Flow in a tunnel
 

Hello to everyone. I'm new with CFD and I need your help.
I try to simulate the flow around a train in a tunnel and for the beginning I created a long rectangular tunnel and imposed the boundary conditions: a velocity-inlet (40 m/s) and a pressure-outlet (0 pascal). (I used transient solver and k-w turbulent model). Can somebody explain me why the pressure near the velocity-inlet reach the maximum and decrease till reach the minimum near pressure outlet?
I also put a box (which represent the train) in the tunnel and the pressure is still maximum near velocity-inlet and decrease moving to pressure outlet. Also the minimum value reached by the pressure is in the back of my train. I expected a maximum value of the pressure in front of the train, but I got the maximum value for pressure near velocity inlet. Can anybody explain me why the pressure is maximum near the velocity-inlet and not in front of the train?
(To evaluate the pressure I used surface integral)

Hello! I do not believe that you are supposed to impose what the pressure should be at the outlet. That will be calculated by the software. All you will have to do it to have your Cabinet open on all sides and apply the velocity BC on the inlet. The purpose of the CFD simulation is to calculate for you the N-S equations, which are pressure and velocity entities...

Regards,
Adriana D.
Aavid Thermalloy
San Jose, CA

Thank you for your replay. If I impose the velocity BC on the inlet what kind of BC should I impose on the other side of my Cabinet? I thought to impose an outflow condition, can it work in your opinion?

I think you can try pressure-far field BC

What software do you use? If you use Icepak to simulate this, then you would have an opening on your cabinet at Inlet and Outlet (with velocity being applied at the opening at the inlet). All the other sides of your cabinet would be Default, or walls. If you use Fluent, then you apply velocity at the inlet, zero pressure on all sides except for the outlet. At the outlet you apply the "outlet" boundary conditions. I hope this helps.

Adriana

Aavid Thermalloy
San Jose, CA

In theory it is correct as you are giving 0 pressure at outlet, inlet pressure should be more than 0 as there will be pressure loss in the system.

*sorry gave reverse info in last post :P

Guys please...

For the setup you described so far (velocity inlet, pressure outlet) the results you obtained are physically valid. The reason is fluid dynamics.
Fluids flow from locations of high pressure to locations of low pressure. So if you impose a fluid flow from the inlet towards the outlet, pressure will be highest at the inlet.

I dont feel like I understand enough what you did after this first step to comment on the results. But I dont think that the setup you describe models the physical process correctly.
First of all, you should not impose a velocity. The flow velocity in the tunnel will be a result of the movement of the train. As a first approach, I would use pressure inlet and pressure outlet boundary conditions.
Then for the movement of the train it is not enough to impose a velocity on the walls of the train. You will have to model the movement with some kind of sliding mesh interface or a remeshing approach. Not an expert on this though so I cant give any further advice here.

Thank you for replys.
Flotus1 my interest is to calculate the difference of pressure between the front and the back of my train in tunnel. I can't use the sliding mesh technique because my computer can't handle a fine mesh.(I used a box of influence to create a fine mesh only near the train while for the far field I imposed the bigger size for mesh's elements) So to simulate the motion of the train in tunnel I considered a train at rest and a moving air. To move the air I imposed a velocity inlet and a pressure outlet but I'm not sure that these boundary condition can simulate, in the first approach, my case. Also the tunnel is quite long and the train is far from the inlet and the outlet. Can somebody tell me why these BC don't simulate correctly my case?
P.S. I use fluent.

Best regards.
Artur A.

Try not to apply zero pressure boundary conditions at the outlet. I think your approach of moving air is good enough for what you want to simulate.

Adriana
Aavid Thermalloy
San Jose, CA

Is "pressure-far field" a better outlet condition in this case? And why?
Thanks

Best regards
Artur A.

I don't think pressure far field is a good bc for this problem; I would set a velocity inlet (air velocity=train velocity) and a pressure outlet (0 Pa gauge if the tunnel is open to atmosphere), or outflow BC.
If you don't use sliding mesh you are simulating a fixed train, like in a sort of a wind tunnel, where are moves instead of the train.
So, if this is your approach, and you want to study the train, I would increase the size of the domain and create a wind tunnel.
The bottom of the tunnel will be wall (no-slip velocity), side walls can be wall (no-slip velocity or zero shear stress)-->you can set also simmetry if side walls are far enough from the train.
Outlet should be far enough from the train, to let the vortices disappear and to not have reversed flow.
What Alex says is 100% correct, your pressure will be higher at inlet than outlet and you will have high pressure in the front of the train caused by the dynamic pressure.

Please post some pictures of pressure contour of the train.

PS: there are a lot of tutorials of cars in wind tunnels, you can read them and replicate your problem.

I think this will be not usefull if boundaries are velocity inlet/pressure outlet: pressure values will be only shifted.

Daniele

To create my model I followed these steps:
1)created the train
2)created an enclosure (air)
3)created a boolean
so at this step I got just the air and the walls of te train inside it. After, when I created the mesh a used a box of influence to create a relative fine mesh near the train.
4)I created the walls for tunnel
5)I created the mesh for air and another one for the tunnel's walls

In this simulation I used sliding mesh tecnique but I tink that this is not very useful becaus me mesh is not fine enough. Is it correct? (You can see on the picture the dimensions of wall elements and air near the walls of the tunnel)
The setups are:
a)BC->velocity inlet = 40 m/s
b)BC->pressure outlet = gauge pressure = 0 pascal
c)BC->everything else are: stationary walls, interior (I got a symmetry plane in the midle of the tunnel to reduce the number of mesh elements)
d)cell zone condition ->walls of tunnel-> mesh motion 50m/s (so the walls of sliding mesh move at 50 m/s in the same direction of the velocity of the air)
e)the solver is : type=pressure based, time=transient
f)viscous model = standard k-omega
g)run calculation-> time step size= 1e-05
h)run calculation-> max iterations/time steps= 20
I used also a second order upwind for: momentum, turbulent kinetic energy, specific dissipation rate, tranient dormulation

On the pictures You can see the pressure contour on the train, symmetry plane, and planes just in front and back the train.
please give me your opinion.

best regards
Artur.Ant

You're setup is not correct.
Why for sliding mesh you are moving air and wall tunnel???
Only the Wall train and the mesh sourrounding the train must move, and you need to create interface between the mesh sourrounding the train and the external mesh.
However the problem is that you probably don't know for this configuration the boundary condition for air inlet.
I suggest to simulate the wind tunnel.

The problem of train in tunnel is very similar to the problem of the elevator, and I found a document of a simulation of a high speed elevator. The author of that document said this about the sliding mesh: [quote: The “Sliding mesh” scheme has the elevator fixed in space while the walls and the fluid pass by at a constant speed]
So I decided to make the same thing.
Also the author of that document don't advise to make a dynamic mesh with the train in motion and the fluid and walls at rest because [quote: It is more numerically laborious and the scheme is only a first order accurate]

Can you confirm that a sliding mesh requires a very fine mesh?
Because in my document I want to report all the reason why I don't use a sliding mesh.
Thank you.

Best regards
Artur.Ant

Then, I don't get what is the advantage of using this type of sliding mesh instead of a static simulation with moving air.
Sliding mesh doesn't require particular fine mesh.
Are you sure the static elevator is using sliding mesh?Or it only moves air and wall tunnel without moving the mesh?In other words, is there only a fluid zone or are there 2 fluid zones connected by interface?

For the simulation of a wind tunnel in your opinion Should I keep the transient solver or not?

The author of that document report this advantage of a sliding mesh [ quote: This model has the correct relationship between the fluid and the hoistway walls]
The authors don't give any other explanation. They call it sliding mesh but don't give any explanation how they created the model or about the setups they used...

With sliding mesh transient solver is a must (obviously): for sliding mesh I mean 2 zones connected by interface, and one or both zones (meshes) are moving, sliding the interface.
For a wind tunnel static mesh I would try with steady solver; if the solution doesn't converge well (maybe because of vortices unsteadiness) I would switch to the transient solver from the steady solution.

Ok thank you.
I will try a simulation of a wind tunnel with a static mesh.
So to sum up:
I can use pressure outlet (pressure gauge 0) and velocity inlet (air velocity = train velocity) with the train and walls of the tunnel at rest.
Is it ok in your opinion?

Yes, it sounds ok to me.