • How does sedimentation vary with the size of the sediment particles?

  • How rapidly do enzyme-coated beads move in a bioreactor?

Louisiana Tech University

Ruston, LA 71272


Symmetry of the geometry

Symmetry of the Geometry

The flow will be symmetric with respect to .

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Components of the incident flow

Components of the Incident Flow

Component of incident velocity in the radial direction,

Incident Velocity

Component of incident velocity in the  - direction,

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Creeping momentum equation

Creeping Momentum Equation

To see how creeping flow simplifies the momentum equation, begin with the equation in the following form (Assume a Newtonian fluid):

For small v, 2nd term on the left is small. It is on the order of v2. (v appears in the right hand term, but only as a first power).

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Ruston, LA 71272


Convective term in spherical coordinates

Convective Term in Spherical Coordinates

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Reynolds number

Reynolds Number

The Reynolds number describes the relative importance of the inertial terms to the viscous terms and can be deduced from a simple dimensional argument.

Louisiana Tech University

Ruston, LA 71272


Reynolds number1

Reynolds Number

Different notations are used to express the Reynolds number. The most typical of these are Re or Nr.

Also, viscosity may be expressed as kinematic ( ) or dynamic () viscosity, so the Reynolds number may be

In the case of creeping flow around a sphere, we use v for the characteristic velocity, and we use the sphere diameter as the characteristic length scale. Thus,

Louisiana Tech University

Ruston, LA 71272


Boundary conditions b c s for creeping flow around a sphere

Boundary Conditions (B.C.s) for Creeping Flow around a Sphere

There is symmetry about the  -axis. Thus (a) nothing depends on , and (b) there is no  - velocity.

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Ruston, LA 71272


Summary of equations to be solved

Summary of Equations to be Solved

We must solve conservation of mass and conservation of momentum, subject to the specified boundary conditions.

Conservation of mass in spherical coordinates is:

Which takes the following form in spherical coordinates (Table 3.1):

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Summary of equations momentum

Summary of Equations (Momentum)

Because there is symmetry in , we only worry about the radial and circumferential components of momentum.

(Incompressible, Newtonian Fluid)

Which takes the following form in spherical coordinates (Table 3.4):

Radial

Azimuthal

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Simplified differential equations

Simplified Differential Equations

Yikes! You mean we need to solve these three partial differential equations!!?

Conservation of Mass

Conservation of Radial Momentum

Conservation of Azimuthal Momentum

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Comments

Comments

Three equations, one first order, two second order.

Three unknowns ( ).

Two independent variables ( ).

Equations are linear (there is a solution).

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Stream function approach

Stream Function Approach

We will use a stream function approach to solve these equations.

The stream function is a differential form that automatically solves the conservation of mass equation and reduces the problem from one with 3 variables to one with two variables.

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Stream function cartesian

Stream Function (Cartesian)

Cartesian coordinates, the two-dimensional continuity equation is:

If we define a stream function, y, such that:

Then the two-dimensional continuity equation becomes:

Louisiana Tech University

Ruston, LA 71272


Summary of the procedure

Summary of the Procedure

  • Use a stream function to satisfy conservation of mass.

    • Form of is known for spherical coordinates.

    • Gives 2 equations (r and  momentum) and 2 unknowns ( and pressure).

    • Need to write B.C.s in terms of the stream function.

  • Obtain the momentum equation in terms of velocity.

  • Rewrite the momentum equation in terms of .

  • Eliminate pressure from the two equations (gives 1 equation (momentum) and 1 unknown, namely ).

  • Use B.C.s to deduce a form for  (equivalently, assume a separable solution).

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Ruston, LA 71272


Procedure continued

Procedure (Continued)

6. Substitute the assumed form for  back into the momentum equation to obtain an ordinary differential equation.

7. Solve the equation for the radial dependence of .

8. Insert the radial dependence back into the form for  to obtain the complete expression for .

9. Use the definition of the stream function to obtain the radial and tangential velocity components from .

10. Use the radial and tangential velocity components in the momentum equation (written in terms of velocities, not in terms of ) to obtain pressure.

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Procedure continued1

Procedure (Continued)

11. Integrate the e3 component of both types of forces (pressure and viscous stresses) over the surface of the sphere to obtain the drag force on the sphere.

Louisiana Tech University

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Stream function

Stream Function

Recall the following form for conservation of mass:

Slide 22

If we define a function (r,) as:

then the equation of continuity is automatically satisfied. We have combined 2 unknowns into 1 and eliminated 1 equation.

Note that other forms work for rectangular and cylindrical coordinates.

Louisiana Tech University

Ruston, LA 71272


Exercise

With:

Exercise

Rewrite the first term in terms of y.

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Exercise1

With:

Exercise

Rewrite the second term in terms of y.

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Momentum eq in terms of

Momentum Eq. in Terms of 

Use

and conservation of mass is satisfied (procedure step 1).

Substitute these expressions into the steady flow momentum equation (Slide 23) to obtain a partial differential equation for  from the momentum equation (procedure step 2):

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Elimination of pressure

Elimination of Pressure

The final equation on the last slide required several steps. The first was the elimination of pressure in the momentum equations. The second was substitution of the form for the stream function into the result. The details will not be shown here, but we will show how pressure can be eliminated from the momentum equations. We have:

We take the curl of this equation to obtain:

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Ruston, LA 71272


Elimination of pressure1

Elimination of Pressure

But it is known that the curl of the gradient of any scalar field is zero (Exercise A.9.1-1). In rectangular coordinates:

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Elimination of pressure2

Elimination of Pressure

Alternatively:

So, for example, the e1 component is:

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Exercise elimination of pressure

Exercise: Elimination of Pressure

One can think of the elimination of pressure as being equivalent to doing a Gaussian elimination type of operation on the pressure term.

This view can be easily illustrated in rectangular coordinates:

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Ruston, LA 71272


Elimination of pressure3

Elimination of Pressure

This view can be easily illustrated in rectangular coordinates:

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Exercise 4 th order equation

Exercise: 4th order equation

With:

What is the momentum equation:

in terms of y?

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Exercise 4 th order equation1

Exercise: 4th order equation

Answer:

or

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Elimination of pressure4

Elimination of Pressure

Fortunately, the book has already done all of this work for us, and has provided the momentum equation in terms of the stream function in spherical coordinates (Table 2.4.2-1). For vf=0:

Admittedly this still looks nasty. However, when we remember that we have already eliminated all of the left-hand terms, the result for the stream function is relatively simple.

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Momentum in terms of y

Momentum in terms of y

If:

How does this simplify for our problem?

Recall:

Steady state

Low Reynolds number

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Stream function creeping flow

Stream Function, Creeping Flow

When the unsteady (left-hand side) terms are eliminated:

This equation was given on slide 35.

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Boundary conditions in terms of

Boundary Conditions in Terms of 

From

and

Exercise: Write these boundary conditions in terms of y.

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Boundary conditions in terms of1

Boundary Conditions in Terms of 

From

must be zero for all  at r=R. Thus, 

must be constant along the curve r=R. But since it’s constant of integration is arbitrary, we can take it to be zero at that boundary. I.e.

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Question

Question

Consider the following curves. Along which of these curves must velocity change with position?

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Comment

Comment

A key to understanding the previous result is that we are talking about the surface of the sphere, where r is fixed.

As r changes, however, we move off of the curve r=R, so ycan change.

y does not change as q changes.

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Boundary conditions in terms of2

Boundary Conditions in Terms of 

From

(See Slide 14)

Thus, in contrast to the surface of the sphere, ywill change with q far from the sphere.

Louisiana Tech University

Ruston, LA 71272


Boundary conditions in terms of3

Boundary Conditions in Terms of 

From

which suggests the  -dependence of the solution.

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Comment on separability

Comment on Separability

For a separable solution we assume that the functional form of y is the product of one factor that depends only on r and another that depends only on q.

Whenever the boundary conditions can be written in this form, it will be possible to find a solution that can be written in this form. Since the equations are linear, the solution will be unique. Therefore, the final solution must be written in this form.

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Comment on separability1

Comment on Separability

In our case, the boundary condition at r=R is:

and the boundary condition at r is:

Both of these forms can be written as a function of r multiplied by a function of q. (For r=R we take R(r)=0). The conclusion that the q dependence like sin2q is reached because these two boundary conditions must hold for all q. A similar statement about the r-dependence cannot be reached. I.e. we only know about two distinct r locations.

Louisiana Tech University

Ruston, LA 71272


Separability

Separability

Again, at r=R:

and at r :

For a separable solution, we look for a form:

Because the q-dependence holds for all q, but the r-dependence does not, we must write:

Louisiana Tech University

Ruston, LA 71272


Momentum equation

Momentum Equation

The momentum equation:

is 2 equations with 3 unknowns (P, vr and v). We have used the stream function to get 2 equations and 2 unknowns (P and ). We then used these two equations to eliminate P.

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Substitute back into momentum

Substitute Back into Momentum

With

(slide 45) becomes:

Note the use of total derivatives.

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Exercise substitute

Exercise: Substitute

into

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Exercise substitute1

Exercise: Substitute

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Exercise so we now need

Exercise: So we now need

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Substitute back into momentum1

Substitute Back into Momentum

The student should recognize the differential equation as an equidimensional equation for which:

Substitution of this form back into the equation yields:

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Equidimensional equation

Equidimensional Equation

The details are like this:

This is a 4th order polynomial, i.e. there are 4 possible values for n which happen to turn out to be -1, 1, 2 and 4.

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Solution for velocity components

Solution for Velocity Components

Once the boundary conditions are evaluated, the solution is:

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Pressure

Pressure

To obtain pressure, we return to the momentum equation:

This form was 2 equations with 3 unknowns, but now vr and v have been determined. Once the forms for these two velocity components are substituted into this equation, one obtains:

Integrate to get P.

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Pressure1

Pressure

The result of this exercise is:

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Force

Force

To obtain force on the sphere, we must remember that force is caused by both the pressure and the viscous stress.

Used to get the z3 component.

z3 is the direction the sphere is moving relative to the fluid.

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Ruston, LA 71272


Potential flow

Potential Flow

Potential flow derives from the viscous part of the momentum equation.

If we write:

Then the viscous part of the momentum equation will automatically be zero.

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Potential flow1

Potential Flow

The continuity equation:

Becomes:

Therefore potential flow reduces to finding solutions to Laplace’s equation.

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