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


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



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:



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



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



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



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)



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


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)


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.


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


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


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


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)


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


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


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.


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


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


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


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


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)


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)


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


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


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


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


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



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



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



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)



1

2

0.20 0.10 0.5

0.10 0.112 0.24( 20 0.012)

Solving: 2.5806 0.1613 20T Φ



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



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



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



*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


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


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)


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,



( )( ) ( )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



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



( ) 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


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)



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,



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


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


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


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


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



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)



( ) 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



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



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



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



( ) ( ) ( ) ( )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)



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



( )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



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



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


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


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


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


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


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*


CASE STUDYCASE STUDY

Node Temperature (C)

1 5.861

2 5.832

3 5.764

4 5.697

5 5.669

Surface temperatures:

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

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