of 42
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HEAT TRANSFER
Prepared by
M. Santhosh Kumar , AP-
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2/42
DERIVATION OF TEMPERATURE FUNCTION (T) AND SHAPE FUNCTION (N) FO
CONDUCTION ELEMENT
xaaT 10
1
01
a
axT
1.1
Consider a bar element with nodes 1 and 2 as shown in figure.T1 and T2 are the tempe
respective nodes. Therefore T1 and T2 are the dof for this bar element.
l
1 2
T2T1
k
Let the temperature function be
Writing Eq 1.1 in matrix form
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1TT 0x
2TT lx
laaT
aT
102
01
1
0
2
1
101
aa
lTT
1.3
1.2
1.4
At node 1
At node 2 ,
Sub the above values in EQ 1.1
Assemble EQ 1.2 & 1.3 in matrix form
, TfunctioneTemperatur
l
xN
l
xN
2
1 1
2
1
11
011
T
Tl
lxT
Sub the EQ1.4 in EQ 1.1
1
1
We know that
From EQ 1.5&1.6
We can get the shape functio
2
1
1
0
11
01
T
Tl
la
a
2
1
21T
TNNT
2
1
T
T
l
x
l
xlT
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Consider a bar element with nodes 1 and 2 as shown in figure.T1 and T2 are the temprespective nodes and k be the thermal conductivity of the material
Stiffness matrix
We know that
Where
Derivation of Stiffness matrix for 1 D heat conduction eleme
dvBDBK T
v
1.1
2211, TNTNTfunctioneTemperatur
lxN
l
xN
2
1 1
l
1 2
T2T1
k
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Strain Displacement Matrix
In 1-D heat conduction
dx
dN
dx
dNB 21
l
lB
llB
T
1
1
111.2
1.3
1.4
materialtheof
tyconductiviThermalkKD
Sub. EQ 1.2,1.3 &1.4 in EQ 1.1 , we
Stiffness matrix for
heat conduction
l
lK
l
C 1
1
0
v
C dvk
ll
llK **11
11
22
22
l
C Ak
ll
llK0
22
22
**11
11
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l
C dxAk
ll
llK0
22
22
***11
11
22
22
11
11
ll
llAklKC
11
11
l
AkKC
CK
TKF C
Thus is the stiffness matrix for he
A= Area of the elementl = length of the element
The general force equation is given by
2
1
11
11
l
Ak
F
F
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11
11
l
AkKCStiffness matrix for 1-D heat conduction
The convection term contribution to
stiffness matrix is dANNhKT
A
endh
functionShapeN
coefficietransferHeath
1.6
Boundary conditions
(i) 1-D heat conduction with free end convection
Consider a element with node 1 and 2 with temperature T1 and T2respectively with convection is from the right end as shown in fig
l
Conduction
k T2T1
h
Convection
T
x
1 2
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Shape functions
At node 2, x=l
Sub EQ 1.7 in EQ 1.6
l
xN
l
xN
2
1 1
1
0
10
TN
N
21 NNN
1.7
dAhKA
endh10
1
0
10
00hAK
endh
1.8
Stiffness matrix [K] = + CK enhK
Convection from the free end at x=
1
0AThF
endh
2
1
N
NAThF
endh
We know that
TKF C
0
0
11
11
1
0hA
l
AkATh
dAhK
endh
10
00
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11
11
l
AkKC
NNhKT
S
h
NhPK
Tl
h 0
dPdS XWhere
(ii)1-D element with conduction, convection and internal heat gene
Consider a rod with nodes 1 and 2 as shown in fig. This rod is subjected to conductio
internal heat generation
Stiffness matrix for
1-D heat conduction
Heat convection part of
stiffness matrix is
1 2k Q Conduction
T2T1 l
Convection
Thermal
Conductivity
Internal heat generation
l
xl
l
xl
xl
hPK
l
h
0
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dx
l
x
l
x
l
x
l
x
l
x
l
xl
hPK
l
h
0
2
2
2
2
2
22
36
63ll
ll
hP
21
12
6
hPlKh
21
12
611
11 hPl
l
AkK
hC KK KmatrixStiffness
Force matrix due to heat generati
dVQNFT
v
Q
dxQNF
Tl
Q *A**0
Q
1
1
2
lAQFQ
dxNAQFTl
Q 0
dx
l
xl
xl
AQ
l
0
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Force matrix due to convection is given by
PdxNhT T
s
l
TdxNPhT
0
l
dx
l
xl
xlPhT
0
1
1
2
lPhTFh
dSNhTF T
s
h
Force matrix
QFFmatrixForce
1
1
2
PlAQF
2
PhTQAlF
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The general force equation is given by
where
TKF
1
1
2
lPhTQAl
2
1
21
12
611
11
T
ThPl
l
Ak
Substituting the values of [K] & {F} in the above equation we obtain
2
1
21
12
611
11
T
ThPl
l
Ak
1
1
2
lPhTQAl
Q=Heat Generation
P=Perimeter , m
The above equation is the finite element equation for 1-D eleme
is subjected to conduction , convection, internal heat generation
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A furnace wall is made up of three layers, inside layer with thermal conductivity
the middle layer with thermal conductivity 0.25 W/mK, the outer layer with ther
conductivity 0.08 W/mK. The respective thickness of inner, middle and outer lay
25cm,5cm, and 3cm respectively. The inside temperature of the wall is 600 C and
the wall is exposed to atmospheric air at 30
C with the heat transfer coefficient o
Determine the nodal temperatures.
Problem 1:
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For element 2:
Nodes 2 & 3
2
1
1
11
2
1
11
11
T
T
l
kA
F
F
2
1
2
1
11
11
25.0
5.8
T
T
F
F
2
1
2
1
3434
3434
T
T
F
F
For element 1:
Nodes 1& 2
2
22
3
2
11
11
l
kA
F
F
3
2
11
11
05.0
25.0*1
F
F
3
2
55
55
F
F
l1
CONDUCTION
k1T1 T21
l2
CONDUCTION
k2T2 T32
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For element 3:
Nodes 3 & 4
The third element is subjected to both
conduction and convection
4
3
3
33
10
00
11
11
1
0
T
ThA
l
kAATh
4
3
10
001*45
11
11
03.0
08.0*1
1
01*303*45
T
T
4
3
3 450
00
666.2666.2
666.2666.2
10*635.13
0
T
T
4
3
3
666.47666.2
666.2666.2
10*635.13
0
T
T
l3
Conduction
k3
T4T3
h
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Assembling the finite elements we get
4
3
2
1
4
3
2
1
666.2666.200
666.2666.750
053934
003434
F
F
F
F
T
T
T
T
Since there is no heat generation&
convection except from the right end
34
321
10*635.13
0
F
FFF
3
4
3
2
10*635.13
0
0
0873
666.2666.200
666.2666.750
053934
003434
T
T
T
1
From Equation 1 form simultan
equation and solve to get value
Solution:
T2=846.2 K
T3=664.1 KT4=323.2 K
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3
3
2
2
1
1
v
u
v
u
v
u
u
Derivation of shape function for heat transfer in 2D elem
Displacement
Consider a triangular element with nodes 1,2,& 3 with nodal displacements u 1,u2,u3
7/25/2019 Unit II and III
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yxv
yxu
654
321
333213
232212
131211
yxu
yxu
yxu
3
2
1
33
22
11
3
2
1
1
1
1
yx
yx
yx
u
u
u
3
2
1
1
33
22
11
3
2
1
1
1
1
u
u
u
yx
yx
yx
3123
1332
31132332
2312332
1
33
22
11
yx-y(x
)(
1
1
1
1
xxxx
yyyy
yxyx
yyxyxyxyx
yx
yx
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3
2
1
123123
211332
122131132332
2312312332
3
2
1
)yx-y(x)yx-y(x
X)()(
1
u
u
u
xxxxxx
yyyyyy
yxyx
xxyyyxyxyx
33
22
11
1
1
1
2
1
yx
yx
yx
A
(21
312332 yxyxyxA
(2 312332 yyxyxyxA
123123
211332
122131132332
3
2
1 )yx-y(x)yx-y(x
X2
1
xxxxxx
yyyyyy
yxyx
A
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3
2
1
321
321
321
3
2
1
X2
1
u
u
u
ccc
bbb
aaa
A
123312231
213132321
122133113223321
yx-yxyx-yx
xxcxxcxxc
yybyybyyb
aayxyxa
1yxu
3
2
1
333222111
3
2
1
321
321
321
222
X211
u
u
u
A
ycxba
A
ycxba
A
ycxbau
u
u
u
ccc
bbb
aaa
Ayxu
21
21
NNNv
NNNu
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A
ycxbaN
A
ycxbaN
A
ycxbaN
2
2
2
3333
2222
1111
3
3
2
2
1
1
321
321
000
000
),(
),(
v
u
v
u
v
u
NNN
NNN
yxv
yxu
uDisplacement function
N1,N2,N3ARE THE SHAPE FUNCTIONS O
TRIANGULAR ELEMENT
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Stiffness matrix and load vector for heat transfer in 2-D element
y
x
3
21
T1 T2
T3
(x1,y1) (x2,y2)
(x3,y3)
332211),(, TNTNTNyxTfunctioneTemperatur A
cxbaN
A
cxbaN
A
ycxbaN
2
2
2
3333
2222
1111
Shape functions are giv
dvBDBKmatrixStiffnessT
C
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24/42
StrainDisplacement matrix is given by
y
N
y
N
y
Nx
N
x
N
x
N
B321
321
321
321
2
1
ccc
bbb
AB
33
22
11
2
1
cbcb
cb
AB
T
Stress strain matrix is given by
y
x
k
kD
0
0
Assuming a unit thickness, dv=da
Now we get the stiffness matrix by substit
[B],[B]T,[D] in [K]
c
b
Ak
k
cb
cb
cb
AK
y
x
C
1
1
33
22
11
2
1
0
0
2
1
cc
bb
k
k
cb
cb
cb
AK
y
x
C
1
1
33
22
11
2 0
0
4
1
For an isotropic material , kx=ky=k NNNNN 2
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p , x y
Thus the Stiffness matrix for conduction is
2
3
2
332323131
3232
2
2
2
22121
31312121
2
1
2
1
4cbccbbccbb
ccbbcbccbb
ccbbccbbcb
A
kKC
convectionformatrixStiffness
dSNNN
N
N
N
hKS
h 321
3
2
1
dSNNhK TS
h
Now considering edge 1-2, alone and N3
NNNNN
NNNNN
NNNNN
hKS
h
2
33231
32
2
221
31211
dSNNNNNN
hKS
h
000
0
02
221
21
2
1
Substitute N3=0 in Equation 1.1
Sub N1=L1,N2=L2 and N3=L3, along the e
2
1 000
0
02
221
21
2
1s
s
h dSLLL
LLL
hK
Sub eq. 1.2,1.3,1.5 in eq. 1.2We know that
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26/42
sdsLL)!1(
!!21
33*2*1
2*1
!12
!2
2
1
2
1
ssdsL
sdsL
6!111
!1!121
ssdsLL
33*2*1
2*1
!12
!2
2
2
2
2
ssdsL
sdsL
1.3
1.4
1.5
00
36
63
2121
2121
2121
ss
ss
hKh
0
1
2
62121
21 shKh
We know that
Now considering edge 2 3 alone Now considering edge 3 1 alone
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27/42
Now considering edge 2-3, alone
And N1=0 at the this edge
Sub N2=L2,N3=L3 , along the
edge 2-3,N1=L1=0
Sub in 1.1
dSLLL
LLLhKS
h
2
332
32
2
2
0
0
000
360
630
000
3232
3232
3232
ss
sshKh
210
120
000
6
32
32
32 shKh
Now considering edge 3-1, alone
And N2=0 at the this edge
Sub N1=L1,N3=L3 , along the
edge 3-1,N2=L2=0
dSLLL
LLL
hKS
h
2331
31
2
1
0
000
0
606
0006
03
1313
1313
1313ss
ss
hKh
201
000
102
6
1313
13
shKh
Stiff t i f C ti i 000012
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28/42
Stiffness matrix for Convection is
133221
hhhh KKKK
201
000
102
6
210
120
000
6000
021
012
6
1313
32322121
sh
shsh
Kh
Stiffness matrix for 2-D heat transfer element is given by
hC KKK
1
0
2
6
000
021
012
6
4
1313
22121
2
3
2
332323131
3232
2
2
2
22121
31312121
2
1
2
1
sh
hsh
cbccbbccbb
ccbbcbccbb
ccbbccbbcb
A
kK
Force ector for 2 D heat transfer dsNqFT
hTF
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29/42
Force vector for 2-D heat transfer
elements is
dvNqF To1
3
2
1
1
L
L
L
AqF o
By using area coordinate system
AdALLL 2*)!2(
!!!321
1
1
1
31
AqF o
dsNqFs
2
dsNN
N
qF
s
s
2
1
3
2
1
2
dsLL
qF
s
s
2
1
0
2
1
2
0
1
1
2
21212
sqF
hTFs
3
hTF 3
hTF 3
2
213
ThF
Unit thickness
Compute the element matrix and vectors for the element shown in figure, When the e
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30/42
Compute the element matrix and vectors for the element shown in figure, When the e
and 3-1 experience convection heat loss.
405010
508535
103525
CK
15.5645.762.20
45.790.140
62.20025.41
hK
200
200
200
1F
3F
02 F
a) Element matrix
(i) conduction(ii) convection
b) Force vectors
Shape function derivation for fluid mechanics We know that
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p
in 2-D element
Let us consider a three noded triangular element as
shown in fig. Let the nodal potentials are p1,p2 and p3
Let the potential function be
3
2
1
p
p
p
p
yxp 321
131211 yxp
232212 yxp
333213 yxp
W
3
2
1
1
33
22
11
3
2
1
1
1
1
p
p
p
yx
yx
yx
3
2
1
321
321
321
3
2
1
2
1
p
pp
ccc
bbbaaa
A
12213
31132
23321
yxyxa
yxyxa
yxyxa
213
132
321
yyb
yyb
yyb
c
c
c1.1
From eq 1.1
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32/42
q
3
2
1
1
yxp
3
2
1
321
321
321
2
11
p
p
p
ccc
bbb
aaa
Ayxp
AxbaN
xbaN
A
xbaN
2
2
2
333
222
111
3
2
1
333222111
222p
p
p
Aycxba
Aycxba
Aycxbap
21),( NNyxp
-2ofmechanicsfluid
shtheare&, 321 NNN
hPDerivation of 1 - D Heat transfer
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33/42
dx
0x lx
OTQ
Convection
FIN
Q
dx
AreaSectionalCrossA
surfaceateTemperaturT
atmosphereAmbient
/
/ 2
T
Thicknessdx
PerimeterP
ConvectionofefficientCoh
mkK
Kmh
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)( TTdxPhdQQQ
0)( TTdxPhdQ
dx
dTKAQ
LawFourierBy
''
0)(
dxbyequationabovethe
TTdxPhdQ
(
Phdx
dx
dTKAd
7/25/2019 Unit II and III
35/42
0)(
insulatedisEnd:ICase
lQ
endat theretemperatuisT
)()(
atmosphereopen toisEnd:IICase
LWhere
TThAlQ L
)(
TatmaintainedissurfaceEnd:IIICase
TThAQL
Boundary conditions By Ritz method
dxxwGDE )()(
By assuming bar element for he
Wt
Wt
Niseqsecondforfunction.
-1Niseqfirstforfunction.
2
1
TThPdx
dx
dTKAdl
0
Thus we get
d
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36/42
0w(x)w(x)0 0
dxTThPdxdxdx
dTKAdl l
0)(dx
dw)(
000
dxxwTThPdxdx
dTKAxw
dx
dTKA
lll
Thus we arrive at the weak form of 1D heat transfer element
v-uvdvu
dx
dwdu;)(
xwu
dxdx
dTKA
dx
ddv
0)(dx
dw)()(
00
0
dxxwTThPdxdx
dTKAxQxw
lll
N l i fi t ti
7/25/2019 Unit II and III
37/42
Now solving first equation
ldx
xdw
lxw
1)(
X1N)( 1
2
Sub in2 1
011
11)(1
0
21
21
00
dxl
xTT
l
xT
l
xhp
dxl
Tl
xT
l
x
dx
dKAxQ
l
x
l
ll
ll
ll
xQl
xdx
l
xThp
dl
xT
l
xT
l
xhpdx
lT
l
xT
l
x
dx
dKA
00
0
2121
0
)(11
111
1
2
1111 22
dxxx
Txx
hpdxTTKAll
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)0(2
322
2
3
11
0
2
2
0
2
32
1
0
2
2
3
2212
0
Ql
xxhpT
Tl
x
l
xT
l
x
l
xxhpdxT
lT
lKA
l
lll
)0(2
322
2
3
2
2
0
2
32
1
2
2
3
0
2212
Qdxl
llhpT
Tl
l
l
lT
l
l
l
llhpT
l
xT
l
xKA
ll
)0(2
3
11121
Ql
hpT
Tll
lhpTl
Tl
KA
12611 21
2
1
hPT
ThPl
T
T
l
KA
)0(1
1
0
0
21221
0
Qdxl
xhpT
dxll
Tll
hpdxl
Tl
Tl
KA
l
A
For second equation
7/25/2019 Unit II and III
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ll
xNxw
1
dx
dw(x);)( 2 3
Sub in3 1
01
11)(
0
21
21
00
dxl
xTT
l
xT
l
xhp
dxl
Tl
xT
l
x
dx
dKAxQ
l
x
l
ll
ll
ll
xQl
xdx
l
xThp
dxl
xT
l
xT
l
xhpdx
lT
l
xT
l
x
dx
dKA
00
0
2121
0
)(
11
1
11122ll
7/25/2019 Unit II and III
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)(
111
0
0
22
2
12
2
21
0
lQdxl
xhpT
dxTl
xT
l
x
l
xhpdx
lT
lT
lKA
l
)(2
332
11
0
2
0
22
3
12
32
2212
0
lQl
xhpT
Tl
xT
l
x
l
xhpdxT
lT
lKA
l
ll
)(2
332
21
0
2212
lQl
hpT
Tl
Tll
hpTl
xT
l
xKA
l
)(2
32
11121
0
lQlhpT
Tll
hpTl
Tl
KAl
21
611
2
1
2
1h
T
ThPl
T
T
l
KA
B
Combining A and B in matrix form
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CombiningAand B in matrix form
2/
2/
21
12
611
11
2
1
2
1
l
l
hPTT
ThPl
T
T
l
KA
Apply boundary conditions
(i) End is insulated i.e. Q(l)=0
2/
2/
21
12
6
11
11
2
1 Q
l
lhPT
T
ThPl
l
KA
(ii) End is open to atmosphere,
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@ end TL=T2; Q(L)=hA(TL-T)
TL Surface temperature end (x=l)
l
l
hPTT
T
hA
hPl
l
KA
2/
2/
0
00
21
12
611
11
2
1
(iii) End is maintained at temp T,
At (x=l)
(2/
2/
21
12
6
11
11
2
1
hA
Q
l
lhPT
T
ThPl
l
KA
