Chapter 4 - 2D Conduction 2012 - Web

Chapter 4 - 2D Conduction 2012 - Web - CHAPTER 4: STEADY...

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C HAPTER 4: S TEADY S TATE , 2-D C ONDUCTION For many heat transfer problems engineers must solve, the effect of the object’s ends, corners, etc. can be neglected without significant error ( semi-infinite assumption ). A 2D or 3D problem becomes 1D However, other problems cannot be simplified to a 1-D system without incorporating large amounts of error or missing some critical phenomena. 2/3 y 1/3 y A B C D E NOT Insulated
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General Heat Diffusion Equation: q z T k z y T k y x T k x t T c p For 2D conduction at steady state, no internal heat generation and assuming constant properties (constant k ): 0 2 2 2 2 dy T d dx T d LAPLACE EQUATION We can solve the Laplace Equation to find temperature as a function of x and y by one of three methods: Exact (analytical) – only works for very select cases of temperature distributions Empirical (graphical) – isotherms and adiabats Approximation (numerical) – nodes, shape factors, finite elements
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A NALYTICAL S OLUTIONS The Laplace Equation can be solved analytically by the method of separation of variables Basic Trick: Assume that the solution can be expressed as the product of a function of x and a function of y , Since T = X * Y , differentiate: dx dX Y dx dT 2 2 2 2 dx X d Y dx T d dy dY X dy dT 2 2 2 2 dy Y d X dy T d Therefore the PDE becomes, 0 2 2 2 2 dy T d dx T d 0 2 2 2 2 dy Y d X dx X d Y ) ( ) ( y Y x X T
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Rearranging, 0 2 2 2 2 dy Y d X dx X d Y 2 2 2 2 1 1 dy Y d Y dx X d X NOTE: L.H.S. function of x only R.H.S. function of y only Since x and y are independent, the the equation can only hold for all x, y values if L.H.S.=R.H.S.=constant= - 2 Consequently, the solution of the original PDE (Laplace) reduces to the solution of the two following ODEs: 0 2 2 2 X dx X d 0 2 2 2 Y dy Y d The general solutions are: ) sin( ) cos( 2 1 x C x C X y y e C e C Y 4 3 Thus, since T = X * Y    y y e C e C x C x C T 4 3 2 1 ) sin( ) cos(
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C 1 , C 2 , C 3 , and C 4 by applying the appropriate boundary conditions for our problem With 2D heat flow we need 2x2 = 4 boundary conditions – the temperatures at x =0, x =W, y =0, y =H (each of the 4 sides of the object) T=given (BC. 4) T=given (BC. 3) T=given (BC. 2) T=given (BC. 1) Y H Depending on the nature of the four B.C.s., determining the integration constants can be straight-forward, very complicated, or IMPOSSIBLE. W
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Chapter 4 - 2D Conduction 2012 - Web - CHAPTER 4: STEADY...

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