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## Creeping Flows

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**Creeping Flows**Steven A. Jones BIEN 501 Wednesday, March 21, 2007 Start on Slide 53 Louisiana Tech University Ruston, LA 71272**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 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 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**• 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 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**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 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**Nonviscous Flow Creeping Flow Velocity Larger velocity near the sphere is an inertial effect. Louisiana Tech University Ruston, LA 71272**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**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**• 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**• 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**The flow will be symmetric with respect to . Louisiana Tech University Ruston, LA 71272**Components of the Incident Flow**Component of incident velocity in the radial direction, Incident Velocity Component of incident velocity in the - direction, Louisiana Tech University Ruston, LA 71272**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). Louisiana Tech University Ruston, LA 71272**Convective Term in Spherical Coordinates**Louisiana Tech University Ruston, LA 71272**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 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 There is symmetry about the -axis. Thus (a) nothing depends on , and (b) there is no - velocity. Louisiana Tech University Ruston, LA 71272**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): Louisiana Tech University Ruston, LA 71272**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 Louisiana Tech University Ruston, LA 71272**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 Louisiana Tech University Ruston, LA 71272**Comments**Three equations, one first order, two second order. Three unknowns ( ). Two independent variables ( ). Equations are linear (there is a solution). Louisiana Tech University Ruston, LA 71272**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. Louisiana Tech University Ruston, LA 71272**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**• 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). Louisiana Tech University Ruston, LA 71272**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. Louisiana Tech University Ruston, LA 71272**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 Ruston, LA 71272**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**With:**Exercise Rewrite the first term in terms of y. Louisiana Tech University Ruston, LA 71272**With:**Exercise Rewrite the second term in terms of y. Louisiana Tech University Ruston, LA 71272**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): Louisiana Tech University Ruston, LA 71272**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: Louisiana Tech University Ruston, LA 71272**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: Louisiana Tech University Ruston, LA 71272**Elimination of Pressure**Alternatively: So, for example, the e1 component is: Louisiana Tech University Ruston, LA 71272**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: Louisiana Tech University Ruston, LA 71272**Elimination of Pressure**This view can be easily illustrated in rectangular coordinates: Louisiana Tech University Ruston, LA 71272**Exercise: 4th order equation**With: What is the momentum equation: in terms of y? Louisiana Tech University Ruston, LA 71272**Exercise: 4th order equation**Answer: or Louisiana Tech University Ruston, LA 71272**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. Louisiana Tech University Ruston, LA 71272**Momentum in terms of y**If: How does this simplify for our problem? Recall: Steady state Low Reynolds number Louisiana Tech University Ruston, LA 71272**Stream Function, Creeping Flow**When the unsteady (left-hand side) terms are eliminated: This equation was given on slide 35. Louisiana Tech University Ruston, LA 71272**Boundary Conditions in Terms of **From and Exercise: Write these boundary conditions in terms of y. Louisiana Tech University Ruston, LA 71272**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. Louisiana Tech University Ruston, LA 71272**Question**Consider the following curves. Along which of these curves must velocity change with position? Louisiana Tech University Ruston, LA 71272**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. Louisiana Tech University Ruston, LA 71272**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 of **From which suggests the -dependence of the solution. Louisiana Tech University Ruston, LA 71272**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. Louisiana Tech University Ruston, LA 71272