Tutorial 1: Heat conduction in a slab 2007 Cornell University BEE453, Professor Ashim Datta Authored by Vineet Rakesh and Frank Kung Software: COMSOL 3.3
Tutorial 1: Heat conduction in a slab ............................................................................................... 1 Problem Specification .................................................................................................................. 1 Step 1: Specifying the Problem Type .......................................................................................... 2 Step 2: Creating the Geometry .................................................................................................... 4 Step 3: Meshing ........................................................................................................................... 5 Step 4: Defining Material Properties and Initial Condition ........................................................... 7 Step 5: Defining Boundary Conditions......................................................................................... 8 Step 6: Specifying Solver Parameters ......................................................................................... 9 Step 7: Postprocessing .............................................................................................................. 11 Step 8: Save and Exit ................................................................................................................ 15
Problem Specification This is the first of a series of examples intended for a gentle introduction to COMSOL. Our goal here is to compute transient temperatures profiles for 1-D heat conduction in the slab below during the heating process. The temperature in both bottom and top surfaces of the slab are kept constant and equal to 90 0C and 20 0C respectively. The sides of the slab are assumed to be insulated. The thickness of the slab is 4 cm. The initial temperature is 20 C. The density is 900 kg/m3. The thermal conductivity is 0.55 W/mK. The specific heat is 3800 J/kg K.
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Step 1: Specifying the Problem Type The problem in this case is transient heat conduction in a 1D setting.
We will first set
COMSOL up for this type of problem The model you specify determine the Governing Equations that will be used. Starting from the energy transport equation:
∂T ∂T k ∂ 2T Q + u = + 2 ∂t ∂x ρc p ∂x ρc N N
Np transient convection diffusion
(1)
source
For our problem, the temperatures are dependant on time; there is no fluid flow and no source terms. The only mode of heat transfer is by diffusion. So the problem is a transient diffusion problem with no convection and heat source. So the governing equation changes to:
∂T k ∂ 2T = ∂t ρc p ∂x 2
(2)
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1.
Start COMSOL by double clicking
on
Multiphysics
the
COMSOL
icon
on
the
Desktop 2.
Select 2D next to Space Dimension (Note: COMSOL can do 1D problems, however to give you a better understanding of COMSOL we’ll model the problem as 2D)
3.
Single Click on COMSOL Multiphysics >> Heat Transfer >> Conduction >> Transient Analysis. Transient Analysis under conduction is selected as we intend to solve a time dependent conduction problem (Equation 2).
4.
Click on the Settings Tab
5.
Set the Unit system to SI
6.
Click OK. COMSOL Window opens up.
7.
Under File, click on Save as…
8.
Create your own folder using your NetID in the My Documents folder and save your work there. Specify the file name (e.g. cond.mph) and save it as .mph file.
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Step 2: Creating the Geometry The geometry in this case is a 1D slab that is 4 cm high (along y axis). We are modeling the problem as 2D and so we assume the dimension of 2 cm in the other direction (i.e. along x axis). In this example (in contrast with the drug delivery example) we will draw out the slab directly without specifying the grid.
1.
Click on Draw >> Specify Objects >> Rectangle. Rectangle window opens up.
2.
Specify width as 0.02 and height as 0.04. These are the dimensions of the slab in m.
3.
Click on OK.
4.
Click on Zoom Extents to fit the geometry in the window.
The geometry that is created is shown in the figure.
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Step 3: Meshing Meshing is dividing the geometry into small elements. We can mesh the face directly, but in this case, we will mesh the edges first and then the face. This method is used to control the number of elements in certain parts of the geometry like the boundaries and interfaces. In many cases we need a finer mesh near the boundary and so we mesh the edge accordingly and what we get is a non-uniform mesh. However in this case we mesh the geometry with a uniform mesh with a spacing of 0.002 between the nodes.
1. Under Mesh, click on Mapped Mesh Parameters… 2. Click on the Boundary Tab 3. In Boundary Selection, select 1 and 4 by left clicking and holding the Ctrl key. We will specify 20 elements each on the left and right edges. 4. Check the box for Constrained edge element distribution 5. Click
on
Number
of
edge
elements, and type in 20 in the box below. 6. For boundary 2 and 3, use number of edge elements as 10. We specify 10 elements each on the top and bottom edges. 7. Press the ‘Remesh’ Button on the bottom.
The
mesh
that
is
obtained is shown in the figure. 8.
Click ‘Ok’
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9. Your screen should now look like this.
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Step 4: Defining Material Properties and Initial Condition We are solving the energy equation and so we need to provide the solver with the appropriate material property values required for the analysis. These properties are: (i) Density (Ď ): The density of the material of the slab is 900 kgm-3 (ii) Thermal Conductivity (k): The thermal conductivity is 0.55 W(mK)-1 (iii) Specific Heat (cp): The specific heat is 3800 J (kgK)-1 The Slab is initially at a temperature of 200C initially. We will also specify this in the software in this step. 1.
Under
Physics,
click
on
1
Subdomain Settings‌ 2. Click on 1 to select the slab. 3. Left click on the text field next to Thermal Conductivity and type 0.55
4. Left click on the text field next to Density and type 900
5. Left click on the text field next to Heat Capacity and type 3800 6. Click on the Init Tab 7. In the box under Initial Value, fill in 293. The slab is initially at a temperature of 20 0C (=293 K). 8. Click Ok
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Step 5: Defining Boundary Conditions The boundary conditions for the problem are: On the bottom boundary: Temperature = 900C On the top boundary: Temperature = 200C On the left boundary: The left boundary is insulated. Therefore, Heat Flux =0. On the right boundary: The right boundary is insulated. Therefore, Heat Flux =0. We now specify these boundary conditions to the solver. The default boundary condition for this solver is thermal insulation so we simply need to change the boundary conditions for the top and bottom
1. Under Physics, click on Boundary Settings‌ 2. Click on 2 in the Boundary Selection box 3. Change the Boundary Condition to Temperature 4. In T0, change the temperature to 90. The bottom of the slab is at 900C (=363 K). 5. Repeat for side 3 using 293. The top of the slab is at 200C (=293 K). 6. Click on OK
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Step 6: Specifying Solver Parameters We now specify the time integration method for the time dependant problem. We use the default values for solver settings for this problem as well. To select fixed time steps, we need to go the “Time Stepping� Tab in the Solver Parameters window. However, we do not do this for the problem. Variable time stepping is used as the default for this problem.
1. Under Solve, click on Solver Parameters. 2. Under the General Tab, select Transient under Analysis
if
it
is
not
already selected 3. Select Time dependent under Solver. 4. In the Times: box, type in 0:10:5400. This tells the solver to start at 0 seconds, then save the solution every 10 seconds until it reaches 5400 seconds. 5. Click Ok 6. Click on Get Initial Value under Solve. This step initializes the solver with the value provided when the
initial
conditions
were specified (Step 4). 7. Under Solve, click on Solver Manager.
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8. Click on the Solve For tab. 9. Select T for temperature if
it
is
not
already
selected. By selecting T, we
are
directing
the
solver to solve for the temperature. 10. Click on the Output tab. 11. Select T for temperature if
it
is
not
already
selected. By selecting T, we
are
directing
the
program to save the temperature values. 12. Press
Solve.
Once
Solve is pressed, the solver
solves
the
transient heat transfer equation (Equation 2 on page 2). It will take approximately 1-2 min to solve. We are now ready
to
do
post
processing (looking at the results).
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Step 7: Postprocessing Post-processing is viewing the results obtained on running the simulations of the problem. We will generate graphs and charts based upon our simulation.
Displaying the Mesh 1. To display the mesh, simply click on the Mesh Mode button. Mesh mode can also be selected by clicking on Mesh >> Mesh Mode. The mesh is shown in the figure below.
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1.
Plot Temperature vs. Time at a particular coordinate We will now plot the temperature history at point (0.01, 0.01) in the interior of the slab to see how the temperature varies with time at that location. 1. Under Postprocessing, click on Cross-Section Plot Parameters‌ (It
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may take some time to open up). 2. Click on the Point tab, 3. Make sure Temperature is selected in the Predefined quantities section. 4. In Coordinates, type in 0.01 for x and 0.01 for y. 5. Press OK The temperature history plot obtained for the point (0.01, 0.01) is shown in the figure below.
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Obtain the surface plot at the last time step We now plot the temperature contour in the slab at the end time (i.e. after 1.5 hrs or 5400 s). 1.
Under
Postprocessing
1
click on Plot Parameters. 2.
Click on the General Tab
3.
Check the box for Surface under Plot Type.
4.
Next to Solution at time: select 5400.
5.
Click on the Surface Tab.
6.
Select Temperature next to Predefined quantities, if it is not already selected.
7.
Click on OK.
The
contour
plot
that
is
obtained is shown on the next page.
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Step 8: Save and Exit Now, before we end the session we need to save the files for future use. 1. Go to File 2. Click on Save 3. Go to File >> Exit
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