Thread: Presure (Stokes's Equation)

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?

Originally Posted by maibs89

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?

--(equation 1)

Equation one should become --(equation 2)

Since you are given ----(equation 3)

For the 1st term of your equation 2, you plug in the derivative of the component of the equation 3, where .

For the 2st term of your equation 2, you plug in the derivative of the component of the equation 3,where .

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