Steady Vs Unsteady flow over a cylinder

Simulation of flow over the cylinder and study of von karman vortex street

Goals

• Simulate the flow over a cylinder and explain the phenomena of karman vortex street.
• First of all we simulate the flow with the steady and unsteady case and calculate the Strouhal Number for Re= 100 in unsteady case.
• Calculate the coefficient of drag and lift over a cylinder by setting the Reynolds number to 10, 100, 1000, 10000 and 100000. (Run with steady solver)
• Discuss the effect of Reynolds number on the coefficient of drag.
• To plot the velocity contours, pressure contours, drag coefficient, lift coefficient and velocity magnitude.

The phenemena of Von Karman Vortex Street

In fluid dynamic, a karman vortex street is the repeating pattern of swirling vortices, caused by a process of vertex shadding, which is responsible for the unsteady of separation of flow of a fluid around a blunt bodies. It is named after the engineer and fluid dynamicist 'Therefore von karman'. A vortex street will form only at a certain range of flow velocities, specified by a certain range of Reynold numbers(Re), typically above a limiting Re value of about 90. The Reynold number of a flow is a measure of ratio of inertial to viscous forces in the flow of fluid around the body or in channel, and may be defined as the nondimensional parameter of the global speed of the whole fluid.

`R_e=frac{rho V D}{mu}`

Where rho is the density of air , V is the velocity of air , and D is the diameter of the cylinder.

mu is the dynamic viscosity of the air.

In the following figure, we easily observe the vortex street behind the circular cylinder in the air.

Part-I simulations

Steady State and Transient simulations at Re=100

1.Geometry

• The above image is a sketch of the 2-dimensional geometry.
• Now we generate the surface in the above sketch by using the pull tool of space claim.

• we select the inner surface of the circle and delete this.
• Dimensions of the geometry are given below

2. Computational Domain and Boundary Conditions

Mathematical Models

In this project we are taking the flow over the cylinder is laminar. The motion law accords with the Navier-Stokes equation. Continuity and momentum equation are formulated as shown below.

`frac{partial}{partial x_i}(rhou_i)=0`

`frac{partial}{partial t}(rho u_i)+frac{partial}{partial x_j}(rho u_iu_j)=-frac{partialp}{partial x_i}+vfrac{partial}{partial x_j}[frac{partialu_i}{partial x_j}]+frac{partial}{partial x_j}[-rho bar(u_i^primeu_j^prime)]` u_i)`

we discretize the above equations by using the finite volume method.

For the presure velocity coupling we used the SIMPLE scheme.

For the spatial gradient discretization we used the least square cell based scheme.

For the pressure discretization we used the second order schemes.

For the momentum equation discretization we used the second order upwind scheme.

3.Meshing Details

• First of all we generate the mesh by using the default mesh setting to check the mesh quality. Default mesh is quadrilateral mesh. we change this quadrilateral mesh to triangles in the next step.
• Then right click on mesh and go to insert and go to method, choose the triangular method. Choose the element size to 0.25m and generate the mesh.
• Mesh around the cylinder also play an important role that's why we increase the number of points on the cylinder wall by using the edge sizing option in mesh. This will make the circle edge smoother and more close to perfect circle.
• The number of divisions used in edge sizing is 36.
• The region very close to the circle is also very important to perfectly simulate the flow over the cylinder, so we will insert the inflation layer.
• The first layer heigth is 5e-3, no of layers =10, growrh rate is 1.2.
• After all the setup generate the mesh
• No of nodes= 21622, No of elements= 41810

Mesh around the circle

4.Case setup details

• Steady state analysis
• Solver type:pressure based
• Velocity formulation:absolute
• Viscous model: Laminar
• Material:air (Density =1 kg/m^3), Viscosity = 0.02, Reynold number=`frac{rhoVD}{mu}` where rho is the density, D is the diameter of the circle and mu is the viscoity of the air. `Re=frac{rhoVD}{mu}=frac{(1)(1)(2)}{0.02}=100`.
• Boundary Conditions:
• At the inlet boundary we take the velocity=1 m/s.
• At the outlet we take the gauge pressure=0.
• At the wall we take the no slip condition.
• In the solution menue we have to create a point 8 meters behind the cylinder to monitor the vortex average velocity at that point.
• we use the Hybrid initialization for initialization.
• After initialization the solution, we have to set the pressure and velocity contours.
• To clearly observe the vortex average behind the cylinder. we set the solution animation.
• run calculation for 2000 iterations

Transient simulation setting

In transient simulation we do the following changing.

• choose the transient analysis
• time step size =0.1, no of time steps = 2000

5(a).Results of steady state simulation

• Residuals graph

After the 500 iterations, the residuals are started to give almost constant values.

• Vertex average of velocity magnitude(m/s)

A monitor point was created horizontal at a distance of 4 times the diameter of the circle (4*2=8) from the center of the circle. The vertex average of velocity magnitude at the monitor was plotted as shown above and it started to give almost same pattern after the 500 iterations. By analysing both residuals and velocity plot we say that solution is converged after 500 iterations.

• Velocity Contours

The above image of contours are at the last step of iterations, that is at 2000 iteration. The flow separation, velocity distribution among the fluid layers and vortex shedding are clearly visible in these contours.

• Pressure Contours

In the above pressure contours graphs we observe that there is the increase of pressure in front of cylinder and drop of pressure occur behind the cylinder.

• Drag

• Lift

In the lift plot we easily observe that after 500 iterations the lift plot shows the same pattern. The value of lift and drag are shown in the below table.

Velocity animation

Table 1

Case	Reynold number	Density	Viscosity	No of iterations	Drag	Lift	Velocity
Steady State	100	1	0.02	2000	1.3440185	0.13169855	0.59834456

5(b).Results of transient simulation

Residuals graph

This is the residual graph of transient simulation.

Vertex average velocity plot

The above plot has been obtained by taking the monitor point at a distance of 8m(4 times the diameter of the circle) horizontally. In the x-axis we are taking the time in seconds. After 155 second vertex average of velocity magnitude are showing the same behaviour.

Velocity contours

The above image of contours are at the last step of iterations, that is at 200 second. The flow separation, velocity distribution among the fluid layers and vortex shedding are clearly visible in these contours.

Pressure Contours

In the above pressure contours graph we observe that there is the increase of pressure in front of cylinder and drop of pressure occur behind the cylinder.

Lift

In the lift plot we easily observe that the after 155 seconds the lift plot shows the lift same pattern. The value of lift and drag are shown in the below table.

Drag

• Calculation of strouhal number

The strouhal number is a dimensionless number describing the oscillating flow mechanism. The parameter is named after Vincenc Strouhal, a Czech physicist who experimented in 1878 with wires experiencing vortex shedding and singing in the wind.The Strouhal number is an integral part of the fundamentals of fluid mechanism.

The Strouhal number is often given as

`st= frac{fL}{U}`

Where f is the frequency, L is the characteristic length(in our case, we take the diameter of circle which is 2), U is the free stream velocity.

First of all, we manually calculate the frequency by taking the maximum amplitude of cycles divided by the total duration of time taken of this cycle. In the following figure calculation of frequency are given below.

`st=frac{(0.0857)(2)}{1}=0.1714`

Part-II Simulations

1.In the second part of the challenge we calculate the lift, drag at different reynold number 10, 100, 1000, 10,000, 100,000 in steady state case.

2.And also we analysis the effect of reynold number on drag coefficient.

First of all we understand the definition of drag coefficient and lift coefficients.

Drag coefficients

In fluid dynamic, the drag coefficient (denoted by `C_d`) is a dimensionless quantity that is used to quantify the drag or resistance of an object in a fluid environment, such as air or water. It is used in the drag equation in which a lower drag coefficient indicates the object will have less aerodynamic or hydrodynamic drag. The drag coefficient is always associated with a particular surface area. The drag coefficient is defined as C_d

`C_d = frac{2F_d}{rho u^2A}`

Where `F_d` is the drag force, `rho`is the mass density of fluid, u is the velocity, A is the reference area.

The reference area depends on what type of drag coefficient is being measured. For automobiles and many other objects, the reference area is the projected frontal area of the vehicle. This may not necessarily be the cross-sectional area of the vehicle, depending on where the cross-section is taken. For example, for a sphere `A = pir^2` (note this is not the surface area = `4pir^2).

Lift coefficient

Lift is defined as the component of the aerodynamic force that is perpendicular to the flow direction and drag is the component that is parallel to the flow direction.

The lift coefficient is defined as `C_l = frac{2L}{rho u^2A}`

Now we have the setup of five different cases.

• Case 1 10 Reynold number
• Case 2 100 Reynold number
• Case 3 1000 Reynold number
• Case 4 10,000 Reynold number
• Case 5 100,000 Reynold number

• Case 1: Reynold number 10

• Geometry and Mesh Details

We do the follwing settings for the mesh refinement.

• The number of divisions used in edge sizing is 42.
• The region very close to the circle is also very important to perfectly simulate the flow over the cylinder, so we will insert the inflation layer.
• The first layer heigth is 2.5e-3, no of layers =8, growrh rate is 1.2.
• element size=0.1m
• No of nodes= 127991 , No of elements=253620

• Case Setup

• Steady state analysis
• Solver type:pressure based
• Velocity formulation:absolute
• Viscous model: Laminar
• Material:air (Density =1 kg/m^3), Viscosity = 0.2, Reynold number=`frac{rhoVD}{mu}``where rho is the density, D is the diameter of the circle and mu is the viscoity of the air. `Re=frac{rhoVD}{mu}=frac{(1)(1)(2)}{0.2}=10`
• Boundary Conditions:
• At the inlet boundary we take the velocity=1 m/s.
• At the outlet we take the gauge pressure=0.
• At the wall we take the no slip condition.
• In the solution menue we have to create a point 8 meters behind the cylinder to monitor the vortex average velocity at that point.
• we use use the standard initialization for initialization.
• After initialization the solution, we have to set the pressure and velocity contours.
• To clearly observe the vortex average behind the cylinder. we set the solution animation.
• run calculation for 2000 iterations

• Results

• Residuals

In the above graph, we oberve that after 1700 iterations graphs showing the same behaviour.

Vortex shedding behind the cylinder

In the the above graph we easily observe that with reynold 10 there is no vertex shedding happening behind the cylinder.

Drag

Lift

Values of lift and drag are shown in the given table.

Velocity contours

In the above image of velocity contours, we easily see that vertex shedding are not happening here.

Pressure contours

In the above pressure contours graph we observe that there is the increase of pressure in front of cylinder and the drop of pressure occur in the behind of the cylinder.

Video

Table 3

Case	Reynold number	Density	Viscosity	No of iterations	Drag	Lift	Velocity
Steady State	10	1	0.2	2000	3.3524707	-0.000201999	0.44375068

Conclusion

With reynold number 10, we easily observe in the above graphs that there is no vertex shedding happening here.

• Case 2: Reynold number 100

Mesh Details

• First of all we generate the mesh by using the default mesh setting to check the mesh quality. Default mesh is quadrilateral mesh. we change this quadrilateral mesh to triangles in the next step.
• Then right click on mesh and go to insert and go to method, choose the triangular method. Choose the element size to 0.25m and generate the mesh.
• Mesh around the cylinder also play an important role that's why we increase the number of points on the cylinder wall by using the edge sizing option in mesh. This will make the circle edge smoother and more close to perfect circle.
• The number of divisions used in edge sizing is 36.
• The region very close to the circle is also very important to perfectly simulate the flow over the cylinder, so we will insert the inflation layer.
• The first layer heigth is 5e-3, no of layers =10, growrh rate is 1.2.
• After all the setup generate the mesh
• No of nodes= 21622, No of elements= 41810

Case setup details

• Steady state analysis
• Solver type:pressure based
• Velocity formulation:absolute
• Viscous model: Laminar
• Material:air (Density =1 kg/m^3), Viscosity = 0.02, Reynold number=`frac{rhoVD}{mu}` where rho is the density, D is the diameter of the circle and mu is the viscoity of the air. `Re=frac{rhoVD}{mu}=frac{(1)(1)(2)}{0.02}=100`.
• Boundary Conditions:
• At the inlet boundary we take the velocity=1 m/s.
• At the outlet we take the gauge pressure=0.
• At the wall we take the no slip condition.
• In the solution menue we have to create a point 8 meters behind the cylinder to monitor the vortex average velocity at that point.
• we use use the standard initialization for initialization.
• After initialization the solution, we have to set the pressure and velocity contours.
• To clearly observe the vortex average behind the cylinder. we set the solution animation.
• run calculation for 2000 iterations

Transient simulation setting

Results

• Residuals

After 500 iterations, the residuals are started to give almost constant values. And solution is going to converge.

Vertex average velocity

In the above graph of vertex average of velocity magnitude we easily observe that there is vertex shedding happening here. A monitor point was created horizontal at a distance of 4 times the diameter of the circle (4*2=8) from the center of the circle. The vertex average of velocity magnitude at the monitor was plotted as shown above and it started to give almost same pattern after the 500 iterations. By analysing both residuals and velocity plot we say that solution is converged after 500 iterations.

Drag

Lift

In the lift plot we easily observe that the after 500 iterations the lift plot shows thw same pattern. The value of lift and drag are shown in the below table.

Velocity contours

The above image of contours are at the last step of iterations, that is at 2000 iteration. The flow separation, velocity distribution among the fluid layers and vortex shedding are clearly visible in these contours.

Pressure contours

In the above pressure contours graph we observe that there is the increase of pressure in front of cylinder and drop of pressure in the behind of cylinder.

Velocity animation

Table 4

Case	Reynold number	Density	Viscosity	No of iterations	Drag	Lift	Velocity
Steady State	100	1	0.02	2000	1.3440185	0.13169855	0.59834456

• Case 3: Reynold number 1000

Mesh Details

• First of all we generate the mesh by using the default mesh setting to check the mesh quality. Default mesh is quadrilateral mesh. we change this quadrilateral mesh to triangles in the next step.
• Then right click on mesh and go to insert and go to method, choose the triangular method. Choose the element size to 0.25m and generate the mesh.
• Mesh around the cylinder also play an important role that's why we increase the number of points on the cylinder wall by using the edge sizing option in mesh. This will make the circle edge smoother and more close to perfect circle.
• The number of divisions used in edge sizing is 36.
• The region very close to the circle is also very important to perfectly simulate the flow over the cylinder, so we will insert the inflation layer.
• The first layer heigth is 5e-3, no of layers =10, growrh rate is 1.2.
• After all the setup generate the mesh
• No of nodes= 21622, No of elements= 41810

Case setup details

• Steady state analysis
• Solver type:pressure based
• Velocity formulation:absolute
• Viscous model: Laminar
• Material:air (Density =1 kg/m^3), Viscosity = 0.02, Reynold number=`frac{rhoVD}{mu}` where rho is the density, D is the diameter of the circle and mu is the viscoity of the air. `Re=frac{rhoVD}{mu}=frac{(1)(1)(2)}{0.002}=1000`.
• Boundary Conditions:
• At the inlet boundary we take the velocity=1 m/s.
• At the outlet we take the gauge pressure=0.
• At the wall we take the no slip condition.
• In the solution menue we have to create a point 8 meters behind the cylinder to monitor the vortex average velocity at that point.
• we use use the standard initialization for initialization.
• After initialization the solution, we have to set the pressure and velocity contours.
• To clearly observe the vortex average behind the cylinder. we set the solution animation.
• run calculation for 2000 iterations

• Results

Residuals

Vertex shedding graph

Drag

Lift

Velocity contours

Pressure contours

Table 5

Case	Reynold number	Density	Viscosity	No of iterations	Drag	Lift	Velocity
Steady State	1000	1	0.002	2000	1.1142115	-0.62878	0.91794097

Velocity animation

• Case 4: Reynold number 10,000

Mesh Details

• First of all we generate the mesh by using the default mesh setting to check the mesh quality. Default mesh is quadrilateral mesh. we change this quadrilateral mesh to triangles in the next step.
• Then right click on mesh and go to insert and go to method, choose the triangular method. Choose the element size to 0.25m and generate the mesh.
• Mesh around the cylinder also play an important role that's why we increase the number of points on the cylinder wall by using the edge sizing option in mesh. This will make the circle edge smoother and more close to perfect circle.
• The number of divisions used in edge sizing is 36.
• The region very close to the circle is also very important to perfectly simulate the flow over the cylinder, so we will insert the inflation layer.
• The first layer heigth is 5e-3, no of layers =8, growrh rate is 1.2.
• After all the setup generate the mesh
• No of nodes=21513 , No of elements= 41736

Case setup details

• Steady state analysis
• Solver type:pressure based
• Velocity formulation:absolute
• Viscous model: Laminar
• Material:air (Density =1 kg/m^3), Viscosity = 0.02, Reynold number=`frac{rhoVD}{mu}``where rho is the density, D is the diameter of the circle and mu is the viscoity of the air. `Re=frac{rhoVD}{mu}=frac{(1)(1)(2)}{0.0002}=10,000`
• Boundary Conditions:
• At the inlet boundary we take the velocity=1 m/s.
• At the outlet we take the gauge pressure=0.
• At the wall we take the no slip condition.
• In the solution menue we have to create a point 8 meters behind the cylinder to monitor the vortex average velocity at that point.
• we use use the standard initialization for initialization.
• After initialization the solution, we have to set the pressure and velocity contours.
• To clearly observe the vortex average behind the cylinder. we set the solution animation.
• run calculation for 2000 iterations

Results

Residuals

Vertex shedding behind the cylinder

Drag

lift

Velocity contours

Pressure Contours

Table 6

Case	Reynold number	Density	Viscosity	No of iterations	Drag	Lift	Velocity
Steady State	10,000	1	0.0002	2000	1.0505107	-0.22233036	0.8150537

Velocity animation

• Case 5: Reynold number 100,000

Mesh Details

• First of all we generate the mesh by using the default mesh setting to check the mesh quality. Default mesh is quadrilateral mesh. we change this quadrilateral mesh to triangles in the next step.
• Then right click on mesh and go to insert and go to method, choose the triangular method. Choose the element size to 0.25m and generate the mesh.
• Mesh around the cylinder also play an important role that's why we increase the number of points on the cylinder wall by using the edge sizing option in mesh. This will make the circle edge smoother and more close to perfect circle.
• The number of divisions used in edge sizing is 36.
• The region very close to the circle is also very important to perfectly simulate the flow over the cylinder, so we will insert the inflation layer.
• The first layer heigth is 5e-3, no of layers =4, growrh rate is 1.2.
• After all the setup generate the mesh
• No of nodes=21295 , No of elements= 41588

Case setup details

• Steady state analysis
• Solver type:pressure based
• Velocity formulation:absolute
• Viscous model: Laminar
• Material:air (Density =1 kg/m^3), Viscosity = 0.02, Reynold number=`frac{rhoVD}{mu}``where rho is the density, D is the diameter of the circle and mu is the viscoity of the air. `Re=frac{rhoVD}{mu}=frac{(1)(1)(2)}{0.00002}=100,000``
• Boundary Conditions:
• At the inlet boundary we take the velocity=1 m/s.
• At the outlet we take the gauge pressure=0.
• At the wall we take the no slip condition.
• In the solution menue we have to create a point 8 meters behind the cylinder to monitor the vortex average velocity at that point.
• we use use the standard initialization for initialization.
• After initialization the solution, we have to set the pressure and velocity contours.
• To clearly observe the vortex average behind the cylinder. we set the solution animation.
• run calculation for 2000 iterations

Results

Residuals

Drag

Lift

Velocity contours

Pressure contours

Table 7

Case	Reynold number	Density	Viscosity	No of iterations	Drag	Lift	Velocity
Steady State	100,000	1	0.00002	2000	0.94188868	0.21187989	0.68501842

Velocity animation

Conclusion of the results of the the Reynold number 10 to 100,000

In all above cases, we easily observe that except the case with reynold number 10 vertex shedding happening in all cases. All the graphs and contours almost showing the same behaviour physicaly but numerically there is a change occur in the values of drag coefficient(Cd) with different reynold numbers.

In the following table we observe that the values of drag coefficient are changed with respect to different reynold number

Reynold number(`R_e`)	drag coefficient(`C_d`)
10	3.3524
100	1.3440
1000	1.1142
10,000	1.0505
100,000	0.94188

In the above table results we observe that as the reynold number increases values of `C_d` are decreases.

Conclusions

• The flow over the cylinder was successfully simulated for both steady and transient cases with reynold number 100 and the phenomena of von karman vortex street is easily observed.
• The pressure rise at the front side of the cylinder, boundary layer formation, flow separation, vortex formation, vortex propogation were clearly observed with the help of contours and animations.
• The strouhal number was observed 0.167 at reynold 100.
• Six different simuations has been run successfully to observe the behaviour of drag coefficients with respect to the reynold numbers.
• At a reynold number 10 we observe that there is no vortex shedding happening while in other cases we easily observed through the different graphs and contours that vertex shedding are happening.
• All the above cases results are also validated through the data available in the literature.

For the results validation this data is taken from webpage.

References

• Hua Bai, Jiawu Li, "Numerical simulation of flow over a circular cylinder at Low Reynolds numbers. Advanced material Research Vols 255-260(2011) pp-942-946
• B.N.Rajani, A. Kandasamy, Sekhar Majumdar, " Numerical simulation of laminar flow past a circular cylinder. Science Direct, Applied mathematical modelling 33 (2009) 1228-1247.
• Fluid Mechanics, Fundamentals and applications. Yunus A Cengel, John M. Cimbala.