(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 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 , the equations governing incompressible viscous fluid motion can be written ∇ · v = 0, ρ Dv = −∇P + µ ∇ 2 v, Dt () () where the quantity P = p + ρ Ψ, () 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. 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 .) Suppose that there is a uniform effective pressure gradient in the x-direction, so that dP = −G, () 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 Viscous flow between parallel plates. difference to the final result. Suppose that the fluid velocity profile between the plates takes the form () v = v x (y) e x . From Section , this profile automatically satisfies the incompressibility constraint ∇ · v = 0, and is also such that Dv/Dt = 0. Hence, Equation () reduces to ∇P , () ∇ 2v = µ or. taking the x-component, G d 2vx () =− . 2 µ dy If the two plates are stationary then the solution that satisfies the no slip .
