Environmental Engineering Reference
In-Depth Information
Solution:
(a) Mass balance (basis F
=
100 mol
/
hr) on entire column:
+ S
F
=
V D +
B
(overall)
( since S
F
=
B
=
V D )
y S S
x F F
+
=
y D V D +
x B B
y D V D =
0
.
95 x F F
0
.
05 x F F
B
0
.
05 x F F
F
10 3
x B =
=
=
0
.
05 x F =
1
.
5
×
.
Mass balance around control volume:
(0) S
y n S
+
x n 1 B
=
x B B
+
B
S x n 1
B
S x B .
y n =
Minimum S corresponds to the maximum slope of the operating line.
To determine the maximum value of B
/ S , substitute values for y n and x n 1 which
correspond to intersection of operating and equilibrium line. Rearranging the mass
balance equation:
B
S (0
B
S (1
10 3 )
8
.
3 (0
.
03)
=
.
03)
.
5
×
S
B =
S
F =
10 3
0
.
03
1
.
5
×
=
0
.
11 moles steam min /
moles feed
.
0
.
27
(b) Actual steam rate
=
2.5 times the minimum.
11) 100
feed
mol
hr
mol
hr
S
=
2
.
5(0
.
=
27
.
5
B
S
100 mol/hr
27.5 mol/hr =
actual =
3
.
6
.
Actual operating line is:
10 3 )
10 3
.
The line can be graphed by plotting two points. We already know one set of points:
(0, 1.5
y
=
3
.
6 x
3
.
6(1
.
5
×
=
3
.
6 x
5
.
5
×
10 3 ). The other will be at x
×
=
0.03:
10 3
y
=
3
.
6(0
.
03)
5
.
5
×
=
0
.
10
.
From Figure 4.18, three theoretical stages are required.
Side streams (see Figure 4.19)
1 The operating line for the top and bottom sections are still drawn as before.
2 The q -line for the feed is still drawn as before.
3Tolocate the middle operating line x S must be known. The middle operating line begins
at the top operating line at the point x
x S and ends at the point where the q -line for
the feed intersects the bottom operating line.
=
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