1. Non stationary heat transfer at ceramic pots firing Janna Mateeva, MP 0053 Department Of Material Science And Engineering Finite Element Method

2. Table Of Contents Review Of The Problem Review Of The Problem Solution with ANSYS / THERMAL Solution with ANSYS / THERMAL 1.Define Element Type 2.Build Geometry 3.Generate Mesh 4.Define Material Property 5.Specify Loads and Boundary Conditions 6.Specify Solution Criteria 7. Post processing

3. Review Of The Problem The 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. Given Ceramic pots are fired in an electric furnace in a temperature curve on Fig.1 The 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 aim Problem aim: An investigation of the transient heat transfer in the volume of the articles

4. Ceramic pot Geometrical model (the blue part)

5. olution with ANSYS/Thermal Solution with ANSYS/Thermal 1. Define Element Type 1. Define 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

6. 2 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

7.  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

8. 3. Generate Mesh The 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

9. 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=273КТ=1393К Кхх=Куу=Kzz0,609231,18043 с909,0721204,752  1500 Table 1

11. 5. Specify Loads and Boundary Conditions 5.1. Initial conditions Initial temperatures of 20  С 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

12. 5.2 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 1 τ2 ≤ τ ≤ τ3 T = a2 + b2. τ (constant heating rate) τ3 ≤ τ ≤ τ4 T = const 2 τ4 ≤ 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

13.  Regime 1: 0  Regime variable  15000; Result=293+0,042.τ  Regime 2: 15000  Regime variable  16800; Result=923  Regime 3: 16800  Regime variable  21000; Result=923+0,048.τ  Regime 4: 21000  Regime variable  22800; Result=1123  Regime 5: 22800  Regime variable  31800; Result=1123 +0,03.τ  Regime 6: 31800  Regime variable  33600; Result=1393  File/Save as/t

14. 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. Analysis options Main Menu /Solution/ Analysis type/ New analysis/ Transient

15. 6. Specify Solution Criteria 6. Specify Solution Criteria The 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 Options  Main 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 SearchOn  Main Menu> Solution> Load Step Opts> Nonlinear> PredictorProgram chosen 6.2. Output Control Options  Main Menu> Solution> Load Step Opts> Output Ctrls> Solu PrintoutOn  Main Menu> Solution> Load Step Opts> Output Ctrls> DB/Results FileEvery 30 substep  Main Menu> Solution> Load Step Opts> Output Ctrls> Integration PtYes if valid

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

18. 6.3. Solve

19. 7. Postprocesing of the results  Main Menu> General Postprocessor/ Read results/ First step…. Temperature field at 900 sec. of the process time

20. Temperature gradients at 900 sec. of the process time

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

22. Heat flux vectors at 900 sec. of the process time

23. Temperature gradient vectors at 900 sec. of the process time