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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