|
[Sponsors] |
Find dp/dn given dp/dx and dp/dy and geometry |
|
LinkBack | Thread Tools | Search this Thread | Display Modes |
October 24, 2007, 20:46 |
Find dp/dn given dp/dx and dp/dy and geometry
|
#1 |
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 |
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 |
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 |
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. |
|
|
|