Scientific Computing I Module 5: Heat Transfer Discrete and - PowerPoint PPT Presentation

Lehrstuhl Informatik V Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models Michael Bader Winter 2009/2010 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 1

Lehrstuhl Informatik V Part I: Discrete Models Wiremesh Model A Finite Volume Model Time Dependent Model Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 2

Lehrstuhl Informatik V Part II: A Continuous Model – The Heat Equation From Discrete to Contiuous Derivation of the ation Heat Equations Boundary and Initial Conditions Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 3

Lehrstuhl Informatik V Part I Discrete Models Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 4

Lehrstuhl Informatik V Motivation: Heat Transfer • objective: compute the temperature distribution of some object • under certain prerequisites: • temperature at object boundaries given • heat sources • material parameters • observation from physical experiments: q ≈ k · δ T heat flow proportional to temperature differences Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 5

Lehrstuhl Informatik V A Wiremesh Model • consider rectangular plate as fine mesh of wires • compute temperature x ij at nodes of the mesh x i,j+1 x i−1,j x i,j x i+1,j x i,j−1 h y h x Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 6

Lehrstuhl Informatik V Wiremesh Model (2) • model assumption: temperatures in equilibrium at every mesh node • 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 7

Lehrstuhl Informatik V A Finite Volume Model • object: a rectangular metal plate (again) • model as a collection of small connected rectangular cells h y h x • examine the heat flow across the cell edges Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 8

Lehrstuhl Informatik V Heat Flow Across the Cell Boundaries • Heat flow across a given edge is proportional to • temperature difference ( T 1 − T 0 ) between the adjacent cells • 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 9

Lehrstuhl Informatik V Temperature change due to heat flow • in equilibrium: total heat flow equal to 0 • 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 10

Lehrstuhl Informatik V Finite Volume Model • divide by h x h y : − k x � � f ij = 2 T ij − T i − 1 , j − T i + 1 , j h x − 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 11

Lehrstuhl Informatik V Towards a Time Dependent Model • idea: set up ODE for each cell • simplification: no external heat sources or drains, i.e. f ij = 0 • 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 12

Lehrstuhl Informatik V Part II A Continuous Model – The Heat Equation Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 13

Lehrstuhl Informatik V From Discrete to Contiuous • remember the discrete model: − k x � � f ij = 2 T ij − T i − 1 , j − T i + 1 , j h x − k y � � 2 T ij − T i , j − 1 − T i , j + 1 h y • 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 14

Lehrstuhl Informatik V From Discrete to Contiuous (2) • replace k x by k / h x , k y by k / h y , and get: − k � � f ij = 2 T ij − T i − 1 , j − T i + 1 , j h 2 x − k � � 2 T ij − T i , j − 1 − T i , j + 1 h 2 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 15

Lehrstuhl Informatik V Derivation of the Heat Equation • finite volume model, but with arbitrary control volume D • change of heat energy (per time) is a result of • transfer of heat energy across D ’s surface, • heat sources and drains in D (external influences) • resulting integral equation: ∂ � � � k ∇ T · � ρ cT dV = q dV + n dS ∂ 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 16

Lehrstuhl Informatik V Derivation of the Heat Equation (2) • according to theorem of Gauß: � � k ∇ T · � n dS = k ∆ T dV ∂ D D • leads to integral equation for any domain D : � ρ cT t − q − k ∆ T dV = 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) Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 17

Lehrstuhl Informatik V Heat Equations Different scenarios: • vanishing external influence, q = 0: T t = κ ∆ T 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” Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 18

Lehrstuhl Informatik V Boundary Conditions Dirichlet boundary conditions: • fix T on (part of) the boundary T ( x , y , z ) = ϕ ( x , y , z ) Neumann boundary 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 Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 19