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Chapter 6 SOLUTION of VISCOUS-FLOW PROBLEMS

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Chapter 6 SOLUTION of VISCOUS-FLOW PROBLEMS Chapter 6 SOLUTION OF VISCOUS-FLOW PROBLEMS 6.1 Introduction HE previous chapter contained derivations of the relationships for the con- T servation of mass and momentum—the equations of motion—in rectangular, cylindrical, and spherical coordinates. All the experimental evidence indicates that these are indeed the most fundamental equations of uid mechanics, and that in principle they govern any situation involving the ow of a Newtonian uid. Unfortunately, because of their all-embracing quality, their solution in analytical terms is dicult or impossible except for relatively simple situations. However, it is important to be aware of these “Navier-Stokes equations,” for the following reasons: 1. They lead to the analytical and exact solution of some simple, yet important problems, as will be demonstrated by examples in this chapter. 2. They form the basis for further work in other areas of chemical engineering. 3. If a few realistic simplifying assumptions are made, they can often lead to approximate solutions that are eminently acceptable for many engineering purposes. Representative examples occur in the study of boundary layers, waves, lubrication, coating of substrates with lms, and inviscid (irrotational) ow. 4. With the aid of more sophisticated techniques, such as those involving power series and asymptotic expansions, and particularly computer-implemented nu- merical methods, they can lead to the solution of moderately or highly ad- vanced problems, such as those involving injection-molding of polymers and even the incredibly dicult problem of weather prediction. The following sections present exact solutions of the equations of motion for several relatively simple problems in rectangular, cylindrical, and spherical coor- dinates. Throughout, unless otherwise stated, the ow is assumed to be steady, laminar and Newtonian, with constant density and viscosity. Although these as- sumptions are necessary in order to obtain solutions, they are nevertheless realistic in many cases. All of the examples in this chapter are characterized by low Reynolds numbers. That is, the viscous forces are much more important than the inertial forces, and 272 6.1—Introduction 273 are usually counterbalanced by pressure or gravitational eects. Typical applica- tions occur at low ow rates and in the ow of high-viscosity polymers. Situations in which viscous eects are relatively unimportant will be discussed in Chapter 7. Solution procedure. The general procedure for solving each problem in- volves the following steps: 1. Make reasonable simplifying assumptions. Almost all of the cases treated here will involve steady incompressible owofaNewtonian uid in a single coordinate direction. Further, gravity may or may not be important, and a certain amount of symmetry may be apparent. 2. Write down the equations of motion—both mass (continuity) and momentum balances—and simplify them according to the assumptions made previously, striking out terms that are zero. Typically, only a very few terms—perhaps only one in some cases—will remain in each dierential equation. The simpli- ed continuity equation usually yields information that can subsequently be used to simplify the momentum equations. 3. Integrate the simplied equations in order to obtain expressions for the de- pendent variables such as velocities and pressure. These expressions will usu- ally contain some, as yet, arbitrary constants—typically two for the velocities (since they appear in second-order derivatives in the momentum equations) and one for the pressure (since it appears only in a rst-order derivative). 4. Invoke the boundary conditions in order to evaluate the constants appearing in the previous step. For pressure, such a condition usually amounts to a specied pressure at a certain location—at the inlet of a pipe, or at a free surface exposed to the atmosphere, for example. For the velocities, these conditions fall into either of the following classications: (a) Continuity of the velocity, amounting to a no-slip condition. Thus, the velocity of the uid in contact with a solid surface typically equals the velocity of that surface—zero if the surface is stationary.1 And, for the few cases in which one uid (A, say) is in contact with another immiscible uid (B), the velocity in uid A equals the velocity in uid B at the common interface. (b) Continuity of the shear stress, usually between two uids A and B, leading to the product of viscosity and a velocity gradient having the same value at the common interface, whether in uid A or B. If uid A is a liquid, and uid B is a relatively stagnant gas, which—because of its low viscosity— is incapable of sustaining any signicant shear stress, then the common shear stress is eectively zero. 5. At this stage, the problem is essentially solved for the pressure and velocities. Finally, if desired, shear-stress distributions can be derived by dierentiating 1 In a few exceptional situations there may be lack of adhesion between the uid and surface, in which case slip can occur. 274 Chapter 6—Solution of Viscous-Flow Problems the velocities in order to obtain the velocity gradients; numerical predictions of process variables can also be made. Typesofow. Two broad classes of viscous ow will be illustrated in this chapter: 1. Poiseuille ow, in which an applied pressure dierence causes uid motion between stationary surfaces. 2. Couette ow, in which a moving surface drags adjacent uid along with it and thereby imparts a motion to the rest of the uid. Occasionally, it is possible to have both types of motion occurring simultaneously, as in the screw extruder analyzed in Example 6.4. 6.2 Solution of the Equations of Motion in Rectangular Coordinates The remainder of this chapter consists almost entirely of a series of worked examples, illustrating the above steps for solving viscous-ow problems. Example 6.1—Flow Between Parallel Plates Fig. E6.1.1 shows the ow of a uid of viscosity , which ows in the x direction between two rectangular plates, whose width is very large in the z direction when compared to their separation in the y direction. Such a situation could occur in a die when a polymer is being extruded at the exit into a sheet, which is subsequently cooled and solidied. Determine the relationship between the ow rate and the pressure drop between the inlet and exit, together with several other quantities of interest. Top plate Large width Inlet Fluid Narrow gap, 2d y Exit x Bottom plate Coordinate system z Fig. E6.1.1 Geometry for ow through a rectangular duct. The spacing between the plates is exaggerated in relation to their length. Simplifying assumptions. The situation is analyzed by referring to a cross section of the duct, shown in Fig. E6.1.2, taken at any xed value of z. Let the depth be 2d (d above and below the centerline or axis of symmetry y = 0), and the length L. Note that the motion is of the Poiseuille type, since it is caused by the applied pressure dierence (p1 p2). Make the following realistic assumptions about the ow: 6.2—Solution of the Equations of Motion in Rectangular Coordinates 275 1. As already stated, it is steady and Newtonian, with constant density and viscosity. (These assumptions will often be taken for granted, and not restated, in later problems.) 2. There is only one nonzero velocity component—that in the direction of ow, vx. Thus, vy = vz =0. 3. Since, in comparison with their spacing, 2d, the plates extend for a very long distance in the z direction, all locations in this direction appear essentially identical to one another. In particular, there is no variation of the velocity in the z direction, so that ∂vx/∂z =0. 4. Gravity acts vertically downwards; hence, gy = g and gx = gz =0. 5. The velocity is zero in contact with the plates, so that vx =0aty = d. Continuity. Start by examining the general continuity equation, (5.48): ∂ ∂(v ) ∂(v ) ∂(v ) + x + y + z =0, (5.48) ∂t ∂x ∂y ∂z which, in view of the constant-density assumption, simplies to Eqn. (5.52): ∂v ∂v ∂v x + y + z =0. (5.52) ∂x ∂y ∂z But since vy = vz =0: ∂v x =0, (E6.1.1) ∂x so vx is independent of the distance from the inlet, and the velocity prole will appear the same for all values of x. Since ∂vx/∂z = 0 (assumption 3), it follows that vx = vx(y) is a function of y only. Velocity Wall profile Inlet Exit v x y d x p p Axis of 1 2 symmetry d Wall L Fig. E6.1.2 Geometry for ow through a rectangular duct. 276 Chapter 6—Solution of Viscous-Flow Problems Momentum balances. With the stated assumptions of a Newtonian uid with constant density and viscosity, Eqn. (5.73) gives the x, y, and z momentum balances: ∂v ∂v ∂v ∂v ∂p ∂2v ∂2v ∂2v x + v x + v x + v x = + x + x + x + g , ∂t x ∂x y ∂y z ∂z ∂x ∂x2 ∂y2 ∂z2 x ∂v ∂v ∂v ∂v ∂p ∂2v ∂2v ∂2v y + v y + v y + v y = + y + y + y + g , ∂t x ∂x y ∂y z ∂z ∂y ∂x2 ∂y2 ∂z2 y ∂v ∂v ∂v ∂v ∂p ∂2v ∂2v ∂2v z + v z + v z + v z = + z + z + z + g . ∂t x ∂x y ∂y z ∂z ∂z ∂x2 ∂y2 ∂z2 z With vy = vz = 0 (from assumption 2), ∂vx/∂x = 0 [from the simplied continuity equation, (E6.1.1)], gy = g, gx = gz = 0 (assumption 4), and steady ow (assumption 1), these momentum balances simplify enormously, to: ∂2v ∂p x = , (E6.1.2) ∂y2 ∂x ∂p = g, (E6.1.3) ∂y ∂p =0. (E6.1.4) ∂z Pressure distribution.
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