Chapter 5
Numerical Methods in Heat Conduction
584
A uranium plate initially at a uniform temperature is subjected to insulation on one side and
convection on the other. The transient finite difference formulation of this problem is to be obtained, and
the nodal temperatures after 5 min and under steady conditions are to be determined.
Assumptions
1
Heat transfer is onedimensional since the plate is large relative to its thickness.
2
Thermal conductivity is constant.
3
Radiation heat transfer is negligible.
Properties
The conductivity and diffusivity are given to be
k
= 28 W/m
⋅
°C and
α
=
×

12 5
10
6
.
m
/ s
2
.
Analysis
The nodal spacing is given to be
∆
x
= 0.02 m. Then the number of nodes becomes
1
/
+
∆
=
x
L
M
= 0.08/0.02+1 = 5. This problem involves 5 unknown nodal temperatures, and thus we
need to have 5 equations. Node 0 is on insulated boundary, and thus we can treat it as an interior note by
using the mirror image concept. Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the
general explicit finite difference relation expressed as
τ
i
m
i
m
i
m
i
m
i
m
i
m
T
T
k
x
g
T
T
T

=
∆
+
+

+
+

1
2
1
1
2
→
k
x
g
T
T
T
T
i
m
i
m
i
m
i
m
i
m
2
1
1
1
)
2
1
(
)
(
∆
+

+
+
=
+

+
The finite difference equation for node 4 on the right surface
subjected to convection is obtained by applying an energy
balance on the half volume element about node 4 and taking
the direction of all heat transfers to be towards the node under
consideration:
t
T
T
C
x
x
g
x
T
T
k
T
T
h
k
x
g
T
T
T
T
k
x
g
T
T
T
T
k
x
g
T
T
T
T
k
x
g
T
T
T
T
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
∆

∆
=
∆
+
∆

+

∆
+

+
+
=
∆
+

+
+
=
∆
+

+
+
=
∆
+

+
+
=
+
∞
+
+
+
+
4
1
4
0
4
3
4
2
0
3
4
2
1
3
2
0
2
3
1
1
2
2
0
1
2
0
1
1
2
0
0
1
1
1
0
2
2
)
(
:
n)
(convectio
4
Node
)
2
1
(
)
(
:
(interior)
3
Node
)
2
1
(
)
(
:
(interior)
2
Node
)
2
1
(
)
(
:
(interior)
1
Node
)
2
1
(
)
(
:
)
(insulated
0
Node
ρ
or
k
x
g
T
k
x
h
T
T
k
x
h
T
i
i
i
2
0
3
4
1
4
)
(
2
2
2
2
1
∆
+
∆
+
+
∆


=
∞
+
where
C
20
C,
W/m
35
C,
W/m
28
,
W/m
10
m,
02
.
0
2
3
6
0
°
=
°
⋅
=
°
⋅
=
=
=
∆
∞
T
h
k
g
x
, and
6
10
5
.
12

×
=
m
2
/s.
The upper limit of the time step
∆
t
is determined from the stability criteria that requires all primary
coefficients to be greater than or equal to zero. The
coefficient of
i
T
4
is smaller in this case, and thus
the stability criteria for this problem can be expressed as
1
2
2
0
1
2 1
2
1
2


≥
→
≤
+
→
≤
+
h x
k
h x
k
t
x
h x
k
∆
∆
∆
∆
∆
(
/
)
(
/
)
since
=
∆
∆
t
x
/
2
. Substituting the given quantities, the maximum allowable the time step becomes
s
6
.
15
C)]
W/m.
28
/(
m)
02
.
0
)(
C
.
W/m
35
(
1
/s)[
m
10
5
.
12
(
2
)
m
02
.
0
(
2
2
6
2
=
°
°
+
×
≤
∆

t
Therefore, any time step less than 15.5 s can be used to solve this problem. For convenience, let us
choose the time step to be
∆
t
= 15 s. Then the mesh Fourier number becomes
575
h, T
∞
∆
x
Insulated
•
•
•
•
•
0
1
2
3
4
g
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View Full DocumentChapter 5
Numerical Methods in Heat Conduction
τ
α
=
=
×
=

∆
∆
t
x
2
6
2
12 5
10
002
046875
(
.
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 Spring '11
 ENgel
 Heat Transfer, Insulation, finite difference, Trombe wall

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