1. SHROFF S.R. ROTARY INSTITUTE OF CHEMICAL TECHNOLOGY
SUBJECT:-MTO-1
TOPIC:-Choice of Solvent for absorption, material balance and liquid-gas ratio for absorption and stripping.
PREPARED BY:-BHUPENDRASINH SOLANKI
Enrolment no:
GUIDED BY: Mr. YASHWANT BHALERAO

2. INTRODUCTION Absorption:
When a gas mixture is contacted into a liquid for a purpose to preferentially dissolved one component of gas mixture is known as absorption.
Desorption or stripping operation is the reverse of absorption.
Absorption operation is of two types; physical and chemical.

3. Selection of solvent for absorption and stripping
Produce specific Solution
Solvent is specified by nature of product
Example:-Production of HCL
Purpose
To remove certain component from the mixture
Based on some important Properties
Example:-Seperation of H2S from natural gas

4. Selection of solvent for absorption and stripping
(A)Gas Solubility: Good solvent is which in solute is highly soluble.
For ex:-
(B) Volatility: Low volatility or low vapor pressure of the solvent.
(C) Cost: The solvent should be cheaper.
(D) Corrosiveness: Non-corrosive or less corrosive solvent reduces equipment construction cost as well as maintenance cost.
(E)Viscosity: For better absorption, a solvent of low viscosity is required.
(F) Miscellaneous: The solvent should be non-toxic, nonflammable, non-hazardous and should be chemically stable.
Selection of solvent for absorption and stripping

5. Material balance & liquid gas ratio for absorption and stripping
The steady state Countercurrent M.T

6. Notations
G1=Rate of input of phase G at section 1, mol/time G2=Rate of output of phase G at section 2, mol/time Gs=Rate of flow phase G on solute free basis, mol/time G=Rate of flow of phase G at any section y,y1,y2=Mole fraction of solute in phase G respective section Y,Y1,Y2=Mole ratio of solute in phase G respective section L1=Rate of output of phase L at section 1, mol/time L2=Rate of input of phase L at section 2, mol/time Ls=Rate of flow phase L on solute free basis, mol/time L=Rate of flow of phase L at any section x,x1,x2=Mole fraction of solute in phase G respective section X,X1,X2=Mole ratio of solute in phase G respective section

7. The steady state Countercurrent M.T
We know the eq. for mole ratio,
Gs=G(1-y) Y=y/(1-y)
Ls=L(1-x) X=x/(1-x)
From this we get,
The steady state Countercurrent M.T

8. The steady state Countercurrent M.T

9. The steady state Countercurrent M.T
The phase L is heavier and goes down the apparatus and the lighter phase G flows up. The phase L and G naturally invisible.
The continuous contact equipment will be called the top end terminal 1 and bottom end terminal 2.
Considering envelop 1 :
The total rate of input of solute.
The total rate of output of solute.

10. The steady state Countercurrent M.T
At steady state,
Rate of input = Rate of output
The molar flow rate of the phase GS and LS are taken on solute free basis. These are simply the flow rate of carrier gas. GS and LS remain constant along the equipment. Since we have assume that they are naturally invisible concentration X and Y are express in mole ratio

11. The steady state Countercurrent M.T
The above eq. is represent a straight line having slope and passing through X2,Y2.
X2 and Y2 the concentration of two phases where the terminal of equipment, while X and Y at are their concentration at any location within the equipment and this called as eq. of operating line.
The material balance over envelop 2 that enclosed the entire apparatus is yield.
Total rate of input =
Total rate of output =

12. The steady state Countercurrent M.T
At steady state condition,
Rate of input = Rate of output
This is the straight line having slope LS / GS.

13. The steady state Countercurrent M.T
For an absorption the operating line always lies above the equilibrium solubility curve, for stripper the line always below the equilibrium curve.
The steady state Countercurrent M.T M

14. The steady state Countercurrent M.T
The entire can also be done using the concentration of the phases in the mole fraction unit instead of the mole ratio unit.
A M.B over envelope 1
Gy+L2x2=G2y2+Lx
Therefore
Gy-G2y2 =Lx-L2x2

15. The steady state Cocurrent process
Fig. Shows the schematic of cocurrent contacting apparatus.
Taking M.B over envelope 1 is
Gs(Y1-Y)=Ls(X-X1)
The above equation represent a straight line having a slope
–Ls/Gs
A M.B over envelope 2 that encloses the entire apparatus is
Gs(Y1-Y2)=Ls(X2-X1)

16. The steady state Cocurrent process

17. M.T in stage-wise contact of two phases

18. M.T in stage-wise contact of two phases
Any device or combination of devices in which two immiscible phases are brought into intimate contact in order to achieve M.T from one phase to another is called stage.
A group of interconnected stages in which the phases flow from one stage to other , in sequence is called cascade

19. Determination of Number of Stages in Countercurrent contact
Consider a N-tray as shown in fig.
In the case of stagewise contact it is convenient that the subscript of concentration term or flow rate denotes the stage from which the stream concerned comes out.
Accordingly in fig. The concentration of phase G entering the column is denoted by YN+1 (the stream as if comes from hypothetical (N+1)th stage preceding stage N).

20. Determination of number of stages in countercurrent contact
Similarly phase L entering plate 1 at the top is imagined as a stream coming from the hypothetical zero plate. The concentration of this stream is X0
Taking M.B over envelope 1 is
Gs(Yn+1-Y1)=Ls(Xn-X0)
Now the overall M.B for entire column is
Gs(YN+1-Y1)=Ls(XN-X0)

21. Determination of number of stages in countercurrent contact

22. Minimum liquid gas ratio for absorbers
If we reduce the liquid flow rate Ls, what happens the operating line .
If we reduce the liquid flow rate the operating line slope will decrease and come closer to the equilibrium line which means the driving force for mass transfer reduced.
we cannot reduce the liquid flow rate indefinitely.

23. Minimum liquid gas ratio for absorbers
So, on further reductions as we can see at the fixed Y2 and X2 locations the slope of the operating line is further reduced where it touches to the equilibrium line.
So, the liquid rate corresponding to this operating line is known as the minimum liquid rate for absorption operation. This is theoretically possible.

24. Minimum liquid gas ratio for absorbers
Now, in fig. we can see if we further reduce the liquid rate the operating line crosses the equilibrium line, that means in one section of the column absorption is taking place and in other section of the column desorption occurs, which is physically impossible
So, the slope of operating line at minimum liquid rate would be equal to Ls(min) by Gs

25. Minimum liquid gas ratio for absorbers
If we plot the absorption curve instead of mole ratio, if we plot it mole fraction.
If we work on mole fractions then both the operating line and the equilibrium line are curved in nature.
So, to obtain a pinch point in this case is difficult and we have to work with two different curves instead of state operating lines.

26. Minimum liquid gas ratio for absorbers
The equilibrium relationship is curved line, but if the relation is represented in linear in nature
In general the actual liquid rate is equal to Ls minimum into 1.2 to 2 times.

27. Minimum liquid gas ratio for absorbers
If the equilibrium line follows Henry’s law such that Y= m X
If m> 1, the equilibrium line is convex downward. So, the pinch point, point corresponding to the minimum liquid rate occurs at bottom of the tower
If m< 1 then equilibrium curve is convex upward, so the operating line at minimum liquid rate touches the equilibrium line at some intermediate points.

28. REFERENCES https://en.wikipedia.org/wiki/Absorption

29. THANK YOU