Tutorial Part1 Ex6 - Heat 2
7
pages
English
Documents
Le téléchargement nécessite un accès à la bibliothèque YouScribe Tout savoir sur nos offres
7
pages
English
Documents
Le téléchargement nécessite un accès à la bibliothèque YouScribe Tout savoir sur nos offres
MARC Designer Tutorial Part 1 Example 6 One-Dimensional Steady-State Heat Transfer Analysis Problem Statement: A hot gas flows through a long tube as shown in Figure E6-1. The inner osurface of the tube is assumed to be heated to the gas temperature of 250 C. The outer surface of othe tube is assumed to be maintained at 25 C. The tube is made of an isotropic material with o 3 othermal conductivity of 6.5 W/m C (or 6.5 x 10 W/mm C). The long tube with uniform temperature at its internal surface can be analyzed using plane strain condition. Due to the symmetry in both geometry and loading about the x- and y- axes, we only need to consider a quadrant of the tube as our finite element model. We wish to determine the temperature distribution across the wall of the tube when the heat transfer is at a steady-state condition, using MARC Designer software. A thick-walled tube: Internal radius, r = 150 mm, i Wall thickness, t = 100 mm A hot gas ato250 C Figure E6-1 A hot gas flows through a long tube. Start MARC Designer. Select STEADY-STATE HEAT TRANSFER from the menu. 1. MESH GENERATION (MESH GENERATION) ELEMENT CLASS: QUAD(4); RETURN; CURVE TYPE: CENTER/POINT/POINT; RETURN; SURFACE TYPE: RULED; RETURN; (COORDINATE SYSTEM) RECTANGULAR [Note that the coordinate system changes to CYLINDRICAL coordinate]; GRID [To display the grid points on screen]; SET: (GRID) SPACING: Type 50, Enter; Type 50, Enter; SIZE: Type 250, Enter; Type ...
Voir
MARC Designer Tutorial
Part1
Example 6 One-Dimensional Steady-State Heat Transfer Analysis Problem Statement: A hot gas flows through a long tube as shown inFigure E6-1. The inner o surface of the tube is assumed to be heated to the gas temperature of 250C. The outer surface of o the tube is assumed to be maintained at 25C. The tube is made of an isotropic material with o 3o thermal conductivity of 6.5 W/mC (or 6.5 x 10W/mm C). The long tube with uniform temperature at its internal surface can be analyzed usingplane straincondition. Due to the symmetry in both geometry and loading about thex- andy- axes, we only need to consider a quadrant of the tube as our finite element model. We wish to determine the temperature distribution across the wall of the tube when the heat transfer is at asteady-statecondition, using MARC Designer software. A thick-walled tube: Internal radius,ri= 150 mm, Wall thickness,t= 100 mm A hot gas a o 250 C Figure E6-1A hot gas flows through a long tube. Start MARC Designer. Select STEADY-STATE HEAT TRANSFER from the menu. 1. MESHGENERATION (MESH GENERATION) ELEMENT CLASS: QUAD(4); RETURN; CURVE TYPE: CENTER/POINT/POINT; RETURN; SURFACE TYPE: RULED; RETURN; (COORDINATE SYSTEM) RECTANGULAR [Note that the coordinate system changes to CYLINDRICALcoordinate]; GRID [To display the grid points on screen]; SET:(GRID) SPACING: Type 50,Enter; Type 50,Enter; SIZE: Type 250,Enter; Type 250,Enter; RETURN; FILL (Bottom Menu). We will now create two curves using the following steps. (CRVS) ADD: Click the center point of cylindrical coordinate (0, 0, 0); Click the second coordinate point in U direction (150, 0, 0); Click the second coordinate point in the V direction (0, 150, 0); (CRVS) ADD: Click the center point of cylindrical coordinate (0, 0, 0); Click the last coordinate point in U direction (250, 0, 0); Click the last coordinate point in the V direction (0, 250, 0); [Note that two curves have been created];
Page 45
MARC Designer Tutorial1 Part Next, we will create a RULED SURFACE based on the two curves just created. (SRSF) ADD: Click on the inner curve; Click on the outer curve; END LIST (#); (ZOOM BOX) OUT (Click two times);
Figure E6-2ruled surface created from two adjacent curves, representing a A quadrant of the tube, considered as the finite element model for the analysis.
We will now convert the surface into 200quadrilateral(QUAD 4) elements. (COORDINATE SYSTEM) CONVERT: (CONVERT) DIVISION; Type 20,Enter; Type 10,Enter; (GEOMETRY/MESH) SURFACE TO ELEMENTS: Click anywhere on the surface just created; [Notice that the color of the surface changes to yellow]; END LIST (#); RETURN; We need to SWEEP the discretized finite element model to eliminate any duplicated nodes created during the discretization process. Then we mustcheckany UPSIDE DOWN elements andflipthem if there are any. (COORDINATE SYSTEM) SWEEP: (SWEEP) ALL; RETURN; RENUMBER: ALL; RETURN; CHECK: (CHECK ELEMENTS) UPSIDE DOWN; FLIP ELEMENTS; (ALL) EXIST; RETURN;
Page 46
MARC Designer Tutorial
Part1
2. SAVEYOUR FILE FILES (Bottom Menu): (MODEL) SAVE AS: Typeheat_1.mud;Enter; Typey;Enter; [Your model is saved in aC:/Designer/Binsubdirectory]; RETURN; RETURN; You have successfully created a finite element model of a quadrant of a circle and discretize it into 200 quadrilateral elements.
3. BOUNDARYCONDITIONS o We assume that the inner surface of the cylinder is at a temperature of 250C and the outer o surface at 25C, and carry out a 1-Dimensional Steady-State Heat Transfer analysis. FIXED TEMPERATURE: ON; TEMPERATURE; Type 250;Enter; OK; (NODES) ADD: Click on all the nodes along the inner curve; END LIST(#); NEW: FIXED TEMPERATURE: ON; TEMPERATURE; Type 25;Enter; OK; (NODES) ADD:Click on all nodes along the outer curve; END LIST (#); ID APPLYS; SAVE (From the bottom menu); RETURN;
Figure E6-3finite element model, discretized using 4-nodes quadrilateral The (QUAD-4) elements. Also shown are the fixed temperature boundary conditions. The inner surface is maintained at 250°C while the outer surface is at°25 C.
Page 47
MARC Designer Tutorial
Part1
4. MATERIALPROPERTIES 3 o We shall assume that the material has a thermal conductivity,k= 6.5 x 10W/mm C. (THERMAL MATERIAL TYPES) HEAT TRANSFER; CONDUCTIVITY; Type 6.5e3; Enter; OK; (ELEMENTS) ADD: (ALL) EXIST.; SAVE (From bottom menu); RETURN; ID MATERIALS;
5. GEOMETRICPROPERTIES We will carry out our analysis based on the PLANE STRAIN condition, thus we need to consider only a unit thickness of the tube. (ELEMENT DIMENSION) PLANAR: (GEOMETRIC PROPERTY TYPE) PLANAR: THICKNESS NORMAL TO PLANE: Type 1 (One unit length);Enter; OK; (ELEMENTS) ADD: (ALL) EXIST.; SAVE (Bottom Menu); RETURN; ID GEOMETRIES; RETURN;
6. LOADCASES (BOUNDARY CONDITIONS) THERMAL; OK; RETURN;
7. JOBS (LOADCASES & OPTION) HEAT TRANSFER; (LOADCASES) SELECT; Click on lcase1the (AVAILABLE LCS); (ANALYSIS DIMENSION) PLANAR; OK; SAVE under (From the bottom menu); ELEMENT TYPES; PLANAR; (QUAD) (4); Click on 39; OK; (ALL) EXIST.; ID TYPES; ID CLASSES; RETURN; RUN JOB; [Notice from the status window at the bottom of the screen if your analysis is successfully completed]; RETURN;
8. RESULTS (FILE) OPEN DEFAULT; (DEFORM SHAPE) DEF ONLY; (SCALAR PLOT) CONTOUR BANDS; (FILE) MONITOR; You should see the contour of temperature distribution across the wall thickness of the pipe, as shown inFigure E6-4.
Page 48
MARC Designer Tutorial
POSTPROCESSING RESULTS
Part1
Figure E6-4postprocessing result showing a contour of temperature The distribution across the wall thickness of the tube.
Page 49
MARC Designer Tutorial Part1 TWO-DIMENSIONAL STEADY-STATE ANALYSIS The above problem can easily be transformed into a 2-Dimensional heat transfer problem. To do this, first CLOSE the postprocessing result file, RESET the program, and then RESTORE the present finite element model. (POSTPROCESSING RESULTS); (FILE); CLOSE; FILE (From the bottom menu); RESET PROGRAM; (MODEL) RESTORE; RETURN; RETURN; Next, removed the temperature boundary condition from the outer surface of the tube. (BOUNDARY CONDITIONS); NEXT; (NODES) REM; (ALL) EXIST.; (This removes the fixed temperature boundary condition from the outer surface. o Specify the 25C fixed temperature boundary conditon on all the nodes along the horizontal and vertical sides of the tube quadrant. (NODES) ADD: Click on all the nodes along the horizontal and vertical sides of the tube quadrant; END LIST(#); SAVE; (Save your model); RETURN; LOADCASES; THERMAL; make sure you select the newly created boundary condition; RETURN;
Figure E6-5thermal boundary conditions for 2-Dimensional steady-state The heat transfer analysis.
Page 50
MARC Designer Tutorial1 Part Run the analysis. Once the analysis is completed, open the postprocessing result file. JOBS; RUN JOB; RETURN; RESULTS; (RepeatStep 8above); This time. you should see a different temperature distribution contour across the wall of the tube, as shown inFigure E6-6. Good Luck!!
Figure E6-6 Thepostprocessing result showing the contour of temperature distribution across the wall thickness of the tube, resulting from a 2-Dimensional steady-state heat transfer analysis. NAZRI B KAMSAH, Ph.D Department of Thermo Fluid March 31, 2003
Page 51
