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Problem 1.2.4
(a) We consider an arbitrary slice of thickness
D
x located between
x
=
a
and
x
=
b
. If
u
H
x, t
L
is the density
or concentaration per unit volume in that slice and
j
ê
is the mass flux per unit area, then the material balance is
(1)
d
ÅÅÅÅÅ
dt
Ÿ
a
b
u
H
x, t
L
A
H
x
L
„
x
= 
j
ê
ÿ
n
ê
a
A
H
a
L

j
ê
ÿ
n
ê
b
A
H
b
L
where
n
ê
a
is the outward directed normal to the face of the slice, pointing in the negative
x
direction at
x
=
0
;
n
ê
b
is the outward directed normal at the face located at
x
=
b
, pointing in the positive
x
direction;
A
H
x
L
is the
surface area of the slice. Now since
(2)

j
ê
ÿ
n
ê
a
A
H
a
L

j
ê
ÿ
n
ê
b
A
H
b
L
=
j
H
a
L
A
H
a
L

j
H
b
L
A
H
b
L
= 
Ÿ
a
b
ÅÅÅÅÅ
x
H
jA
L
„
x
We can rewrite the above balance as
(3)
Ÿ
a
b
ÅÅÅÅÅ
t
@
u
H
x, t
L
A
H
x
LD
„
x
+
Ÿ
a
b
ÅÅÅÅÅ
x
H
jA
L
„
x
=
0
Since our slice is arbitrary, the integrand under the common integral must vanish to give:
(4)
ÅÅÅÅÅ
t
@
u
H
x, t
L
A
H
x
LD
=
ÅÅÅÅÅ
x
H
j A
L
But,
(5)
j
H
x, t
L
= 
!
u
ÅÅÅÅÅÅÅ
x
Thus our material balance becomes
(6)
ÅÅÅÅÅ
t
@
u
H
x, t
L
A
H
x
LD
=
ÅÅÅÅÅ
x
H
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 Fall '09
 Stroeve

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