# 3D velocity profile for Rectangular inlet

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 May 4, 2014, 12:57 3D velocity profile for Rectangular inlet #1 Senior Member   Astio Lamar Join Date: May 2012 Location: Pipe Posts: 170 Rep Power: 6 Hello every body. I want to use a 3D parabolic velocity profile in an inlet on ZY plan. the velocity will be normal to this surface (- X). I find two equation for X velocity which is vary by both Z and Y. x = -1 + 2*z - 3*z*z x = -1 + 2*y - 3*y*y For 2D, it is not difficult, but I don't know how to do it in 3D: I am appreciate, if anybody can help me. Thanks. Last edited by asal; May 6, 2014 at 11:29.

 June 17, 2014, 05:04 #2 New Member   Join Date: Oct 2012 Posts: 4 Rep Power: 6 have i fixed the problem as i am trying the same thing. I have the profile for xy plane and how do i write it for 3d? i would appreciate your help

 June 17, 2014, 05:39 #3 Senior Member   Join Date: Nov 2013 Posts: 1,038 Rep Power: 14 You first need one equation for the x-velocity. Not two, as in the opening post, that is wrong. u_x = f(y,z). Once you have that, you can implement it in exactly the same way as for the 2d case.

 September 18, 2014, 10:49 #4 New Member   bhatti Join Date: Aug 2013 Posts: 2 Rep Power: 0 Did you get the right equation for 3d case. could you please share that equation.

 September 19, 2014, 04:50 #5 Senior Member   Join Date: Nov 2013 Posts: 1,038 Rep Power: 14 The equation depends on the shape. The easiest is circular, that one you can look up in most CFD books. I think all others shapes are difficult.

 September 19, 2014, 05:46 #6 Senior Member   Astio Lamar Join Date: May 2012 Location: Pipe Posts: 170 Rep Power: 6 Hello! I have found a solution for my case and it worked well for me. The shape is not matter, if you can derive the equation in 2D. Let's assume that you want to impose a 3D velocity profile (normal to X direction) to a quadratic door. First you need to know how your velocity will be change in Z and Y direction. finally you will have two equation same as the following: x = -1 + 2*z - 3*z*z x = -1 + 2*y - 3*y*y then simple add these two equation such as the following: x = (-1 + 2*z - 3*z*z) + (-1 + 2*y - 3*y*y) finally you need to normalize your equation to got the right velocity at the center of the door. Hope it useful.

 September 21, 2014, 05:46 #7 Senior Member   Join Date: Nov 2013 Posts: 1,038 Rep Power: 14 Hi Asal, What you suggest is not correct. First of all, your equations are not really parabolic profiles, they are always negative, but let's assume that you chose the wrong coefficients, and write them as: x = 1-z*z x = 1-y*y The first equation is a parabolic profile that is zero at z=-1 and z=1. The second is a parabolic profile which is zero at y=-1 and y=1. So far so good. But what happens if you add them? x = 2 - z*z - y*y This is zero when z*z+y*y=2, in other words a circle with sphere sqrt(2). In the opening that you are simulating, that is only the 4 corners, all other points of this circle are outside your geometry. More specificly: at the boundary of your inlet, the velocity is generally speaking NOT zero. You can only have such a simple profile when the opening is circular. And then it is the solution that I just showed. When the opening is not circular, you don't have a simple parabolic profile.

 September 21, 2014, 13:56 #8 Senior Member   Astio Lamar Join Date: May 2012 Location: Pipe Posts: 170 Rep Power: 6 Hello! This is just an approximation, since I found no better solution. I have used this equation and got reasonable result. The profile will be parabolic, and the mismatch at the boundaries may have a negligible affect on the results. Although, if you have any better solution, we appreciate to hear it!!