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
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
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
5 / 30
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
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
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
Ω
e
(
B
i
)
T
B
j
d Ω
→
K
e
ij
=
κ
(
B
i
)
T
B
j
Z
Ω
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
Ω
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
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
Ω
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
Ω
e
N
e
i
d Ω
=
1
3
·
h
·
Z
Ω
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
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
Ω
e
(
N
e
)
T
f
d Ω
=
f
Z
Ω
e
N
1
d Ω
Z
Ω
e
N
2
d Ω
Z
Ω
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
∂Ω
e
N
(
N
e
)
T
bq
ds
=
bq
Z
∂Ω
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
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
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
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
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
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
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
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
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
Ω
f
d Ω
+
R
∂Ω
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