2.5 Stokes flow past a sphere

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1 Lecture Notes on Fluid Dynamics.63J/.J) by Chiang C. Mei, MIT 007 Spring -5Stokes.tex.5 Stokes flow past a sphere Refs] Lamb: Hydrodynamics Acheson : Elementary Fluid Dynamics, p. 3 ff One of the fundamental results in low Reynolds-number hydrodynamics is the Stokes solution for steady flow past a small sphere. Applications range widely from the determination of electron charges to the physics of aerosols. The continuity equation reads q = 0.5.) With inertia neglected, the approximate momentum equation is 0 = p ρ + ν q.5.) Physically, the presssure gradient drives the flow by overcoming viscous resistence, but does affect the fluid inertia significantly. Refering to Figure.5 for the spherical coordinate system r, θ, φ). Let the ambient velocity be upward and along the polar z) axis: u,v,w) = 0, 0,W). Axial symmetry demands φ = 0, and q = q rr,θ),q θ r,θ), 0) Using a known formula for the divergence in spherical polar coordinates, Eq..5.) becomes r r r q r ) + θ q θ sin θ) = 0.5.3) An equivalent and physically more revealing way is to write r r q ) + θ rq θ sin θ) = 0.5.4) As in the case of rectangular coordinates, we define the stream function ψ to satisify the continuity equation.5.4) identically q r = ψ r sin θ θ, q θ = ψ r.5.5)

2 z o r y x Figure.5.: The spherical coordinates At infinity, the uniform velocity W along z axis can be decomposed into radial and polar components q r = W cos θ = ψ r sin θ θ, q θ = W sin θ = ψ r, r.5.6) The corresponding stream function at infinity follows by integration Using the vector identity ψ = W r sin θ, r.5.7) q) = q) q.5.8) and.5.), we get q = q) = ζ.5.9) Taking the curl of.5.) and using.5.9) we get ζ) = 0.5.0) After some straightforward algebra given in the Appendix, we can show that ) ψ eφ q =.5.) and ) ψ eφ ζ = q = = e φ ψ r + sin θ r θ sin θ )) ψ θ.5.)

3 3 Now from.5.0) q) = ψ e )] φ = 0 hence, the momentum equation.5.0) becomes a scalar equation for ψ. r + sin θ r The boundary conditions on the sphere are The boundary conditions at is Let us try a solution of the form: θ sin θ )) ψ = 0.5.3) θ q r = 0 q θ = 0 on r = a.5.4) ψ W r sin θ.5.5) ψr,θ) = fr) sin θ.5.6) then f is governed by the equi-dimensional differential equation: d ] f = 0.5.7) dr r whose solutions are of the form fr) r n, It is easy to verify that n =,,, 4 so that fr) = A r + Br + Cr + Dr 4 or ] A ψ = sin θ r + Br + Cr + Dr 4 To satisfy.5.5) we set D = 0,C = W/. To satisfy.5.4) we use.5.5) to get q r = 0 = W + A a 3 + B a = 0, q θ = 0 = W A a 3 + B a = 0 Hence A = 4 Wa3, B = 3 4 Wa Finally the stream function is ψ = W r + a3 r 3ar ] sin θ.5.8)

4 4 Inside the parentheses, the first term corresponds to the uniform flow, and the second term to the doublet; together they represent an inviscid flow past a sphere. The third term is called the Stokeslet, representing the viscous correction. The velocity components in the fluid are: cf..5.5) :.5. Physical Deductions q r q θ = W cos θ + a3 r 3a ] 3 r = W sin θ a3 4r 3a ] 3 4r.5.9).5.0). Streamlines: With respect to the the equator along θ = π/, cos θ and q r are odd while sin θ and q θ are even. Hence the streamlines velocity vectors) are symmetric fore and aft.. Vorticity: ζ = ζ φ e φ = rq θ ) r r r ) q r e φ = 3 θ Wasinθ r e φ 3. Pressure : From the r-component of momentum equation p r = µwa cos θ= µ q)) r 3 Integrating with respect to r from r to, we get 4. Stresses and strains: p = p 3 µwa cos θ.5.) r 3 e rr = q ) r 3a r = W cos θ r 3a3 r 4 On the sphere, r = a, e rr = 0 hence τ rr = 0 and On the other hand Hence at r = a: σ rr = p + τ rr = p + 3 µw a e rθ = r r ) qθ + q r r r θ = 3 Wa 3 sin θ r 4 σ rθ = τ rθ = µe rθ = 3 µw a cos θ.5.) sin θ.5.3)

5 5 The resultant stress on the sphere is parallel to the z axis. Σ z = σ rr cos θ τ rθ sin θ = p cos θ + 3 µw a The constant part exerts a net drag in z direction D = π This is the celebrated Stokes formula. A drag coefficient can be defined as o π adφ dθ sin θσ z == 3 µw o a 4πa = 6πµWa.5.4) C D = D ρw πa = 6πµWa ρw πa = 4 ρwa) µ = 4 Re d.5.5) 5. Fall velocity of a particle through a fluid. Equating the drag and the buoyant weight of the eparticle hence 6πµW o a = 4π 3 a3 ρ s ρ f )g W o = 9 g a in cgs units. For a sand grain in water, ν ρ =.5 ρ f ) ρ a = 7.8 ρ f ν ) ρ ρ f =.5, ν = 0 cm /s W o = 3, 670 a cm/s.5.6) To have some quantitative ideas, let us consider two sand of two sizes : a = 0 cm = 0 4 m : For a water droplet in air, then a = 0 3 cm = 0 5 = 0µm, W o = 3.7cm/s; W o = 0.037cm/s = 7cm/hr ρ ρ f = 0 3 = 03, ν = 0.5 cm /sec W o = 7.8) a.5.7) in cgs units. If a = 0 3 cm = 0µm, then W o =.45 cm/sec.

6 6 Details of derivation Details of.5.). Details of.5.). q = ) ψ e φ = = e r r sin θ q = = r sin θ = e φ r sin θ ) ψ e θ θ ψ e ) φ r ψ r sin θ θ ψ r + sin θ r e r e θ e φ r θ φ 0 0 ψ ) ψ r θ φ ψ 0 r )] e r r e θ e φ sin θ θ sin θ ψ θ

2.6 Oseen s improvement for slow flow past a cylinder

2.6 Oseen s improvement for slow flow past a cylinder Lecture Notes on Fluid Dynamics.63J/.J) by Chiang C. Mei, MIT -6oseen.tex [ef] Lamb : Hydrodynamics.6 Oseen s improvement for slow flow past a cylinder.6. Oseen s criticism of Stokes approximation Is Stokes