Scientific Computing I Part II: A Continuous Model The Heat - PowerPoint PPT Presentation

Scientific Computing I Michael Bader Outlines Part I: Discrete Models Scientific Computing I Part II: A Continuous Model – The Heat Equation Module 5: Heat Transfer – Discrete and Contiuous Models Michael Bader Lehrstuhl Informatik V Winter 2005/2006

Scientific Part I: Discrete Models Computing I Michael Bader Outlines Part I: Discrete Models Part II: A Continuous Model – The Heat Equation Wiremesh Model 1 A Finite Volume Model 2 A Time Dependent Model 3

Scientific Part II: A Continuous Model – The Computing I Michael Bader Heat Equation Outlines Part I: Discrete Models Part II: A Continuous Model – The Heat Equation From Discrete to Contiuous 4 Derivation of the Heat Equation 5 Heat Equations 6 Boundary and Initial Conditions 7

Scientific Computing I Michael Bader Wiremesh Model A Finite Volume Model A Time Dependent Part I Model Discrete Models

Scientific Motivation: Heat Transfer Computing I Michael Bader Wiremesh Model objective: compute the temperature A Finite Volume Model distribution of some object A Time Dependent under certain prerequisites: Model temperature at object boundaries given heat sources material parameters observation from physical experiments: q ≈ k · δ T heat flow proportional to temperature differences

Scientific A Wiremesh Model Computing I Michael Bader consider rectangular plate as fine mesh of Wiremesh Model A Finite Volume wires Model compute temperature x ij at nodes of the mesh A Time Dependent Model x i,j+1 x i−1,j x i,j x i+1,j x i,j−1 h y h x

Scientific Wiremesh Model (2) Computing I Michael Bader Wiremesh Model A Finite Volume model assumption: temperatures in Model equilibrium at every mesh node A Time Dependent Model for all temperatures x ij : x ij = 1 � � x i − 1 , j + x i + 1 , j + x i , j − 1 + x i , j + 1 4 temperature known at (part of) the boundary; for example: x 0 , j = T j task: solve system of linear equations

Scientific A Finite Volume Model Computing I Michael Bader object: e.g. a rectangular metal plate Wiremesh Model A Finite Volume model as a collection of small connected Model rectangular cells A Time Dependent Model h y h x examine the heat flow across the cell edges

Scientific Heat Flow Across the Cell Boundaries Computing I Michael Bader Heat flow across a given edge is proportional Wiremesh Model to A Finite Volume Model temperature difference ( T 1 − T 0 ) between the A Time Dependent adjacent cells Model length h of the edge e.g.: heat flow across the left edge: q ( left ) � � = k x T ij − T i − 1 , j h y ij heat flow across all edges determines change of heat energy: � � � � q ij = k x T ij − T i − 1 , j h y + k x T ij − T i + 1 , j h y � � � � + k y T ij − T i , j − 1 h x + k y T ij − T i , j + 1 h x

Scientific Temperature change due to heat Computing I Michael Bader flow Wiremesh Model A Finite Volume Model in equilibrium: total heat flow equal to 0 A Time Dependent Model but: consider additional source term F ij due to external heating radiation F ij = f ij h x h y ( f ij heat flow per area) equilibrium with source term requires q ij + F ij = 0: � � f ij h x h y = − k x h y 2 T ij − T i − 1 , j − T i + 1 , j � � − k y h x 2 T ij − T i , j − 1 − T i , j + 1

Scientific Finite Volume Model Computing I Michael Bader divide by h x h y : Wiremesh Model A Finite Volume − k x Model � � f ij = 2 T ij − T i − 1 , j − T i + 1 , j h x A Time Dependent Model − k y � � 2 T ij − T i , j − 1 − T i , j + 1 h y again, system of linear equations how to treat boundaries? prescribe temperature in a cell (e.g. boundary layer of cells) prescribe heat flow across an edge; for example insulation at left edge: q ( left ) = 0 ij

Scientific A Time Dependent Model Computing I Michael Bader Wiremesh Model A Finite Volume idea: set up ODE for each cell Model A Time Dependent no external heat sources or drains: f ij = 0 Model change of temperature per time is proportional to heat flow into the cell (no longer 0): κ x ˙ � � T ij ( t ) = 2 T ij ( t ) − T i − 1 , j ( t ) − T i + 1 , j ( t ) h x κ y � � + 2 T ij ( t ) − T i , j − 1 ( t ) − T i , j + 1 ( t ) h y solve system of ODE

Scientific Computing I Michael Bader From Discrete to Contiuous Derivation of the Part II Heat Equation Heat Equations Boundary and A Continuous Model – The Heat Initial Conditions Equation

Scientific From Discrete to Contiuous Computing I Michael Bader remember the discrete model: From Discrete to Contiuous − k x � � f ij = 2 T ij − T i − 1 , j − T i + 1 , j Derivation of the h x Heat Equation − k y Heat Equations � � 2 T ij − T i , j − 1 − T i , j + 1 Boundary and h y Initial Conditions assumption:heat flow accross edges is proportional to temperature difference q ( left ) � � = k x T ij − T i − 1 , j h y ij in reality: heat flow proportional to temperature gradient T ij − T i − 1 , j q ( left ) ≈ kh y ij h x

Scientific From Discrete to Contiuous (2) Computing I Michael Bader replace k x by k / h x , k y by k / h y , and get: From Discrete to Contiuous − k � � Derivation of the f ij = 2 T ij − T i − 1 , j − T i + 1 , j Heat Equation h 2 x Heat Equations − k � � 2 T ij − T i , j − 1 − T i , j + 1 Boundary and h 2 Initial Conditions y consider arbitrary small cells: h x , h y → 0: � ∂ 2 T � ∂ 2 T � � f ij = − k − k ∂ x 2 ∂ y 2 ij ij leads to partial differential equation (PDE): � ∂ 2 T ( x , y ) + ∂ 2 T ( x , y ) � − k = f ( x , y ) ∂ x 2 ∂ y 2

Scientific Derivation of the Heat Equation Computing I Michael Bader finite volume model, but with arbitrary control From Discrete to Contiuous volume D Derivation of the change of heat energy (per time) is a result of Heat Equation Heat Equations transfer of heat energy across D ’s surface, Boundary and heat sources and drains in D (external Initial Conditions influences) resulting integral equation: ∂ � � � k ∇ T · � ρ cT d V = q d V + n d S ∂ t D D ∂ D density ρ , specific heat c , and heat conductivity k are material parameters heat sources and drains are modelled in term q

Scientific Derivation of the Heat Equation (2) Computing I Michael Bader according to theorem of Gauß: From Discrete to Contiuous � � k ∇ T · � n d S = k ∆ T d V Derivation of the Heat Equation ∂ D D Heat Equations leads to integral equation for any domain D : Boundary and Initial Conditions � ρ cT t − q − k ∆ T d V = 0 D hence, the integrand has to be identically 0: T t = κ ∆ T + q κ := k ρ c , ρ c κ > 0 is called the thermal diffusion coefficient (since the Laplace operator models a (heat) diffusion process)

Scientific Heat Equations Computing I Michael Bader Different scenarios: From Discrete to Contiuous vanishing external influence, q = 0: Derivation of the Heat Equation T t = κ ∆ T Heat Equations Boundary and Initial Conditions alternate notation � ∂ 2 T ∂ x 2 + ∂ 2 T ∂ y 2 + ∂ 2 T ∂ T � ∂ t = κ · ∂ z 2 equilibrium solution, T t = 0: 0 = κ ∆ T + q − ∆ T = f − → ρ c “Poisson’s Equation”

Scientific Boundary Conditions Computing I Michael Bader Dirichlet boundary conditions: From Discrete to fix T on (part of) the boundary Contiuous Derivation of the Heat Equation T ( x , y , z ) = ϕ ( x , y , z ) Heat Equations Boundary and Neumann boundary conditions: Initial Conditions fix T ’s normal derivative on (part of) the boundary: ∂ T ∂ n ( x , y , z ) = ϕ ( x , y , z ) special case: insulation ∂ T ∂ n ( x , y , z ) = 0