CHAP 5 FINITE ELEMENTS FOR HEAT
TRANSFER PROBLEMS
FINITE ELEMENT ANALYSIS AND DESIGN
Nam-Ho Kim
1
HEAT CONDUCTION ANALYSIS
• Analogy between Stress and Heat Conduction Analysis
Structural problem
Heat transfer problem
Displacement
Temperature (scalar)
Stress/strain
Heat flux (vector)
Displacement B.C.
Fixed temperature B.C.
Surface traction force
Heat flux B.C.
Body force
Internal heat generation
Young’s modulus
Thermal conductivity
– In finite element viewpoint, two problems are identical if a proper
interpretation is given.
• More Complex Problems
– Coupled structural-thermal problems (thermal strain).
– Radiation problem
2
THERMAL PROBLEM
• Goals:
[K T ]{T }  {Q}
Thermal load
Nodal temperature
Conductivity matrix
– Solve for temperature distribution for a given thermal load.
• Boundary Conditions
– Essential BC: Specified temperature
– Natural BC: Specified heat flux
3
STEADY-STATE HEAT TRANSFER PROBLEM
• Fourier Heat Conduction Equation:
– Heat flow from high temperature to low temperature
dT
qx  kA
dx
Thermal conductivity (W/m/C )
Heat flux (Watts)
• Examples of 1D heat conduction problems
Thigh
Thigh
qx
Tlow
qx
Tlow
4
GOVERNING DIFFERENTIAL EQUATION
• Conservation of Energy
– Energy In + Energy Generated = Energy Out + Energy Increase
Ein  Egenerated  Eout  U
• Two modes of heat transfer through the boundary
– Prescribed surface heat flow Qs per unit area
– Convective heat transfer Qh  h T   T
– h: convection coefficient (W/m2/C )

T

¥
Qs
Qg
qx
A
qx +
dqx
Dx
dx
dx
5
GOVERNING DIFFERENTIAL EQUATION cont.
• Conservation of Energy at Steady State
– No change in internal energy (U = 0)
dq


qx  QsPx  h T   T Px  Qg Ax   qx  x x 
dx




Egen
Ein
– P: perimeter of the cross-section


dqx
 Qg A  hP T   T  QsP,
dx
Eout
0 x L
• Apply Fourier Law
d  dT 
kA
 Qg A  hP T   T  QsP  0,


dx 
dx 


0 x L
– Rate of change of heat flux is equal to the sum of heat generated and
heat transferred
6
GOVERNING DIFFERENTIAL EQUATION cont.
• Boundary conditions
– Temperature at the boundary is prescribed (essential BC)
– Heat flux is prescribed (natural BC)
– Example: essential BC at x = 0, and natural BC at x = L:
T(0)  T0

 dT
 qL
kA dx
x L

7
DIRECT METHOD
• Follow the same procedure with 1D bar element
– No need to use differential equation
• Element conduction equation
–
–
–
–
–
Heat can enter the system only through the nodes
Qi: heat enters at node i (Watts)
Divide the solid into a number of elements
Each element has two nodes and two DOFs (Ti and Tj)
For each element, heat entering the element is positive
1
Q1
N
2
Q2
QN
Q3
i
e
qi(e )
xi
L(e)
Ti
j
q(je )
Tj
xj
8
ELEMENT EQUATION
• Fourier law of heat conduction
(e)
i
q
Tj  Ti 

dT
 kA
 kA
dx
L(e)
• From the conservation of energy for the element
(e)
i
q
q
(e)
j
0
(e)
j
q
 kA
(Tj  Ti )
L(e)
• Combine the two equation
q(e)
 kA
i 
 (e)   (e)
qj  L
 1 1  Ti 
 1 1  T 

 j
Element conductance matrix
– Similar to 1D bar element (k = E, T = u, q = f)
9
ASSEMBLY
• Assembly using heat conservation at nodes
– Remember that heat flow into the element is positive
– Equilibrium of heat flow:
 T1   Q1 
 T  Q 
Ni
   
(e)
Qi   qi
[K T ]  2    2 
e 1
NN   


TN 
 
QN 

– Same assembly procedure with 1D bar elements
• Applying BC
– Striking-the-rows works, but not striking-the-columns because
prescribed temperatures are not usually zero
Q2
1
Element 1
q2(1)
2
q2(2)
3
Element 2
10
EXAMPLE
• Calculate nodal temperatures of four elements
– A = 1m2, L = 1m, k = 10W/m/C
200 C
Q1
T1
T2
1
Q2 = 500W
T3
2
T4
3
Q3 = 0
Q4 = –200W
T5
x
4
Q5 = 0
• Element conduction equation
 1 1  T1  q1(1) 
10 
    (1) 

 1 1  T2  q2 

 1 1 T2  q(2)
2
10 
    (2) 

 1 1  T3  q3 

 1 1 T3  q(3)
3
10 

 T   (3) 

1
1

  4  q4 

 1 1 T4  q(4)
4
10 
    (4) 

 1 1  T5  q5 
11
EXAMPLE cont.
• Assembly
 Q1   q1(1) 
 1 1 0 0 0   T1 
Q   (1)
 1 2 1 0 0  T 
(2) 
q

q
2 
 2   2
  2 


(3)
(2)
Q3   q3  q3   10  0 1 2 1 0  T3 
 

Q  q(3)  q(4) 
1

2
1

0
0
 T4 

 4   4 (4) 4 
 0 0 0 1 1 T5 

Q5   q5
• Boundary conditions (T1 = 200 oC, Q1 is unknown)
1
 1

10  0

0
 0
1 0
0  200   Q1 
T2   500 
2 1 0 0  


 

1 2 1 0   T3    0 

0 1 2 1  T4  200 

 

0 0 1 1  
 T5   0 
0
12
EXAMPLE cont.
• Boundary conditions
– Strike the first row
200 
 1 2 1 0 0  
 500 


  T2  

0

1
2

1
0
0




 T 
10 
 0 0 1 2 1  3  200 


  T4  
  0 
 0 0 0 1 1  
 T5 
– Instead of striking the first column, multiply the first column with
T1 = 200 oC and move to RHS
 2 1 0 0  T2   500  2000 
 1 2 1 0  T   0   0 

 

3

10 
 


 0 1 2 1 T4  200   0 

  
 
 0 

0
0

1
1

 T5   0 
– Now, the global matrix is positive-definite and can be solved for nodal
temperatures
13
EXAMPLE cont.
• Nodal temperatures
{T}T = {200 230 210 190 190 }°C
• How much heat input is required to maintain T1 = 200oC?
– Use the deleted first row with known nodal temperatures
Q1  10T1  10T2  0T3  0T4  0T5  300 W
• Other example
100W
3
4
2
1
50 C
1
200W
2
4
3
Q=0
5
5
x
Q=0
14
GALERKIN METHOD FOR HEAT CONDUCTION
• Direct method is limited for nodal heat input
• Need more advanced method for heat generation and
convection heat transfer
• Galerkin method in Chapter 3 can be used for this purpose
j
• Consider element (e)
i
(e )
e
qi
q(je )
• Interpolation
L(e)
T
T
T(x)  TN
i i (x)  TN
j j (x)
xx
 xx 
Ni (x)   1  (e) i  , Nj (x)  (e) i
L 
L

Ti 
T(x)  N {T}  Ni (x) Nj (x)  
Tj 
xi
j
i
xj
Temperature varies linearly
• Heat flux
dT  1
   (e)
dx  L
1 
{T }  B {T }
(e) 
L 
Heat flow is constant
15
GALERKIN METHOD cont.
• Differential equation with heat generation
d 
dT 
kA
 Qg A  0,


dx 
dx 
0 x L
• Substitute approximate solution
d 
dT 
kA

  AQg  R(x)
dx 
dx 
Residual
• Integrate the residual with Ni(x) as a weight
 d  dT 

x  dx  kA dx   AQg Ni (x)dx  0

i 
xj
• Integrate by parts
xj
xj
xj
dT
dT dNi
kA
Ni (x)   kA
dx    AQgNi (x)dx
dx
dx dx
xi
xi
x
i
16
GALERKIN METHOD cont.
• Substitute interpolation relation
xj
x
j
dNj  dNi
 dNi
x kA  Ti dx  Tj dx  dx dx  x AQgNi (x)dx  q(x j )Ni (x j )  q(xi )Ni (xi )
i
i
• Perform integration
kA
(e)
(e)
T

T

Q

q


i
j
i
i
L(e)
• Repeat with Nj(x) as a weight
kA
T  Ti   Q(e)
 q(e)
j
j
(e)  j
L
xj
Q(e)
  AQgNi (x)dx
i
xi
xj
Q(e)
  AQgNj (x)dx
j
xi
17
GALERKIN METHOD cont.
• Combine the two equations
kA
L(e)
(e)
(e)
 1 1  Ti  Qi  qi 
 1 1  T   Q(e)  q(e) 

  j   j
j 

(e)
(e)
[k (e)
]{
T
}

{
Q
}

{
q
}
T
Similar to 1D bar element
– {Q(e)}: thermal load corresponding to the heat source
– {q(e)}: vector of nodal heat flows across the cross-section
• Uniform heat source
(e)
N
(x)
AQ
L


i
g
{Q(e) }   AQg 
dx 

2
xi
Nj (x)
xj
1

1
– Equally divided to the two nodes
• Temperature varies linearly in the element, and the heat
flux is constant
18
EXAMPLE
• Heat chamber
Wall temperature = 200 C
Uniform heat source inside
the wall Q = 400 W/m3.
Thermal conductivity of the
wall is k = 25 W/mC.
Use four elements through
the thickness (unit area)
Boundary Condition:
T1 = 200, qx=1 = 0.
200 C
Insulated
No heat flow
Wall
200 C
x
1m
No heat flow
x
T1
T2
1
T3
2
T4
3
T5
4
19
EXAMPLE cont.
• Element Matrix Equation
– All elements are identical
 1 1  T1  50  q1(1) 
100 
 T   50    (1) 

1
1

  2    q2 
– Assembly
 Q1   q1(1) 
1
Q   (1)
 1
(2) 
q

q
2 
 2   2

(2)
(3) 
Q3   q3  q3   100  0

Q  q(3)  q(4) 
0
 4   4 (4) 4 
 0
Q5   q5

1 0
0   T1   50 
2 1 0 0  T2  100 
   

1 2 1 0  T3   100 

0 1 2 1 T4  100 
  

0 0 1 1  T5   50 
0
20
EXAMPLE cont.
• Boundary Conditions
– At node 1, the temperature is given (T1 = 200). Thus, the heat flux at
node 1 (Q1) should be unknown.
– At node 5, the insulation condition required that the heat flux (Q5)
should be zero. Thus, the temperature at node 5 should be unknown.
– At nodes 2 – 4, the temperature is unknown (T2, T3, T4). Thus the heat
flux should be known.
Q1
1
2
1
 1

100  0

0
 0
3
1 0
4
5
Q5
0  200  50  Q1 
2 1 0 0   T2   100 

 

1 2 1 0   T3    100 

0 1 2 1  T4   100 

 

0 0 1 1   T5   50 
0
21
EXAMPLE cont.
• Imposing Boundary Conditions
– Remove first row because it contains unknown Q1.
– Cannot remove first column because T1 is not zero.
200 
 1 2 1 0 0  
 100 
T
 0 1 2 1 0   2  100 

  T3   
100 

 0 0 1 2 1 
100
 


  T4  
 0 0 0 1 1 
 50 
 T5 
– Instead, move the first column
to the right.
2
 1
100 
0

0
100( 1 200  2  T2  1 T3 )  100
100(2  T2  1 T3 )  100  20000
1 0
0  T2  100  20000  20100 
2 1 0  T3  100   0   100 
   



1 2 1 T4  100   0   100 

0 1 1  T5   50   0   50 
22
EXAMPLE cont.
• Solution
T1  200 C, T2  203.5 C, T3  206 C, T4  207.5 C, T5  208 C
208
207
FEM
Exact
206
T
205
204
203
202
201
200
0
0.2
0.4
x
0.6
0.8
1
• Discussion
– In order to maintain 200 degree at node 1, we need to remove heat
Q1  50  100T1  100T2  350
 Q1  400 W
23
CONVECTION BC
• Convection Boundary Condition
– Happens when a structure is surrounded by fluid
– Does not exist in structural problems
– BC includes unknown temperature (mixed BC)
Wall
qh
T
T
qh  hS(T  T)
Fluid Temperature
Convection Coefficient
– Heat flow is not prescribed. Rather, it is a function of temperature on
the boundary, which is unknown
• 1D Finite Element
– When both Nodes 1 and 2 are convection boundary
 q1  hAT1  hAT1


q2  hAT2  hAT2
T1
T2
T1
T2
24
EXAMPLE (CONVECTION ON THE BOUNDARY)
• Element
equation
T 1¥
T1
h1
(1)
kA  1 1  T1  q1 
    (1) 


L  1 1  T2  q2 
T2
1
T3
2
h3
T 3¥
(2)
kA  1 1 T2  q2 
    (2) 


L  1 1  T3  q3 
• Balance of heat flow
(1)

– Node 1: q1  h1A(T1  T1 )
(1)
(2)
– Node 2: q2  q2  0
(2)

– Node 3: q3  h3 A(T3  T3 )
• Global matrix equation

 1 1 0   T1   h1A(T1  T1 ) 
kA 

  
1 2 1 T2   
0


L 
 0 1 1  T3  h3 A(T3  T3 )
25
EXAMPLE cont.
• Move unknown nodal temperatures to LHS
kA
 kA

h
A

1
L
L

2kA
  kA

L
L

kA

0


L



  T1   h1AT1 
kA    


T

0
 2 

L   

T
h
AT

 3  3 3 
kA
 h3 A 

L
0
• The above matrix is P.D. because of additional positive terms
in diagonal
• How much heat flow through convection boundary?
– After solving for nodal temperature, use
q1(1)  h1A(T1  T1 )
• This is convection at the end of an element
26
EXAMPLE: FURNACE WALL
• Firebrick
k1=1.2W/m/oC
hi=12W/m2/oC
• Insulating brick
k2=0.2W/m/oC
ho=2.0W/m2/oC
Insulating
brick
Firebrick
Ta = 20 C
Tf = 1,500
C
hi
x
0.12 m
0.25 m
0   T1  18,000 
16.8 4.8
 4.8 6.47 1.67  T    0 


 2 
 0
1.67 3.67  T3   40 
{T } T  {1,411 1,190 552}C Convection
No heat flow
boundary
1,500 C
ho
Convection
boundary
20 C
x
2
q(2)

h
(T

T
)


1054
W/m
3
0
a
3
Tf
T1
hi
T2
1
T3
2
ho
Ta
27
CONVECTION ALONG A ROD
• Long rod is submerged into a fluid
• Convection occurs across the entire surface
• Governing differential equation
d 
dT 

kA

AQ

hP
T
 T  0, 0  x  L
g


dx 
dx 


P  2(b  h)
Convection
Fluid T 
b
qi(e )
j
i
xi
h
q(je )
Convection
xj
28
CONVECTION ALONG A ROD cont.
• DE with approximate temperature
d 
dT 

kA

  AQg  hP T  T  R(x)
dx 
dx 


• Minimize the residual with interpolation function Ni(x)
 d 

dT 

x  dx  kA dx   AQg  hP(T  T) Ni (x)dx  0

i 
xj
• Integration by parts
xj
xj
xj
xj
xj
dT
dT dNi
kA
Ni (x)   kA
dx   hPTNidx    AQgNi (x)dx   hPT Nidx
dx
dx dx
xi
xi
xi
xi
x
i
29
CONVECTION ALONG A ROD cont.
• Substitute interpolation scheme and rearrange
xj
x
j
dNj  dNi
 dNi
i i  TN
j j )Ni dx
x kA  Ti dx  Tj dx  dx dx  x hP(TN
i
i
xj
  (AQg  hPT  )Ni dx  q(x j )Ni (x j )  q(x i )Ni (x i )
xi
• Perform integration and simplify
Tj 
kA
(e)  Ti
(e)
(e)
T

T

hpL


Q

q




i
j
i
i
L(e)
3
6


xj

Q(e)

(AQ

hPT
)Ni (x)dx
i
 g
xi
• Repeat the same procedure with interpolation function Nj(x)
30
CONVECTION ALONG A ROD cont.
• Finite element equation with convection along the rod
 kA
 (e)
L
 1 1 hPL(e)
 1 1   6


(e)
(e)
2 1   Ti  Qi  qi 
 1 2  T   Q(e)  q(e) 

   j   j
j 

(e)
(e)
[k (e)

}  {q(e) }
T ]  [k h ]  T  {Q
• Equivalent conductance matrix due to convection
hPL(e)
k  
6
(e)
h
 2 1
 1 2


• Thermal load vector
(e)
(e) 
Q
AQ
L

hPL
T


i
g
(e)
{Q }    
2
Q j 
1

1
31
EXAMPLE: HEAT FLOW IN A COOLING FIN
• k = 0.2 W/mm/C, h = 2104 W/mm2/C
• Element conductance matrix
0.2  200  1 1 2  10 4  320  40 2 1
[k ]  [k ] 



 1 2
40
6

1
1




(e)
T
(e)
h
• Thermal load vector
Convection
T  = 30 C
2  10  320  40  30 1
{Q } 

2
1
4
(e)
160 mm
• Element 1
330 C
x
1.25 mm
Insulated
120 mm
T1
T2
1
T3
2
T4
3
32
EXAMPLE: HEAT FLOW IN A COOLING FIN cont.
• Element conduction equation
 1.8533 0.5733   T1  38.4  q1(1) 
 
 
– Element 1  0.5733 1.8533  T2  38.4  q(1)
2 

 1.8533 0.5733  T2  38.4  q(2)
2
 
   (2) 
– Element 2 

 0.5733 1.8533  T3  38.4  q3 

1.8533 0.5733  T3  38.4 q(3)

3
– Element 3
 0.5733 1.8533  T   38.4    (3) 

 4 
 q4 
• Balance of heat flow
– Node 1
q1(1)  Q1
– Node 2
(2)
q(1)
2  q2  0
– Node 3
(2)
q(2)

q
3
3 0
– Node 4

q(3)
4  hA(T  T4 )
33
EXAMPLE: HEAT FLOW IN A COOLING FIN cont.
• Assembly
38.4  Q1
0
0   T1  

1.853 .573

 .573 3.706 .573
76.8
0  T2  


   

76.8
 0
.573 3.706 .573  T3  


  

0
.573 1.853  T4  38.4  hA(T  T4 )
 0
• Move T4 to LHS and apply known T1 = 330
0
0  330  38.4  Q1 
1.853 .573
  76.8 
 .573 3.706 .573
T2 
0 






 

 0
.573 3.706 .573   T3   76.8 


0
.573 1.893  
 0
 39.6 

 T4 
 
• Move the first column to RHS after multiplying with T1=330
0  T2  265.89 
3.706 .573
 .573 3.706 .573  T    76.8 


 3 
 0
.573 1.893  T4   39.6 
34
EXAMPLE: HEAT FLOW IN A COOLING FIN cont.
• Solve for temperature
T1  330C, T2  77.57C, T3  37.72C, T4  32.34C
350
300
250
200
150
100
50
0
0
40
80
120
35