problem4-66 - PROBLEM 4.66 KNOWN Log rod of rectangular...

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PROBLEM 4.66 KNOWN: Log rod of rectangular cross-section of Problem 4.53 that experiences uniform heat generation while its surfaces are maintained at a fixed temperature. Use the finite-element software FEHT as your analysis tool. FIND: (a) Represent the temperature distribution with representative isotherms; identify significant features; and (b) Determine what heat generation rate will cause the midpoint to reach 600 K with unchanged boundary conditions. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, and (2) Two-dimensional conduction with constant properties. ANALYSIS: (a) Using FEHT , do the following: in Setup , enter an appropriate scale; Draw the outline of the symmetrical section shown in the above schematic; Specify the Boundary Conditions (zero heat flux or adiabatic along the symmetrical lines, and isothermal on the edges). Also Specify the Material Properties and Generation rate. Draw three Element Lines as shown on the annotated version of the FEHT screen below. To reduce the mesh, hit
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Unformatted text preview: Draw/Reduce Mesh until the desired fineness is achieved (256 elements is a good choice). Continued … PROBLEM 4.66 (Cont.) After hitting Run , C heck and then Calculate , use the View/Temperature Contours and select the 10-isopotential option to display the isotherms as shown in an annotated copy of the FEHT screen below. The isotherms are normal to the symmetrical lines as expected since those surfaces are adiabatic. The isotherms, especially near the center, have an elliptical shape. Along the x = 0 axis and the y = 10 mm axis, the temperature gradient is nearly linear. The hottest point is of course the center for which the temperature is T(0, 10 mm) = 401.3 K. < The temperature of this point can be read using the View/Temperatures or View|Tabular Output command . (b) To determine the required generation rate so that T(0, 10 mm) = 600 K, it is necessary to re-run the model with several guessed values of q . After a few trials, find ± 8 q 1.48 10 W / m = × ± 3 <...
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