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Unformatted text preview: Chapter Three One Dimensional Steady State Conduction Heat diffusion Equation: t T k q z T y T x T ∂ ∂ = + ∂ ∂ + ∂ ∂ + ∂ ∂ α 1 2 2 2 2 2 2 for steady state = ∂ ∂ t T and one dimension (xdirection) 2 2 x T ∂ ∂ =0 or ( 29 = dx dx kdT d Plane wall: ( 29 = dx dx kdT d if k = constant 2 2 = dx T d integrating twice 1 c dx dT = T(x) =c 1 X+c 2 Apply B.c to obtain C 1 &C 2 1 2 1 ) ( T c T x T = ⇒ = = 1 1 2 2 ) ( T L c T T L x T + = ⇒ = = L T T c 1 2 1 = Then 1 1 2 ) ( ) ( T L x T T x T + = Linear temperature distribution. Heat transfer: dx dT kA q x = 1 1 2 c L T T dx dT = = Constant. L T T kA L T T kA q x ) ( ) ( 2 1 1 2 = = Flux ) ( 2 1 T T L k A q q x = = ′ ′ 1 T 1 T 2 X=0 X=L Alternative conduction Analysis: Starting with Fourier law for 1dimension dx dT kA q x = Since q x = constant, integrating the above eq. ∫ ∫ = kAdT dx q x If k= constant ∫ ∫ = 2 1 2 1 T T x x x dT kA dx q If limits are know T1= T(x=x 1 ) & T 2 = T(x=x 2 ) Then q x (x 2x 1 ) =kA (T 2T 1 ) q x = kA (T 2T 1 )/(x 2x 1 ) L T T kA x x T T kA q x ) ( ) ( ) ( 2 1 1 2 2 1 = = T 1 >T 2 Problem 1: Derive expression for q x if k= k (1+βT) ( 29 dx dT A T k q β + = 1 T 1 = T(x=x 1 ) & T 2 = T(x=x 2 ) ( 29  + = ) ( 2 ) ( 2 1 2 2 1 2 1 2 T T T T x x A k q β . Problem 2: Derive expression for q x if A = лD 2 X 2 4 See Example 3.3, p.88 a=.25 dx dT x a k dx dT kA q 2 2 4 π = = D=ax ) ( ) 1 1 ( 4 4 1 1 2 2 2 1 1 T T k x x a q dT k x dx a q T T x x = + ⇒ = ∫ ∫ π π Solve for T(x)  = x x a q T x T 1 1 4 ) ( 1 2 1 π ; solve for ) 1 1 ( 4 ) ( 2 1 2 1 2 x x T T k a q x = π q x =constant. 2 T 1 T 2 L X 1 X 2 T(x) x x Electrical Analogy: a) Conduction Heat transfer L T T kA q s s ) ( 2 1 = Current I= ∆V/R. q analogous to I ∆V analogous to T s1T s2 =∆T, driving force. Then R= L/kA R th = L/kA hen q x = ∆T/R th known as thermal resistance . T s1 b) Convection Heat Transfer: hA T T T Ts hA q s conv 1 ) ( ) ( ∞ = ∞ = th conv R T q ∆ = then R th = 1/hA T ∞ c) ConvectionConductionConvection: At steady state: q conv1 = q cond = q conv2 =q q conv1 = h 1 A 1 (T ∞1T s1 ) q cond= kA(T s1T s2 )/L q conv2 = h 2 A 2 (T s2 T ∞2 ) Then: T ∞1T s1 = q/ h 1 A 1 T s1T s2 = q/ kA/L T s2 T ∞2 =q/ h 2 A 2 T ∞1 See fig3.1 p75. 4 th ed. 3 T s2 q R th = L/kA T s T ∞ h q R th = 1/hA T s T ∞1 h 1 T ∞ h 2 q x=0 x=Ll T s1 T s2 1/h 1 A T s1 T s2 T ∞2 L/kA 1/h 2 A Add above eqs: + + = ∞ ∞ 2 2 1 1 2 1 1 1 A h KA L A h q T T and ( 29 + + = ∞ ∞ 2 2 1 1 2 1 1 1 A h KA L A h T T q ( 29 2 1 ∞ ∞ T T = ∆T overall , 1 1 1 1 , A h R conv th = since A h T T R T T q S th S conv 1 1 1 1 1 1 1 = = ∞ ∞ kA L R cond th = , , since th S S cond R T kA L T T q ∆ = = 2 1 , 1 2 2 2 , A h R conv th = since th S conv R T A h T T q ∆ = = ∞ 2 2 2 2 1 Hence; total th overall R T R T q ∆ = ∆...
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 Spring '08
 allan
 Heat, Steady State, Heat Transfer, dx, ts 2, Rth Rtotal

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