MIT2_092F09_lec11

# MIT2_092F09_lec11 - 2.092/2.093 Finite Element Analysis of...

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Unformatted text preview: 2.092/2.093 -- Finite Element Analysis of Solids & Fluids I Fall `09 Lecture 11 - Heat Transfer Analysis Prof. K. J. Bathe Reading assignment: Sections 7.1-7.4.1 MIT OpenCourseWare To discuss heat transfer in systems, first let us define some variables. (x, y, z, t) S Sq = = = Temperature Surface area with prescribed temperature (p ) Surface area with prescribed heat flux into the body Given the geometry, boundary conditions, material laws, and loading, we would like to calculate the tem perature distribution over the body. To obtain the exact solution of the mathematical model, we need to satisfy the following in the differential formulation: Heat flow equilibrium Compatibility Constitutive relation(s) Example: One-Dimensional Case We can derive an expression for system equilibrium from the heat flow equation. q -q +q B dx = 0 x x+dx Using the constitutive equation, x q q =q + dx x x x+dx x 2 -k +k + dx + q B dx = 0 x x2 x q = -k 1 Lecture 11 Heat Transfer Analysis 2.092/2.093, Fall `09 We obtain the result 2 k 2 + q B = 0 in V x k , =0 = qS x L L x=0 Principle of Virtual Temperatures Clearly: 2 k 2 + qB = 0 x 2 L 2 k + q B dx = 0 k + q B dV = A x2 x2 V 0 Hence, 0 L (A) 2 k 2 + q B dx = 0 x 2 Lecture 11 Heat Transfer Analysis L L L - k dx + k q B dx = 0 x x x 0 0 0 L L q B dx + q S k dx = x x 0 0 x=L 2.092/2.093, Fall `09 (B) In 3D, the equation becomes V k dV = V T T q B dV + x y z Sq T q S dSq ; = x y z (C) k k= 0 0 For example: 0 k 0 0 0 k ; = q S = h e - S convection q S = r - S radiation e = temperature of the environment r = temperature of the radiation source (m) (x, y, z, t) = H (m) Substituting this into (C), we obtain K = Q K = K (m) m ; S(m) = H S(m) (m) = B (m) ; (m) = B (m) ; K (m) = V (m) B (m)T k(m) B (m) dV (m) Q = QB + QS QB = QB m (m) ; QB ; (m) = V (m) H (m)T q B(m) dV (m) HS (m) QS = S is unknown, so QS (m) (m) QS m (m) QS = (m) Sq T e (m) h - S dSq (m) = (m) Sq HS (m) T he(m) dSq (m) - (m) Sq HS (m) T hH S (m) dSq (m) Here we need to sum over all Sq (m) for element (m). 3 Lecture 11 Heat Transfer Analysis 2.092/2.093, Fall `09 Example (x, y) = h1 h2 h3 1 h4 2 3 4 1 x 1+ (1 + y) ; h2 = . . . h1 = 4 2 h1,x h2,x h3,x h4,x x = h1,y h2,y h3,y h4,y y B H S = H y=1 1 HS = 1+ x 2 2 1- x 2 0 0 4 MIT OpenCourseWare http://ocw.mit.edu 2.092 / 2.093 Finite Element Analysis of Solids and Fluids I Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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## This note was uploaded on 02/24/2012 for the course MECHANICAL 2.092 taught by Professor Klaus-jürgenbathe during the Fall '09 term at MIT.

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