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2.5 Stokes Flow Past a Sphere

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2.5 Stokes Flow Past a Sphere 1 Notes on 1.63 Advanced Environmental Fluid Mechanics Instructor: C. C. Mei, 2001 [email protected], 1 617 253 2994 December 1, 2002 2-5Stokes.tex 2.5 Stokes flow past a sphere [Refs] Lamb: Hydrodynamics Acheson : Elementary Fluid Dynamics,p.223ff One of the fundamental results in low Reynolds hydrodynamics is the Stokes solution for steady flow past a small sphere. The apllicatiuon range widely form the determination of electron charges to the physics of aerosols. The continuity equation reads ~q =0 (2.5.1) ∇ · With inertia neglected, the approximate momentum equation is p 0= ∇ + ν 2~q (2.5.2) − ρ ∇ Physically, the presssure gradient drives the flow by overcoming viscous resistence, but does affect the fluid inertia significantly. Refering to Figure 2.5 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 ∂ =0, and ~q =(q (r, θ),q (r, θ), 0) ∂φ r θ Eq. (2.5.1) becomes 1 ∂ 1 ∂ (r2q )+ (q sin θ) = 0 (2.5.3) r2 ∂r r r ∂θ θ As in the case of rectangular coordinates, we define the stream function ψ to satisify the continuity equation (2.5.3) identically 1 ∂ψ 1 ∂ψ qr = ,qθ = (2.5.4) r2 sin θ ∂θ −r sin θ ∂r At infinity, the uniform velocity W along z axis can be decomposed into radial and polar components 1 ∂ψ 1 ∂ψ qr = W cos θ = ,qθ = W sin θ = ,r (2.5.5) r2 sin θ ∂θ − −r sin θ ∂r ∼∞ 2 z r q o y f x Figure 2.5.1: The spherical coordinates The corresponding stream function at infinity follows by integration W ψ = r2 sin2 θ,r (2.5.6) 2 ∼∞ Using the vector identity ( ~q)= ( ~q) 2~q (2.5.7) ∇ × ∇ × ∇ ∇ · −∇ and (2.5.1), we get 2~q = ( ~q)= ζ~ (2.5.8) ∇ −∇ × ∇ × −∇ × Taking the curl of (2.5.2) and using (2.5.8) we get ( ζ~) = 0 (2.5.9) ∇ × ∇ × After some straightforward algebra given in the Appendix, we can show that ψ~e ~q = φ (2.5.10) ∇ × Ãr sin θ ! and ψ~e ~e ∂2ψ sin θ ∂ 1 ∂ψ ζ~ = ~q = φ = φ + (2.5.11) ∇ × ∇ × ∇ × Ãr sin θ ! −r sin θ à ∂r2 r2 ∂θ Ãsin θ ∂θ !! Now from (2.5.9) ψ~e ( ~q)= φ =0 ∇ × ∇ × ∇ × ∇ × ∇ × "∇ × Ã∇ × r sin θ !# 3 hence, the momentum equation (2.5.9) becomes a scalar equation for ψ. ∂2 sin θ ∂ 1 ∂ 2 + ψ = 0 (2.5.12) Ã∂r2 r2 ∂θ Ãsin θ ∂θ!! The boundary conditions on the sphere are qr =0 qθ =0 on r = a (2.5.13) The boundary conditions at is ∞ W ψ r2 sin2 θ (2.5.14) → 2 Let us try a solution of the form: ψ(r, θ)=f(r)sin2 θ (2.5.15) then f is governed by the equi-dimensional differential equation: d2 2 2 f = 0 (2.5.16) "dr2 − r2 # whose solutions are of the form f(r) rn,Itiseasytoverifythatn = 1, 1, 2, 4sothat ∝ − A f(r)= + Br + Cr2 + Dr4 r or A ψ =sin2 θ + Br + Cr2 + Dr4 ∙ r ¸ To satisfy (2.5.14) we set D =0,C = W/2. To satisfy (2.5.13) we use (2.5.4) to get W A B A B qr =0= + + =0,qθ =0=W + =0 2 a3 a − a3 a Hence 1 3 A = Wa3,B= Wa 4 −4 Finally the stream function is W a3 3ar ψ = r2 + sin2 θ (2.5.17) 2 " 2r − 2 # 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 fluidare:(cf.(2.5.4): a3 3a qr = W cos θ 1+ (2.5.18) " 2r3 − 2r # a3 3a qθ = W sin θ 1 (2.5.19) − " − 4r3 − 4r # 4 2.5.1 Physical Deductions 1. Streamlines: With respect to the the equator along θ = π/2, cos θ and qr are odd while sin θ and qθ are even. Hence the streamlines (velocity vectors) are symmetric fore and aft. 2. Vorticity: ~ 1 ∂(rqθ) 1 ∂qr 3 sin θ ζ = ζφ~eφ ~eφ = Wa ~eφ Ãr ∂r − r ∂θ ! −2 r2 3. Pressure :Fromther-component of momentum equation ∂p µW a = cos θ(= µ ( ~q)) ∂r r3 − ∇ × ∇ × Integrating with respect to r from r to ,weget ∞ 3 µW a p = p cos θ (2.5.20) ∞ − 2 r3 4. Stresses and strains: 3 1 ∂qr 3a 3a err = = W cos θ 2 ∂r Ã2r2 − 2r4 ! On the sphere, r = a, err = 0 hence σrr =0and 3 µW τrr = p + σrr = p + cos θ (2.5.21) − − ∞ 2 a On the other hand 3 ∂ qθ 1 ∂qr 3 Wa erθ = r + = 4 sin θ ∂r µ r ¶ r ∂θ −2 r Hence at r = a: 3 µW τrθ = σrθ = µerθ = sin θ (2.5.22) −2 a Theresultantstressonthesphereisparalleltothez axis. 3 µW Σz = τrr cos θ τrθ sin θ = p cos θ + − − ∞ 2 a The constant part exerts a net drag in z direction 2π π 3 µW 2 D = adφ dθ sin θΣz == 4πa =6πµW a (2.5.23) Zo Zo 2 a This is the celebrated Stokes formula. Adragcoefficient can be defined as D 6πµW a 24 24 CD = 1 2 2 = 1 2 2 = ρW (2a) = (2.5.24) ρW πa ρW πa Red 2 2 µ 5 5. Fall velocity of a particle through a fluid. Equating the drag and the buoyant weight of the eparticle 4π 3 6πµWoa = a (ρs ρf )g 3 − hence 2 a2 ∆ρ a2 ∆ρ Wo = g = 217.8 9 à ν ρf ! à ν ρf ! in cgs units. For a sand grain in water, ∆ρ 2.5 1 2 2 = − =1.5, ν =10− cm /s ρf 1 2 Wo =32, 670 a cm/s (2.5.25) To have some quantitative ideas, letusconsidertwosandoftwosizes: 2 4 a =10− cm = 10− m: Wo =3.27cm/s; 3 5 a =10− cm = 10− =10µm, Wo =0.0327cm/s =117cm/hr For a water droplet in air, ∆ρ 1 3 2 = 3 =10, ν =0.15 cm /sec ρf 10− then (217.8)103 W = a2 (2.5.26) o 0.15 3 in cgs units. If a =10− cm = 10µm, then Wo =1.452 cm/sec. Details of derivation Details of (2.5.10). ~er ~eθ r sin θ~eφ ψ 1 ∂ ∂ ∂ ~eφ = ¯ ∂r ∂θ ∂φ ¯ ∇ × Ãr sin θ ! r2 sin θ ¯ ¯ ¯ 00 ψ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 ∂ψ 1 ∂ψ = ~er ~eθ Ãr2 sin θ ∂θ ! − Ãr sin θ ∂r ! 6 Details of (2.5.11). ψ~e φ = ~q ∇ × ∇ × r sin θ ∇ × ~er r~eθ r sin θ~eφ 1 ∂ ∂ ∂ = ¯ ∂r ∂θ ∂φ ¯ r2 sin θ ¯ ¯ ¯ 1 ∂ψ 1 ∂ψ 0 ¯ ¯ r2 sin θ ∂θ sin− θ ∂r ¯ ¯ ¯ ~e ¯∂2φ sin θ ∂ 1 ∂ψ ¯ = θ ¯ + ¯ r sin θ " ∂r2 r2 ∂θ Ãsin θ ∂θ !#.
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