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Chapter 2
Heat Conduction Equation
Applying the other boundary condition at
rr
=
1
,
B. C. at
:
=
1
T
g
k
rC
CT
g
k
r
II
=−
+
→
=
+
&&
44
1
2
22
2
1
Substituting this
C
relation into Eq. (
b
) and rearranging give
2
TrT
g
k
I
wire
wire
()
&
(
=+
−
4
1
)
(
c
)
Plastic layer
The mathematical formulation of heat transfer problem in the plastic can be expressed as
0
=
⎟
⎠
⎞
⎜
⎝
⎛
dr
dT
r
dr
d
with
Tr
and
T
I
1
=
−=
−
∞
k
dT r
dr
hTr
T
[( )
]
2
2
The solution of the differential equation is determined by integration to be
r
dT
dr
C
=
1
→
dT
dr
C
r
=
1
→
C
r C
ln
=
+
12
where
C
and
C
1
2
are arbitrary constants.
Applying the boundary conditions give
r
=
r
1
:
Cr
11 2
2
11
ln
ln
+=
→
=
−
r
=
r
2
:
+
−
∞
k
C
r
hC r C
T
1
2
12 2
[(
ln
)
]
→
C
TT
r
r
k
hr
I
1
2
=
−
+
∞
ln
Substituting
C
1
and
C
2
into the general solution, the variation of temperature in plastic is determined to be
C
r
T
C
r
T
r
r
k
hr
r
r
I
plastic
plastic
ln
ln
−
=
+
−
+
∞
1
2
1
We have already utilized the first interface condition by setting the wire and ceramic layer temperatures
equal to
T
at the interface
. The interface temperature
is determined from the second interface
condition that the heat flux in the wire and the plastic layer at
I
=
1
T
I
=
1
must be the same:
−
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Mass, Heat

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