HW12_Solution - EML 4507 FEA & DESGN Fall 2010 HW12...

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Unformatted text preview: EML 4507 FEA & DESGN Fall 2010 HW12 Solution 2. Consider a heat conduction problem described in the figure. Inside of the domain, heat is generated from a uniform heat source Q g = 10 W/m 3 , and the conductivity of the domain is k = 0.1 W/m/ o C. The cross-sectional area A = 1 m 2 . When the temperatures at both ends are fixed at 0 o C, calculate the temperature distribution using (a) two equal-length elements and (b) three equal-length elements. Plot the temperature distribution along with the exact temperature distribution. Solution: (a) Two elements solution: In this case, each element has length of L ( e ) = 0.5 m. Thus, the element matrix equation becomes Element 1: (1) 1 1 (1) 2 2 .2 .2 2.5 .2 .2 2.5 T q T q - + = - + Element 2: (2) 2 2 (2) 3 3 .2 .2 2.5 .2 .2 2.5 T q T q - + = - + Note that the generated heat is equally divided into two end nodes. After assembly, we have 1 1 2 3 3 .2 .2 2.5 .2 .4 .2 5.0 .2 .2 2.5 T Q T T Q - = + -- = - = + Since the temperatures at Nodes 1 and 3 are known and equal to zero, the first and last rows and corresponding columns are deleted to obtain 2 2 [.4]{ } {5.0} 12.5 C T T = ⇒ = ° The temperature between nodes varies linearly. (b) Three elements solution: In this case, each element has length of L ( e ) = 1/3 m. Thus, the element matrix equation becomes 1 m T = 0 o C T = 0 o C Q g = 10 W/m 3 Insulated Insulated x Element 1: (1) 1 1 (1) 2 2 .3 .3 1.667 .3 .3 1.667 T q T q - + = - + Element 2: (2) 2 2 (2) 3 3 .3 .3 1.667 .3 .3 1.667 T q T q - + = - + Element 3: (3) 3 3 (3) 4 4 .3 .3 1.667 .3 .3 1.667 T q T q - + =...
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This note was uploaded on 06/07/2011 for the course EML 4507 taught by Professor Kim during the Fall '11 term at University of Florida.

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HW12_Solution - EML 4507 FEA & DESGN Fall 2010 HW12...

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