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Math Help - Presure (Stokes's Equation)

  1. #1
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    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?
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  2. #2
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    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
    Last edited by novice; October 18th 2009 at 12:32 PM.
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