Overview. Review Multidimensional PDEs. Review Diffusion Solutions. Review Separation of Variables. More diffusion solutions February 2, 2009

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1 More diffusion solutions February, 9 More Diffusion Equation Solutions arry Caretto Mechanical Engineering 5B Seinar in Engineering Analysis February, 9 Overview Review last class Separation of variables and eigenfunction epansions for initial condition Gradient boundary condition In-class eercise for one of the hoewor probles Convection boundary condition Use variable transfors to get hoogenous boundary conditions required for a Stur-iouville proble Review Diffusion and aplace Partial differential equations related to conservation principles of flues governed by potentials Heat transfer fro teperature gradient Mass diffusion fro concentration gradient Current fro electrostatic potential Magnetic flues Ideal fluid flow fro velocity potential Review Multidiensional PDEs General diffusion equation for three diensions Cartesian Cylindrica l Sphere u r r r u u u u + + y z u α u u u u r + + r r r r θ z + r r aplace (steady) u u u cotφ + + sin φ θ r φ r φ 3 4 Review Diffusion Solutions Governs heat conduction and species diffusion for t and a, is teperature, species concentration Initial condition,) u () Boundaries, u (; a, u R ( Started with, a, Diffusivity, α, is aterial property (length) /(tie) u α 5 Review Separation of Variables Assue, X()T( [ X ( ) T X ( ) u [ X ( ) T X ( ) α α αt Dividing by αx()t( gives f( g() which ust equal a constant X ( ) α T X ( ) 6 ME 5B Engineering Analysis

2 More diffusion solutions February, 9 Review General Solution Solve two ODEs dt t + λ αt T t Ae α ( ) dt d X ( ) + λ X ( ) X ( ) Bsin( λ ) + C cos( λ) d, T e X ( ) Ae [ Bsin( λ) + C cos( λ) [ C sin( λ) + C cos( λ) Can use this as starting point for any boundary or initial conditions 7 Review Hoogenous BCs oo at case where, a, In this case, X() is the solution to a Stur-iouville proble X() Bsin(λ) + Ccos(λ) X() Bsin() + Ccos() C X( a ) Bsin(λ a ) Must have λ a nπ (n an integer) Coplete set of orthogonal eigenfunctions: sin(nπ/ a ) 8 Review Initial Condition General solution: su of all eigenfunctions λ n n π, Cne sin( λn) λn n Eigenfunction epansion gives u ( ),) Cn sin n any initial condition a π u ( )sin d a a π C u a ( )sin π a a sin d a a nπ a d 9 u/u Review Solution for u () U t. t. t t.5 t.8 t.3 t.4 t.5 t.8 t. t.5 t.5 t.75 t.. t / Review Nonzero Boundaries Stur-iouville eigenfunction epansion requires zero boundary conditions For nonzero boundaries,, u and a, u R split solution into two functions, v(, + w() v satisfies diffusion equation with zero boundary conditions: v/ α v/ with v(, v( a, ; v(,),) w() w satisfies ODE d w/d with boundary conditions w() u and w( a ) u R u v + w satisfies PDEs, BC and ICs Review Nonzero Boundaries II Eigenfunction epansion used for zero boundary solution v(,) gives a ur u π C ( ) u u sin d a a a After C n is found, solution is nπ t α a, Cne n v(, a v(,) u () w() nπ ur u sin + u + a w() ME 5B Engineering Analysis

3 More diffusion solutions February, 9 (, - u)/(u - u) t. Review u () U Eaple t.8 t. t.5 t. t. t.3 t.4 t.5 t.8 t. t.5 t. t.3 t (u R - u ) / (U - u ).5 t α(tie)/( a) 3 Other Boundary Conditions Can have boundary conditions on gradients Physical eaning is flu Zero gradient of teperature, ass fraction, etc. eans is zero flu of heat, diffusion, etc. Stur-iouville proble requires zero gradient boundary condition Start with separation of variables solution, T e t λ α X ( ) Ae [ Bsin( λ) + C cos( λ) [ C sin( λ) + C cos( λ) 4 Hoewor Proble for /9 Tet, p 56, prob 3: find, for u α,) f ( ) Show that the solution is t nπ n u t A + α (, ) Ane cos( λn) λn n nπ A f ( ) d An f ( ) cos d Note: Kreyszig uses c α 5 Class Eercise Wor proble on previous chart Start with separation of variables result, T e t λ α X ( ) Ae [ Bsin( λ) + C cos( λ) [ C sin( λ) + C cos( λ) Apply zero gradient boundary conditions to get eigenfunction solution Use eigenfunction epansion for initial conditions 6 Convection Boundary Condition oo at gradient and ied boundary conditions Show how aing differential equation diensionless can lead to Stur- iouville proble in initial forulation Consider convective heat transfer to sides of slab in region Because of syetry solve with zero gradient (syetry) condition at 7 Physical Region T Heat Flu q Syetry Condition at Proble Diagra - Heat Flu q q Solution Region T Conduction-convection heat balance at q h( T T) q q h T T 8 ( ) ME 5B Engineering Analysis 3

4 More diffusion solutions February, 9 Proble Definition Diffusion equation Initial condition T α T(,) f () Syetry boundary condition Convection boundary condition + h( T T ) + ht ht Nonhoogenous9 boundary condition Diensionless Variables We can convert the boundary condition at to a Stur-iouville proble by defining a new variable lie θ T T All other occurrences of T, which are in derivatives, will not change For convenience we typically define the following diensionless variables (based on previous results) T T Θ ξ τ T T Reference value (constant initial value) Substitute Diensionless T α α [( T T ) Θ + T α( T T ) τ α [( T T ) Θ + T ( ξ) [( T T ) Θ + T ( ξ) ( T T ) α ( T T ) ( T T ) Θ τ Θ Θ Θ + h( T T ) + h ( T T ) Θ Diensionless Proble Diffusion T Θ Θ equation α τ Initial T(,) f () f ( ξ) T ξ,) condition T T Syetry boundary Θ condition ξ Convection boundary condition + h( T T ) Θ h + Θξ ξ Hoogenous boundary condition General Solution Start with separation of variables solution in diensionless coordinates τ Θ ξ, τ ) e λ C sin( λξ ) + C cos( λξ ) [ ( Θ λe ξ τ [ C λξ ) C sin( λξ ) cos( Boundary condition of zero gradient at ξ requires C Θ τ λe C cos() C sin() Cλe ξ τ [ 3 Θ Boundary at ξ Substitute solution into diensionless convection boundary condition at ξ ξ h + Θ ξ e τ h τ C sin( λ) + e C cos( λ) Get equation to be solved for λ cos( λ) λ cot( λ) h sin( λ) 4 ME 5B Engineering Analysis 4

5 More diffusion solutions February, 9 Functions Finding λ n f (/h) λ /h ine Cotangent Intersections f cot λ 5 Getting Initial Conditions Solution as su of all eigenfunctions h τ ξ, τ ) Ce λ cos( λξ ) λ cot( λ) Eigenfunction epansion for initial condition Valid because Θ are eigenfunctions for a Stur-iouville proble even though eigenvalues are not evenly spaced h ξ,) Θ ( ξ) C cos( λξ) λ cot( λ ) 6 Getting Initial Conditions II Usual forula for C but λ π C ξ )cos( λξ ) dξ λ ξ )cos( λξ ) dξ cos( λ)sin( λ) + λ cos ( λ ξ ) dξ Constant initial teperature, T, gives Θ C λ cos( λ () cos( λ ξ) dξ )sin( λ ) + λ sin( λ) cos( λ )sin( λ ) + λ T T Θ T T 7 Solution Substitute forula for C into previous solution h τ ξ, τ) Ce λ cos( λξ ) λ cot( λ) sin( λ ) e λ cos( ) (, ) τ λξ h Θ ξ τ λ cot( λ ) cos( λ )sin( λ ) + λ Need root-finding ethod to obtain eigenvalues, λ For typical accuracy, can use only first ter in su if τ / >. 8 Values for λ n Solution for h/ n h/ h/ h/ Diensionless Teperatur tau..9 tau. tau. tau.5.8 tau.5.7 tau.75.6 tau.5.4 tau.5.3 tau. tau 3. tau / 3 ME 5B Engineering Analysis 5

6 More diffusion solutions February, 9 Diensionless Teperatur Solution for h/ tau. tau. tau.5 tau.75 tau.35 tau tau.5 tau.5 tau. tau. tau.5 tau Diffusion Equation Suary Create Stur-ioville proble for nonzero spatial boundary conditions Define, v(, + w() Use u u ref in original equation Solve by separation of variables Product solution X()T( Tie solution, T(, will be eponential X() is sine and cosine for Cartesian Apply boundary conditions to deterine eigenvalues and eigenfunctions / 3 3 Diffusion Equation Suary II Write solution as su of all possible eigenfunctions with individual constants Use eigenfunction epansion to atch initial conditions If a solution for, v(, + w() is used the eigenfunction epansion ust be for u () w() Solution is su of all eigenfunctions with constants deterined fro atching initial conditions. 33 Diffusion Equation Suary III Sine eigenfunctions start at n Cosine eigenfunctions start at n and C equation has factor of / not / Can convert diffusion equation with source ter into hoogenous equation Hoewor proble 7 (K, p 56): Find w() such that, v(, + w(), where v() satisfies diffusion equation with v(, v(, to solve u α + Ne a 34 ME 5B Engineering Analysis 6