Countercurrent Multistage Absorption (Tray Absorber) | Mass Transfer - Chemical Engineering PDF Download

4.7. Counter-current multi-stage absorption (Tray absorber) 
In tray absorption tower, multi-stage contact between gas and liquid takes place. In each tray, the liquid is brought into intimate contact of gas and equilibrium is reached thus making an ideal stage. In ideal stage, average composition of liquid leaving the tray is in equilibrium with liquid leaving that tray. The most important step in design of tray absorber is the determination of number of trays. The schematic of tray tower is presented in figure 4.7. The liquid enters from top of the column whereas gas is added from the bottom. The efficiency of the stages can be calculated as:

                                                      (4.18)

Figure 4.7: Schematic of tray tower.
The following parameters should be known for the determination of “number of stages”
(1) Gas feed rate
(2) Concentration of gas at inlet and outlet of the tower
(3) Minimum liquid rate; actual liquid rate is 1.2 to 2 times the minimum liquid rate.
(4) Equilibrium data for construction of equilibrium curve
Now, the number of theoretic stages can be obtained graphically or algebraically. 
 

(A) Graphical Method for the Determination of Number of Ideal Stages 
Overall material balance on tray tower
G_s(Y_N+1 -Y₁)=L_s(X_N -X₀)                                                                                 (4.19)
This is the operating line for tray tower.

If the stage (plate) is ideal, (X_n, Y_n) must lie on the equilibrium line, Y*=f(X). Top plate is located at P(X₀, Y₁) and bottom plate is marked as Q(X_N, Y_N+1) in X-Y plane. A vertical line is drawn from Q point to D point in equilibrium line at (X_N, Y_N). From point D in equilibrium line, a horizontal line is extended up to operating line at E (X_N-1, Y_N). The region QDE stands for N-th plate (refer Figure 4.8). We may get fraction of plates. In that situation, the next whole number will be the actual number of ideal plates. If the overall stage efficiency is known, the number of real plates can be obtained from Equation (4.18). 

Figure 4.8: Graphical determination of number of ideal stages.

(B) Algebraic Determination of Number of Ideal Stages 
If both operating line and equilibrium lines are straight, number of ideal stages can be calculated algebraically.
Let solute transfers from gas to liquid (Absorption)
Equilibrium line, Y=αX 

Point (X_N, Y_N) lies on the equilibrium line: Y_N=αX_N                        (4.20)
Operating line: 
  
Now Equation 4.22 becomes,
                                                         (4.23)
This Equation is linear first order “difference Equation” (non-homogeneous).

Solution by finite difference method 
Corresponding homogeneous Equation:                     (4.24)
Solution is                                                                                (4.25)
                                                                           (4.26)

                                                                                                       (4.27)
Non-homogeneous Equation has a particular solution, which is constant.
Assuming Y_N=Y_N+1, we have, Y=K₂ 

                                                                                      (4.28)
The complete solution is as follows:
                                         (4.29)
Initial conditions:
N=0; Y₀=αX₀ 
  
                                                   (4.30) 

                                                         (4.31)
When N=N+1;
            (4.32)
Taking logarithm in both the sides we get: 
                                   
                                 (4.33)
When   Equation (4.23) becomes
 → Operating line Equation                 (4.34)
Put N=N, N-1, N-2, ………..3,2,1 and add to get

                                                                            (4.35)

Let solute is transferred from liquid to gas (stripping). 
                             (4.36)
When  Equation (4.23) becomes
                                                                     (4.37)

These four Equations (4.33, 4.35-4.37) are called “Kremser Equations”. 

Example Problem 4.2 : It is desired to absorb 95% of acetone by water from a mixture of acetone and nitrogen containing 1.5% of the component in a countercurrent tray tower. Total gas input is 30 kmol/hr and water enters the tower at a rate of 90 kmol/hr. The tower operates at 27�C and 1 atm. The equilibrium relation is Y=2.53X. Determine the number of ideal stages necessary for the separation using (a) graphical method as well as (b) Kremser analysis method.

Solution 4.2:
Basis: 1 hour
G_N+1=30 kmol
y_N+1=0.015
L₀=90 kmol Moles acetone in = 30�0.015 moles=0.45
moles Moles nitrogen in = (30-0.45)
moles=29.55 moles Moles acetone leaving (95% absorbed) = 0.45�(1-0.95) moles=0.0225 moles
Gs=29.55 moles
Ls=90 moles
α=2.53 [as, Y=2.53X] 

 = 0.015
Rewriting Equation (4.19) (operating line) as

X_N=4.68�10^-3
 

(a) Solution by graphical method 
Construction of operating line PQ:
P(X₀, Y₁)=P(0, 7.61�10^-4 )
Q(X_N, Y_N+1)=Q(4.68�10^-3, 0.015)

Construction of equilibrium line (Y=2.53X):

From graphical construction (Figure 4.9), the number of triangles obtained is more than 7. Hence number of ideal stages is 8.


Figure 4.9: Graphical construction for determination of number of stages

(b) Solution by Kremser analysis method 
As Ā≠1, according to Kremser analysis method:

Number of ideal stages is 8.