2 d conduction - C HAPTER 4: S TEADY S TATE , 2-D C...

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Unformatted text preview: 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 ). b 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 General Heat Diffusion Equation: = + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ q z T k z y T k y x T k x & t T c p ∂ ∂ ρ For 2D conduction at steady state, no internal heat generation and assuming constant properties (constant k ): 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: x Exact (analytical) – only works for very select cases of temperature distributions x Empirical (graphical) – isotherms and adiabats x Approximation (numerical) – nodes, shape factors, finite elements 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, 2 2 2 2 = + dy T d dx T d b 2 2 2 2 = + dy Y d X dx X d Y ) ( ) ( y Y x X T = Rearranging, 2 2 2 2 = + dy Y d X dx X d Y b 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= - λ 2 Consequently, the solution of the original PDE (Laplace) reduces to the solution of the two following ODEs: 2 2 2 = + X dx X d λ 2 2 2 = − Y dy Y d λ The general solutions are: ) sin( ) cos( 2 1 x C x C X λ λ + = y y e C e C Y λ λ 4 3 + = − Thus, since T = X * Y [ ] [ ] y y e C e C x C x C T λ λ λ λ 4 3 2 1 ) sin( ) cos( + ⋅ + = − We can solve for the constants 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 X One set of B.C.s that allows for a (somewhat) straight-forward solution, making the substitution 1 T T − = θ : B.C. 1 y = 0 ⇒ T = T 1 or θ = 0 B.C. 2 x = 0 ⇒ T = T 1 or θ = 0 B.C. 3 x = W ⇒ T = T 1 or θ = 0 B.C. 4 y = H ⇒ T = T 1 + T m *sin( π x/W ) or θ = T m *sin( π x/W ) T m is the sine function amplitude We can now substitute these B.C.s We can now substitute these B....
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This note was uploaded on 01/24/2010 for the course CHEM ENG 2A04 taught by Professor Toddhoare during the Winter '10 term at McMaster University.

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2 d conduction - C HAPTER 4: S TEADY S TATE , 2-D C...

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