Chapter 6. Solution of Viscous-Flow Problems

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 films, and inviscid (irrotational) flow.

4. With the aid of more sophisticated techniques, such as those involving power series and asymptotic expansions, and particularly computer-based numerical methods (as implemented by CFD or computational fluid dynamics software such as FlowLab and COMSOL), they can lead to the solution of moderately or highly advanced problems, such as those involving injection-molding of polymers and even the incredibly difficult 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 coordinates, augmented by a couple of CFD examples. Throughout, unless otherwise stated, the flow is assumed to be steady, laminar, and Newtonian, with constant density and viscosity. Although these assumptions are necessary in order to obtain solutions, they are nevertheless realistic in several instances.

The reader is cautioned that probably a majority of industrial processes involve turbulent flow, which is more complicated, and for this reason sometimes receives scant attention in university courses. The author has seen instances in which erroneous assumptions of laminar flow have led to answers that are wildly inaccurate.

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 are usually counterbalanced by pressure or gravitational effects. Typical applications occur in microfluidics (involving tiny channels—see Chapter 12) and in the flow of high-viscosity polymers. Situations in which viscous effects are relatively unimportant will be discussed in Chapter 7.

Solution procedure. The general procedure for solving each problem involves the following steps:

1. Make reasonable simplifying assumptions. Almost all of the cases treated here will involve steady incompressible flow of a Newtonian fluid 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 differential equation. The simplified continuity equation usually yields information that can subsequently be used to simplify the momentum equations.

3. Integrate the simplified equations in order to obtain expressions for the dependent variables such as velocities and pressure. These expressions will usually 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 first-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 specified 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 classifications:

(a) Continuity of the velocity, amounting to a no-slip condition. Thus, the velocity of the fluid 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 fluid (A, for example) is in contact with another immiscible fluid (B), the velocity in fluid A equals the velocity in fluid B at the common interface.

¹ In a few exceptional situations there may be lack of adhesion between the fluid and surface, in which case slip can occur. Also see Example 12.4, in which electroosmosis gives the illusion of slip.

(b) Continuity of the shear stress, usually between two fluids A and B, leading to the product of viscosity and a velocity gradient having the same value at the common interface, whether in fluid A or B. If fluid A is a liquid, and fluid B is a relatively stagnant gas, which—because of its low viscosity—is incapable of sustaining any significant shear stress, then the common shear stress is effectively 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 differentiating the velocities in order to obtain the velocity gradients; numerical predictions of process variables can also be made.

Types of flow. Two broad classes of viscous flow will be illustrated in this chapter:

1. Poiseuille flow, in which an applied pressure difference causes fluid motion between stationary surfaces.

2. Couette flow, in which a moving surface drags adjacent fluid along with it and thereby imparts a motion to the rest of the fluid.

Occasionally, it is possible to have both types of motion occurring simultaneously, as in the screw extruder analyzed in Example 6.4.

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Example 6.1—Flow Between Parallel Plates

Fig. E6.1.1 shows a fluid of viscosity μ that flows 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 solidified. Determine the relationship between the flow rate and the pressure drop between the inlet and exit, together with several other quantities of interest.

Simplifying assumptions. The situation is analyzed by referring to a cross section of the duct, shown in Fig. E6.1.2, taken at any fixed 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 difference (p₁ – p₂). Make the following fairly realistic assumptions about the flow:

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. Entrance effects can be neglected, so that the flow is fully developed, in which case there is only one nonzero velocity component—that in the direction of flow, υ_x. Thus, υ_y = υ_z = 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 ∂υ_x/∂z = 0.

4. Gravity acts vertically downward; hence, g_y = –g and g_x = g_z = 0.

5. The velocity is zero in contact with the plates, so that υ_x = 0 at y = ±d.

Continuity. Start by examining the general continuity equation, (5.48):

which, in view of the constant-density assumption, simplifies to Eqn. (5.52):

But since υ_y = υ_z = 0:

so υ_x is independent of the distance from the inlet, and the velocity profile will appear the same for all values of x. Since ∂υ_x/∂z = 0 (assumption 3), it follows that υ_x = υ_x(y) is a function of y only.

Momentum balances. With the stated assumptions of a Newtonian fluid with constant density and viscosity, Eqn. (5.73) gives the x, y, and z momentum balances:

With υ_y = υ_z = 0 (from assumption 2), ∂υ_x/∂x = 0 [from the simplified continuity equation, (E6.1.1)], g_y = –g, g_x = g_z = 0 (assumption 4), and steady flow (assumption 1), these momentum balances simplify enormously, to:

Pressure distribution. The last of the simplified momentum balances, Eqn. (E6.1.4), indicates no variation of the pressure across the width of the system (in the z direction), which is hardly a surprising result. When integrated, the second simplified momentum balance, Eqn. (E6.1.3), predicts that the pressure varies according to:

Observe carefully that since a partial differential equation is being integrated, we obtain not a constant of integration, but a function of integration, f(x).

Assume—to be verified later—that ∂p/∂x is constant, so that the centerline pressure (at y = 0) is given by a linear function of the form:

The constants a and b may be determined from the inlet and exit centerline pressures:

leading to:

Thus, the centerline pressure falls linearly from p₁ at the inlet to p₂ at the exit:

so that the complete pressure distribution is

That is, the pressure declines linearly, both from the bottom plate to the top plate, and also from the inlet to the exit. In the majority of applications, 2d ≪ L, and the relatively small pressure variation in the y direction is usually ignored. Thus, p₁ and p₂, although strictly the centerline values, are typically referred to as the inlet and the exit pressures, respectively.

Velocity profile. Since, from Eqn. (E6.1.1), υ_x does not depend on x, ∂²υ_x/∂y² appearing in Eqn. (E6.1.2) becomes a total derivative, so this equation can be rewritten as:

which is a second-order ordinary differential equation, in which the pressure gradient will be shown to be uniform between the inlet and exit, being given by:

A minus sign is used on the left-hand side, since ∂p/∂x is negative, thus rendering both sides of Eqn. (E6.1.13) as positive quantities.

Equation (E6.1.12) can be integrated twice, in turn, to yield an expression for the velocity. After multiplication through by dy, a first integration gives:

A second integration, of Eqn. (E6.1.14), yields:

The two constants of integration, c₁ and c₂, are determined by invoking the boundary conditions:

leading to:

Eqns. (E6.1.15) and (E6.1.18) then furnish the velocity profile:

in which –∂p/∂x and (d² – y²) are both positive quantities. The velocity profile is parabolic in shape, and is shown in Fig. E6.1.2.

Alternative integration procedure. Observe that we have used indefinite integrals in the above solution, and have employed the boundary conditions to determine the constants of integration. An alternative approach would again be to integrate Eqn. (E6.1.12) twice, but now to involve definite integrals by inserting the boundary conditions as limits of integration.

Thus, by separating variables, integrating once, and noting from symmetry about the centerline that dυ_x/dy = 0 at y = 0, we obtain:

or:

A second integration, noting that υ_x = 0 at y = d (zero velocity in contact with the upper plate—the no-slip condition) yields:

That is:

in which two minus signs have been introduced into the right-hand side in order to make quantities in both pairs of parentheses positive. This result is identical to the earlier Eqn. (E6.1.19). The student is urged to become familiar with both procedures, before deciding on the one that is individually best suited.

Also, the reader who is troubled by the assumption of symmetry of υ_x about the centerline (and by never using the fact that υ_x = 0 at y = –d), should be reassured by an alternative approach, starting from Eqn. (E6.1.15):

Application of the two boundary conditions, υ_x = 0 at y = ±d, gives

leading again to the velocity profile of Eqn. (E6.1.19) without the assumption of symmetry.

Volumetric flow rate. Integration of the velocity profile yields an expression for the volumetric flow rate Q per unit width of the system. Observe first that the differential flow rate through an element of depth dy is dQ = υ_xdy, so that:

Since from an overall macroscopic balance Q is constant, it follows that ∂p/∂x is also constant, independent of distance x; the assumptions made in Eqns. (E6.1.6) and (E6.1.13) are therefore verified. The mean velocity is the total flow rate per unit depth:

and is therefore two-thirds of the maximum velocity, υ_xmax, which occurs at the centerline, y = 0.

Shear-stress distribution. Finally, the shear-stress distribution is obtained by employing Eqn. (5.60):

By substituting for υ_x from Eqn. (E6.1.15) and recognizing that υ_y = 0, the shear stress is:

Referring back to the sign convention expressed in Fig. 5.12, the first minus sign in Eqn. (E6.1.28) indicates for positive y that the fluid in the region of greater y is acting on the region of lesser y in the negative x direction, thus trying to retard the fluid between it and the centerline, and acting against the pressure gradient. Representative distributions of pressure and shear stress, from Eqns. (E6.1.11) and (E6.1.28), are sketched in Fig. E6.1.3. More precisely, the arrows at the left and right show the external pressure forces acting on the fluid contained between x = 0 and x = L.

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The problem of Example 6.1 was solved by starting with the completely general equations of motion and then simplifying them. An alternative approach involves a direct momentum balance on a differential element of fluid—a “shell”—as illustrated in Example 6.2.

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Example 6.2—Shell Balance for Flow Between Parallel Plates

Employ the shell-balance approach to solve the same problem that was studied in Example 6.1.

Assumptions. The necessary “shell” is in reality a differential element of fluid, as shown in Fig. E6.2. The element, which has dimensions of dx and dy in the plane of the diagram, extends for a depth of dz (any other length may be taken) normal to the plane of the diagram.

If, for the present, the element is taken to be a system that is fixed in space, there are three different types of rate of x momentum transfer to it:

1. A convective transfer of dy dz in through the left-hand face, and an identical amount out through the right-hand face. Note here that we have implicitly assumed the consequences of the continuity equation, expressed in Eqn. (E6.1.1), that υ_x is constant along the duct.

2. Pressure forces on the left- and right-hand faces. The latter will be smaller, because ∂p/∂x is negative in reality.

3. Shear stresses on the lower and upper faces. Observe that the directions of the arrows conform strictly to the sign convention established in Section 5.6.

A momentum balance on the element, which is not accelerating, gives:

The usual cancellations can be made, resulting in:

in which the total derivative recognizes that the shear stress depends only on y and not on x. Substitution for τ_yx from Eqn. (5.60) with υ_y = 0 gives:

which is identical with Eqn. (E6.1.12) that was derived from the simplified Navier-Stokes equations. The remainder of the development then proceeds as in the previous example. Note that the convective terms can be sidestepped entirely if the momentum balance is performed on an element that is chosen to be moving with the fluid, in which case there is no flow either into or out of it.

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The choice of approach—simplifying the full equations of motion, or performing a shell balance—is very much a personal one, and we have generally opted for the former. The application of the Navier-Stokes equations, which are admittedly rather complicated, has the advantages of not “reinventing the (momentum balance) wheel” for each problem, and also of assuring us that no terms are omitted. Conversely, a shell balance has the merits of relative simplicity, although it may be quite difficult to perform convincingly for an element with curved sides, as would occur for the problem in spherical coordinates discussed in Example 6.8.

This section concludes with another example problem, which illustrates the application of two further boundary conditions for a liquid, one involving it in contact with a moving surface, and the other at a gas/liquid interface where there is a condition of zero shear.

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Example 6.3—Film Flow on a Moving Substrate

Fig. E6.3.1 shows a coating experiment involving a flat photographic film that is being pulled up from a processing bath by rollers with a steady velocity U at an angle θ to the horizontal. As the film leaves the bath, it entrains some liquid, and in this particular experiment it has reached the stage where: (a) the velocity of the liquid in contact with the film is υ_x = U at y = 0, (b) the thickness of the liquid is constant at a value δ, and (c) there is no net flow of liquid (as much is being pulled up by the film as is falling back by gravity). (Clearly, if the film were to retain a permanent coating, a net upwards flow of liquid would be needed.)

Perform the following tasks:

1. Write down the differential mass balance and simplify it.

2. Write down the differential momentum balances in the x and y directions. What are the values of g_x and g_y in terms of g and θ? Simplify the momentum balances as much as possible.

3. From the simplified y momentum balance, derive an expression for the pressure p as a function of y, ρ, δ, g, and θ, and hence demonstrate that ∂p/∂x = 0. Assume that the pressure in the surrounding air is zero everywhere.

4. From the simplified x momentum balance, assuming that the air exerts a negligible shear stress τ_yx on the surface of the liquid at y = δ, derive an expression for the liquid velocity υ_x as a function of U, y, δ, and α, where α = ρg sin θ/μ.

5. Also derive an expression for the net liquid flow rate Q (per unit width, normal to the plane of Fig. E6.3.1) in terms of U, δ, and α. Noting that Q = 0, obtain an expression for the film thickness δ in terms of U and α.

6. Sketch the velocity profile υ_x, labeling all important features.

Assumptions and continuity. The following assumptions are reasonable:

1. The flow is steady and Newtonian, with constant density ρ and viscosity μ.

2. The z direction, normal to the plane of the diagram, may be disregarded entirely. Thus, not only is υ_z zero, but all derivatives with respect to z, such as ∂υ_x/∂z, are also zero.

3. There is only one nonzero velocity component, namely, that in the direction of motion of the photographic film, υ_x. Thus, υ_y = υ_z = 0.

4. Gravity acts vertically downwards.

Because of the constant-density assumption, the continuity equation, (5.48), simplifies, as before, to:

But since υ_y = υ_z = 0, it follows that

so υ_x is independent of distance x along the film. Further, υ_x does not depend on z (assumption 2); thus, the velocity profile υ_x = υ_x(y) depends only on y and will appear the same for all values of x.

Momentum balances. With the stated assumptions of a Newtonian fluid with constant density and viscosity, Eqn. (5.73) gives the x and y momentum balances:

Noting that g_x = –g sin θ and g_y = –g cos θ, these momentum balances simplify to:

Integration of Eqn. (E6.3.4), between the free surface at y = δ (where the gauge pressure is zero) and an arbitrary location y (where the pressure is p) gives:

so that:

Note that since a partial differential equation is being integrated, a function of integration, f(x), is again introduced. Another way of looking at it is to observe that if Eqn. (E6.3.6) is differentiated with respect to y, we would recover the original equation, (E6.3.4), because ∂f(x)/∂y = 0.

However, since p = 0 at y = δ (the air/liquid interface) for all values of x, the function f(x) must be zero. Hence, the pressure distribution:

shows that p is not a function of x.

In view of this last result, we may now substitute ∂p/∂x = 0 into the x-momentum balance, Eqn. (E6.3.3), which becomes:

in which the constant α has been introduced to denote ρg sin θ/μ. Observe that the second derivative of the velocity now appears as a total derivative, since υ_x depends on y only.

A first integration of Eqn. (E6.3.8) with respect to y gives:

The boundary condition of zero shear stress at the free surface is now invoked:

Thus, from Eqns. (E6.3.9) and (E6.3.10) at y = δ, the first constant of integration can be determined:

A second integration, of Eqn. (E6.3.9) with respect to y, gives:

The second constant of integration, c₂, can be determined by using the boundary condition that the liquid velocity at y = 0 equals that of the moving photographic film. That is, υ_x = U at y = 0, yielding c₂ = U; thus, the final velocity profile is:

Observe that the velocity profile, which is parabolic, consists of two parts:

1. A constant and positive part, arising from the film velocity, U.

2. A variable and negative part, caused by gravity, which reduces υ_x at increasing distances y from the film and eventually causes it to become negative.

Exactly how much of the liquid is flowing upwards, and how much downwards, depends on the values of the variables U, δ, and α. However, we are asked to investigate the situation in which there is no net flow of liquid—that is, as much is being pulled up by the film as is falling back by gravity. In this case:

giving the thickness of the liquid film as:

The velocity profile for this case of Q = 0 is shown in Fig. E6.3.2.

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Example 6.4—Transient Viscous Diffusion of Momentum (COMSOL)

This problem is one of the few instances in this book in which we investigate transient or unsteady-state fluid flow. It illustrates how momentum is transferred under the diffusive influence of viscosity.

As shown in Fig. E6.4.1, consider a liquid of density ρ = 1 g/cm³ and viscosity μ (whose values will be given later), contained between two parallel plates AB and CD of length L = 2 cm and separation H = 1 cm. The lower plate CD is stationary, and the upper plate AB can undergo two types of movement:

1. At time t = 0, it is suddenly set in motion with a constant x velocity υ_x = υ_x0. In our problem, υ_x0 = 1 cm/s and μ = 0.5 g/cm s = 0.5 P.

2. At time t = 0, it is oscillated to the right and left in such a way that its x velocity varies with time according to υ_x0 = a sin ωt. In our problem the amplitude a = 1 cm/s, the angular velocity ω = 2π, and the viscosity may be either μ = 0.1 or 0.5 P.

Very similar types of motion occur in concentric-cylinder viscometers, as in Figs. 1.1 and 11.13(b), although with smaller moving-surface separations.

Use COMSOL to investigate how υ_x varies between the midpoints of the lower and upper plates at different times, and interpret the results. If needed, further details of the implementation of COMSOL are given in Chapter 14.

Note that COMSOL will employ the full Navier-Stokes and continuity equations in its solution. However, because the only nonzero velocity component is υ_x, the following familiar diffusion-type equation is effectively being solved:

Solution—Case 1 (Constant Velocity)

1. In COMSOL, choose the sequence New, Chemical Engineering Module, Momentum Balance, and Incompressible Navier-Stokes.

2. Draw the rectangle shown in Fig. E6.4.1, with opposite vertices (0, 0) and (2, 1).

3. Under Options/Constants, set vx0 = 1, rho = 1, and mu = 0.5.

4. Under Physics/Subdomain Settings, indicate that the density is rho and the viscosity is mu.

5. Under Physics/Boundary Settings, establish the boundary conditions shown in Fig. E6.4.1. On the upper plate (an “Inflow/Outflow” boundary), set the x and y velocities to be u0 = vx0 and v0 = 0.

6. Since the problem is a transient one, under Solve/Solver Parameters choose Solver/Time dependent and for General/Time Stepping/Times, insert 0:0.01:0.1 0.2:0.1:0.5 1:0.5:2—meaning that solutions will be available for subsequent plotting at t = 0, 0.01, 0.02, ... 0.1, 0.2, 0.3, 0.4, 0.5, 1.0, 1.5, 2.0 s.

7. Initialize the mesh without refinement and solve the problem. Observe, as the solution progresses, that COMSOL starts cautiously with a very small time step, which it realizes can rapidly be increased without compromising the accuracy of the solution.

8. Finally, plot υ_x against distance y between the midpoints of the lower and upper plates, for selected values of time. Under Postprocessing/Cross Section Plot Parameters, make the following selections:

(a) Under the General tab, for Solutions to use, select stored output times of 0.01, 0.02, 0.05, 0.1, 0.2, 0.5, and 2.0 s. On a Macintosh computer, this is achieved by holding down the Command key (Control key for a PC) and clicking successively on each time.

(b) Under the Line/Extrusion tab, choose Line plot, x-velocity as the predefined quantity, y as the x-axis data. For Cross-section line data, indicate x0 = 1, x1 = 1, y0 = 0, and y1 = 1, followed by OK.

Comments on the results. Fig. E6.4.2 shows how the velocity υ_x varies with distance y from the lower plate at various times. The result is a classical case of diffusion—in this instance, of x momentum (generated by the motion of the upper plate, which moves at 1 cm/s) into the liquid below it. For the smallest time plotted (t = 0.01 s), most of the liquid, from y = 0 to 0.7 cm, is essentially undisturbed, followed by a rising υ_x between y = 0.7 and 1 cm. For early times (or for an infinitely deep liquid), before the influence of the lower plate is felt, there is an analytical solution:

in which z = H – y is the downward distance from the upper plate, ν = μ/ρ is the kinematic viscosity, and erfc denotes the complementary error function.

As time increases, more and more of the liquid is set into motion by the diffusion of x momentum from the upper plate. Eventually—theoretically at t = ∞ but effectively at a time of about 2 s—a steady state is reached, in which the shear stress exerted by the upper plate is counterbalanced by the shear stress in the opposite direction from the stationary lower plate. Since the velocity profile is linear, dυ_x/dy is constant; thus, these shear stresses are equal (and opposite) because they are both given by τ_yx = μdυ_x/dy.

Solution—Case 2 (Oscillating Velocity)

Since the problem is very similar to that of Case 1, only the differences between the two solution procedures will be noted:

1. Under Constants, replace vx0 with the amplitude a = 1, and add omega = 2*pi = 6.283185. To start, set the viscosity to be mu = 0.1.

2. Under Boundary Settings, the velocity of the upper boundary u0 will now be a*sin(omega*t).

3. Under Times, insert 0:2:18 19:0.25:20, so solutions will be available at 2-s intervals up to 18 s and at 0.25-s intervals between 19 and 20 s.

4. For plotting, under Solutions to use, select stored output times of 19, 19.25, 19.5, 19.75 and 20 s. Such large values of time will insure that the cyclical velocities are repeating themselves almost exactly—indeed, they are only marginally different from the solutions between 2 and 3 s.

5. Proceeding as before, obtain a second plot for a viscosity of μ = 0.5 P.

Comments on the results. The results for the two values of the viscosity are shown in Figs. E6.4.3 and E6.4.4. Since ω = 2π, the period of the oscillations is T = 1 s. Thus, velocity profiles separated by 0.25 s will occur over one complete cycle of t = 0, 0.25T, 0.5T, 0.75T, and T. Note two main effects:

(a) Penetration depth. For the lower viscosity, μ = 0.1 P, the surface oscillations do not penetrate as far into the liquid as those for the higher viscosity of μ = 0.5 P. In other words, the lower viscosity does not permit as rapid a diffusion of momentum as the higher viscosity.

(b) Phase change. Note that for t = T, when the upper plate has a zero velocity, much of the liquid has a negative velocity, largely due to the diffusional effects of the negative surface velocity at an earlier time such as t = 0.75T. Similar observations can be made for other values of time.

Both effects have applications in the testing of non-Newtonian fluids, for which the phase difference between stress and strain is illustrated in terms of displacements in Fig. 11.11. Since the velocity υ_x is the derivative of the distance moved, the displacement D of the upper plate is:

For an infinitely deep liquid, there is again an exact solution:

in which z is the depth and:

The attenuation and phase change with depth z are very apparent from Eqn. (E6.4.4).

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Since polymers are generally highly viscous, their flows can be obtained by solving the equations of viscous motion. In this chapter, we cover the rudiments of extrusion, die flow, and drawing or spinning. The analysis of calendering and coating is considerably more complicated, but can be rendered tractable if reasonable simplifications, known collectively as the lubrication approximation, are made, as discussed in Chapter 8. The treatments of injection molding and blow molding are beyond the scope of this book.

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Example 6.5—The Single-Screw Extruder

Because molten polymers are usually very viscous, they often need very high pressures to push them through dies. One such “pump” for achieving this is the screw extruder, shown in Fig. E6.5.1. The polymer enters the feed hopper as pellets, falls into the screw channels in the feed section, and is pushed forward by the screw, which rotates at an angular velocity ω, clockwise as seen by an observer looking along the axis from the inlet to the exit (notice that the screw is left-handed). The heated barrel, together with shear effects, melts the pellets, which then become fluid prior to entering the metering section.

There are three zones in the extruder:

1. In the feed zone the granules are transported into the barrel, where they melt just before entering the compression zone. The large constant depth of the channel in the feed zone means that there will be negligible pressure exerted on the downstream melt in the metering zone.

2. In the compression or “transition” zone the channel depth decreases from the feed zone to the metering zone. Therefore, the melt will gradually increase in pressure as it travels along the compression zone, reaching a maximum pressure at the beginning of the metering zone.

3. In the metering zone, length L₀, the screw channel has a constant shallow depth h (with h ≪ r) and the melt in the channel should now be homogeneous or uniform. Thus, the metering zone of the screw acts like a constant-delivery pump since the screw is rotating at a constant speed. The melt pressure uniformly increases as it passes along the metering zone. Therefore, calculations of extruder output are based on the metering zone of the screw. Extruder output calculations are relatively simple due to the uniformity of conditions existing in the metering zone of the screw.

The preliminary analysis given here neglects any heat-transfer effects in the metering section, and also assumes that the polymer has a constant Newtonian viscosity μ.

The investigation is facilitated by taking the viewpoint of a hypothetical observer located on the screw, in which case the screw surface and the flights appear to be stationary, with the barrel moving with velocity V = rω at a helix angle θ to the flight axis, as shown in Fig. E6.5.2. The alternative viewpoint of an observer located on the inside surface of the barrel is not very fruitful, because not only are the flights seen as moving boundaries, but the observations would be periodically blocked as the flights passed over the observer! The width of the screw channel measured perpendicularly to the vertical sides of the flight flanks is designated W.

Solution

Motion in two principal directions is considered:

1. Flow parallel to the flight axis, caused by a barrel velocity of υ_y = V cos θ = rω cos θ relative to the (now effectively stationary) flights and screw.

2. Flow normal to the flight axis, caused by a barrel velocity of υ_x = –V sin θ = –rω sin θ relative to the (stationary) flights and screw.

In each case, the flow is considered one-dimensional, with “end-effects” caused by the presence of the flights being unimportant. A glance at Fig. E6.5.3(b) will give the general idea. Although the flow in the x-direction must reverse itself as it nears the flights, it is reasonable to assume for h ≪ W that there is a substantial central region in which the flow is essentially in the positive or negative x direction.

1. Motion parallel to the flight axis. The reader may wish to investigate the additional simplifying assumptions that give the y momentum balance as:

Integration twice yields the velocity profile as:

Here, the integration constants c₁ and c₂ have been determined in the usual way by applying the boundary conditions:

Note that the negative of the pressure gradient is given in terms of the inlet pressure p₁, the exit pressure p₂, and the total length L (= L₀/sin θ) measured along the screw flight axis by:

and is a negative quantity since the screw action builds up pressure and p₂>p₁. Thus, Eqn. (E6.5.2) predicts a Poiseuille-type backflow (caused by the adverse pressure gradient) and a Couette-type forward flow (caused by the relative motion of the barrel to the screw). The combination is shown in Fig. E6.5.3(a).

The total flow rate Q_y of polymer melt in the direction of the flight axis is obtained by integrating the velocity between the screw and barrel, and recognizing that the width between flights is W:

The actual value of Q_y will depend on the resistance of the die located at the extruder exit. In a hypothetical case, in which the die offers no resistance, there would be no pressure increase in the extruder (p₂ = p₁), leaving only the Couette term in Eqn. (E6.5.5). For the practical situation in which the die offers significant resistance, the Poiseuille term would serve to diminish the flow rate given by the Couette term.

2. Motion normal to the flight axis. By a development very similar to that for flow parallel to the flight axis, we obtain:

Here, Q_x is the flow rate in the x direction, per unit depth along the flight axis, and must equal zero, because the flights at either end of the path act as barriers. The negative of the pressure gradient is therefore:

so that the velocity profile is given by:

Note from Eqn. (E6.5.10) that υ_x is zero when either z = 0 (on the screw surface) or z/h = 2/3. The reader may wish to sketch the general appearance of υ_z(z).

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Representative values. A COMSOL example follows shortly, and it is appropriate to consider typical measurements and polymer properties for screw extruders²:

² I heartily thank Mr. John W. Ellis, of the Petroleum and Petrochemical College at Chulalongkorn University in Bangkok, who is an expert in polymer processing, for supplying these values and most of the information in Table 6.1. Previously he wrote Polymer Products: Design, Materials and Processing, with co-author David Morton-Jones, published by Chapman and Hall (1986). His latest book, Introduction to Plastic Foams, co-authored by Dr. Duanghathai Pentrakoon (of the Faculty of Science, Chulalongkorn University, Bangkok), is published by the Chulalongkorn University Press (2005).

1. The size of the screw varies from a laboratory diameter of 20 mm up to large production machines having diameters of 150 mm (very common) or much larger—up to 600 mm.

2. The helix angle is almost always θ = tan^–1(1/π) = 17.66°, corresponding to a “square thread” screw profile in which the pitch P equals the outside diameter D = 2r of the screw.

3. For a 100-mm-diameter screw, a typical channel depth (metering section) would be about h = 5 or 6 mm.

4. The screw rpm ranges from a slow speed of about 30 rpm to a high speed of 150 rpm. The large extruders run at a slower rpm. A typical screw surface speed (the barrel velocity relative to the screw) for extruding polyethylene is 0.5 m/s. For a 100-mm screw, a typical screw speed would be 65 rpm.

5. The metering section length L₀ depends on the type of polymer being used—amorphous, semicrystalline, etc. This length is usually expressed as a multiple of the screw diameter, such as L₀ = 6.5D. Typical extruder screws have an overall length of 25D.

6. Polymers generally “shear thin” with increasing shear rate, so the viscosity will depend on the prevailing shear rate (related to rpm) of the melt in the screw channel at the melt temperature being used. But for extrusion processes we can assume a melt viscosity of anything between 200 and 1,000 N s/m² (i.e., Pa s) for extrusion grades of polyolefins (LDPE, HDPE, and PP).

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Example 6.6—Flow Patterns in a Screw Extruder (COMSOL)

Consider the screw extruder just discussed in Example 6.5. Solve the equations of motion to determine as precisely as possible the flow pattern given approximately in Fig. E6.5.3(b), normal to the flight axis. Note that the pressure cannot be allowed to “float,” but must be specified to some arbitrary value at any one point in the cross section shown in Fig. E6.6.1, such as point 4.

Based on the comments at the end of Example 6.5, use the following values, which relate to the extrusion of a polyolefin:

1. Polymer properties: ρ = 800 kg/m³, μ = 500 Pa s (N s/m²).

2. Rotational speed N = 65 rpm, so that ω = 2π × 65/60 = 6.8 rad/s.

3. Channel depth h = 5 mm = 0.005 m.

4. For a square pitch, the pitch P and screw diameter D are identical, taken to be 100 mm = 0.1 m. The distance between flights is W = P cos θ = 0.1 × cos 17.66° = 0.0953 m. For simplicity in our example, we shall round up slightly and take W = 0.1 m.

5. The relative velocity is V = ωr = ωD/2 = 6.8 × 0.1/2 = 0.340 m/s, with x component υ_x = –0.340 × sin θ = –0.340 × sin 17.66° = –0.103 –0.1 m/s.

Solution

1. In COMSOL, choose the sequence New, Chemical Engineering Module, Momentum Balance, and Incompressible Navier-Stokes.

2. Under Options/Constants set rho = 800, mu = 500, and Vx = -0.1.

2. Draw the rectangle shown in Fig. E6.6.1, with opposite vertices (0, 0) and (0.1, 0.005).

3. Under Physics/Subdomain Settings, indicate that the density is rho and the viscosity is mu.

4. Under Physics/Boundary Settings, set boundaries 1, 2, 3 to be no-slip and boundary 4 to have u0 = Vx and v0 = 0.

5. Under Physics/Point Settings, check Point constraint for point 4, at which p0 = 0. At points 1, 2, and 3 there is no such constraint.

6. Initialize and refine the mesh twice, giving 1,024 elements and 4,951 degrees of freedom. Then solve the problem.

7. Under Postprocessing, make appropriate choices in the Plot Parameters pulldown menu to generate the following:

(a) An arrow plot of the x velocities.

(b) A picture that shows 20 streamlines.

(c) A contour plot of the pressure distribution. Obtain values for the pressure at the midpoints of the two ends by first double-clicking on “Snap” at the bottom of the screen, which enables the mouse to point freely and not be constrained to the nearest grid lines. Then click on each midpoint in turn and observe the value of the pressure in the lower left-hand corner of the window.

(d) A cross-section plot of the pressure between the midpoints A (0.05, 0.005) of the upper moving surface and B (0.05, 0) of the lower stationary surface. Here, you will need to choose the Line/Extrusion option in the Cross-Section Plot Parameters window.

Discussion of Results

The COMSOL results are shown in Figs. E6.6.2–E6.6.6. The arrow plot of Fig. E6.6.2 clearly shows that the velocity profile is virtually unchanged over the whole width of the cross section. Near the top surface the flow is induced by the moving surface to be in the negative x direction. Toward the fixed bottom surface there is a flow reversal, so that at any value of x the net flow rate dy is zero.

Fig. E6.6.3, which contains 20 streamlines, now indicates that the essential flow reversal occurs very close to each end of the cross section. To make the flow pattern more clear, the streamlines are repeated in Fig. E6.6.4, in which a larger scale is employed in the y direction—that is, the x/y geometry is no longer to scale.

The pressure distribution is shown in Fig. E6.6.5 as a series of isobars. Note that the isobars are spaced very evenly, and that the pressure increase from right to left is from p = 0 to p = 1.1 × 10⁶ N/m². Recall that the simplified theory gave the pressure gradient as:

Thus, the theoretical increase in pressure from right to left should be:

in good agreement with the more accurate two-dimensional solution of the Navier-Stokes equations shown in Fig. E6.6.5.

Finally, Fig. E6.6.6 shows how υ_x varies from the midpoint A of the top surface to the midpoint B of the bottom surface. As explained in Eqn. (E6.5.7), there are two contributions to this velocity profile:

1. A Couette flow from right to left, driven by the leftward-moving top surface.

2. A Poiseuille flow from left to right, driven by the excess pressure at the left.

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Example 6.7—Flow Through an Annular Die

Following the discussion of polymer processing in the previous section, now consider flow through a die that could be located at the exit of the screw extruder of Example 6.4. Consider a die that forms a tube of polymer (other shapes being sheets and filaments). In the die of length D shown in Fig. E6.7, a pressure difference p₂ – p₃ causes a liquid of viscosity μ to flow steadily from left to right in the annular area between two fixed concentric cylinders. Note that p₂ is chosen for the inlet pressure in order to correspond to the extruder exit pressure from Example 6.5. The inner cylinder is solid, whereas the outer one is hollow; their radii are r₁ and r₂, respectively. The problem, which could occur in the extrusion of plastic tubes, is to find the velocity profile in the annular space and the total volumetric flow rate Q. Note that cylindrical coordinates are now involved.

Assumptions and continuity equation. The following assumptions are realistic:

1. There is only one nonzero velocity component, namely that in the direction of flow, υ_z. Thus, υ_r = υ_θ = 0.

2. Gravity acts vertically downward, so that g_z = 0.

3. The axial velocity is independent of the angular location; that is, ∂υ_z/∂θ = 0.

To analyze the situation, again start from the continuity equation, (5.49):

which, for constant density and υ_r = υ_θ = 0, reduces to:

verifying that υ_z is independent of distance from the inlet, and that the velocity profile υ_z = υ_z(r) appears the same for all values of z.

Momentum balances. There are again three momentum balances, one for each of the r, θ, and z directions. If explored, the first two of these would ultimately lead to the pressure variation with r and θ at any cross section, which is of little interest in this problem. Therefore, we extract from Eqn. (5.75) only the z momentum balance:

With υ_r = υ_θ = 0 (from assumption 1), ∂υ_z/∂z = 0 [from Eqn. (E6.7.1)], ∂υ_z/∂θ = 0 (assumption 3), and g_z = 0 (assumption 2), this momentum balance simplifies to:

in which total derivatives are used because υ_z depends only on r.

Shortly, we shall prove that the pressure gradient is uniform between the die inlet and exit, being given by:

in which both sides of the equation are positive quantities. Two successive integrations of Eqn. (E6.7.3) may then be performed, yielding:

The two constants may be evaluated by applying the boundary conditions of zero velocity at the inner and outer walls,

giving:

Substitution of these values for the constants of integration into Eqn. (E6.7.5) yields the final expression for the velocity profile:

which is sketched in Fig. E6.7. Note that the maximum velocity occurs somewhat before the halfway point in progressing from the inner cylinder to the outer cylinder.

Volumetric flow rate. The final quantity of interest is the volumetric flow rate Q. Observing first that the flow rate through an annulus of internal radius r and external radius r + dr is dQ = υ_z2πr dr, integration yields:

Since r ln r is involved in the expression for υ_z, the following indefinite integral is needed:

giving the final result:

Since Q, μ, r₁, and r₂ are constant throughout the die, ∂p/∂z is also constant, thus verifying the hypothesis previously made. Observe that in the limiting case of r₁ → 0, Eqn. (E6.7.11) simplifies to the Hagen-Poiseuille law, already stated in Eqn. (3.12).

This problem may also be solved by performing a momentum balance on a shell that consists of an annulus of internal radius r, external radius r + dr, and length dz.

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Example 6.8—Spinning a Polymeric Fiber

A Newtonian polymeric liquid of viscosity μ is being “spun” (drawn into a fiber or filament of small diameter before solidifying by pulling it through a chemical setting bath) in the apparatus shown in Fig. E6.8.

The liquid volumetric flow rate is Q, and the filament diameters at z = 0 and z = L are D₀ and D_L, respectively. To a first approximation, neglect the effects of gravity, inertia, and surface tension (this last effect is examined in detail in Problem 6.26). Derive an expression for the tensile force F needed to pull the filament downward. Assume that the axial velocity profile is “flat” at any vertical location, so that υ_z depends only on z, which is here most conveniently taken as positive in the downward direction. Also derive an expression for the downward velocity υ_z as a function of z. The inset of Fig. E6.8 shows further details of the notation concerning the filament.

Solution

It is first necessary to determine the radial velocity and hence the pressure inside the filament. From continuity:

Since υ_z depends only on z, its axial derivative is a function of z only, or dυ_z/dz = f(z), so that Eqn. (E6.8.1) may be rearranged and integrated at constant z to give:

But to avoid an infinite value of υ_r at the centerline, g(z) must be zero, giving:

To proceed with reasonable expediency, it is necessary to make some simplification. After accounting for a primary effect (the difference between the pressure in the filament and the surrounding atmosphere), we assume that a secondary effect (variation of pressure across the filament) is negligible; that is, the pressure does not depend on the radial location (see Problem 6.27 for a more accurate investigation). Note that the external (gauge) pressure is zero everywhere, and apply the first part of Eqn. (5.64) at the free surface:

The axial stress is therefore:

It is interesting to note that the same result can be obtained with an alternative assumption.³ The axial tension in the fiber equals the product of the cross-sectional area and the local axial stress:

³ See S. Middleman’s Fundamentals of Polymer Processing, McGraw-Hill, New York, 1977, p. 235. There, the author assumes σ_rr = σ_θθ = 0, followed by the identity: p = –(σ_zz + σ_rr + σ_θθ)/3.

Since the effect of gravity is stated to be insignificant, F is a constant, regardless of the vertical location.

At any location, the volumetric flow rate equals the product of the cross-sectional area and the axial velocity:

A differential equation for the velocity is next obtained by dividing one of the last two equations by the other, and rearranging:

Integration, noting that the inlet velocity at z = 0 is , gives:

so that the axial velocity increases exponentially with distance:

The tension is obtained by applying Eqns. (E6.8.7) and (E6.8.10) just before the filament is taken up by the rollers:

Rearrangement yields:

which predicts a force that increases with higher viscosities, flow rates, and draw-down ratios (υ_zL/υ_z0), and that decreases with longer filaments.

Elimination of F from Eqns. (E6.8.10) and (E6.8.12) gives an expression for the velocity that depends only on the variables specified originally:

a result that is independent of the viscosity.

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Example 6.9—Analysis of a Cone-and-Plate Rheometer

The problem concerns the analysis of a cone-and-plate rheometer, an instrument developed and perfected in the 1950s and 1960s by Prof. Karl Weissenberg, for measuring the viscosity of liquids, and also known as the “Weissenberg rheogoniometer.”⁴ A cross section of the essential features is shown in Fig. E6.9, in which the liquid sample is held by surface tension in the narrow opening between a rotating lower circular plate, of radius R, and an upper cone, making an angle of β with the vertical axis. The plate is rotated steadily in the φ direction with an angular velocity ω, causing the liquid in the gap to move in concentric circles with a velocity υ_φ. (In practice, the tip of the cone is slightly truncated, to avoid friction with the plate.) Observe that the flow is of the Couette type.

⁴ Professor Weissenberg (1893–1976) once related to the author that he (Prof. Weissenberg) was attending an instrument trade show in London. There, the rheogoniometer was being demonstrated by a young salesman who was unaware of Prof. Weissenberg’s identity. Upon inquiring how the instrument worked, the salesman replied: “I’m sorry, sir, but it’s quite complicated, and I don’t think you will be able to understand it.”

The top of the upper shaft—which acts like a torsion bar—is clamped rigidly. However, viscous friction will twist the cone and the lower portions of the upper shaft very slightly; the amount of motion can be detected by a light arm at the extremity of which is a transducer, consisting of a small piece of steel, attached to the arm, and surrounded by a coil of wire; by monitoring the inductance of the coil, the small angle of twist can be obtained; a knowledge of the elastic properties of the shaft then enables the restraining torque T to be obtained. From the analysis given below, it is then possible to deduce the viscosity of the sample. The instrument is so sensitive that if no liquid is present, it is capable of determining the viscosity of the air in the gap!

The problem is best solved using spherical coordinates, because the surfaces of the cone and plate are then described by constant values of the angle θ, namely β and π/2, respectively.

Assumptions and the continuity equation. The following realistic assumptions are made:

1. There is only one nonzero velocity component, namely that in the φ direction, υ_φ. Thus, υ_r = υ_θ = 0.

2. Gravity acts vertically downward, so that g_φ = 0.

3. We do not need to know how the pressure varies in the liquid. Therefore, the r and θ momentum balances, which would supply this information, are not required.

The analysis starts once more from the continuity equation, (5.50):

which, for constant density and υ_r = υ_θ = 0 reduces to:

verifying that υ_φ is independent of the angular location φ, so we are correct in examining just one representative cross section, as shown in Fig. E6.9.

Momentum balances. There are again three momentum balances, one for each of the r, θ, and φ directions. From the third assumption above, the first two such balances are of no significant interest, leaving, from Eqn. (5.77), just that in the φ direction:

With υ_r = υ_θ = 0 (assumption 1), ∂υ_φ/∂φ = 0 [from Eqn. (E6.9.1)], and g_φ = 0 (assumption 2), the momentum balance simplifies to:

Determination of the velocity profile. First, seek an expression for the velocity in the φ direction, which is expected to be proportional to both the distance r from the origin and the angular velocity ω of the lower plate. However, its variation with the coordinate θ is something that has to be discovered. Therefore, postulate a solution of the form:

in which the function f(θ) is to be determined. Substitution of υ_φ from Eqn. (E6.9.4) into Eqn. (E6.9.3), and using f ^′and f ^″ to denote the first and second total derivatives of f with respect to θ, gives:

The reader is encouraged, as always, to check the missing algebraic and trigonometric steps, although they are rather tricky here!⁵

⁵ Although our approach is significantly different from that given on page 98 et seq. of R.B. Bird, W.E. Stewart, and E.N. Lightfoot, Transport Phenomena, Wiley & Sons, New York, 1960, we are indebted to these authors for the helpful hint they gave in establishing the equivalency expressed in our Eqn. (E6.9.5).

By multiplying Eqn. (E6.9.5) through by sin²θ, it follows after integration that the quantity in brackets is a constant, here represented as –2c₁, the reason for the “–2” being that it will cancel with a similar factor later on:

Separation of variables and indefinite integration (without specified limits) yields:

To proceed further, we need the following two standard indefinite integrals and one trigonometric identity⁶:

⁶ From pages 87 (integrals) and 72 (trigonometric identity) of J.H. Perry, ed., Chemical Engineers’ Handbook, 3rd ed., McGraw-Hill, New York, 1950.

Armed with these, Eqn. (E6.9.7) leads to the following expression for f:

in which c₂ is a second constant of integration.

Implementation of the boundary conditions. The constants c₁ and c₂ are found by imposing the two boundary conditions:

1. At the lower plate, where θ = π/2 and the expression in parentheses in Eqn. (E6.9.11) is zero, so that f = c₂, the velocity is simply the radius times the angular velocity:

2. At the surface of the cone, where θ = β, the velocity υ_φ = rωf is zero. Hence f = 0, and Eqn. (E6.9.11) leads to:

Substitution of these expressions for c₁ and c₂ into Eqn. (E6.9.11), and noting that υ_φ = rωf, gives the final (!) expression for the velocity:

As a partial check on the result, note that Eqn. (E6.9.14) reduces to υ_φ = rω when θ = π/2 and to υ_φ = 0 when θ = β.

Shear stress and torque. Recall that the primary goal of this investigation is to determine the torque T needed to hold the cone stationary. The relevant shear stress exerted by the liquid on the surface of the cone is τ_θφ—that exerted on the under surface of the cone (of constant first subscript, θ = β) in the positive φ direction (refer again to Fig. 5.12 for the sign convention and notation for stresses). Since this direction is the same as that of the rotation of the lower plate, we expect that τ_θφ will prove to be positive, thus indicating that the liquid is trying to turn the cone in the same direction in which the lower plate is rotated.

From the second of Eqn. (5.65), the relation for this shear stress is:

Since υ_θ = 0, and recalling Eqns. (E6.9.4), (E6.9.6), and (E6.9.13), the shear stress on the cone becomes:

One importance of this result is that it is independent of r, giving a constant stress and strain throughout the liquid, a significant simplification when deciphering the experimental results for non-Newtonian fluids (see Chapter 11). In effect, the increased velocity differences between the plate and cone at the greater values of r are offset in exact proportion by the larger distances separating them.

The torque exerted by the liquid on the cone (in the positive φ direction) is obtained as follows. The surface area of the cone between radii r and r + dr is 2πr sin βdr, and is located at a lever arm of r sin β from the axis of symmetry. Multiplication by the shear stress and integration gives:

Substitution of (τ_θφ)_θ=β from Eqn. (E6.9.16) and integration gives the torque as:

Problem 6.15 reaches essentially the same conclusion much more quickly for the case of a small angle between the cone and the plate. The torque for holding the cone stationary has the same value, but is, of course, in the negative φ direction.

Since R and g(β) can be determined from the radius and the angle β of the cone in conjunction with Eqn. (E6.9.13), the viscosity μ of the liquid can finally be determined.

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Fig. P6.1 shows a film of a viscous liquid held between two bars spaced a distance L apart. If the film thickness is uniform, and the total volume of liquid is V, show that the force necessary to separate the bars with a relative velocity dL/dt is:

2. Wire coating—M. Fig. P6.2 shows a rodlike wire of radius r₁ that is being pulled steadily with velocity V through a horizontal die of length L and internal radius r₂. The wire and the die are coaxial, and the space between them is filled with a liquid of viscosity μ. The pressure at both ends of the die is atmospheric. The wire is coated with the liquid as it leaves the die, and the thickness of the coating eventually settles down to a uniform value, δ.

Neglecting end effects, use the equations of motion in cylindrical coordinates to derive expressions for:

(a) The velocity profile within the annular space. Assume that there is only one nonzero velocity component, υ_z, and that this depends only on radial position.

(b) The total volumetric flow rate Q through the annulus.

(c) The limiting value for Q if r₁ approaches zero.

(d) The final thickness, δ, of the coating on the wire. (Here, only an equation is needed from which δ could be found.)

(e) The force F needed to pull the wire.

3. Off-center annular flow—D (C). A liquid flows under a pressure gradient ∂p/∂z through the narrow annular space of a die, a cross section of which is shown in Fig. P6.3(a). The coordinate z is in the axial direction, normal to the plane of the diagram. The die consists of a solid inner cylinder with center P and radius b inside a hollow outer cylinder with center O and radius a. The points O and P were intended to coincide, but due to an imperfection of assembly are separated by a small distance δ.

By a simple geometrical argument based on the triangle OPQ, show that the gap width Δ between the two cylinders is given approximately by:

Δ a – b – δ cos θ,

where the angle θ is defined in the diagram.

Now consider the radius arm b swung through an angle dθ, so that it traces an arc of length bdθ. The flow rate dQ through the shaded element in Fig. P6.3(b) is approximately that between parallel plates of width bdθ and separation Δ. Hence, prove that the flow rate through the die is given approximately by:

Q = πbc(2α³ +3 αδ²),

in which:

Assume from Eqn. (E6.1.26) that the flow rate per unit width between two flat plates separated by a distance h is:

What is the ratio of the flow rate if the two cylinders are touching at one point to the flow rate if they are concentric?

4. Compression molding—M. Fig. P6.4 shows the (a) beginning, (b) intermediate, and (c) final stages in the compression molding of a material that behaves as a liquid of high viscosity μ, from an initial cylinder of height H₀ and radius R₀ to a final disk of height H₁ and radius R₁.

In the molding operation, the upper disk A is squeezed with a uniform velocity V toward the stationary lower disk B.

(a) Derive expressions for H and R as functions of time t, H₀, and R₀.

(b) Consider the radially outward volumetric flow rate Q per unit perimeter, crossing a cylinder of radius r, as in Fig. P6.4(b). Obtain a relation for Q in terms of V and r.

(c) Assume by analogy from Eqn. (E6.1.26) that per unit width of a channel of depth H, the volumetric flow rate is:

(d) Ignoring small variations of pressure in the z direction, derive an expression for the radial variation of pressure.

(e) Prove that the total compressive force F that must be exerted downward on the upper disk is:

(f) Draw a sketch that shows how F varies with time.

5. Film draining—M. Fig. P6.5 shows an idealized view of a liquid film of viscosity μ that is draining under gravity down the side of a flat vertical wall. Such a situation would be approximated by the film left on the wall of a tank that was suddenly drained through a large hole in its base.

What are the justifications for assuming that the velocity profile at any distance x below the top of the wall is given by:

where h = h(x) is the local film thickness? Derive an expression for the corresponding downward mass flow rate m per unit wall width (normal to the plane of the diagram).

Perform a transient mass balance on a differential element of the film and prove that h varies with time and position according to:

Now substitute your expression for m, to obtain a partial differential equation for h. Try a solution of the form:

and determine the unknowns c, p, and q. Discuss the limitations of your solution.

Note that a similar situation occurs when a substrate is suddenly lifted from a bath of coating fluid.

6. Sheet “spinning”—M. A Newtonian polymeric liquid of viscosity μ is being “spun” (drawn into a sheet of small thickness before solidifying by pulling it through a chemical setting bath) in the apparatus shown in Fig. P6.6.

The liquid volumetric flow rate is Q, and the sheet thicknesses at z = 0 and z = L are Δ and δ, respectively. The effects of gravity, inertia, and surface tension are negligible. Derive an expression for the tensile force needed to pull the filament downward. Hint: Start by assuming that the vertically downward velocity υ_z depends only on z and that the lateral velocity υ_y is zero. Also derive an expression for the downward velocity υ_z as a function of z.

7. Details of pipe flow—M. A fluid of density ρ and viscosity μ flows from left to right through the horizontal pipe of radius a and length L shown in Fig. P6.7. The pressures at the centers of the inlet and exit are p₁ and p₂, respectively. You may assume that the only nonzero velocity component is υ_z, and that this is not a function of the angular coordinate, θ.

Stating any further necessary assumptions, derive expressions for the following, in terms of any or all of a, L, p₁, p₂, ρ, μ, and the coordinates r, z, and θ:

(a) The velocity profile, υ_z = υ_z(r).

(b) The total volumetric flow rate Q through the pipe.

(c) The pressure p at any point (r, θ, z).

(d) The shear stress, τ_rz.

8. Natural convection—M. Fig. P6.8 shows two infinite parallel vertical walls that are separated by a distance 2d. A fluid of viscosity μ and volume coefficient of expansion β fills the intervening space. The two walls are maintained at uniform temperatures T₁ (cold) and T₂ (hot), and you may assume (to be proved in a heat-transfer course) that there is a linear variation of temperature in the x direction. That is:

The density is not constant, but varies according to:

where is the density at the mean temperature , which occurs at x = 0.

If the resulting natural-convection flow is steady, use the equations of motion to derive an expression for the velocity profile υ_y = υ_y(x) between the plates. Your expression for υ_y should be in terms of any or all of x, d, T₁, T₂, , μ, β, and g.

Hints: In the y momentum balance, you should find yourself facing the following combination:

in which g_y = –g. These two terms are almost in balance, but not quite, leading to a small—but important—buoyancy effect that “drives” the natural convection. The variation of pressure in the y direction may be taken as the normal hydrostatic variation:

We then have:

and this will be found to be a vital contribution to the y momentum balance.

9. Square duct velocity profile—M. A certain flow in rectangular Cartesian coordinates has only one nonzero velocity component, υ_z, and this does not vary with z. If there is no body force, write down the Navier-Stokes equation for the z momentum balance.

One-dimensional, fully developed steady flow occurs under a pressure gradient ∂p/∂z in the z direction, parallel to the axis of a square duct of side 2a, whose cross section is shown in Fig. P6.9. The following equation has been proposed for the velocity profile:

Without attempting to integrate the momentum balance, investigate the possible merits of this proposed solution for υ_z. Explain whether or not it is correct.

10. Poisson’s equation for a square duct—E. A polymeric fluid of uniform viscosity μ is to be extruded after pumping it through a long duct whose cross section is a square of side 2a, shown in Fig. P6.10. The flow is parallel everywhere to the axis of the duct, which is in the z direction, normal to the plane of the diagram.

If ∂p/∂z, μ, and a are specified, show that the problem of obtaining the axial velocity distribution υ_z = υ_z(x, y), amounts to solving Poisson’s equation—of the form ∇²φ = f(x, y), where f is specified and φ is the unknown. Also note that Poisson’s equation can be solved numerically by the COMSOL computational fluid dynamics program introduced in Chapter 14 (see also Example 7.4).

11. Permissible velocity profiles—E. Consider the shear stress τ_yx; why must it be continuous—in the y direction, for example—and not undergo a sudden step-change in its value? Two immiscible Newtonian liquids A and B are in steady laminar flow between two parallel plates. Which—if any—of the velocity profiles shown in Fig. P6.11 are impossible? Profile A meets the interface normally in (b), but at an angle in (c). Any apparent location of the interface at the centerline is coincidental and should be ignored. Explain your answers carefully.

12. “Creeping” flow past a sphere—D. Figure P6.12 shows the steady, “creeping” (very slow) flow of a fluid of viscosity μ past a sphere of radius a. Far away from the sphere, the pressure is p_∞ and the undisturbed fluid velocity is U in the positive z direction. The following velocity components and pressure have been proposed in spherical coordinates:

Assuming the velocities are sufficiently small so that terms such as υ_r(∂υ_r/∂r) can be neglected, and that gravity is unimportant, prove that these equations do indeed satisfy the following conditions, and therefore are the solution to the problem:

(a) The continuity equation.

(b) The r and θ momentum balances.

(c) A pressure of p_∞ and a z velocity of U far away from the sphere.

(d) Zero velocity components on the surface of the sphere.

Also derive an expression for the net force exerted in the z direction by the fluid on the sphere, and compare it with that given by Stokes’ law in Eqn. (4.11).

Note that the problem is one in spherical coordinates, in which the z axis has no formal place, except to serve as a reference direction from which the angle θ is measured. There is also symmetry about this axis, such that any derivatives in the φ direction are zero. Note: The actual derivation of these velocities, starting from the equations of motion, is fairly difficult!

13. Torque in a Couette viscometer—M. Fig. P6.13 shows the horizontal cross section of a concentric cylinder or “Couette” viscometer, which is an apparatus for determining the viscosity μ of the fluid that is placed between the two vertical cylinders. The inner and outer cylinders have radii of r₁ and r₂, respectively. If the inner cylinder is rotated with a steady angular velocity ω, and the outer cylinder is stationary, derive an expression for υ_θ (the θ velocity component) as a function of radial location r.

If, further, the torque required to rotate the inner cylinder is found to be T per unit length of the cylinder, derive an expression whereby the unknown viscosity μ can be determined, in terms of T, ω, r₁, and r₂. Hint: You will need to consider one of the shear stresses given in Table 5.8.

14. Wetted-wall column—M. Fig. P6.14 shows a “wetted-wall” column, in which a thin film of a reacting liquid of viscosity μ flows steadily down a plane wall, possibly for a gas-absorption study. The volumetric flow rate of liquid is specified as Q per unit width of the wall (normal to the plane of the diagram).

Assume that there is only one nonzero velocity component, υ_x, and that this does not vary in the x direction, and that the gas exerts negligible shear stress on the liquid film. Starting from the equations of motion, derive an expression for the “profile” of the velocity υ_x (as a function of ρ, μ, g, y, and δ), and also for the film thickness, δ (as a function of ρ, μ, g, and Q).

15. Simplified view of a Weissenberg rheogoniometer—M. Consider the Weissenberg rheogoniometer with a very shallow cone; thus, referring to Fig. E6.9, β = π/2 – α, where α is a small angle.

(a) Without going through the complicated analysis presented in Example 6.9, outline your reasons for supposing that the shear stress at any location on the cone is:

(b) Hence, prove that the torque required to hold the cone stationary (or to rotate the lower plate) is:

(c) By substituting β = π/2 – α into Eqn. (E6.9.13) and expanding the various functions in power series (only a very few terms are needed), prove that g(β)= 1/(2α), and that Eqn. (E6.9.18) again leads to the expression just obtained for the torque in part (b) above.

16. Parallel-disk rheometer—M. Fig. P6.16 shows the diametral cross section of a viscometer, which consists of two opposed circular horizontal disks, each of radius R, spaced by a vertical distance H; the intervening gap is filled by a liquid of constant viscosity μ and constant density. The upper disk is stationary, and the lower disk is rotated at a steady angular velocity ω in the θ direction.

There is only one nonzero velocity component, υ_θ, so the liquid everywhere moves in circles. Simplify the general continuity equation in cylindrical coordinates, and hence deduce those coordinates (r?, θ?, z?) on which υ_θ may depend.

Now consider the θ-momentum equation, and simplify it by eliminating all zero terms. Explain briefly: (a) why you would expect ∂p/∂θ to be zero, and (b) why you cannot neglect the term ∂²υ_θ/∂z².

Also explain briefly the logic of supposing that the velocity in the θ direction is of the form υ_θ = rωf(z), where the function f(z) is yet to be determined. Now substitute this into the simplified θ-momentum balance and determine f(z), using the boundary conditions that υ_θ is zero on the upper disk and rω on the lower disk.

Why would you designate the shear stress exerted by the liquid on the lower disk as τ_zθ? Evaluate this stress as a function of radius.

17. Screw extruder optimum angle—M. Note that the flow rate through the die of Example 6.7, given in Eqn. (E6.7.11), can be expressed as:

in which c is a factor that accounts for the geometry.

Suppose that this die is now connected to the exit of the extruder studied in Example 6.5, and that p₁ = p₃ = 0, both pressures being atmospheric. Derive an expression for the optimum flight angle θ_opt that will maximize the flow rate Q_y through the extruder and die. Give your answer in terms of any or all of c, D, h, L₀, r, W, μ, and ω.

For what value of c would the pressure at the exit of the extruder approach its largest possible value p_2max? Derive an expression for p_2max.

18. Annular flow in a die—E. Referring to Example 6.7, concerning annular flow in a die, answer the following questions, giving your explanation in both cases:

(a) What form does the velocity profile, υ_z = υ_z(r), assume as the radius r₁ of the inner cylinder becomes vanishingly small?

(b) Does the maximum velocity occur halfway between the inner and outer cylinders, or at some other location?

19. Rotating rod in a fluid—M. Fig. P6.19(a) shows a horizontal cross section of a long vertical cylinder of radius a that is rotated steadily counterclockwise with an angular velocity ω in a very large volume of liquid of viscosity μ. The liquid extends effectively to infinity, where it may be considered at rest. The axis of the cylinder coincides with the z axis.

(a) What type of flow is involved? What coordinate system is appropriate?

(b) Write down the differential equation of mass and that one of the three general momentum balances that is most applicable to the determination of the velocity υ_θ.

(c) Clearly stating your assumptions, simplify the situation so that you obtain an ordinary differential equation with υ_θ as the dependent variable and r as the independent variable.

(d) Integrate this differential equation, and introduce any boundary condition(s), and prove that υ_θ = ωa²/r.

(e) Derive an expression for the shear stress τ_rθ at the surface of the cylinder. Carefully explain the plus or minus sign in this expression.

(f) Derive an expression that gives the torque T needed to rotate the cylinder, per unit length of the cylinder.

(g) Derive an expression for the vorticity component (∇ × v)_z. Comment on your result.

(h) Fig. P6.19(b) shows the initial condition of a mixing experiment in which the cylinder is in the middle of two immiscible liquids, A and B, of identical densities and viscosities. After the cylinder has made one complete rotation, draw a diagram that shows a representative location of the interface between A and B.

20. Two-phase immiscible flow—M. Fig. P6.20 shows an apparatus for measuring the pressure drop of two immiscible liquids as they flow horizontally between two parallel plates that extend indefinitely normal to the plane of the diagram. The liquids, A and B, have viscosities μ_A and μ_B, densities ρ_A and ρ_B, and volumetric flow rates Q_A and Q_B (per unit depth normal to the plane of the figure), respectively. Gravity may be considered unimportant, so that the pressure is essentially only a function of the horizontal distance, x.

(a) What type of flow is involved?

(b) Considering layer A, start from the differential equations of mass and momentum, and, clearly stating your assumptions, simplify the situation so that you obtain a differential equation that relates the horizontal velocity υ_xA to the vertical distance y.

(c) Integrate this differential equation so that you obtain υ_xA in terms of y and any or all of d, dp/dx, μ_A, ρ_A, and (assuming the pressure gradient is uniform) two arbitrary constants of integration, say, c_1A and c_2A. Assume that a similar relationship holds for υ_xB.

(d) Clearly state the four boundary and interfacial conditions, and hence derive expressions for the four constants, thus giving the velocity profiles in the two layers.

(e) Sketch the velocity profiles and the shear-stress distribution for τ_yx between the upper and lower plates.

(f) Until now, we have assumed that the interface level y = d is known. In reality, however, it will depend on the relative flow rates Q_A and Q_B. Show clearly how this dependency could be obtained, but do not actually carry the analysis through to completion.

21. Room humidifier—M. Fig. P6.21 shows a room humidifier, in which a circular impeller rotates about its axis with angular velocity ω. A conical extension dips into a water-bath, sucking up the liquid (of density ρ and viscosity μ), which then spreads out over the impeller as a thin laminar film that rotates everywhere with an angular velocity ω, eventually breaking into drops after leaving the periphery.

(a) Assuming incompressible steady flow, with symmetry about the vertical (z) axis, and a relatively small value of υ_z, what can you say from the continuity equation about the term:

(b) If the pressure in the film is everywhere atmospheric, the only significant inertial term is , and information from (a) can be used to neglect one particular term, to what two terms does the r momentum balance simplify?

(c) Hence prove that the velocity υ_r in the radial direction is a half-parabola in the z direction.

(d) Derive an expression for the total volumetric flow rate Q, and hence prove that the film thickness at any radial location is given by:

where ν is the kinematic viscosity.

22. Transport of inner cylinder—M (C). As shown in Fig. P6.22, a long solid cylinder of radius r_c and length L is being transported by a viscous liquid of the same density down a pipe of radius a, which is much smaller than L. The annular gap, of extent a – r_c, is much smaller than a. Assume: (a) the cylinder remains concentric within the pipe, (b) the flow in the annular gap is laminar, (c) the shear stress is essentially constant across the gap, and (d) entry and exit effects can be neglected. Prove that the velocity of the cylinder is given fairly accurately by:

where Q is the volumetric flow rate of the liquid upstream and downstream of the cylinder. Hint: Concentrate first on understanding the physical situation. Don’t rush headlong into a lengthy analysis with the Navier-Stokes equations!

23. Flow between inclined planes—M (C). A viscous liquid flows between two infinite planes inclined at an angle 2α to each other. Prove that the liquid velocity, which is everywhere parallel to the line of intersection of the planes, is given by:

where z, r, θ are cylindrical coordinates. The z axis is the line of intersection of the planes and the r axis at θ = 0 bisects the angle between the planes. Assume laminar flow, with:

and start by proposing a solution of the form υ_z = rⁿg(θ)(–∂p/∂z)/μ, where the exponent n and function g(θ) are to be determined.

24. Immiscible flow inside a tube—D (C). A film of liquid of viscosity μ₁ flows down the inside wall of a circular tube of radius (λ + Δ). The central core is occupied by a second immiscible liquid of viscosity μ₂, in which there is no net vertical flow. End effects may be neglected, and steady-state circulation in the core liquid has been reached. If the flow in both liquids is laminar, so that the velocity profiles are parabolic as shown in Fig. P6.24, prove that:

where Δ is the thickness of the liquid film, and α is the distance from the wall to the point of maximum velocity in the film. Fig. P6.24 suggests notation for solving the problem. To save time, assume parabolic velocity profiles without proof: u = a + bx + cx² and υ = d + ey + fy².

Discuss what happens when λ/Δ → ∞; and also when μ₂/μ₁ → 0; and when μ₁/μ₂ → 0.

25. Blowing a polyethylene bubble—D. For an incompressible fluid in cylindrical coordinates, write down:

(a) The continuity equation, and simplify it.

(b) Expressions for the viscous normal stresses σ_rr and σ_θθ, in terms of pressure, viscosity, and strain rates.

A polyethylene sheet is made by inflating a cylindrical bubble of molten polymer effectively at constant length, a half cross section of which is shown in Fig. P6.25. The excess pressure inside the bubble is small compared with the external pressure P, so that Δp ≪ P and σ_rr – P.

By means of a suitable force balance on the indicated control volume, prove that the circumferential stress is given by σ_θθ = – P + aΔp/t. Assume pseudo-steady state—that is, the circumferential stress just balances the excess pressure, neglecting any acceleration effects.

Hence, prove that the expansion velocity υ_r of the bubble (at r = a) is given by:

and evaluate it for a bubble of radius 1.0 m and film thickness 1 mm when subjected to an internal gauge pressure of Δp = 40 N/m². The viscosity of polyethylene at the appropriate temperature is 10⁵ N s/m².

26. Surface-tension effect in spinning—M. Example 6.8 ignored surface-tension effects, which would increase the pressure in a filament of radius R approximately by an amount σ/R, where σ is the surface tension. Compare this quantity with the reduction of pressure, μdυ_z/dz, caused by viscous effects, for a polymer with μ = 10⁴ P, σ = 0.030 kg/s², L = 1 m, R_L (exit radius) = 0.0002 m, υ_z0 = 0.02 m/s, and υ_zL = 2 m/s. Consider conditions both at the beginning and end of the filament. Comment briefly on your findings.

27. Radial pressure variations in spinning—M. Example 6.8 assumed that the variation of pressure across the filament was negligible. Investigate the validity of this assumption by starting with the suitably simplified momentum balance:

If υ_z is the local axial velocity, prove that the corresponding increase of pressure from just inside the free surface (p_R) to the centerline (p₀) is:

where β = υ_zL/υ_z0. Obtain an expression for the ratio ξ = (p_R – p₀)/(μdυ_z/dz), in which the denominator is the pressure decrease due to viscosity when crossing the interface from the air into the filament, and which was accounted for in Example 6.6.

Estimate ξ at the beginning of the filament for the situation in which μ = 10⁴ P, L = 1 m, R_L (exit radius) = 0.0002 m, υ_z0 = 0.02 m/s, and υ_zL = 2 m/s. Comment briefly on your findings.

28. Condenser with varying viscosity—M. In a condenser, a viscous liquid flows steadily under gravity as a uniform laminar film down a vertical cooled flat plate. Due to conduction, the liquid temperature T varies linearly across the film, from T₀ at the cooled plate to T₁ at the hotter liquid/vapor interface, according to T = T₀ + y(T₁ – T₀)/h. The viscosity of the liquid is given approximately by μ = μ*(1 – αT), where μ* and α are constants.

Prove that the viscosity at any location can be reexpressed as:

where μ₀ and μ₁ are the viscosities at temperatures T₀ and T₁, respectively.

What is the expression for the shear stress τ_yx for a liquid for steady flow in the x direction? Derive an expression for the velocity υ_x as a function of y. Make sure that you do not base your answer on any equations that assume constant viscosity.

Sketch the velocity profile for both a small and a large value of α.

29. Rotating sphere—M. A sphere of radius R, immersed in a liquid of viscosity μ of infinite extent, rotates about its z axis (see Fig. 5.9(b)) with a steady angular velocity ω in the φ direction. You may assume that the only nonzero liquid velocity component is that in the φ direction, and is of the form υ_φ = c sin θ/r².

(a) Verify that the proposed form for υ_φ satisfies the appropriate momentum balance, and determine c in terms of R and ω.

(b) Prove that the shear stress at any radial location in the liquid is τ_rφ = –3μc sin θ/r³, and carefully explain the minus sign.

(c) Obtain an expression for the torque T needed to rotate the sphere, as a function of R, ω, and μ. You should need the indefinite integral:

30. Moving plate—M. Similar to the first part of Example 6.4, consider a viscous fluid of infinite extent and kinematic viscosity ν with a long flat plate in its surface. At t = 0, the plate suddenly starts moving with a steady velocity υ_x0 in the x direction. Prove, by checking ∂υ_x/∂t = ν∂²υ_x/∂z² and the boundary conditions, that the fluid velocity at a depth z is subsequently given by:

You may assume that:

31. Oscillating plate—M. Similar to the second part of Example 6.4, consider a viscous fluid of infinite extent and kinematic viscosity ν with a long flat plate in its surface. The plate oscillates backwards and forwards in the x direction with velocity υ_x0 = a sin ωt. After any initial transients have died out, prove, by checking ∂υ_x/∂t = ν∂²υ_x/∂z² and the boundary conditions, that the fluid velocity at a depth z is subsequently given by:

in which

32. True/false. Check true or false, as appropriate:

(a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates.

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(b) For horizontal flow of a liquid in a rectangular duct between parallel plates, the boundary conditions can be taken as zero velocity at one of the plates and either zero velocity at the other plate or zero velocity gradient at the centerline.

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(c) For horizontal flow of a liquid in a rectangular duct between parallel plates, the shear stress varies from zero at the plates to a maximum at the centerline.

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(d) For horizontal flow of a liquid in a rectangular duct between parallel plates, a measurement of the pressure gradient enables the shear-stress distribution to be found.

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(e) In fluid mechanics, when integrating a partial differential equation, you get one or more constants of integration, whose values can be determined from the boundary condition(s).

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(f) For flows occurring between r = 0 and r = a in cylindrical coordinates, the term ln r may appear in the final expression for one of the velocity components.

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(g) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile.

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(h) Natural convection is a situation whose analysis depends on not taking the density as constant everywhere.

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(i) A key feature of the Weissenberg rheogoniometer is the fact that a conical upper surface results in a uniform velocity gradient between the cone and the plate, for all values of radial distance.

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(j) If, in three dimensions, the pressure obeys the equation ∂p/∂y = –ρg, and both ∂p/∂x and ∂p/∂z are nonzero, then integration of this equation gives the pressure as p = –ρgy + c, where c is a constant.

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(k) If two immiscible liquids A and B are flowing in the x direction between two parallel plates, both the velocity υ_x and the shear stress τ_yx are continuous at the interface between A and B, where the coordinate y is normal to the plates.

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(l) In compression molding of a disk between two plates, the force required to squeeze the plates together decreases as time increases.

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(m) For flow in a wetted-wall column, the pressure increases from atmospheric pressure at the gas/liquid interface to a maximum at the wall.

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(n) For one-dimensional flow in a pipe—either laminar or turbulent—the shear stress τ_rz varies linearly from zero at the wall to a maximum at the centerline.

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(o) In Example 6.1, for flow between two parallel plates, the shear stress τ_yx is negative in the upper half (where y > 0), meaning that physically it acts in the opposite direction to that indicated by the convention.

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