In general, a geometrically separated flow region can be laminar or turbulent, we cannot say a priori what happens, apart from some well known flow configurations.

In you case you should first consider the flow over a 3D configuration using a laminar time dependent formulation on a reasonable grid size to see if this separation is unsteady and the appearence of a wide range of frequancies.

Hi Jenson Prabhu

How do you calculate Reynolds number?

Hi @Nazaruddin Sinaga

The CFD software calculates the Reynolds number while the solution initiation. I believe it calculates from the input cfm value and the space constriction in the fluid domain.

Dear Jenson Prabhu

What I mean is how you calculate the Reynolds number.

Re = Rho. V. L/mu = 100

What are V and L when you count the Re?

@Filippo Maria Denaro

I understand that I have to run a transient setup to study the flow formation. How can I judge that the flow is unsteady once I did that. Also, if the flow seems to be unsteady is it a good strategy to run the setup again considering a steady state turbelence model?

I ran the simulation considering steady state laminar and turbulence model seperately. With laminar model, the processor temperature is 80 deg. Celsius and the temperature didn't normalise even after 1000 iterations. It is oscillating like a sine wave around 80°C with deviation of 2°C.

But, with turbulence model the solution converged so well and I got the processor temperature as 70°C. Thus the difference between the setups is nearly 10°C and I think turbelence model assumes some heat diffusion to the fluid film near the heat sink fins that enhances the local heat transfer rate. Any thoughts on this?

What you are doing is not correct. The equations you solve for the steady laminar solution and the steady turbulent solution are totally different and the meaning of the resolved variables is different. If the laminar solution does not converge to a steady state that means that the deterministic solution has an unsteady nature. This nature can be still laminar (low frequency oscillation) or develop in a correlated turbulent regime (large range of frequencies). Conversely, the steady turbulent solution is only in statistical sense as a fully developed energy equilibrium regime is assumed. But if the flow is actually laminar and unsteady, solving the steady state modelled equations for turbulence can produce a steady solution that is physically not correct since the model does not vanish for laminar conditions. The turbulence model has generally a dissipative effect, like reducing the Re number.

Again, run the time-dependent 3d solution and start descovering if the nature of the solution is unsteady and the range of frequencies that appear.

Dear Filipo,

Thanjs for the comprehend answer.

Regards

If Reynolds number is correct(1000) the flow is laminar and you need no turbulence model.From other hand flow separation in general(laminar or turbulent flow) has boundary conditions to be exist like adverse pressure gradient should be greater than velocity gradient.If your experience is not enough to choose the computational treatment( i mean the flow is considered as laminar or turbulent) you should have experimental measurements to validate your thoughts.

Nazaruddin Sinaga

I know the mass flow of air entering the space between the heat sink fins. Thus dividing it by the area of the fin gap I can get the velocity. I input this velocity value in the Re number formula. For the characteristic length, I considered the length of the fin (25.4 mm).

By hand calculation I got a value around 1100. The software estimated value is around 980 as it accounts for the momentum losses in the 3-D space. So, I think the Reynolds number is reasonable.

Filippo Maria Denaro

Thank you for elaborating the concept, I will start studying the transient flow pattern as you suggested. I think I should learn more about this and thanks again for getting me to a good starting point.

Ahmed Abdelhameed

Thanks Ahmed Yes, I should request a prototype to experiment and better understand the results.

Dear Jenson Prabhu

The choice for both the velocity and characteristic length depends on which object or problem you want to analyze. Please have a look at the case of annular finned tube bundle. When we want to analyze the overall Nusselt number, normally, we use the maximum velocity (between the tubes), and tube diameter as the characteristic length....

The criteria for flow to be laminar or turbulent is Reynolds number and for flow separation is flow stream pressure gradient. So one should be careful not to confuse between them.

Horimek Abderrahmane

Thank you. I think SST-w model will befit my case. Accordingly, for flow in between closely spaced heat sink fins it will assume k-w model and for other far regions and vorticity behind square blocks it will solve with k-e model correct?

Also, I ran transient simulations and the flow in between the heat sink fins seem to be laminar. But there are flow circulation and vorticity behind some square blocks placed nearly 3 inches in front of the heat sink. So, if I use the SST-w model won't it under predict the temperature at the heat sink area deliberately assuming the boundary layer to be turbulent?

Thanks you all for the answers. I also found this article much related to my question here. So, just sharing it for reference.

https://iopscience.iop.org/article/10.1088/1742-6596/525/1/012002/pdf

Horimek Abderrahmane

" because turbulence models are extensions to laminar flow. we add some terms (of fluctuation) to the mean flow (laminar). "

No, not at all...this idea is wrong..

The statistically mean field does not converge to a laminar field, just consider the example of a flow in a plane channel. Set a value for the Reynolds number for which the mean velocity profiles are computed from DNS data and then compute the Poiseulle solution at the same Reynolds. Now compare the profile...you will see that the two solutions are comparable only at very low values of y+

hi,

Usually for internal flow k-w turbulence model is good . But in your case , sice Re is less laminar is fine , try solving without any turbulence model and compare the result with some existing literature . This method should be fine.

Good luck!

Horimek Abderrahmane

No, but let me extend the issue. From the DNS you get the velocity field v(x,t) as numerical solution. But from this field one performs the statistical averaging so that one gets the mean velocity = V(x). Then you simply compute also v'(x,t)=v(x,t)-V(x).

Now you can compare the profile of the mean velocity V(x) to that of the laminar solution.

The RANS formulation works to solve the statistically averaged equations in order to compute directly V(x). But this is not a laminar solution!

Horimek Abderrahmane : You may try solving the unsteady equation without any models, as it is done in Implicit LES (ILES). By successive refinement you should converge to a solution, whose statistics will become invariant with respect to grid refinement. Our experience suggests this to work for transitional and turbulent flows which are evolving.

Horimek Abderrahmane

" we suppose that V(x,t)=V(x)+V'... the first is the mean velocity (the same if the flow was laminar), the second is the fluctuation. "

no, that is not true, you are doing confusion in this concept...What you are supposing this way is that in a turbulence field the averaged velocity V(x) would satisfy the Navier-Stokes equation for a laminar case (no turbulence model) at the same Re number and that is not true. You can try to check if the turbulent averaged velocity profile in a channel is as same as the quadratic profile in the laminar solution.

The fluctuations acts as a residual to reconstruct the turbulence field, V(x) cannot be a simple laminar solution but is the result of the statistical effect of turbulence.

I give you a simple example of the action of fluctuation on the average field. Consider the laminar unsteady solution for the flow over a cylinder (laminar vortex shedding). You now can still define V(x) as the time averaging and, this way, defining a fluctuation velocity field also in a laminar condition. Of course, V(x) does not satisfy the original NSE.

Horimek Abderrahmane I think you are confused between the mean flow and Laminar flow. If Turbulent vortex shedding mean V(x) = Laminar flow itself. Then how can you explain the Laminar vortex shedding mean ?

They both are not same.