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Chapter 10 Absorption Page 10-1
Chapter 10 Absorption
1. Mass transfer in an NH4 scrubber
G
al
1
1mG /Lln ylml
yomxo
(1) From Eq. 10.3-12, Ky =
2. Mass transfer in a Cl2 scrubber
The basic equation is Eq. 10.3-12
(a) G = 14/60 kg/s
(30 g/mol)*(30 cm)2 = 2.7510 mol/cm -s
-3 2
(b) KG = 1
(1/DG + mcG/DLCL)
3. Stripping of H2S
This needs the use of Fig. 10.2-4.
First we calculate the capacity factor: (G')2F0.2
GLgc
4. Air stripping of chloroform
(a) Operating line = Eq. 10.3-5,
Equilibrium line = Eq. 10.3-6
5. Repacking a column
This needs reference to Fig. 10.2-4
at 20°C, G = 28.84
82.05*293*62.43 = 0.0749 lb/ft , = 0.99821*62.43 = 62.32 lb/ft
3L3
Chapter 10 Absorption Page 10-3
6. Absorption SO2
To obtain a constant driving force, the equilibrium and operating lines must be parallel and
therefore have the same slope. To find the driving force we can use any point in the column; we’ll
use the top since we know both concentrations at that point. The height is given by:
Chapter 10 Absorption Page 10-4
7. Absorption with amines
At minimum liquid flow, the exiting liquid will be in equilibrium with the entering gas, i.e. the
operating and equilibrium line will intersect. Using the equilibrium line, we can find this point:
The slope of the operating line is:
From a mass balance on the column using twice the minimum liquid flow, we have:
8. Hydrogen sulfide absorption
The minimum liquid flow rate occurs when the liquid is in equilibrium with the gas, i.e. the
operating line and equilibrium line are parallel. For twice the minimum liquid flow we have:
A mass balance gives:
To find the height of the column, we need the HTU and NTU:
2
16.8 0.083
mol
GGmGm ms
The length of the column is:
9. Methylene chloride absorption
a) The column length required is given by:
Chapter 10 Absorption Page 10-6
Using the fact that m is small and writing Ky in terms of KG, we have:
Since our column is larger than 3.6 m, it should be able to get the separation desired.
b) From the Kremser Equation we have:
For ten stages:
10. Isobutanol stripping
We need to find the concentrations yℓ and x0. A mass balance will give us one equation and the
other must come from a rate equation. First we perform an overall mass balance:
V, y
L
Chapter 10 Absorption Page 10-7
Now we calculate the number of transfer units our column has:
,
yo
To find m, we use the fact that the vapor pressure of isobutanol at 1 atm is 400 mm Hg.
We can now use the equation for absorption to get our second relationship between yℓ and x0:
NTU 1
ln ylmxl
yomxo
Solving our two equations simultaneously we have:
The percent removal is:
11. Fragrance recovery
a) A mass balance gives:
Chapter 10 Absorption Page 10-8
The fraction in the oil is:
12. Natural gas upgrade
x
For this problem:
A mass balance gives:
ℓ
= 0
x
0
2
y
0
= 0.1
y
ℓ
= 0.01
yo=0.012 x0=0
Chapter 10 Absorption Page 10-9
0.1 4 0.0203
ln 0.012 0 5.84
13. Mercaptan removal
a) Given:
Since the equilibrium and operating lines have the same slope, y – y* is constant.
At the top of the column, we can evaluate Δy:
Chapter 10 Absorption Page 10-10
2 ln 100 9.2mm
14. Hydrogen sulfide separation
a) First we perform a mass balance on the column:
The NTU is given by:
b) The height of the tower is given by:
15. Improved H2S separation
G, y
0
= 630 ppm L, x
0
b)
First we solve for the NTU:
The NTU is given by:
Solving for G/L and substituting 1 for m and 0 for xo we have:
We know everything except x
. We can express x
in terms of G/L by writing a mass balance:
We now have:
