Steady and transient flow simulations of flow over a cylinder by use of Fluent

Title :  Steady and transient flow simulations of flow over a cylinder.

Objective

1.  To simulate the flow over a cylinder and explain the phenomenon of Karman vortex street.

2. To simulate the flow with the steady and unsteady case and calculate the Strouhal Number for Re= 100.

3. To calculate the coefficient of drag and lift over a cylinder by varying the Reynolds number from 50 to 300.

4. To discuss the effect of Reynolds number on the coefficient of drag.

Theory:

           For given challange simulation done for flow over a cylinder. There is use of steady & unsteady method for simultion.When there is fluid flow over a cylinder ther is karman vortex street is generated & which responsible for unsteady  sepration of flow. steady state simulation is run 900 iteration & find out the coefficent drag & lift.

Karman vortex street

In fluid mechanics a karaman vortex street is a repating pattern of swirling vortices , caused by a process known as vortex shedding , which is responsible for the unsteady separation of flow of a fluid around the blunt bodies. its named as karman vortex street.

                             

Strouhal No:

          The Strouhal number represents the ratio of inertial force due to the local accelration of the flow to the inertial forces due to the convective acceleration.In flows characterized by a periodic motion, the Strouhal number is associated with the oscillations of the flow due to the inertial forces relative to the changes in velocity due to the convective acceleration of the flow field. Thus, an alternative definition of the Strouhal number better suited to periodic flows is given by the following expression.

                                St= F *d/U

where F= Vortex shedding frequency

         d= diameter of cylinder

         U = Flow velocity

Given data

1. Inlet velocity =1m/sec

2. Density = 1kg/m3

3. Visocity vary for Re=50 to 300

4. Cylinder diamter = 2m

5. Cylinder length =1m

6. Depth of cylinder =1m

7. Rectangular plate length =60m

8. Rectangular plate Height=20m

9. Distance from inlet of cylinder= 20m

10.Distance from cylinder centre to outlet=40m

11. Element size use= 0.25

Geometry Details

PART- I: Simulate the flow with the steady and unsteady case and calculate the Strouhal Number for Re= 100.

Steady state:

steady state simulation run for 800 iteration & viscosity 0.02 kg-m/s for Re=100 & following result is obtained

Residual plot

Velocity vertex Average

Coefficient drag

Coefficient Lift

 

From the above graph of coefficient of lift the & vertex aevrage velocity after 500 iteration the solution is constant & so that we can say that solution is converged.

following plot of velocity & pressure contour at Re=100

Velocity contour

Pressure contour

From the following video we can see that there is generation karman vertex shedding  generated at Re=100

Velocity Animation

Pressure animation

For the steady state we can not calculate the strouhal no .,because steady state gives us solution for final specific time. But for findng strouhal there is need to generation of frequency of oscillation for flow-time second. so, we can not calculate the strouhal no. for stedy state. 

Unsteady state

Unsteady state simulation run for 1700 time step, time step size=0.01 & viscosity 0.02 kg-m/s for Re=100 & following result is obtained

Residual plot

Velocity vertex Average

Coefficient of Drag

Coefficient of lift

From the above graph of coefficient of lift the & vertex aevrage velocity after  flow- time 120 sec soltion is constant , so that there is solution converged

Strouhal No

from the graph we can see that strouhal no is 0.20. we can calculate the strouhal no manually by use of coefficiet of lift graph.

 there are 5 peaks for 50 second which gives the frequency 5/50=0.1 oscillation/sec. so that

f= frequency of oscillation =0.1

d=diameter of cylinder=2m

U=inlet velocity=1m/s

                                        St (Stouhal no)= F *d/U

                                            St= (0.1*2)/1

                             Strouhal No=0.2 for Re=100 by use of unsteady state method

following plot of velocity & pressure contour at Re=100

Velocity contour

 

Pressure contour

From the following video we can see that there is generation karman vertex shedding  generated at Re=100

Velocity animation

 

Pressure animation

 

 

PART-II:Calculate the coefficient of drag and lift over a cylinder by varying the Reynolds number 

As unsteady method take to much time get the solution , so that there is use of steady method. there is vary the reynold no. in step of 50 , so its 50, 100,150,200,250 ,300. for variation of reynold no. there is calculate the coefficent lift & drag.  steady state simulation for 900 iteration.

Reynold No=50

Residual plot

Coefficient Drag

 

 

Coefficient lift

 

 

Average vetex velocity

From the above graph of coefficient of lift the & vertex aevrage velocity after 500 iteration the solution is constant & so that we can say that solution is converged.

Velocity contour

for Re=50 there is not generated the vortex shedding behind th cylinder.

Pressure contour

Reynold No=100

Residual plot

Velocity vertex Average

Coefficient Drag

Coefficient lift

From the above graph of coefficient of lift the & vertex aevrage velocity after 500 iteration the solution is constant & so that we can say that solution is converged.

velocity contour

for Re=100 there is generated the vortex shedding behind th cylinder.

Pressure contour

Reynold No=150

Residual plot

Coefficient Drag

Coefficient lift

From the above graph of coefficient of lift the & vertex aevrage velocity after 400 iteration the solution is constant & so that we can say that solution is converged.

velocity contour

for Re=150 there is generated the vortex shedding behind th cylinder.

Pressure contour

Reynold No=200

Residual plot

Velocity vertex Average

Coefficient Drag

Coefficient lift

From the above graph of coefficient of lift the & vertex aevrage velocity after 325 iteration the solution is constant & so that we can say that solution is converged.

Velocity contour

vortex shedding generated behind cylinder for  Re=200

Pressure contour

Reynold No=250

Residual plot

Velocity vertex Average

coefficient Drag

 

coefficient lift

From the above graph of coefficient of lift the & vertex aevrage velocity after 280 iteration the solution is constant & so that we can say that solution is converged

velocity contour

vortex shedding behind the cylinder for Re=250

pressure contour

Reynold No=300

Residual plot

Velocity vertex Average

Coefficient Drag 

Coefficient lift

From the above graph of coefficient of lift the & vertex aevrage velocity after 220 iteration the solution is constant & so that we can say that solution is converged

Velocity contour

vortex shedding behind the cylinder for Re=300

Pressure contour

 

Effect of Reynold No on Coefficient Drag

From the above graph we can say that as reynold is increases there is reduction of Coefficient of drag

Effect of Reynold No on Coefficient Lift

From the above graph we can say that as reynold is increases there is increase of Coefficient of lift

Coefficent Drag & Lift at Diffrent Reynold No.

the above value of Cd, & Cl got  for varing the reynold no from 50 to 300 by use of steady state method.

 Overall conclusion

1. For the steady state we can calculate the strouhal no .,because steady state gives us solution for final specific time, as strouhal there is need to generation of frequency of oscillation for flow-time second. 

2. for unsteady state the strouhal no 0.2 is got.

3. There is generation of karman vortex street for steady & unsteady state.

4. As genration of karman vortex street is depend upon the Reynold no, for reynold no=50  dont find the karman vortex street.

5. As reynold is increases there is reduction of Coefficient of drag.

6. As reynold is increases there is increase of Coefficient of lift.