Temperature Expansion Problem

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Brainbustling
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Temperature Expansion Problem

Post by Brainbustling » 13 Feb 2020, 17:05

Hello :),

maybe someone can help me with a problem solving the temperature expansion of a Rod (diameter: 5mm, length: 100mm)
The rod is fastened on one side.
Delta temperature is 100K with an coefficient of thermal expansion of 11.1e-6.
Of my hand bill i´ve got an expansion of 0,111mm.
The Result of the Simulation states, that the expansion is nearly half of that with 0,058mm (0,58e-5m).
I´ve got two boundary conditions. One for the fastened side with an temperature of 100K, the other one for the rest of the rod with just the temperature.
I did vary the temperature and the expansion coefficient(seperatly) with the results did vary proportionally to the variations.
I think i´m missing some options.

thanks for your attention.

here is my sif file:

Header
CHECK KEYWORDS Warn
Mesh DB "." "."
Include Path ""
Results Directory ""
End

Simulation
Max Output Level = 5
Coordinate System = Cartesian
Coordinate Mapping(3) = 1 2 3
Simulation Type = Steady state
Steady State Max Iterations = 1
Output Intervals = 1
Timestepping Method = BDF
BDF Order = 1
Coordinate Scaling = 0.001
Solver Input File = case.sif
Post File = case.vtu
End

Constants
Gravity(4) = 0 -1 0 9.82
Stefan Boltzmann = 5.67e-08
Permittivity of Vacuum = 8.8542e-12
Boltzmann Constant = 1.3807e-23
Unit Charge = 1.602e-19
End

Body 1
Target Bodies(1) = 1
Name = "TEMP1"
Equation = 1
Material = 1
End

Solver 1
Equation = Linear elasticity
Procedure = "StressSolve" "StressSolver"
Variable = -dofs 3 Displacement
Exec Solver = Always
Stabilize = True
Bubbles = False
Lumped Mass Matrix = False
Optimize Bandwidth = True
Steady State Convergence Tolerance = 1.0e-5
Nonlinear System Convergence Tolerance = 1.0e-7
Nonlinear System Max Iterations = 20
Nonlinear System Newton After Iterations = 3
Nonlinear System Newton After Tolerance = 1.0e-3
Nonlinear System Relaxation Factor = 1
Linear System Solver = Iterative
Linear System Iterative Method = BiCGStab
Linear System Max Iterations = 1000
Linear System Convergence Tolerance = 1.0e-10
BiCGstabl polynomial degree = 2
Linear System Preconditioning = ILU0
Linear System ILUT Tolerance = 1.0e-3
Linear System Abort Not Converged = False
Linear System Residual Output = 10
Linear System Precondition Recompute = 1
End

Solver 2
Equation = Heat Equation
Procedure = "HeatSolve" "HeatSolver"
Variable = Temperature
Exec Solver = Always
Stabilize = True
Bubbles = False
Lumped Mass Matrix = False
Optimize Bandwidth = True
Steady State Convergence Tolerance = 1.0e-5
Nonlinear System Convergence Tolerance = 1.0e-7
Nonlinear System Max Iterations = 20
Nonlinear System Newton After Iterations = 3
Nonlinear System Newton After Tolerance = 1.0e-3
Nonlinear System Relaxation Factor = 1
Linear System Solver = Iterative
Linear System Iterative Method = BiCGStab
Linear System Max Iterations = 1000
Linear System Convergence Tolerance = 1.0e-10
BiCGstabl polynomial degree = 2
Linear System Preconditioning = ILU0
Linear System ILUT Tolerance = 1.0e-3
Linear System Abort Not Converged = False
Linear System Residual Output = 10
Linear System Precondition Recompute = 1
End

Equation 1
Name = "Heat"
Calculate Stresses = True
Active Solvers(2) = 1 2
End

Material 1
Name = "Steel_1"
Heat expansion Coefficient = 11.1e-6
Heat Conductivity = 37.2
Sound speed = 5100.0
Heat Capacity = 461.0
Density = 7850.0
Poisson ratio = 0.3
Youngs modulus = 210.0e9
End

Boundary Condition 1
Target Boundaries(2) = 1 3
Name = "Temperature"
Temperature Condition = 1
Temperature = 100
End

Boundary Condition 2
Target Boundaries(1) = 2
Name = "Fastened"
Displacement 3 = 0
Displacement 2 = 0
Displacement 1 = 0
Temperature Condition = 1
Temperature = 100
End
Attachments
case.sif
(3.03 KiB) Downloaded 1 time

kevinarden
Posts: 435
Joined: 25 Jan 2019, 01:28
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Re: Temperature Expansion Problem

Post by kevinarden » 13 Feb 2020, 23:05

The governing equation is

k * Lo * Delta T = 11.1E-6 *100 mm * 100 = 0.111 mm

I just ran this case in Abaqus and the expansion is 0.111 mm

Will check elmer, however, I do not see in your sif how the temperature is a delta T off 100?

I see the boundary is 100, but there is no initial condition, what is and how is the initial temperature set?

kevinarden
Posts: 435
Joined: 25 Jan 2019, 01:28
Antispam: Yes

Re: Temperature Expansion Problem

Post by kevinarden » 14 Feb 2020, 02:55

This case produces the correct result in Elmer. Model is in meters so no scaling is used.
The equation being used for comparison is if the whole body has raised 100 delta.
Note that you are putting temperature in on the surface which does not make the whole body 100 until it gets to steady state. It appears that multiple steady state iterations are needed to allow the surface temperature to make the whole body 100. In my sif I increased the number of steady state iterations. The body heat load is just to produce some heat, but the surface boundary temperatures cause the whole body to go to 100 at steady state (transient would be different). Delta T is caused by setting the initial conditions to 0.
expansion.png
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cylinder.zip
(240.39 KiB) Downloaded 4 times

Brainbustling
Posts: 5
Joined: 11 Feb 2020, 21:32
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Re: Temperature Expansion Problem

Post by Brainbustling » 14 Feb 2020, 11:28

Ahhh yes. You´re right.
I didn´t think of the initial conditions to be set.
Thank you for your explanation. this improves my understanding of the programm.
A pitty that there is no body condition with temperature.
To increase the iterations, so that the whole body gets the temperature is a great idea.
Thank you very much :) :)

raback
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Re: Temperature Expansion Problem

Post by raback » 17 Feb 2020, 13:49

Hi

Dirichlet conditions may also be defined unintuitively in "Body Force" section. And if you're not really solving for temperature there are ways to circumwent that too bu just defining the temperature field with epressions.

-Peter

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