Technology & Engineering
Couette Flow
Couette flow is a type of viscous fluid flow between two parallel plates, where one plate is stationary and the other is in constant motion. This flow is characterized by the shearing motion of the fluid, with the velocity of the fluid increasing linearly from the stationary plate to the moving plate. Couette flow is an important concept in fluid dynamics and has applications in various engineering and scientific fields.
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7 Key excerpts on "Couette Flow"
eBook - PDFMicro- and Nanoscale Fluid Mechanics
Transport in Microfluidic Devices
- Brian J. Kirby(Author)
- 2010(Publication Date)
- Cambridge University Press(Publisher)
2.1.1 Couette Flow Couette Flow is the flow of fluid in between two infinite parallel flat plates driven by the motion of one or more of the plates. Consider steady flow between two infinite parallel moving plates with a uniform pressure field. Consider plates located at y = ± h . The top plate moves in the x direction with speed u H , and the other plate moves in the x direction with speed u L . A schematic of this flow is shown in Fig. 2.1 . The governing equations are the Navier–Stokes equations without body forces: ∂ u ∂ t + u · ∇ u = −∇ p + ∇ 2 u . (2.1) Simplified equation. We solve this flow by making two simplifying approximations. The resulting solution satisfies the boundary condition and governing equation and thus vali-dates these approximations. We begin by assuming that the fluid motion is in the x direc-tion only. Because the flow is in the x direction only, we can replace u with u , resulting in ∂ u ∂ t + u ∂ u ∂ x = − ∂ p ∂ x + ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 + ∂ 2 u ∂ z 2 . (2.2) 41 42 Ch. 2. Unidirectional Flow Fig. 2.1 Couette Flow between two parallel, infinite plates. Our next simplifying approximation is that the velocity profile is independent of x . The first three terms of the equation can therefore be assumed zero, in turn, because (a) the flow is assumed steady, (b) u is assumed independent of x , and (c) the pressure is assumed uniform. Thus the simplified governing equation for this flow is 0 = ∂ 2 u ∂ y 2 . (2.3) In this case, we have chosen a problem statement and geometry that allow much of the Navier–Stokes equation to be ignored; in so doing, we have a system that is much simpler mathematically. EXAMPLE PROBLEM 2.1 Show that, if is uniform and flow is in the x direction only such that u = ( u , 0 , 0), then ∇ · ∇ u = ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 + ∂ 2 u ∂ z 2 , 0 , 0 .
eBook - PDF- Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Robertson(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
y L s U FIGURE 9.3 Couette Flow involves two flat plates, each large enough that the dimensions of the plates can be idealized as infinite. The lower plate is stationary and the upper plate is moving with a constant velocity U. The gap L is small—for example, a fraction of a millimeter. The equation for the pictured velocity profile is derived in the current section. Poiseuille Flow in a Channel 295 Solving for the Velocity Field To solve for the velocity field in Couette Flow, take the following steps. Step 1: Apply the Navier-Stokes equation. In Couette Flow, the pressure gradient in the stream-wise direction is zero (dp/ ds = 0) and the streamlines are in the horizontal direction, which means that dz/ ds = 0. Therefore, the right side of Eq. (9.8) is zero, and this equation reduces to d 2 u dy 2 = 0 (9.9) To solve this differential equation, integrate twice using the method known as “sepa- ration of variables.” u = C 1 y + C 2 (9.10) Step 2: Write the boundary conditions. To solve for the two constants in Eq. (9.10), apply the following two boundary conditions: u = 0 at y = 0 (9.11) u = U at y = L (9.12) Step 3: Apply the boundary conditions. Combine Eq. (9.10) with Eqs. (9.11) and (9.12) to give an equation for the velocity field: u = Uy/ L (9.13) The velocity profile (Fig. 9.3) reveals a linear velocity profile. Deriving Working Equations To develop an equation for the shear stress field, apply τ = μ(du/ dy) to Eq. (9.13): τ = μU/ L (9.14) Eq. (9.14) reveals that the shear stress is a constant at every point. To develop an equation for mean velocity, substitute Eq. (9.13) into V = (1/ A) ∫ udA. After integration, the result is V = U/ 2 . 9.3 Poiseuille Flow in a Channel The conduit considered in this section is a rectangular channel (Fig. 9.4). A channel is a flow passage between two parallel plates, when each plate is wide enough so that the end effects caused by the side walls of the channel can be neglected.
eBook - PDF- W.G. Hoover(Author)
- 2012(Publication Date)
- Elsevier Science(Publisher)
Finally, 201 these stable stationary flows provide much-needed checks for the numerical approaches which are necessary for most real applied problems. After describing these elementary examples we outline numerical techniques which can be applied to more complex and more realistic hydrodynamic problems. 8.2 Plane Couette Flow A viscous fluid confined between two coaxial rotating cylinders can exhibit a variety of unstable unsteady size-and-shape-dependent two- and three-dimensional flows as the relative rotation rate is increased, but if the sample size and rotation rates are both sufficiently small a stable laminar two-dimensional flow can result in which fluid viscosity smoothly dissipates the momentum imparted to the fluid by the moving cylinders. If both cylinder radii greatly exceed the size of the gap between the cylinders, then the simplest special case, plane Couette Flow results, with the angular velocity increasing linearly with radius. See Figure 8.1. In a Cartesian coordinate system, with the radial direction replaced by y and the tangential direction by x, this prototypical flow has the form u x = ey, where e is the strain rate and the x velocity component u x matches the wall velocity at limiting values of y, here chosen equal to ±L/2. We wish to consider only the laminar-flow case in which there is no motion in the z direction. We therefore describe the motion in the xy plane. The two-dimensional Navier-Stokes equations of motion pu = p(3u/3t) + pu-Vu = W-o are satisfied within the flow with the stream velocity u = (u x ,0), with u x proportional to y, and a constant stress tensor o, which includes the viscous contribution a X y = a y x = e: p(3u x /3t) = - pu x (au x /ax) - pu y (au x /3y) + (da xx /dx) + (do yx /dy) = -0 -0 + 0 + 0 = 0; p(3uy/3t) = - pu x (3u y /3x) - puy(3u y /3y) + (dG X y/dx) + (3a yy /3y) = -0 -0 + 0 + 0 = 0.
eBook - PDF- Joseph Katz(Author)
- 2010(Publication Date)
- Cambridge University Press(Publisher)
(5.2c) Here we neglected the body forces. However, if gravitational force is present, its effect can be reintroduced in the pressure-gradient term. 5.3 Laminar Flow between Two Infinite Parallel Plates – The Couette Flow Let us start with the simplest example, the flow between two parallel (infinite) plates, as shown in Fig. 5.1. The fluid is considered viscous and incompressible (such as water or oil), and for this case we neglect the time derivatives and the body forces. We also assume that the flow is laminar, and this statement will be clarified toward the end of the chapter. Let us use a Cartesian coordinate system attached to the lower plate, as shown in Fig. 5.1, and the governing equations for this case are summarized by Eqs. (5.1) and (5.2). Observing Fig. 5.1, we can assume that the lower plate is station-ary and therefore the velocity near the lower wall is zero (at z = 0). We may specu-late about the shape of the velocity distribution (as shown); however, it is clear that the velocity at z = h is equal to the velocity of the upper plate, U . This flow is called the Couette Flow after Maurice Marie Alfred Couette (1858–1943), a well-known French physicist. The model is 2D and there are no changes in the y direction. Also there is no velocity in the z direction, and we summarize this as v = w = 0 . (5.3) 144 Viscous Incompressible Flow: Exact Solutions Now, substituting this into continuity equation (5.1) results in ∂ u ∂ x = 0 . (5.4) The conclusion therefore is that the velocity profile is a function of z and it is the same at any x station: u = u ( z ) . (5.5) Recall that the plates extend to infinity ( −∞ < x < ∞ ) and therefore it is obvious that the velocity profile u ( z ) is the same at any x station. Next we apply all the previous assumptions to momentum equations (5.2), which now reduce to 0 = − 1 ρ ∂ p ∂ x + μ ρ ∂ 2 u ∂ z 2 , (5.6a) 0 = − 1 ρ ∂ p ∂ y , (5.6b) 0 = − 1 ρ ∂ p ∂ z .
eBook - PDFProgress in Heat and Mass Transfer
Selected Papers of the 1970 International Seminar
- W. R. Schowalter, A. V. Luikov, W. J. Minkowycz(Authors)
- 2013(Publication Date)
- Pergamon(Publisher)
Of particular interest are the results for Couette Flow between flat plates oriented at an arbitrary inclination relative to the external field, as in Fig. 11. This inclination may be specified by means of the angle y between the direction of the external field E and the undisturbed vorticity vector °. t That the no-slip boundary condition applies to a suspension is shown by Cox and Brenner [18] to be valid to at least the first order in ^,which is the order of validity of the present theory. 116 H. BRENNER The principal result of this calculation is that the viscosity of the suspension is given by the expression The last term represents the effect of the external field. The dimensionless function F is a complicated function [51] of the inclination y and the nondimensional parameter ED λ = z ; (83) specifying the ratio of the strength of the dipole couple, ED, tending to prevent rotation FIG. 11. Orientation of a Couette Flow apparatus relative to a uniform external field. of the sphere, to the strength of the hydrodynamic couple, 4πμ 0 α 3 ο, tending to cause rotation. Here, E = |E| is the magnitude of the external field and G is the macroscopic shear rate in the Couette apparatus. Because of the dependence of λ upon G, it follows that the suspension viscosity is shear-rate dependent. The quantitative nature of this dependence is depicted in Fig. 12 for a specified apparatus orientation y. The suspension displays shear thinning. At low rates of shear the external field is dominant, resulting in a strongly hindered rate of particle rotation. As the shear rate is increased indefinitely the viscosity asymptotically approaches the Einstein limit, corresponding to free or unhindered particle rotation. In addition to a shear-dependent viscosity, there also exists a complex stress system. No normal stresses exist in the suspension, but the stress is antisymmetric in the sense that t u Φ ϊ β .
eBook - PDF- Olga Moreira(Author)
- 2019(Publication Date)
- Arcler Press(Publisher)
COMBINED HAGEN-POISEUILLE AND COU -ETTE FLOWS We will consider the lower plate is stationary and to plate moving with a uniform velocity U. We consider a velocity gradient as in case of (dp/dx) in the direction of flow. The solution is obtained by superpose the two linear equations: 1) Flow between two plates with X-axis at lower plate. This is, u = (1/μ) x (dp/dx) x {(y 2 /2) – ay} 2) Couette Flow ; – This is, u = (u/2a) x y Then we have: , u = = (1/μ) x (dp/dx) x {(y 2 /2) – ay} + (u/2a) x y; q = discharge, q = ʃ u x dy On integrating and assigning limits as 0, 2a,we get q = [Ua – {(dp/dx) x (2a) 3 }]/(12 x μ); Velocity distribution is: (u/U) = (y/2a) + [K x (y/2a) x {1-(y/2a)}] where K = {(2a) 2 /(2 μU)} x (-dp/dx); The velocity pattern for different values of K are shown in the graph Figure 8.6. When you have pressure gradient is positive, it indicates flow due to gradient opposes the Couette Flow . The average velocity, V av = (q/2a) x ʃ u x dy limits 0 to 2a, V = {(1/2) + (K/6)} x U and q = 2 x a x V av . In a specific case when {(1/2) + (K/6)} = 0, the average velocity and the discharge will become zero. There is no net flow n the negative direction. Theory and Problems of Fluid Dynamics 156 When {(1/2) + (K/6)} = 0; K = {(2a) 2 /(2 μU)} x (-dp/dx) = –3, We have dp/dx = 6μU/(2a) 2 ; , du/dy = (1/μ) x (dp/dx) x (y-a) + (U/2a); u is maximum when du/dy = 0; When y = a[1 + (1/K)]; Shear stress = f x = μ(du/dy); Figure 8.6 In the graph P = K of our description. (y/b) = (y/h); (u/V) = (u/U); 8.8. HAGEN-POISEUILLE FLOW IN CYLINDRICAL TUBES The flow is steady and uniform. Let us find the solution by applying N-S equation. N-S equation in cylindrical coordinates, in our problem there is not tangential and radial flow, only axial. We need to consider the axial flow equations . Besides, in axial direction the flow is uniform and steady. Also velocity will not change with angle θ. On simplification we get, [ Note : (δw/δt) + u(δw/δr) + (v/r)(δw/δθ) + w(δw/δz)
eBook - PDF- G. Böhme(Author)
- 2012(Publication Date)
- North Holland(Publisher)
On the other hand if one represents the torque M as a function of the angular veloci ty w, then this graph gives the relationship between T and y, hence the flow function F(y) as in equation (2.4), because M is proportional to T and w to y. We now consider a device in which the fluid is located between a flat plate and a cone placed on it (Fig. 1.12). We denote the radius of the circular top of the cone by rO. If the angle between the cone and the plate is sufficiently small, the overall shear rate in the liquid has the same value, y = w la, in which w is the angular veloci ty of the plate relative to the stationary cone. Thus the shear stress in the field is also constant. Its amount is found from the torque on the cone. (3.3) One obtains an analogous formula for the viscosi ty of the fluid as for Couette viscometers: 3fJ M '1=-_.-21T r~ w (3.4) 3.1 Rotational viscometer 71 This again makes i t possible, as described above, to determine the viscosity at given rotation speeds by measuring the torque which the motion creates. For practical applications equation (3.2) needs a correction to take into account the end effect. Every Couette viscometer naturally has a finite length. The cylindrical housing at the bottom end is blanked off by a plate, in the vicinity of which in general no shear flow occurs. This unwanted end effect at the base can be completely counteracted if one makes the inner cylinder conical below and places its tip on the plate (Fig. 3.1). Hence the Couette arrangement is combined with the cone-and-plate arrangement, and hence a shear flow arises not only between the cylindrical container and the cylinder, but also at the base, which has a constant shear rate. It is advantageous to choose the cone angle so that the shear rate and therefore the shear stresses in both regions match each other: S = h/rO.
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