Civil Engineering Reference
In-Depth Information
72
F
180
F
40
F
4 in.
(a)
1
2
3
4
5
q
h
(
T
5
40)
1
2
3
4
T
180
F
(b)
Figure 7.4
Example 7.3: (a) Cylindrical pin.
(b) Finite element model.
The exterior surface of the pin is in contact with moving air at
72
◦
F
. The physical data are
given as follows:
D
=
0.5 in.,
L
=
4 in.,
k
x
=
120 Btu/(hr-ft-
◦
F
),
50 Btu/(hr-ft
2
-
◦
F
),
100 Btu/(hr-ft
2
-
◦
F
)
h
air
=
h
water
=
Use four equal-length, two-node elements to obtain a finite element solution for the
temperature distribution across the length of the pin and the heat flow rate through the pin.
■
Solution
Figure 7.4b shows the elements, node numbers, and boundary conditions. The boundary
conditions are expressed as follows
At node 1:
T
1
=
180
◦
F
5
=−
q
5
=−
h
(
T
5
−
40)
k
x
d
T
d
x
At node 5:
Element geometric data is then
1963
in.
2
L
e
=
1 in.,
P
=
(0
.
5)
=
1
.
5708
in.,
A
=
(
/
4)(0
.
5)
2
=
0
.
The leading coefficients of the conductance matrix terms are
120
0
.
1963
144
k
x
A
L
e
=
=
1
.
9630
Btu/(hr-
◦
F)
1
12
50
1
1
12
.
5708
12
h
air
PL
e
6
=
0
.
0909
Btu/(hr-
◦
F)
=
6
where conversion from inches to feet is to be noted.
