Presure (Stokes's Equation)

maibs89 · #1 $Old$ October 11th, 2009, 08:50 PM

grad (p) = μ(grad^2 )u

This is the Stokes' Equation/Momentum Equation
How would I find the pressure? How would I integrate this equation to do so?

u is given as
u= { U[1- (3a/2r) + (a^3)/(2r^3)]cosθ, -U[1 - (3a/4r) - (a^3)/(4r^3)]sinθ }

which was previously worked out in spherical coordinates.

How would I begin?

novice · #2 $Old$ October 18th, 2009, 10:07 AM

Quote:

Originally Posted by maibs89 $View Post$

grad (p) = μ(grad^2 )u

This is the Stokes' Equation/Momentum Equation
How would I find the pressure? How would I integrate this equation to do so?

u is given as
u= { U[1- (3a/2r) + (a^3)/(2r^3)]cosθ, -U[1 - (3a/4r) - (a^3)/(4r^3)]sinθ }

which was previously worked out in spherical coordinates.

How would I begin?

$\bigtriangledown\cdot p=\mu\bigtriangledown^{2}U=\mu(\frac{\partial^2U}{\partial^2x}\overrightarrow i+\frac{\partial^2U}{\partial^2y}\overrightarrow j)$ --(equation 1)

Equation one should become $p=\mu \frac{\partial U}{\partial x} \overrightarrow i +\mu \frac{\partial U}{\partial x}\overrightarrow j$ --(equation 2)

Since you are given $U=\mu \cos \theta(1-\frac{3a}{2r}+\frac{a^{3}}{2r^{3}}) \overrightarrow i+ \mu \sin \theta (1-\frac{3a}{4r}-\frac{a^{3}}{4r^{3}}) \overrightarrow j$ ----(equation 3)

For the 1st term of your equation 2, you plug in the derivative of the $\overrightarrow i$ component of the equation 3, where $r=\sqrt{x^2+y^2}$ .

For the 2st term of your equation 2, you plug in the derivative of the $\overrightarrow j$ component of the equation 3,where $r=\sqrt{x^2+y^2}$ .

The rest is algebraic work. For derivative with respect to x, hold y as constant, and for derivative with respect to y, hold x as constant