1 finite element method fem for heat transfer problems for readers of all backgrounds g. r. liu and...

1

FFinite Element Methodinite Element Method

FEM FOR HEAT TRANSFER PROBLEMS

for readers of all backgroundsfor readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 12:

2Finite Element Method by G. R. Liu and S. S. Quek

CONTENTSCONTENTS

FIELD PROBLEMSWEIGHTED RESIDUAL APPROACH

FOR FEM1D HEAT TRANSFER PROBLEMS2D HEAT TRANSFER PROBLEMSSUMMARYCASE STUDY

3Finite Element Method by G. R. Liu and S. S. Quek

FIELD PROBLEMSFIELD PROBLEMS

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

2 2

2 20x yD D g Q

x y

(Helmholtz equation)

For 1D problems:

2

20

dD g Q

dx

4Finite Element Method by G. R. Liu and S. S. Quek

FIELD PROBLEMSFIELD PROBLEMS

Heat transfer in 2D fin

x z

y

t

f

Heat convection on the surface

f

f

f

f

t

2 2

2 2

Heat supplyHeat convectctionHeat conduction

22( ) ( )f

x y

hhD D q

x y t t

22, fhh

g Q qt t

Note:

5Finite Element Method by G. R. Liu and S. S. Quek

FIELD PROBLEMSFIELD PROBLEMS

Heat transfer in long 2D body

x

z y

Representative plane

t=1

2 2

2 2

Heat supply

Heat conduction

0x yk k qx y

Note:

Dx = kx, Dy =tky,

g = 0 and Q = q

6Finite Element Method by G. R. Liu and S. S. Quek

FIELD PROBLEMSFIELD PROBLEMS

Heat transfer in 1D fin

0

x

P

A

2

2

Heat supplyHeat convectoinHeat conduction

0f

dkA hP hP q

dx

Note:, , fD kA g hP Q q hP

7Finite Element Method by G. R. Liu and S. S. Quek

FIELD PROBLEMSFIELD PROBLEMS

Heat transfer across composite wall

x

y Heat convection Heat convection

2

2

Heat supplyHeat conduction

0d

kA qdx

Note:

, 0, D kA g Q q

8Finite Element Method by G. R. Liu and S. S. Quek

FIELD PROBLEMSFIELD PROBLEMS

Torsional deformation of bar2 2

2 2

1 12 0

G x G y

Note:

Dx=1/G, Dy=1/G, g=0, Q=2

Ideal irrotational fluid flow 2 2

2 20

x y

2 2

2 20

x y

Note:

Dx = Dy = 1, g = Q = 0

( - stress function)

( - streamline function and - potential function)

9Finite Element Method by G. R. Liu and S. S. Quek

FIELD PROBLEMSFIELD PROBLEMS

Accoustic problems

2 2 2

2 2 20

P P wP

x y c

Note:

2

2

wg

c , Dx = Dy = 1, Q = 0

P - the pressure above the ambient pressure ;

w - wave frequency ;

c - wave velocity in the medium

10Finite Element Method by G. R. Liu and S. S. Quek

WEIGHTED RESIDUAL WEIGHTED RESIDUAL APPROACH FOR FEMAPPROACH FOR FEM

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

2 2

2 20x yD D g Q

x y

( ( , )) 0f x y

( ( , ))d d 0AWf x y x y

Approximate solution:

Weight function

(Strong form)

(Weak form)

11Finite Element Method by G. R. Liu and S. S. Quek

WEIGHTED RESIDUAL WEIGHTED RESIDUAL APPROACH FOR FEMAPPROACH FOR FEM

1 2

3

Discretize into smaller elements to ensure better approximation

In each element,

Using N as the weight functions

( )( , ) ( , )h ex y x y N Φ

where 1 1 dnN N N N

Galerkin method

( ) ( ( , ))d de

e T

Af x y x yR N Residuals are then assembled

for all elements and enforced to zero.

12Finite Element Method by G. R. Liu and S. S. Quek

1D HEAT TRANSFER 1D HEAT TRANSFER PROBLEMPROBLEM

1D fin1D fin

2

2

Heat supplyHeat convectoinHeat conduction

0f

dkA hP hP q

dx

00 (Specified boundary condition)

fbhAdx

dkA

(Convective heat loss at free end)

k : thermal conductivityh : convection coefficientA : cross-sectional area of the finP : perimeter of the fin : temperature, and

f : ambient temperature in the fluid

0 x

13Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

1 2

(i)

i j

(1)

x

xi xj

(i) 0

0 x

(n)

n+1

2

2

2

2

dj

i

j j

i i

hxe T h

x

hx xT T h

x x

dD g Q x

dx

dD Q dx g dx

dx

R N

N N

Using Galerkin approach,

where

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

14Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D finIntegration by parts of first term on right-hand side,

d d dj

j j j

i i i

i

xh T hx x xe T T T h

x x xx

d d dD D x Q x g x

dx dx dx

N

R N N N

( )( ) ( )h ex x N ΦUsing

( )( )

( ) ( )

d d d( d )

d d d

( d ) ( d )

j

j

i

i

eeD

j j

i i

e egQ

xh Txe eT

xx

x x eT T

x x

D D xx x x

Q x g x

kb

kf

N NR N Φ

N N N Φ

15Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

( ) ( ) ( ) ( )[ ]e ee e e eD g Q R b k k Φ f

where

( ) d dd d

d d

j j

i i

Tx xe T

D x xD x D x

x x

N Nk B B

( ) dj

i

xe Tg x

g xk N N

( ) dj

i

xe TQ x

Q xf N

( ) d

d

j

i

xhe T

x

Dx

b N

(Thermal conduction)

(Thermal convection)

(External heat supplied)

(Temperature gradient at two ends of element)

d

dx

NB (Strain matrix)

16Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

For linear elements,

( ) j ii j

x x x xx N N

l l

N (Recall 1D truss element)

d d 1 1

d dj i

x x x x

x x l l l l

NB

Therefore,

( )

11 11 1

d1 1 1

j

i

xeD x

kAl D xl l l

l

k(Recall stiffness matrix of truss element)

AE

lfor truss element

17Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

( ) 2 1d

1 26

j

i

j

x je ig x

i

x xx x x x hPllg x

l lx x

l

k(Recall mass matrix of truss element)

A l for truss element

( )

Heat supply Heat convection

1 1d ( )

1 12 2

j

i

j

xeQ fx

i

x xQl llQ x q hP

x x

l

f

18Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

( ) ( )

( )

d0ddd

d dd d

d0d

ji

i

i

j

j e eL R

xx xe T

x x

xx x

x x

kAx kA

xDkAx

xkAx

b b

b N

( ) ( ) ( )e e eL R b b bor

(Left end) (Right end)

i1

(n)

i i+1

(n1)

i1 i (n1)

d

dix x

Dx

1

d

dix x

Dx

i i+1 (n)

1

d

dix x

Dx

d

dix x

Dx

0

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.

19Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

When there is heat convection at boundary,

fbhAdx

dkA

0 0 0e

Rj fb f hA hAhA

bE.g.

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

b = j

Therefore,

( ) ( )

( )00 0

0e e

M s

ieR

j fhAhA

k f

b

20Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin( ) ( ) ( ) ( )e e e eR M s b k Φ f

where ( ) 0 0

0e

M hA

k , ( )0

es

fhA

f

For convection on left side,

( ) ( ) ( ) ( )e e e eL M s b k Φ f

where ( ) 0

0 0e

M

hA

k , ( )

0fe

s

hA

f

21Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

Therefore,

( ) ( )

( ) ( ) ( ) ( ) ( )[ ]e e

e ee e e e eD g M Q S

k f

R k k k Φ f f

( ) ( )e ee e R k Φ f

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

Same form for static mechanics problem

22Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

Direct assembly procedure

( ) ( )e ee e R k Φ f

or

1 11 12 1 1

2 21 22 2 2

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

e e e e e

e e e e e

R k k f

R k k f

(1) (2)

1 2 3

Element 1:

1 11 1 12 1

(1) (1) (1) (1)2R k k f

2 21 1 22 2

(1) (1) (1) (1)2R k k f

23Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

Direct assembly procedure (Cont’d)

Element 2:

1 11 12 1

(2) (2) (2) (2)2 3R k k f

2 21 22 2

(2) (2) (2) (2)2 3R k k f

1 11 1 12 1

(1) (1) (1) (1)20 : 0R k k f

Considering all contributions to a node, and enforcing to zero

(Node 1)

2 21 1 22 11 12 2 1

(1) (1) (1) (1) (2) (2) (1) (2)1 2 30 : ( ) ( ) 0R R k k k k f f (Node 2)

2 21 2 22

(2) (2) (2) (2)3 20 : 0R k k f (Node 3)

24Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

Direct assembly procedure (Cont’d)

In matrix form:

11 12 1 1

21 22 11 12 2 1

21 22 2

(1) (1) (1)

(1) (1) (2) (2) (1) (2)2

(2) (2) (2)3

0

0

k k f

k k k k f f

k k f

(Note: same as assembly introduced before)

25Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

Worked example: Heat transfer in 1D fin

1 2 3 4 5

8 cm

(1 ) (2 ) (3 ) (4 )

0 .4 cm

1 cm Ccm

Wk

03

Ccm

Wh

021.0

Cf

020

8 0 C

Calculate temperature distribution using FEM.

4 linear elements, 5 nodes

26Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

1 2 3 4 5

8 cm

(1 ) (2 ) (3 ) (4 )

0 .4 cm

1 cm Ccm

Wk

03

Ccm

Wh

021.0

Cf

020

8 0 C

Element 1, 2, 3: ( )eMk ,

( )eSf not required

Element 4:( )eMk ,

( )eSf required

0

3 0.40.6

2

kA W

l C

0

0.1 2.8 20.093

6 6

hPl W

C

00.1 0.4 0.04

WhA

C

0.1 2.8 20 25.6

2 2fhPl

W

0.1 0.4 20 0.8fhA W

27Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

For element 1, 2, 3

1 1 2 1

1 1 1 26e kA hPl

l

k

1

12e fhPL

f

1,2,3 0.786 0.507

0.507 0.786

k

1,2,3 5.6

5.6

f

For element 4

1 1 2 1 0 0

1 1 1 2 06e kA hPL

hAL

k

01

12e f

f

hPL

hA

f

4 0.786 0.507

0.507 0.826

k

4 5.6

6.4

f

( ) ( )e e eD g k k k

( ) ( ) ( )e e e eD g M k k k k

( )e eQf f,

, ( ) ( )e e eQ S f f f

28Finite Element Method by G. R. Liu and S. S. Quek

1D fin1D fin

*1

2

3

4

5

0.786 0.507 0 0 0 5.6

0.507 1.572 0.507 0 0 11.2 0

0 0.507 1.572 0.507 0 11.2 0

0 0 0.507 1.572 0.507 11.2 0

0 0 0 0.507 0.826 6.4 0

Q

Heat source(Still unknown)

*1

2

3

4

5

800.786 0.507 0 0 0 5.6

0.507 1.572 0.507 0 0 11.2 80 0.507

0 0.507 1.572 0.507 0 11.2

0 0 0.507 1.572 0.507 11.2

0 0 0 0.507 0.826 6.4

Q

1 = 80, four unknowns – eliminate Q*

Solving: {80.0 42.0 28.2 23.3 22.1}T Φ

29Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

(1) (2) (3)

x

y

2

2

Heat supplyHeat conduction

0d

kA qdx

fbhAdx

dkA

fbhAdx

dkA

Convective boundary:

at x = 0

at x = H

All equations for 1D fin still applies except ( )e

Gk and ( )eQf

Therefore, 1 1

1 1e e

M

kA

l

k k , ( ) ( )e eSf f

vanish. Recall: Only for heat convection

30Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

Worked example: Heat transfer through composite wall

5 c m 2

Cf05

Ccm

Wk

006.0

C020

Ccm

Wk

02.0

Ccm

Wh

021.0

( 1 ) ( 2 )

1 2 3

Calculate the temperature distribution across the wall using the FEM.

2 linear elements, 3 nodes

31Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

5 c m 2

Cf05

Ccm

Wk

006.0

C020

Ccm

Wk

02.0

Ccm

Wh

021.0

( 1 ) ( 2 )

1 2 3

For element 1,

C

W

L

kA0

1.02

12.0

C

WhA

01.011.0

WhA f 5.0511.0

1 0.2 0.1

0.1 0.1

k

1 0.5

0S

f

1 1 0 0

1 1 0e kA

hAL

k

( ) ( )

0fe e

s

hA

f f

32Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

5 c m 2

Cf05

Ccm

Wk

006.0

C020

Ccm

Wk

02.0

Ccm

Wh

021.0

( 1 ) ( 2 )

1 2 3

For element 2,

0

0.06 10.012

5

kA W

L C

1 1

1 1e kA

L

k 2 0.012 0.012

0.012 0.012

k

Upon assembly,

1

2

3

0.20 0.10 0 0.5 0

0.10 0.112 0.012 0 0

0 0.012 0.012 ( 20) 0 Q

(Unknown but required to balance equations)

33Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

1

2

0.20 0.10 0.5

0.10 0.112 0.24( 20 0.012)

Solving: 2.5806 0.1613 20T Φ

34Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

Worked example: Heat transfer through thin film layers

Raw material

0.2 mm

h=0.01 W/cm 2/ C f = 150C

0.2mm

Heater

k=0.1 W/cm/ C 2mm Glass

Iron Platinum

k=0.5 W/cm/C

k=0.4 W/cm/C

Substrate

Plasma

1

2 3 4

(1)

(2) (3)

300C

35Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

1

2 3 4

(1)

(2) (3)

For element 1,

0

0.1 10.5

0.2

kA W

L C

1 0.5 0.5

0.5 0.5

k 1 1

1 1e kA

L

k

For element 2,

0

0.5 125

0.02

kA W

L C

1 1

1 1e kA

L

k 2 25 25

25 25

k

36Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

For element 3,

0

0.4 120

0.02

kA W

L C

00.01 1 0.01

WhA

C

0.01 1 150 1.5fhA W

3 20 20

20 20.01

k

3 0

1.5S

f

1 1 0 0

1 1 0e kA

hAL

k

( ) ( )0

e es

fhA

f f

37Finite Element Method by G. R. Liu and S. S. Quek

Composite wallComposite wall

*1

2

3

4

0.5 0.5 0 0

0.5 25.5 25 0 0

0 25 45 20 0

0 0 20 20.01 1.5

Q

2

3

4

25.5 25 0 150

25 45 20 0

0 20 20.01 1.5

T

T

T

Since, 1 = 300°C,

Solving: 300.0 297.1 297.0 296.9T Φ

* 20.5 300 0.5 297.1 1.45 W/cmQ

38Finite Element Method by G. R. Liu and S. S. Quek

2D HEAT TRANSFER 2D HEAT TRANSFER PROBLEMPROBLEM

Element equationsElement equations2 2

2 20x yD D g Q

x y

1 2

3

For one element,

2 2( )

2 2( )d

e

h he T h

x yAD D g Q A

x y

R N

Note: W = N : Galerkin approach

39Finite Element Method by G. R. Liu and S. S. Quek

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

2 2( )

2 2( )d

e

h he T h

x yAD D g Q A

x y

R N

2

2

TT T

x x x x x

N

N N

2

2 2d ( )d d

e e e

TT T

x x xA A AD A D A D A

x x x x x

N

N NTherefore,

( )d cos de e

T T

AA

x x x

N NGauss’s divergence theorem:

2

2d cos d d

e

TT T

x x xA AD A D D A

x x xx

N

N N

(Product rule of differentiation)

40Finite Element Method by G. R. Liu and S. S. Quek

Element equationsElement equations

2nd integral:2

2d sin d d

e

TT T

y y yA AD A D D A

y y y y

N

N N

( ) cos sin de

h he T

x yD Dx y

R N

de

T h T h

x yAD D A

x x y y

N N

d de e

T h T

A Ag A Q A N N

Therefore,

41Finite Element Method by G. R. Liu and S. S. Quek

Element equationsElement equations

( )( ) ( )h ex x N Φ

( )

( ) cos sin de

e

h he T

x yD Dx y

b

R N

( )

( )de

eD

T Te

x yAD D A

x x y y

k

N N N NΦ

( ) ( )

( )( d ) de e

e eg Q

T e T

A Ag A Q A

k f

N N Φ N

42Finite Element Method by G. R. Liu and S. S. Quek

Element equationsElement equations

( ) ( ) ( ) ( ) ( ) ( )[ ]e e e e e eD g Q R b k k Φ f

where ( ) cos sin de

h he T

x yD Dx y

b N

( ) de

T Te

D x yAD D A

x x y y

N N N Nk

( ) de

e Tg A

g Ak N N

( ) de

e TQ A

Q Af N

43Finite Element Method by G. R. Liu and S. S. Quek

Element equationsElement equations

( ) de

T Te

D x yAD D A

x x y y

N N N Nk

0

0x

y

D

D

DDefine ( ) ( )e ex x

y y

N

Φ BΦN,

1 2

1 2

d

d

n

n

NN N

x x x xNN N

y y y y

N

BN

(Strain matrix)

T TT

x yD Dx x y y

N N N N

B DB ( ) d

e

e TD A

Ak B DB

44Finite Element Method by G. R. Liu and S. S. Quek

Triangular elementsTriangular elements

1( ) ( )

1 2 3 2

3

[ ]e e N N N

NΦ

x,

y,

1 (x1, y1) 1

2 (x2, y2) 2

3 (x3, y3) 3

Ae

i i i iN a b x c y 1

( )2

1( )

2

1( )

2

i j k k je

i j ke

i k je

a x y x yA

b y yA

c x xA

( ) d de e

e T T TD eA A

A A A k B DB B DB B DB

2 2

( ) 2 2

2 24 4

i i j i k i i j i kye x

D i j j j k i j j j k

i k j k k i k j k k

b b b b b c c c c cDD

b b b b b c c c c cA A

b b b b b c c c c c

k

Note: constant strain matrix

(Or Ni = Li)

45Finite Element Method by G. R. Liu and S. S. Quek

Triangular elementsTriangular elements

1( )

2 1 2 3

3

21 1 2 1 3

21 2 2 2 3

21 3 2 3 3

d [ ]d

d

e e

e

e Tg A A

A

N

g A g N N N N A

N

N N N N N

g N N N N N A

N N N N N

k N N

Apnm

pnmALLL pn

A

m 2)!2(

!!!d321

Note:

(Area coordinates)

E.g. 1 1 01 2 1 2 3

1!1!0! 1d d 2 2

(1 1 0 2)! 4 3 2 1 12e eA A

AN N A L L L A A A

( )

2 1 1

1 2 112

1 1 2

eg

gA

kTherefore,

46Finite Element Method by G. R. Liu and S. S. Quek

Triangular elementsTriangular elements

1

( )2

3

1

d 13

1e e e

i

e TQ jA A A

k

N LQA

Q A Q N A Q L A

N L

f N d d

Similarly,

Note: b(e) will be discussed later

47Finite Element Method by G. R. Liu and S. S. Quek

Rectangular elementsRectangular elements

1

2( ) ( )1 2 3 4

3

4

[ ]e e N N N N

NΦ

x

y

1 (x1, y1)

2 (x2, y2)

3 (x3, y3)

2a

4 (x4, y4)

2b

1 ( 1, 1)

2 (1, 1)

3 (1, +1)

2a

4 ( 1, +1)

2b

)1)(1(

)1)(1(

)1)(1(

)1)(1(

41

4

41

3

41

2

41

1

N

N

N

N

, byax

1 1 1 1

1 1 1 1

1 1 1 1 1 1 1 1

0 0 0 0

0 0 0 0a a a a

b b b b

a a a ab b b b

B

48Finite Element Method by G. R. Liu and S. S. Quek

Rectangular elementsRectangular elements1 1( )

1 1 d d d

2 2 1 1 2 1 1 2

2 2 1 1 1 2 2 1

1 1 2 2 1 2 2 16 6

1 1 2 2 1 1 1 2

e

e T TD A

yx

A ab

D aD b

a b

k B DB B DB

1 1( )

1 1

21 1 2 1 3 1 4

21 2 2 2 3 2 3

21 3 2 3 3 3 4

21 4 2 4 3 4 4

d d d

d d

e

e

e T Tg A

A

g A abg

N N N N N N N

N N N N N N Nabg

N N N N N N N

N N N N N N N

k N N N N

( )

4 2 1 2

2 4 2 1

1 2 4 236

2 1 2 4

eg

gA

k

49Finite Element Method by G. R. Liu and S. S. Quek

Rectangular elementsRectangular elements

1

2( )

3

4

1

1d

14

1

e e

e TQ A A

N

N QAQ A Q A

N

N

f N d

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

50Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

( ) ( ) ( )e e eI B b b b

Internal Boundary

i

j

k

p

m 1 2

3 n

q

r

(2)Ib

(2)Ib

(3)Bb

(3)Ib

Vanishing of bI(e)bB

(e) needs to be evaluated at boundary

51Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

1

2

Essential boundary is known

Natural boundary: the derivatives of is known

n

x

n

x

Need not evaluate ( )e

Bb

Need to be concern with bB

(e)

52Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

( ) cos sin de

h he T

x yD Dx y

b N

cos sinh h

x y bD D M Sx y

on natural boundary 2

cos sinh h h

x y bD D k M Sx y n

Heat flux across boundary

1

2

Essential boundary is known

Natural boundary: the derivatives of is known

n

x

n

x

53Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

Insulated boundary:

M = S = 0 ( ) 0eB b

Convective boundary condition:

qc

n

n

k

h b f

q kn

q h

fb hhn

k

M

S

kq kn

h b fq h

b fM

S

k h hn

, fM h S h

54Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

Specified heat flux on boundary:

n

n

k

s

q kn

q

sk qn

qk

qs M=0 S= qs

0 b sM S

k qn

0, sM S q

Positive if heat flows into the boundary

Negative if heat flows out off the boundary

0 insulated

S

55Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

For other cases whereby M, S 0

2

2

( ) cos sin d

( + )d

h he T

B x y

Tb

D Dx y

M S

b N

N

eb NΦ

2

2 2

( ) ( )

( ) ( )

( )

( + )d

( d ) d

e eM B

e T eB

T e T

M S

M S

b N NΦ

N N Φ N

k f

56Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

( ) ( ) ( ) ( )e e e eB M S b k Φ f

where2

( ) de TM M

k N N ,

2

( ) de TS S

f N

x

y

1 (x1, y1)

2 (x2, y2)

3 (x3, y3)

2a

4 (x4, y4)

2b

For a rectangular element,

1 2

1

1 2( )

13

4

de TS

N

NS d S a

N

N

f N

1

1

11

11d

02 0

00

eS

SaSa

f (Equal sharing between nodes 1 and 2)

57Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

Equal sharing valid for all elements with linear shape functions

,2 3

0

1

1

0

eS Sb

f ,3 4

0

0

1

1

eS Sa

f ,1 4

1

0

0

1

eS Sb

f

x,

y,

1 (x1, y1) 1

2 (x2, y2) 2

3 (x3, y3) 3

Ae

Applies to triangular elements too

12,1 2

1

12

0

eS

SL

f 23,2 3

0

12

1

eS

SL

f 13,1 3

1

02

1

eS

SL

f

58Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

( )eMk for rectangular element

2

21 1 2 1 3 1 4

2( ) 1 2 2 2 3 2 4

21 3 2 3 3 3 4

21 4 2 4 3 4 4

deM

N N N N N N N

N N N N N N NM

N N N N N N N

N N N N N N N

k

21 1 2

21( ) 2 1 2

,1 2 1

0 0

0 0d

0 0 0 0

0 0 0 0

eM

N N N

N N NaM

k

59Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

21 12

11 1

1 2d d

4 3N

1 1

1 21 1

1 1 2d d

4 6N N

21 12

21 1

1 2d d

4 3N

( ),1 2

2 16 6

1 26 6

2 1 0 0

1 2 0 02

0 0 0 06

0 0 0 0

0 0

0 0(2 )

0 0 0 0

0 0 0 0

eM

aM

aM

k

Shared in ratio 2/6, 1/6, 1/6, 2/6( ),2 3

0 0 0 0

0 2 1 02

0 1 2 06

0 0 0 0

eM

M b

k

( ),3 4

0 0 0 0

0 0 0 02

0 0 2 16

0 0 1 2

eM

M a

k ( ),1 4

2 0 0 1

0 0 0 02

0 0 0 06

1 0 0 2

eM

M b

k

60Finite Element Method by G. R. Liu and S. S. Quek

Boundary conditions and vector Boundary conditions and vector bb((e)e)

Similar for triangular elements

( ),

2 1 0

1 2 06

0 0 0

ijeM i j

ML

k ( ),

0 0 0

0 2 16

0 1 2

jkeM j k

ML

k

( ),

2 0 1

0 0 06

1 0 2

e ikM i k

ML

k

61Finite Element Method by G. R. Liu and S. S. Quek

Point heat source or sinkPoint heat source or sink

Preferably place node at source or sink

Q* i j

k

1

2

*

F

f

f

j Q

62Finite Element Method by G. R. Liu and S. S. Quek

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

( ) de

e TQ A

Q Af N

(X0, Y0)

Q*

i

j

k

x

y

00 YyXxQQ (Delta function)

( )0 0 d d

e

i

eQ jA

k

N

Q N x X y Y x y

N

f

Point source/sink

0 0

0 0

0 0

,

,

,

ie

Q j

k

N X Y

Q N X Y

N X Y

f

63Finite Element Method by G. R. Liu and S. S. Quek

SUMMARYSUMMARY

2 2

2 2

2 2

0fx y

h hk t k

y

t

xD D g Qx y

02

2

2

2

QGy

Dx

D yx

*

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )e e

e e e e e e e e eD g M Q SQ

k f

R b k k k Φ f f f

( ) c o s s i n d

h fh

bM S

e Tx yD D

x y

b N

y

n

x

S o u r c e a n d s i n k

O n l y f o r e l e m e n t s o n t h e d e r i v a t i v e b o u n d a r y

64Finite Element Method by G. R. Liu and S. S. Quek

CASE STUDYCASE STUDY

6 cm

2 cm

4 cm

Heating cables

Representative section to be analyzed

Insulation

f= 0C

Road surface heated by heating cables under road surface

65Finite Element Method by G. R. Liu and S. S. Quek

CASE STUDYCASE STUDYHeat convectionM=h=0.0034S=f h=-0.017

Repetitive boundary no heat flow acrossM=0, S=0

Repetitive boundary no heat flow acrossM=0, S=0

InsulatedM=0, S=0

fQ*

66Finite Element Method by G. R. Liu and S. S. Quek

CASE STUDYCASE STUDY

Node Temperature (C)

1 5.861

2 5.832

3 5.764

4 5.697

5 5.669

Surface temperatures: