斯托克斯力.pdf

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 + ν∇2 ~q ρ (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 ∂ = 0, and ~q = (qr (r, θ), qθ (r, θ), 0) ∂φ Eq. (2.5.1) becomes 1 ∂ 2 1 ∂ (2.5.3) (r qr ) + (qθ sin θ) = 0 2 r ∂r r ∂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 , q = − (2.5.4) θ 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 qr = qr = W cos θ = 1 ∂ψ r2 sin θ ∂θ , qθ = −W sin θ = − 1 ∂ψ , r sin θ ∂r r∼∞ (2.5.5) 2 z q r o f y x Figure 2.5.1: The spherical coordinates The corresponding stream function at infinity follows by integration W 2 2 r sin θ, 2 ψ= r∼∞ (2.5.6) Using the vector identity ∇ × (∇ × ~q) = ∇(∇ · ~q) − ∇2 ~q (2.5.7) ∇2 ~q = −∇ × (∇ × ~q) = −∇ × ζ~ (2.5.8) and (2.5.1), we get Taking the curl of (2.5.2) and using (2.5.8) we get ~ =0 ∇ × (∇ × ζ) (2.5.9) After some straightforward algebra given in the Appendix, we can show that Ã ψ~eφ ~q = ∇ × r sin θ ! (2.5.10) and Ã ψ~eφ ζ~ = ∇ × ~q = ∇ × ∇ × r sin θ ! ~eφ =− r sin θ Ã ∂ 2 ψ sin θ ∂ + 2 ∂r2 r ∂θ Ã 1 ∂ψ sin θ ∂θ Now from (2.5.9) " Ã ψ~eφ ∇ × ∇ × (∇ × ~q) = ∇ × ∇ × ∇ × ∇ × r sin θ !# =0 !! (2.5.11) 3 hence, the momentum equation (2.5.9) becomes a scalar equation for ψ. Ã sin θ ∂ ∂2 + ∂r2 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 The boundary conditions at ∞ is ψ→ (2.5.13) W 2 2 r sin θ 2 (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 2 − dr2 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) = or A + Br + Cr2 + Dr4 r ∙ ¸ A + Br + Cr2 + Dr4 ψ = sin θ 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 2 qr = 0 = W A B + 3 + = 0, 2 a a qθ = 0 = W − 1 A = W a3 , 4 3 B = − Wa 4 Hence Finally the stream function is " # W 2 a3 3ar ψ= r + − sin2 θ 2 2r 2 A B + =0 a3 a (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 qθ " # a3 3a = W cos θ 1 + 3 − 2r 2r " # a3 3a = −W sin θ 1 − 3 − 4r 4r (2.5.18) (2.5.19) 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 ! 3 1 ∂(rqθ ) 1 ∂qr sin θ ~eφ = − W a 2 ~eφ − r ∂r r ∂θ 2 r 3. Pressure : From the r-component of momentum equation µW a ∂p = 3 cos θ(= −µ∇ × (∇ × ~q)) ∂r r Integrating with respect to r from r to ∞, we get p = p∞ − 4. Stresses and strains: 3 µW a cos θ 2 r3 (2.5.20) Ã ! τrr = −p + σrr = −p∞ + 3 µW cos θ 2 a 1 ∂qr 3a 3a3 err = = W cos θ − 2 ∂r 2r2 2r4 On the sphere, r = a, err = 0 hence σrr = 0 and (2.5.21) On the other hand ∂ erθ = r ∂r µ ¶ qθ 1 ∂qr 3 W a3 =− + sin θ r r ∂θ 2 r4 Hence at r = a: 3 µW sin θ 2 a The resultant stress on the sphere is parallel to the z axis. τrθ = σrθ = µerθ = − Σz = τrr cos θ − τrθ sin θ = −p∞ cos θ + (2.5.22) 3 µW 2 a The constant part exerts a net drag in z direction D= Z 2π adφ o Z π o dθ sin θΣz == 3 µW 4πa2 = 6πµW a 2 a (2.5.23) This is the celebrated Stokes formula. A drag coeﬃcient can be defined as D 6πµW a 24 24 = 1 = ρW (2a) = 2 2 2 2 Red ρW πa ρW πa 2 2 µ CD = 1 (2.5.24) 5 5. Fall velocity of a particle through a fluid. Equating the drag and the buoyant weight of the eparticle 4π 3 a (ρs − ρf )g 6πµWo a = 3 hence Ã 2 a2 ∆ρ Wo = g 9 ν ρf ! Ã a2 ∆ρ = 217.8 ν ρf ! in cgs units. For a sand grain in water, ∆ρ 2.5 − 1 = 1.5, = ρf 1 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; −3 −5 a = 10 cm = 10 = 10µm, Wo = 0.0327cm/s = 117cm/hr For a water droplet in air, ∆ρ 1 = −3 = 103 , ρf 10 ν = 0.15 cm2 /sec then Wo = (217.8)103 2 a 0.15 (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 ~eφ r sin θ = ~er Ã ! ¯ ¯ ~ e 1 ¯¯ ∂r ¯ = 2 r sin θ ¯¯ ∂r 0 1 ∂ψ 2 r sin θ ∂θ ! − ~eθ Ã ¯ ~eθ r sin θ~eφ ¯¯ ∂ ∂θ ∂ ∂φ 0 ψ 1 ∂ψ r sin θ ∂r ! ¯ ¯ ¯ ¯ 6 Details of (2.5.11). ψ~eφ = ∇ × ~q r sin¯θ ¯ ¯ ¯ ~ e r~ e r sin θ~ e r θ φ ¯ 1 ¯¯ ¯ ∂ ∂ ∂ ¯ ¯ = 2 ∂r ∂θ ∂φ ¯ r sin θ ¯¯ 1 ∂ψ −1 ∂ψ ¯ 0 r2 sin θ ∂θ sin θ ∂r ∇×∇× " ~eθ ∂ 2 φ sin θ ∂ = + 2 r sin θ ∂r2 r ∂θ Ã 1 ∂ψ sin θ ∂θ !#