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

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Scientific Computing I

Module 5: Heat Transfer – Discrete and Contiuous Models

Michael Bader

Winter 2009/2010

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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

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Part II: A Continuous Model – The Heat Equation

From Discrete to Contiuous Derivation of the ation Heat Equations Boundary and Initial Conditions

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Part I Discrete Models

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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 · δT heat flow proportional to temperature differences

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A Wiremesh Model

• consider rectangular plate as fine mesh of wires
• compute temperature xij at nodes of the mesh

xi,j xi−1,j xi+1,j xi,j+1 xi,j−1 hx hy

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Wiremesh Model (2)

• model assumption: temperatures in equilibrium at every mesh

node

• for all temperatures xij:

xij = 1 4

• xi−1,j + xi+1,j + xi,j−1 + xi,j+1
• temperature known at (part of) the boundary; for example:

x0,j = Tj

• task: solve system of linear equations

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A Finite Volume Model

• object: a rectangular metal plate (again)
• model as a collection of small connected rectangular cells

hx hy

• examine the heat flow across the cell edges

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Heat Flow Across the Cell Boundaries

• Heat flow across a given edge is proportional to
• temperature difference (T1 − T0) between the adjacent cells
• length h of the edge
• e.g.: heat flow across the left edge:

q(left)

ij

= kx

• Tij − Ti−1,j
• hy
• heat flow across all edges determines change of heat energy:

qij = kx

• Tij − Ti−1,j
• hy + kx
• Tij − Ti+1,j
• hy

+ ky

• Tij − Ti,j−1
• hx + ky
• Tij − Ti,j+1
• hx

Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 9

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Temperature change due to heat flow

• in equilibrium: total heat flow equal to 0
• but: consider additional source term Fij due to
• external heating
• radiation
• Fij = fijhxhy (fij heat flow per area)
• equilibrium with source term requires qij + Fij = 0:

fijhxhy = −kxhy

• 2Tij − Ti−1,j − Ti+1,j
• −kyhx
• 2Tij − Ti,j−1 − Ti,j+1
• Michael Bader: Scientific Computing I

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Finite Volume Model

• divide by hxhy:

fij = −kx hx

• 2Tij − Ti−1,j − Ti+1,j
• −ky

hy

• 2Tij − Ti,j−1 − Ti,j+1
• 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)

ij

= 0

Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 11

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Towards a Time Dependent Model

• idea: set up ODE for each cell
• simplification: no external heat sources or drains, i.e. fij = 0
• change of temperature per time is proportional to heat flow into

the cell (no longer 0): ˙ Tij(t) = κx hx

• 2Tij(t) − Ti−1,j(t) − Ti+1,j(t)
• +

κy hy

• 2Tij(t) − Ti,j−1(t) − Ti,j+1(t)
• solve system of ODE

Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 12

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Part II A Continuous Model – The Heat Equation

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From Discrete to Contiuous

• remember the discrete model:

fij = −kx hx

• 2Tij − Ti−1,j − Ti+1,j
• −ky

hy

• 2Tij − Ti,j−1 − Ti,j+1
• assumption:heat flow accross edges is proportional to

temperature difference q(left)

ij

= kx

• Tij − Ti−1,j
• hy
• in reality: heat flow proportional to temperature gradient

q(left)

ij

≈ khy Tij − Ti−1,j hx

Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 14

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From Discrete to Contiuous (2)

• replace kx by k/hx, ky by k/hy, and get:

fij = − k h2

x

• 2Tij − Ti−1,j − Ti+1,j
• − k

h2

y

• 2Tij − Ti,j−1 − Ti,j+1
• consider arbitrary small cells: hx, hy → 0:

fij = −k ∂2T ∂x2

• ij

− k ∂2T ∂y2

• ij
• leads to partial differential equation (PDE):

−k ∂2T(x, y) ∂x2 + ∂2T(x, y) ∂y2

• = f(x, y)

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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:

∂ ∂t

• D

ρcT dV =

• D

q dV +

• ∂D

k∇T · n dS density ρ, specific heat c, and heat conductivity k are material parameters

• heat sources and drains are modelled in term q

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Derivation of the Heat Equation (2)

• according to theorem of Gauß:
• ∂D

k∇T · n dS =

• D

k∆T dV

• leads to integral equation for any domain D:
• D

ρcTt − q − k∆T dV = 0

• hence, the integrand has to be identically 0:

Tt = κ∆T + q ρc , κ := k ρc

• 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

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Heat Equations

Different scenarios:

• vanishing external influence, q = 0:

Tt = κ∆T alternate notation ∂T ∂t = κ · ∂2T ∂x2 + ∂2T ∂y2 + ∂2T ∂z2

• equilibrium solution, Tt = 0:

0 = κ∆T + q ρc − → −∆T = f “Poisson’s Equation”

Michael Bader: Scientific Computing I Module 5: Heat Transfer – Discrete and Contiuous Models, Winter 2009/2010 18

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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