# Which boundary conditions in flow past circular cylinder domain to use?

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 June 4, 2020, 02:01 Which boundary conditions in flow past circular cylinder domain to use? #1 Senior Member   Hector Redal Join Date: Aug 2010 Location: Madrid, Spain Posts: 243 Rep Power: 17 Hi, Which boundary conditions shall I use in the flow past circular cylinder problem? Right now, I am using two set of boundary conditions: A Case: - Inlet -> V=(Vx,Vy) = (1,0) - Outlet -> dVx/ dx = dVy/dx= 0, P=0 - Uppper / lower boundaries -> dVy/dy = dVx/dy = dP/dy = 0 B Case: - Inlet -> V=(Vx,Vy) = (1,0) - Outlet -> dVx/ dx = dVy/dx= 0, P=0 - Uppper / lower boundaries -> V=(Vx,Vy)= (Vx, 0) // Vx is free, but Vy is set to zero. In the first case (A Case), the simulation blows up. That is, fails to converge. In the second case (B Case), the simulation works, and I obtain an stationary simulation (without vortex shedding pattern). This is for a low reynolds number. But, this simulation is some kind of a confined flow, whereas I want to simulate an unconfined problem. Question: Is case A ill-conditioned? Am I using a wrong set of conditions? The problem is stationary, and in that sense, far away from the cylinder, the conditions for the derivaties (dVx/dx, etc.) must be true. Am I missing anything? Thanks for your support. Best regards, Hector.

 June 4, 2020, 02:58 #2 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,817 Rep Power: 73 You cannot set simultaneously both velocity and pressure

June 4, 2020, 03:32
#3
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Hector Redal
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Quote:
 Originally Posted by FMDenaro You cannot set simultaneously both velocity and pressure

Hi,

Do you mean I cannot set P and V in the outlet interface?
Or in the upper / lower boundaries?

Because in the Case B, I am setting simultaneously P, dVx/dx and dVy/dx in the outlet and it works, and only Vy in the upper and lower boundaries.

BR,

June 4, 2020, 03:42
#4
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Filippo Maria Denaro
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Quote:
 Originally Posted by HectorRedal Hi, Do you mean I cannot set P and V in the outlet interface? Or in the upper / lower boundaries? Because in the Case B, I am setting simultaneously P, dVx/dx and dVy/dx in the outlet and it works, and only Vy in the upper and lower boundaries. BR,

Are you solving the incompressible formulation? Thus, if you prescribe a BC for the velocity (everywhere the boundary is), you have an implicit prescription also for the pressure. Do not confuse the physical set of BCs from the numerical set of BCs. Just think about your inlet condition, you prescribed only the velocity.

When you write the momentum equation you have

Consequently if you prescribe a condition for the normal velocity you should prescribe the resulting condition for the pressure gradient.

June 4, 2020, 08:26
#5
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Hector Redal
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Quote:
 Originally Posted by FMDenaro Are you solving the incompressible formulation? Thus, if you prescribe a BC for the velocity (everywhere the boundary is), you have an implicit prescription also for the pressure. Do not confuse the physical set of BCs from the numerical set of BCs. Just think about your inlet condition, you prescribed only the velocity. When you write the momentum equation you have dv/dt +Grad p = r Consequently if you prescribe a condition for the normal velocity you should prescribe the resulting condition for the pressure gradient.
Yes, I am solving the incompressible formulation.

I have double checked my code, and when setting the V in the upper / lower boundary, I am implictly imposing the gradP as you comment.

Then when imposing the dVx/dx and dVy/dx, I have to impose the gradP consequently, as I understand from your comment.

June 4, 2020, 08:28
#6
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Quote:
 Originally Posted by HectorRedal Yes, I am solving the incompressible formulation. I have double checked my code, and when setting the V in the upper / lower boundary, I am implictly imposing the gradP as you comment. Then when imposing the dVx/dx and dVy/dx, I have to impose the gradP consequently, as I understand from your comment.
Yes, you have to derive the condition for the pressure BC after you set a condition on the normal derivative of the velocity

June 5, 2020, 02:31
#7
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Hector Redal
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Quote:
 Originally Posted by FMDenaro Yes, you have to derive the condition for the pressure BC after you set a condition on the normal derivative of the velocity

So, imposing this type of boundary conditions (dVx/dy = 0 and dVy/dy = 0) implies that the flow is fully developed in those boundaries.

But, what type of boundary condition to use when the condition of fully developed flow is not fullfiled?
You cannot use neither vy=0 (since the flow may enter or exit the domain) nor dVx/dy=0 and dVy/dy=0 (since the flow is not fully developed. The flow can enter or exit the domain as well).

In this case, I don't know which boundary condition to use.

Maybe, that's the reason why the simulation is not converging, since my asumption of fully developed flow in the upper and lower boundary is not fulfilled.

/Hector.

June 5, 2020, 03:25
#8
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Filippo Maria Denaro
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Quote:
 Originally Posted by HectorRedal So, imposing this type of boundary conditions (dVx/dy = 0 and dVy/dy = 0) implies that the flow is fully developed in those boundaries. But, what type of boundary condition to use when the condition of fully developed flow is not fullfiled? You cannot use neither vy=0 (since the flow may enter or exit the domain) nor dVx/dy=0 and dVy/dy=0 (since the flow is not fully developed. The flow can enter or exit the domain as well). In this case, I don't know which boundary condition to use. Maybe, that's the reason why the simulation is not converging, since my asumption of fully developed flow in the upper and lower boundary is not fulfilled. /Hector.

Top and bottom boundaries must be enough far from the cylinder.

June 5, 2020, 04:45
#9
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Hector Redal
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Quote:
 Originally Posted by FMDenaro Top and bottom boundaries must be enough far from the cylinder.

That would suffice when the flow is parallel with the upper and lower boundaries. And this is how I have managed to simulate more of my problems.

But the problem is that now I am trying to simulate a flow past circular cylinder where the upper and lower boundaries are parallel to the inflow, but the gravity force acts vertically. Additionally the cylinder is at higher temperature than the flow, so buoyancy effects happen. In that case, when the flow passes the cylinder, it is heated and travels in oblique direcction.

So, it is not valid the assumption that far away from the cylinider the flow will be parallel. Right? Or am I wrong?

What can I do?

Thanks.

Best regards,
Hector.

June 5, 2020, 05:25
#10
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Filippo Maria Denaro
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Quote:
 Originally Posted by HectorRedal That would suffice when the flow is parallel with the upper and lower boundaries. And this is how I have managed to simulate more of my problems. But the problem is that now I am trying to simulate a flow past circular cylinder where the upper and lower boundaries are parallel to the inflow, but the gravity force acts vertically. Additionally the cylinder is at higher temperature than the flow, so buoyancy effects happen. In that case, when the flow passes the cylinder, it is heated and travels in oblique direcction. So, it is not valid the assumption that far away from the cylinider the flow will be parallel. Right? Or am I wrong? What can I do? Thanks. Best regards, Hector.

Provide a sketch of your exact flow problem, you haven't mentioned in your original post that is not the standard flow around a cylinder.

June 5, 2020, 10:05
#11
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Hector Redal
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Quote:
 Originally Posted by FMDenaro Provide a sketch of your exact flow problem, you haven't mentioned in your original post that is not the standard flow around a cylinder.

Yes, you are right.
I am uploading a drawing of the problem I plan to simulate.

U_inf -> Velocity in the inlet

Tw -> Temperature of the cylinder
T_inf -> Temperature of the bulk fluid
g1 -> gravity force acting horizontally
g2 -> gravity force acting vertically

rho, nu and k -> Properties of the fluid (density, viscosity and thermal conductivity).
Attached Files
 FlowPastCylinderPhysicalRepresentationVortexShedding.pdf (18.1 KB, 14 views)

 June 5, 2020, 14:25 #12 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,817 Rep Power: 73 what do you mean for two different gravity forces? The second case is the inflow directed along the gravity direction with opposite versus? However, what about the difference in the temperature? The flow problem seems dominated by forced convection, I doubt you can expect a relevant bouyancy effect in the vortex shedding. What about the non dimensional parameters?

 June 5, 2020, 17:49 #13 Senior Member   Hector Redal Join Date: Aug 2010 Location: Madrid, Spain Posts: 243 Rep Power: 17 I plan to test several different scenarios: - Horizontal gravity -> Aiding / Opposing buoyancy to the flow (depending on the value of the gravity: positive or negative). - Vertical gravity -> Transverse buoyancy to the flow. The value of non-dimensional parameters are: - Reynolds number in the range of 10 -> 50. - Richarson number in the range of 0.5 -> 1.5. Small Richardson numbers characterize a flow dominated by forced convection. Richardson numbers higher than Richarson=16 indicate that the flow problem is pure natural convection and the influence of forced convection can be neglected. So, I agree with you that the flow is dominated by forced convection. But despite this, I can observe that the vortex shedding moves in oblique direction. I have taken a look at this reference: https://www.sciencedirect.com/scienc...17931008005991

June 5, 2020, 17:57
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Quote:
 Originally Posted by HectorRedal I plan to test several different scenarios: - Horizontal gravity -> Aiding / Opposing buoyancy to the flow (depending on the value of the gravity: positive or negative). - Vertical gravity -> Transverse buoyancy to the flow. The value of non-dimensional parameters are: - Reynolds number in the range of 10 -> 50. - Richarson number in the range of 0.5 -> 1.5. Small Richardson numbers characterize a flow dominated by forced convection. Richardson numbers higher than Richarson=16 indicate that the flow problem is pure natural convection and the influence of forced convection can be neglected. So, I agree with you that the flow is dominated by forced convection. But despite this, I can observe that the vortex shedding moves in oblique direction. I have taken a look at this reference: https://www.sciencedirect.com/scienc...17931008005991

Seems a case with different heat conditions. However, do you get a correct solution in the isothermal case?

June 6, 2020, 02:57
#15
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Hector Redal
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Quote:
 Originally Posted by FMDenaro Seems a case with different heat conditions. However, do you get a correct solution in the isothermal case?

Yes, I get a correct solution in the isothermal case For Re=100, I get a Strouhal number = 0.1667 which matches the result obtained in all the refereces for the incompressible fluid flow problem against a circular cylinder. The drag and drag coefficient also match the reference.

But the point is that I am looking for the boundary conditions to use in the non-isothermal case. Not clear to me.

Do you have a hint?

June 6, 2020, 03:21
#16
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Filippo Maria Denaro
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Quote:
 Originally Posted by HectorRedal Yes, I get a correct solution in the isothermal case For Re=100, I get a Strouhal number = 0.1667 which matches the result obtained in all the refereces for the incompressible fluid flow problem against a circular cylinder. The drag and drag coefficient also match the reference. But the point is that I am looking for the boundary conditions to use in the non-isothermal case. Not clear to me. Do you have a hint?

First of all, what about the domain size for the isothermal case ? Have you also computed the residual of div v in each cell and ensured that is small?

Are you using Bousinnesq model? Remember that it works for small temperature difference,I am not sure you can see a relevant buoyancy effect compared to the corced convection.

However, if the buoyancy is really relevant in your problem you should consider to enlarge the distance of the top boundary from the cylinder. That should avoid that the top boundary would require an outflow-like condition.

June 6, 2020, 04:25
#17
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Hector Redal
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Quote:
 Originally Posted by FMDenaro First of all, what about the domain size for the isothermal case ? Have you also computed the residual of div v in each cell and ensured that is small?
The domain size is Height = 160 times the diameter of the cylinder and the length = 50 times the diameter of the cylinder.
The divergence of v at all points of the domain is neglibible.

Quote:
 Originally Posted by FMDenaro Are you using Bousinnesq model? Remember that it works for small temperature difference,I am not sure you can see a relevant buoyancy effect compared to the corced convection.
Yes, I am using the Boussinesq approximation, with a variation in temperature less than 5% (so the Boussinesq approximation is valid).

Quote:
 Originally Posted by FMDenaro However, if the buoyancy is really relevant in your problem you should consider to enlarge the distance of the top boundary from the cylinder. That should avoid that the top boundary would require an outflow-like condition.
That was exactly what I did, but since the Reynolds number is very low (10-40) the diffusion effects due to boundary are even larger than in the isothermal incompressible problem with Re=100.
So, I have come to the conclusion that the domain boundaries should be located far far away even. The reference I copied above, use a blockage ratio of 1/25, and I am using 1/160.

That's the reason why I am very dubious about the results obtained in that reference with such high blockage ratio. The efects of the boundary in the flow pattern are very high, which invalidates the results.

June 6, 2020, 05:07
#18
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Filippo Maria Denaro
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Quote:
 Originally Posted by HectorRedal The domain size is Height = 160 times the diameter of the cylinder and the length = 50 times the diameter of the cylinder. The divergence of v at all points of the domain is neglibible. Yes, I am using the Boussinesq approximation, with a variation in temperature less than 5% (so the Boussinesq approximation is valid). That was exactly what I did, but since the Reynolds number is very low (10-40) the diffusion effects due to boundary are even larger than in the isothermal incompressible problem with Re=100. So, I have come to the conclusion that the domain boundaries should be located far far away even. The reference I copied above, use a blockage ratio of 1/25, and I am using 1/160. That's the reason why I am very dubious about the results obtained in that reference with such high blockage ratio. The efects of the boundary in the flow pattern are very high, which invalidates the results.

Try to check if the top boundary can manage an outflow condition, you can test the pure buoyancy case (no forced convection) and see what happens on the top.

June 8, 2020, 18:25
#19
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Hector Redal
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Quote:
 Originally Posted by FMDenaro Try to check if the top boundary can manage an outflow condition, you can test the pure buoyancy case (no forced convection) and see what happens on the top.

I simulated a pure convection problem, but not for the cylinder flow but for a vertical chanel in between two heated walls at different temperature, and it worked.

I don't think this time is going to work, since the flow have to enter from the bottom of the domain. Otherwise the continuity equation won't be satisfied.

June 8, 2020, 18:41
#20
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Filippo Maria Denaro
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Quote:
 Originally Posted by HectorRedal I simulated a pure convection problem, but not for the cylinder flow but for a vertical chanel in between two heated walls at different temperature, and it worked. I don't think this time is going to work, since the flow have to enter from the bottom of the domain. Otherwise the continuity equation won't be satisfied.
Yes, if the top boundary is outflow you must have the bottom boundary able to let the flow enter into. I suggest to check the pure buoyancy case with the cylinder heated