# Find dp/dn given dp/dx and dp/dy and geometry

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 October 24, 2007, 20:46 Find dp/dn given dp/dx and dp/dy and geometry #1 CF Guest   Posts: n/a Hi, I'm trying to get dp/dn, which is the derivative of pressure normal to a wall (or grid line) in 2d. Supposed I know dp/dx,dp/dy at that location. In that case, is dp/dn=(dp/dx)*cos(theta) + (dp/dy)*sin(thata), where theta is the angle made with the horizontal ? Thanks!

 October 25, 2007, 00:23 Re: Find dp/dn given dp/dx and dp/dy and geometry #2 nobody Guest   Posts: n/a You have to multiply the pressure gradient (dp/dx,dp/dy) with the normal vector and not the tangent vector of the line in order to get the normal derivative dp/dn. If the tangent vector of the line is (cos(theta),sin(theta)) you have to calculate dp/dn = - (dp/dx)*sin(theta) + (dp/dy)*cos(theta)

 October 25, 2007, 03:11 Re: Find dp/dn given dp/dx and dp/dy and geometry #3 CF Guest   Posts: n/a Oh I meant that theta is the angle the normal made with the horizontal. If that's the case, is the original eqn correct?

 October 25, 2007, 03:24 Re: Find dp/dn given dp/dx and dp/dy and geometry #4 Mayur Guest   Posts: n/a ur original eqn was correct... dp/dn=(grad P).n=(dp/dx*i + dp/dy*j)(Nx*i+Ny*j) Nx=cos theta and Ny is sin theta in ur case... expand and u get ur eqn.

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