8 lect8 2013 stud id 46719 Nieznany (2)

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Steady State Heat Conduction — Example

FE Analysis in 2D

Małgorzata Stojek

Cracow University of Technology

April 2013

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

1 / 30

Example — Steady State Heat Transfer

T=500 [K]

q=100

m

2

W

[ ]

m

[ ]

W

3

f=120

1

1

1

κ

=

1

h

W

m deg

i

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

2 / 30

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FE Discretization

1

2

3

1

2

4

3

5

no elem.

node 1

node 2

node 3

1

1

2

4

2

4

3

1

3

3

4

5

T

3

T

2

T

1

T(x,y)

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

3 / 30

Element Shape Functions

Fact

1

ζ

1

N =

1
2

4

1

4

3

1

2

3

1

0

N

1

+

N

2

+

N

3

=

1

N

1

N

2

N

3

=

1

2A

det

x

2

x

3

y

2

y

3

y

2

y

3

x

3

x

2

det

x

3

x

1

y

3

y

1

y

3

y

1

x

1

x

3

det

x

1

x

2

y

1

y

2

y

1

y

2

x

2

x

1

1

x
y

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

4 / 30

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Shape Functions - Element no 1

First Node

(0,0)

2

3

x

1

y

(0,1)

(1,0)

1

N =y

1

2

3

y

x

B

1

= ∇

N

1

=

N

1

x

N

1

y

!

=

0

1

!

B

1

= ∇

N

1

=

N

1

x

N

1

y

!

=

1

2A

y

2

y

3

x

3

x

2

!

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

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Element Shape Functions - Element no 1

Third Node

(0,0)

2

3

x

1

y

(0,1)

(1,0)

3

N =x

1

2

3

y

x

B

3

= ∇

N

3

=

N

3

x

N 3

y

!

=

1

0

!

B

3

= ∇

N

3

=

N

3

x

N 3

y

!

=

1

2A

y

1

y

2

x

2

x

1

!

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

6 / 30

background image

Element Shape Functions - Element no 1

Second Node

(0,0)

2

3

x

1

y

(0,1)

(1,0)

2

N =1-x-y

y

1

2

3

x

N

2

=

1

N

1

N

3

N

2

=

a

0

+

a

1

x

+

a

2

y

(

0, 0

)

1

=

a

0

(

0, 1

)

0

=

1

+

a

2

a

2

= −

1

(

1, 0

)

0

=

1

+

a

3

a

3

= −

1

B

2

= ∇

N

2

=

N

2

x

N

2

y

!

=

1

1

!

B

2

= ∇

N

2

=

N

2

x

N

2

y

!

=

1

2A

y

3

y

1

x

1

x

3

!

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

7 / 30

Element Shape Functions - Element no 1

Second Node - Third Method

edge

13

:

y

=

ax

+

b

0

= −

y

ax

+

b

edge

13

plane

z

(

x, y

) = −

y

ax

+

b for

z

(

x

1

, y

1

) =

0

z

(

x

3

, y

3

) =

0

N

2

z

(

x, y

) = −

y

ax

+

b (compact support &

z

(

x

2

, y

2

) =

1

)

N

2

(

x, y

) = −

y

ax

+

b

y

2

ax

2

+

b

(0,0)

2

3

x

1

y

(1,0)

(0,1)

y=-x+1

N

2

=

y

x

+

1

−(

0

) − (

0

) +

1

=

1

x

y

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

8 / 30

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Element Stiffness Matrix I

(0,0)

2

3

x

1

y

(0,1)

(1,0)

N

1

N

2

N

3

B

1

B

2

B

3

y

1

x

y

x

0

1

!

1

1

!

1

0

!

K

e

ij

=

κ

R

e

(

B

i

)

T

B

j

d Ω

K

e

ij

=

κ

(

B

i

)

T

B

j

Z

e

d Ω

=

κ

2

(

B

i

)

T

B

j

K

e

11

=

κ
2

0 1

0

1

!

=

κ
2

K

e

12

=

κ
2

0 1

1

1

!

= −

κ
2

. . .

K

e

=

κ
2

1

1

0

1

2

1

0

1

1

NOTE:

X

T

X

=

X

·

X

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

9 / 30

Element Stiffness Matrix II

K

e

=

κ

R

e

(

B

e

)

T

B

e

d Ω

B

e

=

B

1

B

2

B

3

=

0

1 1

1

1 0

K

e

=

κ
2

0

1

1

1

1

0

0

1 1

1

1 0

K

e

=

κ
2

1

1

0

1

2

1

0

1

1

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

10 / 30

background image

Element Stiffness Matrix III

B

e

=

B

1

B

2

B

3

=

1

2A

y

2

y

3

y

3

y

1

y

1

y

2

x

3

x

2

x

1

x

3

x

2

x

1

no elem.

node 1

node 2

node 3

1

1

2

4

2

4

3

1

3

3

4

5

1

2

3

1

2

4

3

5

K

e

=

κ

Z

e

(

B

e

)

T

B

e

d Ω

=

κ

A

(

B

e

)

T

B

e

K

e1

=

K

e3

B

e2

= −

B

e1

=⇒

K

e2

=

K

e1

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

11 / 30

Element Load Vector

General Remarks

F

e

=

F

e

f

+

F

e

BC

1

N

1

2

3

2

N

1

2

3

3

N

1

2

3

Fact

Z

e

N

e

i

d Ω

=

1
3

·

h

·

Z

e

d Ω

=

1
3

·

h

·

A

e

=

1
3

A

e

( tetrahedron’s volume)

Fact

one edge L

12

:

I

nod 2

nod 1

N

e

i

ds

=

1
2

·

h

·

L

12

=

1
2

L

12

i

=

1, 2

0

i

=

3

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

12 / 30

background image

Element Load Vector

Internal Heat Supply (Constant Rate)

f

=

120

W

m

3

,

e

i

A

e

i

=

1
2

m

2

F

e

f

=

Z

e

(

N

e

)

T

f

d Ω

=

f

Z

e

N

1

d Ω

Z

e

N

2

d Ω

Z

e

N

3

d Ω

=

f

A

e

1
3
1
3
1
3

F

e1

f

=

F

e2

f

=

F

e3

f

=

120

·

1
2

·

1
3
1
3
1
3

=

20
20
20

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

13 / 30

Element Load Vector

Constant Inward Heat Flux

bq

=

100

W

m

2

=⇒

F

e

BC

=

Z

e

N

(

N

e

)

T

bq

ds

=

bq

Z

e

N

(

N

e

)

T

ds

(1,0)

1

(0,1)

(0,0)

2

3

1

2

(1,1)

(1,0)

(0,1)

3

1

(2, 0)

3

(1, 1)

2

(1, 0)

100

1
2

·

1

1
2

·

1

0

100

0

1
2

·

1

1
2

·

1

100

1
2

·

2

0

1
2

·

2

F

e1

BC

=

50
50

0

F

e2

BC

=

0

50
50

F

e3

BC

=

70.7

0

70.7

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

14 / 30

background image

Element Load Vector

load vector

elem 1

elem 2

elem 3

F

e

f

20
20
20

20
20
20

20
20
20

F

e

BC

50
50

0

0

50
50

70.7

0

70.7

F

e

=

F

e

f

+

F

e

BC

70
70
20

20
70
70

90.7

20

90.7

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

15 / 30

Assembly Process

Global Stiffness Matrix & Load Vector

no elem.

global DOFs

1

1, 2, 4

2

4, 3, 1

3

3, 4, 5

K

=

0

5

×

5

(+)

K

e1

3

×

3

(+)

K

e2

3

×

3

(+)

K

e3

3

×

3

F

=

0

5

×

1

(+)

F

e1

3

×

1

(+)

F

e2

3

×

1

(+)

F

e3

3

×

1

Before assembly:

K

=

0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0

,

F

=

0
0
0
0
0

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

16 / 30

background image

Assembly Process

First Element

no elem.

global DOFs

1

1, 2, 4

K

e

1

=

κ
2

1

1

0

1

2

1

0

1

1

F

e

1

=

70
70
20

K

=

κ

2

1

1

0

0

0

1

2

0

1

0

0

0

0

0

0

0

1

0

1

0

0

0

0

0

0

F

=

70
70

0

20

0

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

17 / 30

Assembly Process

Second Element

no elem.

global DOFs

2

4, 3, 1

K

e

2

=

κ
2

1

1

0

1

2

1

0

1

1

F

e

2

=

20
70
70

K

=

κ

2

1

+

1

1 0

+

(−

1

)

0

+

0

0

1

2

0

1 0

0

+

(−

1

)

0

0

+

2

0

+

(−

1

)

0

0

+

0

1 0

+

(−

1

)

1

+

1

0

0

0

0

0 0

F

=

70

+

70

70

0

+

70

20

+

20

0

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

18 / 30

background image

Assembly Process

Third Element

no elem.

global DOFs

3

3, 4, 5

K

e

3

=

κ
2

1

1

0

1

2

1

0

1

1

F

e

3

=

90.7

20

90.7

K

=

κ

2

2

1

1

0

0

1

2

0

1

0

1

0

2

+

1

1

+

(−

1

)

0

+

0

0

1

1

+

(−

1

)

2

+

2

0

+

(−

1

)

0

0

0

+

0

0

+

(−

1

)

0

+

1

F

=

140

70

70

+

90.7

40

+

20

0

+

90.7

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

19 / 30

Assembly Process

Global Matrix Equations

Kd

=

F

1
2

2

1

1

0

0

1

2

0

1

0

1

0

3

2

0

0

1

2

4

1

0

0

0

1

1

T

1

T

2

T

3

T

4

T

5

=

140

70

160.7

60

90.7

NOTE:

det K

=

0,

rank

(

K

) =

4

.

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

20 / 30

background image

Essential Boundary Conditions

1

2

3

1

2

4

3

5

T

2

=

T

4

=

T

5

=

500

T

1

T

2

T

3

T

4

T

5

T

1

500

T

3

500
500

1
2

2

1

1

0

0

1

2

0

1

0

1

0

3

2

0

0

1

2

4

1

0

0

0

1

1

T

1

500

T

3

500
500

=

140

70

160.7

60

90.7

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

21 / 30

Solution

1
2

2

1

1

0 0

1

0

3

2 0

T

1

500

T

3

500
500

=

140

160.7

1
2

2

1

1

3

T

1

T

3

=

140

160.7

1
2

1

0 0

0

2 0

500
500
500

1.0

0.5

0.5

1.5

T

1

T

3

=

390

660.7

Solution is

:

T

1

T

3

=

732
685

d

=

T

1

500

T

3

500
500

=

732

500

685

500
500

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

22 / 30

background image

Postprocessing

General Remarks

RECALL:

no elem.

global DOFs

1

1

,

2

,

4

2

4

,

3

,

1

3

3

,

4

,

5

,

d

=

732
500
685
500
500

temperature field

T

e

(

x, y

) =

N

e

d

e

heat flux rate

q

e

(

x, y

) = −

κ

T

= −

κB

e

d

e κ

=

1

= −

B

e

d

e

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

23 / 30

Postprocessing

Element Data

N

e

B

e

= ∇

N

e

elem 1

y

1

x

y

x

0

1 1

1

1 0

elem 2

1

y

x

+

y

1 1

x

0 1

1

1 1

0

elem 3

y

2

y

x

x

1

0

1 1

1

1 0

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

24 / 30

background image

Postprocessing

First Element

no elem.

global DOFs

1

1

,

2

,

4

,

d

=

732

500

685

500

500

temperature field

[

deg

]

T

e

1

(

x, y

) =

N

e

1

d

e

1

=

y

1

x

y

x

732

500

500

=

232y

+

500

heat flux rate

W

m

2

q

e

1

(

x, y

) = −

B

e

1

d

e

1

= −

0

1 1

1

1 0

732

500

500

=

0

232

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

25 / 30

Postprocessing

Second Element

no elem.

global DOFs

2

4

,

3

,

1

,

d

=

732

500

685

500

500

temperature field

[

deg

]

T

e

2

(

x, y

) =

1

y

x

+

y

1 1

x

500

685

732

=

185y

47x

+

547

heat flux rate

W

m

2

q

e

2

(

x, y

) = −

0

1

1

1 1

0

500

685

732

=

47

185

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

26 / 30

background image

Postprocessing

Third Element

no elem.

global DOFs

3

3

,

4

,

5

,

d

=

732
500

685

500

500

temperature field

[

deg

]

T

e

3

(

x, y

) =

y

2

y

x

x

1

685

500

500

=

185y

+

500

heat flux rate

W

m

2

q

e

3

(

x, y

) = −

0

1 1

1

1 0

685

500

500

=

0

185

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

27 / 30

Temperature Field

500

500

500

732

708.5

685

592.5

q

q

q

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

28 / 30

background image

Heat Flux Rate

Heat Exchange

1 8 5 [W/m]

1 8 5 [W/m]

2 3 2 [W/m]

1 8 5 [W/m]

[0, -232]

2

4

1

3

[47, -185]

[0, -185]

5

T=500 [K]

q=100

m

2

W

[ ]

m

[ ]

W

3

f=120

1

1

1

Q

exact

L

25

= −

R

f

d Ω

+

R

N

bq

ds

= − (

180

+

341

) = −

521

Q

app.

L

25

= −

232

185

= −

417

e

L

25

=

521

417

521

100%

=

20%

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

29 / 30

Temperature at Element Centroid

T

3

T

2

T

1

(x ,y )

c c

T

(

x

c

, y

c

) =

T

1

+

T

2

+

T

3

3

since

T

(

x

c

, y

c

) =

N

1

(

x

c

, y

c

)

N

2

(

x

c

, y

c

)

N

3

(

x

c

, y

c

)

T

1

T

2

T

3

=

=

1
3

1
3

1
3

T

1

T

2

T

3

=

1
3

T

1

+

1
3

T

2

+

1
3

T

3

MS

(L-53 CUT)

Heat Conduction in 2D

04/2013

30 / 30


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