Flag General Physics> stoke's law...
question mark

How can drive stokes law?

joun gee , 12 Years ago
Grade
anser 1 Answers
Aman Bansal

Last Activity: 12 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

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...