Home / Examples / Fluid Analysis [Bernoulli] / Example 5: Transient Analysis of Flow around Cylinder

Example 5: Transient Analysis of Flow around Cylinder


General

 

Analysis Space

Item

Settings

Analysis Space

2D

Model Unit

mm

 

Analysis Conditions

Item

Tab

Settings

Solver

Solver Selection

Fluid Analysis [Bernoulli]

Analysis Type

Fluid Analysis

Transient Analysis

Laminar Flow/Turbulent Flow

Fluid Analysis

Select Laminar flow

Layer Mesh Setting for Wall Surface (General Settings)

Fluid Analysis

Specify mesh height of 1st layer

Height of 1st Layer Mesh: 0.5 [mm]
Growth Rate: 1.2
Number of Layers: 5

Timestep

Transient Analysis

Timestep: Specify

Number

Calculation Steps

Output Interval

Timestep [s]

1

1000

1

0.05

 

Meshing Setup

Mesh

General Mesh Size: 3 [mm]

Model

The model is a rectangular sheet body and the material is Air (000_Air). The boundary conditions of inlet and outlet are set on the left edge and the right edge, respectively.

The slip wall outer boundary condition is applied to the top and bottom edges where the boundary condition is not set.

The model is a circle sheet body and the material is iron (007_Fe).

The edges surrounding a circle is a boundary of solid and fluid. Solid wall is automatically set to them.

 

Body Attributes and Materials

Body Number/Type

Body Attribute Name

Material Name

0/Solid

Air

000_Air(*)

1/Solid

Column

007_Fe *

* Available from the material DB

Boundary Condition

Boundary Condition Name/Topology

Tab

Boundary Condition Type

Settings

Inlet/Edge

Fluid

Inlet

Forced Inflow
Specify flow velocity
0.05 [m/s]

Outlet/Face

Fluid

Outlet

Natural Outflow

Outer Boundary Condition

Fluid

Slip Wall

-

 

If Reynolds number exceeds 100, the time dependency of the turbulence occurs. It makes calculation difficult in the steady-state analysis.

The Reynolds number calculated with this model form, material property, and flow velocity is about 126.0. Use transient analysis instead.

 

Viscosity: μ=1.816e-5 [Pa s]

Density: ρ=1.144 [kg/m3]

Kinematic Viscosity: v=μ/ρ=1.816e-5/1.144=1.587e-5 [m2/s]

Flow Velocity: V=0.05 [m/s]

Diameter of Cylinder: D=0.04 [m]

Reynolds Number: Re = V*L/ν=0.05*0.04/1.587e-5 = 126.0

Results

The vectors of the flow velocity distribution around the cylinder at time 48 [s] and 50 [s] are shown below.

The vectors are adjusted to the same length in the graphics setup.

It is known that at a Reynolds number of 126, Karman vortices are shed from the alternating upper and lower sides at the back of the cylinder.
In the diagram below, a vortex appears on the upper side of the cylinder at 48 [s] and the lower side at 50 [s].

 

48 [s]
50 [s]

 

 

The figure below is the contour of Y component of the vorticity at 50 [s]. The color scale is minimum: -2 [/s] and maximum: 2 [/s].

The vortex strength in the Y direction is shown. The clockwise vortices are shown in the positive value, and the counter-clockwise vortices in the negative value.

You can observe the vortices are shed downstream from the cylinder.

 

 

The force on the wall face is shown by the table.

[Column] shows the force which the cylinder receives from the fluid.

 

 

The result below shows only Column/X component (drag F: force in the flowing direction).

(Double click the graph to show the plot option. Deselect [Display Y-axis (primary axis)] except Column/X component. Change the color of column/X component to red.)

The range of the scale for the vertical axis is changed to 0.9 x 10^-7 ~1.0 x 10^-7.

The drag force near time 50[s] fluctuates, centering around a central value of 9.8 x 10^-8 [N] , with a cycle of about 2 [s].

 

 

 

The result below shows only Column/Z component (lift: force in the direction perpendicular to the flow).

(Double click the graph to show the plot option. Deselect [Display Y-axis (primary axis)] except Column/X component. Change the color of column/X component to blue.)

You can see that the lift is fluctuating at a constant cycle of about 4 [s].

 

 

 

 

The drag coefficient CD can be calculated from the drag F.

 

Thickness in Depth Direction: t = 1 [mm]

Cross Sectional Area: S = D * t = 0.04 * 0.001 = 4e-5 [m2]

Dynamic Pressure: Pk = 0.5 * ρ*V^2 = 0.5 * 1.144 * 0.05 * 0.05 = 1.430e-3 [Pa]

Drag Coefficient: CD = F / Pk / S = 9.8e-8 / 1.430e-5 /4e-5 = 1.71