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**FEM FOR HEAT TRANSFER PROBLEMS**

Finite Element Method for readers of all backgrounds G. R. Liu and S. S. Quek CHAPTER 12: FEM FOR HEAT TRANSFER PROBLEMS

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**CONTENTS FIELD PROBLEMS WEIGHTED RESIDUAL APPROACH FOR FEM**

1D HEAT TRANSFER PROBLEMS 2D HEAT TRANSFER PROBLEMS SUMMARY CASE STUDY

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FIELD PROBLEMS General form of system equations of 2D linear steady state field problems: (Helmholtz equation) For 1D problems:

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FIELD PROBLEMS Heat transfer in 2D fin Note:

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**FIELD PROBLEMS Heat transfer in long 2D body Note: Dx = kx, Dy =tky,**

g = 0 and Q = q

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FIELD PROBLEMS Heat transfer in 1D fin Note:

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FIELD PROBLEMS Heat transfer across composite wall Note:

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**FIELD PROBLEMS Torsional deformation of bar**

Note: Dx=1/G, Dy=1/G, g=0, Q=2q ( - stress function) Ideal irrotational fluid flow Note: Dx = Dy = 1, g = Q = 0 ( - streamline function and - potential function)

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**FIELD PROBLEMS Accoustic problems**

P - the pressure above the ambient pressure ; w - wave frequency ; c - wave velocity in the medium Note: , Dx = Dy = 1, Q = 0

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**WEIGHTED RESIDUAL APPROACH FOR FEM**

Establishing FE equations based on governing equations without knowing the functional. (Strong form) Approximate solution: (Weak form) Weight function

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**WEIGHTED RESIDUAL APPROACH FOR FEM**

Discretize into smaller elements to ensure better approximation In each element, Using N as the weight functions where Galerkin method Residuals are then assembled for all elements and enforced to zero.

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**1D HEAT TRANSFER PROBLEM**

1D fin k : thermal conductivity h : convection coefficient A : cross-sectional area of the fin P : perimeter of the fin : temperature, and f : ambient temperature in the fluid (Specified boundary condition) (Convective heat loss at free end)

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1D fin Using Galerkin approach, where D = kA, g = hP, and Q = hP

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1D fin Integration by parts of first term on right-hand side, Using

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**1D fin (Strain matrix) where (Thermal conduction) (Thermal convection)**

(External heat supplied) (Temperature gradient at two ends of element)

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**1D fin For linear elements, (Recall 1D truss element) Therefore,**

for truss element (Recall stiffness matrix of truss element)

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1D fin for truss element (Recall mass matrix of truss element)

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**1D fin or (Left end) (Right end)**

At the internal nodes of the fin, bL(e) and bL(e) vanish upon assembly. At boundaries, where temperature is prescribed, no need to calculate bL(e) or bL(e) first.

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**1D fin When there is heat convection at boundary, E.g.**

Since b is the temperature of the fin at the boundary point, b = j Therefore,

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1D fin where , For convection on left side, where ,

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1D fin Therefore, Residuals are assembled for all elements and enforced to zero: KD = F Same form for static mechanics problem

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1D fin Direct assembly procedure or Element 1:

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**1D fin Direct assembly procedure (Cont’d) Element 2:**

Considering all contributions to a node, and enforcing to zero (Node 1) (Node 2) (Node 3)

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**1D fin Direct assembly procedure (Cont’d) In matrix form:**

(Note: same as assembly introduced before)

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**1D fin Worked example: Heat transfer in 1D fin**

Calculate temperature distribution using FEM. 4 linear elements, 5 nodes

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1D fin Element 1, 2, 3: , not required Element 4: , required

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1D fin For element 1, 2, 3 , For element 4 ,

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**1D fin Heat source (Still unknown)**

1 = 80, four unknowns – eliminate Q* Solving:

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**Composite wall Convective boundary: at x = 0 at x = H**

All equations for 1D fin still applies except Recall: Only for heat convection and vanish. Therefore, ,

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**Composite wall Worked example: Heat transfer through composite wall**

Calculate the temperature distribution across the wall using the FEM. 2 linear elements, 3 nodes

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Composite wall For element 1,

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**Composite wall For element 2, Upon assembly,**

(Unknown but required to balance equations)

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Composite wall Solving:

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**Composite wall Worked example: Heat transfer through thin film layers**

Raw material 0.2 mm h =0.01 W/cm 2 / C f = 150 0.2mm Heater k =0.1 W/cm/ 2mm Glass Iron Platinum =0.5 W/cm/ =0.4 W/cm/ Substrate Plasma 1 3 4 (1) (2) (3) 300C

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Composite wall For element 1, 1 2 3 4 (1) (2) (3) For element 2,

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Composite wall For element 3,

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Composite wall Since, 1 = 300°C, Solving:

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**2D HEAT TRANSFER PROBLEM**

Element equations For one element, Note: W = N : Galerkin approach

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**Element equations Therefore, Gauss’s divergence theorem:**

(Need to use Gauss’s divergence theorem to evaluate integral in residual.) (Product rule of differentiation) Therefore, Gauss’s divergence theorem:

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Element equations 2nd integral: Therefore,

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Element equations

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Element equations where

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Element equations Define , (Strain matrix)

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Triangular elements Note: constant strain matrix (Or Ni = Li)

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Triangular elements Note: (Area coordinates) E.g. Therefore,

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Triangular elements Similarly, Note: b(e) will be discussed later

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Rectangular elements

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Rectangular elements

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Rectangular elements Note: In practice, the integrals are usually evaluated using the Gauss integration scheme

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**Boundary conditions and vector b(e)**

Internal Boundary bB(e) needs to be evaluated at boundary Vanishing of bI(e)

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**Boundary conditions and vector b(e)**

Need not evaluate Need to be concern with bB(e)

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**Boundary conditions and vector b(e)**

on natural boundary 2 Heat flux across boundary

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**Boundary conditions and vector b(e)**

Insulated boundary: M = S = 0 Convective boundary condition:

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**Boundary conditions and vector b(e)**

Specified heat flux on boundary:

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**Boundary conditions and vector b(e)**

For other cases whereby M, S 0

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**Boundary conditions and vector b(e)**

where , For a rectangular element, (Equal sharing between nodes 1 and 2)

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**Boundary conditions and vector b(e)**

Equal sharing valid for all elements with linear shape functions Applies to triangular elements too

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**Boundary conditions and vector b(e)**

for rectangular element

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**Boundary conditions and vector b(e)**

Shared in ratio 2/6, 1/6, 1/6, 2/6

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**Boundary conditions and vector b(e)**

Similar for triangular elements

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**Point heat source or sink**

Preferably place node at source or sink

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**Point heat source or sink within the element**

Point source/sink (Delta function)

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SUMMARY

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CASE STUDY Road surface heated by heating cables under road surface

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**CASE STUDY Heat convection M=h=0.0034 S=ff h=-0.017 fQ***

Repetitive boundary no heat flow across M=0, S=0 Repetitive boundary no heat flow across M=0, S=0 Insulated M=0, S=0

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**CASE STUDY Surface temperatures: Node Temperature (C) 1 5.861 2 5.832**

5.764 4 5.697 5 5.669

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