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F inite Element Method. for readers of all backgrounds. G. R. Liu and S. S. Quek. CHAPTER 12:. FEM FOR HEAT TRANSFER PROBLEMS. CONTENTS. FIELD PROBLEMS WEIGHTED RESIDUAL APPROACH FOR FEM 1D HEAT TRANSFER PROBLEMS 2D HEAT TRANSFER PROBLEMS SUMMARY CASE STUDY. FIELD PROBLEMS.

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F inite element method

Finite Element Method

for readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 12:

FEM FOR HEAT TRANSFER PROBLEMS


Contents

CONTENTS

  • FIELD PROBLEMS

  • WEIGHTED RESIDUAL APPROACH FOR FEM

  • 1D HEAT TRANSFER PROBLEMS

  • 2D HEAT TRANSFER PROBLEMS

  • SUMMARY

  • CASE STUDY


Field problems

FIELD PROBLEMS

  • General form of system equations of 2D linear steady state field problems:

(Helmholtz equation)

  • For 1D problems:


Field problems1

FIELD PROBLEMS

  • Heat transfer in 2D fin

Note:


Field problems2

FIELD PROBLEMS

  • Heat transfer in long 2D body

Note:

Dx = kx, Dy =tky,

g = 0 and Q = q


Field problems3

FIELD PROBLEMS

  • Heat transfer in 1D fin

Note:


Field problems4

FIELD PROBLEMS

  • Heat transfer across composite wall

Note:


Field problems5

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)


Field problems6

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


Weighted residual approach for fem

WEIGHTED RESIDUAL APPROACH FOR FEM

  • Establishing FE equations based on governing equations without knowing the functional.

(Strong form)

Approximate solution:

(Weak form)

Weight function


Weighted residual approach for fem1

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.


1d heat transfer problem

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)


1d fin

1D fin

Using Galerkin approach,

where

D = kA, g = hP, and Q = hP


1d fin1

1D fin

Integration by parts of first term on right-hand side,

Using


1d fin2

1D fin

(Strain matrix)

where

(Thermal conduction)

(Thermal convection)

(External heat supplied)

(Temperature gradient at two ends of element)


1d fin3

1D fin

For linear elements,

(Recall 1D truss element)

Therefore,

for truss element

(Recall stiffness matrix of truss element)


1d fin4

1D fin

for truss element

(Recall mass matrix of truss element)


1d fin5

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.


1d fin6

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,


1d fin7

1D fin

where

,

For convection on left side,

where

,


1d fin8

1D fin

Therefore,

Residuals are assembled for all elements and enforced to zero: KD = F

Same form for static mechanics problem


1d fin9

1D fin

  • Direct assembly procedure

or

Element 1:


1d fin10

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)


1d fin11

1D fin

  • Direct assembly procedure (Cont’d)

In matrix form:

(Note: same as assembly introduced before)


1d fin12

1D fin

  • Worked example: Heat transfer in 1D fin

Calculate temperature distribution using FEM.

4 linear elements, 5 nodes


1d fin13

1D fin

Element 1, 2, 3:

not required

,

Element 4:

,

required


1d fin14

1D fin

For element 1, 2, 3

,

For element 4

,


1d fin15

1D fin

Heat source

(Still unknown)

1 = 80, four unknowns – eliminate Q*

Solving:


Composite wall

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,

,


Composite wall1

Composite wall

  • Worked example: Heat transfer through composite wall

Calculate the temperature distribution across the wall using the FEM.

2 linear elements, 3 nodes


Composite wall2

Composite wall

For element 1,


Composite wall3

Composite wall

For element 2,

Upon assembly,

(Unknown but required to balance equations)


Composite wall4

Composite wall

Solving:


Composite wall5

Heater

300C

·

1

(1)

°

2mm

k

=0.1 W/cm/

C

Glass

Substrate

·

2

Iron

(2)

°

0.2mm

°

k

=0.5 W/cm/

C

3

·

°

Platinum

(3)

0.2 mm

°

k

=0.4 W/cm/

C

4

·

°

2

f

°

= 150

C

h

=0.01 W/cm

/

C

f

Plasma

Raw material

Composite wall

  • Worked example: Heat transfer through thin film layers


Composite wall6

·

1

(1)

·

2

(2)

3

·

(3)

4

·

Composite wall

For element 1,

For element 2,


Composite wall7

Composite wall

For element 3,


Composite wall8

Composite wall

Since, 1 = 300°C,

Solving:


2d heat transfer problem

2D HEAT TRANSFER PROBLEM

Element equations

For one element,

Note: W = N : Galerkin approach


Element equations

Element equations

(Need to use Gauss’s divergence theorem to evaluate integral in residual.)

(Product rule of differentiation)

Therefore,

Gauss’s divergence theorem:


Element equations1

Element equations

2nd integral:

Therefore,


Element equations2

Element equations


Element equations3

Element equations

where


Element equations4

Element equations

Define

,

(Strain matrix)


Triangular elements

Triangular elements

Note: constant strain matrix

(Or Ni = Li)


Triangular elements1

Triangular elements

Note:

(Area coordinates)

E.g.

Therefore,


Triangular elements2

Triangular elements

Similarly,

Note: b(e) will be discussed later


Rectangular elements

Rectangular elements


Rectangular elements1

Rectangular elements


Rectangular elements2

Rectangular elements

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


Boundary conditions and vector b e

Boundary conditions and vector b(e)

Internal

Boundary

bB(e) needs to be evaluated at boundary

Vanishing of bI(e)


Boundary conditions and vector b e1

Boundary conditions and vector b(e)

Need not evaluate

Need to be concern with bB(e)


Boundary conditions and vector b e2

Boundary conditions and vector b(e)

on natural boundary 2

Heat flux across boundary


Boundary conditions and vector b e3

Boundary conditions and vector b(e)

Insulated boundary:

M = S = 0 

Convective boundary condition:


Boundary conditions and vector b e4

Boundary conditions and vector b(e)

Specified heat flux on boundary:


Boundary conditions and vector b e5

Boundary conditions and vector b(e)

For other cases whereby M, S 0


Boundary conditions and vector b e6

Boundary conditions and vector b(e)

where

,

For a rectangular element,

(Equal sharing between nodes 1 and 2)


Boundary conditions and vector b e7

Boundary conditions and vector b(e)

Equal sharing valid for all elements with linear shape functions

Applies to triangular elements too


Boundary conditions and vector b e8

Boundary conditions and vector b(e)

for rectangular element


Boundary conditions and vector b e9

Boundary conditions and vector b(e)

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


Boundary conditions and vector b e10

Boundary conditions and vector b(e)

Similar for triangular elements


Point heat source or sink

Point heat source or sink

Preferably place node at source or sink


Point heat source or sink within the element

Point heat source or sink within the element

Point source/sink

(Delta function)


Summary

SUMMARY


Case study

CASE STUDY

Road surface heated by heating cables under road surface


Case study1

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


Case study2

Node

Temperature (C)

1

5.861

2

5.832

3

5.764

4

5.697

5

5.669

CASE STUDY

Surface temperatures:


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