Challenge

Flow over a cylinder

AIM:

  • To run a baseline simulation and then calculate the Reynolds number.
  • To increase the Reynolds number by a factor of 20%, 40%, and 100% then run the transient flow simulation.

INTRODUCTION:

A common external flow configuration involves the circular cylinder or tube in crossflow, where the flow is normal to the axis of the cylinder.

As the fluid approaches the front side of the tube, the fluid pressure rises from the freestream value to the stagnation point value. The high pressure forces the fluid to move along the tube surface and boundary layers develop on both sides. The pressure force is counteracted by viscous forces and the fluid cannot follow the tube surface to the rear side but separates from both sides of the tube and form two shear layers. The innermost part of the shear layers are in contact with the tube surface and moves slower than the outermost part. As a result, the shear layers roll up. The flow pattern is dependent on the Reynolds number.

Reynolds number:

Re = rho*Ut*D/mu

D = cylinder diameter

mu = dynamic viscosity

rho = density

MODEL SETUP:

Diameter= 0.25m

Length of cylinder=0.5m

PROCEDURE

  • Initially the model of a cylinder with diameter of 0.25m and length of 0.5m is created.
  • In the the flow simulation, velocity in x direction=20m/s, fluid=Air was setup
  • Select the external flow and time dependent.
  • Computational domain is automatically created.
  • Mesh automatically set as level 3.
  • Run the simulation.

Calculations:

Re = rho*Ut*D/mu

Re = 1.225*20*0.25/(1.81*10^(-5))

Re = 338397.79

Now increasing the Reynolds number by 20%, 40% and 100%

At 120% of Reynolds number:

Re = 1.2*338397.79 = 406077.34

Therefore,

Ut = 406077.34 *(1.81*10^(-5))/1.225*0.25

Ut = 24m/s

At 140% of Reynolds number:

Re = 1.4*338397.79 = 473756.90

Therefore,

Ut = 28m/s

At 200% of Reynolds number:

Re = 2*338397.79 = 676795.58

Therefore,

Ut = 40m/s

CALCULATING SIMULATION END TIME

velocity=distance of computation domain/time

computation domain=1.375+0.875=2.25m

  1. Flow Through Time (FTT)=2.25/10=0.225sec

End Time=2*FTT

=0.45 sec

  1. Flow Through Time (FTT)= 2.25/12=0.1875sec

End Time=2*FTT

=0.375sec

  1. Flow Through Time (FTT)= 2.25/14=0.16sec

End Time=2*FTT

=0.32sec

  1. Flow Through Time (FTT)= 2.25/20=0.1125sec

End Time=2*FTT

= 0.225sec

Results:

At initial Reynolds number, Cut plot at middle of the cylinder.

At 20% more than initial Reynolds number, Cut plot at middle of the cylinder.

At 40% more than initial Reynolds number, Cut plot at middle of the cylinder.

At 100% more than initial Reynolds number, Cut plot at middle of the cylinder.

Conclusion:

There is a slight noticeable difference between these plots, where Reynolds number is increased which intern increased velocity of the flow over the cylinder. When speed of air is low, the wake zone is more whereas in case of high speed, it is vice versa.

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