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

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!

Re: Find dp/dn given dp/dx and dp/dy and geometry
 

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)

Re: Find dp/dn given dp/dx and dp/dy and geometry
 

Oh I meant that theta is the angle the normal made with the horizontal. If that's the case, is the original eqn correct?

Re: Find dp/dn given dp/dx and dp/dy and geometry
 

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.