Example 6.1—Flow Between Parallel Plates

Fig. E6.1.1. Geometry for flow through a rectangular duct. The spacing between the plates is exaggerated in relation to their length.

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.

Fig. E6.1.2. Geometry for flow through a rectangular duct.

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, v_x. Thus, v_y = v_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 ∂v_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 v_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 v_y = v_z =0:

so v_x is independent of the distance from the inlet, and the velocity profile will appear the same for all values of x. Since ∂v_x/∂z = 0 (assumption 3), it follows that v_x = v_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 v_y = v_z = 0 (from assumption 2), ∂v_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), v_x does not depend on x, ∂²v_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 dv_x/dy =0 at y = 0, we obtain:

or:

A second integration, noting that v_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 v_x about the centerline (and by never using the fact that v_x = 0 at y = −d), should be reassured by an alternative approach, starting from Eqn. (E6.1.15):

Application of the two boundary conditions, v_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 = v_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, v_xmax, which occurs at the centerline, y =0.

Fig. E6.1.3. Pressure and shear-stress distributions.

Shear-stress distribution

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

By substituting for v_x from Eqn. (E6.1.15) and recognizing that v_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.

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.

Fig. E6.2. Momentum balance on a fluid element.

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 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 v_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 v_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.

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 v_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.)

Fig. E6.3.1. Liquid coating on a photographic film.

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 v_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 v_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 v_z zero, but all derivatives with respect to z, such as ∂v_x/∂z, are also zero.

3. There is only one nonzero velocity component, namely, that in the direction of motion of the photographic film, v_x. Thus, v_y = v_z =0.

4. Gravity acts vertically downwards.

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

But since v_y = v_z = 0, it follows that

so v_x is independent of distance x along the film. Further, v_x does not depend on z (assumption 2); thus, the velocity profile v_x = v_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 v_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:

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, v_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 v_x at increasing distances y from the film and eventually causes it to become negative.

Fig. E6.3.2. Velocity profile in thin liquid layer on moving photographic film for the case of zero net liquid flow rate.

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.

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.

Fig. E6.4.1. Region of liquid being studied, with the four COMSOL boundary conditions.

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 v_x = v_x0. In our problem, v_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 v_x0 = a sin ωt. In our problem the amplitude a = 1 cm/s, the angular velocity ω =2π s⁻1, 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 v_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. All mouse-clicks are left-clicks (the same as Select) unless specifically denoted as right-click (R).

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

Solution—Case 1 (Constant Upper-Plate Velocity)

Select the Physics

1. Open COMSOL and L-click Model Wizard, 2D, Fluid Flow (the little rotating triangle is called a “glyph”), Single-Phase Flow, Laminar Flow, Add. Note the symbols used by COMSOL—u, v, w for velocity components and p for pressure.

2. L-click Study, Time Dependent, Done.

Define the Parameters

3. R-click Global Definitions and select Parameters. In the Settings window to the right of the Model Builder window enter the parameter vx0 in the Name column and its value and units, 1 [cm/s], in the Expression column. Similarly, define L = 2 [cm], H = 1 [cm], rho1 = 1 [g/cm³], and mu1 = 0.5 [g/cm/s].

Establish the Geometry

4. R-click Geometry and select Rectangle. Note the default is based on the lower left corner being at (0, 0). Enter the values for Width and Height as L and H. L-click Build Selected.

Define the Fluid Properties

5. R-click the Materials node within Component 1 and select Blank Material. Note that the new material will have the required material parameters based on the physics of the problem. Enter the parameters rho1 and mu1 as the values for Density and Dynamic viscosity.

Define the Boundary Conditions

6. L-click Laminar Flow, Wall 1 and note the default No Slip boundary conditions that are highlighted.

7. R-click the Laminar Flow node and select Inlet to define an inlet boundary condition, and L-click the left boundary (#1) to select it as the inlet. Note that clicking on a boundary will change its color from red to blue. Within the Settings window L-click the dropdown box in the Boundary Condition node to select Pressure (equal to zero), deselect Suppress backflow, and leave Normal flow checked. R-click the Laminar Flow node and select Outlet, and L-click the right boundary (#4) to select it as the outlet, with zero pressure, Normal flow checked, and Suppress backflow unchecked. R-click the Laminar Flow node and L-click Wall. Select the top boundary (again changing from red to blue), and within the Settings window change the Boundary Condition from No Slip to Sliding wall. Enter vx0 as the value for the tangential velocity Uw.

Create the Mesh and Solve the Problem

8. L-click the Mesh node and change the Element Size in the Settings window to Fine. L-click Build All to construct the mesh. R-click Mesh 1 and Statistics will show that the mesh contains approximately 4,000 elements.

9. L-click the Study 1 little triangular glyph, and then select the Step 1 Time Dependent node and define the output range segments using the range function. For the requested output times, enter the three ranges as: range(0.0, 0.01, 0.1) range(0.2, 0.1, 0.5) range(1.0, 0.5, 2.0).

10. R-click the Study node and select Compute, which gives (after about 15 seconds) a surface plot of the magnitudes of the velocities—from low near the bottom (blue), medium at the middle (turquoise), and high near the top (red).

11. Save your results occasionally—for example, in the file Ex6.4-11.mph.

Display the Results (Case 1)

12. Create a data set at the midpoint in the y-direction. R-click the Data Sets node within the Results tree and select Cut Line 2D. In the Settings window enter the values (L/2, 0) and (L/2, H) for Points 1 and 2. L-click Plot near the top of the Settings window to display the cut line.

13. R-click Results and L-click 1D Plot Group. In the Settings window, Left-click the dropdown dialog and change the data set to Cut Line 2D 1, which was created above. Change the Time selection to “From list” and select the desired individual plot times by holding down the CTRL key (Command on a Mac) and then L-clicking on the individual values 0.01, 0.02, 0.05, 0.1, 0.2, 0.5, and 2.0.

Fig. E6.4.2. Plot of v_x (m/s) against distance y (m) from the midpoint of the lower plate to the midpoint of the upper plate, for the indicated values of time (s).

14. R-click the 1D Plot Group 3 and select Line Graph. Within the Settings window enter the variable for the x-component of velocity, u, in the Expression field.

15. Expand the Coloring and Style section by L-clicking the triangular glyph to the left. Change the Color option to Black by L-clicking the dropdown dialog box.

16. L-click Plot in the Settings window to update the plot. Save in Ex6.4-16.mph or similar.

Discussion of Case 1 (Constant Upper-Plate Velocity)

Fig. E6.4.2 shows how the velocity v_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 v_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 downwards 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, dv_x/dy is constant; thus, these shear stresses are equal (and opposite) because they are both given by τ_yx = μdv_x/dy.

Solution—Case 2 (Oscillating Upper-Plate Velocity)

17. Now solve the oscillatory surface time-dependent problem by first saving the model under a new name, such as Ex6.4-17.mph (or a file name of your choice).

18. Follow Step 3 above (Parameters under Global Definitions) to add the parameters for a and omega with values 1.0 [cm/s] and 2*pi [1/s]. Also update the viscosity by changing the value for mu1 to 0.1 [g/cm/s].

19. Update the top wall boundary condition, Wall 2. L-click Laminar Flow, Wall 2 and enter the tangential velocity value a*sin(omega*t).

20. Update the requested output times. L-click the Study 1, Step 1 Time Dependent node and enter the ranges as: range(0, 2, 18) range(19, 0.25, 20).

21. R-click the Study 1 node and select Compute. The solution took about 100 seconds on the author’s computer. The result is a surface plot showing velocity magnitudes, not needed here.

Fig. E6.4.3. Plot of v_x (m/s) against distance y (m), for one complete cycle of the upper plate. Viscosity μ =0.1 P.

Fig. E6.4.4. Plot of v_x (m/s) against distance y (m), for one complete cycle of the upper plate. Viscosity μ =0.5 P.

22. Update the requested times by selecting 1D Plot Group 3. In the Settings window, change the Time selections to t = 19, 19.25, 19.5, 19.75, and 20 s by L-clicking the first value 19 and concluding the selection with Shift + L-click on the value 20, giving five values altogether. L-click Plot.

23. Save the model, in Ex6.4-23.mph for example.

24. Use Step 3 to update the value of mu1 to 0.5 [g/cm/s].

25. R-click the Study node and select Compute.

26. L-click on the 1D Plot Group 3, observe the changes to the solution, and make a final file save.

Discussion of Case 2 (Oscillating Upper-Plate Velocity)

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 times 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 x-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 v_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).

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.

Fig. E6.5.1. Screw extruder.

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 .

Fig. E6.5.2. Diagonal motion of barrel relative to flights.

Solution

Motion in two principal directions is considered:

1. Flow parallel to the flight axis, caused by a barrel velocity of V_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 V_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).

Fig. E6.5.3. Fluid motion (a) along, and (b) normal to the flight axis, as seen by an observer on the screw.

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:

Note from Eqn. (E6.5.10) that v_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 v_z(z).

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

Fig. E6.6.1. Cross-section of the screw extruder normal to the flight axis. The types of boundary conditions are indicated. The pressure is specified to be zero at the midpoint B of the bottom wall. Eventually, a velocity cross-plot will be made between midpoints A and B of the top and bottom surfaces. The transverse coordinate has been changed from z to y to conform to the COMSOL convention.

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 the midpoint B.

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 V_x = −0.340 × sin θ = −0.340 × sin 17.66^° = −0.103 . = −0.1 m/s.

6. After solving for the flow pattern, make the following plots:

(a) A picture that shows 10 streamlines, one with the true aspect ratio and another with the y-axis “stretched” by a factor of five.

(b) An arrow plot of the x-velocities.

(c) A contour plot of the pressure distribution.

(d) A cross-section plot of the x-velocity between the midpoints A (0.05, 0.005) of the upper moving surface and B (0.05, 0) of the lower stationary surface.

Solution

See Chapter 14 for more details about COMSOL Multiphysics. All mouse-clicks are left-clicks (the same as Select) unless specifically denoted as right-click (R).

Select the Physics

1. Select Model Wizard, 2D, Fluid Flow (the little rotating triangle is called a “glyph”), Single-Phase Flow, Laminar Flow. Add. Note COMSOL’s symbols for pressure and velocity. Study, Stationary, Done.

Define the Parameters

2. R-Global Definitions, Parameters. Enter rho1 = 800 [kg/m³], mu1 = 500 [Pa*s], Vx = −0.1 [m/s] under the Name/Expression columns. (The names rho and mu are reserved by COMSOL for its own use, and so are unavailable to us.)

Establish the Geometry

3. R-Geometry, Rectangle. Note Corner is at x = 0, y = 0. Width = 0.1, Height = 0.005. Build Selected. R-Geometry, Point (this is where we shall specify p = 0). Enter 0.05 and 0.00 for x and y. Build Selected.

Define the Fluid Properties

4. Under Component1, R-Materials, Blank Material. Enter rho1 and mu1 as values. To see the equations being solved, R-click Laminar Flow and select the Equations glyph in the Settings window.

Define the Boundary Conditions

5. R-Laminar Flow, Wall, Wall 2. Hover over the top wall (turns red), L-click to select. Boundary Condition, Sliding Wall. Enter Vx for Uw. Set pressure constraint with R-Laminar Flow, Points, Pressure Point Constraint. Hover over bottom center point and select with L-click (note Pressure is zero). Point 1 should appear in the Active Window.

Create the Mesh and Solve the Problem

6. Mesh 1, Element size, Coarse, Build All. R-Mesh 1, Statistics and note 14,695 elements. To see some of the elements, use the Zoom icon three items, then Zoom Extents to revert.

7. R-Study, Compute (takes about 4 seconds). Gives velocity magnitude, not of interest here.

Display the Streamlines

8. R-Results, 2D Plot Group. R-2D Plot Group 3, Streamline. Settings window, Positioning, Start point controlled, 20 points, yields Fig. E6.6.2.

Fig. E6.6.2. Streamlines, in which the aspect ratio is the same as in the problem statement. The two arrows have been added to emphasize the counter-clockwise direction of flow.

Fig. E6.6.3. Streamlines, in which the height, from y = 0 to 0.005 m has been “stretched” to show more detail. The two arrows have been added to emphasize the counter-clockwise direction of flow.

9. Stretch the y-axis to improve visibility by R-Definitions (under Component 1), View. L-click the glyph at the left of View 2 to expand its tree, Axis. Settings, View Scale, Manual, y-scale factor = 5, Update, Zoom Extents. Streamline 1 then gives Fig. E6.6.3.

Display the Arrow Velocity Plot

10. View 1. R-click Results, 2D Plot Group. R-click 2D Plot Group 4, Arrow Surface 1. Plot. Scale Factor, move slider to right to give Scale Factor 0.034, Color black.

11. In the Graphics window, zoom in and out by either: (a) simultaneously holding the middle mouse button down and moving the mouse forwards and backwards, or (b) clicking on Zoom above the Graphics window and holding the R mouse button down to move the graphics around, giving Fig. E6.6.4.

Fig. E6.6.4. Arrow plot, in which the arrow length is proportional to the local velocity. To improve legibility, only the left portion has been reproduced and enlarged.

Display the Isobars

12. R-click Results and L-click 2D Plot Group.

13. L-click the dropbox (under Definitions for Component 1) for View and select View 1.

14. R-click the new 2D Plot Group 5 and L-click Contour.

Fig. E6.6.5. Isobars, with the pressure p = 0 at the midpoint, varying from p . = −5.9 × 10⁵ Pa at the midpoint of the right-hand end to p = 5.9 × 10⁵ Pa at the left-hand end. Note that except for the two upper corners, where there is a conflict in velocities between the moving and no-slip boundaries, the pressure depends only on the horizontal coordinate x.

15. In Settings, Expression field, enter pressure, p. Levels field, change the Entry Method from Number of Levels to Levels. In the Levels field, enter the pressure range as range(−0.6e6, 0.1e6, 0.6e6), Plot; “range” is a function whose first and last parameters denote the start and end; the middle parameter is the increment.

16. By comparing the isobars with the color bar, or by selecting Level Labels, note that the pressure varies from approximately −5.9 × 10⁵ Pa at the midpoint of the right end to approximately 5.9 × 10⁵ Pa at the midpoint of the left end, and zero in the middle.

17. Change Coloring to Uniform and the Color to black to generate Fig. E6.6.5.

Construct the Velocity Cross-Plot

18. Section the data set at the midpoint in the y-direction: R-click Data Sets within the Results tree and select Cut Line 2D. In the Settings window enter (0.05, 0) and (0.05, 0.005) for Points 1 and 2. L-click Plot to display the cut line.

19. R-click Results and L-click 1D Plot Group 6. In the Settings window, L-click the dropdown dialog and change the data set to Cut Line 2D 1.

20. R-click the 1D Plot Group 6 and select Line Graph 1.

21. Within Settings, Expression field, enter u, the x-velocity component. Expand the Coloring and Style section by L-clicking the triangular glyph at the left.

Fig. E6.6.6. A cross-plot that shows how the horizontal velocity v_x varies from the midpoint A of the top surface, where it is strongly negative, to the midpoint B of the bottom surface, where it is zero.

22. Change the Color option to Black by L-clicking the dropdown dialog box.

23. L-click Plot at the top of the Settings window to update the plot. There is a Recovery option at the bottom of the Settings window under the Quality section. Select Recover Within Domains, which does polynomial smoothing, giving Fig. E6.6.6.

Discussion of Results

The COMSOL results are shown in Figs. E6.6.2–E6.6.6. The arrow plot of Fig. E6.6.4 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. Towards the fixed bottom surface there is a flow reversal, so that at any value of x the net flow rate is zero.

Fig. E6.6.2, 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. E6.6.3, 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 (at the end mid-points) is from −5.9 × 10⁵ Pa to 5.9 × 10⁵ Pa, or 1.18 × 10⁶ Pa total.

Recall that the simplified theory gave the pressure gradient as:

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

in complete 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 v_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.

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 6.6. 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.

Fig. E6.7. Geometry for flow through an annular die.

Assumptions and continuity equation

The following assumptions are realistic:

1. There is only one nonzero velocity component, namely that in the direction of flow, v_z. Thus, v_r = v_θ =0.

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

3. The axial velocity is independent of the angular location; that is, ∂v_z/∂θ =0. To analyze the situation, again start from the continuity equation, (5.49):

which, for constant density and v_r = v_θ = 0, reduces to:

verifying that v_z is independent of distance from the inlet, and that the velocity profile v_z = v_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 v_r = v_θ = 0 (from assumption 1), ∂v_z/∂z = 0 [from Eqn. (E6.7.1)], ∂v_z/∂θ = 0 (assumption 3), and g_z = 0 (assumption 2), this momentum balance simplifies to:

in which total derivatives are used because v_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 = v_z2πr dr, integration yields:

Since r ln r is involved in the expression for v_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.

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 v_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 v_z as a function of z. The inset of Fig. E6.8 shows further details of the notation concerning the filament.

Fig. E6.8. “Spinning” a polymer filament, whose diameter in relation to its length is exaggerated in the diagram.

Solution

It is first necessary to determine the radial velocity and hence the pressure inside the filament. From continuity: 1 r

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

(E6.8.2) But to avoid an infinite value of v_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: 1 v_z

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 (v_zL/v_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:

(E6.8.13)

a result that is independent of the viscosity.

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 v_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.”

Fig. E6.9. Geometry for a Weissenberg rheogoniometer. (The angle between the cone and plate is exaggerated.)

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, v_ϕ. Thus, v_r = v_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 v_r = v_θ = 0 reduces to:

verifying that v_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 v_r = v_θ = 0 (assumption 1), ∂v_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 v_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 sin² θ, 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,

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 v_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 v_ϕ = rωf, gives the final (!) expression for the velocity:

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