EquilStageNotes

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Equilibrium Stage OperationsBGHiggins/UCDavis//Feb1, 2008

IntroductionConsider a gas phase containing a solute of concentration y0 that is mixed with an immiscible liquid phase atconstant T and P. If the solute is miscible in both the liquid and gas phases, it will partition between the two

phases as the system attempts to attain equilibrium, thereby reducing the concentration of the solute in the gas

phase. At equilibrium, the concentration of solute in the liquid phase is related to the concentration in the gas

phase by an equilibrium relation that may be taken to be linear at sufficiently low concentrations:

(1)y = K x

where K is the equilibrium constant. Since the concentration y in the gas phase at equilibrium is less thany0, this

suggests a way for removing an unwanted solute from a gas stream; the process is called gas absorption. A

schematic of such a process is shown below in Figure 1

Figure 1

Single Stage

L, x0

L, x

V , y0 V , y

The contacting unit is called an equilibrium stage if the streams leaving the unit are in equilibrium; that is, the

solvent in the gas stream leaving the unit is given byy = K x.

If the concentration of solute in any given stream is sufficiently low then it is permissible to assume that the

molar (or mass) flow rate of any given phase is constant from stream to stream. With this assumption, a solute

balance over the equilibrium stage gives ( see Figure 1)

(2)L x0 + V y0 = L x + V y

We can represent this equation on a y-x plot by writing (2) as

(3)y = y0 +L

VHx0 - xL


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Thus the solute balance can be represented by a straight line that passes through the point Hx0, y0Land has slope-L/V. According to the mass balance any {x,y} pair of values that satisfies Eq. (3) is possible. However, if the

exit streams are in equilibrium, then this line must intersect the equilibrium line y = K x. which determines the

minimum value of y that can be reached in an equilibrium stage when L/V is fixed, see Figure 2

Figure 2

y0

y

y0

*

x x

-LV

Substituting x=y/K into (3) and definingA =L V Kand then solving for y gives,

(4)yH1 + AL = y0 +L

Vx0

whereA is called the absorption factor. A further rearrangement of (4) gives

(5)y

y0=

1

H1 +AL 1 +

L

V

x0

y0

The ratio y y0 represents the fraction of solute not extracted or absorbed from the gas phase. Hence, thefraction of solute absorbed into the liquid phase H1 -y y0L increases with increasing absorption factor A. To

appreciate the limiting behavior when A becomes small or large, it is convenient to eliminate x0 from Eq. (5) by

defining a hypothetical value y0* , which is the concentration of solute that would be in equilibrium with the inlet

liquid stream (see Figure 2). This strategem gives

(6)

y

y0=

1

H1 +AL 1 +

L

V K

K x0

y0

=1

H1 + AL 1 + A

y0*

y0

where we have used the relationship y0*= K x0 to define the hypothetical value y0

* . There are two limits that are

of interest. The first limit of interest is when A 0. The RHS of Eq. (6) is then 1, which means that y y0 and

there is then no absorption of solute. The second limit is when A . The RHS of (6) approaches y0* y0,

which implies that y y0*. This limit represents the minimum solute concentration that can be achieved in an

equilibrium stage.

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Example

A gas stream containing acetone and air is to be passed through an equilibrium stage so that the acetone can be

absorbed in a water stream. The inlet gas contains 1.0 mol % acetone, and has a molar flow rate of 30 kg mol/h.

The pure water stream has a molar flow rate of 90 kg mol/h. The process is to operate isothermally and at

constant pressure. The equilibrium relationship for the acetone in the gas liquid system is y = 2.53 x. You may

assume that the water and air are immiscible and the molar flow rates entering and leaving the equilibrium stage

are constant.

(i) Determine the absorption factor for any equilibrium stage

(ii) Determine the concentration of acetone in the exit streams

Solution

Solution

(i) Since the air and water can be taken as inert streams, we can determine the absorption factor by noting that

(7)A=L

V K=

90

H30LH2.53L= 1.18577

(ii) The mole fraction leaving the equilibrium stage is given by (5)

(8)y

y0=

1

H1 + AL 1 +

L

V

x0

y0

where

(9)L

V=

90

30= 3, A=

L

V K=

3

2.53= 1.18577, x0 = 0

Substituting these values into (8) gives

(10)y =y0

H1 + AL=

0.01

2.18577= 0.00457505

Recall that the acetone in the water stream is in equilibrium with acetone in the leaving gas stream. Thus

(11)x =y

K=

0.00457505

2.53= 0.00180832

Efficiency of Multiple Countercurrent StagesIn the previous section we saw that the fraction of solute absorbed or extracted from the V-phase depended on

the absorption factor A. Suppose now we added an additional equilibrium stage in series with the previous stage,

as shown in Fig 3. We would like to know is how the additional stage affects the absorption of the solute.

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Figure 3

V , y1

V , y2

V , y3

L, x0

L, x1

L, x2

Stage 1

Stage 2

Intuitively we expect the absorption of the solute to depend on the absorption factor A. To show this mathemati-

cally, we consider control volumes for each stage. In the previous section we showed that the solute species

balance (in terms of mole fraction ratios) around Stage 1 gives

(12)y1H1 + AL = y2 +L

Vx0

which we rewrite as

(13)y2 = H1 + ALy1 -L

Vx0

Similarly, solute balance over stage 2 gives

(14)y3 =L

Vx2 + y2 -

L

Vx1

The terms involving x2, x1 in (14) can be expressed in terms of y2, y1 using the equilibrium relation

yj = K xj to give

(15)y3 = H1 + ALy2 -A y1

Using (13) to eliminatey2 gives

(16)

y3 = H1 + ALH1 + ALy1 - A y1 - H1 +ALL

Vx0

= I1 + A+ A2My1 - H1 +ALL

Vx0

The above expression can be expressed more compactly as



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(17)yN+1 = bNy1 - bN-1

x0, with N = 2

where

(18)bN =i=0

N

Ai, with N = 2

If we have more than two stages we can show that Eq. (17) holds when N>2. Thus we can use the above expres-

sion to determine the fraction of solute not absorbed in the L-phase using N multiple stages. Consider the case

when N=2; then from Eq. (16) we have

(19)

1 = I1 + A+ A2My1

y3- H1 + AL

L

V

x0

y3

y1

y3=

1

b2

+b1

b2

x0

y3

It follows from the definition ofbN that

(20)H1 - AL bN = H1 - ALI1 +A+A2 + + ANM = I1 - AN+1Mwhich means we can express bN as

(21)bN =I1 - AN+1M

1 -A

We can use this result to determine the limiting value of the solute fraction as N . There are two cases to

consider.

Case I: When A< 1, the limiting values for the functions depending on b are

(22)

LimitN->

1

bN

1 - A,

LimitN->

bN-1

bN

1

so that

(23)y1

yN+1 1 - A+

L

V

x0

yN+1as N

Hence if there is no solute in the entering L-phase (x0 = 0), then limiting value of the solute fraction not

absorbed in the L-phase is

(24)y1

yN+1 1 - A as N , with A< 1

Case II: When A> 1, the limiting values for the corresponding b functions are

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LimitN->

1

bN

0,

Limit

N->

bN-1

bN

1

Aso that

(26)y1

yN+1

1

A

L

V

x0

yN+1as N , with A> 1

The last result shows that ifx0 = 0, and A> 1, we can absorbed all of the solute in the V-phase as N . One

can show further that counter current operations are more efficient than co-current operations, and for this reason

they are widely used in the chemical process industry. Keep in mind that the above result assumes K is a con-

stant (i.e. independent of the stage), which is a good approximation if the solute concentration is sufficiently

dilute at each stage.


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