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Copy file name to clipboardExpand all lines: _tutorials/design_features/Inc_Turbulent_Bend_Opt/Inc_Turbulent_Bend_Opt.md
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@@ -135,7 +135,7 @@ As you can see, the row of cells just above the symmetry plane has collapsed ont
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### constrained FFD deformation
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The problem here is that the mesh nodes on the symmetry plane are not allowed to move vertically, they only move horizontally outward inside the symmetry plane. Unfortunately, the FFD deformation is such that mesh nodes just above the symmetry plane are moved down, almost on the symmetry plane. To improve the quality of the mesh deformation, we disallow the vertical movement of the FFD box nodes on the nodes in the bottom plane of the FFD box, with j-index 0 and vertical displacement (0,1,0). In Figure (3), the plane with index j=0 is the bottom plane, indicated in yellow. So we remove entries in the **DV_PARAM** list of the form *(BOX, i_Ind, 0, k_Ind, 0,1,0)*. Additionally, we also disallow the vertical movement of the nodes in the plane j=1.
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Since we disallow vertical movement in 2x(6x6)=72 nodes in the planes with $$j=0$ and $j=1$$, The total degrees of freedom is then $$648 - 72 = 576$$ d.o.f.
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Since we disallow vertical movement in 2x(6x6)=72 nodes in the planes with $$j=0$$ and $$j=1$$, The total degrees of freedom is then $$648 - 72 = 576$$ d.o.f.
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| iteration |$$\Delta P$$, total|$$\Delta P$$, bend| gain, total| gain, bend |
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| --------|--------|--------|--------|--------|
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|0|89.0 [Pa]|20.7 [Pa]| -| - |
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|1|84.9 [Pa]|16.6 [Pa]|4.6 \%|19.8 \%|
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|2|82.1 [Pa]|13.8 [Pa]|7.8 \%|33.3 \%|
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|3|81.6 [Pa]|13.3 [Pa]|8.3 \%|35.7 \%|
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|4|79.9 [Pa]|11.6 [Pa]|10 \%|44 \%|
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|1|84.9 [Pa]|16.6 [Pa]|4.6 %|19.8 %|
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|2|82.1 [Pa]|13.8 [Pa]|7.8 %|33.3 %|
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|3|81.6 [Pa]|13.3 [Pa]|8.3 %|35.7 %|
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|4|79.9 [Pa]|11.6 [Pa]|10 %|44 %|
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We see that the global pressure drop between the inlet and the outlet reduces from 89 Pa to 80 Pa, a reduction of more than 10 \%. However, since we are optimizing only the bend, we should subtract the pressure drop of the straight parts and only consider the pressure drop of the bend for a more fair comparison. In paraview, we can integrate the pressure in a 2D slice at the start and end of the bend. The pressure drop of the bend comes to $$\Delta P = 20.7 Pa$$. That means that the reduction of pressure drop in the bend is actually 44 \% !
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We see that the global pressure drop between the inlet and the outlet reduces from 89 Pa to 80 Pa, a reduction of more than 10 %. However, since we are optimizing only the bend, we should subtract the pressure drop of the straight parts and only consider the pressure drop of the bend for a more fair comparison. In paraview, we can integrate the pressure in a 2D slice at the start and end of the bend. The pressure drop of the bend comes to $$\Delta P = 20.7 Pa$$. That means that the reduction of pressure drop in the bend is actually 44 % !
Copy file name to clipboardExpand all lines: _tutorials/incompressible_flow/Inc_Von_Karman/Inc_Von_Karman.md
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@@ -33,11 +33,11 @@ The resources for this tutorial can be found in the [incompressible_flow/Inc_Von
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### Background
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When the Reynolds number $$Re=\rho \cdot V \cdot D / \mu$$ is low (Re < 40), the flow around a circular cylinder is laminar and steady. At around Re=49, the flow becomes unsteady and a periodic shedding of vortices forms in the wake of the cylinder, known as vortex shedding. The frequency of this vortex shedding is usually expressed in terms of the Strouhal number $$St= f \cdot D / U_{\infty}$$, with f the shedding frequency, D the diameter of the cylinder and U the far-field velocity. Important experimental work can be found in the paper of Williamson, *Vortex Dynamics in the Cylinder Wake*, Annual Review of Fluid Mechanics (1996) [doi](https://doi.org/10.1146/annurev.fl.28.010196.002401). After around Re=180, a second frequency is observed experimentally, in the longitudinal direction. This frequency can only be observed in a 3D simulation. The increase in number of frequencies continues until after around Re=1000 the flow is considered fully turbulent.
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When the Reynolds number $$Re=\rho \cdot V \cdot D / \mu$$ is low (Re < 40), the flow around a circular cylinder is laminar and steady. At around Re=49, the flow becomes unsteady and a periodic shedding of vortices forms in the wake of the cylinder, known as vortex shedding. The frequency of this vortex shedding is usually expressed in terms of the Strouhal number $$St= f \cdot D / U_{\infty}$$, with f the shedding frequency, D the diameter of the cylinder and U the far-field velocity. Important experimental work can be found in the paper of Williamson, *Vortex Dynamics in the Cylinder Wake*, Annual Review of Fluid Mechanics (1996) [doi](https://doi.org/10.1146/annurev.fl.28.010196.002401). After around Re=180, a second frequency is observed experimentally, in the longitudinal direction. This frequency can only be observed in a 3D simulation. The increase in number of frequencies continues until after around Re=1000 the flow is considered fully turbulent.
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### Problem Setup
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The configuration is a circular cylinder of 5 mm surrounded by a far field at $$L = 30 D$$ and a rectangular wake region of $$X = 150 D$$. The far-field velocity is $$U_{\infty} = 0.12 m/s$$. With a viscosity of $$\mu=1.0 \cdot 10^{-5}$$ and a density of $$\rho = 1 kg/m3$$, the Reynolds number is $$Re=120$$.
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The configuration is a circular cylinder of diameter $$D=0.01 m$$surrounded by a far field at $$L = 30 D$$ and a rectangular wake region of $$X = 150 D$$. The far-field velocity is $$U_{\infty} = 0.12 m/s$$. With a viscosity of $$\mu=1.0 \cdot 10^{-5}$$ and a density of $$\rho = 1 kg/m3$$, the Reynolds number is $$Re = 120$$.
Figure 2: Computational domain for the von Karman vortex shedding.
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```
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When we compare the Strouhal number with the experimental data from Williamson, we see in Figure 4 that the frequency is underpredicted. We will vary some numerical settings to investigate if we can improve the prediction of the Strouhal number.
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When we compare the Strouhal number with the experimental data from Williamson, we see in Figure 4 that the frequency is slightly overpredicted. We will vary some numerical settings to investigate the impact on the prediction of the Strouhal number.
Figure (5): Comparison of different numerical settings
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Figure (5): Comparison of different numerical settings.
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In Figure 5, we see the effect of different numerical settings on the prediction of the Strouhal number. The second order scheme predicts a Strouhal number of $$St = 0.1734$$, slightly over predicting the experimental value of $$St_{exp} = 0.170$$. Note that our predictions of the Strouhal frequency depends on the number of samples and sampling rate that we provide to the FFT. We took 2500 timesteps of 0.01 s which contains enough cycles for an accurate frequency prediction using an fft.
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When switching from second order in time to first order, the Strouhal number is under predicted by 6 \% compared to the experimental value. Also, when increasing the time step from 0.01 s to 0.02 seconds, the St decreases by 2 \%. When increasing the time step even further to $$ \Delta t = 0.04 s%%, St is under predicted by 8 \%. The period of the dimensional frequency is $$f \approx 0.5 s$$, so with a timestep of 0.01 s we have 50 time steps per period, we have 25 time steps when $$\Delta t = 0.02 s$$, and only 12 time steps when $$\Delta t = 0.04 s$$. It is clear that 12 time steps per period is not sufficient.
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When switching from second order in time to first order, the Strouhal number is under predicted by 6 % compared to the experimental value. Also, when increasing the time step from 0.01 s to 0.02 seconds, the St decreases by 2 \%. When increasing the time step even further to $$ \Delta t = 0.04 s$$, St is under predicted by 8 %. The period of the dimensional frequency is $$f \approx 0.5 s$$, so with a timestep of 0.01 s we have 50 time steps per period, we have 25 time steps when $$\Delta t = 0.02 s$$, and only 12 time steps when $$\Delta t = 0.04 s$$. It is clear that 12 time steps per period is not sufficient.
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It is also known that the size of the computational domain influences the results, so we reduce the domain by half, $$L = 15 D$$ and $$X = 75 D$$. The Strouhal then increases to $$St = 0.1768$$, an increase of 2 \%. It seems that a far-field that is 15D away from the cylinder is sufficient.
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It is also known that the size of the computational domain influences the results, so we reduce the domain by half, $$L = 15 D$$ and $$X = 75 D$$. The Strouhal then increases to $$St = 0.1768$$, an increase of 2 %. It seems that a far-field that is 15D away from the cylinder is sufficient.
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As a final test, the testcase can be executed for varying Reynolds numbers, ranging from Re=60 to Re=180, giving the result in Figure (6).
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The paraview statefile to create the movie can be found here: [statefile_with_particles.pvsm](https://github.com/su2code/Tutorials/blob/master/incompressible_flow/Inc_Von_Karman/statefile_with_particles.pvsm)
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<p>Computational analysis tools have revolutionized the way we design engineering systems, but most established codes are proprietary, unavailable, or prohibitively expensive for many users. The SU2 team is changing this, making multiphysics analysis and design optimization software freely available and involving everyone in its creation and development.</p>
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<p>Find a detailed description of the code philosophy, components, and implementations in the SU2 <ahref="https://arc.aiaa.org/doi/abs/10.2514/1.J053813">AIAA Journal article</a>. Whether it's <ahref="documents/AIAA-2016-3518.pdf">discrete adjoints</a>, <ahref="documents/2017_J._Phys.3A_Conf._Ser._821_012013.pdf">non-ideal compressible CFD</a>, <ahref="http://dx.doi.org/10.1016/j.compfluid.2016.02.003">high-performance computing</a>, or <ahref="documents/AIAA-2018-3111.pdf"> incompressible flows with heat transfer</a>, SU2 has something for you.</p>
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