Flow over a cylinder
Aim :- To simulate the flow over a cylinder in solidworks
Objective :-
- To simulate the flow over a cylinder for the conditions shown in the video
- To Run a baseline simulation and then calculate the Reynolds number. Then increase the Reynolds number by a factor of 20%, 40%, and 100% then run the transient flow simulation.
- To create animations for pressure-velocity for all the three Reynolds number.
Formula / Governing eqn :-
- using the renolds no we can findout the inlet velocity by the given formula
| `R_e =(rho. V.D) / mu` |
Where
- `R_e`= Reynolds No
- `rho` = density of fluid
- V = Velocity of fluid
- D = dia of pipe
- `mu` = dynamic viscosity
- Input values are as given
- Dia of pipe (D) = 0.02m
- Length of pipe (l) = 0.05m
- Here the value for rho & mu are considered for the air at 25deg Celsius ie air at room temp.
- `rho` = 1.184
- `mu` = 1.849 x 10^-5
- Value for V is considered from the video as the starting/initial value from that Reynolds no is calculated and then increase the Reynolds number by a factor of 20%, 40%, and 100% we get three values as follows
- V = 10 m/s
- `R_e` = ?
We get Re as
Re = 15368.2
Re = 17929.6
Re = 25613.8
- From Re values we have calculated the velocity for the inlet and is as follows
V = 12
V = 14
V =20
- Now Using the velocity Values and length of domain we can calculated the time using the formula
`V = D/T` `ie` `T = D/V`
3D Part Design :-
- Pipe design is shown below for dia=0.02 & L=0.05
- In this Project the Automatic mesh generation was used and the level was set to 4 and is shown below
- An Overall view
- Animation view for the case 1 where cutplot1 represent the velocity flow and the video for it is provided below
Explanation :-
- Software used solid works 2016
- Start by designing the circle of 0.02m dia and using extrude command to extrude for a length of 0.05m
- Now as we want the flow Outside the pipe set new flow simulation
- Considering the velocity as 10m/s and running the baseline simulation to find out the Reynolds no
- Then increase the Reynolds number by a factor of 20%, 40%, and 100% then run the transient flow simulation, thus three Reynold’s no are obtained
- Now by using the formula of Reynold’s no we get the three input velocity values
- Simulation end time should be calculated properly.
- Now it’s time to start the flow simulation
- Using the new wizard, provide the time, velocity values for the flow
- Now Run the study and wait for the calculations been done for the iterations once its done we can obtain the simulation
- Now to create new simulation for the flow over the pipe for velocity & pressure and create an animation of it.
Result :-
- Thus we have studied the flow over a cylinder
- Using the values from baseline simulation calculated three Reynold’s no and from it the three velocity values for the simulation
- Also obtained the animation for the Pressure-Velocity for the three different Reynold’s no & inlet Velocity values
Flow Simulation :-
- The animations for pressure-velocity for all the three Reynolds number & for all the three Velocity values is shown below
- For CASE-1 ( Re = 15368.2 & V = 12 )
- For CASE-2 ( Re = 17929.6 & V = 14 )
- For CASE-3 ( Re = 25613.8 & V = 20 )
Observation & Conclusion :-
- It is seen that the flow changes for increase in the velocity values
- Also the pressure variation is seen to be a bit different in cases 1, 2 & 3
- As velocity values for case2 & 3 increases the end time also changes that results in a smooth flow
- If solver solves for more mesh values and the iterations values, a better looking flow is obtained