Civil Engineering Reference
In-Depth Information
Eliminating the first equation while taking care to include the effect of the specified tem-
perature at node 1 on the remaining equations gives
4
.
2896
−
1
.
8571
376
.
2530
39
0
0
T
2
T
3
T
4
T
5
−
.
.
−
.
.
1
8721
4
2896
1
8721
0
2750
=
0
−
1
.
8721
4
.
2896
−
1
.
8721
39
.
2750
0
0
−
1
.
8721
2
.
2812
25
.
0917
Solving by Gaussian elimination, the nodal temperatures are obtained as
T
2
=
136
.
16
◦
F
T
3
=
111
.
02
◦
F
T
4
=
97
.
23
◦
F
T
5
=
90
.
79
◦
F
The heat flux at node 1 is computed by back substitution of
T
2
into the first equation:
2
.
1448 (180)
−
1
.
8721 (136
.
16)
=
19
.
6375
+
Aq
1
Aq
1
=
111
.
5156
Btu/hr
111
.
5156
0
.
1963
/
144
≈
81, 805
Btu/hr-ft
2
q
1
=
Although the pin length in this example is quite small, use of only four elements rep-
resents a coarse element mesh. To illustrate the effect, recall that, for the linear, two-node
element the first derivative of the field variable, in this case, the temperature gradient, is
constant; that is,
d
T
d
x
=
T
x
=
T
L
e
Using the computed nodal temperatures, the element gradients are
◦
F
in
.
d
T
d
x
=
136
.
16
−
180
1
Element 1:
=−
43
.
84
◦
F
in
.
d
T
d
x
=
111
.
02
−
136
.
16
1
Element 2:
=−
25
.
14
◦
F
in
.
d
T
d
x
=
97
.
23
−
111
.
02
1
Element 3:
=−
13
.
79
◦
F
in
.
d
T
d
x
=
90
.
79
−
97
.
23
1
Element 4:
=−
6
.
44
where the length is expressed in inches for numerical convenience. The computed gradi-
ent values show significant discontinuities at the nodal connections. As the number of
elements is increased, the magnitude of such jump discontinuities in the gradient values
decrease significantly as the finite element approximation approaches the true solution.
