Simple shear flow dissipation

Fig. 9. Average efficiency of stretching of material elements (e.) in a simple shear flow with random reorientation after an average length stretch ym. pgives the with of the distribution of length stretch about the mean value (ym). Results for a random distribution (top) and a normal distribution (bottom) of length stretch are shown. The maximum in the efficiency corresponds to the maximum length stretch for a fixed amount of energy dissipated and occurs at an average stretch of about 5 per period (Khakhar and Ottino, 1986a).

Thus, only the normal Reynolds stresses (i = j) are directly dissipated in a high-Reynolds-number turbulent flow. The shear stresses (i / j), on the other hand, are dissipated indirectly, i.e., the pressure-rate-of-strain tensor first transfers their energy to the normal stresses, where it can be dissipated directly. Without this redistribution of energy, the shear stresses would grow unbounded in a simple shear flow due to the unbalanced production term Vu given by (2.108). This fact is just one illustration of the key role played by 7 ., -in the Reynolds stress balance equation. 

For V2 < V3 a simple shear flow in a perpendicular alignment causes less dissipation than in a parallel alignment. The next step is to study the stability of these alignments in the linear regime. Following the standard procedure (as described above) we find a solvability condition of the linearized equations which does not depend on the shear rate 7 ... 

As an illustration, we will derive the viscous dissipation terms in the energy balance using a simple shear flow system such as the one shown in Fig. 5.6. 

Derive the equation for the steady state temperature profile in a simple shear flow with viscous dissipation. Assume a Newtonian viscosity model. 

C. The Effect of Viscous Dissipation on a Simple Shear Flow... 

C. THE EFFECT OF VISCOUS DISSIPATION ON A SIMPLE SHEAR FLOW... 

Consider steady simple shear flow between infinite parallel plates as in Example 2.2.1b, but do not neglect viscous dissipation. 

Velocity field for simple shear flow (a) isothermal and (b) viscous dissipation. 

Friction is generated at the microscopic level by relative motion between polymer molecules in flow. In a simple shear flow such as that introduced in Section 7.2.1, the relative velocity between a polymer molecule and its surroundings can be estimated as Vc = yf g, where is the polymer coil radius in the melt. If the rate of shear is so small that it does not perturb the polymer shape, the rate of viscous dissipation per molecule q is therefore given by... 

In this section, we return to the analysis of simple, unidirectional shear flow that was considered in Section B of Chap. 3, but instead of neglecting viscous dissipation altogether, we consider its influence when the Brinkman number is small, but nonzero. The starting point is Eqs. (3-34) and (3-35), which are reproduced here for convenience ... 

One application of the solutions (4-55)-(4-61) is to evaluate the effect of viscous dissipation in the use of a shear rheometer to measure the viscosity of a Newtonian fluid. In this experiment, we subject the fluid in a thin gap between two plane walls to a shear flow by moving one of the walls in its own plane at a known velocity and then measuring the shear stress produced at either wall (by measuring the total tangential force and dividing by the area). In the absence of viscous dissipation, the velocity profile is linear and the shear rate is simply given by the tangential velocity U divided by the gap width d. Now, the constitutive equation, (2-87), for an incompressible Newtonian fluid applied to this simple flow situation takes the form... 

It is evident that by adding particles in a flow the amount of energy dissipated will be increased, since the work done by the shearing stresses is increased because of the addition of the solid boundaries associated with the particles. The particles plus liquid, which we term the suspension phase, might therefore be looked upon as a Newtonian fluid but with a coefficient of viscosity larger than that of the pure liquid. To understand the relation between the particle and fluid characteristics, Einstein (1956) set himself the problem of a dilute suspension in a simple Couette flow viscometer and asked what would be the measured viscosity. 

Correspondingly, the parameter has been found analyzing simple turbulent shear flows where is the only non-zero mean velocity gradient. Applying the ife-equation (1.407) to this flow situation we recognize that the dissipation and production terms are approximately equal. In statistical turbulence analysis (e.g., [67]), this property is said to be denoting an equilibrium shear flow. The simplified -equation yields ... 

Very dense shearing flows involve multiple and/or repeated collisions that violate the assumption of instantaneous, uncorrelated, binary collisions on which simple kinetic theory is based. Because of the practical importance of very dense shearing flows in a gravitational field, we describe an extension of the simple kinetic theory that maintains much of the structure of the simpler theory. We phrase and solve a boundary value problem for a dense, inclined flow of dissipative spheres using this extension of the kinetic theory. The numerical solution of this problem reproduces the features of such flows observed in numerical simulations. Finally, we compare the profiles that result from a numerical solution with those of a simple, analytic, approximation theory. 

Simple shear, in which u = u y, with u, T, and v constants, is exceptional. It occurs when the diameter and solid volume fraction of the flow spheres, the diameter and separation of the wall hemispheres, and the two coefficients of restitution are such that in the interior and at the boundaries, the rate at which fluctuation energy is produced by the shear stress working through the mean velocity gradient or through the slip velocity is precisely equal to the local rate at which it is dissipated in collisions. We first treat this homogeneous flow, noting that the apparent rate of shear 2U/L differs from u by 2v/L because of slip at the boundary. 

---