Steady Vs Unsteady flow over a cylinder
Steady Vs Unsteady flow over a cylinder
Part- 1
Aim: To simulate the flow over a cylinder and explain the phenomenon of Karman Vortex Street.
To Find:
- To simulate the flow with steady and unsteady case and calculate the Strouhal Number for Re = 100.
- To calculate the coefficient of drag and lift over a cylinder by setting the Reynolds number to 10,100,1000,10000 & 100000. (Run with steady solver)
- To Discuss the effect of Reynolds number on the coefficient of drag. [ Results should be validated with any standard literature and error should be within 5 %]
Introduction:
- Analysing the flow over a cylinder helps to study the effects of vortex shedding in steady state and in transient state simulations.
- Vortex shedding is an unsteady flow that takes place in a special flow velocities (according to size and shape of the cylindrical body).
- In this flow, vortices are created at the back of the body and detach periodically from either side of the body.
- Vortex shedding is an effect of : viscosity and Reynolds number. It mainly depends up on velocity of the fluid, type of fluid, and shape of the respective objective.
- In this flow, vortices are created at the back of the body and detach periodically from either side of the body.
- Generally vortex shedding occurs above a particular value of Reynolds number, where Reynolds number is represented by Re=(ρ*u*D)/μ). Here ρ = density, u = velocity, D = diameter, μ = absolute viscosity.
- Vortex shedding is basically given by Strouhal Number(st) i.e. st = (f*D)/u, where f = frequency of shedding, D = diameter, u = velocity of fluid.
Procedure :-
Geometry:
- Creating the geometry or import the existing geometry into Space Claim. As per the below drawing the geometry is created.
Meshing:
- Initially when the geometry opens in Meshing, it shows a quadrilateral mesh. In this case consider a triangular mesh.
- Mesh of element size 0.25m is considered to get finer results.
- Once mesh is generated zoom towards the circle, we can see that circle is not in exact shape. Hence flow deviation occurs.
- To capture the flow it should be accurate, for this use Sizing. Sizing helps to control the Mesh.
- To make the circumference accurate click on the circleand enter the number of divisions =36, which makes the circle of cylinder more finer.
- As flow around the cylinder is important, apply a six layers of inflation with reference to the actual circle. To check this zoom in to the circumference of the circle we find six inflation layers.
- Now the mesh follows the body structure i.e. called as body fitted.
- Update and then mesh generates.
- Name the selections of faces as Inlet, Outlet, Wall and Symmetry.
| Details of Mesh | Value |
| Element type (Mesh) | Triangular |
| Element size | 0.25m |
| No of nodes | 20697 |
| Elements | 40502 |
| Maximum Inflation layers | 6 |
| Growth rate (Inflation layers) | 1.2 |
Setup:
Setting up physics:
Solver > Steady state / Transient state, Pressure based, Velocity Formulation=Absolute.
Model > Viscous = Laminar
Material > create a USER-MATERIAL i.e. Density = 1kg/m3, Viscosity = 0.02kg/m-s (Reynolds no. = 100)
Boundary conditions > inlet velocity = 1m/s, outlet- guage pressure = 0 (1 atm)
Setting up solution:
Steady state - HYBRID Initialization
Transient state - STANDARD Initialization
Post processing or Results:
- Create a point where x-axis=8m and y-axis=0m. (As we need point at 4*dia of cylinder)
- Setup contours for both pressure and pressure and velocity.
Solution:
- Go to Definitions and select create a Vertex Average of VELOCITY MAGNITUDE at the point which wee previously created.
- Go to ACTIVITIES, create and create a solution animation for velocity.
- Now INITIALIZE(t=0) and then enter the time step size as well as number of time steps and run the calculations.
- As we refined the mesh finer we need to run for more number of time steps to get better results.
Post processing or Results:
- We can see the Residuals plot, vertex average of velocity magnitude plot as well as pressure and velocity contour plots.
Observations/Results :
Steady State Analysis at Reynolds No. = 100
Residual plot :
Vertex Average of Velocity-Magnitude plot :
Pressure Contour :
Velocity Contour :
coefficient of drag :
Coefficient of Lift :
Animation of Velocity field over a cylinder representing the Vortex Shedding in Steady State:
In Steady State Analysis we cannot calculate time and hence we cannot calculate the frequency. Without frequency wee cannot obtain the Strouhal Number. To find the frequency and Strouhal Number we next use Transient State Analysis.
Transient state Analysis at Reynolds No. = 100
Residual plot :
Vertex Average of Velocity-Magnitude plot :
Pressure Contour :
Velocity Contour :
Animation of Velocity field over a cylinder representing the Vortex Shedding in Transient-State:
Coefficient of Drag and Coefficient of Lift plots for Re= 100(Transient case) :
Calculating the Strouhal Number for Re= 100:
- Vortex shedding is basically given by Strouhal Number(st) i.e. st = (f*D)/u.
where f = frequency of shedding/oscillations
D = diameter,
u = velocity of fluid.
- Strouhal Number is directly praportional on the frequency of Vortex Shedding.
- In steady state analysis we cannot find the frequency as there is no time parameter involved , where as in case of transient Analysis we can find the frequency.
- So the frequency(Hz) can be determined by using FFT Analysis(Fast Fourier Transform).
- For this, go to DEFINITIONS > NEW > FORCE REPORT > select LIFT. Initialize and run the simulation. Once the simulation is completed. Write the generated data and save in a file.
- Now go to FFT >load the input file(lift plot values)> select Y-AXIS function= POWER SPECTRAL DENSITY and X-AXIS function= FREQUENCY > plot FFT >once solution is done Write FFT to file.
- From the saved file we can easily find the value of the frequency. Using this value of frequency we can find the Strouhal Number i.e.
>> st = (f*D)/u.
where f = 0.085 Hz
D = 2m
u = 1m/s
>> st = (0.085*2)/1
>> st = 0.17
- Hence the Strohal Number = 0.17.
> In the vertex average of velocity magnitude plot it is observed that initially the trend was smooth but later it showed an oscillating behaviour after few iterations. This is due to the formation of vortex shedding. This doesnot change after few iterations and hence we can say that the flow has converged.
> In the steady state flow analysis we couldnot actually predict the exact result during the flow but in from the unsteady flow analysis shows the accurate result at any time interval of the flow. Hence the result that is obtained during the unsteady state analysis can be used to calculate the Strouhal number.
Part- 2
Calculating the coefficient of drag and lift over a cylinder by setting the Reynolds number to 10,100,1000,10000 & 100000. (Run with steady solver)
Coefficient of drag - As per fluid dynamics, the drag coefficient is a dimensionless quantity that is used to quantify the drag or the resistance of an object in a fluid environment(air/water). A low drag coefficient is that the streamline shape of the body is such as to enable it to move easily through the surrounding viscous air with the minimum resistance, whereas a high drag coefficient is caused by poor streamlining of the body profile so that there is a high air resistance when the body is in motion. Cd depends upon the orientation of the object to the flow . The coefficient of drag is given by the following equation:
where,
Fd- drag force, ρ= density of fluid, A=Area
Coefficient of Lift - As per fluid dynamics, the drag coefficient is a dimensionless quantity that co relates the lift generated by a lifting body to the fluid density around the body. The component of the aerodynamic force that is perpendicular to the flow direction and the drag is parallel to the flow direction. he coefficient of lift is given by the following equation:
where,
L- lift force, ρ= density of fluid, u=flow speed, A=Area
Geometry,Mesh, case setup considerations :-
- Geometry is same as in the above cases.
- Mesh type= Triangular mesh
- First layer height=5.e-003m
- Max. inflation layers=6
- Growth rate=1.2
- Element size= 0.25m
- Nodes=20697
- No. of Elements=40502
- Steady State analysis- pressure based solver- viscus model= LAMINAR(RE=10,100,1000) and Viscous model = k-omega SST model (Re=10000,100000).
- viscosity = 0.02 for all the cases .
- Inlet velocity is varied to change the reynolds number.
- For finding the right values of drag coefficient and lift coefficient use reference values .
Re=10 :
- Reynolds Number > Re=(rho*u*D)/(mu)
Re=10, rho=1kg/m^3, u=0.1m/s, D=2m, viscosity= 0.02 , No of iterations=1500.
Residual plot :
Vertex Average of Velocity-Magnitude plot :
Coefficient of Drag plot :
Here the Coefficient of Drag = 3.32427
Coefficient of Lift plot :
Here the Coefficient of Lift = 0.0009267
Velocity Coontour Plot :
Pressure Contour Plot :
Re=100 :
- Reynolds Number > Re=(rho*u*D)/(mu)
Re=100, rho=1kg/m^3, u=1m/s, D=2m, viscosity= 0.02 , No of iterations=1000.
Residuals Plot :
Vertex Average of Velocity-Magnitude plot :
Coefficient of Drag plot :
Here the Coefficient of Drag = 1.33652
Coefficient of Lift plot :
Here the Coefficient of Lift = 0.0164
Velocity Contour Plot :
Pressure Contour Plot :
Re=1000 :
- Reynolds Number > Re=(rho*u*D)/(mu)
Re=1000, rho=1kg/m^3, u=10m/s, D=2m, viscosity= 0.02 , No of iterations=1000.
Residuals Plot :
Vertex Average of Velocity-Magnitude plot :
Coefficient of Drag plot :
Here the Coefficient of Drag = 0.75896
Coefficient of Lift plot`
Here the Coefficient of Lift = 0.13587
Velocity Contour Plot :
Pressure Contour Plot :
>> As we know if the Reynold number is less than 2000, then flow in a pipe is Laminar (i.e Re=10, Re=100, Re=1000). Hence we use Viscous model = Laminar. Where as if the Reynold number is greater than 2000, the flow is turbulent (i.e Re=10000 and Re=100000). For this case turbulence model = k-omega SST model.
Re=10,000 :
- Reynolds Number > Re=(rho*u*D)/(mu)
Re=10000, rho=1kg/m^3, u=100m/s, D=2m, viscosity= 0.02 , No of iterations=1000.
Residuals Plot :
Vertex Average of Velocity-Magnitude plot :
Coefficient of Drag plot :
Here the Coefficient of Drag = 0.71922
Coefficient of Lift plot :
Here the Coefficient of Lift = 0.33536
Velocity Contour Plot :
Pressure Contour Plot :
Re=1,00,000 :
- Reynolds Number > Re=(rho*u*D)/(mu)
Re=100000, rho=1kg/m^3, u=1000m/s, D=2m, viscosity= 0.02 , No of iterations=1500.
Residuals Plot :
Vertex Average of Velocity-Magnitude plot :
Coefficient of Drag plot :
Here the Coefficient of Drag = 0.55103
Coefficient of Lift plot :
Here the Coefficient of Lift = 0.28657
Velocity Contour Plot :
Pressure Contour Plot :
Conclusions for the above obtained results - Reynolds Number 10 to 100000 :
From the results obtained it is observed that the formation of vortex shedding is happening in all the cases except Re=10. It is also noted that there is a change in the values of drag coefficient(Cd) with varying the Reynolds Numbers.
The below table shows the variation of drag coefficient on changing the Reynolds Number:
| Reynold's Number | Drag Coefficient (Cd) |
| 10 | 3.32427 |
| 100 | 1.33652 |
| 1000 | 0.75896 |
| 10000 | 0.71922 |
| 100000 | 0.55103 |
The table shows that as the reynolds number increases the value of cd decreases. And hence we can see the variation of drags by changing reynold number. The drag on the object depends on the shape,size and flow conditions.
Conclusion :
- In the first case, the flow over a cylinder is simulated for both Steady and unsteady cases with Re=100 and with the results generated the phenomena of Von Karman Vortex Street is also observed.
- Necessary plots are plotted in order to observe that the the flow is converged and the vortex shedding is happening.
- Contours are generated in order to observe the flow over a cylinder and the animations show the vortex propagation.
- By plotting Power Spectral Density vs Frequency , the Strouhal number is calculated and observed as 0.17 at Re=100.
- In the second case,
six different simulations i.e (Re = 10,100,1000,10000,100000,1000000) are simulated to observe the behaviour of the drag and lift coefficients by changing the reynold number.
- In order to observe the behaviour of the drag coefficients for the simulations, the velocity of the flow is varied to change the Reynolds number and rest of the parameters are kept constant.
- As we know if the Reynold number is less than 2000, then flow in a pipe is Laminar. Hence we use Viscous model = Laminar. Where as if the Reynold number is greater than 2000, the flow is turbulent. For this case turbulence model = k-omega SST model is used for simulations.
- For Re=10 , it is observed that there is no formation of vortex shedding and in rest of the cases it is clearly observed.
- As per the basic conditions and parameters the results are validated from the literature.
- As we refine the mesh we need to run more number of iterations.
Reference Materials :
- Numerical simulation of laminar flow past a circular cylinder. Science Direct, Applied mathematical modelling.
- Numerical Study of the Drag coefficient of an Infinite Circular Cylinder at Low and Moderate Reynolds Numbers using the FLUENT Code. (Berge DJEBEDJIAN and Jean-Pierre RENAUDEAUX,ESCPI, Conservatoire National des Arts et Métiers, Paris, FRANCE).
- Numerical Simulation of Flow over a Circular Cylinder at Low Reynolds
Number. By Hua Bai. Advanced Materials Research Vols 255-260(2011) - https://floringh.gitbooks.io/computational-fluid-dynamics/von_karman_vortex_street_[simflow].html
- http://www.mate.tue.nl/mate/pdfs/2743.pdf