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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 + + = + = = = & & 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 ) ( 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 ) ( ) ( + + = + + + = + + = + + = & & & & &
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
) 3 ( 16 4 16 3 4 24 11 2 ) 0 ( ) , ( 24 11 ) ( 2 ) 0 ( 2 4 2 2 + + + = = + R r r R kR q k qR x c m R q T x r T k qR x T x c m R q T b S b & & π c. From the above results, we have: ) ( 11 24 24 11 ) ( ) ( b S S b T T R k q k qR x T x T = = (4) Nusselt number is defined as hD/k for circular pipes. In this problem, the transfer coefficient, 364 . 4 11 48 11 48 = = = = k hD Nu D k h d. With the given parameter values, we can compute the Reynolds number as: 2300 1648 ) 10 03 . 1 )( 015 . 0 ( ) / 02 . 0 ( 4 4 4 / Re 3 2 < = × = = = = s Pa m s kg D m D m D vD μ ρ & & Hence the flow is laminar, and the temperature in the previous parts is valid. Using Eqs. 3 and 4, and m W R q / 50 2 = , K m x m r T m m m m m mK W m m W m kgK J s kg m W K m r m x T R r r R kR q x c m R q T r x T R r r R kR q k qR x c m R q T r x T S b 04 . 361 ) 20 , 006 . 0 ( ) 0075 . 0 ( 16 ) 006 . 0 ( 4 ) 006 . 0 ( 16 ) 0075 . 0 ( 3 ) 0075 . 0 )( / 12 . 0 ( ) 015 . 0 /( ) / 50 ( 4 ) 20 ( ) / 1990 )( / 02 . 0 ( / 50 350 ) 006 . 0 , 20 ( 16 4 16 3 4 2 ) 0 ( ) , ( 16 4 16 3 4 24 11 2 ) 0 ( ) , ( 2 4 2 2 2 4 2 2 2 4 2 2 = = = + + = = = + + = + + + = & & e. Using Eqs. 3 and 4, the temperature distribution is shown as below:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/06/2011 for the course MECHANICAL 2.006 taught by Professor Blah during the Spring '08 term at MIT.

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online