Environmental Engineering Reference
In-Depth Information
The total homogeneous solution becomes:
k
y
n
n
=
C
i
α
i
.
i
=
1
If two roots are equal (
α
j
, for example):
y
n
=
α
j
(
C
1
n
+
C
2
)
.
D.2
Application to separation process
Now, let us apply the mathematical approach to an equilibrium-stage process illustrated
schematically in Figure D.1.
A mass balance around stage (
n
+
1), assuming
L
and
V
are constant, gives:
Vy
n
+
2
+
Lx
n
=
Vy
n
+
1
+
Lx
n
+
1
.
We further assume:
y
n
+
1
=
m
x
n
+
1
(linear equilibrium relation).
The mass balance equation becomes:
L
mV
.
y
n
+
2
−
(1
+
A
)
y
n
+
1
+
Ay
n
=
0;
A
=
n
.
Solution is of form:
y
n
=
C
α
Substituting:
n
+
2
n
+
1
n
C
α
−
(1
+
A
)
C
α
+
AC
α
=
0
.
n
:
Dividing by
C
α
2
α
−
(1
+
A
)
α
+
A
=
0
.
The two values of which satisfy this equation are
α
=
1
,
A
.
C
1
A
n
Therefore:
y
n
=
C
2
.
The constants
C
1
and
C
2
are evaluated from boundary conditions on the equilibrium-stage
process. They remain constant for each stage.
+
D.2.1
Example D.1
Problem:
Derive the Kremser equation, Equation (3.49). Refer to Figure 3.27 for a diagram of the
equilibrium-stage process.
V
,
y
n
V
,
y
n
+1
n
L
,
x
n
L
,
x
n
+1
Figure D.1
Schematic of equilibrium-stage process.
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