Civil Engineering Reference
In-Depth Information
According to the Fourier law introduced in Chapter 4 we can write
w
u
w
u
q
(5.1)
0
0 and
w
x
w
y
k
Solving the differential equations for
u
, the temperature at a point
Q
with coordinates
x,y
is obtained as
q
(
a
)
0
u
Q
y
C
(5.2)
k
If we assume the temperature at the centre of the circle to be zero, then C= 0.
We now place a cylindrical isolator in the flow and compute how the flow pattern and
temperature distribution changes. The isolator prevents flow to occur in a direction
perpendicular to its boundary, which is computed by
w
u
w
u
w
u
t
k
kn
(
n
)
0
(5.3)
x
y
w
n
w
x
w
y
Where
n
{
n
x
,
n
y
}is the vector normal to the boundary of the isolator (
outward
normal
). Note that the positive direction of this vector is pointing from the infinite
domain into the isolator. For the solution (5.2) just obtained, we find that this condition
is not satisfied, because the flow in the direction normal to the isolator boundary
(marked with a dotted line in Figure 5.1a) is computed as:
()
a
(5.4)
t
n q
q
0
sin
I
y
0
If we want to find out how the isolator changes the flow/temperature distribution,
then we can think of the problem as divided into two parts: the first being the trivial one,
whose solution we just obtained, the second being one where the solution is obtained for
the following boundary condition:
()
b
()
a
(5.5)
t
t
q
0
sin
I
If the two solutions for the flow normal to the boundary of the isolator are added
then:
(5.6)
()
b
()
a
tt
t
0
i.e. the boundary condition that no flow occurs normal to the isolator is satisfied. The
final solution for the temperature is therefore
(
a
)
(
b
)
(5.7)
u
Q
u
Q
u
Q
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