College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 501B Seminar in Engineering Analysis Spring 2009 Class 14443 Instructor Larry Caretto February 16 Homework Solutions 1 Solve the diffusion equation in a cylinder with a convection boundary condition for the potential u r t for 0 r R and t 0 The problem is defined as follows u 1 u r t r r r u r 0 u 0 r u h u r R u r r R u is finite at r 0 k In this problem is the usual diffusivity h is the convection coefficient u is the free stream potential and k is the conductance The parameters h u and k are constant The approach used for solving solution to this problem is a combination of approaches used for two problems done in class 1 diffusion in a cylinder with the boundary condition that u R t uR and 2 diffusion in Cartesian direction with the convection boundary condition The solution is similar to the one found for solving the diffusion equation in a cylinder namely u r t Cm e 2m t R2 m 1 following equation f m r J 0 m u R In this equation m are the roots of the hR J 0 m m J 1 m 0 m 1 k Obtain an equation for Cm for the general initial condition u r 0 u0 r and for the special case where u0 r U0 a constant Use the following integral to find Cm R r J 0 m r dr 2 0 R2 J 0 m R 2 J 1 m R 2 2 We want to solve the diffusion equation in a cylindrical coordinate system 0 r R for the potential u r t with the initial condition that u r 0 u0 r and the convection boundary condition at r R u 1 u r t r r r k u r r R h u r R u 1 We do not have a Sturm Liouville problem because the boundary condition at r R is not homogenous If we define a new temperature variable u u we will have a homogenous set of equations and boundary conditions Jacaranda Engineering Room 3333 Email lcaretto csun edu Mail Code 8348 Phone 818 677 6448 Fax 818 677 7062 1 r t r r r k h r R 0 r r R 2 As usual we seek a solution for r t using separation of variables We postulate a solution that is the product of two functions T t a function of time only and P r a function of the radial coordinate r only With this assumption our solution becomes r t P r T t 3 We substitute equation 3 for u in equation 2 Since P r is a function of r only and T t is a function of t only we obtain the following result 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 4 If we divide the final equation through by the product P r T t we obtain the following result 1 1 T t 1 1 P r r 2 T t t P r r r r 5 The left hand side of equation 5 is a function of t only the right hand side is a function of r only The only way that this can be correct is if both sides equal a constant As before we choose the constant to be equal to 2 This gives us two ordinary differential equations to solve dT t 2 2 T t has the general solution T t Ae t The dt d dP r r 2 rP r 0 The general second differential equation in 5 can be written as dr dr The first equation becomes solution to this equation is P r BJ0 r CY0 r where J0 and Y0 are the Bessel functions of the first and second kind with zero order Thus our general solution for u r t P r T t becomes 2 r t T t P r Ae t BJ 0 r CY0 r 2 e t C1 J 0 r C2Y0 r 6 As r 0 Y0 r to keep the solution finite we require that C2 0 To satisfy the boundary condition on at r R given by equation 2 we must satisfy the following equation for the solution P r dP r h P R 0 dr r R k dJ 0 r h J 0 R 0 dr k r R 7 Since the derivative dJ0 z dz J1 z we can write this boundary condition as follows where we have defined m R J 1 R h J 0 R 0 k Jacaranda Engineering Room 3333 Email lcaretto csun edu f m m J1 m Mail Code 8348 hR J 0 m 0 k 8 Phone 818 677 6448 Fax 818 677 7062 Equation 8 defines a transcendental equation for m R that gives the eigenvalues for this solution A plot of f m showing the first four zeros of this function for hR k 1 is given below A calculation of the first four zeros from equation 8 for hR k 1 gives the following values 1 1 25578371 2 4 079477709 3 7 155799184 and 4 10 27098536 This solution process can be continued to obtain a large number of eigenvalues that might be required for the solution of the series solution in equation 9 For any value of hR k we can solve equation 8 for a sequence of values m that give the eigenvalues m mR The solution to our radial diffusion problem is the sum of all possible solutions from equation 6 with C 2 0 with different values of m This gives r t C m e 2m t R2 m 1 r J0 m R 9 hR 1 k We still have to satisfy the initial condition that u r 0 u 0 r We can do this by using an eigenfunction expansion for the solution to our Sturm Liouville statement of the problem r t Setting t 0 in equation 9 sets the exponential term to one and gives the following result r 0 u r 0 u u0 r u C m J 0 m r 10 m 1 The values of Cm are found from the general equation for orthogonal eigenfunction expansions which includes a weighting function Here the weighting function p r is equal to r For any initial condition u0 r we find the values of Cm from the following equation In the second equation below the integral in the denominator has been evaluated using integral tables R Cm 0 R rJ 0 m r u0 r u dr R r J 0 m r 0 2 dr rJ 0 m r u0 r u dr 2 0 2 R J 0 m R 2 R J1 m R 2 2 2 11 …