Mass Transfer and Mass Transfer Operations - Erden Alpay and Mustafa Demircioğu

MASS TRANSFER and MASS TRANSFER OPERATIONS

Erden ALPAY Mustafa DEMİRCİOĞLU

Ege University-Engineering Faculty-Chemical Engineering Department Bornova-İzmir-TURKEY

September-2006

TABLE OF CONTENTS (You can click subject name in order to reach the related page)

Preface vii Introduction 1

Chapter-1 : MASS TRANSFER BY MOLECULAR DIFFUSION 7

1.1 Introduction 7 1.2 Mass Transfer by Molecular Diffusion in the Gases 9

Integration of the General Flux Equation 9 Equimolar Counter Transfers of A and B 9 A Transfers through Non-Transferring B 10 The Relationship between NA and NB is Fixed by Reaction Stoichiometry 11 The Relationship between NA and NB is Given by the Latent Heats of Vaporization 12 Mass Transfer with Varying Cross-Sectional Area 12 Mass Transfer by Molecular Diffusion in Multi-Component Gas Mixtures 14 Determination of Binary Diffusivities of Gases 14 Experimental Determination 14 Prediction of Binary Gas Diffusivities 18 Estimation of Binary Gas Diffusivities 20 Effect of Temperature and Pressure on Gas Diffusivity 22

1.3 Mass Transfer by Molecular Diffusion in Liquids 22 Determination of Molecular Diffusivities in Liquids 23 Experimental Method 23

Estimation of Liquid Diffusivities 25 Molecular Diffusivity in Concentrated Liquid Solutions 29

Molecular Diffusivities in Electrolytic Solutions 29 Effect of Temperature on Molecular Diffusivity in Liquids 30 Molecular Diffusivity in Multi-Component Liquid Solutions 30

1.4 Continuity Equation for a Binary Mixture 30 1.5 Mass Transfer by Molecular Diffusion in Solids 33

Diffusion that is Independent of the Nature of the Solid 33 Steady-State Diffusion 33 Unsteady-State Diffusion 34 Diffusion that is Dependent on the Nature of the Solid 37 Diffusion of Liquids in Solids 37 Diffusion of Gases in Solids 38

The Relationship between the Fluxes at the Diffusion of Gases in Solids 42 Molecular Diffusivity of Gases in Solids and the Permeability 42

Chapter-2 : MASS TRANSFER BY TURBULENT DIFFUSION and MASS TRANSFER COEFFIENTS 46

2.1 Introduction 46 2.2 A and B Transfer under Equimolar Counter Transfer Conditions 48 2.3 A Transfers through Non-Transferring B 49 2.4 Mass Transfer Coefficients in Laminar Flow 49

Mass Transfer from a Gas into a Liquid Film in Laminar Flow 51

2.5 Mass Transfer Correlations 56 2.6 Mass Transfer Theories 59

Film Theory 59 Penetration Theory 60

2.7 Determination of Effective Concentration Difference for Calculation of Average Flux 61

Chapter-3 : MASS TRANSFER BETWEEN TWO PHASES 68

3.1 Introduction 68 3.2 Equilibrium between Phases 68 3.3 Mass Transfer between Two Phases 69 3.4 Mass Transfer Flux 70 3.5 Overall Mass Transfer Coefficients and Overall Driving Forces 72 3.6 The Relationships between Individual and Overall Mass Transfer Coefficients

73

Chapter-4 : GAS ABSORPTION 78

4.1 Introduction 78 4.2 Gas-Liquid Equilibrium 79

Ideal Solutions 80 Real Solutions 81

4.3 Selection of Solvent 83

4.4 Absorption Operations 85

Gas Absorption in Continuous Contact Type of Equipment 85 Wetted-Wall Column 85 Spray Column 88 Packed Column 88 Gas Absorption in Packed Column 95 Calculation of Number of Transfer Units 100 Calculation of Individual Heights of Transfer Units 104

Determination of Diameter of a Packed Column 106 Stage-Wise Contact Type of Absorption 110 Gas Absorption in Plate Columns 110

4.5 Non isothermal Absorption 126 4.6 Gas Absorption with Chemical Reaction 130

Gas Absorption with Instantaneous Chemical Reaction 133 Calculation of Height of Packing when Chemical Reaction is Instantaneous 135 Calculation of Height of Packing when Chemical Reaction is Slow 138

Calculation of Height of Packing when Rates of Diffusion and Chemical Reaction are Comparable 141

Chapter-5 : DISTILLATION 150

5.1 Introduction 150

5.2 Liquid-Vapor Equilibria 150

Ideal Solutions 153 Deviation from Ideality: Real and Azeotropic Solutions 155 Partial Solubility and Insolubility of the Components in Liquid Phase 158 Volatility and Relative Volatility 161 K-Values 162 Bubble Point Temperature 164 Dew Point Temperature 164 Enthalpy-Composition Diagrams 166

5.3 Methods of Distillation 168 Equilibrium or Flash Distillation 168 Equilibrium or Flash Distillation under Constant Pressure 168

Flash Distillation by Reducing the Pressure of the Heated Liquid 171 Simple or Differential Distillation 175 Rectification or Fractionation 182 Continuous Rectification in a Plate Column 185 Rectification in Batch Operation 219 Rectification in Packed Column 229 Rectification of Azeotropic Solutions 233

5.4 Internal Design of Plate Columns for Liquid-Gas (Vapor) Contact 237 Column Diameter 239 Plate Spacing (P.S.) 240 Liquid Entrainment 241 The Holes 241 Weir 242 Pressure Drop in the Gas along the Plate 242 Design Steps 245

Chapter-6 : LIQUID-LIQUID EXTRACTION 254

6.1 Introduction 254

6.2 Liquid-Liquid Equilibria 255

6.3 Selection of Solvent 258

6.4 Extraction Operations 259 Stage-Wise Operations 259 Single Stage Extraction 260 Cross-Current Multi-Stage Extraction 262

Counter-Current Multi-Stage Extraction 265 Counter-Current Multi-Stage Extraction under Reflux 269 Design of Mixer-Settler Units 272

Appendices 276 References 284

Preface

This CD-book covers the subject-matter on mass transfer and mass transfer operations. The text is primarily intended for undergraduate students in chemical engineering, however the first three chapters may also be used by biochemical, food and environmental engineering students. Practicing process engineers may find the book useful to refresh themselves. The first three chapters of the book provide the students with background necessary to understand the mass transfer operations, dealt with in the subsequent three chapters. In Chapter1, molecular diffusion in gases , liquids and solids is discussed in some details, including determination and prediction of diffusion coeffients. Chapter 2 covers the mass transfer by turbulent diffusion and introduces the student to mass transfer coefficients and to their use. In Chapter 3, mas transfer between two insoluble phases is dealt with and concept of mass transfer resistance is emphasized. The following three chapters deal with macroscopic separation operations based on interface mass transfer, which are gas absorption, distillation and liquid-liquid extraction. Operations are discussed in both plate and packed columns including their internals and sizing. This CD-book was first published in 2004 and corrected and expanded in the years 2005 and 2006. Another CD, prepared for the teachers, contains the animated power point presentations of the topics and solved problems of this CD book. It is consisted of 1387 slides ( 695 in English, 692 in Turkish) and is sent to the interested colleagues. Erden Alpay Mustafa Demircioğlu [email protected] [email protected] September, 2006-İZMİR

VI

Introduction

MASS TRANSFER and MASS TRANSFER OPERATIONS

Mass transfer and mass transfer operations play important roles in chemical engineering practice. These are also important for bioengineering, environmental engineering and food engineering practices. What is mass transfer and how does it take place? In order to understand this, let us take a glass of water. Although the water in the glass looks stagnant to us, it is well known that individual water molecules are not stagnant but move randomly in all directions (Brownian motion). During motions, molecules collide with each other and change directions. Nevertheless, these motions of the molecules do not result in net transfer of mass, because of the fact that, the number of the molecules leaving a region is balanced with the number of the molecules coming to this region. Hence, we cannot mention of mass transfer in a medium consisting of single type of molecules (pure components). Now, let us hang a potassium permanganate (KMO4) crystal into the glass as shown in Fig.1. As we know, potassium permanganate can dissolve in water and has a violet colour. Within a very short time period, we notice that the water surrounding the crystal turns into purple colour. This is the indication of the dissolution of crystal. As the time passes, purple coloured water expands showing the movement of KMO4 molecules. We understand from this that dissolved KMO4 molecules (in reality ions rather than molecules) do not remain stagnant but diffuse from dissolved region to the regions, which do not contain these molecules. After a very long time, a uniform purple colour throughout the whole glass is finally obtained. Uniform colour means that number of the KMO4 molecules in each ml of the solution (concentration of the solution)) at every point of the glass is the same. We understand from this experiment that a net transfer of mass (KMO4) had taken place: the dissolved KMO4 molecules moved from the dissolved region to the regions where they did not exist. This phenomenon is named as mass transfer. As the mass transfer resulted from the movement of the individual molecules, this type of mass transfer is known as mass transfer by

KMO4

Fig.1 Dissolution of KMO4 crystal and mass transfer by molecular diffusion

molecular diffusion. In reality, we cannot follow the movements of the individual molecules by blank eye, as they are very tiny particles. But in this experiment, because of the colour associated with molecules, we can realise the mass transfer by noticing the colour change. If we repeat this experiment, for example with table salt (NaCl), we

No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 1

cannot realise mass transfer with blank eye, although we may follow the dissolution of colourless crystal. Normally, we follow and understand mass transfer by measuring the concentration of the solution at different points. If we return back to KMO4 experiment, we can say that KMO4 molecules transferred from higher concentration regions to the lower concentration regions (how can we say that?). It follows from this that the reason or driving force for a mass transfer is the concentration difference or concentration gradient. Since, because of concentration difference mass transfer takes place, mass transfer can only be met in the solutions. If at two different points of a solution, a concentration difference exists for a component (in the example this component is KMO4), this component transfers from higher concentration region to the lower concentration region to remove this difference. As long as this difference prevails, the mass transfer continues. When the concentration uniformity is obtained, transfer of mass stops, although the movements of the molecules of both water and KMO4 do not stop. But these movements do not bring about net transfer of mass. This is an example of mass transfer by molecular diffusion in a liquid phase. Now, look at another example: Suppose a teacher has a perfume can and by pressing the knob at the top of the can he lets the perfume liquid to vaporize into air of a class-room whose door and windows are closed. Students, sitting in the immediate vicinity of the teacher, smell the nice smell almost instantaneously, but the students, sitting at the back seats, smell the perfume much later. Here, vaporized perfume molecules mix with the molecules of still air near the teacher. Now, the concentration of perfume molecules in the gaseous mixture near the teacher is higher than in the gas at the rest of the room. Thus, reason for mass transfer has been created. Under this concentration difference, perfume molecules start transferring from teacher’s vicinity towards the students. As the air in the room is still, mass transfer is affected by molecular diffusion. This is an example of mass transfer by molecular diffusion in a gas phase. It follows from these two experiments that, rate of mass transfer is rather slow, because the times required for the glass to obtain uniform colour and for the students sitting at the back seats to smell the perfume are both rather long. Now, if we repeat the first experiment by stirring the content of the glass by a stirrer as soon as we hang the KMO4 crystal, we see that the time required for uniform colour attainment is very short. The answer to the question: “what has happened?” is: “stirring has created turbulence”. Turbulence is characteristic of creating molecule groups, called Eddy. These eddies contain large number of both solvent and solute molecules (in the smallest eddy this is higher than 1016) and move rapidly. This type of mass transfer is known as mass transfer by turbulent diffusion. The rate of mass transfer by turbulent diffusion, which depends largely on the intensity of the turbulence, is much greater than the rate of mass transfer by molecular diffusion. In the second experiment mass transfer by turbulent diffusion is affected by switching on a fan as soon as the can’s knob is pressed. As expected, in this case the time required for the students to smell the scent of the perfume is much shorter compared with the experiment without fan. The creation of turbulence can be accomplished by various ways. For example consider a solid horizontal plate as shown in Fig.2, made of a soluble solid or coated with it, along which a liquid flows parallel to the plate (e.g. solid is benzoic acid,


liquid is water). If the average velocity of flow xu is small, the liquid flows as layers slipping one on the other. The velocity of the first layer adjacent to solid is zero and

solute A

x

solid plate

liquid B

(b) (a)

Fig.2 Mass transfer (a) in laminar, (b) in turbulent flow

z

other layers have different velocities, which increase in z-direction. This type of flow is known as streamline or laminar flow. In this type of flow, there is not velocity component in the z-direction (uz=0), and only individual molecules can pass from one layer to another. If the velocity of liquid is increased, after a critical velocity, liquid loses its orderly flow, and an appreciable velocity component in z-direction forms. This type of flow is known as turbulent flow. Disorderly flow is due to the formation of eddies, which are characteristic of turbulent flow. The eddies move rapidly in both x- and z-directions. As the plate is coated with dissolving solid A, a concentration gradient for component A in z-direction sets up, as soon as the flow of liquid B starts. The concentration of solute A in the first liquid layer adjacent to solid plate corresponds to its solubility at the prevailing temperature and this is the highest concentration. Due to this concentration change in z-direction, mass transfer in z-direction must take place. In the laminar flow, since only the movement of single molecules in z-direction is possible, molecules of solute A diffuse from one layer to the next layer in z-direction. In other words, mass transfer can only take place by molecular diffusion. In the turbulent flow, since the movement of group of molecules (eddy) is permissible, transfer of solute A from liquid next to solid surface to the bulk liquid is by turbulent or eddy diffusion. So, whenever a concentration gradient exists for a component of a solution, mass transfer takes place. The transfer mechanism depends on the condition of medium: if the medium is stagnant or flows in laminar regime, transfer of mass is by molecular diffusion. On the other hand, if the medium is mixed or flows in turbulent regime, transfer of mass is by turbulent or eddy diffusion. Although the mass transfer by turbulent diffusion is much more rapid than the mass transfer by molecular diffusion, this is not an instantaneous process but it is a rate process. Interphase mass transfer and mass transfer operations: So far, we have seen mass transfer within a single phase. We consider now the mass transfer between two phases. These two phases may be: liquid-liquid, liquid-gas, liquid-solid and gas-solid. Gas-gas system is not possible, as gases mix with each other in any proportion. As an example, take liquid-gas system. Suppose we contact a gaseous mixture of ammonia (A)-nitrogen (C) with water (S) at room conditions. At these conditions, only ammonia dissolves in water and water does not evaporate into gas. As soon as we have the contact, some of the ammonia molecules in the gas phase next to the interface pass the


interface and dissolve in water. As a result of this, a concentration difference for the ammonia molecules in z- direction sets up, as the number of the ammonia molecules in the unit gas volume next to the interface is less than the number of the ammonia molecules in unit volume in the bulk gas. Under this concentration difference, ammonia transfers from bulk gas to the interface and from there into the liquid phase. Now, look at the liquid phase. At the beginning, water does not contain any ammonia. As soon as the ammonia molecules cross the interface, the liquid next to the interface contains more ammonia molecules than the bulk liquid. So, in the liquid phase a concentration gradient for the ammonia in z-direction sets up as well. As a result of this concentration difference, ammonia transfers from liquid interface to the bulk liquid. If we look at the whole process, we see that ammonia is transferring from the gas phase into the liquid phase. This transfer can proceed until the two phases reach in equilibrium. The equilibrium concentration of ammonia in the liquid phase is the solubility of ammonia in water under the prevailing partial pressure of ammonia and temperature. Once the equilibrium is attained, mass transfer stops and concentration in both phases are uniform throughout. At the equilibrium, concentrations of ammonia in two phases are not equal.

Bulk liquid Bulk gas

z

Liquid phase S

A

inte

rfac

e

Gas phase A+C

This interphase mass transfer can be by molecular or turbulent diffusion depending upon the conditions and the flow characteristics of the phases. If both phases are stagnant or flow in laminar regimes, then the mass transfer is by molecular diffusion in both phases. If both phases are mixed thoroughly by stirrers or they flow in turbulent flow regimes, then the mass transfer in both phases is by turbulent diffusion. Of course, depending upon hydrodynamics of the phases, mass transfer in one phase may be by molecular diffusion and in the other by turbulent diffusion. Look at the net result of this interphase mass transfer. What has happened? With interphase mass transfer we make changes in the concentrations of the phases. Take the gas phase: at the beginning we have a gaseous mixture rich in ammonia, by transferring part of the ammonia to the water we obtained a gaseous mixture lean in ammonia. We can even transfer all the ammonia in the gas into liquid phase by taking necessary measures, leaving almost pure nitrogen behind. So, by applying interphase mass transfer we can separate or even purify a phase. In the example, ammonia is separated from nitrogen by contacting gaseous mixture with water. This operation is known as gas absorption. Gas absorption is used to separate or purify gaseous mixtures. For the separation or purification of liquid mixtures, distillation or extraction or both are used. To all these operations and to some more are given a general name: Mass transfer operations. So, mass transfer operations are the separation or purification operations that depend on interphase mass transfer. What is the importance of separation or purification in chemical process industries? In the chemical process industries, separation is very important. Almost in all the chemical reactions, mixtures, not pure substances are obtained. For sale or subsequent use of a product, certain degree of purity is required. Thus separation of any reactor exit is almost dictated upon us. In some cases, even the reactants


themselves must be purified prior to the entry to a reactor. As an example, consider the production of benzoic acid from toluene by oxidation with air oxygen. Toluene + O2 Benzoic acid Although the reaction can be written in this simple way, industrial production is rather complicated, as it requires large number of separation steps: Reactant toluene, which is a petro-chemical, is generally found mixed with benzene and xylenes and is separated from these by using a series of distillation operation, before pumping it to the reactor. Other reactant oxygen is again obtained by liquefaction and distillation of air. The reactor outlet contains not only the main product benzoic acid, but also un-reacted toluene (conversion is kept at 35-40 %), and the side-products benzaldehyde, acetic acid, benzylbenzoate, etc. This mixture is then sent to a train of separation units, which contains distillation and stripping columns, to separate and purify the benzoic acid. The separation and purification operations used in this production are all mass transfer operations. In the total investment cost of the plant, the investment cost of the reactor is much smaller than the investment cost of the separation units. From this, we may understand the importance of mass transfer operations in chemical engineering practices. Are all the separation operations mass transfer operations? In chemical process industry, various separation operations are used. Not all these separation operations are mass transfer operations. Only the separation operations based on interphase mass transfer are the mass transfer operations. For example, filtration, which is a separation operation, is not a mass transfer operation, since it does not involve interphase mass transfer. It is based on physical separation of a solid from a solid-liquid mixture with the help of a media, such as filter paper or cloth. Sieving is also a separation operation but it is not a mass transfer operation. Because, in the sieving a solid mixture having different sizes of material, is separated into uniform size fractions by using a sieve set. Again repeating, mass transfer operations are the separation operations based on mass transfer between the phases. Here are some industrially important mass transfer operations: gas absorption, stripping or desorption, distillation, liquid-liquid extraction, leaching, adsorption, membrane separation, etc. Almost all the separation operations, used in chemical industry, whether are based on interphase mass transfer or not, are physical in nature and hence they are unit operations. Don’t we use mass transfer in processes other than separation operations? We also make use of mass transfer knowledge in various processes other than mass transfer operations; for example, at the production of ammonia or hydrochloric acid solution. In these cases, we contact the gas ammonia or gas hydrogen chloride with water and by taking the necessary measures we can produce ammonia or hydrochloric acid solution at the desired concentrations. In a fermentation process, substrate dissolved in the solution, diffuses to the microorganism to react there. Here, the rate of mass transfer directly affects the fermentation process. In a catalytic chemical reaction, the reactants diffuse from the bulk of gas phase to the catalyst surface, where reaction takes place. Gaseous products diffuse from the surface of the catalyst to the bulk gas. In this case, rate of mass transfer to or away from the surface affects the production


rate. In drying processes, the moisture in the bulk solid, first diffuses through the solid phase to the surface of the solid, there it evaporates into hot gas at solid-gas interface and continues its diffusion in the gas phase from interface gas to the bulk gas. The rate of drying of the solid may depend on the rate of mass transfer in solid or in gas phase or both.


Chapter-1

MASS TRANSFER BY MOLECULAR DIFFUSION 1.1 Introduction: It was shown in the introduction that mass transfer takes place by molecular diffusion in stagnant media or laminar flow regimes. In any mass transfer process, it is very important to calculate the rate of mass transfer. Now, we will try to write this rate equation. Concentration difference causing mass transfer and the rate can be expressed in molar or mass units. Throughout this chapter molar units will be used. According to Fick, when a concentration gradient exists for component A in z-direction in a binary mixture of A+B, molar diffusional flux of A JAz,is written as:

JAz= -DAB dzdcA

(1-1)

This is known as Fick’s first law equation. Since the flux is defined as quantity of A transferring per unit time per unit area normal to the transfer direction, in SI unit system this is then: k-molA/m2s. dcA/dz is the concentration gradient causing the mass transfer as k-molA/m3 m. (-) sign emphasizes that diffusion always occurs in the direction of decreasing concentration. DAB is called as molecular diffusivity or diffusion coefficient of component A in component B (m2/s). This is a true physical property of the system and hence is characteristic of A-B pair and depends on pressure, temperature and concentration of the mixture. If a concentration gradient exists for the other component B of the mixture, similar equation for this component is written as:

JBz= -DBA dz

dcB (1-2)

Rate of molecular diffusion, zAJ (k-molA/s) can be related to the diffusional flux, by : AzJ

m

AzAz S

JJ = (1-3)

where, Sm (m2) is the mass transfer area which is normal to the transfer direction. If mass units instead of molar units are used, diffusional flux of component A, jAz (kg A/m2s) is written as:

jAz=-DABdzdρA

(1-4)

where (kgA/mAρ 3 ) is the mass concentration of component A. As it has been stated above, these fluxes are the fluxes of molecules. At some cases, the mixture itself moves in the diffusion direction of the components, which is known as bulk or convective flow. In reality, when a concentration gradient exists for one component of a binary mixture in any direction, there must be a concentration difference for the other component of the mixture in the opposite direction. As a result of these concentration differences, both components of the mixture diffuse in the opposite directions. If the rates of these diffusions are not equal in molar units, then the mixture itself drifts in the direction of the component whose molar diffusional rate

No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

7

is greater. So, it is obvious that total molar flux of each component for a fixed observer will be different than the diffusional fluxes of the components. Let us consider this in more details: Let us show the velocities of each type of molecules in z-direction as uAz and uBz. Then, the total molar flux of each component relative to the fixed coordinates are NAz= cA uAz and NBz= cB uBz, where cA and cB are molar concentrations of A and B (k-mol/m3) . On the other hand, since the total molar concentration of the mixture is c=cA+cB, the molar average velocity of the mixture, (m/s) is then: zu&

= zu&c

ucuc BzBAzA + =c

NN BzAz + (1-5)

As the molar diffusional flux of component A is due to the movement of the molecules relative to the molar average velocity of the mixture, then; JAz= cA(uAz- ) = czu& A uAz - zAuc & (1-6) can be written. After substitutions from the equations above finally,

NAz= JAz + c

cA ( NAz + NBz ) (1-7)

is obtained. This equation is known as general flux equation and it relates the molar diffusional flux to total molar flux. As it is seen, total molar flux of a component is the sum of the molar diffusional flux of this component and the flux of this component due to the bulk flow of the mixture. Repeating once more, diffusional flux of a component in a mixture is the flux, which is relative to the average velocity of the mixture, if any. Total flux of the same component is the flux of this component with respect to fixed coordinate system. If there is no bulk flow of the mixture in the direction of diffusion, then total and diffusional fluxes will be the same. Total flux is very important in the design of the equipment in which mass transfer occurs. Similar equation for the other component B will be:

NBz = JBz + c

cB ( NAz + NBz ) (1-8)

The relationship between total molar flux and total molar rate ( )N Az (k-mol A/s) can be written as:

m

AzAz S

NN = (1-9)

Similarly, if instead of molar units, the mass units are used; total mass flux of component A, nAz (kgA/m2s) is written as:

)n(nρρjuρjn BzAz

AAzzAAzAz ++=+= (1-10)

where uz is the mass average velocity of the mixture and given as: / =(( )BzBAzAz uρuρu += ρ BzAz nn + )/ (1-11) ρ


8

1.2 MASS TRANSFER BY MOLECULAR DIFFUSION IN THE GASES 1.2.1. Integration of the General Flux Equation: Equation (1-7) can be integrated under various conditions. Let us consider its integration at steady-state with DAB and Sm are constant. First, substitute JAz from equation (1-1) and then separate the variables. With the limits of the integrals: at z=z1 cA= cA1 and at z = z2 cA=cA2 ;

⎮⌡⌠

+−−=∫

2A

1A

2

1

c

c BAAA

Az

zAB )NN(ccN

dcdz

cD1 is obtained. For the sake of simplicity

subscript z is omitted. Upon integration,

)NN(cNc)NN(cNc

lnNN

1Dc

zz

BA1AA

BA2AA

BAAB

12

+−

+−

+=

− is found. If both sides of this equation are

multiplied by NA and z = z2-z1 is taken, finally;

NA =

cc

NNN

cc

NNN

lnz

Dc.NN

N1A

BA

A

2A

BA

A

AB

BA

A

−+

−+

+ (1-12)

is obtained. This equation can also be written in different forms. From the ideal gas

law, cRTRTVnP == and RTcRT

Vn

p AA

A == can be written, where P and

pA are the total and partial pressure of component A, respectively. Substituting these into equation (1-12),

NA =

Pp

NNN

Pp

NNN

lnRTz

PD.

NNN

1A

BA

A

2A

BA

A

AB

BA

A

−+

−+

+ (1-13)

is obtained. From Dalton’s law yA= pA/P, where yA is the mole fraction of component A in the gas. Substituting this into the equation above, another form of equation (1-12) is found as:

NA = 1A

BA

A

2ABA

A

AB

BA

A

yNN

N

yNN

N

lnRTz

PD.

NNN

−+

−+

+ (1-14)

Any one of these three equations may be used to calculate the total molar flux of component A. But, to use any one of these equations, the relationship between NA and NB must be known. This can be easily obtained from the condition of the system. Let us see below the most commonly encountered cases for the interrelationships of NA and NB. 1.2.2 Equimolar Counter Transfers of A and B: This is a situation, which is frequently met in distillation of binary solutions of equal latent heats of vaporization. So, in this case, NA=-NB = constant and equation (1-12) is indeterminate. Then, we go back to the equation (1-7), with NB=-NA this equation becomes:


9

[ ] 0dz

dcD)N(N

cc

dzdc

DN AABAA

AAABA +−=−++−=

if this equation is integrated under the conditions at which equation (1-7) was integrated.

( 2A1AAB

A ccz

DN −= ) (1-15)

is obtained. This equation is the special solution of equation (1-12) for this case. Similarly, the special solutions of equations (1-13) and (1-14) for this case are:

( 2A1AAB

A ppRTzD

N −= ) (1-16)

and

( 2A1AAB

A yyRTz

PDN −= ) (1-17)

As it is seen, at this special case NA = JA . It follows from this that there can’t be bulk flow in the transfer direction, when equimolar counter transfer occurs. Since in this case NA+NB=0, this case is also known as “mass transfer under zero-net-flux”. It is obvious from the equations above that concentrations of the components change linearly with transfer path as shown in Fig.1.1a. 1.2.3 A Transfers through Non-Transferring B: This occurs in gas absorption operations. In this special case, since component B is not absorbed, NB=0 and then equations (1-12), (1-13) and (1-14) simplify to:

,pp

lnRTz

PDN,

cc

lnz

DcN

1B

2BABA

1B

2BABA ==

1B

2BABA y

yln

RTzPD

N = (1-18)

Although these equations can be used in these forms, we prefer to write them in the form of “flux of a component is proportional to the concentration difference of the same component”. Since, cA1 – cA2 = cB2 – cB1, pA1 – pA2 = pB2 – pB1 and yA1 – yA2 = yB2 – yB1 , the right hand sides of the equations given in (1-18) are first multiplied with the appropriate equation below,

,1cccc

1B2B

2A1A =−− 1

pppp

1B2B

2A1A =−− 1

yyyy

1B2B

2A1A =−−

then by definitions of logarithmic means, which are given below,

( )lnB

1B

2B

1B2B c

cc

ln

cc=

− , ( )lnB

1B

2B

1B2B p

pp

ln

pp=

− , ( )lnB

1B

2B

1B2B y

yy

ln

yy=

−

finally;

( ) )cc(cz

cDN 2A1A

lnB

ABA −= , ( ) )pp(

pRTzPD

N 2A1AlnB

ABA −= ( ) )yy(

yRTzPD

N 2A1AlnB

ABA −=

(1-19) are obtained. As it is seen, in this special case, relationships between concentrations and transfer paths are not linear as shown in Fig.1.1b. Furthermore, it is seen from the figure that (-dpB/dz ) is not zero, and under this concentration gradient component B


10

diffuses also in the opposite direction of component A. But its diffusional flux is exactly balanced with the bulk flow flux of B. As a result of this, to a fixed observer, component B looks non-transferring. So, when only one component of a binary gas mixture is absorbed into a liquid, there is always a bulk flow of the mixture in the transfer direction. This increases the flux of the absorbed component.

P P

pA1

pB1 pA2

pB2

A B

P

pB2

pA2

pA1

pB1

A

pres

sure

s,pA,p

B,P

pres

sure

s,pA,p

B,P

z1 z2 z1 distance,z z2 z (a) (b) 1.2.4. Thsome casplace neasurface, Accordintransfer tfrom thesubstitutiis calcula

A gaseousto the stoic

Ft

Across a measured aCalculate tAt the ope

No part of t

ig.1.1 Concentration profiles (a) in equimolar counter transfer and (b) in A ransfers through non-transferring B

e Relatioes, relatiorby. Co

g to theowards c catalyst sng NB = -ted.

chemical hiometry g

gas film os 0.40 andotal molar rating temp

his CD-book

distance,

nship between NA and NB is Fnship between the fluxes is fi

nsider the gas phase reaction b

m A→ n B

stoichiometry of the reactionatalyst surface at the same timurface to the bulk gas. So, n N (n/m) NA into any one of the e

Example-1.1) Calculation o

reaction is taking place on a solid caiven below:

2A B

f 2 mm thickness adjacent to the s 0.10 bars. fluxes of component A and B. erature and pressure DAB= 4.10-5 m2/

may be multiplied for commercial purp

P

ixed by Reaction Stoichiometry: In xed by the chemical reaction, taking elow taking place on a solid catalyst

, when m moles of component A e n moles of component B transfer A = - m NB can be written. Hence,

quations (1-12), (1-13) and (1-14) NA

f Molar Fluxes

talyst surface at 2 bars and 25 oC according

olid, partial pressures of component A are

s.

oses. E.Alpay & M.Demircioğlu 11

Solution : Equation(1-13) can be used.

z B

A catalystt

25.01

1N5.0N

NNN

N

AA

A

BA

A =−

=−

=+

21

)2/4.0(2)2/1.0(2ln

)m10*2)(K25273)(Kmolk/mbar083.0()bar2)(s/m10*4()2(N

33

25

A −−

+−=

−

−

NA = (3.23 * 10-3) ln (1.083) = 2.58 *10-4 k-mol A/m2s NB = - (0.5) NA = - 1.29*10-4 k-mol B/m2s 1.2.5 The Relationship between NA and NB is Given by the Latent Heats of Vaporization: In the rectification of a binary liquid solution, mass transfer takes place both in liquid and vapor phases. The more volatile component transfers from liquid to vapor and the less volatile from vapor to liquid. As the vapor and liquid are both saturated at the operating conditions, an energy equal to the latent heat of vaporization is needed for the vaporization of more volatile component. This energy is given by the condensation of less volatile component. Thus, the quantities of more and less volatile components that will vaporize and condense are interrelated by λA NA = - λB NB, where λA and λB are latent heats of vaporization of the corresponding components (kJ/k-mol). In this case substituting NB = - (λA/λB ) NA into one of the equations (1-12), (1-13) and (1-14) the fluxes can easily be calculated. Note that if λA = λB, equimolar counter transfers of A and B occurs. 1.2.6 Mass Transfer with Varying Cross-Sectional Area: If the mass transfer area Sm does not remain constant and changes with z, the fluxes in the equation (1-7) do not remain constant even at steady-state. Since, in this case AzN and BzN are constant, NA

and NB in the equation (1-7) are first replaced with m

Az

SN and

m

Bz

SN and Sm is

expressed as function of z and then the integral is performed. By dividing so-obtained ( )N Az with the Sm value calculated at the specified z, NA at this z is obtained.

Example-1.2) Calculation of Molar Flux in Varying Cross-Section

SO2 (A) is transferring at steady-state through non-transferring O2 (B) in a metal conduit, 2.0 meter long, at 10 bars and 598 K. The cross-section of the conduit is rectangular and tapers uniformly from an area of 300 mm by 400 mm to an area of 300 mm by 200 mm. The partial pressure of SO2 is measured as 0.22 bar and 0.055 bar at two ends.


12

Calculate the total molar flux of SO2 at mid-point of the conduit. The molecular diffusivity of SO2 in O2 at 1 bar and 25 oC is 2.25*10-5 m2/s.

Solution :

m = b * a = 0.3 * a (m2)

0.03 z (m2) From eqn.(1-7) for NB=0 and

S

)m(z1.04.0z2

2.04.04.0zaaa 211 −=

−−=

−−=

la

dzdp

RTDJ aAB

A −= Sm = 0.3(0.4 - 0.1 z) = 0.12 -

After substituting for Sm and separating the variables,

dzdp

RTD)

Pp1(N AABA

A −=−

AABA dpPDN

Then from equation above, At z = l / 2 (Sm)l /2= 0.12 – 0.03 (1.0) = 0.09 m2

sm/molAk10*23.109.010*11.1N

)N( A2/A ==l )S(

289

2/m−= −

−

l

Am pPRTdz

S −−=

∫−=−

2Ap AB

oAdp

RTPD

z03.012.0dzN

l A∫ −1Ap

ApP

[ ] [ ] 055.0220A

20

A pPlnPD

z03.012.0lnN

−=−− AB.RT03.0

s/m10*61.7298598

10110*25.2D 26

75.1

5AB

−− =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

ttanconss/Amolk10*11.1N 9A =−= −

l

l/2 z

b=300 mm a1 =400 mm

Sm2pA2

a S

b=300 mm

NB= 0

NA

pA1

(SSm1

a2 =200 mm


13

1.2.7 Mass Transfer by Molecular Diffusion in Multi-Component Gas Mixtures: In many engineering applications of mass transfer, the mixture contains more than two components and all the components of the mixture may diffuse under the available oncentration gradients. The total molar flux of component A in this case can also be c

calculated from equations (1-12, (1-13) and (1-14) by making two modifications in the equations. First, all the NA+ NB terms in the equations are replaced with ∑

=

n

AiiN second,

an effective diffusivity defined by equation (1-20) is taken in the place of DAB.

∑

∑

=

=

−

−=

n

AiiAAi

Ai

n

AiiAA

Aef

)NyNy(D1

NyND (1-20)

With these changes, the equation (1-12) becomes:

ccNzN 1A

nA

n

Aii

A

−∑= N

cc

N

N

lnDcNN

Aii

2An

Aii

A

AefA

−

=

∑

∑

=

= (1-21)

-20), effective diffusivity Dy diffusivities with each of the other components, depends on the

concentration and hence may vary considerably from one end of the diffusion path to the other, and a linear variation with distance can be usually assumed. In some cases

As it is seen from equation (1 Aef, which can be synthesized from its binar

all N’s except NA is zero. In this case equation (1-20) simplifies to:

∑=

ion of Binary the equations above that to use any one of these, binary diffusivity, DAB must be known at the operating conditions. It was shown that effect of concentration on the diffusivity is egligible. The diffusivities can be determined experimentally or may be predicted

ntation conditions. The experimental set-up as shown in Fig.1.2, consists of

−= n

Bi Ai

i

AAef

Dy

y1D (1-22)

1.2.8 Determinat Diffusivities of Gases: It is obvious from

nfrom the kinetic theory of gases or can be estimated from various empirical correlations. Depending upon the physical state of the components at the experimentation conditions, one of the three experimental techniques given below, can be used. 1.2.8.1 Experimental Determination: Three main methods are available. Depending upon the state of the components at experimental conditions, one of these can be used. Two-Bulb Method: This method is used when both components are gas at the experimetwo chambers with V1, V2 volumes, connected with a capillary tube whose length and cross-section are l and S. This set-up can be made of glass or metal depending upon experimental pressure. The chambers are first completely evacuated, then the valve on the capillary tube is closed and chamber-1 is charged with pure A and the chamber-2 with pure B at the same pressure. At θ = 0, the valve on the capillary tube is opened and mass transfer is allowed to take place. After a certain time, the valve is closed and


14

hamber-2 is thoroughly mixed and analysed for component A. By iolume of the capillary tube and assuming uniform concentrations in bothny time of the

2

cva diffusion, equation

ldz

DJ 1A2AABAABA

)cc(Ddc −−=−= (1-2

can be written. The accumulation of component A in chamber-2 is due transfer of this component from chamber-1. Thus,

ld

)c(cDSJSdcV A2A1ABm

AmA2

2−

==θ

is written. On the other hand, the average concentration of component A

(1-24

system is found from any one of the equations beloc are molar concentrations of com n2A1A

o2A

o1A ,c,c,c pone t A in chamb

( ) o2A2

o1A1A21 cVcVcVV +=+ (1-2

( ) 2A21A1A21 cVcVcVV +=+ (1-2

beginning and at the end f nd substituted into equation 1-

chamber-2 at the o the experiment. If cA1 is equation (1-26) a ( 24) ,

∫θ

θ=⎮⌡ − oc 2AA )cc(po

2A

⌠c

2Adc2A

d is obtained. Where, p is defined as /S(l

After performing integral, substituting p and solving for D

p =

AB finally,

ABD

A2A

oA2A21 ccVV −l

21mAB cc

ln)V(VS

D−θ+

=

is obtained. Since V1 , V2 , Sm , l are all known, and Ac can be found fr(1-25), by measuring the cA2 after an experimentation time of θ, D is cal

Winkelmann Method: This method is used when one of the component

to the tube mouth. During this flow, liquid A vaporizes a

ABequation (1-27).

the experimentation conditions. As shown in Fig.1.3, a narrow tube is filleA to a certain depth and the gas component B is made to flow through a lachannel attached in the gas in z-direction by molecular diffusion, as its concentration at thinterface (point-1) is higher than that at the tube mouth (point-2). Mass tran

1

V1 V2

l

z cA2

Sm

o2Ac

cA1

oA1c

Fig.1.2 Experimental set-up for two bulb method


gnoring the chambers at

3)

to the mass

)

( )c in the Aw, where

and er-15) 6)

ed from solv

( )21m V)V21 V+ V

(1-27)

om equation culated from

s is liquid at

nd transfers

d with liquid rge diameter

e liquid-gas sfer takes

15

place under the conditions of “A transfers through non-transferring B”, as component B is not absorbed into liquid A. So, any one of the equations (1-19) can be written.

)pp()p(RTz

N 2A1AlnB

A −= (1-19)

Here, p

PDAB

A1 is the partial pressure of component A at the liquid-gas interface and hence equals the vapor

the liqui oAp ) at the epressure of xperimentation d (

temperature. The partial pressure of A at the tube mouth (pA2) can be taken zero, by making volumetric flow rate of the gas in the channel very high. By measuring the decrease in the liquid level in the tube after an experimentation time of θ, the evaporation flux of A, which is equal to the mass transfer flux of A in the z-direction, can be written as:

θρ

−=d

dVSM

N A

mA

AA , where

eof the tube,VA= Sm(l -z) and d

into equation (1-19):

VA, MA and ρA are volume, molecular weight and

ly. On the other hand, S

1

density of the liquid respectiv m being the cross-sectional area VA = -S dz can be written. If all these are substituted

m

o

lnB

AB

A

A p)p(RTz

PDdtdz

M=

ρ

and

A

∫∫ =θθ

θρ o

A

oAAAB

z

0zd

)(RTpPMD

dzz and finally, lnBp

θρθ

lnB

is obtA

oAAAB2

0 )p(TRpPMD2zz =− (1-28)

ained. By measuring the liquid levels at the beginning (z0) and at the end (zθ ) of the experimentation time θ, DAB can be calculated from this equation.

Example-1.3) Measurement of Molecular Diffusivity

chloride till 10 mm elow its mouth is subjected to air flow. After 55 d in the ube falls 73.8 mm. W apor pressure and the

3

2

The molecular diffusivity of carbon tetrachloride (A) in air (B) will be measured by using

inkelmann’s method at 1 atm. and 50 Wb

oC. The tube filled with liquid carbon tetrahours and 29 minutes the level of the liqui

hat is the value of molecular diffusivity? At 50 td

oC, vensity of carbon tetrachloride are 282 mmHg and 1 500 kg/m . MA= 154.

Solution:

From the givens, zo=10 mm and zt =10 +73.8 = 83.8 mm. mmHg1.608)282760(760)p( =

−−=

)]282760/(760ln[lnB −θ = (55)(3 600) + (29)(60) = 199 740 sec. Th uation (1-28), en, from eq s/m10*63.9

)740199)(282)(013.1)(154(2)1.608)(50273)(083.0)(5001(10*])10()8.83[( 622 −

DAB26−=

+−=

is found.

2

pA1= oAp

pA2≈0

Liquid A

gas B

NA

Fig.1.3 Winkelmann nt

z

l

experime


16

Vaporization of Liquid Drops or Sublimation of Solid Spheres: For the application xperiment, one of the components must be either liquid or sublimable solid at

Fig.1.4 component A, either as liquid

non-transferring B”, and hence flux e

of this ethe experimentation conditions. As shown indrop or solid sphere in rp diameter, is hanged in a large diameter channel through

which gas B flows at low velocity at the experimentation temperature and pressure. Liquid drop or solid sphere vaporizes or sublimes into the gas at liquid (solid)-gas interface (point-1) and from there it transfers into the bulk gas (point-2), because of the concentration difference exists in radial direction. Here, again mass transfer takes place under the conditions of “A transfers through quation,

A B 2

1

Fig.1.4 Vaporization of liquid drops

)pp()p(rRT

PD2A1A

lnB

ABA − (1-19a) N =

written. Again here, pA1 is the partial pressure of A at liquid (solid)-gas interface, at operating temperature. p

hich can be assumed zero, as th of

iswhich is equal to the vapor pressure of A A2 is the partial pressure of A at bulk gas, w e volumetric flow rate gas B is large. Vaporization or sublimation flux of component A, which is also equal to the mass transfer flux of component A in radial direction, can be written as,

θρ

−=d

dVSM

N A

mA

AA , where VA, MA and ρA are volume, molecular weight and density

of the liquid drop or solid sphere respectively. From the sphere geometry, Sm = 4π r2

and VA = 34 π r3 can

1

be written. If the flow of gas B continues until the drop or sphere

evaporates completely, by inserting all the above equations into equation (1- 9a), the equations;

∫∫ −=θρ

θ o

rolnBA

oAABA

p)p(RTpDM drrdP and

2r

)p(RTpPDM 2

p

lnBA

oAABA =θ

ρare obtained. O

n the other hand, from the definition of logarithmic mean,

[ ])pP(/Plnppp)p(

A2B

2BlnB

−

−=

into equation above and by rearranging, finally

po

oA1B =

pln

1B

is written. Substituting this

[ ])pP/(PlnPM2RTr

DoAA

2pA

AB −θρ

= (1-29)

ion, DAB is equation above. imentally measured values of molecular diffusivities of some gas

is found. By measuring the time θ for complete vaporization or sublimatcalculated from theIn Table.1.1, experpairs at 1 atm. pressure and at various temperatures are given. It follows from the table that the values for most gas pairs are at the order of 10-4 m2/s. It can easily be shown that for gaseous mixtures DAB = DBA.


17

Example-1.4) Measurement of Molecular Diffusivity

Molecular diffusivity of iodine (A) in air (B) will be measured at 760 mmHg and 25 oC. For this, solid

dine cast in spherical shape with 4 mm dioA

iameter is suspended in a duct through which air flows. fter 22 hours and 54 m molecular

y? o 3

inutes, iodine sphere completely disappears. What is the value of diffusivitVapor pressure and density of solid iodine at 25 C are 1.07 mmHg and 4 930 kg/m . MA=254.

Solution:

θ = (22)(3 600) + (54)(60) = 82 440 sec. From equation (1-29), s/m10*16.8

)]07.1760/(6025273(*−7ln[*013.1*44082*254*2

)083.0*)10*2(*9304D 2623

AB−

−

=+

=

emperat s identical with the value given in

iction of Binary Gas Diffusivities: The kinetic theory of the gases is

tz derived the theoretical equation below, which can be used to predict the inary diffusivities of non-polar gas pairs or of a polar gas with a non-polar gas, at the

is found. It is seen that this value, after t ure correction, iTable.1.1. 1.2.8.2 Predfairly well developed. Starting from the kinetic theory of the gases, Hirschfelder-Bird-Spobabsence of experimentally found value.

5.0

AB,D2AB

5.0

BA

234

AB11

rP

M1

M1249.0084.1T10

D ⎟⎞

⎜⎛

+Ω

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=

−

(1-30) BA MM ⎟⎠

⎜⎝

In the equation above, DAB is molecular diffusivity in m2/s. T and P are absolute

rature in K, and pressure in N/m2 respectively. MA and ar weights of A and B in kg/k-mol. rAB, which is defined as r =(rA+rB)/2 , is collision

dius in nm. , which is given as function of (kT/

tempe MB are moleculAB

ra AB,D ABΩ ε ) in Fig.1.5, is known as collision function, where k is very well known Boltzmann constant. ABε , which is obtained from the force constants (ε/k) of the components by k/k/ BAAB εε=ε , is known as energy of molecular attraction. The collision radius and force constants for some molecules are given in Table.1.2.

Example-1.5) Prediction of Molecular Diffusivity Predict molecular diffusivity of methane in air at 1 tm. and 0 oC from Hirschfelder-Bird-Spotz Equation:

Table.1-2.

a

From ε/k r (nm) M Methane 148.6 0.3758 A 16 Air 78.6 0.3711 B 29


18

Table 1-1. E imentally det bi iffu es atm. pressure

Gas Temp. Temp. 2/s

xper ermined nary gas d siviti at 1

(Vapor) Pair (oC) DAB*104 m2/s Gas (Vapor) Pair (oC) DAB*104 m

Acetic acid-Air 0 0.1064 Ethyl formate-Air 0 0.337 Acetic acid-CO2 0 0.0716 Ethylene-H2 0 0.486 Acetic acid-H 0.416 Formic acid-Air 0.131 2 0 0 Acetone-Air 0 0.109 H -SO2 2 200 1.23 Air-n-Butanol 59 0.104 H -Su 2 5 5.5 1.121 Air-n-Butanol 2 5.9 0.087 Helium-Nitrogen 20 0.705 Ammonia-Air 0 0.198 n-Heptane-CH4 38 0.066 Ammonia-H2 25 0.784 n-Hexane-H2 15 0.290 Ammonia-H2 85 1.093 Iodine-Air 0 0.07 Ammonia-N2 85 0.328 Mercury-Air 0 0.07 Ammonia-N2 25 0.230 Methanol-Air 0 0.132 Aniline-Air 0 0.061 Methyl acetate-H2 0 0.303 Aniline-Air 2 e-Air 5.9 0.074 Methyl format 0 0.0872 Aniline-Air 59 0.09 Naphthalene-Air 0.0513 0 Argon-Nitrogen 20 0.194 Nitrogen-H2 25 0.784 Benzene-Air -10 0.104 0 0.077 Nitrogen-SO2Benzene-CO2 0 0.0528 3 Nitrogen-Water 4.5 0.256 Benzen-H2 0 0.306 Nitrogen-CO2 25 0.165 CH -H4 2 15 0.694 O -Air 2 0 0.178 CH -O4 2 500 1.1 O -H2 2 0 0.697 CO2-Air 44 0.177 O2-Nitrogen 0 0.181 CO2-Air e-H20 0.138 n-Octan 30 0.271 CO2-H2 0 0 ne-O 0.0705 .55 n-Octa 2 30 CO2-Helium 25 0.612 Phosgene-Air 0 0.095 CO2-O2 0 0.139 n-Propanol-Air 0 0.085 CO2-SO2 70 0.108 n-Propanol-CO2 0 0.0577 CO2-Water 5 5.4 0.211 n-Propanol-H2 0 0.315 CO2-Water 3 4.4 0.202 Toluene-Air 0 0.076 CO-Nitrogen 100 0.318 Toluene-Air 30 0.088 n-Decane-H2 290 0.306 i-Valeric acid-H 0 0.212 Ethanol-Air 0 0.102 Water-Air 2 5.9 0.258 Ethanol-CO2 0 0.0685 Water-O2 59 0.305 Ethanol-CO2 67 0.106 Water-O2 450 1.3


19

Fig. 1.5 Collision function

0.2

0.4

0.6

0.8

1

1.2

1.4

0.1 1 10 100 10000.2

0.3

0.4

0.5

AB,DΩ AB,DΩ

kT/ ABε

)K(074.108)6.78)(6.148(kkkBAAB ==εεε

)nm(37345.02

3711.03758.02

rrr BAAB =

+=

+= =

53.2074.108

273kT

AB

==ε

From Fig.1.5 ΩD,AB = 0.5 is read.

311.01111 05.0

=⎥⎤

⎢⎡ +=⎥

⎤+

2916MM

5.

BA ⎦⎣⎦⎢⎣

⎡

y substituting all these into equation(1-30) B

[ ] s/m10*2)5.0()37345.0)(10*013.1)(1( 25AB

)311.0()311.0)(249.0(084.1D 252

−=−

=

obtained

1.2.8.3 Estimation of Binary Gas Diffusivities: There is large number of empirical developed by analysing large number of experimental results. In the absence

of experimentally obtained values, the suitable ones of these may be used to estimate

)273(10 /34−

is

equations

the binary gas diffusivities. In using these equations, one must be very careful with the units of the terms contained in the equations and with the application range of the equations. The equation given by Fuller-Schettler-Giddings can be successfully used

to estimate the binary diffusivities of polar and non-polar gas pairs with ± % 10 deviation.

( ) ( )

2/111T100.1D

75.19⎟⎞

⎜⎛

+−∗

= (1-31) BMAM2

PAB

3/1B

3/1A vv

⎟⎠

⎜⎝

⎥⎦⎤

⎢⎣⎡ + ∑∑


20

Table 1-2. T n r some molecules he collision radius and force consta ts fo

Gas (Vapor) r (nm) ε/k (K) Gas (Vapor) r (nm) ε/k (K) Acetone 0.4600 560.2 Hydrogen iodide 0.4211 288.7 Acetylene 0.4033 231.8 Hydrogen peroxide 0.4196 289.3 Air 0.3711 78.6 ydrogen sulfide 0.3623 301.1 HAmmonia 0.2900 558.3 Iodine 0.5160 474.2 Argon 0.3542 93.3 Krypton 0.3655 178.9 Arsine 0.4145 259.8 Mercury 0.2969 750.0 Benzene 4 686.2 0.5349 12.3 Mercury dibromide 0.5080 bi-Butane 0.5278 330.1 Mercury dichloride 750.0 0.4550 bn-Butane 5 diiodide 695.6 0.4687 31.4 Mercury 0.5625 Boron trichloride 0.5127 337.7 Methane 0.3758 148.6 Boron trifluoride 0.4198 186.3 Methanol 0.3626 481.8 Bromine 0.4296 507.9 Methyl acetate 0.4936 469.8 Carbon dioxide 0.3941 195.2 Methyl acetylene 0.4761 251.8 Carbon monoxide 0.3690 91.7 Methyl bromide 0.4118 449.2 Carbon sulfide 0.4483 467.0 Methyl chloride 0.4182 350.0 Carbon tetrachloride 0.5947 322.7 Methyl ether 0.4307 395.0 Carbon tetrafluoride e 0.4662 134.0 Methylene chlorid 0.4898 356.3 Chlorine 0.4217 316.0 Neon 0.2820 32.8 Chloroform 0.5389 340.2 Nitrogen 0.3798 71.4 Cyanogen 0.4361 348.6 Nitrogen monoxide 116.7 0.3492 Cyclo –propane 0.4807 248.9 Nitrogen oxidul 0.3828 232.4 Cyclo-hexane ane 341.1 0.6182 297.1 n-Pent 0.5784 2,2 Dimethylpropane 1 lcohol 0.6464 93.4 n-Propyl a 0.4549 576.7Ethane 20.4443 15.7 Oxygen 0.3467 106.7 Ethanol 0.4530 3 .5 62.6 Phosphine 0.3981 251Ethyl acetate 0.5205 521.3 Propane 0.5118 237.1 Ethyl chloride 0.4898 300.0 Propylene 0.4678 298.9 Ethyl ether 0.5678 313.8 Silicon hydride 0.4084 207.6 Ethylene fluoride 0.4163 224.7 Silicon tetra 0.4880 171.9 Fluorine 0.3357 112.6 Stannous bromide 0.6388 563.7 Helium 0.2551 10.2 Sulfur dioxide 0.4112 335.4 n-Hexane 0.5949 399.3 Sulfur hexafluoride 0.5128 222.1 Hydrogen 0.2827 59.7 Trimethyl borate 0.5503 396.7 Hydrogen bromide ide 0.3353 449.0 Uranium hexafluor 0.5967 207.6 Hydrogen chloride 30.3339 44.7 Water 0.2641 809.1 Hydrogen cyanide 0.3630 569.1 Xenon 0.4047 231.0 Hydrogen fluoride 30.3148 30.0

I ve s ular vity in m /s, T and P are absolute temperature in K and pressure in atm., Σv is the diffusional volum m e in m an h from the atomic diffusional volumes and the

ructural contribution of the molecule, in the absence of the value for the molecule.

n the equation abo , DAB i molec diffusi 2

e of the olecul3/k-mol, which c be synt esized

stTable1.3 gives the diffusional volumes of some atoms and molecules.


21

Chen and Othmer derived an equation using critical volumes and critical temperatures instead of diffusional volumes.

( ) ( )

2/181.16 T105.1 ⎞⎛∗ −

BAcBV+

ere again, temperature, pressure and volumes should be taken in d in m3/k-mol respectively. DAB will be obtained in m2/s.

decreases with pressure. This is nderstandable as the kinetic energy hence the mobility of the molecules raises with

cAcBcAAB M

1M

1

VTTPD

24.04.01405.0 ⎟⎟⎠

⎜⎜⎝

+= (1-32)

H K, in atm. an

1.2.9 Effect of Temperature and Pressure on Gas Diffusivity: In the gases, molecular diffusivity increases with temperature bututhe increasing temperature, and the number of the molecules per unit volume increases with rising pressure. These dependences are expressed as;

( ) ( )75.1

1

2

2

1T,PABT,PAB T

TPPDD

1122 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟

n, its value at pressure can easily be calculated from the equation above. In

the equation, temperature T is taken in K.

he integration of general flux equation (1-7) requires that DAB and c are constant. ay

ary considerably with concentration. Nevertheless, due to the lack of knowledge of

⎠

⎞⎜⎜⎝

⎛= (1-33)

So, if molecular diffusivity at one temperature and pressure is knowanother temperature and

1.3 MASS TRANSFER BY MOLECULAR DIFFUSION IN LIQUIDS: TThis is almost so for binary gas mixtures but not so for binary liquids, where both mvthese changes, equation (1-7) is also used in liquids with average values of DAB and c. So, the integration of equation (1-7) at steady-state, with constant S results in :

)NN/(Nx

lnzNN

NBAA1A

BAA2A

BA

AA +−+= (1-)NN/(NxcDN AB +− 34)

where DAB and c are the arithmetic means of the molecular diffusivities and total entrations evaluated at the two ends of diff al p is

defined as , is the mole fraction of component A in the liquid . molar conc usion ath. xA, which

c/cx AA = When mass transfer takes place under the condition of “equimolar counter transfer of A and B”, equation (1-34) simplifies to:

)cc(z

)xx(z

N 2A1A2A1AA −=−= (1-35)

On the other hand, if the transfer is unde

DcD ABAB

r the conditions of “A transfers through non-ation (1-34) simplifies to: transferring B”, equ

)xx()x(zx1

lnz

N 2A1AlnB1A

A −=−

cDx1cD AB2AAB −= (1-36)

where, x is given as B )x/xln(xx 1B2B)x(

1B2BlnB

−=


22

Table 1-3. D

iffusional volumes of some atoms and molecules

Atom Diffusional volume,v m3/kg atom *103

Atom Diffusional volume,v m3/kg atom *103

Carbon 16.5 Oxygen 5.48 Chlorine 19.5 Sulfur 17.0 Hydrogen 1.98 Aromatic ring -20.2 Nitrogen 5.69 Heterocyclic ring -20.2 Molecule Diffusional volume,∑ v

m3/k-mol *103 Molecule Diffusional volume, ∑v

m3/k-mol *103

Air 20.1 He 2.88 Ar 16.1 Kr 22.8 Br2 67.2 N2 17.9 Cl2 37.7 Ne 5.59 CO 18.9 NH3 14.9 CO2 26.9 N2O 35.9 CCl2F2 14.8 O2 16.6 D2 6.70 SF2 69.7 H2 7.07 SO2 41.1 H2O 12.7 Xe 37.9

1.3.1 Determination of Molecular Diffusivities in Liquids: Molecular interactions in a liquid are much greater than that of a gas and the formulation of these interactions is very difficult and as a result of this, the kinetic theory of the liquids is not so developed as the gases. Hence, there is no reliable theoretical equation for the prediction of binary diffusivities in liquids. 1.3.1.1 Experimental Method: For the experimental determination of binary diffusivities of liquids, two-cell method is frequently used. As it is shown in Fig.1.6, this equipment consists of two cells of volume V each, separated with a porous disc of thickness δ, and equipped with stirrers. Two different strengths of dilute binary solutions are prepared and put in each cell, and then the stirrers are set on motion. If

> , component A transfers by molecular diffusion from cell-2 to cell-1 through the stagnant liquid filled the porous disc. Assuming steady-state, the concentration

gradient at any moment can be written as:

o2Ac o

1Ac

δKcc

zc A1A2A −

=∂∂

, where constant K, which

is greater than 1, emphasizes the fact that diffusion path is always greater than δ. Since the solutions are dilute, by neglecting bulk flow contribution,

δKcc

DεdzK

dcDεJN A1A2

ABA

ABAA−

=−== (1-37)

can be written, where shows the fraction of the area which is open to mass transfer in the porous disc.

ε

No part of this e-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

23

Fig.1.6 Experimental set-up for two-cell method

δz

Vc

The component A transferring from cell-2 to cell-1 causes an accumulation in this cell. Hence,

)c(cAB

δKDSε

d A1A2mA1 −=

θ (1-38)

written, where Sm is the cross-sectional area of the disc. Similarly, component A ving the cell-2 causes a depletion in this cell and hence, equation,

dcV

islea

)c(cδKDSε

ddcV A1A2

ABmA2 −=θ

− (1-39)

is also written. If we add equation (1-38) to the equation (1-39),

)c(cδKdd A1A2θDSε2 ABm −

)cd(cVdcdcV A1A2A2A1 =−

−=−θ

(1-40)

is obtained. If the variables are separated and integrated,

∫θ

−

−θ=⎮⌡

⌠−−

−θθ

oABm

cc

cc A1A2

A1A2 dVδKDSε2

cc)c(cdA1A2

oA1

oA2

and finally,

θθ −−

θ=

A1

oA1

oA2

mAB cc

cclnSε2

VδKD (1-41)

is obtained, where 2A1Ao

2Ao

1A c,c,c,c trations of component A in the cells at the beginning (θ =0) and at the end (θ = θ) of the experiment. The constants of the experimental set-up can be collected under one constant, )SεV/2δ(Kα

A2

are molar concen

m= , and this constant can easily be determined by conducting an experiment with a liquid pair

r diffusivity is already known. Experimentally measured diffusivities of some binary liquids are given in Table.1.4.

lution. By comparing Table.1.4 with Table.1.1, it can be realized at values of molecular diffusivities of binary gases are about ten thousand times

higher than the values of molecular diffusivities in binary liquids.

whose molecula

As it was stated before, molecular diffusivities in liquids are also dependant on the concentration of soth

cA2

A1

V

porousdisc

stirrer 1

2


24

Example-1.6) Measurement of Molecular Diffusivity in Liquids

Molecular diffusivity of methanol (A oC in a two-cell experimental set-up whose cell constant is not known. To the cell-1 and cell-2 pure water and methanol solution with 0.085 k-molA/m3 strength are charged. After 63 hours and 36 minutes, methanol concentrations in cell-1 and cell-2 are measured as 0.0195 k-molA/m3 and 0.0655 k-molA/m3 respectively. In order to find the cell constant, an experiment with form system whose molecular diffusivity at 25 oC is known DAB=1.52*10-9 m ducted with initial concentrations of formic acid in cell-1 and cell-2 being as 0.0 and 0.12 k- /m3 and these changing to 0.025 k-molA/m3 and to 0.095 k-molA/m3 after 48 hours and inutes experimentation time. What is the molecular diffusivity of methanol in water at 25 oC and within the concentration range used?

First, find the cell constant α. From the givens: θ = (48)(3 600) + (28)(60) = 174 480 sec.

) in water (B) will be measured at 25

ic acid(A)- water(B) 2as /s is conmolA

28 m

Solution:

32A

31A

3o2A

o1A m/molAk095.0c;m/molAk025.0c;m/molAk12.0c;0.0c −=−=−== θθ

Then, from equation (1-41); 24 m10*92.4

)]025.0095.0/()0.012.0ln[(−=

−−=α

is found. Now this constant can be used to calculate

9 )480174)(10*52.1( −

the molecular diffusivity of methanol in water. 0) + (36)(60) = 228 960 sec.

0.0c;m/molAk0195.0c;m/olAFrom the givens : θ = (63)(3 60

3o2A

o1A m/molAk655mk085.0c;0.0c −2A

31A

3 =−=−== θ θ

Then, from equation (1-41) ; s/m10*32.1

0195.00655.00.0085.0ln

96022810*92.4D 29

4

AB−

−

=−−

=

is obtained. As it is seen value given in Table.1.4.

this value is only 6.5% greater than the

number of empirical

ts and the mixture. Most of them are for dilute so application limitations.

sed to estimate the binary diffusivities in liquids at a temperature range of 5-40 osolute A.

1.3.1.2 Estimation of Liquid Diffusivities: There is large equations derived by various scientists in terms of physical variables of the componen lutions and have

The equation given by Wilke and Chang can be uC, when the solution is dilute for

µ

φ∗=

−

6.0

5.016

A

BBoAB

VT)M(1017.1D (1-42)

In the equation, is the viscosity of the solution in kg/m s, T is absolute temperature ar volume of solute A a

µin K, VA is the mol t normal boiling temperature in m3/k-molA, DAB is molecular diffusivity of solute A in solvent B in m2/s, MB is the molecular weight of solvent B in kg/k-mol B, Bφ is the association factor of s ich is 2.6 for water, 1.9 for methanol, 1.5 for ethanol, 1 for benzene, ether and non-polar solvents such as aliphatic hydro-carbons. Super

olvent B, wh

script (o) on DAB means that solution is dilute for solute A. Molar and atomic volumes of some components at normal boiling temperature are given in Table.1.5. The equation above gives the molecular diffusivities of the solutions whose solvent is water, with ± % 10-15 deviation, and with up to ± % 25 deviation for other solutions. This equation is recommended for


25

solutes whose te A is water molar volumes are below 400*103 m3/k-molA. When soluthe value obtained from the equation must be multiplied with 0.453. Reddy and Doraiswamy obtained a pair of equations, which can be used when the association factor of the solvent is not available.

For (VB/VA) < 1.5 3/1BA

BoAB

)VV(TM1010D

µ

∗= (1-43)

5.017−

5.017For (VB/VA)≥ 1.5 3/1

BA

BoAB

TM105.8D ∗=

− (1-44)

)VV(µ

The accuracy of these equations is almost the same with equation (1-42).

ve an equation which is very simple and used only when the

. Othmer and Thaker gasolvent B is water

6.01.1ABD = (1-45) AVµ

13o 1011.1 ∗ −

In the equations (1-43), (1-44) and (1-45) , µ is the viscosity of the solution in kg/ms, volumes at normal boiling temperature in m3/k-mol, T is the

ture in K and is the molecular diffusivity of solute A in solvent

ker Equation

hanol (A) CH3-CH2-OH 7) + (1)(0.0074) = 0.0592 m3/k-m

o

VA and VB are the molaroABDabsolute tempera

B, when the solution is dilute for solute A.

Example-1.7) Estimation of Liquid Diffusivities Estimate the molecular diffusivity of dilute ethanol in water at 15 oC from;

a) Wilke-Chang Equation b) Othmer-Tha

From Table.1-5 Atomic volumes of Carbon, Hydrogen and Oxygen are read as 0.0148, 0.0037, 0.0074 m3/kg atom. Hence for etVA =(2)(0.0148) + (6)(0.003 ol A For water φB = 2.6 and at 15 C µB = 1.14 cP From Wilke-Chang Equation;

[ ]s/m10*10.1

)10*14.1)(1()0592.0()15273()18)(6.2()10*17.1(

D 2936.0

5.016oAB

−−

− +=

From Othmer-Thaker Equation;

[ ]

=

s/m10*05.1)0592.0()10*14.1)(1(

10*11.1D 29

6.01.13

13oAB

−

−

−

==

are found. As it is seen both values are quite close to the value given in Table.1.4.


26

Table 1-4. Experimentally determined molecular diffusivities in liquids at 1 atm.

oSolute (A) Solvent (B) Temperature( C) Concentration DAB *109 m2/s (k-mol A/m3)

Acetic acid Acetone 25 0 3.31 Acetic acid Benzene 25 0 2.09 Acetic acid Water 0.05 0.769 9.7 Acetic acid Water 0.05 1.26 25 Acetone Water 25 0 1.28 Acetylene Water 25 0 1.78 Ammonia Water 1.24 5 3.5 Ammoni Water 1.64 a 12 1.0 Ammonia Water 20 1.0 2.30 Benzene Chloroform 15 0 2.51 Benzene Water 0 1.0 25Benzoic acid Acetone 25 0 2.62 Benzoic acid Benzene 0 1.38 25Benzoic acid Water 25 0 1.21 n-Butanol Water 15 0 0.77 Butyric acid Water 25 0.05 0.92 Carbon dioxide Wa 25 2.0 ter 0 Chloroform Ethanol 20 1.25 2 Ethanol Benzene 15 0 2.25 Ethanol Chloroform 15 0 2.20 Ethanol Water 10 3.75 0.50 Ethanol Water 10 0.05 0.83 Ethanol Water 10 0 0.84 Ethanol Water 15 0 1.0 Formic acid Benzene 2.28 25 0 Formic acid W 1.52 ater 25 0 Glycerin Water 25 0 0.94 HCl Water 10 9 3.3 HCl Water 10 2.5 2.5 Hydr 25 0 6.3 ogen Water Iodine Benzene 25 0 1.98 KCl Ethylene glycol 25 0.05 0.119 KCl Water 25 0.05 1.87 Methanol Water 25 0 1.24 Methanol Water 15 0 1.28 Oxalic acid Water 25 0 1.61 Oxygen Water 18 0 1.98 Oxygen Water 25 0 2.41 Phenol Ethanol 25 0 0.89 n-Propanol Water 15 0 0.87 Propionic acid Water 25 0.05 1.01 Tartaric acid Water 25 0 0.80 Toluene n-Hexane 25 0 4.21 Water Ethanol 25 0 1.13 Water Glycerin 25 0 0.021


27

Table 1.5 Molar and atomic volumes at normal boiling temperature

Atom ic 0 atom×10

Atomm3/100

Volume 3

Bromine 27.0 Carbon 14

ne HCl-R) 24.6 R-Cl) 21

ium 27.4 ine 8ogen 3.7 e 37

46 ury 19.0

ogen bond 15 amine 10ry amine 12

below) 7.4 le bond in carbony 7 to other elements hyde and ketons 7.4 yl esters 9.1 yl ethers 9.9

ers and ether 9her ester and ethe 11ds (-OH) 12.0 nded to S, P, N 8

sphorou 27.0

members -6members -8

embers -11.5 embers -16.0

thalene -30.0 ene -47.5

Sulfur 25.6 42

ium 35.7 20

le ar Volume, V/k-mol ×103

Molecule Molar Volume, V m3/k-mo 3

.8 Chlori (as R-C

om(end as .6

ChrFluor .7 HydrIodin .0 Lead

c.5-50.1

MerNitr 1 double .6

1 primary1 seconda

.5

.0 Oxygen (outside 1 doub l .4 bonded alde

meth meth ethyl est s .9 hig rs .0 aci bo .3 Pho s Rings 1 three .0 1 four .5

1 five m 1 six m

1 naph Silicon 32.0

1 anthrac

Tin .3 TitanZinc

lecu.4

Mo Mol3m l ×10

Air 29.9 2S 32H .9 Br2 Cl

53.2 7148.4 2 31.2 30.7 H3 25.8 34.0 O 23

S 1.5 2O 36.4 2 14.3 O2 44.8 2O 18.8 2 2

I2N

.5 2

CO NCO2 N .6 CO 5 NH SH O 5.6

Atomim

c Vatom*10

olume 33/kg


28

1.3.2 Molecular Diffusivity in Concentrated Liquid Solutions: The molecular diffusivity in concentrated liquid solutions differs from that in dilute solutions, because of the changes in viscosity with concentration and also because of the changes in the degree of non-ideality of the solutions. Leffler and Cullinan oposed the following empirical equation by analysing the experimentally measured diffusivities in concentrated solutions.

pr

( ) ( ) ⎟⎠

⎞⎜⎝

⎛

∂∂

+=A

AxA

oBA

xB

oABmAB lnx

lnδ1µDµDµD AB (1-46)

This equation is used to estimate the molecular diffusivity of solute A in concentrated solutions whose solvent B does not associate. In the equation, is the molecular diffusivity at infinite dilution in B and is the molecular diffusivity of B at infinite dilution in A. are viscosities of the solution, solute A and the solvent B respectively.The activity coefficient can be obtained from liquid-vapor equilibrium data as the ratio of real to ideal partial pressure of A in vapor in equilibrium with a liquid of concentration xA. for ideal solutions (Raoult’s law)

for non-ideal solutions

oABD

oBAD

BAm µ,µ,µ

Aδ

AxoApAp =

Axo

Ap

Aδ

Ap =

AoA

AA xp

Pyδ =

A graph of log versus logxA is first constructed, then the derivative dlog /dlogxA at any xA, is obtained from the slope of this graph.

1.3.3 Molecular Diffusivities in Electrolyte Solutions: As it is known, molecule dissociates in cations and anions in the electrolyte solution. As the sizes of the ions are smaller than the size of the molecule, mobility of the ions is quite different than the molecule. It is expected that smaller ions will diffuse at higher rates than the larger ions. But this is not so, because of the fact that electrical charge cannot be separable. Nernst has given the following equation for molecular diffusivity of ionic solute A in infinitely dilute solution with solvent B.

Aδ Aδ

−+

−+

−+

−+−

−+

−+

−+

−++

+=

++

=ZZZZ

λλλλT10*8.91

ZZZZ

λλλλ

FRTD

oo

oo

14

oo

oo

2

o

AB (1-47)

where F is the Faraday constant in A s/g-equiv., is molecular diffusivity at infinite dilution in m2/s, and are conductances of anion and cation in (A/cm2)(cm/V)(cm3/g-ekiv.), and Z+ and Z- are the valences of cation and anion, and T is absolute temperature in K. Ionic conductances of some ions at infinite dilution are given in Table.1.6.

Example-1.8) Molecular Diffusivity in Electrolyte Solution Calculate the molecular diffusivity of sulphuric acid ( H2SO4) in water at 20 oC when the solution is ilute for sulphuric acid. emember that sulphuric acid is ionized in water as H+ and . Hence Z+ = 1 and Z_= 2.

oABD

oλ +oλ−

d.R =

4SO


29

From Table.1.6 oλ + = 349.8 and oλ− = 80. Substituting all these into equation (1-47),

)2)(1()21(.

)808.349()80)(8.349()293)(10*91.8(D 14

AB

++

= −

DAB= 2.55*10-9 m2/s is found.

Table 1-6. Ionic conductances of some ions at infinite dilution

Cation Anion o−λ o

+λ Ag+ 61.9 Br- 78.4H+ 349.8 Cl- 76.35Li+ 38.7 Cl −

3O 64.6 Na+ 50.1 −

4ClO 67.6K+ 73.5 F- 55.4

−4 73.6 I- 76.8NH

Ca+2 59.5 −3NO 71.46

Cu+2 56.6 OH 198.6-

Mg+2 53.0 69.623CO−

Zn+2 52.8 80.024SO−

1.3.4 Effect of Temperature on Molecular Diffusivity in Liquids: The change of molecular diffusivity in liquids with temperature is given by the equation below:

1T

AB

2T

ABT

DT

D⎟⎠⎞

⎜⎝⎛ µ

=⎟⎠⎞

⎜⎝⎛ µ (1-48)

Since the viscosity of the solution decreases with increasing temperature, nothing can be said about the change of diffusivity with temperature. 1.3.5 Molecular Diffusivity in Multi-Component Liquid Solutions: Mass transfer flux of solute A in liquid solutions containing more than two components can be calculated from ∑ the flux equations derived for binary solution by using

=

of NA+ NB and by replacing DAB with effective diffusivity, DAef. Perkin and

solu

1.4 CONTINUITY EQUATION FOR A BINARY MIXTURE:

general equation for mass transfer of a component of a binary mixture of A+B can

n

AiiN in place

Geankoplis have given the following equation for the effective diffusivity of solute A, which is dilute in the strong tions of B and C. 0.80.80.8

CoACCB

oABBm

oAef µDxµDxµD += (1-49)

CBm µandµ,µ are the viscosities of solution, components B and C .

Abe derived by considering all the possible changes such as molecular diffusion, bulk


30

flow, chemical reaction at unsteady-state. Let us consider a differential volume, as shown in Fig. 1.7, within a ary m re write the mass balance for component A for this differential v lu Mass rate of inflow of A - m ate tflow f ass rate of formation of A = mass rate of accumulation of A. Let us find the mathematical equivalents of these expressions and substitute them into quation. Assume that component A enters this differential volume through three faces

zAz

bin fluid ixtu and o me:

ass r of ou o A + m

ewith a common corner at T and leaves it through other three faces. Hence, rate of mass flow of A in:

yx)n(zx)n(zy)n( yAyxAx ∆∆+∆ and rate of mass flow of A out: ∆+∆∆yx)n(zx)n(zy)n( zzAzyyAyxxAx ∆∆+∆∆+∆∆ ∆+∆+∆+ , where is the total mass Axn

flux of A in x-direction and subscripts x and x + x∆ outside the brackets show its values at points x and x + x∆ respectively. If component A forms from component B with a chemical reaction and its mass formation rate in unit volume is rA (kg A/m3 s), then its mass rate of formation is zyxrA ∆∆∆ . On the other hand, since total mass of component A in the volume is Azyx ρ∆∆∆ , where Aρ (kg A/m3) is the mass density of component A, mass rate of accumulation of component A in differential volume is then, )/(zyx A θ∂ρ∂∆∆∆ , where θ(s) shows the time. If all these values are substituted into the above-given mass balance expression:

[ ] [ ] [ ]

∆z∆y∆xA)/Aρ(∆z∆y∆x

∆y∆x)(n)Ax(nz∆xx

rt =∂∂

+−+

)

Az(n)Az(n∆z∆x)Ay(n)Ay(n∆z∆y)Ax ∆zzy∆yyx−+−+

++

(1-50is obtained. If both sides of this equation are divided with ∆x ∆y ∆z and then limits of each term is taken as ∆x, ∆y and ∆z go to zero;

Fig.1.7 Differential volume ix in m ture

y

(nAx)x (nAx)x+∆x

∆x

∆z

∆y

x

z

T


31

Ar=θ∂

∂+⎟⎟⎠

⎞∂∂

+∂∂

+ AAzAyAx ρz

ny

n (1-51) ⎜⎜

⎝

⎛∂∂

xn

obtained. Similar analyses for component B gives:

is

BBBzByBx r

zn

yn

xn

=θ∂ρ∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂ (1-52)

these two equations ar ed,

If e summ

=θ∂

ρ∂+

∂

+∂+

∂

+∂ +

∂

+∂

zz

)B

nA

n(

yy

)nA

n(

xx

)B

nA

n(0 (1-53)

found. This gives total mass balance for the mixture. Note that and BA =+ , where ρ is the mass density of the mixture.

follows from equation (1-11) that u)nn(

B

BA ρρρ +=0r

isrIt zzBA ρ=+ . If the derivative of this with

spect to z is taken,re

zu

zu

z)nn(

zzzBA

∂∂

+∂∂

=∂+ ρρ is obtained. If similar equations for x- and y-directions

are written and then derivatives of these are taken with respect to y and z;

∂

n(xxx x ∂∂∂

uxxBA += ρ and u)nn( ∂∂+ ρ∂yyy y ∂∂

uu)n yyBA ∂

+∂ + ∂

=∂

ρρ are also obtained.

tuted into equation (1-53) finally to, Now, if all these equations are substi

0z

uy

ux

uzyx zyx

zyx =θ∂

+∂

+∂

+∂

+⎟⎟⎠

⎜⎜⎝ ∂

+∂

+∂

ρ (1-54)

is reached. This equation is known as “continuity equation for the mixture”. All the derivatives of ρ is zero, when ρ is constant, then equation (1-54) reduces to:

uuu ρ∂ρ∂ρ∂ρ∂⎞⎛ ∂∂∂

0=∂∂

+∂

∂+

∂∂

zu

yu

xu zyx

This equation is known as “continuity equation for inco le .

(1-55)

mpressib fluids”. Now, return back to equation (1-51) From equation (1-10),

zu

zu

zj

zn A

zz

AAzAz

∂∂

+∂∂

+∂∂

=∂∂ ρ

ρ is obtained. If similar equations to the equation (1-10)

are written for x- and y-directions a uations are differentiated with respect to x and y, and all these three equation ed into equ ion (1-51)

nd these eqs are substitut at

Arz

Azj

y

Ayj

x

Axj

z

Azu

y

A

x

A

z

zu

y

yu

x

xuA

A =⎟⎟⎠

⎞⎜⎜⎝

⎛

⎠

⎞⎜⎜⎝

⎛∂

∂+

∂

∂+

∂

∂+

∂

ρ∂+

∂

ρ∂

∂

ρ∂

∂

∂

∂

∂+

∂

∂ρ+

θ∂

ρ∂ (1-56)

finally obtained. When the density of the mixture is

yuxu⎟⎟ +++

constant, this equation mplifies to:

issi

AAzAyAxA

zA

yA

xA r

zj

yj

xj

zu

yu

xu =⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂ρ∂

+∂ρ∂

+∂ρ∂

+θ∂ρ∂ (1-57)

omponent By dividing both sides of this equation with MA, the molecular weight of cA, and by noting that; AA cM =/Aρ , AAA RM/r = and AA JM =/jA ;

AAzAyAxA

zA

yA

xA R

zJ

yJJcccc ⎛ ∂∂∂∂∂∂

xzu

yu

xu =⎟⎟

⎠

⎞⎜⎜⎝ ∂

∂+

∂+

∂+

∂+

∂+

∂+

θ∂ (1-58)


32

is obtained. Since )x/c(DJ AABAx ∂∂−= then; 2Ax ∂∂ If this and

) and ()x/c(D AAB ∂∂−= .

equivalents of (x/J 2

y/JAy ∂∂ z/JAz ∂∂ ) are all substituted into equation (1-58) finally,

A2A

2AA

xA R

zccu

xcuc

+⎟⎟⎞

∂+

∂∂

+θ∂

∂ 22

2A

2

ABA

zA

y yxcD

zcu

yc

⎠⎜⎜⎝

⎛ ∂+

∂∂

+∂∂

=∂∂

+∂∂ (1-59)

to the equation below, when the velocity of the mixture is zero and no chemical reaction takes place:

is found. This equation simplifies

⎟⎟⎠

⎜⎝ ∂∂∂θ∂ 222AB zyx

ion is known as Fick’s second law equation.

⎞⎜⎛ ∂

+∂

+∂

=∂ A

2A

2A

2A cccDc (1-60)

This equat

mass transferspecified initial and boundary conditions gives the concentration

profile of component A, fro

1.5 MASS TRANSF LECULAR DIFFUSION

portant roles in some engineering applications. As example leaching, adsorption, drying of solids,

of solutions by solid membranes and chemical reactions taking place in the

ll be used e of the

solid, at the others this is strongly influenced by the nature of the solid. So, these two

when the solute dissolves andiffusion mechanisms are qu

he diffusivity is independent of the concentration, which are realistic cases in many engineering applications,

Equation (1-59), which is continuity equation for component A in a mixture of A+B, is also known as “general equation for ”. Solution of this equation for any geometry under

m which mass transfer flux of this component can easily be calculated.

ER BY MO IN SOLIDS:

Transfer of gas, liquid and solid by molecular diffusion in solids play im

separationpores of solid catalysts can be cited. Transfers of solutes in solids proceed under several ways. Although the transfer mechanisms in the solids are not simple as in the solutions of gases and liquids, in the majority of them Fick’s law can stisuccessfully. While in some solids diffusion is independent of the structur

situations will be dealt with separately. 1.5.1 Diffusion that is Independent of the Nature of the Solid: This situation is met,

d forms a homogeneous solution with the solid. Although ite different for many systems, Fick’s first law equation

can be written to calculate the flux of the solute A at steady-state operations. 1.5.1.1 Steady-State Diffusion: At the absence of bulk flow, and when t

dz

DN AABA −= (1-61)

can be written

dc

. Integration of this equation under steady-state conditions with constant S yields to:

z

NA)c(cD A2A1AB −

= (1-62)

When the mass transfer area changes with z, NA is first replaced with mA/SN and then the integration is performed. As an example, at the transfer of solute A in radial direction (z = r) through the wall of a cylindrical pipe,


33

)c)/r(rln A2

12

− (1-63)

is obtaine

(cD2πN A1AB

A =l

d, where r1, r2 are the inner and outer radii and l is the length of the pipe.

.5.1.2 Unsteady-State Diffusion: Many mass transfer processe

e in the solid. This equation is geometries with given initial an bo

2a, 2b, 2c dimensions: the x and y-directions, are

1 s occurring in solidsproceed under unsteady-state conditions. In this case concentration of solute A not only depends on location but also on the time. For that, Fick’s second law equation can be used to determine the concentration profile of solutsolved for various d undary conditions. Diffusion through a slab of Let us first assume that small surfaces of the slab, which are the surfaces normal toinsulated against mass transfer. Assume that this solid, whose initial solute concentration is uniform throughout at Aoc , is dropped into a solution that can dissolve the solute A, at θ =0. The concentration of solute A at the surfaces of the solid remains constant at cAs (this concentration is the concentration that the solid would have had, if the solid ha is the

solid

d remained in the solution infinitely long time). Thus, AscAoc −

measure of the solute that can be removed from the solid at the operating conditions. Ifthe solid is kept in the solution for a θ time, the concentration of solute A in thedrops to the uniform concentration of Ac . Thus, Ac - cAs is the measure of the solute that remained in the solid and potentially can still be removed, if the contact time is

prolonged. Hence, the ratio of ( Ac - cAs)/( AscAoc − ) shows the fraction of solute A

o θ. On that has not transferred and remained in the s lid at the end of time the other hand, subtraction of this ratio from 1, gives the fraction of solute A that has already transferred from solid to the solution. With these, the equation (1-60) simplifies to:

cAs

Ac

Aoc

-a a o

cA

ncen

tratio

n

A A

Co

Distance from the center

z

Fig.1.8 Concentration profiles in the slab at unsteady-state mass transfer


34

2A

2

ABA

zcDc

∂∂

=θ∂

∂ (1-64)

his equation must be solved under the following initial and boundary conditions: T

θ < 0 -a < z < a cA= Aoc = constant 0 < θ < ∞ z = ± a cA = cAs = constant θ > 0 -a < z < a cA = Ac There are different techniques for the solution of partial differential equations such as above. Method using Fourier transformations yields to:

amβmβmβ

22AB

AsAo

AsA F.........)e251e

91(e

π8

aDf

ccccF 111 =+++=⎟

⎠⎞

⎜⎝⎛ θ

=−−

= −−− (1-65)

where m and β1 are m = π2/4 and β1 = DABθ/a2. Function Fa is given in Fig.1.9. If mass transfer is allowed from the small surfaces of the slab, equations similar to the

0.001

0.01

0.1

1

0 0.1 0. 0.3 0.5 0.6 0.72 0.4

FaFbFcFr Fs

β1= 2AB

aD θ β2= 2

AB

bD θ β3= 2

AB

cD θ

Fig.1.9 Values of Functions Fa, Fb, Fc, Fs and Fr

Fa , Fb , Fc (slab)

Fr (cylinder)

Fs (sphere)

2a

2b

2c

2a

2c

2axy

z


35

equation (1-65) are found es. Finally, the equation for mass transfefrom all the surfaces of the

for these surfac r slab becomes;

⎟⎠

⎜⎝

=−

=2a

fcc

F ⎞⎛ θ− ABAsA Dcccba22

FFFc

fb

f =⎟⎠

⎜⎝

⎟⎠

⎜⎝

(1-ABAB DD ⎞⎛ θ⎞⎛ θ 66) AsAo

where β2 = 2

AB32

AB

cβ DD θθ

= . Functions Fb and Fc a e also given b

r in Fig.1.9

l the solid, Fick’s second law

ation is first written in the cylindrical coordinates and then solved with suitable ial and boundary conditions. The function Fr, which gives the concen le

n n by

Diffusion through a cylindrical solid: If the solute A diffuses through a cylindricasolid of radius a, length 2c, to the solution surroundingequinit tration profiof solute A in radial direction is thus obtai ed. The profile in x-direction is givethe same function obtained for the slab. Thus, for mass transfer from all the surfaces ofylinder, c

rc2AB

2AB

AsAo

AsA FFa

D fc

DfccccF =⎟

⎠⎞

⎜⎝⎛ θ′⎟

⎠⎞

⎜⎝⎛ θ

=−−

= (1-67)

written. Function Fr was also added to Fig.1.9.

iffusion through a solid sphere of radius a: When the solute A transfers from a lid sphere into a solution surrounding the solid, Fick’s second law equation is first ritten in the spherical coordinates and then solved with the appropriate initial and oundary conditions. Result is;

is Dsowb

s2AB

AsAo

AsA Fa

D fccccF =⎟

⎠⎞

⎜⎝⎛ θ′′=

−−

= (1-68)

unction Fs is also plotted in Fig.1.9. he equations given above can also be used, when solute A transfers from solution to e solid. the slab and cylinder geometries mass transfer is assumed to take place from the

oth faces in a direction. If this is not the case (mass transfer takes place from only one f the faces in one direction), to the denominator of β in this direction 4 is added, ecause length of the diffusion path, in this case, is twice.

Example-1.9) Unsteady-State Diffusion in Solid

otton seeds which contain 18 kg oil per m3 of solid and are pressed in the form of slabs with 12 mm 8 mm x 3 mm dimensions, are immersed in hexane to extract the oil. Calculate their oil content ter 1 hour of contact. The solution is well mixed so that resistance to mass transfer is only in the lid phase. Molecular diffusivity of oil in solid at the operating conditions is 4.1*10-10 m2/s.

Gi Asked :

FTthInbob

Cxafso vens :

solidofm/oilkg18 3Ao = cAs = 0 ?cAo = c

2θ

a = 3 mm, 2b = 8 mm = 1 hour

, 2c = 12 mm, DAB = 4.1*10-10 m2/s


36

For mass transfer from all the surfaces of the slab,

⎟⎠⎞

⎜⎝⎛ θ

=−−

=Ao

A

cF

2ABAs

aD

fccc

cba2AB

2AB FFF

cDD θθ

As

fb

f =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ (1-66)

656.0)10*5.1(

)3600)(1)(10*1.4(Dβ23

10AB ==θ

=−

−

From Fig.1-9 Fa = 0.16 a 21

093.0)10*4(

)3600)(1)(10*1.4(b

D θ From Fig.1-9 F = 0.63 β23

10

2AB

2 ===−

−

b

041.0)10*6(

)3600)(1)(10*1.4(c

Dβ23

10

2AB

3 ==θ

=−

−

From Fig.1-9 Fc = 0.78

0786.0)78.0)(63.0)(16.0(0180cA ==

−−

solidofm/oilkg42.1cA = 3 is found.

sion that is Dependent on the Nature of the Se used for the calculation of mass

ansfer flux, when the diffusion is independent of the nature of the solid. On the other and, in some cases, the mass transfer is strongly influenced from the type and the size f the pores in the solid, and the terms showing the effects of these appear in the flux

ch are cA1 and cA2, are different at two faces such as (c > c ). As a result of this, solute A transfers from face-1 to face-2, through the

id. flow is

1.5.2 Diffu olid: As it has been seen, the Fick’s first and second law equations can btrhoequations. 1.5.2.1 Diffusion of Liquids in Solids: Let us assume that the pores of a solid, as shown in Fig.1.10, is filled with a binary liquid solution of A+B and molar concentrations of solute A, whi

A1 A2

A A+B A+B

A

z 1 2

pores of the solid. It is obvious from the figure that the transfer path is greater than z2-z1 and the mass transfer area is smaller than Sm, total area of each face of the solWith these facts in consideration, total molar flux of A, neglecting the bulk

Fig.1.10 Diffusion of liquids through porous solid

written as:

12t zzk −A2A1

AB2Acc

Dε

N−

= 9 (1-6 )


37

where ε and kt, which must be found experimentally, are the void fraction of the solid nd the tortuousity factor. Notice that mass transfer area in this case, is εSm. One of e tortuousity factors accounts for the fact th ath is greater than

z2-z1 and corrects it, the other one emphasizes the fact that pores at the faces make angle with the surfaces and correct diffusivity of solute A in solvent B, is not influenced from the solid. This D may be grouped with ε and kt in a way given by equation y is known as

DAef=(ε ) DA

1.5.2.2 Diffusion of Gases in Solids iameter of the pores in the solid and the mean free path of the gas molecules, transfer

of gases in the porous solids take place at different mechanisms. The mean free path of the gas molecules is given by;

ath at true mass transfer p

s it. DAB, which is the molecularAB

(1-70) that the resultant diffusiviteffective diffusivity for the system. / 2

tk B (1-70) : Depending upon the relationship between the

ds

0.5

A

A

P=A M2π

RTµ3.2λ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ (1-71)

A1 A2

KA

where λA(m) is mean free path of the gas molecules, which is defined as the distance the molecule travels before colliding with another molecule on its way, µA is the viscosity of the gas(kg/m s), P (N/m2) is total pressure, T(K) is absolute temperature, and R is the general gas constant, which is 8314 J/k-mol K. The mechanisms that solute A molecules may encounter during their transfers in a gaseous mixture of A+B through a porous solid under the concentration difference of (p - p ) at a constant total pressure of P are shown schematically in Fig.1.11 Knudsen diffusion: If the mean free path of the molecules is much greater than the diameter of the pores, the molecules collide with the wall of the pores rather than colliding with each other. This type of diffusion is known as Knudsen diffusion and

nudsen diffusion coefficient D (m2/s) is given as; K

0.5

AK A 48=

MT

d.5D ⎟⎠

⎞⎜⎝

⎛ (1-72)

here, d(m) is the diameter of the pore, which can be calculated from w

(a)

(b)

(c)

pA1 pA2

NA

NA

NA

Fig.1.11 Mass transfer along the pore (a) by Knudsen diffusion, (b) by Molecular diffusion, (c) by Transition diffusion


38

bk ρSε4

d= (1-73)

In the equation above, ρb (kg/m3), is the bulk density of the solid, Sk(m2/kg) is the surface area in unit mass of solid and ε is the void fraction. Hence at steady-state, total molar flux of solute A through a pore of length l (m), is obtained from:

)y(yRTdzRTdz A2A1KAA

l

PDdpDdcDN KAA −=−= KAA −= (1-74)

Kn =

As it is seen, transfer of A through the pore is not influenced by the existence of B, as molecules A collide with the wall of the pore rather than with the molecules of B. Knudsen diffusion occurs when the Knudsen number, defined by

dλ N

is greater than 10.

(1-75)

Molecular diffusion: If the diameter of the pore is much greater than the mean free path of the molecules, molecule-molecule collision rather than molecule-wall collision dominates throughout the pores, and as a result of this, pores do not have any effects on the transfer of solute A in a gas mixture of A+B through the pores of a solid. Hence, in this case, for the total molar flux of A equation (1-7) can be used. This equation can be written as,

)N(NydzRT

N BAAAAB

A ++−= (1-76)

By defining N

dyD

R=NA/(NA+NB) , the equation (1-76) can also be written as;

dz)/Ny(1dz)/Ny(1

N ABABA −

−=−

−=dpDdyPD AA

RARA (1-77)

Upon integration of this equation at steady-state with S=constant, finally;

A1R

A2RABRA yN

yNlnRT

P−

DNN −= (1-78)

l

Transition (mixed) diffusion: If the Knudsen number is greater than 0.01 but smaller than 10, both molecule-molecule and molecule-wall collisions are important in the pores. This case is known as transition or mixed diffusion, and a flux equation for this case can be derived as follows: The loss itravel dz distance in the pores and collide with the walls of the pores is found by multiplying dpA, obtained from equation (1-74) with Ac, which is cross-sectional area of the pores;

is obtained. Equation (1-78) is valid, when Knudsen number < 0.01.

n the momentum of A molecules, when they

cKA

AcKA AdzDRT

NA)(dp =− (1-79)

Similarly, momentum loss due to the molecule-molecule collision is found from equation (1-77) as;

cAdz)/Ny(1D RA

ABAcMA

RTNA)(dp −=− (1-80)


39

Hence, total momentum is obtained by summing the two equationsabove,

loss of the gas

-(dpA) Ac = cKA

A AdzDRT

N + cRAAB

A Adz)/Ny(1DRT

N −

with rearranging,

dz

dyRTP

D1/)/D/Ny(11

N A

KAABRAA ⎥⎦

⎤⎢⎣

⎡

+−−= (1-81)

this equation at steady-state finally, and upon integration of

⎥⎦

⎤⎢⎣

⎡+−+−

=KAABRA1

KAABRA2ABRA /DD/Ny1

/DD/Ny1lnRT

PDNNl

(1-82)

is obtained. This equation reduces to the equation (1-74) at low pressures, and to the s .

ygen. Molecular diffusivity of oxygen in carbon monoxide at 1 atm. and 20 oC is 7.84*10-5 m2/s and

at operating conditions is 0.026 cP. From eqn.(1-71)

is obtained.

equation (1-78) at high pres ures

Example-1.10) Diffusion of Gases in Solid

Oxygen (A) is diffusing in a mixture of oxygen+carbon monoxide at 0.05 atm. and 400 K in the pores of a solid catalyst. The pores are cylindrical with 4 microns diameter and 80 mm length. At the two ends of the pores mole fractions of oxygen are 0.10 and 0.02. The flux ratio is NA= - 2NB. Calculate the total molar flux of ox

viscosity of oxygen

m10*11.2)32)(2(

)400)(8314()10*013.1)(

From eqn.(1-75)

As 0.01<NKn<10 , mass transfer takes place in the transition regime,

From equation(1-82)

05.0(A =λ)10*1)(026.0)(2.3( 6

5

3−

−

=⎥⎦

⎤⎢⎣

⎡π

5.0

528.0)10*1)(4(

(NKn =)10*11.2

6=

−

6−

2N)5.0(N

NNN

NNAA

A

BA

AR =

−+=

+=

s/m10*86.632400)10*1)(4)(5.48(D 24

5.0

6KA

−− =⎟⎠⎞

⎜⎝⎛=

s/m10*71.220273

4001) 23

75.1

−=⎟⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎛10*84.7(D 5−=

05.0⎝AB

95.310*86.6(D )4

Kn

AB ==−

)10*71.2(D 3−

sm/molAk10*4.8)95.3()2/10.0(1)95.3()2/02.0(1ln

)10*80)(400)(083.0()013.1)(05.0)(10*71N

3

3

A −=⎥⎦

⎤⎢⎣

⎡+−+−

=−

−.2)(2( 27−


40

Table 1-7. Permeability and Diffusion Coefficients in Some Solids

SoluteA Solid B Temp., oC

Permeability PM

)cm/.atm(scm)P.T.N(Acm

2

3

Solubility,s

.atmBcm)P.T.N(Acm

3

3

Molecular

Diffusivity,DAB (m2/s)

H2 Vulca 0 8.55 10nized rubber 25 0.342 10-6 0.04 -10

O2 25 0.152 10-6 0.070 2.17 10-10

N2 25 0.054 10-6 0.035 1.54 10-10 CO 25 1.003 10 0.900 1.11 10-10-6

2 H Vulc 5 0.37 102 anized

neoprene 0 - 0.06 0-1

17 0.051 1.03 10-10

27 0.053 1.80 10-10

46.5 0.050 4.81 10-10

H2S Nylon 30 2.60 10-9 5.20 5.0 10-14

Cellulose acetate

30 2.56 10-9 16.0 1.6 10-14

60 4.64 10 9.1 5.1 10-9 -14

Polyvinyl butyral 0 2.25 10-8 15.0 1.5 10-13

30 5.04 10-8 8.0 6.3 10-13

2.04 10 0.3 6.8 10-13-9 Polyvinyl 30 trifluoroacetate

60 5.80 10-9 0.2 2.9 10 2-1

Mylar A 0 1.43 10-10 11.0 1.3 10-15

60 1.74 10-9 2.8 6.2 10-14

Saran 30 2.69 10-10 2.8 9.6 10-15

45 1.13 10-9 3.9 2.9 10-15

75 7.80 10-9 1.3 6.0 10-13

Air Newspaper 25 0.357 Air Leather 25 0.152-0.684 H O Paraffin 23 0.160 10-6 2 H2O Cellophane 38 0.91-1.82 10-6 He Pyrex Glass 0 2.81 10-11 20 4.86 10-11 100 20.06 10-11 Air Porcelain 25 0.81 10-11 He SiO2 20 0.01 2.4-5.5 10-14

H2 Ni 85 0.202 1.16 10-12

125 0.194 3.4 10-12

165 0.192 10.5 10-12

H2 Fe 20 2.59 10-13

CO Ni 950 4.0 10-12

Bi Pb 20 1.1 10-20

Cd Cu 20 2.7 10-19

Al Cu 20 1.3 10-34


41

1.5.3 The Relationship between the Fluxes at the Diffusion of Gases in Solids: At the transfer of gases through the pores of a solid in a system which is closed to atmosphere and kept under constant pre lw ion .

caus act th ends of the pore total pressure i astem s to: N - NB. Hence, it follows from this that in a closed

system at constant pressure mass transfer always takes place under “equimolar counter sfer c h er open ther the wing relationship betw e f dless er ma takes e by Knudsen, molecula ansi on.

i √a bin his gi A√ NB √ MB tated erent

effects in other cases may govern the relationships between the fluxes.

.5.4 Mo cular Diffusivity of Gases in Solids and th bility n the iffusion f gases through s he ermeability, instead of molecular diffusivity, is

The , PM is defined as the volume of the gas as cm at 1 tm. and th h 1 c of the 1 cm thick at 1 second

gas in th (cm3 as/cm3 s at ip a eability cular di d the olubilitys; M = AB s

ence, if ility of a in so perme of the solid for this gas re know olecular diffusivit gas thr his soli y be alculated the equatio ove. .7, the lities, th lities nd molec iffusivities f me g irs are

PROBLEMS

oss hicknes , c B e diffusing in opposite directions at and steady cond olar fl x of A is twice that of B. Partial

res o e of the film are 0 .10 bar. alculate the tota molar fluxes of A nd B. Molecular diffusivit A 0 oC and 1 bar is 2*10-5 m2/s. Ans. NA= - N =

uimo nter transfers o nd CO are occurring by m r diffusion in a 0.11 ng. The partial pressures of N2 at each en mmH .

) Calculate the total fluxe the components as at 298 atm. ) Repea the calculations 473 K and 1 a the rease?

epeat th culations at 298 3 atm Do the fluxes chan [A . a) NA = 7.36* m2s ]

a bin aminar gas mixture, equimo r counter transfers of A and B occurs at 1 bar olar ntrations of A at one a point two, which are 2 mm apart, a

3 and 0.08 k-mol/m3 respectively. Temperature in the gas hase does not r nt and es wi tance as T = T1e where T1 is the temperature at point 1 and i 5 K, z

distance in m. alculate the total molar flux of A. Molecular diffusivity at 1 bar and 305 K is 2.5*10-5 m2/s.

ssure, there is a ays the relat ship Σ Ni = 0 This is be

binary sye of the f this lead

at at both s constant. For A =

tran onditions”. On t e oth hand in an system e is alwaysfollo een th luxes, regar of wheth ss transfer plac r or tr tion diffusi ΣN Mi = 0 For ary system t ves N MA = - . As s before, diff

1 le e Permea : Usually, id o olids, t pused. permeability 3, measur dea 0 oC, that passes roug m2 surface solid ofunder a pressure difference of 1 atm. If g

the solubility of the mong pe

e solid is s ffusivity olid atm.) the rel

s given ionsh rm , mole an

s ia

P 104 D (1-83)

H the solub

the m gas lid and the ability

a n, y of this ough t d can easilc from

lar dn ab In Table.1 solubi e permeabi

a u or so as-solid pa given.

1.1 Acrs

a gas film of to

s 0.2 mm omponents A andm

ar2 barressu

25 C. UnderA at each sid

-state itions, total upa

f .40 bar and 0in B

C l y of at

[ B

1.2 Eq lar cou f N2 a olecula at steady-state tube m lo d are 80 g and 10 mmHga s of kg/m2s K and 1b t at tm. Do

? fluxes inc

c) R

e cal

K and

.

gens

1.3 In 10-7 k-mol A/

ary l la pressure. re measured as 0.04 The m conce point nd

k-mol/m p emain constachang th dis -200z , ts value is 30isC


42

[Ans. NA = 1.4*10-4 k-mol A/m2s ] 1.4 Water at 25 oC is flowing in a covered irrigation pipe below ground. In every 30 m there is a vent line 25 mm inside diameter and 300 mm long to the atmosphere which is at 760 mmHg and 25 oC. There are 10 vents in 300 m pipe. The outside air can be assumed to be dry. Calculate daily water loss as kg. The vapor pressure of water at 25 oC is 23.8 mmHg. [ Ans. 8.55*10-4 kg water / day ] 1.5 Water vapor is diffusing at steady-state through stagnant air in a conical channel at 54 oC and 1 bar. The height of channel is 10 meters and the diameter of it changes uniformly from 0.8 m at the bottom to 0.4 m at the top. The partial pressures of water vapor at the bottom and at the top of channel are 2 500 Pa and 500 Pa, respectively. If air is non-transferring, calculate: a) The molar rate of transfer of water vapor as kg/h, b) The molar fluxes of water vapor at the top and at the bottom of the channel. [Ans. a) h/waterkg10*29.4NA = 5− ]

ases back through the same gas film to the bulk gas composed of equimolar A and

A .

h

0oC and 2 atm total pressure, the mole percents of the components are 50, 25, and 25

1.6 A gaseous mixture consisting of A+B is contacted with a liquid solvent S, which can dissolve only component A and itself does not evaporate into the liquid at the operating conditions. The two phases flow counter-currently in laminar flow regimes. Considering molecular diffusions and bulk flows of the components write the appropriate expressions on the arrows.

1.7 In a catalytic reactor, the dimerization reaction 2A A2 is being carried out. Each catalyst particle is surrounded by a laminar gas film of 0.16 mm thickness through which gas A diffuses in order to arrive at catalyst surface. Re ction occurs instantaneously on the catalyst surface and product A2 diffu

2a) Calculate the total molar fluxes of A and A2 at 140 oC and 1 bar pressure. b) Find the distance from the catalyst surface where t e mole fraction of A is 0.25. Molecular diffusivity at the operating conditions is 2.4*10-5 m2/s. [ Ans. a) NA = 2.55*10-3 k-mol A /m2s, b) 0.074 mm ] 1.8 Hydrogen is being oxidized on a solid catalyst surface according to the reaction below: 2H2+O2 2H2O H2 and O2 diffuse through a laminar gas film of 0.2 mm thickness to the catalyst surface, react there instantly with complete conversion, and H2O diffuses back through the same film into the bulk gas. At the bulk gas at 30

Gas phase(A+B) Liquid phase(S+A) Interface

A

for H2(A), O2(B) and H2O(C) respectively.


43

Calculate: a) The effective diffusivity of H2 in O2 - H2O mixture. b) Total molar fluxes of H2 , O2 and H2O. The binary diffusivities at 200 oC and 1 atm are DAB=1.8*10-4 m2/s -4 2, D =2.1*10 m /s and

8 k-mol/m2s ]

at the left and right side of the gas film adjacent to the solid

y y

ACDBC=5.9*10-4 m2/s. [Ans. a) DAef =1.47*10-4 m2/s, b) NA = 0.01 1.9 The following reaction takes place on a solid catalyst surface at 4.2 bar and 323 oC.

A(g) + 3 B(g) 4 C (g) The mole fractions of the componentscatalyst surface are given as;

yA B C

Left side 0.20 0.60 0.20 Right side 0.05 0.15 0.80

Calculate the total molar fluxes of each component by assuming that thickness of the film is 0.1 mm. Binary diffusivities at 1 bar and 25 oC are given as DAB =1*10-5 m2/s, DAC = 2*10-5 m2/s, DBC= 1.5*10-5 m2/s. 1.10 In the diffusion system shown on the Figure, pure liquid A evaporates into a as B at constant

der steady-state conditions, the position of liquid-gas interface is kept fixed at z = z1, and the mole fraction of A is yA1, which corresponds to equilibrium with the liquid at the interface. The solubility of B in A is negligible. At the tube mouth (z = z2) mole fraction of A remains constant at yA2a) Derive the following equation for the total molar flux of A at any z

gg an experiment unpressure and temperature. Durin

.

1A

A

1

ABA y1

y1ln

zzcD

N−−

−=

b) Derive the following equation for the concentration profile of B in the gas phase.

A

B A+B

z2

z1 1

2

1z2z1zz

1B

2B

1B

B

yy

yy −

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

For the case t= 23 C, P = 1 bar, Do -6 2 o

rd-Spotz

++ . AB BA

AB = 8.35*10 m /s, z2-z1= 150 mm and vapor pressure of A at 23 C 100 mmHg, calculate: c) The total molar flux of A, d) The distance where the partial pressure of A is the half of that at the liquid-gas interface. Due to the large flow rate of gas B, the partial pressure of A at the tube mouth may be taken zero. 1.11 Predict the molecular diffusivity of CO in Cl at 1 bar and 25 oC from2 2 Hirschfelder-Biequation. 1.12 Calculate the molecular diffusivity of copper sulfate ( CuSO4) in water at 25 oC, when the solution is dilute for copper sulfate. Remember that copper sulfate ionizes in water as Cu and

.13 Prove that in a binary gas mixture D = D .

=4SO

1 1.14 Derive equation (1-20). Hint : Write total molar flux equations for binary and multi-component gas mixture. 1.15 Derive equation (1-22) starting with equation (1-20).


44

1.16 Starting with Fick’s first law equation show that for steady-state equimolar counter diffusion of A and B, in which T is proportional to zn, the

gas A through a non-isothermal medium consisting oftotal molar flux of gas A is given by

])D/Tz()D/Tz[(R 1AB2AB

Az −)pp)(2/n1(N 2A1A −−

=

where pA is the partial pressure of component A, R is the gas constant, T is absolute temperature, z is c and subscripts 1 and 2 refer to the two

1.17 a).Show that for steady-state diffusion of compon A fr a cylindrical surface through a non-transferring gas film of B , the tot ux p A linder surface is given by

the distance in the direction of diffusion, n is a onstant, boundaries between the diffusion takes place. Assume that DAB is proportional to T1.5 and n = 2.

ent omal molar fl of com onent at the cy

)a/1ln(RTa χ+

where P is the total pressure, a the radius of the cylinder,

)p 1B/lN 1A

pn( 2BPDAB=

χ the thickness of the gas film, and pB1 and pB2 the partial pressure of component B at the cylinder surface and at the other side of the gas film.

) A cylindrical brass rod, 12 mm in diameter, is given a protective coat of lacquer. After application

one per cm of rod surface. The vapor pressure of fficient of acetone through air may be taken

. Assume that flow rate of B is sufficiently high so ]

bin an acetone solvent, the lacquer is dried in a chamber through which acetone-free air is passed at 70 oC and atmospheric pressure. If the temperature at the surface of the coating is assumed to remain constant at the wet-bulb value of 10 oC until drying is complete and acetone is assumed to diffuse from the rod through a stagnant gas film of 6 mm thick, how long does the lacquer take to dry?

2

that pA2 ≈ 0. [Ans.: b) 147 sec.

The original lacquer coating contains 10 mg of acetacetone at 10oC is 147 mbar and molecular diffusion coeas 8.0*10-6 m2/s


45

Chapter-2

MASS TRANSFER BY TURBULENT DIFFUSION and

MASS TRANSFER COEFFIENTS 2.1 Introduction: In the discussion of mass transfer, so far the emphasis has been on molecular transport in fluids that were stagnant or in laminar flow. In much of the mass transfer applications, the rate of mass transfer by molecular diffusion is too small and more rapid transfer is necessary. To speed up the transfer, the fluid velocity is increased so that flow regime changes from laminar to turbulent and as a result of this mass transfer takes place by turbulent or eddy diffusion. To have a fluid in turbulent flow requires this fluid to be flowing past another immiscible fluid or a solid surface. An example of this was already given in Fig.2 in the introduction. When a fluid flows past a surface under turbulent flow conditions, the actual velocity of small particles (eddies) of fluid cannot be given mathematically as in laminar flow. In laminar flow fluid flows in smooth streamlines whose behaviours can be described mathematically. In turbulent flow there are no orderly streamlines and no equation to describe the behaviour. There is large number of eddies which move rapidly and randomly in all the directions. When a solute is dissolving from a solid surface and being transferred into the fluid, there is a high concentration of this solute in the fluid adjacent to the solid and its concentration decreases as the distance from the surface increases.

A typical concentration profile, which is explained below, is shown in Fig.2.1. Adjacent to the surface, a thin laminar film is present, in which the mass transfer occurs by molecular diffusion, since no eddies can penetrate into this layer. Since mass transfer by molecular diffusion is a slow process, large concentration change or drop in concentration occurs across this laminar film. Adjacent to this layer is the transition or buffer layer to which some eddies can penetrate and mass transfer is by the sum of the molecular diffusion and by turbulent

diffusion. In this layer there is a gradual change from almost pure molecular diffusion at one end to the turbulent diffusion at the other end. For that, concentration decrease is much less in this layer. In the true turbulent core adjacent to buffer layer most of the transfer is by turbulent or eddy diffusion. Molecular diffusion still occurs, but its contribution is little to the over-all mass transfer in this layer. The concentration decrease is very small here, since the rapid movement of eddies evens out any gradient tending to exist. As the average concentration of solute A, Ac is very close to the minimum concentration, cA2, the thickness of laminar sublayer is very small compared

cA2

A cA1

Ac

Distance from surface, z 0Mol

ar c

once

ntra

tion

of so

lute

A, c

A

Laminar sublayer

Buffer layer

True turbulent core

cA=f(z)

Fig.2.1 Typical concentration profile in turbulently flowing fluids

to the thickness of the true turbulent core. Although the thickness of the laminar sublayer decreases sharply with an increase in turbulence, this can never be made zero.

46No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 46

It was stated that even the motions of the particles of a pure fluid in turbulent flow cannot be described mathematically with today’s knowledge of us. It can then easily be realized that the situation will be more complicated when a second component and hence the mass transfer exist. For that reason, the flux equation for turbulent diffusion can only be written by analogy to the flux equation for molecular diffusion.

dz

dc)D(J AMABA ε+−= (2-1)

where JA is the molar diffusional flux of A in a turbulently flowing mixture of A+B as (k-molA/m2s), (dcA/dz) is the concentration gradient causing this diffusion. Mε is known as turbulent or eddy diffusion coefficient for mass transfer. Although it has the same dimension with DAB and looks counter part of DAB in turbulent diffusion, it is not a true physical property as it depends not only on the components but more than this on the flow conditions of the fluid. Hence, Mε is strongly dependent on z, distance from the surface of the solid. Although this dependence cannot be given by an equation yet, it is obvious that =0 at z=0, and it increases with increasing z. If the mixture itself moves in the transfer direction (bulk flow) the flux given by equation

Mε

(2-1) is the flux of component A relative to the molar average velocity of the mixture defined by equation (1-5). A question may be asked: “ why was DAB added to the equation (2-1)?” The answer is very simple; “not only in the laminar and buffer layers but also in the true turbulent core molecular diffusion still exist”. For the integration of equation (2-1) )z(fM =ε is required. It was already said that this is not known yet. By defining an average Mε , which is independent of z, integration can easily be performed at constant Sm and at steady-state with boundary conditions; at z = 0 cA=cA1 and at z = z cA=cA2;

)cc(z

DJ 2A1AMAB

A −ε+

= (2-2)

In general; cA1=cAi, which is equilibrium solubility of component A in B at the prevailing conditions, and from the above-given explanations A2A cc ≅ . In the equation (2-2) z and Mε are both influenced by the flow conditions and they must be determined experimentally. Since DAB must also be found experimentally, then all these three variables are collected under one coefficient, ck′ (m/s), which is known as mass transfer coefficient and is determined experimentally.

z

Dk MABc

ε+≡′ (2-3)

It is obvious that all the parameters effecting DAB, Mε and z will effect the . So- defined is only the mass transfer coefficient of the eddies and molecules diffusing under the available concentration difference. If any bulk flow exists in the transfer direction, flux arising from this flow is excluded from the flux obtained by using this coefficient. Since we are normally interested in the total flux rather than diffusional flux alone, in this case equation (1-7) can be written as;

ck′

ck′

)NN(c

cdz

dc)D(N BAAA

MABA ++ε+−= (2-4)

From the integration of this equation at steady-state with the same boundary conditions used in the integration of equation (2-1) ;


c1Ac

c2Ac

c

BNANAN

BNANAN

lnZ

MABD

BNAN

ANAN

−+

−++

+=

ε (2-5)

is obtained. Substituting from equation (2-3) gives; ck′

c1Ac

c2Ac

BNANAN

BNANAN

BNAN

ANAN lncck

−+

−+

+= ′ (2-6)

We prefer to write this equation in the form of “Flux equals the product of mass transfer coefficient and concentration difference causing this transfer” and hence,

)cc(kN 2A1Ac

A −β′

= (2-7)

is written. Here β accounts for the bulk flow contribution to the mass transfer and is therefore known as bulk flow contribution term or drift factor. β is 1, when no bulk flow in the transfer direction exists. By equating equations (2-6) and (2-7) an expression is obtained for the calculation of β.

)cc(kc/cNc/cNlnckN 2A1A

c

1AR

2ARcR −

β′

=−−′

( ) ( )( )( )

( )R

lnAR

1AR

2ARR

1AR2ARN

c/cN

c/cNc/cNlnN

c/cNc/cN −=

−−

−−−=β (2-8)

where, NR is defined by NR=NA/(NA+NB). Now, let us see what the β will be at some special cases, which were already considered in Chapter-1. But before this, note that β is almost 1, when the solution is dilute for component A (cA/c ), whatever the relationship between the fluxes is. 0≅ 2.2 A and B Transfer under Equimolar Counter Transfer Conditions: Since NA= - NB, NR is indeterminate and then we go to equation (2-4). If NB is substituted with -NA and then integrated with the boundary conditions at which equation (2-2) was integrated, NA= (2-9) )cc(k 2A1Ac −′

is obtained. Comparing this equation with the equation (2-7) β=1 is seen. This is an expected result, as it was already proved in Chapter-1 that there cannot be bulk flow in “equimolar counter transfer of A and B”. It follows from this that; mass transfer coefficient given by is the mass transfer coefficient for “ equimolar counter transfer of A and B”.

ck′

As the concentrations in the solutions are expressed in different units, the mass transfer coefficients have also different units, because NA has always the same unit. In the gas phase, partial pressure and mole fraction are the most commonly used units next to the molar concentration. Hence, for gas mixtures equation (2-9) can be written as: )yy(k)pp(k)cc(kN 2A1Ay2A1AG2A1AcA −′=−′=−′= (2-10) Interrelationships between three mass transfer coefficients can be found as

, by noting that yckkPk cyG ′=′=′ A= cA/c and cA= pA/P .


Similarly in the liquid solutions equation (2-9) is written as; )xx(k)cc(kN 2A1Ax2A1ALA −′=−′= (2-11) where )M/(kckk LLx ρ′=′=′ can easily be proven. Here, and M are density and molecular weight of liquid solution.

ρ

2.3 A Transfers through Non-Transferring B: In this special case as NB=0 then NR=1, and since from equation (2-8): β= (1-cA/c)ln = (1-xA)ln = (xB)ln veya =(1-yA)ln = (yB)ln finally equation,

)cc(k)cc()x(

kN 2A1Ac2A1A

lnB

cA −=−

′= (2-12)

is obtained. In the gas mixtures this equation can also be written as; )yy(k)cc(k)pp(kN 2A1Ay2A1Ac2A1AGA −=−=−= (2-13) where, kG= , and lnBG )P/(pk′ lnBcc )/(ykk ′= lnByy )P/(pkk ′= The relationships between three mass transfer coefficients are: kG P = ky = kc c In liquid solutions the equation (2-12) is written as; )xx(k)cc(kN 2A1Ax2A1AA L −=−= (2-14) where, kL= and lnBL )/(xk′ lnBxx )/(xkk ′= It follows from the equations above that primless mass transfer coefficients are the mass transfer coefficients for “A transfers through non-transferring B”. Furthermore, while is independent of concentration, kck′ c depends on concentration of the solution. Mass transfer coefficient equals /βkc′ , when solutions are not dilute and when mass transfer takes place under the conditions other than two special cases given above. Often experimental determination of mass transfer coefficients is conducted when β is not 1. To use this coefficient for other values of NR, it is first converted to and then is corrected to the desired value of N

ck′

R. If the mass transfer coefficient is determined experimentally in a mixture, which is very dilute for component A, then regardless of the value of NR used, the experimental value of (kc)exp. equals ck′ and kc, as the bulk flow contribution is negligible. The flux equations and the relationships between various mass transfer coefficients for the two special cases mentioned above were all collected in Table.2.1 2.4 Mass Transfer Coefficients in Laminar Flow: In principle, mass transfer coefficients are not required for laminar flow, since molecular diffusion prevails and equations given in Chapter-1 can be used to calculate mass transfer rates. But, for complex geometrical systems, it is either difficult to describe the laminar flow mathematically or to solve the written equation without making many simplifying assumptions. This then may lead to the oversimplification of the case. For that even for laminar flow, mass transfer coefficients are measured experimentally and correlated. Below, theoretical determination of mass transfer coefficient in laminar flow for a very simple geometry will be described. As it will be seen, even the geometry is very simple; many simplifying assumptions will be made to solve the complex mathematical expressions.


Table 2-1. Mass transfer coefficients and Flux equations

General equations:

)yy(k

)xx(k

)cc(k

N 2A1Ay

2A1Ax

2A1Ac

A −β

′=−

β′

=−β′

=

)xN/()xN(lnN

)xN()xN(

1AR2ARR

1AR2AR−−

−−−=β NR = NA/(NA+NB)

Special cases: A transfers through non- Equimolar counter transfer transferring B Units of mass transfer coefficients β=1 β=(1-xA)ln in SI Gases: )pp(kN 2A1AGA −′= )yy(kN 2A1AyA −′= )cc(kN 2A1AcA −′=

)pp(kN 2A1AGA −= )yy(kN 2A1AyA −= )cc(kN 2A1AcA −= )YY(kN 2A1AYA −=

k-mol A /m2 s bar k-mol A /m2 s k-mol A /m2 s (k-molA /m3) k-molA/m2 s (k-mol A/k-mol B)

Liquids: )xx(kN 2A1AxA −′= )cc(kN 2A1ALA −′=

)xx(kN 2A1AxA −= )cc(kN 2A1ALA −=

k-mol A/m2 s k-mol A/m2 s (k-mol A/m3)

Conversions of mass transfer coefficients:

Gas : lnBGlnB

ccc )p(kRT

)p(kRTPkck ==′=′ = ky yk

P)(p lnB ′=

Liquid: lnBxxLlnBLL )x(kkM

kc)x(kck =′=ρ′==′


2.4.1 Mass Transfer from a Gas into a Liquid Film in Laminar Flow: Let us consider the transfer of solute gas A into a liquid film B in laminar flow down a vertical flat plate of length l, and width b. Our aim is to calculate the liquid phase mass transfer coefficient kL theoretically, which in turn can be used to compute the total flux or rate of the solute A dissolved in liquid B at any point of the plate. As it is seen from Fig.2.2, the flow direction is x, mass transfer direction is z and thickness of

Ac=b*δ

ux=f(z)

x=0

Liquid

cAx=f(z)

Axc

δ

Liquid

A

cAi

y

z

x

Ac

ux,max

xu

b

Solid plate

Liquid film

lAA cc =

AoA cc =

dSm=b*dx

Ac

l

x=l

S

G

dx

Fig.2.2 Mass transfer from a gas into a laminar liquid film the liquid film is constant at . The concentration of solute A in the inlet liquid is uniform and shown with

δAoc . The concentration of solute A in the first liquid layer

adjacent to the gas is cAi, which is the equilibrium solubility of solute A in liquid B at the prevailing temperature and pressure. cAi does not change with x. The solution of the problem is started by writing the continuity equation for component A, which was derived in the previous chapter, and Navier-Stokes equation, which describes the motion of the fluid in x-direction.


A2A

2

2A

2

2A

2

ABA

zA

yA

xA R

zc

yc

xcD

zcu

ycu

xcuc

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

+θ∂

∂ (1-58)

gxP

zu

yu

xu

zuu

yuu

xuuu

2z

2

2

y2

2x

2x

zx

yx

xx ρ+

∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

µ=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+θ∂

∂ρ (2-15)

These two equations must be solved simultaneously. First from equation (2-15) velocity profile, ux=f(z) is obtained, then by substituting this into equation (1-58) concentration profile is found. It is obvious that these equations cannot be solved in these forms and hence some simplifications are required. These are: 1o) There is no chemical reaction between A and B and hence RA = 0. 2o). There are no changes in the conditions in y-direction, as a result of this, all the derivatives with respect to y are set to zero. 3o) Steady-state prevails so that 0/uand0/c xA =θ∂∂=θ∂∂ . 4o) Rate of absorption of solute A in liquid B is small, this leads to 0uz ≅ . 5o) Rate of molecular diffusion of solute A in x-direction is negligible compared to the rate of transfer by bulk motion in this direction, so that can be written. 60)x/c(D 2

A2

AB =∂∂ o) Flow is fully developed and hence . 70x/uand0x/u 2

x2

x =∂∂=∂∂ o) Pressure drop in flow direction is negligible, then 0x/P =∂∂ . 8o) All the physical properties ( ) are constant.

ABD,,µρ

Under these assumptions equation (2-15) simplifies to:

0gdz

ud2x

2=ρ+µ (2-16)

If this equation is integrated with boundary conditions of: at z = δ ux=0 and at z = 0 , 0dz/du x =

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛δ

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛δ

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛δ

−µδρ

=2

max,x

2

x

22

xz1uz1u

23z1

2gu (2-17)

is obtained, where ux and xu are the local and average velocities. The average velocity, which is written as /3µρgδu 2

x = is found from the integration of local velocity over the whole cross-section. ux,max is the maximum velocity which occurs at gas-liquid interface and is obtained as 2/u3u xmax,x = from the equation above by inserting z = 0. As for the equation (1-58), with the assumptions given above this equation also simplifies to:

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

=∂∂

2A

2

ABA

x zcD

xcu (2-18)

If ux is substituted from equation (2-17), this equation becomes

2A

2

ABA

2

max,x zcD

xcz1u

∂∂

=∂∂

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛δ

− (2-19)

Equation (2-19) must be integrated with the following boundary conditions: at z=0 for < x < l c0 A = cAi , at x = 0 for 0 < z < δ cA = Aoc and at z = δ for 0 < x < l

. The equation above is still complicated for integration and may be integrated at two different conditions:

0z/cA =∂∂


1o) Contact time and hence the length of plate is long: In this case solute A can penetrate into all the liquid layers. This condition is met, when the liquid Reynolds number, which is defined as Re = 4Γ/µ, is smaller than 100. Upon the integration under these conditions first cA= f(x,z) is obtained , then by inserting x = l for all the x’s in the equation, the concentration profile in the outlet liquid, cAl = f(z) is found. The average concentration in the outlet liquid, which is independent of z, is then obtained by performing the following integral;

∫=δ

o xA

x

A dzbucb δu

1cll

(2-20)

If all these steps are accomplished, finally the average concentration of the outlet liquid is found as:

.....e0.03599e0.100e0.7857cccc η4105.6η839.3η5.1213

AoAi

AAi +++=−− −−−l (2-21)

where x2

AB u/3δ2Dη l= . Consider the differential volume in the liquid as shown in Fig.2.2. Since solute A transferring through the shaded area will cause an increase in the A content of the solution, dxb)c(ckcdbδu AAiLAx −= can be written. By separating the variables and integrating;

∫∫ ==⎮⌡⌠

−δ

lll

oLo L

c

c AAi

Ax dxkdxk

cccdu

A

Ao

l

l AAi

AoAixL cc

cclnuk−−δ

= (2-22)

is obtained, where and Lk Lk are local and average mass transfer coefficients for liquid phase. On the other hand for Re < 100, the first term of the equation (2-21) will be sufficient. If this term is inserted into equation (2-22),

δ

≅η+δ

=δ

=η

ABx1213.5

xL

D41.3)1213.5241.0(u7857.0

elnukll

(2-23)

or 41.3ShDk

.avAB

L ==δ (2-24)

is finally obtained, where Shav. is average Sherwood number. 2o) Contact time is short or 100 < Re < 1 200: In this case, solute A can only penetrate into the first layers of the liquid, hence ux can be replaced by ux,max. When

this is done 2A

2

ABmax,x

A

zcD

)u/x(c

∂∂

=∂

∂ or by taking x/ux,max = θ, where θ is contact

time of gas-liquid,


2A

2

ABA

zcDc

∂∂

=θ∂

∂ (2-25)

is obtained. This equation is solved with the boundary conditions: at θ= 0 for 0 < z < ∞ cA= Aoc , at θ > 0 for z =0 cA=cAi and at θ 0 for z = ∞ c≥ A= Aoc . As it is seen, because of very short contact time, the thickness of the liquid film is taken as ∞. Laplace transformation technique can be used for the solution. Thus, by taking Laplace transforms of both sides:

₤ ⎟⎠⎞

⎜⎝⎛

θ∂θ∂ ),z(cA = θ⎮⌡

⌠θ∂

∂∞θ− dec

o

sA = =θθθ∂∂∫∞ θ− de),z(c

o

sA AoA c)s,z(cs −

₤ ⎟⎠⎞

⎜⎝⎛

∂θ∂

2A

2

AB z),z(cD = =θ⎮⌡

⌠∂

θ∂ θ−

∞

dez

),z(cD s

o

2A

2

AB 2A

2

ABo

sA2

2

AB dz)s,z(cdDde),z(c

zD =θθ

∂∂∫∞

θ−

are obtained, where s is Laplace parameter. By inserting these transformations into equation (2-25), ordinary differential equation,

AB

AoA

AB2

A2

Dc)s,z(c

Ds

dx)s,z(cd

−=− (2-26)

is obtained, the solution of which is :

s

ceKeK)s,z(c AoD/sz2

D/sz1A

ABAB ++= − (2-27)

where, K1 and K2 are integration constants. In order to evaluate these constants the two remaining boundary conditions, after the Laplace transformations, are used. The transformations of these boundary conditions lead to: ₤[cA(0,θ)]=cA(0,s)=₤(cAi)=cAi/s and ₤[cA(z,θ)]= cA(z,s)= ₤ s/c)c( AoAo = . z→∞

Since from second boundary condition, s

ceK)0(Ks

c Ao21

Ao ++= ∞ is written, K2 = 0 is

found. From the first boundary condition cAi/s = K1 (1) + s/cAo and K1 s/)cc( AoAi −= is written. By substituting all these into equation (2-27) finally,

s

cs

e)cc()s,z(c AoD/sz

AoAiAAB

+−=−

(2-28)

is obtained. Taking the anti-Laplace of this equation with the help of Table.App.2.1, the concentration profile of solute A is found as:

θ

−+=θAB

AoAiAoA D4zerfc)cc(c),z(c (2-29)

where, )D4/z(erf1)D4/z(erfc ABAB θ−=θ is error function complementary. As it

is known, error function, which is defined by erf u = ∫ −

πuo

z dze2 2

, is an infinite series

and its value for various u is given in Table.App.2.2. Returning back to the equation (2-29) and substituting (x/ ux,max) for θ,

max,xAB

AoAiAoA u/xD4zerfc)cc(c)x,z(c −+= (2-30)


is obtained. Again, by inserting x = l in the equation (2-30), the concentration profile of solute A in the outlet liquid, cAl is found, and this is:

max,xAB

AoAiAoA u/D4zerfc)cc(ccl

l−=− (2-31)

Average concentration lAc is found as,

dzu/D4

zerfc)cc(dzb)cc(b

1cco

max,xABo

AoAiAoAAoA ∫∫

∞∞

δ−

=−δ

=−l

ll

π/u4D

δ)c(ccc maxx,ABAoAi

AoA

ll

−=− (2-32)

In this case, by writing solute A balance for the differential volume, an equation similar to equation (2-22) is obtained. When this is done,

l

l AAi

AoAimaxx,L cc

cclnδu

k−−

= (2-33)

is found. In short contact time the change in the A content of the liquid will be very small, as a result of this,

AoAi

AoA

AAi

AoAi

cccc

ccccln

−−

≅−−

l

l

can be approximated. Hence, from the

combination of equations (2-32) and (2-33) finally,

ll πρδΓ6D

πu4D

k ABmaxx,ABL == (2-34)

is obtained, where Γ is the mass flow rate of liquid per unit width of the plate (kg/ms) and is defined by δuρ/bmΓ x== & (2-35) If it is remembered that /3µρgδu 2

x = , from these two equations, the thickness of the liquid film δ; (2-36) 3/12 )g/3( ρΓµ=δis found. As it is seen, after very long mathematical manipulations, two theoretical equations for the calculation of average liquid phase mass transfer coefficient, Lk , when solute A transfers from a gas into a liquid in laminar flow on a vertical plate, have been derived. These average coefficients, as will be proved in section 2.7, are used in the equation below, to compute the average mass transfer flux. lnAAiLA )cc(kN −= (2-37) As stated in the derivations, Lk is computed from equation (2-23), when Reynolds number is smaller than 100, and from equation (2-34), when Reynolds number is between 100 and 1200. In order to obtain reliable results, 8 assumptions made at the start of the derivations must be met. From the comparison of the experimentally determined Lk values with the computed values, it has been found that experimentally measured values are greater than computed values. The deviations are more pronounced at high Reynolds


numbers, as the ripples that form at gas-liquid interface are more stable at these conditions.

Example-2.1 Calculation of Mass Transfer Coefficient Cl2 gas is being absorbed into water film flowing down on a vertical plate which is 1.2 m high and 200 mm wide. Calculate the average liquid phase mass transfer coefficient for a water flow rate of 0.15 m3/h at 25 oC. Density and viscosity of water and molecular diffusivity of Cl2 in water at 25 oC are 977 kg/m3, 0.894 cP and 1.44*10-9 m2/s.

Solution:

Mass flow rate of water, s/kg0415.03600/)997*15.0(m ==& . Mass flow rate of water per unit width of plate, from equation (2-35), kg/ms0.20800.0415/0.2Γ == Thickness of the liquid film from equation (2-36), .m10*85.3)]81.9()997/()208.0*10*894.0*3[( 43/123 −− ==δ Reynolds number, Re = 4Γ/µ = 4*0.208/(0.894*10-3) = 931. Hence, equation (2-34) is applicable. s/m10*52.3

1.2*)10*(3.85*997*π0.208*)10*(1.44*6k 5

4-

-9

L−==

is found. 2.5 Mass Transfer Correlations: As it has been shown above, even for a very simple geometry for the solution of theoretically written mathematical expressions, many simplifying assumptions are made and the solution still requires lengthy calculations. In the end, because of the simplifying assumptions, the values computed from theoretically derived equations might become quite different from experimentally measured values. From this, it can easily be realized that why experimentally measured mass transfer coefficients are preferred even in laminar flow. In the turbulent mass transfer, even writing the mathematical equation describing the process is almost impossible, let alone the solution of the equation. Hence, experimentation is left as only alternative. To keep the number of the experimentations at a reasonable level and also for the formulation of the results, a method which is known as Dimensional Analysis is frequently used in the analysis of many engineering problems, when the differential equation describing the process can not be written and large number of variables are involved. As it can be realized, to find a relationship between one dependent and large number of independent variables with experimentation, one of the independent variables is changed, while the others are kept constant and the effect of this changing independent variable on the dependent variable is observed. This is repeated with all the independent variables systematically. It is obvious then how lengthy will be the experimentation when large number of variables are involved. But not only this, the organizing and expression of the results in the form of a generalized equation become also very difficult. Instead of this, before experimentation, the variables both dependent and independent are grouped such that resultant group is dimensionless. Once grouped, the effect of group is important not the individual variables contained in the group. It is obvious that number of the groups is much smaller than the number of the variables. After grouping, experimentations are conducted and a relationship among the groups is obtained. To some dimensionless groups or numbers the names of the renowned scientists were assigned, such as


Reynolds Number, Schmidt Number, Sherwood Number etc. Before giving an example to the analysis, which will be conducted in this way for mass transfer, let us mention from a theorem, which is known as Buckingham Theorem and enables one to compute the number of the dimensionless groups from number of the variables. The statement of this theorem is as follows: The functional relationship among q quantities, whose units may be given in terms of n fundamental units, may be written as m = q-n dimensionless groups or numbers. Now, as example let us consider mass transfer from a pipe wall to a fluid flowing inside the pipe at steady-state, where A shows the transferring component and B the fluid. Determination of the independent variables that affect the mass transfer coefficient constitutes the first step of the analysis. Let us assume that these are: inside diameter of the pipe D, molecular diffusivity of A in B D

ck′

AB, density and the viscosity of the fluidρ , and finally the average velocity of the fluidµ xu . An error in estimating the variables either in excess or less at this stage yields no result in dimensional analysis. So, )uµ,ρ,,DD,(fk xABc =′ (2-38) can be written, from which q = 6. If the dimension of each term is found from its definition equation n =3, which are length l, mass M and time θ, is obtained. Thus, from m = 6-3 = 3, the number of the dimensionless groups required to give the relationship among the variables is found as 3. Now, let us find these dimensionless groups. Let us assume that the variables in the dimensionless groups are with the powers given in the equation below: [ ] [ ] [ ] [ ] [ ] e

xdcb

ABa

c uDDfk µρ=′ (2-39) If the dimensions of all the variables are substituted, (2-40) eedddc3cbb2a1 MM −−−−−− θθθ=θ llllll

is obtained. As the equation to be found will be homogeneous in dimension, the sum of the powers of each dimension on the left of the equation (2-40) must equal the sum of the powers of the same dimension on the right of the equation. Hence, For l : 1 = a+2b-3c-d+e For M : 0 = c+d For θ : -1 = -b-d-e are written. Since there are 3 equations but 5 unknowns, then, if c and e are kept constant and a, b and d are expressed in terms of c and e ; a = -1+e, b =1+c-e and d = -c are found. From equation (2-39) )uDD(fk e

xccec1

ABe1

c−−++− µρ=′ and from here,

cAB

e

AB

x

AB

c

µρD

DuDf

DDk

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

′ can be written. By setting e and c arbitrarily equal to 1

and multiplying the last two terms, µ

uDρµρD*

DuD xAB

AB

x =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

finally,

γ

AB

β

x

AB

c

Dρµ

µuDρα

DDk

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

′ (2-41)


Table 2-2. Mass transfer correlations for widely used geometries

No. Geometry Equation Application range Charact. dimension

Type of

fluid 1

Inside pipe flow

Sh= 0.023 Re0.83 Sc0.33

Sh= 0.0149 Re0.88 Sc0.33

4 000< Re< 60 000 0.6 <Sc <3 000 10 000< Re<4*105

Sc > 100

Pipe diameter

Fluid

2 Flow parallel to a flat plate

JD = 0.664 5.0xRe−

Sh=0.03725.0

iScSc43.0Sc8.0

xRe ⎟⎟⎠

⎞⎜⎜⎝

⎛

Sh= 0.027 25.0

iScSc43.0ScxRe ⎟⎟

⎠

⎞⎜⎜⎝

⎛

Re < 50 000 5*105<Rex <3*107

0.7<Sc <380 2*104<Rex< 5*105

0.7<Sc <380

Length of plate

Fluid

3 Gas flow Parallel to a flat plate in confined duct

JD = 0.11 Re-0.29

2 600<Re <22 000

Length of plate

Gas

4 Wetted-wall column

Sh = 3.41

Sh= 5.0

ScReh2

3⎟⎠⎞

⎜⎝⎛

πδ

Sh= 1.76*10-5 Re1.506 Sc0.5

Re= 1004<

µΓ

100<Re 1200< 1 300< Re <8 300

Thickness of liquid film

Liquid

5 Flow Perpendicular To cylinder

44.060.0 ScRe281.0Sh =

Sh= (0.35+0.34 Re0.5 + 0.15 Re0.58) Sc0.3

400 <Re< 25 000 0.6<Sc<2.6 0.1<Re< 105

0.7<Sc< 1 500

Diameter of cylinder

Gas Fluid

6 Flow past single sphere

Sh = Sho + 0.347 (Re.Sc0.5)0.62

Sho= 810 ScGr 244.0Sc333.0)ScGr(0254.02

810 ScGr 25.0)ScGr(569.02

>+

<+

1.8<Re.Sc0.5<6*105

0.6<Sc<3 200

Diameter of sphere

Fluid

7 Packed column (*)

JD= 75.5Re06.2 −ε

JD= 815.0Re4.20 −ε

JD= 3/2Re09.1 −ε

JD = 31.0Re250.0 −ε

90< Re< 4 000 Sc=0.6 5 000< Re < 10 300 Sc=0.6 0.0016 < Re <55 168 < Sc < 70 600 5 < Re < 1 500 168 < Sc < 70 600

Diameter of packing

Gas Gas Liquid Liquid

Grashof number, 23g

Gr ⎟⎠⎞

⎜⎝⎛µ

ρ

ρ

ρ∆=

l33.0D ScRe

ShJ = ; ε : Void fraction ; JD- factor ,

(*) In the Reynolds numbers superficial velocities (velocity based on empty column) are used.


is obtained. As the three dimensionless groups are named as Sherwood (Sh), Reynolds (Re) and Schmidt (Sc) numbers respectively, equation (2-41) can also be written as; Sh = (2-42) γβ ScReαwhere the constants α, β and γ must be determined experimentally. This is accomplished by making several experiments with different A and B under different conditions in pipes. So obtained equation is now a general equation and it can be used to compute the mass transfer coefficient of the fluid, when mass transfer takes place from (to) the wall of a pipe to (from) a fluid flowing inside the pipe. For different geometries, similar but not the same equations can be derived. Extensive lists of mass transfer correlations thus obtained, are given in the Tables.5.21-5.28 (pages: 5-59 to 5-77) in Perry’s Handbook. Some of them are shown in Table 2.2 below. In using these empirical correlations, one must be very careful with the four points given below: 1o) Check whether the equation to be used is developed for the geometry you are going to use. 2o) Find out what characteristic dimensions are used in Re and Sh numbers of the equation you are going to use and use the same characteristic dimensions in your calculations. 3o) Be careful with the application ranges of Re and Sc numbers. And never use any equation outside the application ranges. 4o) Check for what type of fluid the equation is valid, as some equations are valid for only liquids or for only gases. 2.6 Mass Transfer Theories : In the design of process equipment generally mass transfer coefficients obtained from empirical correlations are used. However some theoretical methods have been developed to give an explanation of mass transfer coefficients. These are known as mass transfer theories. Only two of them are briefly presented below. 2.6.1 Film theory : The film theory which is the oldest of the mass transfer theories, was first proposed by Whitman in 1923. According to this theory, “There is a thin fictitious laminar film adjacent to the phase boundary in which all the resistance to mass transfer is concentrated. Rest of the fluid is in turbulent motion and does not have any resistance to the transfer. Mass transfer takes place across this film by molecular diffusion under steady-state condition unidirectionally. As the capacity of the film is very small, concentration profile in the film is established instantaneously. Due to low solute concentration in the film, zero net-flux approximation can be made”. It is obvious that thickness of the fictitious film should be greater than the thickness of the laminar sub-layer. With these assumptions, general equation for mass transfer (eqn.1-59) simplifies to the equation below:

0dz

cdD

2A

2

AB = (2-43)

The boundary conditions required for the solution are obtained from the statement of the theory as : B.C. 1) at z = 0 cA = cAi 2) at z = zL cA = cAL where, zL is the thickness of the fictitious film, cAi and cAL are the interface and bulk concentrations which are both constant.


Solution of equation (2-43) with the stated boundary conditions gives the concentration profile as :

zz

cccc

L

ALAiAiA ⎥

⎦

⎤⎢⎣

⎡ −−= (2-44)

Since this is a straight line equation, change of concentration of solute in the film is linear. Then, total molar flux of solute A crossing the interface becomes;

)cc(z

Dz

ccD

dzdc

DN ALAi

L

AB

L

ALAiAB

0z

AABA −=⎥

⎦

⎤⎢⎣

⎡ −−−=⎟

⎠⎞

⎜⎝⎛−=

=

(2-45)

As mass transfer coefficient is defined by

)cc(

NforcedrivingMolarfluxmolarTotalk

ALAi

AL −

== (2-46)

by substituting NA from equations (2-45) into equation (2-46),

L

ABL z

Dk = (2-47)

is obtained. So, according to the film theory, mass transfer coefficient is proportional to .But most of the empirical mass transfer correlations shows that k

0.1ABD

L is proportional to . Later on, some attempts have been made to modify the original film theory so that

is obtained. It follows from this that original film theory oversimplifies the mass transfer mechanism across a phase boundary. However because of its simplicity, original form of the film theory is still used in complex cases such as mass transfer accompanied by a chemical reaction.

3/2ABD

3/2ABL Dk α

2.6.2 Penetration theory : This theory was first proposed by Higbie and modified later by Danckwerts. The statement of the model is as follows: “Hydrodynamic conditions exist which allow for unidirectional mass transfer to take place by unsteady molecular diffusion. A constant concentration is instantaneously established at the interface and significant depth of penetration of solute is smaller than the depth of undisturbed fluid. Solute concentration is low so that zero net flux conditions are approximated.” With these assumptions, general equation for mass transfer (eqn.1-59) simplifies to the equation below:

2A

2

ABA

zc

Dc

∂∂

=∂∂θ

(2-48)

The boundary conditions required for the solution are obtained from the statement of the theory as follows:


B.C. 1) at θ = 0 for 0 < z < ∞ cA = cAo 2) at θ > 0 for z = 0 cA = cAi 3) at θ ≥ 0 for z = ∞ cA = cAo where, cAi and cAo are the solute concentrations at the interface and at the undisturbed fluid. Solution of equation(2-48) can be done by using Laplace transformation technique. The result is:

θABAoAi

AoA

D2zerf1

cccc

−=−−

(2-49)

Then, instantaneous total molar flux of solute A crossing the interface becomes;

)cc(D

dzdc

D)(N AoAiAB

0z

AABA −=⎟

⎠⎞

⎜⎝⎛−=

= πθθ (2-50)

According to regular surface renewal model (Higbie model) average total molar flux;

)cc(D

2N AoAiAB

A −=πτ

(2-51)

and then average mass transfer coefficient,

πτ

ABL

D2k = (2-52)

are obtained. According to random surface renewal model (Danckwerts model) average total molar flux; )cc(sDN AoAiABA −= (2-53) and then average mass transfer coefficient, sDk ABL = (2-54) where τ is the time of penetration of the solute(s), s is the fractional surface renewal rate(sec-1 ). As it is seen, the penetration theory predicts that mass transfer coefficient is proportional to . This is valid in some mass transfer systems, such as mass transfer from gas to liquid in packed column where liquid flows over packing and contact surface is renewed constantly in short periods.

5.0ABD

Other mass transfer theories-some combining film and penetration theories- have been developed which predict a gradual change of the exponent of DAB from 0 to 1.0 depending upon degree of turbulence. 2.7 Determination of Effective Concentration Difference for Calculation of Average Flux: The flux equations given above are used to compute the local fluxes in the equipment, as the mass transfer coefficients and the concentration differences are local values. Since the mass transfer coefficients and the concentration differences change along the equipment, what concentration difference should be taken to


compute the average flux for the equipment? To reply this, let us consider mass transfer from a horizontal solid plate of length l, and width b, as shown in Fig.2.3 to a

1 Ac 2A c

cAi

Liquid B

dSmSm

Ac

x

x= 0 x+dx x=l

A xu

A

cAi b

Dissolving solid

x

z

Q

Fig.2.3 Determination of effective concentration difference liquid which flows over the plate with an average velocity of xu (m/s), and a volumetric flow rate of Q (m3/s). Let us show with A2A1 candc (k-molA/m3) the average concentrations of solute A in the inlet and outlet solutions. As it was stated before, the concentration of solute A in the first liquid layer on the solid plate reaches its solubility value, cAi at the prevailing temperature almost instantaneously and remains at this value throughout the mass transfer. The average concentration of solute A in the liquid Ac , increases with increasing x. The concentration difference, which causes mass transfer in z-direction also changes with x, is being )cc( 1AAi − at the inlet, )cc( AAi − at any x and )cc( 2AAi − at the outlet. So, which one of these should be used to compute the average mass transfer flux? Let us consider the liquid on the differential area of dS, and at steady-state write the mass balance for solute A; The molar rate of dissolution of A from the area dSm = Molar rate of transfer of solute A in z-direction normal to the surface dSm. Hence, mAAiLmAA dS)c(ckdSNcdQ −== (2-55) is written, where kL is local mass transfer coefficient and Acd is the increase in the concentration of liquid as liquid flows a distance of dx. Separating the variables and integrating gives:

A1Ai

A2Ai

L

c

cAAiLo

mm ccccln

kQ

cccd

kQdSS

A2

A1

−−

−=⎮⌡⌠

−== ∫

l

(2-56)

where Lk is the average mass transfer coefficient for the liquid for the whole plate. On the other hand, if the same is repeated for the whole plate, The molar rate of dissolution of A from the area Sm = Molar rate of transfer of solute A in z-direction normal to the surface Sm. By defining an average concentration difference with ( av.AAi )cc − for the whole plate,


av.AAiLA1A2m

A )c(ck)cc(SQN −=−= (2-57)

is written. The value of so-defined ( av.AAi )cc − can easily be found by eliminating

Lk between equations (2-56) and (2-57).

A1Ai

A2Aiav.AAi

mA1A2

m ccccln)c(c

SQ)cc(

SQ

−−

−−=−

From this,

)c(c)c(c

ln

)c(c)c(c)c(c

A2Ai

A1Ai

A2AiA1Aiav.AAi

−−

−−−=− (2-58)

is finally obtained. As it is seen from the equation above, the average concentration difference(driving force for mass transfer) to be used to compute the average molar flux is the logarithmic mean of the concentration differences obtained at the inlet and outlet of the equipment. Hence,

lnAAiav.AAi )c(c)c(c −=− . So, for the average flux, the following equation can be written; lnAAiLA )cc(kN −= (2-59) The average mass transfer coefficient Lk to be used in the equation is the arithmetic mean of the mass transfer coefficients computed at the inlet and outlet conditions of the equipment.

Example-2.2) Calculation of Gas Phase Mass Transfer Coefficient As water flows down as a thin film on the inside surface of a vertical pipe of 100 mm diameter, dry air flows upward at 30 oC and 1 bar pressure. a) Calculate the mass transfer coefficient for the gas phase for an air flow rate of 30 m3/h. b) For 60 % relative humidity of exit air, what must be the length of the pipe? Density and viscosity of air at 30 oC and 1 bar are 0.019 cP and 1.14 kg/m3, and molecular diffusivity of water vapor in air at 0 oC and 1 bar is 2.20*10-5 m2/s.Vapor pressure of water at 30 oC is 0.0424 bar.

Solution :

Water evaporates into air and transfers in z-direction. x2

water (A)

A)

air(B)A)

A l

1

zpipe Average air velocity in x-direction

s/m06.1

)10*100)(4/()3600/30(

AQu

23c

Bx =

π==

−

3606

)10*1)(019.0()10*100)(06.1)(14.1(DuRe

3

3x ==µ

ρ=

−

−

s/m10*64.2

)027330273)10*20.2(

TT)D()D( 25

75.1

5

75.1

1

2TABTAB 12

−− =⎟⎟⎠

⎞⎜⎜⎝

⎛++

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

631.0

)10*64.2)(14.1()10*1)(019.0(

DSc

5

3

AB

==ρµ

=−

−

(Note that in the calculations of Re & Sc numbers density and viscosity of air are taken.)

From Table.2-2 Sh = 0.023 Re0.83 Sc0.33


31.28

PDD)p(RTk

DDkSh

AB

lnBG

AB

c ==′

=Sh= 0.023 (6 360)0.83 (0.631)0.33 = 28.31

barsm/Amolk10*0.3

)30273)(083.0)(10*100()10*64.2)(31.28(

RTD)D)(31.28(

P)p(

k 243

5ABlnB

G −=+

== −−

−

(For dilute gas mixture (pB)ln / P≈1) (b) Relative Humidty of air is defined as : (Partial pressure of water vapor in air/vapor pressure of water at the prevailing temperature)

H A= pA2 / p oA The air leaving section-2 will be at 60 % realative humidity. Then, pA2 = (60/100)(0.0424) = 0.02544 bar Total molar flux of water vapor in z-direction for the whole pipe;

lnAoAGlnAAiGA )pp(k)pp(kN −=−=

Driving force for mass transfer at section-1 : bar0424.0p)pp( oA1A

oA ==−

Driving force for mass transfer at section-2 : bar01696.002544.00424.0)pp( 2AoA =−=−

Average driving force for the pipe: bar0278.0

01696.00424.0ln

02544.0

pppln

)pp(p)pp(

2AoA

oA

2AoA

oA

lnAoA ==

−

−−=−

Assume that gas is dilute in water vapor (to be checked later). Then;

barsm/Amolk10*0.3kP

)p(k 24G

lnBG −== −

NA= (3.0*10-4)(0.0278) = 8.34*10-6 k-mol A /m2 s Rate of evaporation of water into air = Rate of transfer of water vapor into air

)002544.0(

)30273)(083.0()3600/30()pp(

RTQ

)cc(QN 1A2AG

1A2AGA −+

=−=−= = 8.43*10-6 k-mol A /s

(For dilute gas mixture: QG = QB ). Mass transfer area and the length of the pipe needed under these conditions:

.m22.3

)10*100)((01.1

DS

3m =

π=

π=

−l2

6

6

A

Am m01.1

10*34.810*43.8

NNS ===

−

−

Check whether dilute gas mixture assumption is correct or not ! At section-1: bar101pPpandbar9576.00424.01pPp 1AB

oABi =−=−==−=−=

bar979.0

)9576.0/1ln(9576.01

)p/pln(pp)p(

BiB

BiBlnB =

−=

−=

Then; (kG)1 = (3.0*10-4)(1/0.979) = 3.06*10-4 k-mol A /m2s bar At section-2 : bar974.0024544.01pPpbar9576.00424.01pPp 2AB

oABi =−=−==−=−=

bar966.0

)9576.0/974.0ln(9576.0974.0

)p/pln(pp)p(

BiB

BiBlnB =

−=

−=

Then; (kG)2 = (3.0*10-4)(1/ 0.966) = 3.1*10-4 k-mol A /m2s bar Average mass transfer coefficient for the pipe:


barsm/molk10*08.3

210*)10.306.3(k 24

4

G −=+

= −−

Average total molar flux for the pipe :

sm/molAk10*56.8)0278.0)(10*08.3()pp(kN 264lnA

oAGA −==−= −−

This gives l = 3.14 m. As it is seen, this is very close to 3.22 m (only 2.5 % excess), hence dilute gas mixture assumption is valid.

Example-2.3) Calculation of Liquid Phase Mass Transfer Coefficient

100 mm diameter spherical solid was covered with benzoic acid. Water at 35 oC flows past this solid with an average velocity of 0.12 m/s.

Water(B)

Benzoic acid (A) Calculate:

a) Liquid phase mass transfer coefficient, kc b) Amount of benzoic asid dissolved in 1 hour as gr. Viscosity and density of water at 35 oC are 7.2*10-4 kg/ms and 995 kg/m3. Solubility and molecular diffusivity of benzoic acid in water at 35 oC are 0.31 gr B.Ac/100 ml water and 1.25*10-9 m2 /s.

Solution: Since the solubility of benzoic acid in water is very small, dilute solution assumption can easily be done.

58316

)10*2.7()10*100)(12.0)(995(du

Re4

3px ==

µρ

=−

−

9.578

)10*25.1)(995()10*2.7(

DSc

9

4

AB

==ρµ

=−

−

Then, from Table.2.2. Sh = Sho + 0.347 (Re.Sc0.5 )0.62 For Sho Grashof number must be evaluated.

As ∆ρ = 0 then Gr =0 Hence from Table.2-2 Sho = 2 23

pdgGr ⎟⎟

⎠

⎞⎜⎜⎝

⎛µρ

ρρ∆

=

Re Sc0.5 = (16 583)(578.9)0.5 = 398 990 Sh = 2 + 0.347 (398 990)0.62 = 1 032.2

2.0321

Ddk

Ddk

ShAB

pc

AB

pc ==′

=

s/m10*29.1

)10*100()10*25.1)(2.0321(

dD)2.0321(k 5

3

9

p

ABc

−

−

−

===

3

33Ai m/Amolk0254.0m

Amolk)122/1.3(m1

Akg1.3Bml100Agr31.0.c)b −=

−===

sm/molAk10*28.3)00254.0)(10*29.1()cc(kN 275AAicA −=−=−= −−

2232pm m0314.0)10*100)((dS =π=π= −

h/Ac.Bgr52.4)1000)(3600)(122)(0314.0)(10*28.3()1000)(3600(MSNm 7

AmAA === −&


Problems 2.1 Gas phase mass transfer coefficient for the transfer of NH3(A) through non-transferring N2(B) at 760 mmHg and 20 oC has been given as kc=10.4 k-mol/m2h(k-mol/m3). The mixture is dilute for NH3. Convert this coefficient to ky (k-mol/m2h), kG [k-mol/m2h(mmHg)], kY [k-mol/m2h(k-molA/k-molB)]. 2.2 Ethanol flows through a pipe of 50 mm inside diameter, with a velocity of 1.0 m/s at 20 oC. The inside surface of the pipe was covered with naphthalene, whose solubility in ethanol at 20 oC is 9.5 gr naphthalene per 100 ml of solution. a) Calculate the liquid phase mass transfer coefficient, kL. b) What must be the length of the pipe, for an exit ethanol solution containing 1.2 gr naphthalene per liter? Inlet ethanol does not contain naphthalene. Density and viscosity of ethanol at 20 oC are 789 kg/m3 and 2*10-3 kg/ms, and molecular diffusivity of naphthalene in ethanol at 20oC is 0.8*10-9 m2/s. [ Ans. b) ℓ = 7.03 m ] 2.3 In a channel which is 200 mm in diameter and 5 m in length, a gas flows at 2 bars and 20 oC with a velocity of 2 m/s. To this gas solute A transfers from the wall of the channel. Calculate the mass transfer coefficient k'

G as k-mol/m2s bar. Viscosity and density of the gas and the molecular diffusivity at the operating conditions are 0.68*10-3 Ns/m2, 1.51 kg/m3 and 1.25*10-4 m2/s . In the literature, the following mass transfer correlations are given for the channel geometry: i) Sh = 0.041 Re0.60 Sc1/3 Valid for 10 < Re < 25 000; for only liquids (Characteristic dimension is length) ii) JD= 0.068 Re-0.35 Valid for 10 < Re < 10 000; for fluids (Characteristic dimension is diameter) iii) Sh= 0.01 Re-0.44 Sc1/3 Valid for 1 000 < Re < 10 000; for only gases (Characteristic dimension is diameter) [Ans. k'

G = 2.21*10-4 k-mol/m2 s bar ]

2.4 Pure water at 26°C is flowing at a rate of 2*10-3 m3/s through a packed bed of benzoic acid spheres having a mass transfer area of 0.01 m2. The concentration of benzoic acid in the outlet liquid is measured as 3*10-3 k-mol/m3. Calculate:

a) The molar rate of dissolution of benzoic acid, b) The average mass transfer coefficient Lk The solubility of benzoic acid in water at 26 oC is 0.03 k-mol Ben.Ac./m3 of solution. [Ans. b) s/m021.0k L = ] 2.5 Show that in a column filled (packed) with solid spheres of dp diameter, the area per unit volume of packed bed, ap is given as;

p

p d)1(6a ε−

=

where, ε is void fraction of the bed defined as volume of the voids divided by the total volume. 2.6 In order to find the liquid phase mass transfer coefficient, Lk experimentally, water is made to flow at an average velocity of 0.1 m/s through a pipe which is covered with benzoic acid. Diameter and length of the pipe are 100 mm and 4 m. Concentration of benzoic acid in the outlet water is measured as 0.0106 k-mol/m3. At the operating temperature, viscosity and density of water are 0.9*10-3 kg/ms and 990 kg/m3. The solubility and the molecular diffusivity of benzoic acid in water at the given temperature are 0.0205 k-mol/m3 and 7.2*10-7 m2/s. a) Calculate Lk at the experimental conditions,


b) Find the same Lk from the appropriate correlation in Table.2.2 and compare the two values. [Ans. a) 4.55*10-4 m/s ] 2.7 Pure gas B at 2 bars and 20oC is flowing over a solid surface wetted with water (A). The mass transfer coefficient for the system is given as 0.006 k-mol/m2s. If the area of solid surface is 0.1 m2, what is the rate of vaporisation of water as kg/h ?. Vapor pressure of water at 20 oC is 0.023 bar. [Ans. 0.447 kg water / h] 2.8 From experiments conducted at atmospheric conditions in a wetted-wall column using an air velocity of 2 m/s and a peripheral water rate of 1m3/h m, the following values were obtained for the mass transfer coefficients: kG = 2.5 k-mol/m2h bar for the absorption of ammonia in water kL = 0.4 m/h for the absorption of CO2 in water What value of the over-all mass transfer coefficient KG would you expect for the absorption of SO2 in water at atmospheric pressure under the same flow conditions? The equilibrium data for the absorption of SO2 in water, over the operating range employed, are represented by the equation y = 10x, where y and x are weight fractions of SO2 in air and water phases. At the experimental temperatures the molecular diffusivities of NH3 and SO2 in air are 2.32*10-5 and 1.16*10-5 m2/s and of CO2 and SO2 in water 1.5*10-9 and 1.25*10-9 m2/s. Assume that transfer coefficients vary with , and the solutions are dilute. Take the density of water as 1000 kg/m

3/2ABD

3. [Ans.: KG = 0.70 k-mol/h m2 bar]


Chapter-3

MASS TRANSFER BETWEEN TWO PHASES 3.1 Introduction: So far we have seen the mass transfer within a single phase. In most of the mass transfer applications such as in mass transfer operations, however, the two insoluble phases are brought into contact in order to accomplish transfer of one or more components between them. The two phases in contact may be gas-liquid, liquid-liquid, gas-solid or liquid-solid. Important point here is the insolubility of the phases within each other. If one of the phases contains a component, which can dissolve also in the other phase, this component may transfer from the phase to the other. The condition for the transfer is the non-equilibrium of the phases. If the two phases are in equilibrium, transfer is not possible. Concentration gradients exist in the two phases in contact is the indication of mass transfer taking place between the phases. When no concentration gradients exist in the phases, the two phases are said to be in equilibrium. It is obvious from these explanations that equilibrium is very important in the understanding and analysis of mass transfer between two phases. 3.2 Equilibrium between Phases: Equilibrium characteristics of each system are different and peculiar to the system. For example, equilibrium characteristics of a gas-liquid system are quite different than that of a liquid-liquid system. Before investigating the equilibrium concept, let us see the Phase Rule, which was first given by Gibbs and is a very useful tool for finding the independent variables, which can be changed without upsetting the equilibrium. According to this rule, P+F =C+2 is always valid for a system at equilibrium, where P is the number of the phases in the system, C is the total number of the components in the system and F is degree of freedom that is to say, the number of independent variables. Typical independent variables are: Pressure, temperature and concentrations of components in the phases. For example number of independent variables for a system containing two phases and three components, at most two being in each phase, is found as 3 from the equation above. In this case, possible independent variables are: pressure, temperature and the compositions of one component in each phase. If pressure and temperature are fixed, the number of the variables that can be changed without upsetting the equilibrium is left as one. Hence; if the concentration of the component in one phase is specified, the concentration of the same component in the other phase is automatically set. Or, if the temperature and the compositions of one component in two phases are fixed, we cannot say anything about the pressure, as it has been set automatically. Let us return to equilibrium concept and as an example take gas-liquid equilibria. These equilibria are important in gas absorption operations. Suppose we have an ammonia-nitrogen gaseous mixture and bring it in contact with a fixed amount of liquid water in a closed container, which can be kept under constant temperature and pressure. Since the ammonia but not nitrogen is soluble in water, some ammonia molecules will transfer from gas to the liquid crossing the interface. This transfer continues until no more ammonia can dissolve in the liquid. The concentrations within each phase then become uniform by internal diffusion and do not change any more. This condition is known as equilibrium condition. In reality the transfer of ammonia molecules between the phases does not stop (dynamic equilibrium), but this transfer


does not result in net transfer of mass as the number of ammonia molecules going to liquid at any time interval is exactly balanced by the number of the ammonia molecules coming from the liquid phase at the same time interval. By analysing two phases for ammonia, a set of equilibrium concentrations such as x and y (mole fraction of ammonia in liquid and gas phases respectively) are obtained. These values do not

P=cons

t=cons

mole fraction of ammonia in liquid, x

mol

e fr

actio

n of

am

mon

ia in

gas

, y

o o

Fig.3.1 An example for gas-liquid equilibrium

change any more with time. Suppose that we add some more ammonia to the gas phase, as the equilibrium is upset some of the ammonia added to the gas phase transfers to the liquid. This transfer continues until a new equilibrium is attained. Once the equilibrium is reached, transfer of ammonia stops again. When the gas and liquid samples are analysed again for ammonia, higher concentrations in each phase are found. If these additions and analysis are repeated, a complete set of equilibrium relationship is obtained. By plotting x values versus the y values, equilibrium distribution curve for the ammonia as shown in Fig.3.1, is obtained. This curve results irrespective of the amounts of water and gas that we start with and influenced only by the temperature and pressure. It is important to note that at equilibrium the concentrations of ammonia in the two phases are not equal. It may then be asked: “what are equal at the equilibrium”? The answer to this question is: “the chemical potentials of ammonia in each phase”. So, it is the equality of the chemical potentials of the distributed component that stops the mass transfer. If the chemical potentials of the ammonia in liquid and gas are shown with , at equilibrium = . Whenever a component distributes itself between insoluble phases, a dynamic equilibrium is established. Each equilibria are peculiar to particular system. For example replacement of water with toluene or replacement of ammonia with SO

GA

LA µandµ

LAµ G

Aµ

2 will each result in new curves. The temperature and pressure each does affect the equilibrium, resulting with another curves for the same system. Equilibrium relationship for a liquid-liquid system will bear no relation to that for gas-liquid system. 3.3 Mass Transfer between Two Phases: If the two phases in contact are not in equilibrium, mass transfer will certainly take place between the phases. In many engineering applications of mass transfer, the two phases flow at steady-state with


constant velocities in counter directions to each other. As an example, let us consider again a gas-liquid system. Suppose that gaseous mixture consisting of A and C is contacted with a solvent S, that can dissolve only solute A, in a simple equipment such as wetted-wall column, which is shown in Fig.4.2. In this equipment, as liquid flows as a thin film down the inside surface of the column, gaseous mixture flows upward. Gas mixture changes its composition from high to a low solute concentration as it flows upward, while liquid dissolves the solute A and leaves at bottom as solute-rich liquid.

Con

cent

ratio

ns in

liqu

id A+C A+S

Liquid phase

inte

rfac

e

A

yi

y

x

xi

o Distance from interface Distance from interface

Fig.3.2 Concentration profiles in two phases

Gas phase

Con

cent

ratio

ns in

gas

z z

At every section of the column, solute A transfers from gas to liquid. Typical concentration profiles of solute A in each phase at a particular level of the column are shown schematically in Fig.3.2. As it is seen, concentration of solute A in the gas phase drops from its bulk value y to its interface value yi, hence concentration difference or driving force, which causes the transfer of solute A in the gas from bulk to interface is y-yi. In the liquid phase, concentration of the same component decreases from its interface value xi to its bulk value x, hence concentration difference or driving force that causes the transfer of solute A from interface liquid to the bulk liquid is xi-x. As there is mass transfer from gas to liquid, x and y cannot be equilibrium values. It is assumed that there is no resistance to mass transfer (except at some special cases) across the interface and as a result of this yi and xi are equilibrium values related by the system’s equilibrium distribution curve. As they are equilibrium values, which one will be greater, totally depends on the equilibrium relationship of the system. The concentration rise at the interface from yi to xi as shown in Fig.3.2 is not a barrier to mass transfer in the direction from gas to liquid. The determining factor for the direction of mass transfer is neither the relationship between yi and xi nor the difference between y and x, but is the difference between the bulk and interface concentrations in a phase, hence, if y-yi or xi-x is positive, solute A transfers from gas to liquid as shown in the Figure. On the other hand, if these differences are negative, the transfer direction is from liquid to gas. There is no direct effect of y-x difference on the transfer direction. 3.4 Mass Transfer Flux: The most important point in a mass transfer between the phases is to compute the mass transfer fluxes of the transferring components. At


steady-state, since the mass transfer flux of solute A from bulk gas to the interface gas equals the flux of the same component in the liquid from interface liquid to the bulk liquid, the following equations can be written for solute A:

)xx(k

)yy(k

N iix

xi

iy

yA −

β′

=−β

′= (3-1)

where, are the individual mass transfer coefficients for gas and liquid phases and are the bulk flow correction terms for the same phases, which can be given by the following equations:

xy kandk ′′

ixiy βandβ

)yN/()yNln(N

)yN()yN(

iRRR

iRRiy −−

−−−=β and

)xN/()xNln(N)xN()xN(

iRRR

iRRix −−

−−−=β (3-2)

Any one of the two equations in (3-1) can be used to compute the total molar flux of solute A. Since y and x are the bulk concentrations, they can easily be determined with sampling from the phases and analysing them. Individual mass transfer coefficients, on the other hand, can be computed from the appropriate mass transfer correlations by inserting the system’s properties and flow parameters. Bulk flow contribution terms can be found from the equations given in (3-2). But, since it is almost impossible to take samples right through the interface fluids, yi and xi can not be determined experimentally, but they can be computed from simultaneous solution of system’s equilibrium relationship and equations given in (3-1). Since, in the majority of the cases, equilibrium relationship is given in graphical form, the solution is done on the graph. If equation (3-1) is rearranged,

i

i

iyy

ixxxxyy

/k/k

−−

=β′β′

− (3-3)

is obtained. As it is seen, this equation represents a straight line on xy-diagram,

passing through points P(x;y) and M(xi;yi) with a slope of iyy

ixx/k/kβ′β′

− . Since xi and yi

Equilibrium distribution curve

slope=iyy

ixx

/k/kβ′β′

−

y*

y

yi

Mole fraction of solute A in liquid, x

o

E

M

P

R

x*xix

slope= m'

slope= m''

Fig.3.3 Determination of interface concentrations

Mol

e fr

actio

n of

solu

te A

in g

as, y


values are equilibrium values, point M must also lie on the equilibrium distribution curve. As point P and slope are known, by drawing the line starting at P with known slope, point M, hence xi and yi are found. Now, flux can be computed by using any one of the equations in (3-1). In reality, drawing of PM line is not so easy, as the slope contains the sought for xi and yi. Nevertheless, if transfer of solute A in two phases takes place in dilute solutions or under the condition of equimolar counter transfer , then and slope of PM line is simply:

1ixiy =β=β

yx k/k ′′− and hence drawing of PM line is straightforward. But, if solute A transfers through non-transferring C in the gas and through non-transferring S in the liquid as in the example above, and in addition if the solutions are not dilute for solute A, the bulk flow correction terms and hence the slope of PM line

becomes, and ilniln x)(1andy)(1 −−lniy

lnix)y1/(k)x1/(k

−′−′

− , where (1-y)iln and (1-x)iln

are given as:

)xx)/(1ln(1)x(1x)(1

x)(1and)yy)/(1ln(1)y(1y)(1

y)(1i

iiln

i

iiln −−

−−−=−

−−−−−

=− (3-4)

In this case, drawing of PM line can only be done by trial and error. In other cases, first NR is calculated from the relationship between the fluxes, then with the help of equations (3-2) and (3-3) and by trial and error xi and yi are found. 3.5 Overall Mass Transfer Coefficients and Overall Driving Forces: To compute the molar flux of transferring solute A; yi or xi must be known, which are in many cases found by trial and error. In addition to this, mass transfer coefficients are obtained experimentally. At some experimental conditions, the measurements of

, the individual mass transfer coefficients, are very difficult, as the complete elimination of the resistance to mass transfer in one phase is almost impossible and hence overall mass transfer coefficients based on gas or liquid phase

are measured. Computing total molar flux by using these overall coefficients requires the definitions of overall concentration differences or driving forces for the system. It is obvious that (y-x) or (x-y) cannot be taken as overall driving forces, as the real driving force for mass transfer is the difference between the chemical potentials of the transferring component in two phases, not the difference between its concentrations in two phases. It is very well known from thermodynamics that, the relationship between the concentration of solute A in the gas y and its chemical potential is quite different from the relationship of x and its chemical potential Hence, if concentration differences are to be used with overall mass transfer coefficients, the measure of x in terms of gas phase unit must be subtracted from y, not the x itself. If this is shown by , the overall driving force in terms of gas phase concentrations for the system then becomes ( ), where is the equilibrium value of x and hence, point R(x, ) is a point on the equilibrium distribution curve. On the other hand, if the overall driving force is to be expressed in terms of liquid phase concentrations, this should be written as (

xy kandk ′′

xy KandK ′′

Gµ.µ L

∗y∗− yy ∗y

∗y

xx −∗ ), where is the measure of y in terms of liquid phase units and this is equilibrium value of y and hence, point

∗x


E( ;y) lies on the equilibrium distribution curve of the system. In the light of these explanations, for the total molar flux of solute A, which transfers from gas phase to the liquid, in addition to the equations (3-1),

∗x

)xx(K

)yy(K

Nx

x

y

yA −

β=−

β

′= ∗

∗

∗∗

∗ (3-5)

can be written, where xy KandK ′′ are overall mass transfer coefficients based on

gas and liquid phase respectively, are bulk flow correction terms for gas and liquid phases , which are defined by;

∗∗ ββ xy ve

)xN/()xNln(N)xN()xN(and

)yN/()yNln(N)yN()yN(

RRR

RRx

RRR

RRy ∗

∗∗

∗

∗∗

−−−−−

=β−−

−−−=β (3-6)

For dilute solutions and for “equimolar counter transfer” conditions and hence equation (3-5) simplifies to:

1xy =β=β ∗∗

(3-7) )xx(K)yy(KN xyA −′=−′= ∗∗

If solute A transfers in both phases through non-transferring components as in the example above: and hence equation (3-5) is written as:

∗∗∗∗ −=−= lnxlny x)(1βandy)(1β

(3-8) )xx(K)yy(KN xyA −=−= ∗∗

where, . and are defined by:

∗∗ −′=−′= lnxxlnyy x)/(1KKandy)/(1KK ∗− ln)y1( ∗− ln)x1(

)xx)/(1ln(1)x(1x)(1

x)(1and)yy)/(1ln(1)y(1y)(1

y)(1 lnln ∗

∗∗

∗

∗∗

−−−−−

=−−−−−−

=− (3-9)

3.6 The Relationships between Individual and Overall Mass Transfer Coefficients: If the individual mass transfer coefficients are computed from mass transfer correlations, overall mass transfer coefficients can easily be synthesized from them. To derive the relationship between them, first write (y-y*) = (y-yi) + (yi-y*) = (y-yi) + m′ (xİ-x) from Fig.3.3, where m′ is the slope of RM line. Then, substitute the driving forces from equations (3-1) and (3-5) and make the necessary simplifications. The result is:

ixxiyyyy /k

m/k1

/K1

β′′

+β′

=β′ ∗

(3-10)

Similarly, from the same figure, (x*-x) =(x*-xi) + (xi-x) =m"(x*-x) + (xi-x) can be written. Substituting the driving forces from equations (3-1) and (3-5) gives;

ixxiyyxx /k

1/km

1/K1

β′+

β′′′=

β′ ∗ (3-11)

where, is the slope of EM line. If the equilibrium distribution curve is a straight line then, = m.

m ′′mm ′′=′

The term on the left hand side of equation (3-10) gives the total resistance to mass transfer in the system. On the other hand, the first term on the right hand side is the resistance to mass transfer in the gas phase only, and the second term is the resistance to mass transfer in the liquid phase only. Hence, total resistance to mass transfer in the


system is the sum of the resistances lying in each phase. The same can be said by looking at the equation (3-11). It follows from this that multiplication of the ratio of the first terms on the right hand sides to the terms on the left hand sides of the equations (3-10) and (3-11) with 100 gives the percent of resistance lying in the gas phase and subtraction of this from 100 is the percent of resistance in the liquid. This, which is known as resistance analysis, is very useful in deciding at what phase changes must be made in order to increase the rate of mass transfer occurring between the phases. It is obvious that, changes made in the right direction in the phase, which has the greater resistance will increase the rate of mass transfer significantly. In many cases, the magnitudes of individual mass transfer coefficients are almost the same. In these cases then the value of the slope of equilibrium distribution curve determines the values of resistances in each phase. If the slope m′ is very great (this is so when the solubility of A in S is very small) the first term on the right hand side of equation (3-10) can be neglected. Thus,

ixxyy /k

m/K1

β′′

≅β′ ∗

(3-12)

can be written. It follows from this that total resistance to mass transfer in the system is almost in the liquid phase. This special case is known as “mass transfer under the control of liquid phase”. It is obvious then that in order to increase the rate of mass transfer, measures must be taken in the liquid phase not in the gas phase. On the contrary to this, if the slope m′ is very small (this is so when the solubility of A in S is very great) the second term on the right hand side of equation (3-10) can be neglected. Thus,

iyyyy /k

1/K1

β′≅

β′ ∗ (3-13)

can be written. It follows from this equation that total resistance to mass transfer lays almost in the gas phase. This special case is known as “mass transfer under the control of gas phase”. Then, in order to speed up the mass transfer, measures must be taken in the gas phase, not in the liquid phase.

Example-3.1) Interphase Mass Transfer

In a distillation operation, one of the components transfers from a binary vapor mixture of A+B into the liquid , during which other component transfers from liquid to vapor. The molar latent heats of vaporization of components are such that λA = 3λB. At one section of the equipment, the mole fractions of component A in liquid and vapor are measured as 0.25 and 0.32 respectively. a) Find the transfer directions of the components, b) Calculate total molar fluxes of components, c) Calculate the percentage resistances in liquid and vapor phases, d) Plot the concentration profiles of components A and B in both phases qualitatively. The individual mass transfer coefficients for liquid and vapor phases were calculated at the operating conditions from the appropriate correlations as;

.sm/molk10*5.1k 24y −=′ −sm/molk10*3k 24

x −=′ −

The equilibrium relationship within the operating range is linear and expressed as y =1.8 x, where y and x are the mole fractions of component A in vapor and liquid respectively.


Solution:

a) Nothing can be said for the transfer directions by looking at x and y values. From NB = - (λA /λB ) NA = - 3NA Then, 5.0

N3NN

NNN

NAA

A

BA

AA −=

−=

+=

Solution is by trial and error.

slope of PM line; ix

iy

ix

iy4

4

ix

iy

y

x

iyy

ixx 2.10*5.1

10*3.kk

/k/km

β

β−=

β

β=

β

β′′

−=β′β′

−= −

−

(1)As first trial, assume βix = βiy =1 then; m = - 2 From eqn.(3-3) and yi=1.8xi

0.32 – 1.8 xi = -2 (0.25 - xi) xi = 0.216 and yi = 1.8 (0.216) = 0.389 Substitute these into eqn.(3-2)

466.1

)216.05.0()25.05.0(ln)5.0(

)216.05.0()25.05.0(ix =

−−−−

−

−−−−−=β 71.1

)389.05.0()32.05.0(ln)5.0(

)389.05.0()32.05.0(iy =

−−−−−

−−−−−=β

Then, new slope of PM line; 33.2466.171.1)2(m −=−=

New values for xi and yi with this m; 0.32- 1.8 xi = - 2.33 (0.25 – xi) xi = 0.219 and yi = 1.8 (0.219) = 0.394 These are sufficiently close to the previous ones and hence trial is stopped. Final values are xi = 0.219 , yi = 0.394 , βix= 1.466 and βiy= 1.71 As (y-yi) = (0.32-0.394) < 0 then , component A transfers from liquid to vapor, and the other component B transfers from vapor to liquid.

sm/Amolk10*49.6)394.032.0(71.110*5.1)yy(

kN)b 26

4

i

iy

yA −−=−=−

β

′= −

−

or

sm/Amolk10*35.6)25.0219.0(466.110*3)xx(kN 26

4

i

ix

xA −−=−=−

β′

= −−

(-) sign shows that component A transfers from liquid to vapor. NB = - 3 NA = (-3)(-6.42*10-6) = 1.93*10-5 k-mol B/m2 s c) m' = m" = 1.8 Hence, from eqn.(3-10)

b) 19620797840011

466.1/)10*3(8.1

71.1/)10*5.1(1

/K1

44yy

=+=+=β′ −−∗

Percentage resistance in liquid : Percentage resistance in vapor:

%5.56100.1962040011100.

/K/1/k/1

yy

iyy ==β′β′

∗%5.43100.19620

7968100./K/1/k/m

yy

ixx ==β′β′′∗


For check, calculate NA by using over-all values; y* = 1.8 x = (1.8)(0.25) = 0.45 x* = y/1.8 = (0.32)/(1.8) = 0.178

sm/molk10*91.8220111K 25

x

x −==β′ −∗sm/molk10*95.4

196201K 25

y

y −==β

′−

∗

From eqn.(3-5)

sm/Amolk10*44.6)45.032.0)(10*95.4()yy(K

N 265

y

yA −−=−=−

β

′= −−∗

∗

sm/Amolk10*42.6)25.0178.0)(10*91.8()xx(KN 265

x

xA −−=−=−

β= −−∗

∗

∗

d) x = 0.25 y = 0.32 xi = 0.219 yi = 0.394 xB = 1- x = 1- 0.25 = 0.75 yB = 1- 0.32 = 0.68 xBi =1- xi =1- 0.219 = 0.781 yBi = 1- 0.394 = 0.606

interface

xB =f(z)

y, yB

0.0 0.0

z z

0.4

x 0.2

xBi xB

B A 0.8

0.6

0.4

yB

yBi

yi

xi 0.2

0.0

VAPOR (A + B) LIQUID (A + B)

y=f(z)

yB =f(z)

x =f(z) x, xB

0.6

0.8

y

Problems

3.1 A saturated binary vapor mixture consisting of A and B, contains 16 mole percent component A. This mixture is contacted with a saturated binary liquid solution consisting of the same components and containing 11 mole percent component A, at the same pressure. Latent heats of vaporization of both components are the same. Individual mass transfer coefficients at the operating conditions are 2*10-3 k-mol/m2s for liquid and 2.2*10-3 k-mol/m2s for the vapor phase. At the operating pressure and concentration range involved equilibrium distribution curve may be represented by y= 5.5 x, where y and x are mole fractions of component A in vapor and liquid phases respectively.


a) Calculate the interface compositions, b) Find the transfer directions of components A and B, c) Calculate percentage resistance to mass transfer in the liquid phase.

3.2 Air containing 4 mole percent of acetylene is contacted with water containing 0.02 mole percent acetylene in an equilibrium unit at 25 oC and 3.039 bar.

a) Determine the transfer direction of acetylene b) Calculate the interface compositions of acetylene in both phases c) Compute the ratio of the resistance in gas phase to the total resistance of both phases. Equilibrium relationship can be given as pA

= 1.01*106 x, where pA is the partial pressure (mm Hg) of

acetylene in air and x is the mole fraction of acetylene in water. Individual mass transfer coefficients are given as kx = 6.0*10-3 kmol/m2s; ky = 3.0*10-4 kmol/m2s. 3.3 An air-SO2 gaseous mixture, containing 3 % SO2 (A) by volume is in contact with water (S) containing 1 % SO2 by mole at 750 mmHg and 20 oC. At the operating conditions individual mass transfer coefficients are calculated from the appropriate correlations as kx = 5*10-3 k-mol/m2s and ky = 2.5*10-3 k-mol/m2s. The equilibrium relationship of the system is given as pA = 30x, where pA is the partial pressure of SO2 in mmHg and x is the mole fraction of SO2 in the liquid. a) Find the transfer direction of SO2. b) Calculate total molar flux of SO2. c) Calculate percentage resistances to mass transfer in liquid and gas phases, and comment on the result. 3.4 In a wetted-wall column an air-H2S (A) gas mixture is flowing up by a water (S) film which is flowing down on the inner surface at 1.50 atm and 30 oC. The gas phase mass transfer coefficient, kc has been predicted as 9.57*10-4 m/s. At a certain plane, the mole fraction of H2S in the liquid at the gas-liquid interface is 2*10-5 and partial pressure of H2S in the bulk gas is 0.05 atm. Calculate the total molar flux of absorption of H2S. Equilibrium relationship is given as pA= 609 x , where pA is the partial pressure of H2S in atm. and x is the mole fraction of H2S in liquid. [ Ans. a) NA = 1.46*10-6 k-mol H2S/m2s ] 3.5 Benzene and toluene are being separated by distillation in a column at 760 mmHg. At a particular point in the column vapor and liquid phases contain 63 and 50 mole percent benzene respectively and local value of the total molar flux of benzene is 0.05 k-mol/h m2. If 85 percent of the total resistance to mass transfer is in the vapor phase, calculate: a) The interfacial compositions, b) The values of local individual mass transfer coefficients. 3.6 In the distillation of a binary solution of A+B in a packed column, the mole fractions of A in liquid and vapor are measured as 0.28 and 0.26 at one section. If the ratio of mass transfer resistance in the liquid to that in the vapor is 1.5, a) What are the values of interface compositions? b) What are the transfer directions of the components? Vapor liquid equilibrium at the operating conditions may be given as y = 3x, where y and x are the mole fractions of component A in vapor and liquid. [Ans. a) xi = 0.164, yi = 0.492 ]


Chapter-4

GAS ABSORPTION 4.1 Introduction: Gas absorption is a mass transfer operation by which a gaseous mixture is separated into its components by contacting it with a suitable liquid solvent, such that one or more components of the gas phase transfer from gas to liquid phase. In this operation, the original phase is a gas mixture and liquid solvent is brought from the outside. This operation depends on the difference in the solubility of gases in liquids. Operation is carried out at low temperatures so that liquid solvent itself does not evaporate into gas phase. Hence, direction of mass transfer is from gas to liquid only. This operation is not a final separation operation, as upon the absorption, a liquid solution is produced, from which solute A can only be recovered by applying another mass transfer operation, such as stripping or distillation. Gas absorption is used either to recover the valuable components or to remove the noxious substances from the gaseous mixtures. For the latter, it has also found wide applications in the environmental engineering in the recent years. Removal of sulphur oxides from stack gases of a power generator burning coal is a good example for this. Because of large amounts of gas to be processed, absorption towers up to 20-meter diameters are in use today. If solute A is transferred from liquid solution to a gas phase, by contacting the solution with an inert gas C, operation is then called as stripping or desorption. As the transfer direction of mass in this case is from liquid to gas, desorption operation is the reverse of gas absorption operation. As it is seen, in the desorption operation, the original phase is liquid and the second phase, which is brought from the outside, is a gas. Desorption operation may be used alone or with the combination of gas absorption. As an example for gas absorption in chemical industry and the combined use of absorption-desorption, let us consider the separation and purification of coke-oven gases. The gas phase leaving a coke-oven is a mixture of benzene, toluene and xylenes vapors with nitrogen, hydrogen and carbon oxides gases. When this gas phase is contacted with a high molecular weight liquid solution of paraffinic hydro-carbons at low temperature, organic vapors benzene, toluene and xylenes are absorbed into the liquid, leaving back the inert gas mixture of nitrogen, hydrogen and carbon oxides. Thus, the separation of organic vapors from inert gases is accomplished. The liquid solution is then heated up and sent to a desorption tower, where it is contacted with superheated steam. The superheated steam there strips off the absorbed organics from the liquid, leaving behind the solvent liquid, which is, upon cooling, re-circulated back to the absorption column. The gas mixture consisting of steam and organics is then passed to a cooler-condenser, where total condensation occurs, giving two liquid phases in the decanter; water phase being at the bottom, because of insolubility of organics in the water. Thus, recovery of organics from coke-oven gas has been realized with the application of absorption and desorption operations. For the recovery of benzene, toluene and xylenes in pure states from the liquid solution, another mass transfer operation called rectification is used. In the gas absorption and desorption operations, the two phases involved are the same: gas and liquid. Mass transfer takes place between these two phases, only the directions


of transfer being different. Hence, for the understanding of these two operations, gas-liquid equilibria must be known very well. 4.2 Gas-Liquid Equilibria: If a pure gas is brought to equilibrium with a liquid that can dissolve this gas, at a constant pressure and temperature, the concentration of the gas in the solution is known as equilibrium solubility or simply solubility of the gas in this liquid. The solubility is generally obtained experimentally. If the pressure is increased, keeping the temperature constant, the solubility of gas increases. By plotting the solubilities against the pressures, the so-called solubility or equilibrium (distribution) curve for this gas-liquid system is produced. This is shown in Fig.4.1 for

gases of A1, A2 and A3 in a certain liquid. The solubility of a gas in a liquid decreases with increasing temperature. This is because of the fact that the dissolving of a gas in a liquid results in the liberation of heat. This effect is shown in the figure for gas A1. The solubilities of different gases in the same liquid at the same pressure and temperature are different. It is obvious from Fig.4.1 that the solubility of gas A3 is greater than the solubilities of gases A1 and A2. The solubilities of many industrially important gases in water were measured experimentally and some are given in Table.App.4.1-4.4.

A1

A3

A1 A2

t1

x2 x1

t2

t1

t1

t2 > t1

x3 o

t: temperature

o

Pres

sure

s of A

1,A2,A

3 gas

es

Mole fractions of A1,A2,A3 gases in liquid

Fig.4.1 The solubilities of gases in liquid

Let us now consider the solubilities of gas mixtures in the liquids. It has been found that, at two special cases, the solubility of a component in the gas mixture is not influenced by the existences of the other components in the gas, provided that partial pressure of the component, instead of total pressure is taken. One of the special cases occurs, when only one component of the gas mixture is soluble in the liquid considered. For example if this special case is valid, then the solubility curve of gas, say A3 in Fig.4.1 is also the solubility curve of this gas, when this gas is mixed with an insoluble (inert) gas. This enormously reduces the number of the experimentation needed for the determination of solubilities of gas mixtures. The second special case is met, when the binary solutions obtained upon the dissolving of each component of a gas mixture in the liquid considered are ideal. For example, if the solutions produced by dissolving each of the A1, A2 and A3 gases in a certain liquid are ideal, the solubility curve of say A1, when it is mixed with A2 and A3 is the same with the one given in Fig.4.1. This applies of course to the other gases, A2 and A3. But, on the other


hand if the binary solutions produced are not ideal, the solubility curves obtained with pure gases cannot be used for the gas mixtures. New experiments are needed for each composition of the mixture. 4.2.1 Ideal Solutions: It was stated that gas-liquid equilibria are found experimentally. There is an exception to this. This is the ideal solutions case. Because, as first stated by Raoult, in an ideal solution, the partial pressure of solute A in the gas, pA is simply equals the product of its vapor pressure, at the prevailing temperature and its mole fraction in the liquid solution, x. Thus Raoult’s law is written as:

oAp

(4-1) xpp oAA =

As it is seen, any term showing the effect of solvent S does not appear in the equation. But it mustn’t be forgotten that S should be such a solvent that solution of A in S will be ideal. So, it follows from equation (4-1) that solubility of a solute in ideal solution in any solvent is always the same in terms of mole fractions. The vapor pressure of each component is only the function of temperature and experimentally determined relationship is given either in graphical form or by Antoine type equation. Since is only function of the temperature, the equation (4-1) is represented by a straight line passing through the origin at a constant temperature. Hence, the solubility curve for an ideal system is a straight line passing through the origin, which requires only one point to draw. For a solution to be considered ideal, the following four points must be fulfilled at the same time: 1

oAp

o) Attraction and repulsion forces between S-S and A-S molecules must be the same. 2o) The volume of the liquid solution must change linearly with the composition of the solution. 3o) Dissolving of solute A in solvent S must neither release nor absorb heat. Latent heat of condensation associated with the condensing of vapor is excluded from this. 4o) The total pressure of gas must change linearly with gas composition expressed as mole fraction. In reality there is no solution fulfilling all these four conditions at the same time. The closest approach is seen at the solutions of optic isomers, where the size, the structure and the chemical nature of A and S molecules are the same. In practice, many solutions can be considered ideal from engineering point of view, when the deviations from ideality are not so significant. Especially, the solutions of adjacent or near adjacent members of a homologous series of organic chemicals are considered so.

Example-4.1) Ideal Solution Solutions of propane and butane in paraffin oil (M.W.=300) are both ideal. A gaseous mixture consisting of propane-and butane is brought to equilibrium in paraffin oil at 2 bars and 10 oC, at which volume percents of propane and butane in the gas are measured as 80 and 20 respectively. Calculate the equilibrium solubilities of propane and butane at the given conditions as mole and mass percent. Antoine constants of propane and butane are as follows:

a b c

Propane 6.82973 813.200 248.000

Butane 6.83029 945.900 240.000


Antoine equation is , where po (mmHg) and t(oC)

tcbaplog o

+−=

Solution:

vapor pressures of propane and butane at 10oC mmHg03.4762po

1A =678.3

102482.81382973.6plog o

1A =+

−=Propane(1) :

047.3

102409.94583029.6plog o

2A =+

−=mmHg5.1113po

2A =Butane(2) : In the gas mixtures : volume percent = mole percent, then y1=80/100=0.8 and y2=20/100=0.2 Raoult’s law : ; Dalton’s law : xpp o

AA = yPpA = , combining these two : oAp

yPx =

Propane mole fraction in the liquid: 252.0013.1*)760/03.7624(

)2)(8.0(p

Pyxo

1A

11 ===

Butane mole fraction in the liquid: 270.0

013.1*)760/5.1131()2)(2.0(

pPyx

o2A

22 ===

Mole fraction of paraffin oil in the liquid: xB=1-(x1+x2) =1-(0.252+0.270)= 0.478 MA1 = 44, MA2 = 58, MS = 300 In 100 k-mol liquid solution: 25.2 k-mol Propane = (25.2)(44) = 1 108.8 kg Propane 27.0 k-mol Butane = (27.0)(58) = 1 566 kg Butane 47.8 k-mol Paraffin oil = (47.8)(300) = 14 340 kg Paraffin oil Total mass of the liquid solution of 100 k-mol = 17 014.8 kg mole/mole w/w Propane : 25.2 (1 108.8/17 014.8)*100 = 6.52 Butane : 27.0 (1 566/17 014.8)*100 = 9.20 Paraffin oil : 47.8 (14 240/17 014.8)*100 = 84.28

100 100

4.2.2 Real Solutions: The solutions, which are far from fulfilling the conditions given above, are called non-ideal or real solutions. The solubility diagrams of such solutions cannot be represented by straight lines passing through the origin. They are mostly curves. But, at most of them, solubility curve, which is determined experimentally, is almost a straight line at very low solute concentration range and hence at this region, x p AA H ′= (4-2) can be written. This equation is known as Henry’s law equation, as it is first given by Henry. A (bar), which is known as Henry’s law constant, is only function of temperature for a A-S system. Experimentally found Henry’s law constants for many gases dissolved in water at various temperatures are given in Table.4.1. It follows from equation (4-2) that the dimension of Henry’s law constant is pressure. Henry’s law is also written as,

H ′

pA =ΉA cA (4-2b) or as, y = mA x (4-2c)


Here, ΉA(bar m3/k-mol) and mA(-) are also known as Henry’s law constants. Interrelationships between them are; mA= Ή’A/P = ΉA c/P (4-3) where, c (k-mol/m3) is the total molar concentration of liquid solution.

Table 4-1. Values of Henry’s law constants for some gases in water, ( AH ′ * 10-4 bar)

t( oC) Air CO2 CO C2H6 H2 H2S CH4 NO N2 O2 0 4.38 0.0738 3.57 1.28 5.87 0.0272 2.27 1.71 5.36 2.58 10 5.56 0.105 4.48 1.92 6.44 0.0381 3.01 2.21 6.77 3.31 20 6.73 0.144 5.43 2.66 6.92 0.0489 3.81 2.67 8.15 4.06 30 7.81 0.188 6.28 3.47 7.39 0.0617 4.55 3.14 9.36 4.81 40 8.81 0.236 7.05 4.29 7.61 0.0755 5.27 3.57 10.54 5.42 50 9.58 0.287 7.71 5.07 7.75 0.0896 5.85 3.95 11.45 5.96 60 10.23 0.345 8.32 5.72 7.75 0.104 6.34 4.23 12.16 6.37 70 10.64 8.56 6.31 7.71 0.121 6.75 4.44 12.66 6.72 80 10.84 8.56 6.70 7.65 0.137 6.91 4.54 12.76 6.96 90 10.94 8.57 6.96 7.61 0.146 7.01 4.58 12.76 7.08 100 10.84 8.57 7.01 7.55 0.150 7.10 4.60 12.76 7.10 There is a relationship between Henry’s law constant, Ή’A and the heat of absorption, ∆Hs (kJ/k-mol), which is given as; Ή’A

RT/HSeC ∆−= (4-4) where, C is a constant that depends on A-S pair, R(=8.314 kJ/k-mol K) and T(K) are general gas constant and absolute temperature. So, if the values of Henry’s law constants at two different temperatures are known, heat of absorption of the system can easily be computed from equation (4-4).

Example-4.2) Calculation of Heat of Absorption

For the system ammonia-water, which obeys Henry’s law for low concentration of ammonia, the Henry’s law constant has the following values: Temperature(oC) 20 25 30 35 Constant (bar) 0.73 0.96 1.23 1.55 What is the value of heat of absorption (solution) of ammonia in water over this temperature range ?

Solution :

RTH

Cln s∆−=Take ln of both sides of the eqn.(4-4) lnΉ’A

or,

RT303.2H

Clog s∆−= logΉ’A


So, slope of logΉ’A vs. 1 / T line =

R303.2Hs∆

−

t (oC) 20 25 30 35

)K(T1 1− 0.00341 0.00335 0.00330 0.003247

Ή’A (bar) 0.73 0.96 1.23 1.55

logΉ’A -0.137 -0.018 0.090 0.190

8.906110*)243.3345.3(

0.02.0xytgslope 3 −=

−−

−=∆∆

−=α−= −

Then, ∆Hs = - (-1960.8)(2.303)(8.314) = 37 544 kJ / k-mol A absorbed 4.3 Selection of Solvent: As it was explained in the introduction section, in absorption operations solvent S is selected and brought from outside. In many absorption operations, there is large number of potential solvents, (excluding the cases at which main purpose is to produce certain solution such as ammonia solution in water, for which solvent is already fixed), among which, the most suitable one should be selected by the designer. In selecting the most suitable solvent, the following points should be considered: 1o) Solubility of the gas: select the solvent in which the solubility of solute A is the highest, as this will require small quantity of solvent for the operation. In looking at the solubilities, mass units should be used not the mole units, as high molar solubility may mean low mass solubility due to the high molecular weight of the solvent. 2o) Volatility of the solvent: the inert gas leaving the absorber is almost saturated with the solvent vapor at the prevailing temperature. Hence, if the vapor pressure of solvent is high at this temperature, considerable amount of solvent loss will occur. This will not only result in loss of money but may also cause air pollution, as the inert gas is normally let to the atmosphere. So, in order to prevent or

α

∆y

y

∆xx

0.2.

0.1.

0.0.

-0.2.0

-0.1.0

0.0033 0.0034 )K(

T1 1−

logΉ’A H tg θ = -tgα

θ

0.003345 0.003243


reduce these, the solvent, which has the lowest vapor pressure at the operating temperature, should be selected. 3o) Corrosiveness of the solvent: the selected solvent should not be corrosive to the ordinary construction materials, otherwise either the equipment or part of it will be replaced frequently or expensive construction materials, which are resistant to the attack of solvent should be selected. 4o) Viscosity of solvent: viscosity of the liquid is not only important in heat and mass transfer taking place in it, but also important in pumping it. It is seen from mass and heat transfer correlations that high viscosity affects both heat and mass transfer coefficients adversely. Since low transfer coefficients mean low rates of transfer, then the size of equipment should be great. Hence, the viscosity of solvent should be as low as possible. This is also important for the flooding, which will be dealt with later. 5o) Cost of solvent: although the solvent is recovered and re-used in absorption operation, solvent losses are inevitable. So, to keep the initial cost and the cost due to the solvent losses low, the unit purchasing cost of solvent should be low. 6o) Other points: In addition to the points cited above, some other properties of potential solvents should also be considered. These are: toxicity, chemical stability, flammability and freezing point of solvent.

Example-4.3) Selection of Absorption Temperature In the production of maleic anhydride (MAN) from benzene with air oxidation, a new scheme is being considered to recover the MAN from reactor exit gas mixture. For this purpose, the gas mixture containing 3.5 mass percent MAN vapor (remaining may be assumed as air) will be washed with dibutyl phthalate solvent (B.P.=340 oC, MS=278) in an absorber operating at 760 mmHg. Find the solvent losses per ton of MAN recovered at the following 5 absorption temperatures, assuming that all the MAN is recovered and the inert gas(air) leaving the absorber is saturated with solvent vapor. Absorption temperatures are: 25, 50, 100, 150, 200 oC. The vapor pressures of dibutyl phthalate at these temperatures are : 2.5*10-5; 4.8*10-4; 0.044; 1.07 and 11.52 mmHg.

Solution: As the gas entering the absorber contains 3.5 mass percent MAN vapor, the amount of inert gas(air) per ton of MAN is: mc= (96.5)(1 000)/3.5 = 27 571.4 kg or nc= 27 571.4/29 = 950.7 k-mol The amount of dibutyl phthalate vapor (mB) in the absorber exit gas is found as;

PppPpy

Ppy c

oS

cc

oB

S =+==

TcS

c

T

coB

T

S nnnPp

nn

Pp

nn

=+==

From these equations,

,orpP

pnn o

S

oS

cS−

=

oS

oS

oS

oS

oS

oS

cSS p760p6.294264

p760p)7.950)(278(

pPpnMm

−=

−=

−=

is obtained. After calculating mS values at each temperature from the equation above and finding the density of the absorber exit gas from ideal gas law, the following table has been constructed.


As it is seen from the table, at elevated absorption temperatures considerable loss of solvent occurs. This results not only in loss of money but also in air pollution. As the maximum emission of dibutyl phthalate to atmosphere is limited to 5 mg/m3 gas, absorption temperature must be kept below 50 oC.

t(oC) 25 50 100 150 200

)mmHg(poS 2.5*10-5 4.8*10-4 0.044 1.07 11.52

mS (kg dibp/ton MAN) 0.0087 0.167 15.30 372.62 4 067.7

ρG (kg/m3) 1.187 1.10 0.95 0.84 0.84

Dose (mg dibp/m3 gas) 0.375 6.7 527.2 11 252.4 123 928 4.4 Absorption Operations: The absorption operations carried out in chemical industry can be divided in two main groups by looking at the contact type of gas and liquid phases. In one group of operations, the gas and liquid phases are in close contact with each other from the inlet points to the outlet points. Hence, the mass transfer from gas to liquid takes place at every point of the equipment. These types of operations are called as continuous-contact type of operations and the equipment in which these types of operations are carried out are named as continuous-contact type of equipment. Wetted-wall columns, spray columns and packed columns are typical continuous contact type equipment. In other group of operations, gas and liquid phases come in contact with each other at certain compartments of the equipment, which are called stages, and where mass transfer from gas to liquid takes place. After certain time of contact, the phases separate out and gas phase goes to the next stage up, and liquid phase flows to the next stage down through different channels, during which they are not in contact, hence there is no mass transfer between. As these types of operations are called as stage-wise operations, the equipment in which these types of operations are carried out, are named as stage-wise contact type of equipment. Plate (tray) columns are typical stage-wise contact type of equipment. In both type of operations, the two phases flow counter-currently in the equipment, which is normally operated continuously at steady-state. Below, gas absorption operations will be first dealt with in continuous contact type of equipment and later in stage-wise contact type of equipment. 4.4.1 Gas Absorption in Continuous Contact Type of Equipment: As it was explained above, gas absorption operations are carried out in wetted-wall columns, spray columns and packed columns as continuous-contact type of operation. 4.4.1.1 Wetted-Wall Column: This is the simplest equipment used in gas absorption. In a vertical pipe, while liquid solvent flows as a thin film down on inside surface of the pipe, the gas phase is made to flow through the pipe upward, as shown in Fig.4.2.a. Down flowing liquid absorbs solute A from gas at every section of the pipe. Since the mass transfer area is limited with the diameter and the length of the pipe, this equipment has very small mass transfer area per unit volume. For that reason, wetted-wall columns are no longer used in modern plants except the case that when heat of absorption is large and is the main problem in the design of absorber. In these cases wetted-wall columns equipped with cooling jackets are quite useful. By circulating a suitable cooling medium in the jacket, heat of absorption is removed from the system and thus absorption medium is kept at constant temperature. If the heat of absorption is very great, construction of the absorber in the form of shell and tube heat exchanger is


advised, as in this case more heat transfer area (of course also mass transfer area) per unit volume can be stacked. In the design of such wetted-wall absorbers not only the absorption duty but also the heat transfer duty is taken into consideration. If the diameter and length of the tubes are specified, the number of the tubes required for heat and mass transfer are computed independently and greater one is taken for the

Gas A+C

construction. The required mass transfer area (Sm), hence the number of the tubes (n) for a given absorption duty ( AN ) is computed from,

A

Aim N

NdnS =π= l (4-5)

where, di(m) and l(m) are the inside diameter and the length of each tube, NA is the total molar flux of solute A, which is computed from suitable equations given in Chapter-3. The absorption duty, AN can easily be calculated from the given molar flow rate of gas and from the mole fractions of solute A at inlet and outlet gas.

Example-4.4) Design of Wetted-Wall Absorber

A binary gas mixture containing 5 percent solute A by volume is to be scrubbed with a liquid in a wetted- wall absorber operating counter-currently at 35 °C and 1 bar pressure. The wetted-wall

Cooling liquid

Cooling liquid Cooling jacket

Solution

Gas A+C

Liquid solvent

Gas A+C

Liquid film Column

A

A

A

A

S

Cooling liquid

Liquid solvent S

Gas A+C

Solution

Cooling liquid Tubes

Liquid film Column

(b)(a)

Fig.4.2 Wetted-wall columns : (a) Single pipe type, (b) Shell and tube type


column, due to the large heat of absorption, is to be constructed in ‘shell & tube’ form, from 6 m long ∅25*2 mm stainless steel tubes. Calculate the number of the tubes required for the below given values: Gas and liquid flow rates to the absorber are 25k-mol/h and 45 k-mol/h respectively. Exit gas will not contain more than 0.5 percent solute A by volume. Equilibrium relationship for the system at the operating conditions may be represented as y*= 0.52x, where y and x are mole fractions of solute A in gas and liquid respectively. Over-all mass transfer coefficient based on gas phase is calculated as K'

y = 6.02 *10-4 k-mol/m2s. Heat of absorption of solute A in the solvent is 120.103 kJ/k-mol A absorbed. Over-all heat transfer coefficient based on outside tube area is estimated as 140 W/m2 K. Inlet and outlet temperatures of cooling water, which circulates in the shell are 25 °C and 30 oC respectively.

Solution:

First, find G2, L1 and x1

01274.0)0245.0(52.0x52.0y ===∗

)yy(y ∗

11

111 −=∆

otal molar flux of solute A :

riving force for mass tran

Driving force for mass transfer at the top :

Total material balance : G1+ L2 = G2 + L1Solute A balance : G1y1 + L2 x2 = G2 y2 + L1x1Inert C balance : Gs= G2 (1 - y2) = G1(1 - y1)

L1= G1+L2-G2=25+45-23.87= 46.13 k-mol / h (25)(0.05)+(45)(0.0)=(23.87)(0.005) + (46.13) x1 x1 = 0.0245 Now, calculate the required number of the tubes for mass transfer duty;

Smi=n smi

T D sfer at the bottom :

L2= 45 k-mol/h x2=0.0

tco=30 oC

tci=25 oC

y2=0.005

G2

h/molk87.23005.01

)05.01(25y1

)y1(GG2

112 −=

−−

=−−

=

2

l=6 m

liA

Ami dn

NN

S π==1

doy1=0.05 G1=25 k-mol/h

t

smi= π di l

ao= π dol

dii

l

tube

L1,x1

)005.0)(87.23()05.0)(25(yGyGN 2211A −=−=

s10*14.3

h13.1 4 AmolkAmolk −−

== −

ln

y

yA )yy(N ∗

∗−

βK′

=

0373.0)01274.005.0(y =1 −=∆

)yy(y ∗222 −=∆

005.0)0.0005.0(y2 =−=∆0.0)0.0(52.0x52.0y 22 ===∗

0161.0)005.0/0373.0ln()y/yln(

)yy()y(21

21lnln ==

∆∆=−=∆ ∗ 005.00373.0yy −∆−∆


Average driving force for the absorber :

smAmolk10*69.9)0161.0(

110*02.6)yy(

KN

2

64

ln

y

yA

−==−

β

′= −

−∗

∗Total molar flux of solute A :

26

4

A

Ami m42.32

10*69.910*14.3

NN

S ===−

−Then, required mass transfer area;

829.81)6)(021.0(

42.32d

SsSn

i

mi

mi

mi ==π

=π

==l

Then, required number of the tubes; Required heat transfer area ; where, Ao= n ao

lno

o )T(UqA =∆

Heat load : )kW(s/kJ68.37)10*120)(10*14.3(HNq 34sA ===∆= −

Driving force for heat transfer at the bottom and at the top: ∆T1= 35-25 = 10 oC , ∆T2=35-30 = 5 oC

K2.7C2.7)5/10ln(

510)T/Tln(

TT)T( o

21

21ln ==

−=

∆∆∆−∆

=∆Average driving force for heat transfer for the absorber:

2

3

lno

o m38.37)2.7)(140(

10*68.37)T(U

qA ==∆

=Then, required heat transfer area ;

804.79)6)(025.0(

38.37dA

aAn

o

o

o

o ==π

=π

==l

Then, required number of the tubes; Conclusion: An absorber with 82 tubes will be sufficient for the specified heat and mass transfer duties.

4.4.1.2 Spray Column: This simple equipment is essentially a vertical pipe equipped with a liquid-spraying nozzle-group at the top. As the solvent droplets formed at the nozzles rain down in the column, the gas mixture flows upward coming in contact with the liquid droplets, during which solute A transfers from gas to liquid. The surfaces of liquid drops act as mass transfer area; hence creating large number of small diameter drops at the spray nozzles is preferred from mass transfer point of view. But, because of the back-mixing, drops cannot be made very small. In addition to this, drops tend to coalesce by touching with each other, after they travelled certain distance. For that, the height of column cannot be made very long. This means spray columns can only be used when the concentration change required in the gas phase is small. 4.4.1.3 Packed Column: The most commonly used continuous-contact type of equipment is

the packed column. As the name implies, these columns are vertical pipes filled with suitable packing materials. German chemist Dr.F.Raschig, in search for increasing the

Liquid solvent S

liquid drops

Gas A+C

liquid distributor

column

Gas A+C

solution

Fig.4.3 Spray column used in gas absorption


efficiency of wetted-wall columns, filled the empty pipe with inert solid particles and used for the first time the so-obtained columns in mass transfer operations. Later on, use of packed columns has extended rapidly with the inventions of various new packing materials. In packed columns, the liquid solvent, which is introduced to the column at the top through a liquid distributor, flows down through the voids by wetting the surfaces of all the packings as thin films. The gas phase, which enters the column at the bottom, flows upward through the same voids by coming in contact with the liquid films on the packing, during which solute A transfers from gas to liquid films. It follows from this that the surfaces of all the packings act as mass transfer area. For that reasons, the packed columns are the columns, which have the highest mass transfer area per unit volume of the equipment and because of this, they are the most widely used continuous-contact type mass transfer equipment. Packing materials: There is large number of packing materials of different shape and size and improvement of the existing ones and invention of new ones continue without interruption. The common features of the packing materials that are to be looked at for the selection of the best ones for a specific operation are: 1o) Specific surface, aP (m2/m3): it is defined as the surface area per unit volume of packed bed and it must be as great as possible, since these surfaces act as mass transfer area, if of course are wetted properly with the liquid. 2o) Void fraction, ε (m3/m3): the void fraction is defined as the volume of the empty spaces divided by total volume of the packing. As the liquid and gas flow through these voids, whenever possible void fraction should be very large. Small ε results in high pressure drop in the gas phase and also reduces the flooding gas velocity; in turns the first one increases the operating cost, the latter one the fixed capital cost. 3o) Inertness of the packing, which is known as insolubility of the packing material in gas and liquid at the operating conditions, is important for keeping the mass transfer area in the column as unchanged and insuring the solution against contamination. 4o) Robustness: the packing should have sufficient strength against breakages and distortions during filling, emptying and piling. 5o) Cost: as in many cases the cost of packing material is the main factor determining the total cost of column, its unit cost should be as low as possible. As stated above, various types of packing were developed by various companies, and some of them are shown in Fig.4.4. These may be divided in three groups. Ring types: these are hollow cylinders whose diameters are equal to the lengths. Raschig Rings, Lessing Rings and Pall Rings are the most commonly known types. Saddle types: these are more complicated in structure and common types are: Berl, Intalox, Torus, Novalox Saddles . Structured types: they are produced with weaving the thin wires or stripes in different shapes and are used at special applications, as they are rather expensive. Montz-Pak, Koch-Flexipac and Sulzer-Mellapak are the very well known types. All these packings are produced from different materials. Ceramic, carbon, various metals and various plastics are selected as construction material depending upon the corrosiveness of gas and liquid to secure the inertness property of the packing stated above. Ring and saddle types are mostly constructed from ceramic, as this is resistant to the attack of many gases and liquids. Packings are produced in different sizes, which are standardized. The important properties of some of the packings in the first and second groups are given in Table.4.2. More detailed information is obtained from the catalogues of the producers. The packings in the third


group whose aP and ε are much higher than others, are constructed specially in the modules of 0.25 to 0.50 m heights for the specified column diameter. For large

diameter

columns, the modules are produced in segments to facilitate easy

Raschig Rings Lessing Rings Pall Rings

Berl Saddles Novalox Saddles Torus Saddles Intalox Saddles

N

Montz-Pak type: BSH Koch-Flexipac Sulzer-Mellapak

Fig.4.4 Types of some packings

o part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 90

Table 4-2. Properties of some randomly filled packings

S i z e of p a c k i n g, mm Packing 6 9.5 13 16 19 25 32 38 50 76 R a s c h i g R i n g s

Ceramic Wall thickness,mm Cf ε ap , m2/m3 Metal 0.8 mm thickness Cf

ε ap , m2/m3 1.6 mm thickness Cf ε ap , m2/m3

0.8 1 600 0.73 787 700 0.69 774

1.6 1 000 0.68 508 390

2.4 580 0.63 364 300 0.84 420 410 0.73 387

2.4 380 0.68 328 170 29

2.4 255 0.73 262 155 0.88 274 220 0.78 236

3 155 0.73 190 115 0.92 206 137 0.85 186

4.8 125 0.74 148 110 0.87 162

4.8 95 0.71 125 83 0.90 135

6 65 0.74 92 57 0.92 103

9.5 37 0.78 62 32 0.95 68

P a l l R i n g s Plastics Cf ε ap , m2/m3

Metal Cf ε ap , m2/m3

Flexiring Cf ε ap , m2/m3

Hy-pak Cf ε

97 0.87 341 70 0.93 341 78 0.92 341

52 0.90 206 48 0.94 206 45 0.94 213 45 0.96

40 0.91 128 28 0.95 128 28 0.96 131

25 0.92 102 20 0.96 102 22 0.96 115 18 0.97

15 0.97

B e r l S a d d l e s Ceramic Cf ε ap , m2/m3

900 0.60 899

240 0.63 466

170 0.66 269

110 0.69 249

65 0.75 144

45 0.72 105

I n t a l o x S a d d l e s Ceramic Cf ε ap , m2/m3

Plastics Cf ε ap , m2/m3

725 0.75 984

330

200 0.78 623

145 0.77 335

98 0.775 256 33 0.91 207

52 0.81 195

40 0.79 118 21 0.93 108

22 16 0.94 89


diameter columns, the modules are produced in segments to facilitate easy installation. Ring type packings are filled into the column either in stacked or more commonly randomly. It is obvious that the packings in the second group can only be filled randomly. To minimize the breakages of packings in random filling, the column is first filled with water and then the packings are poured in. Column internals: In the construction of a packed column, apart from the packings some ancillary equipment is also used. These are: 1o) Packing support: packing supports are used at the bottoms of each packed section to carry the weight of packing materials. Hence packing supports should be strong enough to carry the weight, which is especially important for heavy material packings. In addition to this, packing support should have a large void area to permit the passages of gas and liquid phases without causing high pressure drops in them, and must not prevent the uniform distribution of the phases across the whole cross-section of the column. Various types of packing supports have been developed to accomplish these points. The bar grid is obtained by welding metal strips at equal spaces in vertical positions to a metal circular strip, which fits the column. In the more sophisticated ones, which are used in large diameter columns, gas and liquid phases pass through separate slots and holes. In Fig.4.5 some of the packing supports, used in industry are shown. 2o) Liquid distributors and re-distributors: It is very important to wet all the surfaces of the packings as otherwise these will not act as mass transfer area. Hence, requirement for a uniform distribution of liquid throughout the whole cross-section of the column is obvious. For that a liquid distributor at the top of the column should be installed to distribute the liquid very well. Various liquid distributors were developed for this purpose. In randomly filled columns, the density of packing is smaller near the wall, and this causes a tendency of liquid to separate and flow near the wall and gas to flow in the centre. This is known as channelling and results in loss of efficiency of the column. Although this can be reduced by making the ratio of packing diameter to the column diameter smaller than 1/8, complete elimination of channelling is only possible by installing liquid re-distributors at interval varying from 3 to 10 times the diameter of column but at least every 6 to 7 meter. The liquid re-distributors collect the liquid near the column wall and re-distribute it throughout the whole cross-section again. In practice column is made in sections of 3 to 10 diameter heights and to the bottom of each section a packing support and a liquid re-distributor are installed. In Fig.4.6 the most commonly used liquid distributors and re-distributors are shown. 3o) Hold-down plates: To prevent the carry over of packing materials of type-1 and type-2 by the gas, a hold-down plate is placed at the top of a packed section (just below the liquid distributor) of the column. This is especially important, when the gas velocity is high. For packings made from heavy materials such as ceramic, hold-down plate may restrain on the packed section freely, but for plastic packings hold -down plate must be clamped to the wall of the column. As shown in Fig. 4.7, bar grid type hold-down plates are generally used for this purpose. 4o) Entrainment eliminators: Especially at high gas velocities, the gas leaving the top of the packing may carry off droplets of liquid. This not only causes the loss of solvent but also may cause pollution of atmosphere. To prevent this, above the liquid distributor a mist eliminator is installed. Various types of mist eliminators are in use. A plate obtained by placing a layer of


about 100 mm thick wire mesh of 98-99 percent voids in a circular metal chamber can collect almost all the droplets in the gas (Fig.4.7 d). Randomly filled about 1 meter high packed bed will do the same duty.

Fig.4.5 Different types of packing supports


a b

e f

c d

g h Fig.4.6 Liquid distributors (a,b,c,d,e,f) and re-distributors (g,h)


Fig.4.7 Hold-down plates (a,b) and Entrainment eliminators (c,d)

s shown in Fig.4.8

on in Packed Column: The calculation of height of packing of a acked absorption column is the most important step of the design. We will derive an

ual to the total material leaving the column,

a b

c d

The view of vertical cut of a randomly packed column containing all these ancillarieis 4.4.2 Gas Absorptipequation now for the calculation of this. We assume that only one component (component A) of the gas mixture is absorbed in isothermally operating column. Consider the schematically drawn column in Fig.4.9, where G and L are the total molar flow rates of gas and liquid as k-mol/s respectively. As the flow rates of gas and liquid change along the column, with G1, L1 at the bottom, with G2, L2 at the top and hence with G, L at any section of the column theirs values are shown. If x and y show the mole fractions of solute A in liquid and gas, similarly with x1, y1 at the bottom, with x2, y2 at the top and with x, y at any section of the column theirs values are shown. Although G and L change along the column, Gs and Ls which show the molar flow rates of inert gas component and liquid solvent respectively as k-mol/s remain constant throughout the column. Let us write total material balance for the whole column at steady-state: As the total material entering the column is eq G1 + L2 = G2 + L1 (4-6) is written.



G2,y2

L2,x2

h=z

dh

1

2

G1,y1

L1,x1

h=0

G,yL,x

gas

liquid

Fig.4.9 Schematic representation of a packed column

Fig.4.8 View of vertical cut of a packed absorption column

Similarly at steady-state, solute A balance for the whole column becomes, G1 y1 + L2 x2 = G2 y2 + L1 x1 (4-7) For the inert components of gas and liquid, Gs = G1(1-y1) = G(1-y) = G2(1-y2) Ls = L1(1-x1) = L(1-x) = L2(1-x2) (4-8) are written. Now, similar equations between any section and top of the column are written as,

G + L2 = G2 + L (4-9)

G y + L2 x2 = G2 y2 + L x (4-10)

If the last equation is solved for y,

GxLyGx

GLy 2222 −

+=

cons.GG21 =≅

(4-11)

is obtained. This equation, which gives the relationship between gas and liquid compositions at any section of the column, is known as operating line equation. So, if this equation is known for the column, by knowing the composition of any phase at any level of the column the composition of the other phase at the same level can easily be computed from this equation. As shown in Fig.4.10 a, equation (4-11) represents a curve on xy-diagram, limited by the points T(x2,y2) and D(x1,y1) and having a variable slope of L/G . If both gas and liquid phases are dilute for solute A, G ≅ and

, and then equation (4-11) becomes, cons.LLL 21 =≅≅

22 xGLyx

GLy −+= (4-12)

This equation is represented with a straight line on xy-diagram, again limited by the points T and D (Fig.4-10 b). As it is seen, operating line is a straight line for dilute solutions. When the solutions are not dilute, operating line as was shown above, is a curve. As the drawing of a curve is not easy, we prefer to write operating line equation in a different way so that its representation on the diagram is still made by a straight line. For this, if G, L, G2, L2 are taken from the material balance equations and substituted into equation (4-11),

)y1/(G)x1/(xL)y1/(yGx

)y1/(G)x1/(Ly

s

22s22s

s

s

−−−−

+−−

= and from here,

2s

s2

s

s XGL

YXGL

−+=Y (4-13)

is obtained, where Y and X which are defined by Y= y/(1-y) and X = x /(1-x) are the mole ratios in the gas and liquid phases respectively. Namely, Y is the moles of solute


A in the gas divided by the moles of inert gas component, and X is the moles of solute A in the liquid divided by the moles of liquid solvent. As it is seen from Fig.4-10 c, the equation (4-13) is represented by a straight line on XY diagram, limited by the points T(X2,Y2) and D(X1,Y1) and with a constant slope of Ls/Gs. So, working with mole ratios instead of mole fractions the operating line is linearized. Summing up, if both solutions are dilute (x ≤ 0.05 and y ≤ 0.05 is sufficient to consider the solutions dilute) equation (4-12), if the solutions are strong, equation (4-13) is used as operating line equation.

y Y For the derivation of an equation from which height of packing can be computed, start with considering a differential volume dv in the packing. Note that height of differential volume, dh is related to this volume by dv = Ac dh, where Ac is the cross-sectional area of the empty column. At steady-state, along this differential volume solute A balance is written as: The loss of solute A by the gas = The gain of solute A by the liquid Hence, -d(G y) = d(L x) = NA dSm (4-14) is obtained, where NA is the total molar flux of A transferring from gas to liquid, dSm is the total mass transfer area available in the differential volume dv. If a mass transfer area per unit volume of packing is defined by av(m2/m3) dSm becomes dSm = av Ac dh. As it is seen, here av instead of ap is taken. This is because of the fact that geometrically available area ap is mass transfer area only when it is wetted thoroughly. Un-wetted part of ap does not act as mass transfer area. It is assumed that so-defined av is constant throughout the column. This is so, when column is filled uniformly with the packing and distribution of gas and liquid phases across the column cross-section is the same at any height. By substituting dSm from equation above and solving for dh,

⎮⌡⌠⌠

1122 xLyGZ L(d)yG(d

=⎮⌡−== ∫

2211 xLvcA

yGvcAo aAN

)xaAN

dhZ (4-15)

is obtained. In reality, this equation is general equation for mass transfer between two phases represented by G and L, in a packed column. Different solutions of equation

y1

equation(4-11)

slope=(L/G)≠cons. y2

x2 x

Y1

y2

y

equation(4-13)

T

x1

D y1

x2 x1

T T

D D

x Xslope=(L/G)=cons. slope=(Ls/Gs) =cons.

equation(4-12)

Y2

X2 X1

(c) (b)(a)

Fig.4.10 Representation of Operating Lines on the graph


(4-15) is obtained for each mass transfer operation, as NA is differently expressed in terms of mass transfer coefficients and driving forces for each operation. If it is remembered from Chapter-3, for “A transferring through non-transferring components” in both phases, as it is the case for gas absorption, NA is given as:

)xx()x1(

K)yy(

)y1(

K)xx(

)x1(k

)yy()y1(

kN

ln

x

ln

yi

lni

xi

lni

yA −

−

′=−

−

′=−

−′

=−−

′= ∗

∗∗

∗(4-16)

On the other hand,

y1dyG

)y1(dyG

y1ydG

y1yG

2s

ss

−=

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

d(G y) = d and

d(L x) = x1

dxL)x1(

dxLx1

xdLx1xL

d 2s

ss

−=

−=⎟

⎠⎞

⎜⎝⎛−

=⎟⎠

⎞⎜⎝

⎛−

0.1xx

0.1yy LK,k,GK,k ∝′′∝′′

cons.KL/,cons.kL/cons.,KG/cons.,kG/ xxyy

(4-17)

can be written. By substituting all these into equation (4-15) and taking constants Ac, av outside the integral and furthermore assuming and hence ≅′≅′≅′≅′

8.0yy GK, ∝′′ 8.0

xx LK,k ∝′′

(If it is looked at mass

transfer correlations k and are seen. But here assuming the powers as 1.0, instead of 0.8 the error involved is negligible from engineering point of view) finally;

⎮⌡⌠

−−−

′=⎮⌡

⌠−−

−′

1

2

1

2

x

xi

lni

cvx

y

yi

lni

cvy )xx)(x1(dx)x1(

AakL

)yy)(y1(dy)y1(

AakG Z =

⎮⌡

⌠

−−

−′

=⎮⌡

⌠

−−

−′ ∗

∗

∗

∗1

2

1

2

x

x

ln

cvx

y

y

ln

cvy )xx)(x1(dx)x1(

AaKL

)yy)(y1(dy)y1(

AaKG (4-18) =

are obtained. In the equations above, the terms outside the integrals are abbreviated with HG, HL, HOG, HOL respectively and the terms under the integral signs with NG, NL, NOG, NOL respectively, Hence;

cvx AakL

′ cvy AaKG

′ cvx AaKL

′cvy AakG

′HG = , HL = , HOG = , HOL = (4-19)

Height of one gas transfer unit

Height of one liquid transfer unit

Height of one overallgas transfer unit

Height of one overallliquid transfer unit

NG=⎮⌡⌠

−−

1

2yi

lni)yy)(y1(

dy)y1( −y

NL=⎮ NOG=⌡⌠

−−−

1

2

x

xi

lni)xx)(x1(

dx)x1(⎮⌡

⌠

−−

−∗

∗1

2

y

y

ln

)yy)(y1(dy)y1(

⎮⌡

⌠

−−

−∗

∗1

2

x

x

ln

)xx)(x1(dx)x1(

NOL=

Number of gas transfer units

Number of liquid transfer units

Number of overallgas transfer units

Number of overallliquid transfer units


and in shorthand writing,

Z = HG NG = HL NL = HOG NOG = HOL NOL (4-20)

is obtained. Depending upon the information available any one of these four equations can be used to compute the height of packing, z. N values are named as number of transfer units and H values as height of one transfer unit. Hence the height of packing required to make a change in the composition of a phase (for gas this is from y1 to y2) is found by multiplying the number of the transfer units needed to accomplish this change (this is NG or NOG for gas) with the height of one transfer unit (this is HG for NG and HOG for NOG). Note that N’s are dimensionless and H’s have the dimensions of length. Equation (4-20) can also be used at the desorption operations, as in these operations solute A transfers through non-transferring components in both phases as well. But, since transfer direction of solute A is from liquid to gas, the driving forces in N’s should be taken as (yi-y), (x-xi), (y*-y) and (x-x*) respectively. 4.4.2.1 Calculation of Number of Transfer Units: N values are obtained by performing the integrals. These integrals however cannot be solved directly, when the

relationships between say y and yi or x and xi are not given functionally as is the case at many applications. Then graphical integration is resorted to. Graphical integration: First, operating line and equilibrium distribution curve are both drawn on a millimetric paper as shown in Fig.4.11. Point P’s are then arbitrarily chosen between T and D and from these points PM lines are drawn by trial and error

y*T

y1

y

yi

y2

x x* xi x1 x2

0

slope= -lniy

lnix

)y1/(k)x1/(k

−′−′

D

M

P

0

Operating line


Mole fraction of solute A in liquid, x

R

E

P=cons. t=cons.

Fig.4.11 Graphical integration of N values

Mol

e fr

actio

n of

solu

te A

in g

as, y


as explained in Chapter-3. For the calculation of NG, yi values and for the calculation of NL , xi values are read from points M and the following table is constructed. In the

select read from graph c a l c u l a t e

y yi (1-y) (1-yi) (y-yi) (1-y)iln (1-y)iln/(1-y)(y-yi)

y1 - - - - - -

- - - - - - -

- - - - - - -

y2 - - - - - -

table, only the values required for the calculation of NG are shown. The values in the last column of the table are taken against the corresponding values in the first column on a millimetric paper. The area under the curve between limits of y1 and y2 gives the value of NG. The accuracy of the calculation depends on the number of the P points selected. The calculations of NOG and NOL are much easier, as in this case; drawing of PM lines is not required. For the calculation of NOG ,the ordinates of points R and for NOL , the abscissa values of points E are needed. Note that at desorption the operating line lies below the equilibrium distribution curve. Dilute solutions: If both solutions are dilute then 1 can be neglected next to x and y and integral terms simplify so that analytical solutions can be derived. Take NOG as an example, with this assumption it is written as,

⎮⌡

⌠−

=∗

1

2

y

y

OGyy

dyN (4-21)

if the equilibrium distribution curve is given by the following linear equation between x1 and x2 (m and n being constants); y* = mx + n (4-22) and if it is remembered that equation (4-12) is the operating line equation for dilute solutions,

22 xGLyx

GLy −+= (4-12)

Then by substituting the equation below, which is obtained by eliminating x between equations (4-12) and (4-22), into equation (4-21)

nmx)yy(L

mGy 22 ++−=∗ (4-23)

⎮⎮

⌡

⌠

−−+−=

⎮⎮

⌡

⌠

−−−−=

1

2

1

2

y

yn2mx2y

LmG)

LmG1(y

dy

y

yn2mx)2yy(

LmGy

dyOGN


n2mx2yL

mG)L

mG1(2y

n2mx2yL

mG)L

mG1(1yln

LmG1

1

−−+−

−−+−

− =

)nmx(y

nmx)yy(L

mGyln

LmG1

1

22

2211

+−

⎥⎦⎤

⎢⎣⎡ ++−−

− =

2)yy(1)yy(ln

LmG1

1∗−

∗−

−

)yy 21∗∗ −

∗2y )yy( 21

∗∗ −

= (4-24)

is obtained. If equation (4-23) is first written at top conditions and then ( is

solved, by noting that [(mx2 + n ) = ]; = )yy(L

mG21 − is obtained. It

follows from this that: )yy(

)yy()yy(yyyy

1L

mG

21

21

21

21−

−−−=

−−

−=−∗∗∗∗

1

when this is substituted into equation (4-24) finally,

2,1ln,

21

)yy(yy

∗−

−

[ ]

NOG = (4-25)

is found. Note that (y-y*)ln,1,2 ,which is given by the equation below, is the logarithmic mean of the overall driving forces for mass transfer calculated at the top and at bottom of the column.

21

212,1ln,

)yy/()yy)yy()yy()yy∗∗

∗∗∗

−−

−−−=−

(ln(

Hence, the number of the overall gas transfer units, NOG required for the creation of a concentration difference in the gas phase, which is (y1-y2), is simply obtained by dividing this concentration difference with the logarithmic mean of the overall driving forces calculated at the top and bottom of the column. If the system obeys Henry’s law, the equilibrium distribution curve is given by y*= mx and then NOG becomes:

⎥⎦

⎤⎡α

+α−−−

=1)/11(

mxmxyln1N

2

21⎢⎣α− y/11 2

OG (4-26)

mG/Lwhere, α is known as absorption factor and given as α = .

Similarly, at the desorption operation for NOL;

α−

⎦⎥⎢

⎣α+α−

−=

1

)1(m/yx

lnN 11

OL

⎤⎡ − m/yx 12

(4-27)

can be derived. Graphical Construction of Transfer Units (Baker’s Method): It follows from equation (4-21) that one overall gas transfer unit results when the change in gas


composition equals the mean overall driving force for mass transfer, causing the change. Baker, considering this fact, developed a graphical method for the calculation of number of transfer units by drawing. According to this method, first, operating line and equilibrium distribution curve are plotted on a millimetric paper as shown in Fig.4.12. Then, if NOG is to be calculated a so-called bisect line, dividing the vertical distances between operating line and equilibrium distribution curve in two equal parts everywhere, is drawn. By starting at any end (in the Figure this is top end) of the column number of the overall gas transfer units are drawn as follows: A horizontal line TF is so drawn that TE=EF and then is continued vertically until cutting the operating line (point R). Now, if it can be shown that HI = FR, then the right angle triangle TFR represent one overall gas transfer unit, as FR is the concentration difference occurred in the gas and HI is the mean overall driving force causing this change. THE and TFR are similar triangles and hence, TE/TF = HE/FR can be written. On the other hand TF=2TE from the drawing and 2HE = HI from the definition of bisect line and hence finally, FR =2HE = HI is obtained. This proves that right angle triangle TFR is an overall gas transfer unit. The drawing in this way is continued until reaching point D. The number of the right angle triangles found gives the required number of the overall gas transfer units. For Fig.4.11 this is slightly smaller than 4.

y

By drawingline and eqthe same stransfer uni

0

yI

y1

yR

yHy2

No part of this

the bisect line which divides the horizontal distances between operating uilibrium distribution curve in two equal parts everywhere, and following teps explained above starting at point D, the number of the overall liquid ts, NOL can be calculated.

K

C

R H

D

Operating line

Equilibrium disribution curve

T E F

I

x2

P=cons.

t=cons.

Bisect line

Draw TE=EF

xx1 0

Fig.4.12 Determination of NOG by Baker’s Method

CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 103

Example-4.5 Calculation of Number of Transfer Units Air, containing 3 % acetone by volume is to be washed with water in a packed column operating counter-currently at 1 bar and 25 oC, to reduce the acetone content to 0.5 % by volume. The flow rates of the gas and acetone-free water to the column are 300 k-mol/h and 690 k-mol/h respectively. a) Write the operating line equation, b) Calculate the concentration of liquid solution leaving the column as mass percent, c) Calculate the percentage recovery of acetone, d) Calculate number of the over-all gas transfer units required. At the operating pressure, temperature and the concentration range the equilibrium distribution of acetone between air and water is given as y*=2.2x, where y and x are the mole fraction of acetone in gas and liquid respectively.

Solution :

Assumption : Solutions are dilute (to be checked later !) y2=0.005 a) Equation (4-12) is the operating line equation for dilute solutions.

2

1

Then; )0.0(300690005.0x

300690y −+=

005.0x3.2y +=

005.0x3.2y 11 +=

x2=0 L=690 k-mol/h

b) Operating line equation at the bottom of the column;

Then, 0109.03.2

005.003.03.2

05.0yx 11 =

−=

−=

h/kg6.85512)58)(51.7()18)(690(L =+=&

As it is seen, dilute solutions assumption is valid.

y1=0.03 G=300 k-mol/h

Absorbed acetone : L(x1-x2)= 690(0.0109-0.0)= 7.51 k-mol/h Amount of solution leaving the column,

x1=?

Then, mass percent of acetone in the solution;

%39.3100.6.85512)58)(51.7(100x1 ==&

%4.83100.)03.0)(300(

51.7=c) Percentage recovery of acetone =

d) Since the solutions are dilute, equation (4-21) or (4-25) can be used. From equation (4-21)

( )⎮⌡⌠

+=

⎮⎮⌡

⌠

−−

=⎮⌡⌠

−=⎮⌡

⌠−

=∗

03.0

005.0

03.0

005.0

03.0

005.0

y

yOG 0048.0y0435.0

dy

3.2005.0y2.2y

dyx2.2y

dyyy

dyN1

2

51.40048.0)005.0)(0435.0(

0048.0)03.0)(0435.0(ln0435.01

=++ = is obtained.

4.4.2.2 Calculation of Individual Heights of Transfer Units: For the calculation of height of packing by using the number of gas or liquid transfer units, the relevant height of one transfer unit is needed. These can only be found experimentally as they contain mass transfer coefficients. Instead of determining individual mass transfer coefficients and mass transfer area per unit volume of packing separately and then substituting them into equations defining H’s, H values are directly measured. For the


experimental determination of HL, sparingly soluble gases in water such as O2, N2 are used as test gas. Sherwood and Holloway have given the so-obtained results by (4-28) 5.0

Lb

LL Sc)/L(aH µ′=where, HL is the height of one liquid transfer unit in (m), L' is the mass flow flux of liquid phase in (kg/m2s), µL is the viscosity of liquid in (kg/ms), and ScL is the Schmidt number of the liquid phase. a and b are the constants that depend on the type and the size of packing. Their values for Raschig rings and Berl saddles for the most commonly used sizes are given in Table.4.3.

Table 4-3. The constants of Sherwood-Holloway equation Packing mmxmm a b Raschig ring 9.5x9.5 0.00032 0.46 12.7x12.7 0.00072 0.35 25x25 0.00235 0.22 38x38 0.00260 0.22 50x50 0.00293 0.22 Berl saddle 12.7x12.7 0.00146 0.28 25x25 0.00128 0.28 38x38 0.00137 0.28 For the determination of HG, the mixtures of very soluble gases with air, such as ammonia are used as test gases. Fellinger has given the following dimensional equation: (4-29) 5.0cb Sc)L( ′G )G(aH ′= Gwhere, HG is the height of one gas transfer unit in (m), G' and L' are the mass flow fluxes of gas and liquid phases respectively in (kg/m2s), and ScG is the Schmidt number of the gas phase. a, b and c are the constants that depend on the type and size of packing. Their values for Raschig rings and Berl saddles for the most commonly used sizes are given in Table.4.4

Table 4-4. The constants of Fellinger equation Packing, mmxmm a b c Raschig ring 9.5x9.5 0.62 0.45 -0.47 25x25 0.56 0.32 -0.51 38x38 0.69 0.38 -0.40 50x50 0.89 0.41 -.045 Berl saddle 12.7x12.7 0.54 0.30 -0.74 25x25 0.46 0.36 -0.24 38x38 0.65 0.32 -0.45


The equations given above should be used cautiously only when experimentally measured value for the system with the specified packing is not available, as they have wide scattering of data. In the literature, experimentally measured values for various systems with different packings are available. Perry’s handbook contains a list of references for various systems. HOG and HOL can easily be synthesized from HG and HL. To show this, take the following equation from Chapter-3:

xyy km

k1

K1

′+

′=

′ (4-30)

If both sides of this equation are multiplied with G/avAc and in addition, the last term of the right hand side is multiplied also with L/L;

cvxcvycvy AakL

LmG

AakG

AaKG

′+

′=

′ is obtained. By comparing this equation with

the equations (4-19),

LGOG HL

mGH +=H (4-31)

is found. Similar analysis for HOL results in,

GLOL HmGLHH +=

∝ 8.1Gu

(4-32)

4.4.2.3 Determination of Diameter of a Packed Column: Determination of the diameter of a packed column constitutes the second step in the design of a packed column. As was explained before, gas and liquid phases flow counter currently through the voids of the packing. The hydrodynamic interaction of gas and liquid in these voids is the most important point influencing the diameter of the column. Assume that a randomly packed column made of glass and equipped with a device to measure the pressure drop of the gas along the packing, is used for experimentation. The lines carrying the gas and liquid to the column contain rotameters to measure the flow rates. In the first experiment let us assume that gas is made to flow through dry packing. A table can be formed by writing the pressure drops in the gas (∆P) against the increasing flow rates of gas. If the logarithms of pressure drops are plotted against the logarithms of superficial gas velocities (uG), which are calculated by dividing the volumetric flow rates of gas by the cross-sectional area of empty column, the line-1 shown in Fig.4.13 is obtained. Calculated slope of this line gives ∆P , which indicates that gas flow through the voids is turbulent. If the experiment is repeated with wet packing, this time the line-2 is obtained, indicating that higher pressure drops at the same gas velocities, as expected, are recorded. In the third experiment, again the pressure drops at various gas flow rates are monitored but unlike the first two, this time liquid flows down at a constant flow rate of L1. In this case, the ∆P-uG relationship has three distinct regions: Up to the point a the change of ∆P with uG is the same, of course with higher ∆P values for the same uG, but after this point, the change becomes steeper up until another point b, from where the change is even more steeper than the previous one.


L3> L2 > L1 3

If we look at the inside of the column during this experiment, we notice that up to the point a, gas and liquid phases flow through the column without affecting each other as in the first two experiments. After point a, it is seen that interaction between liquid and gas starts and so-called liquid hold-up (Ho), which is defined as the volume of the liquid retained per unit volume of the packing, increases with increasing gas flow. Point a is named as loading point. From point b on, liquid has no chance to flow down the column as it is pushed up by the gas and forced out of the column through the exit nozzle of gas. This condition, which is a hydrodynamic crisis, is known as Flooding and the velocity of gas corresponding to point b, uGF is named as flooding gas velocity. Repeating the experiments with higher constant liquid flows (L2 and L3) produces the same phenomena, with loading and flooding points being at lower gas velocities. It is obvious that in the design of packed columns, flooding should be evaded. Hence, always a gas velocity below the flooding gas velocity must be selected. It can easily be realized that flooding phenomena has nothing to do with mass transfer; it is purely a hydrodynamic phenomena. By using various gases, liquids and packing materials and by working under different flow rates and conditions, a generalized equation in graphical form has been developed as shown in Fig.4.14, for calculating flooding gas velocity for any gas-liquid pair in a randomly packed column at any condition. The flooding curve is given as uppermost curve in the Figure. To the same Figure, constant pressure drop curves of gas phase as Pascal per meter of packing are also added, as the pressure drop in the gas phase along the packing is very important in the design of packed columns. Here, are the mass flow rates of liquid and gas as (kg/s) ,

GveL &&

GL ve ρρ are the densities of liquid and gas as (kg/m3), µL is the viscosity of liquid as(kg/ms) and G' is the mass flow flux of the gas phase as (kg/m2s). When reading the ordinate of the flooding curve, G′ should be replaced by . The effect of packing material on the flooding is given by Cf parameter, which is given for various packings in Table.4.2. By dividing

FG′

FG′ with the density of the gas, flooding gas velocity uGF is first obtained, then by taking some percent (in practice usually 65-75 %) of this velocity, operating gas velocity uG is found . Then from,

L2

Log

(∆P)

uGF

a

0

L1 2 L3

1 b bb

a aL=0

Log uG

0

Fig.4.13 Loading and Flooding points


Dc = the diameter is calculated. In some cases, the diameter is calculated according to a specified pressure drop in the gas per unit height of packing. In this case, G

5.0GG )u/G4( πρ&

G/GAc ′= &

5.0LG )/)(G/L( ρρ&&

' is computed from the ordinate of the point, which is obtained by

Pressure drop in gas(Pa/m.packing)

5.0

GL

G

GL

⎟⎟⎠

⎞⎜⎜⎝

⎛ρ−ρ

ρ&

&

Fig.4.14 Flooding and pressure drops in randomly packed columns

Flooding curve

)GL(G

1.0LfC2G

ρ−ρρ

µ′

going vertically up from the calculated abscissa value until cutting the specified pressure drop curve. Then, first cross-sectional area from and later the diameter from Dc = (4Ac/π)0.5

is calculated. The diameter of a column, which is packed with structured packing material given in the third group (page:84), is normally calculated by using the information given in the producer’s catalogues. For example for the calculation of the diameter of a column which will be packed with type: BHS-400 packing material of Montz GmbH (shown in Fig.4.4), first an s parameter is calculated from s = then a x-factor from x = - 1.443 s x5.0

L eρ and finally a flooding factor from FFL = C , where C is given by the producer as 0.115 for type :BHS-400. Finally by taking some percent of FFL (a figure between 70-75 % is recommended by the producer) the load factor Fv and then by using equation uG =Fv / Gρ

cLw aA/L ρ= &

33.0w

operating gas velocity is computed. It has been stated many times that in order to obtain the highest efficiency, the packing should be wetted properly at every point of the column. It has been shown experimentally that the wetting rate, which is defined as L and is

related to the liquid hold-up by the equation Ho = 0.0136 ap L , should not be smaller than 2.10

p

-5 m3/ms for randomly packed columns. If this cannot be secured by changing the size of packing, liquid circulation will be tried. The packing materials given in the third group have the advantages of being wetted thoroughly at very low liquid rates.


Example-4.6) Calculation of Diameter and Height of Packing The column in Example-4.4) will be packed with 32*1.6 mm metal Raschig rings. Calculate:

a) The diameter of the column for a gas side pressure drop of 400 N/m2 per m of packing, b) The height of packing required, c) Check for minimum wetting rate and if it is not satisfied make necessary changes, d) Find out with what percent of the flooding gas velocity the operation is carried out,

Heights of transfer units for acetone absorption into water in a column packed with 32*1.6 mm Raschig rings were found experimentally and given as ;

417.0

395.0

G LG397.1H′′

=22.0

L L345.0H and ′= L& ′ where, HG & HL are in m. G′ are in kg/m2s.

Solution: a) Calculate the diameter at the bottom of the column, where liquid and gas flow rates are the highest.

Average molecular weight of the gas at the bottom; 87.29)58)(03.0()29)(97.0(G =M +=

3G m/kg21.1

)25273)(083.0()87.29)(1(

=+

=

s/kg57.3h/kg6.85512 ==&

Density of the gas; ρ

As the density and viscosity of the 3.39 % acetone solution at 25 oC the density and viscosity of water may be taken ,which are 997 kg/m3 and 0.894 cP . Mass flow rate of liquid, , L

s/kg49.2h/kg9618)87.29)(300(M*G G ====&G Mass flow rate of gas,

05.021.1997

21.149.257.3

GL 5.05.0

GL

G =⎟⎠⎞

⎜⎝⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ρ−ρ

ρ&

& Then from Fig.4.14, for ∆P/z = 400 N/m2 per m.

08.0)(

CG

GLG

1.0Lf

2=

ρ−ρρµ′ is read. From Table.4.2 Cf =110 and ap = 162 m-1. Then,

Cross-sectional area required,

sm/kg33.1)10*894.0)(110(

)21.1997)(21.1)(08.0(G 2

5.0

1.03=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=′

−

2c m872.1

33.149.2

GGA ==′

=&

m55.1872.1*4D5.0

c =⎟⎠⎞

⎜⎝⎛

π=Diameter of the column,

sm/kg872.157.3

AL 2

c

===′&b) 91.1L HL = (0.345)(1.91)0.22 = 0.398 m

m194.1397.1(H417.0

395.0

G ==

m575.1)398.0)(3.2/1)(2.2(194.1HOG

)91.1()33.1()

From equation (4-31), =+=Finally, from equation(4-20), the required height of packing, z = (1.575)(4.51) = 7.10 m

c) Wetting rate, m10*18.1)162)(872.1)(997(

57.3aA

LL 35

pcLW

−===ρ

& ms/

As this figure is less than the recommended value of 2*10-5 m3/ms, either re-circulation of the liquid must be considered or the size of the packing must be increased. When possible the latter choice is preferred to the first one otherwise reduction in the driving force and hence increase in the number of the transfer units is inevitable. If 50*1.6 mm Raschig rings are taken , instead of 32*1.6 mm, Cf =57 and ap =103 m-1 (from Table.4.2). Then,


sm/kg85.157

110)33.1(G 2

5.0

=⎟⎟⎠

⎞⎜⎜⎝

⎛=′

sm/kg65.2346.1/57.3L 2==′

New cros-section, AC = 2.49/1.85 = 1.346 m2 ; new diameter, Dc = [(4*1.346)/π]0.5 = 1.31 m.

m186.1)65.2()85.1()397.1(H 417.0

395.0

G ==, HL = 0.345 (2.65)0.22 = 0.428 m,

HOG= 1.186 + (2.2)(1/ 2.3) (0.428) = 1.595 m. and z = (1.595)(4.51) = 7.19 m. In reality height of packing will increase more than this , as the constants and exponents of HG & HL correlations will be higher for this size of the packing .

ms/m10*58.2)

57.3L 35W

−== . The wetting rate is now satisfactory. 103)(346.1)(997(

05.0G

5.0

GL

G =⎟⎟⎠

⎞⎜⎜⎝

⎛ρ−ρ

ρL&

&d) From flooding curve of Fig.4.14, for

28.0)(

CG

GLG

1.0Lf

2F =

ρ−ρρµ′ sm/kg46.3)85.1()08.0/28.0( 25.0

F ==′ is found. Then, G

%5.53100.46.385.1100.

GG

==′′

Hence, percent of flooding = , which allows for safe operation. F

4.4.3 Stage-Wise Contact Type of Absorption: As was stated before, gas absorption operations are also carried out in stage-wise contact type of equipment. The most commonly used stage-wise contact type of equipment is plate (tray) column. 4.4.3.1 Gas Absorption in Plate Columns: The plate column is a vertical pipe containing horizontal plates at certain distances, which are known as plate spacing. Various types of plates have been developed and are in use in industry, the common properties of all are: They are circular metal plates of the diameter which are slightly smaller than the inside diameter of the column. At large diameter columns, they are divided into segments to facilitate easy installation. A segmental part of the plate is cut off and a vertical metal sheet, which is known as down comer apron, at a length which is 40-50 mm shorter than the plate spacing is attached to the plate to create a channel between column shell and the apron, which is known as down-comer and through which liquid phase flows. By leaving a segmental area equals the area of down comer cross-section at the other side of the plate untouched, holes or channels are opened in the area between, to facilitate gas flow through the plate. When the holes are opened in the size range of 3-15 mm diameters, no additional parts are added to the holes and so-obtained plate is known as sieve or perforated plate. In the other types of plates, the hole diameters are lager (50-150 mm) and some special parts are attached into or onto them to distribute the gas as tiny bubbles in the liquid on the plate, thus valve plate, bubble-cap plate, tunnel-plate etc. are formed. The down comer apron is extended about 25-150 mm above the plate to form a weir, by which a liquid pool on the plate is always secured during the operation. Gas and liquid phases come in contact only in the liquid pools formed on the plates and hence mass transfer between the phases can only take place here. Between the plates; gas and liquid phases flow through separate channels without seeing each other. In Fig. 4.16 (a) the view of vertical cut of a sieve plate absorption column is shown. The column is schematically shown as in Fig.4.16 (b). The plates are numbered from top of the column, and with n and N any plate and the last plate of the column are shown respectively. As the total


flow rates of gas and liquid (G and L) and the concentrations of these phases change from plate to plate, in order to refer to a particular phase and composition, the number of the plate from which this stream comes is given as subscript to this property. Thus, L3 , G3 , x3 , y3 show the flow rates of liquid and gas leaving the third plate and the mole fractions of solute A in these streams respectively. The streams and compositions in the schematically shown column were numbered according to this principle. It is easy to understand why the liquid and gas phases entering the column are shown by Lo

Downcomer

2

Gas

G1,y1

Lo,xo Liquid

Solution

Gas

Plate

Shell

1

L1,x1 G2,y2

L2,x2 G3,y3

n

∇

N

Gn+1,yn+1 Ln,xn

GN+1,yN+1

LN,xN

Down-comer apron

Weir

(b) (a)

Fig.4.16 (a) View of vertical cut of a sieve-plate absorption column, (b) Schematicall showing of a plate column


and GN+1. The compositions of gas and liquid phases entering any plate n, which are xn-1 and yn+1 are not equilibrium values. During the contact, they approach the equilibrium values, as solute A transfers from gas to liquid, and if sufficient contact time is provided they finally reach in equilibrium. Then the compositions of the two phases (Ln and Gn phases) leaving the plate, which are xn and yn become coordinates of a point lying on the equilibrium distribution curve of the system. So-operating plate is named as equilibrium plate, ideal plate or theoretical plate. On the other hand, if the two phases do not remain sufficiently long time in contact, then the two phases leaving the plate will not be in equilibrium and so-operating plate is named as real or actual plate. In practice, the plates operate as real plate, since it is not economical to keep the two phases in contact very long time. From these definitions, it is understood that equilibrium plate accomplishes maximum possible change in the compositions of the phases, whereas real plate does only part of this maximum change. Nevertheless, in the analysis of plate columns; all the plates are first assumed as operating as equilibrium plates and then the number of the equilibrium plates needed for the given separation is computed. Then, by determining the approaches of the plates to the equilibrium plates, this equilibrium plate number is converted to the actual plate number, which really interests us. A concept, which is known as plate efficiency is frequently used to determine the performances of real plates. Plate efficiency can be defined in terms of gas and liquid phases. To the memory of Murphree, who first used these efficiencies, Murphree gas and Murphree liquid efficiencies are defined as:

1nn

1nnML xx

xx

−

∗

−

−−

=∗

+

+

−−

=n1n

n1nMG yy

yyE E (4-33)

It follows from the equations above that the plate efficiency is defined as the ratio of the actual change in the composition of a phase on a plate to the change that would have been obtained, if the two phases leaving the plate were in equilibrium. The changes in the compositions of gas and liquid on an equilibrium and on a real plate are shown in Fig.4.17. The value of the plate efficiencies defined by equation (4-33) may change from plate

y

yn+1

yn

Real PlateEquilibrium Plate


∗nx

yn+1

yn

∗ny

xxxn xn-1 xn-1 xn


Slope=-L/G Slope=-L/G

Fig.4.17 Changes in the compositions of gas and liquid on an Equilibrium and on a Real Plate

y


to plate. Hence, it is more convenient to use over-all column efficiency, Eo in converting the number of the equilibrium plates to the number of the real plates, as this efficiency is defined as the ratio of the number of the equilibrium plates needed to accomplish the desired change in the compositions, to the number of the real plates needed to accomplish the same change. Eo = Nideal/Nreal (4-34) The efficiencies of both plate and column are determined in pilot size columns experimentally. Some experimental results obtained in bubble-cap absorption columns are listed in Table 14-56 of Perry’s Chemical Engineers’ Handbook (4th ed. Page:14-38). The correlation given by O’Connell in graphical form and shown in Fig.4.18 may be used to estimate the over all column efficiency, when experimentally measured values are not available. Although this correlation was obtained with bubble-cap absorber, it can be used for other types of plates for approximate calculations. In the figure below; m(=y/x) is the slope of equilibrium distribution curve, ML, µL(kg/m.s) and ρL(kg/m3) are the molecular weight, viscosity and the density of the liquid phase. By dividing the number of the ideal plates obtained for a given absorption by using

any one of the methods given below, by the over all column efficiency, the number of the real plates is found. Although the number of the ideal plates may be a fractional number, the number of the real plates must always be an integral number.

Ove

r all

colu

mn

effic

ienc

y, E

o

L

LLmMρµ

Fig.4.18 Over all column efficiency for bubble-cap absorber

Calculation of the Number of the Ideal Plates: In the design of a plate type absorber, calculation of the number of the ideal plates constitutes the first step. Below, necessary equations are to be derived. Let us assume again that only one component of the gas mixture is absorbed in an isothermally operating column. By writing total material and solute A balances for the whole column shown in Fig.4.16b, Lo + GN+1 = LN + G1 (4-35) Lo xo + GN+1 yN+1 = LN xN + G1 y1 (4-36)


are obtained. As the flow rates of the inert components of the gas and liquid phases are constant then, Gs = G1(1-y1) = G n+1(1-yn+1) = GN+1(1-yN+1) and Ls = Lo(1-xo) = Ln(1-xn) = LN(1-xN) (4-37) can be written. By writing the same balances between top of the column and any plate (plate n), Lo + Gn+1 = Ln + G1 (4-38) Lo xo + Gn+1 yn+1 = Ln xn + G1y1 (4-39) and by solving yn+1 from the last equation,

1n

oo11n

1n

nG

xLyGxGL

++

−+ yn+1 = (4-40)

is obtained. This equation is the operating line equation, as it gives the relationship between the composition of liquid leaving plate n and the composition of the gas entering the same plate. Operating line in this form is represented by a curve on xy-diagram between the points T(xo,y1) and D(xN,yN+1). When both solutions are dilute, Lo=Ln=LN= L= cons. and GN+1=Gn+1 =G1= G= cons., and hence equation (4-40) becomes:

GxLyx

GL o

1n −+ yn+1 = (4-41)

Operating line, written in this form is represented by a straight line on xy-diagram, bounded by the same points. When the solutions are not dilute, equation (4-40), by substituting the values of G1, Gn+1 , Lo and Ln from equation (4-37), can be written as,

os

s1n

s

s1n X

GLYX

GLY −+=+ (4-42)

As it is seen, operating line equation written in terms of mole ratios rather than mole fractions is represented by a straight line on XY-diagram. Summing up, in dilute solutions equation (4-41) and in strong solutions equation (4-42) is preferred. After writing operating line equation, the calculation of the number of the ideal plates is started. But before this, let us see what are known at the start of the design. These are: molar flow rate of gas to the column GN+1, its composition yN+1, the permissible concentration of the outlet gas y1, or percentage recovery of solute A from the inlet gas and the operating pressure and temperature. The quantity and the composition of liquid solvent (Lo,xo)must also be selected and set at the start of calculations by the designer. Plate to Plate Calculation Method or Lewis Method: First, equilibrium data of the system at operating conditions are obtained. If the solutions are dilute, these are plotted on xy-, if not on XY- coordinates and thus equilibrium distribution curve is obtained. Then, operating line is written either as equation (4-42) or (4-43). The calculation is now started at the top of the column. Since y1 is known, by entering this on the equilibrium distribution diagram, x1 is read. Later, this x1 is inserted into operating line equation and y2 is calculated. Now the calculated y2 value is entered on the equilibrium distribution diagram and x2 is read. Calculation is continued in this way by using equilibrium distribution curve and operating line equation alternately


Y(y)

until yN+1 value is reached. The number, that is found by subtracting 1 from the subscript of the last calculated y value, is the number of the ideal plates required to accomplish the given absorption. Graphical Calculation Method or Mc Cabe-Thiele Method: By drawing the operating line on the same diagram with equilibrium distribution curve, the calculation can be realized on the graph. As it is shown in Fig.4.19, the calculation is started again at the top of the column. By drawing a horizontal line from point T until cutting the equilibrium distribution curve and then a vertical line until cutting the operating line, a right angle triangle is formed. If we look at the values around the triangle, we see that these are the compositions of the streams around the first ideal plate of the column. Hence, this triangle represents the first ideal plate in the column. By continuing the construction of right angle triangles in this way until reaching the point D, and counting the number of the triangles, the number of the ideal plates is found.

Example-4.7) Gas Absorption in Plate Column

A 150 k-mol/h gas consisting of ammonia (A) and air (C) will be washed with water (S) in a plate column operating at 20 oC and 800 mmHg. Gas contains 20 percent ammonia by volume and this will be reduced to 1.96 percent. The water, which is ammonia-free, will be supplied at a flow rate of 2 581 kg/h. Calculate: a) The concentration of liquid solution leaving the column, b) The percentage recovery of ammonia, c) The number of the equilibrium plates needed, d) The quantity of ammonia transferred from gas to liquid on each equilibrium plate.

Solution: Gas is not dilute, Hence work with mole ratios.

Y1(y1)

YN+1=Y4 (yN+1= y4)

Operating line


1

Y2(y2)

Xo(xo) X1(x1) X2(x2) X3(x3) X(x)

Y3(y3)

T

3

2

D

0 0

P=cons. t=cons.

Fig.4.19 Calculation of number of ideal plates by Mc Cabe-Thiele method


First, obtain equilibrium data at 20 oC and P = 800 mmHg, then convert these to X-Y and plot on a mm paper.

y

Gn+1,yn+1

N

2

1

L2,x2

L1,x1

Ln,xn

GN+1= 150 k-mol/h yN+1 = 0.20

G2,y2

LN,xN

xo =0.0

n

G3,y3

y

LN,xN

≈

≈≈

≈

y1=0.0196

h/kg5812Lo =&From App.Table.4.1 Solubility of NH3 in water at 20 oC Mass of NH3 per 100 mass of H2O 20 15 10 7.5 5 4 3 2

Partial press. of NH3 (mmHg) 166 114 69.6 50.0 31.7 24.9 18.2 12.0

Conversion of the data to X and Y

)M/m()M/m(M/m

x1xX−

=

Fig.Example-4.7a

x

BBAA

AA

+=

y1yY−

=Pp

y A=

The first values; With the conversion of other values, Equilibrium Data are obtained as follows :

X 0.212 0.159 0.106 0.079 0.053 0.042 0.032 0.021 Y 0.262 0.167 0.095 0.067 0.042 0.032 0.024 0.015

These are plotted on a millimetric paper as shown in Fig.Example-4.7b

Operating line equation; If we write this equation at the bottom of the column; Then, from here XN = 0.192 and is found. a) or as mass fraction : is obtained. b) Ammonia entering the column : GN+1 yN+1 = Gs YN+1 = (150)(0.20) = 30 k-mol/h Ammonia recovered (absorbed): Gs (YN+1 - Y1) = 120 (0.25 – 0.02) = 27.6 k-mol/h

175.0)/()/(=

+=

18100172017/20x 212.0

175.01X ==

175.0−

262.0208.01

208.0Y =−

=208.0800166y ==

25.020.01

20.0y1

y 0.0x1

xXo

oo =

−=02.0

0196.010196.0

y1yY

1

11 =

−=

−=Y

1N

1N1N =

−=

−=

+

++

G h/Cmolk120)20.01(150)y1(G 1N1Ns −=−=−= ++ oooos L)01(L)x1(LL =−=−=

h/Bmolk4.14318/5812LL os −===

)0.0(120

4.14302.0X120

4.143Y n1n −+=+

020.0X1Y 195. n1n +=+

020.0X195.1Y N1N +=+

161.0192.01

192.0Xx ===

X1 N

NN ++

020.0X195.125.0 N +=

153.00((

())(x())((

x ) =−+

=+

=)18)(161.1)17)(161.0

)17)(161.0M1Mx

M)(x(

BNAN

ANN −&


%92100.0.306.27

== Then, percentage recovery of ammonia c) Locate points T (0.0; 0.02) and D (0.192; 0.25) on the XY-diagram and by joining them draw the operating line. Then, using McCabe-Thiele method find the number of the ideal plates, which is found as 6.

d) Read the X and Y values around each ideal plate from the Fig.Example-4.7b as:

Xo X1 X2 X3 X4 X5 X6

0.0 0.025 0.063 0.115 0.150 0.175 0.192

Y1 Y2 Y3 Y4 Y5 Y6 Y7

0.02 0.050 0.095 0.158 0.20 0.23 0.25

Xo=

X4

Y P = 800 mm Hg, t = 20 oC 0.26

0.02

Y1= 0.02

0.06

0.06

0.1

0.10 0.14

0.14

0.18

0.22

0.22

D Y7

1

2

4

3

5

6

0.192=XN

T

Operating Line

X1 X2 X3 X5 X6

Y2

Y3

Y4

Y5

Y6

YN+1= 0.25


0 X 0 0.18

Fig.Example-4.7b


Loss of ammonia by the gas on plate n = GS(Yn+1-Yn) On the first plate : GS(Y2-Y1) = 120 (0.050-0.02) = 3.6 k-mol /h On the second plate : GS(Y3-Y2) = 120 (0.095-0.05) = 5.4 k-mol /h On the third plate : GS(Y4-Y3) = 120 (0.158-0.095) = 7.56 k-mol /h On the fourth plate : GS(Y5-Y4) = 120 (0.20-0.158) = 5.04 k-mol /h On the fifth plate : GS(Y6-Y5) = 120 (0.23-0.20) = 3.6 k-mol /h On the sixth plate : GS(Y7-Y6) = 120 (0.25-0.23) = 2.40 k-mol /h + Total = 27.6 k-mol NH3/h Or, gain of ammonia by the liquid on each plate = LS(Xn-Xn-1) On the first plate : LS(X1-Xo) = 143.4 (0.025-0.0) = 3.59 k-mol /h On the second plate : LS(X2-X1) = 143.4 (0.063-0.025) = 5.45 k-mol /h On the third plate : LS(X3-X2) = 143.4 (0.115-0.063) = 7.46 k-mol /h On the fourth plate : LS(X4-X3) = 143.4 (0.15-0.115) = 5.02 k-mol /h On the fifth plate : LS(X5-X4) = 143.4 (0.175-0.15) = 3.59 k-mol /h On the sixth plate : LS(X6-X5) = 143.4 (0.192-0.175) = 2.44 k-mol /h + Total = 27.55 k-mol NH3/h In section (b) ammonia recovery was calculated as 27.6 k-mol/h. These are in perfect agreement considering graphical reading accuracy. Absorption Factor Method: It is known that when the solutions are dilute; operating line and the equilibrium relationship both written in terms of mole fractions are represented by straight lines. If in this case, solute A balance and the equilibrium relationship for the first plate are written: L (xo-x1) = G (y1-y2) (4-43) y1 = m x1 (4-44) and, if x1 is eliminated between these two equations,

1)mG/L(x)G/L(yy o2

1 ++

= (4-45)

is obtained. When the absorption factor, which is defined as L/mG = α is substituted into the equation above,

1xmyy o2

1 +αα+

= (4-46)

is found. If the similar steps are repeated for the second plate,

1yy

1xmyy 1313

2 +αα+

=+αα+

= (4-47)

is obtained. Upon substituting y1 from equation (4-46),

1xmy)1(y 2

o2

32 +α+α

α++α= (4-48)

is found. Multiplying the nominator and denominator of the above equation by (α-1)

1mx)1(y)1(y 3

o2

32

2 −α−αα+−α

= (4-49)

is obtained. If similar steps are repeated for 3rd , 4th and finally for the last plate (plate N ) of the column:


1mx)1(y)1(y 1N

oN

1NN

N −α−αα+−α

= ++ (4-50)

is found. On the other hand, the solute A balance for the whole column is written as: L (xo-xN) = G (y1-yN+1) (4-51) Since the equilibrium relationship for the last plate is yN/m = xN , substituting this into equation above and solving for yN gives;

αα+−

= + o11NN

xmyyy (4-52)

As the left hand sides of equation (4-50) and (4-52) are the same, then the right hand sides must also be the same. Hence, by equating and rearranging these equations;

1mxyyy

1N

1N

o1N

11N

−αα−α

=−−

+

+

+

+ (4-53)

s obtained. This equation can be used to compute the change in the composition of a gas at a certain absorption factor in an existing column whose ideal plate number is already known. Cross multiplication and then solution for αN+1

gives;

α+⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

⎟⎠⎞

⎜⎝⎛

α−α

= + 1mxy

mxy1

o1

o1NN1mxy

mxy)1(o1

o1N1N +⎥⎦

⎤⎢⎣

⎡−−

−α=α ++ If both sides are divided by α, α

is found. Finally by taking the logarithm of both sides,

α

⎥⎦

⎤⎢⎣

⎡

α+⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

⎟⎠⎞

⎜⎝⎛

α−α

=

+

log

1mxy

mxy1logN o1

o1N

(4-54)

is obtained. This analytical equation can be used for the calculation of the number of the ideal plates, when the solutions are dilute. In desorption operation, following the same steps;

( )

)/1(log α

1m/yxm/yxlog

N 1NN

1No⎥⎦

⎤⎢⎣

⎡α+α−

−−

= +

+

(4-55)

is found. Determination of Quantity of Liquid Solvent: The selection criteria of the most suitable solvent among the potential solvents have been given in Section.4.3. Once the type of solvent is selected, next the amount of solvent must be decided. The effect of solvent quantity on the absorption will be seen now, by considering a plate type absorber; the analysis applies to the packed columns as well. Let us try to draw the operating line on the graph on which equilibrium distribution curve has already been plotted on XY-coordinates as shown in Fig.4.20a. Since the values at the top of the column are known, point T can easily be located on the graph. Only the ordinate value of point D is known, if this YN+1 is entered on the graph it is then obvious that point D will be on the horizontal line drawn from YN+1 . Gs is calculated from GN+1 and yN+1, but as Lo is not known, Ls can not be calculated. Now let us assume that we select a certain quantity of solvent Lo. Ls and the slope Ls/Gs can now be calculated. If we draw the line from point T with the slope of Ls/Gs ,this will be the operating line for this


selected liquid quantity and the abscissa of point D, which is XN, gives the composition of liquid solution leaving the column. The required number of the ideal plates is then found by drawing the right angle triangles between equilibrium distribution curve and operating line as explained above. Now, if we reduce the quantity of solvent to be used, the slope of the operating line decreases and hence operating line approaches the equilibrium distribution curve, and the bottom condition of the column is shown now with point D1. As it is seen from the figure, the concentration of solution leaving the column is greater now and the number of the equilibrium plates needed for the same absorption is also greater, as the space between the operating line and equilibrium curve became narrower. It follows from here that when the quantity of solvent is decreased keeping everything else constant, the number of the equilibrium plates (in the case of packed column, the height of packing) increases. By decreasing the solvent rate further, a condition is reached at which operating line touches the equilibrium distribution curve. This is shown with point D' on Fig.4.20 (a) and with point K on Fig.4.20(b). An attempt to draw the right angle triangles between operating line and equilibrium distribution curve results in a pinch around point D' or K, indicating that the number of the equilibrium plates (or height of packing) needed to accomplish the specified absorption with this solvent rate is infinite. This solvent rate is then known as minimum solvent rate. In this case, the

concentration of liquid solution leaving the column-as expected- becomes maximum. This maximum value of XN as shown in Fig.4.20(a), is the equilibrium value of YN+1 , when equilibrium distribution curve is concave upward. Desired absorption cannot be accomplished even in an infinitely long column, at solvent rates smaller than minimum solvent rate. It is obvious then that the solvent rate to be selected for an absorption operation should be higher than the minimum solvent rate. The selected solvent rate is generally expressed in terms of minimum solvent rate in the form of Ls = β (Ls)min. β , which is always greater than 1, is determined by considering the economy of the

Y Y

YN+1

slope=(Ls/Gs)min

Xo

D

T

D'D D1

T

XN (XN)maxXo XNX

D'

(XN)max= X ∗N

X

K

Y1

YN+1

Y1



slope=Ls/Gs

slope=(Ls/Gs)min

slope=Ls/Gs

Fig.4.20 Determination of Minimum Solvent Rate in gas absorption

(a) (b)


separation. The β value, which makes the total separation cost minimum, is known as optimum β value and the solvent rate corresponding to this β is also known as optimum solvent rate. The total cost is the sum of the fixed capital cost of the system, which includes either a stripping or a distillation column next to the absorption column, and the operating cost. For the determination of minimum solvent rate, the slope of operating line at minimum solvent rate conditions is written as;

( ) o.maxN

11N

mins

sXX

YYGL

−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ + (4-56)

As Gs, Xo, Y1 and YN+1 are all known, reading the (XN)max. value from the graph, (Ls)min is computed from equation (4-56). When the solutions are dilute, equation (4-56) is written as;

( ) o.maxN

11N

min xxyy

GL

−−

=⎟⎠⎞

⎜⎝⎛ + (4-57)

In the stripping operation, the flow rate and the inlet and outlet concentrations of the liquid phase are known and the designer selects the flow rate of the inert gas. So, in this operation, we face with the problem of finding minimum and optimum gas flow rates. First, point D is located with the known values on the diagram on which equilibrium distribution curve has already been plotted. Notice that in the stripping, operating line is below the equilibrium distribution curve and positions of points D and T are switched. Only the abscissa value of point T, which is Xo is known and its ordinate value Y1 is fixed by the quantity of gas selected. If the equilibrium relationship is linear or concave downward, the intersection point of the vertical from Xo with equilibrium distribution curve gives point T', by which the operating line at minimum gas condition is obtained as shown in Fig.4.21 (a). If the equilibrium distribution curve is concave upward as shown in Fig.4.21(b), the intersection point of the line drawn vertically up from Xo with the line drawn tangent to the curve from point D gives the point T'. The tangent is the operating line at minimum gas rate conditions. Note that only the first of so-determined (Y1)max. is the equilibrium value of

D D

T

T'

XN

K

Xo

Y1

XN Xo X X

Y1 T

T'

slope=(Ls/Gs)max

(Y1)max

(Y1)max

YN+1 YN+1

slope=(Ls/Gs)max

slope=(Ls/Gs)slope=(Ls/Gs)


a) b) Fig.4.21 Determination of minimum gas flow rate in desorption operation


YY


Xo and the slope of operating line at minimum gas rate is a maximum. The minimum gas flow rate is then calculated from the slope of so-obtained operating line, which is given in equation (4-58), by inserting the known values of Ls, Xo, XN ,YN+1 and (Y1)max.

( )No

1Nmax1

.maxs

sXX

YYGL

−−

=⎟⎟⎠

⎞

⎝

+⎜⎜⎛ (4-58)

Example-4.8 Determination of Minimum Solvent Rate (Strong Solution)

Calculate the minimum water rate and hence the β value selected for the absorption operation given in the Example-4.7.

Solution: After plotting the equilibrium curve

Xo=

Y on XY-diagram, a horizontal line from P = 800 mm Hg, t = 20 oC 0.26

0.02

Y1= 0.02

0.06

0.06

0.1

0.10 0.14

0.14

0.18

0.22

0.22

YN+1=0.25 is drawn until cutting the D'

T


YN+1= 0.25 equilibrium distribution curve (point D'), Labsis of which gives (XN)max = 0.206.

Then, from equation (4-56),

0.0206.002.025.0120)L( minS −

−=

= 134.0 k-mol water/h is found. As the selected water rate was 2 581 kg/h, the value of β is;

07.10.13418/2581

==β

∗1x

It follows from this that at the absorption operation in the Example-4.7 , only 7 % excess water from the minimum was used. 206.0)X( maxN = 0

X 0 0.18

Example-4.9) Determination of Minimum Solvent Rate (Dilute Solution)

Find the minimum solvent rate and the β value for the Example-4.5.

Solution: In this case solutions are dilute and equilibrium relationship is given as y*=2.2x. Hence, equation (4-57) after writing for packed column, can be used for the calculation of minimum solvent rate.

0136.02.203.0

2.2y

x 11 ===∗First, is calculated from

h/molk5.5510.00136.0

005.003.0300xxyyGL

21

21min −=

−−

=−−

=∗

Then, from equation (4-57), is obtained.


As in the Example-4.5 690 k-mol/h solvent was used, 25.1

5.551690

==

β is found.

This shows that 25 % excess of the solvent from the minimum was used in the Example-4.5.

Example-4.10) Estimation of Number of Real Plates Estimate the number of the real plates from O’Connell correlation for the absorption operation given in Example-4.7.

Solution:

As the slope of equilibrium curve changes, the geometric mean of the slopes calculated at the top and bottom conditions of the colum is taken. Equilibrium value of Y1= 0.02 is read from Fig.Example-4.7b as X1= 0.025

0244.0025.01

025.0X1

Xx

1

11 =

+=

+=Then, and 0196.0

02.0102.0

Y1Yy

1

11 =

+=

+=

Hence, the slope at the top condition: mT = y1 / x1= (0.0196) / (0.0244) = 0.803 Equilibrium value of X6 = 0.192 is read from the Fig.Example-4.7b as Y6 =0.23 Then, and Hence, the slope at the bottom condition: mD= y6/x6 = 0.187 / 0.161 =1.16 Average slope:; Liquid is water at the top and 15.3 % ammonia solution at the bottom. As average it can be considered as 7.7 % ammonia solution. So, ML, ρL, µL will be taken for 7.7% ammonia solution at 20 oC. From Perry’s Handbook , ρL = 965 kg/m3, but no value for µL is available. Hence take µL= µwater = 1 cP, and ML = Mwater = 18. Then, is obtained. = y6 / x6

With this value as absis from Fig.4.18 4 Eo = 0.48 is read. Hence; NReal= 13 is found.

m Example-4-11 Stripping in Plate Column

Propane will be stripped from a non-volatile oil by steam in a plate column operating at 138 oC and 2.5 bars absolute pressure. The flow rate of liquid to the column is 120 k-mol /h and it contains 18.1 mole percent propane. The stripped oil should not contain more than 1.96 mole percent propane. Steam flow rate to the column will be 1.25 times the minimum steam rate. a) Calculate the steam rate as kg/h, b) Find the composition of the gas phase leaving the column, c) Find the percentage recovery of the propane, d) Calculate the number of ideal plates needed, e) Estimate the number of real plates from O’Connell correlation,

187.023.01

23.0Yy === 161.0192.01

192.0Xx6 ===

X1 6

6

++Y1 6

66 ++

965.0)16.1)(803.0()m)(m(m DT ===

53

L

LL 10*8.1965

)10*1)(18)(965.0(mM −−

=ρµ

=

5.1248.6NN

o

IdealalRe ===

0E


Equilibrium relationship at 138 oC and 2.5 bars is given as : Y= 34 X, where Y and X are mole ratios in gas and liquid phases. The molecular weights of oil and propane are 300 and 44, and density and viscosity of oil at 138 o C are 900 kg/m3 and 0.5 cP . Solution:

y1A : propane, S : oil, C : steam G1,y1

As the liquid solution is not dilute ,work with mole ratios.

Gn+1,yn+1

N

2

1

L2,x2

L1,x1

Ln,xn

GN+1

y1

yN+1

G2,y2

LNxN= 0.0196

Lo=120 k-mol/hxo= 0.181

n

G3,y3

GSLN,xN

≈

≈≈

≈

22.0

181.01181.0

x1x

02.00196.01

0196.0x1

xXN

NN =

−=

−=

Xo

oo =

−=

−=

Y 0.0

LS=Lo(1-xo)=120(1-0.181)= 98.28 k-mol B/h 1N =+

o1 ===∗

Y 48.7)22.0)(34(X34

First, plot Equilibrium Line ( Y=34X) on X-Y coordinates with these in mind. Operating line equation,

D= y6 / x6 o

S

S1n

S

S1n X

GLYX

GLY −+=+

Then, a) is obtained. b) Operating line equation, Operating line equation written at the bottom of the column: Then; 0 = (29.87)(0.02)+Y1-6.57 from which Y1= 5.97 and or, as mass fraction; is obtained. c) Propane entered : Loxo = (120)(0.181) = 21.72 k-mol/h Propane recovered(stripped) : GS(Y1-YN+1) = 3.29(5.97- 0) =19.64 k-mol/h Then, percentage recovery of propane : (19.64 / 21.72)*100 = 90.4 %

)X(YYandGL)( ⎟

⎞⎜⎛

At G o1

maxS

SminS =⎟

⎠⎜⎝

∗

h/steammolk63.24.3728.98

4.37L)G( S

minS −===4.3702.022.00.048.7

XXYY

GL

No

1N1

maxS

S =−−

=−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

∗

h/steammolk29.3)63.2)(25.1()G(G minSS −===β

h/steamkg59(G N ==== )18)(29.3(M)29.3G B1S +&& 2.

)22.0(29.328.98YX

29.328.98Y 1n1n −+=+

57.6YX87.29Y 1n1n −+=+

57.6YX87.29Y 1N1N −+=+

857.097.51Y1

y ===97.5Y

1

11 ++

94.0)18)(857.01()44)(857.0(

)44)(857.0(M===

M)y1(Myy

yC1A1

A11 −+−+&


d) As point T( 0.22 ; 5..97) and D ( 0.02 ; 0.0 ) are known, the operating line is plotted by joinning T with D. Then, the number of the ideal plates is found by McCabe-Thiele method as shown in Fig.Example-4.11 as; 3.6

mm30mm9

6acab6N ideal =+=+=

Or, since both equilibrium and opcalculation of required number of Here; m = Y*/ X =34 Very close agreement between tw e) At the top of the column : Y X1=Y1/m= 5.97/ 34 = 0.176 and At the bottom of the column : xN YN =34 XN = (34)(0.02) =0.68 an

T'

D

2

1

3

6

T

Xo

XN YN+1

X1

Y

0.02

1.0

2.0

0.06

3.0

0.10 0.14

4.0

0.18

5.0

0.22

6.0

7.0

X

8.0 P = 2.5 bars t =138 oC

48.7Y1 =∗

Y=34X Y

Operating Line at slope =

maxs

s

GL

⎟⎟⎠

⎞

⎝⎜⎜⎛

Operating Line

X2X3

4

5

Y1=5.97

D

6

0.03

0.2

0.01

0.4

0.05

0.6

0.8

1.0

0 X

Y6

7 X6

b

0

a c

X7

XXlog

N⎢⎣

⎡

=

.0

.0logN

⎢⎣⎡

=

No part of this CD-book may be mul

Fig.Example-4.11

erating lines are straight, equation (4-55) can be used for the the ideal plates after replacing y and x with Y and X.

(4-55)

and

is obtained.

o methods is obvious.

1=5.97 or y1 = 0.857

mT = y1 /x1 = 0.857 / 0.15 = 5.71

= 0.0196 or XN = 0.02

d mD =yN /xN=0.405/ 0.0196 =20.66

( )

)/1(log

1m/Ym/Y

1NN

1No

α

⎥⎦

⎤α+α−

−−

+

+

879.034

87.29m

G/L SS ===α

( )13.6

)879.0/1(log

879.0879.0134/0.00234/0.022

=⎥⎦⎤+−

−−

15.176.

176.0X

1 ++0

01X1x 1

1 ===

405.068.01

0Y1

Yy N ===68.

N

N

++

tiplied for commercial purposes. E.Alpay & M.Demircioğlu 125

Average slope :

is CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 126

Then, From Fig. 4.18 Eo = 0.13 is read.

86.10)66.20)(71.5()m)(m(m DT ===

3

3 )(mM==

µ

L

LL 10*81.1900

10*5.0)(300)(86.10 −−

ρ Finally, is obtained. 5.48

13.03.6

ENN

o

IdealalRe === NReal = 49

4.5. Non-isothermal Absorption: In all the analysis above, existence of isothermal conditions are assumed. But, on the other hand it is very well known that absorption of gases into liquids releases heat. So, in order to have constant temperature in an absorption column, either the heat released must be removed or the heat of absorption and/or rate of absorption must be very low. Cooling in plate type columns is easier than in packed columns, as cooling coils can readily be attached to the plates as shown in Fig.4.22, and through which cooling liquid is circulated. In some other applications, the liquid solution is taken to the outside of the column from time to time and cooled in an external heat exchanger and returned back to the column. When the cooling is not applied and the heat of absorption is great, the rise of temperature in the column is inevitable. This causes reduction in the capacity of the column. In order to prevent capacity loss, higher liquid flow rates are used, which in turn, results in excessive dilution of the solution leaving the column. In the design of an adiabatically operating absorption column, not only mass balance but also enthalpy balance equation is needed and the solution can only be made by trial and error. It follows from the thermodynamics that molar specific enthalpy of a liquid solution at temperature tL (oC) is written as;

Fig.4.22 A bubble-cap plate containing cooling pipes

(4-59) sRLL H)tt(ch ∆+−=where tR is the reference temperature in (oC), Lc

sH

; is the specific heat of the liquid solution in (kJ/k-moloC), and calculated at the arithmetic mean of tL and tR temperatures and ∆ is the heat of mixing or integral heat of solution at given temperature and liquid composition in (kJ/k-molA). The specific heat of liquid solution is computed from the pure component’s specific heats, , LScLAc by using the following equation,

No part of th

(4-60) LSLAL c)x1(cxc −+=When this equation is substituted into equation (4-59),

[ ] +−−+= )tt(c)x1(cx RLLSLA sHh ∆ (4-61) is obtained. Similarly, the molar specific enthalpy of a binary gas mixture of A+C at temperature tG is given by

[ ] [ ] sRGGCARGGA H)tt(c)y1()tt(cyH ∆+−−+λ+−=

Aλ

0A =λ

11NN1N1Noo HGhLHGhL +=+ ++

1n1nNN1N1Nnn HGhLHGhL ++++ +=+

Aλ sH∆

1NH +

(4-62) where, is the latent heat of vaporization of component A at reference temperature in (kJ/k-mol A), which is taken into consideration when component A is liquid at reference temperature, otherwise . If total material, solute A and enthalpy balances are written for a plate type absorption column shown in Fig.4.23; Lo + GN+1 = LN + G1 (4-63) Lo xo + GN+1 yN+1 = LN xN + G1y1 (4-64)

(4-65) If the same balances are written between column bottom and any plate n in the column. Ln + GN+1 = LN + Gn+1 (4-66) Ln xn + GN+1 yN+1 = LN xN + Gn+1yn+1 (4-67)

(4-68) are obtained. The knowns at the start of calculations are: GN+1, yN+1 , tG,N+1 , y1 , Lo , xo , tLo . In addition to these

t1 1

2

n

N

Ln,hn,xn

G1,H1,y1,tG1

GN+1HN+1 yN+1tG,N+1

tN

GN, HN, yN

Gn+1,yn+1

Lo,hO xo,tLo

, , cLA=f(t), cLS =f(t), cGA = f(t), cGC =f(t) and y*= m x, m= f(t) must be found from the relevant literature. Then, the calculations are started at the bottom of the column. First, from equation (4-61) ho and from equation (4-62) are calculated. Then, from Gs = G1(1-y1) = GN+1(1-yN+1) G1, Gs and from equations (4-63) and (4-64) LN and xN are calculated. Now an assumption for the temperature of the first plate of the column t1 is made (it is obvious that tL1=tG1=t1). As t1 is now known, H1 from equation (4-62) and later hN from equation (4-65) are computed. Hence, all the streams around the column and their compositions and specific enthalpies have been determined. From equation (4-61) tLN and at this tLN mN are found. Then, from yN=mN xN and GN =Gs/(1-yN) yN, GN and from equation (4-62) HN are computed. Hence the gas phase leaving plate N, has been specified. By taking n=N-1 we come to the plate above. First, from equation (4-66) LN-1 , from equation (4-67) xN-1 , from equation (4-68) hN-1 and from equation (4-61) tN-1 are computed. Then, at this tN-1 mN-1 is read. From yN-1=mN-1 xN-1 and GN-1=Gs/(1-yN-1) yN-1 and GN-1 , and then from equation (4-62) HN-1 are calculated. By taking n=N-2 from equation (4-66) LN-2 , from equation (4-67) xN-2 , and from equation (4-68) hN-2 are computed. Calculation is continued in this way until y1 value is reached. So-obtained t1 is then compared with the assumed t1. If the difference between assumed and calculated values is acceptable,

LN,hN,xN,tLN

Fig.4.23


ythe calculations and hence the solution are correct. If the difference is great, the calculations are repeated with a new t1 value. Repeating is continued until the difference is reasonable. As it is seen, calculations require trial and error and in many cases this is very lengthy. A short method giving reasonably accurate result has been developed by assuming that the heat liberated in the liquid phase upon the absorption, remains in the liquid. Although with this assumption, the temperature of the liquid hence, the number of the ideal plates (or height of packing) required is found higher than the actual values, this over-design in the number of ideal plates (or in the height of packing) can be taken as design security. If an enthalpy balance is written for a dZ height in the column, assuming L and G are constant; G dHG + ∆Hs L dx = L dhL (4-69) or G cG dtG + ∆Hs L dx = L cL dtL (4-70) is obtained. with the assumption of ∆Hs L dx > G cG dtG , ∫∫ =

2LtLL

2xs dtcdxHLtx

∆ and from here,

∆Hs (x-x2) = )tt(c 2LLL − (4-71) is found, where point 2 represents top of the column. If absorbed solute A is a liquid at reference temperature, ∆Hs then includes the latent heat of vaporization. With the help of this equation, the equilibrium distribution diagram of the adiabatic absorption column is plotted as follows: equilibrium distribution lines at various tL temperatures which are greater than tL2, are drawn on a millimetric paper. At the same time x values corresponding to these temperatures are calculated from equation (4-71). The calculated x values are then marked on the corresponding equilibrium distribution lines and by joining these points, the equilibrium distribution curve at adiabatic condition is obtained as shown in Fig.4.24. This equilibrium curve can now be used at the absorption of solute A in an adiabatically operating packed or plate column.

)Example-4.12 Non-Isothermal Gas Absorption The concentration of ammonia in a gas stream, consisting of ammonia and air is to be reduced from 5 to 0.5 % by volume, by contacting it with water, which is ammonia-free and at 20 oC, in an absorption column at 1 bar pressure.. The molar heat of absoption is 37 544 kJ/k-mole of ammonia absorbed and molar specific heat of liquid is 75.3 kJ/k-mole oC . Over the range of operation, the equilibrium relationships at various temperatures are linear and given as y*= m x, where y and x are the mole fractions of ammonia in gas and liquid respectively. The slope m changes with temperature as follows: Temperature(oC) 20 25 30 35 m 0.73 0.96 1.23 1.55

tL2

tL3

tL4

tL5

tL1Adiabatic equilibrium curve

P=cons.

xx2 x3 x4 x5 x1

Fig.4.24 Equilibrium distribution curve


a) Draw the equilibrium curve for non-isothermal (adiabatic) absorption b) Calculate the minimum solvent rate for a gas rate of 50 k-mol/h, c) Draw the operating line for a solvent rate which is 1.15 times the minimum rate, d) Explain, what would be, if isothermal absorption at 20 oC were assumed ?

Solution:

a) The short method can be used. Hence, from equation (4-71),

is written. )20t(002.0)20t(

544375.730.0)tt(

Hc

xx LL2LL

s

L2 −=−+=−+=

∆ If the x values are calculated at the given temperatures:

tL 20 25 30 35

x 0.0 0.01 0.02 0.03

is obtained. After drawing y*= m x lines at the given temperatures, corresponding x values are marked on these lines and by joining them, the equilibrium distribution curve at adyabatic condition is obtained as shown in Fig.Example-4.12. b) For dilute solutions operating line is given by the equation (4-12). At minimum liquid flow rate this is written as,

2min21min1 x)GL(yx)

GL(y −+= ∗

From which, is obtained. is read from

21

21

min xxyy

GL

−−

=⎟⎠⎞

⎜⎝⎛

∗ x1 =∗ 032.0

the diagram by drawing a horizontal line from y1 = 0.05 until cutting the adiabatic equilibrium curve. Then, is found. h/molk700050L −= c) L = 1.15 (70.31) = 80.86 k-mol/h

31.0.0032.0

005.05.min −

−=

Then, operating line, )0.0(

5086.80005.0x

5086.80y −+=

y = 1.62 x + 0.005 By writing operating line at the bottom of the column, is found. 028.0

62.1005.005.0

62.1005.0yx 1

1 =−

=−

= With the help of this, point D is located and by joining D with T operating line is drawn. d) Extend the horizontal line from y1= 0.05 until cutting the equilibrium line drawn at 20 oC and read x1 =

∗ 069.0 Then, minimum solvent rate and operating solvent h/molkL −= 61.32

0.0069.0005.005.050min −

−=

rate, L = 1.15 (32.61) = 37.5 k-mol/h are obtained. Hence, operating line is: y = 0.75 x + 0.005 and x1= 0.06. With the help of this value, point D is located and the operating line for isothermal absorption at 20 oC condition is drawn by joining D with T. As it is seen from the figure, this line cuts the adiabatic equilibrium curve at point K, whose temperature is about 28 oC and x value is 0.014. This means that if -with isotermal absorption at 20 oC assumption- a solvent rate of 37.5 k-mol/h is


selected, absorption stopes when the temperature of the liquid reaches 28 oC at which time mol fraction of ammonia in the liquid is 0.014, much below the 0.028 value.

Adiabatic equilibrium curve

4.6 Gas Absorption with Chemical Reaction : In absorption operations as the concentration of solute A in the liquid phase increases, rate of absorption of A slows down due to the equilibrium back pressure of solute. This results in an increase in the size of absorber or else excessive amount of solvent must be used. In order to reduce this equilibrium back pressure and hence to keep the rate of absorption at a reasonable level, the absorbed solute A may be reacted with a reactant B added into the solvent S, according to the stoichiometry given below, to form a product P. aA + bB pP (in solvent S) Reaction of absorbed solute A with reactant B may enhance the rate of absorption considerably. But, if solute A is to be recovered, the above given reaction must be reversible. For example reaction between ammonia and aqueous sulphuric acid proceeds toward the right at low temperature, so that back pressure of ammonia is almost zero.

2 NH3 + H2SO4 (NH4)2SO4 But at elevated temperature, reaction drives to the left giving ammonia and sulphuric acid back (regeneration).

y1= 0.05

T

0 0.01 0.02 0.03 0.04 0.05

Mole fraction of ammonia in liquid,x

30 oC

20 oC 25 oC

35 oC

x2 = 0

0.04

0.02

0.06

Mol

e fr

actio

n of

am

mon

ia in

gas

,y

y2 =0.005

D

x1=0.028= x1=0.06= 032.0x1 =∗

D

Operating Line

Operating Line for Isothermal Absorption at 20 oC

D' D'

P=1 bar

D

K

x =0.014=

0.06= 0.07=

069.0x1 =∗

Fig.Example-4.12


In some other cases, the production of component P may be the main objective of the process such as the absorption of NO2 in water for the production of HNO3 or absorption of SO3 in H2SO4 for the manufacture of Oleum. In some cases, absorbed solute A may react directly with solvent S. An example from chemical industry to this is the absorption of maleic anhydride vapor from air into water. Maleic anhydride upon absorption reacts with part of the water to form maleic acid which can dissolve in water. Overall rate of absorption of solute A is then influenced by both the mass transfer resistance and the rate of chemical reaction, both in the liquid phase. From the values of these two, overall rate is determined. In some cases, chemical reaction may be very rapid (fast reaction) so that rate of mass transfer controls the overall rate of absorption. In some other cases, rate of reaction is very small (slow reaction) and hence it controls the overall absorption rate. Between these two extremes, rate of mass transfer is comparable with the rate of chemical reaction and both must be considered in computing the overall rate of absorption, which is the most difficult case to analyze. In the analysis of gas absorption with chemical reaction, film or penetration theory may be taken as model. In many analysis results obtained from these two theories are very close to each other. Hence, in the analysis below the validity of film theory due to its simplicity is to be assumed. Typical concentration profiles of solute A in gas and liquid phases when the chemical reaction taking place is slow are shown in Fig.4.25., along with the profiles at pure physical absorption. As it is seen -due to the slow reaction-the absorbed solute A can not be depleted completly in the liquid film with the reaction and concentration profile of it in the liquid phase is no longer linear.

But when the reaction is fast enough, absorbed solute A is completly consumed in the liquid film with the reaction and the concentration profile of it along with that of the rectant B in the liquid may be depicted as in Fig.4.26.

Fig.4.25 Concentration profiles in two phase at physical absorption and at absorption with chemical reaction

oAc

oAic

pA

pAi

oAip

Liquid filmGas film

G a s p h a s e L i q u i d p h a s e

cAi

cA

I

Mol

ar c

once

ntra

tion

of so

lute

A

Parti

al p

ress

ure

of so

lute

A

A + C A

A+B+P+S

0 zLzG

: Absorption+chemical reaction: Physical absorption


In the part of the liquid film to the left of reaction zone only solute A diffuses from interface toward reaction zone, and in the part of the liquid film right to the reaction zone only reactant B diffuses from bulk liquid toward reaction zone. In the reaction zone itself diffusions of both components and chemical reaction take place simultaneously.

pA

pAi

Liquid filmGas film


cAi

cB

I

Parti

al p

ress

ure

of so

lute

A

Mol

ar c

once

ntra

tions

0 zLzG

Fig.4.26 Concentration profiles in two phase at absorption+fast chemical reaction in liquid phase

B

Reaction zone

AA + C

One extreme case of fast reaction is the instantaneous reaction. In this case reaction zone reduces to a reaction plane as shown in Fig.4.27. As there is no simultaneity between reaction and diffusion, this case is the most easy to analyze

pA

pAi

Liquid filmGas film


cAi

I

AA + C

zG 0 zLzR

B

Reaction plane

Parti

al p

ress

ure

of so

lute

A

cB

Mol

ar c

once

ntra

tions

Fig.4.27 Concentration profiles in two phase at absorption with instantaneous chemical

reaction in liquid phase


4.6.1 Gas Absorption with Instantaneous Chemical Reaction: From film theory representation; for the diffusion of solute A in the part of the liquid film shown in Fig.4.27:

)0c(zD

N Ai

R

ASA −= (4-72)

can be written. By multiplying right hand side with zL/zL and remembering kL=DAS/zL from film theory, this equation becomes;

Ai

R

LLAi

R

L

L

ASA c

zz

kczz

.z

DN == (4-73)

Similarly, for the diffusion of reactant B in the other part of the liquid film:

)z/z(1c

.DD

k)z/z(1

c.

zD

.DD

)0c(zz

DN

LR

B

AS

BSL

LR

B

L

AS

AS

BSB

RL

BSB −

−=−

−=−−

−= (4-74)

On the other hand, from the stoichiometry of the reaction, bNA +aNB = 0 (4-75) can be written. If, first zL/zR is solved from equation(4-73) and substituted in equation(4-74) and then so-obtained NB is substituted into equation (4-75) finally,

)cDD

.bac(kN B

AS

BSAiLA += (4-76)

is found. [At physical absorption this was : NA=kL(cAi-cA)] With the elimination of cAi, equation(4-76) can be written in more appropriate form as follows: At steady-state, in the gas phase; NA=kG(pA-pAi) (4-77) can be written. Assuming that Henry’s law equation is valid for the gas-liquid equilibrium,

pAi= ΉA cAi (4-78) If pAi is eliminated between last two equations;

)kN

p(1cG

AAAi −= (4-79)

By substituting this into equation (4-76); ΗA

⎥⎦

⎤+− B

AS

BS

G

AA c

DD

.ba)

kN

p⎢⎣

⎡=

1kN LA ( (4-80)

LG

L

B

AS

BSA

kkk

cp=

GkDD

ba

+

+

ΗA

ΗA

ΗA

On the other hand,


LGG kk1

K1

=+=

If this is substituted into equation above finall

( += pKN AGA ΗA

is obtained. Position of reaction plane in the liquid film cB. If the first three terms are kept constant, zR ch[which is known as critical value and shown wwhich means that interface becomes reaction are to be as shown in Fig.4.28. As it is seen from the Figure, at cB ≥ (cB)crit. eq

NA=kG pA So, it follows from these that, for the calculati

when, cB < (cB)crit equation(4-81when, cB ≥ (cB)crit equation(4-82)

(cB)crit value for any given pA is found as folloFrom Fig.4.26, at or above (cB)crit ,

L

B zD

N −=AS

BS

L

AS.critB

BS

DD

.z

D)c( −=

is obtained. If this and NA from equation(4-82 b kG pA - a kL(DBS / DAS)(cB)crit.

A

L

G

BS

AS.critB p

kk

.DD

.ab)c( =

pA LiquidGas film

G a s p h a s e L i

Parti

al p

ress

ure

of so

lute

A

0

Reap

A

I

ΗA

zR

A + C

zG

Fig.4.28 Concentration profiles in two phase at reaction when cB

No part of this CD-book may be multiplied for commercial

Η

LG

LG

kkkk +

y,

)B

AS

BS cDD

.ba (4-81)

is dependant on (DAS/DBS), (a/b), pA and

anges only with cB. There is a value of cB ith (cB)crit.] that at or above which zR =0,

plane. Concentration profiles in this case

uation(4-77) becomes, (4-82)

on of total flux, ), is used . ws;

critB

AS

BSL.critB )c(

DD

k)c( −= (4-83)

) are substituted into equation(4-75) = 0 from which,

(4-84)

film

q u i d p h a s e

(cB)crit.

Mol

ar c

once

ntra

tions

zL

ction lane

B

A

absorption with instantaneous chemical ≥ (cB)crit.

purposes. E.Alpay & M.Demircioğlu 134

is found. Or for a given value of cB; (pA)crit. is:

BLBS c

k.

D

GAS

.critA kD.

ba)p( = (4-84b)

4.6.2 Calculation of Height of Packing when Chemical Reaction is Instantaneous:: With the assumption of dilute solutions, solute A balance along dh differential height in the packed section;

-G(dpA/P) = NA dSm (4-85) = (a/b)LvdcB (4-85b)

where, G is total molar flow rate(k-mol/s) of gas and Lv is volumetric flow rate(m3/s) of liquid, both are constant. By defining a mass transfer area per unit volume of packing with av(m2/m3), total mass transfer area in the differential volume becomes, dSm= avAcdh where Ac(m2) is the cross-sectional area of the empty colum. Upon substitution of this into equation(4-85) and integrating from bottom to the top of the packing;

⎮⌡⌠== ∫

1Ap

2Ap A

A

cv

z

o Ndp

PAaGdhZ (4-86)

Fig.4.29 is obtained.

1

2

dh

h=0

h=z

Lv

cB2

G

Lv

G pA2

NA

cB pA

pA1

cB1

pA

pAx

pA1

pA2

cB1 cB2cBxcB

Case-1

Case-2

Case-3

(cB1)crit.(cB2)crit.

1

1

2

12

2

(pA)crit.= f(cB)

Fig.4.30 Stoichiometric operating lines for three different cases


Depending upon the values of cB1 and cB2 , three different cases are encountered at which NAs to be substituted into equation(4-86) are different. These cases are shown schematically in Fig.4.230. Case-1) cB1 ≥ (cB1)crit and cB2 > (cB2)crit In this case NA=kGpA and then,

2A

1AG

1Ap

2Ap A

A

Gcv pp

lnHp

dpkPAa

GZ =⎮⌡⌠= (4-87)

cvG

G PAakGH =is obtained. Where (m) is height of one gas transfer unit.

Case-2) ) cB1 < (cB1)crit and cB2 ≤ (cB2)crit In this case NA value must be substituted from equation(4-81).

⎮⌡ +

2Ap A

OG

BBS pc

DD

.ba

⎮⌠

=⎮⎮

⌡

⌠

+=

1Ap

B

AS

BS

A

1Ap

2ApAS

A

A

Gcv cDD

.ba

dpH

p

dpKPAa

GZ (4-88) ΗAΗA

cvG

OG PAaKH =

G (m) is the height of one overall gas transfer unit. is found. Where,

By integrating equation(4-85b) between bottom and any section of the column, -G(pA1-pA)/P = (a/b)Lv(cB1-cB) from which, cB = cB1+(b/a)(G/LvP)(pA1-pA) (4-89) is obtained. This equation is known as “stoichiometric operating line equation” If this cB is substituted into equation (4-88) and integrated;

(pA + ΗA 1B

AS

BS )cD

.b

Da

(pA + ΗA 2B

AS

BS )cDD

.ba

is found. Where )G/L( v=α is absorption factor.

Case-3) cB1 < (cB1)crit and cB2 > (cB2)crit In this case between top and any level(level-x) the solution of case-1 and between this level and bottom of the column solution of case-2 apply. Hence;

(pA + ΗA 1BBS )c

DD

.ba

ln

DD

.1α

−1H

Z

AS

BS

OG= (4-90)

ΗA / P

OGHp

No pa
AS
(pA + ΗA xB

AS

BS )cDD

.ba

ln

D.1

α−

D1plnHZ

AS

BS2A

AxG += (4-91)

rt of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 136

pAx and cBx are found by first writing equations (4-84) and(4-89) at level-x and then solving;

L

G

BS

ASvAx

kk

.DD

.G

PL1

Gbp+

=1B

v1A c

PL.ap +

(4-92)

GPL

kk

.DD

c

ac

v

G

L

AS

BS

1B

Bx

+= G

PL.

bap

.bv

1A + (4-93)

Example-4.13) Calculation of Height of Packing when Reaction is Instantaneous

An ammonia-air gaseous mixture is to be washed with aqueous sulfuric acid solution containing 0.25 k-mol acid per m3 of solution, in a packed column operating counter-currently at 1 bar and 20 oC, to reduce the partial pressure of ammonia from 0.05 bar to 0.001 bar. The flow rates of gas and liquid to the column are 150 k-mol/h and 25 m3/h respectively and cross-sectional area of the column is 1 m2. An instantaneous reaction between ammonia and sulfuric acid in the liquid phase takes place according to the stoichiometry below:

2 NH3 + H2SO4 (NH4)2SO4 At the operating conditions, volumetric mass transfer coefficients are given as: kGav=50 k-mol/m3h bar and kLav = 2.0 h-1 , and the value of Henry’s law constant is 0.73 bar. Calculate the required height of packing. Solution: pA2=0.001 bar

GFirst find cB1 from equations(4-85) and (4-85b); -(150)(0.05-0.001)/1= (2/1)(25)(cB1-0.25) cB1= 0.103 k-mol/m3

Molecular diffusivity of ammonia in water from Table.1.4, DAS=2.3*10-9 m2/s. The molecular diffusivity of sulphuric acid in water is taken from Example-1.8) DBS= 2.55*10-9 m2/s. Now, we can find which case is applicable for the solution. From equation(4-84), (cB)crit. =(1/2)(2.3 /2.55)(50/2) pA =11.3 pA Then, at column bottom; (cB1)crit. = (11.3)(0.05)= 0.565 k-mol/m3

Hence, (cB1)crit. > cB1 At column top ; (cB2)crit. = (11.3)(0.001) = 0.0113 k-mol/m3

Hence, (cB2)crit. < cB2 It follows from here that case-3 is applicable. Then, from equation (4-92);

bar0177.0)2/50)(55.2/3.2](150/)1*25[(1

)103.0](150/)1*25)[(1/2(05.0pAx =+

+=

From equation (4-93);

Lv=25 m3/h

1

2cB2=0.25 k-mol/m3

z

G=150 k-pA1=0.05 bar

Lv cB1


3

Bx m/molk20.0]150/)1*25[()50/2)(3.2/55.2()103.0](150/)1*25)[(1/2(05.0)2/1(c −=

++

=

As the solution is dilute in ammonia and density of water at 20 oC is 998 kg/ m3 c=998/18=55.44 k-mol/m3 and from equation(4-3)ΉA = Ή’A /c=(0.73)/(55.44) = 0.0132 bar m3/k-mol.

molk/mhbar0266.02

0132.0501

aK1 3

vG

−=+=m3=)1)(1)(50(

150H G =

m99.3)1)(1(

)0266.0)(150(PA c

==aKGH

vG

OG =

63.121/0132.0

150/25==hence; α

Substituting all these into equation (4-91);

[ ])20.0)(3.2/55.2).(1/2)(0132.0(0177.0

)103.0)(1/1)(3.2/55.2)(1/2)(0132.0(05.0ln)3.2/55.2.(

63.1211

99.3001.0

0177.0ln)3(Z+

+

−+=

∗A

Z = 8.62 + 3.55 = 12.17 m is found. 4.6.3 Calculation of Height of Packing when Chemical Reaction is Slow: When the chemical reaction is slow, only part of the absorbed solute A reacts with reactant B and G

1

2

dh

h=0

h=Z

Lv

cB2

G

Lv

pA

pA1

cB

NA

ZP

ZR cA

cA1

cA2=0

cB1

pA2unreacted part of it remains in the leaving solution. Again, with dilute solutions assumption; along dh differential height in the packed section: -G(dpA/P) = NA dSm = NA av Ac dh (4-94) = consumption of A by chemical reaction + increase in content of A in bulk liquid = -RAHoAc dh - Lv dcA (4-94b) = (a/b) Lv dcB - Lv dcA (4-94c) can be written.Where Ho (m3 liquid /m3 packing) is the specific hold-up. Considering gas phase; NA =KG (pA- p )= KG (pA - ΗA cA) (4-95) can be written.After substituting this into equation (4-94) and integrating from bottom to the top of the column;

⎮⌡⌠=⎮⌡

⌠−

== ∫1Ap

2Ap AA

AOG

1Ap

2Ap AA

A

CvG

z

0 cpdp

Hcp

dpPAaK

GdhZ (4-96) −ΗA ΗA

is obtained. Note that in this case; cA = f (pA,cB) [ Remember: in physical absorption, cA = f(pA)]


In order to perform the integral, the following model may be assumed. Since incoming liquid does not contain any solute A and the chemical reaction in the liquid phase is directly proportional to the concentration of solute A, a zone at the top part of the column may be assumed in which only physical absorption of solute A occurs, followed by another zone in which-due to the sufficient concentration build-up of solute A-chemical reaction is more significant than the absorption. Hence, in the physical absorption zone from equation(4-94) and (4-94b) with the assumption of Lv dcA ≫ RAHoAc dh ; -G(dpA/P) = Lv dcA and from the integration of this between the top and any level within the physical absorption zone with cA2 = 0.

)pp(PL 2AA

v

−GcA = (4-97)

is found. Substituting this into equation (4-96) and taking pA1= pAx (partial pressure of solute A at the end of physical absorption zone), height of physical absorption zone;

⎮⌡⌠

+−=

Axp

AOGP /p)/11(p

dpHZ

αα

2Ap 2AA

⎥⎦

⎤⎢⎣

⎡+−

−=

αα

α1

pp

)/11(ln/11

HZ

2A

AxOGP (4-98)

is obtained. In the chemical reaction zone, from equations(4-94) and (4-94b) with the assumption of RAHoAc dh ≫ Lv dcA ; NA av Ac dh = - RAHoAc dh and from which, NA =-RAHo / av is found. Substituting this into equation (4-95); (pA - ΗA cA)= -RAHo/KG av (4-99) and this into equation (4-96) and taking pA2 = pAx ; the height of chemical reaction zone;

⎮⌡⌠−=

1Ap

Axp A

A

o

vGOGR R

dpH

aKHZ (4-100)

is obtained. Then, total height of the packing becomes; Z = ZP + ZR (4-101) For each specific reaction rate equation, there are special solutions of equations(4-98) and (4-100). Special case-1) Reaction between A and B is 0th order with respect to A; RA= - ko Then from equation(4-100);

)pp(H

kdp

HaK

HZ Ax1A

o

OG1Ap

Axp o

A

o

vGOGR −=⎮⌡

⌠=β

(4-102)


is found. Where, vGooo aK/Hk=βAt h=ZR equations(4-97) and(4-99) become;

)pp(PL

Gc 2AAx

v

Ax −=

(pAx - ΗA cAx)= koHo/KG av =βo

ααβ

/11/p

p 2AoAx −

−=From these two; is obtained. Substituting this into

equations(4-98) and (4-102);

)p/ 2Aoln(/11

HZ OG

p βα−

= (4-103)

[ ]o2A1A

o

OGR )/p(p)/11(

)/11(H

Z βααβα

−+−−

= (4-104)

are found. Special case-2) Reaction between A and B is first order with respect to A; RA= - k1cA . Then from equation(4-100)

(4-105) ΗA

⎮⌠=⎮

⌠ 1Ap

AOG1Ap

AvGOG

dpHdpaKH

β ⌡⌡=

Axp A1Axp A1o

R cckHZ

Where, From equation(4-99) by noting the value of β1 ;

1

AA 1

pβ+

= is found. After substitution into equation(4-105) and integration; c

ΗA vL

o1

vG

o11 aK

HkaK

Hk==β

ΗA

Ax

1A

1

1OGR p

pln

1HZ

ββ+

= (4-106)

is obtained. At h = ZR equations(4-97) and(4-99) become;

)pp(PL

Gc 2AAx

v

Ax −=

(pAx - ΗA cAx)= k1 cA Ho/KG av = cAx β1 ΗA

)1/(1p

p1

2AAx βα +−=From these two; is obtained.. Substituting this into

equations(4-98) and (4-106);

αββ

α −+−=

1

1OGp 1

ln/11

ZH

(4-107)


⎥⎦

⎤⎢⎡

+−

+=

2A

1AOG

1R p

p)

11(lnH

1Z

βα

ββ

(4-108) ⎣ 11

are found. Special case-3) Reaction between A and B is first order with respect to A and B , RA= - k2cAcB. In this case solution is done by stepwise integration. Hence equations (4- 94) must be written as finite difference:. -G(dpA/P) = KG(pA- cA)av. avAc∆h (4-109) = -RAHoAc ∆h - Lv ∆cA (4-109b)

ΗA

= (a/b) Lv ∆cB - Lv ∆cA (4-109c) These equations can be written for ∆hj height which is bounded with the planes j and j-1, as: G(pAj-pAj-1)/P = KG(pA- cA)av. avAc∆hj (4-110) = k2 (cAcB)av.HoAc∆hj+Lv(cAj-cAj-1) (4-110b) =(a/b)Lv(cBj-1-cBj) + Lv(cAj-cAj-1) (4-110c) Calculations may be started at the top of the column, where pAj-1, cAj-1 and cBj-1 are all known. By selecting a pAj value, as a first approximation, (pA- cA)av. = (pA- cA)j-1 and (cAcB)av. = (cAcB)j-1 can be written. Then, from equation (4-110); ∆hj = HOG(pAj-pAj-1)/(pA- cA)j-1 (4-111) From equation(4-110b), cAj = cAj-1 + (G/LvP)(pAj-pAj-1)-k2(cAcB)j-1 HoAc∆hj/Lv (4-111b) From equation(4-110c), cBj = cBj-1 –(b/a)(G/LvP)(pAj-pAj-1)+(b/a)(cAj-cAj-1) (4-111c) As a second approximation; (pA- cA)av. = (1/2)[(pA- cA)j + (pA- cA)j-1] (4-112) (cAcB)av. =(1/2)[(cAcB)j + (cAcB)j-1] (4-112b) with these values ∆hj, cAj and cBj are re-calculated from equations (4-111). These are repeated until practically no difference between the last and previous calculations is observed. After then, to the second step of the calculations is leaped. The steps are continued until reaching the bottom of the column. Summation of calculated ∆hj s gives the required height of packing. 4.6.4 Calculation of Height of Packing when Rates of Diffusion and Chemical Reaction are comparable: In this case, basic equation with Film theory representation becomes,

0R A2A

2

AS =+dz

cdD (4-113)

This equation must be solved with the following boundary conditions:

∆hj

Lv

cB2

G

Lv

G pA2

cA2=0

pAj

pA1

cBj-1

cAj-1 pAj-1

cBj

cAj

ΗA

ΗA ΗAcB1 cA1

ΗA

ΗA ΗA ΗA


1o) at z = 0 cA =cAi = constant 2o) at z = zL cA = cAL where cAL is the concentration of solute A in the bulk liquid. Special case-1) Reaction is 0th order with respect to solute A, RA=-ko

0Dk

dzcd

AS

o2A

2

=−Then, (4-114)

Double integration of this equation results in

212

AS

oA IzIz

D2k

c ++= (4-115)

Integration constants I1 and I2 are evaluated from the boundary conditions as;

L

and noting that, kL =DAS/zL

Ai2LASoAL

1 zcz)D2/k(c

I−−

= and I2 = cAi Substituting these into equation(4-115)

2

AS

o

L

oALAi

AS

LAiA z

D2k

zk2k

)cc(Dk

cc +⎥⎦

⎤⎢⎣

⎡+−−= (4-116)

is obtained. Total molar flux of solute A crossing the interface becomes;

L

oASALAiL

0z

AASAi k2

kD)cc(k

dzdc

DN +−=⎟⎠⎞

⎜⎝⎛−=

=

(4-117)

Or, if cAi is solved from NAi=kG(pA- cAi) and substituted into equation above; ΗA

⎥⎦

⎤⎢⎣

⎡−−= )

k2kD

cpKN2L

oASALAGAi (4-118) (ΗA

is found. Calculation of the height of packing in this case is accomplished by substituting the above given value for NA in equation (4-110) and then conducting stepwise integration as explained in the previous section. Special case-2) Reaction is 1st order with respect to solute A, RA=-k1cA Then, from equation(4-113);

0cDk

dzcd

A

AS

12A

2

=−

qz2

qz1A eIeI −+=

(4-119)

This equation must be solved with the boundary conditions given above. By taking q2=k1/DAS this equation can readily be integrated. Result is: c (4-120) Integration constants can be evaluated by applying the boundary conditions. These are:

)qzsinh(cec)qz

L

ALLqz

AiL +−2

sinh(c2I Ai

1 = )qzsinh(2cec

L

ALLqz

Ai2

−= I

By substituting these into equation (4-120),


)qzsinh()]zz(qsinh[c)qzsinh(c

cL

LAiAL

A

−+= (4-121)

is obtained. Total molar flux crossing the interface;

)cosh

cc()k()

coshc

dzdc

(DN ALAiRL

ALASAi ββ

−=−−= (4-122) c(Hak) AiL0zA ==

is found. Where β=qzL , Ha is Hatta number defined by Ha=β/tghβ and (kL)R is liquid phase mass transfer coefficient when chemical reaction exist and defined by (kL)R=kLHa . Eliminating cAi with the help of NAi=kG(pA- cAi), equation (4-122) can also be written as;

ΗA

⎟⎟⎠

⎞⎜⎜⎝

⎛− ALARG c

coshp)

βΗA (4-123) =Ai K(N

where,

Hakk)k(k)K( LGRLGRG

+=+=111 ΗA ΗA

(4-124)

Criteria for fast and slow reaction;

2ecosh

2

12

)2/1()2/1(2

e 22 βββββββ

+=β+−−++

=+ −

=

2/1)2)2/1()2/1(

eeetgh

2

22

β/1()2/1(e 22

βββββββββ

ββ

ββ

+=

β +−++++−−++

=−

=−

+ −

21

tgh

2ββ

β+==Then, Ha is found.

The reaction is slow if β2/2≪1 (or β ≤ 0.2). Then, as Ha =1 and. cosh β=1, equation (4-122) becomes; NAi = kL (cAi-cAL) This is the flux equation for physical absorption. As expected, in this case contribution of chemical reaction on the absorption is negligible.Height of packing is calculated by using equations given in section 4.6.3. The reaction is fast if β is great (β≫3). So, in this case;

22eee

2eecosh

1βββββ

β =+

=+

=−

0e2

cosh1

≈=ββ

and

1ee

=+

=−ββ

βeetgh − −ββ

and Ha = β

And flux equation in this case from equations(4-122) and(4-123); NAi= kL βcAi = (KG)R pA (4-125) Where,


AS1GRG Dkk1

)K(1

+=ΗA

(4-126)

For the calculation of height of packing; -G dpA/P = (KG)R

avAcpAdh

2Ap1A

ROG

pln)H(Z =can be written. From this (4-127)

is obtained. Where;

RL

v

GROG )H(L

H)H( +=G (4-128)

and AS1c

v

DkAL

vcvRL

vRL aAa)k(

L)H( == (4-129)

In Fig.4.31 conditions for slow and fast reaction are summarized.

L

RL )k/(11=

k/1Ha

Fig.4.31 Values of Ha number and β at slow and fast reaction

Problems

4.1 Plot equilibrium curves of ammonia(A)-air(C)-water (S) system at 20 oC and 800 mmHg in the concentration units shown in the table below:

logβ 0 0.2

0.5

30

1.0

10

Slow reaction

Fast reaction

Diffusion and reaction are comparable

Mass ratio of AMass ratio of A

Mole ratio of AMole ratio of A

Mass fraction of AMass fraction of A

Mole fraction of AMole fraction of A

Mass fraction of APartial pressure of A (mm Hg)

Mole fraction of APartial pressure of A (mm Hg)

Molar concentration (k-mol A/ m3solution) Partial pressure of A (mm Hg)

Concentration unit in the liquid Concentration unit in the gas

ΗA


Solubilities of ammonia in water are given in the Table. App. 4.1. Densities of aqueous ammonia solutions are listed in Perry’s Handbook. 4.2 With the help of Table.4.1, compare the solubilities of CO2 and H2S in water at 1 bar and 20 oC. 4.3 A 20 k-mol/h binary gas mixture containing 4 percent solute A by volume is to be scrubbed with a 40 k-mol/h liquid in a wetted- wall absorber operating counter-currently at 38 °C and 1 bar pressure, to reduce the solute A content of the gas to 0.3 percent by volume. The wetted-wall column, due to the large heat of absorption, is to be constructed in ‘shell & tube’ form, from 3 m long Ø30×2 mm stainless steel tubes. Heat transfer calculations have shown that 170 tubes will be sufficient. Check whether this tube number is sufficient for mass transfer. Give your conclusion. Equilibrium relationship for the system at the operating conditions may be represented as y*= 0.5 x, where y and x are mole fractions of solute A in gas and liquid respectively. Over-all mass transfer coefficient based on gas phase is calculated as K'y = 4.3*10-4 k-mol/m2s. 4.4.Show that solution of the equation below after replacing logarithmic mean with arithmetic mean

⎮⌡⌠

−−−

=

2yi

lniG )yy)(y1(

dy)y1(N

1y

1

21y

2y i

G y1y1

ln21

yydyN

−−

+⎮⌡⌠

−=gives

4.5 99 percent of the ammonia is to be recovered from 10 mole percent ammonia-air mixture by scrubbing with water in a packed column operating counter-currently at 20 oC and 800 mmHg. The mass flow rates of the gas and liquid to the column are 0.95 kg/s and 0.65 kg/s respectively and the cross-sectional area of the empty column is 1 m2. If the over-all volumetric mass transfer coefficient KGav = 0.1 k-mol/m3s bar, find the necessary height of packing. 4.6 Air, containing 3% acetone vapor by volume is to be washed with water in a packed column, operating counter-currently at 1 bar and 25 oC. Permissible acetone in the exit gas is 0.5% by volume, and the flow rate of the gas to the column is 300 k-mol /h. Water, which is acetone-free, will be supplied at a rate of 1.25 times the minimum water flow rate. The column is to be packed with 32*1.6 mm metal Raschig rings: a) Calculate the flow rate of water as kg/h, b) Write the operating line equation, c) Calculate the concentration of liquid solution leaving the column as mass percent, d) Calculate the percentage recovery of acetone, e) Calculate the diameter of the column, for a gas side pressure drop of 400 N/m2 per meter height of packing, f) Find out with what per cent of the flooding gas velocity the operation is carried out, g) Calculate the height of packing required. At the operating pressure and temperature and within the concentration range involved, equilibrium distribution of acetone between air-water is given as y*=2.2x, where y and x are mole fractions of acetone in gas and liquid respectively. Height of transfer units for acetone absorption into water in a column packed with 32*1.6 mm Raschig rings were found experimentally and given as: HG =1.397[G' 0.395/L' 0.417]; HL = 0.395L' 0.22, where HG and HL are in m., G' and L' are in kg/m2 s 4.7. A packed absorption column is to be designed to reduce the solute A content of a gas mixture from 1.5 mole percent to 0.01 mole percent by contacting it with a liquid solvent, which is solute-free. Molar ratio of liquid to gas is constant at 9 throughout the column. The equilibrium relationship at the operating conditions and the concentration range involved may be expressed as y*= 6x, where y and x are the mole fractions of solute A in gas and liquid respectively. Calculate the number of the overall gas transfer units needed, starting from the definition equation of it. [Ans. : 11.77]


4.8 An absorption column, packed with 38 mm ceramic Berl saddles, is to be designed to contact a gas phase of 1.0 kg/s with a liquid phase of 7.23 kg/s , counter-currently. a) Calculate the diameter of the column for a gas side pressure drop of 200 N/m2 per meter of packing. b) If this diameter is selected, determine with what percent of the flooding gas velocity the column operates. At the operating conditions, the viscosity of liquid and the densities of gas and liquid are: 1.5 cP, 2.9 kg/m3, 950 kg/m3. [Ans. : a) 0.966 m, b) % 53.6 ] 4.9 Carbon disulfide CS2 and nitrogen N2 gas mixture, containing 4.5 % CS2 by volume is to be scrubbed with a hydrocarbon oil in a plate column, operating counter-currently at 760 mmHg and 24 oC, to reduce CS2 content to 0.5 % by volume. The entering oil is CS2-free. Using Mc Cabe-Thiele method: a) Calculate the minimum liquid/gas ratio. b) For a liquid/gas ratio of 1.33 times the minimum, determine the number of the ideal plates needed. This system obeys Raoult’s law and the vapor pressure of CS2 at 24 oC is 342 mmHg. 4.10 An absorption column packed with 25 mm plastic Pall rings is to be designed to recover solute A at 1 bar and 27 oC. The flow rates of the gas and liquid phases are 14.4 k-mol/h and 32.8 k-mol/h respectively. a) Calculate the diameter of the column for a gas velocity of 60 % of the flooding gas velocity. b) What will be the pressure drop per meter of packing ? Density and viscosity of the liquid at 27 oC are 910 kg/m3 and 1.2 cP. The molecular weights of liquid and gas phases are 55 and 31 kg/k-mol. [Ans. : a) 0.312 m, b) 400 N/m2] 4.11 A column of 350 mm inside diameter, packed with 25 mm plastic Pall rings to a depth of 4.0 m, is already available. This column is now being considered for a new absorption operation. The flow rate of the gas to be processed is 7.2 k-mol/h and the solute content of this gas should be reduced from 2% to 0.1% by volume. The operation will be carried out counter-currently at 27 oC and 1 bar. The liquid solvent, which is solute-free, will be introduced at a flow rate which is 1.5 times the minimum solvent rate. At these conditions: height of one over-all gas transfer unit, HOG is calculated as 0.56 m. Gas fan connected to the column can develop a pressure difference of 400 N/m2 along 4.0 m packed height. Equilibrium relationship of the system may be represented by y*=1.6x, where y and x are the mole fractions of solute in gas and liquid phases respectively. Find out whether this column can be used for the new operation or not under the given conditions. If your answer is not, what alterations would you recommend? Density and viscosity of the liquid at 27 oC are 910 kg/m3 and 1.2 cP. The molecular weights of liquid and gas are 55 and 31 kg/k-mol. 4.12 A packed column is to be designed to absorb benzene vapor from air by contacting it with a hydrocarbon oil at 25 oC and 760 mmHg. The partial pressure of benzene will be reduced from 6 mmHg to 0.5 mmHg. The entering hydrocarbon oil is benzene-free. Calculate: a) The minimum value of L/G ratio, b) The number of the over-all gas transfer units needed, for an operating L/G ratio which is 1.25 times the minimum L/G ratio. Equilibrium relationship at 25 0C and 760 mmHg is y*= 0.12 x, where y*and x are the mole fractions of benzene in gas and liquid respectively. [ Ans. b) NOG = 6.86 ] 4.13 Solute A will be absorbed from an inert gas containing 3 mole percent solute A into a solvent S in a packed column operating counter-currently at 20 oC and 2.26 bar. The gas leaving the column will not contain more than 0.05 mole percent solute A. The flow rates of solvent and gas to the column are 110 k-mol/h and 100 k-mol/h respectively. The column will be packed with 50 mm ceramic Berl saddles. a) Calculate the required number of the overall gas transfer units NOG by using any suitable method.


b) For a gas side pressure drop of 200 N/m2 per meter height of packing, calculate the diameter of the column. At the operating conditions equilibrium relationship of the system is given as y =0.5x, where x and y are the mole fractions of solute A in liquid and gas. At 20 oC density and viscosity of liquid are 860 kg/m3 and 0.5 cP ; molecular weights of gas and liquid are 28 and 47 kg/k-mol. [Ans.: a) 6.42, b) 0.70 m ] 4.14 Two gas streams will be scrubbed with water at 20 oC and 800 mmHg in a packed column. The first stream which contains 4.8 mole percent ammonia and 95.2 mole percent air will be introduced at the bottom of the column at a rate of 126 k-mol/h. The second stream containing 2.5 mole percent ammonia and 97.5 mole percent air will be fed at the proper point at a flow rate of 133 k-mol/h. The mole fraction of ammonia in the leaving gas should not be more than 0.005. The water, which is ammonia free will be supplied at a rate of 3 600 kg/h. Calculate the total packed height required and the introduction point of the second stream. The maximum gas flow flux at any level in the column is to be 120 k-mol/h m2 based on empty column cross-section. For the packing used, over-all volumetric mass transfer coefficient

,where Kyav is in k-mol/m( ) 57.0

vy G5.6aK ′= 3h and G' is in k-mol/m2h. Assume dilute solutions and hence take the flow rates of the streams entering the column constant throughout the column. 4.15 In a certain absorption process conducted at 20 oC and atmospheric pressure, 1360 kg/h of gas of molecular weight 30 is to be contacted with water in a packed column of 3.65 m packing height. Ceramic Raschig rings are selected as the packing material. Liquid/gas mass ratio of 2 is to be employed. For a gas operating velocity of 60 % of the flooding gas velocity: a) Determine the column diameter for the best packing size among 25, 50 and 76 mm sizes, b) Calculate the pressure drop along the packing in mbar. Density and viscosity of liquid phase may be taken as 1000 kg/m3 and 1 cP. 4.16 An existing packed column, 3.3 m high, is to be employed for countercurrent absorption of a solute from a carrier gas into a liquid solvent. The inlet gas contains 3.8 % by volume of the solute and solvent enters the column solute-free. What will be the solute content of the outlet gas if the liquid/gas molar ratio employed is 1.2 ? Under the selected operating conditions the individual heights of the transfer units are known to be 0.23 m for the gas phase and 0.40 for the liquid phase. The equilibrium relation is y* = 0.96 x, where y and x are the mole fractions of solute in gas and liquid. 4.17 A solute will be recovered from a binary gas mixture in a plate column by contacting it with a solvent at 25 oC and 1 atm. The gas contains 4 mole percent solute and percentage recovery will be at least 90. The maximum molar ratio for L/G is to be 1.7 during the operation. There are two alternative solvents which can be used. The equilibrium relationships with these solvents at the operating conditions are y* = 120 x2 (for solvent I) and y* = 1.36 x (for solvent II) , where y and x are the mole fractions of solute in gas and liquid phases respectively. a) Which solvent is suitable for the operation? b) What is the number of the ideal plates needed with this solvent? 4.18 A 20 k-mol/h air-SO2 gaseous mixture containing 3 % SO2 by volume is contacted with a 30 k-mol/h water containing 0.1 % SO2 by mole in a stage (on a plate). The leaving gas contains 2 % SO2 by volume. a) Calculate the composition of the leaving liquid, b) Find the Murphere gas and Murphere liquid phase efficiencies. At the operating conditions, gas-liquid equilibrium is given as y*=1.2x, where y and x are the mole fractions of SO2 in gas and liquid. [Ans. : a) x=0.0077, b) EMG = 0.48, EML = 0.43 ]


4.19 A packed laboratory column, 2 m long, is used to remove a solute from air by countercurrent absorption in water. The mole fraction of solute in the inlet gas is 0.08 and this is reduced to 0.05 at outlet. The equilibrium relation is given by y* = 0.75 x, where y and x are the mole fractions of solute in the gas and liquid. The column is provided with a number of liquid sampling points and the following results were obtained in a trial run: z(m) 0 0.2 0.5 0.8 1.2 1.6 1.8 2.0 x 0.075 0.066 0.054 0.043 0.027 0.011 0.05 0 where z is the distance up the column measured from the bottom of the packing and x is the mole fraction of solute in the liquid. From the above results, plot a curve of 1/HOL versus z, where HOL is the local height of one overall liquid transfer unit. Use vertical axis for z. 4.20 A liquid solvent containing 5.3 mass percent solute A is contacted with a solute-free gas in a counter-current stripping tower isothermally to reduce its solute content to 0.5 mass percent. The flow rate of solute-free gas is selected as 1.4 times its minimum value. Calculate the concentration of outlet gas and percentage recovery of solute. The equilibrium data at the operating conditions are given as;

x 0.005 0.01 0.02 0.03 0.04 0.05 0.055 y 0.0006 0.00125 0.0027 0.0044 0.0065 0.009 0.0113

where x and y are the mass fractions of the solute A in liquid and gas. [Ans. : 0.0060 (mass fraction] 4.21 A thirty-plate column is to be used to recover n-pentane from solvent oil by steam stripping. The liquid contains 6 k-mol of n-pentane per 100 k-mol of oil and it is desired to reduce the solute content to 0.1 k-mol per 100 k-mol of oil. Assuming isothermal operation and an over-all column efficiency of 30 percent; a) Find the required steam rate per 1 k-mol of liquid rate, b) Calculate the ratio of steam rate to the minimum steam rate, c) Find the number of the plates needed if this ratio were selected as 2. The equilibrium relationship for the system at the operating conditions may be given as Y=3 X, where Y and X are the mol ratios in the gas and liquid phases respectively. [Ans.: a) 0.414, b) 1.34] 4.22 An existing packed column, 3.9 m high, is to be employed for countercurrent desorption by live steam (steam stripping) of a solute from a liquid solvent. The mole fraction of solute in the inlet liquid is 0.045 and the steam enters solute-free. The process can be regarded as isothermal and the column is well insulated so that no condensation occurs inside the column. If the recommended gas/liquid molar ratio is 0.6, calculate the mole fraction of solute in the outlet solvent. The equilibrium relation for the system is y* = 2 x, where y and x are the mole fractions of solute in the gas and liquid. Under the operating conditions the heights of individual transfer units are 0.3 m for the gas phase and 0.4 m for the liquid phase. 4.23 The concentration of solute A in a gas stream is to be reduced from 5 to 0.5 % by volume, by contacting it with a solvent in an absorption column at 1 bar. Solvent is available as solute-free at 20 oC. The molar heat of absoption is 42 000 kJ/k-mole of A absorbed and molar spesific heat of liquid is 84 kJ/k-mole oC . Over the range of operation, the equilibrium relationships at various temperatures are linear and given as y*= m x. The slope m changes with temperature as follows:

t(oC) 20 30 40 50

m 0.52 0.68 0.84 1.0

a) Draw the equilibrium curve for non-isothermal absorption, b) Calculate the minimum solvent rate , for a gas rate of 50 k-mol/h, c) Draw the operating line for a solvent rate which is 1.15 times the minimum rate,


d) Explain, what would be, if isothermal absorption at 20 oC were assumed ? In the solution use “the short method”. 4.24 Neutralization reaction between ammonia and phosphoric acid shown below is instantaneous. 3NH3 + H3PO4 (NH4)3PO4 An air-ammonia gaseous mixture containing 10 percent ammonia by volume is to be washed with aqueous phosphoric acid solution, contaning 0.37 k-mol acid per m3 of solution at 1 bar and 20 oC in a packed column, to reduce the ammonia content of the air to 0.1 mole percent. The cross-sectional area of the empty column is 1.6 m2. Flow rates of the gas and liquid to the column are 150 k-mol/h and 22.5 m3/h respectively. Volumetric mass transfer coefficients at the operating conditions were estimated as kGav = 40 k-mol/m3h bar and kLav = 4.8 1/h. Henry’s law constant for ammonia water system at 20oC is 0.0132 bar m3/k-mol. Molecular diffusivity of phosphoric acid in water at 20oC may be taken as 2.85*10-9 m2/s. Calculate the height of packing needed. [ Ans. Z = 3.78 m ] 4.25 Reaction between chlorine gas and aqueous sodium hydroxide is slow and first order, Cl2 + NaOH NaOCl + HCl Air containing 6 percent Cl2 gas by volume is to be scrubbed with aqueous NaOH solution in a packed column operating counter-currently at 1 bar and 20 oC, to reduce the volume percent of Cl2 to 0.1. Calculate the height of packing required for the following case: [Ans. Z = 3.36 m ] Volumetric flow rate of liquid : 10 m3/h Molar flow rate of gas : 150 k-mol/h Cross-sectional area of the column : 1 m2

Rate constant of the reaction : 0.15 1/s Volumetric mass transfer coefficient for the gas phase : 0.125 k-mol/m3sbar Volumetric mass transfer coefficient for the liquid phase : 0.006 1/s Liquid hold-up : 0.04 m3liquid/m3 of packing Henry’s law constant : 0.04 bar m3/k-mol

4.26 Solute A is absorbed at atmospheric pressure by countercurrent contact with a liquid solvent containing reactant B. The rection between A and B is instantaneous and proceeds according to the stoichiometry below; A + B P (in solvent S) The partial pressure of A in the inlet gas is 50 mbar and the concentration of B in the inlet liquid is 0.15 k-mol/m3. If the height of packing is 0.9 m, calculate the partial pressure of A in the outlet gas. The height of overall gas transfer unit for physical absorption is 1.0 m, the absorption factor is 1.3, the Henry’s law constant is 0.25 bar m3/k-mol, the gas and liquid phase mass transfer coefficients are 3.0 k-mol/m2h bar and 0.1 m/h respectively. The liquid phase diffusivities of A and B may be taken as equal.


Chapter-5

DISTILLATION

5.1 Introduction: Distillation operation is a mass transfer operation aimed at separating a liquid solution into its components. The second phase, which is vapor, is not brought from the outside as in other mass transfer operations, but is created from the original liquid phase by partial vaporization. Then, in distillation operation, the two phases involved are vapor and liquid and mass transfer taking place between these two phases enriches the phases in the components. The principle, on which separation with distillation is based, is the difference in the volatilities of the components of the solution, which are all volatile but at different levels. Distillation differs hence from Evaporation, which is an important unit operation, in that in evaporation only the solvent of the liquid solution is volatile at the operating temperature and the solute, which is practically non-volatile, does not transfer to the vapor phase. For example, the separation of K2SO4 from its aqueous solution is an evaporation operation as K2SO4 solid is not volatile at the evaporation temperature of the solution. But, on the other hand, separation of a methanol/ethanol solution is a distillation operation, as the vapor obtained by partial vaporization contains both of the components. But, since the vapor thus formed is richer in the more volatile methanol than the remaining liquid, a certain degree of separation is thus achieved. Now, if so-obtained vapor mixture is partially condensed, the remaining vapor is further enriched in methanol as more ethanol than the methanol condenses. So, “partial vaporization of a liquid solution and partial condensation of a vapor mixture always enrich the vapor in the more volatile component, and the liquid in the less volatile component”. With the repeating of partial vaporizations and partial condensations many times, finally a vapor is produced, which is almost pure in the more volatile component, and a liquid which is almost pure in the less volatile component; thus giving a total separation of liquid solution. As it is seen, distillation gives complete separation of the solution. For that, it is the most commonly used separation operation in chemical engineering practice, when liquid solutions are to be separated. In order to understand distillation operation, the equilibrium between liquid and vapor must be known very well. 5.2 Liquid-Vapor Equilibria: Let us start with considering the behaviour of a binary solution of A+B, which is kept under constant pressure in a closed container. Let us assume that component A is more volatile than the component B. When the vapor pressure of a component at all the temperatures is greater than that of other component, this component is then named as more volatile component (MVC in short), while the other is being named as less volatile (LVC in short). Assume that x and y are the mole fractions of component A in liquid and vapor phases respectively, and tA and tB show the boiling point temperatures of pure A and pure B at the prevailing pressure. Typical liquid-vapor equilibrium of this solution is shown in Fig.5.1. Consider the liquid solution whose temperature and composition are tK and xk and represented by point K. If this solution, which is a cold liquid, is heated slowly, the first vapor bubble forms at temperature tH and its composition yJ is read from point J. As expected, yJ is greater than xk. xk and yJ are the compositions of liquid and vapor phases which are in both mass and thermal equilibrium. Temperature tH is known as


the bubble point temperature of the liquid solution whose composition is xk. The line HJ, which relates two equilibrium phases, is named as tie-line. If more heat is added to the mixture, both temperature of the mixture and the quantity of the vapor in the mixture increase at the expense of liquid. For example, point M represents the vapor-liquid mixture at temperature tL, which separates in equilibrium liquid and vapor phases represented by the points L and G, the line LG being a tie-line. As it is seen, the composition of liquid drops from xk to xL ,and that of vapor from yJ to yG with increasing vaporization, but yG is again greater than xL. The ratio of the moles of vapor

Mole fraction of the MVC in vapor, y

yR= xk

Te

mpe

ratu

res,

t

1

1

0

tABubble point temperature curve

tH

tL

tS

tB

tR

L

H

K

M G

NS

R

P=cons.

0

xH =xK

Dew point temperature curve

Tie-lines

xL xS

yG yJ

JTem

pera

ture

s, t

Mole fraction of the MVC in liquid, x

Fig.5.1 Typical temperature-composition diagram (t-xy) of a binary solution

to the moles of liquid in the mixture is given by the lines ratio of LM/GM, which is known as inverse lever-arm rule. If the mixture is further heated, both the temperature and vapor quantity increase further, and finally to a point N is reached at which the last liquid drop, whose composition is read as xs from point S vaporizes at temperature ts. As it is noticed, the liquid solution has started boiling at temperature tH and ended up at temperature ts. It follows from this that, liquid solutions unlike the pure liquids have no constant boiling point temperatures but have boiling temperature ranges, which is tH-ts for a liquid solution of composition xk. Hence, the curve given by the points tB-S-L-H-tA is named as bubble-point-temperature curve, as it shows the saturated liquid solutions. If the vapor represented by point N is further heated, it becomes superheated vapor, as in this case only its temperature increases. Let us consider the phenomena in the reverse direction. If the superheated vapor mixture whose temperature and composition are tR and yR and represented by point R is cooled slowly, no phase change is observed until the temperature drops to tN=ts, at which first liquid drop (dew) forms, whose composition is read as xs from point S. By further cooling, the quantity of liquid increases at the expense of vapor during which both liquid and vapor compositions alter. For example, when point M is reached, the


compositions of the equilibrium liquid and vapor are read from points L and G as xL and yG. As it is seen, while the composition of liquid increases from xs to xL, the composition of vapor rises from yR to yG. If the cooling is further continued, when the temperature tH is reached, the last vapor bubble whose composition is yj, condenses giving a saturated liquid mixture represented by point H. Further cooling sub- cools the liquid bringing it back to the condition represented by point K. As it is seen, superheated vapor represented by point R first becomes saturated vapor at point N (temperature tN) then starts condensing, and condensation continues until the temperature drops to tJ (point J) during which its composition enriches in the MVC and finally the last vapor bubble, which is richest in the MVC condenses at point J with composition yJ. The condensation process takes place at a temperature range of tN-tJ ,not at a constant temperature. Since the curve given by the points tB-N-G-J-tA represents saturated vapor mixtures (vapor mixtures at their dew points), it is known as dew-point-temperature curve. Hence, any point between bubble-point-temperature

curve and dew-point-temperature curve represents a liquid-vapor mixture, which upon separation gives two equilibrium phases: saturated liquid and saturated vapor, which are located on the bubble-point-temperature curve and on the dew-point-temperature curve by drawing the tie-line through the point representing the mixture. Their amounts are found by applying the inverse lever-arm rule to the tie-line. Any point just on the bubble-point-temperature curve represents saturated liquid, and on the dew-point-temperature curve represents saturated vapor. Points above dew-point-temperature curve and below the bubble-point-temperature curve represent superheated vapor and sub-cooled liquid

respectively.

(L,G) (H,J)

(S,N) tH

xL

P=cons.

tL

tS

xH

yN

xS

yG

yJ

y= x

0 0

1

1 mole fraction of the MVC in liquid, x

mol

e fr

actio

n of

the

MV

C in

vap

or, y

equilibrium curve

Fig.5.2 xy-diagram of a binary mixture

If the liquid represented by point K is heated in an open vessel, it starts boiling at temperature tH represented by point H. Since the vapor escapes from the vessel, the remaining saturated liquid becomes poorer in the MVC; hence its temperature and composition shift toward left along the bubble-point-temperature curve with the vaporization. The equilibrium or t-xy diagram shown in Fig.5.1 is peculiar to each binary mixture and it is only obtained experimentally for the real solutions. Various experimental techniques have been developed for the measurements, most of which are given by E.R.Gilliland and C.S. Robinson in their book “Elements of Fractional Distillation” Mc Graw-Hill, 1950. The experimentally measured vapor-liquid equilibria for many binary solutions are given in the literature. The book written by Hirata et. al contains large number of binary solutions’ vapor-liquid equilibria (“Computer Aided Data Book of Vapor-Liquid Equilibria” Elsevier Scientific Publishing Company, 1975).


Sometimes, the vapor-liquid equilibria are given on simple xy-diagrams as shown in Fig.5.2. Although the relationship between liquid and vapor compositions can easily be seen, temperature information is not normally given in these diagrams (to the Figure these were added for the explanation purpose). 5.2.1 Ideal Solutions: Vapor-liquid equilibria of ideal solutions can be computed directly from the vapor pressures of the pure components without resorting to the experimentation. The criteria for ideality have already been given in Section 4.2.1 For ideal solutions, Raoult’s law and for ideal gas (vapor) mixtures Dalton’s law are applicable. Raoult’s law for a binary solution of A+B is written as: xpp o

AA =

pres

sure

P-y

P-x

xpB −

x,y

xp A −

0 (B)

oAp

t = cons.

1 (A)

Fig.5.3 Ideal solution

oBp

(5-1) x)(1pp o

BB −=By summing up these two equations;

(5-2) x)(1pxpppP oB

oABA −+=+=

is obtained. From these, it is understood that partial and total pressures change linearly with the mole fraction as shown in Fig.5.3. By solving x from these two equations,

oB

oA

oB

pppP

x−

−= (5-3)

and from the combination of Raoult and Dalton laws,

xP

pP

py

oAA == (5-4)

are obtained. Now, with the help of these two equations, liquid-vapor equilibrium can be computed as follows: arbitrary temperatures between the boiling point temperatures of pure components at the specified total pressure are selected and vapor pressures of the components at these temperatures are calculated from the Antoine equations of the components. Then, from equation (5-3) x, and from equation (5-4) y are computed. With the so-calculated x and y values, t-xy and xy-diagrams are plotted. The accuracy of the plots depends on the number of the temperatures selected.

Example-5.1) Plotting of xy- and t-xy Diagrams of an Ideal System

Solutions of benzene- toluene obey Raoult’s law. a) Compute the vapor-liquid equilibrium of this system at 760 mmHg. b) Plot t-xy and xy- diagrams at 760 mmHg for this system. Vapor pressure of benzene and toluene can be calculated from Antoine equation which is written as:

oi

oi pwhere

tcbaplog+

−= in mmHg, and t in oC. Antoine constants a, b and c for benzene

and toluene are given in the table below:


Component tbp (oC) a b c t range ( oC)

Benzene(A) 80.1 6.90565 1211.033 220.79 7-135

Toluene(B) 110.6 6.95334 1343.943 219.38 7-135

Solution :

Select t = 85 oC, then ;

mmHg7.881p9453.2

8579.220033.121190565.6plog o

AoA ==

+−=

mmHg1.345p538.2

8538.219943.134395334.6plog o

BoB ==

+−=

773.0

345.1881.7345.1760x =

−−

=From equations (5-3) and (5-4) 897.0)773.0( ==760

881.7y

By selecting various temperatures and repeating the steps given above, the following table is obtained.

Select C a l c u l a t e

t (oC) (mmHg) oAp (mmHg) o

Bp x y

80.1 760.0 292.3 1.0 1.0

85.0 881.7 345.1 0.773 0.897

90.0 1021.0 406.8 0.575 0.773

95.0 1176.8 476.9 0.405 0.626

100.0 1350.5 556.4 0.256 0.455

105.0 1543.2 646.0 0.127 0.258

110.6 1783.4 759.5 0.0 0.0

With the data in the table above t-xy and xy-diagrams of the system can easily be plotted on the millimetric papers as shown below.

t-xy Diagram

75

85

95

105

115

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

mole fraction of benzene,x,y

tem

pera

ture

(oC

)

xy-Diagram

00.10.20.30.40.50.60.70.80.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

mole fraction of benzene in liquid,x

mol

e fr

actio

n of

ben

zene

in

vapo

r,y


5.2.2 Deviation from Ideality: Real and Azeotropic Solutions: In real solutions, the tal pressure measured over the solution may be greater or smaller than the total

e measured pressure is higher than the alculated pressure, this solution is said to deviate from Raoult’s law in positive irection, if the reverse happens, deviation is said to be in negative direction. The artial pressures of the components in such systems do no longer change linearly with e composition. In Fig.5.4, the changes of total and partial pressures with the

e direction and the degree of deviation of the solution 1, the lution is ideal, as expected and when they are

s positive, a d smaller than 1, the deviation is negative. If

ctivity coeffic to 1, devia

oeff nts with osit , and threat and the diffee total pressinimum. This ty

topressure calculated from equation (5-2). If thcdpthcompositions are shown for positive and negative deviations. As it is seen from the figures, system approaches ideal system when the mole fraction of the component in the liquid nears 1. For the real (actual) solutions; xpγp o

AAA =

x)(1pγp oBBB −= (5-5)

are written, where γ and γ are the liquid phase activity coefficients of the A B

components and they show th are from ideality. When they so

greater than 1, deviation i n

P-x

the alosec

c iciep ivelygth uremboiling solutionsazeotropism, whazeotropism. Conand maximum boi

oBp

xp A −

0 (B)

oAp

t=cons. oApP-x

op

pres

sure

s

Fig

No part of this CD-boo

P-y

ients, whiction fromcompositioe o r dere betw

p

thence

versus me of solu

. Passingereas passtant temp

ling azeotr

x,y

B

P

(a) .5.4 Deviat

k may be mult

h are de idealitns aviating een the

re

ole frations a

througsing therature

opic sys

xpB −

1 (A

ion from

iplied for

pendant by is sma

r different negatively. When the de vapor pressures of the tw

oth on temperatll. In Fig.5.5, t

shown fo two

ction curve may pass tre said to form azeotroph a maximum results rough a minimum giv, constant pressure and xtems are all shown in Fig.

)

xpB −

0 (B)

idealty (a (+), (b) in (-) d

pres

sure

s

) in

commercial purposes. E.Alpay &

-y

s mviationo com

ure anhe ch

yste

hroughic solu

in mes m

y-diagr5.6. As

x,y

(b) irectio

M.Demi

s, one from ideponents is

maxim or c

d compositanges of a

de

ationsinimum aximum ams of mi are seen f

xp A −

1 (A

n

rcioğlu

t =cons.

ality is small, um or onstant

ion arectivity viating

boiling boiling nimum rom

)

155

Fig.5.5 Change of activity coefficients with composions (a) at positive, (b) at negative deviation

(a) propanol-water (b) acetone-chloroform

the figures, at point Az saturated vapor and liquid curves touch each other. The liquid solution corresponding to this composition boils at constant temperature as if it were a ure liq formed has the same co id. While a omponent is more volatile on the left of point Az, it becomes less volatile on the right

ehaving in this way cannot be separated completely by using classical distillation ethods. Minimum boiling azeotropism is more common than the maximum boiling

where in mmHg and t in C. Solution :

Vapor pressure of acetone at 65.9 C :

rom equation (5-5) :

as γA greater th ution is not m Raoult’s law.

p uid, and the vapor mposition with the liqucf the point, hence the volatilities of the components change at point Az. The solutions o

bmazeotropism, and there is large number of binary and ternary solutions showing azeotropism. The books written by L.E. Horsly contain almost all the azeotropic systems (Azeotropic Data-I,II, Am. Chem. Society, 1952-1962). In some systems azeotropic point, hence the azeotropic composition may change or even completely disappears with changing pressure. Ethanol-water solution, which is an industrially important solution, has a minimum boiling azeotropism, which occurs at 78.2 oC with a composition of 95.6 mass percent ethanol at 1 atm. The azeotropism of this system disappears at pressures below 70 mmHg.

Example-5.2) Check for Ideality

Equilibrium partial pressure of acetone in vapor over a liquid solution of acetone-methanol, containing 30 mole percent acetone and saturated at 65.9 oC is measured as 398 mmHg. Find out whether this solution is ideal or not. State the direction of deviation from Raoult’s law, if exist. The vapor pressure of acetone is given as: o o

Apt5.237

24.7+

−=9.1279plog o

A

o

F

an 1, the sol ideal and positively deviates fro

0215.39.655.237 +

mmHg7.105010p 0215.3oA

9.24.7plog =−=1279o

A

==

26.1A )30.0)(7.1050(398

==γ


Az

Az

oBp

Az

o


Ap

pB-x

Az

x,y

Az

P-x

pA-x

P-y

pB-x

0 (B)

1.0(A)

x,y

pre

ssur

es

tA

Az

pA-x

P-x oAp

opB

P-y

1.0(A)

0 (B)

t =cons. t =cons.

tA

tB

tB

L

x 0 (B)

0 (B)

1.0(A)

1.0(A)

GG

G

G

L

L

L

y

x

x x xy y

y > x

y

y > x

y < x

y < x

P= cons.

x0

0 0

0 1.0

y

1.01.0

1.0

tem

pera

ture

s

y = x y = x

y

x,y x,y

P= cons.

P= cons. P= cons.

Fig.5.6 (a) Minimum and (b) maximum boiling azeotropism

a b

t-x

t-y

t-y

t-x

pre

ssur

este

mpe

ratu

res

5.2.3 Partial Solubility and Insolubility of the Components in Liquid Phase: In some special cases, components A and B are partially soluble or even completely insoluble within each other. Vapor-liquid equilibria of such systems are quite different than the normal systems’ vapor-liquid equilibria. Let us consider first the system that forms two liquid phases because of incomplete solubility of the components. In such a system each compone erts its own vapor pressure in the vapor phase and then the total pressure is written as,

(5-6

If such a liquid mixture is heated in an open vessel, it starts boiling when the pressure given by equation (5-6) reaches the outside pressure and the temperature and pressure remain constant until the MVC of the mixture is completely consumed up (first period). As soon as the MVC is removed completely, the pressure of the system drops to the vapor pressure of the LVC at the prevailing temperature. If heating is continued, the boiling starts again when the vapor pressure of the LVC rises to the outside pressure and then the temperature remains again constant until this component vaporizes completely (second period). The mole fractions of the more and less volatile compo ts in the vapor at the first period at w e two components vaporize together are calculated from the equations below,

nt ex

oB

oA ppP += )

nen hich th

oB

oAp

oA

oA

pp

Pp

y+

== (5-7)

and

oB

oA

op oBp

+B

ppPy1 ==− (5-8)

These values remain constant throughout the first period. This special case forms the basis of am stillation, which is frequently used to separate a high boiling organic material, which is insoluble in water, from the n volatile impurities. For that, water is added to the mixture and then heating is started. The mixture starts boiling at relatively low temperature (less than 100 oC for atmospheric pressure) due to the high vapor pressure of water and as long as wate i he mixture, temperature does not change. Steam consumption per unit mass of organic material distilled can be com uted from,

ste dion-

r ex sts in t this

p

oBB

oA

BB

A

pMp18

y)(1My18

mm

vapororganickgsteamkg

=−

== (5-9)

where, y is the mole fraction of water vapor (steam) in vapo B is the molecular weight of vaporizing organic material. Hence, energy required qs as kJ per mB distilled is given by;

the r mixture, M

kg of organic material

[ ] [ ]BoLBAoLAoBB

oA

B

s λ)t(tcλ)t(tcpM

p18mq

+−++−= (5-10)


where, to and t are the initial and boiling point temperatures as oC, λA and λB are latent heats of vaporization of water and organic material at temperature t as (kJ/kg), LAc and

LBc are the mean specific heats of water and organic material as ( kJ/kg oC). As the water and organic material do not dissolve within each other, they separate in two phases in a decanter upon the condensation of vapor. By using steam distillation the organic materials, which normally boil at high temperatures and thus have the risk of decomposition, can be separated at low temperatures without decomposition. But

ly for low due to the h

d) The loss of ortho-nitrotoluene with water, e) The energy requirement for direct vaporization of ortho-nitrotoluene.

used to purify th re? distillation.

o 62.3 99 99.5

the cost of separation by steam distillation may be very high especialmolecular weight organics, igh rate of steam requirement.

Example-5.3) Steam Distillation

500 kg ortho-nitrotoluene (M.W=137; b.p.= 222 oC) at 25 oC, will be purified from very small quantity of non-volatile impurities, with steam distillation in a jacketed vessel equipped with a condenser and a decanter. Water at 25 oC will be continuously supplied to the vessel to maintain a liquid water level. Distillation is to be carried out at 760 mm Hg pressure.

Calculate: a) The temperature at which distillation will proceed, b) The amount of water to be used, c) The heating steam requirement. Steam is saturated at 40 kN/m2 gauge pressure and

condenses in the jacket.

f) What other way may be e ortho-nitrotoluene at the same temperatuCompare this with steam

Temperature ( C) 60 100 110 123.5 222op (mmHg) - 8 760 1060 - -A

- 733.5 746. Water

cLA(kJ/kgK) 4.182 (A)

λA(kJ/kg) 2267 2238 oBp (mmHg) 12.8 13.1 13.4

cLB(kJ/kgK) 1.57 1.75

o-Nitro toluene (B) λB(kJ/kg) 389.6 340

Solution:

a) When mmHg760Ppp oB

oA ==+ distillation starts.

Assume, t = 99 oC , from table above,

oC

Then, distillation starts and proceeds at 99.5 oC

c) The steam requirment is obtained from equation(5-10).

mmHg760mmHg3.7468.125.733pp oB

oA <=+=+

Assume, t = 99.5

b) From equation (5-9)

,attakenbemustcandc LBLA

===)1.13)(137(

)8.746)(18)(500(pMp18mm

oBB

oAB

A waterkg7453

C3.622

5.99252

tt oo ≅+

=+

[ ] [ 1.82019=+−++−= 389.625)(99.5(1.57)226725)(99.5(4.182)13.1)(137)(mB

s ](746.8)18q

g9.7591.138.746pp oB

oA mmH760mmHg ≅=+=+


Then, required heat energy : qs= (19 820.1) mB= (19 820.1)(500) = 9.91*106 kJ Absolute pressure of the heating steam ; Ps=Poutside + Pgauge= 760+760(40/101.3) =1 060 mmHg

the table ; t

From

e) Direct vaporization at 760

s = 110 oC and λs =2 238 kJ/kg Then, required heating steam amount : d) In the Perry’s Handbook, the solubility of o-nitrotoluene in water at 30 oC is given as : 0.07 kg o-nitrotoluene per 100 kg water. Then, loss of o-nitrotoluene =

mmHg takes place at 222 oC , BLc is taken at (25+222)/2=123.5 oC. Then, energy requirment:

istillation would be enough

f) As other me temperature, which corresponds to 13,1 mmHg absolute pre r In this case; In othenough.

ng ene s com w e ene qui r cr vac ecision is

much t at compone an B hav only so ith each her in st he r-li equ libria o such ar terestin with rtial solubility range. The vapor-liquid

libri /n butano em hich h usthis way 7. The area below points D and E s ows th concen n range at which partial solubility in liquid p rs. The binary mixture represented

nt in this area is heterogeneous and separates in two insoluble liquid phases. or example, the mixture which has 82 mole percent water and at 85 oC and shown by

rium nt e percent n-butanol and being represented by point C, the other

rcent water and 2 mole percent n-butanol and being represented ses are found by drawing the liquid tie line

assing through point M, and the compositions of the phases change with temperature. presented by a point in this region is heated up slowly, when the

mperature reaches boiling point temperature, which is about 93 oC, vapor forms. The by point Az, remains constant at

zeotropic composition of 77 mole percent water and 23 mole percent n-butanol until ne of the components is depleted. In the regions outside this region (left of point D

and right of point E) liquid solutions are homogeneous and behave as normal systems.

As it is seen, only of the energy for steam din this case.

way, vacuum evaporation at the sassu e (760-13.1≈747 mmHg vacuum), may be considered.

er words; only of the energy for steam distillation would be

Saving in the heati

en. rgy i pared ith th rgy re red fo eating uum. D

then giv Some systems deviate from the ideality so h nts A

uid d e

partial s

lubility w inin

otg

liquidin pa

ate. T vapo q i fsystem e ratherequi um of water - l syst , w as ind trial importance and behaves in

, is shown in Fig.5. h e tratiohase occu

by a poiFpoint M separates in two equilib liquid phases one containing 58 mole percewater and 42 molontaining 98 mole pec

by point K. These equilibrium liquid phapIf a liquid mixture retecomposition of this vapor, which is representedao

==λ

=2238

10*91.9qm

6

s

ss kg4284

=100

)07.0)(7453( kg62.2

[ ]BoLB λ)t(tc +−= Bs mq [ ]=+−= 34025)(2221.75)500( kJ10*43.3 5

%46.3100.10*43.36

5

=10*91.9

[ ]=+−= 389.625)(99.51.57)500(qs *53.2 kJ105

%55.2100.10*91.9 6

=10*53.2 5


For example, if the liquid solution represented by point F is heated up, it starts boiling at point H and the first vapor bubble formed, is represented by point J. As the vaporization proceeds, the composition of liquid shifts toward S, and the composition of vapor toward N. 5.2.4 Volatility and Relative Volatility: The volatilities of A and B in a binary solution are defined as;

x

pα A

A = x1

pα B

B −= (5-11)

The relative volatility of A to B is then given as,

x)/(1p

/xpαα

αB

A

B

A

−== (5-12)

From Dalton’s law Py)(1pandPyp BA −== can be written. If these are substituted into equation (5-12);

y)(1xx)(1y

α−−

= (5-13)

D

NS

P =cons.

1.0

two liquid phase

mole fraction of water in liquid and vapo yr, x,

y =x

AZ

AZ

tB

KM

E

D

C

E

tA

F

HJ

t-y

t-y

t-xt-x

H,J

liquid

vapor

0 0

0

1.0

1.0

mole fraction of water in liquid, x

mol

e fr

actio

n of

in

vap

or,

wat

er

y

tem

pera

ture

,t

Fig.5.7 Vapor-liquid equilibrium diagrams of water / n-butanol


and solving for y,

x1)(α1 −+

xαy = (5-14)

is obtained. This equation is used to e equilibria of binary solutions in functional way, when the rechange much within the operating temperature range. For ideal system, equation (5-14) reduces to equation below after substituting from Raoult’s law, (5-15) Although the vapor pressures change with temperature, their ratio remains almost constant in ideal solutions. Thus, equation (5-14) can be used to express vapor-liquid equilibriums in functional way in ideal solutions. In many real solutions, the change of relative volatility within the temperature rang is not so great. Thus, equation (5-14) can still be u nce between the relative volatilities calculated at the top and bottom conditions of the column is not greater than 15%, in which case geometric mean of the two relative volatilities is recommended for equation (5-14). Table.5.1 contains the relative volatilities of some binary solutions calculated at the boiling pothe solutions. For the separation of a binary solution by the classical distillation methods, the relative volatility of the solution should nowhere be 1, and greater the value of α easier the separation of solution by distillation. Note that at azeotropic composition α =1.

Example-5.4) Calculation of Relative Volatility

Calculate the relative volatilities of benzene to toluene at x = 0.1 , 0.3 , 0.5 , 0.7 and 0.9 and compare the values. From xy-diag sponding y values are read and f he table below;

x 0.1 0.3 0.5 0.7 0.9

xpress the vapor-liquid lative volatility is a constant or does not

oB

oA p/pα =

e involvedsed in the real solutions, when the differe

int temperatures of the pure components of

ram of benzene/toluene at 760 mmHg plotted in Example-5.1, correrom the equation (5-13) α values are computed. Results are given in t

y 0.205 0.51 0.72 0.855 0.96

α 2.32 2.43 2.57 2.53 2.67

As expected, α values do not change much with the composition.

relationship between liquid and vapor compositions of component-i in a binary or a multi-component solution at equilibrium may be

yi = Ki xi (5-16)

where, K is known as the K value or distribution coefficient of component-i. This eful for the comput ibria of multi-

For ideal solutions, Ki here that K value of a component depends on

5.2.5 K-Values: The

expressed as,

i

equation is especially us ation of vapor-liquid equilcomponent solutions. = /Ppi and for real solutions K

o

i = /Pγp ioi . It follows from


temperature and pressure. The K values for many components were measured experimentally and are given either in the classical diagrams such as shown in Fig.5.8, or in nomographific diagrams developed by C.L. DePriester. Table 5-1. The relative volatilities of some binary solutions computed at the boiling points of the pure components of the solutions

Solution Boiling p. of component-1 α

Boiling p. of component-2

(o (oC) α

C) Benzene-ethy 83.48 1.109 lene dichloride 80.1 1.113 Benzene-toluene 80.1 2.61 110.7 2.315 n-Butyl chloride-n-Butyl bromide 77.5 2.08 101.6 1.87 Chloroform-carbon tetra chloride 61.1 1.71 76.6 1.60 Ethanol-isopropanol 78.3 1.18 82.3 1.17 Ethanol-propanol 78.3 2.18 97.2 2.03 Ethyl chloride-Ethyl bromide 12.5 3.23 38.4 2.79 Ethyl ether-benzen 34. 80.2 3.95 e 6 5.16 Ethylene dibromide-propilen dibromide 31. 141.5 1.30 1 7 1.30 Ethylene dichloride-trichloro 3. 7 2.33 ethane 8 5 2.52 113.n-Heptane-methylcyclohexane 98.4 1.058 100.3 1.056 n-Hexane-n-heptan 9. 98.4 2.33 e 6 0 2.613 Methanol-ethanol 64.7 1.73 78.1 1.64 Methanol-isobutanol 64.6 6.1 107.5 4.4 Methanol-propanol 64.6 3.89 97.2 3.15 Methyl acetone-ethyl acetate 56.8 2.036 77.1 1.923 Phenol-o-cresol 181.2 1.30 190.6 1.275 Phenol-m-cresol 181.2 1.768 201.5 1.699 Phenol-p-cresol 181.2 1.793 202.2 1.728 Toluene-benzyl chloride 110.7 7.75 178.0 4.45 Toluene-chlorotoluene 110.7 4.76 162.0 3.65 Water-ethylene glycol 100.0 49.8 197.0 13.2 Water-ethylene glycol(150 mm Hg) 60.1 98.0 150.2 21.0 Water-ethylene glycol (50 mm Hg) 38.1 76400 202 244

Fig.5.8 K-Values of Some Hydrocarbons at 1 atm.

(K)


5.2.6 Bubble Point Temperature: In order to find the bubble point temperature and the composition of the first bubble to be formed from a known composition of a liquid solution at a specified pressure, the equation, ∑ yi = 1 =∑ Ki xi (5-17) can be used. Since P and xi are known, an arbitrary temperature is selected and at this temperature and the specified pressure the K value of each component is found and then product of Kx for each component is computed. If the summation of these products is 1 or very close to 1, the selected temperature is the bubble point temperature of the solution and the product Kx for any component gives its mole fraction in the first vapor bubble formed. If the summation deviates considerably from unity, a new temperature is selected and steps are repeated with this temperature. The temperature selection is repeated until the summation gives 1. 5.2.7 Dew Point Temperature: The dew point temperature of a known composition vapor mixtu t liquid drop to be formed, can be computed from;

re at constant pressure, and the composition of the firs

i

ii K

1x Σ==Σ (5-18)

The tion is done by trial and error by selecting a temperature. At this temperature and given pressure the K value of each component is read and then the ratio of y/K is found. If the summation of these ratios is 1 or very close to 1, the selected temperature is the dew point temperature of the vap

y

calcula

or mixture and y/K ratio r each component gives its mole fraction in the first liquid drop that will form. If the

-5.5) Calculation of De erature of a Ternary Mixture

T luene/o-xylene are ideal. e dew point temperature of a vapor, containing 20 mole per nzene ole p o-xylene at mHg total pressure. he first liqu plet form erature range at which all the vapor will condense (condensation range) ? A nents are given table bel

a b t )

fosummation deviates substantially from 1, with new temperature selections the steps arerepeated until the summation can be accepted as 1.

Example w Point Temp

ernary solutions of benzene/to a) Calculate th cent be , 30 mercent toluene and 50 mole percent 760 m b) What is the composition of t id dro ed? c) What is the temp

ntoine constants for the compo in the ow:

Component tbp (oC) c range ( Co

Benzene(A) 80.1 6.90565 11.033 .79 12 220 7-135

Toluene(B) 110.6 6.95334 43.943 .38 7-135 13 219

o-Xylene(C) 144.4 6.99891 74.679 .69 14 213 7-145

Solution

F xA + xB + xC =

:

rom equation (5-18); 1=p

PyPyoC

CoB

+ppo

A

yP BA +


Antoine equations; lcualated. Then;

o ew point temperature of the vapor

at 129 oC. As the ondensation proceeds the temperature drops. The condensation temperature of the last vapor bubble

is the bubble point temperature of the liquid, who e composition is equal to the composition of the

Then, find the tur q co n is : .20 ; xB=0.30 ; xC=0.50. For t

oCp

)(50.0())2.0(++=

oBpo

Ap)760760)(760)(0 30.0(

ooB

oA p

380p

228p

152++=

Solution is by trial & error : Assume td =110 oC

C

2446.311079.220

033.121190565.6plog oA =

+−=

As this is greater than 1, the selected temperature is too low.

New assumption : t =132 oC. Then, fromd

are ca As the summation is smaller than 1, the assumed temperature is high. New assumption : td =129 oC. At this temperature,

are calculated. Then; a) As this is almost 1, the selected temperature 129 C is the dmixture. b) The composition of the first liquid droplet formed is :

c) The condensation range of the vapor (td - tb) : The vapor starts condensingc

svapor at the start.

bubble pointhis, equation (5-17) is use

tempera e of the lid.

uid, whose mpositio xA = 0

8731.211038.219

943.134395334.6plog oB =

+−=

4431.211069.213

679.147499891.6plog oC =

+−=

mmHg4.1756poA =

mmHg7.746poB =

mmHg4.277poC =

762.14.277

3807.746

2284.1756

152xxx CBA =++=++

mmHg6.1344p;mmHg1.2971p oB

o =A =

mmHg8.540poC =

924.08.540

3806.1344

2281.2971

152xxx CBA =++=++

mmHg4.1246p;mmHg4.2776p oB

oA ==

mmHg2.496poC =

004.13802281522.4964.12464.2776CBA xxx =++=++

055.04.2776

x A

152== 183.0

4.1246x B

228== 766.0380

2.496x C ==

0760(p o

CoB

o

B

2( 4ooo

.10

)50.=

p760

)30.0(+

p760

20.0(A +yC =+yyA +

)p58.6p9.3p63. CBA 5 0.110*++= − =


Assume tb =110 oC, then ,

4)]*10-4 = 0.94 s the summation is less than 1, the assumed temperature is low, but it is rather close to the bubble oint temperature. ew assumption ; tb=113 oC, then ;

yA + yB + yC =[(2.63)(1894.6)+(3.95)(812.7)+(6.58)(305.4)]*10-4 = 1.02 ince this is almost 1, the assumed temence, the condensation range

Then; yA + yB + yC = [(2.63)(1756.4)+(3.95)(746.7)+(6.58)(277.ApN S perature is the bubble point temperture. H : 129 - 113 oC

In some distillation operations, the energy ifference between vapor and liquid phases is important and is to be known. Enthalpy-

composition diagrams, which are obtained by plotting saturated liquid and saturated vapor enthalpies against the compositions, are in this case proved to be very useful. The specific enthalpy, h (kJ/k-mol) of a liquid solution is give

5.2.8 Enthalpy-Composition Diagrams: d

n as: soLtL ∆H)t(ch +−= (5-19)

where, ∆Hs (kJ /k-mol solution) is heat of solution at to temperature and at given concentration, to is reference temperature, tLcalculated and

is the temperature at which enthalpy is Lc (kJ/k-mol oC) is the molar heat capacity of the solution calculated at

the arithmetic mean of to and tL. If temperature tL equals the bubble point temperature of the solution, then h is the specific e

the heat capacities of pure components by nthalpy of the saturated solution. The heat

capacity of the solution is related to LBLAL cx)(1cxc −+= Hence, specific ent d liquid soluti

(5-20) on can be written as:

h = [halpy of the saturate

LBLA cx)(1cx −+ ] (tb- to) +∆ s (5-21) he sign of ∆Hs is negative, if heat is released during mixing. For ideal solutions, H = 0.

btained by assuming that heated up to the dew point temperature of

ixtur d w ly and mixed with the other component. ence,

HT∆ sThe specific enthalpy of a saturated vapor mixture, H is oeach component of the mixture is separately the m e, t h totalere it is vaporized H [ ] [ ]BodLBAodLA λ)t(tcy)(1λ)t(tcyH +−−++−= (5-22) an be written.

shown in Fig.5.9. As it is seen, xy-diagram is also plotted with this diagram. ific enthalpies of the saturated liquid

solutions and the upper curve the specific enthalpies of saturated vapor mixtures. Any point above the saturated vapor enthalpies curve, such as point R represents uperheated vapor, whereas the points belo the saturated liquid solution enthalpies

rial and error with the help of xy-diagram below.

mmHg4.1756poA = mmHg7.746po

B = mmHg4.277poC =

mmHg4.305p;mmHg7.812p;mmHg6.1894p oC

oB

oA ===

cA typical enthalpy-composition diagram obtained by plotting equations (5-21) and (5-22) isThe lower curve in the diagram shows the spec

s wcurve such as point K show the cold liquid solutions. The point between two curves such as point M represents a vapor-liquid mixture, which upon separation, gives a saturated liquid solution shown by point L and a saturated vapor mixture represented by point G, which are in equilibrium, and locations of them are found by drawing the tie-line passing through point M by t


Similarly, the saturated vapor mixture G1, which will be in equilibrium with saturated liquid solution represented by point L1, can easily be found by drawing the tie line

ely. characteristic of enthalpy-composition diagrams is tion of

n and subtraction rules along with so-called inverse lever-arm rule.

passing through point L1. The vertical distance between two curves gives the energy required to vaporize the saturated liquid solution totally. Hence, the vertical distances at x = 0 and x = 1 give the latent heats of vaporization of the pure components B and A respectivAn important the applicagraphical additioIf a liquid solution given by point K is added to a vapor mixture shown by R, the resultant mixture is always represented by a point on the straight line (point M) joining the two original mixtures (graphical addition), and its exact location depends on the relative amounts of the phases added and is found by applying inverse lever-arm rule to the line as: Amount of liquid * length of KM line = Amount of vapor * length of RM line Since amounts of liquid and vapor are both known, by measuring the length of KR line and dividing it with the ratio of (amount of liquid/ amount of vapor), point M can easily be located. Similarly, if a mixture shown by point R is subtracted from a mixture given by point M, the mixture that will be left behind is always on the

L1

liquid-vapor mixture

y =x

cold liquid

Spec

ific

thal

en

pies

h,H

0

R G

G1

M

KL

y1x

HG

hM

x1xkzM

HR

hL

L,G

L1,G1

yyR

hK

1.0

x1

y1

y

x

x,y

Fig.5.9 A typical enthalpy-composition diagram

0

0

x

1.0

saturated vaporsuperheated vapor

P =cons.

saturated liquid

y

1.0


extension and behind t t M of the line joining the two original mixtures (graphical subtraction). This is shown by point K in the figure. The position of point K is now found by appl g aga , point K being the support point. The proof of these can be made as follows: Total material balance : K + R = M (5-23) Component A balance : K xk + R yR = M zEnthalpy balance : K hk + R HR= (5-25) If M is eliminated between equations (5-23) and (5-24),

he poin

yin in the inverse lever-arm rule

M (5-24) M hM

kMz= MR zy

R −−

Similarly by eliminating M between equations (5-23)

Kx

and (5-25)

kM

R

hhhHK

−M

R−

= is obtained. From he two equations,

t se

kM

MRyxz

−−

= kM

MR

hhhH

−−

z (5-26)

is found. This equation represents a straight line on enthalpy-composition diagram passing through the oints K k) , M( zM;hM) and R R R). 5.3 Methods of Distillation: After seeing liquid-vapor equilibria, now the distillation methods used in practice can be considered. There are three basic methods of distillation used in industry, from simple to complicated these are:

- Equilibrium or Flash Distillation - Simple or Differential Distillation - Rectification or Fractionation

Let us consider these s, the solution is assumed to be consisting of A and B. However, from time to time the multi-

(xk;h (y ;H p

methods in some details. In the calculation

component mixtures are to be considered too. 5.3.1 Equilibrium or Flash Distillation: The simplest of all the three distillation methods is known as flash or equilibrium distillation. This method is carried out continuously as a single stage operation in two different ways in practice: 5.3.1.1 Equilibrium or Flash Distillation under Constant Pressure: As shown schematically in Fig.5.10, the feed liquid, F is continuously pumped through a heat

vapor

vapor-liquidseparator partial vaporizer

top product D, yD, HD, P, t

liquid

bottom product

liquid

feed

liquid-vapor F, xF, hF, P, tF

P, t, hF1

qS W, xw, hw, P, t

Fig.5.10 Flow diagram of continuous flash distillation under constant pressure


exchanger, where part of it, is vaporized by adjusting the energy qs(kJ/k-mol) given. The vapor-liquid mixture then passes to a vapor-liquid separator from where vapor is taken as top and liquid as bottom products, which are in equilibrium with each other. The top product vapor later may be condensed in a condenser, if it is needed. As heat exchanger, double-pipe or shell and tube type heat exchanger is used. When the temperature is high, the partial vaporizer is placed into a direct- fired oven.

is obvious hich is a

(5-27) F = Dy w (5-28)

: FhF + qS = DHD + Whw (5-29)

eed , top and bottom yD, xW are the and

bottom products respectively and, hF, HD and hW show their specific enthalpies (kJ/k-mol).

ation is ned by ε

A simple cyclone type separator may be used as vapor-liquid separator. It that the top product vapor is richer in the MVC than the bottom product, wliquid. For the whole system: Total material balance : F = D + W MVC balance : Fx D + WxEnthalpy balance = FhF1 can be written, where F, D and W are the molar flow rates of the fproducts (k-mol/s), xF, mole fractions of the MVC in the feed, top

If a fractional vaporiz defi ≡ D/F, then with the elimination of W between equations (5-27) and (5-28) and between equations (5-27) and (5-29),

WDWD hHxy −

SWFWF /Fq)h(hxxε +−=

−=

−is obtained. The left hand side of this equation can also be written as;

(5-30)

ε

xxεε1y F

WD +−

−= (5-31)

As it i e ip between the compositions of top and bottom een these two is given by the equilibrium

For that, the line given by equation

needed for the operation can easily be calculated as shown in Fig.5.11. For the determination of temperature of the mixture, the t-xy diagram of the system at operating pressure is also needed. It is obvious that if the equilibrium distribution curve of the system can be expressed in functional way such as given by equation (5-14), xW and yD are then calculated directly from equations (5-14) and (5-31). In any mass transfer operation, the recovery of the product next to the purities is also important. Hence, percentage recovery of the MVC into the top product from the feed is calculated from;

P.R.=

s s en, this equation gives the relationsh products, another relationship betw

distribution curve of the system, as these two streams are in equilibrium. Equation (5-31) is a straight line on xy-diagram and passes through point F(xF;xF) for a fixed є. Computation of yD and xW is generally made on the graph as the equilibrium relationship is usually given in graphical form. (5-31) is drawn on the diagram, which already contains the equilibrium distribution curve. From the intersection point Q(xW;yD) xW and yD are read. If, in addition, the enthalpy-composition diagram of the system is available, the heat energy

100xyε100

xFyD

F

D

F

D = (5-32)


As it is seen from Fig.5.11, the purity of the top product, which is shown by yD decreases with increasing fractional vaporization, but the percentage recovery of the MVC increases.

Example-5.6) Flash Distilation at

Constant Pressure An aqueous acetone solution, containing 40 mole percent acetone is to be flash distilled at 760 mmHg pressu

P =cons.

at constant pressure

D HD

re. Calculathe compositions of top and bottoproducts and percentage recovery

te m of

acetone for the following fractional , 0.66. ater system at le.App.5.1.

By substituting the x and є values into

are obtained. All these lines, when drawn on through point F(0.40;0.40). Another points are found as

.0;0.8),(0.0;0.6) for lines. With the help

of these points, the lines are drawn and yD values are shown below.

vaporizations: 0.2, 0.4, 0.5xy-diagram of acetone-w760 mmHg is given in Tab

Fequation(5-31); For ε =0.20 yD = - 4 xw+ 2 For ε =0.40 yD = - 1.5 xw+ 1 For ε =0.50 yD = - xw+ 0.8 For ε =0.66 yD = - 0.515 xw+ 0.6

xy-diagram, pass

(0.25;1.0), (0.0;1.0), (0the first and following

from the Q points xw andread from the diagram asPercentage recoveries are then computed

from equation (5-32). The results are given in the table below.

Fig.5.11 Flash distillation

F

M

W

N

Q

hF1=hF+qs/F

xF

spec

ific

enth

alpi

es h

,H

hw

F hF

x yx

xw

w F

yD

x,yD

y =x

0

1.0

0

0 1.0

1.0

y

x

Vapor-Liquid Equilibrium of Aqueous Acetone at 760 mmHg

1

0

0.1

0.2

4

0.5

0.6

0.7

0.8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Mol

e fr

actio

f ace

tone

in v

ap

0.3

0.n o

0.9

or,y

Mole fraction of acetone in liquid,x

yD= - 4xw + 2

yD= - w + 11.5x

yD= - 0.8 xw +

yD=- 0.515xw+0.60

xF

F

ε=0.5

ε =0.4

ε=0.2

ε=0.66


ε

yD

xw

0.20 0.80 0.30 40.0

0.40 0.76 0.16 76.0 0.50 0.72 0.075 90.0 0.66 0.58 0.04 95.7

As it is seen, when fractional vaporization is increased, purity of top product decre

No part of this 171CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

ases but the creases.

:

vaporization takes place. The vapor-liquid mixture is then directed

to a vapor-liquid separator from where top product as vapor, bottom product as liquid are withdra n. This type of operation is more com e operation in practice and the name “flash distillation” is o type of operation than the first type of operation, as the su q t the outlet of the pressure reducing valve. As in this c a also not known, the solution requires trial and errordiagram, t-xy diagram of the system is also needed. As shown in Fig.5.13, first a temperature is selected and the composition of top and bottom products, xw, yD are read from t-xy diagram and from the first term on the right hand side of equation (5-30) є is calculated. Then these compositions are carried to the enthalpy-composition diag a diagram; and substituting these values into the second term on the right hand side of equation (5-30), the є is once more calculated. If the two є are the same, the selected temperature and the calculations are correct. If the difference between two rea ulations are repeated with the new temperature selections until the є s are almost the same. As it is seen from the figures, although certa ion is achieved by using flash distillation, obtaining comple separation and hence pure products is almost impossible, although the purities are the relative volatility is great. In practice, the flash distillation is generally used to divide the multi-

percentage recovery of the MVC in 5.3.1.2 Flash Distillation by Reducing the Pressure of the Heated Liquid Solution In this type of operation as shown in Fig.5.12, the liquid solution, which is heated up to an elevated temperature in a heat exchanger, is passed through a pressure reducing valve where flash

w mon than th first type of more appr priate for this perheated li uid really flashes a

se fraction l vaporization є isa. Next to the enthalpy-composition

ram and hW, HD nd e read from this hF1 ar

є is g t, then the calc

in degree of separatte

relatively high when

bottom product W, xw, hw, P, t

Fig.5.12 Flow diagram of continuous flash distillation at reducing pressure

liquid

feed

liquid-vapor

F tF , xF, hF, P1,

Liquid-vaporseparator

top product D, yD, HD, P, t

liquid

vapor

P1, tF1 hF1H e a t e r

qS

liquid

P , t

pressure red valve

ucing

%100.xy.R.P

F

D =ε=

component feed into two rough products, which are then purified by more sophisticated distillation techniques. In some cases, working in this way is more

economic than using sophisticated distillation methods directly.

H

P =c ns.o For multi-componenfollowing equations can be written:

t solutions, the

Total material balance: F = D + W (5-27) Component-i balance: FxFi = DyDi + Wx (5-33) Enthalpy balance:

Fi Di Wiis obtained from equations (5-27)

then,

D

M

W

D

hF1=hF+qs/F

wi

FhF+qs =FhF1= DHD + Whw (5-29) x = ε y + (1-ε) x

and (5-33), by noting that ε = D/F. If the vapor-liquid equilibrium is given as: yDi = Ki xWi (5-34)

ε)(1Kε i +xx Fi

Wi −= (5-35)

Between equations (5-27) and (5-29), hF1 = ε HD + 6)

can also be written in terms of latent heat of

is found.

(1-ε) hw (5-3

is obtained. This equation

vaporization, heat capacities and temperatures as, tcε)(1t)cλ(εtc wLGoF1FL −++= (5-37) where, reference temperature has been taken zero, oλ (kJ/k-mol) is the latent heat of vaporization of the mixture at reference temperature, FLc and wLc are the heat capacities of the liquid feed and bottom product (kJ/k-mol K) and Gc is the heat capacity of vapor mixture (kJ/k-mol K). All the heat capacities are calculated at average temperatures. Note that temperature tF1 is the temperature of the feed after the heater. In a flash distillation of a multi-component solution ε, P and xF are known. The solution is done as follows: a temperature t is selected and the K value for each component is read from the graph at this temperature and given pressure. Then the mole fraction of each component in the bottom product is computed from equation (5-35). If the summation of these mole fractions is one, then the selected temperature is the temperature of the vapor-liquid mixture after flashing and the mole fraction of each component in the top product is then calculated from equation (5-34). If the summation is different than one, with new temperatures the calculations are repeated.

spec

ific

enth

alpi

es h

, H

F hF

hw

xw xF

x,yyD 1.0

xF xw yD 0 1.0

tem

pera

ture

t

tB

tA

x,y

DW M

h-x

H-y

t-x

t-y

Fig.5.13 Flash distillation by pressure reduction


If the flash distillation is conducted by pressure reduction, equation (5-37) must also be used for the solution, as in this case ε is also unknown. Solution is again started with a temperature selection. With the help of this temperature first ε is computed from equation (5-37). Then the K values are read and the mole fractions in the bottom product are calculated from equation (5-35). If the summation of xwi values gives 1, then the selected temperature is the temperature of the vapor-liquid mixture after

t, the calculations are repeated by satisfied.

flashing and the calculations are correct; if noselecting new temperatures until this condition is

Example-5.7) sh Distil ion by

The bottom product from a distillation column is a bina sition a is supplied continuously at 135 oC and ar p nal vap

ospheric sure apor anproducts. The latent heat of vaporization at 0 oC i 000 feed and bottom product is 250 kJ/k-mole K, the m r sp 80 kJ/k-mole K and the relative volatility is 3. The system obeys Rao s law LVC with temperature is as follows:

t(oC) 60 65 70

Fla lat Pressure Reduction

ry liquid solut 0.5 moleorization

ion of compofr ction. Iton adiabatic pressure reduction to atm

5 bpress 38

olault’

ressure. Calculate its fractio and the compositions of v d liquid kJ/k-mole, molar specific heat of the liquid

ecific heat of vapor is 1 and the variation of vapor pressure of the

75 80 85 90 280 0 4 6 340 41 86 570 70 780

Solution

)barpoB m(

Top product

liquid-vapor

x =0.5

(vapor)

F

Bottom product (liquid)

vapor-liquidseparator

D,yD,HD,P,t

W,xw,hw,P,t

P= 013 bar1.

t

t∆

Distillation column

1 =5 bar P

hK mol-kJ/k 250cFL =

Kmolk/kJ180 −=Gc

Bottom product ary) (bin K mol-kJ/k 250cw =L

tF1 35 oC Press ducingve

=1 ure reval

From equation (5-30) wD

w

WD

WF

xyx-0.5

xyxxε

−=

−−

= (1)

2)

No

From equation (5-37) ( ) )t135(250ttc 1FwL −− (
( ) t)180250(00038tcc GwLo −−
=−−λ

=ε

For ideal solutions equations (5-3), (5-4) and (5-15) are also valid. Solution is by trial and error.

S

p

elect t = 78 oC. Then from the given table mbar533poB = is read.

art of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 173

mbar1599)533)(3(pp oB

oA ==α=

71.0)45.0(1013

==== wD xPP

y

Then from equation (1) 192.00.450.71

0.45-0.5ε =−

=

From equation (2) 438.0)78)(70(38000

)78135(250t)180250(00038

)t135(250−

=−

=−−

−=ε

N

From equation (5-15)

Since no check is available, selected temperature is not correct.

Select t = 81 oC. From table above, mbar590poB =

BA ==α= mbar1770)590)(3(pp oo
From equation (5-15)
From equations (5-3) and (5-4) 45.053315995331013

=−−

=−

−= o

BoA

oB

w pppP

x

1599oAA pp

From equations (5-3) and (5-4) 359.059017705903101

pppPx

oB

oA

oB

w =−−

=−−

=

1770pp oAA
627.0)359.0(
1013x w =

PPyD ===

0.359-0.5
Then from equation (1) 526.0= 0.3590.627
ε−

=

From equation (2) 418.0)81)(70(38000

)81135(250t)180250(00038

)t135(250=

−−

=−−

−=ε

So, t is between 79 and 81 C. Select t = 80 oC. From table above o mbar570poB =

From equation (5-15) mbar1710)570)(3(poB

oA ==α= p

From equations (5-3) and (5-4) 389.057017105701013pP

oo

oB

ppx

BA

w =−−

=−−

=

657.0)389.0(10131710x

Pp

Ppy w

oAA

D ====

0.389-0.5

174

Then from equation (1) 414.00.3890.657

ε =−

=

o part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

tttacawltAfao

5ocF

)80135(250)t135(250 −−

N

424.0)80)(70(38000t)180250(00038

=−

=−−

=ε From equation (2)

Since two є are almost equal, trial and error is stopped. Hence flash temperature is 80 oC, fractional vaporization є =(0.414+0.424)/2= 0.419, mole fractions of the MVC in the vapor and

9. liquid are 0.657 and 0.38

hrough which heating medium is circulated. Vessel is charged with liquid solution up o 70-75 % of its volume and then heating medium is set to circulation. When the emperature of the solution reaches its bubble point temperature, vaporization starts nd the vapor formed rises through the pipe connecting boiler and condenser and upon ondensing there the liquid, which is called top product, flows into a receiver. By djusting the heat energy in the boiler, slow and constant rate vaporization is achieved, hich results in attaining equilibrium at any time between rising vapor and remaining

iquid during the operation. Since the vapor formed is always richer in the MVC than h becomes poorer in the MVC as the distillation proceeds. s a result of this, the forming vapor steadily becomes poorer also in the MVC. It

ollows from this that the vapor formed at the start of the vaporization is the richest nd the vapor formed at the end of vaporization is the poorest in the MVC. The peration is stopped when the amount of liquid in the vessel or its composition drops

.3.2 Simple or Differential Distillation: Simple distillation, which is a single-stage peration, is conducted batch-wise. The equipment, which is named as batch-still, onsists mainly of a boiler and a condenser connected to each other as shown in ig.5.14. The boiler of batch-still is a fairly large vessel containing heating coils

e remaining liquid, liquid

Boiler

Co densen r

Receiver

Charge

Bottoms

Heating medium

Top product

Cooling medim Vapor

Fig.5.14 Batch-still

o part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 175

to a predetermined value. Then, heating is stopped and the vessel content is discharged (bottom product) and the vessel is now ready for a new charge with fresh batch of solution. It is important to note that during the operation although the rising vapor is always in equilibrium with the remaining liquid, the liquid, which is collected in the receiver (top product) at the end of operation, is not in equilibrium with the bottom product. As the comra the MVC than the cuts collected in the subsequent receivers. The analysis of simple

position of the vapor steadily changes, using more than one eceiver and hence collecting more than one top product, which are named as cuts are lso possible. It is obvious that the cut collected in the first receiver will be richer in

distillation was first made by Rayleigh as follows: Suppose that the liquid charged to the vessel consists of A and B and quantity of the initial charge is F k-mole and the mole fraction of the MVC in this charge is xF, and at the end of operation W k-mole liquid is left in the vessel whose composition is xw. Consider the MVC balance at any moment during the vaporization, at which quantity of liquid in the vessel, is L k-mole, the mole fraction of the MVC in this liquid is x, the amount of vapor rising is dL k-mole and its composition is y*, y* dL = d (L x) (5-38) Where (*) shows that the rising vapor is in equilibrium with the remaining liquid, whose composition is x. Then; y* dL = L dx + x dL dL (y*-x) = L dx

⎮⌡⌠

−=⎮

⌡⌠

∗

FxF

xydx

LdL

wxW

and finally, =WF

ln ⎮⌡⌠

−∗

F

w

x

xxy

dx (5-39)

is obtained. With the help of this equation, any one of the F and xw variables, of which any three are known at the start of the operation, can be calculated. It is obvious that in order to perform the integral, y* must be expressed in terms of x. In most of the cases this relationship, which is equilibrium relationship of the system, is given in graphical form and hence, the solution of the problem is then accomplished by graphical integration. For that, as shown in Fig.5.15, y* values at arbitrarily selected x values, which are between xF and xw, are read from the equilibrium diagram and then the calculated 1/( y*-x) values are carried against the corresponding x values on a mill metric paper. The area under the curve between the limits of xw and xF gives the value of the integral. If the relationship between y* and x can be expressed in functional way, the integration can be conducted directly. For example, when the relative volatility of the system is constant, y* can be substituted from equation (5-14) and upon integration,

, W, xF

F

w

Fw

wF

x1x1

ln)x(1x)x(1x

ln1α

1WF

ln−−

+−−

−= (5-40)


or

)x(1W

)x(1Flnα

xWxF

lnw

F

w

F

−−

= (5-41)

is obtained. In addition to these equations, Total material balance : F = D + W Total material balance : F = D + W (5-42)

VC balance : F xF = D xD,av. + W xw (5-43) an also be written from which the average composition of the top product

D,av. can be com the MVC is given

P.R. =

, (5-42)

VC balance : F xF = D xD,av. + W xw (5-43) an also be written from which the average composition of the top product

D,av. can be com the MVC is given

P.R. =

McMc(composited composition), x puted. Percentage recovery of as, (composited composition), x puted. Percentage recovery of as,

100xF

xD

F

av.D, (5-44)

ulti-comp nent ide t’s say component j) is selected as key component and the relative volatilities are defined to If the solution is a m o al solution, one of the components (le

this component. Hence, αij shows the relative volatility of component i to the component j. In this case equation (5-41) is written as,

wj

Fjij

wi

Fi

xWxF

lnαxWxF

ln = (5-45)

If this equation is to be used to compute the composition of the bottom product, it must be solved simultaneously with the equation below by trial and error,

∑ =n

1wi 1x (5-46)

The heat energy requirement in the simple distillation can be calculated from, λD)t(tcFqq q FFLLST +−=+= (5-47) where, qT (kJ) is the total heat energy required, qS (kJ) is the sensible heat needed to bring the solution from initial temperature tF (oC) to the average distillation

oC), qL (kJ) is the latent heat requitemperature t ( red to distill D k-mol distillate, FLc ( kJ/k-m e solution calculated at the arithmetic ol oC), is the molar heat capacity of th

0

Area= WFln

xy1−∗

xw xF xx

x

y

y

y =x

00

1.0

0

P =cons.

Fig.5.15 Graphical integration

1.0


mean of F, and t and t λ (kJ/k-mol) is the latent heat of vaporization of the mixture calculated at the mean boiling point temperature. The time for one batch operation can be calculated by adding the heating (sensible heat transfer) and distillation (latent heat transfer) times to the time required for charging and discharging the batch-still. Heating of the solution from initial temperature tF to the average distillation temperature t is an unsteady-state operation and the time needed is given by,

(sec)cF/UAttttln

FL

H

FH

S

⎥⎦

⎤⎢⎣

⎡−−

=θ (5-48)

for isothermal heating fluid such as saturated steam at tH (oC) temperature, and by

(sec)

K1K

cFcm

ttln F1H

⎥⎢ −tt

FL

HH

1HS

⎟⎠⎞

⎜⎝⎛ −

⎦

⎤

⎣

⎡ −

=θ&

(5-49)

al heating fluid whose inl

for non-isotherm et temperature is tH1. K is given by HHcmeK &= .

UA

In the equations above, (kg/s), and cH (kJ/kg oC) are the mass flow eating fluid, A is the heat transfer area in the batch-still

Hm&rate and mass specific heat of h(m2) and U is the overall heat transfer coefficient (kW/m2 oC). For the distillation time,

(sec)qL=θ (5-50) )tt(AUL −

lnHb

can be used. Where, Ub is the overall heat transfer coefficient of boiling solution 2 o

operation;

er solution a conta ing 20 ole percent a tone w stopped when the m le fraction of acetone in the

tillate to be produced,

=

(kW/m C). Then, total time for one batch θBT = θS + θL + θCD (sec) (5-51) where, θCD is the time (sec) for charging and discharging the batch-still. Although a better separation is achieved in the simple distillation than the flash distillation operating at the same conditions, a complete separation is still not possible. However, high relative volatility may give high purity top products as shown in theexamples below.

Example-5.8) Simple Distillation

123.1 k-mole acetone-wat at 20 oC nd in m ce ill be distilled in a batch still at 760 mmHg. Distillation will be obatch still drops to 0.04. Calculate: a) The amount of dis b) The mole fraction of acetone in the distillate (purity), c) The percentage recovery of acetone, d) Estimate the time of one batch operation for the conditions: Heat transfer area in the still is A 10 m2, heating is supplied by the saturated steam condensing inside the tubes of the still at 1.05 bar gauge


pressure. Over-all heat transfer coeffients for the heating and vaporization periods are estimated as U=200 W/m2K and Ub= 1 000 W/m2K. e) If the same fractional vaporization is realized in a single stage by flash distillation at the same pressure, what will be the purity and percentage recovery of acetone? Molar specific heat of the solution at mean temperature is 83 kJ/k-mol oC, and molar latent heat of vaporizations for acetone and water at 71.4 oC are 28 880 kJ/k-mol and 42 000 kJ/k-mol respectively.

First, plot xy- and t-xy diagrams of the system and then evaluate the integral given by eqn.(5-39) by

values between xF and xW and s as shown in table below:

select read c a l c u l a t e

Vapor-liquid equilibrium for acetone-water system at 760 mmHg is given in Table.App.5.1.

Solution:

selecting arbitrary x reading the corresponding y* value

x y* xy

1−∗

⎮⌠

20.0 dx⌡ −∗

04.0 xy

0.04 0.55 1.961 -

0.08 [(1. 0.71 1.587 961+1.587)/2]*(0.08-0.04) = 0.07096

0.12 0.742 1.608 [(1.587+1.608)/2]*(0.12-0.08) = 0.0639

0.16 0.765 1.653 [(1.608+1.653)/2]*(0.16-0.12) = 0.06522

0.20 0.780 1.724 [(1.653+1.724)/2]*(0.20-0.16) = 0.06754

(The value of integral) Total = 0.2676 W = 94.19 k-mol

44), d) As it is seen from the t-xy diagram oC and ends at 78 oC. Then, take the mean temperature e.

a) From equation (5-42), D = 123.1 - 94.19 = 28.91 k-mol b) From equation (5-43), (123.1)(0.20) = (28.91) xD,av + (94.19)(0.04) xD,av = 0.721 c) From equation (5-

of the system, distillation starts at 64.8 71.4 oC as the distillation temperatur

From equation (5-47), the sensible heat required, qS = (123.1)(83)(71.4 - 20) = 525 170 kJ

As λ = (28 880)(0.721) +( 42 000)(1 - 0.721) = 32 540 kJ/k-mol, The latent heat required, q = (28.91)(32 483) = 940 732 kJ

(5-48), the tLime required to heat the solution from 20 oC to 71.4 oC ; From equation

2676.0dxFln =⎮⌠=

∗

20.0

xyW 04.0⌡ −307.1e1.123 2676.0 ==

W

%7.84100.)20.0)(1.123()721.0)(91.28(100.

Fx.R.P

F

av,D ===Dx

[ ] min07h1.sec ≈9973)83)(41.)4.71121(/)20121(ln

S =135(/)10)(200.0(

−−=θ


Vapor-Liquid Equilibrium of Aqueous Acetone at 760 mmHg t-xy diagram of Aqueous Acetone at 760

mmHg

100

C)

To account for heating of the batch-still itself (metal), F is taken as (1.1)(123.1) =135.41 k-mol . From steam tables, temperature of the saturated steam at Ps=1.05+1.013 = 2.063 bar absolute pressure is read as t o

From ( e ti So, f uat - 51); req h operation, with 1 h charging and discharging time; θ 07 mi e) Fractional vaporization,

H =121 equation

C. 5-50), th me required for distillation;

rom eq ion (5 the time uired for one batc

BT = 1 h n + 32 min + 1 h = 2 h 39 min

235.01.123

91.28==ε

Then, from ) equation (5-31235.0

85.

20.05 x235.0

23.0y WD +−

−=

0xy WD

1

26.3 +−= is obtained. By drawing this line on ed.

rom equation (5-32)

s it is seen, simple distillation gives higher purity and higher P.R. than the flash distillation under

Example-5.9) Simple Distillation: Effect of Relative Volatility on Purity

.1 k-mole (12 000 kg) chloroform-toluene solution at 20 oC and containing 20 mole percent

T

) Estimate the tim ea in the still is A=10 2, heating is supplied by the saturated steam condensing inside the tubes of the still at 1.05 bar gauge

ressure. Over-all heat transfer coeffients for the heating and vaporization periods are estimated as =200 W/m2K and Ub= 1 000 W/m2K. Compare the purity of distillate with the Example-5.8 and comment on the results.

Assume that specific heat and latent heat of solution remain constant at 140 kJ/k-mol oC and

xy-diagram, and reading the coordinates of point Q ; yD = 0.67, xw = 0.065 are obtain F Aidentical operation conditions.

123chloroform will be distilled in a batch still at 760 mmHg. Distillation will be stopped when the mole fraction of chloroform in the batch still drops to 0.04. Calculate: a) he amount of distillate to be produced, b) he mole fraction of chloroform in the distillate (purity), T c) The percentage recovery of chloroform, d) Heat energy requirement for the operation. e e of one batch operation for the conditions: Heat transfer armpUf)

min32sec8971 ≈=)4.71121)(10)(.1(

732940−

=0Lθ

00.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.20.30.40.50.60.70.8

y

0.91

x

50te

60

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

m

70

pera

t

80

90

ure,

t (o

Mole fraction of acetone, x,y

%7.78100.)20.0()67.0()235.0(.R.P ==


32 000 kJ/k-mol. Mchloroform= 119.4 , Mtoluene= 92 Vapor-liquid equilibrium of the system at 760 mmHg is given in Table.App.5.2. Solution: a) First, plot xy- and t-xy diagrams of the system and evaluate the integral given by equation (5-39) by selecting arbitrary x values between xW and xF as in the previous example.

select read c a l c u l a t e

x y*

0.04 0.155 8.696 -

0.08 0.275 5.128 [(8.696+5.128)/2]*(0.08-0.04) = 0.27648

0.12 0.358 4.202 [(5.128+4.202)/2]*(0.12-0.08) = 0.1866

0.16 0.430 3.704 [(4.202+3.704)/2]*(0.16-0.12) = 0.15812

0.20 0.500 3.333 [(3.704+3.333)/2]*(0.20-0.16) = 0.14074

Total = 0.76194

ol

) From equation (5-43), + (57.47)(0.04) xD,av = 0. 34

) From equation (5-44),

e 100.5 oC as the distillation temperature. rom equation (5-47), the sensible heat required, q = (123.1)(140)(100.5 - 20) = 1 387 337 kJ

The latent heTotal heat required, q = 3 487 497 kJ

metal), F is taken as (1.1)(123.1) =135.41 k-mol. team at P =1.05+1.013 = 2.063 bar absolute pressure

W = 57.47 k-m From equation (5-42), D= 123.1 – 57.47 = 65.63 k-mol b (123.1)(0.20) = (65.63) xD,av c d) It is seen from the t-xy diagram of the system, distillation starts at 94.5 oC and ends at 106.5 oC Then, take the mean temperaturF S

at required, qL= (65.63)(32 000) = 2 100 160 kJ T

e) From equation (5-48), the time required to heat the solution from 20 oC to 100.5 oC ; To account for heating of the batch-still itself (From steam tables, temperature of the saturated s sis read as tH=121 oC. From equation (5-50), the time required for distillation;

min51h2sec24510

)5.100121)(10)(0.1( −

1601002L =θ ≈=

xy −∗

1⎮⌡ −∗

04.0 xy⌠

20.0 dx

76194.0xyW 04.0

⎮⌡ −∗

dxF 20.0⌠ln == 142.2e

W==

1.123 76194.0

%6.

[ ]

90100.)20.0)(1.123()34.0)(63.65(.R.P ==

min12h4.sec11615)5.100121(/)20121(ln≈=

−−)401)(41.135(/)10)(200.0(S =θ


So, f uation (5-51); requ charging and discharging time; = 4 n + 2 f) As it is y of the top product is m duct of the previous exampl er t and th ystems. From xy-diagrams of the two systems, the relative volatility of acetone to water and the relative volatility of chloro to to er t ntrati lp of equation

x 0.04 0.08 0.12 0.16 0.20

rom eq the time ired for one batch operation, with 1 h

θBT h 12 mi h 51 min + 1 h = 8 h 03 min

seen, the purit uch less than the purity of the top prois, look at the relative volatilities of the two se. In ord o underst

form luene ov he conce on range involved were calculated with the he(5-13) as follows: αAcetone-Water 29.33 28.16 21.09 17.09 14.18 3 3.96 4.0 αChloroform-Toluene 4.4 4. 6 4.09 As olatilities of the two system. Flash and imple distillations can produce high purity top products only when the relative volatil

it is seen, there is great difference between the relative vss

ity of the these examples that flash or simple distillation techniques can

tions such as chloroform -toluene whose relative volatilities are not great.

iques r distillation method, which is

he previous nes, must be used for high purity products and high percentage recoveries. For that, e rectificatio mical industry.The

quipment used is multi-stage column, which may be either packed or plate type. lthough continuous operation is more common, batch-wise operation is not also very

is a

) is introduced centrally to the column. As the liquid flows down, it tes where mass transfer between the

phases takes place in both directions. The vapor and liquid contacting on a plate try to or that, part of the LVC passes from vapor to liquid by partial

ystem is great. So, it is obvious fromnot be used to separate soluAs will be considered in the next sections, rectification is the only technique to separate such solutions. 5.3.3 Rectification or Fractionation: None of the ab

solution. Anotheove given distillation techn

produces complete separation of the known as rectification and which is more complicated and expensive than toth n is the most widely used separation method in cheeAseldom. A typical rectification column is shown in Fig.5.16. Although this columnplate column it could also be a packed column. As it is seen, the liquid solution to be separated (the feedcomes in contact with rising vapor on the pla

come equilibrium. F

Mole fraction of chloroform, x,y

Tem

pera

ture

, t (O

C)

t-x

t-y

t-xy diagram of chloroform-toluene at 760 mmHg xy-diagram of chloroform-toluene at 760

mmHg

00.10.20.3

0.91

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

tB

0.40.50.60.70.8

0x

y

tA


condensation. This enriches the liquid in the LVC and the remaining vapor in the MVC. The heat of nsation given by the condensing LVC is taken up by the MVC and part of it vaporizes. While this enriches the liquid in LVC further, the vapor is also enriched in the MVC further. As a result of these, while the liquid leaving the plate and flowing to the plate below becomes richer in the LVC than the liquid entering the plate, the vapor leaving the plate and going to the plate above becomes richer in the MVC than the vapor incoming the plate. These partial vaporizations and

partial condensations are repeated on each plate. As a result of these, the liquid solution flowing down from plate to plate continuously enriches in the LVC. By providing sufficient number of the plates (or packing

made almost pure in the LVC. Part of this liquid is

mass transfer is formed in the column. Notice that as this vapor is

rmed liqui h is ost p LVC s also

most the . The

column; this vapor enriches in the

y

conde

height) in this region of the column, the liquid reaching bottom of the column can be

vaporized in a reboiler and returned back to the bottom of the column as reflux vapor, thus a vapor phase, which is required for interphase

fo from a d, whicalm ure in the , itself ial pure in LVCremaining part of the liquid is withdrawn as bottom product. This liquid is the liquid, which is richest in the LVC in the column. If we follow the vapor given back to the

MVC as it passes from plate to plate up by repeated partial vaporizations and condensations taking place on the plates as explained above. It has the highest MVC composition at leaving the plate to which the feed is introduced (the feed plate).

However, although this vapor is enriched in the MVC considerably, it is not pure in the MVC, as it still contains some LVC. This vapor must be brought in contact with a liquid, which is richer in the MVC in order to remove the remaining LVC. This contact is provided on the plates (or in the packing) placed above the feed plate. As the rising vapor comes in contact with a liquid which is richer in the MVC on a plate, the LVC of the vapor passes from vapor to liquid by partial condensation and the MVC of

e liquid passes from liquid to vapor by taking this heat of condensation. Bth

liquid

condenser

v apor

qC

1 Reflux liquid Top

product Enriching section

Bottom product

vapor

Feed

Stripping section

N

reboiler

Fig.5.16 A plate rectification column

qs


providing sufficient number of plates (or packing height) above the feed plate, the vapor leaving the column can be made almost pure in the MVC. The vapor, leaving the column is directed to a condenser, where it condenses totally and part of the

condensate (liquid) is returned back to the top of the column as reflux liquid to provide the liquid phase rich in the MVC on the plates above the feed plate. The

ining part of the condensate is drawn as top product or distillate. e look at the liquid and vapor phases ing inside the column, we see that apor, which enters the column at the m as almost pure in the LVC, turns VC as it rises in the column and

es the column as almost pure in the C at the top. The liquid, which is n at the top of the column as almost in the MVC, turns to the LVC as it s down and with joining the feed at feed plate leaves the column at the m as almost pure in the LVC. As it een, all the streams leaving the mn at the top and bottom are not n as products, but parts of them, after e changes, are returned back to the mn as reflux. If this is not done, uction of pure products is not ible. The liquids in the column are ys at their bubble point temperatures the vapors are at their dew point

erature in the column is always at op and the highest temperature is at ottom. As it is seen, the feed divides

column in two sections. The section

ed as stripping section of the mn as in this section the rising vapor s off the MVC from the liquid. The

section above the feed plate is known as enriching section of the column as the vapor in this section is further enriched in the MVC. The purities of the top and bottom products do not only depend on the number of the plates in each section but

Feed

G1,y1,H1

reboiler

F,zF,hF

Top product

condenser

G2,y2 H2

D,xD,h

remawithIf wflowthe v

oM

leavMVgivepureflowthe bottois scolutakephascoluprodpossalwaand temperatures. Hence, the lowest tempthe tthe bthe below and including the feed plate is namcolustrip

bottto

Bottom product

D

1

m

F

ig.5.17 Notations in a continuously operating plate rectification column

L1,x1 h1

n

2

N

L2,x h2

2 G3,y3 H3

Gn+1,yn+1 Hn+1

Ln,xn hn

Gm+1,ym+1 Hm+1

Lm,xm hm

GN+1,yN+1

Lo,xo ho

qC

Ln-1

,xn-1 hn-1

Gn,yn Hn

LN,xN HN+1 hN

qS

W,xw,hw


also on the ratio of liquid to vapor in each section. Typical rectification system as shown in Figure does not consist only of the column, but it contains also a reboiler and a condenser as inseparable part of the system. In many applications, in order to reduce

bottom (sometimes top) product and the e products to storage temperatures, are

e insulated against the heat losses. The may be plate or packed column. Below, on column will be given. Column: The first step in the design of a the numbers of the plates required in s in the column are first assumed as ideal s absorption, although they are not, and ach section are computed. By using plate

oretical plates are then converted to the mbered from top to the bottom and n and ng section respectively. N is the last plate flow rates of the liquid and vapor phases mole fraction of the MVC in liquid and ate, the number of the plate from which nce, L

the heating cost, a heat exchanger between feed is placed. oolers, which cool thfrequently added to the product lines. The column itself and all the hot lines arcolumn, which is the main part of the systemfirst the design principles of a plate rectificati5.3.3.1 Continuous Rectification in a Plateplate rectification column is to determine enrichin d str g s ons e plateor equilibrium or theoretical s, as in gathen numbers of the ideal plates required in eor colu ffic es, numbers of the thenumbers of the real plates. The plates are num show any plate in the enriching and strippiof the column. L and G are the total molar respectively as k-mol/s. x and y show the vapor. Since these change from plate to plthese streams come, i given as subscript. Hethe 5th e, m raction of the MVC inspecific enthalpy of the vapor leaving the 7

lar flow rates of the feed, top and bottomtion of the MVC in the feed, top and bo

p and bottomThe total and MVC balances and the enthalpysteady-state are written as, Total material balance : Gn+1 + Ln-1 = Gn + Ln MVC ba ce Gn+1 yn+1 + Ln-1 xn-1= Gn yn + LnEnthalpy balance :

Gn+1 Hn+1+ Ln-1 hn-1 = Gn Hn+ Lwhere, H a e the specific enthalpliquid. equations (5-21) and (5-22), Hn+1 =

The c

g an ippin ecti . All thplate

mn e ienci

5, y4 and H7 show the liquid leaving the vapor leaving the 4 plate and the plate respectively. F, D and W are the

products and z

sth

thplat ole f

F, xD and xw are the mole ttom products and h

moF, hD and hw are the

products respectively. balance equations around the plate n at

(5-52)

fracspecific enthalpies of the feed, to

lan : xn (5-53)

n hn (5-54) ies of the saturated vapor and saturated nd hn arn+1

From Bλ)y(1λy)t(t 1nA1no1nLB +++ −c)y(1)t(tcy o1nLA1n ++ −+−

++−1n+

(tc)x(1)t(tcxh nnonLAnn s∆H)toLB +−−+−= can be written. In many cases, t ≅ d

BA λ,λ > 1nt + an ∆ Hs and hence ∆ Hs can be equation from the first,

Bλ

n

neglected and then by subtracting t nd 1nn1n )y1yhH +++

he seco1nA (λ −+=− (5-55) th

hen λA ≈ λB(

is obtained. It is known that the molar (not e mass) latent heats of vaporization of

=some liquids are almost equal. T λ ) and it follows from equation (5-55) Hn+1-hn = λ = constant


This means that the heat liberated upon the condensation of 1 mole of vapor on a plate is just enough to vaporize only 1 mole of liquid. It follows from here that Gn+1= Gn and from equation (5-52) Ln-1 = Ln . Hence, if the molar latent heats of vaporization of the components forming the solution are almost equal (a deviation up to 25% is assumed equal from engineering calculations point of view) the molar (not the mass) flow rates of liquid and vapor remain constant throughout, provided that there is no stream entering and leaving the column in the section considered. In this special case, which is known as “constant molar over-flow case” the subscripts are not needed to be used and hence with G and L in the enriching section and with LandG in the stripping section the molar flow rates of vapor and liquid can be shown. As a stream (the feed) is added to the column in somewhere, the flow rates of vapor and liquid are not expected to be the same in the enriching and stripping sections. The calculation of the number of the ideal plates (and also the height of packing) is simplified considerably when constant molar over-flow case applies, as in this case the knowing of xy-diagram of the system at the operating conditions is sufficient. On the contrary, if constant molar over-flow case does not apply, enthalpy-concentration diagram next to the xy-diagram is also needed as in this case flow rates of vapor and liquid, even expressed in molar units, do not remain constant and change from plate to plate. Below, rectification under constant molar over-flow is to be first considered. Rectification under Constant Molar Over-Flow: At steady-state, the following equations between top and any plate of the enriching section (envelope-1) of the column, as shown schematically in Fig.5.18, can be written: Total material balance : G = L + D (5-56) MVC balance : G yn+1 = L xn + D xD (5-57) The last equation can also be written as:

GxD

xL

y D+= (5-58)G n1n+

g the plate. Hence, this equation is nothing but operating line equation. As

s the perating line equation. It is represented by a straight line passing

through poin tio of

most important operating parameter. The ) is then preferred written in terms of reflux ratio. By dividing both

or and denominator of equation (5-58) by D and by noting that RD = L/D,

This equation relates the compositions of the liquid leaving and the vapor enterinsame numbered this equation was written in the enriching section then the equation (5-58) ienriching section o

t D(x ;x ) on xy-diagram. In any rectification operation, the raD Dliquid rate given back to column at the top to the rate of top product, which is known as reflux ratio and shown by RD , is theequation (5-58nominat

D

nD

D1n R1

xR1

Ry

++

+=+ (5-59) Dx

his equation is represented by

is obtained. T a straight line passing through point D(xD;xD) with a slope of RD/(1+RD) on xy-diagram. Similarly, in the stripping section of the column between any plate and bottom of thecolumn (envelope-2);


Total material balance : WGL += (5-60)

MVC balance : w1mm xWyGxL += + (5-61)

are written. The last equation can also be written as,

GxW

xGL

y w1m −=+ (5-62)

This equation gives the relationship between vapor entering and liquid leaving the same numbered plate. Hence, it is an operating line equation. Since it was written in the stripping section, it is then stripping section operating line equation. This line cuts the diagonal at point W(x

m

F D wand percentage recovery of the MVC

:

w;xw). For the whole column; the following equations can also be written: Total material balance :F = D + W (5-63) MVC balance F z = D x + W x (5-64)

.100zF

P.R.F

D= (5-65) xD

Let us now find the intersection point of the two operating lines,

which is given as Q(xq;yq). Since this point lies both on enriching and stripping section operating lines, the coordinates of this point must satisfy the equations (5-57) and (5-61). Hence, Dqq xDxLyG += wqq xWxLyG −= are written. By subtracting the second from the first, wDqq xWxDx)L(Ly)G(G ++−=− (5-66) is obtained. If the value of D xD + W xw is substituted from equation (5-64) and then both sides of the equation is divided by F:

Fqq zxF

yF

+⎠⎝

=⎠⎝

(5-67)

is found. On the other hand, total material balance ar

LLGG⎟⎞

⎜⎛ −

⎟⎞

⎜⎛ −

ound the feed plate is written as GLGLF +=++ . By rearranging this equation,

L

G L

G

F

xN

Envelope-1

F,zF

D,xD

1

m

Fig.5.18 Constant molar over-flow rectification

n

2

N

L G

xn

G,y1

L,xo

W,xw

qC

yn+1

xm ym+1

L G

qS

yN+1 G

L

Envelope-2


1F

GGF

LL+⎟

⎠⎞

⎜⎝⎛ −

=⎟⎠⎞

⎜⎝⎛ −

is obtained. Upon substitution of

(5-68)

)G(G − /F from this equation into equation (5-67),

Fz qq xF

LLy

FLL

1 +⎟⎠⎞

⎜⎝⎛ −

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

− (5-69)

is found. If a q-par ter is defined by

ame

FLL

q−

≡

finally, equation (5-71) is obtained from equation

(5-70)

s (5-69) and (5-70)

1qz

x1q

qy F

qq −−

−=

This equation, which is known as q-line equatpassing thro point F(zF;zF) w of operating lines cut each other always on this linintersection po s of the two operating linesparameter defined by equation (5 0) must be knthat the value of q-parameter depends on the thermal condition of may be introduced to the column in one of e 5 liquid; as in this case the feed is below s satcondensation of the part of the vapor coming from stripping section on the feed plate in order to become saturated liquid. The chan aphases around the feed plate is schematically shoseen. An equation for the calculation of exact va

(5-71)

ion, is represented by a straight line q/(q-1) on xy-diagram. Since the two e, equation (5-71) is then the locus of . For the drawing of q-line, the q-own. It follows from equation (5-70)

ugh ith a slope

int-7

the feed. The feed thermal conditions. The th

itse are: a) Cold

uration temperature, it causes to the

ge of quantities of v por and liquid wn in Fig.5.19 a, from which q > 1 is lue of q, which will be greater than 1,

L

GL

G

F

L

GL

G

F

L

G LGL

G

F

L G

L

G L

G

F

F

(a) (b)

(d) (e)

(c)

Fig.5.19 The change of quantities of vapor and liquid around the feed plate according to ermal conditions of feed the th


can easily be derived by considering enthalpy balance around the feed plate as will be done later. b) Saturated liquid : since the feed is saturated liquid solution, it simply

ming from the enrichi A mixture of va

n in Fig. 5.19 c. q-param

ply the mole fraction of the liquid from stripping section

q-parameter equals to zero. e) Super-heated se the su

tan zero.

ditions of ematically shown in F

joins the liquid co ng section of the column as shown in Fig.5.19 b, and q in this case equals to 1. c) por-liquid: in this case vapor part of the feed joins the vapor coming from stripping section and liquid part joins the liquidcoming from the enriching section as show eter, which is 0 <q <1, is sim in the feed mixture. d) Saturated vapor: as the feed is saturated vapor, it joins the vapor comingas shown in Fig.5.19 d and the value of vapor: in this ca per-heatedness of the feed vapor is given up to the liquid coming from the enriching section on the feed plate and as a resul of this, part of the liquid vaporizes as shown in Fig.5.19 e and the value of q is smaller thThe appearances of the q-lines on xy-diagram at these 5 different thermal conthe feed are sch ig.5.20. An enthalpy balance around the feed plate gives, GLGLF GHhLHGLhFh +=++ (5-72) Assuming LL hh ≅ and GG HH ≅ this becomes L)L(hG)G(HFh LGF −=−+ By dividing both sides of this equation with F and substituting the values from equations (5-68) and (5-70) finally,

λ−

=−−

= FG

LG

FG hHhHhH

q (5-73)

is obtained. This equation is another definition of q-parameter in terms of enthalpies. Since HG and hL are the enthalpies of saturated vapor and liquid, the difference between them is simply latent heat of vaporization. For the cold liquid feed, λ)t(tchFH FFbFLG +−=− can be written, where tF and tFb are the entering and bubble point temperatures of the feed and FLc is the molar heat capacity of the feed liquid, calculated at the arithmetic mean of tF and tFb. By substituting this into equation (5-73),

λ

)t(tc1q FFbFL −

+= (5-74)

is obtained. For superheated vapor feed, )tt(chHG FFdFGF −=− can be written, where, tF and tFd are the entering and the dew point temperatures of the feed vapor and FGc is the molar heat capacity of the feed vapor calculated at arithmetic mean of tF and tFd . By entering this into equation (5-73),

λ

)t(tcq FFdFG −

= (5-75)

is found.

GH,G

L,hL

F,hF

G,HG

Lh,L


After seeing the operating and q-lines, we can now start with the calculation of the

number of the ideal platthe know

es required for a given separation. But before this, let us write ns. They are: F, z , thermal condition of the feed, xD and xw (or percentage

e operating pressure and obtains the gner also decides on the reflux ratio

he operation by considering the criteria, which will be given later. Of eck whether the consta e ap heats of vaporization t th

hen q-line is drawn on the same the q value from the ther e

Since the condenser is a total condenser (there r operates as partial co ll b

o D composition of the vapor y1 composition of the liquid x1 leaving the same plate is read from

this y1 into the diagra

2compared with the known xq to check whether the feed plate

t calculated x val hanues the stripping section operating line equation is used instead

ng section operating line equation, as we have entered the stripping section of the column. Calculation is continued in the stripping section in the same way as in the enriching section; once using equilibrium curve and once using operating line

Frecovery of the MVC). The designer then selects thxy-diagram of the system at this pressure. The desito be used in tcourse, he must also ch nt molar over-flow cas plies or not by obtaining the latent of the components a e operating conditions. Plate to Plate Calculation Method or Lewis-Sorel Method: First xy-diagram of the system is plotted at the operating pressure, and tdiagram after calculating mal condition of the feed. Th operating lines for stripping and enriching section are then written by substituting the known values into equations (5-59) and (5-62) and by drawing the enriching section operating line on the same diagram point Q hence xq is found. The calculations are now started at the top of the column.are cases where condense ndenser, this case wi e dealt with later), y = x = x . Hence, the leaving the first plate is 1known then theequilibrium curve by entering m. Now by substituting this x1 into enriching section operating line equation y2, the composition of the vapor leaving the second ideal plate is computed. Returning back to the equilibrium curve x2 is read from the diagram by entering the y value. After reading each x value from the equilibrium diagram, it is is reached or not. If the las ue is equal to or smaller t xq then for the calculation of y valof enrichi

y=x(a)

0<q<1

(d)

q< 0

(e)

zF

q>1q=1

q= 0

F

(b)

(c) y

1.0

x

Fig.5.20 The appearances of q-lines on xy-diagram

00

1.0

P=cons.


equation, but this time comparing the calculated x values with the given xw. When the last calculated x value is equal to or smaller than xw, the calculation is stopped as we have reached the bottom of the column and the subscript on the last x value gives the total number of the ideal plates that must be in the column to achieve the specified separation. Graphical Calcul In the calculation of the number of the ideal pl uilibrium curve and operating lines are alt n. McCabe and Thiele, seeing t he op s on the diagram on which equilibrium curve and q-line already been plotted, and then they have drawn the right angle triangles between equilibrium curve and operating lines starting at the top conditions of the column as shown in Fig.5.21. Each triangle represents an ideal plate in the column, as the compositions around which are the compositions around each ideal plate. Around q-line a switch from enriching section operating line to the stripping section operating line is done. T numbers ngle triangles above and below q-line es in enriching and stripping sections.

ation Method or McCabe-Thiele Method: ates by plate to plate calculation method, eqernately used, starting at the top of the colum

his, have drawn thad

erating line

he of the right agive the required numbers of the ideal plat

1.0

The Importance of Reflux Ratio: It was stated above that the reflux ratio is an important operating parameter, and the designer does its selection. Let us consider the effect of reflux ratio on the number of the ideal plates. It follows from Fig.5.21 that if the reflux ratio is decreased keeping everything else constant, the operating lines approach the equilibrium curve and as a result of this, more right angle triangles are drawn between them. This means that more ideal plates are needed to make the same separation than before. It is understood from this that the number of the ideal plates is inversely proportional to the reflux ratio. If the reflux ratio is further decreased as

Fig.5.21 Determination of number of ideal plates by McCabe-Thiele method

D

W

F

Q

K

q-line

Enric.sec.op.line

Strip.sec.op.line

y=x

D

D

R1x+

00

P=cons.

zF xD=xo 1.0

y2

y1

y3

y4

2

1

3

4 y5

y7

y6

y8

6

7

8

5

x3 x4 x5 x6 x2 x1 x7 xw x8

x

y


his CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu 192

nce the ideal plates is infinite at this reflux ratio which is named as inimum reflux ratio and shown by RDm. So, the minimum reflux ratio is defined as

condition of the feed for a given system. It is obvious that reflux ratios greater than minimum reflux ratio are to be selected by the designer. In practice operating reflux ratio, which is always greater th m m r ux rat in terms of minimum reflux ratio in the form of RD= β R . The value of β, which is always greater than 1, is selected with economic considerations as will be explained later. At minimum reflux ratio, the enriching section operating line is written as,

shown in Fig.5.22a, point Q further approaches the equilibrium curve, finally coming on to it (point Q'). In other words, the two operating lines and the q-line cut each other on the equilibrium curve. Construction of right angle triangles starting at the top conditions results in a pinch at or around the Q' point showing that number of the triangles, hemthe reflux ratio at which desired separation can only be done in a column containing infinite number of the ideal plates. In some systems, the enriching section operating line may cut the equilibrium curve before cutting the q-line as shown in Fig.5.22b. In this case, the tangent drawn from point D is taken as enriching section operating line at minimum reflux condition. The minimum reflux ratio depends on the thermal

y=x

P=cons.

Q'

an inimu efl io, is given Dm

Dmn

Dm1n R1R1 +++

Hence, it can easily be calculated either by equating the slope of DQ' line (for Fig.5.22 b the slope of tangent) to R

DDm xRxy += (5-76)

Dm/(1+RDm) or equating its intercept value (point K') to xD/(1+RDm). From the first,

qq

qDDm xy

yxR

′−′

′−= (5-77)

is obtained.

Q'

F F

D

W

D

W

Q Q

0

Strip.sec.op.line at m reflux in. ratio

q-line q-line

Enr.sec.op.line at min. reflux ratio

000

T

K'

K'

Dm

D

R1x

+

xD 1.0 1.0

1.0 1.0

zF zF xw xD xw qx′

qy′

qx′

qy′

y

y=x

x x

yP=cons.

Dm

D

R1x

+

(a) (b)

Fig.5.22 Determination of minimum reflux ratio

Strip.sec.op.line at min.ref ratio

Enr.sec.op.line at min. reflux ratio

lux

No part of t

Underwood has derived a series of analytical equations from which minimum reflux ratio can be computed without resorting to drawing. He first found the coordinates of the intersection point of enriching section operating line and q-line at minimum reflux conditions as,

Dm

DmFDq Rq

)R(1z1)(qxx

+++−

=′ qR

zRxqyDm

FDmDq +

+=′

then he substituted these coordinates into the definition equation of relative volatility

written at point Q' which is )x/(1x)y/(1y

αqq

qq

′−′

′−′= and obtained the following equation,

[ ]

)x(11)(q)z(11)(R)xq(1)z(1R DDm

DmFD

DFDm

DFDm

−−+−+=

−+− (5-78)

RDm can be computed from this equation only b trial and e or a given value of q. However, equation (5-78) simplifies to the following equations for

q =1

1)(Rz1)(qxαxqzR ++−+

F

y rror f

⎥⎦

⎤⎢⎣ −

−−

−=

)z(1)x(1α

z1α F

D

FDm -79)

q =0

⎡ x1 D (5R

⎟⎠

⎞⎜⎝

⎛

−−

−−

=F

D

F

DDm z1

x1zxα

1α1

R (5-80)

The α in the equations must be taken at the intersection p t Q') of the two operating lines. Let u on er now another limit of reflux ratio, which is known as total reflux. The reflux ratio, which is defined by R = /D becom finite at D = 0. When no top product is withdrawn from n, no bottom product must be withdrawn and hence no feed is introduced. Hence, certain quantity of solution rged to the column is vaporized in the reboiler and then condensed in the condenser and returned back to the column. In this ty under total reflux, the slopes of both operating lines become al between points D

flux

an analytical equation for e nutive volatility is constant or near constant.

oint (poin

s c sidD L es in

the columcha

pe of operation, which is known as operation1 and hence the diagon

and W becomes operating line, as no feed is introduced to the column only one operating line exists. As shown in Fig.5.23, the number of the ideal plates needed for a given separation is a minimum under total reflux. It is obvious that operation under total reflux cannot be a normal operation, but the columns operate under total reflux for a certain time to reach the steady-state, when they are first taken in operation, after which is passed to normal operation slowly. The operation time under total redepends on both the size of the column and the system. Fenske has derived the calculation of th mber of the ideal plates under total reflux, when the relaFrom the definition equation of relative volatility for the last plate of the column,

N

NN xα

1y

=−

(5-81) N x1y −

al reflux y = x n b ome

N-1 and then this equatio ec s; can be written. But at tot N

N

N

1N

1N

x1x

αx1

x−

=− −

− (5-82)


Similar equation for the plate (N-1) is,

N

Nx (5-83) 2

1N

1N

1N

1N

x1α

x1x

αy1

y−

=−

=− −

−

−

−

are written for the other plates pIf similar equations until reaching the first late,

N

N

1 x1α

y1 −=

− (5-84)

is obtained. For

N1 xy

a column equipped with a total condenser y1 = xD, and if the bottom product is taken from the last plate of the column, xN = xw. Then, equation (5-84) becomes,

w

wN

D

D

x1x

αx1

x−

=−

(5-85)

By taking the logarithm of both sides and solving for N,

αlog

)x(1x)x(1x

logN Dw

wD

m−−

= (5-86)

is finally obtained. As in this case the required ideal plate number is minimum, N = Nm is written. If relative volatility changes along the column, but if this change is not great, geometric mean of the α values calculated at the top and at the bottom of the column is used in the equation above. Optimum Reflux Ratio: The total cost of separation by rectification is the sum of the fixed capital cost of the system and the operating cost, as in many other engineering operations. The fixed capital cost of the system consists mainly of the costs of column, reboiler, condenser, coolers and other heat exchangers and the pumps. The column

e greatest of all, is proportional to the diameter and the height (or number of the plates) of the column. Since at RD=RDm the column cost is

he cost of op in the heig

lthough with incre ce

cost, which is probably th

infinite (why?), the fixed capital cost is also infinite. With increasing RD, tcolumn decreases due to the dr ht, and hence the fixed capital cost of the system also decreases. A asing RD the sizes and hen the costs of

5

1

W

F

D

xw zF x

xD

4

3

2

y =x

P=cons.

y

0

1.0

0 1.0

Fig.5.23 Determination of ideal plates under total reflux


reboiler and condenser increase (why?), these are much offset by the savings in the column cost due to the sharp drop in the height of the column at RDs near the RDm. But after certain value of RD, this tendency reverses and the fixed capital cost of the system increases with increasing RD. This can be explained as follows: At RDs which are fairly greater than RDm, drop in the height of column, although exists, is not significant, but increase in the diameter is rather high due to the increased vapor and liquid flows in the column, and as a result of this the cost of column increases with increasing RD in this region. Not only this, but the costs of reboiler and condenser rise also due to the increase vapor loads. Hence fixed capital cost of the system increases with increasing RD in this region as shown qualitatively in Fig.5.24. As for the operating cost, this cost arises when the system operates. The operating cost in a rectification operation consists mainly of the costs of heating and cooling duties (or loads) of reboiler and condenser which are given by qs = and qc =λG λG respectively. As it is seen, these costs are directly proporti

es with increasing RD. o

onal to the vapor flow rates in the column.

A material balance aro G = D(1+Rund the condenser gives D), which shows that cost of condenser load increas On the other hand, since f r a saturated liquid feed )RD(1GG D+== can be written, the cost of reboiler duty also increases with RD. It follows from here that operating cost increases always with reflux ratio as shown in Fig.5.24. The total cost is obtained by adding these two costs as shown in Fig.5.24. The minimum of this total cost curve gives the optimum value of the reflux ratio. The optimum reflux ratio does not only change from system to system but also from time to time for a system. The best example of the change with time was lived at the petrol crisis in 1973. Suddenly soaring petrol prices brought the RDopvalues much nearer to the RDm.values almost overnight. Many operating plants then tried to balance the sharply rising operating cost by adding more plates to their columns and operating them under reflux ratios, which are much closer to the minimum reflux ratios than before. For many systems the optimum reflux ratio lies somewhere around 1.25RDm

min.

Total cost

Ann

ual c

ost,

YTL

Operating cost

Fixed capital cost

RDm RD (RD)op

Fig.5.24 Determination of optimum reflux ratio


Optimum Feed Plate: The plate to which feed is introduced is known as feed plate and it is the first plate of the stripping section. As shown in Fig.5.25, in the drawing of

right angle triangles, which represent the plates, starting from point D it cannot be passed from enriching section operating line to the stripping section operating line before point E. Similarly, it must be passed from enriching section operating line to the stripping section operating line at last at point R. So, it follows from this that from one operating line to the other can be passed at anywhere between E and R. The best passing point is the poin which th umber of the ideal plates needed for a given separation is a

t at e n

minimum. If it is noticed the figure, it is seen that the maximum enrichments in the vapor phases (hence maximum mass transfer

between the phases) occur on the plates, which are close to the intersection point of two operating lines. It follows from this that m ber of the plates is obtained, if the switch from enriching section operating line to the stripping section operating line is done at or near the intersection point (point Q), although the column operates at any feed introduction point between points E and R. So, the optimum feed plate is at or near the intersection point, Q. In the feed plate, the feed and the vapor and liquid coming from stripping and enriching sections mix together and adjust themselves to the new conditions. Hence, the enrichment that will be obtained on the feed plate is usually ignored as design safety and one additional ideal plate is added to the number of the ideal plates calculated for stripping section. Notice the locations of feed introduction nozzles in the column for cases of liquid, vapor or mixture of vapor-liquid feed. For possi ight occur in future due to the changes in the

D

F

W

Q R

E

xD

q-line

xw zF

y

x

P =cons.

0 0

1.0

1.0

y = x

Fig.5.25 Optimum feed plate

inimum num

ble changes in the feed plate location that mcomposition and/or in the thermal condition of the feed,

opening of additional introduction nozzles during the construction stage of the column and closing them with blind-flanges are strongly advised (why?). Condensers: The vapor leaving the column must be condensed. Condensation is carried out in a shell and tube type heat exchanger usually located outside the column. Vapor coming from the column is condensed either in the tubes or in the shell depending upon the conditions. The cooling medium is then circulated in the shell in the first case or in the tubes in the latter case. In some cases, shell and tube condenser is placed inside the column vertically as shown in Fig.5.26a. In this type of condenser, liquid returns back as the vapor rises through the tubes and hence, flooding of the condenser must be prevented by observing the following equation in the design of the condenser: [ ]0.25

L0.5L

0.25G

0.5G ρuρu + < 0.6 [ ]0.25

GLi )ρ(ρdg − (5-87) where, uG and uL are the velocities of vapor and liquid in the tubes as m/s, based on empty cross sectional area of the tubes, ρG and ρL are the densities of vapor and liquid


Top produc

(a)

Condenser

(b)

1.plate

( c)

Receiver

Receiver

enser

Condenser

Column

Cond

Column

Column

Reflux liquid

t

Top product

Top product

Cooling liquid

1.plate

Vent

Vent

Cooling liquid

Cooling liquid

slope=-L/D

xD= yD

yD

xo

D

y1

x1

Enr sec.op.line

C

1

x 1.0

Reflux liquid

1.plate

y

(d)

y = x

Fig.5.26 Types of condensers : (a) built-in condenser, (b) total condenser, (c) partial ondenser, (d) enrichm the partial condens c ent in er



condenser, which is placed outside the column, must have sufficient elevation to permit the return of the condensate to the column top under the gravity. If this is not observed, the condens d with condensate partly or totally and it does not fulfil the design duty. In la located at or near ground level and the reflux liquid is pumped to the top of the column. Wherever the condenser is located, only the latent heat of vaporization of the vapor must b n it and hence, the condensate produced must be a saturated liquid. If it is sub-cooled, y the cooling medium is wasted but also undue condensation of some vapor on the first plate occurs. The part of the condensate, which will be taken as top product, is cooled in a cooler before flowing into storage vessel. In some special cases, only the part of the vapor, which will be returned back to the column as reflux liquid, is condensed and remaining part is taken as top product in vapor form. This is resorted to, when the capacity of the available condenser is not ent or the top product is used in vapor

in a subsequent process. This condenser is then known as partial condenser and if the partial condensation takes place slowly, the vapor and liquid reach in equilibrium, as they will remain in contact long time. As in this case partial condenser makes enrichment equals one ideal plate as shown in Fig.5.26d, the number of the ideal plates in the enriching section may be decreased one. But, in practice this is not done with the assumption that the equilibrium will not be attained in the condenser. In the ondenseusually the cooling water coming from the cooling tower is used as cooling medium. At lo ation temperatures cooling medium may be ch er or cooled ethylene glycol solution. If the condensation temperature is sufficiently high, atmospheric air can also be used as cooling medium. Condenser load qc(kW); in a total condenser is calculated f

as kg/m3, di is the inside diameter of the tubes in m. In small capacity systems, the

er is fillerge capacity plant, the condenser is

e removed i not onl

sufficiform

c r,

w condens illed wat

rom qc=G λ and in a partial condenser from qc=(G-D) λ = L λ , where λ (kJ/k-mol) is the average latent heat of condensation. Reboilers: The reboiler, which is an important part of the rectification systeessentially a heat exchanger and, depending upon the heat load and the properties of the liquid solution, in di s are construct ifferent modes are operated. It is generally placed outside the column as a separate piece of equipment. But when heat load is low and the fouli , it can be located in the column as shown in Fig.5.27a. For larg ties, kettle-type (b) and vertical thermo-syphon types (c, d) are widely used. To prevent the fouling of the tubes, thermo-syphon type reboil erally operated on partial vaporization principle; the mixture issuing from the reboiler comprises both liquid and vapor. For heat sensitive componen mo-syphon type reboilers take the liquid from the last plate of the column. This type is known as once-through type (d). But as in this type of reboilers heat transfer is by natural convection, when the viscosity of solution is still high at the operating temperature, over-all heat transfer coefficient becomes rather low. In order to increase the heat transfer coefficient, a pump between the column and reboiler is installed and this reboiler is then called as forced circulation reboiler. As in hes in eq ler becomes equivalent to one ideal plate. Hence, the number of the ideal plates in the stripping section can be decreased one. However, in practice this is taken as design safety and plate number is not reduced in the stripping section. The diameter of kettle-type reboiler must be so

the kettle-type reboiler, vaporization is slow and low, the vapor formed reacuilibrium with the liquid in the reboiler and then reboi

ts, ther

ers are gen

ng is not a probleme capaci

ed and at dfferent type

m, is

Fig.5.27 Types of reboilers : (a) built-in reboiler, (b) kettle-type reboiler, (c) thermo- syphon type reboiler, (d) once-through thermo-syphon type reboiler

bottom product

∇

last plate

Column

heatingmedium

Reb

oile

r

∇

last plate

Column

heatinmedium

g

∇

∇

Column Column

last plate last plate

heating medium

heating medium

Reboiler

Reboiler

bottom product bottom product

(a) (b)

Reb

oile

r

bottom product

(c) (d)


chosen that the velocity of vapor at liquid-vapor interface, uG (m/s) satisfies the following equation;

0.5

G

GLG ρ

ρρ0.2 ⎢⎡

<u ⎥⎦

⎤

⎣trainment cannot be evaded.

− (5-88)

Otherwise, excessive liquid enIn the reboilers, the follo media, which are listed in the order of increasing temperature, can be used: hot water, steam, hot oil and Dowtherm vapor. The reboiler load or duty, qs ( kW) is calculated from qs =

wing heating

λG , where λ ( kJ/k-mol) is the latent heat of vaporization of solution.

Example-5.10) Rectification under Constant Molar Over-flow

A 15 000 kg/h methanol/n-propanol solution at 20 OC, containing 22.3 mass percent ol will be rectified in a plate c mHg to produce top and bottom products containing 91 mass percent methanol and 97.8 mass percent n-propanol respectively. A reflux ratio 1.87 times the minimum reflux ratio is selected for the operation. The column is equipped with a total condenser and a kettle-type reboiler. a) Calculate the flow rates of top and bottom products, b) Write the q-line and the operating line equations for enriching and stripping section of the column. Calculate : c) Number of the ideal plates in each section by McCabe - Thiele method, d) The amounts of methanol and n-propanol transferred between liquid and vapor phases on the 2nd and 6th ideal plates. e) Condenser and Reboiler loads.

Vapor-liquid equilibriu system at 760 mmHg is given in Table.App.5.3 and some physical data for the components are given below.

Methanol ol n-Propan

Boiling point, b.p. (oC) at 760 mmHg 64.5 97.2

Molecular weight 32 60

Latent heat of vaporization, λ (kJ/kg) at b.p. 1 100 693

Specific heat, cL (kJ/kg oC) at 50 o 2.70 2C .65

Solution : First, check whether “Constant Molar Over-flow” approximation is applicable For methanol : λA= (1 100)*(32) = 35 200 kJ/k-mol For n-propanol : λB = (693)*(60) = 41 580 kJ/k-m

As the difference is not much, “Constant Molar Over-flow” case is applicable. a) Then work with molar units,

h/molk78.29860

)223.01)(00015(32

)223.0)(00015(F −=−

+=

35.060/)223.01()32/223.0(

z F =−+

=)32/223.0(

95.060/)91.01()32/91.0(

x D =−+

=)32/91.0(

ol

m of the

olumn at 760 mmethan


From equations (5-63) and (5-64), W = 298.78 - 101.78 = 197 k-mol/h b) To find the the must be known. Plot t-xy diagram and read Fb

rmal condition of the feed, bubble point temperature of the feed (tFb) the bubble point temperature of the feed as t = 80.5 oC.

As tF= 20 oC , the feed is sub-cooled liquid and for the calculation of q-parameter eqn.(5-74) must be used. Mean temperature for the FLc is 50 oC. Then, molar specific heat of the feed and average latent heat of vaporization;

molk/kJ390382/)5804120035( +=λ −=

21.139038

)205.80)(6.133(1q =−

+=

nt Q' is marked. By joining point D with Q' and extending e line until cutting y-axis K' = 0.56 is read.

Then from minimu Opera = (1.87)(0.696) = 1.30

667.1x762.5

Then, from eqn.(5-71);

y After drawing this line on xy-diagram, poi

qq −=

04.0)60/978.0(32/)978.01(

32/)978.01(xw =+−

−=

h/molk78.101)04.095.0(

)04.035.0(78.298)xx()xz(FD

wD

wF −=−

−=

−−

=

th

, m reflux ratio is computed.

ting reflux ratio, RD

6.0156.0

1 9695.0Kx DR Dm =−=−

′=

t-xy diagram of methanol/n-pr

600 0.1 0.2 0.3 0.4

mole fraction of me

at)

opanol at 760 mmHg

70tem

80ures

900C

100

0.5 0.6 0.7 0.8 0.9 1

thanol in liquid and vapor,x,y

per

, t (

tB

t-y

t-x

tAzF=0.35

tFb=80.5 C o

Cmolk/kJ6.133)60)(65.2)(35.01()32)(70.2)(35.0(c oFL −=−+=


L = D RD= (101.78)(1.30) = 132.3 k-mol/h and G = D(1+RD) = (101.78)(1+1.30) = 234,1 k-mol/h Then, from section operating line operating line;

c) Mark K = 0.413 on y-axis and by joining it with point D draw the enriching section operating line. Mark point Q and by joining it with point W draw stripping section operating line. Then, step off the right angle triangles between equilibrium curve and enriching section operating line starting at point D, and do the same after point Q between equilibrium curve and stripping section operating line until reaching point W. Total number of the ideal plates needed is found as 9( a close examination shows that exact number is 8.5), 3 needed in the enriching section, 6 needed in the stripping section. Although the reboiler is kettle-type, as design safety provide 6 ideal plates in the stripping section of the column.

eqn.(5-59) enriching section And from eqn.(5-62) stripping;

h/molk84.493)78.298)(21.1(3.132qFLL −=+=+= h/molk84.29619784.493WLG −=−=−=

1.3010.95x

1.3011.30y n1n +

++

=+

413.0565.0 +=+ n1n xy

296.84(0.04)x

296.84493.84y m1m

)197(−=+

0266.0664.1 −=+ m1m xy

xy-diagram of methanol/n-propanol at 760 mmHg

0.1

0.4

0.7

0.8

0.4 0.5 0.6 0.7 0.8 0.9 1

mol

e fr

act

p

1

0

0 0.1 0.2 0.3

9

8

1

2

3

6

D

F

0.2

0.3

0.5

n of

0.6

met

han

0.9

mole fraction of methanol in liquid,x

iool

in v

aor

,y

7

4

5

Q

K=0.413

W

yq= 5.762 xq + 1.667y1

y2

y3

y4

y5

y6

y7

y8

y9y10

zF xD

x1x2x6 x3x5 x4x7

x9 xw

y=x413.0565.0 +=+ n1n xy

0274.0685.1 −=+ m1m xy

Q'

K'=0.56

x8


d) 2nd ideal plate is in the enriching section, and the composition of the liquid and vapor phases

Fo lar over-flow” rectification; on ea“Moles of met nol transferred f oles of n-propanol transferred from vapor to liquid” is validThen, n-propanol transferred from vapor to liquid = 22.29 k-mol/h = (22.29)(60) =1 337.4 kg /h 6th ideal plate is in the stripping section, and x5=0.288 , x6=0.195 , y7 =0.300 , y6 =0.455 are read from the diagram. Then ,

entering and leaving this plate are read from the diagram as, x1=0.834 ,x2=0.665 ,y3 =0.79, y2 =0.885. Then, L (x1 - x2) = (132.2)(0.834-0.665) = 22.34 k-mol methanol/h Or G (y2 - y3) = (234.1)(0.885-0.79) = 22.24 k-mol methanol/h As it is seen these are vey close to each other, take the average value 22.29 k-mol methanol/h In mass units, this makes = (22.29)(32)= 713.3 kg methanol/h

r “ Constant mo ch plate, rom liquid to vapor = mha

.

=− )x 6x(L 5 (493.84)(0.288-0.195) = 4 k-mol methanol/h Or,

5.93=− )yy(G 76 (296.84)(0.455-0.300) = 46 -mol methanol/h.

Take the mean value = 45.97 k-mol methanol/h As it is seen, g quantities (rat f mass transfers) are quite different even on mole basis on the 2nd and 6th

e) As the condenser is total condenser; 2 496.4 kJ/s (=kW) Reboiler load, 3 165.5 kJ/s (= kW) Use of Open Steam: Slightly superheated steam at column pressure may be directly injected to the bottom of the column eliminating the reboiler, when the bottom product is water. In this case, although the component balance along the column does not change, al ma rial alanc chan s and comes,

.01 k

transferrin e o plates.

( )=⎟⎟⎠

⎞⎜⎜⎝

⎛== 39038

60031.234Gq c λ

the tot te b e ge beWDGF +=+ (5-89)

as seen from Fig.5.28 a. Here, G is e flow rate of steam as k-mol/s. As yN+1 in this case, point W lies on the x-axis and the required number of the ideal plates

th = 0

Bottom product

strip.sec.op.line

zF

N-1

N

xw xN

F

x

L , xw =xN

Q

xN-1 yN+1

y

yN-1

yN

G , yN+1 =0

yN-1

q-line

xN-1 yN

xN Steam

N-1

N

y = x

∇

equilibrium curve

Fig.5.28 Use of open steam in the column

L

( )=⎟⎟⎠

⎞⎜⎜⎝

⎛== 39038

600384.296q S G λ

(a) (b)

W


increases in the stripping section as shown in Fig.5.28 b. This increase is mostly limited with one ideal plate and hence a certain saving in the fixed capital cost of the system is obtained as the cost of one ideal plate is generally much smaller than the cost

steam to the boiler house for reuse and,

steam. Hence, by comparing the

treams: In e top and bottom products uid product. Although this

e fraction n in Fig.5.29. As it is ns: upper, middle and lower section. In the

iddle section, the flow rates of liquid, vapor and any plate are shown with Ls, Gs and n e be ween the top and any l ne equati

ritten as,

of reboiler. But use of open steam is much costlier than the use of closed-loopedas in the latter case condensed steam is returned back demineralized water is used for the production of saving that will be obtained by elimination of reboiler with the cost increase due to the open steam use, a decision is made whether the use of open steam should be preferredor not. Side S some applications, a product other than this also withdrawn from the column, which is usually a liq

mponent rectification, it is also used in some application is more common in multi-cobinary rectifications. Let us show the flow rate of intermediate product with S as k-mol/s and the mol of the MVC in it with xs as showseen, the column is divided in three sectiomk respectively. By writing the M C bala c t of the columnlate in the upper section, the operating i on for the upper section can be

Vpw

D

Dn

D

D1n R1

xx

R1R

y+

++

=+ (5-59)

Between any plate in the middle section and top of the column, Total material balance : (5-90) Gs = Ls + S + D

GMVC balance : syk+1 = Lsxk + Sxs + DxD (5-91) can be written. From the last equation,

s

Dsk

s

s1k G

DxSxx

GL

y+

+=+ (5-92)

is obtained. This is operating line equation the middle section of the column and the

coordinates of its intersection with the diagonal is found as U(

of

)DSDxSx

;DSDxSx DsDs

++

++

with the he f equation (5-90). If in this case a qS-parameter is defined by

lp o

SLLq S

S

−= (5-93)

qS-line, which is the locus of intersection nts of upper ection operating lines is obtained as;

poi and middle s

1qxx

1qqy

S

Sq

S

Sq SS +

++

= (5-94)

This is a straight line passing through point S( S S). For a saturated liquid side product, qS = -1 and the qS-line is a vertical line from point S, and for a saturated vapour side product, qS = 0 and the qS-line is a horizontal line from point S. In this case q-parameter is defined by,

x ;x


F

LLq −= S (5-95)

passing through point F(zF;zF) is then the ts of middl

the lower section becomes;

and the q-line given by equation (5-71) andlocus of intersection poin e and lower section operating lines. The operating line equation of

GxWL w−= x

Gy m1m+ (5-62)

Considering the whole column, three more equations as;

(5-97)

Total material balance : F = D + W + S (5-96) MVC balance : FzF = DxD + WxW + SxS

Fig.5.29 Withdrawal of side product

xk

m

k

n

1

F,zF

LS

N

LG

xn

W,xw

GS

S, xS

Top product

yn+1

xm ym+1

L G

Bottom product

Side product

yk+1

L

D,xD

L,xo

qC

qS

xN

yN+1 G

Upper section

(a)

Lower section

D

U

S

F

op.line for lower sec.

q-line

Q

W

qs-line

op.line for middle sec.

op.line for upp.sec.

K

D

DxR1+

0 0

1.0

Feed

y

xs zFxw

P=cons.

x xD

Middle section

(b)


Percentage recovery of the MVC : 100.Fz

SxDx.R.PF

SD += (5-98)

can be written. The required number of the ideal plates can now be easily obtained by first locating points F, D, W and S, then drawing qS-, q- and three operating lines as shown in Fig.5.29b, and finally constructing the right angle triangles between equilibrium curve and the operating lines starting at point D (For simplicity sake this is not shown in the

Example-5.1 al of Side P

An aqueous methanol solution containi ole percent methanol will be re n a plate columnequipped with a total condenser, at 760 mmHg to pr uce a 10 l/h saturated liquid side produccontaining ole en thanol , a top product containing 95 mole percent methanol and a bottom product containing 4 mole pe n e feed, which is at 20 oC, will be introduced at a flow rate of 173 k-m h. Operatin flux ratio is selected as 1.235. Assuming constant molar

n. number of the ideal plates needed and the locations of the feed and

Vapor - liquid equilibrium for methanol-water system at 760 mmHg is in TaMolar specific heat of the feed at mean temperature and the latent heat of vaporization of the solution are 76.5 kJ/k-mol oC and 38 000 kJ/k-mol respectively.

Solution :

First, plot t-xy and xy-diagrams as sho elow. Read bub e point temperature of the feed from t-xy diagram as tFb= 81 oC. Hence the feed is subcooled then calculate the q-parameter freqn.(5-74)

Figure). 1) Withdraw roduct

ng 25 m ctified i t od k-mo

60 m pec t merceg re

t methanol. Thol/ over-flow;

a) Write the operating line equations for each section of the columCalculate the side product

ble.App.5.4. b)

wn b the blliquid, om

123.100038

)2081)(5.76(1q =−

+=

q-line from equation (5-71); 1123.1

25.0x1123.1

123.1yq = q −−

−

yq = 9.13 xq – 2.03 From equations (5-94) and (10)(0.60) + (0.04)(173-10-D) D = 33.77 k-mol/h W= 173-10-33.77 = 129.23 k-mol/h

1.71 k-mol/h k-mol/h

ol/h

(5-95); (173)(0.25) = (0.95)D +

L = DRD = (33.77)(1.235) = 4 G = L + D = 41.71 + 33.77 = 75.48As side product S is saturated liquid, GS = G = 75.48 k-mol/h From equation (5-90); LS = 75.48 – 10 - 33.77 = 31.71 k-mFrom equation (5-97); L = 31.71 + (1.123)From equation (5-60);

(173) = 225.99 k-mol/h W− = 225.99-1LG = 29.23 = 96.76 k-mol/h

line from equation (5-59);235.11

95.0x235.1y +Then, upper section operating 235.11 n1n ++

=+

yn+1 = 0.553 xn+ 0.425

Middle section operating line from equation (5-92); 48.5

x48.75 k1k ++ 7

)10( +

yk+1 = 0.42 xk + 0.505

)95.0)(77.33(60.0)(71.31y =


Lower section operating line from equation (5-62); 76.96

)04.0)(23.129(x

76.9699.225y m1m −=+

ym+1 = 2.43 xm - 0.0535

W, D, S on xy-diagram. Then, draw q-line and qS-line. By marking K1=0.425 n y-axis and joining it with D, upper section operating line is first drawn and point E is located. Then,

First, mark the points F,o

W

mol

e fr

actio

n of

met

hano

l in

vapo

r, y

F

S

E

D

zF xw

xS

xD

qS-line 23

45

67

8

9

03.2x13.9y qq −=

K2

mole fraction of methanol in liquid, x

xy-diagram of aqueous methanol solution at 760 mmHg

K1

Q

1

10

Mole fractions of methanol, x,y

Tem

pera

ture

(o C)

tFb

zF

t-xy diagram of aqueous methanol solution at 760 mmHg


by marking K2=0.505 on y-axis and joining it with E, middle section operating line is drawn. Finally, by joining W with Q, lower section operating line is drawn. Upper section operating line has meaning between D and E, middle section operating line between E and Q, and lower section operating line between Q and W. Stepping off the right angle triangles between equilibrium curve and the operating lines gives the required number of the equilibrium plates, which is found as 10. It follows from the figure that, feed must be introduced on the 6th , and the side product must be withdrawn from the 4th ideal plate.

High Purity Products: When the purities of the products are high, the values of xD and xw are very close to 1.0 and 0.0 respectively, and the calculation of number of the ideal plates according to McCabe and Thiele method cannot be done accurately near the points D and W. The equilibrium curves of almost all the systems are linear near the x =0 and x= 1. Let us show the slopes of these linear parts of the curve with K1 and K2 at the bottom and top respectively. In this case, drawing of right angle triangles is started at the feed plate and is continued in both sections until accurate drawings can be done and numberings are started from the plate above the feed plate for enriching section and from the feed plat r stripping section. Suppose n and m show the last drawn plates in enriching and strip After plates n and m, the numbers of the plat uations given below. The enriching section of the column may be considered as an absorption olumn for the LVC and hence an equation similar to the equation (4-54) can be ritten

e foping section respectively.

es in the remaining parts are calculated from the eq

cw for the calculation of number of the ideal plates, ne in this part.


αloge

ααK/)x1()y1(n 2o1 ⎦⎣ ⎠⎝−−−

=

α 11K/)x1()y1(log 2on⎡

+⎟⎞

⎜⎛ −−−−

⎥⎤

⎢ (5-99)

where, α is absorption factor defined by α = L/K2G, yn is the mole fraction of the MVC in the vapor leaving the last drawn plate (plate n) in the enriching section. Number of the total ideal plates in the enriching section then becomes NE = n + ne. Similarly, since the stripping section of the column works as a str per (dethe MVC, an equation similar to equation (4-55) can be written for the calculation of the number of the ideal plates after plate m in the stripping section of the column.

ip sorber) for

( )

)/1 α

αα

(low

g

1/Kx

xxlog

m 1

1wm

s

⎥⎦

⎤⎢⎣

⎡+−

−

= (5-100)

w e

/Kx w −

her , 1KG/L=α and xm is the mole fraction of the MVC in the liquid leaving the last drawn plate (plate m) in the stripping section. The number of the total ideal plates in the stripping section of the column is then obtained from Ns = m + ms. Rectification under Varying Over-flow: When the latent heats of vaporization of the components are not equal and/or the heat of solution is very great, the flow rates of the phases do not remain constant even expressed in molar units. In this case, the methods given above cannot be used for the calculation of the number of ideal plates. Enthalpy balances next to the to pute the flow rates of tal and MVC balances are required to com

vapors and liquids, which change from plate to plate. By considering the plate column shown in Fig.5.30a, the following equations can be written between any plate (plate n) in the enriching section and the top of the column (envelope-1): Total material balance : Gn+1 = Ln + D (5-101) MVC balance : Gn+1 yn+1 = Ln xn + D xD (5-102) Enthalpy balance : Gn+1 Hn+1 = Ln hn + D hD + qc (5-103) f the energy removed per unit mole of top product in the condenser is shown by

the enriching section, there is a net flow of MVC in the upward direction, which is constant and equals the quantity of the MVC in the top product. Similarly, if the enthalpy of the liquid leaving plate n is subtracted from the enthplate, G H - L h = ∆ h (5-107)

8)

In the stripping section, between any plate (plate m) and bottom of the column (envelope-2);

lance : Lm = Gm+1 + W MVC balance : Lmxm = Gm+1ym+1 + Wxw (5-110)

nthalpy balance : L h + q = G H +Wh (5-111) ritten.

(5-112)

IQCD=qc/D, the last equation can also be written as; Gn+1 Hn+1 = Ln hn + D (hD + QCD) (5-104) Now let us imagine that we subtract the quantity of the liquid leaving plate n from the quantity of the vapor entering the same plate and show this difference with ∆e then, Gn+1- Ln = ∆e (5-105) is written. So-defined ∆e is named as difference flow or total net flow. If equations (5-101) and (5-105) are compared, it is seen that ∆e = D, hence there is a total net flow in the enriching section which is in upward direction, constant and equals the flow rate of the top product. Similarly, if we imagine the subtraction of the quantity of the MVC in the liquid leaving plate n from the quantity of the MVC in the vapor entering the same plate we write, Gn+1yn+1- Lnxn = ∆e x∆e (5-106) where, x∆e shows the mole fraction of the MVC in the ∆e. By comparing equation 5-106) with equation (5-102) x∆e= xD is obtained. Hence, it is said that in

alpy of the vapor entering the same

n+1 n+1 n n e ∆ecan be written. Where, h∆e is the specific enthalpy of the ∆e. Comparison of this equation with equation (5-104) gives, h = h + Q (5-10∆e D CDSince both hD and QCD are positive h∆e is also positive. Hence, there is an enthalpy net flow in the enriching section in the upward direction, which is constant and equals the sum of the enthalpy of the top product and energy removed in the condenser.

Total material ba (5-109)

E m m s m+1 m+1 wcan be w By defining Qsw=qs/W , equation (5-111) becomes, Lmhm - Gm+1Hm+1 = W (hw- Qsw) Now let us assume that we subtract the quantity of vapor entering the plate m from the quantity of the liquid leaving the same plate and show this difference with ∆s then, Lm- Gm+1 = ∆s (5-113)


is written. So-defined ∆s is named as difference flow or total net flow. If equations (5-109) and (5-113) are compared, it is seen that ∆s = W, hence there is a total net flow in the stripping section which is in downward direction, constant and equals the flow rate of the bottom product. Similarly, if we subtract the quantity of the MVC in vapor ntering plate m from the quantity of the MVC in the liquid leaving the same plate we

e h is constant and

equa

, w ic he

nden

(5-121) i (5-1

ewrite, Lmxm- Gm+1ym+1 = ∆s x∆s (5-114)

where, x∆s shows the mole fraction of the MVC in the ∆s. By comparing equation (5-110) with equation (5-114) x∆s= xw is obtained. Hence, it is said that in the stripping

dsection, there is a net flow of MVC in the ownward dir ction, whicequals the quantity of the MVC in the bottom product. Similarly, if the enthalpy of the vapor entering plate m is subtracted from the enthalpy of the liquid leaving the same plate, Lmhm - Gm+1Hm+1 = ∆sh∆s (5-115) can be written. Where, h∆s, is the specific enthalpy of ∆s stream. If this equation is compared with equation (5-112), h∆s = hw – Qsw (5-116) is obtained. Both Qsw and hw are positive and since Qsw is always greater than hw, h∆s is then always negative. Hence, in the stripping section, there is a constant enthalpy net flow in upward direction ls in magnitude the value, which is obtained from the

subtraction of enthalpy removed by bottom product from the energy given in thereboiler. Summarizing, in the enriching section of the column total net flow, component net flow and the enthalpy net flow are all in upward direction. But, in the stripping section while total net flow and component net flow are in downward direction, the enthalpy net flow is in upward direction. So, it can be said that in a rectification column, there is always an enthalpy net flow h h carries the MVC from reboiler to t condenser of the column. Now, let us write total material, the MVC and enthalpy balances for the whole column: Total material balance : F = D + W (5-63)

∆or : F = ∆e + s (5-117) MVC balance : F zF = D xD + W xw (5-64) Enthalpy balance : FhF + qs = DhD + Whw+ qc (5-118) or : FhF = ∆e h∆e + ∆s h∆s (5-119) In a rectification operation, the reflux ratio, which is defined as (RD=Lo/D) is an important operating parameter and the relationship between RD and h∆e must be

en al oknown. By writing total material and th py balances around the c ser, this can be found as follows:

G1 = Lo + D (5-120) G1H1 = Loho + D hD + qcIf the value of G1 is substituted from equation (5-120) nto equation 21) and the

edefinitions of RD , QCD and h∆ are remembered then,

O1

1∆eD hH

HhR

−−

= (5-122)

is obtained.


G1,y1,H1

With the help of the equations above, the number of the ideal plates required for a

Condenser

G1

L4

L6

Fig.5.30 Varying over-flow rectification

Reboiler

F,zF,hF

Top product

qC

Bottom product

G2,y2 H2

D,xD,hD

Feed

1

m

L1,x1 h1

n

2

N

L2,x2 h2

G3,y3 H3

Gn+1,yn+1 Hn+1

Ln,xn hn

Gm+1,ym+1 Hm+1

Lm,xm hm

GN+1,yN+1 HN+1

LN,xN hN

Lo,xo ho

W,xw,hw

qS

Ln-1,xn-1 hn-1

Gn,yn Hn

G6 G5

G3 G2

H1

hDD

h∆e

L1 L2

∆e

L5

∆S

L3

W

4 3 2 1

P =cons.

65

x5 xD=y1 x2 x1zF x4 x3 x6

xW

h∆s

F

G4

hF

Spe

cific

ent

ha,h

lpie

s, H

x,y

H-y

h-x

1.00

y

x0

y=x

Envelope-1

Envelope-2

(b)

(a)


varying over-flow rectification can be calculated by using Ponchon and Savarit Method, which involves the graphical drawing of the ideal plates on enthalpy-composition diagram of the system as shown in Fig.5.30 b. At the start of the design F, zF, hF, xD and xw (or percentage recovery) are known. The operating pressure and reflux ratio are se cted by the designer and then enthalpy-composition; xy- and t-xy diagrams are obtained at the operating pressure. If the

is a total condenser then y1= xo= xD and ho = hD. Points F, W, D and G1 are first located on the enthalpy-composition diagram as their compositions and thermal conditions ar all known (reme r that a ors and liquids leaving the plates and the top and bottom duc are ted vapors and liquids). Then point ∆e, which will be on the vertical from x = xD is located by calculating its enthalpy coordinate h∆e equation (5-122). H1 and ho, which are required for the calculation are read from enthalpy-composition diagram. Equation (5-117) indicat tha oints F, ∆e and ∆s must lie on a straight line. Thus, by joining points ∆e and F and extending it until cutting the vertical from x = xw, ∆s is located. Now the drawi the l plates on en py-com ition diagram is started. Since the composition of the vapor leaving the f i l plate is known, t com o the liquid leaving the same plate can be und by carrying the y1 value to the xy-diagram below. Then, this x1 composition is carried to the enthalpy-composition diagram and point L1 is located. The tie line joining G1 and L1 represents the first ideal plate in the column. Equation (5-105) indicates that L1, G2 and ∆e must lie on t s e s a ht ne. In dit n, Gshould b e saturated vapor curve. Thus, L1 joined with e ll locate the 2, the vapor leaving the second ideal plate. Then the composition of the liquid leaving the second plate, x2 is read from xy-diagram below by carrying the y2 value from diagram above. With the help of this x2, point L2 is located on enthalpy-composition diagram. The tie line joining L2 i G2 shows the second ideal plate in the column. G3 is then easily located by joining L2 ∆e according to equation (5-105). The calculation is thus continued by alternate drawing of tie lines and operating lines issuing from point ∆e ∆eF∆s line, ∆e losses its meaning as into the stripping section of the column is entered. From now on, point ∆s is joined with points L’s in order to find points G’s. The calcula the same way until reaching x = x e of the tie line, which cuts the ∆eF∆s line gives the ideal plate number to which the feed should be introduced (in Fig.5-30 b this is the 4th ideal plate). While the number of the tie lines on the right of ∆eF∆s line gives the number of the ideal plates that must be in the enriching section, the number of the tie lines on the left of the ∆eF∆s line, including the feed gives the number of the ideal plates in the stripping section of the column. The lines issuing from ∆e and ∆s points are known as operating lines. It follows from here that th ine of each plate is different as the quantities of both liquid and vapor are different for each plate. The simultaneous solution of total material, component and enthalpy balances around each plate gives the quantities of liquid and vapor entering and leaving the plate. The condenser and reboileFrom equation (5-108) QCD = qc/D = h∆e- hD (5-123)

le

condenser

e mbe ll the vap pro ts satura

fromes t p

ng of idea thalirst fo

posdea he p sition of

he am tr ig∆

li ad ioG

2 wie on th

w th and

as explained above. On the left of

tion is continued inw. Th number

e operating l

r duties are calculated,

and from equation (5-116) Qsw = qs/W = hw- h∆s (5-124)


As it is seen, RD, qc and qs are all interrelated. If any one of these three is specified, the other two are automatically fixed. If, instead of RD, qc is selected, ∆e is located with the help of equation (5-123). When qs is selected, then first ∆s is located for which h∆s is calculated from equation (5-116). Minimum Reflux Ratio: When a tie line coincides with an operating line during the drawing, this means that required number of the ideal plates at this reflux ratio is infinite. Hence, this reflux ratio is known as minimum reflux ratio. To find the minimum reflux ratio, arbitrary tie lines between points F and D are drawn and extended until cutting the vertical from x = xD. The tie line, which when extended, cuts the vertical at the farthest point from the diagram gives ∆em. Hence, by inserting the ordinate value of this point h∆em into equation (5-122), RDm is found as;

O1

1∆emDm hH

HhR

−−

= (5-125)

The tie line which gives the minimum reflux ratio is usually the tie line of which extension passes through point F as shown in Fig.5.31. Total Reflux: at total reflux no products are withdrawn from the column. Since RD = ∞ at D = 0, it follows from equation (5-122) that h is also infinite. As a result ∆eof these, both ∆e and ∆s go to infinity. Hence, determination of the number of the ideal plates under total reflux, which is a minimum, by drawing on enthalpy-composition diagram is done as shown in Fig.5.32. As it is seen from the figure, the operating lines in this case are parallel to ordinate.

93Sp

ecifi

c en

thal

pies

, H,h

∆em

∆e ∞

∆sm

D

W

h∆em

h∆sm

h-x

zF xw 1.0

1.0

1.0 0

0

0

1.0 0

0

0

1.0

x

x,y

H-y

y

y = x

y

x

P=cons.

1

3

D

xD

∆s - ∞

H-y

h-x

xw

W

x,y1.0

2

y = x

P=cons.

Fig..5.31 Determination of Minimum Reflux Ratio Fig.5.32 Number of Ideal Plates at Total Reflux

xD

Spec

ific

enth

alpi

es, H

,h

4


Partial Condenser: if the condenser is a partial condenser and the reflux liquid and the top product vapor reach in equilibrium, the condenser then makes an enrichment which is equivalent to one ideal plate as shown in Fig.5.33. Since in this case, top product D is a saturated vapor, point D lies on saturated vapor curve and then its

∆e = HD+QCD. Use of Open Steam: when the bottom product of the column is water, slightly superheated steam at column pressure can be directly injected to the bottom of the column eliminating the use of reboiler. In this case for the whole column; Total material balance : F + GN+1 = D + W (5-126) MVC balance : F zF + GN+1(0) = D xD + W xw (5-127) Enthalpy balance : F hF + GN+1 HN+1 = D hD + qc + W hw (5-128)

can be written. Similarly, for the stripping section of the column;

specific enthalpy is shown by HD. Hence, the enthalpy coordinate of point ∆e is then given by h

Equilibrium curve

Fig.5.33 Partial condenser

G1,y1,H1

D,y ,HD D

L,xo,ho

qC

1

1.0slope=-Lo/D

xD= yD

yD

xo

D

y1

x1

enr.sec.op.line

C

1

x 1.0

y

y = x

H1

yD

C 1

h∆e∆e

xo y1 x1 1.0

x,y

D

G1

HD

H-y

Lo

L1

h-x

(a)

(b)

Top product


F

N-1

y = xN-1

spec

ific

e

(5-129)

MVC balance : LNxN- GN+1(0) = Wxw- GN+1(0) = ∆sx s (5-130) Enthalpy balance : LNhN- GN+1H N+1HN+1 = ∆sh∆s (5-131) are written. First points F, D, W, G1 and GN+1 are located on enthalpy-com osition diagram. Then with the help of known RD, point ∆e is located. Then the points G nd points ∆e and F are joined and extended. The intersection point of these two lines gives point ∆s. Then starting at point G1, the ideal plates are drawn in the known way. In Fig.5.34 b, only the last two plates of the column are shown. The quantity of open steam GN+1 required is calculated fro y first reading the value of x∆s from the diagram above.

-5.12 Rectification under Varying Over-flow

c) Total net flow, ammonia net flow and enthalpy net flow in each section, ates,

Total material balance : LN- GN+1 = W- GN+1 = ∆s ∆

N+1 = Whw- G

pN+1 and W a

m the equations above b

Example Aqueous ammonia solution containing 25 mass percent ammonia will be rectified in a plate column at 10 bars pressure to produce a top and a bottom product with 95 and 4 mass percent ammmonia. The feed, which is saturated liquid, will be supplied at a flow rate of 5 500 kg/h. The reflux ratio to be used is 0.758. The column is equipped with a total condenser and kettle-type reboiler. Calculate:

a) Flow rates of top and bottom products and P.R. of ammonia, b) Number of the ideal plates needed,

d) Flow rates and composition of each stream entering and leaving the pl

nth.

,H,h

(a)

∇

LN=W

GN

GN-1

LN-1

LN GN+1

N+1=0 y

xN=xw

hN=h

Steam N

N-1LN=W

GN+1

w

GN-1

GN

LN-1 LN-2

∆S

∆e

yN+1=00

0

xw x∆S

N

N

h∆S

x,y

y

strip.sec.op.line

h-x

H-y

P=cons.

hN=hw

xN

zF

E

(b)

Fig.5.34 Use of open steam

Bottom product

x


e) Enthalpy

First plot ent a) From 5-64)

Condenser and reboiler duty -composition data of the system at 10 bars are given in Table.App.5.12.

Solution:

halpy-composition and xy-diagram of the system.

equations(5-63) and (

h/kg2.269104.095.004.025.0)5005(

xxz

FDwD

F =−−

=−

= W= 5 500-1269.2 = 4 0.8 kg/h

From equat (5-65)

x w− 23

%ion 7.87100.)(5500(

)95.0)(2.1269(.R.P ==

b) From enthalp mposition diagra 17 = 20 400 kJ/k-mol and λB= 2 000*18 = 36 000 kJ/k-mol are found. So, va er-flow case applies. Then, calculation can be conducted in mass units. From the diagram hD=ho= 280 kJ/kg and H1= 1600 kJ/kg are read. Hence f uation(5-122) h∆e= 1 600 + 0.758(1 600 600 kJ/kg is calculated. With the help of this point ∆e is located. By joinin this with point F a ding the line point ∆s is located. Then, starting at point G1 tie lines and operating lines are drawn as outlined above. Number of the ideal plates needed is found as 4, and the feed must be introduce on the second plate. Allthough kettle reboiler makes a separation equal to one ideal plate, for de fety 4 ideal plates are ended r the column. c) From the diagram, hw = 720 kJ/kg and h∆S = - 320 kJ/kg are read. Total net flow in the enriching section : ∆e = D = 1 269.2 kg/h Total net flow in the stripping section : ∆s = W = 4 230.8 kg/h Ammonia net flow in the enriching section : ∆ x =Dx =(1 269.2)(0.95) = 1 205.7 kg NH3/h Ammonia net flow in the stripp 0.8)(0.04) = 169.2 kg NH3/h

Enthalpy net flow in the enriching section : ∆h∆e = (1 269.2/3 600)(2 600) = 916.6 kJ/s 4

1 =1205.7, L1 = 395.5 kg/h and G2 = 1 664.7 kg/h are calculated.

mpirical Correlations for Estimating Number of Ideal Plates: First Gilliland later mate of the number

f the ideal plates for a given continuous rectification. Both empirical correlations as

equ o

)25.0y-co m λA= 1 200*

rying ov

rom eq-280)= 2nd exden

dsign sa recomm fo

e ∆e Ding section : ∆sx∆s=Wxw=(4 23

Enthalpy net flow in the stripping section : ∆sh∆s= ( 230.8/3 600)(-320) = - 376.1 kJ/s d) L =D R = (1269.2)(0.758) =962.1 kg/h, G = 962.1 + 1269.2 = 2 231.3 kg/h o D 1From the graph, x1= 0.445 and y2 = 0.83 are read.Then, from G2-L1=1269.2 and 0.83G2-0.445LFrom the graph, x2=0.195 and y3=0.615 are read. Then, from L2-G3 = 4230.8 and 0.195L2-0.615G3 =169.2, L2= 5 792.2kg/h and G3= 1 561.4 kg/h are found. From the graph, x3=0.09 and y4=0.22 are read. Then, from L3-G4= 4230.8 and 0.09L3-0.22G4=169.2, L3= 5 858.3 kg/h and G4= 1 627.5 kg/h are found. e) Condenser duty, qc = D(h∆e-hD) = (1 269.2/3 600)(2 600-280) = 817.9 kJ/s Reboiler duty, qs = W(hw-h∆s) = (4 230.8/3 600)[720-(-320)] = 1 222.2 kJ/s EErbar and Maddox have given empirical correlations for quick estioshown in Fig. 5.35 and Fig.5.36 require only minimum reflux ratio and minimum number of the ideal plates at total reflux. For quick estimate of the required number of the ideal plates from these figures, minimum reflux ratio may be obtained from

equation and the minimum number of the ideal plates from Fenske Underwood ati n.


Enthalpy-composition diagram of ammonia-water at 10 bars

0

0.1

0.2

0.3

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

mol

fra

-4800

-4400

-4000

-3600

0.4

ion

0.5mon

i

0.6

a in

0.7r,y

0.8

0.9

1 -80

1.1

1.2

1.3

1.4

400

800

ntha

lpy

of s

olu1.5 1200

1.6 1600 ,h

1.7

1.8

molfraction of ammonia in liquid,x

ctm

-3200

-2800

-2400

-2000

-1600

-1200

0

-400

0

2000

2400

2800

3200

mol fraction of ammonia,x,y

spec

ific

etio

n, H

(kJ/

kg s

olut

ion)

1.9

2

of a

vap

o

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

D

xD

hD

H1

h∆e∆e

G1

G2

G3

G4

L1

L2L3

L4 W

F

∆S

zF

xw

4

3

2 1

hw

h∆s

Enthalpy-concentration diagram of ammonia-water at 10 bars Mass fraction of ammonia, x,y

Mass fraction of ammonia in liquid, x

ctio

n of

am

mon

ia in

vap

or,

Mas

s fra

y


F

F

No part of this CD-book may be multip

ig.5.35 Gilliland correlation

ig.5.36 Erbar - Maddox correlation

lied for commercial purposes. E.Alpay & M.Demircioğlu 218


Example-5.13) Estimation of Number of Ideal Plates from Empirical Correlations

Estimate the number of the ideal plates needed for the Example-5.10 from; a) Gilliland correlation, b) Erbar-Maddox correlation.

Compare the results with the result of Example-5.10.

Solution: From xy-diagram, at Q' 73.0y,415.0x qq =′=′ are read. Then relative volatility at this point, from equation (5-13) : α = 0.73(1-0.415)/[0.415(1-0.73)] = 3.81 RDm is calculated from Underwood equation [equation (5-78)] by trial and error, Try RDm= 0.696;

)95.01)(121.1()35.01)(1696.0()]1696.0(35.0)121.1(95.0[81.3

)95.01)(21.1()35.01)(696.0()95.0)(21.1()35.0)(696.0(

−−+−+++−

=−+−

+

2.716 ≈ 2.715 So, the assumed RDm is correct. For the calculation of number of the ideal plates at total reflux from Fenske equation average α is

eeded. At the top of the column, y = 0.95 , x = 0.834 , αT = 0.95(1-0.834)/[0.834(1-0.95)] = 3.78 t the bottom of the column, y = 0.12 , x = 0.04 , αB = 0.12(1-0.04)/[0.04(1-0.12)] = 3.27

nA

52.3)27.3)(78.3(BTAV ==αα=α ,

then from equation (5-86),

87.4)52.3log(

)95.01(04.0)04.01(95.0log

Nm =−−

=

a) 263.0

30.11696.030.1

R1RR

D

DmD =+−

=+− Then, from Fig.5.35 4.0

2NNN m =

+− is read. Finally,

number of the ideal plates, 45.94.01

8.087.4N =−

+= is found. A close look at McCabe-Thiele diagram

of Example-5.10 reveals that 8.5 ideal plates are needed. Then Gilliland correlation estimates the number of the ideal plates about [(9.45-8.5)/8.5]*100 = 11 % higher than the actual value.

b) 41.0

696.01696.0

R1R

Dm

Dm =+

=+

and 57.030.11

30.1R1

R

D

D =+

=+

then, from Fig.5.36 Nm/N=0.72 is read.

Finally, number of the ideal plates, N = 4.87/0.72 = 6.76 is found. As it is seen, the Erbar-Maddox correlation stimates the required number of the ideal plates about [(6.76-8.5)/8.5]*100 = -20.5 % less than the actual value. In this particular case, Gilliland correlation gives better result than the Erbar-Maddox correlation. 5.3.3.2 Rectification in Batch Operation: When the quantity of the solution to be separated is small or products at different purities are required, then the rectification operation is carried out batch-wise. A plate type batch rectification column is shown in Fig. 5.37. The volume of boiler is selected such that it takes one batch of solution at 70% fullness. Column operates first under total reflux until the top product reaches the desired composition, from there on top product is withdrawn. During the operation, no bottom product is withdrawn. The purity of top product depends on the number of the plates available in the column and the composition of the liquid in the boiler. The

quid in the boiler becomes poorer in the MVC as the distillation proceeds; as no feed given to the column cher in the MVC. As a

e

liis and as the vapor formed is always ri

result of this, top product beco the MVC as the time passes. If the

steadily in the course of distillation. It follows from here t

mes also poorer inpurity of the top product is to be kept constant, then the reflux ratio must be increased

hat batch rectification can be reased

ine,

carried out in two different ways: either under constant reflux ratio or under increflux ratio. As the column consists of only enriching section, there is only one operating lwhich is given as;

yn+1=D

Dn

D

D

R1x

xR1

R+

++

(5-132)

While in operations under constant reflux ratio the slope of this line remains constant but its intercept changes; in operations under increasing reflux ratio, the slope of the line changes but the intercept remains constant. Batch Rectification under Constant Reflux Ratio: In this case, xD continuously decreases as the reflux ratio is kept constant. A MVC balance at any moment during the rectification can be written as; Lx = (dL)xD+(L-dL)(x-dx) (5-133)

Charge

Bottoms

Heatimediu

ng m

Fig.5.37 Batch rectification column

G,y1

n

1

N

L G

xn

Column

yn+1

Boiler

Top product

Reflux

L,xo D,xD

qC

Condenser


Where, L and dL are the quantities of the liquid in the boiler and the vapor rising from the boiler as k-mole respectively at nsidered, and x and xD are the mole fraction of the MVC in the boiler liquid and in the rising vapor. If the product (dL)(dx) is ignored,

the time co

xxD −

dxL

dL= (5-134)

is obtained from equation (5-133). If this eend of rectification;

quation is integrated from the start to the

⎮=F

⌡

⌠

−=⎮

⌡⌠

F

w

x

xD

F

Wxx

dxW

lnL

dL (5-135)

is found. Where, F and W are the quantities of the charge bottoms as k-mole, and xF and xw are the mole fractions of the MVC in the charge and bottoms. As it is seen, this equation is similar to the Rayleigh equation, but here xD is not the equilibrium value of x. Total material and the MVC balances can be written at the end of operation as; F = D + W (5-136) FxF = DxD,av. + Wxw (5-137) (5-138) With the help of equatio 35), (5-136) and (5-137), the quantity of top product and its average composition at the end of operation for which F, xF, RD, xw and N (number of ideal plates) are known, can be calculated as follows: first the slope of operating line is calculated from equation (5-132) and a line with this slope is drawn on xy-diagram. Then starting at point F, the given number of right angle triangles (N) is drawn between this line and e line up and down. Final line is the operating line at the start of distillation and thus xDi is found. Then

D D

F wl given by equation (5-135) from which W is then found. Finally,

5-137) D and xD,av. uted The heat energy required in the operation can be calculated from;

and the

100.Fx

Dx.R.P

F

av,D=

ns (5-1

equilibrium curve by shifting th

arbitrary xD values, which are smaller than xDi are chosen and parallel lines to the first operating line from the points having these xD s as abscissa are drawn (each line is an operating line corresponding to the selected xD). The given numbers of the right angle triangles are again located between these lines and the equilibrium curve starting this time at points whose abscissa are the selected x . With thus found x values, 1/(x -x) values are calculated and then x values are carried against calculated 1/(xD-x) values on a millimetric paper. The area under the curve between the limits of x and x gives the value of integrafrom equations (5-136) and ( are comp .

λ)DR1()t(tcFqqq DFFLLST ++−=+= (5-139) where, qs (kJ) is the sensible heat needed to heat up the solution from initial

otemperature tF ( C) to the average distillation temperature, t ( C) and qoL (kJ) is the

latent heat required to produce GT (k-mol) vapor which is given as GT=D(1+RD), FLc (kJ/k-mol oC) is the molar heat capacity of the solution calculated at the arithmetic


mean of t and tF and λ (kJ/k-mol) is the molar latent heat of vaporization of the solution calculated at mean temperature. The time for one batch operation can be calculated by adding the heating (sensible heat transfer) and distillation (latent heat transfer) times to the time required for charging

tion from ini

me, equation (5-50) and for the total batch time, equation (5-51) can be used.

Example-5.14) Batch Rectification under Constant Reflux Ratio

n will be

) Estimate the time of on he conditions: Heat transfer area in the still is A=10 2, heating is supplied by the saturated steam condensing inside the tub till at 1

ressure. Over-all heat transfer coeffients for the heating and vaporizatio are es2 2

slope on xy-diagram and locate 3 right ang

In th

and discharging the batch-still. Heating of the solu tial temperature tF to the average distillation temperature t is an unsteady-state operation and the time needed is given by the equations (5-48) and (5-49). For the distillation ti

A 12 000 kg chloroform-toluene solution containing 24.6 mass percent chloroform will be rectified in a plate type batch column, which is equivalent to 3 ideal plates for this system. The solutiocharged to the still at 20 oC and distillation at a reflux ratio of 3.125 and 760 mmHg will be continued until the mole fraction of chloroform in the still drops to 0.04. Calculate : a) Quantity and average composition of the distillate to be produced, b) Percentage recovery of chloroform, c) Heat Energy requirement for the operation. d e batch operation for tm es of the s .05 bar gauge

n periods timated as pU=200 W/m K and Ub= 1 000 W/m K. Assume that specific heat and latent heat of solution remain constat at 140 kJ/k-mol oC and 32 000 kJ/k-mol. Mchloroform= 119.4 , Mtoluene= 92 Vapor-liquid equilibrium of the system at 760 mmHg is given in Table.App.5.2. Solution : a) First, plot t-xy and xy-diagrams of the system. Calculate the slope of operating line from eqn.(5-132),

molk1.12392/)246.01)(00012(4.119/)246.0)(00012(F −=−+=

20.0

92)246.01(

4.119246.0

4.119246.0

x F =−

+=

758.0125.31

125.3R1

R D =+

=+

D

Draw a line with this le triangles between this line and equilibrium curve, starting at point F(0.20;0.20), by shifting this line up and down. This line is the operating line at the start of the distillation and xDi is read from the diagram. Select an arbitrary xD value and draw the operating line from this xD value and starting at this point locate 3 right angle triangles and then read the corresponding x value. Repeat this step untill reaching or exceding the xW.

e figure below only initial and final operating lines are shown.


xD x

0.90 0.20 1.429

0.80 0.12 1.471 0.5(1.429+1.471)(0.20-0.12) = 0.116

0.70 0.085 1.626 0.5(1.471+1.626)(0.12-0.085) = 0.0542

0.50 0.05 2.222 0.5(1.626+2.222)(0.085-0.05) = 0.06734

0.40 0.04 2.778 0.5(2.222+2.778)(0.05-0.04) = 0.025

Then, from equat

Total = 0.2625 ion (5-135);

xx1

D − ⎮⌠

20.0 dx⌡ − xx04.0 D

2625.0WFln = 30.1e

WF 2625.0 == molk69.94

30.11.123W −==

tA

t-xy diagram of chloroform-toluene at 760 mmHg

tB

Mole fraction of chloroform, x,y xF=0.20 xW = 0.04

t-x

t-y

at

ure,

t (

Tem

per

OC

)

ti = 94.5

tf = 106.5

Mole fraction of chloroform in liquid,x

Mo

form

in v

a

xy-diagram of chloroform-toluene at 760 mmHg

por,

y

le fr

actio

n of

chl

oro

xF=0.20

xW=0.04

xDi=0.90

Di

xDf =0.40

Operating lines

Df


Then from equ n ( ); = F- W ol From equation.(5-137) ; (123.1)(0.20) = (28.41) xD,av= 0 b) From equation (5-138) ; c) From equation (5- e atThe averag g p p

atio 5-136 D = 123.1- 94.69 = 28.41 k-mxD,av+ (94.69(0.04)

.733

139) ; Soint tem

nsible heerature

required = e boilin

%6.84100.1(

)733.0)(2(.R.P ==)20.0)(1.23

41.8

ofFrom t-xy d o te 4.Sensible heat required = (123.1)(140)( 100.5- 20) = 1 387 337 kJ

atent heat required = = (1+ 3.125)(28.41)(32 000) = 3 750 120 kJ

) m 20 C is an unsteady e needed is alculated from equation (5-48). o account for heating of the batch-still itself (metal) , take F= (1.1)(123.1) =135.41 k-mol.

From the stread as ts

For the ca distillation time, equation (5-50) is used: By considering additional 1 h time for charging the feed to the still and discharging the bottom product from the still at the end of operation, total batch time is found as ; 12 min + 5 h 36 in. Batch Rectification at Constant Top Product Composition: As was said before, if the composition of the top product is to be kept constant, the reflux ratio must be increased steadily. In this type of operation, generally N, F, xF, xD and xw (or percentage recovery) are known. First D and W are calculated from equations (5-136) and (5-137). Then o ry line towards F is drawn and until the N right angle triangles are located between equilibrium curve and

ux ratio. From its slope or intercept the initial reflux ratio, RDi is calculated. Then from the same D, another line but in this case with greater slope is drawn and the slope of this line again is changed until the given number of the right angle triangles is located between equilibrium curve and this line between the limits of xD and xw. The reflux ratio, RDf calculated from the slope of this line gives the reflux ratio at which the operation will be stopped. It is obvious that the reflux ratio will be increased continuously from the initial value to the last value. The heat energy requirement in this case is calculated from;

solution t can be calculated from; t=(ti +tf)/2 iagram f the sys m, ti = 9 5 oC, tf =106.5 oC are read.

)t(tcF FFL −

L 7+ 3 750 120 = 5 137 457 kJ

λD)R D+1(

Total heat required = 1 387 33 d oC to 100.5 o Heating of solution fro -state operation and the timcT

eam tables, temperature of the saturated steam at Ps=1.05+1 =2.05 bar absolute pressure is =121 oC.

lculation o

[ ] min12h4.sec11615)140)(41.135(/)10)(200.0(S ≈=)5.100121(/)20121(ln −−

f

=θ

min36h5sec12220)5.100121)(10)(0.1(

)1207503)(1.1(≈=

−=θL

θBT = 1 h + 4 h min = 10 h 48 m

from point D wh se abscissa is xD, an arbitra

this line between the limits of xD and xF,, the slope of this line is changed. The line satisfying this is the operating line corresponding to the initial refl

λ+ TF G)t −=+= FLLST (tcFqqq (5-140) GT (k-mol) which i al a ount por to be formed, can be calculated as follows;

s the tot m of va


Total material and the MVC balances at any time θ during the rectification can be

(5-141) F (5-142)

here, D is the quantity of top product collected until this time. By eliminating L

written as; F = Dθ + L

θxF = Dθ xD + Lx

Wbetween the two equations,

⎥⎢=θ FD F ⎤⎡ − xx

⎦⎣ − xx D

(5-143)

hen, = dD)R1(dGG (5-144) example with the help of equation

is obtained. θθ DGT

∫∫ +T =0 D0T

his equation may be written in different ways, for T(5-143), or noting that [ ])G/L(1/1R1 D −=+ ;

[ ]⎮⌠−=⎮

⌠ +−= D dx)xx(FdxR1)xx(FG

⌡ −−⌡ −WW x

2D

FDx

2D

FDT )xx()G/L(1)xx(he evaluation o

Fx

(5-145)

trary operating lines ween R and RDf are d awn and then t e giv

ated between the equiiangles into equation (5-143)

orresponding D s are computed. Then θ

ime required for heating of the solution from initial temperature tF to the average ) or from (5-49).

or the calculation of distillation time;

Fx

T f GT may be done as follows: from point D arbifor reflux ratios bet Di r h en number of right ngle triangles are loc librium curve and the operating lines. By a

substituting the x values read from the last trθc s and from the slope of the operating lines RD

carrying (1+RD) values against the corresponding D s on a millimetric paper and calculating the area under the curve between the limits of RDi and RDf, gives the value of GT. Tdistillation temperature t is calculated either from equation (5-48F

(sec)GGT

L =θ (5-146)

where, G is the molar flow rate of vapor (k-mol/s), which remains constant during the distillation and can be calculated from the heat transfer considerations as follows;

λ

−=

λ= lnHb )tt(AUqG (5-147)

Then, total time for one batch operation is calculated from equation (5-51).

Example 5-15) Batch Rectification at Constant Top Product Composition

The batch of chloroform-toluene solution given in Example 5-14 will be rectified in the same column but this time under varying reflux ratio. The solution will be charged to the still at 20 oC and distillation will be continued, keeping the mole fraction of chloroform in the distillate constant at 0.733, at 760 mmHg until the mole fraction of chloroform in the still drops to 0.04. Calculate : a) The initial and final reflux ratio to be used, b) Percentage recovery of chloroform, c) Heat energy requirement for the operation. d) Estimate the time of one batch operation for the conditions given in Example-5.14. e) Compare the result with Example-5.9 and 5.14.


Assume that specific heat and latent heat of solution remain constat at 140 kJ/k-mol oC and 32 000 kJ/k-mol. Solution : a) First, points F, D and W are located on the xy-diagram of the system. A line from point D is drawn and slope of this line is changed until 3 right angle triangles are located between this line and equilibrium curve within the points D and F. This line is the operating line at the start of distillation.

sfrom point D with greater slope is drawn and the slope of this line is changed until 3 right angle

his time between point D and W fit. This line is the operating line at the end of the slope of this line is the

9 and RDf = 25.18

o qS = (123.1)(140)( 100 - 20) = 1 387 337 kJ

the calculation of latent heat required, first GT must be calculated. For this various

between initial and final are drawn (not shown in the figure) and after locating three right angle

Reflux ratio calculated from the slope of this line i the initial reflux ratio (RDi). Then another line

triangles but tdistillation. Reflux ratio calculated from final reflux ratio (RDf). These are found as RDi = 0.92b) As F, xF , D and xD are the same with the previous example, P.R. = 84.6 %. b c) Sensible heat required is calculated fr m equation (5-140) .5

For operating lines

triangles between each and equilibrium curve starting at point D, x values are read from the diagram. By calculating RD s from the slopes of these operating lines and Dθs from equation (5-143), and then carrying the (1+RD) values against corresponding Dθ values, and calculating the area under the curve between the limits of D = 0 and Dθ = 28.41 GT is obtained.

x Dθ 1+RD

0.20 0 1,929 - ∫ +=

41.28

0 DT dD)R1(G θ

0.16 8.59 2.443 0.5(1.929+2.443)(8.59-0) = 18.78

0.135 13.38 2.932 0.5(2.443+2.932)(13.38-8.59) = 12.87

0.1 19.45 5(2.932+3.665)(19.45-13.38) = 19.69 3.665 0.

0.092 20.74 4.887 0.5(3.665+4.887)(20.74-19.45) = 5.52

0.072 23.84 7.33 0.5(4.887+7.33)(23.84-20.74) = 18.94

0.04 28. (7.33+26.18)(28.41-23.84) = 76.57 41 26.18 0.5

GT = 152.37 k-mol

Then, latent heat required, qL = (152.37)(32 000)= 4 875 840 kJ

2 min. n of distillation time, equation (5-146) is used along with equation (5-147).

Total heat required, qT = 1 387 337+ 4 875 840 = 6 263 177 kJ d) Heating of the solution from 20 oC to the distillation temperature 100.5 oC is an unsteady-state operation and the time required is the same with the Example-5.12, which was 4 h 1For the calculatio min16h7sec16326 ≈

)5.100121)(10)(0.1()8408754)(1.1(

=L −=θ

(10 % excess energy was taken to account for the heating of the column itself).


So, one batch time with 1 hour extra time for charging and discharging of the still is found from equation (5-51); θBT = 12 h 28 min e) Comparison of the results of Example-5.9, Example-5.14 and Example-5.15

Example-5.9 Example-5.14 Example-5.15

Operation mode Simple distillation Batch distillation Batch distllation tant with RD= constant with xD= cons

Purity of top product le fra

34

(mo ction) 0. 0.733 0.733 P.R. of the MVC (%)

90.6

84.6

84.6

Total heat energy (kJ) 3 348 497 5 137 457 6 263 177 One batch time 8 h 03 min 10 h 48 min 12

h 28 min

It follows from the table that simple distillation is not suitable for this system as the relative volatility of chloroform to toluene is not great. Batch distillation under constant reflux ratio requires leand batch time than the batch distillation under varying reflux ratio. Hence this mode of op

ss energy eration is

mber of the ideal

preferred for this system. Calculation of Number of Real Plates: After determining the nulates needed for a given separation, the number of the real plates can be found by p

M ction uid, x ole fra of chloroform in liq

xy-diagram of chloroform-toluene at 760 mmHg M

ole

frac

tion

of c

hlor

ofor

m in

vap

or, y

D

xD = 0.733 xF = 0.20

xW = 0.04

Initial operating line

Final operating line


either applying plate or over-all column efficiency defined in Chapter-4. These efficiencies are found experimentally in correlation giving by O’ d to calculate over-all column efficiency for rectification when experimentally obtained value is not available. Although this correlation was obtained for bubble-cap columns, it can also

tive c the column, the relative volatility calculated at

ed. The liquid viscosity is taken as the molar viscosities of pure

omponents (kg/ms) at column average temperature.

Example-5.16 Estimation of Number of Real Plates

Estimate the numbers of the real plates needed in eac lu ple 5-10 from O’CViscositi anol and at 81 OC are 0.298 cP and 0. ctively : From t-x at the t mn, fo ead as 67 oC, and at the bottom of the column, for x9 =0.04 is read as 95 C. Then, average column temperature is 81 0C. At this tem -xy diagram nd of

the feed liquid at 81oC,

pilot size plate columns. The graphical Connell and shown in Fig.5.38 may be use

α µFL

Ove

r-al

l col

umn

effic

ienc

y, E

o

Fig.5.38 Ove r-all column efficiency at rectification

be used for sieve and valve tray columns for approximate calculations. If the relavolatility of the system hanges along he column average temperature is ust

average viscosity of the feed liquid, which is calculated from thec

h section of the co mn given in Exam

onnell correlation. es of liquid meth n-propanol 61 cP respe

Solution

y diagram of the system, temperature op of the coluo

r y1=0.95 is r

p rature, from t : x = 0.342 a y =0.658 are read. Then relative volatilityemethanol to n-propanol from equation (5-13),

olar average viscosity of

7.3)658.01)(342.0()342.01)(658.0(

=−−

=α

M Then, αµFL= (3.7)(0.50*10-4) = 1.85*10-4

ms/kg10*50.0)10*61.0)(35.01()10*298.0)(35.0( 433FL

−−− =−+=µ


Now enter this value into Fig.5.38 and read Eo=0.74.

Number of the real plates needed in the enriching section = 574.03

=

Number of the real plates needed in the stripping section = 974.06

=

Total number of the real plates needed in the column = 14

5.3.3.3 Rectification in Packed Column: Rectification operation can also be carried

t in continuous contact type of equipment. Especially, when the flow rate of solution be rectified is small, the packed column is preferred as in this case construction of te column is rather difficult due to the small diameter, although the possibility of

rtridge type construction exists. For rectification at fine vacuum, packed columns are many cases preferred because of the fact that special structured packings, developed ently have very low pressure drop along the packing at even fairly high liquid and

por loads. To give an idea, for Montz-Pak, type: A3-500 packing element, pressure p per meter of packing is given as function of liquid and vapor loads in Fig.5.39.

Another case, at which the packed column is preferred to the plate column, is the rectification of heat sensitive solution. As the liquid hold-up of a packed column is

uch smaller than the plate column, the residence time of the solution in the packed composition of heat sensitive components of

olution is rather low. Except these special cases, the choice of column is done by

the diameter of a packed rectification column accomplished by us ding or pressure drop per unit eight of packing as for packed absorption column (Fig.4.14), as gas and vapor hydro-

por and liquid loads are generally quite eter calculations

ust be done for each section separately.

outo placain recvadro

mcolumn is low and as a result of this, deslooking at the economy of the operation. In the construction of packed rectification column, all the criteria seen in the construction of packed absorption column are valid. For example, if the feed is a liquid, a liquid distributor must be installed at the feed introduction section. The calculation ofis ing the same correlation for floohdynamically are the same. But, as the vaifferent in enriching and stripping sectionsd of the column, the diam

m

GGu ρ Pa0.5

L m3/m2h

∆pG m

bar/m

.

Fig.5.39 Pressure drop at Montz-Pak, type: A3-500


The calculation of the height of packing required for a giving rectification is done by n. But here again the enriching and

stripping section must be considered separately. If the constant molar over-flow case is pplicable, then xy-diagram of the system is sufficient for the calculation, otherwise

o requ ed. The total material, omponent and enthalpy balance equations written for rectification in a plate column

as in the packed column the denoting the plates

d(Gy) = -d(Lx) = NAavAcdhe (5-149)

tant molar over-flow = - N and then from Chapter-3 for N ,

)x(xKy)(yK)x(xky)(yk xyixiyA∗∗ −′=−′=−′=−′= (5-150)

can be written. By substituting all these into equation (5-148) finally;

using the same procedure as used in gas absorptio

aenthalpy-composition diagram of the system is als ircare also valid for rectification in packed column. But, ontact of two phases is continuous, the subscripts in the equationsc

must be omitted. If the MVC balance is written for a differential height, dhe in the enriching section of the column as shown in Fig.5.40a d(Gy) = -d(Lx) = NA dSm (5-148) where, NA is the total molar flux of the MVC, dSm is the mass transfer area available in the differential volume dv. If we define an av as the mass transfer area per unit volume of the packing then, dSm = avAcdhe can be written, where Ac is the cross-sectional area of the empty column in enriching section. By substituting this into equation (5-148)

is obtained. G and L remain approximately constant even consdoes not apply. Hence NA B A

N

⎮⌡⌠

−′=⎮⌡

⌠−′

=⎮⌡⌠

−′=⎮⌡

⌠−′

=h=∗∗∫

2)Lx(

q)Lx(cvx

2)Gy(

q)Gy(cvy

2)Lx(

q)Lx(icvx

2)Gy(

q)Gy(icvy

eZ

0 ee )xx(AaK)Lx(d

)yy(AaK)Gy(d

)xx(Aak)Lx(d

)yy(Aak)Gy(d

dZ

is obtained. (5-151) In the case of varying over-flow case, the integral terms are evaluated as follows: as example consider side. From point ∆e arbitrary lines (operating lines) are drawn on the right hand side of ∆ F∆ line

the evaluation of first integral on the right hande s

(enriching section) and points G and L are marked and corresponding y and x values are transferred to the xy-diagram above (Fig.5.40c). By joining the points having these x and y values as coordinates, the enriching section operating line of the column is plotted on the xy-diagram of the system as shown in Fig.5.40c. Then arbitrary P points on the operating line are chosen and x and y values of these points are read and with the help of ∆e=G-L and ∆ex∆e= Gy-Lx, the G and L values are computed at the sections considered. Then by drawing the lines from points P with slopes (- vyvx ak/ak ′′ ), points M on the equilibrium curve are marked and corresponding yi values are read. The area under the curve, which is plotted by taking the 1/ y)(yAak icvy −′ values against the Gy, between the limits of (Gy)q and (Gy)2 gives the value of integral hence the ze.


The evaluation of equations (5-151) is much simpler in constant molar over-flow case as in this case G and L are constant through out the enriching section and consequently;

cvyG Aak

GH

′=

cvxL Aak

LH

′=

cvxL Aak

LH

′=

cvxOL AaK

H′

= (5-152)

can be taken outside the integral signs with the names of height of one transfer unit and the remaining integral terms under the names of number of transfer u

L

nits simplify to:

y=x

W

P q-line

Q

G1,y1

Top product

qCG2,y2

F,zF

Fig.5.40 Rectification in packed column

D,xD

L2,x2

∆S

qS

1.0

y

D

hs=o

F

Strip.sec.op.line

Enric

hing

sec.

Enr.sec.op.line

y2

0 0

xi zF xw=x1

(b)

xD=x2

M

x

yiy

yq

xq

y1

1.0

x

he=o

hs=zs

dhs

dhe

he=ze

Gq,yq Lq,xq

L1,x1

W xw=x1

Bottom product

Strip

ping

sec.

x,y

(a)

(c)

zF

h-x

Spec

ific

enth

alpi

es, H

, h

∆ e

L

GG

GG

LLL F

W

D

xD=x2 xw=x1

P=cons. H-y

1.0 0

L L


⎮⌡⌠

−=

2

q

x

xi

L xxdx

N ⎮⌡⌠

−=

2

q

y

yi

G ⎮⌡⌠

yydy

N −∗ yy

=

2

q

y

y

OGdy

N ⎮⌡x

⌠−

= ∗

2

q

x

OL xxdx

N (5-153)

The solutions of these integrals are again a plished by using graphical integration technique by selecting arbitrary P points on e enriching section operating line, which is a straight line in this case, and drawi the lines with slopes (- ) and locating the M points on the equilibrium curve. If t teps from equation (5- pe the stripping section of the column, the height of packing of this section, (zs) ound. he integramu e taken as (Gy)q - (Gy)1 and (Lx)q - (Lx)1. In t se of v ying ve flow, to plot the operating line on xy-diagram, the required values of x and y are found by drawing arbitrary straight lines this time fro point ∆s on the left hand side of ∆ F∆

ccom th

ng

atedis f

vyvx ak/ak ′′

he s

st b

148) are re for T l limits in this section

he ca ar o r-

m e s and line. For constant molar over-flow case, G L must be use n the eq ations

52) a wi e cross-sectional area of the stripping section, if this is different than the area of enriching section. Heights of transfer units are determined experimentally as in the gas absorption. For this, pilot size columns filled with the same packing are used. As an example, HOG values obtained by Furna and Taylor at atmospheric rectification of ethanol-water system are given in Table.5.2 for two different packing types.

Table OG values for ethanol-water system

Another method, which is used at the calculation of height of packing, is the height

or Ponchon-Savarit method depending upon the conditions. If the height of selected packing, wh a separation equivalent to one theoretical plate under the operating conditions, is known; by multiplying this height with the

n the vapor is above 17 mm liquid head per meter of packing.

Packings Depth of packing(m)

Liquid flow flux(kg/m2s) (m)

d i u (5-1 long th th

s

5-2. H

HOG

50 m Raschig rings 3.0 1.06 0.670 m 25 mm Raschig rings 3.0 1.0 2 0.3669.5 mm Raschig rings 2.44 0.416 0.396 9.5 mm Raschig rings 2.4 0.780 0.305 4 25 mm Berl saddles 2.75 0.195 270.4 25 mm Berl saddles 2.75 1.25 0.335 12.5 mm Berl saddles 3.0 0.25 0.457 12.5 mm Berl saddles 3.0 1.196 0.274

equivalent to a theoretical plate (H.E.T.P.). As it has been shown, the number of the theoretical plates required for a given rectification can be calculated by using McCabe-Thiele

ich makes

number of the theoretical plates, the required height of packing is found. H.E.T.P. values at a rectification column filled with pall rings of 25, 38 and 50 mm sizes were found in the ranges of 0.4-0.5 m., 0.60-0.75 m. and 0.75-1.0 m. respectively, when the pressure drop i


A3-500

The producers of special packing materials give generally the number of the theoretical plates that will be achieved per meter of packing (reciprocal of H.E.T.P.) as

t types of packing produced by Montz GmbH are

ration of the solution is not possible. Depending upon the compositions

pe, one of the components may be obtained in pure state. As an example consider the binary solution having

% MVC, as Fig a 10 -mol/h feed,

containing 30% C, bjected to rectification, the top product can approach to th otropic sitio course, if sufficient num he or packing height is provid enriching section) and with the assumption that all the MVC appears

top p , a product of 0x30/0.65 5 k-m withdrawn

nd 100-46.15 = 53.85 k-mol/h bottom product which is

he MVC.

function of flow factor, which is defined as uG√ρG. As an example, in Fig.5.41 the separation powers of two differenshown. 5.3.3.4 Rectification of Azeotropic Solutions: Binary solutions having minimum or maximum boiling azeotropic points may be subjected to rectification. But as the azeotropic composition cannot be exceeded, a complete sepa

of the feed and azeotro

minimum boiling azeotrope at 65shown in .5.42. If 0 k

MV is su

e aze compo n (of ber of t plates ed in the

in the roduct top D=(10from the colum

)= 46.1n, leaving behi

ol/h is

almost pure in the LVC. As it is seen, in this case a pure product consisting of LVC can be obtained, the top product being an azeotrope. Now, let us assume that the feed flow is again 100 k-mol/h, but it contains this time 75% MVC. In this case, the top product may again approach the azeotropic composition. With the assumption that almost all the LVC is in the top

product, a top product of D = (100x25)/(1-0.65) = 71.43 k-mol/h is obtained, leaving behind 100-71.43 = 28.57 k-mol/h bottom product which is almost pure in t

Fig.5.42 Rectification of a minimum boiling azeotrope

zF

F

D

Az

Az

tA

tB t-y

t-x t

zF

zF

zF 0.2

0.60.40.2

1.0

0.8 1.0

0

0

x,y

0.4 0.6 0.8

x

y

1.0

y=x

Fig.5.41 The separation powers of two different type of packing

5.0GG Pau ρ

Nid

eal/m

. pac

king

5.0GG Pau ρ

Nid

eal/m

. pac

king


As it is seen, in this case a product, which is almost pure in the MVC is produced, ile the other product is not pure. The azeotropes having maximum b point ave similarly, but in this case bottom products are azeotropes and the top products pure products. It follows from these that eotropic solutions cannot be separated pure products by ordinary rectification. As it will be considered later, only by ing a third component to the solution, com aration is made possible. But in

this case, another column to recover the added component is necessary. However, some binary azeotropic solutions can be separated completely without adding a third component to the solution. These solutions either form hetero-azeotrope

por forms two quid phas ed as hetero-zeotropism. The reason for hetero-azeotropism is the limited solubilities of the

b point ave similarly, but in this case bottom products are azeotropes and the top products pure products. It follows from these that eotropic solutions cannot be separated pure products by ordinary rectification. As it will be considered later, only by ing a third component to the solution, com aration is made possible. But in

this case, another column to recover the added component is necessary. However, some binary azeotropic solutions can be separated completely without adding a third component to the solution. These solutions either form hetero-azeotrope r their azeo por forms two quid phas ed as hetero-zeotropism. The reason for hetero-azeotropism is the limited solubilities of the

whbehareintoadd

behareintoadd

oilingoiling

az

plete sep

az

plete sep

oor their azeotropic composition change with pressure. If an azeotropic vaes upon condensation, this type of azeotropism is calltropic composition change with pressure. If an azeotropic vaes upon condensation, this type of azeotropism is calllili

aacomponents in liquid state. Water/n-butanol is a hetero-azeotropic system, for liquid water and n-butanol dissolve limitedly within each other as shown in Table.5.3. Table 5-3. The solubilities of water/n-butanol binary

components in liquid state. Water/n-butanol is a hetero-azeotropic system, for liquid water and n-butanol dissolve limitedly within each other as shown in Table.5.3. Table 5-3. The solubilities of water/n-butanol binary

Temperature(oC) Butanol layer (mass % of alcohol)

Water layer (mass % of alcohol)

Temperature(oC) Butanol layer (mass % of alcohol)

Water layer (mass % of alcohol)

80.38 9.95 5 10 80.33 8.92 15 80.14 8.21 20 79.93 7.81 25 79.73 7.35 30 79.38 7.08 35 78.94 6.83 40 78.59 6.60 50 77.58 6.46 60 76.38 6.52 70 74.79 6.73

Wa r/n-butanol solut ns hence can bcontaining two columns as shown in Fdecanter, which is loc d beneath the of the decanter is pumped to the top introduced to the top of another columstripping sections and there is a comcom g fr f c n ave 23 mol % butanol) and upon co densiwhere they separate in two insoluble reboiler of the water column stripes ocolumn, giving it the water. By provi

te io e completely separated in a distillation system ig.5.43. As it is seen, the feed is introduced to a condenser. While the bottom layer (water layer) of a column, the upper layer (butanol layer) is n. As it is seen, both columns consist of only

mon condenser for both columns. The vapors almost azeotropic compositions (approximately ng in this condenser, they flow to the decanter, liquid phases. The water vapor rising from the ff the butanol from the liquid flowing down the ding sufficient number of the plates or packing

height, the vapor at the top of the column can be brought to the azeotropic composition. The liquid leaving the column at the bottom is almost pure water and part of it is vaporized in a reboiler and returned back to the column, while the remaining

ate

in om top o the olum s hn


part is discarded. In this column, open steam may also be used with the elimination of the reboiler. The vapor, rising from the reboiler of the other column and which is almost pure in n-butanol, stripes off the water from down-flowing liquid, giving it the

azeotrope azeotrope

Water column

y = x

WI

slope=LI/GI

1

Az

condenser

n-Butanol

N

1

yDI ≈0.77

N

1

decanter

GI GII

Water

reboilers

Water column n-Butanol column Feed

xoI

yDII ≈0.77 qc

xoII

xwI xwIIWI WII

GI GIILI

LII LIILI

GIIGI

LII

qsI

F, zF

qsII

LI

0.2 0 0.4 0.6 0.8

1

N

N

n-Butanol column

xoI xoII

yDI

yDII

xwI xwII

WII

slope=LII/GII

1.0 P =760 mm Hg

0

Mole fraction of water in liquid, x

Mol

e fr

actio

n of

wat

er in

1.0

vap

or, y

I II

760 mmHg

760 mmHg

Fig.5.43 Rectification of n-butanol/water solution


n-butanol. Hence with locating sufficient number of the plates or providing sufficient height of packing in this column, the composition of the vapor is brought to the azeotropic composition, when it reaches top of the column, and the liquid is made to reach almost pure butanol composition, when it leaves the column at the bottom. The design of both columns er they are plate or packed, can be made with the methods given aIn another group of azeotropic solutions, the composition of azeotrope changes with pressure. In this case, by using two co , each ope t different pressure, complete separation of azeotropic mixture is accomplished. Aceto nitrile/water solution is a typical example of this group. While binary azeotrope contains 73 mole percent acet ni le t 76 g, this rises to 84 mole percent at 150 mmHg. Aqueous aceto nitrile solution, containing approximately 30 mole percent aceto nitrile, is mi the top product of the second column and is fed column, which es at 150 mmHg pressure as shown in Fig.5.44. While the azeotropic mixture is withdrawn at the top of the column, the excess water is taken from the bottom and discarded. The top product of this column is then fed to the second column, wh operates at 760 m Hg. The top product of this column, which is an azeotrope with 73 mole percent aceto nitrile, is mixed with the aqueous solution coming from the plant to make the feed to the first column as mention before. The excess aceto-nitrile leaves the second column as bottom product in almost pure state. The columns, which y be plate or packed can easily be designed by using the methods giving above.

ma

mich

to the firstxed withoperat

a 0 mmHtrio

rating alumns

bove. wheth

xoI

F, zF

Aceto nitrile

N

1

N

1

Water

Reboilers

Water column Aceto nitrile column

ED

xoII

WII I , xwI

xDII

FII =DII

Fig.5.44 Rectification of aceto nitrile/water solution

qsII

qcI

qsI

qcII

P=760 mmHgP=150 mmHg

I II

DII

zFII = xDII

FI , zFI

Azeotrope-I Azeotrope-II

WI ,xwI

FE


5.4 Internal Design of Plate Columns for Liquid-Gas (Vapor) Contact: The plate columns are widely used in gas absorption, desorption and rectification operations. The first step in the design of these columns is to find the number of the plates needed for a given separation for which the ratio of the amounts of two phases (not the absolute values) flowing inside the column along with the equilibrium relationship of the system is sufficient. Next step constitutes the calculation of the diameter of the

re. The

Fig.5.45 Types of plates There are different types of plates, which are used in today’s industry. Some of them are shown in Fig.5.45. The most commonly used ones are sieve (perforated) plate and the valves tray (plate). The bubble cap plates, which were once used widely, are now used in some special cases due to the high cost. The sieve plates are now almost standard in use, because of very well developed design equations and low cost. The only disadvantage of these plates is to be prone to the weeping at reduced which causes the loss of efficiency. If during the operation, vapor load is to be subjected to low values from time to time, it is then the use of valves trays are recommended as they adjust the holes area automatically depending upon the vapor load and hence they operate in a wide range of vapor loads wi sing the efficiency. Below, design principles of a sieve plate column are to be given as an example. Although the equations for other types a similar but they are not identical. A sieve plate is essentially a circular metal plate on part of it, which is called active area, Aa small diameter holes are drilled for gas flow. Part of the plate area, which is called down-comer area, Ad is cut off to make a segmental channel for liquid flow as shown in Fig.5.46. A metal sheet, called down comer apron is fixed to the cut part of the plate vertically to form the channel and is extended above the plate to create a weir, which is required to keep id pool te. On the area of the plate, which is projected area of the down-comer above, holes are not drilled, because no vapor flow through the down-comer is desired. Thus, the active area of a plate is Aa=Ac-2Ad, whe

column and the design of plates themselves. The diameter is directly related to the amounts of the phases flowing inside the column. In the determination of the diameter of a plate column, the flooding criterion is used as in the packed columns. Of course flooding relationship in plate columns is quite different and complicated than the packed columns as the design of the plates themselves play an important role hedesign is accomplished by compromising various opposing tendencies, which require large number of trial and error.

Bubble Cap Plate Sieve Plate Valves Tray Valves Tray

vapor loads,

thout lo

of plates re

a liqu on the pla

re Ac is the total cross-sectional area of the plate.


In order to keep plate efficiency as high as possible, gas or vapor is dispersed as small bubbles in the liquid pool formed on the plate. It is obvious that great depth of liquid on the plate means long contact time between gas and liquid which in turns, means high plate efficiency. But great liquid depth causes high pressure drop in the gas rising through the liquid. This results in high operating cost in gas absorbers and in high temperatures at the bottom of rectifiers, which are both not desired. It follows from here that height of weir must be kept at a reasonable level, neither too high nor too

low. On the other hand, gas must be dispersed in bubbles as small as possible and their velocities through the liquid pool on the plate must be as high as possible for a high plate efficiency, as the first increases the mass transfer area and the latter the mass transfer coefficient, due to the increased turbulence. But high gas velocity through the liquid pool again means high pressure drop in the gas side. Not only this, high gas velocity also causes liquid entrainment which in turns decreases the plate efficiency.

and which reduces

So, it follows from these explanations that up to a certain value, increase in gas velocity increases plate efficiency due to the increased turbulence, but after thendecreases it due to the excessive liquid entrainment. The pressure drop in the gas side influences the liquid level in the down-comer above; high gas side pressure drop means high liquid level (back-up) in the down-comer. If the liquid level in the down-comer reaches the liquid level of the plate above, the whole column is then filled with the liquid and the gas phase passes through the liquid in fluctuating pulses and the plate efficiency rapidly drops to a very low value. This condition is known as flooding. So, the gas velocity in the column must be so selected that no flooding should be observed. On the other hand, low gas velocity through the holes results in liquid raining down through the holes which is known as weeping

Projected down-comer area

hw

how h

Ad

d

holes

Aa

LW

liquid

Weir

Plate

Vapor

Down-comer

Column

hdc

(a)

Weir

(b)

Plate (Ac) Net area (An)

Ad

Active area

Down-comer apron

Down-com area

er

Fig.5.46 (a) Vertical cut of a sieve plate column, (b)view of a sieve plate from the top


the plate efficiency sharply. It follows from the above explanations that there is a safe operating region for the column depending upon the gas and liquid flow rates as shown in Fig.5.47. The conning is observed at low liquid flow rates, in which gas pushes the liquid around the hole away and it passes as inverted cone through the liquid. This reduces the plate efficiency drastically as the gas-liquid contact becomes very poor. The design of plates in the satisfactory operation region is of prime importance and requires many trial and errors, which are explained below.

5.4.1 Column meter: The diameter of column Dc (m) is com ted from equation,

Con

ning

Satisfactory operating region

Flooding

Liquid entrainment

Weeping

Limitation of d mer own-co F

low

rate

of g

as

Flow rate of liquid

Fig.5.47 Satisfactory operating region of sieve-plate column

Dia pu

0.5

GGc a)(1uπρ

G4D ⎥⎦⎢⎣ −

= (5-1⎤⎡ &

54)

where, ρG (kg/m3) is density of gas (vapor), uG (m/s) is gas velocity based on net area (An) and a is the ratio of Ad/Ac. The net area is the area through which gas phase flows between the plates. Hence, it is given as An=Ac-Ad. Gas velocity uG is taken as some percent of flooding gas velocity uGF, which was found dependent on

ahLG /AAσ,P.S.,,ρ,ρ,L,G && . Here, L& (kg/s) is the liquid flow rate, ρL (kg/m3) is the density of the liquid, P.S.(m.) is plate spacing, σ (N/m) surface tension of liquid and Ah (m2) is total hole area. The first four of these variables are collected under one group, which is known as flow parameter, as follows,

L

GLG ρ

ρGL

F &

&= (5-155)

Dependence of uncorrected capacity factor, K1 on this flow parameter at various plate spacings has been determined experimentally and the results are given in Fig.5.48. The capacity factor obtained from this diagram is valid for σ = 0.02 N/m and Ah/Aa ≥ 0.1.


When these values are different, K1 is multiplied by f.(σ/0.02)0.2 and corrected capacity factor K1c is found. Factor f is obtained from the table below.

Ah/Aa ≥ 0.1 0.08 0.06f 1.0 0.9 0.8

Then from equation below the gas velocity, which will cause the flooding, is alculated.

c

G

GL1cGF K=

ρρρ

u−

(5-156)

y taking som al operating gas

commonly used range in th s.

B e percent of this velocity (usually 60-80 %), normvelocity uG (m/s) is found. Finally, inserting this value into equation (5-154) the diameter of the column is obtained. It is the property of a circle that once the ratio of Lw/Dc is specified then a = Ad/Ac is automatically fixed. This relationship is given in the table below for the most

e design

Lw/Dc 0.690 0.695 0.700 0.705 0.710 0.715 0.720 0.725 0.730 a = Ad/Ac 0.0833 0.0855 0.0878 0.090 0.0922 0.0945 0.0968 0.0992 0.1016

5.4.2 Plate Spacing (P.S.)(m): The plate spacing, which is the distance between two

is the most ihence it must be kept as small as possible, if there are no other reasons such as onstruction of manholes etc. In practice, plate spacing is selected 0.45m or smaller for

consecutive plates, mportant parameter fixing the height of a column and

ccolumns having diameters less than 1.2m and higher plate spacings are employed for larger diameter columns, although this is not a rule. At cryogenic rectifications such as the rectification of air, plate spacings as low as 0.15-0.20 m are used.

0.6 0.4 0.080.060.04 0.02 0.2 0.01 0.1 1.0 0.8 2.0 5.0

0.02

0.01

Cap

acity

fact

or, K

1

0.04

0.08 0.06

0.1

0.2 Plate spacing, m

Flow parameter, 5.0

GLF ⎟⎟⎞

⎜⎜⎛ ρ

=&

L

LG G ⎠⎝ ρ&

Fig.5.48 Capacity factor


5.4.3 Liquid Entrainment: The liquid entrainment must be kept at a low value as it reduces the plate efficiency. The fractional entrainment, which is defined as,

E = eL +&

e (5-157)

must be kept below 0.1 for a sat ac e I on above, e is the liquid

ent of flooding gas velocity and these relationships found experimentally

eters of the holes, h echange from 3 to 15 mm, the 5 mm being the most widely used one. They are drilled at the corners of either a

or an equilateral triangle, the pitch (distance between the hole centres), PT (mm) being 2.5 to 5 hole diameters. While smaller pitch causes the coalescence of the gas bubbles issuing from the holes, greater pitch makes the diameter of the column unduly large. Between down-comer and the first ro f les a lso between the last row of holes and the weir 100 mm’s un-drilled spaces must be left (why?). By observing these, the holes must be

uted evenly in the active area. Total holes area accounts for generally

isf tory d sign. n the equatias kg/s carried back to the plate above by the gas and L is the flow rate of liquid as kg/s. So-defined fractional entrainment was found to be a function of flow parameter and the perc

&

are given in Fig.5.49. 5.4.4 The Holes: The diam d (mm) punched in th active area

1.0

0.1

0.01

002

0.6

0.04

0.02

0.06

0.4

0.2

0.08

0.008 0.006

0.004

Percent of floodingsquare

w o ho nd a

distrib

between 5-15% of the active area. The ratio of total holes area to the active area is function of the ratio of hole diameter to the hole pitch, the constant being dependent on the holes

configuration. For equilateral hole configuration the relationship is ; 2

T

h

P7⎢⎣a

ion,

h d0

A⎥⎦

⎤⎡ )

n r o

0.9=A

(5-158

a d for squa e hole c nfigurat

2

T

h

a

h

Pd

0.785AA

⎥⎦

⎤⎢⎣

⎡= (5-159)

are given. As it is seen, more holes can be drilled in equilateral configuration than the square configuration. The total number of the holes drilled, n is then calculated from,

62h

h 104dπn

A −= (5-160)

0.001

0.

Flow parameter, FLG

Fra

ctio

nal e

ntra

inm

ent,

E

0.2 0.01 0.1 0.06 0.04 0.02 1.0

Fig.5.49 Liquid entrainment

0.8


5.4.5 Weir: As explained before, great weir height, hw (mm) increases the plate efficiency because of the long contact time between gas and liquid but, it also

u mm

(

increases the pressure drop in the gas side. Hence, an optimum height m st be found for the weir. This is as low as 20 for vacuum operations and as high as 75 mm for high-pressure operations and mostly 50 mm for atmospheric operations. The length of weir, LW (m) is selected such that liquid weir crest, how mm liquid) is not excessive. For this, it is recommended that volumetric flow rate of liquid per unit length of weir is to be less than 0.025 m /sm.

3

Liquid weir crest: liquid crest over the weir is proportional to the volumetric flow rate of liquid and inversely proportional to the length of weir and this relationship after

perimentation was given as; ex2/3

wLw Lρ

L750 ⎥⎦

⎤⎢⎣

⎡=

& (mm) (5-161)

r a reasonable plate efficiency, h

oh

Fosh5.4thphcadrtheprpaplakncaho

w+how ould be no smaller than 50 mm. .6 Pressure Drop in the Gas along

e Plate: The pressure drop that the gas ase will encounter as it passes the plate, n be divided in two parts: dry pressure op and the pressure drop that occurs as gas passes through aerated liquid. Dry

essure drop that occurs during the ssage of gas through the holes of dry te can easily be found from very well own orifice equation, which for this se can be written in terms of liquid head (mm) as;

2h

L

G2

ou

ρρ

C1

51 ⎟⎠

⎞⎜⎝

⎛= (5-162)

ere, u

o

wh

h

h (m/s) is the velocity of gas through the holes which can be calculated

from,

Orif

ice

cons

0 5 10 15 20

(Ah/Aa)*100

and less

Fi

0.95 0.90

85

0.80 0.75 0.70 0.65

Plate Thickn hess/Hole diameter= χ/d

0.

g.5.50 Orifice constant

tant

, Co

hGh Aρ

Gu

&= (5-163)

tion of the ratio of plate thickness to the diameter of hole and the ratio of total holes area to the active area and this relationship is given in

The pressure drop that is due to the passage of gas through aerated liquid, h is given

)

tion factor and

The orifice coefficient, Co is a func

Fig.5.50. a

as mm liquid head by, ha = Qp(hw+how) (5-164 where, Qp is known as aera can be found from Fig.5.51.


Hence, the total pressure drop in the gas side per plate, hT as mm liquid head is calculated from, hT = ho + ha (5-165) Weeping: the weeping, which is known as the liquid coming down from the holes must be prevented or at least must be kept at a reasonably low level. To secure this, the velocity of the gas through the holes, uh must always be higher than the minimum hole velocity, uhm, which is giving by,

uhm = 0.5G

h2

ρ)d0.90(25.4K −−

(5-166)

ship between them is

+how) and the relationgive in Fig.5.52. Dow -com vel in

as and liquid phases as they flow, to e liquid level on the plate, which is (hw+how). Total pressure drop in the gas side was

d as hT. The pressure drop in the liquid side as it flows through t r

ed as the velocity of the liquid and the distance it travels, are very small. But

e restriction at the plate entrance is created deliberately by

as bubbles through the down-comer. Generally an opening corresponding to hdc -comer apron.

Hence, the cross-sectional area through which liquid flows at the plate entrance

where, K2 is a constant dependent on (hwnn er back-up: the liquid le the down-comer, which is also known as

32 31 30 29 28 27

K2

down-comer back-up, is very important during the operation. This level is obtained by adding the pressure drops that will occur in the gthobtaine he down-comeand through the contraction formed at the plate entrance can be calculated by using the hydraulics equations. The pressure drop due to the flow in the down-comer can readily be neglectthe pressure drop encountered during the flow through the restriction at plate entrance may be significant. Thimmersing the down-comer apron into the liquid on the plate to prevent the escape of g(mm) = hw-10 is left on the plate between plate deck and tip of down

Fig.5.52 Constant K2

Qp

0.5

GaGa A

GFρ

=&

1.5 1.0 2.5 3.0 2.0 0.0

Fig.5.51 Aeration factor

0.4

hw+how , mm

1.0 0.8 0.6

0 20 40 60 80 100 120


becomes Ape= Lw* hdc*10-3 (m2) and the pressure drop in the liquid side due to the flow through this restriction, hpe (mm liquid) is given as;

hpe= 1662

peL AL

⎥⎥⎦

⎤

⎢⎢⎣

⎡

ρ

& (5-167)

This pressure drop reflects itself in the down-comer back-up too. The liquid flowing into the down-comer from the previous plate is an aerated liquid and hence, by taking this into account for design sa iven as,

fety, finally liquid level in the down-comer is g

p

peTowwda Q

hhhhh

+++=

So-calculated level must always be smaller than P.S. to evade the flooding. In practice rather great safety margins are used and columns are designe

3

(5-168)

d such that

hda ≤ ½ (P.S)*10 (5-169) Residence time in down-comer: The liquid flowing into the down-comer is an aerated liquid. This liquid must be freed from the gas bubbles before it enters the plate below. The experiments have shown that a residence time of 5 second will be enough for the bubbles to coalesce and escape back to the gas phase above. The residence time of liquid in the down-comer can be calculated from the equation below;

L

ddr /ρL

)(P.S.A&

=θ (5-170)

Above; the equations, which are required for the calculation of diameter and some important parameters of a sieve plate column for gas-liquid contact were listed. Below, the sequence of the use of these equations under the name of design steps is given.

how

hw

hd

liquid

Weir

Plate

Vapor

Down-comer

Column

ho

how

hw

ha

hda

hpe

Fig.5.53 Down-comer back-up


5.4.7 Design Steps: 1o) First calculate as kg/s, which are no longer changed. 2o) Calculate or estimate ρL, ρG and σ at the operating conditions. 3o) Assume a plate spacing, P.S. 4o) Make a plate lay-out and assume a percent of flooding. From plate lay-out, selection of Lw/Dc, dh T and hw are understood. These can be changed at any time. 5o) Calculate the diameter of the column as explained above. 6o) Calculate fractional entrainment, E from Fig.5.49. If the value is greater than 0.1, then reduce the percent of flooding, which w s assumed at step 4. 7o) Check for weeping by calculating uh and uhm. If weeping occurs, then return back to step 4 and re-assume the relevant ones in the plate lay-out. 8) Calculate the total pressure drop along a plate. If this is too high, return back to step 4 and re-assume the relevant ones in plate lay-out. 9o) Calculate the down-comer back-up. If this is greater than half of the plate spacing, first go to step 4 and renew the relevant as . After trying all the possible changes in step 4, if down-comer back-up is still greater than half of the plate spacing, then go to step 3 and increase the plate spacing. 10o) Calculate the down-comer residence time. If this is smaller tep 4 and change the relevant ones in the assume

al design of a sieve plate column is accomplished by first making checking themn and deso is rea

ction where are the greatest. As for the rectification columns, diameter is

de different.

Example-5.17) Internal Design of a Sieve-Plate Rectification Column

te column with at the

GandL &&

, P

a

sumptions

than 5 s., then go back to sd plate lay-out.

Do not forget the fact that you should re-calculate all the steps down from the step to which you have returned. As it is seen, internsome assumptions and then one by one. In the design of absorptio rption columns, the design lized at the

GandL &&secalculated at the top and at the bottom of the column. If no great difference is found, then the greater diameter is selected for both sections, but the plate lay-outs in each section may be different, if necessary. If the difference between the diameters of two

ctions is quite great and/or the construction material of the column is expensive, then sethe diameters and the plate lay-outs of each section maybe ma

For the separation of methanol/n-propanol solution given in Example-5.10 , a sieve-pla

m plate thickness will be used. Design the stripping section of the column with the values 4 mbottom plate.

Solution:

From Example-5.10, ,h/molk84.493L −= h/molk84.296G −= , tN= 95 oC, xN=0.022 (mass ac.), x = 0.04 (mol frac.), y = 0.12 (mol frac.) fr

Mass flow ratN

e of liquid, N

s/kg1.8)]60)(96.0()32)(04.0)[(6003/84.493(MLL L =+==& Mass flow rate of vapor, s/460)(88.0()32)(12.0)[(6003/84.G +& kg7.)]296(MG G ===

t 95 oC , A 3methL m/kg716=ρ 3prop

L m/kg735=ρ , m/N016.0meth =σ , m/N018.0prop =σ 3

L m/kg6.734)735)(978.0()716)(022.0( =+=ρ , m/N018.0)018.0)(978.0()016.0)(022.0( ≅+=σ

3GbottomG m/kg0.2

)95273)(082.0()64.56)(07.1(

RTMP

=+

==ρ (Pressure at the bottom was assumed as 1.07 atm)

Assumptions: P.S.=0.60 m, Lw/Dc=0.69, hw=50 mm, dh=5 mm, PT =2.5 dh(∆), Percent of flooding =70


Flow parameter, from equation(5-155) 09.06.734

0.27.41.8FLG == , Then, uncorrected capacity factor

from Fig.5.48 for P.S.=0.60 m is read as K1=0.1 From equation (5-158) Ah/Aa= 0.907(1/2.5)2 = 0.145. Then, as f = 1, corrected capacity factor, K1c = (0.1)(1.0)(0.018/0.02)0.2 = 0.098 Flooding vapor velocity from equation(5-156); uGF = 0.098[(734.6-2.0)/2.0]0.5 = 1.88 m/s Operating vapor velocity, uG = (0.70)(1.88) = 1.32 m/s For Lw/Dc = 0.69 a = Ad/Ac = 0.0833. Then, column diameter from equation (5-154),

m573.1)0833.01)(32.1)(0.2)((

)7.4)(4(D5.0

c =⎥⎦

⎤⎢⎣

⎡−

=π

. Take Dc = 1.60 m.

Total column area, Ac = (0.785)(1.60)2 = 2.01 m2

Down-comer area, Ad = (0.0833)(2.01) = 0.167 m2

Net area, An = 2.01-0.167 = 1.843 m2

Active area, A = 1.843-0.167 = 1.676 ma2

Total hole area, A = (0.145)(1.676) = 0.243 mh2

Length of weir, L = (0.69)(1.60) = 1.10 m

w

-Check for liquid entrainment, From Fig.5.49 for FLG = 0.09 and 70 % of flooding , E = 0.021 is read. As this value is smaller

how = 50 +

than 0.1, liquid entrainment is acceptable. -Check for weeping, Vapor velocity through the holes from equation (5-153), uh = 4.7/(2.0)(0.243) = 9.67 m/s Liquid weir crest from equation (5-161), how = 750 [8.1/(734.6)(1.10)]2/3 = 34.87 mm hw + 34.87 = 84.87 mm. From Fig.5.52 K2 = 30.8 is read. Minimum vapor velocity through the holes from equation (5-166),

s/m8.80.2

)54.25(90.08.30u hm =−−

=

Although uh > uhm the difference is very small. It is advisable to increase the uh . This can easily be accomplished by drilling less holes in the active area. For this, change PT/dh from 2.5 to 2.8. With this new value, Ah/Aa = 0.907 (1/2.8)2 = 0.116 is found. As f is still 1, K1c and hence Dc

will not change. Only, A will have new value. This is : A = (0.116)(1.676) = 0.194 m2. Now, vapor velocity is

much secure than the previous one to prevent the weeping.

a)*100 = 11.6 and with χ/d = 4/5 = 0.8, from Fig.5.50 C = 0.80 is read. Then, from

equation (5-162);

h h velocity through the holes rises to uh = 4.7/(2.0)(0.194) = 12.11 m/s. This

-Check for pressure drop in the vapor along the plate; Dry pressure drop at the vapor side along the plate,

For (Ah/A h o

headliquidmm82.31)11.12(6.734

0.280.0151h 2

2

o =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

FGa = 4.7/[(1.676)(2.0)0.5] = 1.98 ≈ 2. Fro aeration factor, Qp = 0.61 is read. Pressure drop in the vapor through the aerated liquid, from equation (5-164);

1 (84.87) = 51.77 mm liquid head. late,

id head. rresponds to (83.59)(734.6)/13 600 = 4.52 mm Hg, which can be e for atmospheric rectification column.

re drop in the liquid plate entrance; nce at plate entrance, hdc = 50-10 = 40 m at plate entrance,

Ape = (1.10)(40*10-3) = 0.044 m2. Hence pressure drop in the liquid at plate entrance from equation (5-167), hpe = 166[8.1/(734.6)(0.044)]2 = 10.43 mm liquid head. -Check for down-comer back-up; Down-comer back-up from equation(5-168),

m Fig.5.51

ha = 0.6 Total pressure drop in the vapor through the p h = 31.82 + 51.77 = 83.59 mm liquT This pressure drop co considered reasonabl Pressu at Cleara m. Then, flow area


mm3.29361.0

43.1059.8387.3450hda =+++

=

As h < ½ (P.S de are valid. -Check for down-comer residence time from equa

da .)*103 , assumptions mation (5-170);

s1.9)60.0)(167.0(==θ

)6.34

07 atm) at the start of the design is quite close to this.

Finalised Design

7/1.8(dr

As this value is greater than 5 s, is acceptable. -Check for pressure at the bottom of the column, Pbottom= Ptop+ NReal hT = 1 + (14)(4.52)/760 = 1.083 atm. So, the assumed value (1.

:

Percent of flooding = 70 E = 0.021

uh = 12.11 m/s uhm = 8.8 m/s

θdr = 9.1 s = 1.60 m Pbottom = 1.083 atm = 1.10 m h = 293.3 mm

8 = 14

or pressure with temperature for n-hexane and n-heptane is as follows:

P.S. = 0.60 m h = 50 mm w d = 5 mm h χ (thickness of plate) = 4 mm PT (∆) = 14 mm Dc L W da h = 40 mm dc n (number of the holes/plate) = 5 90 N Real

Problems 5.1 The variation of vap po(mbar) 13.3 26.7 53.3 80 133 267 533 1013 2026

hexane - 25.0 - 14.1 - 2.3 5.4 15.8 31.6 49.6 68.7 93.0 t ( oC) heptane - 2.1 9.5 22.3 30.6 41.8 58.7 78.0 98.4 124.8

equilibrium data for the system at 1 atmAssuming ideal behaviour, calculate and plot the vapor-liquid .

on containing 0.30 mole fraction of the MVC

b) Wc) F equilibrium liquid and vapor phases at 94 oC and 1 bar The system obeys Raoult’s and Dalton’s laws and its relative volatility is 3.0. The variation of vapor pres

5.2 a) Calculate the bubble point of a binary liquid solutiat 1 s e bar pres ur .

hat is the dew point of the vapor mixture containing 30 mol percent MVC at 1 bar. ind the compositions of

sure of the less volatile component with temperature is given by the expression

1750T

o

620.7plog oB −=

where Bp is in mbar and T in Kelvin. [Ans.: a) 90 0C, b) 98 oC, c) x = 0.206, y = 0.438]

erent binary solutions at 760 mmHg, ositions were calculated as follows:

5.3 From the vapor-liquid equilibrium measurements of three diffthe relative volatilities at two different liquid comp

system αAB at x = 0.3 αAB at x = 0.8 1 Ethylacetate(A)-Ethanol(B) 1.49 0.66 2 Methanol(A)-Water(B) 4.63 2.69 3 Etyhlenedichloride(A)-Benzene(B) 0.898 0.902


Which of these three system

s shows azeotropism between the given concentrations, why?

urified from very small

C will be continuously supplied to the vessel to maintain a l benzoate is insoluble in ater. Distillation is to be carried out at 760 mm

ated at 40 kN/m2 gauge pressure and condenses in the cket.

5.4 1 200 kg n-propyl benzoate (M.W.= 164 ; b.p.=231 oC) at 20 oC, will be pn a jacketed vessel equipped with a quantity of non-volatile impurities, with steam distillation i

ocondenser and a decanter. Water at 20 liquid water level. N-propy wHg pressure. Calculate: a) The temperature at which distillation will proceed, b) The amount of water to be used, c) The heating steam requirement. Steam is saturja

m rTe peratu e ( oC) 97.93 98.62 99.31 99.66 99.83 110

oAp (mmHg) 705.69 723.44 741.56 751.47 755.40 1060 Water

(A) λA (kJ/kg)

2267 2238

B (mmHg) op 7.89 8.17 8.46 8.53 8.68 - n-Propyl benzoate

λ (kJ/kg) - (B) B

- -

352.2 -

-

O4

ver the temperature range of operation specific heats of water and n-propyl benzoate may be taken as .18 kJ/kgK and 1.87 kJ/kgK.

5.5 An acetone-water solution containing 60 % acetone by weight is separated by flash distillation at tmospheric pressure in two equilibrium stages. Half weight) is vaporized in the

ture is

ine the tem re firs e, a e fr l vaporisation in the second, together o n o u ach .

is n o th ne tw or products together? quilibrium -w a he ssu giv ow e x show the

a of the solution (byfirst stage, the remaining liquid solution being passed to the second stage, where its tempera

ised to 80 oC. raa) Determ

cperatu in the t stag nd th actiona

with theb) What

mpositio s of vap r and liq id in e stagethe perce data for

tage recacetone

very of ater at

e acetotmosp

in theric pre

o vapre is E en bel , wher and y

mass fractions of acetone in liquid and vapor respectively. [Ans.: a) 67.5 oC, ε2 = 0.28, b) 94 %]

t(oC) 100 88.2 80 71.9 67.6 65.2 63.4 62.0 60.8 59.7 58.2 56.4x 0 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 y 0 0.623 0.784 0.876 0.91 0.925 0.933 0.94 0.946 0.953 0.963 1.0

5.6 A binary liquid solution containing 36 mole percent MVC is to be flash-distilled at constant pressure. The flow rate of the feed is 55 k-mol/h and 47 percent of this will be vaporized. The relative

5 at the operating concentration range. Calculate: nd bottom products,

.7 Cumen is an important precursor for the production of phenol and acetone. It is produced by the

Propane 40.0 mole percent l mol mol mol

volatility of the MVC to the LVC is constant at 12.a) The compositions of top ab) The percentage recovery of the MVC. [ Ans. a) yD = 0.63, xw = 0.12 ] 5catalytic reaction between benzene and propylene. Composition of the mixture leaving the catalytic reactor is given below;

Propylene 2.0 mo e percent Benzene 27.0 percent Cumen 30.0 e percent Less volatile organics 1.0 e percent


In order to separate cumen from the others, first flash distillation is used to remove the propane and ropylene, later benzene is recovered and recycled.

Temperature,oC 37.8 65.6 93.3

pCalculate the loss of benzene and cumen if the above-given mixture is flash-distilled at 2.4 bar constant pressure to remove the 90 mole percent of the propane. Assume that Raoult’s and Dalton’s laws are applicable. The vapor pressures of the components as atm are given at three different temperatures in the table below:

Propan 12.8 23.3 39 Propylene 15.3 27.6 46 Benzene 0.217 0.62 1.45 Cumen 0.013 0.05 0.16

5.8 A 60 k wh 2 ycol (EG) and 68 mol percent water (W). This solution will

-mol/h liquid solution ich is at 25 oC, contains 3 mol percent ethylene glbe flash- distilled at 1.013 bar constant pressure.

For a fracti ori tioa) The flash te eratub) The compositions of top and bottom products, c) The perc ery c p 5.9 A 40 k iqui o ing ol nt ac d L) 55 mol percent ethanol (ET) will be subjected to differential (simple) distillation at 1.013 bar pressure. Operation will

e stopped when the mol fraction of AL in the still drops to 0.15.

sition of distillate, ) Find the percentage recovery of AL.

onal vamp

p za n of 0.50 calculate: re,

entage recov of EG. (EG is re overed in bottom roduct)

-mol l d solution c ntain 45 m perce etaldehy e (A and

ba) Find the temperature of the still at the end of the operation, b) Calculate the average compoc 5.10 a) Derive the following equation for simple (differential) distillation of a binary solution, whose vapor-liquid equilibrium may be given as y*= mx (m is a constant, x,y are mole fractions) , starting

1mFxF −

⎟⎟⎞

⎜⎜⎛

= whewith Rayleigh Equation.

1

re, F and W are the total moles of the feed

) A 80 k-mol binary solution containing 23 mole percent MVC will be subjected to simple distillation at 7 io l b pe e o ct t C e st ps to 0.03. Ca late o u il d e e o) If the same solution is subjected

lation at the same

wxW ⎠⎝(charge) and bottom product, xF and xw are the mole fraction of the MVC in the feed and bottom product respectively. b

60 mmHg. Distillat n wil e stop d wh n the m le fra ion of he MV in th batch- ill dro

lcu the average compositi n of topto flash dis

prod ct (disttillation under

late) an perc the same co

ntage r covery nditions with the same

f the MVC. cpercent of vaporization, what will be the purity of the top product and the percentage recovery of the MVC? d) Compare the results in b) and c) and comment on!

liquid equilibrium at 760 mmHg may be given as Within the concentration range involved vapor-y*=6.5x, where x and y are mole fractions. 5.11 60 k-mole acetone-water solution, containing 40 mole percent acetone will be distilled in a batch still at 760 mmHg. Distillation will be stopped when the mole fraction of acetone in the batch still drops to 0.05. Calculate: a) The amount of distillate to be produced, b) The mole fraction of acetone in the distillate (purity), c) The percentage recovery of acetone, d) If the same percent of vaporization is affected in a single stage by flash distilpressure, what will be the purity and percentage recovery of acetone?


5.12 It is desired to obtain an ortho-cresol product in 95 % purity by mole by simple distillation of a solution containing 65 mole percent ortho-cresol and 35 mole percent phenol. Show that the ortho-

relation at the operating pressure and over range of concentration is presented by the equation: y = 0.015 x + 0.05 where x and y are mole fractions of phenol in liquid

and vapor. 5.13 A binary liquid sol oC containing 4 perc enzene and 55 mole percent heptane is to be fed to a rectification colum i e at eric pressure. It is required to produce a top product g not mo nt h

) Calculate the minimum reflux ratio,

ol. The erature range is 185 kJ/k-mol oC.

g 60 mole percent benzene is ic pressure. The feed is

of the recirculated liquid is to be operate the column at a reflux ratio 1.3

the composition of vapor entering the condenser and of the reflux liquid entering the column,

[Ans.: a) xw = 0.046, b) y = 0.905, x = 0.85 c) y = 0.092, x = 0.072]

5.15 An ethyl acetate-acetic acid solution, containing 35 mole percent ethyl acetate will be rectified in

d will be introduced as liquid-vapor mixture in which mole fraction of liquid is

as operating reflux ratio.

the number of the ideal plates by using McCabe-Thiele Method. as kW.

.16 A plate rectification column is to be designed to separate a 12 500 kg/h ethyl acetate-acetic acid

ratio which is 1,50 times the minimum is to be selected for the operation. The column ed with a total condenser and a kettle-type reboiler.

solution remains constant at 28 500 kJ/k-mol. Liquid heat kJ/kg oC and 2.21 kJ/kg oC respectively.

ottom products as kg/h and k-mol/h and their compositions as

uation and compare them.

cresol recovery is trebled by using a partial condenser to condense 75 % of the vapor and continuously returning this back to the still (reflux liquid), rather than using conventional batch-still. Assume that the hold-up in the condenser is negligible, and that condensate is in equilibrium with the remaining vapor. The equilibriumre

ution at 15 5 mole ent bn which w ll operat atmosph

containin re than 5 mole perce eptane. ab) What will be the composition of the bottom product, if the top product flow rate is 40 % of the feed rate?

00 kJ/k-mLatent heat of vaporization of the solutions may be considered constant at 23 0e feed over the required tempaverage specific heat of th

5.14 A solution of benzene and methylisobutylketone (MIBK) containin

be separated by rectification in a sieve-plate column operating at atmospherto40 mol percent vapor 60 mole percent liquid and 97 % of its benzene content is to be recovered in the top product, which should contain 95 mole percent benzene. The column is to be equipped with a partial condenser and a recirculation reboiler in which 25 % vaporised. The feed rate is 400 k-mol/h and it is suggested to times its minimum value. Determine: a) the composition of the bottom product,

)bc) the composition of vapor entering the column and of the liquid leaving the bottom plate, d) the heat loads of the reboiler and condenser. The molar latent heat of vaporization is constant at 32 000 kJ/k-mol

1 o N+1 N

a plate column operating at 760 mmHg. The column will be equipped with a total condenser and a total reboiler. The fee0.65. 96 percent of the ethyl acetate in the feed will be recovered into the top product which will contain 95 mole percent ethyl acetate. A reflux ratio of 1.50 times the minimum reflux ratio will be selected For a feed flow rate of 220 k-mol/h : a) Calculate the flow rates of top and bottom products. b) Write the equations of q-line, and enriching and stripping section operating lines. c) Find d) Calculate the condenser and reboiler loads (duties) Molar latent heats of vaporization of both components are the same at 28 500 kJ/k-mol. 5solution, which is at 25 oC and contains 44 mass percent ethyl acetate. The column will operate at 760 mmHg top pressure. 91 percent of the ethyl acetate in the feed is to be recovered in 96.5 mass percent purity. A refluxwill be equippAverage latent heat of vaporization of the capacities of ethyl acetate and acetic acid at 60oC are 2.06a) Calculate the flow rates of top and bmass and mole percent. b) Find the minimum reflux ratio from xy-diagram and Underwood eq


c) Find the minimum number of the ideal plates needed from xy-diagram and Fenske equation and compare the two values. d) Find the number of the ideal plates needed by using Mc Cabe-Thiele method. e) Estimate the number of the ideal plates needed from Gilliland and Erbar-Maddox correlations and

emperature of each ideal plate and write these on the schematic column above. ) Calculate the quantities of ethyl acetate and acetic acid transferred between liquid and vapor phases

s as kW.

continuous rectification column is used to separate a binary solution. Enriching and stripping

y = 0.739 x + 0.226

nser, it condenses only the part of the vapor required .05 k-mol/h, W = 40.95 k-mol/h]

.18 An existing column providing 7 equilibrium plates and equipped with a kettle-type reboiler and a

, r production in the reboiler as k-mol/h,

at the same purity if reboiler capacity were

vaporation. But during the evaporation components,

aturated vapor coming from the evaporator is sent to a rectification column where essence is

r k-mol of feed to the column.

y of methyl antranilate and water vapor, and relative volatility

compare these with the value in section d). f) Find the compositions of liquid and vapor phases around each ideal plate and show them on a schematically drawn plate column. g) Find the thon each ideal plate as kg/h and make a table with them. i) Estimate the number of the real plates from O’Connell’s correlation. At column average temperature viscosities of ethyl acetate and acetic acid are 0.21 cP and 0.47 cP. j) Calculate condenser and reboiler load 5.17 Asection operating lines for this operation are given as; n+1 n ym+1 = 1.28 xm - 0.0143 where x and y are the mole fractions of the MVC in liquid and vapor phases and, n and m show any plate in the enriching and stripping sections respectively. Feed flow rate is 75 k-mol/h and it contains 42 mole percent MVC. Heat added in the reboiler (reboiler load) is 1 300 kJ/s. Constant molar over-flow is applicable and latent heat of vaporization of the solution is 32 000 kJ/k-mol. Calculate: a) The mole fraction of the MVC in the top and bottom products and their flow rates, b) The thermal condition of the feed, c) Condenser load. Condenser is a partial condeas reflux liquid. [Ans. : a) xD = 0.866, xw = 0.051, D = 34 5total condenser is being considered for the separation of 100 k-mol/ h saturated aqueous methanol solution containing 40 mole percent methanol. Operation will be carried out at atmospheric pressure and 95 percent of the methanol in the feed is to be recovered in 98 mole percent purity. The feed can be introduced at any point and the column will be operated at whatever the reflux ratio is required to produce both the purity and the recovery specified. Assuming constant molar over-flowa) Calculate the necessary rate of vapob) What would be the percentage recovery of methanol limited to 90 k-mol vapor/h. 5.19 Fruit juice concentrates are prepared commercially by eflavoring volatile components are lost with the vapor. In order to recover these flavoringswithdrawn at the top as liquid. The most important flavor component in the vapor coming from grape juice concentrating evaporator is methyl anthranilate, which is known to occur at 2*10-7 mole fraction in the vapor. 90 percent of this component is to be recovered in a plate rectification column as top product with 99 mole percent purity. Assuming constant molar over-flow; a) Calculate minimum steam generation in the reboiler as k-mol peb) If the steam generation in the reboiler is 40 % greater than the minimum steam generation, what is the number of the equilibrium plates needed for the specified recovery? Due to the low concentration of the other volatile flavor components, feed to the column may be treated as binary system consisting onlof methyl antranilate to water at high dilution is 3.5. [Ans.: a) 12F/G = ]


5.20 An ethanol- propanol solution is to be separated in a plate rectification column at atmospheric

l. 95 % of the ethanol

e reboiler, both of which have Murphree vapor phase

mum value, calculate:

t at 47 000 kJ/k-mol.

.21 It is required to design a rectification column to separate methanol from water at atmospheric

Stream Quality Flow rate, k-mol/h Methanol mol fraction

pressure. The feed is 40 mole percent vapor and contains 45 mole percent ethanoshould be recovered in the top product, which should be 92 % pure by mole. The column is provided with a partial condenser and a kettle-typefficiencies of 100 %. If the feed rate is 250 k-mol/h, the overall column efficiency is 60 % and the reflux ratio is set at 1.4 times its minia) The composition of the bottom product, b) The number of the real plates required in the column, assuming that feed plate is located optimally, c) The duty of the condenser in kW. The system may be assumed to have a constant relative volatility of 2.1 at the operating pressure. Latent heat of vaporization of the solutions may be assumed constan 5pressure. The following table gives the design requirements for the feeds and products;

Feed-1 Saturated vapor 400 0.50 Feed-2 Saturated liquid 200 0.30 Top product Saturated liquid 150 0.96 Bottom product Saturated liquid 0.04 Side product Saturated liquid 0.70

The column has a total condenser and a reboiler. Assuming constant molar over-flow; a) Calculate the minimum reflux ratio,

which the side

g, the mole fractions of methanol in the liquid samples taken from the exit of the 1st, 2nd and 3rd

n a batch column at 760 mmHg, to produce a top The operation will be carried out keeping the top product

fraction of SCl in the ter

corresponding to the 0.15 mole fraction of SCl2 in the batch-still. alculate:

oloC

b) For a reflux rate of 400 k-mol/h to the top plate, find the total number of the equilibrium plates needed, the number of the plates to which each feed should be introduced and that fromstream should be withdrawn and vapor flow rate to the column. 5.22 In the continuous rectification of methanol/n-propanol solution in a plate column operating 760 mmHplates are measured as 0.88, 0.70 and 0.45 respectively under the total reflux conditions. Calculate the Murphree liquid efficiencies of the three plates. The mole fraction of methanol in the vapor going to the condenser is 0.95. 5.23 A batch of 10 000 kg sulfur dichloride(SCl2)- carbon tetra chloride(CCl4) solution at 25 oC and containing 40 mass percent SCl2, will be rectified iproduct of 90 mole percent SCl . 2composition constant and increasing the reflux ratio steadily until the mole 2batch-still drops to 0.15. The reflux ratio at the last moment of the distillation will be 20 % greathan the minimum reflux ratioCa) The number of the equilibrium plates needed. b) The reflux ratio at the start of the distillation. c) The total heat energy requirement. d) The time of duration of one batch operation, for the conditions : heat transfer area in the batch-still is 5 m2, over-all heat transfer coefficients for the heating and vaporization periods are 300 W/m2K and 750 W/m2K, and the heating is affected by the condensation of saturated steam at 1 atm absolute pressure in the coils. Specific heat and latent heat of vaporization of the solution may be taken constant at 125 kJ/k-mand 30 000 kJ/k-mol over the operating temperature range. 5.24 A packed rectification column is to be designed to separate an equimolar solution of benzene and toluene to obtain products of 95 percent purity at 760 mmHg. The feed is to be introduced as saturated


liquid. For the conditions stated H = 0.5 m and H =1.0 m. For a reflux ratio 25 % greater than the

ed as follows: P.S. = 0.45 m, dh = 6 mm,

cur or not. If yes, what would you change in the assumed plate

ensities of liquid and gas and surface tension of the liquid are: 850 kg/m , 2.2 kg/m and 0.40 N/m

.26 Design the enriching section of the column given in Problem.5.16) with the values at the top plate. The thickness of the plates will be 5 mm sities and surface ten the ethyl acetate and acetic acid are 825 kg/m , .017 N/m and 0.022 N/m respe

G Lminimum and assuming constant molar over-flow calculate; a) NL, NG and NOG, b) The heights of packing required for each section of the column. 5.25 A Plate lay-out for the internal design of a sieve-plate gas absorption column for gas and liquid flow rates of 4.2 kg/sec and 6.1 kg/sec respectively, is assum PT = 3 dh , hw = 50 mm, Lw/Dc= 0.705. a) Calculate the diameter of the column for this lay-out, assuming a percent of flooding of 70. b) Find out whether weeping will oclay-out? c) Calculate the total pressure drop in the gas per plate.

3 3Drespectively. 5

. At the temperature3

of the top plate the densions of 984 kg/m3, 0

ctively.


Chapter-6

LIQUID-LIQUID EXTRACTION 6.1 Introduction: Liquid-liquid extraction or solvent extraction, which is a mass transfer operation, is used to separate liquid solutions. In this operation, the solute A, which is dissolved in solvent C, is transferred into a new solvent B, which is either insoluble or partially soluble in solvent C. Solvent C is named as raffinate solvent and solvent B as extract solvent or simply the solvent. The limited solubilities of solvents constitute the basis of this operation. Thus formed two liquid layers, each containing all the components to different extend, in the case of partial solubility, separate from each other due to the density difference. The layer, which contains extract solvent more than raffinate solvent is called as extract phase and the other layer, which contains the raffinate solvent more than extract solvent, is called as raffinate phase. Raffinate or extract phase may be denser depending upon the conditions. The best condition is attained when the extract and raffinate solvents are completely insoluble, but this is a rather rare case. In most of the cases, the two solvents are partially soluble within each other and hence all the components exist in each phase. If the contact time of the two phases is long and/or the intensity of mixing is high, the two phases reach in equilibrium. At equilibrium, the ratio of A to C in the extract phase is greater than that of in the raffinate phase and as a result of this the separation of solute A from raffinate solvent C occurs at certain degree. Repeating the contact with solvent B, separation can be further increased. As it is seen, the liquid-liquid extraction is not a final separation operation since the solute A must be separated from the new solvent B, which is mostly done by rectification. It may be then asked: “why the solution of A+C is not separated by direct application of rectification”. The answer to this question is: “because of either physical impossibilities or economic considerations”. Physical impossibilities occur, when the solution of A+C has azeotropic point. In this case, by changing the solvent of the solution, azeotropism is removed. Another example for physical impossibilities is the solution of heat sensitive components. For example, if solute A is a heat sensitive material and the boiling point of solvent C is high and the solution is dilute in A, the boiling point of the solution will be inevitably high, which causes the decomposition of solute A at direct rectification of the solution. Instead of rectifying the solution of A+C, rectifying the solution of A+B, after replacing solvent C by new solvent B, which has relatively low boiling point, prevents the decomposition of solute A during rectification. At some cases, the solvent extraction followed by rectification may be more economic than the direct rectification of the solution. This occurs, when the solution A+C is dilute in component A, component C is the MVC and the latent heat of vaporization of which is relatively high. In this case solvent C is first replaced with a new solvent B by solvent extraction, which is more volatile than A and has a very low latent heat of vaporization and then solute A is separated from solvent B by rectification. Although another separation operation is added, the cost of recovery of unit amount of solute A is lower than that of direct rectification due to the reduced energy requirement. Liquid-liquid equilibria are required to understand and to design any solvent extraction operation as the mass transfer in this operation is between two insoluble liquid phases.


6.2 Liquid-Liquid Equilibria: Certain quantities of liquids A, B and C are mixed thoroughly until the equilibrium is attained and then the samples are taken from the two layers, which form after stopping the mixing. By analysing the samples; x, xB and xC ,which are the mass fractions of components A, B and C respectively in the raffinate phase and y, yB and yC which are the mass fractions of the same components in the extract phase are determined. By repeating the experiments at different proportions of components A, B and C in the same way, a table is formed as shown below. (Notice that the mass fractions having no subscript are the mass fractions of solute A). As the solubilities change with temperature, the temperature is kept constant R a f f i n a t e p h a s e E x t r a c t p h a s e x xB xC y yB yC - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - during the experiments and this temperature is reported with the experimental results. The graphical representation of the results can be done on xy-type diagrams, when the two solvents are completely insoluble within each other (in this case all the xB and yC

values are zero). This type of representation is obtained by carrying

values against Ycx/xX =′ ' = y/yB values on a millimetric paper as shown in Fig.6.1. Notice that X′ and Y′ are weight ratios. If all the three components exist in both phases, xy-type diagrams cannot be used for the representation of equilibrium. Ternary diagrams may be used. The right angle triangular diagrams are preferred to the equilateral triangular diagrams, as for which special millimetric papers are required and enlargement of part of it is rather difficult. As shown in Fig.6.2, the apexes of the triangle represent pure components, the sides the binary liquid mixtures of the components represented by

the two apexes and the points in the triangle, the ternary liquid mixtures. As xC = 1-(x+xB) and yC = 1-(y+yB), there is no need for the representation of the mass fractions of component C on the diagram. On these diagrams, graphical additions and subtractions can be done. For example, if a ternary liquid solution represented by point L is added to another ternary liquid solution given by point G, the resultant solution, which is represented by point M is always on the straight line joining L and G. Point M can be located by applying inverse lever-arm rule to the LG line, which is written as

Weight ratio in raffinate phase, X'=x /1-x

t = cons.

Wei

ght r

atio

in e

xtra

ct p

hase

, Y' =y

/1-y

0.0 0.0

Fig.6.1 Liquid-liquid equilibrium when B and C are insoluble within each other


A

∆

CB

M G1

G

L1 L

Mass fractions of B in raffinate aextract phases, x

nd B,yB

1.0

1.0 0.0

0.0

0.5

0.5

1.0

Mas

s fra

ctio

n of

A in

ext

ract

pha

se,y

0.5

1.0 0.5 Mass fraction of A in raffinate phase, x

0.00.0

Mas

s fra

ctio

ns o

f A in

raff

inat

e

and

extra

ct p

hase

s, x,

y

Fig.6.2 (a) Right angle triangular diagram, (b) distribution diagram

MG*GML*L = , where L and G shows the quantities of liquids as kg. Similarly, if from a ternary liquid solution represented by point L1, a ternary liquid solution represented by point G1 is subtracted, the remaining liquid solution, which is represented by point ∆ is always on the extension and beyond the point L1 of the straight line joining L1 and G1. Inverse lever-arm rule written for line G1∆, point ∆ being the support point, locates the ∆ point. The solubility diagram of the system is obtained by carrying x values against xB and y values against yB from the table. Next to this diagram, the distribution diagram of the system, which is plotted by carrying the values of x against the values of y, and required to find the extract and raffinate phases at equilibrium is also plotted. By plotting the experimentally found values of various ternary systems, two types of solubility and distribution diagrams shown in Fig.6.3 and 6.4 were obtained. In the ternary system shown in Fig.6.3, only one pair (pair of B-C) has partial solubility. The other two pairs (pairs of A-B and A-C) dissolve within each other at any proportions. Any point under the solubility curve such as point M shows heterogeneous ternary mixtures and any point outside the curve shows homogeneous ternary solutions. Accordingly, point J represents a binary heterogeneous mixture, which separates in two binary homogeneous solutions represented by points R and T, whose compositions can be directly read from the solubility diagram and their quantities are calculated by applying inverse lever-arm rule to the line RJT. Point R gives the maximum solubility of B in C, and point T the maximum solubility of C in B. The heterogeneous ternary mixture represented by point M gives two homogeneous ternary solutions represented by points G and L upon settling. Since they will be in equilibrium, they are located by drawing the tie line passing through point M by trial and error with the help of distribution diagram. Since the solution represented by point L is richer in the raffinate solvent, it is called raffinate phase and the other phase, which is richer in the extract solvent, is called extract phase. While the compositions of extract and raffinate phases obtained are directly read from the solubility diagram, their amounts are found by


applying inverse lever-arm rule to the line LMG. The part of the solubility curve bounded by RLP represents raffinate phases; the remaining part represents extract phases. Point P, at which the compositions of both phases are the same, is known as plait point. The tie line reduces to a point at plait point. As it is seen from Fig.6.3b, y/x ratio which is given by K and known as distribution coefficient is greater than 1. This means that solute A prefers solvent B rather than solvent C. In the ternary system shown in Fig.6.4, two pairs (pairs of A-B and B-C) are partially

1.0

y

BC

P

(a)

Fig.6.3 (a) Solubility, (b) Distribution diagram for one pair partially soluble system

L

xB,yB

P M

R

G

J T

y

x

G,L

xP=yP

x

x

y

yP

x=y

yB

xB

0.00.0 0.0

0.0

1.0 1.0

Tie line

(b)

xp

Extract phase curveRaffinate phase curve

Extract phase curve

x=y

1.0

1.0

y

BC

U

L

xB,yB

V M

R

G

J T

y

x

G,L

x

x

y

yB xB

0.00.0 0.0

0.0

1.0

Tie line

Raffinate phase curve

Distribution curve

A 1.0

Distribution curve

A 1.0

x,y

x

x,y

x

(b)(a)

Fig.6.4 (a) Solubility, (b) Distribution diagram of two pairs partially soluble system


soluble and the other pair (pair of A-C) is soluble at any proportions. Points U and V show the maximum solubilities of A and B within each other. Any point between two curves represents heterogeneous mixtures and any point outside the curves gives homogeneous solutions. The tie line passing through point M locates the equilibrium extract and raffinate phases, which will be obtained upon the settling of ternary heterogeneous mixture represented by point M. RLU curve represents raffinate phases and the TGV curve the extract phases. In this case, distribution coefficient K is smaller than 1, which means preference of solute A is at raffinate solvent side. The liquid-liquid equilibria are also represented on so-called solvent-free coordinated diagrams, which involves first expressing of compositions of extract and raffinate phases on solvent-free bases and then plotting them on xy-type diagram. This type of diagrams is especially useful for two pairs partially soluble systems, as their extract phase curves, which are rather short in right angle triangular diagrams, can be extended to any value on these coordinates. For this, first the compositions of components A, B and C are expressed by excluding solvent B in the denominators as x'= x/(x+xC), = xBx′ B/(x+xC) and Cx′ = xC/(x+xC) in the raffinate phase and as y'= y/(y+yC), = yBy′ B/(y+yC) and Cy′ = yC/(y+yC) in the extract phase and then by carrying values against xBx′ ' values and By′ values against y' values, raffinate and extract phase curves are plotted as shown in Fig.6.5. Below this diagram, by plotting x' values against y' values another diagram is also drawn, which is used to draw the tie

lines on the diagram above. Any point between extract and raffinate curves, such as point M represents heterogeneous ternary mixture, which upon settling gives equilibrium extract and raffinate phases, which are located on the diagram by drawing the tie line passing through point M by trial and error with the help of diagram below. While their compositions are directly read from the diagram, their amounts are calculated by applying inverse lever-arm rule to the line LMG.

No part of this CD-book may be multiplied for commercial pu

0

G

M

L

By′

Raffinate phase curve

x′

yyB ′−′

B

B

xy′′

Bx′

L,G

y′

xx B ′−′

y′

x′

y =x

0

0

x′1.0

1.0

y′

1.0

y,x ′′

Extract phase curve

Tie line

R

T

U

V

t =cons.

Fig.6.5 Liquid-liquid equilibrium on solvent-free coordinates

6.3 Selection of Solvent: In many cases, there is more than one solvent that can be used for a specific extraction operation. The following properties of the solvents are looked at to select the best one among the potential solvents: 1o) Selectivity of solvent (β): It is defined as the ratio of y/yC to x/xC at equilibrium, and is the separation power of solvent B of solute A from raffinate solvent C. Whatever is the α in rectification is

rposes. E.Alpay & M.Demircioğlu 258

the β in extraction. If β =1, this solvent has no selectivity for solute A and hence it cannot be used for the extraction. Higher the value of β easier is the separation of solute A from solvent C.

C

C

x/xy/y

β= (6-1)

2o) Distribution coefficient (K): It is defined as K= y/x and it should be as great as possible to accomplish the given extraction with small amount of solvent usage, although it is not necessary that it should be greater than 1. 3o) Insolubility of solvent: the selected solvent should have no or low solubility in raffinate solvent to minimize the solvent losses, otherwise another rectification column is to be used to recover the solvent dissolved in raffinate solvent. 4o) Separation of solvent from solute: as it is stated before, the solution after extraction is separated by rectification. Hence, selected solvent should not form azeotropic solution with solute A and when it is possible, the relative volatility of this solution should be high. 5o) Density of solvent: for the separation of extract and raffinate phases, density difference should exist between the phases. Hence, the density of selected solvent should be different than the density of raffinate solvent. 6o) Interfacial tension: in the extraction operation, one of the phases is dispersed first in the other phase as small droplets by intensive mixing and then by allowing the settling, the coalescence occurs. Interfacial tension between the liquids plays an important role in dispersion and coalescence. Low interfacial tension means easy dispersion but difficult coalescence. As short coalescence time is preferred, selected solvent should have high interfacial tension with the raffinate solvent. 7o) Chemical stability: as the solvent is recovered by rectification and re-used many times and during which it is heated and cooled repeatedly, it should be chemically stable so that no decomposition should occur. 8o) Other properties: whenever possible selected solvent should have low viscosity and vapor pressure, should be non-toxic, should not catch fire easily and should have low unit cost. 6.4 Extraction Operations: Extraction operations are carried out either as stage-wise contact type of operation or continuous contact type of operation in practice. 6.4.1 Stage-Wise Operations: In the stage-wise contact type of operations, the two phases come in contact for a certain time, during which mass transfer takes place

between the phases and then the two phases separate out. The mixer-settler unit shown in Fig.6.6 is a typical stage-wise contact type of equipment used in extraction operations. The equipment, which consists of two main sections, one being mixing and the other settling section, can be constructed as single piece of equipment as shown in the Figure or, as two separate units. In the mixing section, the two phases are mixed thoroughly with the help of a stirrer to disperse one of the phases as small droplets in the other

Feed F, xF

Settler

Mixer

Raffinate phase,L,x

Extract phase,G,y

Solvent,S,ys

Fig.6.6 Mixer-settler unit


Main interface

Exit of light phase

Main interface

Up-comer

Perforated plate

Perforated plate

Light phase

Light phase

Down-comer

Coalesed light phase

Dispersed light phase

Dispersed heavy phase

Coalesed heavy phase

Exit of heavy phase

Heavy phase

Exit of heavy phase

Heavy phase

Exit of light phase

(a) (b)

Fig.6.7 Perforated plate extraction column : a) light phase dispersed, (b) heavy phase dispersed

phase during which main part of mass transfer occurs between the phases. If the mixing time is enough, the two phases reach in equilibrium and the unit is then known as equilibrium unit or equilibrium stage. The heterogeneous mixture then flows into settling section of the unit, where coalescence of the drops occurs and the two phases separate out and leave the unit through different channels. Operation is carried out in continuous way, at which the feed and solvent are continuously pumped to the mixing section and the extract and raffinate phases are withdrawn from the settling section continuously. Mixer-settler units are the only extraction equipment for single stage or cross-current multi-stage operations. In the multi-stage counter-current operations perforated-plate columns can also be used next to the mixer-settler units. These columns are similar to the sieve-plate columns used in gas absorption and rectification operations but they don’t have weirs. The light or heavy phase may be dispersed depending upon the conditions. In Fig.6.7, two perforated plate extraction columns are shown in which light and heavy phases are dispersed. 6.4.1.1 Single Stage Extraction: The amounts of raffinate and extract products and their compositions can be calculated by simultaneous solution of the equilibrium


relationship and the total and components balance equations which can be written along the equilibrium mixer-settler unit shown in Fig.6.8 as; Total material balance: F+S = M1 = L1+G1 (6-2) Solute A balance: FxF + Sys = M1xM1 = L1x1 +G1y1 (6-3) Solvent B: FxFB+ SysB = M1xM1B = L1x1B

+ G1y1B (6-4) where F and S are the amounts of feed and solvent as kg, if the operation is carried out

batch-wise, and the flow rates of the feed and solvent as kg/s, if the operation is conducted continuously. When the solvent does not contain solute A then ys=0. Points F and S can easily be located on the diagram as their compositions are known. It follows from Fig.6.9 that the feed is a binary solution of A and C and the solvent contains small amount of solute A. Then either by applying inverse lever-arm rule to the line FS or by solving the left hand side of equation (6-3) point M1 is

located. By drawing the tie line passing through point M1 by trial and error, points L1 and G1 and hence raffinate and extract products are located. While the compositions of

these products are directly read from the diagram, their amounts are calculated either by applying inverse-lever-arm-rule to the line L1M1G1 or by simultaneous solution of the right hand sides of equations (6-2) and (6-3). As it is seen, in the calculations the stage is assumed as equilibrium stage.

Raffinate product, L1,x1

1

Solvent, S,ys

Extract product, G1,y1

Feed, F,xF

Fig.6.8 An Equilibrium Stage

xM1

S

1.0

1.0

x=y

BC

P

(a)

L1

xB,yB

M1

F G1

D G1,L1

x1

x1

y1

yS 0.0

0.0 0.0 0.01.0

E

Gm

Lm

y1

x1

xM1B x1B y1B ySB

xF

(b)

P

A 1.0

y

x,y

x

Fig.6.9 Solution of single stage extraction

For extraction operation, point M1 must lie always in the heterogeneous region. By taking this fact into consideration, minimum and maximum values of solvent for a


given extraction can be calculated as follows: If the amount of solvent is reduced, point M1 shifts toward point F. Hence point M1, coming to point D corresponds to the minimum solvent quantity for the given extraction at which infinitely small amount of extract phase shown by Gm is obtained. On the contrary, point M1 approaches to point S with the increase in solvent amount. Hence point M1, coming to point E corresponds to maximum solvent quantity for the given extraction at which infinitely small amount of raffinate phase shown by point Lm is produced. The percentage recovery of solute A into extract phase is given by;

.100Fx

xLFx.R.PF

11F −= (6-5)

The percentage recovery of solute A into extract phase, in many cases, is rather small in a single stage extraction. To increase this, more than one stage is connected in series to form multi-stage cascades. As the connections of stages are done in two different ways, two different types of operation are possible. 6.4.1.2 Cross-Current Multi-Stage Extraction: In this type of operation, the solvent is divided in parts and each part is mixed in a stage with the raffinate phase coming from the previous stage to recover the solute, which is not recovered in the previous stage. The flow diagrams in real and schematic showing are given in Fig.6.10 for three-stage operation. As it is seen, the stages are numbered from left to right and the phases and their compositions leaving any stage are shown by writing the number of the stage as subscript. The analysis of cross-current multi-stage extraction is easily done by repeating the equations written for single stage operation, for each stage. Hence, total material and solute A balances for any stage (stage n) in the cascade;

Raffinate product

Extract product

L3,x3

G3

y3

1 1 2 2

(a)

3 3

F,xF

F,xF

G1

y1

G1

y1 G2

y2

G2

y2 G3

y3

G3

y3

L1,x1

G, y Extract product

G, y

L2,x2

L1,x1

(b)

Feed

Raffinate product

L3,x3

L2,x2

S,ys

S,ys S1

S1 S2

S2

S3

S3

Solvent

SolventFeed

Fig.6.10 Flow diagram of cross-current multi-stage extraction

Total material balance: Ln-1 + Sn = Mn = Ln + Gn (6-6) Solute A balance: Ln-1xn-1 + Snys = MnxMn = Lnxn + Gnyn (6-7) By substituting Lo = F and xo = xF, these equations for the first stage (n=1) becomes,

F + S1 = M1 = L1 + G1FxF + S1ys = M1xM1 = L1x1 +G1y1

Points F and S1 can be located on the diagram as their compositions are known. Then, as explained above, points M1, L1 and G1 are located and the amounts of L1 and G1 are calculated. By taking n=2, equations (6-6) and (6-7) can be written for stage two as;


L1 + S2 = M2 = L2 + G2L1x1 + S2ys = M2xM2 = L2x2 + G2y2

From which it is seen that points L1, M2 and S2 should be on the same straight line. Hence, after joining point L1 with point S2, point M2 is located on this line as explained above. From the right hand sides of the equations above, it is understood that points L2, M2 and G2 should be on the same straight line and in addition, points L2 and G2 must be on the solubility curve. Hence, by drawing the tie line passing through point M2, points L2 and G2 are located and the amounts of L2 and G2 are calculated. The solution is continued in this way until reaching the last stage as shown in Fig.6.11.

x1

x=y

1.0

1.0

y

BC

P

(a)

L3

xB,yB

P

G2

S

y1

yS

0.00.0 0.0

0.01.0

G1

x1x3 x2

G3

L2

L1

M2

M3

M1

2

3

1 F

y1

x1

x2 x2

y2 y2

xF

y3

x3 x3

y3

(b)

x,y

A 1.0

x

Fig.6.11 Solution of three-stage cross-current extraction on right angle triangular diagram

The percentage recovery of solute A is calculated from,

P.R. = .100Fx

xLFx

F

NNF − (6-8)

If the solution is done on solvent-free coordinated diagram; for any stage n; Total material balance: nGLMSL nnn1n ′+′=′=′+′ − (6-9) Solute A balance: nyGxLxMySxL nnnMnnsn1n1n ′′+′′=′′=′′+′′ −− (6-10) Solvent B balance: nBnnBnMnBnsBn1)B(n1n yGxLxMySxL ′′+′′=′′=′′+′′ −− (6-11) where L', G', S', M' and F' are the mass flow rates of the streams on solvent-free basis. These equations for the first stage (n=1) by taking Foo xxandFL ′=′′=′ can be written as;

1111 GLMSF ′+′=′=′+′ 1111M11s1F yGxLxMySxF ′′+′′=′′=′′+′′

1B11B1M1B1sB1FB yGxLxMySxF ′′+′′=′′=′′+′′


Points are first located on the diagram with the help of their known compositions as shown in Fig.6.12. Then point M

SandF

1 is located on the line joining these two points in the known way. By drawing the tie line passing through point M1, points L1 and G1 are located. The compositions of these phases are read from the diagram and the amounts of are calculated in the known way. Equations (6-9), (6-10) and (6-11) are then written for the second and third stages and the amounts of phases leaving the stages and their compositions are obtained in the same way as explained above. The percentage recovery of solute A is calculated from;

11 GandL ′′

P.R.= 100.xF

xLxF

F

NNF

′′′′−′′

(6-12)

If the raffinate and extract solvents are completely insoluble, the solution is much simplified, as the flow rates of raffinat and extract solvents, C and B, do not change from stage to stage. A component A balance around stage n is written as;

nYnBnXCsYnB1nXC ′+′=′+−′ (6-13)

From which,

n1n

ns

n XXYY

BC

′−′′−′

=−−

(6-14)

is obtained. This equation represents a straight line on X'Y'-diagram passing through points ( )Y;X s1n ′′ − and ( with a slope of (-C/B

)nn ′′

solution is done as follows: first, equilibrium relat

Y;Xn), which is known

as operating line for stage n. Remember that the equilibrium relationship in this case is also given on X'Y'-diagram. Hence, ionship is drawn on X'Y'-diagram,

then equation (6-14) is written for the first stage by taking Fo XX ′=′ as;

F

G1

M1

L1

x'1

B

B

xy′′

y'1x'

F

x',y'

y =x

0

0

x'

1.0

1.0

y'

1.0

t =cons.

x'B-x'

y'1B

y'S

x3' x2

' x'1

y'3

y'2

y'

G2

y'2

M2

M3

y'3 x2

'

2 3

x'3

G3

L3

x'3B

L2

F

S

y'B-y'

1

y'SB

Fig.6.12 Solution of cross-current three-stage extraction on solvent-free coordinated diagram

1F

1s

1 XXYY

BC

′−′′−′

=−

with the known values, point

)Y;X( sF ′′ is located on the diagram and the line with slope of (–C/B1) passing through this point is drawn. The coordinates of intersection


point of this line with equilibrium curve give 11 YandX ′′ . Then by writing equation (6-14) for the second and subsequent equilibrium stages, the solution is continued in the same way. In Fig.6.13, solution for a cascade containing three equilibrium stages is shown. The percentage recovery of solute A is computed from the equation below;

P.R. = 100.X

XX.100XC

XX

F

NF

F

NF

′CC′

′ ′−′=

−′ (6-15)

Y'

6.4.1.3 Counter-Current Multi-Stage Extraction: The stages may also be connected in another way such that the two phases flow in counter directions as shown in Fig.6.14. The counter-current operation has advantages over the cross-current operation. For a fixed percentage recovery; number of the equilibrium stages required is smaller at the same solvent rate or at a constant number of equilibrium stages

Total material and solute A balances along the cascade are written as;

required solvent rate is less. n shows any stage, N shows the last stage of the cascade.

(6-16) FxF

Y'S

Y'3

Y'2

Y'1

X'3 X'

2 X'1 X'

FX'

1

0.0 0.0

2

3

Equilibrium curve

Fig.6.13 Determination of equilibrium stages at cross-current multi-stage extraction when the solvents are insoluble

slope= -C/B1

slope = -C/B2

slope = -C/B3

Solvent

Raffinate product

Gn+1, yn+1

LN, xN F, xF

Feed LN-1, xN-1 Ln-1, x n-1 L1, x1 L2, x2 Ln, x n

GN, yN G2, y2

1 2 n N

Gn, yn G3, y3 S, yS G1, y1

Extract product

∆ (+)

Fig.6.14 Flow diagram of counter-current multi-stage extraction

Total material balance: F + S = M = LN + G1 Solute A balance: +Sys= MxM = LNxN+ G1y1 (6-17)


These equations indicate that point M lies on the FS and LNG1 lines, hence at the intersection point of these two lines. Equation (6-16) can also be written as; F - G1 = LN - S = ∆s (6-18)

here, ∆s is the total net flow stant along the cascade, hence

F - G1 = Ln - Gn+1 = LN - S = ∆s (6-19)

ith the help of equations given above the number of the equilibrium stages required

w which remains conequation (6-18) can be generalized to; If total net flow is from left to right as shown in Fig.6.14, F > G1 or LN > S, the otherwise it is from right to left. It is seen from equation (6-19) that point ∆s will be the intersection point of the lines FG1 and LNS.

A

Fig.6.15 Solution of four-stage counter-current extraction on right angle triangular diagram

y

xF

x

x=y

BC

P

xB,yB

S

x2

x,y

0.0 0.0 0.0

0.0

1.0

1.0 1.0

L2

G3 L1

F

y1

x3

y3

x4 x1

x4

x3

G4

LN=L4

3

∆S

M

G1

G2

L3

2 1

4

x1

y2

y4

x2

Wfor a given extraction can be calculated on the equilibrium diagram of the system as follows: Points F, S and LN are first located on the solubility diagram in Fig.6.15 as their compositions are known . Notice that point LN should be on the solubility curve. By joining points F and S, point M is located on this line in the known way. Then point LN is joined with point M and is extended until cutting the solubility curve, which gives point G1. Point G1 is joined with point F and extended; point S is joined with point LN and extended. The intersection point is ∆s. Now drawing of equilibrium stages can be started. By drawing the tie line passing through point G1, point L1 is located; this tie line shows the first equilibrium stage in the cascade. According to equation (6-19) points ∆s, L1 and G2 must be on the same straight line; in addition point G2 must lie on the solubility curve, hence by joining points ∆s and L1 and extending the line, point G2 is found. The raffinate phase L2, which is in equilibrium with extract phase represented by point G2 can be found by drawing the tie line passing through point G2. Then returning back to the point ∆s and joining it with point L2 and extending the line, point G3 is located. The drawing is continued in this way until reaching the point LN; once drawing the tie line once drawing the operating line


(the lines drawn from point ∆s are known as operating lines). The number of the tie lines drawn gives the required number of the equilibrium stages for the given extraction. After computing the amount of extract product G1 by applying inverse lever-arm rule to the line LNMG1, the percentage recovery of solute A is calculated from;

FFx xP.R. .100N NL

FFx−

= (6-20)

For the solution of the problem on the solvent-free coordinated diagram; total

F' + S = M =

material balance on solvent-free basis and the solute A balance along the cascade can be written as; ' '

N1 LG ′+′ (6-21)

F yGxF NNMs xLxMyS 11 +′′ ′′=′′=′′+′′quation (6-21) can also be written as,

∆SLG

(6-22) E n1 LGF −′=′−′ sN1n ′=′−′=′ + (6-23)

his means that total net flow on solvent-free basis re

= F and if the solvent is pure, S = 0,

T mains constant along the cascade. If the fee ' 'd does not contain any solvent then FS'

sy′ = 0, F' = M', MXXF ′=′ . The solution on solvent-free

M

x∆SB'

x'3

y'S

x'N

0

G1 B

B

xy′′

x',y'

y =x

0

0

x'

1.0

1.0

y'

1.0

G2

y'2

y'3

x'1 x2

' x3'

G3

∆S

Sy'SB

y'1B

x'B-x'

y'B-y'

L3 LN

Fx'

F

1 2

3

y'1y'

2

L1 L2

y'3

coordinated diagram of the system is done as follows: first, points F, S and LN are located on the diagram with the help of known compositions (in Fig.6.16). Points F and S are joined and point M is located on this line in the known way. By joining points LN and M and extending this line, point G1 is located. According to equation (6-23), point ∆s is obtained by drawing FG1 and SLN lines. Then, drawing of equilibrium stages is started. By drawing the tie line passing through point G1, point L1 is obtained. Equation (6-23) indicates that points ∆s, L1 and G2 should be on the same straight line; hence by joining ∆s with L1 and extending the line, point G2 is found. The tie line passing through point G2 gives point L2. The solution is continued by drawing once an operating line (the lines drawn from point ∆s are known as operating lines) once a tie line until Fig.6.16 Solution of counter-current extraction on

solvent-free coordinated diagram


reaching the point LN. The number of the tie lines drawn gives the number of the equilibrium stages required for the given extraction. The percentage recovery of solute A is calculated from;

F

F

F xP.R. .100F x

N NL x′ ′ ′ ′−=

′ ′ (6-24)

If raffinate and extract solvents are completely insoluble, the solution is simplified as the amounts of raffinate and extract solvent do not change from stage to stage. C and B showing the mass flow rates of the solvents, the solute A balance between any stage n and the last stage, N can be written as; sn1nN YBXCYBXC ′+′=′+′ + (6.25) If this equation is solved for , 1nY +′

BXC

YXBC

Y Nsn1n

′−′+′=′+ (6-26)

is obtained. This equation represents a straight line on X'Y'-diagram passing through points ( with a slope of (C/B). The number of the equilibrium stages is then found as follows: first, equilibrium relationship of the system is plotted on a X

)Y;X(and)Y;X 1FsN′′′′

'Y'-diagram, then the operating line given by equation (6-26) is drawn on the same diagram as the values of C, B, Ns XY ′′ , and FX′ are all known at the start of the design. The number of the right angle triangles located between equilibrium curve and operating line, starting at one end and stopping at the other end of the cascade, gives the number of the equilibrium stages required for the given extraction as shown in Fig.6.17 (McCabe-Thiele method).

The percentage recovery of solute A into the extract phase is given by;

F

F

XP.R. .100X

NX′ ′−=

′ (6-27)

Minimum solvent rate: In counter-current multi-stage extraction if the solvent rate S is reduced, point M shifts toward F and point ∆s comes closer to the diagram as seen from Fig.6.15. The closer ∆s means more equilibrium stages for the same extraction. Hence, for the same extraction, less number of equilibrium stages is required, if the solvent rate is increased; and more number of equilibrium stages is needed, if the solvent rate is decreased. If one of the operating lines coincides with a tie line during the calculation, the required number of the

equilibrium stages then becomes infinite. The solvent rate corresponding to this condition is known as minimum solvent rate. Of course, the solvent rate that is to be selected for the operation is to be greater than the minimum solvent rate. Hence, the minimum solvent rate for a given extraction is found as follows: arbitrary tie lines between points F and LN are drawn and they are extended until cutting the extension of

Operating line slope= C/B

Equilibrium curve 1

2

3

4

X'X'

N

Y'

Y'

Y's

X'F

Fig.6.17 Determination of number of the equilibrium stages in counter-current extraction, when solvents are insoluble


SLN line. The tie line, which when extended cuts the SLN line at the farthest point from the diagram, gives ∆sm, which is the ∆s point at minimum solvent rate. Usually this is the tie line whose extension, passes through point F as shown in Fig.6.18. Then point ∆sm is joined with point F and extended until cutting the solubility curve, which is

applying inverse lever-arm rule to line FM

shown by point (G1)m. Intersection point of lines (G1)mLN and FS locates point Mm. By

.4.1.4 Counter-Current Multi-Stage Extraction under Reflux: It is obvious that,

mS, minimum solvent rate Sm is calculated. Operating solvent rate S is always selected as S=β (Sm), β being always greater than 1. Its exact value is obtained by making an economic analysis around the combined system of extraction and rectification units. Notice that point ∆s, which corresponds to operating solvent rate, is always behind the point ∆sm.

yA

6the extract product leaving a multi-stage cascade operating counter-currently may be enriched up to a level that it is in equilibrium with the feed. No more enrichment is possible in counter-current contact. For further enrichment of extract product, reflux must be used at the extract product end. There is no need for reflux at the raffinate product end. Flow diagram of a multi-stage cascade using reflux is shown in Fig.6.19.

Fig.6.18 Determination of minimum solvent rate

x

x=y

BC

P

xB,yB

S

x,y

0.00.0 0.0

0.0

1.0

1.0 1.0

LN

F

( (G1)m y1)m

Mm

∆S ∆Sm

2solvent separat

G,y Lo xo

G1,y1

Se,ye

L1,x1

G ,y3 3 G2,y2 Gn,yn

L2,x2 Ln-1,xn-1

n f m N 1

Gn+1,yn+1 Gf,yf Gf+1,yf+1 Gm,ym

Ln,xn Lf-1,xf-1 Lm,xm Lf,xf

E n r i c h i n g s e c t i o n

Gm+1,ym+1

Lm-1,xm-1

GN,yN

LN-1,xN-1 LN,xN

S,yS

D, xD F, xF

S t r i p p i n g s e c t i o n

Fig.6.19 Flow diagram of multi-stage extraction under reflux

Raf

FeedExtract product

finatereflux

solvent

product


As it is seen, the feed is introduced at somewhere between the first and last stages of the cascade. In the enriching section, the solute A content of the extract phase is increased by contacting this phase with a raffinate phase which is rich in solute A. The raffinate phase, rich in solute A, is obtained by refluxing the part of the extract product, which is produced by separating the solvent of the extract phase leaving the first stage of cascade in a solvent separator, which is usually a distillation column. In the stripping section, solute A is stripped off the raffinate phase by the extract phase flowing in counter direction. Calculation of number of the equilibrium stages needed for a given extraction is done on solvent-free coordinated diagram. In the enriching section, total material balance between any stage n and the first stage is written as; DLG n1n

′+′=′+ (6-28)

This equation can also be written as, en1n ∆DLG ′=′=′−′

+ (6-29) It follows from this equation that in the enriching section of the cascade there is a constant total net flow, which is from right to left and equals the amount of extract product on solvent-free basis. Similarly solute A balance in this section yields to: Dnn1n1n xDxLyG ′′+′′=′′

++ (6-30) which can also be written as; ∆eeDnn1n1n y∆xDxLyG ′′=′′=′′−′′

++ (6-31) It follows from this equation that in the enriching section of the cascade there is a constant solute A flow, which is from right to left and equals the amount of solute A removed by the extract product. From the solvent B balance in this section, ∆eBeDBnBn1)B(n1n y∆xDxLyG ′′=′′=′′−′′

++ (6-32) can also be written. In the stripping section, total material balance on solvent-free basis between any stage m and the last stage N gives; N1mm LGSL ′+′=′+′ + (6-33) which can be written also as, sN1mm ∆SLGL ′=′−′=′−′ + (6-34) This equation indicates that there is a total net flow in the stripping section of the cascade which is given by . s∆′Solute A balance in this section becomes; NN1m1msmm xLyGySxL ′′+′′=′′+′′ ++ (6-35) which can also be written as, ∆sssNN1m1mmm x∆ySxLyGxL ′′=′′−′′=′′−′′ ++ (6-36) This means that in the stripping section there is a net flow of solute A whose value equals . ∆ssx∆ ′′From the solvent B balance in this section, ∆sBssBNBN1)B(m1mmBm x∆ySxLyGxL ′′=′′−′′=′′−′′ ++ (6-37) can also be written. Total material balance for the whole cascade;


F' + S' = D' + NL′ (6-38) ith the help of equations (6-29) and (6-34) this eW quation can also be written as;

s∆∆F e ′+′=′ (6-39) Determination of number of the equilibrium stages is done as follows: point F and S are first located, as their compositions are known. Then, with the help of

( )D o ∆ex x y′ ′ ′= =

and Nx′ , points, D(Lo) and LN are located. Point LN must lie on the raffinate phase curve but not the point Lo. Points S and LN are joined and extended

vertical from beyond LN. A( )∆eoD yxx ′=′=′ fixes the

point G1, from whic 1Byh ′ is read. The values of 1By′ , oBx′ and selected RD are substituted i equation (6-40) and ∆eBy′ is solved. With the he

nto

lp of this, point ∆e is located.

oB1BD xyD

R′−′

=′

= (6-40)

By joining point ∆

1B∆eBo yyL ′−′′

e with point F and extending the line until cutting the extension of SLN line, point ∆s is found. Then drawing of equilibrium stages is started. By drawing the tie line passing through point G1, point L1 is located. Equation (6-29) indicates that points G2, L1 and ∆e must be on the same straight line (operating line), hence by joining the

points L1 and ∆e , point G2 is found. Solution is then continued by drawing the tie line passing through point G2. On the left hand side of point F, point ∆s is used instead of point ∆e to draw the operating lines. While total number of the tie lines drawn gives the number of the equilibrium stages required for the given extraction, the number of the tie line cutting ∆ F

Nx′

LN L5

Fx′

0

B

B

xy′′

y,x ′′

y =x

0

0

x′

1.0

1.0

y′1.0

4x′

4y′

2y′

3′y

1x′2x′3x ′

G3

∆S

SSBy′

B1y′

yyB′−′

G4

F

SBx∆′

5y′

2 1 3 4 5

G1G2

G5

L1L2 L3

L4

∆e

Lo,D

eD yx ∆′=′

5x′

eBy∆′

Sy′

S∆x′

xx B′−′

1y′

Fig.6.20 er reflux on solvent-free coordinated diagram Solution of multi-stage extraction und

e ∆s line gives the feed stage. The reflux ratio is selected by the designer. If the reflux ratio is decreased, points ∆e and ∆s approach the diagram, which means more stages are required for the given extraction. On the contrary, if the reflux ratio is increased, points ∆e and ∆s go away from the diagram, which means less number of stages is enough for the given extraction. If during the drawing, one of the operating lines coincides with a tie line,


then infinite number of the stages is required for the given extraction. The reflux ratio corresponding to this case is known as minimum reflux ratio. For the determination of minimum reflux ratio, arbitrary tie lines between point F and D are drawn and extended until cutting the vertical drawn at ( )∆eoD yxx ′=′=′ . The tie line, which when extended cuts the vertical at the farthest point from the diagram, gives point ∆em from which value of is read and finally from, ∆emBy′

oB1B

1B∆emBDm xy

yyR

′−′′−′

= (6-41)

RDm is calculated. Selected RD for the extraction should always be greater than RDm at which, point ∆e lies always beyond point ∆em. 6.4.2 Design of Mixer-Settler Units: The mixing section of a mixer-settler unit is generally made as a closed vessel equipped with turbine type impeller. Vessel does not contain baffles when it is operated full, which is a general practice. The two phases enter the vessel through separate channels and are mixed usually less than 60 seconds at which equilibrium or near equilibrium condition is attained. Turbine impellers are generally flat-blade type and the ratio of impeller diameter (di) to the vessel diameter (Dv) is kept between 0.25 and 0.33. The required power for this type of vessels, which are full and do not contain baffles, can be found from Fig.6.21, where n is the revolution of impeller per second, P is required power as Watt, ρM (kg/m3) and µM (kg/ms) are the density and the viscosity of the mixture, which are obtained from the following equations,

dφdρsφsρMρ +=

)µµφµ6

(1φµ

µow

oo

w

wM +

+= , > 0.4 (6-42) wφ

)µµφµ1.5

(1φµ

µow

ww

o

oM +

+= , < 0.4 wφ

In the equations above, the subscripts s and d show the continuous and disperse phases and w and o water and organic phases respectively. φ shows the volume fraction of the component in the mixture, indicated at its subscript. The volume fraction of dispersed phase in the vessel , is usually smaller than its value in the feed ( ) which is giving as;

dφ

dFφ

)q/(qqφ sFdFdFdF += (6-43)

The ratio of to is given as, dφ dFφ

0.0987

s

d0401

4s

s30.430

s0.427

2sdF

3s

0.247

3L

2sdF

dF

d

µµ

µ9.81ρσ

∆ρρ

σρqµ

σVµqP

3.39φφ ⎟

⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛= (6-44)

where, qdF and qsF are the volumetric flow rates of dispersed and continuous phases at vessel entrance as (m3/s), VL is the volume of liquid in the vessel as (m3), which is defined as [VL= 0.785 ], σ is the interfacial tension between two liquid phases as (N/m) and z is the height of liquid level in the vessel as (m).

zD2v


0.1

1

10

100

1 10 100 1000 10000 100000

M

M2i ndReµρ

=

0.1

1

10000 100000 1000000

M35

i ndPρ

Fig.6.21 Power requirment in un-baffled vessels equipped with turbine type impeller

Re

The average diameter of dispersed phase liquid drops in the vessel, dp (m) and average specific surface (mass transfer area per unit volume of mixture), av (m2/m3) are calculated from;

0.274

s

0.204

ML

0.0473

s

sφ0.7322.066p ρ

σρV

Pρµ

10d d ⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=

−+− (6-45)

p

dv d

φ6a = (6-46)

Individual mass transfer coefficient of continuous phase, kLs can be estimated from the following equation, which was obtained for mass transfer from small solid particles,

0.36s

0.17

v

i

0.622/3

s

s1/3

L

4/3p

As

pLss Sc

Dd

µρ

VP

d0.472D

dkSh ⎟

⎠

⎞⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛+== (6-47)

where DAs is the molecular diffusivity of solute A in continuous phase as (m2/s), which may be component B or C and Scs is Schmidt number for continuous phase, which is given as Scs = µs/ρs DAs. Individual mass transfer coefficient of dispersed phase kLd is calculated from equation;

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ θ−

θ= ∑

∞

12p

Adn2n

pLd d

D64λexpT83ln

6d

k (6-48)

where θ (s) is the residence time in the vessel, which is calculated from θ = VL/(qsF+qdF). DAd (m2/s) is the molecular diffusivity of solute A in the dispersed


component, which can be component B or C. The values of Tn and λn are given in

Table.6.1. Hence over-all mass transfer coefficient based on dispersed phase is obtained as;

LssdLdLd km

1k1

K1

+= (6-49)

where, msd is distribution coefficient defined as the ratio of molar concentration of solute A in the continuous phase cAs (k-molA/m3 continuous phase) to the molar concentration of solute A in dispersed phase cAd (k-molA/m3 dispersed phase). Murphree stage efficiency is then based on the number of transfer units. The number of over-all transfer units for dispersed phase in molar concentration units assuming dilute solution, can be written as;

⎮⌡⌠

−=

∗

1Adc

2Adc AdAd

Ad

ccdc

odN (6-50)

where it is assumed that molar concentration of solute A in dispersed phase changes from cAd1 to cAd2 and is the molar concentration of solute A in dispersed phase, when it reaches in equilibrium with the continuous phase. By assuming constant exit concentration for thoroughly mixed vessel, the equation above is written as;

∗Adc

∫ ∗∗ −−

=−

=1Adc

2Adc2Ad2Ad

2Ad1AdAd cc

ccdcAd2Ad2

od cc1N (6-51)

Table 6-1. The values of Tn

and λn in equation (6-48)

kLsdp/DAs λ1 λ2 λ3 T1 T2 T33.20 0.262 0.424 1.49 0.107 5.33 0.386 8.00 0.534 10.7 0.680 4.92 1.49 0.300 16.0 0.860 5.26 1.48 0.382 21.3 0.982 5.63 1.47 0.428 26.7 1.082 5.90 15.7 1.49 0.495 0.205 53.3 1.324 7.04 17.5 1.43 0.603 0.298 107 1.484 7.88 19.5 1.39 0.603 0.384 213 1.560 8.50 20.8 1.31 0.588 0.396 320 1.600 8.62 21.3 1.31 0.583 0.391 ∞ 1.656 9.08 22.2 1.29 0.596 0.386

On the other hand from definition equation,

vLddod

od aK/uz

HzN == (6-52)

can be written. Where, Hod (m) is the height of one over-all dispersed phase transfer unit, ud (m/s) is the velocity of dispersed phase in the vessel, which is calculated from


[ ud= qdF /0.785 ]. 2vD

As the Murphree stage efficiency for the dispersed phase is,

∗−−

=Ad2Ad1

Ad2Ad1Md cc

ccE (6-53)

Then,

1)cc/()cc()cc/()cc(

)cc()cc(ccE

2Ad2Ad2Ad1Ad

2Ad2Ad2Ad1Ad

2Ad2Ad2Ad1Ad

2Ad1AdMd +−−

−−=

−+−−

= ∗

∗

∗

and finally with the help of equation (6-51);

1N

NEod

odMd += (6-54)

is obtained. By substituting the Nod value, which is obtained from equation (6-51) into equation (6-54) the efficiency of mixer-settler unit is computed. The mixture then flows into settling section of the unit as emulsion, which means one of the phases is dispersed finely into another phase. The behaviour of the emulsion is very important for the separation into two phases in the settler. The stable emulsions, at which the droplet diameters are between 1-1.5 µm, the density difference of the phases and the interfacial tension between the phases is small, cannot be separated easily. When the diameter of the drops exceeds 1 mm., the separation quickens. The high viscosity of continuous phase and the dust particles, which usually accumulate at the interface, hinder the coalescence. An unstable emulsion settles and coalesces rapidly as soon as it flows into settler and a sharp interface forms quickly, which is known as primary break. However, one of the phases, usually that in majority, remains clouded by very fine fog of the dispersed phase. The cloud will eventually settle and leave the clouded phase clear, which is known as secondary break. In continuous multi-stage operations the wait for secondary break between the stages usually is not economic and it is contended with the primary break, which is a matter of minutes. The velocity of emulsion from mixer to settler must be sufficiently low so that it does not disturb the already separated phases in the settler. In sizing the settler, the settling times measured in laboratory are used. In the absence of experimental measurements, diameter of settler Dd (m), which is normally taken as ¼ of the length, is estimated from; Dd = 8.4 (qs+ qd)0.5 (6-55) The emulsions containing very fine droplets may be passed through a coalescer in order to increase the size of the droplets and hence reduce their settling time, before entering the settler. Cotton fibres, glass wool, fibres glass, steel wool, polypropylene cloth etc. can be used for this purpose.


APPENDICES

276

Table.App. 2.1 Laplace Transforms

No. Function f(θ)

Transform

₤(s) ∫∞

θ− θθ=0

s d)(fe

1 1 s1

2 θ 21s

3 )!1n(

1n

−θ −

.,..3,2,11 =nsn

4 )(

1πθ

s

1

5 πθ2 2/3

1s

6 θae as −1

7 ⎟⎠⎞

⎜⎝⎛

θ−

πθ 4kexp

)3(2

k 2

0>− ke sk

8 ⎟⎠

⎞⎜⎝

⎛θ2

kerfc 01 ≥− kes

sk

9 ⎟⎟⎠

⎞⎜⎜⎝

⎛

θ−

πθ 4kexp

)(1 2

01 ≥− kes

sk

10 ⎟⎠

⎞⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

θ−

θ−

πθ

2kerfck

4kexp2

2 02

3≥−−

kes sk

11 ⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛θ

+θ

+θθ−2

kerfc2

kaerfc)aexp()akexp( 2

0)(

≥+

−k

sasae sk

12 ⎟⎠

⎞⎜⎝

⎛θ

+θθ2

kaerfc)aexp()akexp( 2 0)(

≥+

−k

sase sk

13 ⎟⎠⎞

⎜⎝⎛ θ

k2erf 0)()exp(1 22 >kkserfcsk

s

277

Table. App. 2.2 Error Function

∫u

o

z dze2erf2−

π=u ;

dzdu.e2)uerf(

dzd 2u−

π−= ; erf(-u) =-erf u; erf(o)=o;

erf(∞)=1

u erf u u erf u 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

0.0 0.056372 0.112463 0.167996 0.222703 0.276326 0.328627 0.379382 0.428392 0.475482 0.520500 0.563323 0.603856 0.642029 0.677801 0.711156 0.742101 0.770668 0.796908 0.820891 0.842701

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0

0.880205 0.910314 0.934008 0.952285 0.966105 0.976348 0.983790 0.989091 0.992790 0.995322 0.997020 0.998137 0.998857 0.999311 0.999573 0.999764 0.999866 0.999925 0.999959 0.999978

278

Table. App. 4.1 Solubility of ammonia in water

Partial pressure of ammonia, mmHg kg NH3 ----------- 100 kg H2O

0oC 10oC 20oC 30oC 40oC 50oC 60oC

100 947 90 785 80 636 987 70 500 780 60 380 600 945 50 275 439 686 40 190 301 470 719 30 119 190 298 454 692 25 89.5 144 227 352 534 825 20 64 103.5 166 260 395 596 834 15 42.7 70.1 114 179 273 405 583 10 25.1 41.8 69.6 110 167 247 361 7.5 17.7 29.9 50.0 79.7 120 179 261 5 11.2 19.1 31.7 51.0 76.5 115 165 4 16.1 24.9 40.1 60.8 91.1 129.2 3 11.3 18.2 29.6 45.0 67.1 94.3 2 12.0 19.3 30.0 44.5 61.0 1 15.4 22.2 30.2

Table. App. 4.2 Solubility of sulphur dioxide in water

Partial pressure of sulphur dioxide, mmHg kg SO2---------- 100 kg H2O 0oC 7oC 10oC 15oC 20oC 30oC 40oC 50oC 20 646 657 15 474 637 726 10 308 417 474 567 698 7.5 228 307 349 419 517 688 5.0 148 198 226 270 336 452 665 2.5 69 92 105 127 161 216 322 458 1.5 38 51 59 71 92 125 186 266 1.0 23.3 31 37 44 59 79 121 172 0.7 15.2 20.6 23.6 28.0 39.0 52 87 116 0.5 9.9 13.5 15.6 19.3 26.0 36 57 82 0.3 5.1 6.9 7.9 10.0 14.1 19.7 - - 0.1 1.2 1.5 1.75 2.2 3.2 4.7 7.5 12 0.05 0.6 0.7 0.75 0.8 1.2 1.7 2.8 4.7 0.02 0.25 0.3 0.3 0.3 0.5 0.6 0.8 1.3

279

280

Table. App. 4.3 Solubility of carbondioxide in 15.3% Monoethanolamine solution

k-mol CO2/ k-mol amine Partial pressure of CO2,mmHg

40oC 60oC 80oC 100oC 120oC 140oC 1 0.383 0.096 5 0.438 0.152 10 0.471 0.412 0.194 30 0.518 0.459 0.379 0.265 50 0.542 0.482 0.405 0.299 70 0.558 0.498 0.422 0.322 0.200 100 0.576 0.516 0.442 0.347 0.227 0.109 200 0.614 0.552 0.481 0.393 0.281 0.162 300 0.639 0.574 0.505 0.423 0.314 0.194 400 0.657 0.591 0.523 0.442 0.336 0.219 500 0.672 0.605 0.538 0.458 0.355 0.237 600 0.686 0.615 0.550 0.472 0.370 0.254 760 0.705 0.631 0.566 0.489 0.390 0.275 1000 0.727 0.650 0.584 0.509 0.413 0.300 2000 0.702 0.637 0.562 0.476 0.366 3000 0.669 0.596 0.513 0.408 5000 0.712 0.641 0.562 0.464 7000 0.742 0.672 0.597 0.500

Table. App. 4.4 Solubility of H2S in 15.3% Monoethanolamine solution

k-mol H2S / k-mol amine Partial pressure of H2S, mmHg 40oC 60oC 80oC 100oC 120oC 140oC 1 0.128 0.029 3 0.212 0.137 0.050 0.025 0.016 5 0.271 0.171 0.065 0.036 0.025 10 0.374 0.240 0.141 0.091 0.056 0.040 30 0.579 0.386 0.243 0.160 0.101 0.072 50 0.683 0.472 0.314 0.203 0.139 0.091 70 0.750 0.534 0.364 0.238 0.153 0.106 100 0.802 0.600 0.422 0.279 0.182 0.124 200 0.890 0.722 0.545 0.374 0.256 0.167 300 0.931 0.790 0.617 0.439 0.312 0.200 400 0.949 0.836 0.666 0.490 0.357 0.226 500 0.959 0.871 0.706 0.536 0.393 - 600 0.970 0.900 0.738 0.575 0.426 700 0.980 0.921 - 0.607 0.453 800 - 0.942 - 0.636 -

Table.App.5.1 Vapor-liquid equilibrium of acetone-water system at 760 mmHg t(oC) 100 74.8 68.5 64.8 63.1 61.7 60.5 59.4 58.4 57.5 56.7 56.2

x 0.0 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0

y 0.0 0.64 0.73 0.78 0.80 0.82 0.84 0.86 0.88 0.90 0.94 1.0

Table.App.5.2 Vapor-liquid equilibrium of chloroform-toluene system at 760 mmHg

t (oC) 110.7 105.2 101.1 94.5 89.0 84.0 79.3 74.9 70.8 67.2 63.9 61.1

x 0 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

y 0 0.187 0.317 0.500 0.632 0.735 0.816 0.878 0.925 0.959 0.983 1.0

Table.App.5.3 Vapor-liquid equilibrium of methanol-n-propanol system at 760 mmHg

t(oC) 97.2 94.4 91.8 86.9 82.7 78.9 75.6 72.7 70.2 68.0 66.1 65.3 64.5

x 0.0 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95 1.0

y 0.0 0.148 0.272 0.467 0.61 0.716 0.795 0.856 0.902 0.939 0.970 0.984 1.0

Table.App.5.4 Vapor- liquid equilibrium for methanol-water system at 760 mmHg

t (oC) 100 93.7 89.2 83.2 79.1 76.1 73.6 71.5 69.6 67.8 66.1 64.5

x 0 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y 0 0.243 0.390 0.562 0.664 0.735 0.791 0.839 0.882 0.923 0.962 1.0

Table.App.5.5 Vapor-liquid equilibrium for ethyl acetate - acetic acid system at 760 mmHg

t(oC) 118 113,4 109,3 102,5 97,0 92,6 88,9 85,9 83,3 81,1 79,1 77,2x 0,0 0,05 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,0 y 0,0 0,18 0,32 0,51 0,64 0,74 0,80 0,86 0,90 0,93 0,97 1,0

Table.App.5.6 Vapor-liquid equilibrium for benzene- heptane system at 760 mmHg

t(oC) 97.4 96.3 94.1 92.0 90.0 88.1 86.3 84.6 83.1 81.6 80.8 x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 y 0.105 0.187 0.337 0.457 0.552 0.638 0.714 0.782 0.85 0.922 0.958

Table.App.5.7 Vapor-liquid equilibrium for benzene- methylisobutylketone system at 760 mmHg

t(oC) 116.1 110.3 105.2 100.8 96.8 93.3 90.2 87.3 84.8 82.4 80.1x 0.0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 y 0.0 0.267 0.455 0.598 0.69 0.778 0.838 0.89 0.933 0.968 1.0

281

282

Table.App.5.8 Vapor-liquid equilibrium for acetaldehyde-ethanol system at 760 mmHg

t(oC) 78.3 79.3 79.8 77.1 69.1 58.4 48.1 39.7.1 33.1 28.1 24.1 21.1x 0 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 y 0 0.030 0.100 0.358 0.650 0.841 0.931 0.969 0.986 0.994 0.998 1.0

Table.App.5.9 Vapor-liquid equilibrium for water-ethylenglycole system at 760 mmHg

t(oC) 197.1 178.9 165.8 148.1 136.2 127.4 120.5 115.0 110.3 106.3 103.0 100

x 0.0 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y 0.0 0.464 0.675 0.852 0.921 0.954 0.973 0.983 0.990 0.995 0.998 1.0

Table.App.5.10 Vapor-liquid equilibrium for benzene-toluene system at 760 mmHg

t(oC) 110.68 107.96 105.59 101.52 98.0 94.85 91.95 89.26 86.75 84.4 82.19 80.12

x 0.0 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 y 0.0 0.12 0.22 0.38 0.51 0.62 0.71 0.79 0.85 0.91 0.96 1.0

Table.App.5.11 Vapor-liquid equilibrium for sulfur dichloride-carbon tetrachloride system at 760 mmHg

t(oC) 76.7 74.5 72.4 70.5 68.6 66.9 65.2 63.6 62.1 60.7 59.3

x 0.0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.0 y 0.0 0.18 0.32 0.43 0.54 0.65 0.73 0.81 0.89 0.95 1.0

x and y values at the above-given tables are the mole fractions of the first written components

Table.App.5.12 Enthalpy-composition data for aqueous ammonia at 10 bar

x, mass fraction of ammonia in liquid

Specific enthalpy of saturated liquid, h (kJ/kg)

y, mass fraction of ammonia in vapor

Specific enthalpy of saturated vapor,H (kJ/kg)

0 790 0 2 810 0.0951 600 0.622 2 280 0.20 420 0.824 1 960 0.30 310 0.903 1 775 0.40 200 0.94 1 650 0.5 140 0.958 1 580 0.6 100 0.97 - 0.7 140 0.975 - 0.8 160 0.98 1 530 0.9 240 0.99 - 1.0 310 1.0 1 500

283

Table.App.5 Periodic Table

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Ia IIa IIIb IVb Vb VIb VIIb VIIIb Ib IIb IIIa IVa Va VIa VIIa VIIIa Group

Period

1 1 H 1.01

2 He 4.00

2 3 Li 6.94

4 Be 9.01

5 B 10.81

6 C 12.01

7 N 14.01

8 O 15.99

9 F 19.00

10 Ne 20.18

3 11 Na 22.99

12 Mg 24.31

13 Al 26.98

14 Si 28.09

15 P 30.97

16 S 32.07

17 Cl 35.45

18 Ar 39.95

4 19 K 39.10

20 Ca 40.08

21 Sc 44.96

22 Ti 47.88

23 V 50.94

24 Cr 52.00

25 Mn 54.94

26 Fe 55.85

27 Co 58.93

28 Ni 58.70

29 Cu 63.55

30 Zn 65.39

31 Ga 69.72

32 Ge 72.61

33 As 74.92

34 Se 78.96

35 Br 79.90

36 Kr 83.80

5 37 Rb 85.47

38 Sr 87.62

39 Y 88.91

40 Zr 91.22

41 Nb 92.91

42 Mo 95.94

43 Tc 98.91

44 Ru 101.07

45 Rh 102.91

46 Pd 106.42

47 Ag 107.87

48 Cd 112.41

49 In 114.82

50 Sn 118.71

51 Sb 121.75

52 Te 127.60

53 I 126.90

54 Xe 131.29

6 55 Cs 132.91

56 Ba 137.33

57 La 138.91

* 72 Hf 178.49

73 Ta 180.95

74 W 183.85

75 Re 186.21

76 Os 190.20

77 Ir 192.22

78 Pt 195.08

79 Au 196.97

80 Hg 200.59

81 Tl 204.38

82 Pb 207.20

83 Bi 208.98

84 Po (209)

85 At (210)

86 Rn (222)

7 87 Fr (223)

88 Ra 226.03

89 Ac (227)

** 104 Rf (261)

105 Ha (262)

106 Sg (263)

107 Ns (262)

108 Hs (265)

109 Mt (266)

110 Uun

111 Uuu

112 Uub

113 114 115 116 117 118

Lanthanoide * 58 Ce 140.12

59 Pr 140.91

60 Nd 144.24

61 Pm (145)

62 Sm 150.36

63 Eu 151.96

64 Gd 157.25

65 Tb 158.93

66 Dy 162.50

67 Ho 164.93

68 Er 167.26

69 Tm 168.93

70 Yb 173.04

71 Lu 174.97

Actinoide ** 90 Th 232.04

91 Pa 231.04

92 U 238.03

93 Np 237.05

94 Pu (244)

95 Am (243)

96 Cm (247)

97 Bk (247)

98 Cf (251)

99 Es (254)

100 Fm (257)

101 Md (258)

102 No (259)

103 Lr (260)

REFERENCES 1- Mass Transfer Operations. 3rd ed. R.E.Treybal, McGraw-Hill, 1980. 2- Mass Transport Phenomena. C.J.Geankoplis. Holt, Rinehart and Winston, 1972. 3- Chemical Engineering. Vol.1 (2nd ed.) and Vol.2 (3rd ed.) J.M. Coulson and J.F. Richardson. Pergamon Press, 1970-1978. 4- Perry’s Chemical Engineers’ Handbook. 7th ed. D.W.Green (edit.), McGraw-Hill, 1998. 5- Transport Processes and Unit Operations. 3rd ed. C.J.Geankoplis, Prentice-Hall, 1993. 6- Distillation Engineering. R.Billet, Chemical Pub. Co., 1979. 7- Mass Transfer (Turkish). E.Alpay, Ege Univ. Eng. Fac. Pub. No.3, 1984. 8- Stoffaustausch einschliesslich chemischer Reaktionen. H.Brauer, Sauerlander AG, Aarau, 1971. 9- Mass Transfer. T.K.Sherwood, R.L.Pigford and C.R.Wilke, McGraw-Hill, 1975. 10-Diffusional Mass Transfer. A.H.P Skelland, John Wiley, 1974. 11-Fundamentals of Transport Phenomena. R.W.Fahien, McGraw-Hill, 1983. 12-Mass Transfer. A.L.Hines and R.N.Maddox, Prentice-Hall, 1985. 13-Vapor-Liquid Equilibria. M.Hirata, S.Ohe and K.Nagahama, Elsev. Sci. Pub., 1975. 14-Elements of Fractional Distillation. E.R.Gilliland and C.S.Robinson, McGraw-Hill, 1950. 15-Azeotropic Data. Vol.1-2. L.E.Horsley. Am.Chem.Soc., 1952-1962. 16-Process Heat Transfer. D.Q.Kern, McGraw-Hill, 1950. 17-Gas-Liquid Reactions. P.V.Danckwerts, McGraw-Hill, 1970.

284

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