Step change in the boundary condition of conduction problems Narasimha Murthy 1772037 ME535 Final Project

Introduction Many engineering problems in heat transfer involves discontinuous boundary conditions i.e. an abrupt change in the boundary condition along a smooth curve of line. These type of problems are called mixed or split boundary value problems Heat conduction on a plate whose top surface is both partially insulated and partly exposed to a fluid medium with constant temperature is an example of a mixed boundary value problem.

Introduction Heat conduction between bodies with wavy surfaces is an example of mixed boundary value problem

Introduction These problems are usually solved by artificial interface methods. Here we introduce an interface separating boundaries and, solutions are then found for both regions and matched at the artificial interface in order to find the series coefficients. . In solving numerically, discontinuities can cause serious difficulty.  It has been established that Finite difference methods require a very fine mesh near the discontinuity We hope that using a non uniform grid to analyse the problem numerically will give us more precise results and decrease the computational time

Introduction Problem Statement The fluid has a large thermal diffusivity, so the Peclet number of the flow is small. .

Background 𝑈 𝜕𝑇 𝜕𝑋 =𝛼[ 𝜕 2 𝑇 𝜕 𝑋 2 + 𝜕 2 𝑇 𝜕 𝑌 2 ] 𝑈 𝜕𝑇 𝜕𝑋 =𝛼[ 𝜕 2 𝑇 𝜕 𝑋 2 + 𝜕 2 𝑇 𝜕 𝑌 2 ] 𝑃𝑒 𝜕𝜃 𝜕𝑥 =[ 𝜕 2 𝜃 𝜕 𝑥 2 + 𝜕 2 𝜃 𝜕 𝑦 2 ] If Peclet number is large we can ignore the axial conduction For the case of our problem, we cannot ignore the axial conduction

Non-Uniform grid The non-uniform grid we used here 𝑋=𝐷𝑠+(𝐴 𝑋𝑐−𝐿𝑠 1−𝑠 𝑠) For our problem, D = 400, Xc = 200 and s contains 400 equidistant points.

Non-Uniform Grid Finite Difference equations f ′ Xi = f X i+1 −f X i−1 X i+1 − X i−1 +( X i+1 − X i 2 − X i − X i−1 2 2! X i+1 − X i−1 f ′′ Xi )+O h 2 𝑇𝑟𝑢𝑛𝑐𝑎𝑡𝑖𝑜𝑛 𝑒𝑟𝑟𝑜𝑟 = 1−𝑟 𝑓 ′′ 𝑋𝑖 2 ℎ 𝑖 𝑟= ℎ 𝑖 ℎ 𝑖−1

Non-Uniform Grid Finite Difference Equations According to Lagrange interpolation, we can represent any function f(x) locally as a polynomial Using Lagrange interpolation, 𝑓 ′ 𝑥 𝑖 =( − ℎ 𝑖+1 ℎ 𝑖 ℎ 𝑖+1 + ℎ 𝑖 )𝑓 𝑥 𝑖−1 +( ℎ 𝑖+1 − ℎ 𝑖 ℎ 𝑖+1 ℎ 𝑖 )𝑓 𝑥 𝑖 +( ℎ 𝑖 ℎ 𝑖+1 ( ℎ 𝑖+1 + ℎ 𝑖 ) )𝑓 𝑥 𝑖+1 𝑓 ′ ′ 𝑥 𝑖 = 2 ℎ 𝑖 ℎ 𝑖+1 + ℎ 𝑖 𝑓 𝑥 𝑖−1 + −2 ℎ 𝑖+1 ℎ 𝑖 𝑓 𝑥 𝑖 +( 2 ℎ 𝑖+1 ℎ 𝑖+1 + ℎ 𝑖 )𝑓 𝑥 𝑖+1 ℎ 𝑖 = 𝑋 𝑖 − 𝑋 𝑖−1

Numerical Solution & Discussion Consider Pe = 0.01, T1 = 300K, T2 = 350K, L = 10

Numerical Solution & Discussion Comparison with analytical solution

Numerical Solution & Discussion Comparison with analytical solution

Conclusion Lagrange interpolation has been used to get a finite difference equations of the order of h2. The resulting solution matches with analytical solution quite well in the vicinity of discontinuity, but at the discontinuity the numerical solution does not match the analytical solution.

Future Work