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fem 2 dimensional heat conduction problems
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2D- HEAT CONDUCTION (PROBLEM)A long bar of rectangular cross sections (f 0, f 0)
having thermal conductivity of 1.5 W / m c is subjected to boundary conditions as shown fig. Two opposite sides are maintained at uniform temperat
Q CV
2
ure of T=180 .One side is insulated and other
side is subjected to convection process with T =25 c, h=50W/mDetermine the temperature distribution in the bar.
C
C
T=180 C
0q
0.4m
2h=50W/m C
T =25 c
T=180 C
0.3
Discretize the domain into 5-node,3-element
0q
2
3
1
45
0.4m
0.3m2
1
3
Local
Global
exx yy cvB cv cv cvB qs qsB QK K K K a f f f f f
0 0 0 01:Element
i j
k
1(0,0) 2(0.4,0)
3(0.4,0.15)
2
:1 0 0
1/ 2 1 0.4 01 0.4 0.15
1(0.4 0.15) (1/ 2)
0.3
Kxx
A
m
ij
k
x y
1
111
11/ 2 1 0.4 0
1 0.4 0.15
i J J
K K
x yN x y
x y
x yA
j
k
ki
111
11/ 2 1 0.4 0.15
1 0 0
j k k
i i
x yN x y
x y
x yA
221 21 22 21 23
222 21 22 22 23
223 21 23 22 23
xx
m m m m mK KAt m m m m m
m m m m m
111
11/ 2 1 0 0
1 0.4 0
k i i
j j
x yN x y
x y
x yA
ij
11 21 31
22 32 12
33 13 23
,1, 2.5, 02.5, 6.666, 06.666, 0
by solvingm m mm m mm m m
2( 2.5) 6.25 00.495 6.25 6.25 0
0 0 0
0.28125 0.28125 00.28125 0.28125 0
0 0 0
xxK
231 31 32 31 33
2 232 31 32 32 33
233 31 33 32 33
,
yy
similarly
m m m m mK KAt m m m m m
m m m m m
2
2
0 0 00.045 0 ( 6.666) ( 6.666) 6.666
0 ( 6.666) 6.666 6.666
0 0 00 1.999 1.9990 1.999 1.999
yyK
0 0 0/ 6 0 2 1
0 1 2
0 0 01.25 0 2 1
0 1 2
0 0 00 2.5 1.250 1.25 2.5
cvB B jkK h tl
0/ 2 1
1
093.75 1
1
093.7593.75
cvB B jk aBf h tl T
1
2
3
1:
0.28125 0.28125 0 00.28125 4.7808 0.7495 93.75
0 0.7495 4.4995 93.75
Element equation for elemen
TTT
1T 2T 3T
exx yy cvB cv cv cvB qs qsB QK K K K a f f f f f
2 :Element0 0 0 0 0 0 0
221 21 22 21 23
222 21 22 22 23
223 21 23 22 23
T
xxN NK Kt dxdyx xm m m m m
KAt m m m m mm m m m m
2
:1 0 0.3
1/ 2 1 0 01 0.4 0.15
0.3(1 0.4) (1/ 2)
0.06
Kxx
A
m
ijk
x y
1(0,0)
5(0,0.3)
2
i
j
k3(0.4,0.15)
111
11/ 2 1 0 0
1 0.4 0.15
i J J
K K
x yN x y
x y
x yA
jk
ki
111
11/ 2 1 0.4 0.15
1 0 0.3
j k k
i i
x yN x y
x y
x yA
111
11/ 2 1 0 0.3
1 0 0
k i i
j j
x yN x y
x y
x yA
ij
11 21 31
12 22 32
13 23 33
,0, 1.25, 3.3331, 1.25, 3.3330, 2.5, 0
by solvingm m mm m mm m m
21.25 1.5625 3.1250.09 1.562 1.562 3.125
3.125 3.125 6.25
0.140 0.140 0.28120.140 0.140 0.2812
0.2812 0.2812 0.562
xxK
2
,
3.33 11.108 00.09 0 11.108 0
0 0 0
0.998 0.999 00 0.999 00 0 0
yy
similarly
K
5
1
3
2 :
1.138 1.138 0.2812 00.140 11.248 0.2812 0
0.2812 0.2812 0.562 0
Element equation for elemen
TTT
5T 1T 3T
exx yy cvB cv cv cvB qs qsB QK K K K a f f f f f
2 :Element
0 0 0 0
2
:1 0 0.3
1/ 2 1 0.4 0.31 0.4 0.15
1(0.4 0.15) 0.3(0.4 0.4) (1/ 2)
0.03
Kxx
A
m
ijk
x y
111
11/ 2 1 0.4 0.3
1 0.4 0.15
i J J
K K
x yN x y
x y
x yA
5(0,0.3)
3(0.4,0.15)
i j
k
4(0.4,0.3)
3
j
k
ki
111
11/ 2 1 0.4 0.15
1 0 0.3
j k k
i i
x yN x y
x y
x yA
111
11/ 2 1 0 0.3
1 0.4 0.3
k i i
j j
x yN x y
x y
x yA
11 21 31
12 22 32
13 23 33
,1, 2.5, 0
2, 2.5, 6.662, 0, 6.66
by solvingm m mm m mm m m
ij
22.5 6.25 00.045 6.25 6.25 0
0 0 0
0.2812 0.2812 00.2812 0.2812 0
0 1.995 1.995
xxK
,0 0 0
0.045 0 44.355 44.3550 44.355 44.355
0 0 00 1.995 1.9950 1.995 1.995
yy
similarly
K
0 0 0/ 6 0 2 1
0 1 2
0 0 0(50 1 0.15) / 6 0 2 1
0 1 2
0 0 00 2.5 1.250 1.25 2.5
cvB B jkK h tl
0/ 2 1
1cvB B jk aBf h tl T
5
4
3
3:
0.2812 0.2812 0 00.2812 4.7762 0.745 0
0 0.745 4.495 0
Element equation for elemen
TTT
5T 4T 3T
02.8125 1
1
02.81252.8125
,
(0.28124 11.248) 0.28125 (0 0.2812) 0 0.1400.2812 4.7808 0.74915 0 00.2812 0.7495 (4.4995 4.495 0.562) 0.745 0.2812
0 0 0.745 4.7762 0.28121.138 0 0.2812 0.2812 (1.138 0.2812)
Assembling all theelement equations
1
2
3
4
5
093.75
93.75 93.7593.75
0
TTTTT
1T 2T 3T 4T 5T
1 2 5 4
1
2
3
4
5
,180
11.529 0.28125 0.2812 0 0.1400.2812 4.7808 0.74915 0 00.2812 0.7495 9.556 0.745 0.2812
0 0 0.745 4.7762 0.28121.138 0 0.2812 0.2812 1.419
Applyboundry conditionsT T T T C
TTTTT
093.75
187.5093.75
0
1T 2T 3T 4T 5T
1 2 3 4 5
1 2 5 4
3
3
0.2821( ) 0.749( ) 9.556( ) 0.745( ) 0.2812( ) 187.50' 180
9.556( ) 187.50 268.9047.76
3 ,
T T T T TApplybc sT T T T C
TT C
Tempratureat thenode in thebar element
3 47.76T C
Re :sult
