Counter-current cascade of ideal stages

Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik

3

2

1

Let’s consider three ideal stages.

The solvent stream enters the cascade by the first stage. The solvent can be pure or can have a certain concentration of solute, represented by xo.

The gas enters the cascade by the last stage and thus, flows in counter-current to the solvent flow.

The operating line is defined by points of compositions of gas and liquid phases at the same level of the process. Points on the operating line are:

L, x0

x1

x3 G, y4

y1

y2

y3 x210 , yx 21 , yx 32 , yx 43 , yx

Because the stages are ideal, the equilibrium is reached in each of them. This means that the gas and the liquid phase going out from a stage are at equilibrium, and thus lie on the equilibrium line.

11 , yx 22 , yx 33 , yx

Local partial mass balance

The equilibrium equation:

Dividing by G:

The operating line is:

The slope of the operating line is positive (L/G)

jj xLyGxLyG 101

xmy

jj xG

Lyx

G

Ly 101

Mass balances

L, x0y1

1

y2

2

x1

yj

j

xj-1

yj+1 xj

n

G, yn+1 xn

dotted area

solute

101 yx

G

Lx

G

Ly jj

Data are the initial gas composition, yn+1, as well as the initial solvent composition, xo. On the diagram:

When the final gas composition, y1, is known because it is restricted to be less than a certain minimum value (specification), the first point of the operating line can be set.

L, x0y1

1

y2

2

x1

yj

j

xj-1

yj+1 xj

n

G, yn+1 xn

y = m x

y

x

Yn+1

x0

Y1

For the specific slope (L/G), we can set the operating line:

L, x0y1

1

y2

2

x1

yj

j

xj-1

yj+1 xj

n

G, yn+1 xn

y = m x

y

x

Yn+1

x0

Y1

L/G

101 yx

G

Lx

G

Ly jj

The graphical resolution consists on a step by step construction.

L, x0y1

1

y2

2

x1

yj

j

xj-1

yj+1 xj

n

G, yn+1 xn

y = m x

y

x

Yn+1

x0

Y1

We know that x1 is at equilibrium with y1. It is then easy to set x1 on the diagram.

x1

We also know that x1 and y2 are the same point on the operating line, since they are both at the same level in the cascade.

Y2

And so on, until the specification is reached...

Y3

Y4

x2 x3 x4

A sequential resolution is also possible without the graphical representation, alternating the use of the operating line equation and the equilibrium equation.

As we have seen in the case of the cross-current cascade, a mathematical resolution can also be done.

Introducing the absorption factor in the mass balance equation obtained before...

01101 yAyyAyx

G

Lx

G

Ly jjj

Comparing this equation and the general Difference Partial equation, we get

01 yAyA

And applying the general solution we get

A

yAyA

A

yAyyy j

j

110101

0 1 A

And for the n stage:

A

yAyA

A

yAyyy n

n 1101101

01

A

AyAyAyy

nn

n

1

1 1

011

01

The fractional absorption can now be calculated, by substituting the value of yn+1:

1

11

101

11

n

n

n

A

A

yy

yy 1 A

This is the Kremser equation.

When A=1, the Kremser equation cannot be used. The fraction of absorption is then calculated:

Ayy

yy

n

n

1

11

01

11 1 A

And solving for n, the number of stages can be calculated:

ALn1

A1Ln

n