2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 15 - Field problems
Prof. K.J. Bathe
MIT OpenCourseWare
Heat transfer, incompressible/inviscid/irrotational flow, seepage flow, etc.
Reading:
Sec. 7.2-7.3
• Differential formulation
• Variational formulation
• Incremental formulation
• F.E. discretization
15.1
Heat transfer
Assume V constant for now:
S = Sθ ∪ Sq
θ(x, y, z, t) is unkown except θ|Sθ = θpr . In addition, q s
|Sq is also prescribed.
15.1.1
Differential formulation
I. Heat flow equilibrium in V and on Sq .
∂θ
II. Constitutive laws qx = −k ∂x
.
∂θ
qy = −k
∂y
∂θ
qz = −k
∂z
(15.1)
(15.2)
III. Compatibility: temperatures need to be continuous and satisfy the boundary conditions.
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MIT 2.094
15. Field problems
Heat flow equilibrium gives
�
�
�
�
�
�
∂
∂θ
∂
∂θ
∂
∂θ
k
+
k
+
k
= −q B
∂x
∂x
∂y
∂y
∂z
∂z
(15.3)
where q B is the heat generated per unit volume. Recall 1D case:
unit cross-section
dV = dx · (1)
(15.4)
q |x − q |x+dx + q B dx = 0
�
∂qx
q|x − q|x +
dx + q B dx = 0
∂x
(15.5)
�
−
�
∂
∂x
∂
∂x
−k
�
k
∂θ
∂x
∂θ
∂x
�
�
(15.6)
� + qB �
�=0
dx
dx
�
(15.7)
= −q B
(15.8)
We also need to satisfy
∂θ
= qS
∂n
k
(15.9)
on Sq .
15.1.2
�
θ
Principle of virtual temperatures
∂
∂x
�
∂θ
k
∂x
�
+ ··· + q
B
�
=0
(15.10)
�
( θ�S = 0 and θ to be continuous.)
θ
�
�
θ
V
∂
∂x
�
�
�
∂θ
k
+ · · · + q B dV = 0
∂x
(15.11)
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MIT 2.094
15. Field problems
Transform using divergence theorem (see Ex 4.2, 7.1)
�
�
�
�T
Sq
�
B
θ
kθ dV =
θq dV +
θ q S dSq
����
V
V
heat flow
⎛
∂θ
∂x
⎞
⎜
⎜
θ =⎜
⎜
⎝
∂θ
∂y
⎟
⎟
⎟
⎟
⎠
�
(15.12)
Sq
(15.13)
∂θ
∂z
⎡
k
k=⎣ 0
0
0
k
0
⎤
0
0 ⎦
k
(15.14)
Convection boundary condition
�
�
qS = h θe − θS
(15.15)
where θe is the given environmental temperature.
Radiation
�
� �4 �
4
q S = κ∗ (θr ) − θS
�
� �2 � � r
��
�
2
= κ∗ (θr ) + θS
θ + θS θr − θS
�
�
= κ θr − θS
(15.16)
(15.17)
(15.18)
where κ = κ(θS ) and θr is given temperature of source. At time t + Δt
�
�
�
� T t+Δt t+Δt �
S
t+Δt B
θ
k
θ dV =
θ
q dV +
θ t+Δtq S dSq
V
V
t+Δt
θ = tθ + θ
Let
t+Δt (i)
or
θ
t+Δt (0)
with
θ
=
t+Δt (i−1)
θ
(15.19)
Sq
(15.20)
+ Δθ
(i)
(15.21)
t
= θ
(15.22)
From (15.19)
�
�T
(i)
θ t+Δtk(i−1) Δθ � dV
V
�
�
�T
t+Δt � (i−1)
=
θ t+Δtq B dV −
θ t+Δtk(i−1)
θ
dV
V
V
�
�
��
�
(i−1)
(i)
S
+
θ t+Δth(i−1) t+Δtθ e − t+Δtθ S
+ ΔθS
dSq
(15.23)
Sq
where the ΔθS
(i)
term would be moved to the left-hand side.
We considered the convection conditions
�
�
�
S
θ t+Δth t+Δtθ e − t+Δtθ S dSq
(15.24)
Sq
The radiation conditions would be included similarly.
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MIT 2.094
15. Field problems
F.E. discretization
θ = H1x4 · t+Δtθ̂4x1
t+Δt
t+Δt �
θ2x1
t+Δt S
for 4-node 2D planar element
(15.25)
= B2x4 · t+Δtθ̂4x1
S
θ =H ·
(15.26)
t+Δt
θ̂
(15.27)
For (15.23)
�
⎛
⎞
�
gives
(i)
θ t+Δtk(i−1) Δθ � dV =⇒ ⎝ ����
B T t+Δtk(i−1) ����
B dV ⎠ Δ ����
θ̂ (i)
� �� �
V
�T
V
�
4x2
θ t+Δtq B dV ⇒
V
�
2x4
2x2
(15.28)
4x1
H T t+Δtq B dV
(15.29)
V
�
θ
� T t+Δt (i−1) t+Δt � (i−1)
k
θ
��
dV ⇒
V
B T t+Δtk(i−1) BdV
�
V
�
θ
S T t+Δt (i−1)
h
�
t+Δt e
θ −
�
t+Δt S (i−1)
θ
t+Δt (i−1)
�
+ ΔθS
(i)
��
dSq
θ̂
��
known
(15.30)
�
=⇒
Sq
⎛
�
⎞⎞
(15.31)
T
Sq
15.2
⎛
S t+Δt (i−1)
h
H S ⎝ t+Δtθ̂ e − ⎝ t+Δtθ̂ (i−1) +Δ ����
H
θ̂ (i) ⎠⎠ dSq
� �� �
����
� �� �
� �� �
4x1
1x4
4x1
4x1
4x1
Inviscid, incompressible, irrotational flow
2D case: vx , vy are velocities in x and y directions.
�·v =0
∂vx
∂vy
+
=0
∂x
∂y
∂vx
∂vy
−
=0
∂y
∂x
or
Reading:
Sec. 7.3.2
(15.32)
(incompressible)
(15.33)
(irrotational)
(15.34)
Use the potential φ(x, y),
vx =
⇒
∂φ
∂x
vy =
∂φ
∂y
(15.35)
∂2φ ∂2φ
+ 2 = 0 in V
∂x2
∂y
(15.36)
(Same as the heat transfer equation with k = 1, q B = 0)
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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