Flow over a cylinder

Aim : - Simulate the flow over a cylinder with different reynolds number

Reynolds number

The Reynolds number is the ratio of inertial forces to viscous forces and is a convenient parameter for predicting if a flow condition will be laminar or turbulent. It can be interpreted that when the viscous forces are dominant (slow flow, low Re) they are sufficient enough to keep all the fluid particles in line, then the flow is laminar. Even very low Re indicates viscous creeping motion, where inertia effects are negligible. When the inertial forces dominate over the viscous forces (when the fluid is flowing faster and Re is larger) then the flow is turbulent.

• It is defined as:
  
  • `Re = (rhoVD)/mu` ------- equation (1)
  • Where
    • V is the flow velocity,
    • D is a characteristic liner dimension (whitch is equal to the diameter of cylinder)
    • `rho` is fluid density
    • `mu`is dynamic viscosity
      

Calculations : -

Cylinder diameter = 0.04m

Cylinder length = 0.05m

Density of air at 20 degree celsius is 1.205kg/m^3

dynamic viscosity at 20 degree celsius is 1.82 *10^-5 kg/ms

Baseline

Inlet velocity 10m/s

Using above reynolds number formulae we get reynolds number equal to 26483

length of domain is 0.22+0.14 = 0.36m

Flow through time = 0.36/10 = 0.036 sec

Simulation end time = 2*Flow through time = 0.072sec

20% increase in reynolds number

After 20% increase in baseline reynolds number we get new reynolds number equal to 31780

Put the value of new reynolds number in equation(1) keeping other factor constant we get new input velocity as 12m/s

Length of domain = 0.36m

Flow throught time = 0.36/12 = 0.03 sec

Simulation end time = 0.06 sec

40% increase in reynolds number

After 40% increase in baseline reynolds number we get new reynolds number equal to 37076

Put the value of new reynolds number in equation(1) keeping other factor constant we get new input velocity as 14m/s

Length of domain = 0.36m

Flow throught time = 0.36/14 = 0.0257 sec

Simulation end time = 0.0514 sec

100% increase in reynolds number

After 100% increase in baseline reynolds number we get new reynolds number equal to 52966

Put the value of new reynolds number in equation(1) keeping other factor constant we get new input velocity as 20m/s

Length of domain = 0.36m

Flow throught time = 0.36/20 = 0.018

Simulation end time = 0.036

Baseline Simulation (Reynolds number = 26483 or Input velocity = 10m/s)

Velocity variation

• Animation
  • https://youtu.be/x4F0j4xwAAE

Pressure variation

• Animation
  • https://youtu.be/s1oUSG6m4xE

20% increase in reynolds number (Reynolds number = 31780 or Input velocity 12m/s)

Velocity variation

• Animation
  • https://youtu.be/SufD6NfLCm4

Pressure variation

• Animation
  • https://youtu.be/vcJGhjpzIPg

40% increase in reynolds number (Reynolds number = 37076 or Input velocity 14m/s)

Velocity variation

• Animation
  • https://youtu.be/LLgF-iN9qXo

Pressure variation

• Animation
  • https://youtu.be/hcA3OvDh2Xk

100% increase in reynolds number (Reynolds number = 52966 or Input velocity 20m/s)

Velocity

• Animation
  • https://youtu.be/ngcTOvgy-EQ

Pressure

• Animation
  • https://youtu.be/fgbkQZmeFVc

Observation and conclusion

• At stagnation point when the air touch the surface of cylinder the velocity of air is zero then their is a sideward flow of air at the speed higher than the input velocity and at the back of cylinder the velocity is again zero .
• At the stagnation point when the air touch the surface of cylinder the pressure is at maximum because all the kinetic energy of air converted into the pressure head , In side of of cylinder the pressure is lower than the atmospheric pressure beacuse of high kinetic head .
• When we increase the reynolds number the input velocity also increase so the flow throught time and simulation end time decrease and the simulation reached the study state in less time .