Civil Engineering Reference
In-Depth Information
Second, element 3 is subjected to convection on both lateral surfaces as well as the two
edges defined by nodes 8-9 and 6-9. Consequently, three integrations are required as
follows:
2
H
(
e
)
h
{
T
}−
T
a
)d
A
(
e
)
=
h
([
N
]
+
h
([
N
]
{
T
}−
T
a
)d
A
8
−
9
A
(
e
)
A
8
−
9
+
h
([
N
]
{
T
}−
T
a
)d
A
6
−
9
A
6
−
9
where
A
(
e
)
is element area in the
xy
plane and the multiplier in the first term (2) accounts
for both lateral surfaces.
Transforming the first integral to normalized coordinates results in
1
1
1
1
1
I
1
=
2
hab
([
N
]
{
T
} −
T
a
)
d
r
d
s
=
2
hab
[
N
]
d
r
d
s
{
T
} −
2
habT
a
d
r
d
s
−
1
−
1
−
1
−
1
−
1
1
1
2
hA
4
=
[
N
]
d
r
d
s
{
T
} −
2
hAT
a
−
1
−
1
Therefore, we need integrate the interpolation functions only over the area of the element,
as all other terms are known constants. For example,
1
1
1
1
1
1
(1
−
r
)
2
2
(1
−
s
)
2
2
1
4
(1
1
4
N
1
d
r
d
s
=
−
r
)(1
−
s
)d
r
d
s
=
1
=
1
−
−
1
−
1
−
1
−
1
−
1
An identical result is obtained when the other three functions are integrated. The integral
corresponding to convection from the element lateral surfaces is then
2
hA
T
(3)
+
T
(3)
2
+
T
(3)
3
+
T
(3)
4
1
I
1
=
−
T
a
4
The first term in the parentheses is the average of the nodal temperatures, and this is a
general result for the rectangular element. Substituting numerical values
111
.
982
+
90
.
966
+
89
.
041
+
106
.
507
4
68
2(50)(1)
2
144
I
1
=
−
=
21
.
96 Btu/hr
Next, we consider the edge convection terms. Along edge 8-9,
I
2
=
h
([
N
]
{
T
}−
T
a
)d
A
8
−
9
A
8
−
9
and, since
r
=
1
along that edge,
d
A
8
−
9
=
tb
d
s
,
and the integral becomes
1
I
2
=
htb
([
N
r
=
1
]
{
T
}−
T
a
d
s
)
−
1
1
1
1
4
[0
=
htb
1
−
s
1
+
s
0] d
s
{
T
}−
htbT
a
d
s
−
1
−
1
