Dr Saad Al-ShahraniChE 334: Separation Processes Absorption of Dilute Mixtures Graphical...

Dr Saad Al-ShahraniChE 334: Separation Processes

Absorption of Dilute Mixtures

Graphical Equilibrium Stage Method for Trayed Tower

Consider the countercurrent-flow, trayed tower for absorption (or stripping) operating under:

1. Isobaric, isothermal, continuous, steady-state flow conditions.

2. Phase equilibrium is assumed to be achieved at each of the trays between the vapor and liquid streams leaving the tray.

V

L


Each tray is treated as an equilibrium stage. Assume that the only component transferred from one phase to other is the solute. For application to an absorber

let:L' = molar flow rate of solute-free absorbent

V' = molar flow rate of solute-free gas (carrier gas)

X = mole ratio of solute to solute-free absorbent in the liquid = moles of solute/ mole of solute free-absorbent (solvent)

Y = mole ratio of solute to solute-free gas in the vapor = moles of solute / moles of solute-free gas (inert gas)



With these definitions, values of L' and V' remain constant through the tower, assuming that:

no vaporization of absorbent into carrier gas.

no absorption of carrier gas by liquid.

E.g. (air + NH3) + H2O(solvent)

carrier solute

no H2O evaporate with air

no absorption of air in H2O.

For the solute at any equilibrium stage n, the K-value is given in terms of X and Y as:

n

yn

xn

(1)



where Y = y/(1 - y) and X = x/(1- x ) .

The balances are written around one end of the tower and an arbitrary intermediate equilibrium stage n.

For the absorber (solute balance):

(2)

L`Xn

V`Yn+1

Yn+1

Y1

X0 XN



or, solving for Yn+1

(3)

The terminal points of this line represent the conditions at the top and bottom of the towers.

Equation (3). which is called operating-line equations, is plotted as shown in the last Figure.

The operating line is above the equilibrium line because, for a given

solute concentration in the liquid, the solute concentration in the gas is

always greater than the equilibrium value (driving force from mass

transfer of solute from the gas to the liquid).



The operating line is straight line with a slope of L`/V`.

The terminal point of the operating line at the top of the tower is fixed at X0 by the amount of solute.

The terminal point of the operating line at the bottom of the tower depend on Yn+1 and the slope of the operating line and, thus, the flow rate, L` of solute-free absorbent.



Minimum Absorbent Flow Rate Operating lines for four different absorbent flow rates shown in Figure 6.9

(Seader) as shown below, where each operating line passes through the terminal point, (Y1, X0), at the top of the column and corresponds to a different liquid absorbent rate and corresponding slope, L'/ V'.

X0

(liquid in)XNmin

(for L min)

Moles solute/mole solute-free liquid, X



To achieve the desired value of Y1 for given YN+I , X0, and V', the solute-

free absorbent flow rate L', must lie in the range of (operating line 1) to L`min (operating line 4)

Solute balance for the entire absorber

VYLXVYLX NN 110

N= number of stage in the absorber

(6)

(7)

Notes:

The operating line can terminate the equilibrium line, as operating line 4, but can not cross it because that would be a violation of the second law of thermodynamics (it is possible to transfer a mass from the low concentration to the higher concentration)



The value of L`min corresponds to a value of XN (leaving the bottom of

the tower) in equilibrium with YN+1, the solute concentration in the feed

gas.

It takes an infinite number of stages for this equilibrium to be achieved.

For stage N, eqn. (1) becomes, for the minimum absorbent

)1/(

)1/( 11

NN

NNN XX

YYK

(8)



Solving (8) for XN and substituting the result into (7) gives:

011

11min )1(/

)(

XKKYY

YYVL

NNNN

N

For dilute-solute conditions, where Y y and X x, (9) approach

(9)

01

11min )/(

)(

xKy

yyVL

NN

N(10)

(fraction of solute absorbed)NKVL min (11)

)(

)(

/)(

)(

)/(

)(

01

11

01

11

01

11min

NN

NN

NNN

N

NN

N

Kxy

yyKV

KKxy

yyV

xKy

yyVL



This equation is reasonable because it would be expected that L`min

would increase with increasing V`, K-value, and fraction of solute

absorbed.

The selection of the actual operating absorbent flow rate is based on

some multiple of L`min, typically from 1.1 to 2. A value of 1.5

corresponds closely to the value of 1.4 for the optimal absorption factor

mentioned earlier



NUMBER OF EQUILIBRIUM STAGES

As shown in Figure (a). the operating line relates the

Solute concentration in the vapor passing upward between

two stages to the solute concentration in the liquid passing

downward between the same two stages.

operation

equilibrium Figure (b) illustrates that the equilibrium curve relates the

solute concentration in the vapor leaving an equilibrium

stage to the salute concentration in the liquid leaving the

same stage.

(a)

(b)



X0

X1

X2

X3

Y1

Y2

Y3

Y4

(1)

(2)

(3)E. L

O. L

Start from the top of the tower (at the bottom of the Y-X diagram) and move to the bottom of the tower (at the top of the Y-X diagram) by constructing a staircase alternating between the operating line and the equilibrium line.

TOP

Bottom

N = 3



Example:

When molasses is fermented to produce a liquor containing ethyl alcohol, a CO2-rich vapor containing a small amount of ethyl alcohol is evolved. The alcohol can be recovered by absorption with water in a sieve-tray tower. For the following conditions, determine the number of equilibrium stages required for countercurrent flow of liquid and gas, assuming isothermal, isobaric conditions in the tower and neglecting mass transfer of all components except ethyl alcohol. The entering liquid flow rate is 1.5 times the minimum value.

Entering gas:

180 kmol/h, 98% CO2, 2% ethyl alcohol: 30°C, 110 kPa

Entering liquid absorbent:

100% water; 30°C, 110 kPa

Required recovery (absorption) of ethyl alcohol: 97%



Solution

Assume that the exiting absorbent will be dilute alcohol, whose K-value is determined from a. modified law,

K = Psat/P.

The vapor pressure of ethyl alcohol at 30oC= 10.5 kpa.

At infinite dilution in water activity coefficient of ethyl alcohol is taken as 6.0

therefore, K = 6*10.5/110 = 0.57

NKVL min (fraction of solute absorbed)

V`= 180*0.98 = 176.4 kmol/ h



L`min = (176.4)(0.57)(0.97) = 97.5 kmol/h

The actual solute-free absorbent rate are 1.5 times the minimum

L` = (1.5) L`min = (1.5)(97.5) = 146.2 kmol/h

The amount of ethyl alcohol transferred from the gas to the liquid acid is 97% of the amount of alcohol in the entering gas, or

(0.97)(0.02)(180)=3 .49 kmo/h

the amount of ethyl alcohol remaining in the exiting gas is:

(1-0.97)(0.02)(180) = 0.11 kmol/h

we now compute the alcohol mole ratios at both ends of the operating line as follows:



top X0 =0, Y1 =(0.11)/ (176.4) = 0.0006

bottom YN+1 = (0.11+3.49)/(176.4)= 0.0204 XN =(3.49)/ (146.2) = 0.0239

The equation for the operating line with X0 = 0 is

The equilibrium curve for ethyl alcohol can be determined using the value of K = 0.57 computed above.



For this dilute system in ethyl alcohol, the maximum error in Y is 1.0% if

Y is taken simply as Y = KX = 0.57X

The equilibrium curve, which is almost straight in this example.

Solving for Y, we obtain



● ●

●

Equilibrium curve

Operating lin

e

213

4

5

6

7

Mo

les

of

alco

ho

l/m

ole

of

alco

ho

l fr

ee g

as.

Y

Moles of alcohol/mole of alcohol free liquid. X


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Dr Saad Al-ShahraniChE 334: Separation Processes Absorption of Dilute Mixtures Graphical...

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