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MARC Designer Tutorial Part 1

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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

surface of the tube is assumed to be heated to the gas temperature of 250

o

C. The outer surface of the tube is assumed to be maintained at 25oC. The tube is made of an isotropic material with

thermal conductivity of 6.5 W/moC (or 6.5 x 10

3W/mm

oC).

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 onlyneed 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.

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 toCYLINDRICAL coordinate];

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 Vdirection (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 ];

A thick-walled tube:

Internal radius, r i = 150 mm,

Wall thickness, t = 100 mmA hot gas at

250 oC




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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 E6-2 A ruled surface created from two adjacent curves, representing a

quadrant of the tube, considered as the finite element model for the analysis.

We will now convert the surface into 200 quadrilateral (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 must check any UPSIDE DOWN elements

and flip them if there are any.

(COORDINATE SYSTEM) SWEEP: (SWEEP) ALL; RETURN; RENUMBER: ALL;RETURN; CHECK: (CHECK ELEMENTS) UPSIDE DOWN; FLIP ELEMENTS; (ALL)

EXIST; RETURN;




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2. SAVE YOUR FILE

FILES (Bottom Menu): (MODEL) SAVE AS: Type heat_1.mud ; Enter ; Type y; Enter ; [Your

model is saved in a C:/Designer/Bin subdirectory]; RETURN; RETURN;

You have successfully created a finite element model of a quadrant of a circle and discretizeit into 200 quadrilateral elements.

3. BOUNDARY CONDITIONS

We assume that the inner surface of the cylinder is at a temperature of 250oC and the outer

surface at 25oC, 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 E6-3 The finite element model, discretized using 4-nodes quadrilateral

(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.




8

4. MATERIAL PROPERTIES

We shall assume that the material has a thermal conductivity, k = 6.5 x 103

W/mmoC.

(THERMAL MATERIAL TYPES) HEAT TRANSFER; CONDUCTIVITY; Type 6.5e3;

Enter ; OK; (ELEMENTS) ADD: (ALL) EXIST.; SAVE (From bottom menu); RETURN; ID

MATERIALS;

5. GEOMETRIC PROPERTIES

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

lcase1 under the (AVAILABLE LCS); (ANALYSIS DIMENSION) PLANAR; OK; SAVE(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 in Figure E6-4.




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POSTPROCESSING RESULTS

Figure E6-4 The postprocessing result showing a contour of temperature

distribution across the wall thickness of the tube.




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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 removesthe fixed temperature boundary condition from the outer surface.

Specify the 25oC 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 E6-5 The thermal boundary conditions for 2-Dimensional steady-state

heat transfer analysis.




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Run the analysis. Once the analysis is completed, open the postprocessing result file.

JOBS; RUN JOB; RETURN;

RESULTS; (Repeat Step 8 above);

This time. you should see a different temperature distribution contour across the wall of the tube,as shown in Figure E6-6. Good Luck!!

Figure E6-6 The postprocessing 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

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