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F inite Element Method

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Finite Element Method

for readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 12:

FEM FOR HEAT TRANSFER PROBLEMS

- FIELD PROBLEMS
- WEIGHTED RESIDUAL APPROACH FOR FEM
- 1D HEAT TRANSFER PROBLEMS
- 2D HEAT TRANSFER PROBLEMS
- SUMMARY
- CASE STUDY

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

(Helmholtz equation)

- For 1D problems:

- Heat transfer in 2D fin

Note:

- Heat transfer in long 2D body

Note:

Dx = kx, Dy =tky,

g = 0 and Q = q

- Heat transfer in 1D fin

Note:

- Heat transfer across composite wall

Note:

- 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)

- Accoustic problems

P - the pressure above the ambient pressure ;

w - wave frequency ;

c - wave velocity in the medium

Note:

, Dx = Dy = 1, Q = 0

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

(Strong form)

Approximate solution:

(Weak form)

Weight function

- 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 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)

Using Galerkin approach,

where

D = kA, g = hP, and Q = hP

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

Using

(Strain matrix)

where

(Thermal conduction)

(Thermal convection)

(External heat supplied)

(Temperature gradient at two ends of element)

For linear elements,

(Recall 1D truss element)

Therefore,

for truss element

(Recall stiffness matrix of truss element)

for truss element

(Recall mass matrix of truss element)

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.

When there is heat convection at boundary,

E.g.

Since b is the temperature of the fin at the boundary point,

b = j

Therefore,

where

,

For convection on left side,

where

,

Therefore,

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

Same form for static mechanics problem

- Direct assembly procedure

or

Element 1:

- Direct assembly procedure (Cont’d)

Element 2:

Considering all contributions to a node, and enforcing to zero

(Node 1)

(Node 2)

(Node 3)

- Direct assembly procedure (Cont’d)

In matrix form:

(Note: same as assembly introduced before)

- Worked example: Heat transfer in 1D fin

Calculate temperature distribution using FEM.

4 linear elements, 5 nodes

Element 1, 2, 3:

not required

,

Element 4:

,

required

For element 1, 2, 3

,

For element 4

,

Heat source

(Still unknown)

1 = 80, four unknowns – eliminate Q*

Solving:

Convective boundary:

at x = 0

at x = H

All equations for 1D fin still applies except

Recall: Only for heat convection

and

vanish.

Therefore,

,

- Worked example: Heat transfer through composite wall

Calculate the temperature distribution across the wall using the FEM.

2 linear elements, 3 nodes

For element 1,

For element 2,

Upon assembly,

(Unknown but required to balance equations)

Solving:

Heater

300C

·

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

- Worked example: Heat transfer through thin film layers

·

1

(1)

·

2

(2)

3

·

(3)

4

·

For element 1,

For element 2,

For element 3,

Since, 1 = 300°C,

Solving:

Element equations

For one element,

Note: W = N : Galerkin approach

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

(Product rule of differentiation)

Therefore,

Gauss’s divergence theorem:

2nd integral:

Therefore,

where

Define

,

(Strain matrix)

Note: constant strain matrix

(Or Ni = Li)

Note:

(Area coordinates)

E.g.

Therefore,

Similarly,

Note: b(e) will be discussed later

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

Internal

Boundary

bB(e) needs to be evaluated at boundary

Vanishing of bI(e)

Need not evaluate

Need to be concern with bB(e)

on natural boundary 2

Heat flux across boundary

Insulated boundary:

M = S = 0

Convective boundary condition:

Specified heat flux on boundary:

For other cases whereby M, S 0

where

,

For a rectangular element,

(Equal sharing between nodes 1 and 2)

Equal sharing valid for all elements with linear shape functions

Applies to triangular elements too

for rectangular element

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

Similar for triangular elements

Preferably place node at source or sink

Point source/sink

(Delta function)

Road surface heated by heating cables under road surface

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

Node

Temperature (C)

1

5.861

2

5.832

3

5.764

4

5.697

5

5.669

Surface temperatures: