Separation Process Technology

Course by Prof. Marco Mazzotti

Summary by Michael Ehrenstein

December 23, 2014

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Contents

1 Fundamentals 61.1 Liquid-vapour equilibrium (LVE) . . . . . . . . . . . . . . . . 6

1.1.1 General expression for the iso-fugacity condition . . . 61.2 Binary systems representation . . . . . . . . . . . . . . . . . . 71.3 Gibbs Phase Rule . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.1 Single component systems . . . . . . . . . . . . . . . . 91.3.2 Binary systems . . . . . . . . . . . . . . . . . . . . . . 10

1.4 Raoults law (ideal systems) . . . . . . . . . . . . . . . . . . . 101.5 Distribution coefficients . . . . . . . . . . . . . . . . . . . . . 111.6 Real systems and Raoults law . . . . . . . . . . . . . . . . . 121.7 Constant of relative volatility (CRV) . . . . . . . . . . . . . . 13

1.7.1 Binary systems . . . . . . . . . . . . . . . . . . . . . . 131.7.2 Multicomponent systems . . . . . . . . . . . . . . . . 14

1.8 Bubble and dew point . . . . . . . . . . . . . . . . . . . . . . 141.8.1 Bubble point . . . . . . . . . . . . . . . . . . . . . . . 141.8.2 Dew point . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.9 Henrys law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.10 Ternary systems representation . . . . . . . . . . . . . . . . . 161.11 Azeotropic systems . . . . . . . . . . . . . . . . . . . . . . . . 191.12 Fundamental principles of mass transfer . . . . . . . . . . . . 221.13 Ficks law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.14 Film model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.14.1 Steady state ordinary molecular diffusion . . . . . . . 231.14.2 One-film model . . . . . . . . . . . . . . . . . . . . . . 241.14.3 Two-film model . . . . . . . . . . . . . . . . . . . . . . 251.14.4 The overall mass transfer coefficient (MTC) . . . . . . 26

1.15 Numerical methods - principals . . . . . . . . . . . . . . . . . 271.16 Half interval method . . . . . . . . . . . . . . . . . . . . . . . 271.17 Newton method . . . . . . . . . . . . . . . . . . . . . . . . . . 281.18 Lever-arm rule . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2 Contactors 302.1 Gas-liquid contactors . . . . . . . . . . . . . . . . . . . . . . . 30

2.1.1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . 302.1.2 Equipment overview . . . . . . . . . . . . . . . . . . . 30

2.2 Tray columns . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.2.1 Description . . . . . . . . . . . . . . . . . . . . . . . . 312.2.2 Tray types . . . . . . . . . . . . . . . . . . . . . . . . 332.2.3 Operative conditions . . . . . . . . . . . . . . . . . . . 35

2.3 Packed columns . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3.1 Description . . . . . . . . . . . . . . . . . . . . . . . . 362.3.2 Packing types . . . . . . . . . . . . . . . . . . . . . . . 37

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2.3.3 Comparison between packed columns and tray columns 382.4 Liquid-liquid contactors . . . . . . . . . . . . . . . . . . . . . 39

3 Flash distillation 393.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.1.1 Definitions and aims . . . . . . . . . . . . . . . . . . . 393.1.2 Process description . . . . . . . . . . . . . . . . . . . . 40

3.2 Binary flash . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.2.1 Problem description . . . . . . . . . . . . . . . . . . . 403.2.2 Data, specifications, unknowns . . . . . . . . . . . . . 413.2.3 Process design . . . . . . . . . . . . . . . . . . . . . . 423.2.4 Process verification . . . . . . . . . . . . . . . . . . . . 45

3.3 Multicomponent flash . . . . . . . . . . . . . . . . . . . . . . 453.3.1 Problem description . . . . . . . . . . . . . . . . . . . 453.3.2 Data, specifications, unknowns . . . . . . . . . . . . . 453.3.3 Split factor . . . . . . . . . . . . . . . . . . . . . . . . 483.3.4 Bubble point . . . . . . . . . . . . . . . . . . . . . . . 493.3.5 Dew point . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.4 Nonideal flash . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.4.1 Process design . . . . . . . . . . . . . . . . . . . . . . 49

3.5 Flash cascades . . . . . . . . . . . . . . . . . . . . . . . . . . 513.5.1 Flash operations at constant temperature . . . . . . . 513.5.2 Flash operations at constant pressure . . . . . . . . . 52

4 Absorption and stripping 544.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.1.1 Definitions and aims . . . . . . . . . . . . . . . . . . . 544.1.2 Process description . . . . . . . . . . . . . . . . . . . . 55

4.2 Gas plant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.2.1 Solvent selection . . . . . . . . . . . . . . . . . . . . . 564.2.2 Industrial schemes . . . . . . . . . . . . . . . . . . . . 57

4.3 Absorption in a single equilibrium stage . . . . . . . . . . . . 594.3.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . 594.3.2 The liquid-vapour equilibrium . . . . . . . . . . . . . . 604.3.3 Single equilibrium stage . . . . . . . . . . . . . . . . . 624.3.4 Co-current cascade configuration . . . . . . . . . . . . 664.3.5 Cross-current cascade configuration . . . . . . . . . . 704.3.6 Counter-current cascade configuration . . . . . . . . . 74

4.4 Absorption operations in tray columns . . . . . . . . . . . . . 794.4.1 Mass balance . . . . . . . . . . . . . . . . . . . . . . . 794.4.2 The operating line . . . . . . . . . . . . . . . . . . . . 804.4.3 Problem description . . . . . . . . . . . . . . . . . . . 824.4.4 Process design and verification . . . . . . . . . . . . . 834.4.5 Process design - linear case . . . . . . . . . . . . . . . 85

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4.4.6 Process design - nonlinear case . . . . . . . . . . . . . 884.4.7 Process verification - linear case . . . . . . . . . . . . 914.4.8 Process verification - nonlinear case . . . . . . . . . . 94

4.5 Stripping operations in tray columns . . . . . . . . . . . . . . 974.5.1 Comparison: absorption vs. stripping . . . . . . . . . 974.5.2 Process design . . . . . . . . . . . . . . . . . . . . . . 984.5.3 Process verification . . . . . . . . . . . . . . . . . . . . 101

4.6 Nonideal operations . . . . . . . . . . . . . . . . . . . . . . . 1024.6.1 Absorption with efficiency stage . . . . . . . . . . . . 103

4.7 Multicomponent operations . . . . . . . . . . . . . . . . . . . 1074.7.1 Design of a multisolute absorption . . . . . . . . . . . 107

4.8 Non-isothermal operations . . . . . . . . . . . . . . . . . . . . 1114.8.1 Design of a non-isothermal operation . . . . . . . . . . 111

4.9 Absorption operations in packed columns . . . . . . . . . . . 1194.9.1 Problem description . . . . . . . . . . . . . . . . . . . 1194.9.2 Design based on the HTU/NTU concept . . . . . . . . 120

4.10 Chemical absorption . . . . . . . . . . . . . . . . . . . . . . . 1274.10.1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . 1274.10.2 Case study: H2S absorption with Diethanolamine . . . 128

5 Distillation 1315.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

5.1.1 Definitions and aims . . . . . . . . . . . . . . . . . . . 1315.1.2 From flash... . . . . . . . . . . . . . . . . . . . . . . . 1315.1.3 ... to distillation . . . . . . . . . . . . . . . . . . . . . 1345.1.4 Industrial distillation . . . . . . . . . . . . . . . . . . . 135

5.2 Process description . . . . . . . . . . . . . . . . . . . . . . . . 1365.2.1 The column streams . . . . . . . . . . . . . . . . . . . 1365.2.2 The distillation column . . . . . . . . . . . . . . . . . 1395.2.3 The working principles: thermodynamics . . . . . . . 140

5.3 Binary distillation design . . . . . . . . . . . . . . . . . . . . 1415.3.1 Problem definition . . . . . . . . . . . . . . . . . . . . 1415.3.2 External balances: mass balance around the column . 1435.3.3 Constant Molar Overflow (CMO) assumption . . . . . 1435.3.4 Operating lines . . . . . . . . . . . . . . . . . . . . . . 1465.3.5 Number of stages: McCabe Thiele procedure . . . . . 1495.3.6 Feed quality . . . . . . . . . . . . . . . . . . . . . . . . 1525.3.7 The reflux ratio . . . . . . . . . . . . . . . . . . . . . . 1545.3.8 Heat balances . . . . . . . . . . . . . . . . . . . . . . . 1565.3.9 Final design . . . . . . . . . . . . . . . . . . . . . . . . 157

5.4 Multicomponent distillation design . . . . . . . . . . . . . . . 1575.4.1 Problem definition . . . . . . . . . . . . . . . . . . . . 1575.4.2 The external balances issue . . . . . . . . . . . . . . 1585.4.3 (C 2) assumptions to solve the external balances . . 159

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5.4.4 Approximate shortcut methods . . . . . . . . . . . . . 1615.4.5 Fenske equation . . . . . . . . . . . . . . . . . . . . . . 1625.4.6 Underwood equations . . . . . . . . . . . . . . . . . . 1655.4.7 Gilliland correlation . . . . . . . . . . . . . . . . . . . 1695.4.8 Position of feed stage . . . . . . . . . . . . . . . . . . . 170

5.5 Non-standard distillation . . . . . . . . . . . . . . . . . . . . 1705.5.1 Partial condenser . . . . . . . . . . . . . . . . . . . . . 1705.5.2 Total reboiler . . . . . . . . . . . . . . . . . . . . . . . 1725.5.3 Open steam (or direct steam) . . . . . . . . . . . . . . 1735.5.4 Side streams . . . . . . . . . . . . . . . . . . . . . . . 1745.5.5 Multiple feeds . . . . . . . . . . . . . . . . . . . . . . . 177

5.6 Azeotropic column . . . . . . . . . . . . . . . . . . . . . . . . 1835.6.1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . 1835.6.2 Extractive distillation . . . . . . . . . . . . . . . . . . 1845.6.3 Azeotropic distillation . . . . . . . . . . . . . . . . . . 1865.6.4 Pressure swing distillation . . . . . . . . . . . . . . . . 189

6 Liquid-liquid extraction 1916.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

6.1.1 Definitions and aims . . . . . . . . . . . . . . . . . . . 1916.1.2 Process description . . . . . . . . . . . . . . . . . . . . 191

6.2 Single stage operations . . . . . . . . . . . . . . . . . . . . . . 1936.2.1 Single stage: problem definition . . . . . . . . . . . . . 1936.2.2 Single stage: process design . . . . . . . . . . . . . . . 1946.2.3 Single stage: solvent range . . . . . . . . . . . . . . . . 195

6.3 Multistage operations: cross-current . . . . . . . . . . . . . . 1956.3.1 Cross-current: cascade configuration . . . . . . . . . . 1956.3.2 Cross-current: process design . . . . . . . . . . . . . . 1966.3.3 Cross-current: effect of the number of stages on sepa-

ration . . . . . . . . . . . . . . . . . . . . . . . . . . . 1986.4 Multistage operations: counter-current . . . . . . . . . . . . . 200

6.4.1 Counter-current: column configuration . . . . . . . . . 2006.4.2 Counter-current: stream definition . . . . . . . . . . 2016.4.3 Counter-current: process design I . . . . . . . . . . . . 2026.4.4 Counter-current: solvent rate and position: Smin

and Dmin . . . . . . . . . . . . . . . . . . . . . . . . . 2046.4.5 Counter-current: process design II . . . . . . . . . . . 206

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This document contains most things found on the hyper-TVT site. It alsocontains some errors, and is probably incomplete.

1 Fundamentals

1.1 Liquid-vapour equilibrium (LVE)

Thermal equilibrium: Heat-transfer stops,Tliq = Tvap (1)

Mechanical equilibrium: Forces are balanced, usually that means:Pliq = Pvap (2)

Phase equilibrium: rate of condensation = rate of evaporation, nochange in composition. If T and P are constant, equal rates require aminimum of free energy, and:

li = vi (3)

If the chemical potentials are the same, the fugacities (because di =RTd ln(fi) are also:

fLi (x, T, P ) = fvi (y, T, P ) (4)

1.1.1 General expression for the iso-fugacity condition

In the most general form, the iso-fugacity condition can be expressed as

xii(x, T, P )fLi,pure(T, P ) = yii(y, T, P )P (5)

with i =aixi

and i =fiPi

being the activity and fugacity coefficients, respec-tively, and a being the activity. i refers to a specific component, and, in themost general case, the activity and fugacity coefficients depend on temper-ature, pressure and all compositions (making x and y vector compositions).Equation (5) can be further simplified for most practical cases, using thefollowing assumptions:

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The liquid fugacity of the pure component i can be expressed via itsvapour pressure:

fLi,pure(T, P ) = Pvi (6)

The pressure has a negligible effect on the activity coefficient:i(x, T, P ) = i(x, T ) (7)

The gas is ideal, which means that the fugacity coefficient for thecomponent i is equal to 1:

i(y, T, P ) = 1 (8)

With these simplifications, the isofugacity condition can be rewritten:

xii(x, T )Pvi (T ) = yiP (9)

Activity coefficients can be calculated using different correlations, e.g. thecorrelations of Margules.

1.2 Binary systems representation

A liquid-vapour equilibrium of a binary system can be graphically repre-sented in different ways:

T vs. x P vs. x y vs. x H vs. x

One of the most convenient is mole fraction in gas phase vs. mole fractionin liquid phase (y vs. x, usually of the more volatile component). The equi-librium curve can refer to either constant temperature or constant pressure,with the other value changing along the equilibrium line (which representstwo phases in equilibrium):

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In the temperature vs. composition diagram, there are two lines representingT vs. x and T vs. y: The liquid equilibrium line and the gas equilibriumline, respectively. There are 3 regions: At high temperatures, only vapouris present, at low temperatures, only liquid is present. Finally, there is aregion representing a liquid-vapour equilibrium, with different compositionsyi and xi of the vapour and liquid phase.

At T1, only vapour is present. Moving to T2 in the biphasic region, a certainamount of vapour condenses. The same kind of diagram is obtained for a Pvs. x representation.Finally, there is the enthalpy vs. composition diagram.

Isotherms can be drawn between the saturated vapour and saturated liquidcurves, these are called tie-lines. Example:

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1.3 Gibbs Phase Rule

The degrees of freedom describe the numbers of variables that can be arbi-trarily specified without modifying the system:

degrees of freedom = # variables# constraints (10)

For non-reacting systems, this relationship can be expressed via the Gibbsphase rule (with F = degrees of freedom, C = number of components andP = number of phases):

F = C P + 2 (11)So, for a binary system with 2 phases, there are 2 degrees of freedom. There-fore, 2 variables can be set to uniquely identify the system (e.g. T and P,T and x, P and x); there are 2 independent variables, while the third isdefined by the other two. The Gibbs phase rule refers to intensive variables(independent of the size of the system). Examples of extensive variables arevolume, flow rate and number of moles.

1.3.1 Single component systems

Applying the Gibbs phase rule to the following example, we get 2 degrees offreedom for point A. So, 2 variables can be changed (pressure, temperature)without changing the phase. The same would happen for point B. For pointC, however, the Gibbs phase rule only returns 1 degree of freedom. Only oneof the variables can be chosen at will, while the other is defined by the firstone. In doing so, one would move along the liquid-vapour equilibrium linewhile keeping the same state of the system. Finally, we have the point T:

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There are no degrees of freedom left, no matter which variable is changed,the state of the system would change. This point is called the triple point.The temperature and pressure corresponding to this point are the criticalvalues, Tc and Pc.

1.3.2 Binary systems

Using the diagram from earlier,

point A would give us 3 degrees of freedom. Since we are at constant pres-sure, we can still change 2 variables (temperature and composition) and stillhave the state of the system remain unchanged. On point B, we would have2 degrees of freedom, meaning that apart from the already fixed pressure,we can still change temperature or composition, the remaining variable thenbeing defined by the former. The same holds for points C and D.

1.4 Raoults law (ideal systems)

Raoults law states, that The partial pressure of a component is equal toits mole fraction in the liquid multiplied by its vapour pressure. Raoults

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law is the isofugacity condition for ideal vapour and liquid phases. For abinary system, it can be expressed as follows:

PA = xAPvA (12)

PB = xBPvB (13)

Since it holds, that PA = yAP , PB = yBP and P = PAPB, it follows that:

yA = xAP vAP

(14)

yB = xBP vBP

(15)

Raoults law can be shown in a P vs. x diagram:

1.5 Distribution coefficients

Raoults law can also be expressed using the distribution coefficients kA andkB:

kA =yAxA

=P vAP

(16)

kB =yBxB

=P vBP

(17)

They are strongly dependent on the temperature, since P vA and PvB are also,

according to the Antoine equation (with A, B and C being constants forpure compounds easily found in the literature):

ln(P vi ) = AB

T (K) C (18)

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1.6 Real systems and Raoults law

In reality, the pressure-composition correlation is not linear, as expressed byRaoults law, but is much rather represented by curves (green):

To apply Raoults law to real gases, the activity coefficient i can be in-cluded:

PA = AxAPvA (19)

PB = BxBPvB (20)

where A and B depend on temperature, pressure and composition:

i = i(T, P, xi) (21)

For i > 1, the deviation from Raoults law is positive and the curves bendon top, as seen in the previous figure.For i < 1, the deviation from Raoults law is negative and the curves bendbelow, as seen in the next figure.

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For i = 1, we get an ideal system. At high values of xA, one can use theideal version of Raoults law, since the deviation from the real case becomesvery small.

1.7 Constant of relative volatility (CRV)

1.7.1 Binary systems

The constant of relative volatility is the ration of the distribution coefficientsof the two components:

AB =kAkB

(22)

which means, that:

AB =P vAP vB

(23)

These equations hold for binary, ideal systems following Raoults law. It ispossible to write the following system of equations:

yA = kAxA (24)

1 yA = kB(1 xA) (25)And, combining them:

yA =ABxA

1 + (AB 1)xA (26)

The relative volatility is often just a weak function of temperature, and cantherefore be assumed constant in small temperature ranges. Equation (26)is the equation of a curve and can be drawn in an y vs. x diagram:

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1.7.2 Multicomponent systems

The most general expression for the distribution coefficient of component iis:

ki =if

Li,pure

iP(27)

For most practical cases, it simplifies to (using (6) and (8)):

ki =iP

vi

P(28)

Or, assuming an ideal mixture when Raoults law is valid:

ki =P viP

(29)

For multicomponent systems, it is useful to define the constant of relativevolatility with respect to a reference component r:

i =kikr

(30)

withkr =

yrxr

(31)

So we get: i

yi = kri

ixi (32)

kr =1i ixi

(33)

yi =ixii ixi

(34)

1.8 Bubble and dew point

1.8.1 Bubble point

The bubble point is the temperature, at which a liquid mixture starts to boil.At a defined pressure and composition, bubble and dew point are fixed. Thebubble point TBP fulfils the following equations:

i

yi = 1 (35)

yi = ki(TBP )xi (36)

It can be calculated iteratively. In order to do this, one must first estimate avalue for TBP . From this, ki(TBP , P, xi) is calculated (using the known totalpressure and Antoine equation constant). Finally, the iteration continuesuntil:

i

kixi = 1 (37)

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1.8.2 Dew point

The condensation temperature TDP fulfils the following system of equations:i

xi = 1 (38)

xi =yi

ki(TDP )(39)

Like the bubble point, it can be determined iteratively, the one differencebeing the exit condition:

i

yiki

= 1 (40)

1.9 Henrys law

As we saw earlier in the pressure-composition diagram, we can use Raoultslaw for high compositions, since the linear behaviour coincides with thetangent to the curve. For xi

HiP

= mi (43)

where mi is the slope of the equilibrium line in the y vs. x - diagram:

The larger the slope, the more willing a component is to leave the liquidphase for the gas phase. This corresponds to the temperature dependenceof Hi:

Hi(T ) = H0 exp( ERT

) (44)

So, Hi increases with temperature, as does the slope. Absorption is thereforefavoured at low temperatures and high pressures.

1.10 Ternary systems representation

Ternary systems are usually represented using either right triangular or equi-lateral triangular diagrams. Any point in the following example representsa different composition of the system:

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In particular:

The points at the three apexes A, B and C represent the three purecompounds.

Any point on a side of the triangle represents a binary mixture of thetwo points that enclose the side.

Any point in the triangle (e.g. S, R) represents a mixture of the threecomponents A, B and C.

The sum of the lengths of the perpendicular lines drawn from any pointin the diagram is equal to the height of the triangle. So, if the height isscaled from 0 to 100, the length of the line perpendicular to (for example)the side opposite of A corresponds to the percentage of A in the mixture.The contents of B and C can be determined in the same way.The degrees of freedom can be determined using the Gibbs phase rule. Atthe point S (outside the immiscibility region), we have 4 degrees of freedom.Since pressure and temperature are set, this leaves us with 2 compositionswe can chose. The third one is then determined. A point in the immiscibilityregion, like R, only has 1 degree of freedom left (after subtracting pressureand temperature). The compositions of the two phases are determined bythe end points of the tie line (M, P).The curve LMNOPQ represents the saturation curve for the system, alsocalled solubility envelope. Herein, the three components are not fully mis-cible, and any system inside this immiscibility region (like R) will split into

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two phases in equilibrium to both ends of the corresponding tie line. PointO is called the plait point, the two phases have the same composition andbecome one (the tie line becomes a point). The solubility envelope changeswith different temperatures and pressures. For example, it becomes smallerfor higher temperatures:

When the process design of a ternary system uses graphical methods, it isoften more convenient to employ a right triangular diagram, e.g.:

A diagram typically doesnt contain all tie lines that are needed for thegraphical solution of problems. Additional tie lines can be constructed givenan auxiliary line.

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One starts with the given blue auxiliary line, B-K. Then, choosing any pointP on this line, parallels through P to A-C and B-C are drawn. The resultingintersections of these two lines with the edge of the immiscibility region givesone tie line. Reversely, given a couple (the more lines, the more precise) tie-lines, one can reconstruct the auxiliary line. The same procedure can befollowed with right right triangular diagrams.

1.11 Azeotropic systems

Sufficiently non-ideal systems with relatively close boiling points often showazeotropic behaviour (B and C):

Not all non-ideal systems have to show azeotropic behaviour. For example,systems can have only positive ( > 1)

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or negative ( < 1) deviations from ideality.

When substances have a very large positive deviation from ideality (verylarge ), the liquid mixture has a minimum boiling point at the azeotropiccomposition:The liquid mixture boils, but the composition of the liquid and gas phasesdo not change any further

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The vapour and liquid compositions are identical, and the distribution co-efficients and k-values of all components are equal to 1.

The three previous and next figures represent homogeneous azeotropes, atthe azeotropic composition L, there is only one liquid phase.Negative deviation from ideality leads to maximum boiling point azeotropes:

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Heterogeneous azeotropes exists when the two components are only par-tially miscible as liquid. A vapour mixture at the azeotropic compositionL, condenses into two different liquid phases at equilibrium compositionsrespectively K and M.

Heterogeneous azeotropes are always minimum boiling, because of the verylarge positive deviation from ideality which is necessary to cause the sep-aration of the liquid phases. All presented systems are binary, so no morethan one heterogeneous azeotrope can exist (one vapour phase can coexistwith a maximum of two liquid phases).

1.12 Fundamental principles of mass transfer

Mass transfer is described as the movement of a component from one loca-tion to another due to disequilibrium, between phases or in the same phase.There are two basic mechanisms for mass transfer, molecular diffusion andturbulent diffusion, where molecular diffusion is a lot slower than its tur-bulent counterpart. The first one happens due to spontaneous microscopicmotion of individual molecules, while the second one is a random, macro-scopic fluid motion. Here, only molecular diffusion due to concentration gra-dients will be considered, focussing on binary systems. Other driving forces

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other than the aforementioned concentration gradient will not be consid-ered. Most systems treated in this course have mass transfer taking placeacross a two-phase interface. Therefore, it is important to choose the equip-ment in such a way, that the interface between the two exchanging phasesis maximised.

1.13 Ficks law

Assuming a system of homogeneous fluid composition is divided by a wall.Adding particles to one of the two divisions, then removing the wall, shows aconcentration gradient (after some time). A disequilibrium had been createdwithin the same phase, and the particles started moving in direction of lowerparticle concentration z.

This phenomenon, molecular diffusion, stops, when the red particles areuniformly distributed in the phase. The flux of components can be describedby Ficks law:

Ni = Didcidz

[mol s1 cm2

](45)

The orders of magnitude for the diffusion coefficient:

Liquids: Di = 105 cm2 s1

Gases: Di = 101 cm2 s1

Solids: Di = 1010 1013 cm2 s1

1.14 Film model

1.14.1 Steady state ordinary molecular diffusion

In separation processes, it is very important to consider the mass transferacross an interface between a gas and a liquid (absorption, stripping, flashevaporation, distillation) or between two liquids (liquid liquid extraction).The key concept of the film model is that all resistances to mass transferbetween two phases reside in the phases themselves, there is no resistanceto crossing the actual interface. The resistances occur in small regions toboth sides of the interface.

23

The film model can be viewed as a membrane separating two bulk phases ofuniform compositions. Concentration differences exist between the sides ofthe membrane and act as the driving force of mass transfer. The membranerepresents the liquid film, and not the interface!

1.14.2 One-film model

In the one-film model, only the film on the liquid side is considered forthe mass transfer resistance. We assume to have a pure gas of componentsA diffusing (absorbing) into a nonvolatile liquid B. Therefore, there is noresistance on the gas side and at the interface, the equilibrium condition isgiven by Henrys law:

cA,i = HAPA (46)

The assumption is made that all molecules of A diffusing into the liquid filmwill diffuse into the bulk of B.

Due to the film being very thin, this assumption is valid. Bulk flow of A isneglected, so all transfer of A is due to molecular diffusion.Applying the assumption of a steady-state to Ficks law (45), we get:

dNAdz

=d(DAB(dcAdz ))

dz= 0 (47)

24

d2cAdz2

= 0 (48)

or, using the mole fractions xA =cAc

NA = DAB dxAdz

(49)

and applying the material balance

dNAdz

= 0 with xz=0 = xAi and xz= = xA,bulk (50)

we get:

NA = DAB c xA,i xA,bulk

(51)

The ratio DAB can be replaced by the mass transfer coefficient kc to give:

NA = kc c (xA,i xA,bulk) (52)

NA = kc (cA,i cA,bulk) (53)Therefore, the one-film model postulates a linear concentration gradientwithin the film. The rate of mass transfer, NA is directly proportional to theconcentration difference, which acts as the driving force for mass transfer.

1.14.3 Two-film model

The two-film model assumes that resistances occur in both a small film onthe liquid side of the interface as well as in a small film on the gas side of theinterface. Again, the interface adds nothing to the resistances, and we canassume that the liquid and gas phases are at equilibrium at the interface

yi = feq.(xi) (54)

and, if Henrys law can be applied, that the equilibrium can be assumedlinear:

yi = mxi (55)

At steady state, with no accumulation possible, we can say that

N = kG cG (y yi) = ky(y yi) (56)

N = kL cL (xi x) = kx(xi x) (57)for the gas and liquid side, respectively, where ky and kx are the locallyapplicable mass transfer coefficients. Since equations (56) and (57) must beequal, it is possible to write:

ky(y yi) = kx(xi x) (58)

25

The ratiokxky

=(y yi)(xi x) (59)

is the slope connecting the operation line to the equilibrium line, or, in otherwords, the line connecting equilibrium composition with bulk composition.

Therefore, if we know the local mass transfer coefficients, we could calculatethe rate of mass transfer. Unfortunately, it is impossible to get close enoughto the interface to experimentally determine these coefficients. Therefore, adifferent expression for N has to be found.

1.14.4 The overall mass transfer coefficient (MTC)

In order to get rid of the interface concentrations, we express the drivingforce via. equilibrium concentrations using Henrys law:

y = y y = y mx (60)N = Ky(y y) (61)

With Ky being the overall mass transfer coefficient. It can be related to thelocal mass transfer coefficients via the following steps:

N = Ky((y yi) + (yi y)) (62)N = Ky((y yi) +m(xi x)) (63)

It still holds, that:

y yi = Nky

(64)

xi x = Nkx

(65)

and, finally:1

Ky=

1

ky+m

kx(66)

26

1Kx=

1

mky+

1

kx(67)

If the gas diffusing into the liquid is very soluble, absorption is favoured,the slope of the equilibrium line, m, is very small, and the second term inequation (66) also becomes very small. It then holds, that:

1

Ky=

1

ky(68)

For the opposite case, if the gas is not soluble, then the slope of m becomesvery large. The first term in equation (67) can be neglected and we get:

1

Kx=

1

kx(69)

1.15 Numerical methods - principals

It is often required to find the root(s) for a problem of the form

f(x) = 0 (70)

since all analytical expressions can be written in this form. Following aretwo numerical methods for solving problems of this kind.

1.16 Half interval method

This method is generally used when there is no analytical expression forthe derivative of f(x) with respect to x available. The algorithm works asfollows:

27

Of course, choosing a and b for each subsequent iteration uses xnew as eithera or b. Also, a requirement for a and b is:

f(a)f(b) < 0 (71)

1.17 Newton method

This algorithm is applied if an analytical expression for the derivative of thefunction is available, it is significantly faster than the half interval method.First, a starting value xi is chosen (x0 in the graph below). The derivativef (x) is then calculated. The condition is, that

y = f(xi) + f(xi)(x xi) = 0 (72)

from which a value for the next iteration can be found:

xi+1 = xi f(xi)f (xi)

(73)

This equation can also be obtained via the equations for the tangent.

28

1.18 Lever-arm rule

The lever-arm rule is based on the same principle as the see-saw moment-arm balance. In equilibrium, it has to hold that:

(weight F1) distance(F1M) = (weight F2) distance(MF2) (74)

The lever-arm rule can be applied to mixing calculation on enthalpy vs.composition as well as on ternary diagrams for extraction. In the secondcase, the lever-arm rule can be used to easily solve overall mass balances fortwo mixing streams.

Doing the overall mass balance, it becomes apparent that F1, F2 and M lieon the same straight line.

29

The mass of the two streams F1 and F2 is then proportional to their respec-tive distance to the point M from the point representing the other stream(shorter distance MF2 means heavier F1).

2 Contactors

2.1 Gas-liquid contactors

2.1.1 Principles

Gas-liquid mass transfer takes place across an interface at the phase bound-ary over a certain contact time. Any equipment used for this kind of oper-ation should provide a high interface area. Additionally, proper mixing isalso a requirement to be considered in the design of a good contactor.

2.1.2 Equipment overview

Many types of column designs are available: Tray columns, packed columns,spray towers, bubble columns, centrifugal contactors, ... . The mainly differin the choice of continuous phase, which can be either the gas or the liquid,and in the mode of flow, which can be co-current, counter-current or cross-current. In the end, they all must enable a sufficient rate of mass transferby providing adequate mixing and contact time.

Tray columns are cylindrical vessels, which operate in either cross- orcounter-current operation and can be operated under pressure. The masstransfer takes place in separate trays.

30

Packed columns are vertical cylindrical vessels containing packing mate-rial between sections 1 and 2 in the following figure. The liquid flows downthe column via gravity, while the gas rises from the bottom. Mass transfertakes place between the liquid film coating the packing area (thereby greatlyincreasing the contact area) and the gas bubbling through from the bottom.Packed columns can be operated under pressure.

Spray columns are cylindrical vessels filled with gas through which theliquid is sprayed. They can be advantageous when the solute is very solublein the liquid phase.

Bubble columns are cylindrical vessels filled with liquid through whichthe gas phase is bubbled.

Tray and packed columns are the devices most used in separation technologyusing gas-liquid mixtures.

2.2 Tray columns

2.2.1 Description

Tray columns can be operated under pressure and consist of a cylindricalvessel in which gas and liquid flow either flow in a cross-current or counter-current fashion to each other. Mass transfer occurs on a series of metal traysor plates.

31

In counter-current operation, gas and liquid flow through the same perfo-rations, the plates take up the whole cross-section of the column. In cross-current operation, the liquid flows over the plate and down a downcomer,and only part of the cross-section is taken up by the plate. The operatingrange and liquid flow pattern may be controlled for a better mass transferefficiency via the area of the downcomer, which leads to cross-current sys-tems being more widely used. The number of downcomers is not limited toone. In a double pass plate, for instance, two downcomers are arranged, sothat the horizontal direction of liquid flow changes with every plate.

On the other hand, the more downcomers are employed, the lower the areaavailable for gas-dispensers (perforations in the plate) becomes.Due to plate disposition and stream flow pattern, different zones are presentaround a plate, with different regimes.

32

In the above figure, zone A is the downcomer from the above plate. Liquidenters the tray here as a continuous phase and flows through the zone aeratedby the upflowing gas stream (tray), before departing to the next tray onthe right hand side (D). Zone B is a continuous liquid directly above theperforations of the plate. There is a froth (bubbly, or aerated, mixture ofvapour and liquid, with observable height) over the tray. Liquid dropletsare ejected into the gas phase above the froth. Under specific circumstances(high volumetric vapour to liquid flow), this zone can invert to a continuousvapour spray, meaning that the vapour continuous phase is extended and itsliquid content will decrease with height (distance to the tray). A properlydesigned and functioning sieve tray operates in froth contacting mode if atall possible. Zone C describes the continuous vapour phase containing liquiddroplets. The next downcomer is shown as D, with the only difference to Abeing the location of entry (right vs. left). Finally, zone E describes the gasflow to the plate.

2.2.2 Tray types

The most common and used plates are:

Sieve plate

33

A sieve plate is usually a plate with simple round perforations. Up-flowing gas prevents the liquid from flowing through these perfora-tions, although with low gas rates, this can still occur (weeping phe-nomenon). Different kinds of perforations can be used to increaseperformance.

Valve plate

The valve plate-design minimises the phenomenon of weeping. It con-sists of a cap covering the perforation and legs to limit the verticalrise of said cap. The lower the gas flow, the lower the rise of the valve,although this also depends on pressure. This allows for control of theamount of liquid draining to the plate below and therefore also of theliquid retention time on the plate.

Bubble- (or bell-) cap plate

34

A bubble-cap plate consists of fixed caps mounted over the perfora-tions. The caps have rectangular or triangular slots cut around itsside. The gas flows up through the perforations, rises, turns aroundand passes out through the slots of the cup directly into the liquid,producing a froth regime. The caps are well sealed to the platesurface, preventing the liquid from draining at low gas flow rate.

Tray comparison Plates types can be compared on the basis ofcost, pressure drop, efficiency, vapour capacity and flexibility.

Usually, sieve plates are preferred because of their low cost, unlessflexibility is required. Because of the high associated pressure drop,bubble caps are rarely used, although they have been widely used inthe past, especially for cross-flow plates. They have now largely beenreplaced by valve plates.

2.2.3 Operative conditions

Four regimes can be distinguished in a tray column, depending on flow ratesand continuous phase:

1. Bubbles regime

35

The liquid is quiescent and the gas bubbles through it. This is thecase at low gas flow rates. Due to the poor mixing of gas and liquid,this case exhibits low stage efficiency. This low efficiency, coupledwith the low gas flow rate, makes the bubbles regime undesirable forcommercial applications.

2. Foam regime

At higher gas flow rates compared to the bubbles regime, the gas risesfaster and a foam is formed on top of the liquid. The stage efficiency isincreased by a higher interfacial area (more bubbles), though, shouldthe foam become too much and too stable, the phenomenon of en-trainment becomes excessive: Liquid is carried to the tray above, thestage efficiency drops and the column may even become flooded (filledwith liquid) and inoperative. An additional chemical antifoam agentis therefore required.

3. Froth regime

The liquid is the continuous phase and the gas passes through it asjets or bubbles, with the surface of the liquid boiling and splashing. Avery high mass transfer due to a very good liquid-gas contact makesthis the most common and favoured regime.

4. Spray regime

At very high gas flow rates, the gas is the continuous phase and theliquid is sprayed into it as fine droplets. Because of the poor liquidmixing, mass transfer rates are usually low in this regime. A shiftfrom froth to spray regime can result in a significant reduction instage efficiency (e.g. from 65% to 40%).

2.3 Packed columns

2.3.1 Description

Packed columns are cylindrical vessels containing one or more sections ofpacking material. The liquid moves downward through gravity, while thegas flows upward in a counter-current fashion. The packing material aims to

36

maximise the gas-liquid interfacial area to enhance mass transfer. Packingcan be either structured (a.) or random (d.). The packing material restsbetween a lower support plate and an upper hold-down plate (b.), whichaim at avoiding packing movement. The liquid distributer (c.) is situatedabove the hold-down plate, and ensures even distribution of the liquid flow.

2.3.2 Packing types

The types of packing material can be split into random packing and struc-tured (aka. ordered, stacked or arranged) packing. A good packing materialshould have a high surface area to volume ratio, since mass transfer takesplace between the liquid layer coating the packing material and the gasflowing through the packing elements. Additionally, a high liquid flow ca-pacity through the packing layer is required to uniformly coat the packingmaterial with liquid (good wettability) as well as good resistance to pres-sure (high strength) as well as a low pressure drop. The larger the size ofthe packing material units, the lower the pressure drop, but also the masstransfer. Optimal packing materials compromise between the two. Metaland plastic packing material has largely replaced the ceramic of the old days.

For random packing, the old rashig rings (d3) and berl saddles (d2, yel-low pieces) are substituted by pall rings, intalox saddles, mini cascade ring,and a broad range of other different shaped elements (d1, d2). However,ceramic elements are still used when dealing with corrosive materials andgood wettability is required. Metal elements provide good wettability and

37

high strength. Plastic elements, on the other hand (usually polypropylene)have poor wettability and fail at high temperatures. Due to their low cost,they are nevertheless widely used.

Structured packing is a lot more cost intensive than random packing, buthas the advantage of higher mass transfer as well as a very low pressuredrop.

2.3.3 Comparison between packed columns and tray columns

Packed columns offer a lower operating range:

Too low liquid flow rate: Wetting may not be sufficient. Too hight liquid flow rate: Plate columns are often more economic. If solid particles are present in the liquid: Cleaning of the packing is

very expensive and complicated compared to plate columns.

Working under stressed temperature and pressure conditions: Packingelements can easily break.

On the other hand, packing can also be the more favourable option:

Costs: Packed columns are usually less expensive for small columndiamenters (< 0.6 m).

Handling of corrosive chemicals: Cheap ceramic or other chemicallyresistant packing can easily be used.

Foaming liquid: Foaming liquid can be handled more easily in packedcolumns because of the lower agitation of the liquid by the gas.

38

2.4 Liquid-liquid contactors

Liquid-liquid contactors employ agitation of the liquid in order to enhancemass transfer by increasing the dispersion of one phase in the other andincreasing the interfacial area. The agitators create mixing zones, whichalternate with settling zones along the length of the column.

Various commercially available column designs for liquid-liquid contactorsmainly differ in the design of the agitators and stator disks. (Respectivelya and b in the above figure). The goal of the extraction process is to purifyan aqueous stream (heavy, dispersed phase) by extracting a solute with awater-immiscible organic solvent (light, continuous phase). The pollutant,having a higher affinity for the organic phase, is transferred to the latteralong the length of the column.

3 Flash distillation

3.1 Introduction

3.1.1 Definitions and aims

Flash evaporation is one of the simplest unit operations. A liquid stream ispartially vaporised in a flash drum at a certain pressure and temperature.The result is vapour, which is richer in the more volatile component than

39

the remaining liquid. The opposite operation to flash distillation (partialevaporation) is partial condensation (with a complete vapour feed). Knowl-edge of the dew point of the mixture is very important.

Usually, flash distillation cannot achieve a large degree of separation, and istherefore employed as an auxiliary operation to prepare streams for furtherprocessing. In some cases however, like the desalination of sea water, com-plete separation can be achieved.

If only two components are present in the feed, the flash is called binary,while more than two components in the feed define a multicomponent flash.

3.1.2 Process description

The flash can be seen as a distillation with only one equilibrium stage. Theoperation stops, when the liquid and vapour streams reach the equilibriumcompositions defined by temperature and pressure, and the two streams caneasily be separated.

The incoming liquid is first heated and pressurised, before being fed into thedrum. Due to the large pressure drop, the liquid evaporates very quickly(hence flash).

3.2 Binary flash

3.2.1 Problem description

In the case of a binary flash, the feed consists only of two components: Aand B. One of the key assumptions here is, that the flash drum acts as asingle equilibrium stage, so liquid and vapour are in equilibrium at the end

40

of the flash operation.

The goal is now either to design a new flash unit for a new separation or toverify that an existing flash drum is unit is good to carry out an existingone.

3.2.2 Data, specifications, unknowns

Using the Gibbs phase rule, we receive 2 degrees of freedom for the system,which should translate to 7 equations and 5 variables. Let us check that forthe dashed line in the figure.

Usually, the design of a flash drum consists of determining the temperatureand pressure at which the flash takes place, the amount of heat providedand finally the liquid and vapour stream compositions and flow rates. Thisgives us:

Unknowns: Tdrum, Pdrum, QH , V, y, L, x

For the sake of simplicity, x and y refer to the more volatile component.

The data of the feed mixture is usually known:

Data: TF , PF , F, zFFor this system, the following equations can bewritten:

Phase equilibrium equations (using eq. (12), Raoult):

yP = xP vA(T ) (75)

(1 y)P = (1 x)P vB(T ) (76)

41

These equations can also be written using the distribution coefficients (viaeq. (29):

y = xkA(T ) (77)

(1 y) = (1 x)kB(T ) (78)

Material balance equations:

F = L+ V (79)

Fz = Lx+ V y (80)

Energy balance equations:

FhF +QH = LhL + V HV (81)

So, 5 equations and 7 variables confirms our 2 degrees of freedom.

In order to solve our system of equations, the two degrees of freedom mustbe specified, meaning that 2 of the 7 unknowns must be fixed. This choicedepends on the availability of the data. In the aforementioned cases of pro-cess design and process verification, the availability of starting data differs.Additionally, in real life industrial situations there are often limitations onoperating conditions, and therefore temperature and pressure are often nota free choice. Usually, one of the two specifications in a flash drum is thepressure. For the last degree of freedom, several approaches can be consid-ered, depending on process design or process verification.

It is important to note, that the equilibrium and mass balances containvariables, which do not appear in the energy balance equation and the otherway around. In this case, the two sets of equations are decoupled. In thesequential procedure of solving the system, we will go through the first groupof equations (equilibrium and mass balance) and solve the system for L, V ,x, y and after this, the energy balance would be solved to get QH .

3.2.3 Process design

In this case, the given task is to design a new flash unit for a requiredseparation. Given the following system data:

Unknowns: Tdrum, QH , V, y, L, x

Data: TF , Pdrum, PF , F, zF

42

We assume that the pressure of the flash drum is known. The system nowhas one degree of freedom left, and we need to specify this variable to un-equivocally determine the system. The following possibilities, which variableto specify, can be considered:

a. Vapour mole fraction y

b. Liquid mole fraction x

c. V/F : Fraction of vaporised feed, also indicated as .

d. L/F : Fraction of feed remaining liquid.

e. Temperature in the drum, Tdrum

1st case: x or y is known. The solving procedure is an iterative methoduntil convergence:

1. Assume a temperature Tdrum and calculate y (or x) from one of theequilibrium equations.

2. Verify, if the found composition value also fulfils the second equilibriumequation.

3. Find L and V from the mass balance equations.

4. Solve the energy balance equation for QH .

2nd case: V/F or L/F is known. With this, and the overall massbalance equations

F = L+ V (79)

Fz = Lx+ V y (80)

it is possible to draw the operating line

y = LVx+

F

Vz (82)

43

and find x and y graphically.

The temperature can now be found from the ratio:

y

x=P vA(T )

P(83)

The energy balance is an independent equation, with the unknown QH . hLand HV can be found from an enthalpy-composition diagram of the binarymixture,

while hf must be calculated from the following equation (since the feed isneither a saturated liquid, nor has equilibrium composition):

hf = xACp,A(T Tref) + xBCp,B(T Tref) (84)

3rd case: Tdrum is known. If Tdrum is known, PVA (T ) and P

VB (T ) can

be calculated. Using P , we can now calculate x and y using the phaseequilibrium equations

yP = xP vA(T ) (75)

(1 y)P = (1 x)P vB(T ) (76)

44

and, subsequently, L and V can be calculated via the mass balance equa-tions. QH can be calculated from the independent energy balance.

Very often, the distribution coefficients kA(T ) and/or kB(T ) are given. Theyare related to the vapour pressures via equations (16) and (17):

kA =yAxA

=P vAP

(16)

kB =yBxB

=P vBP

(17)

3.2.4 Process verification

The purpose of the verification problem is to check, how good a separationcan be achieved in an existing flash unit. The following system is given:

Data: TF , PF , F, zF , Pdrum, QH

In this case, QH is a known value. This is often true, when the flash drumis insulated and process is assumed to be adiabatic, i.e. QH = 0.

Unknowns: Tdrum, V, y, L, x

The solving procedure in this case follows an iterative path:

1. Assume a temperature Tdrum and calculate x and y from the equilib-rium equations (75) and (76).

2. Calculate L and V from the mass balance equations (79) and (80).

3. Verify, if the found values fulfil the energy balance equation (81).

3.3 Multicomponent flash

3.3.1 Problem description

If the feeding stream has more than 2 components, an analytical procedure isneeded. The key assumption, that the flash drum acts as a single equilibriumstage, implies that the vapour and liquid streams leaving the drum haveequilibrium composition.

3.3.2 Data, specifications, unknowns

The system scheme looks like the one for the binary system:

45

For this system, we can write:

Unknowns: Tdrum, Pdrum, QH , V, yi, L, xi = 2C + 5 in total, wherei = 1, 2, ..., C are the components.

The data of the feed is known:

Data: TF , PF , F, ziand the following equations can be written:

Phase equilibrium equations: C independent equations.

yi = kixi with i = 1, ..., C (85)

The phase equilibrium equations are usually written in terms of dis-tribution coefficients:

ki = ki(Tdrum, Pdrum, xi) (86)

For an ideal system, ki does not depend on the mole fractions and theabove equation reduces to:

ki = ki(Tdrum, Pdrum) (87)

Stoichiometric rules: 2 independent equations.Ci=1

yi = 1 (88)

Ci=1

xi = 1 (89)

46

Material balance equations: C independent equations.F = L+ V (90)

Fzi = Lxi + V yi with i = 1, ..., C 1 (91)

Energy balance equations: 1 independent equation.FhF +QH = LhL + V HV (92)

Combining equations (88), (89), (110), (91) and (92), we get a total of(2C + 3) independent equations. This confirms, that a multicomponentflash (like a binary flash) has 2 degrees of freedom [(2C + 5) unknowns -(2C + 3) equations].

As discussed for the binary case, one of these two degrees of freedom istaken by the pressure of the flash drum, Pdrum, while for the last remain-ing degree of freedom, the choice depends on the available data, which inturn depends on the kind of problem (design vs. verification).

The following procedure can be used to solve the system of equations:

First, use the phase equilibrium equations (85)

yi = kixi (85)

and replace yi in the mass balance equation (91):

Fzi = Lxi + V kixi (93)

solving for xi:

xi =Fzi

L+ V ki(94)

and substituting L by the mass balance equation (110):

xi =Fzi

F V + V ki (95)

It is common to divide the numerator and denominator by the feed rate F,we get expressions for xi and yi:

xi =zi

1 + (ki 1)VFyi =

kizi

1 + (ki 1)VF(96)

The ratio VF is usually denoted by the Greek letter , because it is boundedbetween 0 and 1. Applying the stoichiometric relations (88) and (89)

Ci=1

xi = 1

Ci=1

yi = 1 (97)

47

equations (96) can be rewritten as:

Ci=1

zi1 + (ki 1) = 1

Ci=1

kizi1 + (ki 1) = 1 (98)

and, subtracting equations (98) from each other:

Ci=1

kizi1 + (ki 1)

Ci=1

zi1 + (ki 1) = 1 1 (99)

Ci=1

(ki 1)zi1 + (ki 1) = 0 (100)

In order to solve equation (100), two cases can be considered:

ki are known: Find the root to the function f() = 0 to find .Once is known, equations (96) are used to calculate xi and yi.

is given: Equation (100) is used to find Tdrum, following an iterativeprocedure:

1. Assume an initial value T drum.

2. Calculate ki with the assumed Tdrum.

3. Calculate the root of the function f(Tdrum) = 0.

4. If the root coincides with the assumed T drum, this is the value forTdrum. If not, start again at point 1.

For the analytical search of the root of a function f(x) = 0, one of thenumerical methods, presented in the basics section, can be used.

3.3.3 Split factor

If the feeding stream contains more than 2 components, an analytical pro-cedure is needed.

The split factor is defined as:

Si =V yiLxi

(101)

Using the well-known equilibrium equation

xiPvi = yiPi (102)

48

this can be rewritten as

Si =V P viLPi

(103)

With the overall mass balance, we get:

Fzi = Lxi + V yi (104)

Dividing by Lxi, the split factor appears:

FziLxi

= 1 +V yiLxi

= 1 + Si (105)

This can be rearranged to

Lxi =Fzi

1 + SiV yi =

FziSi1 + Si

(106)

from which we can obtain:

Ci=1

V yi = VCi=1

Lxi = L (107)

3.3.4 Bubble point

It is sometimes important to know the bubble point of the liquid feed. Theflash drum is used to bring the liquid to its boiling point (the very first dropbecomes vapour). At this point, no vapour is produced yet and = VF = 0.Of the two degrees of freedom that we know this system has, one is thepressure (as usual), while the other one will be defined by . zi now coincideswith xi, since everything remains in the liquid phase. Once we know xi, wecan make an assumption for TBP and follow an iterative procedure.

3.3.5 Dew point

Analogous to the bubble point. In this case, the feed is entirely in the vapourphase and the purpose is to bring it to the dew point. No liquid is produced,and = 1, and therefore also zi = yi. Knowing xi, we again assume TBPand proceed iteratively.

3.4 Nonideal flash

3.4.1 Process design

In case of a real system, the previous assumed independence of the distri-bution coefficients from the mole fractions is no longer valid:

ki = ki(Tdrum, Pdrum, xi) (86)

49

Using the isofugacity condition

fLi (x, T, P ) = fvi (y, T, P ) (4)

liquid fugacity = vapour fugacity (108)

and assuming an ideal gas, we can write:

xii(P, T, x)Pvi (T ) = yiP (9)

with i being the activity coefficient which can be obtained from the Mar-gules correlations.

So, for a multicomponent, ideal flash, we get:

ki(P, T, xi) =yixi

= i(P, T, x)P vi (T )

P(109)

We can write the following equations:

Material balance equations: C independent equations.F = L+ V (110)

Fzi = Lxi + V yi with i = 1, ..., C 1 (111)

Phase equilibrium equations: C independent equations.

ki =yixi

with i = 1, ..., C (112)

Stoichiometric rules: 2 independent equations.Ci=1

yi = 1 (113)

Ci=1

xi = 1 (114)

As with the ideal case, everything can be rewritten in the form ofequation (100):

Ci=1

(ki 1)zi1 + (ki 1) = 0 (100)

Again, this system is solved iteratively:

1. Assume an initial value for the mole fraction xn. Usually, oneassumes an ideal system for this.

50

2. From this mole fraction, calculate ki(P, T, xn).

3. With this, use equation (100) to find .

4. From the mass and equilibrium balances, recalculate xn.

5. If xn is close enough to xn, it is the true value. If not, the iteration

is repeated from step 1 until convergence.1

3.5 Flash cascades

3.5.1 Flash operations at constant temperature

As noted previously, flash operations can be seen as distillations with onlyone equilibrium stage. For this reason, high degrees of separation usuallycannot be achieved. Therefore, a flash is usually the first process in in-dustrial processes in order to achieve a crude separation before feeding thestream to a distillation or other separation/purification process.

Due to the parallels between flash and distillation, it is possible to describethe latter through a series of units of the former.

In this figure, all units operate at the same temperature but different pres-sure. Towards the top, the stream becomes richer in the most volatile com-ponent until the purified stream V1 is obtained. On the bottom, the opposite

51

happens, and the least volatile component is collected in L5. Additionally,other undesired streams (Ln6=5, Vn6=1) are produced and can be recycled (af-ter repressurising) to the previous flash stage in order to reduce waste.

For a binary system, this flash cascade can be described in a P vs. x diagram.The composition of the liquid (or vapour) stream fed to the cascade of stages(green points in the figure below) is very close to the composition of therecycled streams to the same stage. Obviously, the total feed composition(not shown) will lie somewhere between these two, depending on the flowrates of the two streams.

The flash cascade can be a useful operation. However it becomes unattrac-tive when several stages are needed to reach the specifications on operatingand economical considerations. Every stream of the flash cascade must bepressurised to reach the conditions for the following flash operation, and notneeded outlets must be re-pressurised back when recycled to the previousstage.

3.5.2 Flash operations at constant pressure

The other option that presents itself would be to keep the pressure constantwhile changing the temperature, shown in the following T vs. x diagram.

52

Again, the composition of the liquid or vapour stream fed to the cascade isindicated by the green points, and the total feed composition (not shown)will lie somewhere between the compositions of recycle and feed, dependingon the flow rates. In this case, feeds leaving the individual flash units haveto be heated up (in the case of liquid streams ) or cooled down (in the caseof vapour streams).

53

Again, the flash cascade allows for a better degree of separation, while onthe other hand the constant re-heating can make it very inconvenient.

4 Absorption and stripping

4.1 Introduction

4.1.1 Definitions and aims

Absorption is a unit operation in which one or more selected components,the solutes, are removed from a gas mixture (solute + carrier gas) by contactwith a nonvolatile liquid stream, the solvent. The carrier gas is assumed tobe insoluble in the solvent, so only the solute moves to the liquid phase.

Mass transfer of solute to the solvent is due to its greater affinity towardsthe liquid phase compared with the gas phase. The transport of the soluterequires that the two phases are in contact for a certain contact-time.

The affinity can be of physical or chemical nature. In the latter case a re-action is present between solute and solvent termed as chemical absorption.An example of this is the removal of CO2 or H2S removal by NaOH throughreversible reaction. An irreversible reaction occurs when using MEA (mo-noethanolamine) instead of NaOH. In the case of irreversible reactions, theresulting solvent must be disposed of, whereas in reversible reactions, thesolvent can be regenerated. Therefore, reversible reactions are often pre-ferred.

In physical absorption only physical interactions are present between soluteand solvent. An example is the transport of acetone from an air stream toa water stream. The acetone dissolves into the water while the air passes out.

Stripping is the opposite operation of absorption. In this unit operation,one or more components of a liquid mixture are removed by evaporationinto an insoluble gas stream. The same assumptions as in the absorptioncase are valid here. i.e. the carrier liquid is nonvolatile and the gas is insolu-ble into the solvent. Only the solute moves from the liquid to the gas stream.

54

The aim of these operations is the separation of the inlet stream (gas forabsorption and liquid for stripping) from one or more components mixed init. This may be needed for one of the following reasons:

These components (the solutes) are pollutants and they have to beremoved from the gas or liquid stream before it is released to theenvironment.

They are valuable components and they have to be recuperated beforethe stream is wasted.

They are intermediate products and they must be separated from thestream to be used in further productions.

However it is interesting to note that in these operation units, while therequired stream is being purified, another one is being polluted with theseparated components, which moved into it. For this reason, as we see inthe next section, stripping and absorption are two very linked processes.

4.1.2 Process description

The most typical configuration for an absorption/stripping column puts thetwo streams in counter-current. The gas enters the column from the bottomwith a solute concentration of yn+1 and a flow rate of G. The liquid entersat the top, flowing down the column by gravity, having an initial soluteconcentration of x0 and flow rate of L.

This configuration has proved to most successful, since it doesnt require alot of energy for the flow transport. The two streams exchange matter alongthe length of the column, before exiting on the opposite sides. In absorp-tion, the solute moves from the gas to the liquid stream, while the opposite

55

happens in stripping.

The gas leaves the column at the top with a solute concentration of y1 andthe same flow rate G, while the liquid leaves at the bottom with a soluteconcentration xn and the same flow rate L. If the two streams exchangesolute, for the case of absorption, it holds that y1

reversible (under certain conditions such as at an elevated tempera-ture). Otherwise the solvent cannot be regenerated.

b) Selectivity The selectivity is an important factor when the processdeals with the separation of one or more components from a mixtureof gases. Only when the solvent has more affinity for the componentsto be separated, the process can be successful.

c) Solvent regeneration process Since the solvent is usually recycled,regeneration should be taken into account: The relative ease of regen-eration may determine the selection of the solvent.

4.2.2 Industrial schemes

If a gas stream has been purified by a suitable solvent in an absorption col-umn, what happens to the polluted solvent? How expensive would it be touse fresh solvent in every feed?

An appropriate approach to this would be regeneration of the liquid solventand recycle it to the absorption column. This is done by a stripping opera-tion. Therefore, most of the time an absorption operation is coupled with astripping counterpart.

Different plant configurations are possible depending on the nature and diffi-culty of the solvent regeneration. Thermodynamic aspects (desorption tem-perature or decomposition temperature of the solvent), solvent volatility,and chemical/physical aspects like corrosivity, viscosity, toxicity, as well ascosts, are taken into account when choosing the solvent for a specific ab-sorption process keeping regeneration in mind. When solvent volatility isvery low, i.e. solvent is not present in the gas stream, a simple regenerationprocess by heating is adequate. The system then looks very simple, andfollowing are three possible configurations:

57

In the first figure, the liquid absorbent to be recycled enters the strippingcolumn at the top and comes into contact with the stripping vapour. Thiscan be a steam or any inert gas with the right thermodynamic conditionsto strip the polluted solvent. The clean absorbent is then recycled to theabsorption column.

This second option has the to be recycled absorbent enter a re-boiled strip-ping column. The stripping vapour consists of the liquid solvent itself. Partof the recovered solvent is continuously recycled to the absorber.

58

Lastly, a distillation column can also be used for regeneration. The pollutedabsorbent is fed to the distillation column. In the bottom, the solvent iscollected and sent back to the absorber.

4.3 Absorption in a single equilibrium stage

4.3.1 Assumptions

In order to approach the study of the absorption/stripping operation, acertain number of assumptions are needed in order to simplify the designand easily understand the basics concepts:

In reality, most of the time, the above assumptions about gas and liquidstreams are true. However different options can be considered for the so-

59

lute, the equilibrium, and the temperature, i.e. multicomponent, nonlinearequilibrium, nonisothermal operation.

4.3.2 The liquid-vapour equilibrium

Linear equilibrium

Typically in an absorption or stripping problem the solute, which hasto be removed, is present in the liquid or in gas phase at very lowconcentration (1% ), the relationship in between y and x is not linearany more but generally curved, expressed as

yA = f(xA) (117)

which can be visualised as:

60

Mole ratios vs. mole fractions in the equilibrium representa-tion

Besides the mole fraction, there are others ways to express the gas andthe liquid composition. In some cases it can be more useful to expressit with the mole ratios, which refer to the moles of carrier instead ofthe total number of moles as a reference:

Xi =Ni,liquidNn,liquid

=moles of solute i in the liquid

moles pure carrier liquid (component n)(118)

Yi =Ni,gasNn,gas

=moles of solute i in the gas

moles pure carrier gas (component n)(119)

Molar fractions and molar ratios are linked with the following rela-tionships:

xi =Xi

1 +Xiyi =

Yi1 + Yi

(120)

Xi =xi

1 xi Yi =yi

1 yi (121)(122)

If the equilibrium is a straight line y = mx in the x vs. y diagram, itwill be a curve Y = f(X) in the X vs. Y counterpart:

61

However in most of the cases, as we have already pointed out, we are inthe condition of infinite dilution of the solute (very low concentration).In this case the following simplification can be made:

xi Xi yi Yi (123)

which allows us to assume a linear equilibrium in the x-y and X-Yplane.

As we can see, this is of a key importance to simplify the application ofthe McCabe-Thiele graphical methods for the design of an absorptionor a stripping process.

4.3.3 Single equilibrium stage

During the absorption operation, the gas phase and the liquid phase mustbe in contact. Before considering the different possible configurations, weconsider the thermodynamical aspects of this contact between phases.

The solute contained in G transfers to the liquid phase L. The concentrationin the gas decreases while the concentration in the liquid increases. Thepairs of points (concentration of solute in the gas and liquid phases) at eachmoment constitute the operating line.

62

Mass transfer ends when equilibrium is reached. Graphically, it means thatthe operating line crosses the equilibrium line. The driving force for absorp-tion can be qualitatively seen as the distance between the equilibrium andoperating line. Therefore, when the two lines cross, the driving force is zero.

The gas at the inlet has a certain concentration of solute, represented by themolar fraction, y0. The solvent is put in contact with the gas phase. Thesolvent can be pure or can have a certain concentration of solute, representedby x0.

The equilibrium between phases can be represented in an x vs. y diagram fora given temperature and pressure. Sometimes the equilibrium is a straightline, y = mx. The inlet compositions of the two phases can be representedin the diagram.

63

After contact of the inlet gas stream with the solvent, the concentration ofthe solute in the gas phase decreases: y < y0. Likewise, the solvent is nowcharged with solute: x > x0. Because this is an ideal equilibrium stage,equilibrium between the phases is reached. The mass balance equation is:

L x0 +G y0 = L x+G y (124)total mass of solute at inlet = total mass of solute at outlet

This can be solved with respect to y:

y =

(y0 +

L

G x0) LG x (125)

The slope of the operating line is then LG :

Because this is an equilibrium stage, the equilibrium is reached and theoperating line touches the equilibrium line:

64

The equilibrium line represents a barrier and determines a region of values(x, y) that cannot be reached for any ratio LG .

Using the equilibrium and operating line, a value for y corresponding to theintersection can be found:

y =

(y0 +

L

G x0) LG x y = mx (126)

y =

(y0 +

L

G x0) LG ym

(127)

y =

(y0 +

L

G m x0

m

) LG ym

(128)

Per definition, the ratio

A =L

G m (129)is called absorption factor A.

y =

(y0 +A m x0

) LG ym

(130)

Using the equilibrium equation, the product m x0 can be replaced by y0,i.e. the composition of a gas at equilibrium with a liquid of composition x0(we do this, because we want to get all our xs out of the equation. Theusual equilibrium, y0 = mx0 doesnt work here, because the two variablescome from different streams).

y =

(y0 +A y0

) LG ym

(131)

65

This finally gives for y:

y =y0 +A y0

1 +A(132)

The fraction of absorption is defined as:

=y0 yy0 y0

= 1 11 +A

(133)

Here, the numerator is the distance between y0 and y, while the denominatoris the distance between y0 and the composition of a gas at equilibriumwith x0. This represents the maximum solute exchange. The fraction ofabsorption can also be written in terms of the absorption factor A.

A 0 0 y y0A 1 y y0

y0 represents the minimum concentration of solute in the gas phase, so themaximum absorption (A tends to infinity).

4.3.4 Co-current cascade configuration

The assumptions remain the same as for the last case. When the gas flow isparallel to the solvent flow in a cascade of ideal stages, only the first stageis effective. As we saw when we studied the single ideal stage, the solventflow is at equilibrium with the gas at the outlet. This means that the con-centration of solute in the solvent is maximal and, at the same conditionsof temperature and pressure, the absorption operation cannot be more ef-fective. There is no driving force at equilibrium.

66

A cascade of single stages, all of them at the same conditions, is then equiv-alent to one single stage.

First, we consider two ideal stages. The solvent is put in contact with thegas phase in the first stage. The solvent can be pure or can have a certainconcentration of solute, represented by x0.

The equilibrium between phases can be represented in an x-y diagram fora given temperature and pressure. Again, lets consider it a straight liney = mx.

The inlet compositions of the two phases can be represented in the diagram(x0, y0).

The content of solute in the gas after the first stage is now lower than atthe entrance: y1 < y0, and the solvent is now charged with solute, x1 > x0.Because this is an ideal equilibrium stage, the equilibrium between the twophases is reached.

Repeating what we saw before in the case of a single ideal stage, the changein composition in the two phases can be represented in the following diagram:

67

The equation for y from before, with the operating line LG :

y =

(y0 +

L

G x0) LG x (134)

For the single ideal stage, we derived the equation for y1

y =y0 +Ay

0

1 +A(135)

with A being the absorption factor. Now, we have a second ideal stage,which can be handled analogously, following the same derivation as for thesingle equilibrium stage:

y2 =y1 +Ay

1

1 +A(136)

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or, for any stage:

yi =yi1 +Ay0

1 +A(137)

The aim of a second stage would be to have y2 smaller than y1, so that weimprove the operation by decreasing the concentration of pollutant in thegas stream.

But since the equilibrium represents a barrier and determines a region ofvalues (x, y) that cannot be reached for any ratio LG and any number ofstages

we have already reached equilibrium and can move no further.

One can also arrive at this conclusion mathematically: Starting at

y2 =y1 +Ay

1

1 +A(138)

we reach equilibriumy1 = y1 (139)

so we get

y2 =y1(1 +A)

(1 +A)(140)

and finallyy2 = y1 (141)

The composition equation for the last stage mirrors the one for the first

yn =y0 +Ay

0

1 +A(142)

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as does the fractional absorption:

=yn+1 y1yn+1 y0

= 1 11 +A

(143)

4.3.5 Cross-current cascade configuration

Again, the assumptions remain unchanged.

In this case the solvent stream is divided in n streams before entering then stages of the cascade. The solvent flows in cross-current configuration tothe gas flow.

At every stage the solvent that comes in is fresh. The solvent coming out isat equilibrium with the gas stream and is not driven to another stage butcollected and mixed with the other solvent outlets.

That way, the performance of the operation is improved with respect to asingle stage or a co-current cascade.

Each step has a different operating line, since the conditions of the concen-tration in the gas change. The slope of the operating line is constant, sincewe consider variations on the gas flow not being significant and the amountof solvent is kept constant.

We consider three ideal stages.

70

The solvent stream is divided and each new stream is driven to a stage. Thesolvent can be pure or can have a certain concentration of solute, representedby x0. Every stage is receiving solvent with the same composition x0. Everystage also receives the same amount of solvent, (L/3). The equilibriumbetween phases can again be represented in an x-y diagram, for a giventemperature and pressure, as a straight line y = mx.

Again, the inlet compositions can be represented in the diagram; the contentof solute in gas is lower than at the entrance, while the concentration ofsolute in the liquid is higher. The equilibrium between phases is also reached.Compared to what we saw before, there is a subtle change in the expressionfor y:

y =

(y0 +

L/3

G x0) L/3

G x (144)

Therefore, the slope of our operating line is now ( L3G), and, in terms ofabsorption factor A:

y1 =y0 + (A/3)y

0

1 + (A/3)(145)

71

Looking at the second ideal stage, the gas at the entrance has a concentra-tion equal to y1. Because the solvent has the same composition as before,the new operating point must be located on the same vertical.

The outlet of the second stage is again at equilibrium, and because gas andsolvent flow are approximately constant, the new operating line has the sameslope: ( L3G).

Since the equilibrium is reached, we can easily draw the new operating line.For the concentration in terms of A, we get (material balance, divide by G,

replace (x2 =y2m ), replace with A, replace (x0 =

y0m ), replace

LGm with A,

solve for y2):

y2 =y1 + (A/3)y

0

1 + (A/3)(146)

Other stages work analogously. We find y0 in the numerator for every stage,since the fed solvent always has the same concentration x0.

72

Sometimes, the solvent flows are chosen to be different at each stage. Thismakes sense, since the concentration of solute in the gas decreases in eachstage and thus the driving force decreases as well as the solvent requirements.Then, the slopes of the operating lines are different and must be calculated.

L1G

L2G

L2G

(147)

In principle, the solvent flow rate will increase and thus the absolute valueof the slope will also increase.

Looking at the expressions for the final composition of the gas coming outof the last stage (yn), which is of primary interest, we see that we have tosolve the equation:

yi =yi1 + (A/N)y0

1 + (A/N)(148)

This can be rearranged to give

yi =1

1 + (A/N)yi1 +

(A/N)

1 + (A/N)y0 (149)

which is a First Order Difference Equation. A First Order Difference equa-tion of the form

yi = yi1 + (150)

has the solution

yi = y0i +

1 i1 (151)

We see, that in our case the coefficients are

= y0(A/N)

1 + (A/N) =

1

1 + (A/N)(152)

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and the general solution therefore

yi = y01

(1 +A/N)i+ y0

(1 1

(1 +A/N)i

)(153)

So, for the last stage, where i = n:

yn = y01

(1 +A/N)n+ y0

(1 1

(1 +A/N)n

)(154)

and for the fractional absorption:

=y0 yny0 y0

= 1 1(1 +A/N)n

(155)

4.3.6 Counter-current cascade configuration

Once again, the assumptions remain the same.

In the counter-current configuration, the gas flows in the opposite directionto the solvent. The gas, rich in pollutant at the entrance, is put in contactwith the solvent already charged with pollutant (outlet). On the contrary,on the other side, the gas now poor in pollutant is put in contact with theclean solvent.

This is the characteristic of any counter-current process, and is convenientbecause the driving force is nearly constant along the cascade (the opera-tion). On the diagram the driving force is represented qualitatively by thedistance between the equilibrium and the operating lines.

Consider three ideal stages. The solvent stream enters the cascade by thefirst stage. It can be pure or can have a certain concentration of solute,represented by x0. The gas enters the cascade by the last stage and flows incounter-current to the solvent flow.

74

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

(x0, y1) (x1, y2) (x2, y3) (x3, y4)

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

(x1, y1) (x2, y2) (x3, y3)

We now do a local (dotted area in the next figure) partial (only for thesolute) mass balance:

75

Gyj+1 + Lx0 = Gy1 + Lxj (156)

Again, with the equilibrium equation y = mx and after dividing by G, weget the operating line

yj+1 =L

Gxj + (L

Gx0 + y1) (157)

which gives us a positive slope LG for our operating line.

Known data in this case are the initial gas composition, yn+1, as well as theinitial solvent composition, x0. When the final gas composition, y1, is known,because it is restricted to less than a certain minimum value (specification),the first point of the operating line can be set.

76

For the specific slope (LG), we can set the operating line through that point.

The graphical resolution consists of a step by step construction. We knowthat x1 is at equilibrium with y1. This lets us set x1 on the diagram. Wealso know that x1 and y2 are on the same operating line, since they are atthe same levelin the cascade. And so on and so forth, until the specificationis reached.

77

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

Introducing the absorption factor in the mass balance equation obtainedbefore gives us

yj+1 = Ayj + (y1 Ay0) (158)Comparing this equation to the general Difference Partial equation, we get

= A = y1 Ay0 (159)Applying the general solution, we find

yj =

(y0

y1 Ay01A

)Aj +

y1 Ay01A A 6= 1 (160)

This can be transformed (j = n+ 1) to

yn+1 = y0A

n+1 + (y1 Ay0)1An+1

1A (161)

for the (n+ 1)th stage.

The fractional absorption can now also be calculated by substituting thevalue of yn+1:

=yn+1 y1yn+1 y0

= 1 A 1An+1 1 A 6= 1 (162)

This is the Kremser equation. Solving the Kremser equation for n, thenumber of stages to the specified purity can be calculated:

n =

ln

(1/A1

)lnA

(163)

78

When A = 1, the Kremser equation cannot be used. The fraction of absorp-tion is then calculated via

=yn+1 y1yn+1 y0

= 1 11 +A

A = 1 (164)

4.4 Absorption operations in tray columns

4.4.1 Mass balance

Lets consider an absorption case. In the most typical configuration the gasand the liquid enter and leave the column at opposite ends (counter-currentconfiguration).

The gas enters from the bottom with a flow rate of G and a compositionof yn+1, while the liquid enters from the top with a flow rate of L and acomposition of x0.

The gas leaves the column at the top with the same flow rate of G and acomposition of y1, while the liquid leaves the column at the bottom withthe same flow rate of L and a composition of xn.

79

This gives us the known overall mass balance:

Lx0 +Gyn+1 = Lxn +Gy1 (165)

For the mass balance for a section between stage 1 and the generic stage i,we can write:

Lx0 +Gyi+1 = Lxi +Gy1 (166)

Solved for yi+1, this equation gives:

yi+1 =L

Gxi + (y1 L

Gx0) (167)

4.4.2 The operating line

The operating line is the points of compositions (xi, yi+1), where xi andyi+1 are respectively the compositions of liquid and gas stream exchangingmatter in the generic stage i, as shown in the figure below:

(xi, yi+1) are correlated by the mass balance at the stage (i+ 1):

yi+1 =L

Gxi + (y1 L

Gx0) (168)

80

In the x-y diagram this correlation can be represented by a straight line withslope LG and it is known as operating line.

In general the ratio LG is not constant since during the process the solutemoves from one phase to another, hence modifying the flow rates of boththe liquid and the gas phase.

However in the condition of infinite dilution of the solute (very low concen-tration), the amount of absorbed solute is so small that its passage from onestream to the other doesnt significantly impact the flow rates of the twoinvolved streams.

Moreover the following assumptions have been made:

1. Solvent is not volatile

2. Carrier gas is not soluble

Therefore the liquid and gas flow rates can be considered constant and theequation is represented in the x-y plan as a straight line:

If for any reason, the assumption of infinite dilution is not true, only the inertpart of the liquid and the gas phase can be considered and their compositionswould be expressed as mole ratios. In this case it is:

1. L0 = inert part of liquid (solvent without solute)

2. G0 = carrier gas (without solute)

3. Xi = liquid composition in mole ratio =LxiL0

4. Yi = gas composition in mole ratio =GyiG0

81

From this, we get mass balance

L0X0 +G0Yi+1 = L0Xi +G0Y1 (169)

from which we get the operating line

Yi+1 =L0G0

Xi + (Y1 L0G0

X0) (170)

which is again a straight line in the X - Y plane.

More in general the mole ratio is not very often used because of the difficultyof representing the equilibrium with a linear correlation in the (X-Y) plane.

Whenever it would be possible the assumption of very diluted solution willbe made, allowing for the use of the mole fractions to express the gas andliquid compositions.

4.4.3 Problem description

A gas phase containing a pollutant with a concentration of yn+1 must bepurified. To this aim, a suitable liquid solvent is used. The initial liquidsolute concentration x0 is very low or zero.

The two streams exchange solute during the passage in the column.

The outlet gas is now cleaned and contains a pollutant concentration withinspecification y1.

82

The pollutant is removed with the outlet liquid which now contains a con-centration of pollutant xn.

The column under examination is a rather standard tray column, where theliquid and gas stream are brought into counter-current contact.

The operating conditions of the column, like temperature T and pressure P ,are selected in order to favour the absorption process in terms of thermody-namic equilibrium, and to optimise the separation process.

Once again, the following assumptions are made:

The assumption, whether a linear or non-linear equilibrium is assumed, de-pends on the solute concentration in the liquid feed, xsolute. I