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How can drive stokes law?

How can drive stokes law?

Grade:

1 Answers

Aman Bansal
592 Points
9 years ago

Dear Student,

Flow around a sphere

For the case of a sphere in a uniform far field flow, it is advantageous to use a cylindrical coordinate system ( r , φ , z ). The z–axis is through the centre of the sphere and aligned with the mean flow direction, while r is the radius as measured perpendicular to the z–axis. The origin is at the sphere centre. Because the flow is axisymmetric around the z–axis, it is independent of the azimuth φ.

In this cylindrical coordinate system, the incompressible flow can be described with a Stokes stream function ψ, depending on r and z:

   v = -\frac{1}{r}\frac{\partial\psi}{\partial z},   \qquad   w = \frac{1}{r}\frac{\partial\psi}{\partial r},

with v and w the flow velocity components in the r and z direction, respectively. The azimuthal velocity component in the φ–direction is equal to zero, in this axisymmetric case. The volume flux, through a tube bounded by a surface of some constant value ψ, is equal to 2π ψ and is constant.

For this case of an axisymmetric flow, the only non-zero of the vorticity vector ω is the azimuthal φ–component ωφ

   \omega_\varphi = \frac{\partial v}{\partial z} - \frac{\partial w}{\partial r}     = - \frac{\partial}{\partial r} \left( \frac{1}{r}\frac{\partial\psi}{\partial r} \right) - \frac{1}{r}\, \frac{\partial^2\psi}{\partial z^2}.

The Laplace operator, applied to the vorticity ωφ, becomes in this cylindrical coordinate system with axisymmetry:

\nabla^2 \omega_\varphi = \frac{1}{r}\frac{\partial}{\partial r}\left( r\, \frac{\partial\omega_\varphi}{\partial r} \right) + \frac{\partial^2 \omega_\varphi}{\partial z^2} - \frac{\omega_\varphi}{r^{2}} = 0.

From the previous two equations, and with the appropriate boundary conditions, for a far-field uniform-flow velocity V in the z–direction and a sphere of radius R, the solution is found to be

   \psi = - \frac{1}{2}\, V\, r^2\, \left[      1      - \frac{3}{2} \frac{R}{\sqrt{r^2+z^2}}      + \frac{1}{2} \left( \frac{R}{\sqrt{r^2+z^2}} \right)^3\;   \right].

The viscous force per unit area σ, exerted by the flow on the surface on the sphere, is in the z–direction everywhere. More strikingly, it has also the same value everywhere on the sphere:

\boldsymbol{\sigma} = \frac{3\, \eta\, V}{2\, R}\, \mathbf{e}_z

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Thanks

Aman Bansal

Askiitian Expert

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