CP302 Separation Process Principles
Mass Transfer - Set 7
Course content of
Mass transfer section
L
T
A
Diffusion
Theory of interface mass transfer
Mass transfer coefficients, overall coefficients
and transfer units
04
01
03
Application of absorption, extraction and
adsorption
Concept of continuous contacting
equipment
04
01
04
Simultaneous heat and mass transfer in gasliquid contacting, and solids drying
04
01
03
03 Nov 2011
Prof. R. Shanthini
1
Example 6.10 of Ref 2
Experimental data have been obtained for air containing
1.6% by volume of SO2 being scrubbed with pure water in a
packed column of 1.5 m2 in cross-sectional area and 3.5 m
in packed height. Entering gas and liquid flow rates are
0.062 and 2.2 kmol/s, respectively. If the outlet mole fraction
of SO2 in the gas is 0.004 and column temperature is near
ambient with KSO2 = 40, calculate the following:
a) The NOG for absorption of SO2
b) The HOG in meters
c) The volumetric, overall mass-transfer coefficient, Kya for
SO2 in kmol/m3.s
03 Nov 2011
Prof. R. Shanthini
2
Solution to Example 6.10 of Ref 2
Data:
yin = 0.016
Treated gas
Gout, yout
Inlet solvent
Lin, xin
xin = 0
S = 1.5 m2
G
y
Z = 3.5 m
L
x
dz
G = 0.062 kmol/s
G
y+dy
L = 2.2 kmol/s
L
x+dx
Z
z
yout = 0.004
KSO2 = 40
Inlet gas
Gin, yin
03 Nov 2011
Prof. R. Shanthini
Spent solvent
Lout, xout
3
Solution to Example 6.10 of Ref 2
Data:
yin = 0.016
Operating line:
xin = 0
y = (L / G) x + yout - (L / G) xin
S = 1.5 m2
= (2.2/0.062) x + 0.004
Z = 3.5 m
G = 0.062 kmol/s
L = 2.2 kmol/s
yout = 0.004
KSO2 = 40
03 Nov 2011
Equilibrium line:
y=Kx
= 40 x
Prof. R. Shanthini
4
Solution to Example 6.10 of Ref 2
0.02
0.018
0.016
Bottom of the column (xout = ?, yin = 0.016)
0.014
y
0.012
Operating line
0.01
Equilibrium line
0.008
0.006
Top of the column (xin = 0, yout = 0.004)
0.004
0.002
0
0
0.0001
0.0002
0.0003
0.0004
0.0005
x
03 Nov 2011
Prof. R. Shanthini
5
Solution to Example 6.10 of Ref 2
a) NOG could be calculated using (83)
1
NOG =
(1 - KG/L)
(1 - KG/L) (yin - K xin)
ln
+ KG/L
yout - K xin
where KG / L = KSO2G / L = 40*0.062 / 2.2 = 1.127
1
(1 – 1.127) (0.016 - 0)
ln
NOG =
+ 1.127
(1 – 1.127)
0.004 - 0
= 3.78
03 Nov 2011
Prof. R. Shanthini
6
Solution to Example 6.10 of Ref 2
b) HOG could be calculated using (80)
HOG ≡
G
KyaS
=
0.062 kmol/s
(Kya)(1.5 m2)
Cannot continue this way since the value of Kya is not known.
Use (82) instead to calculate HOG since Z and NOG are known.
HOG = Z / NOG = 3.5 m / 3.78
03 Nov 2011
Prof. R. Shanthini
= 0.926 m
7
Solution to Example 6.10 of Ref 2
c) Kya can be calculated using (80)
HOG ≡
0.926 m
=
G
KyaS
0.062 kmol/s
(Kya)(1.5 m2)
0.062 kmol/s
Kya =
(0.926 m)(1.5 m2)
03 Nov 2011
= 0.044 kmol/m3.s
Prof. R. Shanthini
8
Example 6.11 of Ref 2 (modified)
A gaseous reactor effluent consisting of 2 mol% ethylene
oxide in an inert gas is scrubbed with water at 30oC and 20
atm. The total gas feed rate is 2500 lbmol/h, and the water
rate entering the scrubber is 3500 lbmol/h. The column, with
a diameter of 4 ft, is packed in two 12-ft-high sections with
1.5 in metal Pall rings. A liquid redistributer is located
between the two packed sections. Under the operating
conditions for the scrubber, the K-value for ethylene oxide is
0.85 and estimated values of kya and kxa are 200 lbmol/h.ft3
and 2643 lbmol/h.ft3 , respectively. Calculate the following:
a) Kya
b) HOG and NOG
c) Yout and xout
03 Nov 2011
Prof. R. Shanthini
9
Solution to Example 6.11 of Ref 2 (modified)
Treated gas
Gout, yout
Data:
yin = 0.02
Inlet solvent
Lin, xin
xin = 0
G = 2500 lbmol/h
G
y
L = 3500 lbmol/h
dz
S = π (4/2) ft2 = 12.6 ft2
G
y+dy
Z = 2 x 12 ft = 24 ft
L
x
L
x+dx
Z
z
K = 0.85
kya = 200 lbmol/h.ft3 and
kxa = 165 lbmol/h.ft3
Inlet gas
Gin, yin
03 Nov 2011
Prof. R. Shanthini
Spent solvent
Lout, xout
10
Solution to Example 6.11 of Ref 2 (modified)
a) Kya can be calculated using (78).
1
1
=
+
Kya
kya
K
kxa
=
1
200
+
0.85
165
Kya = 98.5 lbmol/h.ft3
b) HOG can be calculated using (80).
HOG ≡
G
KyaS
=
2500 lbmol/h
(98.5 lbmol/h.ft3)(12.6 ft2)
= 2.02 ft
Use (82) to calculate NOG since Z and HOG are known.
NOG = Z / HOG = 24 ft / 2.02 ft = 11.88
03 Nov 2011
Prof. R. Shanthini
11
Solution to Example 6.11 of Ref 2 (modified)
c) yout could be calculated using (83)
1
NOG =
(1 - KG/L)
(1 - KG/L) (yin - K xin)
ln
+ KG/L
yout - K xin
where KG / L = 0.85 * 2500 / 3500 = 0.607
1
(1 – 0.607) (0.02 - 0)
ln
11.88 =
+ 0.607
y
0
out
(1 – 0.607)
(84)
yout = 0.00007 = 0.007%
03 Nov 2011
Prof. R. Shanthini
12
Solution to Example 6.11 of Ref 2 (modified)
RHS of (84)
Enlarge it.
18
16
14
12
10
8
6
4
2
0
0.00%
0.50%
1.00%
1.50%
2.00%
2.50%
yout
03 Nov 2011
Prof. R. Shanthini
13
Solution to Example 6.11 of Ref 2 (modified)
RHS of (84)
yout = 0.01 mol%
18
17
16
15
14
13
12
11
10
9
8
7
6
0.00%
0.01%
0.02%
0.03%
0.04%
0.05%
yout
03 Nov 2011
Prof. R. Shanthini
14
Solution to Example 6.11 of Ref 2 (modified)
xout can be determined from the mass balance.
xout = (G/L) yin - (G/L) yout + xin
= (25/35) 0.02 - (25/35) 0.0001 + 0
= 0.0142 = 1.42%
03 Nov 2011
Prof. R. Shanthini
15
Determining the minimum liquid flow rate:
Calculate (L/G)min for the removal of 90% of the ammonia from
a 3540 mol/min feed gas containing 3% ammonia and 97% air.
The inlet liquid is pure water and the temperature and pressure
are 293 K and 1 atm, respectively. The equilibrium ratio of mole
fraction of ammonia in air to mole fraction of ammonia in water
at the column condition can be taken as 0.772.
Since a small liquid flow rate results in a high tower (which is
costly), and a large liquid flow rate requires a large diameter
tower (which is also costly), the optimum liquid flow rate of 1.2
to 1.5 times the minimum flow rate is used in practice.
Assuming the liquid flow rate to be 1.5 times the minimum,
determine NTU, HTU and the height of tower required. Take Kya
= 82 mol/m3.s and the cross-sectional area of the tower as 1.5
m 2.
03 Nov 2011
Prof. R. Shanthini
16
Solution to determining the minimum liquid flow rate:
Treated gas
Gout, yout
Data:
yin = 0.03
Inlet solvent
Lin, xin
xin = 0
G = 3540 mol/min
G
y
Kammonia = 0.772
L
x
dz
yout can be determined from the data
provided as shown in the next page.
G
y+dy
L
x+dx
Z
z
We need to calculate (L/G)min which
can be done graphically as shown in
the following pages.
Dilute mixtures are assumed.
03 Nov 2011
Prof. R. Shanthini
Inlet gas
Gin, yin
Spent solvent
Lout, xout
17
Solution to determining the minimum liquid flow rate:
Ammonia in the inlet stream
= 0.03 x 3540 mol/min
Air in the inlet stream
= 0.97 x 3540 mol/min
Ammonia removed
= 0.9 x 0.03 x 3540 mol/min
Ammonia in the outlet stream
= 0.1 x 0.03 x 3540 mol/min
Air in the outlet stream
= 0.97 x 3540 mol/min
yout
= 0.1 x 0.03 x 3540 / (0.1 x 0.03 x 3540 + 0.97 x 3540)
= 0.1 x 0.03 / (0.1 x 0.03 + 0.97)
= 0.003083
≈ 0.003
03 Nov 2011
Prof. R. Shanthini
18
Solution to determining the minimum liquid flow rate:
Operating line with yout = 0.003 and xin = 0:
y = (L / G) x + yout - (L / G) xin
= (L / G) x + 0.003
Since L/G is not known, we
cannot plot the operating line.
Equilibrium line:
y=Kx
= 0.772 x
Equilibrium line can be plotted.
03 Nov 2011
Prof. R. Shanthini
19
Solution to determining the minimum liquid flow rate:
0.04
0.035
Bottom of the column (xout = ?, yin = 0.03)
0.03
y
0.025
0.02
Equilibrium line
0.015
0.01
Top of the column (xin = 0, yout = 0.003)
0.005
0
0
0.01
0.02
0.03
0.04
0.05
0.06
x
03 Nov 2011
Prof. R. Shanthini
20
Solution to determining the minimum liquid flow rate:
0.04
0.035
Bottom of the column (xout = ?, yin = 0.03)
0.03
y
0.025
Operating line should
pass through this
point.
0.02
0.015
Equilibrium line
Top of the column
0.01
Top of the column (xin = 0, yout = 0.003)
0.005
0
0
0.01
0.02
0.03
0.04
0.05
0.06
x
03 Nov 2011
Prof. R. Shanthini
21
Solution to determining the minimum liquid flow rate:
0.04
0.035
Bottom of the column (xout = ?, yin = 0.03)
0.03
y
0.025
Operating line with the
slope (L/G)min should
connect these two points.
0.02
0.015
Equilibrium line
0.01
Top of the column (xin = 0, yout = 0.003)
0.005
0
0
0.01
0.02
0.03
0.04
0.05
0.06
x
03 Nov 2011
Prof. R. Shanthini
22
Solution to determining the minimum liquid flow rate:
0.04
0.035
Bottom of the column (xout = ?, yin = 0.03)
0.03
y
0.025
0.02
Equilibrium line
0.015
Operating line at minimum (L/G)
0.01
Top of the column (xin = 0, yout = 0.003)
0.005
0
0
0.01
0.02
0.03
0.04
0.05
0.06
x
03 Nov 2011
Prof. R. Shanthini
23
Solution to determining the minimum liquid flow rate:
0.04
0.035
Bottom of the column (xout = ?, yin = 0.03)
0.03
y
0.025
(L/G)min = (0.03-0.003)/xout
0.02
xout = 0.03/0.772 = 0.0389
0.015
(L/G)min = (0.03-0.003) / 0.0389
0.01
= 0.695 mol of water/mol of air
Top of the column (xin = 0, yout = 0.003)
0.005
0
0
0.01
0.02
0.03
0.04
0.05
0.06
x
03 Nov 2011
Prof. R. Shanthini
24
Solution to determining the minimum liquid flow rate:
(L/G)min = 0.695 mol of water/mol of air
G = 3540 mol/min (given)
Lmin = 0.695 x 3540 = 2460 mol/min
(L/G)operating = 1.5 x (L/G)min = 1.5 x 0.695
= 1.042 mol of water/mol of air
Loperating = 1.042 x 3540 = 3688 mol/min
03 Nov 2011
Prof. R. Shanthini
25
Solution to determining the minimum liquid flow rate:
0.04
0.035
Bottom of the column (yin = 0.03)
0.03
xout = ??
xout = 0.0389
y
0.025
0.02
Equilibrium line
0.015
Operating line at minimum (L/G)
Operating line at operating (L/G)
0.01
Top of the column (xin = 0, yout = 0.003)
0.005
0
0
0.01
0.02
0.03
0.04
0.05
0.06
x
03 Nov 2011
Prof. R. Shanthini
26
Solution to determining the minimum liquid flow rate:
0.04
0.035
Bottom of the column (yin = 0.03)
0.03
xout = ??
y
0.025
(L/G)operating = (0.03-0.003)/xout
0.02
Xout = (0.03-0.003)/1.042
0.015
= 0.0259
0.01
Top of the column (xin = 0, yout = 0.003)
0.005
0
0
0.01
0.02
0.03
0.04
0.05
0.06
x
03 Nov 2011
Prof. R. Shanthini
27
Solution to determining the minimum liquid flow rate:
NTU (NOG) could be calculated using (83)
1
NOG =
(1 - KG/L)
(1 - KG/L) (yin - K xin)
ln
+ KG/L
yout - K xin
where KG / L = 0.772 / 1.042 = 0.741
1
(1 – 0.741) (0.03 - 0)
ln
NOG =
+ 0.741
(1 – 0.741)
0.003 - 0
= 4.65
03 Nov 2011
Prof. R. Shanthini
28
Solution to determining the minimum liquid flow rate:
HTU (HOG) could be calculated using (80)
HOG =
=
G
KyaS
=
3540 mol/min
(82 mol/m3.s )(1.5 m2)
3540/60 mol/s
82 x 1.5 mol/m.s
= 0.48 m
Height of the tower (Z) can be calculated as follows:
Z = NOG x HOG = 4.65 x 0.48 m = 2.23 m
03 Nov 2011
Prof. R. Shanthini
29
Summary: Equations for Packed Columns for dilute solutions
The packed height is given by:
yin
dy
G
Z =
y – y*
KyaS
yout
Treated gas
Gout, yout
∫
HOG
Inlet solvent
Lin, xin
G
y
dz
NOG
HOL
G
y+dy
NOL
L
x
L
x+dx
Z
z
Z =
L
KxaS
xout
dx
x* – x
∫
Inlet gas
Gin, yin
xin
03 Nov 2011
Prof. R. Shanthini
Spent solvent
Lout, xout
30
Summary: Equations for Packed Columns for dilute solutions
Distributed:
Photocopy of Table 16.4 Alternative mass transfer
coefficient groupings for gas absorption
from
Henley EJ and Seader JD, 1981, Equilibrium-Stage
Separation Operations in Chemical Engineering, John
Wiley & Sons.
03 Nov 2011
Prof. R. Shanthini
31
Gas absorption, Stripping and Extraction
Gas absorption:
NOG and HOG are used
Stripping:
NOL and HOL are used
Extraction:
NOL and HOL are used
Humidification:
NG and HG are used.
03 Nov 2011
Prof. R. Shanthini
32