Chapter 3
Example 3.2-5. ---------------------------------------------------------------------------------Sulfur dioxide produced by the combustion of sulfur in air is absorbed in water. Pure SO2 is
then recovered from the solution by steam stripping. Make a preliminary design for the
absorption column. The feed will be 5000 kg/hr of gas containing 8 mole percent SO2. The
gas will be cooled to 20oC. A 95 percent recovery of the SO2 is required2.
Solution ----------------------------------------------------------------------------------------Operation is at atmospheric pressure as the solubility of SO2 in water is high. The feed water
temperature will be taken as 20oC.
Table E-1. Equilibrium data for SO2 at 1 atm and 20oC.
0
.000564
.000842
.001403
.001965
.00279
0
.0112
.01855
.0342
.0513
.0775
x
y
At 95 percent recovery of SO2
xin = 0
.00420
.121
yout
yout = (0.05)(0.08) = 0.004
Slope of equilibrium line: y* = mx
G = 5000 kg/hr
0.0775 = m(0.00279) ⇒ m = 27.8
To decide the most economic water flow rate,
the stripper should be considered together with the
absorption design. For this example, the absorption
design will be considered alone.
xout
yin = 0.08
The number of gas transfer unit may be estimated from
NOG =
*


y − yout
1
ln (1 − mG / L) in
+ mG / L
*
y out − yout
1 − mG / L 

Where G =
5000
= 0.1055 lbmol/s
(0.454)(3600)( 29)
yin = 0.08, yout = 0.004, y*out = mxin = (27.8)(0) = 0
Evaluate as NOG a function of mG/L
mG/L
0.2
NOG
3.5
L,lbmol/s 14.66
0.3
3.8
9.78
0.4
4.2
7.33
0.5
4.7
5.87
0.6
5.4
4.89
3-25
0.7
6.3
4.19
0.8
7.8
3.67
0.9
10.6
3.26
0.99
17.4
2.96
The optimum will be between mG/L = 0.6 to 0.8. Below 0.6 there is only a small
decrease in the number of stages required with increasing liquid rate and above 0.8, the
number of stages increases rapidly with decreasing liquid rate (Figure E-1)
18
16
L (lbm ol/s), NOG
14
12
10
L
NOG
8
6
4
2
0.2
0.3
0.4
0 .5
0.6
0.7
0 .8
0.9
1
m G/L
Figure E-1. Optimum liquid flow rate for Sulfur Dioxide absorption.
Check the liquid outlet composition at mG/L = 0.6 and at mG/L = 0.8. Assuming dilute
solution, the material balance is
L(xout – xin) = G(yin – yout) ⇒ xout = G(yin – yout)/L + xin
xout = (0.08 – 0.004)
mG
27.8 L
At mG/L = 0.6, xout = 1.64×10-3
At mG/L = 0.8, xout = 2.19×10-3
Use mG/L = 0.8 as the higher concentration will favor the stripper design and operation
without significantly increasing the number of stages needed in the absorber. Therefore
NOG = 8
3-26
Estimate column diameter:: 2 methods.
First design method: chooses the pre
pressure
ssure drop per unit length of packing then use
Figure 13.41 to evaluate G'(lb/s
'(lb/s⋅ft2). Note: G in the ordinate of Figure E-2 is actually G'.
lb 
 lbmol 
G
 M .W .vapor

s 
lbmol 
Area = 
lb
G'
s ⋅ ft 2
Check the percent of flooding (65 to 90%).
Figure E-2 Generalized flooding and pressure drop correlation.
10
Second design method: Use the flooding curve of Figure E-2
2 or the Eq. (3.2-21)
(3.2
to
determine G'flood and operate at some percentage of G'flood.
G' = (.65 to .90) G'flood
Once G'' is known the column cross
cross-sectional area A can be determined.
10
Wankat, P. C., Equilibrium Staged Separations
Separations, Elsevier, 1988, pg. 420
3-27
 G ' 2 Fψµ 0.2 
log10 
= − 1.6678 − 1.085 log10(Flv) − 0.29655[log10(Flv)]2

 ρG ρ L gc 
1/ 2
L ρ 
L'  ρ G 
= m  G
where
Flv =


G'  ρL 
Gm  ρ L 
Gm are mass flow rates in this expression.
(3.2-11)
1/ 2
= the abscissa of Figure E--2. Note: Lm and
The first design method will be applied. The physical properties of the gas can be taken
as those for air as the concentration of SO2 is low.
G=
5000
= 0.1055 lbmol/s
(0.454)(3600)( 29)
At mG/L = 0.8, L = mG/0.8
0.8 = (27.8)(0.1055)/0.8 = 3.666 lbmol/s
Packing: Choose 1.5" Ceramic Intalox saddle (picture below) with packing parameters F =
32, α = 0.13, β = 0.15. Intalox saddle is one type of dumped packings.
 G ' 2 Fψµ 0.2 
F is the parameterr in the ordinate of Figure E
E-2 
 . α and β are the
 ρG ρ L gc 
parameters to determine the pressure drop ∆p in inches of water per foot of packing given by
 G' 2 

∆p = α(10 ) 
ρ
 G
βL'
Air (gas) density at 20oC: ρG =
(3.2-22)
(29)( 273)
= 0.0753 lb/ft3
(359)( 293)
Liquid density: ρL = 62.3 lb/ft3, liquid viscosity = 1 cp.
The column will be designed for a pressure drop of 0.5 in of water/ft of packing. Table
E-22 shows the recommended design values.
3-28
Table E-2. Recommended design value for ∆p (inches of water per foot of packing)
∆p (inches of water per foot of packing)
0.2 – 0.6
0.5 – 1.0
Application
Absorber and stripper
Distillation (atmospheric & moderate
pressure)
Vacuum columns
0.1 – 0.4
If very low bottom pressures are required, structured packings or special low pressure drop
dumped packings should be considered (Hyperfil, Multifil, or Dixon rings).
The column area may be estimated from the pressure drop ∆p using either Eq. (3.2-22)
or Figure E-2. The procedure for using Figure E-2 will be discussed
ψ=
Density of
Density of
water
=1
liquid
1/ 2
L ρ 
= m  G
Gm  ρ L 
L'  ρ G 
Flv =
 
G'  ρL 
1/ 2
1/ 2
(3.67)(18)  0.0753 
=


(0.1055)( 29)  62.3 
= 0.75
From Figure E-2 at Flv = 0.75 and ∆p = 0.5 in water/ft of packing
 G ' 2 Fψµ 0.2 
 ρ ρ g  = 0.018
 G L c 
 G ' 2f Fψµ 0.2 
At flooding using the flooding line or Eq. 3.2-21: 
 = 0.0294
 ρ G ρ L g c 
 G'

 G'
 f
2

G'
 = 0.018 = 0.612 ⇒
= 0.78 (O.K., between 65 and 90 %)

G
'
0
.
0294
f

 0.018 ρ G ρ L g c 
G' = 

Fµ 0.2


1/ 2
 (0.018)(0.0753)(62.3)(32.2) 
= 

(52)(1)


1/ 2
= 0.229 lb/s⋅ft2
Note: µ in the above expression is the liquid viscosity in centipoises.
Gas mass flow rate Gm = (0.1055)(29) = 3.06 lb/s = AcG'
3.06
 (4)(13.38) 
Ac =
= 13.38 ft2 ⇒ Dc = 

0.229
π


3-29
1/ 2
= 4.13 ft
Check packing size: Recommend size ranges are
Column diameter
< 1 ft
1 to 3 ft
> 3 ft
Use packing size
1 in.
1 to 1.5 in.
2 to 3 in.
In general, the largest size of packing that is suitable for the size of column should be used,
up to 2 in. Small sizes are appreciably more expensive than the larger sizes. Above 2 in., the
lower cost does not normally compensate for the lower mass transfer efficiency. Use of too
large a size in a small column can cause poor liquid distribution. Since 1.5 in. ceramic
Intalox saddle is used in this example, a larger size could be considered.
The height of packing may be determined from the following formula
hp = NOGHOG
The height of overall gas transfer unit, HOG, may be evaluated from the height of gas transfer
unit, HG, and the height of liquid transfer unit, HL
HOG = HG +
mG
HL
L
The correlation for HG is
HG =
Where
ψ ( D ' col ) b1 ( h p / 10)1/ 3 ( Scv )1/ 2
[(3600) L' ( µ L / µV ) −1.25 (σ L / σ V ) −0.8 ]b 2
b1 = 1.11 for saddles, b2 = 0.50 for saddles
D'col = lesser of column diameter in ft or 2
hp = height of packed bed in ft
Scv = Schmidt number for vapor = µv/ρvDv
L' = mass flux of liquid, lb/s⋅ft2
σ = surface tension of liquid (L) or water (W)
ψ = packing parameter given by Figure E-3.
3-30
0
20
40
60
80 90
Figure E-3. Packing parameter ψ (ft) as a function of percent flood11
Ceramic Berl saddles are used to make conservative estimate of packing parameter for
Intalox saddles since mass transfer efficiency of Intalox saddles is higher than that of the
equivalent size Berl saddles.
µL
ρ
σ
=1; L =1; L =1
µW
ρW
σW
102
103
104
105
Figure E-4. Packing parameter φ (ft) as a function of L'
Figure E-5. Vapor load coefficient
The correlation for HL is
HL = φCfL(hp/10)0.15(ScL)1/2
where
φ = packing parameter given in Figure E-412
CfL = vapor load coefficient given in Figure E-512
ScL = Schmidt number for liquid = µL/ρLDL
Estimate hp by assuming a value for so HOG that HG and HL can be evaluated. Let HOG =
2.2 ft, then
11
12
Wankat, P. C., Equilibrium Staged Separations
Separations, Elsevier, 1988, pg. 652
Wankat, P. C., Equilibrium Staged Separations
Separations, Elsevier, 1988, pg. 654
3-31
hp = NOG HOG = (8)(2.2) ≈ 18 ft
At 78% flooding and ∆p = 1.5 in, Figure E-3 gives a packing value ψ = 65 ft. From the
PROP program (T.K. Nguyen Website)
Diffusivity of SO2 in water at 20oC:
Diffusivity of SO2 in air at 20oC and 1 atm:
Viscosity of gas (air) at 20oC:
DL = 7.5×10-6 cm2/s = 8.07×10-9 ft2/s
Dv = 0.122 cm2/s = 1.31×10-4 ft2/s
µv = 1.82×10-5 kg/m⋅s = 1.22×10-5 lb/ft⋅s
Scv = µv/ρvDv = (1.22×10-5)/[(7.53×10-2)(1.31×10-4)] = 1.237
ScL = µL/ρLDL = (6.72×10-4)/[(62.3)(8.07×10-9)] = 1337
L' = (3.666)(18)/(13.38) = 4.93 lb/s⋅ft2 = 1.78×104 lb/hr⋅ft2
Flooding ratio = 0.78 ⇒ CfL ≈ 0.70 (Figure 19-8)
ψ ( D ' col ) b1 ( h p / 10)1/ 3 ( Scv )1/ 2
(65)(2)1.11 (18 / 10)1/ 3 (1.237)1 / 2
HG =
=
= 1.42 ft
[(3600) L' ( µ L / µV ) −1.25 (σ L / σ V ) −0.8 ]b 2
[(3600)( 4.93)]1 / 2
L' = 1.78×104 lb/hr⋅ft2 ⇒ φ ≈ 0.1 (Figure E-4)
HL = φCfL(hp/10)0.15(ScL)1/2 = (0.1)(0.7)(1.8)0.15(1337)1/2 = 2.80 ft
HOG = HG +
mG
HL = 1.42 + (0.8)(2.80) = 3.66 ft
L
hp = NOG HOG = (8)(3.66) = 29.3 ft
Repeat the calculation with hp = 30 ft
(65)(2)1.11 (30 / 10)1 / 3 (1.237)1/ 2
 3 
HG =
=

1/ 2
[(3600)(4.93)]
 1 .8 
 3 
HL = (0.1)(0.7)(3.0)0.15(1337)1/2 = 

 1.8 
HOG = HG +
1/ 3
(1.42) = 1.69 ft
0.15
(2.80) = 3.02 ft
mG
HL = 1.69 + (0.8)(3.02) = 4.1 ft
L
hp = NOG HOG = (8)(4.1 = 32.8 ft
3-32