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Chemical Engineering 150B Fall 2005
Problem Set #11
Solutions
(165 point assignment)
Problem 1.
(40 Points) A distillation operating will separate a mixture of 8 components with the
following feed rates at 700 kPa:
Compound
f
i
(kmol /hr)
C
3
2500
iC
4
400
nC
4
600
iC
5
100
nC
5
200
nC
6
40
nC
7
50
nC
8
40
It is known that the heavy key is iC
5
, which will have a distillate rate of 15 kmol/hr, and the light key is
nC
4
, which will have a bottoms rate of 6 kmol/hr.
Furthermore, assume for all parts of this problem that
only the light key and heavy key distribute; all other components do not distribute.
Determine the
following:
(a)
the minimum number of equilibrium stages and the distribution of nonkey components by the
Fenske equation.
(b)
what class of separation is this?
(c)
the minimum external reflux rate and distribution of nonkey components at minimum reflux by the
appropriate Underwood equation (see class notes from Wed, November 23 for handling
distributing keys and nondistributing other components) if the feed is a bubblepoint liquid at
column pressure.
Hint: You may find the following equation useful:
3
12
61
2
ln
ln
P
TT
TP
a
aa
Ka
a
P
P
=+++
+
,
where P is in psia, T is in
o
R, and the constants are the following:
Compound
a
T1
a
T2
a
T6
a
P1
a
P3
C
3
970,688.5626
0
7.15059
0.76984 6.90224
iC
4
1,166,846
0
7.72668
0.92213
0
nC
4
1,280,557
0
7.94986
0.96455
0
iC
5
1,481,583
0
7.58071
0.93159
0
nC
5
1,524,891
0
7.33129
0.89143
0
nC
6
1,778,901
0
6.96783
0.84634
0
nC
7
2,018,803
0
6.52914
0.79543
0
nC
8
0
7,646.81641
12.48457 0.73152
0
In addition, you might find MathCAD useful to find the bubblepoint temperature of the feed, bottoms,
and distillate streams, so that you can determine the Kvalue of each component in each stream.
This will
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allow the calculation of volatility of each stream for various components to help you use the Fenske and
Underwood equations.
Part a:
To use the Fenske equation, we need the geometric mean of the relative volatilities of the distillate and
bottoms stream.
In order to get a rough estimate of these values, we will use an extremely crude estimate
for the separation.
We will assume that anything lighter than nC
4
(which is iC
4
and C
3
) only appears in
the distillate, while anything heavier than iC
5
(which include nC
5
, nC
6
, nC
7
, and nC
8
) only appear in of the
bottoms stream.
nC4 and iC5 will have the flow rates specified in the problem statement.
Therefore, the
distillate stream will have the following flow rates:
Compound
d
i
(kmol /hr)
z
i
(kmol /hr)
C
3
2500
0.712
iC
4
400
0.114
nC
4
594
0.169
iC
5
15
0.005
nC
5
0
0
nC
6
0
0
nC
7
0
0
nC
8
0
0
and the bottoms flow rates are the following:
Compound
b
i
(kmol /hr)
z
i
(kmol /hr)
C
3
0
0
iC
4
0
0
nC
4
6
0.014
iC
5
85
0.202
nC
5
200
0.475
nC
6
40
0.095
nC
7
50
0.119
nC
8
40
0.095
Using these flow rates, we can calculate the mole fraction of each stream, which is shown to the right on
the tables.
These numbers will be useful in order to estimate the relative volatility between species in the
distillate and bottoms streams.
First of all, we must calculate the temperature of the above streams by
doing a bubble point temperature calculation at a given pressure, which we know is 700 kPa (101.5 psia).
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 Spring '08
 Bell
 Distillation

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