Flow over a cylinder

Aim :

1. Simulation the flow over a cylinder.
2. Run a baseline simulation and calculate the Reynolds number. Then increase the Reynolds number by a factor of 20%, 40% and 100% and then run the simulation.
3. Create animations for pressure-velocity for all the Reynolds number.

Introduction :

A fluid flow is classified as being internal or external flow. The flow of an unbounded fluid over a surface is external flow. The flow in a pipe or duct is internal flow if the fluid is completely bounded by solid surface.

The flow of a fluid is transient or unsteady if its flow parameters (i.e. velocity and pressure) are depend not only on the position in the coordinate system used to describe the field of the flow , but also on time.

The Reynolds number (Re) is an important dimensionless quality in fluid mechanics used to help predict flow patterns in different fluid flow situation. At low Reynolds numbers, flows tend to be dominated by laminar (sheet like) flow, while at high Reynolds numbers turbulence results from differences in the fluid’s speed and direction, which may sometimes intersect or even move counter to the overall direction of the flow.

The Reynolds number is defined as,

`Re= ρuL/μ=uL/v`

Where,

ᵨ is the density of the fluid.

u is the velocity of the fluid with respect to the object.

L is the characteristics linear dimension.

μ is the dynamic viscosity of the fluids.

v is the kinematics velocity of the fluid.

Geometry of the model :

In this project, a cylinder is created in Solidworks CAD software. The design parameters are as follows

Diameter of the cylinder = 0.04m

Length of the cylinder = 0.05m

Initially a baseline velocity of air is considered as 0.03651m/s. on the basis of initial velocity, Reynolds number is calculated.

Inputs	Values
Analysis type	External flow
Fluid	Air
Diameter	0.04m
Initial velocity	0.03651m/s
Density of the air	1.225kg/m^3
Dynamic viscosity	17.89*10^-6 Pa.s

So,

By substituting the above values we get the Reynolds number,

Re = 100

Similarly, Reynolds number at 20%, 40% and 100% is calculated,

At 20% Re = 120

At 40% Re = 140

At 100% Re = 200

For length of the domain = 0.22+0.14 = 0.36m (in x-axis direction)

so, time = distance(length of the domain)/velocity.

For Re = 100

Velocity = 0.03651m/s

Flow through time = 0.36/0.03651 = 9.86sec

So end time = 2 * flow through time = 2 * 9.86 = 19.72sec

• Pressure plot and animation :

• velocity plot and animation :

Results :

Case 1 : increased Reynolds number by 20%

Then, Re = 120

`Re= ρuL/μ`

so, Velocity, v = 0.04381m/s

Flow through time = length of the domain/velocity = 0.36/0.04381 = 8.22sec

End time = 2*8.22=16.44sec

• Pressure plot and animation :

• Velocity plot and animation :-

Case 2 : increased Reynolds number by 40%

Then, Re = 140

`Re= ρuL/μ`

so, Velocity, v = 0.05111m/s

Flow through time = length of the domain/velocity = 0.36/0.05111 = 7.04sec

End time = 2*7.04=14.08sec

• Pressure plot and animation :

• Velocity plot and animation :-

Case 3 : increased Reynolds number by 100%

Then, Re = 200

`Re= ρuL/μ`

so, Velocity, v = 0.07302m/s

Flow through time = length of the domain/velocity = 0.36/0.07302 = 4.93sec

End time = 2*4.93=9.86sec

• Pressure plot and animation :-

• Velocity plot and animation :-

Conclusion :

1. For velocity distribution, we can say that the velocity of the air just before striking the cylinder is low but after striking the cylinder, it passes over the cylinder with high velocity and then in back velocity decreases.
2. For pressure distribution, if the velocity greater then pressure will be lesser. Low pressure region is created where the velocity is high and high pressure region is created where velocity is low. But in this project as the velocity is very low, so it is very difficult to identify the pressure region.
3. We can see that as velocity increases the flow through time decreases.