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PS5_sol_S08

# PS5_sol_S08 - Problem Set 5(2.006 Spring 08 Problem 1 T 1...

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Problem Set 5 (2.006, Spring 08) Problem 1 a. Given x T v r T r r r = 1 α (1) For a fully developed laminar flow in circular tube, = 2 2 1 2 R r v v ave where 2 R m v ave ρπ & = If we assume T(x,r) = T b (x) + Θ (r), we will have c m R q x T & π 2 = Substitute into Eq. (1), ) ( ) ln( ) ( 16 4 4 ) , ( ) ( 4 2 4 1 4 1 2 1 2 1 2 1 2 4 2 1 2 4 2 2 2 2 2 2 x C r x C R r r kR q r x T x C R r r kR q r T r R r kR q r T r r r c m R q R r R m r T r r r c k + + = + = = = & & π ρπ ρ Since T(x,r) is finite everywhere including at r=0, hence, C 1 = 0. We have the boundary condition of T(x, r=R) = T S (x). Hence, ) 2 ( 16 4 16 3 4 ) ( ) , ( 16 3 4 ) ( ) ( ) ( 16 4 4 ) ( 2 4 2 2 2 2 2 2 4 2 + = = + = R r r R kR q x T r x T R kR q x T x C x C R R R kR q x T S S S b. The bulk temperature is defined as: c m r x cT r v rdr x T R b & = 0 ) , ( ) ( 2 ) ( ρ π By using the result from part a, R S b R S b R S b R o S b R r R r r R r r R r kR q x T x c m R q T rdr R r R r r R r r R kR q x T x c m R q T rdr R r r R R r kR q x T x c m R q T c m R r r R kR q c R r R m rdr x T x T 0 4 8 2 6 4 2 6 4 2 2 3 0 4 6 2 4 2 2 4 2 2 3 0 2 4 2 2 2 2 3 2 4 2 2 2 2 2 128 24 64 3 96 16 32 3 16 ) ( 2 ) 0 ( 16 4 16 3 16 4 16 3 16 ) ( 2 ) 0 ( 16 4 16 3 1 16 ) ( 2 ) 0 ( 16 4 16 3 4 1 2 2 ) ( ) ( + + = + + + = + + = + + = & & & & & π π π ρπ ρ π

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