ChE 120B
6 - 1
Conduction Separation of Variable
Consider a rod of finite length insulated at one end
Same basic equation, but quite a different physical problem.
2
2
T
T
t
x
α
∂
∂
=
∂
∂
We now have a length scale to work with so non-dimensionalize:
0
s
s
T
T
T
T
θ
−
=
−
→
Note that this
θ
is a little different than before
we want boundary conditions to equal zero.
2
2
/
p
kt
t
x
L
C
L
L
α
η
τ
ρ
=
=
So we have
2
2
θ
θ
τ
η
∂
∂
=
∂
∂
Boundary conditions:
0
at
0
for all
1
for all
for
0
0
for all
1
θ
η
τ
θ
η
τ
θ
τ
η
η
=
=
=
=
∂
=
=
∂
Assume that a solution of the form
( )
(
)
(
)
,
Y
X
τ
η
θ τ η
=
exits.
This is called a separation of variables approach.
This
will only work if the boundary conditions are of a certain type.
Check our boundary conditions.
0 at
0;
0 at
d
L
dx
θ
θ
η
η
=
=
=
=
We’ll find out why these are the right type soon.
Insert our new variables into the heat
conduction equation:
2
2
Y
X
X
Y
τ
η
∂
∂
=
∂
∂
.
Divide both sides by
XY
to get:
2
λ
−
≡
1
dY
Y
d
τ
=
2
2
1
d X
X
d
η
}
can replace
∂
by
d
because
Y
and
X
are
functions of only one
variable
can only
be true if
equal to a
constant
=
function of
time only
=
function of
position

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