《力学》课程教学资源（参考资料）斯托克斯力 Stokes flow past a sphere

1 Notes on 1.63 Advanced Environmental Fluid Mechanics Instructor:C.C.Mei,2001 ccmei@mit.edu,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.223 ff 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 7.q=0 (2.5.1) With inertia neglected,the approximate momentum equation is 0=-2+v2g (2.5.2) 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,0,o).Let the ambient velocity be upward and along the polar (z)axis:(u,v,w)=(0,0,W).Axial symmetry demands 86=0,andg=(g(,9),9o(,9),0) Eq.(2.5.1)becomes 182gr)+ 18 r2Or ra0 (gosine)=0 (2.5.3) As in the case of rectangular coordinates,we define the stream function to satisify the continuity equation (2.5.3)identically 10b 1∂b 4r=r2sin0∂9' 90= (2.5.4) rsin0 Or At infinity,the uniform velocity W along z axis can be decomposed into radial and polar components 10b 18b 4-Wcos0=2sin 00 e--W sin0-- rsin98r,r~∞ (2.5.5)

1 Notes on 1.63 Advanced Environmental Fluid Mechanics Instructor: C. C. Mei, 2001 ccmei@mit.edu, 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. 223 ff 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 0 = −∇p ρ + ν∇2 ~q (2.5.2) 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 ∂ ∂φ = 0, and ~q = (qr(r, θ), qθ(r, θ), 0) Eq. (2.5.1) becomes 1 r2 ∂ ∂r (r2 qr) + 1 r ∂ ∂θ (qθ sin θ) = 0 (2.5.3) As in the case of rectangular coordinates, we define the stream function ψ to satisify the continuity equation (2.5.3) identically qr = 1 r2 sin θ ∂ψ ∂θ , qθ = − 1 r sin θ ∂ψ ∂r (2.5.4) At infinity, the uniform velocity W along z axis can be decomposed into radial and polar components qr = W cos θ = 1 r2 sin θ ∂ψ ∂θ , qθ = −W sin θ = − 1 r sin θ ∂ψ ∂r , r ∼ ∞ (2.5.5)

2 Q Figure 2.5.1:The spherical coordinates The corresponding stream function at infinity follows by integration 2 in0. 2.5.6) Using the vector identity 7×(V×=7(7·)-7d (2.5.7) and (2.5.1),we get g=-V×(7×=-7×6 (2.5.8) Taking the curl of(2.5.2)and using (2.5.8)we get V×(7×G)=0 (2.5.9) After some straightforward algebra given in the Appendix,we can show that g-7× (2.5.10) r sin 0 and (=7×g=V×V× 山eo 包过 82b sin8 1 aψ X (2.5.11) r sin 0 rsin r288 sin0 80 Now from (2.5.9) vx×v×=vxxx(x】=0

2 z y x r o
Figure 2.5.1: The spherical coordinates The corresponding stream function at infinity follows by integration ψ = W 2 r2 sin2 θ, r ∼ ∞ (2.5.6) 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 ~q = ∇ × Ã ψ~eφ r sin θ ! (2.5.10) and ~ζ = ∇ × ~q = ∇ × ∇ × Ã ψ~eφ r sin θ ! = − ~eφ r sin θ Ã ∂2ψ ∂r2 + sin θ r2 ∂ ∂θ Ã 1 sin θ ∂ψ ∂θ !! (2.5.11) Now from (2.5.9) ∇ × ∇ × (∇ × ~q) = ∇ × ∇ × " ∇ × Ã ∇ × ψ~eφ r sin θ !# = 0

3 hence,the momentum equation(2.5.9)becomes a scalar equation for 2 sin 1 2 ψ=0 (2.5.12) 8r2 r280 sin00 The boundary conditions on the sphere are gr=0 go=0 on r=a (2.5.13) The boundary conditions at oo is ψ→ 22n20 (2.5.14) Let us try a solution of the form: (r,)=f(r）sin20 (2.5.15) then f is governed by the equi-dimensional differential equation: d 272 dr2-2 f=0 (2.5.16) whose solutions are of the form f(r)o r",It is easy to verify that n=-1,1,2,4 so that )+Br+Cr+Dr or =m29[2+Br+c2+D 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 0三，+元女。=09如=0=w-A+B=0 a Hence A=-Wa, 1 Finally the stream function is W「 a3 3ar = sin20 2 2+2r 2 (2.5.17) Inside the parentheses,the first term corresponds to the uniform flow,and the second term to the doublet;together they represent an inviscid fow past a sphere.The third term is called the Stokeslet,representing the viscous correction. The velocity components in the fluid are:(cf.(2.5.4): Wcos 1+ 3a1 gr i 23-27 (2.5.18) ge =-Wsin0 a3 3a 1- 43 (2.5.19) 4r

3 hence, the momentum equation (2.5.9) becomes a scalar equation for ψ. Ã ∂2 ∂r2 + sin θ r2 ∂ ∂θ Ã 1 sin θ ∂ ∂θ!!2 ψ = 0 (2.5.12) The boundary conditions on the sphere are qr = 0 qθ = 0 on r = a (2.5.13) The boundary conditions at ∞ is ψ → W 2 r2 sin2 θ (2.5.14) 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 dr2 − 2 r2 #2 f = 0 (2.5.16) whose solutions are of the form f(r) ∝ rn, It is easy to verify that n = −1, 1, 2, 4 so that f(r) = A r + Br + Cr2 + Dr4 or ψ = sin2 θ ∙A r + Br + Cr2 + Dr4 ¸ To satisfy (2.5.14) we set D = 0, C = W/2. To satisfy (2.5.13) we use (2.5.4) to get qr =0= W 2 + A a3 + B a = 0, qθ =0= W − A a3 + B a = 0 Hence A = 1 4 W a3 , B = −3 4 W a Finally the stream function is ψ = W 2 " r2 + a3 2r − 3ar 2 # sin2 θ (2.5.17) 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. (2.5.4) : qr = W cos θ " 1 + a3 2r3 − 3a 2r # (2.5.18) qθ = −W sin θ " 1 − a3 4r3 − 3a 4r # (2.5.19)

4 2.5.1 Physical Deductions 1.Streamlines:With respect to the the equator along 0=/2,cos 0 and gr are odd while sin 0 and go are even.Hence the streamlines(velocity vectors)are symmetric fore and aft. 2.Vorticity: 5-SoEo 18(rge)18gr sin r08 >Wa- 3.Pressure From the r-component of momentum equation Op uWa Or cos(=-V×(V×) Integrating with respect to r from r to oo,we get 3 uWa p=p∞一 23c0s9 (2.5.20) 4.Stresses and strains: 1 8gr =W cos0 3a3 Or 2r2 2r4 On the sphere,r=a,err =0 hence orr=0 and Trr =-p+Orr=-Poo+ 3uW cos0 (2.5.21)） 2a On the other hand 9 10g,3Wa -sin0 r00 2r4 Hence at r=a: 3uW T0=00=er0=-2a sin (2.5.22) The resultant stress on the sphere is parallel to the z axis. ∑：=Trr COS9-Tr8sin0=-P Cos0+ 3uw 2a The constant part exerts a net drag in z direction D= de sin :==a -4ra2=6πμWa (2.5.23) This is the celebrated Stokes formula. A drag coefficient can be defined as D 6πuWa 24 24 Co-pWiza=ipWira =eW(2a)= (2.5.24) Rea

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: ~ζ = ζφ~eφ Ã 1 r ∂(rqθ) ∂r − 1 r ∂qr ∂θ ! ~eφ = −3 2 W asin θ r2 ~eφ 3. Pressure : From the r-component of momentum equation ∂p ∂r = µW a r3 cos θ(= −µ∇ × (∇ × ~q)) Integrating with respect to r from r to ∞, we get p = p∞ − 3 2 µW a r3 cos θ (2.5.20) 4. Stresses and strains: 1 2 err = ∂qr ∂r = W cos θ Ã 3a 2r2 − 3a3 2r4 ! On the sphere, r = a, err = 0 hence σrr = 0 and τrr = −p + σrr = −p∞ + 3 2 µW a cos θ (2.5.21) On the other hand erθ = r ∂ ∂r µqθ r ¶ + 1 r ∂qr ∂θ = −3 2 W a3 r4 sin θ Hence at r = a: τrθ = σrθ = µerθ = −3 2 µW a sin θ (2.5.22) The resultant stress on the sphere is parallel to the z axis. Σz = τrr cos θ − τrθ sin θ = −p∞ cos θ + 3 2 µW a The constant part exerts a net drag in z direction D = Z 2π o adφ Z π o dθ sin θΣz == 3 2 µW a 4πa2 = 6πµW a (2.5.23) This is the celebrated Stokes formula. A drag coefficient can be defined as CD = D 1 2 ρW2πa2 = 6πµW a 1 2 ρW2πa2 = 24 ρW(2a) µ = 24 Red (2.5.24)

5 5.Fall velocity of a particle through a fluid.Equating the drag and the buoyant weight of the eparticle 6iwa=行en.-pn9 hence 2a2△ 217.8 "Sp v Pf in cgs units.For a sand grain in water, =1=1.5,v=10-2cm2/5 △2=2.5-1 W。=32,670a2cm/s (2.5.25) To have some quantitative ideas,let us consider two sand of two sizes a=10-2cm=10-4m:W。=3.27cm/s; a=10-3cm=10-5=10um,W。=0.0327cm/s=117cm/hr For a water droplet in air, 2=1 10-=103,v=0.15cm2/sec then -0 (2.5.26) in cgs units.If a 10-3 cm 10um,then Wo =1.452 cm/sec. Details of derivation Details of(2.5.10). 1 eo rsin beo r2 sin0 品 品 0 0 1. ∂b 1 =e r2 sin 0 00 rsin0 Or

5 5. Fall velocity of a particle through a fluid. Equating the drag and the buoyant weight of the eparticle 6πµWoa = 4π 3 a3 (ρs − ρf )g hence Wo = 2 9 g à a2 ν ∆ρ ρf ! = 217.8 à a2 ν ∆ρ ρf ! in cgs units. For a sand grain in water, ∆ρ ρf = 2.5 − 1 1 = 1.5, ν = 10−2 cm2 /s Wo = 32, 670 a2 cm/s (2.5.25) To have some quantitative ideas, let us consider two sand of two sizes : a = 10−2 cm = 10−4 m : Wo = 3.27cm/s; a = 10−3 cm = 10−5 = 10µm, Wo = 0.0327cm/s = 117cm/hr For a water droplet in air, ∆ρ ρf = 1 10−3 = 103 , ν = 0.15 cm2 /sec then Wo = (217.8)103 0.15 a2 (2.5.26) in cgs units. If a = 10−3 cm = 10µm, then Wo = 1.452 cm/sec. Details of derivation Details of (2.5.10). ∇ × Ã ψ r sin θ ~eφ ! = 1 r2 sin θ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ~er ~eθ r sin θ~eφ ∂ ∂r ∂ ∂θ ∂ ∂φ 0 0 ψ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ~er à 1 r2 sin θ ∂ψ ∂θ ! − ~eθ à 1 r sin θ ∂ψ ∂r !

6 Details of(2.5.11). Vxvx wea =V×q rsin 1 rsin beo 8 8 8 r2 sin0 86 地 -10地 T2 sin 0 80 sin0 Or 0 、 ee 「ao sin 8 10 rsin Or2 r2 sin00

6 Details of (2.5.11). ∇ × ∇ × ψ~eφ r sin θ = ∇ × ~q = 1 r2 sin θ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ~er r~eθ r sin θ~eφ ∂ ∂r ∂ ∂θ ∂ ∂φ 1 r2 sin θ ∂ψ ∂θ −1 sin θ ∂ψ ∂r 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ~eθ r sin θ " ∂2φ ∂r2 + sin θ r2 ∂ ∂θ Ã 1 sin θ ∂ψ ∂θ !#