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Unsteady Flow Past a Cylinder
Created using ANSYS 13.0
Problem Specification
Consider the unsteady state case of a fluid flowing past a cylinder, as illustrated above. For this tutorial we will use a Reynolds Number of 120. In order to simplify the computation, the diameter of the cylinder is set to 1 m, the x component of the velocity is set to 1 m/s and the density of the fluid is set to 1 kg/m^3. Thus, the dynamic viscosity must be set to 8.333x10^-3 kg/m*s in order to obtain the desired Reynolds number.
Compared to the steady case, the unsteady case includes an additional time-derivative term in the Navier-Stokes equations:
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{include: FLUENT Google Analytics} {panel} Author: John Singleton and Rajesh Bhaskaran {color:#ff0000}{*}Problem Specification{*}{color} [1. Pre-Analysis & Start-Up|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Pre-Analysis & Start-Up] [2. Geometry|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Geometry] [3. Mesh|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Mesh] [4. Setup (Physics)|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Setup (Physics)] [5. Solution|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Solution] [6. Results|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Results] [7. Verification and Validation|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Verification & Validation] [Exercises|FLUENT 12.1 - Unsteady Flow Past a Cylinder - Exercises] {panel} {note}Under Construction\!\!{note} h2. Problem Specification \\ !prob_spec_v2.png!\\ \\ Consider the unsteady state case of a fluid flowing past a cylinder, as illustrated above. For this tutorial we will use a Reynolds Number of 120. In order to simplify the computation, the diameter of the cylinder is set to 1 m, the x component of the velocity is set to 1 m/s and the density of the fluid is set to 1 kg/m^3. Thus, the dynamic viscosity must be set to 8.333x10^-3 kg/m*s in order to obtain the desired Reynolds number. For this Unsteady Case, the governing equation becomes non linear due to the addition of a time derivative term: {latex} \begin{eqnarray} \frac{\partial \vec{u}}{\partial t} + \rho (\vec{u}\cdot \triangledown)\vec{u} = -\triangledown p + \mu \triangledown^{2} \vec{u} \end{eqnarray} {latex} |
The
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methods
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implemented
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by
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FLUENT
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to
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solve
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a
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time
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dependent
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system
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are
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very
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similar
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to
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those
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used
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in
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a
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steady-state
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case.
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In this case,
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the
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domain
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and
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boundary
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conditions
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will
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be
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the
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same
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as
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the
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Steady
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Flow
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Past
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a
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Cylinder.
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However,
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because
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this
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is
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a
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transient
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system,
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initial
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conditions
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at
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t=0
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are
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required.
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To solve the system,
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we
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need
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to
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input
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the
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desired
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time
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range
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and
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time
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step
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into
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FLUENT.
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The program will then compute a solution for the first time step, iterating until convergence or a limit of iterations is reached, then will proceed to the next time step, "marching" through time until the end time is reached.
Go to Step 1: Pre-Analysis & Start-Up