Radial diffusion equation February 4 2009 Outline Diffusion Equations in Cylindrical Coordinates Review last class Gradient and convection boundary condition Larry Caretto Mechanical Engineering 501B Seminar in Engineering Analysis February 4 2009 Diffusion equation in radial coordinates Solution by separation of variables Result is form of Bessel s equation Review Bessel functions Eigenfunction expansion in Bessel functions 2 Review Homework Problem Review Homework Problem II Page 561 problem 5 find u x t for 0 x L General solution is sum of all eigenfunctions u u 2 t x u x 2 u x x 0 0 u x 0 f x x L u x t Ane n t cos n x n 0 2 0 for cosines General orthogonal relationship for An An Start with separation of variables solution C1 sin x C2 cos x u x t e 2 t C1 cos x C2 sin x u x e 2 t L n n L n x dx L L 2 n x cos L dx 0 f x cos 0 Final cosine result has A0 L L 1 2 n x A0 f x dx An f x cos dx L0 L0 L For u x x 0 0 C1 0 For u x x L 0 n L for integer n 3 Review Convection Problem T Heat Flux qx L Heat Flux qx L x L Symmetry Condition x 0 T x x L 0 x 0 T Conduction convection heat balance at x L T h Tx L T q x L k x x L x t 2 L L ME 501B Engineering Analysis Review Dimensionless Problem Diffusion equation Solution Region Physical Region 4 T T T0 5 T T 2T 2 t x 2 2 Initial T x 0 f0 x 0 condition T Symmetry boundary x 0 x 0 condition Convection boundary condition k T x h Tx L T 0 x L f 0 L T T0 T 0 0 hL 1 0 1 k Homogenous6 boundary condition 1 Radial diffusion equation February 4 2009 Review Finding n Review Initial Conditions 25 Usual formula for Cm but m m k hL 20 15 k hL 1 10 Functions 5 Cm Line Cotangent Intersections 0 1 1 0 cos m d 0 1 cos 2 m d 2 m 0 cos m d 0 cos m sin m m 0 5 Constant initial temperature T0 gives 0 1 10 1 15 cot 20 Cm 25 0 2 4 6 8 10 12 14 16 18 20 x 2 m 1 cos m d 0 cos m sin m m 2 sin m cos m sin m m 7 8 Review Solution Cm General solution 1 m 1 cos m sin m m 0 1 e m cos m 2 0 8 Solution for T x 0 T0 0 1 2m 2 sin m e cos m m 1 cos m sin m m Need root finding method to obtain eigenvalues m hL k cot m tau 1 Dimensionless Temperatur 0 1 0 0 1 0 2 0 3 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1 10 Also gets constants C1 and C2 which determines functions that are in solution tau 1 5 0 tau 3 tau 4 Apply boundary conditions to determine eigenvalues tau 0 75 0 2 tau 2 Time solution will be exponential tau 0 5 0 3 tau 1 5 0 3 Solve by separation of variables tau 0 35 0 5 0 4 0 4 Define u x t v x t w x Use u uref in original equation 0 7 0 6 0 5 Create Sturm Lioville problem for nonzero spatial boundary conditions 3 00 0 tau 5 00 0 01 0 0 02 tau 1 Review Summary 0 0 5 tau 0 75 0 6 x L u ta ta u tau 0 2 tau 0 5 0 u ta ta u tau 0 02 0 1 tau 0 1 tau 0 05 0 7 0 1 Review Solution for hL k 10 0 9 tau 0 1 0 2 9 0 8 tau 0 2 0 9 Dimensionless Temperatur 2 m 0 cos m d Review Solution for hL k 1 0 4 0 5 0 6 x L ME 501B Engineering Analysis 0 7 0 8 0 9 1 11 12 2 Radial diffusion equation February 4 2009 Review Summary II Cylindrical Diffusion Equation Write solution as sum of all possible eigenfunctions with individual constants Use eigenfunction expansion to match initial conditions If a solution for u x t v x t w x is used the eigenfunction expansion must be for u0 x w x Solution is sum of all eigenfunctions with constants determined from matching initial conditions General diffusion equation for three dimensions Cartesian Cylindrica l Sphere 2u 2u u 2u t 2u 2u 2u x 2 y 2 z 2 2u One dimensional radial equation 1 u 1 2u 2u r r r r r 2 2 z 2 1 2 u 1 2u 1 2u cot u 2 r 2 2 2 r r r r sin 2 r 2 2 r 13 Radial Diffusion Equation New Variable for u R t uR u r t v r t uR Governs diffusion heat conduction in cylinder for t 0 and 0 r R u r t is temperature species concentration Initial condition u r 0 u0 r No boundary condition at r 0 except that u 0 t is finite u r r 0 0 by symmetry Diffusivity is material property length 2 time 14 u 1 u r t r r r v r t satisfies diffusion equation v R t 0 and v 0 t is finite Gives a Sturm Liouville problem for radial function Since uR is a constant u r t is a solution to the original problem It satisfies the differential equation and the boundary conditions u v u R 1 u 1 v u R r r t t r r r r r r 15 Separation of Variables Solve ODEs to Get v r t Assume v r t P r T t u P r T t T t P r t t t 1 u 1 P r T t 1 P r r r T t r r r r r r r r r r Divide by P r T t 1 1 T t T t t Have exponential ODEs in time and Bessel s equation for radial function dT t 2 t 2 T t T t Ae dt This is form of d dP r r 2 rP r 0 Bessel s equation dr dr Solution of Bessel s equation is P r BJ0 r CY0 r 1 1 P r r 2 P r r r r J0 and Y0 are order zero Bessel functions of first and second kind respectively 17 ME 501B Engineering Analysis 16 18 3 Radial diffusion equation February 4 2009 Review Bessel s Equation Gamma Functions d 2 y x 1 dy x x 2 2 …