Chapter 13 Staged Distillation Page 13-1
Chapter 13 Staged Distillation
1. Pentane-heptane distillation
F, xF = 0.3,
q = 0.5
D, xD= 0.95,
B, xB= 0.1
From Eq. 10.1-1 and 10.1-2,
0.3F = 0.1B + 0.95D
We found that B = 0.765F, D = 0.235F
The feed line is from Eq. 10.3-12,
We need to find the minimum RD. The feed line intercepts with
the equilibrium line at (0.126, 0.474). Thus the minimum RD is
0.474 – 0.126 = 1.37 => R = 1.3R = 1.78
The operating line of rectifying section is Eq. 10.1-8,
Which intercepts with the feed line at (0.157, 0.443). This gives the operating line of
stripping section as the line passing through (0.157, 0.443) and (0.1, 0.1):
0
0.8
1
0 0.2 0.4 0.6 0.8 1
x
(a) The plot shows there are 5 stages.
n xn yn
(b) The feed stage is the 4th plate.
(c) For = 0.6, the equilibrium curve is moved
downward, as the broken line shows.
2. Distillation with two feeds
a) To find the distillate flow rate, we simply write overall and species balances:
Chapter 13 Staged Distillation Page 13-2
BDFF
21
s
D3.85
85.0
b) Based on the concentration and state of the feeds, one can quickly see that the 100 mol/s feed
must be fed higher than the 70 mol/s feed. To find the minimum reflux ratio, we construct a q
line for the larger feed stream. Since it is a saturated liquid, the q line is a vertical line at a
liquid concentration of 0.6. We then draw an operating line between the point (xD, xD) and the
intersection of the q line and the equilibrium line. Extending this line to the y axis gives an
intercept of about 0.675. A mass balance on the top of the column gives:
F1, sat. liq.
G
G
L
L + F1
L + F1
B, xB
D, xD
G
G
G
Manipulation gives:
D
D
D
R
R
RL
D
L
DL
DL
L
G
L
1
1
1
1
1
11
Substituting, we have:
D
D
D
D
x
R
R
Solving for the reflux ratio, we have:
675.0
675.0
min, D
D
Since we want 1.5 times the minimum, our reflux ratio will be:
The rectifying operating line is therefore:
c) The stripping line is harder. A mass balance on the bottom of the column and the reboiler
gives:
Manipulation gives:
D
D
L
D
DL
Substituting we have:
22
FRD
D
D
From above:
s
Chapter 13 Staged Distillation Page 13-4
The operating line is therefore:
7.8405.0
1005.03.85
0.6
0.7
d) Given the operating lines above, we simply connect the two endpoints to form a third
operating line and step off the stages. This separation requires eight stages including the
reboiler.
e) The plot shows that the 100 mol/s feed should be on stage three and the 70 mol/s feed should
be on stage seven.
Chapter 13 Staged Distillation Page 13-5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
y, y
n+1
3. Methanol-water distillation using an old still
First we calculate the minimum RD. Assuming the feed is saturated liquid, the feed line
intercepts with the equilibrium line at (0.32, 0.681). Thus
From Eq. 10.1-8, the operating line for the rectifying section is
1 + 0.913 + 0.913
1 + 0.913 xn = 0.470 + 0.477 xn
This line intercepts with the feed line at (0.32, 0.623). The operating line for the stripping
section is from Eq. 10.1-13, however, because xB is not given, we cannot write it out at this
moment. This can be done by trial-and error method. In the following figure, we assume xB
= 0.2, then B = 994 mol/hr, and D = 206 mol/hr. The operating line for the stripping section is
Chapter 13 Staged Distillation Page 13-6
then
yn+1 = -0.505 + 3.527 xn
The first 3 stages have = 0.5, thus the
equilibrium line is lower and plotted as the
broken in the figure.
0.6
0.8
1
x
n xn yn
4. Acetone-ethanol distillation using an old still
The distillation uses high RD, therefore the
operating line is the diagonal line. Knowing
that xB = 0.1, the concentration of each stage
can be obtained graphically, from bottom to the
top, as shown in the figure.
0.4
0.6
0.8
1
y
n xn yn
(b) From mass balance,
5. Acetone-acetic acid distillation
First we calculate the Murphree efficiency from Eq. 10.5-8. The mass transfer coefficient
Chapter 13 Staged Distillation Page 13-7
From the equilibrium data, 0.07 < m < 3.24,
From mass balance,
Thus B = 61.5 lb-mol/hr, D = 38.5 lb-mol/hr.
Hence, G = (1 + RD)D = 55.4 lb-mol/hr, L = RDD = 16.9 lb-mol/hr.
From Eq. 10.5-15, (Note that the unit of G in Eq. 10.5-15 is mol/cm2-s)
G = 55.4*453.6/3600
(15)2 = 0.01 mol/cm -s
2
=> = 1 – exp(-Kyal/G) = 0.58 ~ 0.91
The operating line for the rectifying section is
and the operating line for the stripping section is
The following table lists the concentration of
each stage.
0.8
1
n xn yn
The distillation needs 10 stages. The 3rd stage is the feed stage.
6. Methanol-chlorobenzene separation
(a) No refluxing case:
Based on a mass balance, we have
Chapter 13 Staged Distillation Page 13-8
(b) One stage case:
Because the reflux ratio is large, the operating line is close to the diagonal:
The equilibrium line is linear at this range and passes through (0.18, 0.82) and (1, 1):
y* = 0.8 + 0.2x
(c) Three-stage case
For multiple stages, we use the Kremser Equation, Eq. 10.4-12
=> y1 = 0.99856