Tray Efficiency

Absorption – Calculating the Number of Stages Now we know what an absorption column looks like and we have established the nomenclature for the flows in and out of the stages in the column.

Ln-1 xn-1

Vn yn

Vn+1 yn+1

Ln xn

We also established that the mass balances for a section of column from the top of the column to stage n was (where the mole fractions refer to the solute material): La

Overall: Component La x a

Vn
Vn


1

1

 L V n a

yn


1

 L x V y n n a a

We can rearrange this equation to give us an expression for yn+1 yn


1

 L n x n  V a y a  L a x a . . . . . .(1) Vn
1 Vn
1

(if we use n = 0 we should find y1= ya ,because x0 is xa) To apply this in absorption we need to know something about how Ln and Vn+1 are related. As we move from the base of the column Vn+1 will decrease, as some of the gas stream is absorbed by the liquid. At the same time Ln will be increasing for the same reason. Assumptions – the non-solute part of the gas is insoluble in the liquid.  the liquid solvent is not volatile. This means that the molar flowrate of the insoluble gas is constant, we will call this V' and the molar flowrate of the liquid, which we will call L' is also constant, throughout the column. At any point in the column the mole fraction of solute in the gas is (where m is the molar flowrate of solute): y

m V' m



m 1

so y

m

yV ' my

yV '

m 

y 1

V'

y

The total vapour flowrate at this point in the column is then V' + m. V

so

V' 1 y

Similarly L 

L' 1 x

and specifically Ln  

L' 1  xn 

V'  and V n
1  1  y n
1

. . . . . (2)

Substituting (2) into (1) gives



yn


1

 L ' 1  y n
1  V ' 1  xn 



V' y  1  ya  a

L' x 1  xa  a

V'  1  yn
1 

solvent is initially pure. Rearranging gives: 

L' 1  y n
1   yn
1  xn V ' 1  xn 







V ' ya

1

1

ya 

remembering that L' =La if the



yn
1  L' x a 1  y n
1    V' 1  xa  V'

Which finally simplifies to: L'

{

xa

1  xa



xn

1  xn

6 { V'

1

ya ya

 

1

yn 

1

yn  1 

6

. . . . .(3)

This is the Operating Line Equation, which represents the relationship between the mole fraction of solute in the liquid draining from one stage (xn) and the vapour rising from the stage below (yn+1). It is often used by plotting it on a graph of yn+1 vs xn, which is constructed from the equation. Generally the problems we are trying to solve are where we know the separation we require ie ya; the vapour flow rate at the base Vb and its composition yb;, the purity of the solvent feed xa; and either La or xb. We will also have equilibrium data. We want to find the number of stages required.

Method of solution 1. ya = y1 so we can use equilibrium data to find x1 (assuming that the stage is ideal then the liquid and vapour leaving it will be at equilibrium. 2. Substitute this value into equation (3) to find y2 ( so n = 1 and n+1 = 2) 3. Use the equilibrium relationship to find x2 repeat until yn+1 ≥ yb Graphical Method of Solution – essentially the same but plotted on a graph showing the operating line and the equilibrium line on the same axes.

Operating line yn+1 vs xn yb y

y2 Equilibrium relationship yn vs xn

ya = y1

xa

x1

x

xN

x2

Only two theoretical stages needed. These theoretical stages are 100% efficient ie each reaches equilibrium. If the real stages were say, 33% efficient there would be six required. Here the operating line is above the equilibrium line, ie at every point in the column the vapour phase composition is richer than the composition of a vapour in equilibrium with the liquid. In other words there is a “driving force” for mass transfer from the vapour to the liquid. If the operating line were below the equilibrium line the driving force would act in the other directing and the solute would be “stripped” from the liquid into the vapour. If the operating line and the equilibrium line cross this represents a “pinch”in the column, a limiting concentration. When the lines cross the driving force for mass transfer becomes zero.

an infinite number of stages is required to reach the pinch concentration.

We can turn the operating line, equation (3) into a straight line by writing it in terms of yn  1 x Xn  and Y n  1  , the mole ratios of solute:solvent and 1 yn  1 1  xn  solute:insoluble gas, respectively. So equation (3) becomes





L ' X a  X n   V 'Y a  Y n ! 1  Using this graphically requires recalculating the equilibrium data, but this is often required anyway. The straight line is much simpler to plot, since only two points are required (or the gradient and one point).

increased L' Y

Slope L'/V'

Yb

Ya Xa

X1

X2

Xb

X If we increase L'while holding V'constant we need a) less stages for the same separation or get b) better separation in the same number of stages. So the performance of an absorber can be controlled by controlling the flowrate of solvent (which is common sense?).

What is the minimum solvent rate? The minimum solvent rate is the flowrate which just produces a pinch in the system ie the operating line crosses the equilibrium line at yb. Hence at the limiting solvent flow (the minimum solvent rate) the number of theoretical stages required is infinite. We always therefore work at some multiple of the minimum solvent rate.

Y

Yb

pinch

Ya Xa The slope of the operating line is

X L 'min V'

. If the solvent flowrate is changed the

gradient of the line changes (the vapour flow is generally set by process requirements. This is only applicable when using mole ratio coordinates, because in mole fraction coordinates the operating line is a curve unless the y' s are very small, for example if y 1 " $ 0.005 in which case it is possible to use y = 0.005 then Y " 1# y 1 # 0.005 %'& mole fractions in the the constructions above. Stripping (desorption) Here a liquid stream rich in solute transfers that solute to a stream of initially clean gas. The method of determining the number of stages is the same as for absorption except for the fact that the operating line lies below the equilibrium line. Since the transfer of solute to the gas is favoured by high temperature strippers are generally operated at as high a temperature as possible while absorbers are operated at as low a temperature as possible. Increasing the temperature moves the equilibrium line up.