Non stationary heat transfer at ceramic pots firingNon stationary heat transfer at ceramic pots firingJanna Mateeva, MP 0053

Department Of Material Science And Engineering

Finite Element Method

Table Of ContentsTable Of Contents Review Of The ProblemReview Of The Problem

Solution with ANSYS / THERMALSolution with ANSYS / THERMAL

1.1. Define Element Type Define Element Type

2.2. Build GeometryBuild Geometry

3.3. Generate MeshGenerate Mesh

4.4. Define Material PropertyDefine Material Property

5.5. Specify Loads and Boundary ConditionsSpecify Loads and Boundary Conditions

6.6. Specify Solution CriteriaSpecify Solution Criteria 7. Post processing7. Post processing

Review Of The ProblemReview Of The ProblemThe pots are fired in a chamber furnace

according a temperature curve. The heat transfer between the pot’s bottom and the furnace floor can be neglected. The energy equation is solved at temperature boundary conditions, which reflect the temperature change at the given heating rate. The thermal effects due to the phase forming reactions can in the raw material can be neglected –there are low contents of chemical reagents in the ceramic mass. Because of the existing of an axis symmetry of the heating and the temperature field in the volume of the pot, the solution can be made 2D in cylindrical coordinates. But for a good visualization of the problem, the solution is made 3D.

GivenGivenCeramic pots are fired in an electric furnace in a temperature curve on Fig.1The pot has a cone form with sizes:- Top: inner diameter 24 cm, external diameter 25 cm.- Bottom: inner diameter 14 cm, external diameter 15 cm- Height of the pot: 20 cm

Problem aimProblem aim: An investigation of the transient heat transfer in the volume of the articles

Ceramic pot

Geometrical model(the blue part)

Solution with ANSYS/Thermalolution with ANSYS/Thermal

1.1. Define Element TypeDefine Element Type Preferences >Thermal.

The finite elements SOLID 70 are suitable for a problem solution. Main Menu /Preprocessor/ Element type/ Add, Edit, Delete/

Add/Thermal Mass/ Solid/ Solid 70

2 Building of the geometrical model2 Building of the geometrical model It’s enough to be investigating a part of the pot due to existing of thermal and geometrical symmetry. We create only a part of the pot – 30 degree .

We create two volume cones – outer and inner. Then we cut the inner cone from the outer volume with the command overlap

The geometrical model is made in the menu: Main Menu /Preprocessor/ Modeling/ Create/ Volumes/ Cone/By Dimension /bottom radius=0.075; top radius=0.125; z1=0; z2=0.2; starting angle=0; ending angle=30

Main Menu /Preprocessor/ Modeling/ Create/ Volumes/ Cone/By Dimension /bottom radius=0.065; top radius=0.125; z1=0.01; z2=0.2; starting angle=0; ending angle=30

Main Menu /Preprocessor/Modeling/Operate/Booleans/Overlap

3. Generate Mesh3. Generate MeshThe model is discretizated with the finite elements with sizes 0.002m:

Main Menu /Preprocessor/ Mеshing/ Meshtool/ Global Set/Element length 0.002 Main Menu /Preprocessor/ Mеshing/ Meshtool/ Areas Set/Element length 0.002 Main Menu /Preprocessor/ Mеshing/ Meshtool/ Shape –tetragonal Main Menu /Preprocessor/ Mеshing/ Meshtool/ Mesh /Select the pot/OK

Finite elements mesh

4. Define Material Property 4. Define Material Property The physical properties of the pot’s materials are given below: Specific heat capacity: c = 0,9 + 0,000167.t kJ/kg.K Thermal conductivity: Кхх=Куу=Kzz = 837 + 0,264. t W/(m.K) Density: = 1500 kg/m3

These properties are applied in the menu: Main Menu /Preprocessor/ Material properties/Material

models/Thermal/ The values, determinate by the above expressions (table 1) can be

used for that aim.

Т=273К Т=1393К

Кхх=Куу=Kzz 0,60923 1,18043

с 909,072 1204,752

1500 1500

Table 1

5. Specify Loads and Boundary Conditions5. Specify Loads and Boundary Conditions5.1. Initial conditions5.1. Initial conditions

Initial temperatures of 20С are specified at all nodes: Main Menu /Preprocessor/ Loads/ Define loads/ Apply/ Initial

conditions/Define/Pick all nodes Main Menu /Solution/ Define loads/ Apply/ Initial conditions/Define//Pick all nodes

5.2 Boundary conditions5.2 Boundary conditions

The transient thermal act on the pots in the furnace can be model with applying the temperature change to the surfaces according the temperature curve

Mathematical descriptions of the boundary conditions:T = f(τ)

0 ≤ τ ≤ τ1 T = a1 + b1. Τ (constant heating rate)τ1 ≤ τ ≤ τ2 T = const 1τ2 ≤ τ ≤ τ3 T = a2 + b2. τ (constant heating rate)τ3 ≤ τ ≤ τ4 T = const 2τ4 ≤ τ ≤ τ5 T = a3 + b3. τ (constant heating rate)τ5 ≤ τ ≤ τ6 T = const 3

The coefficients a, b and the constants are calculated according the different heating rates of the temperature curve.

The temperature change rate is specified at the steps:

1) Definition of the function

Main Menu /Solution/ Define loads/ Apply/Function/Define, Edit/Multivalued function, based on regime variable

Regime variable = Time

Regime 1: 0 Regime variable15000; Result=293+0,042.τ Regime 2: 15000 Regime variable16800; Result=923 Regime 3: 16800 Regime variable21000; Result=923+0,048.τ Regime 4: 21000 Regime variable22800; Result=1123 Regime 5: 22800 Regime variable31800; Result=1123 +0,03.τ Regime 6: 31800 Regime variable33600; Result=1393

File/Save as/t

2) Creation of the table according the function Main Menu /Solution/ Define loads/ Apply/Function/Read file/t/

Open Table parameter name/ t /OK

3) Applying the heat generation rate on the nodes Main Menu /Solution/ Define loads/ Apply/Thermal/ Heat Generation

Rate /On nodes / Pick all/ Existing table /%t%

5.3. 5.3. AnalAnalyyssiis optionss options

Main Menu /Solution/ Analysis type/ New analysis/ Transient

6. Specify Solution Criteria6. Specify Solution CriteriaThe calculations can be made at automatic calculated or user definite time steps. The convergence is better when the time steps are small. In that exercises the calculations are made at time steps 60s. The results are written to the results file every 30 steps.

Main Menu> Solution> Load Step Opts> Time/Frequenc> Time/TimeStep/Time of end of load step = 31800; Time step size=60

6.1. Nonlinear OptionsMain Menu> Solution> Load Step Opts> Nonlinear> Equilibrium IterNo of

equilibrium iteration =7 Main Menu> Solution> Load Step Opts> Nonlinear> Convergence CritMain

Menu> Solution> Load Step Opts> Nonlinear> Criteria to StopDo not stop Main Menu> Solution> Load Step Opts> Nonlinear> Line SearchOnMain Menu> Solution> Load Step Opts> Nonlinear> PredictorProgram

chosen 6.2. Output Control Options

Main Menu> Solution> Load Step Opts> Output Ctrls> Solu PrintoutOnMain Menu> Solution> Load Step Opts> Output Ctrls> DB/Results

FileEvery 30 substep Main Menu> Solution> Load Step Opts> Output Ctrls> Integration PtYes if

valid

Main Menu> Solution> Load Step Opts> Output Ctrls> Integration Pt/Yes if valid

6.3. Solve6.3. Solve

7. Postprocesing of the results7. Postprocesing of the resultsMain Menu> General Postprocessor/ Read results/ First step….

Temperature field at 900 sec. of the process time

Temperature gradients at 900 sec. of the process time

Heat flux (scalar field) at 900 sec. of the process time

Heat flux vectors at 900 sec. of the process time

Temperature gradient vectors at 900 sec. of the process time