DERIVATION OF STOKE’S EQUATION IN VISCOUS FLUID
I mean that the equation is,
F = 6πƞrv; where, ƞ is the coefficient of viscosity, r is the radius of sphere which flows through the fluid, v is the terminal velocity and F is the force acted on the fluid for the viscous drag.
I am sorry to convey that I couldn’t find the proper derivation of Stoke’s equation. But here I wish to say about the magical power of our mind, for past one month I searched for the derivation of Stoke’s equation, but I couldn’t find it anywhere at last I could derive it by myself. But it is not correct
We have the Newton’s equation for viscosity,
F = ƞA (dv/dx) --------------------------- (1)
‘A’ is the area of fluid displaced and (dv/dx) velocity gradient or it is written as “div v” here (dv/dr), since the quantities varies radial.
In this figure we can see that only half of the sphere is push the fluid, therefore A = 2πr2 because, this figure 1 misunderstands you actually the fluid motion is considering from bottom to half of the sphere, but go on half to top the flow will be turbulent due to downward motion of sphere, so we cannot use Newton’s equation on the another half face.We have the equation for continuity
(dβ/dt) + V (grad β) + β (div v) = 0; OR
Where β is the density of fluid, here β is independent of time because fluid undergoes terminal velocity that is no force for compressing the liquid. Therefore the equation be
(dv/dr) = - {v/β}(dβ/dr)
Therefore equation (1) becomes
F = -ƞ2πr2{v/β}(dβ/dr)-----------------------------------------(2)
Let ‘M’ be the mass flows out by the sphere, And using Archimedes principle, the volume displaced is equal to the volume of sphere , So β = (3M/4πr3)
And (dβ/dr) = -(9M/4πr4)
These two equations are substitute in equation (2) we get the Stoke’s equation
| I.e. F = 6πƞrv |
