Seminar 9: Batch rectification
Example 1: Batch rectification with constant reflux.
20kmol of a methanol solution in water (40 mol.% methanol) is distillated by differential
distillation in a column with 4 theoretical stages including still. The process is finished when
the mol friction of methanol decreases to 0.03. The reflux ratio is constant and has a value of
R=3. The column works at normal pressure. Calculate:
1. amount of distillate and bottoms
2. average composition of distillate
3. the total amount of vapors condensed in total condenser.
Equilibrium data of methanol-water system at 101.3 kPa:
x.100
y.100
0
0
3.2
19.0
5.2
29.4
5.9
30.8
7.5
35.2
8.8
39.0
Solution:
1. Scheme and material balance:
D
W
F
Q
At the end of the process:
nF=nW+nD (1)
nFxF=nDxD+nWxW (2)
Rayleigh equation:
x
n F F dxW
(3)
ln

nW xW y D  xW
2. draw x- y diagram
15.4
49.0
18.2
55.2
22.5
59.3
29.0
64.3
34.9
70.3
59.0
83.0
81.3
91.8
91.8
96.3
100
100
100
90
80
70
60
50
40
30
20
10
0
0
20
40
60
80
100
2. Choose different composition of distillate
3. Calculate xD/R+1 for each xD
4. Draw operating line than 4 steps between the equilibrium curve and operating line
starting from xD
5. Take of the corresponding xw
100
95
90
85
80
75
70
65
60
55
50
45
40
35
30
25
20
15
10
5
0
0
5
10
15
20
25
Example for xD=0.90
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
xD
xD/R+1
Received
xW
1/xD-xW
0.6
0.15
0.03
0.75
0.1875
0.036
0.80
0.2
0.045
0.90
0.225
0.1
0.92
0.230
0.140
0.94
0.235
0.230
0.96
0.240
0.400
1.75
1.4
1.33
1.25
1.28
1.408
1.79
6. Calculate the right side of equation 3 using a numerical or graphical method.
2
1.8
1.6
1/xD-xW
1.4
1.2
1
0.8
J=0.536
0.6
0.4
0.2
0
0
0.05
0.1
0.15
0.2
0.25
xW
For example using trapezium rule
xF
f ( x) i 1  f ( x) i
dxW
xi 1  xi  =0.563
x y D  xW = i
2
W
7. calculation of nW, nD, xD, and nV
n
ln F  0.563
nW
nF
 1.702  nW  11.75kmol
nW
n D  n F  nW  8.25kmol
n F x F  nW xW
 0.927
nD
nV =nD(R+1) = 33 kmol
xD 
0.3
0.35
0.4
0.45
Example 2: Batch rectification with constant distillate composition
A mixture of benzene and toluene in an amount of 1200 kg and mass friction of benzene
0.317 is distillated by differential rectification into a distillate with 93 wt% of benzene
(During the process the distillate composition is constant). The rectification is finished when
the friction of benzene in the reboiler decreases under 3.4 wt.%. At this composition of the
bottoms the R/Rmin =2. and average efficiency of the column is 70 %. The pressure is 101.3
kPa. Calculate:
1. The amounts of distillate and bottom
2. The Number of real stages in the column
3. Cooling water requirement in total condenser, if its temperature is changed by 38 oC.
4. Requirement of vapor with a pressure of 0.2 Mpa.
Equilibrium data:
x.
y.
t/oC
0
0
110.6
0.1
0.208
106.1
0.2
0.372
102.2
0.3
0.507
98.6
0.397
0.618
95.2
Heats of evaporation of pure components:
t/oC
20
60
ΔvhB [kJ/kg] 435.3
408.1
ΔvhT [kJ/kg] 407.3
388.5
0.489
0.710
92.1
0.592
0.789
89.4
0.7
0.853
86.8
100
378.8
368.4
Solution:
Scheme and material balance:
D
W
F
Q
At the end of the process:
mF=mW+mD
mFwF=mDwD+mWwW
m ( w  wW )
mD  F F
 379kg
wD  wW
mW  mF  mD  821kg
Conversion of mass basis parameters to mol basis parameters:
0.803
0.914
84.4
0.903
0.957
82.3
0.95
0.979
81.2
140
345.8
343.7
1
1
80.2
x FA
wFA
MA

 0.354 , by the same way xDA= 0.940, and xWA=0.04
wFA wFB

MA MB
nD 
mD
mD

 4.80kmol
M D M A xD A  M B xD B
M A  78kg / kmol
M B  92kg / kmol
nF 
mF
mF

 13.79kmol
M F M A xF A  M B xF B
nW  nF  nD  8.99kmol
Calculation of Rmin:
- Draw x-y diagram
- Draw the operating line for Rmin and xD at the end of the process
1
0.95
0.9
0.85
0.8
0.75
0.7
0.65
0.6
y
0.55
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
x
From the picture above
xD
Rmin  1
 0.05
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
xD
 1  17.8
0.05
Calculation of R:
R  2.Rmin  35.6
Rmin 
xD
 0.026
R 1
Draw the operating line for R and xD at the end of the process
Draw steps between operating line and equilibrium curve.
The operating line section on axis y is
1
0.95
0.9
0.85
0.8
0.75
0.7
0.65
0.6
y
0.55
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
x
The number of steps=7.8
NTS=7.8-1=6.8
Number of real stages
NTS
NRS 
 10

Calculation of total amount of vapor produced in the still:
xF
R  1 dx (1)
nV  n F x D  x F  
n 1
2
xw  x D  x n 1 
Calculation steps:
x
1. choose a number of R , D , respectively.
R 1
xD
2. Calculate
, R respectively.
R 1
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
3. For each R, draw the calculated number of steps between the equilibrium curve and
operating line (in our case it is 8 steps) and determine xn+1
R  1 , for each R
4. Calculate the under integral function
xD  xn1 2
5. Using numerical integration (for example trapezium method ) calculate the integral
xF
R  1
x xD  xn1 2 dxn1
w
6. calculate nV in equation (1)
1
0.026
0.1
0.15
0.2
0.3
2
xD
R 1
R
35.6
8.4
5.25
3.69
2.13
3
xn+1
0.04
0.105
0.158
0.212
0.331
45.2
13.5
10.2
8.85
8.44
R  1
4
xD  xn1 
2
50
45
R+1/(xD-xn+1)
40
35
30
25
20
15
10
5
0
0
0.05
0.1
0.15
0.2
xn+1
xF
5.
J= 
xw
R  1
xD  xn1 2
dxn 1 = 3.82
xF
6. nV  n F x D  x F  
xw
R  1
x D  xn1 2
dxn1 = 30.87 kmol.
Calculation of cooling water requirement:
0.25
0.3
0.35
Assuming only the condensation of vapor in condenser and negligible heat losses the
condenser heat duty can be calculated by:
Qc  nV  v hD
The distillate temperature can be obtained from the t-x,y diagram tD=81.5 oC. Graf
The condensation heat of individual components at this temperature are:
 v h A =392.35 kJ/kg =30 600 kJ/ kmol
 v h B =377.7 kJ/kg=34750 kJ/kmol
 v hD   v h A x A   v hB x B =30850 kJ/kmol
Qc  nV  v hD =952308 kJ=952.3 MJ
cooling water requirement mwater:
Qc
m water 
=5995.4 kg.
t water c p water
Steam requirement calculation:
Assuming only the condensation of steam in reboiler and negligible heat losses, the heat duty
of reboiler can be calculated by:
QW  nV  v hW
Assuming that nv is constant and perfect thermal insulation: Qw=Qc
Than for steam requirement we can write:
Qc
0.2 Mpa
=432.44 kg.,  v hsteam
=2202.2 kJ/kg.
msteam 
 v hsteam