1. Unit Operation 1
By: Dr. Amer Aziz
Dr. Farah Talib
Absorption
Lect. No. 2 Part 1 and 2
2. Example (1):- Absorption column receiving about 130 kmol/hr of feed gas containing 9 mol% of solute. It
is required to remove 93% of solute using 150 kmol/hr liquid solvent. The feed solvent has 0.4% of solute
in it. The overall tray efficiency is to be 45%.
Equilibrium data are: x 0.013 0.033 0.049 0.064 0.074 0.093 0.106
y 0.01 0.026 0.043 0.06 0.073 0.1 0.126
Determine the No. of trays required ?
Solution:- A) To solve graphically:-
xT = 0.004 , yT = ? , yB = 0.09 , xB = ?
To remove 93% from 9% initial:
yT = 0.09 ( 1 – 0.93 ) → yT = 0.0063
to find xB make over all M.B :- Gs (yB - yT) = Ls ( xB – xT ) 130 (0.09 – 0.0063 ) = 150 ( xB – 0.004)
→
xB = 0.0765
3. to find the operating line equation:- make M.B. between top of column and any section:-
→
Gs (y - yT) = Ls ( x – xT ) → 130 ( y – 0.0063 ) = 150 ( x – 0.004)
Y = 1.15 x – 0.0017 ….. Operating line equation
1- Plot equilibrium line from the table.
2- operating line between ( 0.004 , 0.0063 ) , ( 0.0765 , 0.09 ).
From the Fig. :-
No. of theoretical trays = 8 trays
No. of actual trays = 8/0.45 = 18 trays
To draw:-
4.
5.
6.
7.
8. B) To solve analytically:-
G
*
m
L
A
m = 1.1 , A = 150 / (1.1 * 130) = 1.049
N =
𝐿𝑜𝑔 {
0.09−1.1∗0.004
0.0063−1.1∗0.004
1−
1
1.049
+
1
1.049
}
𝐿𝑜𝑔 1.049
N =
𝐿𝑜𝑔 3.2
𝐿𝑜𝑔 1.049
→
N = 25 No. of theoretical trays.
No. of actual ( real ) trays = 25 / eff. = 25 / 0.45 ,
No. of actual ( real ) trays = 56 trays
9. C) Using Kremser – Chart
A = 1.049 ~ 1.05
yt − 𝑚 ∗ xt
yB − 𝑚 ∗ xt
→
0.0063 − 1.1 ∗ 0.004
0.09 − 1.1 ∗ 0.004
= 0.02
From chart …. Find ( N ) from ( 1.05 ) and ( 0.02 ) :-
No. of theoretical trays = 25 trays
No. of actual ( real ) trays = 25 / eff. = 25 / 0.45 ,
No. of actual ( real ) trays = 56 trays
11. Example (2):- It is required to remove 99% of the solute (C) from a solution of (C) in gas (G) by using
a pure solvent liquid (L) in a counter – current cascade. The feed gas containing 8 mol% C in the
mixture enters the column at the bottom at a rate of 5500 kg/hr. the solvent enters at the top at a
rate of 7685 kg/hr.
- Write down the equation of the operating line.
- Determine the number of trays required to perform the separation if overall efficiency is to be
40%.
- The equilibrium relation y = 1.32 x.
Solution:- A) To solve graphically:-
xT = 0 (free solvent ) , yT = ? , yB = 0.08 , xB = ?
To remove 99 % from 8 % initial:
yT = 0.08 ( 1 – 0.99 ) → yT = 0.0008
12. to find xB make over all M.B :-
Gs (yB - yT) = Ls ( xB – xT ) → 5500 (0.08 – 0.0008 ) = 7685 ( xB – 0)
xB = 0.056
To find the operating line equation:-
Gs (y - yT) = Ls ( x – xT )
→ Make M.B. between top of column and any section:-
5500 ( y – 0.0008 ) = 7685 ( x – 0)
→
Y = 1.39 x – 0.0008 ….. Operating line equation
To draw:-
1- Plot equilibrium line after assume values of x
and applied it in the equilibrium equation…….
y = 1.32 x
Y
X
0
0
0.0132
0.01
0.0264
0.02
0.0396
0.03
0.0528
0.04
0.066
0.05
0.0792
0.06
Always a little more than xB →
13. 2- operating line
between ( 0 , 0.0008 ) ,
( 0.056 , 0.08 ).
From the Fig. :-
No. of theoretical trays =
33 trays
No. of actual trays =
33/0.4 = 83 trays
14.
15.
16.
17.
18. B) To solve analytically:-
G
*
m
L
A
m = 1.32 , A = 7685 / (1.32 * 5500) = 1.058 ,
N =
𝐿𝑜𝑔 {
0.08−0
0.0008−0
1−
1
1.058
+
1
1.058
}
𝐿𝑜𝑔 1.058
N =
34
0.4
N = 34 No. of theoretical trays.
No. of actual ( real ) trays = 34 / eff. = 34 / 0.4 ,
No. of actual ( real ) trays = 85 trays
19. C) Using Kremser – Chart
A = 1.058 ~ 1.06
yt − 𝑚 ∗ xt
yB − 𝑚 ∗ xt
→
0.0008 − 0
0.08 − 0
= 0.01
From chart …. Find ( N ) from ( 1.06 ) and ( 0.01 ) :-
No. of theoretical trays = 35 trays
No. of actual ( real ) trays = 35 / eff. = 35 / 0.4 ,
No. of actual ( real ) trays = 88 trays
20. Example (3):- A sieve tray tower is being designed for a gas absorption process. The entering gas contains
1.8 % (molar) of (A) which must be absorbed. The gas should leave the tower containing no more than 0.1
% (molar) of (A). The liquid to be used as absorbent initially contains 0.01 % of (A). The system obeys
Henry's law with m = 1.41. At the bottom of the tower, the molar liquid – to – gas ratio = 2.115, and at the
other extreme is (2.326). For these conditions, it has been seen that EMGE = 0.65 . Find No. of trays?
Solution:-
GT
*
m
LT
A1 5
.
1
A1
1.41
2.115
→ GB
*
m
LB
A2
𝐴 = 𝐴1 ∗ 𝐴2 → 𝐴 = 1.5 ∗ 1.65 A = 1.57
yB = 0.018 , yT = 0.001 , xT = 0.0001 , m = 1.41
21. No. of theoretical trays = 4.64 ~ 5 trays.
Eo=
ln[1+EMG
1
𝐴
−1 ]
ln(
1
𝐴
)
Eo = 0.596
No. of actual ( real ) trays = 5 / eff. = 5 / 0.596
No. of actual ( real ) trays = 8.3 ~ 8 trays.
