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(BQ) Part 2 book "Theoretical fluid mechanics" has contents: Incompressible viscous flow, waves in incompressible fluids, terrestrial ocean tides, equilibrium of compressible fluids, one dimensional compressible inviscid flow, two - dimensional compressible inviscid flow. | 10 Incompressible Viscous Flow 10.1 Introduction This chapter investigates incompressible flow in which viscosity plays a significant role throughout the bulk of the fluid. Such flow generally takes place at relatively low Reynolds number. From Section 1.14, the equations governing incompressible viscous fluid motion can be written ∇ · v = 0, ρ Dv = −∇P + µ ∇ 2 v, Dt (10.1) (10.2) where the quantity P = p + ρ Ψ, (10.3) which is a combination of the actual fluid pressure, p, and the gravitational potential energy per unit volume, ρ Ψ , is known as the effective pressure. Here, ρ is the fluid mass density, µ the fluid viscosity, and Ψ the gravitational potential. More information on this topic can be found in Batchelor 2000. 10.2 Flow Between Parallel Plates Consider steady, two-dimensional, viscous flow between two parallel plates that are situated a perpendicular distance d apart. Let x be a longitudinal coordinate measuring distance along the plates, and let y be a transverse coordinate such that the plates are located at y = 0 and y = d. (See Figure 10.1.) Suppose that there is a uniform effective pressure gradient in the x-direction, so that dP = −G, (10.4) dx where G is a constant. Here, the quantity G could represent a gradient in actual fluid pressure, a gradient in gravitational potential energy (due to an inclination of the plates to the horizontal), or some combination of the two—it actually makes no 287 288 Theoretical Fluid Mechanics y=d vx(y) −∇P y x y=0 Figure 10.1 Viscous flow between parallel plates. difference to the final result. Suppose that the fluid velocity profile between the plates takes the form (10.5) v = v x (y) e x . From Section 1.18, this profile automatically satisfies the incompressibility constraint ∇ · v = 0, and is also such that Dv/Dt = 0. Hence, Equation (10.2) reduces to ∇P , (10.6) ∇ 2v = µ or. taking the x-component, G d 2vx (10.7) =− . 2 µ dy If the two plates are stationary then the solution that satisfies the no slip .