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Covers topics like multicomponent distillation and liquid liquid extraction
Typology: Assignments
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Question Paper
Answer Booklet
Question 1
One hundred kilograms of a 50% solution of C in A is equilibrated with 70 kg of solvent B
containing 2% of C. At equilibrium, the raffinate phase has a mass of 80 kg and has 52% A and
8% B in it. What is the selectivity or separation factor of C? (10)
Solution:
β
C , A
y
C
y
A
x
C
x
A
At equilibrium:
R = 80 kg x
RA
=0.52 x
RC
Total balance on the system:
E = 90 kg
Component balance on C
x
Fc
F + S y
SC
= E y
EC
RC
0.5 × 100 + 70 × 0.02= 90 × y
EC
y
EC
Calculating amount of A in extract via solvent balance:
Amount of solvent initially fed = 70 × 0.98=68.6 kg
Amount of solvent ∈ raffinate = 80 × 0.08=6.4 kg
Amount of solvent ∈ theextract =68.6−6.4=62.2 kg
Amount of A ∈ the extract = 90 −62.2−( 0.215 × 90 )=8.45 kg
y
A
β
C , A
Q 1.2 (provide graph paper to students)
500 kg/h of a 40%wt mixture of DPH in Docosane is to be extracted in a counter current system
with 500 kg/h of solvent containing 98% wt of Furfural and 2% DPH to produce a raffinate that
contains only 5%wt DPH. The equilibrium and tie-line data is given below.
Equilibrium data (mass%)
C , F
C , S
C , M
C , M
C , M
Point M is located on FS line and E 1
by joining R N
M to cut equilibrium line. The operating point
(∆) is then located by extending lines FE 1
and R
N
S to intersection point. R
n
located using tie-
lines. Graphical solution shown below: [12]
From the graph, 4 stages are required. [2]
Question 2
a) A mixture of hydrocarbons containing 32 mole% of n-hexane, 38 moles% n-heptane and 30
moles% n-octane is to be distilled to obtain a distillate product with 0.05 mole fraction of n-
heptane and a bottom product of 0.05 mole fraction of n-hexane. The column operating
pressure and temperature are 1.2 atm and 102
o
C respectively. 40
%
of the feed is to be
vaporized. Calculate the product composition and the minimum number of ideal plates. NB:
The K i
values for n-hexane, n-heptane and n-octane are 2.08, 0.92 and 0.42 respectively at
o
Show your working clearly and be logically.
First identify the key and non-key components
Light Key (LK) Heavy Key (HK) Heavier than Heavy Key (HHK)
n-hexane n-heptane n-octane
Assumptions: Non-distributing assumption, thus (1) no n-octane in the distillate (2) 0.
mole fraction of n-hexane is in the distillate. Assume basis: 100 moles/h of the feed rate.
Overall mass balance: F = D+W
Component mass balance (Light key-n-hexane): Fx lk
= Dx lk
+ Wx lk
The composition of the bottom product can be calculated using the conditions given. These
conditions include that all n-octane, all but 0.05 D of the n-heptane, 0.05 mole fraction
of n-hexane.
In turn the following compositions can be deduced:
Keys Components
LK 32 28.5 0.95 3.5 0.05 2.
HK 38 1.5 0.05 36.5 0.521 0.
HHK 30 0 0 30 0.429 0.
Total - 100 30 1 70 1 -
And now use the Fenske Equation (below) to obtain the minimum number of plates
min
=ln ¿ ¿
min
ln (
ln (2.26)
=6.5 7 ( Including thereboiler )
b) The actual number of stages and minimum reflux ratio of the above separation in determined
to the 9 and 1.58 respectively. Using the correlation below determine the actual reflux ratio
Calculate (
min
) using the value of the actual number of stages provided i.e.
Interpolate the value of
min
R
s
Solve the above equations simultaneously to get
s
=3.9 4 stages below feed tray , therefore N
R
