2.092/2.093 — Finite Element Analysis of Solids & Fluids I
Fall ‘09
Lecture 11 - Heat Transfer Analysis
Prof. K. J. Bathe
MIT OpenCourseWare
Reading assignment: Sections 7.1-7.4.1
To discuss heat transfer in systems, first let us define some variables.
θ(x, y, z, t)
Sθ
Sq
=
=
=
Temperature
Surface area with prescribed temperature (θp )
Surface area with prescribed heat flux into the body
Given the geometry, boundary conditions, material laws, and loading, we would like to calculate the tem­
perature distribution over the body. To obtain the exact solution of the mathematical model, we need to
satisfy the following in the differential formulation:
• Heat flow equilibrium
• Compatibility
• Constitutive relation(s)
Example: One-Dimensional Case
We can derive an expression for system equilibrium from the heat flow equation.
�
�
�
�
q � −q �
+q B dx = 0
x
x+dx
Using the constitutive equation,
∂θ
∂x
�
�
∂q ��
�
�
q�
=q� +
� dx
∂x x
x+dx
x
�
�
∂θ
∂θ
∂2θ
−k
+k
+
dx + q B dx = 0
∂x ∂x2
∂x
q = −k
1
Lecture 11
We obtain the result
Heat Transfer Analysis
2.092/2.093, Fall ‘09
∂2θ
k 2 + q B = 0 in V
∂x
�
�
∂θ ��
�
�
k
, θ�
=0
� = qS �
∂x L
L
x=0
Principle of Virtual Temperatures
Clearly:
� 2
�
∂ θ
θ¯ k 2 + q B = 0
∂x
� 2
�
�
�
� L � 2
∂ θ
∂ θ
B
B
¯
¯
θ k 2 +q
θ k 2 +q
dx = 0
dV = A
∂x
∂x
V
0
Hence,
�
0
L
� 2
�
∂ θ
θ¯ k 2 + q B dx = 0
∂x
2
(A)
Lecture 11
Heat Transfer Analysis
�L � L �
�
� L
∂θ¯ ∂θ
∂θ ��
¯
¯ B dx = 0
−
k
dx
+
θk
θq
∂x ∂x
∂x �0
0
0
�
� L � ¯� � �
� L
� S� �
∂θ
∂θ
¯ B dx + θq
¯
�
θq
k
dx =
�
∂x
∂x
0
0
x=L
2.092/2.093, Fall ‘09
(B)
In 3D, the equation becomes
�
�T
�
�
θ kθ dV =
V
�
T B
⎡
k
k=⎣ 0
0
⎤
0
0 ⎦
k
0
k
0
θ̄T q S dSq
θ̄ q dV +
V
(C)
Sq
⎡
∂θ
∂x
∂θ
∂y
∂θ
∂z
⎢
⎢
; θ� = ⎣
⎤
⎡
⎥
⎥
⎦
⎢
�
; θ =⎢
⎣
∂θ̄
∂x
∂θ̄
∂y
∂θ̄
∂z
⎤
⎥
⎥
⎦
For example:
�
�
q S = h θe − θS → convection
�
�
q S = κ θr − θS → radiation
θe = temperature of the environment
θr = temperature of the radiation source
θ(m) (x, y, z, t) = H (m) θ
θ
�
(m)
= B (m) θ
; θS(m) = H S(m) θ
; θ̄
�
(m)
= B (m) θ¯
Substituting this into (C), we obtain
Kθ = Q
K = ΣK (m)
m
; K (m) =
�
V
B (m)T k(m) B (m) dV (m)
(m)
Q = QB + QS
(m)
QB = ΣQB
m
QS =
(m)
ΣQS
m
(m)
; QB
;
(m)
QS
�
H (m)T q B(m) dV (m)
=
V (m)
�
HS
=
(m)
T
(m)
Sq
� � e
��(m)
h θ − θS
dSq (m)
θS is unknown, so
(m)
QS
=
�
(m)
Sq
HS
(m)
Here we need to sum over all Sq
(m)
T
(m)
hθe(m) dSq
for element (m).
3
−
��
(m)
Sq
HS
(m)
T
hH S
(m)
(m)
dSq
�
θ
Lecture 11
Heat Transfer Analysis
2.092/2.093, Fall ‘09
Example
⎡
θ(x, y) =
�
h1
h2
h3
⎤
θ1
� ⎢ θ2 ⎥
⎥
h4 ⎢
⎣ θ3 ⎦
θ4
1�
x�
1+
(1 + y) ; h2 = . . .
h1 =
4
2
⎤
⎡
⎤ ⎡
∂θ
h
h
h
h
2,x
3,x
4,x
⎣ ∂x ⎦ = ⎣ 1,x
⎦θ
∂θ
h
h
h
h
1,y
2,y
3,y
4,y
∂y
�
��
�
B
�
H S = H �y=1
→
�
1� �
HS =
1 + x2
2
4
�
1−
x
2
�
0
0
�
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2.092 / 2.093 Finite Element Analysis of Solids and Fluids I
Fall 2009
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