Environmental Engineering Reference
In-Depth Information
Solution:
The boundary conditions are:
n
=
0;
x
=
x
0
n
=
N
+
1;
y
=
y
N
+
1
.
C
1
A
n
+
1
The general solution is
y
n
=
+
C
2
.
In Chapter 2, we defined:
y
1
=
mx
0
(value of
y
which would be in equilibrium with
x
0
)
=
y
0
y
1
y
0
=
=
C
1
+
C
2
C
1
A
n
+
1
y
N
+
1
=
+
C
2
.
We can immediately write:
A
n
+
1
C
1
(
A
n
+
1
y
N
+
1
−
y
1
−
A
)
A
−
y
1
=
1)
=
A
n
+
1
.
C
1
(
A
n
+
1
y
N
+
1
−
−
1
−
This is a similar equation to that given in Section 3.7.
Example D.2: distillation, separation of H
2
18
O from H
2
16
O
D.2.2
Problem:
Water contains 0.002 mole fraction H
2
18
O. Using the Fenske equation, estimate the number
of stages required to separate H
2
18
O from H
2
16
O and compare to the result from the Smoker
equation.
Overall mass balance on entire column section (Figure D.2):
Vy
n
−
1
=
Lx
n
+
Dx
D
.
K
A
K
B
=
y
(1
−
x
)
Relative volatility:
α
=
y
)
, assumed constant.
x
(1
−
V
D
,
x
D
L
V
,
y
n
− 1
L
,
x
n
Figure D.2
Enriching section of distillation column.
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