# Y+

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 November 8, 2006, 05:58 Y+ #1 Ralph Guest   Posts: n/a Sponsored Links Hi, Could anyone give me a definition of Y+ and why it has to be within a certain range to model turbulent flows? I'm pretty new to CFD and struggling with this concept. I know it's to do with the distance between the wall and the first node, and the log law of the wall, but I can't really find a good (or indeed understandable) explanation. Thanks a lot, Ralph

 November 8, 2006, 11:26 Re: Y+ #2 sach Guest   Posts: n/a u can get on net, or in books like fox and macdermott but its non dimentional no as a function of a cell center height and the distance from the start of fully deleloped boundary layer for k-e model range of y+ wil be 30 to 150for k-w sst it needs Y + = 1 to 5 normaly keep 1 mm height of the first prism layer from the boundary this will solve most of the cases

 November 8, 2006, 11:52 Re: Y+ #3 Paolo Lampitella Guest   Posts: n/a Hi Ralph, the question about y+ is more simple than could appear. The y+ is defined as: y+ = y*Ut/v where y is the distance from the wall, v is the cinematic viscosity of the fluid and Ut, also known as friction velocity, is defined as: Ut = (Tw/rho)^(-1/2) where rho is the density of the fluid and Tw is the friction at the wall. As you can see, y+ is just a local transverse Reynolds number based on the distance from the wall and the relative characteristic velocity Ut which, from dimensional analysis, can be defined only as stated above. Now that we have defined such a Re(y) we can go over. The question about the simulation of turbulent flows arise because, when you set up such a simulation, the only thing you can say about the velocity near the wall is that it is 0 at wall, nothing else, while at the first node away from the wall you will have, as a result of your simulation, a certain velocity. As you know, the above defined Tw, the force that the fluid locally exercise on the wall, is given by: Tw = rho*v*(du/dy) so it depends on the wall derivative of the velocity profile and the same is for the vorticity that you introduce in the flow, which also depends by the wall derivative of the velocity profile. So a lot of quantities that strongly affects the solution, expecially in the boundary layers, depends by this derivative. If now you consider the numerical aspect, in it's simpler form such a derivative can be expressed by: (du/dy)w approx.= (u(i,2)-u(i,1))/dx = u(i,2)/dx where u(i,1) is the velocity at the wall and is 0. So it depends exclusively by the velocity in the first cell near wall and its distance from the wall. Here arise the question about the y+ because, if we want to use such a numerical approximation (it must works) we have to be sure that locally the linear approximation used for the derivative is real (we cannot put the first node at 10 miles from the wall and say that the velocity profile varies linearly between the wall and the point). The only way to be sure of this is that the first point of the grid is in the viscous sub-layer where we know that the velocity profile is linear in y+ (and so in y) and so the velocity in the first node is well resolved. One way to determine an equation for y+ is to assume a law for the velocity profile at wall and use an empirical correlation of the tipe Tw=f(Re) for the skin friction. Substituting these in the equation for y+ and setting y+=1 is possible to obtain the necessary distance from the wall for the first node as a function of the Reynolds number and the distance along the wall. I hope i've been useful.

 November 8, 2006, 13:14 Re: Y+ #4 Paolo Lampitella Guest   Posts: n/a I made a mistake writing Ut, obviously it is defined as: Ut = (Tw/rho)^(1/2)

 December 7, 2006, 09:36 Re: Y+ #5 Igor Guest   Posts: n/a I think your first suggestion was right. Ut = (Tw*rho)^(1/2) since y+=y*((Tw*rho)^(1/2))/v Regards, Igor

 December 9, 2006, 14:01 Re: Y+ #6 Paolo Lampitella Guest   Posts: n/a I think that the last was right, because, by dimensional analysis(SI): [Tw]=Kg*(m/s^2)*(1/m^2) [rho]=kg/m^3 and so, to be [Ut]=m/s it must be: Ut=(Tw/rho)^0.5 I think it's right. Regards. Paolo

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