6.1 Steady, One-Dimensional Rectilinear Flows Steady, Spherically Symmetric Radial Flows 42

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1 Contents 6 UNIDIRECTIONAL FLOWS Steady, One-Dimensional Rectilinear Flows 6. Steady, Axisymmetric Rectilinear Flows Steady, Axisymmetric Torsional Flows Steady, Axisymmetric Radial Flows Steady, Spherically Symmetric Radial Flows Transient One-Dimensional Unidirectional Flows Steady Two-Dimensional Rectilinear Flows Problems References 81 i

2 ii CONTENTS

3 Chapter 6 UNIDIRECTIONAL FLOWS Isothermal, laminar, incompressible Newtonian ow is governed by a system of four scalar partial dierential equations (PDEs) these are the continuity equation and the three components of the Navier-Stokes equation. The pressure and the three velocity components are the primary unknowns, which are, in general, functions of time and of spatial coordinates. This system of PDEs is amenable to analytical solution for limited classes of ow. Even in the case of relatively simple ows in regular geometries, the nonlinearities introduced by the convective terms rule out the possibility of nding analytical solutions. This explains the extensive use of numerical methods in Fluid Mechanics [1]. Computational Fluid Dynamics (CFD) is certainly the fastest growing branch of uid mechanics, largely as a result of the increasing availability and power of computers, and the parallel advancement of versatile numerical techniques. In this chapter, we study certain classes of incompressible ows, in which the Navier-Stokes equations are simplied signicantly to lead to analytical solutions. These classes concern unidirectional ows, that is, ows which have only one nonzero velocity component, u i. Hence, the number of the primary unknowns is reduced to two: the velocity component, u i, and pressure, p. In manyows of interest, the PDEs corresponding to the two unknown elds are decoupled. As a result, one can rst nd u i,by solving the corresponding component of the Navier-Stokes equation, and then calculate the pressure. Another consequence of the unidirectionality assumption, is that u i is a function of at most two spatial variables and time. Therefore, in the worst case scenario of incompressible, unidirectional ow one has to solve a PDE with three independent variables, one of which is time. The number of independent variables is reduced to two in (a) transient one-dimensional (1D) unidirectional ows in which u i is a function of one spatial independent variable and time and (b) steady two-dimensional (D) unidirectional ows in which u i is a function of two spatial independent variables. 1

4 Chapter 6. Unidirectional Flows The resulting PDEs in the above two cases can often be solved using various techniques, such as the separation of variables [] and similarity methods [3]. In steady, one-dimensional unidirectional ows, the number of independent variables is reduced to one. In these ows, the governing equation for the nonzero velocity component is just a linear, second-order ordinary dierential equation (ODE) which can be solved easily using well-known formulas and techniques. Such ows are studied in the rst three sections of this chapter. In particular, in Sections 1 and, we study ows in which the streamlines are straight lines, i.e., one-dimensional rectilinear ows with u x =u x (y)andu y =u z =0 (Section 6.1), and axisymmetric rectilinear ows with u z =u z (r) andu r =u =0 (Section 6.). In Section 6.3, we study axisymmetric torsional (or swirling) ows, with u =u (r) andu z =u r =0. In this case, the streamlines are circles centered at the axis of symmetry. In Sections 6.4 and 6.5, we discuss briey steady radial ows, with axial and spherical symmetry, respectively. An interesting feature of radial ows is that the nonzero radial velocity component, u r =u r (r), is determined from the continuity equation rather than from the radial component of the Navier-Stokes equation. In Section 6.6, we study transient, one-dimensional unidirectional ows. Finally, in Section 6.7, we consider examples of steady, two-dimensional unidirectional ows. Unidirectional ows, although simple, are important in a diversity of uid transferring and processing applications. As demonstrated in examples in the following sections, once the velocity and the pressure are known, the nonzero components of the stress tensor, such as the shear stress, as well as other useful macroscopic quantities, such as the volumetric ow rate and the shear force (or drag) on solid boundaries in contact with the uid, can be easily determined. Let us point out that analytical solutions can also be found for a limited class of two-dimensional almost unidirectional or bidirectional ows by means of the potential function and/or the stream function, as demonstrated in Chapters 8 to 10. Approximate solutions for limiting values of the involved parameters can be constructed by asymptotic and perturbation analyses, which are the topics of Chapters 7 and 9, with the most profound examples being the lubrication, thin-lm, and boundarylayer approximations. 6.1 Steady, One-Dimensional Rectilinear Flows Rectilinear ows, i.e., ows in which the streamlines are straight lines, are usually described in Cartesian coordinates, with one of the axes being parallel to the ow direction. If the ow is axisymmetric, a cylindrical coordinate system with the z-axis

5 6.1 Steady, One-Dimensional Rectilinear Flows 3 coinciding with the axis of symmetry of the ow is usually used. Let us assume that a Cartesian coordinate system is chosen to describe a rectilinear ow, with the x-axis being parallel to the ow direction, as in Fig. 6.1, where the geometry of the ow in a channel of rectangular cross section is shown. Therefore, u x is the only nonzero velocity component and From the continuity equation for incompressible u y = u z =0: (6.1) =0 we nd =0 which indicates that u x does not change in the ow direction, i.e., u x is independent of x: u x = u x (y z t) : (6.) Flows satisfying Eqs. (6.1) and (6.) are called fully developed. Flows in tubes of constant cross section, such astheoneshown in Fig. 6.1, can be considered fully developed if the tube is suciently long so that entry and exit eects can be neglected. Due to Eqs. (6.1) and (6.), the + x + x + u z is reduced x = + = u u u x + x! u x + x : (6.3) If now the ow is steady, then the time derivative in the x-momentum equation is zero, and Eq. (6.3) becomes u u x + x =0: (6.4) The last equation which describes any steady, two-dimensional rectilinear ow in the x-direction is studied in Section 6.5. In many unidirectional ows, it can be assumed u

6 4 Chapter 6. Unidirectional Flows y FLOW W x H Figure 6.1. Geometry of ow in a channel of rectangular cross section. z and u x can be treated as a function of y alone, i.e., u x = u x (y) : (6.5) With the latter assumption, the x-momentum equation is reduced to: + d u x dy + g x =0: (6.6) The only nonzero component of the stress tensor is the shear stress yx, yx = du x dy (6.7) in terms of which the x-momentum equation takes the form + d yx dy + g x =0: (6.8) Equation (6.6) is a linear second-order ordinary dierential equation and can be integrated directly if =const: Its general solution is given by u x (y) = ; g x y + c 1 y + c : (6.10)

7 ;;;;; ;;;;; ;;; ;;; 6.1 Steady, One-Dimensional Rectilinear Flows 5 V - ;;;;; ;;; ; ;;;; ;;;; ;;; - -; ;- ; ;; - u x = VH y - ;;;;; ;;; ; ;;;; ;;;; ;;; ;; ;;; ;; 6 H? ;; ;;; ;; y 6 - x Figure 6.. Plane Couette ow. Therefore, the velocity prole is a parabola and involves two constants, c 1 and c, which are determined by applying appropriate boundary conditions for the particular ow. The shear stress, yx = xy, is linear, i.e., yx = du x dy ; g x y + c 1 : (6.11) Note that the y- and z-momentum components do not involve the velocity u x since u y =u z =0, they degenerate to the hydrostatic pressure expressions + g y =0 and + g z =0: (6.1) Integrating Eqs. (6.9) and (6.1), we obtain the following expression for the pressure: p x + g yy + g z z + c (6.13) where c is a constant of integration which may be evaluated in any particular ow problem by specifying the value of the pressure at a point. In Table 6.1, we tabulate the assumptions, the governing equations, and the general solution for steady, one-dimensional rectilinear ows in Cartesian coordinates. Important ows in this category are: 1. Plane Couette ow, i.e., fully-developed ow between parallel at plates of innite dimensions, driven by the steady motion of one of the plates. (Such a ow is called shear-driven ow.) The geometry of this ow is depicted in Fig. 6., where the upper wall is moving with constant speed V (so that it remains in the same plane) while the lower one is xed. The pressure gradient iszero everywhere and the gravity term is neglected. This ow is studied in Example

8 6 Chapter 6. Unidirectional Flows Assumptions: u y = u z =const. Continuity: x-momentum: y-momentum: z-momentum: General =0 =) u x = u x (y) + d u x dy + g x =0 + g y =0 + g z =0 u x = ; g x y + c 1 y + c yx = xy ; g x y + c 1 p x + g yy + g z z + c Table 6.1. Governing equations and general solution for steady, one-dimensional rectilinear ows in Cartesian coordinates.. Fully-developed plane Poiseuille ow, i.e., ow between parallel plates of innite width and length, driven by a constant pressure gradient, imposed by a pushing or pulling device (a pump or vacuum, respectively), and/or gravity. This ow is an idealization of the ow in a channel of rectangular cross section, with the width W being much greater than the height H of the channel (see Fig. 6.1). Obviously, this idealization does not hold near the two lateral walls, where the ow is two-dimensional. The geometry of the plane Poiseuille ow is depicted in Fig This ow is studied in Examples 6.1. to 6.1.5, for

9 6.1 Steady, One-Dimensional Rectilinear Flows 7 dierent boundary conditions. 3. Thin lm ow down an inclined plane, driven by gravity (i.e., elevation dierences), under the absence of surface tension. The pressure gradient is usually assumed to be everywhere zero. Such a ow is illustrated in Fig. 6.8, and is studied in Example All the above ows are rotational, with vorticity generation at the solid boundaries, i j k! = ruj w = ; k 6= 0 w u x 0 0 w The vorticity diuses away from the wall, and penetrates the main ow at a rate (d u x =dy ). The extensional stretching or compression along streamlines is zero, i.e., _ =0 Material lines connecting twomoving uid particles traveling along dierent streamlines both rotate and stretch, where stretching is induced by rotation. However, the principal directions of strain rotate with respect to those of vorticity. Therefore, strain is relaxed, and the ow is weak. Example Plane Couette ow Plane Couette ow, 1 named after Couette who introduced it in 1890 to measure viscosity, is fully-developed ow induced between two innite parallel plates, placed at a distance H apart, when one of them, say the upper one, is moving steadily with speed V relative to the other (Fig. 6.). Assuming that the pressure gradient and the gravity inthex-direction are zero, the general solution for u x is: u x = c 1 y + c : For the geometry depicted in Fig. 6., the boundary conditions are: u x =0 at y =0 (lower plate is stationary) u x = V at y = H (upper plate is moving): By means of the abovetwo conditions, we nd that c =0 and c 1 =V=H. Substituting the two constants into the general solution, yields u x = V H y: (6.14) 1 Plane Couette ow is also known as simple shear ow.

10 ;;; ;;; 8 Chapter 6. Unidirectional Flows The velocity u x then varies linearly across the gap. The corresponding shear stress is constant, yx = V H : (6.15) A number of macroscopic quantities, such as the volumetric ow rate and the shear stress at the wall, can be calculated. The volumetric ow rate per unit width is calculated by integrating u x along the gap: Q W Z H = u x dy = 0 Z H 0 V ydy =) H Q W = 1 HV : (6.16) The shear stress w exerted by the uid on the upper plate is w = ; yx j y=h = ; V H : (6.17) The minus sign accounts for the upper wall facing the negative y-direction of the chosen system of coordinates. The shear force per unit width required to move the upper plate is then Z F L W = ; w dx = V H L where L is the length of the plate. ;;;;; ;;;;; ;;;;; ;;;;; 0 V - ;;; ;- ;;;; ;;;; ;;; ;; ;;; ;; u x =V H - -? ;;; ; ;;;; ;;;; - ;;; ;; ;;; ;; V y 6 - x Figure 6.3. Plug ow. Finally, let us consider the case where both plates move with the same speed V, as in Fig By invoking the boundary conditions u x (0) = u x (H)=V

11 6.1 Steady, One-Dimensional Rectilinear Flows 9 we nd that c 1 =0 and c =V, and, therefore, u x = V: Thus, in this case, plane Couette ow degenerates into plug ow. Example Fully-developed plane Poiseuille ow Plane Poiseuille ow, named after the channel experiments by Poiseuille in 1840, occurs when a liquid is forced between two stationary innite at plates, under constant pressure gradient =@x and zero gravity. The general steady-state solution is u x (y) = y + c 1 y + c (6.18) and yx y + c 1 : (6.19) u x =u x (y) yx H y x Figure 6.4. Plane Poiseuille ow. By taking the origin of the Cartesian coordinates to be on the plane of symmetry of the ow, as in Fig. 6.4, and by assuming that the distance between the two plates is H, the boundary conditions are: yx = du x =0 at y =0 (symmetry) dy u x =0 at y = H (stationary plate) : Note that the condition u x =0 at y=;h may be used instead of any of the above conditions. By invoking the boundary conditions at y=0 and H, we nd that c 1 =0 and c = ; H :

12 10 Chapter 6. Unidirectional Flows The two constants are substituted into the general solution to obtain the following parabolic velocity prole, u x = ; (H ; y ) : (6.0) If the pressure gradient is negative, then the ow is in the positive direction, as in Fig Obviously, the velocity u x attains its maximum value at the centerline (y=0): u x max = ; 1 The volumetric ow rate per unit width H : Q W = Z H ;H u x dy = Z H 0 ; (H ; y ) dy =) Q = ; H 3 W: (6.1) As expected, Eq. (6.1) indicates that the volumetric ow rate Q is proportional to the pressure gradient, =@x, and inversely proportional to the viscosity. Note also that, since =@x is negative, Q is positive. The average velocity, u x,inthe channel is: u x = Q WH = ; 3 The shear stress distribution is given by H : = y i.e., yx varies linearly from y=0 to H, being zero at the centerline and attaining its maximum absolute value at the wall. The shear stress exerted by the uid on the wall at y=h is w = ; yx j y=h = H: Example Plane Poiseuille ow with slip Consider again the fully-developed plane Poiseuille ow of the previous example, and assume that slip occurs along the two plates according to the slip law w = u w at y = H

13 6.1 Steady, One-Dimensional Rectilinear Flows 11 where is a material slip parameter, w is the shear stress exerted by the uid on the plate, w = ; yx j y=h and u w is the slip velocity. Calculate the velocity distribution and the volume ow rate per unit width. w =; yx =u w u w u x =u x (y) H y x Figure 6.5. Plane Poiseuille ow with slip. Solution: We rst note that the ow is still symmetric with respect to the centerline. In this case, the boundary conditions are: yx = du x =0 at y =0 dy w = u w at y = H: The condition at y=0 yields c 1 =0. Consequently, and yx u x = 1 Applying the condition at y=h, we obtain u w = 1 w =) u x (H) =; y + c y =) w = H =) H + c = ; H:

14 1 Chapter 6. Unidirectional Flows Consequently, and c = ; 1 u x = H + H H + H ; y : (6.3) Note that this expression reduces to the standard Poiseuille ow prole when!1. Since the slip velocity isinversely proportional to the slip coecient, the standard no-slip condition is recovered. An alternative expression of the velocity distribution is u x = u w ; H ; y which indicates that u x is just the superposition of the slip velocity u w to the velocity distribution of the previous example. For the volumetric ow rate per unit width, we obtain: Q W Z H = u x dy =u w H ; 0 H 3 =) Q = ; H W: (6.4) H Example Plane Couette-Poiseuille ow Consider again fully-developed plane Poiseuille ow with the upper plate moving with constant speed, V (Fig. 6.6). This ow is called plane Couette-Poiseuille ow or general Couette ow. In contrast to the previous two examples, this ow is not symmetric with respect to the centerline of the channel, and, therefore, having the origin of the Cartesian coordinates on the centerline is not convenient. Therefore, the origin is moved to the lower plate. The boundary conditions for this ow are: u x =0 at y =0 u x = V at y = a where a is the distance between the two plates. Applying the two conditions, we get c =0 and V = a + c 1 a =) c 1 = V a ; a

15 6.1 Steady, One-Dimensional Rectilinear Flows 13 V V u x =u x (y) a y x Figure 6.6. Plane Poiseuille ow with the upper plate moving with constant speed. respectively. Therefore, The shear stress distribution is given by u x = V a y ; (ay ; y ) : (6.5) yx = V a ; 1 (a ; y) : It is a simple exercise to show that Eq. (6.5) reduces to the standard Poiseuille velocity prole for stationary plates, given by Eq. (6.0). (Keep in mind that a=h and that the y-axis has been translated by a distance H.) If instead, the pressure gradient is zero, the ow degenerates to the plane Couette ow studied in Example 1.6.1, and the velocity distribution is linear. Hence, the solution in Eq. (6.5) is the sum of the solutions to the above two separate ow problems. This superposition of solutions is a result of the linearity of the governing equation (6.6) and boundary conditions. Note also that Eq. (6.5) is valid not only when both the pressure gradient and the wall motion drive the uid in the same direction, as in the present example, but also when they oppose each other. In the latter case, some reverse ow {in the negative x direction{ can occur when =@x >0. Finally, let us nd the point y where the velocity attains its maximum value. This point is a zero of the shear stress (or, equivalently, of the velocity derivative, du x =dy): 0= V a ; (a ; y ) =) y = a + V :

16 14 Chapter 6. Unidirectional Flows The ow is symmetric with respect to the centerline, if y =a=, i.e., when V =0. The maximum velocity u x max is determined by substituting y into Eq. (6.5). Example Poiseuille ow between inclined plates Consider steady ow between two parallel inclined plates, driven by both constant pressure gradient and gravity. The distance between the two plates is H and the chosen system of coordinates is shown in Fig The angle formed by the two plates and the horizontal direction is. y x H ;g cos j g sin i u x (y) g Figure 6.7. Poiseuille ow between inclined plates. The general solution for u x is given by Eq. (6.10): u x (y) = ; g x y + c 1 y + c : Since, g x = g sin we get u x (y) = ; g sin y + c 1 y + c :

17 6.1 Steady, One-Dimensional Rectilinear Flows 15 Integration of this equation with respect to y and application of the boundary conditions, du x =dy=0 at y=0 and u x =0 at y=h, give u x (y) = 1 The pressure is obtained from Eq. (6.13) as + g sin (H ; y ) : (6.7) p x + g y y + c =) p = x + g cosy + c Example Thin lm ow Consider a thin lm of an incompressible Newtonian liquid owing down an inclined plane (Fig. 6.8). The ambient air is assumed to be stationary, and, therefore, the ow is driven by gravity alone. Assuming that the surface tension of the liquid is negligible, and that the lm is of uniform thickness, calculate the velocity and the volumetric ow rate per unit width. Solution: The governing equation of the ow is with general solution d u x + g dy x =0 =) d u x dy u x g sin = ; y + c 1y + c : = ;g sin As for the boundary conditions, we have no slip along the solid boundary, u x =0 at y =0 and no shearing at the free surface (the ambient air is stationary), yx = du x dy =0 at y = : Applying the above two conditions, we nd that c =0 and c 1 =g sin=(), and thus u x = g sin y ; y! : (6.9)

18 16 Chapter 6. Unidirectional Flows y Stationary air ( yx =0 at y=) x u x (y) ;g cos j g sin i Thin lm g Figure 6.8. Film ow down an inclined plane. The velocity prole is semiparabolic, and attains its maximum value at the free surface, g sin u x max = u x () = : The volume ow rate per unit width is Q W = Z 0 u x dy = g sin 3 3 (6.30) and the average velocity, u x,over a cross section of the lm is given by u x = Q W = g sin 3 : Note that if the lm is horizontal, then sin=0 and u x is zero, i.e., no ow occurs. If the lm is vertical, then sin=1, and u x = g! y ; y (6.31)

19 6.1 Steady, One-Dimensional Rectilinear Flows 17 and Q W By virtue of Eq. (6.13), the pressure is given by = g3 3 : (6.3) p = g y y + c = ;g cos y + c: At the free surface, the pressure must be equal to the atmospheric pressure, p 0,so p 0 = ;g cos + c and p = p 0 + g ( ; y) cos: (6.33) Example Two-layer plane Couette ow Two immiscible incompressible liquids A and B of densities A and B ( A > B ) and viscosities A and B ow between two parallel plates. The ow is induced by the motion of the upper plate which moves with speed V, while the lower plate is stationary (Fig. 6.9). V H B Fluid B u B x =ub x (y) H A Fluid A u A x =u A x (y) y x Figure 6.9. Two-layer plane Couette ow. The velocity distributions in both layers obey Eq. (6.6) and are given by Eq. (6.10). Since the pressure gradient and gravity are both zero, u A x = c A 1 y + c A 0 y H A u B x = c B 1 y + c B H A y H A + H B

20 18 Chapter 6. Unidirectional Flows where c A 1, c A, c B 1 and c B are integration constants determined by conditions at the solid boundaries and the interface of the twolayers. The no-slip boundary conditions at the two plates are applied rst. At y=0, u A x =0 therefore, At y=h A + H B, u B x =V therefore, The two velocity distributions become c A =0: c B = V ; CB 1 (H A + H B ) : u A x = c A 1 y 0 y H A u B x = V ; c B 1 (H A + H B ; y) H A y H A + H B : At the interface (y=h A ), we have two additional conditions: (a) the velocity distribution is continuous, i.e., u A x = ub x at y = H A (b) momentum transfer through the interface is continuous, i.e., A yx = B yx at y = H A =) du A x du B x A = B dy dy From the interface conditions, we nd that c A 1 = B V A H B + B H A and c B 1 = Hence, the velocity proles in the two layers are u A x = at y = H A : A V A H B + B H A : B V A H B + B H A y 0 y H A (6.34) u B A V x = V ; (H A + H B ; y) H A y H A + H B : (6.35) A H B + B H A If the two liquids are of the same viscosity, A = B =, then the two velocity proles are the same, and the results simplify to the linear velocity prole for onelayer Couette ow, u A x = u B x = V H A + H B y:

21 Sec. 6.. Steady, Axisymmetric Rectilinear Flows Steady, Axisymmetric Rectilinear Flows Axisymmetric ows are conveniently studied in a cylindrical coordinate system, (r z), with the z-axis coinciding with the axis of symmetry of the ow. Axisymmetry means that there is no variation of the velocity with = 0 : (6.36) There are three important classes of axisymmetric unidirectional ows (i.e., ows in which only one of the three velocity components, u r, u and u z, is nonzero): 1. Axisymmetric rectilinear ows, inwhich only the axial velocity component, u z, is nonzero. The streamlines are straight lines. Typical ows are fullydeveloped pressure-driven ows in cylindrical tubes and annuli, and open lm ows down cylinders or conical pipes.. Axisymmetric torsional ows, in which only the azimuthal velocity component, u, is nonzero. The streamlines are circles centered on the axis of symmetry. These ows, studied in Section 6.3, are good prototypes of rigid-body rotation, ow in rotating mixing devices, and swirling ows, such as tornados. 3. Axisymmetric radial ows, in which only the radial velocity component, u r, is nonzero. These ows, studied in Section 6.4, are typical models for radial ows through porous media, migration of oil towards drilling wells, and suction ows from porous pipes and annuli. As already mentioned, in axisymmetric rectilinear ows, The continuity equation for incompressible ow, becomes (ru r)+ 1 r u r = u @u =0: =0 From the above equation and the axisymmetry condition (6.36), we deduce that u z = u z (r t) : (6.38)

22 0 Chapter 6. Unidirectional Flows Due to Eqs. (6.36)-(6.38), the z-momentum z z + u r + z + u r = ; @u z + 1 u + g z is simplied = g z : (6.39) For steady ow, u z =u z (r) and Eq. (6.39) becomes an ordinary dierential equation, + 1 d r dr r du z dr + g z =0: (6.40) The only nonzero components of the stress tensor are the shear stresses rz and zr, for which we have is rz = zr = du z dr (6.41) + 1 d r dr (r rz) +g z =0: (6.4) When the pressure gradient =@z is constant, the general solution of Eq. (6.39) u z = 1 ; g z r + c 1 ln r + c : (6.43) For rz,weget rz = ; g z r + c 1 r : (6.44) The constants c 1 and c are determined from the boundary conditions of the ow. The assumptions, the governing equations and the general solution for steady, axisymmetric rectilinear ows are summarized in Table 6.. Example Hagen-Poiseuille ow Fully-developed axisymmetric Poiseuille ow, or Hagen-Poiseuille ow, studied experimentally by Hagen in 1839 and Poiseuille in 1840, is the pressure-driven ow in innitely long cylindrical tubes. The geometry of the ow is shown in Fig Assuming that gravity is zero, the general solution for u z is u z = 1 r + c 1 ln r + c :

23 6. Steady, Axisymmetric Rectilinear Flows 1 Assumptions: u r = u =const. =0 =) u z = u z (r) z-momentum: r-momentum: + r 1 dr d r du z dr + g r =0 + g z =0 -momentum: General solution: ; 1 + g =0 u z = 4 ; g z r + c 1 ln r + c rz = zr = ; g z r + c 1 r p z + c(r ) [ c(r )=const. when g r =g =0 ] Table 6.. Governing equations and general solution for steady, axisymmetric rectilinear ows. The constants c 1 and c are determined by the boundary conditions of the ow. Along the axis of symmetry, the velocity u z must be nite, Since the wall of the tube is stationary, u z nite at r =0: u z =0 at r = R:

24 Chapter 6. Unidirectional Flows u z =u z (r) rz R r z Figure Axisymmetric Poiseuille ow. By applying the two conditions, we getc 1 =0 and and, therefore, c = ; 1 R u z = ; 1 R ; r (6.45) which represents a parabolic velocity prole (Fig. 6.10). The shear stress varies linearly with r, rz = r and the shear stress exerted by the uid on the wall is w = ; rz j r=r = ; R: (Note that the contact area faces the negative r-direction.) The maximum velocity occurs at r=0, For the volume ow rate, we get: u z max = ; 1 R : Q = Z R 0 u z r dr = Z R 0 (R ; r )rdr =) Q = ; R4 : (6.46)

25 6. Steady, Axisymmetric Rectilinear Flows 3 Note that, since the pressure gradient =@z is negative, Q is positive. Equation (6.46) is the famous experimental result of Hagen and Poiseuille, also known as the fourth-power law. This basic equation is used to determine the viscosity from capillary viscometer data after taking into account the so-called Bagley correction for the inlet and exit pressure losses. The average velocity, u z, in the tube is u z = Q R = ; 1 R : Example 6... Fully-developed ow in an annulus Consider fully-developed pressure-driven ow of a Newtonian liquid in a suciently long annulus of radii R and R, where <1(Fig. 6.11). For zero gravity, the general solution for the axial velocity u z is u z = 1 r + c 1 ln r + c : R r u z =u z (r) R z Figure Fully-developed ow in an annulus. Applying the boundary conditions, u z =0 at r = R u z =0 at r = R we nd that c 1 = ; 1 R 1 ; ln(1=)

26 4 Chapter 6. Unidirectional Flows and c = ; 1 R ; c 1 ln R: Substituting c 1 and c into the general solution we obtain: " u z = ; 1 r R 1 ; + 1 ; R ln(1=) ln r R # : (6.47) The shear stress is given by " rz = 1 R r R ; 1 ; ln(1=) R # : (6.48) r The maximum velocity occurs at the point where rz =0 (which is equivalent to du z =dr=0), i.e., at " # 1 r ; 1= = R : ln(1=) Substituting into Eq. (6.47), we get ( u z max = ; 1 R 1 ; 1 ; ln(1=) For the volume ow rate, we have Q = Z R 0 u z r dr = ; Q = ; R Z R 0 " 1 R4 " 1 ; 4 ; The average velocity, u z, in the annulus is u z = " 1 ; ln 1 ; ln(1=) #) r # + 1 ; R ln(1=) ln r rdr =) R ; # 1 ; ln(1=) " Q R ; (R) = ; 1 1+ R ; : : (6.49) ; # 1 ; ln(1=) Example Film ow down a vertical cylinder A Newtonian liquid is falling vertically on the outside surface of an innitely long cylinder of radius R, in the form of a thin uniform axisymmetric lm, in contact :

27 6. Steady, Axisymmetric Rectilinear Flows 5 Q r z R u z (r) Stationary air Figure 6.1. Thin lm ow down a vertical cylinder. with stationary air (Fig. 6.1). If the volumetric ow rate of the lm is Q, calculate its thickness. Assume that the ow is steady, and that surface tension is zero. Solution: Equation (6.43) applies =0: u z = ; 1 4 g z r + c 1 ln r + c Since the air is stationary, the shear stress on the free surface of the lm is zero, rz = du z dr =0 at r = R + =) c (R + ) 1 = g : At r=r, u z =0 consequently, c = 1 4 gr ; c 1 ln R: Substituting into the general solution, we get u z = 1 4 g R ; r +(R + ) ln r R : (6.50)

28 6 Chapter 6. Unidirectional Flows For the volume ow rate,q, we have: Q = Z R+ R u z r dr = g Z R+ R R ; r +(R + ) ln r rdr: R After integration and some algebraic manipulations, we nd that Q = (4 8 gr4 1+ R 4 ln 1+ ; " + 3 R R R 1+ R ; 1 #) : (6.51) When the annular lm is very thin, it can be approximated as a thin planar lm. We willshow that this is indeed the case, by proving that for R 1 Eq. (6.51) reduces to the expression found in Example for a thin vertical planar lm. Letting leads to the following expression for Q, = R Q = 8 gr4 n 4(1+) 4 ln (1 + ) ; ( + ) h 3(1+) ; 1 io : Expanding ln(1 + ) into Taylor series, we get ln(1 + ) = ; ; O(5 ) : Thus (1 + ) 4 ln(1 + ) = ( ) " ; ; O(5 ) # Consequently, or Q = 8 gr4 4 = O( 5 ) O( 5 ) ; ( ) Q = 8 gr ; O( 5 ) :

29 6. Steady, Axisymmetric Rectilinear Flows 7 Keeping only the third-order term, we get Q = gr4 =) 3 R Q R = g3 3 : By setting R equal to W, the last equation becomes identical to Eq. (6.3). Example Annular ow with the outer cylinder moving Consider fully-developed ow of a Newtonian liquid between two coaxial cylinders of innite length and radii R and R, where <1. The outer cylinder is steadily translated parallel to its axis with speed V, whereas the inner cylinder is xed (Fig. 6.13). For this problem, the pressure gradient and gravity are assumed to be negligible. V R r u z =u z (r) R z Figure Flow in an annulus driven by the motion of the outer cylinder. The general solution for the axial velocity u z takes the form u z = c 1 ln r + c : For r=r, u z =0, and for r=r, u z =V. Consequently, c 1 = V ln(1=) Therefore, the velocity distribution is given by u z and c = ;V ln(r) ln(1=) : = V ln r R ln(1=) : (6.5)

30 8 Chapter 6. Unidirectional Flows Let us now examine two limiting cases of this ow. (a) For!0, the annular ow degenerates to ow in a tube. From Eq. (6.5), we have ; u z = lim V ln r R!0 ln(1=) = V lim 1+ ln r R = V:!0 ln(1=) In other words, we have plug ow (solid-body translation) in a tube. (b) For!1, the annular ow is approximately a plane Couette ow. To demonstrate this, let = 1 ; 1=1 ; and R = R ; R =(1; )R =) R = R : Introducing Cartesian coordinates, (y z), with the origin on the surface of the inner cylinder, we have y = r ; R =) Substituting into Eq. (6.5), we get u z Using L'H^opital's rule, we nd that ; lim V ln 1+ y!0 ln(1 + ) R = V ln ; 1+ y R ln(1 + ) = lim!0 V y R r R =1+ y R : : (6.53) y R = V y R : Therefore, for small values of, that is for!1, we obtain a linear velocity distribution which corresponds to plane Couette ow between plates separated by a distance R. 6.3 Steady, Axisymmetric Torsional Flows In axisymmetric torsional ows, also referred to as swirling ows, u r = u z =0 (6.54) and the streamlines are circles centered at the axis of symmetry. Such ows usually occur when rigid cylindrical boundaries (concentric to the symmetry axis of the

31 6.3 Steady, Axisymmetric Torsional Flows 9 ow) are rotating about their axis. Due to the axisymmetry =@=0, the continuity equation for incompressible ow, (ru r)+ =0 is automatically satised. Assuming that the gravitational acceleration is parallel to the symmetry axis of the ow, g = ;g e z (6.55) the r- and z-momentum equations are simplied as follows, u r (6.56) + g =0: Equation (6.56) suggests that the centrifugal force on an element of uid balances the force produced by the radial pressure gradient. Equation (6.57) represents the standard hydrostatic expression. Note also that Eq. (6.56) provides an example in which the nonlinear convective terms are not vanishing. In the present case, however, this nonlinearity poses no diculties in obtaining the analytical solution for u. As explained below, u is determined from the -momentum equation which is decoupled from Eq. (6.56). By assuming =0 and by integrating Eq. (6.57), we get p = ;g z + c(r t) consequently, =@r is not a function of z. Then, from Eq. (6.56) we deduce that u = u (r t) : (6.58) Due to the above assumptions, the -momentum equation reduces (ru ) : (6.59) For steady ow, we obtain the linear ordinary dierential equation d 1 d dr r dr (ru ) =0 (6.60)

32 30 Chapter 6. Unidirectional Flows the general solution of which is u = c 1 r + c r : (6.61) The constants c 1 and c are determined from the boundary conditions of the ow. Assumptions: u r = u =0, g = ;g e z Continuity: -momentum: Satised identically d dr 1r d dr (ru ) =0 + g =0 r-momentum: General solution: u r u = c 1 r + c r =) u = u (r) r = r = ; c c p = 1 r r +c 1 c ln r ; c r ; g z + c Table 6.3. Governing equations and general solution for steady, axisymmetric torsional ows. The pressure distribution is determined by integrating Eqs. (6.56) and (6.57): p = Z u r dr ; g z =)

33 6.3 Steady, Axisymmetric Torsional Flows 31! p = c 1r +c 1 c ln r ; c ; g z + c (6.6) r where c is a constant ofintegration, evaluated in any particular problem by specifying the value of the pressure at a reference point. Note that, under the above assumptions, the only nonzero components of the stress tensor are the shear stresses, r = r = r d dr u r in terms of which the -momentum equation takes the form (6.63) The general solution for r is d dr (r r )=0: (6.64) r = ; c r : (6.65) The assumptions, the governing equations and the general solution for steady, axisymmetric torsional ows are summarized in Table 6.3. Example Steady ow between rotating cylinders The ow between rotating coaxial cylinders is known as the circular Couette ow, and is the basis for Couette rotational-type viscometers. Consider the steady ow of an incompressible Newtonian liquid between twovertical coaxial cylinders of innite length and radii R 1 and R, respectively, occurring when the two cylinders are rotating about their common axis with angular velocities 1 and, in the absence of gravity (Fig. 6.14). The general form of the angular velocity u is given by Eq. (6.61), The boundary conditions, u = c 1 r + c r : u = 1 R 1 at r = R 1 u = R at r = R The time-dependent owbetween rotating cylinders is much more interesting, especially the manner in which it destabilizes for large values of 1, leading to the generation of axisymmetric Taylor vortices [4].

34 3 Chapter 6. Unidirectional Flows 1 R 1 R r Figure Geometry of circular Couette ow. result in c 1 = R ; R 1 1 R ; R 1 and c = ; R 1 R R ; ( ; 1 ) : R 1 Therefore, u = 1 R ; R 1 (R ; R 1 1 ) r ; R 1R ( ; 1 ) 1 : (6.66) r Note that the viscosity does not appear in Eq. (6.66), because shearing between adjacent cylindrical shells of uid is zero. This observation is analogous to that made for the plane Couette ow [Eq. (6.14)]. Also, from Eqs. (6.6) and (6.65), we get and 1 1 p = (R ; R 1 ) (R ; R 1 1) r +R 1 R (R ; R 1 1)( ; 1 )lnr r = ; 1 R4 1R 4 ( ; 1 ) 1 r + c (6.67) R 1 R (R ; R 1 ) ( ; 1 ) 1 r : (6.68) Let us now examine the four special cases of ow between rotating cylinders, illustrated in Fig

35 6.3 Steady, Axisymmetric Torsional Flows 33 R 1 R r (a) The inner cylinder is xed ( 1 =0) u = R R ; R r ; R 1 r 1 For p, see Eq. (6.70) (b) 1 = = R 1 R u = r p= 1 r + c r (Rigid-body rotation) (c) No inner cylinder R u = r p= 1 r + c r (Rigid-body rotation) (d) No outer cylinder 1 R 1 u =R r r p=; 1 R r + c Figure Dierent cases of ow between rotating vertical coaxial cylinders of innite height.

36 34 Chapter 6. Unidirectional Flows (a) The inner cylinder is xed, i.e., 1 =0. In this case, u = R R ; R 1 r ; R 1 r! (6.69) and! R 4 p = r (R ; R 1 ) +R 1 ln r ; R4 1 + c: (6.70) r The constant c can be determined by setting p=p 0 at r=r 1 accordingly, " R 4 p = r ; R 1 (R ; +R R 1 ) 1 ln r ; R4 1 1 # R 1 r ; 1 + p R 0 : (6.71) 1 For the shear stress, r,weget r = R 1 R 1 R ; R 1 r : (6.7) The shear stress exerted by the liquid to the outer cylinder is R 1 w = ; r j r=r = ; R ; R 1 : (6.73) In viscosity measurements, one measures the torque T per unit height L, at the outer cylinder, T L =R (; w ) =) T L = 4 R 1R R ; : (6.74) R 1 The unknown viscosity of a liquid can be determined using the above relation. When the gap between the two cylinders is very small, circular Couette ow can be approximated as a plane Couette ow. Indeed, letting r=r 1 +r, we get from Eq. (6.69) u = R R ; R 1 When R 1! R,r=R 1 1 and, therefore, + r R 1 1+ r R 1 r: u = R (R ; R 1 ) r = R R ; R 1 r

37 6.3 Steady, Axisymmetric Torsional Flows 35 which is a linear velocity distribution corresponding to plane Couette ow between plates separated by a distance R -R 1, with the upper plate moving with velocity R. (b) The two cylinders rotate with the same angular velocity, i.e., 1 = =: In thic case, c 1 = and c =0. Consequently, u =r (6.75) which corresponds to rigid-body rotation. This is also indicated by the zero tangential stress, r = ; c r =0: For the pressure, we get p = 1 r + c: (6.76) (c) The inner cylinder is removed. In thic case, c 1 = and c =0, since u (and r ) are nite at r=0. This ow is the limiting case of the previous one for R 1!0, u = r r =0 and p = 1 r + c: (d) The outer cylinder is removed, i.e., the inner cylinder is rotating in an innite pool of liquid. In this case, u!0asr!1, and, therefore, c 1 =0. At r=r 1, u = 1 R 1 which gives c = R 1 1 : Consequently, u = R r (6.77) 1 r = ;R 1 1 r (6.78) and p = ; 1 1 R c: (6.79) r The shear stress exerted by the liquid to the cylinder is w = r j r=r1 = ; 1 : (6.80)

38 36 Chapter 6. Unidirectional Flows The torque per unit height required to rotate the cylinder is T L =R 1 (; w)=4 R 1 1 : (6.81) In the previous example, we studied ows between vertical coaxial cylinders of innite height ignoring the gravitational acceleration. As indicated by Eq. (6.6), gravity has no inuence on the velocity and aects only the pressure. In case of rotating liquids with a free surface, the gravity term should be included if the top part of the ow and the shape of the free surface were of interest. If surface tension eects are neglected, the pressure on the free surface is constant. Therefore, the locus of the free surface can be determined using Eq. (6.6). Example Shape of free surface in torsional ows In this example, we study two dierent torsional ows with a free surface. First, we consider steady ow of a liquid contained in a large cylindrical container and agitated by avertical rod of radius R that is coaxial to the container and rotates at angular velocity. If the radius of the container is much larger than R, one may assume that the rod rotates in an innite pool of liquid (Fig. 6.16). z R p=p 0 z 0 r Figure Rotating rod in a pool of liquid. From the results of Example 6.3.1, we have c 1 =0 and c =R. Therefore, u = R 1 r

39 6.3 Steady, Axisymmetric Torsional Flows 37 and p = ; 1 R4 1 ; g z + c: r With the surface tension eects neglected, the pressure on the free surface is equal to the atmospheric pressure, p 0. To determine the constant c, we assume that the free surface contacts the rod at z=z 0.Thus, we obtain c = p R4 1 R + g z 0 and p = 1 R4 1 R ; 1 r ; g (z ; z 0 )+p 0 : (6.8) Since the pressure is constant along the free surface, the equation of the latter is 0=p ; p 0 = 1 1 R4 R ; 1 ; g (z ; z r 0 ) =) z = z 0 + R g 1 ; R r! : (6.83) The elevation of the free surface increases with the radial distance r and approaches asymptotically the value z 1 = z 0 + R : g This ow behavior, known as rod dipping, is a characteristic of generalized-newtonian liquids, whereas viscoelastic liquids exhibit rod climbing (i.e., they climb the rotating rod) [5]. Consider now steady ow of a liquid contained in a cylindrical container of radius R rotating at angular velocity (Fig. 6.17). From Example 6.3.1, we know that this ow corresponds to rigid-body rotation, i.e., u =r: The pressure is given by p = 1 r ; g z + c: Letting z 0 be the elevation of the free surface at r=0, and p 0 be the atmospheric pressure, we get c = p 0 + g z 0

40 38 Chapter 6. Unidirectional Flows z p=p 0 z 0 r R Figure Free surface of liquid in a rotating cylindrical container. and thus p = 1 r ; g (z ; z 0 )+p 0 : (6.84) The equation of the free surface is 0= p ; p 0 = 1 r ; g (z ; z 0 ) =) z = z 0 + g r (6.85) i.e., the free surface is a parabola. Example Superposition of Poiseuille and Couette ows Consider steady ow of a liquid in a cylindrical tube occurring when a constant pressure gradient =@z is applied, while the tube is rotating about its axis with constant angular velocity (Fig. 6.18). This is obviously a bidirectional ow, since the axial and azimuthal velocity components, u z and u, are nonzero. The ow can be considered as a superposition of axisymmetric Poiseuille and circular Couette ows, for which we have: u z = u z (r)= ; 1 (R ; r ) and u = u (r)= r: This superposition is dynamically admissible, since it does not violate the continuity equation, which is automatically satised.

41 6.3 Steady, Axisymmetric Torsional Flows =const. R r z Figure Flow in a rotating tube under constant pressure gradient. Moreover, the governing equations of the ow, i.e., the z- and -momentum equations, (ru ) are linear and uncoupled. Hence, the velocity for this ow is given by u = u z e z + u e = ; 1 4 (R ; r ) e z +r e (6.86) which describes a helical ow. The pressure is obtained by integrating the r-momentum equation, u r taking into account that =@z is constant. It turns out that p z + 1 r + c (6.87) which is simply the sum of the pressure distributions of the two superposed ows. It should be noted, however, that this might not be the case in superposition of other unidirectional ows.

42 40 Chapter 6. Unidirectional Flows 6.4 Steady, Axisymmetric Radial Flows In axisymmetric radial ows, u z = u =0: (6.88) Evidently, the streamlines are straight lines perpendicular to the axis of symmetry (Fig. 6.19). z y x r Figure Streamlines in axisymmetric radial ow. For the sake of simplicity, we will assume that u r, in addition to being axisymmetric, does not depend on z. In other words, we assume that, in steady-state, u r is only a function of r: u r = u r (r) : (6.89) A characteristic of radial ows is that the non-vanishing radial velocity component is determined by the conservation of mass rather than by the r-component of the conservation of momentum equation. This implies that u r is independent of the viscosity of the liquid. (More precisely, u r is independent of the constitutive equation of the uid.) Due to Eq. (6.88), the continuity equation is (ru r)=0 (6.90) which gives u r = c 1 r (6.91) where c 1 is a constant. The velocity u r can also be obtained from a macroscopic mass balance. If Q is the volumetric ow rate per unit height, L, then Q = u r (rl) =)

43 6.4 Steady, Axisymmetric Radial Flows 41 u r = Q L r (6.9) which is identical to Eq. (6.91) for c 1 =Q=(L). Assumptions: Continuity: u z = u =0 u r = u r (r), g = ;g e z d dr (ru r)=0 =) u r = c 1 r r-momentum: u r du r dr = + g =0 -momentum: General =0 =) p = p(r z) u r = c 1 r rr = ; c 1 r, = c 1 r p = ; c 1 r ; g z + c Table 6.4. radial ows. Governing equations and general solution for steady, axisymmetric Letting g = ;g e z (6.93) the r-component of the Navier-Stokes equation is simplied to u r du r dr = : (6.94)

44 4 Chapter 6. Unidirectional Flows Note that the above equation contains a non-vanishing nonlinear convective term. The z- and -components of the Navier-Stokes equation are reduced to the standard hydrostatic expression, + g =0 and =0 (6.96) respectively. The latter equation dictates that p=p(r z). Integration of Eqs. (6.94) and (6.95) gives p(r z) = ; Z u r du r dr dr ; g z + c Z 1 = c 1 dr ; g z + c =) r3 p(r z)=; c 1 ; g z + c (6.97) r where the integration constant c is determined by specifying the value of the pressure at a point. In axisymmetric radial ows, there are two non-vanishing stress components: rr = du r dr = ; c 1 r (6.98) = u r r = c 1 r : (6.99) The assumptions, the governing equations and the general solution for steady, axisymmetric radial ows are summarized in Table Steady, Spherically Symmetric Radial Flows In spherically symmetric radial ows, the uid particles move towards or away from the center of solid, liquid or gas spheres. Examples of such ows are ow around a gas bubble which grows or collapses in a liquid bath, ow towards a spherical sink, and ow away from a point source. The analysis of spherically symmetric radial ows is similar to that of the axisymmetric ones. The assumptions and the results are tabulated in Table 6.5. Obviously,

45 6.5 Steady, Spherically Symmetric Radial Flows 43 Assumptions: Continuity: u = u =0 u r = u r (r), g = 0 d dr (r u r )=0 =) u r = c 1 r r-momentum: -momentum: -momentum: General solution: u r du r =0 u r = c 1 r = rr = ;4 c 1 r 3, = = c 1 r 3 p = ; c 1 r 4 + c Table 6.5. Governing equations and general solution for steady, spherically symmetric radial ows. spherical coordinates are the natural choice for the analysis. In steady-state, the radial velocity component is a function of the radial distance, while the other two velocity components are zero: u r = u r (r) (6.100) u = u =0: (6.101) As in axisymmetric radial ows, u r is determined from the continuity equation

46 44 Chapter 6. Unidirectional Flows as u r = c 1 r (6.10) or u r = Q 4r (6.103) where Q is the volumetric ow rate. The pressure is given by p(r) =; c 1 + c: (6.104) r4 (Note that, in spherically symmetric ows, gravity is neglected.) Finally, there are now three non-vanishing stress components: rr = du r dr = ;4 c 1 r 3 (6.105) = = u r r = c 1 r 3 : (6.106) Example Bubble growth in a Newtonian liquid Boiling of a liquid often originates from small air bubbles which grow radially in the liquid. Consider a spherical bubble of radius R(t) in a pool of liquid, growing at a rate dr = k: dt The velocity, u r, and the pressure, p, can be calculated using Eqs. (6.10) and (6.104), respectively. At rst, we calculate the constant c 1.Atr=R, u r =dr=dt=k or c 1 R = k =) c 1 = kr : Substituting c 1 into Eqs. (6.10) and (6.104), we get and u r = k R r p = ;k R4 r 4 + c: Note that the pressure near the surface of the bubble may attain small or even negative values, which favor evaporation of the liquid and expansion of the bubble.

47 Sec Transient One-Dimensional Unidirectional Flows Transient One-Dimensional Unidirectional Flows In Sections 6.1 to 6.3, we studied three classes of steady-state unidirectional ows, where the dependent variable, i.e., the nonzero velocity component, was assumed to be a function of a single spatial independent variable. The governing equation for such a ow is a linear second-order ordinary dierential equation which is integrated to arrive at a general solution. The general solution contains two integration constants which are determined by the boundary conditions at the endpoints of the one-dimensional domain over which the analytical solution is sought. In the present section, we consider one-dimensional, transient unidirectional ows. Hence, the dependent variable is now a function of two independent variables, one of which is time, t. The governing equations for these ows are partial dierential equations. In fact, we have already encountered some of these PDEs in Sections , while simplifying the corresponding components of the Navier- Stokes equation. For the sake of convenience, these are listed below. (a) For transient one-dimensional rectilinear ow in Cartesian coordinates with u y =u z =0 and u x =u x (y = u + g x : (6.107) (b) For transient axisymmetric rectilinear ow withu r =u =0 and u z =u z (r t), z = + = u g z + g z : (6.108) (c) For transient axisymmetric torsional ow with u z =u r =0 and u =u (r (ru + ; 1 r u : (6.109)

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