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

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Old   October 24, 2007, 20:46
Default Find dp/dn given dp/dx and dp/dy and geometry
  #1
CF
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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!
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Old   October 25, 2007, 00:23
Default Re: Find dp/dn given dp/dx and dp/dy and geometry
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nobody
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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)
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Old   October 25, 2007, 03:11
Default Re: Find dp/dn given dp/dx and dp/dy and geometry
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CF
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Oh I meant that theta is the angle the normal made with the horizontal. If that's the case, is the original eqn correct?
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Old   October 25, 2007, 03:24
Default Re: Find dp/dn given dp/dx and dp/dy and geometry
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Mayur
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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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