Steady Vs Unsteady flow over a cylinder

                                                    FLOW OVER CYLINDER # challenge 3

Aim

To Simulate the flow with the steady and unsteady case and calculate the Strouhal Number for Re= 100. Also to Calculate the coefficient of drag and lift over a cylinder by setting the Reynolds number to 10,100,1000,10000 & 100000. And to discuss the effect of Reynolds number on the coefficient of drag.

Introduction

Flow around a circular cylinder still remains a challenging problem in fluid mechanics, where intensive investigations are continued even today to understand the complex unsteady dynamics of the cylinder wake flow. Cross-flow normal to the axis of a stationary circular cylinder and the associated problems of heat and mass transport are encountered in a wide variety of engineering applications, Such as important to study the flow past bluff bodies are offshore and marine engineering, marine pipe lines etcetra.Both experimental measurements and numerical computations have confirmed the onset of instability of the wake flow behind a cylinder beyond a critical Reynolds number, leading finally to a kind of periodic flow identified by definite frequencies, well-known as the Von Karman vortex street.

Theory

The flow usually separates at one or more sharp corners of the cross-section geometry itself, forming a system of vortices in the wake on either side of the mid symmetry plane.

As the flow Reynolds number (Re=UD/ν) based on the free stream velocity (U), cylinder diameter (D) and kinematic viscosity ν of the fluid, changes from a creeping laminar flow value of the order of 0.1 to a turbulent flow of the Re of a million or even higher.Steady laminar flow exists at Reynolds number between 5 and 40 with a pair of symmetric counter-rotating vortices formed behind the cylinder.Beyond a critical value of Re, depending on the other flow disturbances, a transverse oscillation sets in with loss of flow symmetry and vortices are shed from the cylinder surface.

Solving and Modelling

1.Create the geometry

2.Mesh the geometry

3.Pre processing 

4.Post-processing

For solving the problem first we should construct the model through Spaceclaim.With Dimensions 20m*60m in area of boundary and 2m as the radius of the 2D cylinder.Using staedy and unsteady case we will ca;culate strauhals number from the velocity plot,this is done by calculating the frequency of the plot.And also using the steady solver we will find the coefficient of drag and lift for specified reynolds number.

Preprocessing

The element size is changed from default to 0.25 ,The meshed model is open for pre-processing. For pre-processing we configure the Fluent launcher with double precision and in serial, also with steady state  and trancient state Pressure-based solver with absolute velocity formulation. Afterwards initial boundary conditions are assigned i.e. with inlet velocity and other parameters. By selecting the turbulence model we initialise the solver and input the number of iteration we desire and calculate. The plots including residual plots are generated, from the graph we can determine the convergence, the velocity, the coefficient of drag and lift  also which model prevails.

The meshed part with element size of 0.25m

 the total number of elements generated                                The mesh aspect ratio

Solution model                                                                   Fluent launcher set up

Results (Part 1)(steady and unsteady case and calculate the Strouhal Number for Re= 100)

1. Steady state (350 iteration)

Residuals(figure a)

Velocity plot

Coefficient of drag plot

Coefficient of lift plot

velocity contour

Pressure contour

Result(steady state)

For steady state analysis for the flow over the cylinder we found from the velocity plots generated that the frequency to be 0.1314 hz,By applying the strouhals number equation Sn=(frequency * diameter of cylinder)/velocity of the inlet. We can confirm from this that straouhls number for steady state as 0.2628.Running for 350 iterations ,While it has attained convergence in between 300 and 350 iteration.We can see this because the amplitude of velocity and residuals remained constant.

2.Transient state(1000 time steps with size of 0.1s)

Residuals(figure b)

velocity average

drag coefficient

coefficient of lift

Countor plots

contour plot for pressure

Result (transient)

from the velocity plot the formation of vertex started to happen at 28 s and gets converged around  75 to 80 s.While the strouhal number for transient state will be 0.25 with frequency as 0.125 and velocity 1m/s and diameter of cylinder as 2m.

Inference and Results

By comparing the plots of steady state and transient we can conclude that both have converged

Steady state  (figure a)                                                                                              Transient (figure b)

In steady state the solution has converged within 300 iteration while similarly the transient state also converged in 75s

strohaul number for steady state=0.2628

strohaul number for transient state=0.25

Conclusion

Considering the real world vortex shedding situations it is better to choose transient solution over steady state considering accurate predition but comes with a trade off that it is computationaly expensive.Also from the Pressure plot we can see that the pressure is maximum on the front side indicating the impact effect of the fluid.

Part 2.

 coefficient of drag and lift over a cylinder by setting the Reynolds number to 10,100,1000,10000 & 100000 with steady state analysis.provided the viscosity to be a constant(0.05 kg/m-s) also density constant(1 kg/m3).The reynolds number is varied by changing the inlet velocity

For re 10 inlet velocity 0.25m/s  

Lift Coefficient(0.000928)                                                                                               Drag coefficient(3.3242)

For re 100 inlet velocity 2.5m/s 

Lift coefficient         (0.0892)                                                                      Drag coefficient(1.33407)

for re 1000 inlet velocity 25m/s 

 Lift coefficient    (0.40663)                                                                          Drag coefficient(0.71076)

for re 10000 inlet velocity 250m/s  

    Lift coefficient    (0.5634)                                                                                Drag coefficient (0.7502)

for re 100000 inlet velocity 2500m/s  

   lift coefficient(0.5942)                                                                     Drag coefficient(0.69188)

Conclusion 

Comparing the drag coefficients and Reynolds number from the above plots results we can conlude that the drag coefficients reduces with increasing Reynolds number.The graph given below shows how the Coefficient of Drag decreases with increase in Reynolds number For a cylinder.By Comparing the Results from The plot generated and the graph from below(figure2) we can confirm that our solution is correct and we are on the right path.From figure 1 we can see that the drag for cylinder is 0.8 at reynolds number 10^4 ,Finalising from our result at reynolds number 10^4 we obtained around 0.75as coefficient of drag.

Table and plot 

Coeff lift/drag v/s Reynolds number This plot confirms the theory and experiment that coefficient of drag decreases with increase in Reynolds number.

      figure 1                                                                         figure 2

Reference

• Nasa aeronautics and Space admnistration    https://www.grc.nasa.gov/www/k-12/airplane/dragsphere.html#:~:text=For%20most%20aerodynamic%20objects%2C%20the,large%20range%20of%20Reynolds%20numbers.&text=In%20all%20of%20the%20cases,to%20increase%20the%20Reynolds%20number.
• DRAG COEFFICIENT OF A CYLINDER author Giancarlo Bruschi http://sv.20file.org/up1/916_0.pdf
• Wiki pedia https://en.wikipedia.org/wiki/Drag_coefficient
• Numerical simulation of laminar flow past a cylinder ,Author links open overlay panelB.N.Rajani^aA.Kandasamy^bSekharMajumdarhttps://www.sciencedirect.com/science/article/pii/S0307904X08000243