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

1.3 Mass Transfer by Molecular Diffusion in Liquids Determination of Molecular Diffusivities in Liquids

22 22 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

Stage-Wise Contact Type of Absorption Gas Absorption in Plate Columns

106 110 110

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 Equilibrium or Flash Distillation under Constant Pressure

168 168

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 [email protected]

Mustafa Demircioğlu [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 classroom 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,

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

2

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

x

solute A solid plate (a)

(b)

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

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 zdirection 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


3

interface

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 Liquid phase volume next to the interface is less than the number Gas phase A+C S of the ammonia molecules in unit volume in the A bulk gas. Under this concentration difference, ammonia transfers from bulk gas to the interface Bulk gas Bulk liquid 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 z 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 zdirection 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. 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


4

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 unreacted 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


5

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.


6

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 zdirection in a binary mixture of A+B, molar diffusional flux of A JAz,is written as: JAz= -DAB dc A

(1-1)

dz

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 dc B

(1-2)

dz

Rate of molecular diffusion, J A z (k-molA/s) can be related to the diffusional flux, J Az by : J (1-3) J Az = Az Sm 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=-DAB

dρ A dz

(1-4)

where ρ A (kgA/m3 ) 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 7 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

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, u& z (m/s) is then: u& z =

c A u Az + c B u Bz N Az + N Bz = c c

(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- u& z ) = cA uAz - c A u& z can be written. After substitutions from the equations above finally, NAz= JAz +

cA ( NAz + NBz ) c

(1-6) (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 +

cB ( NAz + NBz ) c

(1-8)

The relationship between total molar flux and total molar rate ( N Az ) (k-mol A/s) can be written as: N (1-9) N Az = Az Sm Similarly, if instead of molar units, the mass units are used; total mass flux of component A, nAz (kgA/m2s) is written as:

nAz = j Az + ρ A u z = j Az + where

ρA (nAz + nBz ) ρ

(1-10)

uz is the mass average velocity of the mixture and given as:

u z = ( ρAu Az + ρBuBz ) / ρ =( nAz + nBz )/ ρ

(1-11) 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 ; 1 cD AB

∫

z2 z1

c

A2 dc A ⌠ dz = − ⎮ ⌡c A1 cN A − c A ( N A + N B )

is obtained. For the sake of simplicity

subscript z is omitted. Upon integration,

c N A − c A2 (N A + N B ) z 2 − z1 1 ln is found. If both sides of this equation are = c D AB N A + N B c N A − c A1 ( N A + N B )

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

NA c − A2 NA + NB c ln NA c − A1 NA + NB c

NA c D AB . NA + NB z

(1-12)

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

n RT = cRT V

and

pA =

nA RT = c A RT V

can be written, where P and

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

NA D P . AB N A + N B RTz

ln

NA p − A2 NA + NB P NA p − A1 NA + NB P

(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 =

D P NA . AB N A + N B RTz

ln

NA − y A2 NA + NB NA − y A1 NA + NB

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

N A = − D AB

dc A c A + [N A + (− N A )] = − D AB dc A + 0 dz c dz

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

D AB (c A1 − c A 2 ) z

(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: NA =

D AB RTz

NA =

D AB P RTz

( p A1 − p A 2 )

(1-16)

and

( y A1 − y A 2 )

(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: NA =

c c D AB ln B2 , c B1 z

NA =

p D AB P ln B2 , p B1 RTz

NA =

D AB P y ln B2 RTz y B1

(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”. pA1 – pA2 = pB2 – pB1 and yA1 – yA2 = yB2 – yB1 , Since, cA1 – cA2 = cB2 – cB1, the right hand sides of the equations given in (1-18) are first multiplied with the appropriate equation below, c A1 − c A 2 =1, c B2 − c B1

p A1 − p A 2 =1 p B2 − p B1

y A1 − y A 2 =1 y B2 − y B1

then by definitions of logarithmic means, which are given below, c B2 − c B1 = (c B )ln , c B2 ln c B1

p B2 − p B1 = (p B )ln , p B2 ln p B1

y B2 − y B1 = (y B )ln y B2 ln y B1

finally; NA =

D AB c D AB P D AB P ( c A1 − c A 2 ) , N A = ( p A1 − p A 2 ) N A = ( y A1 − y A 2 ) z (c B ) ln RTz (p B )ln RTz (y B ) ln

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

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

A

pB2

B pA1

pressures,pA,pB,P

pressures,pA,pB,P

P

A

P pB2

pB1 pA1

pA2

pB1 z1

distance,z

pA2 z1

z2

distance,z

(a)

z2

(b)

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

1.2.4. The Relationship between NA and NB is Fixed by Reaction Stoichiometry: In some cases, relationship between the fluxes is fixed by the chemical reaction, taking place nearby. Consider the gas phase reaction below taking place on a solid catalyst surface, m A→ n B According to the stoichiometry of the reaction, when m moles of component A transfer towards catalyst surface at the same time n moles of component B transfer from the catalyst surface to the bulk gas. So, n NA = - m NB can be written. Hence, substituting NB = - (n/m) NA into any one of the equations (1-12), (1-13) and (1-14) NA is calculated. Example-1.1) Calculation of Molar Fluxes A gaseous chemical reaction is taking place on a solid catalyst surface at 2 bars and 25 oC according to the stoichiometry given below: 2A

B

Across a gas film of 2 mm thickness adjacent to the solid, partial pressures of component A are measured as 0.40 and 0.10 bars. Calculate total molar fluxes of component A and B. At the operating temperature and pressure DAB= 4.10-5 m2/s.


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

t

A

NA NA 1 = = =2 N A + N B N A − 0.5 N A 1 − 0.5

catalyst

B

z

1

N A = ( 2)

2

( 4 *10−5 m 2 / s)(2 bar ) 2 − (0.1 / 2) ln 3 −3 (0.083 bar m / k − mol K )(273 + 25 K )(2 *10 m) 2 − (0.4 / 2)

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 (112), (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 N Az and N Bz are constant, NA N Az N and Bz and Sm is Sm Sm 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.

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

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.


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 :

b=300 mm

Sm1

NA (S

a1 =400 mm

S

a

Sm2

a2 =200 mm

pA2

NB= 0

pA1

b=300 mm l/2

z

l Sm = b * a = 0.3 * a (m2) a = a1 −

a1 − a 2 0.4 − 0.2 z = 0. 4 − z = 0.4 − 0.1 z (m) l 2

Sm = 0.3(0.4 - 0.1 z) = 0.12 - 0.03 z (m2) From eqn.(1-7) for NB=0 and J A = − D AB dp a RT dz

N A (1 −

pA D dp A ) = − AB P RT dz

NA D P dp A dz = − AB Sm RT P − p A NA

−

l

dz

∫o 0.12 − 0.03 z

= −

After substituting for Sm and separating the variables,

D AB P RT

p A2

∫p

A1

dp A P − pA

NA D P 055 ln [0.12 − 0.03 z ]02 = AB ln [P − p A ]00..22 0.03 RT D AB

⎛ 1 ⎞ ⎛ 598 ⎞ = 2.25 *10 ⎜ ⎟ ⎜ ⎟ ⎝ 10 ⎠ ⎝ 298 ⎠ −5

Then from equation above,

At z = l / 2

(N A ) l / 2 =

1.75

= 7.61*10 −6 m 2 / s N A =1.11*10−9 k − mol A / s = cons tan t

(Sm)l /2= 0.12 – 0.03 (1.0) = 0.09 m2

NA 1.11 * 10 −9 = = 1.23 * 10 −8 k − molA / m 2 s (S m ) l / 2 0.09


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 concentration gradients. The total molar flux of component A in this case can also be calculated from equations (1-12, (1-13) and (1-14) by making two modifications in the n

equations. First, all the NA+ NB terms in the equations are replaced with ∑ N i second, i=A

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

D Aef =

NA − yA ∑ Ni i=A

1 ( yi N A − yA Ni ) ∑ i = A D Ai n

(1-20)

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

NA =

N A c D Aef ln n z ∑ N

i=A

i

∑

i=A

Ni

NA

−

cA2 c

c − A1 n c ∑ Ni

(1-21)

i =A

As it is seen from equation (1-20), effective diffusivity DAef, which can be synthesized from its binary 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 all N’s except NA is zero. In this case equation (1-20) simplifies to: D Aef =

1− yA n y ∑ i i = B D Ai

(1-22)

1.2.8 Determination of Binary Diffusivities of Gases: It is obvious from 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 negligible. The diffusivities can be determined experimentally or may be predicted from 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 experimentation conditions. The experimental set-up as shown in Fig.1.2, consists of two 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

1

c oA1

Sm

2 c oA 2 V2

V1

l

cA1 z

cA2

Fig.1.2 Experimental set-up for two bulb method

chamber-2 is thoroughly mixed and analysed for component A. By ignoring the volume of the capillary tube and assuming uniform concentrations in both chambers at any time of the diffusion, equation dc D (c − c ) J A = −D AB A = − AB A 2 A1 (1-23) dz l can be written. The accumulation of component A in chamber-2 is due to the mass transfer of this component from chamber-1. Thus, S m D AB (c A1 − c A2 ) dc V2 A2 = S m J A = (1-24) dθ l is written. On the other hand, the average concentration of component A (c A ) in the system is found from any one of the equations below, where c oA1 , c oA 2 , c A1 , c A 2 are molar concentrations of component A in chamber-1 and

(V1 + V2 ) cA = V1 c oA1 + V2 c oA 2

(V1 + V2 ) c A = V1 c A1 + V2 c A 2

(1-25) (1-26)

chamber-2 at the beginning and at the end of the experiment. If cA1 is solved from equation (1-26) and substituted into equation (1-24) , c θ D (V + V2 ) dc A 2 ⌠ p = AB 1 = ∫o dθ is obtained. Where, p is defined as ⎮ (l/S m )V1 V2 ⌡c p ( cA − c A 2 ) A2

o A2

After performing integral, substituting p and solving for DAB finally, l V1 V2 c − co D AB = ln A A2 S m (V1 + V2 ) θ c A − c A2

(1-27)

is obtained. Since V1 , V2 , Sm , l are all known, and c A can be found from equation (1-25), by measuring the cA2 after an experimentation time of θ, DAB is calculated from equation (1-27). Winkelmann Method: This method is used when one of the components is liquid at the experimentation conditions. As shown in Fig.1.3, a narrow tube is filled with liquid A to a certain depth and the gas component B is made to flow through a large diameter channel attached to the tube mouth. During this flow, liquid A vaporizes and transfers in the gas in z-direction by molecular diffusion, as its concentration at the liquid-gas interface (point-1) is higher than that at the tube mouth (point-2). Mass transfer takes 15 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

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 gas B equations (1-19) can be written. 2

NA =

pA2≈0 NA

1

pA1=

(1-19)

Here, pA1 is the partial pressure of component A at the liquid-gas interface and hence equals the vapor pressure of the liquid ( p oA ) at the experimentation 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:

z

l

D AB P ( p A1 − p A 2 ) RTz (p B ) ln

p oA

Liquid A

Fig.1.3 Winkelmann experiment

ρ A dVA , where VA, MA and ρA are volume, molecular weight and M A S m dθ density of the liquid respectively. On the other hand, Sm being the cross-sectional area of the tube,VA= Sm(l -z) and dVA = -Sm dz can be written. If all these are substituted NA = −

into equation (1-19): and

∫

zθ z0

ρ A dz D AB P = p oA M A dt RTz (p B ) ln

D AB M A P p oA z dz = ρ A RT ( p B ) ln

∫

θ

dθ

and finally,

o

2 D AB M A P p oA θ (1-28) ρ A R T (p B ) ln is obtained. 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. z θ2 − z 02 =

Example-1.3) Measurement of Molecular Diffusivity

The molecular diffusivity of carbon tetrachloride (A) in air (B) will be measured by using Winkelmann’s method at 1 atm. and 50 oC. The tube filled with liquid carbon tetrachloride till 10 mm below its mouth is subjected to air flow. After 55 hours and 29 minutes the level of the liquid in the tube falls 73.8 mm. What is the value of molecular diffusivity? At 50 oC, vapor pressure and the density of carbon tetrachloride are 282 mmHg and 1 500 kg/m3. MA= 154. Solution: From the givens, zo=10 mm and zt =10 +73.8 = 83.8 mm. (p B ) ln =

760 − (760 − 282) = 608.1 mmHg ln[760 /(760 − 282)]

θ = (55)(3 600) + (29)(60) = 199 740 sec. Then, from equation (1-28), D AB =

[(83.8) 2 − (10) 2 ] *10 −6 (1 500)(0.083)(273 + 50)(608.1) = 9.63 *10 −6 m 2 / s 2(154)(1.013)(282)(199 740)

is found.


Vaporization of Liquid Drops or Sublimation of Solid Spheres: For the application of this experiment, one of the components must be either liquid or sublimable solid at the experimentation conditions. As shown in Fig.1.4 component A, either as liquid drop or solid sphere in rp diameter, is hanged in a large diameter channel through which gas B flows at low velocity at the 2 experimentation temperature and pressure. Liquid drop or solid sphere vaporizes or B A 1 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 Fig.1.4 Vaporization of liquid drops direction. Here, again mass transfer takes place under the conditions of “A transfers through non-transferring B”, and hence flux equation, NA =

D AB P ( p A1 − p A 2 ) RT r (p B ) ln

(1-19a)

is written. Again here, pA1 is the partial pressure of A at liquid (solid)-gas interface, which is equal to the vapor pressure of A at operating temperature. pA2 is the partial pressure of A at bulk gas, which can be assumed zero, as the volumetric flow rate of 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, ρ A dVA , where VA, MA and ρA are volume, molecular weight and density NA = − M A S m dθ of the liquid drop or solid sphere respectively. From the sphere geometry, Sm = 4π r2 and VA =

4 3 π r can be written. If the flow of gas B continues until the drop or sphere 3

evaporates completely, by inserting all the above equations into equation (1-19a), the equations; o rp2 M A D AB P p oA M A D AB P p oA θ dθ = − ∫r r dr and θ= 2 ρ A RT (p B ) ln ∫o ρ A RT (p B ) ln are obtained. On the other hand, from the definition of logarithmic mean, p

p B2 − p B1 p oA (p B ) ln = = p B2 ln P / (P − p oA ) ln p B1

[

]

is written. Substituting this into equation above and by rearranging, finally ρ A rp2 RT D AB = (1-29) 2 M A θ P ln[P /( P − p oA )] is found. By measuring the time θ for complete vaporization or sublimation, DAB is calculated from the equation above. In Table.1.1, experimentally measured values of molecular diffusivities of some gas pairs 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

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 iodine cast in spherical shape with 4 mm diameter is suspended in a duct through which air flows. After 22 hours and 54 minutes, iodine sphere completely disappears. What is the value of molecular diffusivity? Vapor pressure and density of solid iodine at 25 oC are 1.07 mmHg and 4 930 kg/m3. MA=254. Solution: θ = (22)(3 600) + (54)(60) = 82 440 sec. From equation (1-29), D AB =

4 930 * (2 *10 −3 ) 2 * 0.083 * (273 + 25) = 8.16 *10−6 m 2 / s 2 * 254 * 82 440 *1.013 * ln[760 /(760 − 1.07)]

is found. It is seen that this value, after temperature correction, is identical with the value given in Table.1.1.

1.2.8.2 Prediction of Binary Gas Diffusivities: The kinetic theory of the gases is fairly well developed. Starting from the kinetic theory of the gases, HirschfelderBird-Spotz derived the theoretical equation below, which can be used to predict the binary diffusivities of non-polar gas pairs or of a polar gas with a non-polar gas, at the absence of experimentally found value. 10 D AB =

−4

0.5 ⎞ ⎛ ⎛ ⎞ 1 1 ⎜ ⎟⎟ ⎟ + T 2 1.084 − 0.249 ⎜⎜ ⎜ 0.5 ⎝ M A M B ⎠ ⎟⎠ ⎛ 1 1 ⎞ ⎝ ⎜⎜ ⎟⎟ + 2 Ω D,AB P rAB ⎝ MA MB ⎠ 3

(1-30)

In the equation above, DAB is molecular diffusivity in m2/s. T and P are absolute temperature in K, and pressure in N/m2 respectively. MA and MB are molecular weights of A and B in kg/k-mol. rAB, which is defined as rAB=(rA+rB)/2 , is collision radius in nm. Ω D,AB , which is given as function of (kT/ ε AB ) in Fig.1.5, is known as collision function, where k is very well known Boltzmann constant. ε AB , which is obtained from the force constants (ε/k) of the components by ε AB / k = ε A ε B / k , 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 atm. and 0 oC from Hirschfelder-Bird-Spotz Equation: From Table.1-2. Methane Air

ε/k 148.6 78.6

r (nm) 0.3758 0.3711

A B

M 16 29


Table 1-1. Experimentally determined binary gas diffusivities at 1 atm. pressure

Gas (Vapor) Pair Acetic acid-Air Acetic acid-CO2 Acetic acid-H2 Acetone-Air Air-n-Butanol Air-n-Butanol Ammonia-Air Ammonia-H2 Ammonia-H2 Ammonia-N2 Ammonia-N2 Aniline-Air Aniline-Air Aniline-Air Argon-Nitrogen Benzene-Air Benzene-CO2 Benzen-H2 CH4-H2 CH4-O2 CO2-Air CO2-Air CO2-H2 CO2-Helium CO2-O2 CO2-SO2 CO2-Water CO2-Water CO-Nitrogen n-Decane-H2 Ethanol-Air Ethanol-CO2 Ethanol-CO2

Temp. D *104 m2/s (oC) AB 0 0.1064 0 0.0716 0 0.416 0 0.109 59 0.104 25.9 0.087 0 0.198 25 0.784 85 1.093 85 0.328 25 0.230 0 0.061 25.9 0.074 59 0.09 20 0.194 0 0.077 0 0.0528 0 0.306 15 0.694 500 1.1 44 0.177 0 0.138 0 0.55 25 0.612 0 0.139 70 0.108 55.4 0.211 34.4 0.202 100 0.318 90 0.306 0 0.102 0 0.0685 67 0.106

Temp. DAB*104 m2/s (oC) Ethyl formate-Air 0 0.337 Ethylene-H2 0 0.486 Formic acid-Air 0 0.131 H2-SO2 200 1.23 H2-Su 55.5 1.121 Helium-Nitrogen 20 0.705 n-Heptane-CH4 38 0.066 n-Hexane-H2 15 0.290 Iodine-Air 0 0.07 Mercury-Air 0 0.07 Methanol-Air 0 0.132 Methyl acetate-H2 0 0.303 Methyl formate-Air 0 0.0872 Naphthalene-Air 0 0.0513 Nitrogen-H2 25 0.784 Nitrogen-SO2 -10 0.104 Nitrogen-Water 34.5 0.256 Nitrogen-CO2 25 0.165 O2-Air 0 0.178 O2-H2 0 0.697 O2-Nitrogen 0 0.181 n-Octane-H2 30 0.271 n-Octane-O2 30 0.0705 Phosgene-Air 0 0.095 n-Propanol-Air 0 0.085 n-Propanol-CO2 0 0.0577 n-Propanol-H2 0 0.315 Toluene-Air 0 0.076 Toluene-Air 30 0.088 i-Valeric acid-H2 0 0.212 Water-Air 25.9 0.258 Water-O2 59 0.305 Water-O2 450 1.3 Gas (Vapor) Pair


1.4

0.5

1.2 1

0.4

Ω D ,AB 0.8

Ω D ,AB

0.6

0.3

0.4 0.2

0.2 0.1

1

10

100

1000

kT/ ε AB Fig. 1.5 Collision function

rAB =

rA + rB 0.3758 + 0.3711 = = 0.37345 (nm) 2 2

kT 273 = = 2.53 ε AB 108.074

From Fig.1.5

⎡ 1 1 ⎤ + ⎢ ⎥ ⎣MA MB ⎦

0.5

ε AB = k

εA εB = k k

(148.6)(78.6) = 108.074 (K )

ΩD,AB = 0.5 is read.

1⎤ ⎡1 =⎢ + ⎥ 16 29 ⎣ ⎦

0.5

= 0.311

By substituting all these into equation(1-30)

10 −4 ( 273) 3 / 2 [1.084 − (0.249)(0.311)](0.311) = 2 * 10 −5 m 2 / s (1)(1.013 * 10 5 )(0.37345) 2 (0.5) is obtained D AB =

1.2.8.3 Estimation of Binary Gas Diffusivities: There is large number of empirical equations 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 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. D AB =

1.0 ∗ 10−9 T1.75

1/ 2 ⎛ 1 1 ⎞ ⎜ ⎟ + 2 ⎜⎝ M A M B ⎟⎠ P ⎡(∑ v )1A/ 3 + (∑ v )1B/ 3 ⎤ ⎢⎣ ⎥⎦

(1-31)


Table 1-2. The collision radius and force constants for some molecules Gas (Vapor) Acetone Acetylene Air Ammonia Argon Arsine Benzene bi-Butane bn-Butane Boron trichloride Boron trifluoride Bromine Carbon dioxide Carbon monoxide Carbon sulfide Carbon tetrachloride Carbon tetrafluoride Chlorine Chloroform Cyanogen Cyclo –propane Cyclo-hexane 2,2 Dimethylpropane Ethane Ethanol Ethyl acetate Ethyl chloride Ethyl ether Ethylene Fluorine Helium n-Hexane Hydrogen Hydrogen bromide Hydrogen chloride Hydrogen cyanide Hydrogen fluoride

r (nm) 0.4600 0.4033 0.3711 0.2900 0.3542 0.4145 0.5349 0.5278 0.4687 0.5127 0.4198 0.4296 0.3941 0.3690 0.4483 0.5947 0.4662 0.4217 0.5389 0.4361 0.4807 0.6182 0.6464 0.4443 0.4530 0.5205 0.4898 0.5678 0.4163 0.3357 0.2551 0.5949 0.2827 0.3353 0.3339 0.3630 0.3148

ε/k (K) 560.2 231.8 78.6 558.3 93.3 259.8 412.3 330.1 531.4 337.7 186.3 507.9 195.2 91.7 467.0 322.7 134.0 316.0 340.2 348.6 248.9 297.1 193.4 215.7 362.6 521.3 300.0 313.8 224.7 112.6 10.2 399.3 59.7 449.0 344.7 569.1 330.0

Gas (Vapor) Hydrogen iodide Hydrogen peroxide Hydrogen sulfide Iodine Krypton Mercury Mercury dibromide Mercury dichloride Mercury diiodide Methane Methanol Methyl acetate Methyl acetylene Methyl bromide Methyl chloride Methyl ether Methylene chloride Neon Nitrogen Nitrogen monoxide Nitrogen oxidul n-Pentane n-Propyl alcohol Oxygen Phosphine Propane Propylene Silicon hydride Silicon tetrafluoride Stannous bromide Sulfur dioxide Sulfur hexafluoride Trimethyl borate Uranium hexafluoride Water Xenon

r (nm) 0.4211 0.4196 0.3623 0.5160 0.3655 0.2969 0.5080 0.4550 0.5625 0.3758 0.3626 0.4936 0.4761 0.4118 0.4182 0.4307 0.4898 0.2820 0.3798 0.3492 0.3828 0.5784 0.4549 0.3467 0.3981 0.5118 0.4678 0.4084 0.4880 0.6388 0.4112 0.5128 0.5503 0.5967 0.2641 0.4047

ε/k (K) 288.7 289.3 301.1 474.2 178.9 750.0 686.2 750.0 695.6 148.6 481.8 469.8 251.8 449.2 350.0 395.0 356.3 32.8 71.4 116.7 232.4 341.1 576.7 106.7 251.5 237.1 298.9 207.6 171.9 563.7 335.4 222.1 396.7 207.6 809.1 231.0

In the equation above, DAB is molecular diffusivity in m2/s, T and P are absolute temperature in K and pressure in atm., Σv is the diffusional volume of the molecule in m3/k-mol, which can be synthesized from the atomic diffusional volumes and the structural contribution of the molecule, in the absence of the value for the molecule. Table1.3 gives the diffusional volumes of some atoms and molecules.


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

1/ 2

1.5 ∗ 10 −6 T1.81 P (TcA TcB )

0.1405

(V

0.4

cA

)

0.4 2

+ VcB

⎛ 1 1 ⎞ ⎟⎟ ⎜⎜ + ⎝ MA MB ⎠

(1-32)

Here again, temperature, pressure and volumes should be taken in K, in atm. and in m3/k-mol respectively. DAB will be obtained in m2/s. 1.2.9 Effect of Temperature and Pressure on Gas Diffusivity: In the gases, molecular diffusivity increases with temperature but decreases with pressure. This is understandable as the kinetic energy hence the mobility of the molecules raises with the increasing temperature, and the number of the molecules per unit volume increases with rising pressure. These dependences are expressed as;

(D AB )P ,T 2

2

1.75

⎛ P ⎞⎛ T ⎞ = (D AB )P ,T ⎜⎜ 1 ⎟⎟⎜⎜ 2 ⎟⎟ 1 1 ⎝ P2 ⎠⎝ T1 ⎠

(1-33)

So, if molecular diffusivity at one temperature and pressure is known, its value at another temperature and pressure can easily be calculated from the equation above. In the equation, temperature T is taken in K. 1.3 MASS TRANSFER BY MOLECULAR DIFFUSION IN LIQUIDS:

The integration of general flux equation (1-7) requires that DAB and c are constant. This is almost so for binary gas mixtures but not so for binary liquids, where both may vary considerably with concentration. Nevertheless, due to the lack of knowledge of these 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 : NA =

D AB c NA x − N A /( N A + N B ) ln A 2 NA + NB z x A1 − N A /( N A + N B )

(1-34)

where DAB and c are the arithmetic means of the molecular diffusivities and total molar concentrations evaluated at the two ends of diffusional path. xA, which is defined as x A = c A / c , is the mole fraction of component A in the liquid . When mass transfer takes place under the condition of “equimolar counter transfer of A and B”, equation (1-34) simplifies to: NA =

D AB c D ( x A1 − x A 2 ) = AB (c A1 − c A 2 ) z z

(1-35)

On the other hand, if the transfer is under the conditions of “A transfers through nontransferring B”, equation (1-34) simplifies to: D AB c 1 − x A2 D AB c ln = ( x A1 − x A 2 ) z 1 − x A1 z( x B ) ln x − x B1 where, xB is given as ( x B ) ln = B2 ln(x B2 / x B1 ) NA =

(1-36)


Table 1-3. Diffusional volumes of some atoms and molecules

Atom Carbon Chlorine Hydrogen Nitrogen Molecule Air Ar Br2 Cl2 CO CO2 CCl2F2 D2 H2 H2O

Diffusional volume,v m3/kg atom *103 16.5 19.5 1.98 5.69 Diffusional volume,∑ v m3/k-mol *103 20.1 16.1 67.2 37.7 18.9 26.9 14.8 6.70 7.07 12.7

Atom Oxygen Sulfur Aromatic ring Heterocyclic ring Molecule He Kr N2 Ne NH3 N2O O2 SF2 SO2 Xe

Diffusional volume,v m3/kg atom *103 5.48 17.0 -20.2 -20.2 Diffusional volume, ∑v m3/k-mol *103 2.88 22.8 17.9 5.59 14.9 35.9 16.6 69.7 41.1 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 c oA 2 > c oA1 , 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 ∂c c − c A1 gradient at any moment can be written as: A = A2 , where constant K, which ∂z Kδ 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, dc A c − c A1 = ε D AB A2 (1-37) N A = J A = − ε D AB K dz Kδ can be written, where ε shows the fraction of the area which is open to mass transfer in the porous disc.

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

V cA1

1

stirrer δ

z

porous disc

2 cA2

V

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

The component A transferring from cell-2 to cell-1 causes an accumulation in this cell. Hence, dc ε Sm D AB (1-38) (c A2 − c A1 ) V A1 = dθ Kδ is written, where Sm is the cross-sectional area of the disc. Similarly, component A leaving the cell-2 causes a depletion in this cell and hence, equation, dc ε Sm D AB (1-39) − V A2 = (c A2 − c A1 ) dθ Kδ is also written. If we add equation (1-38) to the equation (1-39), d(c A2 − c A1 ) 2 ε S m D AB dc − dc A2 = −V = (c A2 − c A1 ) (1-40) V A1 dθ Kδ dθ is obtained. If the variables are separated and integrated, c θA2 −cθA1

⌠ −⎮ ⌡c

o A2

−c A1 o

d (c A2 − c A1 ) 2 ε S m D AB = c A2 − c A1 KδV KδV D AB = 2 ε Sm θ

θ

∫o dθ ln

and finally,

c oA2 − c oA1 c θA2 − c θA1

(1-41)

is obtained, where c oA1 , c oA 2 , c A1 , c A 2 are molar concentrations 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, α = (K δ V/2 ε S m ) , and this constant can easily be determined by conducting an experiment with a liquid pair whose molecular diffusivity is already known. Experimentally measured diffusivities of some binary liquids are given in Table.1.4. As it was stated before, molecular diffusivities in liquids are also dependant on the concentration of solution. By comparing Table.1.4 with Table.1.1, it can be realized that values of molecular diffusivities of binary gases are about ten thousand times higher than the values of molecular diffusivities in binary liquids. 24 No part of this e-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

Example-1.6) Measurement of Molecular Diffusivity in Liquids

Molecular diffusivity of methanol (A) in water (B) will be measured at 25 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 kmolA/m3 respectively. In order to find the cell constant, an experiment with formic acid(A)- water(B) system whose molecular diffusivity at 25 oC is known as DAB=1.52*10-9 m2/s is conducted with initial concentrations of formic acid in cell-1 and cell-2 being as 0.0 and 0.12 k-molA/m3 and these changing to 0.025 kmolA/m3 and to 0.095 k-molA/m3 after 48 hours and 28 minutes experimentation time. What is the molecular diffusivity of methanol in water at 25 oC and within the concentration range used? Solution: First, find the cell constant α. From the givens: θ = (48)(3 600) + (28)(60) = 174 480 sec. o o c A1 = 0.0 ; c A 2 = 0.12 k − molA / m 3 ; c θA1 = 0.025 k − molA / m 3 ; c θA 2 = 0.095 k − molA / m 3 Then, from equation (1-41); (1.52 * 10 −9 )(174 480) α= = 4.92 * 10 −4 m 2 ln[(0.12 − 0.0) /(0.095 − 0.025)] is found. Now this constant can be used to calculate the molecular diffusivity of methanol in water. From the givens : θ = (63)(3 600) + (36)(60) = 228 960 sec. c oA1 = 0.0 ; c oA 2 = 0.085 k − molA / m 3 ; c θA1 = 0.0195 k − molA / m 3 ; c θA 2 = 0.0655 k − molA / m 3 Then, from equation (1-41) ; 4.92 * 10 −4 0.085 − 0.0 D AB = ln = 1.32 * 10 −9 m 2 / s 228 960 0.0655 − 0.0195 is obtained. As it is seen this value is only 6.5% greater than the value given in Table.1.4.

1.3.1.2 Estimation of Liquid Diffusivities: There is large number of empirical equations derived by various scientists in terms of physical variables of the components and the mixture. Most of them are for dilute solutions and have application limitations. The equation given by Wilke and Chang can be used to estimate the binary diffusivities in liquids at a temperature range of 5-40 oC, when the solution is dilute for solute A. D oAB =

1.17 ∗10 −16 (φ B M B ) 0.5 T VA0.6 µ

(1-42)

In the equation, µ is the viscosity of the solution in kg/m s, T is absolute temperature in K, VA is the molar volume of solute A at 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, φ B is the association factor of solvent B, which 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. Superscript (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


solutes whose molar volumes are below 400*103 m3/k-molA. When solute A is water the 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

D oAB =

For (VB/VA) ≥ 1.5

D oAB =

10 ∗ 10 −17 M 0B.5 T µ (VA VB )1/ 3 8.5 ∗10 −17 M 0B.5 T µ (VA VB )1 / 3

(1-43)

(1-44)

The accuracy of these equations is almost the same with equation (1-42). Othmer and Thaker gave an equation which is very simple and used only when the solvent B is water. D oAB =

1.11 ∗ 10 −13 µ1.1 VA0.6

(1-45)

In the equations (1-43), (1-44) and (1-45) , µ is the viscosity of the solution in kg/ms, VA and VB are the molar volumes at normal boiling temperature in m3/k-mol, T is the absolute temperature in K and D oAB is the molecular diffusivity of solute A in solvent 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-Thaker Equation 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 ethanol (A) CH3-CH2-OH VA =(2)(0.0148) + (6)(0.0037) + (1)(0.0074) = 0.0592 m3/k-mol A For water φB = 2.6 and at 15 oC µB = 1.14 cP From Wilke-Chang Equation;

D oAB =

(1.17 *10 −16 )[(2.6)(18)]0.5 (273 + 15) (0.0592)

0.6

−3

= 1.10 *10 −9 m 2 / s

(1)(1.14 *10 )

From Othmer-Thaker Equation; 1.11 * 10 −13 = 1.05 * 10 −9 m 2 / s D oAB = −3 1.1 0.6 (1)(1.14 * 10 ) (0.0592)

[

]

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


Table 1-4. Experimentally determined molecular diffusivities in liquids at 1 atm. Solute (A)

Solvent (B)

Acetic acid Acetic acid Acetic acid Acetic acid Acetone Acetylene Ammonia Ammonia Ammonia Benzene Benzene Benzoic acid Benzoic acid Benzoic acid n-Butanol Butyric acid Carbon dioxide Chloroform Ethanol Ethanol Ethanol Ethanol Ethanol Ethanol Formic acid Formic acid Glycerin HCl HCl Hydrogen Iodine KCl KCl Methanol Methanol Oxalic acid Oxygen Oxygen Phenol n-Propanol Propionic acid Tartaric acid Toluene Water Water

Acetone Benzene Water Water Water Water Water Water Water Chloroform Water Acetone Benzene Water Water Water Water Ethanol Benzene Chloroform Water Water Water Water Benzene Water Water Water Water Water Benzene Ethylene glycol Water Water Water Water Water Water Ethanol Water Water Water n-Hexane Ethanol Glycerin

Temperature(oC) 25 25 9.7 25 25 25 5 12 20 15 25 25 25 25 15 25 25 20 15 15 10 10 10 15 25 25 25 10 10 25 25 25 25 25 15 25 18 25 25 15 25 25 25 25 25

Concentration (k-mol A/m3) 0 0 0.05 0.05 0 0 3.5 1.0 1.0 0 0 0 0 0 0 0.05 0 2 0 0 3.75 0.05 0 0 0 0 0 9 2.5 0 0 0.05 0.05 0 0 0 0 0 0 0 0.05 0 0 0 0

DAB *109 m2/s

3.31 2.09 0.769 1.26 1.28 1.78 1.24 1.64 2.30 2.51 1.0 2.62 1.38 1.21 0.77 0.92 2.0 1.25 2.25 2.20 0.50 0.83 0.84 1.0 2.28 1.52 0.94 3.3 2.5 6.3 1.98 0.119 1.87 1.24 1.28 1.61 1.98 2.41 0.89 0.87 1.01 0.80 4.21 1.13 0.021


Table 1.5 Molar and atomic volumes at normal boiling temperature Atom Bromine Carbon Chlorine

AtomcicVVolume Atomi olume 3 m33/100 /kg atom*10 m 0 atom×103 27.0 14.8 24.6 21.6 27.4 8.7 3.7 37.0 46.5-50.1 19.0

(as R-CHCl-R) (end as R-Cl)

Chromium Fluorine Hydrogen Iodine Lead Mercury Nitrogen

Oxygen

1 double bond 1 primary amine 1 secondary amine (outside below) 1 double bond in carbonyl bonded to other elements aldehyde and ketons methyl esters methyl ethers ethyl esters and ethers higher ester and ethers acids (-OH) bonded to S, P, N

15.6 10.5 12.0 7.4 7.4 7.4 9.1 9.9 9.9 11.0 12.0 8.3 27.0

Phosphorous Rings 1 three members 1 four members 1 five members 1 six members 1 naphthalene 1 anthracene Silicon Sulfur Tin Titanium Zinc Molecule Air Br2 Cl2 CO CO2 COS H2 H2O

Molar Volume, V m3/k-mol ×103 29.9 53.2 48.4 30.7 34.0 51.5 14.3 18.8

Molecule H2S I2 N2 NH3 NO N2O SO2 O2

-6.0 -8.5 -11.5 -16.0 -30.0 -47.5 32.0 25.6 42.3 35.7 20.4 Molar Volume, V m3/k-mol ×103 32.9 71.5 31.2 25.8 23.6 36.4 44.8 25.6


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 proposed the following empirical equation by analysing the experimentally measured diffusivities in concentrated solutions. ⎛ ∂ lnδ A ⎞ ⎟ D AB µ m = (D oAB µ B ) x B (D oBA µ A ) x A ⎜1 + (1-46) ⎝ ∂lnx A ⎠ This equation is used to estimate the molecular diffusivity of solute A in concentrated solutions whose solvent B does not associate. In the equation, D oAB is the molecular

diffusivity at infinite dilution in B and D oBA is the molecular diffusivity of B at infinite dilution in A. µ m , µ A , µ B are viscosities of the solution, solute A and the solvent B respectively.The activity coefficient δ A 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) p = po x A A A for non-ideal solutions p = δ po x A A A A y P δ A = oA pA x A A graph of log δ A versus logxA is first constructed, then the derivative dlog δ A /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. λ λ Z +Z RT λ λ Z + Z = 8.91*10 T D = (1-47) F λ +λ ZZ λ +λ ZZ o

AB

2

o

o

+

−

+

o

o

+

−

−

+

−

−14

o

o

+

−

+

o

o

+

−

−

+

−

D oAB

where F is the Faraday constant in A s/g-equiv., is molecular diffusivity at infinite dilution in m2/s, λ o+ and λ o− 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 dilute for sulphuric acid. .Remember that sulphuric acid is ionized in water as H+ and SO =4 . Hence Z+ = 1 and Z_= 2.


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

D AB = (8.91 * 10 −14 )(293)

(349.8)(80) (1 + 2) . (349.8 + 80) (1)(2)

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

Table 1-6. Ionic conductances of some ions at infinite dilution Cation Ag+ H+ Li+ Na+ K+ NH −4

Ca+2 Cu+2 Mg+2 Zn+2

λo+

Anion λo− 61.9 Br- 78.4 349.8 Cl- 76.35 38.7 Cl O 3− 64.6 50.1 ClO −4 67.6 73.5 F55.4 73.6 I 76.8 − 59.5 NO 3 71.46 56.6 OH- 198.6 53.0 CO 3−2 69.6 52.8 SO −42 80.0

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: ⎛ D AB µ ⎞ ⎛D µ⎞ ⎜ ⎟ = ⎜ AB ⎟ ⎝ T ⎠ T 2 ⎝ T ⎠ T1

(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 n

calculated from the flux equations derived for binary solution by using ∑ N i in place i=A

of NA+ NB and by replacing DAB with effective diffusivity, DAef. Perkin and Geankoplis have given the following equation for the effective diffusivity of solute A, which is dilute in the strong solutions of B and C. o o 0.8 0.8 D oAef µ 0.8 (1-49) m = x B D AB µ B + x C D AC µ C µ m , µ B and µ C are the viscosities of solution, components B and C . 1.4 CONTINUITY EQUATION FOR A BINARY MIXTURE:

A general equation for mass transfer of a component of a binary mixture of A+B can be derived by considering all the possible changes such as molecular diffusion, bulk


z

(nAx)x

(nAx)x+∆x

∆z

y

T

∆y

∆x

x Fig.1.7 Differential volume in mixture

flow, chemical reaction at unsteady-state. Let us consider a differential volume, as shown in Fig. 1.7, within a binary fluid mixture and write the mass balance for component A for this differential volume: Mass rate of inflow of A - mass rate of outflow of A + mass rate of formation of A = mass rate of accumulation of A.

Let us find the mathematical equivalents of these expressions and substitute them into equation. Assume that component A enters this differential volume through three faces with a common corner at T and leaves it through other three faces. Hence, rate of mass flow of A in: and rate of mass flow of A out: (n Ax ) x ∆ y ∆ z + (n Ay ) y ∆ x ∆ z + (n Az ) z ∆ x ∆ y (n Ax ) x + ∆x ∆ y ∆ z + (n Ay ) y + ∆y ∆ x ∆ z + ( n Az ) z + ∆z ∆ x ∆ y , where n Ax is the total mass 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 rA ∆x ∆y ∆z . On the other hand, since total mass of component A in the volume is ∆ x ∆ y ∆ z ρ A , where ρ A (kg A/m3) is the mass density of component A, mass rate of accumulation of component A in differential volume is then, ∆ x ∆ y ∆ z ( ∂ρ A / ∂ θ) , where θ(s) shows the time. If all these values are substituted into the above-given mass balance expression:

{[

(n

Ax

)

x + ∆x

− (n

Ax

)

x

]∆y ∆z + [(n Ay )

y + ∆y

− (n

Ay

)

y

]

[

∆x ∆z + (n

Az

)

z + ∆z

− (n

Az

)

z

]∆x ∆y}+

∆x ∆y ∆z ( ∂ρ /∂t ) = r ∆x ∆y ∆z A A

(1-50) is 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;


⎛ ∂n Ax ∂n Ay ∂n Az ⎞ ∂ρ A ⎟+ ⎜⎜ = rA + + ∂z ⎟⎠ ∂θ ∂y ⎝ ∂x is obtained. Similar analyses for component B gives: ⎛ ∂n Bx ∂n By ∂n Bz ⎞ ∂ρ B ⎟⎟ + ⎜⎜ = rB + + ∂ θ x y z ∂ ∂ ∂ ⎠ ⎝ If these two equations are summed, ∂(n

A

(1-51)

(1-52)

∂(n + n ) ∂(n + n ) +n ) A B y B x + A B z + ∂ρ = + ∂z ∂θ ∂y ∂x

0

(1-53)

is found. This gives total mass balance for the mixture. Note that ρ = ρ A + ρ B and rA + rB = 0 , where ρ is the mass density of the mixture. It follows from equation (1-11) that ( n A + n B )z = ρ u z . If the derivative of this with respect to z is taken, ∂( n A + n B )z ∂u ∂ρ = ρ z + uz is obtained. If similar equations for x- and y-directions ∂z

∂z

∂z

are written and then derivatives of these are taken with respect to y and z; ∂( n A + n B )y ∂u y ∂( n A + n B )x ∂u ∂ρ ∂ρ and =ρ are also obtained. + uy = ρ x + ux ∂x

∂x

∂x

∂y

∂y

∂y

Now, if all these equations are substituted into equation (1-53) finally to, ∂u ⎛ ∂u ∂u ⎞ ∂ρ ∂ρ ∂ρ ∂ρ ρ⎜⎜ x + y + z ⎟⎟ + u x =0 (1-54) + + uz + uy ∂z ⎠ ∂y ∂z ∂θ ∂y ∂x ⎝ ∂x 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: ∂u x ∂u y ∂u z =0 + + ∂z ∂y ∂x

(1-55)

This equation is known as “continuity equation for incompressible fluids”. Now, return back to equation (1-51). From equation (1-10), ∂ρ A ∂u ∂n Az ∂j Az is obtained. If similar equations to the equation (1-10) + ρ A z + uz = ∂z

∂z

∂z

∂z

are written for x- and y-directions and these equations are differentiated with respect to x and y, and all these three equations are substituted into equation (1-51) ∂j ⎛ ∂j ⎞ ∂ρ ∂ρ ∂ρ ∂ρ ∂j ⎛ ∂u x ∂u y ∂u z ⎞ Ay Az ⎟ A ⎜ Ax A A A ⎟ +ρ ⎜ + + + ux + uy + uz + + =r (1-56) + A⎜ ⎟ ⎜ ∂x ⎟ A ∂θ ∂ ∂ ∂ ∂ ∂ ∂ ∂ y ∂ z z y x z x y ⎝ ⎠ ⎝ ⎠ is finally obtained. When the density of the mixture is constant, this equation simplifies to: ∂j ⎛ ∂j ∂j ⎞ ∂ρ ∂ρ ∂ρ ∂ρ A (1-57) + u x A + u y A + u z A + ⎜⎜ Ax + Ay + Az ⎟⎟ = rA ∂z ⎠ ∂y ∂z ⎝ ∂x ∂y ∂x ∂θ By dividing both sides of this equation with MA, the molecular weight of component A, and by noting that; ρ A / M A = c A , rA / M A = R A and j A / M A = J A ; ∂J ⎛ ∂J ∂J ⎞ ∂c ∂c ∂c ∂c A + u x A + u y A + u z A + ⎜⎜ Ax + Ay + Az ⎟⎟ = R A ∂z ⎠ ∂y ∂z ⎝ ∂x ∂y ∂x ∂θ

(1-58)


is obtained. Since J Ax = −D AB (∂c A / ∂x ) then; ∂J Ax / ∂x = −D AB (∂ 2 c A / ∂x 2 ) . If this and equivalents of ( ∂J Ay / ∂y ) and ( ∂J Az / ∂z ) are all substituted into equation (1-58) finally, ⎛ ∂ 2c ∂ 2cA ∂ 2cA ⎞ ∂c ∂c ∂c ∂c A ⎟ + RA (1-59) + + u x A + u y A + u z A = D AB ⎜⎜ 2A + 2 2 ⎟ x y z ∂ ∂ ∂ ∂z ∂y ∂x ∂θ ⎠ ⎝ is found. This equation simplifies to the equation below, when the velocity of the mixture is zero and no chemical reaction takes place: ⎛ ∂ 2cA ∂ 2cA ∂ 2cA ⎞ ∂c A ⎟ (1-60) + = D AB ⎜⎜ 2 + ∂z 2 ⎟⎠ ∂y 2 ∂θ ⎝ ∂x This equation is known as Fick’s second law equation. Equation (1-59), which is continuity equation for component A in a mixture of A+B, is also known as “general equation for mass transfer”. Solution of this equation for any geometry under specified initial and boundary conditions gives the concentration profile of component A, from which mass transfer flux of this component can easily be calculated. 1.5 MASS TRANSFER BY MOLECULAR DIFFUSION IN SOLIDS:

Transfer of gas, liquid and solid by molecular diffusion in solids play important roles in some engineering applications. As example leaching, adsorption, drying of solids, separation of solutions by solid membranes and chemical reactions taking place in the pores 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 still be used successfully. While in some solids diffusion is independent of the structure of the solid, at the others this is strongly influenced by the nature of the solid. So, these two situations will be dealt with separately. 1.5.1 Diffusion that is Independent of the Nature of the Solid: This situation is met, when the solute dissolves and forms a homogeneous solution with the solid. Although diffusion mechanisms are quite 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 the diffusivity is independent of the concentration, which are realistic cases in many engineering applications, dc N A = − D AB A (1-61) dz can be written. Integration of this equation under steady-state conditions with constant S yields to: D (c − c ) N A = AB A1 A2 (1-62) z When the mass transfer area changes with z, NA is first replaced with N A /S m 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,


NA =

2π l D AB (c A1 − c A2 ) ln (r2 /r1 )

(1-63)

is obtained, where r1, r2 are the inner and outer radii and l is the length of the pipe. 1.5.1.2 Unsteady-State Diffusion: Many mass transfer processes occurring in solids proceed 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 solute in the solid. This equation is solved for various geometries with given initial and boundary conditions. Diffusion through a slab of 2a, 2b, 2c dimensions: Let us first assume that small surfaces of the slab, which are the surfaces normal to the x and y-directions, are insulated against mass transfer. Assume that this solid, whose initial solute concentration is uniform throughout at c Ao , 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 is the the solid had remained in the solution infinitely long time). Thus, c − c Ao

As

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

Concentration

A

A

cAo

cA

cA cAs

z -a

a

o

Distance from the center

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

prolonged. Hence, the ratio of ( c A - cAs)/( c

Ao

−c

As

) shows the fraction of solute A

that has not transferred and remained in the solid at the end of time θ. On 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:


∂ 2cA ∂c A = D AB ∂z 2 ∂θ

(1-64)

This equation must be solved under the following initial and boundary conditions: θ<0 -a < z < a cA= cAo = constant 0<θ<∞ z=±a cA = cAs = constant θ>0 -a < z < a cA = cA There are different techniques for the solution of partial differential equations such as above. Method using Fourier transformations yields to: c − c As 1 −mβ 1 8 ⎛D θ⎞ = f ⎜ AB2 ⎟ = 2 (e −mβ + e −mβ + e + .........) = Fa (1-65) F = A 25 9 π cAo − c As ⎝ a ⎠ 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 1

1

1

1 2a

Fa , Fb , Fc (slab)

2c 2b

Fr (cylinder)

0.1 2a 2c

Fa Fb Fc Fr Fs

Fs (sphere)

0.01

x

2a

y z

0.001 0

0.1

β1=

0.2

0.3

D AB θ a2

β2=

0.4

D AB θ b2

0.5

β 3=

0.6

0.7

D AB θ c2

Fig.1.9 Values of Functions Fa, Fb, Fc, Fs and Fr 35 No part of this e-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

equation (1-65) are found for these surfaces. Finally, the equation for mass transfer from all the surfaces of the slab becomes; c − c As ⎛D θ⎞ ⎛D θ⎞ ⎛D θ⎞ (1-66) F = A = f ⎜ AB2 ⎟ f ⎜ AB2 ⎟f ⎜ AB2 ⎟ = Fa Fb Fc cAo − c As ⎝ a ⎠ ⎝ b ⎠ ⎝ c ⎠ D θ D θ β 3 = AB2 . Functions Fb and Fc are also given in Fig.1.9 where β2 = AB2 b c Diffusion through a cylindrical solid: If the solute A diffuses through a cylindrical solid of radius a, length 2c, to the solution surrounding the solid, Fick’s second law equation is first written in the cylindrical coordinates and then solved with suitable initial and boundary conditions. The function Fr, which gives the concentration profile of solute A in radial direction is thus obtained. The profile in x-direction is given by the same function obtained for the slab. Thus, for mass transfer from all the surfaces of cylinder, c − c As ⎛D θ⎞ ⎛ D θ⎞ (1-67) = f ⎜ AB2 ⎟ f ′ ⎜ AB2 ⎟ = Fc Fr F = A cAo − c As ⎝ c ⎠ ⎝ a ⎠ is written. Function Fr was also added to Fig.1.9.

Diffusion through a solid sphere of radius a: When the solute A transfers from a solid sphere into a solution surrounding the solid, Fick’s second law equation is first written in the spherical coordinates and then solved with the appropriate initial and boundary conditions. Result is; c − c As ⎛D θ⎞ (1-68) F = A = f ′′ ⎜ AB2 ⎟ = Fs cAo − c As ⎝ a ⎠

Function Fs is also plotted in Fig.1.9. The equations given above can also be used, when solute A transfers from solution to the solid. In the slab and cylinder geometries mass transfer is assumed to take place from the both faces in a direction. If this is not the case (mass transfer takes place from only one of the faces in one direction), to the denominator of β in this direction 4 is added, because length of the diffusion path, in this case, is twice. Example-1.9) Unsteady-State Diffusion in Solid

Cotton seeds which contain 18 kg oil per m3 of solid and are pressed in the form of slabs with 12 mm x 8 mm x 3 mm dimensions, are immersed in hexane to extract the oil. Calculate their oil content after 1 hour of contact. The solution is well mixed so that resistance to mass transfer is only in the solid phase. Molecular diffusivity of oil in solid at the operating conditions is 4.1*10-10 m2/s. Givens : c Ao = 18 kg oil / m 3 of solid 2a = 3 mm, 2b = 8 mm, 2c = 12 mm, θ = 1 hour

cAs = 0 DAB = 4.1*10-10 m2/s

Asked : c Ao = ?


For mass transfer from all the surfaces of the slab, c A − c As ⎛ D θ⎞ ⎛ D θ⎞ ⎛ D θ⎞ = f ⎜ AB2 ⎟ f ⎜ AB2 ⎟f ⎜ AB2 ⎟ = Fa Fb Fc c Ao − c As ⎝ a ⎠ ⎝ b ⎠ ⎝ c ⎠ D θ ( 4.1*10 −10 )(1)(3600) From Fig.1-9 β 1 = AB2 = = 0.656 a (1.5 *10 −3 ) 2 D θ (4.1*10 −10 )(1)(3600) From Fig.1-9 β 2 = AB2 = = 0.093 b (4 *10 −3 ) 2

F =

β3 =

D AB θ (4.1 *10 −10 )(1)(3600) = = 0.041 c2 (6 *10 −3 ) 2

From Fig.1-9

(1-66) Fa = 0.16 Fb = 0.63

Fc = 0.78

cA − 0 = (0.16)(0.63)(0.78) = 0.0786 18 − 0

c A = 1.42 kg oil / m 3 of solid

is found.

1.5.2 Diffusion that is Dependent on the Nature of the Solid: As it has been seen, the Fick’s first and second law equations can be used for the calculation of mass transfer flux, when the diffusion is independent of the nature of the solid. On the other hand, in some cases, the mass transfer is strongly influenced from the type and the size of the pores in the solid, and the terms showing the effects of these appear in the flux equations. 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, which are cA1 and cA2, are different at two faces such as (cA1> cA2). As a result of this, solute A transfers from face-1 to face-2, through the

A A+B A+B A z 1

2

Fig.1.10 Diffusion of liquids through porous solid

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 solid. With these facts in consideration, total molar flux of A, neglecting the bulk flow is written as: c −c ε N A = 2 D AB A1 A2 (1-69) z 2 − z1 kt


where ε and kt, which must be found experimentally, are the void fraction of the solid and the tortuousity factor. Notice that mass transfer area in this case, is εSm. One of the tortuousity factors accounts for the fact that true mass transfer path 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 corrects it. DAB, which is the molecular diffusivity of solute A in solvent B, is not influenced from the solid. This DAB may be grouped with ε and kt in a way given by equation (1-70) that the resultant diffusivity is known as effective diffusivity for the system. DAef=(ε/ k 2t ) DAB (1-70) 1.5.2.2 Diffusion of Gases in Solids: Depending upon the relationship between the diameter of the pores in the solid and the mean free path of the gas molecules, transfer of gases in the porous solids takes place at different mechanisms. The mean free path of the gas molecules is given by; 3.2 µ A λA = P

⎛ RT ⎜⎜ ⎝ 2π M A

⎞ ⎟⎟ ⎠

0.5

(1-71)

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 (pA1 - pA2) 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 Knudsen diffusion coefficient DKA(m2/s) is given as; ⎛ T ⎞ ⎟ (1-72) D KA = 48.5 d ⎜ ⎝ MA ⎠ where, d(m) is the diameter of the pore, which can be calculated from 0.5

pA1

pA2

NA (a)

NA (b)

NA (c) Fig.1.11 Mass transfer along the pore (a) by Knudsen diffusion, (b) by Molecular diffusion, (c) by Transition diffusion 38 No part of this e-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

4ε (1-73) Sk ρ b 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: dc D dp A D P = KA (y A1 − y A2 ) (1-74) N A = − D KA A = − KA dz RT dz RTl d=

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 NKn =

λ d

(1-75)

is greater than 10. 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, D dy A N A = − AB + y A (N A + N B ) (1-76) RT dz By defining NR=NA/(NA+NB) , the equation (1-76) can also be written as; D AB P dy A D AB dp A NA = − =− (1-77) (1 − y A /N R ) dz (1 − y A /N R ) dz Upon integration of this equation at steady-state with S=constant, finally; N − y A2 D P (1-78) N A = N R AB ln R N R − y A1 RTl is obtained. Equation (1-78) is valid, when Knudsen number < 0.01. 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 in the momentum of A molecules, when they travel 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; RT − (dp A ) K A c = N A dz A c (1-79) D KA Similarly, momentum loss due to the molecule-molecule collision is found from equation (1-77) as; RT − (dp A ) M A c = N A (1 − y A /N R ) dz A c (1-80) D AB 39 No part of this e-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

Hence, total momentum loss of the gas is obtained by summing the two equations above, RT RT -(dpA) Ac = N A dz A c + N A (1 − y A /N R ) dz A c D KA D AB with rearranging, ⎤ P dy A ⎡ 1 (1-81) NA = − ⎢ ⎥ ⎣ (1 − y A /N R )/D AB + 1/ D KA ⎦ RT dz and upon integration of this equation at steady-state finally, ⎡1 − y A2 /N R + D AB /D KA ⎤ D AB P NA = NR ln ⎢ (1-82) ⎥ RTl 1 y /N D /D − + A1 R AB KA ⎦ ⎣ is obtained. This equation reduces to the equation (1-74) at low pressures, and to the equation (1-78) at high pressures.

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 oxygen. Molecular diffusivity of oxygen in carbon monoxide at 1 atm. and 20 oC is 7.84*10-5 m2/s and viscosity of oxygen at operating conditions is 0.026 cP. From eqn.(1-71) 0 .5

(3.2)(0.026)(1*10 −3 ) ⎡ (8314)(400) ⎤ λA = = 2.11 *10 − 6 m (0.05)(1.013 *10 5 ) ⎢⎣ (2π)(32) ⎥⎦

From eqn.(1-75) N Kn =

(2.11*10 −6 ) = 0.528 (4)(1 *10 − 6 )

As 0.01
NA NA = =2 N A + N B N A + (−0.5) N A 0.5

⎛ 400 ⎞ −4 2 D KA = (48.5)(4)(1*10 −6 )⎜ ⎟ = 6.86 *10 m / s ⎝ 32 ⎠ 1.75 ⎛ 1 ⎞ ⎛ 400 ⎞ −5 −3 2 D AB = (7.84 * 10 ) ⎜ ⎟ = 2.71 * 10 m / s ⎟⎜ 0 . 05 273 20 + ⎠ ⎠⎝ ⎝

D AB (2.71 *10 −3 ) = = 3.95 D Kn (6.86 *10 − 4 )

From equation(1-82) NA =

(2)(2.71 *10 −3 )(0.05)(1.013) ⎡1 − (0.02 / 2) + (3.95) ⎤ −7 2 ln ⎢ ⎥ = 8.4 *10 k − molA / m s (0.083)(400)(80 *10 −3 ) 1 ( 0 . 10 / 2 ) ( 3 . 95 ) − + ⎦ ⎣

is obtained.


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

SoluteA H2 O2 N2 CO2 H2

H2S

Solid B Vulcanized rubber

Vulcanized neoprene

Nylon Cellulose acetate Polyvinyl butyral Polyvinyl trifluoroacetate Mylar A Saran

Air Air H2O H2O He

Newspaper Leather Paraffin Cellophane Pyrex Glass

Air He H2

Porcelain SiO2 Ni

H2 CO Bi Cd Al

Fe Ni Pb Cu Cu

Temp., o C 25 25 25 25 0

Permeability PM cm3 A( N.T.P) cm 2 s (atm. / cm) 0.342 10-6 0.152 10-6 0.054 10-6 1.003 10-6 -

Solubility,s

Molecular

cm 3 A( N.T.P) Diffusivity,DAB (m2/s) cm 3 B atm. 0.040 0.070 0.035 0.900 0.065

8.55 10-10 2.17 10-10 1.54 10-10 1.11 10-10 0.37 10-10 1.03 10-10 1.80 10-10 4.81 10-10 5.0 10-14 1.6 10-14

17 27 46.5 30 30

2.60 10-9 2.56 10-9

0.051 0.053 0.050 5.20 16.0

60 0 30 30

4.64 10-9 2.25 10-8 5.04 10-8 2.04 10-9

9.1 15.0 8.0 0.3

5.1 1.5 6.3 6.8

10-14 10-13 10-13 10-13

0.2 11.0 2.8 2.8 3.9 1.3

2.9 1.3 6.2 9.6 2.9 6.0

10-12 10-15 10-14 10-15 10-15 10-13

60 0 60 30 45 75 25 25 23 38 0 20 100 25 20 85 125 165 20 950 20 20 20

5.80 10-9 1.43 10-10 1.74 10-9 2.69 10-10 1.13 10-9 7.80 10-9 0.357 0.152-0.684 0.160 10-6 0.91-1.82 10-6 2.81 10-11 4.86 10-11 20.06 10-11 0.81 10-11

0.01 0.202 0.194 0.192

2.4-5.5 10-14 1.16 10-12 3.4 10-12 10.5 10-12 2.59 10-13 4.0 10-12 1.1 10-20 2.7 10-19 1.3 10-34


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 pressure, there is always the relationship Σ Ni = 0. This is because of the fact that at both ends of the pore total pressure is constant. For a binary system this leads to: NA = - NB. Hence, it follows from this that in a closed system at constant pressure mass transfer always takes place under “equimolar counter transfer conditions”. On the other hand in an open system there is always the following relationship between the fluxes, regardless of whether mass transfer takes place by Knudsen, molecular or transition diffusion. ΣNi √Mi = 0 For a binary system this gives NA√ MA = - NB √ MB. As stated before, different effects in other cases may govern the relationships between the fluxes. 1.5.4 Molecular Diffusivity of Gases in Solids and the Permeability: Usually, in the diffusion of gases through solids, the permeability, instead of molecular diffusivity, is used. The permeability, PM is defined as the volume of the gas as cm3, measured at 1 atm. and 0 oC, that passes through 1 cm2 surface of the solid of 1 cm thick at 1 second under a pressure difference of 1 atm. If the solubility of the gas in the solid is s (cm3 gas/cm3 solid atm.) the relationship among permeability, molecular diffusivity and the solubility is given as; PM = 104 DAB s (1-83)

Hence, if the solubility of a gas in solid and the permeability of the solid for this gas are known, the molecular diffusivity of this gas through this solid can easily be calculated from the equation above. In Table.1.7, the solubilities, the permeabilities and molecular diffusivities for some gas-solid pairs are given.

PROBLEMS 1.1 Across a gas film of thickness 0.2 mm, components A and B are diffusing in opposite directions at 2 bars and 25 oC. Under steady-state conditions, total molar flux of A is twice that of B. Partial pressures of A at each side of the film are 0.40 bar and 0.10 bar. Calculate the total molar fluxes of A and B. Molecular diffusivity of A in B at 0 oC and 1 bar is 2*10-5 m2/s. [ Ans. NA= - NB = 1.2 Equimolar counter transfers of N2 and CO are occurring by molecular diffusion at steady-state in a tube 0.11 m long. The partial pressures of N2 at each end are 80 mmHg and 10 mmHg. a) Calculate the total fluxes of the components as kg/m2s at 298 K and 1 atm. b) Repeat the calculations at 473 K and 1 atm. Do the fluxes increase? c) Repeat the calculations at 298 K and 3 atm. Do the fluxes change? [Ans. a) NA = 7.36*10-7 k-mol A/ m2s ] 1.3 In a binary laminar gas mixture, equimolar counter transfers of A and B occurs at 1 bar pressure. The molar concentrations of A at point one and point two, which are 2 mm apart, are measured as 0.04 k-mol/m3 and 0.08 k-mol/m3 respectively. Temperature in the gas phase does not remain constant and changes with distance as T = T1e-200z , where T1 is the temperature at point 1 and its value is 305 K, z is distance in m. Calculate the total molar flux of A. Molecular diffusivity at 1 bar and 305 K is 2.5*10-5 m2/s.


[Ans. NA = 1.4*10-4 k-mol A/m2s ] 1.4 Water at 25 C 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) N A = 4.29 *10 −5 kg water / h ] o

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.

Interface Liquid phase(S+A)

Gas phase(A+B)

A

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. Reaction occurs instantaneously on the catalyst surface and product A2 diffuses back through the same gas film to the bulk gas composed of equimolar A and A2. a) 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 the 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 300oC and 2 atm total pressure, the mole percents of the components are 50, 25, and 25 for H2(A), O2(B) and H2O(C) respectively.


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, DAC=2.1*10-4 m2/s and DBC=5.9*10-4 m2/s. [Ans. a) DAef =1.47*10-4 m2/s, b) NA = 0.018 k-mol/m2s ] 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 components at the left and right side of the gas film adjacent to the solid catalyst surface are given as; yA yB yC 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 gas B at constant pressure and temperature. During an experiment under steady-state conditions, the position of liquidgas 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 yA2. a) Derive the following equation for the total molar flux A+B of A at any z B

NA =

cD AB 1 − y A ln z − z 1 1 − y A1

b) Derive the following equation for the concentration profile of B in the gas phase. z − z1

y B ⎛ y B 2 ⎞ z 2 − z1 ⎟ =⎜ y B1 ⎜⎝ y B1 ⎟⎠

2 1

z2

z1

A

For the case t= 23 oC, P = 1 bar, DAB = 8.35*10-6m2/s, z2-z1= 150 mm and vapor pressure of A at 23 oC 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 CO2 in Cl2 at 1 bar and 25 oC from Hirschfelder-Bird-Spotz equation. 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 SO =4 . 1.13 Prove that in a binary gas mixture DAB = DBA. 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).


1.16 Starting with Fick’s first law equation show that for steady-state equimolar counter diffusion of gas A through a non-isothermal medium consisting of A and B, in which T is proportional to zn, the total molar flux of gas A is given by

N Az =

(1 − n / 2)(p A1 − p A 2 ) R[(Tz / D AB ) 2 − (Tz / D AB )1 ]

where pA is the partial pressure of component A, R is the gas constant, T is absolute temperature, z is the distance in the direction of diffusion, n is a constant, and subscripts 1 and 2 refer to the two boundaries between the diffusion takes place. Assume that DAB is proportional to T1.5 and n = 2. 1.17 a).Show that for steady-state diffusion of component A from a cylindrical surface through a nontransferring gas film of B , the total molar flux of component A at the cylinder surface is given by

D AB P ln(p B 2 / p B1 ) RTa ln(1 + χ / a ) where P is the total pressure, a the radius of the cylinder, χ the thickness of the gas film, and pB1 and N A1 =

pB2 the partial pressure of component B at the cylinder surface and at the other side of the gas film. b) A cylindrical brass rod, 12 mm in diameter, is given a protective coat of lacquer. After application in an acetone solvent, the lacquer is dried in a chamber through which acetone-free air is passed at 70 o C 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? The original lacquer coating contains 10 mg of acetone per cm2 of rod surface. The vapor pressure of acetone at 10oC is 147 mbar and molecular diffusion coefficient of acetone through air may be taken as 8.0*10-6 m2/s. Assume that flow rate of B is sufficiently high so that pA2 ≈ 0. [Ans.: b) 147 sec.]


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

Molar concentration of solute A,

cA

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 A explained below, is shown in Fig.2.1. Adjacent Laminar sublayer cA1 to the surface, a thin laminar film is present, in which the mass transfer occurs by molecular Buffer layer diffusion, since no eddies can penetrate into this True turbulent core layer. Since mass transfer by molecular cA diffusion is a slow process, large concentration cA2 cA=f(z) change or drop in concentration occurs across this laminar film. Adjacent to this layer is the 0 Distance from surface, z transition or buffer layer to which some Fig.2.1 Typical concentration profile eddies can penetrate and mass transfer is by the in turbulently flowing fluids 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, c A is very close to the minimum concentration, cA2, the thickness of laminar sublayer is very small compared 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. J A = −(D AB + ε M )

dc A dz

(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. ε M 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, ε M 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 ε M =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 (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) ε M = f ( z ) is required. It was already said that this is not known yet. By defining an average ε M , 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; JA =

D AB + εM (c A1 − c A 2 ) z

(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 c A 2 ≅ c A . In the equation (2-2) z and ε M 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, k ′c (m/s), which is known as mass transfer coefficient and is determined experimentally. k ′c ≡

D AB + ε M z

(2-3)

It is obvious that all the parameters effecting DAB, ε M and z will effect the k ′c . Sodefined k ′c 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; N A = − (D AB + ε M )

dc A c A + (N A + N B ) dz c

(2-4)

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


47

N NA =

NA NA + NB

D

AB

+

εM

Z

N

c

ln

A +N

A

A

cA 2

−

cA1

B

N N

−

A +N

B

c

(2-5)

c

is obtained. Substituting k ′c from equation (2-3) gives; N NA =

NA NA + NB

k ′c c ln

N

A +N

A

A

cA 2

−

cA1

B

N N

−

A +N

B

c

(2-6)

c

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, NA =

k ′c (c A1 − c A 2 ) 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. N R − c A 2 / c k ′c = (c A1 − c A 2 ) β N R − c A1 / c (N R − c A 2 / c) − (N R − c A1 / c) = (N R − c A / c)ln β= (N − c A 2 / c ) NR N R ln R (N R − c A1 / c) N R k ′c c ln

(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 ≅ 0 ), whatever the relationship between the fluxes is. 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, (2-9) NA= k ′c (c A1 − c A 2 ) 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 k ′c is the mass transfer coefficient for “ equimolar counter transfer of A and B”. 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: N A = k ′c (c A1 − c A 2 ) = k ′G (p A1 − p A 2 ) = k ′y ( y A1 − y A 2 ) (2-10) Interrelationships between three mass transfer coefficients can be found as k ′G P = k ′y = k ′c c , by noting that yA= cA/c and cA= pA/P . 48No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

48

Similarly in the liquid solutions equation (2-9) is written as;

(2-11) N A = k ′L (c A1 − c A 2 ) = k ′x ( x A1 − x A 2 ) where k ′x = k ′L c = k ′L (ρ / M ) can easily be proven. Here, ρ and M are density and

molecular weight of liquid solution. 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, NA =

k ′c (c A1 − c A 2 ) = k c (c A1 − c A 2 ) ( x B ) ln

(2-12)

is obtained. In the gas mixtures this equation can also be written as; N A = k G (p A1 − p A 2 ) = k c (c A1 − c A 2 ) = k y ( y A1 − y A 2 )

(2-13)

where, kG= k ′G P/(p B ) ln , k c = k ′c /(y B ) ln and k y = k ′y P/(p B ) ln The relationships between three mass transfer coefficients are: kG P = ky = kc c In liquid solutions the equation (2-12) is written as; (2-14) N A = k L (c A1 − c A 2 ) = k x ( x A1 − x A 2 ) where, kL= k ′L /(x B ) ln and k x = k ′x /(x B ) ln 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 k ′c is independent of concentration, kc depends on concentration of the solution. Mass transfer coefficient equals k ′c /β , 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 k ′c and then is corrected to the desired value of NR. 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 k ′c 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.


49

Table 2-1. Mass transfer coefficients and Flux equations General equations:

NA =

β=

k ′y k ′c k′ (c A1 − c A 2 ) = x ( x A1 − x A 2 ) = ( y A1 − y A 2 ) β β β

( N R − x A 2 ) − ( N R − x A1 ) N R ln ( N R − x A 2 ) /( N R − x A1 )

NR = NA/(NA+NB)

Special cases:

Equimolar counter transfer β=1

A transfers through nontransferring B β=(1-xA)ln

Units of mass transfer coefficients in SI

Gases:

N A = k ′G (p A1 − p A 2 )

N A = k G (p A1 − p A 2 )

k-mol A /m2 s bar

N A = k ′y ( y A1 − y A 2 )

N A = k y ( y A1 − y A 2 )

k-mol A /m2 s

N A = k ′c (c A1 − c A 2 )

N A = k c (c A1 − c A 2 )

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

N A = k Y ( YA1 − YA 2 )

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

N A = k ′x ( x A1 − x A 2 )

N A = k x ( x A1 − x A 2 )

k-mol A/m2 s

N A = k ′L (c A1 − c A 2 )

N A = k L (c A1 − c A 2 )

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

Liquids:

Conversions of mass transfer coefficients:

Gas :

Liquid:

k ′c c = k ′c

(p ) P (p ) = k c B ln = k G ( p B ) ln = ky B ln = k ′y RT RT P

k ′L c = k L ( x B ) ln c = k ′L

ρ = k ′x = k x ( x B ) ln M


50

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 Liquid

z y x

Ac=b*δ b

x=0

c A = c Ao

dx

δ

Liquid film

G A

Solid plate cA

l

dSm=b*dx

S

ux,max u x ux=f(z)

c Ax

cAi Ac

cAx=f(z)

x=l

c A = cAl Liquid

Fig.2.2 Mass transfer from a gas into a laminar liquid film

the liquid film is constant at δ . The concentration of solute A in the inlet liquid is uniform and shown with c Ao . 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.


51

⎛ ∂ 2cA ∂ 2cA ∂ 2cA ⎞ ∂c A ∂c A ∂c A ∂c A ⎟ + RA + ux + uy + uz = D AB ⎜⎜ 2 + + (1-58) ∂θ ∂x ∂y ∂z ∂y 2 ∂z 2 ⎟⎠ ⎝ ∂x ⎛ ∂ 2 u x ∂ 2 u y ∂ 2 u z ⎞ ∂P ⎛ ∂u x ∂u x ∂u x ∂u x ⎞ ⎟− ⎟ = µ⎜⎜ 2 + (2-15) ρ⎜⎜ + ux + uy + uz + + ρg ∂x ∂y ∂z ⎟⎠ ∂y 2 ∂z 2 ⎟⎠ ∂x ⎝ ∂θ ⎝ ∂x 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 ∂c A /∂θ = 0 and ∂u x /∂θ = 0 . 4o) Rate of absorption of solute A in liquid B is small, this leads to u z ≅ 0 . 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 D AB (∂ 2 c A / ∂x 2 ) = 0 can be written. 6o) Flow is fully developed and hence ∂u x /∂x = 0 and ∂ 2 u x /∂x 2 = 0 . 7o) Pressure drop in flow direction is negligible, then ∂P / ∂x = 0 . 8o) All the physical properties ( ρ, µ, D AB ) are constant. Under these assumptions equation (2-15) simplifies to: µ

d 2u x dz 2

+ ρg = 0

(2-16)

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

ρg δ 2 2µ

⎡ ⎛ z ⎞2 ⎤ 3 ⎢1 − ⎜ ⎟ ⎥ = u x ⎢⎣ ⎝ δ ⎠ ⎥⎦ 2

⎡ ⎛ z ⎞2 ⎤ ⎡ ⎛ z ⎞2 ⎤ ⎢1 − ⎜ ⎟ ⎥ = u x , max ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ δ ⎠ ⎥⎦ ⎢⎣ ⎝ δ ⎠ ⎥⎦

(2-17)

is obtained, where ux and u x are the local and average velocities. The average velocity, which is written as u x = ρgδ 2 /3µ 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 u x , max = 3u x / 2 from the equation above by inserting z = 0. As for the equation (1-58), with the assumptions given above this equation also simplifies to: ux

⎛ ∂ 2c A ⎞ ∂c A ⎟ = D AB ⎜ ⎜ ∂z 2 ⎟ ∂x ⎝ ⎠

(2-18)

If ux is substituted from equation (2-17), this equation becomes ⎡ ⎛ z ⎞ 2 ⎤ ∂c ∂ 2c A u x , max ⎢1 − ⎜ ⎟ ⎥ A = D AB ∂z 2 ⎢⎣ ⎝ δ ⎠ ⎥⎦ ∂x

(2-19)

Equation (2-19) must be integrated with the following boundary conditions: at z=0 for 0 < x < l cA = cAi , at x = 0 for 0 < z < δ cA = cAo and at z = δ for 0 < x < l ∂c A / ∂z = 0 . The equation above is still complicated for integration and may be integrated at two different conditions: 52No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

52

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 1 cAl = c Al u x b dz (2-20) u x δ b ∫o If all these steps are accomplished, finally the average concentration of the outlet liquid is found as: c Ai − cAl = 0.7857 e −5.1213 η + 0.100 e −39.3 8 η + 0.03599 e −105.6 4 η + ..... (2-21) c Ai − cAo where η = 2D ABl/3δ 2 u x . 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, u x δ b d cA = k L (c Ai − cA ) b dx can be written. By separating the variables and integrating; c Al

l l d cA ⌠ u xδ ⎮ = ∫o k L dx = k L ∫o dx ⌡ c c Ai − cA Ao

kL =

ux δ c −c ln Ai Ao l c Ai − cAl

(2-22)

is obtained, where k L and k L 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), kL = or

ux δ e5.1213 η u δ D ln = x (0.241 + 5.1213η) ≅ 3.41 AB δ l 0.7857 l

kL δ = Sh av. = 3.41 D AB

(2-23)

(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

∂c A ∂ 2c A = D AB ∂ ( x / u x , max ) ∂z 2

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

time of gas-liquid,


53

∂ 2cA ∂c A = D AB (2-25) ∂z 2 ∂θ is obtained. This equation is solved with the boundary conditions: at θ= 0 for 0 < z < ∞ cA= cAo , at θ > 0 for z =0 cA=cAi and at θ ≥ 0 for z = ∞ cA= c Ao . 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: ∞ ∂ ∞ ⎛ ∂c A (z, θ) ⎞ ⌠ ∂c A −sθ e dθ = ₤⎜ c A (z, θ) e −sθ dθ = s c A (z, s) − c Ao ⎟=⎮ ∫ o ∂θ ⎝ ∂θ ⎠ ⌡ o ∂θ ∞

2 2 2 ∞ ∂ 2 c A (z, θ) ⎞ ⎛ ⌠ ∂ c A (z, θ) e −sθ dθ = D ∂ c (z, θ)e −sθ dθ = D d c A (z, s) D ₤ ⎜ D AB = ⎟ AB ⎮ AB A AB ∂z 2 ⌡ ∂z 2 ∫o dz 2 ∂z 2 ⎠ ⎝ o

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

−

s D AB

c A ( z, s ) = −

c Ao D AB

(2-26)

is obtained, the solution of which is : c A ( z, s ) = K 1 e

− z s / D AB

+ K 2 ez

s / DAB

+

c Ao s

(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)= ₤ ( cAo ) = cAo / s . z→∞ c Ao c = K1 (0) + K 2 e ∞ + Ao is written, K2 = 0 is s s found. From the first boundary condition cAi/s = K1 (1) + cAo / s and K1 = (c Ai − c Ao ) / s

Since from second boundary condition,

is written. By substituting all these into equation (2-27) finally, c A (z, s) = (c Ai − cAo )

e

− z s / D AB

s

+

cAo s

(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: z c A (z, θ) = cAo + (c Ai − cAo ) erfc (2-29) 4D ABθ where, erfc (z / 4D AB θ ) = 1 − erf (z / 4D AB θ ) is error function complementary. As it is known, error function, which is defined by erf u =

2 u −z2 ∫ e dz , is an infinite series π o

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, c A (z, x ) = c Ao + (c Ai − c Ao ) erfc

z 4D AB x / u x , max

(2-30)


54

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: z (2-31) c Al − c Ao = (c Ai − c Ao ) erfc 4D AB l / u x ,max Average concentration cAl is found as, (c Ai − cAo ) 1 ∞ c Al − c Ao = ( c − c ) b dz = l A Ao ∫ δb o δ

∫

∞

o

erfc

z dz 4D AB l / u x ,max

(c Ai − cAo ) 4D ABl/u x,max (2-32) δ π 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, u δ c −c k L = x,max ln Ai Ao (2-33) l c Ai − cAl is found. In short contact time the change in the A content of the liquid will be very small, as a result of this, c −c c −c ln Ai Ao ≅ Al Ao can be approximated. Hence, from the c Ai − cAl c Ai − cAo combination of equations (2-32) and (2-33) finally, 4D AB u x,max 6D AB Γ kL = = (2-34) πl πρδ l is obtained, where Γ is the mass flow rate of liquid per unit width of the plate (kg/ms) and is defined by & /b = ρ u x δ Γ =m (2-35) cAl − cAo =

If it is remembered that u x = ρgδ 2 /3µ , from these two equations, the thickness of the liquid film δ; δ = (3µΓ / ρ 2 g )1 / 3 (2-36) is found. As it is seen, after very long mathematical manipulations, two theoretical equations for the calculation of average liquid phase mass transfer coefficient, k L , 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. N A = k L (c Ai − cA ) ln (2-37) As stated in the derivations, k L 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 k L 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 55No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

55

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, m & = (0.15 * 997) / 3600 = 0.0415 kg / s . Mass flow rate of water per unit width of plate, from equation (2-35), Γ = 0.0415/0.20 = 0.208 kg/ms Thickness of the liquid film from equation (2-36), δ = [(3 * 0.894 *10−3 * 0.208) /(997) 2 (9.81)]1 / 3 = 3.85 *10−4 m. Reynolds number, Re = 4Γ/µ = 4*0.208/(0.894*10-3) = 931. Hence, equation (2-34) is applicable. kL =

6 * (1.44 *10 -9 ) * 0.208 = 3.52 * 10 −5 m / s π * 997 * (3.85 *10 -4 ) *1.2

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 56No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

56

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 k ′c 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 DAB, density and the viscosity of the fluid ρ , µ and finally the average velocity of the fluid u x . An error in estimating the variables either in excess or less at this stage yields no result in dimensional analysis. So, k ′c = f ( D, D AB , ρ, µ, u x ) (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: a b c e k ′c = f { [D] [D AB ] [ρ] [µ] d [u x ] } (2-39) If the dimensions of all the variables are substituted, lθ −1 = l a l 2 b θ − b M c l −3c M d l −d θ − d l e θ −e (2-40) 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. e −1+ e 1+ c − e c −c and from here, From equation (2-39) k ′c = f (D D AB ρ µ u x ) ⎛ Dux k ′c D = f ⎜⎜ D AB ⎝ D AB

⎞ ⎟⎟ ⎠

e

⎛ D AB ρ ⎞ ⎜⎜ ⎟⎟ ⎝ µ ⎠

c

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

and multiplying the last two terms,

⎛ Dux ⎜⎜ ⎝ D AB

⎞ ⎛ D AB ρ ⎞ ρ D u x ⎟⎟ * ⎜⎜ ⎟⎟ = µ µ ⎝ ⎠ ⎠

finally, ⎛ ρ D ux k ′c D = α ⎜⎜ D AB ⎝ µ

⎞ ⎟⎟ ⎠

β

⎛ µ ⎞ ⎜⎜ ⎟⎟ ρ D ⎝ AB ⎠

γ

(2-41)


57

Table 2-2. Mass transfer correlations for widely used geometries No.

Geometry

Equation

Application range

Sh= 0.023 Re0.83 Sc0.33

1

4 000< Re< 60 000 0.6 100 Re < 50 000 Length of 5*105
Inside pipe flow Sh= 0.0149 Re0.88 Sc0.33 2

Flow parallel to a flat plate

3

JD = 0.664 Re −x0.5 0.25 ⎞ ⎟ ⎟ ⎠ 0.25 ⎛ Sc ⎞ ⎟ Sh= 0.027 Re x Sc 0.43 ⎜⎜ ⎟ ⎝ Sc i ⎠ ⎛ Sc ⎝ Sc i

Sh=0.037 Re 0x.8 Sc 0.43 ⎜⎜

Gas flow Parallel to a flat plate in confined duct

2 600
Sh = 3.41

Re=

Wetted-wall column ⎛ 3δ ⎞ Re Sc ⎟ ⎝ 2πh ⎠

Sh= ⎜

Type of fluid

Fluid

Fluid

2*104
JD = 0.11 Re-0.29

4

Charact. dimension

4Γ < 100 µ

100
0.5

Length of plate

Gas

Thickness of liquid film

Liquid

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

1 300< Re <8 300

Sh = 0.281 Re 0.60 Sc 0.44

400
Sh= (0.35+0.34 Re0.5 + 0.15 Re0.58) Sc0.3

0.1
Sh = Sho + 0.347 (Re.Sc0.5)0.62

1.8
5

Flow Perpendicular To cylinder

Gas Diameter of cylinder Fluid

6

Flow past single sphere

Sho=

{2 + 0.569(Gr Sc) 0.25

{

2 + 0.0254 (Gr Sc)

0.333

Gr Sc < 10 Sc

0.244

Fluid

8

Gr Sc > 10

8

0.6
7

Packed column (*)

JD=

2.06 Re − 5.75 ε

90< Re< 4 000 Sc=0.6

Gas

JD=

20.4 Re − 0.815 ε

5 000< Re < 10 300 Sc=0.6 Diameter of packing 0.0016 < Re <55 168 < Sc < 70 600

Gas

JD=

JD =

Grashof number, Gr =

1.09 − 2 / 3 Re ε 0.250 Re − 0.31 ε

3 gl ∆ρ ⎛ ρ ⎞

⎜ ⎟ ⎝µ⎠

2

; JD- factor , J D =

5 < Re < 1 500 168 < Sc < 70 600

Liquid

Liquid

Sh ; ε : Void fraction Re Sc 0.33

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


58

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; (2-42) Sh = α Re β Sc γ where 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: d 2cA =0 (2-43) D AB dz 2 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. 59No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

59

Solution of equation (2-43) with the stated boundary conditions gives the concentration profile as : ⎡ c − c AL ⎤ (2-44) c A = c Ai − ⎢ Ai ⎥z z ⎣ ⎦ L 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; ⎡ c − c AL ⎤ D AB ⎛ dc ⎞ (c Ai − c AL ) N A = − D AB ⎜ A ⎟ = − D AB ⎢− Ai ⎥= z z ⎝ dz ⎠ z = 0 ⎣ ⎦ L L

(2-45)

As mass transfer coefficient is defined by

kL =

Total molar flux NA = Molar driving force (c Ai − c AL )

(2-46)

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

kL =

D AB zL

(2-47)

is obtained. So, according to the film theory, mass transfer coefficient is proportional to D 1AB.0 .But most of the empirical mass transfer correlations shows that kL is proportional to D 2AB/ 3 . Later on, some attempts have been made to modify the original film theory so that k L α D 2AB/ 3 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. 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: ∂ 2cA ∂c A (2-48) = D AB ∂z 2 ∂θ The boundary conditions required for the solution are obtained from the statement of the theory as follows:


60

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: c A − c Ao z = 1 − erf (2-49) c Ai − c Ao 2 D AB θ B.C.

Then, instantaneous total molar flux of solute A crossing the interface becomes; D AB ⎛ dc ⎞ (2-50) N A (θ) = − D AB ⎜ A ⎟ = (c Ai − c Ao ) πθ ⎝ dz ⎠ z = 0 According to regular surface renewal model (Higbie model) average total molar flux; D AB NA = 2 (c Ai − c Ao ) (2-51) πτ and then average mass transfer coefficient, D AB kL = 2 (2-52) πτ are obtained. According to random surface renewal model (Danckwerts model) average total molar flux; (2-53) N A = D AB s (c Ai − c Ao ) and then average mass transfer coefficient, (2-54) k L = D AB s 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 D 0AB.5 . 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.

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


61

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 z x

ux Liquid B

cA

A

Q A

c A1 cAi

Sm

dSm x

x= 0

x+dx

cA 2

cAi Dissolving solid

b x= l

Fig.2.3 Determination of effective concentration difference

liquid which flows over the plate with an average velocity of u x (m/s), and a volumetric flow rate of Q (m3/s). Let us show with cA1 and cA2 (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 c A , increases with increasing x. The concentration difference, which causes mass transfer in z-direction also changes with x, is being (c Ai − c A1 ) at the inlet, (c Ai − c A ) at any x and (c Ai − c A 2 ) 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,

Q d cA = N A dSm = k L (c Ai − cA ) dSm (2-55) is written, where kL is local mass transfer coefficient and d c A is the increase in the concentration of liquid as liquid flows a distance of dx. Separating the variables and integrating gives: l

Sm = ∫ dSm o

c A2

Q ⌠ dc Q c − cA2 =− ln Ai = ⎮ k L ⌡ c Ai − cA kL c Ai − cA1

(2-56)

c A1

where k L 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 ( c Ai − cA ) av. for the whole plate,


62

Q (cA2 − cA1 ) = k L (c Ai − cA ) av. (2-57) Sm is written. The value of so-defined ( c Ai − cA ) av. can easily be found by eliminating k L between equations (2-56) and (2-57). c − cA2 Q Q (cA2 − cA1 ) = − (c Ai − cA ) av. ln Ai Sm Sm c Ai − cA1 From this, (c − cA1 ) − (c Ai − cA2 ) (2-58) (c Ai − cA ) av. = Ai (c Ai − cA1 ) ln (c Ai − cA2 ) 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, (c Ai − cA ) av. = (c Ai − cA ) ln . So, for the average flux, the following equation can be written; N A = k L (c Ai − cA ) ln (2-59) The average mass transfer coefficient k L 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. NA =

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. Average air velocity in x-direction

water (A)

x

A)

2

pipe

A

ux =

QB (30 / 3600) = =1.06 m / s A c ( π / 4)(100 *10 −3 ) 2

z l 1

ρu D (1.14)(1.06)(100 *10 −3 ) Re = x = = 6 360 µ (0.019)(1 *10−3 ) 1.75

⎛T ⎞ (D AB ) T = (D AB ) T ⎜⎜ 2 ⎟⎟ ⎝ T1 ⎠ 2

Sc =

1

air(B)

A)

1.75

⎛ 273 + 30 ⎞ ⎟ = (2.20 *10 −5 )⎜⎜ 273 + 0) ⎟⎠ ⎝ −3

= 2.64 *10 −5 m 2 / s

µ (0.019)(1*10 ) = = 0.631 (Note that in the calculations of Re & Sc numbers ρD AB (1.14)(2.64 *10 −5 ) density and viscosity of air are taken.)

From Table.2-2

Sh = 0.023 Re0.83 Sc0.33


63

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

kG

Sh =

k′c D k G RT (p B ) ln D = = 28.31 D AB D AB P

(p B ) ln (28.31)(D AB ) (28.31)(2.64 * 10 −5 ) = = = 3.0 * 10 − 4 k − mol A / m 2 s bar −3 P D RT (100 * 10 )(0.083)(273 + 30) (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) o

H A= pA2 / pA 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; N A = k G (p Ai − p A )ln = k G (poA − p A )ln Driving force for mass transfer at section-1 : (p oA − p A1 ) = p oA = 0.0424 bar Driving force for mass transfer at section-2 : (p oA − p A 2 ) = 0.0424 − 0.02544 = 0.01696 bar Average driving force for the pipe: p oA − ( p oA − p A 2 ) 0.02544 = = 0.0278 bar o 0.0424 p ln ln o A 0.01696 pA − pA2 Assume that gas is dilute in water vapor (to be checked later). Then; (p oA − p A ) ln =

kG

(p B ) ln = k G = 3.0 * 10 −4 k − mol A / m 2 s bar P

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 N A = QG (c A 2 − c A1 ) =

QG (30 / 3600) (0.02544 − 0) (p A2 − p A1 ) = (0.083)(273 + 30) RT

= 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: Sm =

N A 8.43 *10 −6 = = 1.01 m 2 N A 8.34 *10 −6

l=

Sm 1.01 = = 3.22 m. πD (π)(100 *10 −3 )

Check whether dilute gas mixture assumption is correct or not ! At section-1: p Bi = P − p oA = 1 − 0.0424 = 0.9576 bar and p B = P − p A1 = 1 − 0 = 1 bar (p B )ln =

p B − p Bi 1 − 0.9576 = = 0.979 bar ln(p B / p Bi ) ln(1 / 0.9576)

Then; (kG)1 = (3.0*10-4)(1/0.979) = 3.06*10-4 k-mol A /m2s bar At section-2 : p Bi = P − p oA = 1 − 0.0424 = 0.9576 bar p B = P − p A 2 = 1 − 0.024544 = 0.974 bar

(p B ) ln =

p B − p Bi 0.974 − 0.9576 = = 0.966 bar ln(p B / p Bi ) ln(0.974 / 0.9576)

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:


64

kG =

(3.06 + 3.10) *10 −4 = 3.08 *10 − 4 k − mol / m 2 s bar 2

Average total molar flux for the pipe : N A = k G (p oA − p A ) ln = (3.08 *10 −4 )(0.0278) = 8.56 *10 −6 k − molA / m 2 s

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. Benzoic Calculate: acid (A) a) Liquid phase mass transfer coefficient, kc b) Amount of benzoic asid dissolved in 1 hour as gr. Water(B) 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. ρu d (995)(0.12)(100 *10 −3 ) Re = x p = =16 583 µ (7.2 *10− 4 ) Sc =

µ (7.2 *10−4 ) = = 578.9 ρ D AB (995)(1.25 *10−9 )

Gr =

g d 3p ∆ρ ⎛ ρ ⎞ ⎜ ⎟ ρ ⎜⎝ µ ⎟⎠

2

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

Re Sc0.5 = (16 583)(578.9)0.5 = 398 990 Sh = 2 + 0.347 (398 990)0.62 = 1 032.2 k′ d k d Sh = c p = c p =1 032.2 D AB D AB k c = (1 032.2 )

D AB (1 032.2)(1.25 *10 −9 ) = = 1.29 *10 −5 m / s dp (100 *10 −3 )

0.31 gr A 3.1 kg A (3.1 / 122) k − mol A = = = 0.0254 k − mol A / m 3 100 ml B 1m 3 m3 N A = k c (c Ai − c A ) = (1.29 *10 −5 )(0.0254 − 0) = 3.28 *10 −7 k − molA / m 2 s

b)

c Ai = .

S m = πd 2p = (π)(100 * 10 −3 ) 2 = 0.0314 m 2 & A = N A S m M A (3600)(1000) = (3.28 * 10 −7 )(0.0314)(122)(3600)(1000) = 4.52 gr B.Ac / h m


65

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 [ Ans. b) ℓ = 7.03 m ] naphthalene in ethanol at 20oC is 0.8*10-9 m2/s. 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 k L The solubility of benzoic acid in water at 26 oC is 0.03 k-mol Ben.Ac./m3 of solution. [Ans. b) k L = 0.021 m / s ] 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;

ap =

6(1 − ε) dp

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, k L 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 k L at the experimental conditions,


66

b) Find the same k L 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: for the absorption of ammonia in water kG = 2.5 k-mol/m2h bar 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. 2/3 Assume that transfer coefficients vary with D AB , and the solutions are dilute. Take the density of [Ans.: KG = 0.70 k-mol/h m2 bar] water as 1000 kg/m3.


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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, liquidliquid, 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 gasliquid 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


68

mole fraction of ammonia in gas, y

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

o o

mole fraction of ammonia in liquid, x

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 µ LA and µ GA , at equilibrium µ LA = µ GA . 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 SO2 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

69

y

A+C

Liquid phase A+S A

xi yi

Concentrations in liquid

Concentrations in gas

Gas phase

interface

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.

x z

z Distance from interface o

Distance from interface

Fig.3.2 Concentration profiles in two phases

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


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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: NA =

k ′y

β iy

(y − yi ) =

k ′x (x i − x) β ix

(3-1)

where, k ′y and k ′x are the individual mass transfer coefficients for gas and liquid phases and β iy and β ix are the bulk flow correction terms for the same phases, which can be given by the following equations: β iy =

( N R − y) − ( N R − y i ) N R ln( N R − y) /( N R − y i )

and β ix =

(N R − x) − (N R − x i ) N R ln( N R − x ) /( N R − x i )

(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, −

k ′x / β ix y − yi = k ′y / β iy x − xi

(3-3)

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

Mole fraction of solute A in gas, y

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

slope= − P

y

k ′x / β ix . Since xi and yi k ′y / β iy

k ′x / β ix k ′y / β iy

E Equilibrium distribution curve slope= m''

yi

M

y*

R o

x

slope= m' xi

x*

Mole fraction of solute A in liquid, x

Fig.3.3 Determination of interface concentrations No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

71

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 β iy = β ix = 1 and slope of PM line is simply: − k ′x / k ′y and hence drawing of PM line is straightforward. But, if solute A transfers through non-transferring C in the gas and through nontransferring 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, (1 − y)iln and

(1 − x) iln and −

k ′x /(1 − x ) i ln , where (1-y)iln and (1-x)iln k ′y /(1 − y) i ln

are given as:

(1 − y) − (1 − y i ) (1 − x) − (1 − x i ) and (1 − x)iln = (3-4) ln(1 − y)/(1 − y i ) ln(1 − x)/(1 − x i ) 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 k ′y and k ′x , 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 K′y and K′x 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 µ G is quite different from the relationship of x and its chemical potential µ L . 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 y ∗ , the overall driving force in terms of gas (1 − y)iln =

phase concentrations for the system then becomes ( y − y ∗ ), where y ∗ is the equilibrium value of x and hence, point R(x, y ∗ ) 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 ( x ∗ − x ), where x ∗ is the measure of y in terms of liquid phase units and this is equilibrium value of y and hence, point


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E( x ∗ ;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), NA =

K ′y β ∗y

∗

(y − y ) =

K ∗x β∗x

(x ∗ − x)

(3-5)

can be written, where K′y and K′x are overall mass transfer coefficients based on gas and liquid phase respectively, β∗y ve β∗x are bulk flow correction terms for gas and liquid phases , which are defined by; ( N R − y) − ( N R − y ∗ ) β = N R ln( N R − y) /( N R − y ∗ ) ∗ y

(N R − x) − ( N R − x ∗ ) and β = N R ln( N R − x ) /( N R − x ∗ ) ∗ x

(3-6)

For dilute solutions and for “equimolar counter transfer” conditions β∗y = β∗x = 1 and hence equation (3-5) simplifies to: (3-7) N A = K ′y ( y − y ∗ ) = K ′x ( x ∗ − x ) If solute A transfers in both phases through non-transferring components as in the example above: β∗y = (1 − y)∗ln and β∗x = (1 − x)∗ln and hence equation (3-5) is written as: (3-8) N A = K y ( y − y∗ ) = K x (x ∗ − x) where, K y = K ′y /(1 − y)∗ln

and

K x = K ′x /(1 − x)∗ln . (1 − y) ∗ln and (1 − x ) ∗ln are defined

by: (1 − y) − (1 − y∗ ) (1 − x) − (1 − x ∗ ) ∗ (3-9) (1 − y) = and (1 − x) ln = ln(1 − y)/(1 − y∗ ) ln(1 − x)/(1 − x ∗ ) 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: ∗ ln

1 K ′y / β ∗y

=

1 m′ + k ′y / β iy k ′x / β ix

(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; 1

=

1 1 + m ′′ k ′y / β iy k ′x / β ix

(3-11)

K ′x / β ∗x where, m ′′ is the slope of EM line. If the equilibrium distribution curve is a straight line then, m′ = m′′ = m.

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


73

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 (310) can be neglected. Thus, 1 K ′y / β ∗y

≅

m′ k ′x / β ix

(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, 1 K ′y / β∗y

≅

1 k ′y / β iy

(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;

k′x = 3 *10−4 k − mol / m 2s

k′y =1.5 *10 −4 k − mol / m 2s.

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.


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

NA =

Then,

NA NA = = − 0.5 N A + N B N A − 3N A

Solution is by trial and error. slope of PM line;

m=−

βiy k ′x / βix k ′ βiy 3 *10 −4 βiy =− x . = . = −2 −4 k ′y / βiy k ′y βix 1.5 *10 βix βix

(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

βiy =

yi = 1.8 (0.216) = 0.389

Substitute these into eqn.(3-2)

( −0.5 − 0.32) − ( −0.5 − 0.389) =1.71 (−0.5 − 0.32) (−0.5) ln ( −0.5 − 0.389)

Then, new slope of PM line;

m = ( −2)

βix =

(−0.5 − 0.25) − (−0.5 − 0.216) = 1.466 (−0.5 − 0.25) (−0.5) ln (−0.5 − 0.216)

1.71 = −2.33 1.466

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.

b)

NA =

or

NA =

k ′y β iy

(y − y i ) =

1.5 * 10 −4 (0.32 − 0.394 ) = − 6.49 * 10 − 6 k − mol A / m 2 s 1.71

k′x 3 *10−4 (x i − x) = (0.219 − 0.25) = − 6.35 *10 −6 k − mol A / m 2s βix 1.466

(-) 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

b)

c) m' = m" = 1.8 Hence, from eqn.(3-10)

1 1 1.8 = + =11 400 + 8 797 = 20 196 ∗ −4 K′y / β y (1.5 *10 ) / 1.71 (3 *10 − 4 ) / 1.466 Percentage resistance in liquid :

m′ / k′x / βix 8 796 .100 = .100 = 43.5 % ∗ 1 / K′y / β y 20 196

Percentage resistance in vapor:

1 / k′y / βiy 11 400 .100 = 56.5 % .100 = ∗ 20 196 1 / K′y / β y


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

K′y β∗y

=

x* = y/1.8 = (0.32)/(1.8) = 0.178

K′x 1 = = 8.91*10 −5 k − mol / m 2s ∗ β x 11 220

1 = 4.95 *10 −5 k − mol / m 2s 20 196

From eqn.(3-5)

NA =

K′y ( y − y∗ ) = (4.95 *10−5 )(0.32 − 0.45) = − 6.44 *10 −6 k − mol A / m 2s ∗ βy

NA =

K ∗x ∗ ( x − x ) = (8.91*10 −5 )(0.178 − 0.25) = − 6.42 *10 −6 k − mol A / m 2 s ∗ βx

d)

x = 0.25 xi = 0.219 xB = 1- x = 1- 0.25 = 0.75 xBi =1- xi =1- 0.219 = 0.781

y = 0.32 yi = 0.394 yB = 1- 0.32 = 0.68 yBi = 1- 0.394 = 0.606 interface LIQUID (A + B)

VAPOR (A + B)

B 0.8

yB =f(z)

yB 0.6

A

0.8 xB xBi

xB =f(z)

0.6

yBi

x, xB

y, yB

yi

0.4 y 0.2

0.4

x =f(z)

x 0.2

xi

y=f(z)

0.0

0.0 z

0.0

z

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.


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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 ]


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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 cokeoven 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


78

of transfer being different. Hence, for the understanding of these two operations, gasliquid 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 t2 Pressures of A1,A2,A3 gases

A1

t1

A2

t: temperature t2 > t1

A1

t1

A3

o

o

t1

x2

x1 x3 Mole fractions of A1,A2,A3 gases in liquid

Fig.4.1 The solubilities of gases in liquid

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

79

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, p oA at the prevailing temperature and its mole fraction in the liquid solution, x. Thus Raoult’s law is written as: p A = p oA x (4-1) 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 p oA 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: 1o) 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


80

Antoine equation is , log p o = a −

b c+t

where po (mmHg) and t(oC) Solution:

vapor pressures of propane and butane at 10oC Propane(1) :

log p oA1 = 6.82973 −

813.2 = 3.678 248 + 10

p oA1 = 4762.03 mmHg

Butane(2) :

log p oA 2 = 6.83029 −

945.9 = 3.047 240 + 10

p oA 2 =1113.5 mmHg

In the gas mixtures : volume percent = mole percent, then y1=80/100=0.8 and y2=20/100=0.2 Raoult’s law : p A = p oA x ; Dalton’s law : p A = yP , combining these two : x = yP p oA Propane mole fraction in the liquid:

x1 =

y1 P (0.8)(2) = = 0.252 p oA1 (4 762.03 / 760) *1.013

y2P (0.2)(2) = = 0.270 o p A 2 (1 113.5 / 760) *1.013 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

Butane mole fraction in the liquid:

x2 =

mole/mole Propane Butane Paraffin oil

: : :

25.2 27.0 47.8 100

(1 108.8/17 014.8)*100 (1 566/17 014.8)*100 (14 240/17 014.8)*100

w/w = 6.52 = 9.20 = 84.28 100

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, p A = H A′ x (4-2) can be written. This equation is known as Henry’s law equation, as it is first given by Henry. H 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, (4-2b) pA =ΉA cA (4-2c) or as, y = mA x


81

Here, ΉA(bar m3/k-mol) and mA(-) are also known as Henry’s law constants. Interrelationships between them are; (4-3) mA= Ή’A/P = ΉA c/P 3 where, c (k-mol/m ) is the total molar concentration of liquid solution. Table 4-1. Values of Henry’s law constants for some gases in water, ( H A′ * 10-4 bar) t( oC) 0 10 20 30 40 50 60 70 80 90 100

Air 4.38 5.56 6.73 7.81 8.81 9.58 10.23 10.64 10.84 10.94 10.84

CO2 0.0738 0.105 0.144 0.188 0.236 0.287 0.345

CO 3.57 4.48 5.43 6.28 7.05 7.71 8.32 8.56 8.56 8.57 8.57

C2H6 1.28 1.92 2.66 3.47 4.29 5.07 5.72 6.31 6.70 6.96 7.01

H2 5.87 6.44 6.92 7.39 7.61 7.75 7.75 7.71 7.65 7.61 7.55

H2S 0.0272 0.0381 0.0489 0.0617 0.0755 0.0896 0.104 0.121 0.137 0.146 0.150

CH4 2.27 3.01 3.81 4.55 5.27 5.85 6.34 6.75 6.91 7.01 7.10

NO 1.71 2.21 2.67 3.14 3.57 3.95 4.23 4.44 4.54 4.58 4.60

N2 5.36 6.77 8.15 9.36 10.54 11.45 12.16 12.66 12.76 12.76 12.76

O2 2.58 3.31 4.06 4.81 5.42 5.96 6.37 6.72 6.96 7.08 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 = C e − ∆H / RT (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). S

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) Constant (bar)

20 0.73

25 0.96

30 1.23

35 1.55

What is the value of heat of absorption (solution) of ammonia in water over this temperature range ? Solution : Take ln of both sides of the eqn.(4-4)

lnΉ’A = ln C −

or,

logΉ’A = log C −

∆H s RT ∆H s 2.303 RT


82

So, slope of logΉ’A vs. 1 / T line = − ∆H s 2.303 R t (oC)

20

1 −1 (K ) T

Ή’A (bar) logΉ’A

0.2

25

30

35

0.00335

0.00330

0.003247

0.73

0.96

1.23

1.55

-0.137

-0.018

0.090

0.190

0.00341

.

logΉ’A

H

∆y

. y 0. 0.

tg θ = -tgα

0.1

.0

-0.1 0.003243

-0.2

α

θ

0.0033

x

∆x

0.0034

1 ( K −1 ) T

0.003345

.0

∆y 0 .2 − 0 . 0 = − 1 906 .8 =− ∆x (3.345 − 3.243) *10 −3 Then, ∆Hs = - (-1960.8)(2.303)(8.314) = 37 544 kJ / k-mol A absorbed slope = − tgα = −

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


83

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; p po yS = B yc = c p So + p c = P P P n S p oB = nT P

n c pc = nT P

nS + n c = nT

From these equations, nS = nc

p So

or,

P − p So

mS = M Sn c

p So

= (278)(950.7)

p So

= 264 294.6

p So

P − p So 760 − p So 760 − p So 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.


84

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 -5

-4

100

150

200

2.5*10

4.8*10

0.044

1.07

11.52

0.0087

0.167

15.30

372.62

4 067.7

ρG (kg/m )

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

p So

(mmHg)

mS (kg dibp/ton MAN) 3

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, wettedwall 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


85

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 Gas A+C Liquid solvent S

Liquid solvent S

A Column

A

Cooling liquid Liquid film

Cooling liquid

Tubes

Liquid film

Column

Cooling liquid

A Cooling jacket

Cooling liquid

A

Gas A+C

Gas A+C Solution

Solution

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

construction. The required mass transfer area (Sm), hence the number of the tubes (n) for a given absorption duty ( N A ) is computed from, N Sm = n π d i l = A (4-5) NA 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, N A 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


86

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 G2 Total material balance : G1+ L2 = G2 + L1 Solute A balance : G1y1 + L2 x2 = G2 y2 + L1x1 Inert C balance : Gs= G2 (1 - y2) = G1(1 - y1) G2 =

y2=0.005

G1 (1 − y1 ) 25 (1 − 0.05) = = 23.87 k − mol / h 1 − y2 1 − 0.005

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; S mi =

NA = n π di l NA

L2= 45 k-mol/h x2=0.0

2

tco=30 oC

l=6 m

Smi=n smi tci=25 oC

1

do

y1=0.05 G1=25 k-mol/h smi= π di l

dii

L1,x1

l

N A = G1 y1 − G 2 y 2 = (25)(0.05) − (23.87)(0.005) ao= π dol

=1.13

tube

k − mol A k − mol A = 3.14 *10 −4 s h

Total molar flux of solute A : t

NA =

K′y ( y − y∗ ) ln β∗y

Driving force for mass transfer at the bottom :

y1∗ = 0.52 x1 = 0.52(0.0245) = 0.01274

∆y1 = ( y1 − y1∗ )

∆y1 = (0.05 − 0.01274) = 0.0373

Driving force for mass transfer at the top : y∗2 = 0.52 x 2 = 0.52(0.0) = 0.0

∆y 2 = ( y 2 − y∗2 ) ∆y 2 = (0.005 − 0.0) = 0.005

( ∆y) ln = ( y − y∗ ) ln =

∆y1 − ∆y 2 0.0373 − 0.005 = = 0.0161 ln(∆y1 / ∆y 2 ) ln(0.0373 / 0.005)


87

Average driving force for the absorber :

Total molar flux of solute A : Then, required mass transfer area;

NA =

K′y 6.02 *10−4 k − mol A ∗ ( y y ) (0.0161) = 9.69 *10−6 − = ln ∗ 1 m2 s βy −4 N 3.14 * 10 S mi = A = = 32.42 m 2 N A 9.69 * 10 −6

Smi S mi 32.42 = = = 81.9 = 82 s mi π d i l π (0.021)(6) q Required heat transfer area ; A = where, Ao= n ao o U o (∆T ) ln −4 3 q = N ∆ H = ( 3 . 14 * 10 )( 120 * 10 ) = 37.68 kJ / s (= kW ) Heat load : A s Then, required number of the tubes;

n=

Driving force for heat transfer at the bottom and at the top: ∆T1= 35-25 = 10 oC , ∆T2=35-30 = 5 oC Average driving force for heat transfer for the absorber: Then, required heat transfer area ;

Ao =

Then, required number of the tubes;

n=

(∆T) ln =

10 − 5 ∆T1 − ∆T2 = = 7.2 o C = 7.2 K ln(∆T1 / ∆T2 ) ln(10 / 5)

q 37.68 *103 = = 37.38 m 2 U o (∆T) ln (140)(7.2)

Ao A 37.38 = o = = 79.4 = 80 a o π d o l π (0.025)(6)

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 Gas equipped with a liquid-spraying nozzle-group A+C at the top. As the solvent droplets formed at the nozzles rain down in the column, the gas liquid mixture flows upward coming in contact with distributor the liquid droplets, during which solute A Liquid solvent S transfers from gas to liquid. The surfaces of liquid drops act as mass transfer area; hence liquid column creating large number of small diameter drops drops at the spray nozzles is preferred from mass transfer point of view. But, because of the back-mixing, drops cannot be made very Gas small. In addition to this, drops tend to A+C 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 solution when the concentration change required in the Fig.4.3 Spray column used in gas absorption 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

88

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

89

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

Raschig Rings

Berl Saddles

Pall Rings diameter

Lessing Rings

Novalox Saddles

Torus Saddles

Intalox Saddles

columns, the modules are produced in segments to facilitate easy

Montz-Pak type: BSH

Koch-Flexipac

Sulzer-Mellapak

Fig.4.4 Types of some packings


90

Table 4-2. Properties of some randomly filled packings Packing

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

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

6

0.8 1 600 0.73 787

1.6 1 000 0.68 508

2.4 580 0.63 364

2.4 380 0.68 328

2.4 255 0.73 262

3 155 0.73 190

700 0.69 774

390

300 0.84 420

170

155 0.88 274

115 0.92 206

220 0.78 236 g s

137 0.85 186

29 410 0.73 387 P a l l R i n

Plastics Cf ε ap , m2/m3 Metal Cf ε ap , m2/m3 Flexiring Cf ε ap , m2/m3 Hy-pak Cf ε B e

Ceramic Cf ε ap , m2/m3 Ceramic Cf ε ap , m2/m3 Plastics Cf ε ap , m2/m3

900 0.60 899 I n t 725 0.75 984

a

330

r

200 0.78 623

9.5 6 4.8 65 37 95 0.71 0.74 0.78 62 125 92

110 0.87 162

57 32 83 0.90 0.92 0.95 135 103 68

52 0.90 206

25 40 0.91 0.92 128 102

70 0.93 341

48 0.94 206

20 28 0.95 0.96 128 102

78 0.92 341

45 0.94 213

22 28 0.96 0.96 131 115

S

x

a

d

76

4.8 125 0.74 148

97 0.87 341

l

240 0.63 466 l o

50

45 0.96 d l e

15 18 0.97 0.97 s

170 0.66 269 S a

110 0.69 249 d d

65 45 0.75 0.72 144 105 l e s

145 0.77 335

98 0.775 256

52 40 0.81 0.79 195 118

33 0.91 207


22

16 21 0.93 0.94 108 89

91

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


92

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


93

a

b

c

d

e

f

g

h

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


94

a

c

b

d

Fig.4.7 Hold-down plates (a,b) and Entrainment eliminators (c,d) The view of vertical cut of a randomly packed column containing all these ancillaries is shown in Fig.4.8 4.4.2 Gas Absorption in Packed Column: The calculation of height of packing of a packed absorption column is the most important step of the design. We will derive an equation 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 equal to the total material leaving the column, (4-6) G1 + L 2 = G 2 + L 1 is written.


95

gas liquid G2,y2

L2,x2 2

h=z

dh

L,x

G,y

h=0

1

G1,y1

L1,x1 Fig.4.9 Schematic representation of a packed column

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


96

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

(4-9)

G y + L 2 x 2 = G2 y 2 + L x

(4-10)

If the last equation is solved for y, y=

G y − L2 x 2 L x+ 2 2 G G

(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, G1 ≅ G 2 ≅ G = cons. and L1 ≅ L 2 ≅ L =cons. , and then equation (4-11) becomes,

y=

L L x + y2 − x 2 G G

(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), y=

G y /(1 − y 2 ) − Ls x 2 /(1 − x 2 ) Ls /(1 − x ) x+ s 2 G s /(1 − y) G s /(1 − y) L L Y = s X + Y2 − s X 2 Gs Gs

and from here, (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


97

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

equation(4-11)

y1

D

y1 y2

Y

y

T

slope=(L/G)≠cons.

x2

x1

y2

equation(4-12)

T

equation(4-13)

slope=(L/G)=cons.

x

x x1

x2

Y2

T

slope=(Ls/Gs) =cons.

X2

(b)

(a)

D

Y1

D

X1

X

(c)

Fig.4.10 Representation of Operating Lines on the graph

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 crosssectional 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, Z=

Z

∫ o

G2 y2

⌠ dh = − ⎮ ⌡

G 1 y1

L1 x 1

d (G y ) = NA Ac a v

d(L x ) ⌠ ⎮ N A a ⌡ A c v

(4-15)

L2 x2

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


98

(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: NA =

k ′y

(y − yi ) =

(1 − y) i ln

K ′y K ′x k ′x (x i − x) = (y − y∗ ) = ( x ∗ − x ) (4-16) ∗ ∗ (1 − x ) i ln (1 − y) ln (1 − x ) ln

On the other hand, ⎛G y⎞

⎛ y ⎞

G dy

G dy

s ⎟⎟ = d(G y) = d⎜⎜ s ⎟⎟ = G s d ⎜⎜ = 2 1− y ⎝1 − y ⎠ ⎝ 1 − y ⎠ (1 − y)

and

L dx L dx ⎛ Ls x ⎞ ⎛ x ⎞ ⎟ = L s d⎜ ⎟= s 2 = 1− x ⎝ 1 − x ⎠ (1 − x ) ⎝1− x ⎠

d(L x) = d⎜

(4-17)

can be written. By substituting all these into equation (4-15) and taking constants Ac, av outside the integral and furthermore assuming k ′y , K ′y ∝ G1.0 , k ′x , K ′x ∝ L1.0 and hence G/k ′y ≅ cons., G/K ′y ≅ cons., L/k ′x ≅ cons. , L/K ′x ≅ cons. (If it is looked at mass

transfer correlations k ′y , K ′y ∝ G 0.8 and k ′x , K ′x ∝ L0.8 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; y1

x1

y2

x2

G L ⌠ (1 − y) i ln dy ⌠ (1 − x ) i ln dx = Z= ⎮ k ′y a v A c ⌡ (1 − y)( y − y i ) k ′x a v A c ⎮ ⌡ (1 − x )( x i − x ) y1

=

x1

y) ∗ln

⌠ (1 − ⌠ (1 − x ) ∗ln dx dy L G = ⎮ ⎮ K ′x a v A c ⌡ (1 − x )( x ∗ − x ) K ′y a v A c ⌡ (1 − y)( y − y ∗ ) y2

(4-18)

x2

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; HG =

G , k ′y a v A c

Height of one gas transfer unit

HL =

,

Height of one liquid transfer unit

y1

⌠ (1 − y) i ln dy NG= ⎮ ⌡ (1 − y)( y − y i ) y2 Number of transfer units

L k ′x a v A c

x1

⌠ (1 − x ) i ln dx NL= ⎮ ⌡ (1 − x )( x i − x ) Number of transfer units

G , K ′y a v A c

Height of one overall gas transfer unit

x2 gas

HOG =

y1

⌠ (1 − y) ∗ln dy NOG= ⎮ ∗ ⌡ (1 − y)( y − y ) y2

liquid

Number of overall gas transfer units

HOL =

L K ′x a v A c

(4-19)

Height of one overall liquid transfer unit

x1

⌠ (1 − x ) ∗ln dx NOL= ⎮ ∗ ⌡ (1 − x )( x − x ) x2 Number of overall liquid transfer units


99

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

Mole fraction of solute A in gas, y

D

P=cons. t=cons.

y1

Operating line Equilibrium distribution curve

P

y

E yi y2

M T

y*

slope= -

k ′x /(1 − x ) i ln k ′y /(1 − y) i ln

R

0 0

x2

x

x* x1 xi Mole fraction of solute A in liquid, x Fig.4.11 Graphical integration of N values

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


100

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

y

yi

y1

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

y2

-

-

-

-

-

-

e

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

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, y1

⌠ dy N OG = ⎮ ⌡ y − y∗

(4-21)

y2

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, y=

L L x + y2 − x 2 G G

(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) y∗ = y1 ⌠ N OG = ⎮ ⎮ ⌡ y2

mG ( y − y 2 ) + mx 2 + n L

y1 ⌠ dy = ⎮ mG ⎮ y− ( y − y 2 ) − mx 2 − n ⌡ L y2

(4-23)

dy mG mG y(1 − )+ y 2 − mx 2 − n L L


101

=

=

1 mG 1− L

1 mG 1− L

mG mG )+ y 2 − mx 2 − n L L ln mG mG y 2 (1 − ) + y 2 − mx 2 − n L L y1 (1 −

⎤ ⎡ mG ( y 1 − y 2 ) + mx 2 + n ⎥ y1 − ⎢ ⎦ ⎣ L ln y 2 − (mx 2 + n )

=

( y − y∗ )1 1 ln mG ( y − y∗ ) 2 1− L

(4-24)

is obtained. If equation (4-23) is first written at top conditions and then ( y1∗ − y ∗2 ) is mG ( y1 − y 2 ) is obtained. It L y ∗ − y ∗2 ( y − y ∗ )1 − ( y − y ∗ ) 2 mG 1− = 1− 1 = L y1 − y 2 ( y1 − y 2 )

solved, by noting that [(mx2 + n ) = y ∗2 ]; ( y1∗ − y ∗2 ) = follows from this that:

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

y1 − y 2

(4-25)

( y − y ∗ ) ln,1,2

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. ( y − y ∗ ) ln,1, 2 =

( y − y ∗ )1 − ( y − y ∗ ) 2

[

ln ( y − y ∗ )1 /( y − y ∗ ) 2

]

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: ⎡ y − mx 2 1 1⎤ N OG = ln ⎢ 1 (1 − 1 / α) + ⎥ (4-26) 1 −1/ α α⎦ ⎣ y 2 − mx 2 where, α is known as absorption factor and given as α =

L/G . m

Similarly, at the desorption operation for NOL; ⎤ ⎡ x − y1 / m ln ⎢ 2 (1 − α) + α ⎥ x − y1 / m ⎦ (4-27) N OL = ⎣ 1 1− α 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

102

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 D

y1

K

P=cons. t=cons.

Operating line

Equilibrium disribution curve

yR yH y2 yI

Bisect line

R H T

F

E I

C

Draw TE=EF

0 0

x2

x1

x

Fig.4.12 Determination of NOG by Baker’s Method

By drawing the bisect line which divides the horizontal distances between operating line and equilibrium distribution curve in two equal parts everywhere, and following the same steps explained above starting at point D, the number of the overall liquid transfer units, NOL can be calculated.


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 !) a) Equation (4-12) is the operating line equation for dilute solutions. 690 690 (0.0) x + 0.005 − 300 300 y = 2.3 x + 0.005

y2=0.005

y=

Then;

L=690 k-mol/h 2

x2=0

1

G=300 k-mol/h

b) Operating line equation at the bottom of the column; y1 = 2.3 x 1 + 0.005 y − 0.05 0.03 − 0.005 x1 = 1 = = 0.0109 2.3 2.3

Then,

As it is seen, dilute solutions assumption is valid. Absorbed acetone : L(x1-x2)= 690(0.0109-0.0)= 7.51 k-mol/h Amount of solution leaving the column,

y1=0.03

L& = (690)(18) + (7.51)(58) = 12 855.6 kg / h

x1=? Then, mass percent of acetone in the solution; (7.51)(58) x& 1 100 = .100 = 3.39 % 12 855.6 7.51 c) Percentage recovery of acetone = .100 = 83.4 % (300)(0.03) d) Since the solutions are dilute, equation (4-21) or (4-25) can be used. From equation (4-21) 0.03

0.03 y 0.03 ⌠ dy dy dy ⌠ ⌠ dy ⌠ ⎮ = =⎮ = = ⎮ ⎮ ∗ − y 0 . 005 0 . 0435 y + 0.0048 − − y y y 2 . 2 x ⎮ y − 2.2( ⌡y ⌡0.005 ) ⌡0.005 ⌡0.005 2.3 1 (0.0435)(0.03) + 0.0048 is obtained. = ln = 4.51 0.0435 (0.0435)(0.005) + 0.0048 1

N OG

2

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


104

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 H L = a (L′ / µ L ) b Sc 0L.5 (4-28) 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 Raschig ring

mmxmm

a

b

9.5x9.5 12.7x12.7 25x25 38x38 50x50

0.00032 0.00072 0.00235 0.00260 0.00293

0.46 0.35 0.22 0.22 0.22

12.7x12.7 25x25 38x38

0.00146 0.00128 0.00137

0.28 0.28 0.28

Berl saddle

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: H G = a (G ′) b (L′) c Sc 0G.5 (4-29) ' ' where, 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, Raschig ring

mmxmm

a

b

c

9.5x9.5 25x25 38x38 50x50

0.62 0.56 0.69 0.89

0.45 0.32 0.38 0.41

-0.47 -0.51 -0.40 -.045

12.7x12.7 25x25 38x38

0.54 0.46 0.65

0.30 0.36 0.32

-0.74 -0.24 -0.45

Berl saddle


105

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: 1 1 m = + K ′y k ′y k ′x

(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; G G mG L = + K ′y a v A c k ′y a v A c L k ′x a v A c

is obtained. By comparing this equation with

the equations (4-19), H OG = H G +

mG HL L

(4-31)

is found. Similar analysis for HOL results in, H OL = H L +

L HG mG

(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 ∝ u1G.8 , 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.


106

L3> L2 > L1

3 L2

Log(∆P)

L3 b a

0 0

b

a

a

uGF

2

L1

1 b

L=0

Log uG

Fig.4.13 Loading and Flooding points

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, L& ve G& are the mass flow rates of liquid and gas as (kg/s) , ρ L ve ρ G 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 G ′F . 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 G ′F with the density of the gas, flooding gas velocity uGF is first obtained, then by taking some percent (in practice usually 6575 %) of this velocity, operating gas velocity uG is found . Then from,


107

Dc = (4G& / πρ G u G ) 0.5 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' is computed from the ordinate of the point, which is obtained by

Flooding curve

G ′ 2 C f µ 0L.1 ρ G (ρ L − ρ G )

Pressure drop in gas(Pa/m.packing)

L& ⎛ ρ G ⎜ & ⎜ρ −ρ G G ⎝ L

⎞ ⎟⎟ ⎠

0.5

Fig.4.14 Flooding and pressure drops in randomly packed columns

going vertically up from the calculated abscissa value until cutting the specified pressure drop curve. Then, first cross-sectional area from A c = G& / G ′ 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 & )(ρ / ρ ) 0.5 then a xin Fig.4.4), first an s parameter is calculated from s = (L& / G G L factor from x = - 1.443 s and finally a flooding factor from FFL = C ρ 0L.5 e x , 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 / ρ G 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 w = L& / ρ L A c a p and is related to the liquid hold-up by the equation Ho = 0.0136 ap L0w.33 , should not be smaller than 2.10-5 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.


108

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 ; 2 G ′ 0 .395 and H L = 0.345 L′ 0.22 where, HG & HL are in m. G ′ & L′ are in kg/m s. H = 1 . 397 G

L′

0 . 417

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; M G = (0.97)(29) + (0.03)(58) = 29.87 (1)(29.87) Density of the gas; ρG = =1.21 kg / m 3 (0.083)(273 + 25) 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& = 12 855 .6 kg / h = 3.57 kg / s

Mass flow rate of gas,

& = G * M = (300 )( 29.87 ) = 8 961 kg / h = 2.49 kg / s G G

L& ⎛ ρ G ⎜ & ⎜ρ −ρ G G ⎝ L

0 .5

⎞ 3.57 ⎛ 1.21 ⎞ ⎟⎟ = ⎜ ⎟ = 0.05 2.49 ⎝ 997 − 1.21 ⎠ ⎠

G ′ 2 C f µ 0L.1 = 0.08 ρ G (ρ L − ρ G )

0 .5

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

is read. From Table.4.2 Cf =110 and ap = 162 m-1. Then, 0.5

⎛ (0.08)(1.21)(997 − 1.21) ⎞ ⎟⎟ = 1.33 kg / m 2s G′ = ⎜⎜ − 3 0.1 ⎠ ⎝ (110)(0.894 *10 ) & Cross-sectional area required, A c = G = 2.49 = 1.872 m 2 G ′ 1.33 0.5 ⎛ 4 * 1.872 ⎞ Diameter of the column, Dc = ⎜ ⎟ = 1.55 m π ⎝ ⎠ & L 3 . 57 b) HL = (0.345)(1.91)0.22 = 0.398 m L′ = = = 1.91 kg / m 2 s A c 1.872 (1.33) 0.395 H G = (1.397) = 1.194 m (1.91) 0.417 From equation (4-31), H OG = 1.194 + (2.2)(1 / 2.3)(0.398) = 1.575 m Finally, from equation(4-20), the required height of packing, z = (1.575)(4.51) = 7.10 m L& 3.57 c) Wetting rate, L W = = = 1.18 * 10 − 5 m 3 / ms ρ L A c a p (997)(1.872)(162)

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,


109

0.5

⎛ 110 ⎞ ⎟⎟ G ′ = (1.33)⎜⎜ = 1.85 kg / m 2 s ⎝ 57 ⎠ New cros-section, AC = 2.49/1.85 = 1.346 m2 ; new diameter, Dc = [(4*1.346)/π]0.5 = 1.31 m. L′ = 3.57 / 1.346 = 2.65 kg / m 2s , HL = 0.345 (2.65)0.22 = 0.428 m,

H G = (1.397 )

(1.85) 0.395 = 1.186 m ( 2.65) 0.417

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 . 3.57 LW = = 2.58 * 10 − 5 m 3 / ms . The wetting rate is now satisfactory. (997)(1.346)(103) d) From flooding curve of Fig.4.14, G ′F 2 C f µ 0L.1 = 0.28 is found. ρ G (ρ L − ρ G )

Hence, percent of flooding =

for

Then,

L& ⎛ ρ G ⎜ & ⎜ρ −ρ G G ⎝ L

⎞ ⎟⎟ ⎠

0 .5

= 0.05

G′F = (0.28 / 0.08) 0.5 (1.85) = 3.46 kg / m2s

G′ 1.85 .100 = .100 = 53.5 % , which allows for safe operation. G′F 3.46

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

110

Gas G1,y1

Lo,xo

Liquid Downcomer 1 Weir

G2,y2

L1,x1

G3,y3

L2,x2

Gn+1,yn+1

Ln,xn

2 Down-comer apron

Plate

n Shell

N Gas GN+1,yN+1

∇

LN,xN

Solution (a)

(b)

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

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

111

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:

E MG =

y n +1 − y n y n +1 − y ∗n

E ML =

x n − x n −1 x ∗n − x n −1

(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

y

Real Plate

Equilibrium Plate

Equilibrium distribution curve

yn xn-1

x xn

Slope=-L/G

yn+1

Slope=-L/G

yn+1

yn


y ∗n xn-1

xn x ∗n

x

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


112

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)

Over all column efficiency, Eo

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 bubblecap 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

mM L µ L ρL Fig.4.18 Over all column efficiency for bubble-cap absorber

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. 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)

113

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 and by solving yn+1 from the last equation, yn+1 =

G y − Lo x o Ln xn + 1 1 G n +1 G n +1

(4-39)

(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 xydiagram 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: yn+1 =

L xo L x n + y1 − G G

(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, Yn +1 =

L Ls X n + Y1 − s X o Gs Gs

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

114

Y(y)

YN+1=Y4 (yN+1= y4)

P=cons. t=cons.

D 3

Y3 (y3) Operating line

2

Y2 (y2) Y1 (y1)


1

T

0 0 Xo(xo)

X(x)

X1(x1)

X2(x2)

X3(x3)

Fig.4.19 Calculation of number of ideal plates by Mc Cabe-Thiele method

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.


115

First, obtain equilibrium data at 20 oC and P = 800 mmHg, then convert these to X-Y and plot on a mm paper. From App.Table.4.1 Solubility of NH3 in water at 20 oC

Mass of NH3 per 100 mass of H2O 20 15 Partial press. of NH3 (mmHg)

10

7.5

5

4

3

y1=0.0196 y

y

L& o = 2 581 kg / h xo =0.0

1

2

G2,y2

L1,x1

G3,y3

L2,x2

2

166 114 69.6 50.0 31.7 24.9 18.2 12.0

≈

≈

n

Conversion of the data to X and Y

x=

mA / MA (m A / M A ) + (m B / M B )

Gn+1,yn+1 Ln,xn

X =

≈

x 1 − x

≈

N LN,xN

y Y = 1− y The first values; p y= A P

GN+1= 150 k-mol/h yN+1 = 0.20

LN,xN

0 .175 = 0 .212 1 − 0 .175

x=

20 / 17 = 0 . 175 ( 20 / 17 ) + (100 / 18 )

X=

y=

166 = 0 . 208 800

Y=

Fig.Example-4.7a

0.208 = 0.262 1 − 0.208

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 YN +1 = G

s

y N +1 0.20 = = 0.25 1 − y N +1 1 − 0.20

=G

N +1

Y1 =

(1 − y N + 1 ) = 150 (1 − 0 . 20 ) = 120

y1 0 . 0196 = = 0 . 02 1 − y 1 1 − 0 . 0196

X

o

=

x o 1 − x

= 0 .0 o

L s = L o (1 − x o ) = L o (1 − 0 ) = L o

k − mol C / h

L s = L o = 2 581 / 18 = 143 . 4 k − mol B / h Operating line equation;

Yn +1 =

143.4 143.4 (0.0) X n + 0.02 − 120 120

Yn +1 = 1.195 X n + 0.020 If we write this equation at the bottom of the column;

YN +1 = 1.195 X N + 0.020

Then, 0.25 = 1.195 X N + 0.020 from here XN = 0.192 and xN = XN = 0.192 = 0.161 1+XN 1+0.192 a) or as mass fraction : x& N =

( x N )( M A )

(0.161)(17 ) = 0.153 ( x N )( M A ) + (1 − x N )( M B ) (0.161)(17 ) + (1 − 0.161)(18) =

is found. 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


116

Then, percentage recovery of ammonia =

27.6 .100 = 92 % 30.0

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. Y

P = 800 mm Hg, t = 20 oC

0.26

Y7

D

Y6

6

YN+1= 0.25

0.22 5

Y5

0.18

4 Y4


Operating Line

0.14

3 Y3

0.1

0.06

2 Y2

1 Y1= 0.02

0.192=XN

T

0 Xo= 0

0.02

X3

X2

X1 0.06

0.10

X4

X6

X5 0.18

0.14

0.22

X

Fig.Example-4.7b

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


117

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 and, if x1 is eliminated between these two equations, y1 =

y 2 + (L / G ) x o (L / mG) + 1

(4-44) (4-45)

is obtained. When the absorption factor, which is defined as L/mG = α is substituted into the equation above, y + mα x o y1 = 2 (4-46) α +1 is found. If the similar steps are repeated for the second plate, y + mα x1 y + α y1 y2 = 3 = 3 (4-47) α +1 α +1 is obtained. Upon substituting y1 from equation (4-46), (α + 1) y 3 + mα 2 x o y2 = (4-48) α2 + α +1 is found. Multiplying the nominator and denominator of the above equation by (α-1) (α 2 − 1) y 3 + α 2 (α − 1) mx o y2 = (4-49) α3 − 1 is obtained. If similar steps are repeated for 3rd , 4th and finally for the last plate (plate N ) of the column: No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

118

(α N − 1) y N +1 + α N (α − 1) mx o yN = α N +1 − 1

(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; y − y1 + mα x o y N = N +1 (4-52) α 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; y N +1 − y1 α N +1 − α = N +1 (4-53) y N +1 − mx o α −1 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; ⎡y − mx o ⎤ If both sides are divided by α, α N = ⎛⎜ α − 1 ⎞⎟⎛⎜ y N +1 − mx o ⎞⎟ + 1 α N +1 = (α − 1) ⎢ N +1 ⎥ +1 ⎣ y1 − mx o ⎦

⎝ α ⎠⎜⎝ y1 − mx o ⎟⎠

α

is found. Finally by taking the logarithm of both sides, ⎡⎛ α − 1 ⎞⎛ y N +1 − mx o ⎞ 1 ⎤ ⎟⎟ + ⎥ log ⎢⎜ ⎟⎜⎜ − α⎦ α y mx ⎠ ⎝ o ⎠ ⎝ 1 ⎣ N= (4-54) log α 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; ⎡ x − y N +1 / m ⎤ log ⎢ o (1 − α ) + α ⎥ ⎣ x N − y N +1 / m ⎦ N= (4-55) log (1 / α) 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

119

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 Y

Y

D

YN+1

YN+1

'

D

slope=Ls/Gs

slope=Ls/Gs Equilibrium distribution curve

T

Y1

Y1

XN

(a)

(XN)max= X∗N

X

Equilibrium distribution curve slope=(Ls/Gs)min

K

slope=(Ls/Gs)min

Xo

D'

D

D1

T

Xo

(XN)max

XN

X

(b)

Fig.4.20 Determination of Minimum Solvent Rate in gas absorption

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

120

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; ⎛ Ls ⎞ YN +1 − Y1 ⎟⎟ = ⎜⎜ ⎝ G s ⎠ min (X N )max . − X o

(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; y N +1 − y1 ⎛L⎞ ⎜ ⎟ = ⎝ G ⎠ min (x N )max . − x o

(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 Y

Y (Y1)max



'

T

'

slope=(Ls/Gs)max Y1

T

(Y1)max

T

slope=(Ls/Gs)max Y1

T K

slope=(Ls/Gs)

slope=(Ls/Gs) YN+1

YN+1

D XN

Xo

X

D XN

Xo

X

a) b) Fig.4.21 Determination of minimum gas flow rate in desorption operation No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

121

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.

(Y ) − YN +1 ⎛ Ls ⎞ ⎟⎟ ⎜⎜ = 1 max Xo − X N ⎝ G s ⎠ max .

(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 on XY-diagram, a horizontal line from YN+1=0.25 is drawn until cutting the equilibrium distribution curve (point D'), absis of which gives (XN)max = 0.206. Then, from equation (4-56),

0.25 − 0.02 0.206 − 0.0 = 134.0 k-mol water/h is found. As the selected water rate was 2 581 kg/h, the value of β is; 2581 / 18 β= =1.07 134.0 It follows from this that at the absorption operation in the Example-4.7 , only 7 % excess water from the minimum was used. (L S ) min = 120

Y

P = 800 mm Hg, t = 20 oC

0.26 YN+1= 0.25

D'

L 0.22


0.18

0.14

0.1

0.06

Y1= 0.02

(X N ) max = 0.206

T

0 Xo= 0

0.02

0.06

0.10

0.22

0.18

0.14

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. ∗ y First, x 1 is calculated from x 1∗ = 1 = 0.03 = 0.0136 2 .2 2 .2 Then, from equation (4-57), L min = G y1 − y 2 = 300 0.03 − 0.005 = 551.5 k − mol / h x1∗ − x 2 0.0136 − 0.0


is obtained.

122

X

As in the Example-4.5 690 k-mol/h solvent was used, β=

690 = 1.25 is found. 551.5

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 y1 =

Then,

Y1 0.02 = = 0.0196 1 + Y1 1 + 0.02

and

x1 =

X1 0.025 = = 0.0244 1 + X1 1 + 0.025

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,

y6 =

Y6 0.23 = = 0.187 1 + Y6 1 + 0.23

and

x6 =

X6 0.192 = = 0.161 1 + X 6 1 + 0.192

Hence, the slope at the bottom condition: mD= y6/x6 = 0.187 / 0.161 =1.16 Average slope:; m = (m T )(m D ) = (0.803)(1.16) = 0.965 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,

mM L µ L (0.965)(18)(1 * 10 −3 ) = = 1.8 * 10 − 5 ρL 965

is obtained.

= y6 / x6 With this value as absis from Fig.4.18 Hence;

N Re al =

N Ideal 6 = =12.5 E o 0.48

4 E = 0.48 is read. o

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,


123

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: G1y,y11

A : propane, S : oil, C : steam As the liquid solution is not dilute ,work with mole ratios.

y1 xo 0.181 Xo = = = 0.22 1 − x o 1 − 0.181

xN 0.0196 XN = = = 0.02 1 − x N 1 − 0.0196

YN +1 = 0.0

G2,y2 L1,x1 2

LS=Lo(1-xo)=120(1-0.181)= 98.28 k-mol B/h

Y1 = 34 X o = (34)(0.22) = 7.48

D= y6 / x6 ⎛L ⎞ ⎜ ⎟ At (G ) Then,

Gn+1,yn+1 Ln,xn

≈

LN,xN

GS

LN xN= 0.0196

Y1∗ = Y( X o )

and

⎜G ⎟ ⎝ S ⎠ max

≈

N

L LS X n + Y1 − S X o GS GS

S

min

≈

n

First, plot Equilibrium Line ( Y=34X) on X-Y coordinates with these in mind. Operating line equation,

Yn +1 =

L2,x2

G3,y3

≈

∗

S

Lo=120 k-mol/h xo= 0.181

1

⎛ LS ⎞ Y ∗ − YN +1 7.48 − 0.0 ⎟⎟ = 1 ⎜⎜ = = 37.4 ⎝ G S ⎠ max X o − X N 0.22 − 0.02

(G S ) min =

GN+1 yN+1

L S 98.28 = = 2.63 k − mol steam / h 37.4 37.4

a) G S =β (G S ) min = (1.25)(2.63) = 3.29 k − mol steam / h

& =G & G = (3.29 )M B = (3.29)(18) = 59.2 kg steam / h S N +1 b) Operating line equation,

Yn +1 =

is obtained.

98.28 98.28 (0.22) X n + Y1 − 3.29 3.29

Yn +1 = 29.87 X n + Y1 − 6.57 Operating line equation written at the bottom of the column: Then;

0 = (29.87)(0.02)+Y1-6.57 from which

y1 = as mass fraction; c) Propane entered :

Y1= 5.97 and

Y1 5.97 = = 0.857 1 + Y1 1 + 5.97

y& 1 =

YN +1 = 29.87 X N + Y1 − 6.57

or,

y1M A (0.857)(44) = = 0.94 is obtained. y 1 M A + (1 − y 1 )M C (0.857)(44) + (1 − 0.857)(18) 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 %


124

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;

N ideal = 6 + Y

o

P = 2.5 bars t =138

8.0

Y1 = 7 .48

9 mm ab = 6+ = 6.3 ac 30 mm

C

T'

∗

7.0

6.0

Y=34X

Y T

Y1=5.97

5.0

1

Operating Line at ⎛ L s slope = ⎜⎜ G ⎝ s

4.0

⎞ ⎟⎟ ⎠ max

1.0 Y6 0.8

2

6 0.6

3

3.0

Operating Line

4

2.0

0.4

5

1.0

6

X3

D

YN+1

0.02

0.06

0.10

X2

0.14

X1

Xo

0.18

0.22

b

c

0.2

7

X

X6

0

0 X 0.01 D 7

XN

a

0.03

X

0.05

Fig.Example-4.11

Or, since both equilibrium and operating lines are straight, equation (4-55) can be used for the calculation of required number of the ideal plates after replacing y and x with Y and X.

Here;

⎡ X − YN +1 / m ⎤ (1 − α ) + α ⎥ log ⎢ o ⎣ X N − YN +1 / m ⎦ N= log (1 / α) L /G 29.87 α= S S= = 0.879 m = Y*/ X =34 and 34 m ⎡ 0.22 − 0.0 / 34 log ⎢ (1 − 0.879) + 0.879⎤⎥ − 0 . 02 0 . 0 / 34 ⎣ ⎦ = 6.13 N= log (1 / 0.879)

(4-55)

is obtained.

Very close agreement between two methods is obvious. e) At the top of the column :

Y1=5.97

X1=Y1/m= 5.97/ 34 = 0.176 and x 1 =

or y1 = 0.857

X1 0.176 = = 0.15 1 + X 1 1 + 0.176

At the bottom of the column : xN = 0.0196 YN =34 XN = (34)(0.02) =0.68 and

yN =

or

mT = y1 /x1 = 0.857 / 0.15 = 5.71

XN = 0.02

YN 0.68 = = 0.405 1 + YN 1 + 0.68

mD =yN /xN=0.405/ 0.0196 =20.66


125

Average slope :

m = ( m T )(m D ) = (5.71)(20.66) =10.86

mM L µ L (10.86)(300)(0.5 *10 −3 ) = =1.81*10 −3 ρL 900 N 6.3 N Re al = Ideal = = 48.5 Finally, Eo 0.13

Then,

From Fig. 4.18 Eo = 0.13 is read.

NReal = 49

is obtained.

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 h = c L ( t L − t R ) + ∆H s

(4-59) where tR is the reference temperature in ( C), c L ; is the specific heat of the liquid solution in (kJ/k-moloC), and calculated at the arithmetic mean of tL and tR temperatures and ∆H s 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, cLA , cLS by using the following equation, o


126

cL = x cLA + (1 − x ) cLS (4-60) When this equation is substituted into equation (4-59), h = [x cLA + (1 − x ) cLS ]( t L − t R ) + ∆H s (4-61) is obtained. Similarly, the molar specific enthalpy of a binary gas mixture of A+C at temperature tG is given by H = y[c GA ( t G − t R ) + λ A ] + (1 − y) [c GC ( t G − t R ) ] + ∆H s (4-62) where, λ A is the latent heat of vaporization of G1,H1,y1,tG1 component A at reference temperature in (kJ/k-mol A), which is taken into consideration when component A is liquid at reference temperature, Lo,hO otherwise λ A = 0 . If total material, solute A and xo,tLo t1 enthalpy balances are written for a plate type 1 absorption column shown in Fig.4.23; 2 Lo + GN+1 = LN + G1 (4-63) Lo xo + GN+1 yN+1 = LN xN + G1y1 (4-64) n L o h o + G N + 1 H N + 1 = L N h N + G 1 H1 (4-65) Gn+1,yn+1 Ln,hn,xn If the same balances are written between G , H , y column bottom and any plate n in the column. N N N Ln + GN+1 = LN + Gn+1 (4-66) tN N GN+1 Ln xn + GN+1 yN+1 = LN xN + Gn+1yn+1 (4-67) HN+1 yN+1 L n h n + G N +1H N +1 = L N h N + G n +1H n +1 (4-68) tG,N+1 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 LN,hN,xN,tLN λ A , ∆H s , cLA=f(t), cLS =f(t), cGA = f(t), cGC =f(t) Fig.4.23 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) H N +1 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, No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

127

the calculations and hence the solution are y P=cons. correct. If the difference is great, the calculations are repeated with a new t1 value. Repeating is tL1 Adiabatic equilibrium continued until the difference is reasonable. curve As it is seen, calculations require trial and error tL5 and in many cases this is very lengthy. tL4 A short method giving reasonably accurate t L3 result has been developed by assuming that the t L2 heat liberated in the liquid phase upon the absorption, remains in the liquid. Although with this assumption, the temperature of the liquid x hence, the number of the ideal plates (or height x3 x2 x4 x5 x1 of packing) required is found higher than the actual values, this over-design in the number of Fig.4.24 Equilibrium distribution ideal plates (or in the height of packing) can be curve 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 ,

∆H s ∫ dx = c L x

x2

∫

tL

tL 2

dt L and from here,

(4-71) ∆Hs (x-x2) = c L ( t L − t L 2 ) 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


128

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),

x = x2 +

cL 73.5 (t L − t L2 ) = 0.0 + (t L − 20) = 0.002(t L − 20) is written. ∆Hs 37 544

If the x values are calculated at the given temperatures: tL

20

25

x

0.0

0.01

30

35

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,

y1 = (

L L ) min x 1∗ + y 2 − ( ) min x 2 G G

y − y2 ⎛L⎞ ⎜ ⎟ = 1∗ x1 − x 2 ⎝ G ⎠ min

From which,

is obtained. x 1 = 0.032 is read from ∗

the diagram by drawing a horizontal line from y1 = 0.05 until cutting the adiabatic equilibrium curve.

0.05 − 0.005 = 70.31 k − mol / h 0.032 − 0.0 c) L = 1.15 (70.31) = 80.86 k-mol/h

Then,

L min = 50

is found.

80.86 80.86 x + 0.005 − (0.0) 50 50 y = 1.62 x + 0.005 By writing operating line at the bottom of the column, Then, operating line,

y=

x1 =

y 1 − 0 .005 0 .05 − 0 .005 = = 0 .028 1 .62 1 .62

is found.

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

x 1∗ = 0.069

0.05 − 0.005 = 32.61 k − mol / h and operating solvent 0.069 − 0.0 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 Then, minimum solvent rate

L min = 50


129

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.

Mole fraction of ammonia in gas,y

Adiabatic equilibrium curve 35 oC

P=1 bar

30 oC

0.06 D

y1= 0.05

D

D'

25 oC

20 oC

D

D'

0.04

Operating Line Operating Line for Isothermal Absorption at 20 oC

K

0.02 T

0 x2 = 0

0.01

x1=0.06=

x1=0.028= x ∗ = 0.032 1

x =0.014=

y2 =0.005

0.02

0.04

0.03

0.05

x 1∗ = 0.069

0.06=

0.07=

Mole fraction of ammonia in liquid,x

Fig.Example-4.12

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).


130

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. L i q u i d

Partial pressure of solute A

pA

A+C

p h a s e

Liquid film

Gas film

A A+B+P+S

o

p Ai c oAi

pAi

cAi

c oA cA zG : Physical absorption

0

Molar concentration of solute A

I G a s p h a s e

zL : Absorption+chemical reaction

Fig.4.25 Concentration profiles in two phase at physical absorption and at absorption with chemical reaction

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.


131

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. I

L i q u i d

pA

A+C

p h a s e

Liquid film

Gas film

A

Reaction zone

pAi

B cAi cB zG

Molar concentrations


G a s p h a s e

zL

0

Fig.4.26 Concentration profiles in two phase at absorption+fast chemical reaction in liquid phase

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 I

L i q u i d

pA

A+C

p h a s e

Liquid film

Gas film

A Reaction plane

pAi

B cAi cB zG

0

zR



G a s p h a s e

zL

Fig.4.27 Concentration profiles in two phase at absorption with instantaneous chemical reaction in liquid phase


132

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: D N A = AS (c Ai − 0) (4-72) zR can be written. By multiplying right hand side with zL/zL and remembering kL=DAS/zL from film theory, this equation becomes; D z z N A = AS . L c Ai = k L L c Ai (4-73) zL zR zR Similarly, for the diffusion of reactant B in the other part of the liquid film: D BS D D D cB cB (4-74) NB = − (c B − 0) = − BS . AS . = −k L BS . zL − zR D AS z L 1 − (z R / z L ) D AS 1 − (z R / z L ) 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, a D N A = k L (c Ai + . BS c B ) (4-76) b D AS 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; (4-77) NA=kG(pA-pAi) can be written. Assuming that Henry’s law equation is valid for the gas-liquid equilibrium, (4-78) pAi= ΉA cAi If pAi is eliminated between last two equations; N 1 c Ai = (p A − A ) ΗA kG By substituting this into equation (4-76); ⎤ ⎡1 N a D N A = k L ⎢ (p A − A ) + . BS c B ⎥ kG b D AS ⎦ ⎣ΗA D a p A + ΗA BS c B b D AS = ΗA k G + k L kGkL

(4-79)

(4-80)

On the other hand,


133

ΗA k G + k L 1 1 ΗA = + = KG kG kL kGkL

If this is substituted into equation above finally, D N A = K G (p A + ΗA a . BS c B ) (4-81) b D AS is obtained. Position of reaction plane in the liquid film is dependant on (DAS/DBS), (a/b), pA and cB.

pA

A+C

I

L i q u i d

p h a s e

Liquid film

Gas film

A Reaction plane

(cB)crit. B

zG

0 zR



G a s p h a s e

zL

Fig.4.28 Concentration profiles in two phase at absorption with instantaneous chemical reaction when cB ≥ (cB)crit.

If the first three terms are kept constant, zR changes only with cB. There is a value of cB [which is known as critical value and shown with (cB)crit.] that at or above which zR =0, which means that interface becomes reaction plane. Concentration profiles in this case are to be as shown in Fig.4.28. As it is seen from the Figure, at cB ≥ (cB)crit. equation(4-77) becomes, (4-82) NA=kG pA So, it follows from these that, for the calculation of total flux, when, cB < (cB)crit equation(4-81), when, cB ≥ (cB)crit equation(4-82) is used . (cB)crit value for any given pA is found as follows; From Fig.4.26, at or above (cB)crit , D D D D N B = − BS (c B ) crit . = − AS . BS (c B ) crit . = − k L BS (c B ) crit (4-83) zL z L D AS D AS is obtained. If this and NA from equation(4-82) are substituted into equation(4-75) from which, b kG pA - a kL(DBS / DAS)(cB)crit. = 0 b D k (c B ) crit . = . AS . G p A (4-84) a D BS k L


134

is found. Or for a given value of cB; (pA)crit. is: a D k (4-84b) (p A ) crit . = . BS . L c B b D AS k G 4.6.2 Calculation of Height of Packing when Chemical Reaction is Instantaneous:: With the assumption of dilute solutions, solute G pA2 A balance along dh differential height in the packed section; (4-85) -G(dpA/P) = NA dSm (4-85b) = (a/b)LvdcB 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 2 where Ac(m ) 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;

Lv cB2 h=z

2

NA

dh

cB

pA 1

h=0 G pA1

p A1

G ⌠ dp A Z = ∫ dh = o a v PA c ⎮ ⌡p A 2 N A z

Lv cB1

(4-86)

Fig.4.29 is obtained. pA 1

(pA)crit.= f(cB) Case-3 1

Case-2

pAx

2 2 1

pA1

Case-1

pA2

2

(cB2)crit.

(cB1)crit.

cB1

cBx

cB2

cB

Fig.4.30 Stoichiometric operating lines for three different cases


135

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. In this case NA=kGpA and then, Case-1) cB1 ≥ (cB1)crit and cB2 > (cB2)crit p

A1 p G ⌠ dp A Z= = H G ln A1 (4-87) ⎮ a v PA c k G ⌡p A 2 p A p A2 G is obtained. Where H G = (m) is height of one gas transfer unit. k G a v PA c Case-2) ) cB1 < (cB1)crit and cB2 ≤ (cB2)crit In this case NA value must be substituted from equation(4-81).

p A1

p A1

⌠ ⌠ dp dp A G A ⎮ Z= = H OG ⎮ a D BS a D BS a v PA c K G ⎮ ⎮ cB cB ⌡p A 2 p A +ΗA . ⌡p A 2 p A + ΗA . b D AS b D AS

(4-88)

G (m) is the height of one overall gas transfer unit. K G a v PA c By integrating equation(4-85b) between bottom and any section of the column,

is found. Where, H OG =

from which, -G(pA1-pA)/P = (a/b)Lv(cB1-cB) (4-89) cB = cB1+(b/a)(G/LvP)(pA1-pA) is obtained. This equation is known as “stoichiometric operating line equation” If this cB is substituted into equation (4-88) and integrated; a D (pA + ΗA . BS c B ) 1 H OG b D AS ln Z= (4-90) 1 D a D 1 − . BS (pA + ΗA . BS c B ) 2 α D AS b D AS is found. Where α =

(L v / G ) is absorption factor. ΗA / P

cB2 > (cB2)crit In this case between top and any Case-3) cB1 < (cB1)crit and level(level-x) the solution of case-1 and between this level and bottom of the column solution of case-2 apply. Hence; a D (pA + ΗA . BS c B ) 1 H OG p b D AS ln Z = H G ln Ax + (4-91) 1 D BS p A2 D a BS 1− . (pA + ΗA . cB )x α D AS b D AS


136

pAx and cBx are found by first writing equations (4-84) and(4-89) at level-x and then solving; a L P p A1 + . v c B1 b G p Ax = L P D k 1 + v . AS . G G D BS k L a L P p A1 + . v c B1 b b G c Bx = . a D BS k L L v P + . D AS k G G

(4-92)

(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: First 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 2 Molecular diffusivity of ammonia in water from Table.1.4, -9 2 DAS=2.3*10 m /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 1 Hence, (cB1)crit. > cB1 3 At column top ; (cB2)crit. = (11.3)(0.001) = 0.0113 k-mol/m Hence, (cB2)crit. < cB2 It follows from here that case-3 is applicable. Then, from equation (4-92);

p Ax

0.05 + (2 / 1)[(25 * 1) / 150](0.103) = = 0.0177 bar 1 + [(25 * 1) / 150](2.3 / 2.55)(50 / 2)

pA2=0.001 bar G

Lv=25 m3/h cB2=0.25 k-mol/m3 z

G=150 kpA1=0.05 bar Lv cB1

From equation (4-93);


137

c Bx = (1 / 2)

0.05 + (2 / 1)[(25 * 1) / 150](0.103) = 0.20 k − mol / m 3 ( 2.55 / 2.3)(2 / 50) + [(25 * 1) / 150]

As the solution is dilute in ammonia and density of water at 20 oC is 998 kg/ m3 c=998/18=55.44 kmol/m3 and from equation(4-3)ΉA = Ή’A /c=(0.73)/(55.44) = 0.0132 bar m3/k-mol. HG =

1 1 0.0132 = + = 0.0266 bar h m 3 / k − mol K G a v 50 2

150 =3 m (50)(1)(1)

(150)(0.0266) G = = 3.99 m K G a v PA c (1)(1) hence; α = 25 / 150 = 12.63 0.0132 / 1 H OG =

Substituting all these into equation (4-91);

Z = (3) ln

0.0177 + 0.001

3.99 1 1− .(2.55 / 2.3) 12.63

ln[

0.05 + (0.0132)(2 / 1)(2.55 / 2.3)(1 / 1)(0.103) 0.0177 + (0.0132)(2 / 1).(2.55 / 2.3)(0.20)

]

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 pA2 unreacted part of it remains in the leaving solution. Lv Again, with dilute solutions assumption; along dh cB2 2 differential height in the packed section: cA2=0 h=Z (4-94) -G(dpA/P) = NA dSm = NA av Ac dh = consumption of A by chemical reaction + increase in content of A in bulk liquid ZP (4-94b) = -RAHoAc dh - Lv dcA dh NA = (a/b) Lv dcB - Lv dcA (4-94c) can be written.Where Ho (m3 liquid /m3 packing) is cB pA ZR the specific hold-up. cA Considering gas phase; h=0 (4-95) NA =KG (pA- p ∗A )= KG (pA - ΗA cA) 1 G can be written.After substituting this into equation pA1 (4-94) and integrating from bottom to the top of Lv cB1 cA1 the column; p

p

A1 A1 dp A dp A G ⌠ ⌠ (4-96) = H OG ⎮ Z = ∫ dh = ⎮ 0 K G a v PA C ⌡p A 2 p A −ΗA c A ⌡p A 2 p A −ΗA c A z

is obtained. Note that in this case; cA = f (pA,cB) [ Remember: in physical absorption, cA = f(pA)]


138

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 ; and from the integration of this between the top and any level -G(dpA/P) = Lv dcA within the physical absorption zone with cA2 = 0. G cA = (p A − p A 2 ) (4-97) LvP 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; p

Ax dp A ⌠ Z P = H OG ⎮ ⌡p A 2 p A (1 − 1 / α) + p A 2 / α

ZP =

⎡ H OG p 1⎤ ln ⎢(1 − 1 / α) Ax + ⎥ 1 − 1/ α p A2 α⎦ ⎣

(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); (4-99) (pA - ΗA cA)= -RAHo/KG av and this into equation (4-96) and taking pA2 = pAx ; the height of chemical reaction zone; p

Z R = − H OG

K G a v ⌠ A1 dp A Ho ⎮ ⌡p Ax R A

(4-100)

is obtained. Then, total height of the packing becomes; Z = Z P + 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); p K G a v ⌠ A1 dp A H Z R = H OG = OG (p A1 − p Ax ) (4-102) ⎮ βo H o ⌡p Ax k o No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

139

is found. Where, β o = k o H o / K G a v At h=ZR equations(4-97) and(4-99) become; G c Ax = (p Ax − p A 2 ) LvP (pAx - ΗA cAx)= koHo/KG av =βo β o − p A2 / α 1 − 1/ α equations(4-98) and (4-102);

From these two;

p Ax =

is obtained. Substituting this into

H OG ln(β o / p A 2 ) 1 − 1/ α H OG [(1 − 1 / α) p A1 + (p A 2 / α) − β o ] ZR = (1 − 1 / α) β o Zp =

(4-103) (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) p

Z R = H OG

Where,

p

K G a v ⌠ A1 dp A H OG ⌠ A1 dp A = H ok1 ⎮ β1 ⎮ ⌡p Ax c A ⌡p AxΗA c A

β1 =

(4-105)

k 1H o kH = 1 o ΗA K G a v K L a v

From equation(4-99) by noting the value of β1 ; pA is found. After substitution into equation(4-105) and integration; ΗA c A = 1 + β1 Z R = H OG

1 + β1 p ln A1 β1 p Ax

(4-106)

is obtained. At h = ZR equations(4-97) and(4-99) become; G c Ax = (p Ax − p A 2 ) LvP (pAx - ΗA cAx)= k1 cA Ho/KG av = ΗAcAx β1 p A2 is obtained.. Substituting this into From these two; p Ax = 1 − α /(1 + β 1 ) equations(4-98) and (4-106); H OG β1 Zp = (4-107) ln 1 − 1/ α 1 + β1 − α


140

ZR =

⎡ 1 + β1 α p A1 ⎤ H OG ln ⎢(1 − ) ⎥ β1 1 + β1 p A2 ⎦ ⎣

(4-108)

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- ΗA cA)av. avAc∆h (4-109) G pA2 Lv = -RAHoAc ∆h - Lv ∆cA (4-109b) = (a/b) Lv ∆cB - Lv ∆cA (4-109c) cB2 These equations can be written for ∆hj height which cA2=0 is bounded with the planes j and j-1, as: cAj-1 pAj-1 cBj-1 G(pAj-pAj-1)/P = KG(pA- ΗA 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)

∆hj

pAj

cAj cBj

Calculations may be started at the top of the column, where G pAj-1, cAj-1 and cBj-1 are all known. By selecting a pAj value, pA1 as a first approximation, (pA-ΗA cA)av. = (pA-ΗA cA)j-1 Lv cB1 cA1 and (cAcB)av. = (cAcB)j-1 can be written. Then, from equation (4-110); ∆hj = HOG(pAj-pAj-1)/(pA- ΗA 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- ΗA cA)av. = (1/2)[(pA- ΗA cA)j + (pA- ΗA 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, d 2cA D AS + RA = 0 dz 2 This equation must be solved with the following boundary conditions:


(4-113)

141

1o) at z = 0 cA =cAi = constant o 2 ) 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 k d 2cA (4-114) − o =0 2 dz D AS Double integration of this equation results in k c A = o z 2 + I1 z + I 2 (4-115) 2D AS Integration constants I1 and I2 are evaluated from the boundary conditions as; Then,

c AL − (k o / 2D AS )z 2L − c Ai and I2 = cAi Substituting these into equation(4-115) I1 = zL and noting that, kL =DAS/zL ⎡k k ⎤ ko 2 c A = c Ai − ⎢ L (c Ai − c AL ) + o ⎥ z + z (4-116) 2k L ⎦ 2D AS ⎣ D AS is obtained. Total molar flux of solute A crossing the interface becomes; D k ⎛ dc ⎞ (4-117) N Ai = −D AS ⎜ A ⎟ = k L (c Ai − c AL ) + AS o 2k L ⎝ dz ⎠ z = 0 Or, if cAi is solved from NAi=kG(pA-ΗA cAi) and substituted into equation above; ⎡ D k ⎤ N Ai = K G ⎢p A − ΗA (c AL − AS 2 o )⎥ 2k L ⎦ ⎣

(4-118)

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); d 2cA k − 1 cA = 0 (4-119) 2 dz D AS 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 A = I 1 e qz + I 2 e − qz (4-120) Integration constants can be evaluated by applying the boundary conditions. These are: 2 c Ai sinh(qz L ) − c Ai e qz L + c AL c Ai e qz L − c AL I1 = I2 = 2 sinh(qz L ) 2 sinh(qz L ) By substituting these into equation (4-120),


142

cA =

c AL sinh(qz) + c Ai sinh[q ( z L − z)] sinh(qz L )

(4-121)

is obtained. Total molar flux crossing the interface; dc c c N Ai = −D AS ( A ) z = 0 = k L Ha (c Ai − AL ) = (k L ) R (c Ai − AL ) (4-122) dz cosh β cosh β 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- ΗA cAi), equation (4-122) can also be written as; ΗA ⎛ ⎞ N Ai = (K G ) R ⎜⎜ p A − c AL ⎟⎟ (4-123) cosh β ⎝ ⎠ where, ΗA ΗA 1 1 1 = + = + (4-124) (K G ) R k G (k L ) R k G k L Ha Criteria for fast and slow reaction; cosh β =

e β + e − β (1 + β + β 2 / 2) − (1 − β + β 2 / 2) β2 = = 1+ 2 2 2

e β − e − β (1 + β + β 2 / 2) − (1 − β + β 2 / 2) β = = β −β 2 2 e +e (1 + β + β / 2) + (1 − β + β / 2) 1 + β 2 / 2 β β2 = 1+ Then, Ha = is found. tghβ 2 tghβ =

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; 1 β

1 2 e +e e +e eβ and = β ≈0 cosh β = = = cosh β e 2 2 2 β −β e −e and Ha = β tghβ = β =1 e + e −β And flux equation in this case from equations(4-122) and(4-123); NAi= kL βcAi = (KG)R pA Where, β

−β

β


(4-125)

143

ΗA 1 1 = + (K G ) R k G k 1 D AS For the calculation of height of packing;

can be written. From this

(4-126)

-G dpA/P = (KG)R avAcpAdh p Z = (H OG ) R ln A1 p A2

(4-127)

is obtained. Where; (H OG ) R = H G +

ΗA G (H L ) R Lv

(4-128)

Lv Lv = (k L ) R a v A c a v A c k 1 D AS In Fig.4.31 conditions for slow and fast reaction are summarized. (H L ) R =

and

(4-129)

1.0 Slow reaction

1 1 /(k L ) R = Ha 1/ k L 0.5

Diffusion and reaction are comparable

Fast reaction

0

0

logβ 0.2

3

10

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:

Concentration unit in the gas

Concentration unit in the liquid

Partial pressure of A (mm Hg)

Molar concentration (k-mol A/ m3solution)


Mole fraction of A


Mass fraction of A

Mole fraction of A

Mole fraction of A

Mass fraction of A

Mass fraction of A

Mole ratio of A

Mole ratio of A

Mass ratio of A

Mass ratio of A


144

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 y1

⌠ (1 − y) i ln dy NG = ⎮ ⌡ (1 − y)( y − y i ) y2

1 1 − y2 1 ⌠ dy + ln NG = ⎮ 1 − y1 ⌡y 2 y − y i 2 y

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 solutefree. 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]


145

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 [Ans. : a) 0.966 m, b) % 53.6 ] kg/m3, 950 kg/m3. 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 o C, 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.


146

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 0.57 K y a v = 6.5 (G ′) ,where Kyav is in k-mol/m3h 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 kmol/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 ]


147

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

148

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 ] 3 Volumetric flow rate of liquid : 10 m /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.


149

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


150

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 vaporliquid 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

0

yG

1

yJ

R

tB tS

Temperatures, t

Temperatures, t

P=cons. tR

Dew point N temperature curve S G

tL

L

M

tH

Tie-lines

J

H

tA

Bubble point temperature curve K 0 xS

xL

xH =xK

1

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

151

mole fraction of the MVC in vapor, y

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 1 represents a liquid-vapor mixture, which P=cons. upon separation gives two equilibrium yJ equilibrium (H,J) phases: saturated liquid and saturated vapor, curve yG (L,G) which are located on the bubble-pointy= x temperature curve and on the dew-pointtH yN (S,N) temperature curve by drawing the tie-line tL through the point representing the mixture. Their amounts are found by applying the tS inverse lever-arm rule to the tie-line. Any point just on the bubble-point-temperature curve represents saturated liquid, and on the 0 dew-point-temperature curve represents 0 xS 1 xH xL saturated vapor. Points above dew-pointmole fraction of the MVC in liquid, x temperature curve and below the bubblepoint-temperature curve represent Fig.5.2 xy-diagram of a binary mixture superheated vapor and sub-cooled liquid respectively. 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).


152

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: p A = p oA x t = cons.

P = p A + p B = p x + p (1 − x) o A

o B

(5-2)

p oA

P-x

p oB P-y pressure

p B = p oB (1 − x) (5-1) By summing up these two equations;

pA − x

pB − x

is obtained. From these, it is understood that partial and total pressures change linearly with the mole fraction as shown in Fig.5.3. 1 0 x,y By solving x from these two equations, (A) (B) P − p oB Fig.5.3 Ideal solution x= o (5-3) o pA − pB and from the combination of Raoult and Dalton laws, p po (5-4) y= A = A x P P 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: b o log p io = a − where p io in mmHg, and t in C. Antoine constants a, b and c for benzene c+t and toluene are given in the table below:


153

Component

tbp (oC)

a

Benzene(A)

80.1

6.90565

1211.033

220.79

7-135

Toluene(B)

110.6

6.95334

1343.943

219.38

7-135

b

t range ( oC)

c

Solution : Select t = 85 oC, then ;

log p oA = 6.90565 −

1211.033 = 2.9453 220.79 + 85

log p oB = 6.95334 −

1343.943 = 2.538 219.38 + 85 x=

From equations (5-3) and (5-4)

p oA = 881.7 mmHg

p oB = 345.1 mmHg

760 − 345.1 = 0.773 881.7 − 345.1

y=

881.7 (0.773) = 0.897 760

By selecting various temperatures and repeating the steps given above, the following table is obtained.

Select

C

a

t (oC)

p oA (mmHg)

l

c

u

l

a

t

p oB (mmHg)

e 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

xy-Diagram mole fraction of benzene in vapor,y

temperature (oC)

115

105

95

85

75

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 mole fraction of benzene,x,y

1

mole fraction of benzene in liquid,x


154

5.2.2 Deviation from Ideality: Real and Azeotropic Solutions: In real solutions, the total pressure measured over the solution may be greater or smaller than the total pressure calculated from equation (5-2). If the measured pressure is higher than the calculated pressure, this solution is said to deviate from Raoult’s law in positive direction, if the reverse happens, deviation is said to be in negative direction. The partial pressures of the components in such systems do no longer change linearly with the composition. In Fig.5.4, the changes of total and partial pressures with the compositions 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; p A = γ A p oA x p B = γ B p oB (1 − x)

(5-5)

are written, where γ A and γ B are the liquid phase activity coefficients of the components and they show the direction and the degree of deviation of the solution from ideality. When they are 1, the solution is ideal, as expected and when they are greater than 1, deviation is positive, and smaller than 1, the deviation is negative. If

t=cons.

P-x

t =cons.

p P-y

p oB pA − x

o A

p

P-y

pB − x

pB − x

pA − x

pressures

pressures

p oA

P-x o B

0 (B)

x,y

1 (A)

0 (B)

x,y

1 (A)

(b) (a) Fig.5.4 Deviation from idealty (a) in (+), (b) in (-) direction the activity coefficients, which are dependant both on temperature and composition are

close to 1, deviation from ideality is small. In Fig.5.5, the changes of activity coefficients with compositions are shown for two different systems, one deviating positively, and the other deviating negatively. When the deviation from ideality is great and the difference between the vapor pressures of the two components is small, the total pressure versus mole fraction curve may pass through a maximum or minimum. This type of solutions are said to form azeotropic solutions or constant boiling solutions. Passing through a maximum results in minimum boiling azeotropism, whereas passing through a minimum gives maximum boiling azeotropism. Constant temperature, constant pressure and xy-diagrams of minimum and maximum boiling azeotropic systems are all shown in Fig.5.6. As are seen from


155

(a) propanol-water

(b) acetone-chloroform

Fig.5.5 Change of activity coefficients with composions (a) at positive, (b) at negative deviation

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 pure liquid, and the vapor formed has the same composition with the liquid. While a component is more volatile on the left of point Az, it becomes less volatile on the right of the point, hence the volatilities of the components change at point Az. The solutions behaving in this way cannot be separated completely by using classical distillation methods. Minimum boiling azeotropism is more common than the maximum boiling azeotropism, 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.

1279.9 The vapor pressure of acetone is given as: log p oA = 7.24 − where p oA in mmHg and t in oC. 237.5 + t Solution : 1279.9 log p oA = 7.24 − = 3.0215 Vapor pressure of acetone at 65.9 oC : 237.5 + 65.9 p oA =10 3.0215 = 1050.7 mmHg From equation (5-5) :

γA =

398 =1.26 (1050.7)(0.30)

as γA greater than 1, the solution is not ideal and positively deviates from Raoult’s law.


156

Az

t =cons.

t =cons.

p oA

P-x P-x

p

P-y

o A

Az

p oB P-y

pA-x

pressures

pressures

p

o B

pB-x

0 (B)

pA-x pB-x

0 (B)

1.0 (A)

x,y

1.0 (A)

x,y Az t-y

P= cons.

P= cons.

G

tB

t-x

G L

tB

t-y G t-x

0 x (B)

G L

tA L

Az

y

y x,y

temperatures

temperatures

L

tA

x 1.0 (A)

1.0

0 (B)

x

y

x x,y

1.0 P= cons. y

y
Az

Az

y>x

1.0 (A)

y>x

P= cons.

y

y

y=x

y=x y
0

0 x

0

a

1.0

0

x

1.0

b

Fig.5.6 (a) Minimum and (b) maximum boiling azeotropism


157

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 component exerts its own vapor pressure in the vapor phase and then the total pressure is written as, P = p oA + p oB

(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 components in the vapor at the first period at which the two components vaporize together are calculated from the equations below,

and

p oA p oA y= = o P p A + p oB

(5-7)

p oB p oB 1− y = = o P p A + p oB

(5-8)

These values remain constant throughout the first period. This special case forms the basis of steam distillation, which is frequently used to separate a high boiling organic material, which is insoluble in water, from the non-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 water exists in the mixture, this temperature does not change. Steam consumption per unit mass of organic material distilled can be computed from, mA 18 p oA kg steam 18 y = = = kg organic vapor mB M B (1 − y) M B p oB

(5-9)

where, y is the mole fraction of water vapor (steam) in the vapor mixture, MB is the molecular weight of vaporizing organic material. Hence, energy required qs as kJ per mB kg of organic material distilled is given by; qs 18 p oA [cLA (t − t o ) + λ A ]+ [cLB (t − t o ) + λ B ] = m B M B p oB


(5-10)

158

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), cLA and cLB 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 the cost of separation by steam distillation may be very high especially for low molecular weight organics, due to the high 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. d) The loss of ortho-nitrotoluene with water, e) The energy requirement for direct vaporization of ortho-nitrotoluene. f) What other way may be used to purify the ortho-nitrotoluene at the same temperature? Compare this with steam distillation. Temperature ( oC) Water p oA (mmHg) (A) cLA(kJ/kgK) λA(kJ/kg) o-Nitro p o (mmHg) B toluene cLB(kJ/kgK) (B) λB(kJ/kg)

60 -

62.3 -

99 733.5

99.5 746.8

100 760

12.8

2267 13.1

13.4

110 1060

123.5 -

222 -

4.182 2238

1.57

1.75 389.6

340

Solution: a) When p oA + p oB = P = 760 mmHg distillation starts. Assume, t = 99 oC , from table above,

p oA + p oB = 733.5 + 12.8 = 746.3 mmHg < 760 mmHg

Assume, t = 99.5 oC

p oA + p oB = 746.8 + 13.1 = 759.9 mmHg ≅ 760 mmHg

Then, distillation starts and proceeds at 99.5 oC b) From equation (5-9)

mA =

m B18 p oA (500)(18)(746.8) = = 3 745 kg water M B p oB (137)(13.1)

c) The steam requirment is obtained from equation(5-10). c LA and c LB must be taken at,

t o + t 25 + 99.5 = ≅ 62.3 o C 2 2 qs 18 (746.8) [(4.182) (99.5 − 25) + 2267] + [(1.57) (99.5 − 25) + 389.6]= 19 820.1 = m B (137)( 13.1)


159

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 From the table ; ts = 110 oC and λs =2 238 kJ/kg Then, required heating steam amount :

ms =

q s 9.91 *10 6 = = 4 428 kg λs 2238

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 =

(3 745)(0.07) = 2.62 kg 100

e) Direct vaporization at 760 mmHg takes place at 222 oC , c BL is taken at (25+222)/2=123.5 oC. Then, energy requirment: q s = m B [c LB (t − t o ) + λ B ]

= (500) [1.75 (222 − 25) + 340]= 3.43 * 10 5 kJ

5 As it is seen, only 3.43 *10 .100 = 3.46 % of the energy for steam distillation would be enough in this case. 9.91 *106

f) As other way, vacuum evaporation at the same temperature, which corresponds to 13,1 mmHg absolute pressure (760-13.1≈747 mmHg vacuum), may be considered.

In this case; In other enough.

5 q s = (500) [1.57 (99.5 − 25) + 389.6] = 2.53 * 10 kJ

words;

only

2.53 *105 .100 = 2.55 % of the energy for steam distillation would be 9.91 *106

Saving in the heating energy is compared with the energy required for creating vacuum. Decision is then given.

Some systems deviate from the ideality so much that components A and B have only partial solubility within each other in liquid state. The vapor-liquid equilibria of such systems are rather interesting within partial solubility range. The vapor-liquid equilibrium of water/n-butanol system, which has industrial importance and behaves in this way, is shown in Fig.5.7. The area below points D and E shows the concentration range at which partial solubility in liquid phase occurs. The binary mixture represented by a point in this area is heterogeneous and separates in two insoluble liquid phases. For example, the mixture which has 82 mole percent water and at 85 oC and shown by point M separates in two equilibrium liquid phases one containing 58 mole percent water and 42 mole percent n-butanol and being represented by point C, the other containing 98 mole percent water and 2 mole percent n-butanol and being represented by point K. These equilibrium liquid phases are found by drawing the liquid tie line passing through point M, and the compositions of the phases change with temperature. If a liquid mixture represented by a point in this region is heated up slowly, when the temperature reaches boiling point temperature, which is about 93 oC, vapor forms. The composition of this vapor, which is represented by point Az, remains constant at azeotropic composition of 77 mole percent water and 23 mole percent n-butanol until one 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.


160

P =cons. vapor

tB temperature,t

N

t-y

S

J H t-x F

tA

t-y AZ

D

liquid

t-x

E K

C

M two liquid phase

mole fraction of water in vapor, y

0 mole fraction of water in liquid and vapor, x,y 1.0

1.0 D AZ

E

y =x

H,J

0 0

mole fraction of water in liquid, x

1.0

Fig.5.7 Vapor-liquid equilibrium diagrams of water / n-butanol

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; αA =

pA x

αB =

pB 1− x

(5-11)

The relative volatility of A to B is then given as, α p A /x α= A = (5-12) α B p B /(1 − x) From Dalton’s law p A = y P and p B = (1 − y) P can be written. If these are substituted into equation (5-12); y (1 − x) α= (5-13) x (1 − y) No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

161

and solving for y, αx (5-14) 1 + (α − 1) x is obtained. This equation is used to express the vapor-liquid equilibria of binary solutions in functional way, when the relative volatility is a constant or does not change much within the operating temperature range. For ideal system, equation (514) reduces to equation below after substituting from Raoult’s law, α = p oA / p oB (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 range involved is not so great. Thus, equation (5-14) can still be used in the real solutions, when the difference 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 point temperatures of the pure components of the 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. y=

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-diagram of benzene/toluene at 760 mmHg plotted in Example-5.1, corresponding y values are read and from the equation (5-13) α values are computed. Results are given in the table below; x

0.1

0.3

0.5

0.7

0.9

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.

5.2.5 K-Values: The relationship between liquid and vapor compositions of component-i in a binary or a multi-component solution at equilibrium may be expressed as, yi = Ki xi (5-16)

where, Ki is known as the K value or distribution coefficient of component-i. This equation is especially useful for the computation of vapor-liquid equilibria of multicomponent solutions. For ideal solutions, Ki = p io /P and for real solutions Ki = p io γ i /P . It follows from here that K value of a component depends on


162

(K)

Fig.5.8 K-Values of Some Hydrocarbons at 1 atm.

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 Benzene-ethylene dichloride Benzene-toluene n-Butyl chloride-n-Butyl bromide Chloroform-carbon tetra chloride Ethanol-isopropanol Ethanol-propanol Ethyl chloride-Ethyl bromide Ethyl ether-benzene Ethylene dibromide-propilen dibromide Ethylene dichloride-trichloroethane n-Heptane-methylcyclohexane n-Hexane-n-heptane Methanol-ethanol Methanol-isobutanol Methanol-propanol Methyl acetone-ethyl acetate Phenol-o-cresol Phenol-m-cresol Phenol-p-cresol Toluene-benzyl chloride Toluene-chlorotoluene Water-ethylene glycol Water-ethylene glycol(150 mm Hg) Water-ethylene glycol (50 mm Hg)

Boiling p. of component-1 (oC) 80.1 80.1 77.5 61.1 78.3 78.3 12.5 34.6 131.7 83.5 98.4 69.0 64.7 64.6 64.6 56.8 181.2 181.2 181.2 110.7 110.7 100.0 60.1 38.1

α

1.113 2.61 2.08 1.71 1.18 2.18 3.23 5.16 1.30 2.52 1.058 2.613 1.73 6.1 3.89 2.036 1.30 1.768 1.793 7.75 4.76 49.8 98.0 76400

Boiling p. of component-2 (oC) 83.48 110.7 101.6 76.6 82.3 97.2 38.4 80.2 141.5 113.7 100.3 98.4 78.1 107.5 97.2 77.1 190.6 201.5 202.2 178.0 162.0 197.0 150.2 202


α

1.109 2.315 1.87 1.60 1.17 2.03 2.79 3.95 1.30 2.33 1.056 2.33 1.64 4.4 3.15 1.923 1.275 1.699 1.728 4.45 3.65 13.2 21.0 244

163

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 mixture at constant pressure, and the composition of the first liquid drop to be formed, can be computed from; Σ xi = 1 = Σ

yi Ki

(5-18)

The calculation 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 vapor mixture and y/K ratio for each component gives its mole fraction in the first liquid drop that will form. If the summation deviates substantially from 1, with new temperature selections the steps are repeated until the summation can be accepted as 1. Example-5.5) Calculation of Dew Point Temperature of a Ternary Mixture

Ternary solutions of benzene/toluene/o-xylene are ideal. a) Calculate the dew point temperature of a vapor, containing 20 mole percent benzene, 30 mole percent toluene and 50 mole percent o-xylene at 760 mmHg total pressure. b) What is the composition of the first liquid droplet formed? c) What is the temperature range at which all the vapor will condense (condensation range) ? Antoine constants for the components 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

o-Xylene(C)

144.4

6.99891

1474.679

213.69

7-145

Solution :

From equation (5-18);

xA + xB + xC =

yA P yBP yCP + o + o =1 p oA pB pC


164

=

(0.20)(760) (0.30)(760) (0.50)(760) + + p oA p oB p oC

=

152 228 380 + o + o p oA pB pC

Solution is by trial & error : Assume td =110 oC

log p oA = 6.90565 −

1211.033 = 3.2446 220.79 + 110

p oA = 1756.4 mmHg

log p oB = 6.95334 −

1343.943 = 2.8731 219.38 + 110

p oB = 746.7 mmHg

log p oC = 6.99891 −

1474.679 = 2.4431 213.69 + 110

xA + xB + xC =

152 228 380 + + =1.762 1756.4 746.7 277.4

p oC = 277.4 mmHg

As this is greater than 1, the selected temperature is too low.

p oA = 2971.1 mmHg ; p oB = 1344.6 mmHg

New assumption : td =132 oC. Then, from Antoine equations; are calcualated. Then; p oC = 540.8 mmHg

xA + xB + xC =

152 228 380 + + = 0.924 2971.1 1344.6 540.8

As the summation is smaller than 1, the assumed temperature is high. New assumption : td =129 oC. At this temperature, are calculated. Then; p oC = 496.2 mmHg

xA + xB + xC =

p oA = 2776.4 mmHg ; p oB = 1246.4 mmHg

152 228 380 + + =1.004 2776.4 1246.4 496.2

a) As this is almost 1, the selected temperature 129 oC is the dew point temperature of the vapor mixture. b) The composition of the first liquid droplet formed is :

xA =

152 = 0.055 2776.4

xB =

228 = 0.183 1246.4

xC =

380 = 0.766 496.2

c) The condensation range of the vapor (td - tb) : The vapor starts condensing at 129 oC. As the condensation proceeds the temperature drops. The condensation temperature of the last vapor bubble is the bubble point temperature of the liquid, whose composition is equal to the composition of the vapor at the start. Then, find the bubble point temperature of the liquid, whose composition is : xA = 0.20 ; xB=0.30 ; xC=0.50. For this, equation (5-17) is used.

yA + yB + yC =

p oA (0.20 p oB (0.30) p oC (0.50) + + = 1.0 760 760 760

= (2.63 p oA + 3.95 p oB + 6.58 p oC ) *10 −4 =1.0


165

Assume tb =110 oC, then ,

o p oA = 1756.4 mmHg p B = 746.7 mmHg

p oC = 277.4 mmHg

Then; yA + yB + yC = [(2.63)(1756.4)+(3.95)(746.7)+(6.58)(277.4)]*10-4 = 0.94 As the summation is less than 1, the assumed temperature is low, but it is rather close to the bubble point temperature. o o o New assumption ; tb=113 oC, then ; p A = 1894.6 mmHg ; p B = 812.7 mmHg ; p C = 305.4 mmHg yA + yB + yC =[(2.63)(1894.6)+(3.95)(812.7)+(6.58)(305.4)]*10-4 = 1.02 Since this is almost 1, the assumed temperature is the bubble point temperture. Hence, the condensation range : 129 - 113 oC

5.2.8 Enthalpy-Composition Diagrams: In some distillation operations, the energy difference between vapor and liquid phases is important and is to be known. Enthalpycomposition 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 given as: h = cL (t L − t o ) + ∆H s (5-19) where, ∆Hs (kJ /k-mol solution) is heat of solution at to temperature and at given concentration, to is reference temperature, tL is the temperature at which enthalpy is calculated and cL (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 enthalpy of the saturated solution. The heat capacity of the solution is related to the heat capacities of pure components by cL = x cLA + (1 − x) cLB (5-20) Hence, specific enthalpy of the saturated liquid solution can be written as: (5-21) h = [ x cLA + (1 − x) cLB ] (tb- to) +∆Hs The sign of ∆Hs is negative, if heat is released during mixing. For ideal solutions, ∆Hs= 0. The specific enthalpy of a saturated vapor mixture, H is obtained by assuming that each component of the mixture is separately heated up to the dew point temperature of the mixture, td where it is vaporized totally and mixed with the other component. Hence, H = y [cLA (t d − t o ) + λ A ] + (1 − y) [cLB (t d − t o ) + λ B ] (5-22) can be written. A typical enthalpy-composition diagram obtained by plotting equations (5-21) and (5-22) is shown in Fig.5.9. As it is seen, xy-diagram is also plotted with this diagram. The lower curve in the diagram shows the specific 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 superheated vapor, whereas the points below the saturated liquid solution enthalpies curve 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 trial and error with the help of xy-diagram below.


166

P =cons. superheated vapor

R saturated vapor

HR

G

HG

Specific enthalpies h,H

G1

hM

liquid-vapor mixture

M

hL

cold liquid

hK 0

L K

yR x

zM

xk

L1 saturated liquid y

x1

y1

1.0

x,y 1.0 y1

L1,G1 y =x

y L,G y

0 0

x1

x

1.0

x

Fig.5.9 A typical enthalpy-composition diagram

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 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 respectively. An important characteristic of enthalpy-composition diagrams is the application of graphical addition and subtraction rules along with so-called inverse lever-arm rule. If 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


167

extension and behind the point 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 applying again the inverse lever-arm rule, 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 zM (5-24) Enthalpy balance : K hk + R HR= M hM (5-25) If M is eliminated between equations (5-23) and (5-24), K yR − zM = R zM − x k Similarly by eliminating M between equations (5-23) and (5-25) K H − hM = R is obtained. From these two equations, R hM − hk yR − zM H − hM = R (5-26) zM − x k hM − hk is found. This equation represents a straight line on enthalpy-composition diagram passing through the points K(xk;hk) , M( zM;hM) and R(yR;HR). 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 methods in some details. In the calculations, the solution is assumed to be consisting of A and B. However, from time to time the multicomponent 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 top product vapor

feed

partial vaporizer

P, t, hF1

D, yD, HD, P, t

vapor-liquid separator

F, xF, hF, P, tF liquid

liquid-vapor

qS

liquid

W, xw, hw, P, t bottom product

Fig.5.10 Flow diagram of continuous flash distillation under constant pressure No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

168

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. A simple cyclone type separator may be used as vapor-liquid separator. It is obvious that the top product vapor is richer in the MVC than the bottom product, which is a liquid. For the whole system: Total material balance : F= D+W (5-27) MVC balance : FxF = DyD + Wxw (5-28) Enthalpy balance : FhF + qS = FhF1= DHD + Whw (5-29) can be written, where F, D and W are the molar flow rates of the feed , top and bottom products (k-mol/s), xF, yD, xW are the mole fractions of the MVC in the feed, top and bottom products respectively and, hF, HD and hW show their specific enthalpies (kJ/kmol). If a fractional vaporization is defined by ε ≡ D/F, then with the elimination of W between equations (5-27) and (5-28) and between equations (5-27) and (5-29), x − x W (h F − h W ) + q S /F ε= F = (5-30) yD − xW HD − hW is obtained. The left hand side of this equation can also be written as; x 1− ε (5-31) xW + F ε ε As it is seen, this equation gives the relationship between the compositions of top and bottom products, another relationship between these two is given by the equilibrium 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. For that, the line given by equation (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 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; DyD y 100 = ε D 100 P.R.= (5-32) F xF xF yD = −


169

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. P =cons. D

specific enthalpies h,H

hF1=hF+qs/F

M

hw

W

hF

F 0

xF

xw

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 pressure. Calculate the compositions of top and bottom products and percentage recovery of acetone for the following fractional vaporizations: 0.2, 0.4, 0.5, 0.66. xy-diagram of acetone-water system at 760 mmHg is given in Table.App.5.1.

HD

yD

By substituting the xF and є values into equation(5-31);

1.0

x,y 1.0 y

N Q

For For For For

y =x

yD

ε =0.20 ε =0.40 ε =0.50 ε =0.66

yD = - 4 xw+ 2 yD = - 1.5 xw+ 1 yD = - xw+ 0.8 yD = - 0.515 xw+ 0.6

F

are obtained. All these lines, when drawn on xy-diagram, pass through point F(0.40;0.40). Another points are found as 0 (0.25;1.0), (0.0;1.0), (0.0;0.8),(0.0;0.6) for xF 1.0 xw 0 the first and following lines. With the help x of these points, the lines are drawn and Fig.5.11 Flash distillation at constant pressure from the Q points xw and yD values are read from the diagram as shown below. Percentage recoveries are then computed from equation (5-32). The results are given in the table below.

Vapor-Liquid Equilibrium of Aqueous Acetone at 760 m m Hg

Mole fraction of acetone in vapor,y

1 0.9

ε =0.4

0.8 0.7

ε=0.5

0.6 0.5

ε=0.2

yD= - 4xw + 2 yD= - 1.5xw + 1

ε=0.66

0.4

F

0.3

yD= - xw + 0.8 yD=- 0.515xw+0.60

0.2 0.1

xF

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Mole fraction of acetone in liquid,x


170

P.R. = ε

yD .100 = % xF

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 decreases but the percentage recovery of the MVC increases.

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 vaporization takes place. The vapor-liquid mixture is then directed top product vapor feed

He ater

F, xF, hF, P1, tF

P1, tF1 hF1

P, t

D, yD, HD, P, t

Liquid-vapor separator

liquid

liquid

liquid-vapor

qS

pressure reducing valve

liquid bottom product

W, xw, hw, P, t

Fig.5.12 Flow diagram of continuous flash distillation at reducing pressure

to a vapor-liquid separator from where top product as vapor, bottom product as liquid are withdrawn. This type of operation is more common than the first type of operation in practice and the name “flash distillation” is more appropriate for this type of operation than the first type of operation, as the superheated liquid really flashes at the outlet of the pressure reducing valve. As in this case fractional vaporization є is also not known, the solution requires trial and error. Next to the enthalpy-composition diagram, 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 diagram and hW, HD and hF1 are read from this 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 є is great, then the calculations are repeated with the new temperature selections until the є s are almost the same. As it is seen from the figures, although certain degree of separation is achieved by using flash distillation, obtaining complete separation and hence pure products is almost impossible, although the purities are relatively high when the relative volatility is great. In practice, the flash distillation is generally used to divide the multi-


171

specific enthalpies h, H

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. P =cons. H-y For multi-component solutions, the D following equations can be written: HD Total material balance: F= D+W (5-27) hF1=hF+qs/F Component-i balance: M FxFi = DyDi + Wxwi (5-33) Enthalpy balance: hw W FhF+qs =FhF1= DHD + Whw (5-29) h-x xFi = ε yDi + (1-ε) xWi hF F is obtained from equations (5-27) xw 1.0 y and (5-33), by noting that ε = D/F. xF D x,y If the vapor-liquid equilibrium is given as: (5-34) yDi = Ki xWi

temperature

tB

then,

t-y D

t

W

x Wi =

M

xw

(5-35)

tA

t-x 0

x Fi ε K i + (1 − ε)

xF

yD

1.0

x,y

is found. Between equations (5-27) and (5-29), hF1 = ε HD + (1-ε) hw

(5-36)

Fig.5.13 Flash distillation by pressure reduction

is obtained. This equation can also be written in terms of latent heat of vaporization, heat capacities and temperatures as, c FL t F1 = ε (λ o + cG t) + (1 − ε) c wL t

(5-37)

where, reference temperature has been taken zero, λ o (kJ/k-mol) is the latent heat of vaporization of the mixture at reference temperature, c FL and c wL are the heat capacities of the liquid feed and bottom product (kJ/k-mol K) and cG 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. No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

172

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 flashing and the calculations are correct; if not, the calculations are repeated by selecting new temperatures until this condition is satisfied. Example-5.7) Flash Distillation by Pressure Reduction

The bottom product from a distillation column is a binary liquid solution of composition 0.5 mole fraction. It is supplied continuously at 135 oC and 5 bar pressure. Calculate its fractional vaporization on adiabatic pressure reduction to atmospheric pressure and the compositions of vapor and liquid products. The latent heat of vaporization at 0 oC is 38 000 kJ/k-mole, molar specific heat of the liquid feed and bottom product is 250 kJ/k-mole K, the molar specific heat of vapor is 180 kJ/k-mole K and the relative volatility is 3. The system obeys Raoult’s law and the variation of vapor pressure of the LVC with temperature is as follows: t(oC) p oB ( mbar)

60 280

65 340

70 410

75 486

80 570

85 670

90 780

Top product (vapor)

D,yD,HD,P,t

Distillation column

c G = 180 kJ / k − mol K

P1 =5 bar

vapor-liquid separator

liquidvapor

∆ Bottom product (binary)

xF =0.5 tF1=135 oC

t

c FL = 250 kJ/k - mol K

t

h

W,xw,hw,P,t Bottom product (liquid)

c w L = 250 kJ/k - mol K

P=1.013 bar

Pressure reducing valve

Solution

From equation (5-30)


ε=

ε=

xF − xW 0.5 - x w = yD − x W yD − x w c wL (t F1 − t ) 250(135 − t ) = λ o − (c wL − cG )t 38 000 − (250 − 180) t

(1)

(2)

For ideal solutions equations (5-3), (5-4) and (5-15) are also valid. Solution is by trial and error. Select t = 78 oC. Then from the given table p oB = 533 mbar is read.


173

From equation (5-15) p oA = α p oB = (3)(533 ) =1599 mbar From equations (5-3) and (5-4)

P − p oB

xw =

p oA

−

p oB

=

1013 − 533 = 0.45 1599 − 533

pA po 1599 = A xw = (0.45) = 0.71 P P 1013

yD =

Then from equation (1)

ε=

0.5 - 0.45 = 0.192 0.71 − 0.45

From equation (2)

ε=

250(135 − t ) 250(135 − 78) = = 0.438 38 000 − (250 − 180) t 38000 − (70)(78)

Since no check is available, selected temperature is not correct. Select t = 81 oC. From table above, p oB = 590 mbar

p oA = α p oB = (3)(590 ) =1770 mbar



xw =

P − p oB 1013 − 590 = = 0.359 p oA − p oB 1770 − 590

yD =

p A p oA 1770 = xw = (0.359) = 0.627 P P 1013


ε=

0.5 - 0.359 = 0.526 0.627 − 0.359

From equation (2)

ε=

250(135 − t ) 250(135 − 81) = = 0.418 38 000 − (250 − 180) t 38000 − (70)(81)

So, t is between 79 and 81 oC. Select From equation (5-15)

t = 80 oC. From table above p oB = 570 mbar

p oA = α p oB = (3)(570 ) =1710 mbar



xw =

P − p oB 1013 − 570 = = 0.389 p oA − p oB 1710 − 570

yD =

p A p oA 1710 = xw = (0.389) = 0.657 P P 1013

ε=

0.5 - 0.389 = 0.414 0.657 − 0.389


174

From equation (2)

ε=

250(135 − t ) 250(135 − 80) = = 0.424 38 000 − (250 − 180) t 38000 − (70)(80)

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 liquid are 0.657 and 0.389.

5.3.2 Simple or Differential Distillation: Simple distillation, which is a single-stage operation, is conducted batch-wise. The equipment, which is named as batch-still, consists mainly of a boiler and a condenser connected to each other as shown in Fig.5.14. The boiler of batch-still is a fairly large vessel containing heating coils through which heating medium is circulated. Vessel is charged with liquid solution up to 70-75 % of its volume and then heating medium is set to circulation. When the temperature of the solution reaches its bubble point temperature, vaporization starts and the vapor formed rises through the pipe connecting boiler and condenser and upon condensing there the liquid, which is called top product, flows into a receiver. By adjusting the heat energy in the boiler, slow and constant rate vaporization is achieved, which results in attaining equilibrium at any time between rising vapor and remaining liquid during the operation. Since the vapor formed is always richer in the MVC than the remaining liquid, liquid becomes poorer in the MVC as the distillation proceeds. As a result of this, the forming vapor steadily becomes poorer also in the MVC. It follows from this that the vapor formed at the start of the vaporization is the richest and the vapor formed at the end of vaporization is the poorest in the MVC. The operation is stopped when the amount of liquid in the vessel or its composition drops Condenser

Cooling medim

Vapor

Top product

Charge

Boiler

Heating medium

Receiver

Bottoms

Fig.5.14 Batch-still


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 composition of the vapor steadily changes, using more than one receiver and hence collecting more than one top product, which are named as cuts are also possible. It is obvious that the cut collected in the first receiver will be richer in the MVC than the cuts collected in the subsequent receivers. The analysis of simple 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 kmole 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 F

xF

W

xw

⌠ dx ⌠ dL ⎮ = ⎮ ∗ ⌡ L ⌡y − x xF

and finally,

F ln = W

⌠ dx ⎮ ∗ ⌡y − x

(5-39)

xw

is obtained. With the help of this equation, any one of the F, W, xF 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, x (1 − x w ) 1− xw F 1 ln = + ln ln F (5-40) W α −1 x w (1 − x F ) 1− xF


176

1.0

P =cons.

1 y −x

y

∗

y =x

y

Area= ln

F W

0

0 0

1.0

x

0

xw

xF

x

x Fig.5.15 Graphical integration

or

ln

F xF F (1 − x F ) = α ln W xw W (1 − x w )

(5-41)

is obtained. In addition to these equations,, Total material balance : F = D+W (5-42) (5-43) MVC balance : F xF = D xD,av. + W xw can also be written from which the average composition of the top product (composited composition), xD,av. can be computed. Percentage recovery of the MVC is given as, D x D,av. P.R. = 100 (5-44) F xF If the solution is a multi-component ideal solution, one of the components (let’s say component j) is selected as key component and the relative volatilities are defined to this component. Hence, αij shows the relative volatility of component i to the component j. In this case equation (5-41) is written as, F x Fj F x Fi (5-45) ln = α ij ln W x wi W x wj 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

∑x

wi

=1

(5-46)

1

The heat energy requirement in the simple distillation can be calculated from, q T = q S + q L = F cFL (t − t F ) + D λ (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 temperature t (oC), qL (kJ) is the latent heat required to distill D k-mol distillate, c FL ( kJ/k-mol oC), is the molar heat capacity of the solution calculated at the arithmetic


177

mean of t and tF, and 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, ⎡t − t ⎤ ln ⎢ H F ⎥ t −t ⎦ θS = ⎣ H (5-48) (sec) UA / FcFL for isothermal heating fluid such as saturated steam at tH (oC) temperature, and by ⎡t − t ⎤ ln ⎢ H1 F ⎥ ⎣ t H1 − t ⎦ (5-49) θS = (sec) & HcH ⎛ K − 1 ⎞ m ⎜ ⎟ FcFL ⎝ K ⎠ for non-isothermal heating fluid whose inlet temperature is tH1. K is given by UA

& H (kg/s), and cH (kJ/kg oC) are the mass flow K = e m& c . In the equations above, m rate and mass specific heat of heating fluid, A is the heat transfer area in the batch-still (m2) and U is the overall heat transfer coefficient (kW/m2 oC). For the distillation time, qL (sec) θL = (5-50) U b A( t H − t ) ln can be used. Where, Ub is the overall heat transfer coefficient of boiling solution (kW/m2 oC). Then, total time for one batch operation; (sec) (5-51) θBT = θS + θL + θCD 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 the examples below. H H

Example-5.8) Simple Distillation

123.1 k-mole acetone-water solution at 20 oC and containing 20 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 0.04. 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) 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


178

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. Vapor-liquid equilibrium for acetone-water system at 760 mmHg is given in Table.App.5.1. Solution: First, plot xy- and t-xy diagrams of the system and then evaluate the integral given by eqn.(5-39) by selecting arbitrary x values between xF and xW and reading the corresponding y* values as shown in table below: select x

read

c

y*

1 ∗ y −x

a

l

c

u

l

a

t

e

0.20

dx ⌠ ⎮ ∗ ⌡0.04 y − x

0.04

0.55

1.961

-

0.08

0.71

1.587

[(1.961+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

0.20

ln

F ⌠ dx =⎮ = 0.2676 ∗ W ⌡0.04 y − x

123.1 0.2676 =e = 1.307 W

W = 94.19 k-mol

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 Dx D ,av (28.91)(0.721) .100 = .100 = 84.7 % c) From equation (5-44), P.R. = Fx F (123.1)(0.20) d) As it is seen from the t-xy diagram of the system, distillation starts at 64.8 oC and ends at 78 oC. Then, take the mean temperature 71.4 oC as the distillation temperature. 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, qL = (28.91)(32 483) = 940 732 kJ From equation (5-48), the time required to heat the solution from 20 oC to 71.4 oC ; θS =

ln[(121 − 20) / (121 − 71.4)] = 3 997 sec . ≈ 1 h 07 min (0.200)(10) / (135.41)(83)


179

t-xy diagram of Aqueous Acetone at 760 mmHg

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

tem perat ure,t (o C)

y

Vapor-Liquid Equilibrium of Aqueous Acetone at 760 mmHg

0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

100

1

90 80 70 60 50 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

x

1

Mole fraction of acetone, x,y

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 tH =121 oC. From equation (5-50), the time required for distillation;

θL =

940 732 = 1 897 sec ≈ 32 min (1.0)(10)(121 − 71.4)

So, from equation (5- 51); the time required for one batch operation, with 1 h charging and discharging time; θBT = 1 h 07 min + 32 min + 1 h = 2 h 39 min

e) Fractional vaporization, ε = 28.91 = 0.235 123.1 Then, from equation (5-31) y D = − 1 − 0.235 x W + 0.20 0.235 y D = − 3.26 x W + 0.85

0.235

is obtained. By drawing this line on xy-diagram, and reading the coordinates of point Q ; yD = 0.67, xw = 0.065 are obtained. From equation (5-32)

P.R. = (0.235)

(0.67) .100 = 78.7 % (0.20)

As it is seen, simple distillation gives higher purity and higher P.R. than the flash distillation under identical operation conditions. Example-5.9) Simple Distillation: Effect of Relative Volatility on Purity

123.1 k-mole (12 000 kg) chloroform-toluene solution at 20 oC and containing 20 mole percent chloroform 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) The amount of distillate to be produced, b) The mole fraction of chloroform in the distillate (purity), c) The percentage recovery of chloroform, d) Heat energy requirement for the operation. e) 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. f) 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


180

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

c

read

a

l

c

u

l

a

t

e

0.20

dx ⌠ ⎮ ∗ ⌡0.04 y − x

1 ∗ y −x

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 0.20

ln

F ⌠ dx = 0.76194 =⎮ ∗ W ⌡0.04 y − x

123.1 0.76194 =e = 2.142 W

W = 57.47 k-mol

From equation (5-42), D= 123.1 – 57.47 = 65.63 k-mol b) From equation (5-43), (123.1)(0.20) = (65.63) xD,av + (57.47)(0.04) c) From equation (5-44),

P.R. =

xD,av = 0. 34

(65.63)(0.34) .100 = 90.6 % (123.1)(0.20)

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 temperature 100.5 oC as the distillation temperature. From equation (5-47), the sensible heat required, qS = (123.1)(140)(100.5 - 20) = 1 387 337 kJ The latent heat required, qL= (65.63)(32 000) = 2 100 160 kJ Total heat required, qT= 3 487 497 kJ e) From equation (5-48), the time required to heat the solution from 20 oC to 100.5 oC ;

θS =

ln[(121 − 20) / (121 − 100.5)] = 15 116 sec . ≈ 4 h 12 min (0.200)(10) / (135.41)(140)

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 tH=121 oC. From equation (5-50), the time required for distillation; 2 100 160 θL = =10 245 sec ≈ 2 h 51 min (1.0)(10)(121 − 100.5)


181

t-xy diagram of chloroform-toluene at 760 mmHg

Temperature, t (OC)

tB

y

t-y

t-x

tA

Mole fraction of chloroform, x,y

xy-diagram of chloroform-toluene at 760 mmHg 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x So, from equation (5-51); the time required for one batch operation, with 1 h discharging time; θBT = 4 h 12 min + 2 h 51 min + 1 h = 8 h 03 min

charging and

f) As it is seen, the purity of the top product is much less than the purity of the top product of the previous example. In order to understand this, look at the relative volatilities of the two systems. From xy-diagrams of the two systems, the relative volatility of acetone to water and the relative volatility of chloroform to toluene over the concentration range involved were calculated with the help of equation (5-13) as follows:

x αAcetone-Water αChloroform-Toluene

0.04 29.33 4.4

0.08 28.16 4.36

0.12 21.09 4.09

0.16 17.09 3.96

0.20 14.18 4.0

As it is seen, there is great difference between the relative volatilities of the two system. Flash and simple distillations can produce high purity top products only when the relative volatility of the system is great. So, it is obvious from these examples that flash or simple distillation techniques can not be used to separate solutions such as chloroform -toluene whose relative volatilities are not great. As 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 above given distillation techniques produces complete separation of the solution. Another distillation method, which is known as rectification and which is more complicated and expensive than the previous ones, must be used for high purity products and high percentage recoveries. For that, the rectification is the most widely used separation method in chemical industry.The equipment used is multi-stage column, which may be either packed or plate type. Although continuous operation is more common, batch-wise operation is not also very seldom. A typical rectification column is shown in Fig.5.16. Although this column is a plate column it could also be a packed column. As it is seen, the liquid solution to be separated (the feed) is introduced centrally to the column. As the liquid flows down, it comes in contact with rising vapor on the plates where mass transfer between the phases takes place in both directions. The vapor and liquid contacting on a plate try to come equilibrium. For that, part of the LVC passes from vapor to liquid by partial


182

condensation. This enriches the liquid in the LVC and the remaining vapor in the MVC. The heat of condensation 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 vapor each plate. As a result of these, the liquid solution flowing down from qC plate to plate continuously enriches in the LVC. By providing sufficient condenser number of the plates (or packing height) in this region of the column, the liquid reaching bottom of the 1 Reflux column can be made almost pure in liquid Top the LVC. Part of this liquid is product Enriching vaporized in a reboiler and returned section back to the bottom of the column as reflux vapor, thus a vapor phase, which is required for interphase Feed mass transfer is formed in the column. Notice that as this vapor is formed from a liquid, which is Stripping almost pure in the LVC, itself is also section almost pure in the LVC. The remaining part of the liquid is withdrawn as bottom product. This vapor N liquid is the liquid, which is richest in the LVC in the column. If we qs follow the vapor given back to the column; this vapor enriches in the reboiler MVC as it passes from plate to plate liquid up by repeated partial vaporizations Bottom and condensations taking place on product the plates as explained above. It has the highest MVC composition at Fig.5.16 A plate rectification column 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 the liquid passes from liquid to vapor by taking this heat of condensation. By


183

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 G1,y1,H1 provide the liquid phase rich in the MVC condenser on the plates above the feed plate. The qC remaining part of the condensate is withdrawn as top product or distillate. If we look at the liquid and vapor phases flowing inside the column, we see that 1 Lo,xo the vapor, which enters the column at the ho D,xD,hD L1,x1 G2,y2 bottom as almost pure in the LVC, turns h1 H2 Top to MVC as it rises in the column and 2 product leaves the column as almost pure in the L2,x2 G3,y3 MVC at the top. The liquid, which is h2 H3 given at the top of the column as almost pure in the MVC, turns to the LVC as it Ln-1,xn-1 Gn,yn hn-1 Hn flows down and with joining the feed at n the feed plate leaves the column at the bottom as almost pure in the LVC. As it Ln,xn Gn+1,yn+1 is seen, all the streams leaving the hn Hn+1 column at the top and bottom are not taken as products, but parts of them, after Feed phase changes, are returned back to the F,zF,hF column as reflux. If this is not done, production of pure products is not possible. The liquids in the column are m always at their bubble point temperatures and the vapors are at their dew point Lm,xm Gm+1,ym+1 hm Hm+1 temperatures. Hence, the lowest temperature in the column is always at the top and the highest temperature is at the bottom. As it is seen, the feed divides N the column in two sections. The section GN+1,yN+1 LN,xN below and including the feed plate is hN HN+1 named as stripping section of the reboiler column as in this section the rising vapor strips off the MVC from the liquid. The qS section above the feed plate is known as Bottom enriching section of the column as the product vapor in this section is further enriched in W,xw,hw the MVC. The purities of the top and bottom Fig.5.17 Notations in a continuously operating products do not only depend on the plate rectification column number of the plates in each section but


184

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 the heating cost, a heat exchanger between bottom (sometimes top) product and the feed is placed. The coolers, which cool the products to storage temperatures, are frequently added to the product lines. The column itself and all the hot lines are insulated against the heat losses. The column, which is the main part of the system may be plate or packed column. Below, first the design principles of a plate rectification column will be given. 5.3.3.1 Continuous Rectification in a Plate Column: The first step in the design of a plate rectification column is to determine the numbers of the plates required in enriching and stripping sections. All the plates in the column are first assumed as ideal or equilibrium or theoretical plates, as in gas absorption, although they are not, and then numbers of the ideal plates required in each section are computed. By using plate or column efficiencies, numbers of the theoretical plates are then converted to the numbers of the real plates. The plates are numbered from top to the bottom and n and m show any plate in the enriching and stripping section respectively. N is the last plate of the column. L and G are the total molar flow rates of the liquid and vapor phases respectively as k-mol/s. x and y show the mole fraction of the MVC in liquid and vapor. Since these change from plate to plate, the number of the plate from which these streams come, is given as subscript. Hence, L5, y4 and H7 show the liquid leaving the 5th plate, mole fraction of the MVC in the vapor leaving the 4th plate and the specific enthalpy of the vapor leaving the 7th plate respectively. F, D and W are the molar flow rates of the feed, top and bottom products and zF, xD and xw are the mole fraction of the MVC in the feed, top and bottom products and hF, hD and hw are the specific enthalpies of the feed, top and bottom products respectively. The total and MVC balances and the enthalpy balance equations around the plate n at steady-state are written as, Total material balance : (5-52) Gn+1 + Ln-1 = Gn + Ln MVC balance : (5-53) Gn+1 yn+1 + Ln-1 xn-1= Gn yn + Ln xn Enthalpy balance : (5-54) Gn+1 Hn+1+ Ln-1 hn-1 = Gn Hn+ Ln hn where, Hn+1 and hn are the specific enthalpies of the saturated vapor and saturated liquid. From equations (5-21) and (5-22), Hn+1 = y n+1 cLA (t n +1 − t o ) + (1 − y n +1 ) cLB (t n+1 − t o ) + y n+1 λ A + (1 − y n+1 ) λ B h n = x n cLA (t n − t o ) + (1 − x n ) cLB (t n − t o ) + ∆H s

can be written. In many cases, t n ≅ t n+1 and λ A , λ B > ∆ Hs and hence ∆ Hs can be neglected and then by subtracting the second equation from the first, H n +1 − h n = y n+1 λ A + (1 − y n+1 ) λ B (5-55) is obtained. It is known that the molar (not the mass) latent heats of vaporization of some liquids are almost equal. Then λA ≈ λB(= λ ) and it follows from equation (5-55) Hn+1-hn = λ = constant No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

185

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 G and L 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 xydiagram 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: y n+1 =

D xD L xn + G G

(5-58)

This equation relates the compositions of the liquid leaving and the vapor entering the same numbered plate. Hence, this equation is nothing but operating line equation. As this equation was written in the enriching section then the equation (5-58) is the enriching section operating line equation. It is represented by a straight line passing through point D(xD;xD) on xy-diagram. In any rectification operation, the ratio of liquid 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 the most important operating parameter. The equation (5-58) is then preferred written in terms of reflux ratio. By dividing both nominator and denominator of equation (5-58) by D and by noting that RD = L/D, RD xD (5-59) xn + 1+ RD 1+ RD is obtained. This equation is represented by 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 the column (envelope-2); y n+1 =


186

G,y1 qC

Envelope-1

1

L,xo

LG

D,xD

2 n xn

yn+1

F,zF

L G Envelope-2

m xm

ym+1

G yN+1

N

qS

Total material balance : L= G + W (5-60) MVC balance : L x m = G y m+1 + W x w (5-61) are written. The last equation can also be written as, W xw L xm − y m+1 = (5-62) G G 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(xw;xw). For the whole column; the following equations can also be written: Total material balance :F = D + W (5-63)

MVC balance F zF = D xD + W xw (5-64) and percentage recovery of the MVC W,xw D xD : P.R. = .100 (5-65) F zF Fig.5.18 Constant molar over-flow rectification 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 (561). Hence, G yq = L x q + D x D L xN

G yq = L x q − W x w are written. By subtracting the second from the first, (G − G) y q = (L − L) x q + D x D + W x w (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: ⎛G − G ⎞ ⎛L − L⎞ L ⎜ ⎟ yq = ⎜ ⎟ x + zF G (5-67) ⎝ F ⎠ q ⎝ F ⎠ F is found. On the other hand, total material balance around the feed plate is written as F + L + G = L + G . By rearranging L G this equation, No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

187

⎛L − L⎞ ⎛G − G⎞ ⎟+1 ⎜ ⎟=⎜ ⎝ F ⎠ ⎝ F ⎠

(5-68)

is obtained. Upon substitution of (G − G) /F from this equation into equation (5-67), ⎡ ⎛ L − L ⎞⎤ ⎛L − L⎞ (5-69) ⎢1 − ⎜⎝ F ⎟⎠⎥ y q = ⎜⎝ F ⎟⎠ x q + z F ⎣ ⎦ is found. If a q-parameter is defined by L−L q≡ (5-70) F finally, equation (5-71) is obtained from equations (5-69) and (5-70) q zF yq = xq − (5-71) q −1 q −1 This equation, which is known as q-line equation, is represented by a straight line passing through point F(zF;zF) with a slope of q/(q-1) on xy-diagram. Since the two operating lines cut each other always on this line, equation (5-71) is then the locus of intersection points of the two operating lines. For the drawing of q-line, the qparameter defined by equation (5-70) must be known. It follows from equation (5-70) that the value of q-parameter depends on the thermal condition of the feed. The feed may be introduced to the column in one of the 5 thermal conditions. These are: a) Cold liquid; as in this case the feed is below its saturation temperature, it causes to the condensation of the part of the vapor coming from stripping section on the feed plate in order to become saturated liquid. The change of quantities of vapor and liquid phases around the feed plate is schematically shown in Fig.5.19 a, from which q > 1 is seen. An equation for the calculation of exact value of q, which will be greater than 1, L

L

G

L

G

F

F

L

F

L

G

(a)

L

G

G

(c)

(b)

L

G

G

L

G F

F

L

G (d)

L

G

(e)

Fig.5.19 The change of quantities of vapor and liquid around the feed plate according to the thermal conditions of feed No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

188

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 joins the liquid coming from the enriching section of the column as shown in Fig.5.19 b, and q in this case equals to 1. c) A mixture of vapor-liquid: in this case vapor part of the feed joins the vapor coming from stripping section and liquid part joins the liquid coming from the enriching section as shown in Fig. 5.19 c. q-parameter, which is 0
L,hL

G,HG

L, h L

G, H G

F,hF

HG − h F H − hF = G HG − h L λ

(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, H G − h F = cFL (t Fb − t F ) + λ can be written, where tF and tFb are the entering and bubble point temperatures of the feed and cFL 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), c (t − t ) q = 1 + FL Fb F (5-74) λ is obtained. For superheated vapor feed, H G − h F = c FG ( t Fd − t F ) can be written, where, tF and tFd are the entering and the dew point temperatures of the feed vapor and c FG is the molar heat capacity of the feed vapor calculated at arithmetic mean of tF and tFd . By entering this into equation (5-73), c (t − t ) (5-75) q = FG Fd F λ is found.


189

After seeing the operating and q-lines, we can now start with the calculation of the 1.0 P=cons. 0
y

q>1

q=1 (b)

y=x

(a)

(c) q= 0

(d) F (e)

q< 0 0 0

zF

x

1.0

Fig.5.20 The appearances of q-lines on xy-diagram

number of the ideal plates required for a given separation. But before this, let us write the knowns. They are: F, zF, thermal condition of the feed, xD and xw (or percentage recovery of the MVC). The designer then selects the operating pressure and obtains the xy-diagram of the system at this pressure. The designer also decides on the reflux ratio to be used in the operation by considering the criteria, which will be given later. Of course, he must also check whether the constant molar over-flow case applies or not by obtaining the latent heats of vaporization of the components at the operating conditions. Plate to Plate Calculation Method or Lewis-Sorel Method: First xy-diagram of the system is plotted at the operating pressure, and then q-line is drawn on the same diagram after calculating the q value from the thermal condition of the feed. The 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. Since the condenser is a total condenser (there are cases where condenser operates as partial condenser, this case will be dealt with later), y1 = xo = xD. Hence, the composition of the vapor y1 leaving the first plate is known then the composition of the liquid x1 leaving the same plate is read from equilibrium curve by entering this y1 into the diagram. 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 y2 value. After reading each x value from the equilibrium diagram, it is compared with the known xq to check whether the feed plate is reached or not. If the last calculated x value is equal to or smaller than xq then for the calculation of y values the stripping section operating line equation is used instead of enriching 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


190

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 Calculation Method or McCabe-Thiele Method: In the calculation of the number of the ideal plates by plate to plate calculation method, equilibrium curve and operating lines are alternately used, starting at the top of the column. McCabe and Thiele, seeing this, have drawn the operating lines on the diagram on which equilibrium curve and q-line had 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. The numbers of the right angle triangles above and below q-line give the required numbers of the ideal plates in enriching and stripping sections. 1.0

y

y1 y2

P=cons.

y3

q-line

1 2

D

y=x

3

y4

4

y5

5

Enric.sec.op.line Q

y6 K

6

F

y7

xD 1+ RD

7

Strip.sec.op.line y8 0 0

8

W xw x7 x6 x8

x5

x4

zF x3

x2

x1

xD=xo 1.0

x

Fig.5.21 Determination of number of ideal plates by McCabe-Thiele method

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

191

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, hence the ideal plates is infinite at this reflux ratio which is named as minimum reflux ratio and shown by RDm. So, the minimum reflux ratio is defined as the 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 1.0

1.0

P=cons.

q-line

D

y

P=cons.

q-line

y

D

T

'

y′q

Q

K'

Enr.sec.op.line at min. reflux ratio

Q

y′q

Enr.sec.op.line at min. reflux ratio

Q'

K'

Q

y=x

y=x

xD 1 + R Dm

xD 1 + R Dm

F

F Strip.sec.op.line at min.reflux ratio

Strip.sec.op.line at min.reflux ratio

W

0 0 xw

W

0

x′q

xD 1.0

zF

x

0 xw

x ′q

xD 1.0

zF

x (b)

(a)

Fig.5.22 Determination of minimum reflux ratio

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 than minimum reflux ratio, is given in terms of minimum reflux ratio in the form of RD= β RDm. 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, xD R Dm (5-76) xn + y n+1 = 1 + R Dm 1 + R Dm Hence, it can easily be calculated either by equating the slope of DQ' line (for Fig.5.22 b the slope of tangent) to RDm/(1+RDm) or equating its intercept value (point K') to xD/(1+RDm). From the first, x D − y′q R Dm = (5-77) y′q − x ′q is obtained.


192

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, x (q − 1) + z F (1 + R Dm ) q x D + R Dm z F y′q = x ′q = D q + R Dm R Dm + q then he substituted these coordinates into the definition equation of relative volatility y′q /(1 − y′q ) and obtained the following equation, written at point Q' which is α = x ′q /(1 − x ′q ) R Dm z F + q x D α [x D (q − 1) + z F (R Dm + 1)] = R Dm (1 − z F ) + q(1 − x D ) (R Dm + 1) (1 − z F ) + (q − 1) (1 − x D )

(5-78)

RDm can be computed from this equation only by trial and error for a given value of q. However, equation (5-78) simplifies to the following equations for 1 ⎡ xD α (1 − x D ) ⎤ R Dm = q =1 (5-79) − α − 1 ⎢⎣ z F (1 − z F ) ⎥⎦ 1 ⎛α xD 1− xD ⎞ ⎜ ⎟ (5-80) − α − 1 ⎝ zF 1 − zF ⎠ The α in the equations must be taken at the intersection point (point Q') of the two operating lines. Let us consider now another limit of reflux ratio, which is known as total reflux. The reflux ratio, which is defined by RD = L/D becomes infinite at D = 0. When no top product is withdrawn from the column, no bottom product must be withdrawn and hence no feed is introduced. Hence, certain quantity of solution charged to the column is vaporized in the reboiler and then condensed in the condenser and returned back to the column. In this type of operation, which is known as operation under total reflux, the slopes of both operating lines become 1 and hence the diagonal between points D 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 reflux depends on both the size of the column and the system. Fenske has derived an analytical equation for the calculation of the number of the ideal plates under total reflux, when the relative volatility is constant or near constant. From the definition equation of relative volatility for the last plate of the column, xN yN =α (5-81) 1− xN 1− yN can be written. But at total reflux yN = xN-1 and then this equation becomes; x N−1 xN (5-82) =α 1 − x N−1 1− xN

q =0

R Dm =


193

1.0

P=cons.

1

D

2

y 3

y =x F

4

5

W

0 0

xw

zF

x

xD 1.0

Fig.5.23 Determination of ideal plates under total reflux

Similar equation for the plate (N-1) is, y N−1 x N−1 xN (5-83) =α = α2 1 − y N−1 1 − x N−1 1− xN If similar equations are written for the other plates until reaching the first plate, xN y1 (5-84) = αN 1 − y1 1− xN is obtained. For 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, xw xD (5-85) = αN 1− xD 1− xw By taking the logarithm of both sides and solving for N, x (1 − x w ) log D x w (1 − x D ) (5-86) Nm = log α 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 cost, which is probably the 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 infinite (why?), the fixed capital cost is also infinite. With increasing RD, the cost of column decreases due to the drop in the height, and hence the fixed capital cost of the system also decreases. Although with increasing RD the sizes and hence the costs of


194

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 = G λ and qc = G λ respectively. As it is seen, these costs are directly proportional to the vapor flow rates in the column.

Annual cost, YTL

Total cost

Operating cost min.

Fixed capital cost

RDm (RD)op

RD

Fig.5.24 Determination of optimum reflux ratio

A material balance around the condenser gives G = D(1+RD), which shows that cost of condenser load increases with increasing RD. On the other hand, since for a saturated liquid feed G = G = D(1 + R 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


195

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 1.0 P =cons. plates, starting from point D it cannot be D passed from enriching section operating line q-line E to the stripping section operating line before point E. Similarly, it must be passed from y enriching section operating line to the Q y=x R stripping section operating line at last at point R. So, it follows from this that from one F operating line to the other can be passed at anywhere between E and R. The best passing W point is the point at which the number of the 0 0 xw x zF ideal plates needed for a given separation is a D 1.0 x minimum. If it is noticed the figure, it is seen Fig.5.25 Optimum feed plate 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 minimum number 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 vaporliquid feed. For possible changes in the feed plate location that might occur in future due to the changes in the composition 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:

[u

0.5 G

0.25 ρ G0.25 + u 0.5 L ρ L ] < 0.6 [g d i (ρ L − ρ G )]

0.25

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

196

Top product

Vent Condenser Condenser

Cooling liquid Cooling liquid Receiver Reflux liquid

1.plate Column

Column Top product

( c)

1.plate

1.0

x slope=-L/D

(a)

yD Vent

D

C

y

y1 1

Condenser

xD= yD xo

x1 Enr sec.op.line

Cooling liquid

y=x Receiver (d)

Reflux liquid

1.plate Top product Column

(b) Fig.5.26 Types of condensers : (a) built-in condenser, (b) total condenser, (c) partial condenser, (d) enrichment in the partial condenser


197

as kg/m3, di is the inside diameter of the tubes in m. In small capacity systems, the 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 condenser is filled with condensate partly or totally and it does not fulfil the design duty. In large capacity plant, the condenser is 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 be removed in it and hence, the condensate produced must be a saturated liquid. If it is sub-cooled, not only 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 sufficient or the top product is used in vapor form 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 condenser, usually the cooling water coming from the cooling tower is used as cooling medium. At low condensation temperatures cooling medium may be chilled water 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 from 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 system, is essentially a heat exchanger and, depending upon the heat load and the properties of the liquid solution, in different types are constructed and at different modes are operated. It is generally placed outside the column as a separate piece of equipment. But when heat load is low and the fouling is not a problem, it can be located in the column as shown in Fig.5.27a. For large capacities, kettle-type (b) and vertical thermosyphon types (c, d) are widely used. To prevent the fouling of the tubes, thermosyphon type reboilers are generally operated on partial vaporization principle; the mixture issuing from the reboiler comprises both liquid and vapor. For heat sensitive components, thermo-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 the kettle-type reboiler, vaporization is slow and low, the vapor formed reaches in equilibrium with the liquid in the reboiler and then reboiler 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


198

Column

Column

last plate

last plate

∇

heating medium

Reboiler heating medium

∇ Reboiler bottom product bottom product (a)

(b)

Column

Column

last plate

last plate

∇ heating medium

Reboiler

Reboiler

∇ heating medium

bottom product

bottom product

(c)

(d)

Fig.5.27 Types of reboilers : (a) built-in reboiler, (b) kettle-type reboiler, (c) thermosyphon type reboiler, (d) once-through thermo-syphon type reboiler


199

chosen that the velocity of vapor at liquid-vapor interface, uG (m/s) satisfies the following equation; 0.5

⎡ ρ − ρG ⎤ (5-88) u G < 0.2 ⎢ L ⎥ ρ G ⎦ ⎣ Otherwise, excessive liquid entrainment cannot be evaded. In the reboilers, the following heating 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 = G λ , 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 methanol will be rectified in a plate column at 760 mmHg 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 equilibrium of the system at 760 mmHg is given in Table.App.5.3 and some physical data for the components are given below. Methanol Boiling point, b.p. (oC) at 760 mmHg

n-Propanol

64.5

97.2

32

60

Latent heat of vaporization, λ (kJ/kg) at b.p.

1 100

693

Specific heat, cL (kJ/kg oC) at 50 oC

2.70

2.65

Molecular weight

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-mol As the difference is not much, “Constant Molar Over-flow” case is applicable. a) Then work with molar units,

F= zF =

(15 000 )( 0 . 223 ) (15 000 )(1 − 0 . 223 ) + = 298 . 78 k − mol / h 32 60

(0.223 / 32) = 0.35 (0.223 / 32) + (1 − 0.223) / 60

xD =

(0.91 / 32 ) = 0.95 (0.91 / 32 ) + (1 − 0.91) / 60


200

xw =

(1 − 0.978) / 32 = 0.04 (1 − 0.978) / 32 + (0.978/ 60)

From equations (5-63) and (5-64),

D =

F(z F − x w ) 298 . 78 ( 0 . 35 − 0 . 04 ) = 101 . 78 k − mol / h = (x D − x w ) ( 0 . 95 − 0 . 04 )

W = 298.78 - 101.78 = 197 k-mol/h b) To find the thermal condition of the feed, bubble point temperature of the feed (tFb) must be known. Plot t-xy diagram and read the bubble point temperature of the feed as tFb = 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 cFL is 50 oC. Then, molar specific heat of the feed and average latent heat of vaporization;

c = (0.35)(2.70)(32) + (1 − 0.35)(2.65)(60) =133.6 kJ / k − mol o C λFL= (35 200 + 41 580) / 2 = 38 390 kJ / k − mol q =1+

(133.6)(80.5 − 20) = 1.21 38 390

Then, from eqn.(5-71); y q = 5 . 762 x q − 1 . 667

After drawing this line on xy-diagram, point Q' is marked. By joining point D with Q' and extending the line until cutting y-axis K' = 0.56 is read.

xD 0.95 −1= − 1 = 0.696 K′ 0.56 Operating reflux ratio, RD = (1.87)(0.696) = 1.30 Then from,

R Dm =

minimum reflux ratio is computed.

t-xy diagram of methanol/n-propanol at 760 mmHg 100

temperatures , t (0C)

tB 90

t-y 80

tFb=80.5 oC t-x

70

tA

zF=0.35 60 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

mole fraction of methanol in liquid and vapor,x,y


201

xy-diagram of methanol/n-propanol a t 760 mmHg 1

y1 0.9

1

yq= 5.762 xq + 1.667

y2

D

2 y3

mole fraction of methanol in vapor,y

0.8

3

Q' 0.7

y4

0.6

y5

4 Q

y n +1 = 0.565 x n + 0.413

5

'

K =0.56

y=x

0.5

y6 0.4

K=0.413

0.3

y7

6 F 7

0.2

y m + 1 = 1 . 685 x m − 0 . 0274

y8

8 0.1

y9 y10

9

W x7

x6

x5

0.2

0.3

0

x90 xw x0.1 8

x4 zF

0.4

x3 0.5

x1

x2 0.6

0.7

0.8

0.9

xD

1

mole fraction of methanol in liquid,x

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

L = L + qF = 132.3 + (1.21)(298.78) = 493.84 k − mol / h G = L − W = 493.84 − 197 = 296.84 k − mol / h

Then, from eqn.(5-59) enriching section operating line; y n +1 =

0.95 1.30 xn + 1 + 1.30 1 + 1.30

y n +1 = 0.565 x n + 0.413

And from eqn.(5-62) stripping section operating line; (197)(0.04) 493.84 xm − y m +1 = 296.84 296.84 y m +1 = 1.664 x m − 0.0266

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.


202

d) 2nd ideal plate is in the enriching section, and the composition of the liquid and vapor phases 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 For “ Constant molar over-flow” rectification; on each plate, “Moles of methanol transferred from liquid to vapor = moles of n-propanol transferred from vapor to liquid” is valid. Then, 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 , L ( x 5 − x 6 ) = (493.84)(0.288-0.195) = 45.93 k-mol methanol/h Or, G ( y 6 − y 7 ) = (296.84)(0.455-0.300) = 46.01 k-mol methanol/h. Take the mean value = 45.97 k-mol methanol/h As it is seen, transferring quantities (rate of mass transfers) are quite different even on mole basis on the 2nd and 6th plates. ⎛ 234 . 1 ⎞ ⎟⎟ (38 390 ) = 2 496.4 kJ/s (=kW) e) As the condenser is total condenser; q c = G λ = ⎜⎜ ⎝ 3 600 ⎠

Reboiler load,

⎛ 296 .84 ⎞ ⎟⎟ (38 390 ) = 3 165.5 kJ/s (= kW) q S = G λ = ⎜⎜ ⎝ 3 600 ⎠

L

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, the total material balance changes and becomes, F + G = D+W (5-89) as seen from Fig.5.28 a. Here, G is the flow rate of steam as k-mol/s. As yN+1= 0 in this case, point W lies on the x-axis and the required number of the ideal plates q-line y equilibrium curve yN-1

Q

N-1

xN-1

F

yN-1

yN

y=x

N-1

N

xN

Steam

∇

yN

G , yN+1 =0 yN+1

Bottom product

strip.sec.op.line N

xw xN

xN-1

zF x

W

L , xw =xN (a)

(b)

Fig.5.28 Use of open steam in the column No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

203

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 of reboiler. But use of open steam is much costlier than the use of closed-looped steam as in the latter case condensed steam is returned back to the boiler house for reuse and, demineralized water is used for the production of steam. Hence, by comparing the 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 preferred or not.

Side Streams: In some applications, a product other than the top and bottom products is also withdrawn from the column, which is usually a liquid product. Although this application is more common in multi-component rectification, it is also used in some binary rectifications. Let us show the flow rate of intermediate product with S as kmol/s and the mole fraction of the MVC in it with xs as shown in Fig.5.29. As it is seen, the column is divided in three sections: upper, middle and lower section. In the middle section, the flow rates of liquid, vapor and any plate are shown with Ls, Gs and k respectively. By writing the MVC balance between the top of the column and any plate in the upper section, the operating line equation for the upper section can be written as, xD RD (5-59) xn + y n+1 = 1+ RD 1+ RD Between any plate in the middle section and top of the column, (5-90) Total material balance : Gs = Ls + S + D (5-91) MVC balance : Gsyk+1 = Lsxk + Sxs + DxD can be written. From the last equation, Sx s + Dx D L (5-92) y k+1 = s x k + Gs Gs is obtained. This is operating line equation of the middle section of the column and the Sx + Dx D Sx s + Dx D ; ) coordinates of its intersection with the diagonal is found as U( s S+ D S+ D with the help of equation (5-90). If in this case a qS-parameter is defined by L −L qS = S (5-93) S qS-line, which is the locus of intersection points of upper and middle section operating lines is obtained as; qS x yq = xq + S (5-94) qS + 1 qS + 1 This is a straight line passing through point S(xS;xS). 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, S

S


204

qC

G

L

1

L,xo D,xD

Upper section

Top product

n yn+1

P=cons. 1.0

Side product

xn

S, xS GS

k

qs-line

q-line

D

y

LS

yk+1

xk

U

Middle section Q

op.line for upp.sec. S

Feed

F,zF

G

F

xD 1+ R D

m Lower section

op.line for middle sec.

K

L

ym+1

op.line for lower sec.

xm

N

G yN+1

W

0 0

xw

zF

qS

L xN

xD

xs x

Bottom product

W,xw

(a)

(b)

Fig.5.29 Withdrawal of side product

L − LS (5-95) F and the q-line given by equation (5-71) and passing through point F(zF;zF) is then the locus of intersection points of middle and lower section operating lines. The operating line equation of the lower section becomes; W xw L (5-62) y m+1 = xm − G G q=

Considering the whole column, three more equations as; Total material balance MVC balance

: :

F=D+W+S FzF = DxD + WxW + SxS


(5-96) (5-97)

205

Percentage recovery of the MVC : P.R. =

Dx D + Sx S .100 Fz F

(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 Figure). Example-5.11) Withdrawal of Side Product An aqueous methanol solution containing 25 mole percent methanol will be rectified in a plate column equipped with a total condenser, at 760 mmHg to produce a 10 k-mol/h saturated liquid side product containing 60 mole pecent methanol , a top product containing 95 mole percent methanol and a bottom product containing 4 mole percent methanol. The feed, which is at 20 oC, will be introduced at a flow rate of 173 k-mol/h. Operating reflux ratio is selected as 1.235. Assuming constant molar over-flow; a) Write the operating line equations for each section of the column. b) Calculate the number of the ideal plates needed and the locations of the feed and side product Vapor - liquid equilibrium for methanol-water system at 760 mmHg is in Table.App.5.4. Molar 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 shown below. Read the bubble point temperature of the feed from t-xy diagram as tFb= 81 oC. Hence the feed is subcooled liquid, then calculate the q-parameter from eqn.(5-74) (76.5)(81 − 20) q =1 + = 1.123 38 000 q-line from equation (5-71);

From equations (5-94) and (5-95);

1.123 0.25 xq − 1.123 − 1 1.123 − 1 yq = 9.13 xq – 2.03

yq =

(173)(0.25) = (0.95)D + (10)(0.60) + (0.04)(173-10-D) D = 33.77 k-mol/h W= 173-10-33.77 = 129.23 k-mol/h

L = DRD = (33.77)(1.235) = 41.71 k-mol/h G = L + D = 41.71 + 33.77 = 75.48 k-mol/h As side product S is saturated liquid, From equation (5-90); From equation (5-97); From equation (5-60);

GS = G = 75.48 k-mol/h LS = 75.48 – 10 - 33.77 = 31.71 k-mol/h L = 31.71 + (1.123)(173) = 225.99 k-mol/h G = L − W = 225.99-129.23 = 96.76 k-mol/h 1.235 0.95 Then, upper section operating line from equation (5-59); y n +1 = xn + 1 + 1.235 1 + 1.235 yn+1 = 0.553 xn+ 0.425 Middle section operating line from equation (5-92);

y k +1 =

(10)(0.60) + (33.77)(0.95) 31.71 xk + 75.48 75.48

yk+1 = 0.42 xk + 0.505


206

Temperature (oC)

tFb

zF Mole fractions of methanol, x,y

t-xy diagram of aqueous methanol solution at 760 mmHg

y m +1 =

Lower section operating line from equation (5-62);

(129.23)(0.04) 225.99 xm − 96.76 96.76

ym+1 = 2.43 xm - 0.0535

First, mark the points F, W, D, S on xy-diagram. Then, draw q-line and qS-line. By marking K1=0.425 on y-axis and joining it with D, upper section operating line is first drawn and point E is located. Then,

mole fraction of methanol in vapor, y

1

qS-line

y q = 9.13 x q − 2.03

2 3

D

4 5 6 7

Q

8

K2

E S

K1 9 F 10 xS

W xw

zF

xD mole fraction of methanol in liquid, x

xy-diagram of aqueous methanol solution at 760 mmHg No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

207

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 plate for stripping section. Suppose n and m show the last drawn plates in enriching and stripping section respectively. After plates n and m, the numbers of the plates in the remaining parts are calculated from the equations given below. The enriching section of the column may be considered as an absorption column for the LVC and hence an equation similar to the equation (4-54) can be written for the calculation of number of the ideal plates, ne in this part. ⎡ (1 − y n ) − (1 − x o ) / K 2 ⎛ α − 1 ⎞ 1 ⎤ log ⎢ ⎜ ⎟+ (1 − y1 ) − (1 − x o ) / K 2 ⎝ α ⎠ α ⎥⎦ ⎣ (5-99) ne = log α 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 stripper (desorber) for the 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. ⎡ x − x w /K 1 ⎤ (1 − α ) + α ⎥ log ⎢ m ⎣ x w − x w /K 1 ⎦ (5-100) ms = log (1 / α) where, α = L / GK 1 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 total and MVC balances are required to compute the flow rates of


208

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): (5-101) Total material balance : Gn+1 = Ln + D (5-102) MVC balance : Gn+1 yn+1 = Ln xn + D xD (5-103) Enthalpy balance : Gn+1 Hn+1 = Ln hn + D hD + qc If the energy removed per unit mole of top product in the condenser is shown by QCD=qc/D, the last equation can also be written as; (5-104) Gn+1 Hn+1 = Ln hn + D (hD + QCD) 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, (5-105) Gn+1- Ln = ∆e 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, (5-106) Gn+1yn+1- Lnxn = ∆e x∆e where, x∆e shows the mole fraction of the MVC in the ∆e. By comparing equation 5106) with equation (5-102) x∆e= xD is obtained. Hence, it is said that in 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 enthalpy of the vapor entering the same plate, (5-107) Gn+1 Hn+1- Ln hn = ∆eh∆e can be written. Where, h∆e is the specific enthalpy of the ∆e. Comparison of this equation with equation (5-104) gives, (5-108) h∆e = hD + QCD Since 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. In the stripping section, between any plate (plate m) and bottom of the column (envelope-2); (5-109) Total material balance : Lm = Gm+1 + W (5-110) MVC balance : Lmxm = Gm+1ym+1 + Wxw (5-111) Enthalpy balance : Lmhm + qs = Gm+1Hm+1 +Whw can be written. By defining Qsw=qs/W , equation (5-111) becomes, (5-112) 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, (5-113) Lm- Gm+1 = ∆s


209

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 entering plate m from the quantity of the MVC in the liquid leaving the same plate we write, (5-114) Lmxm- Gm+1ym+1 = ∆s x∆s 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 section, there is a net flow of MVC in the downward direction, which is constant and equals 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, (5-115) Lmhm - Gm+1Hm+1 = ∆sh∆s can be written. Where, h∆s, is the specific enthalpy of ∆s stream. If this equation is compared with equation (5-112), (5-116) h∆s = hw – Qsw 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 equals in magnitude the value, which is obtained from the subtraction of enthalpy removed by bottom product from the energy given in the reboiler. 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, which carries the MVC from reboiler to the 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) (5-117) or : F = ∆e + ∆s (5-64) MVC balance : F zF = D xD + W xw (5-118) Enthalpy balance : FhF + qs = DhD + Whw+ qc (5-119) or : FhF = ∆e h∆e + ∆s h∆s 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 known. By writing total material and enthalpy balances around the condenser, this can be found as follows: (5-120) G1 = Lo + D (5-121) G1H1 = Loho + D hD + qc If the value of G1 is substituted from equation (5-120) into equation (5-121) and the definitions of RD , QCD and h∆e are remembered then, h − H1 (5-122) R D = ∆e H1 − h O is obtained.


210

G1,y1,H1 Condenser qC Envelope-1 1 L1,x1 h1

G2,y2 H2

Lo,xo ho

Top product D,xD,hD

∆e

2

Ln-1,xn-1 hn-1

G3,y3 H3 Gn,yn Hn

n Ln,xn hn

Gn+1,yn+1 Hn+1

P =cons. Specific enthalpies, H,h

L2,x2 h2

G5

G4 G3

G2

G1

H1

6 5

L6

W

4

3

1

2

h-x

L5 L4

hF

L3

F

Feed xW x5

x6

F,zF,hF

h∆e

H-y

G6

x3

x4 zF

L2

L1

x2

x1

hD

D

xD=y1

x,y

h∆s

∆S

m Lm,xm hm

Gm+1,ym+1 Hm+1

LN,xN hN

GN+1,yN+1 HN+1

y

y=x

Envelope-2 N 0

Reboiler

1.0

x

0

(b)

qS

Bottom product W,xw,hw

(a)

Fig.5.30 Varying over-flow rectification

With the help of the equations above, the number of the ideal plates required for a


211

varying over-flow rectification can be calculated by using Ponchon and Savarit Method, which involves the graphical drawing of the ideal plates on enthalpycomposition 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 selected by the designer and then enthalpycomposition; xy- and t-xy diagrams are obtained at the operating pressure. If the condenser 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 are all known (remember that all the vapors and liquids leaving the plates and the top and bottom products are saturated vapors and liquids). Then point ∆e, which will be on the vertical from x = xD is located by calculating its enthalpy coordinate h∆e from equation (5-122). H1 and ho, which are required for the calculation are read from enthalpy-composition diagram. Equation (5-117) indicates that points 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 drawing of the ideal plates on enthalpy-composition diagram is started. Since the composition of the vapor leaving the first ideal plate is known, the composition of the liquid leaving the same plate can be found by carrying the y1 value to the xydiagram 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 the same straight line. In addition, G2 should be on the saturated vapor curve. Thus, L1 joined with ∆e will locate the G2, 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 with G2 shows the second ideal plate in the column. G3 is then easily located by joining L2 and ∆e according to equation (5-105). The calculation is thus continued by alternate drawing of tie lines and operating lines issuing from point ∆e as explained above. On the left of ∆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 calculation is continued in the same way until reaching x = xw. The number 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 the operating line 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 reboiler duties are calculated, From equation (5-108) QCD = qc/D = h∆e- hD and from equation (5-116) Qsw = qs/W = hw- h∆s


(5-123) (5-124)

212

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; h − H1 (5-125) R Dm = ∆em H1 − h O 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∆e is also infinite. As a result of 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. ∆em

h∆em

h-x 9 3

D

4

xw

zF

3

1

W

h-x

D

∆s - ∞ 0

xD 1.0

x,y

∞

2

∆sm

0

∆e

H-y

H-y

W

h∆sm

P=cons.

Specific enthalpies, H,h

Specific enthalpies, H,h

P=cons.

xw

x,y

xD 1.0

1.0

1.0

y

y y=x

y=x

0

0 0

x

1.0

Fig..5.31 Determination of Minimum Reflux Ratio

0

x

1.0

Fig.5.32 Number of Ideal Plates at Total Reflux


213

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

h∆e

H-y G1 G1,y1,H1

D qC

C

1 D,yD,HD

H1 HD

h-x L1

Top product

Lo

L,xo,ho

y1

xo

x1

yD 1.0

x,y

1

x

1.0 1.0

slope=-Lo/D

yD

C

D y

y1 1

(a)

Equilibrium curve

xD= yD xo

x1 enr.sec.op.line

y=x

(b)

Fig.5.33 Partial condenser

specific enthalpy is shown by HD. Hence, the enthalpy coordinate of point ∆e is then given by h∆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; (5-126) Total material balance : F + GN+1 = D + W (5-127) MVC balance : F zF + GN+1(0) = D xD + W xw (5-128) Enthalpy balance : F hF + GN+1 HN+1 = D hD + qc + W hw can be written. Similarly, for the stripping section of the column;


214

specific enth.,H,h

GN-1

GN+1

P=cons. GN GN-1 N-1

hN=hw LN=W

LN-1 L N-2

N-1 GN

LN-1

h∆S

N

∆S

zF

Steam GN+1 yN+1=0

∇

y

x,y

N-1

y=x

Bottom product LN=W xN=xw hN=hw

h-x

F

0 LN

∆e

H-y

N

N

strip.sec.op.line E

yN+1=0 0

xw x∆S xN

x

(b)

(a) Fig.5.34 Use of open steam

(5-129) Total material balance : LN- GN+1 = W- GN+1 = ∆s (5-130) MVC balance : LNxN- GN+1(0) = Wxw- GN+1(0) = ∆sx∆s (5-131) Enthalpy balance : LNhN- GN+1HN+1 = Whw- GN+1HN+1 = ∆sh∆s are written. First points F, D, W, G1 and GN+1 are located on enthalpy-composition diagram. Then with the help of known RD, point ∆e is located. Then the points GN+1 and W and 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 from the equations above by first reading the value of x∆s from the diagram above. Example-5.12 Rectification under Varying Over-flow 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, c) Total net flow, ammonia net flow and enthalpy net flow in each section, d) Flow rates and composition of each stream entering and leaving the plates,


215

e) Condenser and reboiler duty Enthalpy-composition data of the system at 10 bars are given in Table.App.5.12. Solution: First plot enthalpy-composition and xy-diagram of the system.

a) From equations(5-63) and (5-64) z − xw 0.25 − 0.04 = (5 500) = 1 269.2 kg / h W= 5 500-1269.2 = 4 230.8 kg/h D=F F 0.95 − 0.04 xD − xw (1269.2)(0.95) .100 = 87.7 % From equation (5-65) P.R. = (5500)(0.25) b) From enthalpy-composition diagram λA= 1 200*17 = 20 400 kJ/k-mol and λB= 2 000*18 = 36 000 kJ/k-mol are found. So, varying over-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 from equation(5-122) h∆e= 1 600 + 0.758(1 600-280)= 2 600 kJ/kg is calculated. With the help of this point ∆e is located. By joinin this with point F and exdending 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 introduced on the second plate. Allthough kettle reboiler makes a separation equal to one ideal plate, for design safety 4 ideal plates are recommended for 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 = 1 205.7 kg NH3/h Ammonia net flow in the enriching section : ∆ex∆e=DxD=(1 269.2)(0.95) Ammonia net flow in the stripping section : ∆sx∆s=Wxw=(4 230.8)(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 Enthalpy net flow in the stripping section : ∆sh∆s= (4 230.8/3 600)(-320) = - 376.1 kJ/s d) Lo=D RD= (1269.2)(0.758) =962.1 kg/h, G1 = 962.1 + 1269.2 = 2 231.3 kg/h From the graph, x1= 0.445 and y2 = 0.83 are read.Then, from G2-L1=1269.2 and 0.83G2-0.445L1 =1205.7, L1 = 395.5 kg/h and G2 = 1 664.7 kg/h are calculated. From 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

Empirical Correlations for Estimating Number of Ideal Plates: First Gilliland later Erbar and Maddox have given empirical correlations for quick estimate of the number of the ideal plates for a given continuous rectification. Both empirical correlations as shown 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 Underwood equation and the minimum number of the ideal plates from Fenske equation.


216

Enthalpy-concentration diagramofofammonia-water ammonia-water at bars 10 bars Enthalpy-composition diagram at 10 Mass fraction of ammonia, x,y

mol fraction of ammonia,x,y 2

3200 2800

∆e

G4

1.8

h∆e 2400

G3 G2

1.7

2000

4

1.6

G1

H1600 1

3 1.5 1.4

1200

L4

hw

1

2

800

W L3

1.3

L2

400

F xw

1.2

h

0

1.1∆s

D

L1 0.1

0.2

zF

0.3

0.4

hD

0.5

0.6

0.7

0.8

0.9

xD

0 1

∆S

-400

1

molfra fraction ct ion ofofam monia in vapo r,y Mass ammonia in vapor, y

specific enthalpy of solution, H ,h (kJ/kg solution)

1.9

-800

0.9

-1200

0.8

-1600

0.7

-2000

0.6

-2400

0.5

-2800

0.4

-3200

0.3

-3600

0.2

-4000

0.1

-4400

0

-4800 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

molfraction of ammonia in liquid,x Mass fraction of ammonia in liquid, x


217

Fig.5.35 Gilliland correlation

Fig.5.36 Erbar - Maddox correlation No part of this CD-book may be multiplied 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' x ′q = 0.415 , y′q = 0.73 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; (0.696)(0.35) + (1.21)(0.95) 3.81[0.95(1.21 − 1) + 0.35(0.696 + 1)] = (0.696)(1 − 0.35) + (1.21)(1 − 0.95) (0.696 + 1)(1 − 0.35) + (1.21 − 1)(1 − 0.95)

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 needed. 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 At the bottom of the column, y = 0.12 , x = 0.04 , αB = 0.12(1-0.04)/[0.04(1-0.12)] = 3.27 α AV = α T α B = (3.78)(3.27) = 3.52 , then from equation (5-86), log Nm =

0.95(1 − 0.04) 0.04(1 − 0.95) = 4.87 log(3.52)

a) R D − R Dm = 1.30 − 0.696 = 0.263 1+ RD 1 + 1.30

Then, from Fig.5.35

N − Nm = 0.4 is read. Finally, N+2

4.87 + 0.8 = 9.45 is found. A close look at McCabe-Thiele diagram 1 − 0.4 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.

number of the ideal plates,

N=

RD 1.30 b) R Dm = 0.696 = 0.41 and then, from Fig.5.36 Nm/N=0.72 is read. = = 0.57 1 + R Dm 1 + 0.696 1 + R D 1 + 1.30 Finally, number of the ideal plates, N = 4.87/0.72 = 6.76 is found. As it is seen, the Erbar-Maddox correlation estimates 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 liquid in the boiler becomes poorer in the MVC as the distillation proceeds; as no feed is given to the column and as the vapor formed is always richer in the MVC. As a No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

219

G,y1 qC

Condenser Reflux

L

G 1

L,xo

D,xD Top product

n xn

yn+1 Column

Charge

N

Boiler

Heating medium Bottoms

Fig.5.37 Batch rectification column

result of this, top product becomes also poorer in the MVC as the time passes. If the purity of the top product is to be kept constant, then the reflux ratio must be increased steadily in the course of distillation. It follows from here that batch rectification can be carried out in two different ways: either under constant reflux ratio or under increased reflux ratio. As the column consists of only enriching section, there is only one operating line, which is given as; RD xD (5-132) xn + yn+1= 1+ RD 1+ RD 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; (5-133) Lx = (dL)xD+(L-dL)(x-dx) No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

220

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 the time considered, 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, dx dL = (5-134) L xD − x is obtained from equation (5-133). If this equation is integrated from the start to the end of rectification; F

xF

⌠ dx F ⌠ dL ⎮ = ln = ⎮ ⌡L W ⌡ xD − x W

(5-135)

xw

is found. Where, F and W are the quantities of the charge and the 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) Dx D ,av P.R. = .100 (5-138) Fx F With the help of equations (5-135), (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 equilibrium curve by shifting the line up and down. Final line is the operating line at the start of distillation and thus xDi is found. Then 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 xD. With thus found x values, 1/(xD-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 xF and xw gives the value of integral given by equation (5-135) from which W is then found. Finally, from equations (5-136) and (5-137) D and xD,av. are computed. The heat energy required in the operation can be calculated from; q T = q S + q L = F cFL (t − t F ) + (1 + R D )D λ (5-139) where, qs (kJ) is the sensible heat needed to heat up the solution from initial temperature tF (oC) to the average distillation temperature, t (oC) and qL (kJ) is the latent heat required to produce GT (k-mol) vapor which is given as GT=D(1+RD), cFL (kJ/k-mol oC) is the molar heat capacity of the solution calculated at the arithmetic


221

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 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 the equations (5-48) and (5-49). For the distillation time, equation (5-50) and for the total batch time, equation (5-51) can be used. Example-5.14) Batch Rectification under Constant Reflux Ratio 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 solution will be charged 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) 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. 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. 0.246 119 .4 xF = = 0.20 0.246 (1 − 0.246) + 119.4 92 F = (12 000)(0.246) / 119.4 + (12 000)(1 − 0.246) / 92 = 123.1 k − mol

Calculate the slope of operating line from eqn.(5-132), R D 1+ R

= D

3 . 125 1 + 3 . 125

= 0 . 758

Draw a line with this slope on xy-diagram and locate 3 right angle 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. In the figure below only initial and final operating lines are shown.


222

xD

1 xD − x

x

0 . 20

dx ⌠ ⎮ x ⌡ 0 . 04 D − 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 Total = 0.2625

Then, from equation (5-135); ln

F = 0.2625 W

F = e 0.2625 = 1.30 W

W=

123.1 = 94.69 k − mol 1.30

t-xy diagram of chloroform-toluene at 760 mmHg

tB tf = 106.5 ti = 94.5

Temperature, t (OC)

t-y t-x

tA

xW = 0.04

Mole fraction of chloroform, x,y

xF=0.20

Mole fraction of chloroform in vapor, y

xy-diagram of chloroform-toluene at 760 mmHg

Di

Df

Operating lines xF=0.20 xW=0.04

xDf =0.40

xDi=0.90

Mole fraction of chloroform in liquid,x


223

Then from equation (5-136); D = F- W = 123.1- 94.69 = 28.41 k-mol (123.1)(0.20) = (28.41)xD,av+ (94.69(0.04) From equation.(5-137) ; xD,av= 0.733

b) From equation (5-138) ;

P.R. =

(28.41)(0.733) .100 = 84.6 % (123.1)(0.20)

c) From equation (5-139) ; Sensible heat required = F c FL (t − t F ) The average boiling point temperature of solution t can be calculated from; t=(ti +tf)/2 From t-xy diagram of the system, ti = 94.5 oC, tf =106.5 oC are read. Sensible heat required = (123.1)(140)( 100.5- 20) = 1 387 337 kJ Latent heat required = (1 + R D ) D λ = (1+ 3.125)(28.41)(32 000) = 3 750 120 kJ Total heat required = 1 387 337+ 3 750 120 = 5 137 457 kJ

d) Heating of solution from 20 oC to 100.5 oC is an unsteady-state operation and the time needed is calculated from equation (5-48). To account for heating of the batch-still itself (metal) , take F= (1.1)(123.1) =135.41 k-mol. From the steam tables, temperature of the saturated steam at Ps=1.05+1 =2.05 bar absolute pressure is read as ts=121 oC. ln[(121 − 20) / (121 − 100.5)] = 15 116 sec . ≈ 4 h 12 min θS = (0.200)(10) / (135.41)(140) For the calculation of distillation time, equation (5-50) is used:

(1.1)(3 750 120) = 20 122 sec ≈ 5 h 36 min (1.0)(10)(121 − 100.5) 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 ; θL =

θBT = 1 h + 4 h 12 min + 5 h 36 min = 10 h 48 min.

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 from point D whose abscissa is xD, an arbitrary line towards F is drawn and until the N right angle triangles are located between equilibrium curve and 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 reflux 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; q T = q S + q L = F cFL (t − t F ) + G T λ (5-140) GT (k-mol) which is the total amount of vapor to be formed, can be calculated as follows; No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

224

Total material and the MVC balances at any time θ during the rectification can be written as; F = Dθ + L (5-141) FxF = Dθ xD + Lx (5-142) Where, Dθ is the quantity of top product collected until this time. By eliminating L between the two equations, ⎡x − x⎤ Dθ = F ⎢ F (5-143) ⎥ x x − ⎣ D ⎦ is obtained. G T = ∫0 dG θ = ∫0 (1 + R D ) dD θ GT

Then,

D

(5-144)

This equation may be written in different ways, for example with the help of equation (5-143), or noting that 1 + R D = 1 /[1 − ( L / G )] ; dx ⌠ 1+ RD ⌠ G T = F( x D − x F )⎮ dx = F( x D − x F )⎮ (5-145) 2 2 ⌡x [1 − (L / G )]( x D − x ) ⌡x ( x D − x ) The evaluation of GT may be done as follows: from point D arbitrary operating lines for reflux ratios between RDi and RDf are drawn and then the given number of right angle triangles are located between the equilibrium curve and the operating lines. By substituting the x values read from the last triangles into equation (5-143) corresponding Dθ s and from the slope of the operating lines RD s are computed. Then carrying (1+RD) values against the corresponding Dθ s on a millimetric paper and calculating the area under the curve between the limits of RDi and RDf, gives the value of GT. Time required for heating of the solution from initial temperature tF to the average distillation temperature t is calculated either from equation (5-48) or from (5-49). For the calculation of distillation time; G θL = T (sec) (5-146) G 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; q U A( t H − t ) ln (5-147) G= = b λ λ Then, total time for one batch operation is calculated from equation (5-51). xF

xF

W

W

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.


225

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. Reflux ratio calculated from the slope of this line is the initial reflux ratio (RDi). Then another line from point D with greater slope is drawn and the slope of this line is changed until 3 right angle triangles but this time between point D and W fit. This line is the operating line at the end of distillation. Reflux ratio calculated from the slope of this line is the final reflux ratio (RDf). These are found as RDi = 0.929 and RDf = 25.18 b) As F, xF , D and xD are the same with the previous example, P.R. = 84.6 %. b c) Sensible heat required is calculated from equation (5-140) qS = (123.1)(140)( 100.5- 20) = 1 387 337 kJ For the calculation of latent heat required, first GT must be calculated. For this various operating lines between initial and final are drawn (not shown in the figure) and after locating three right angle 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

GT = ∫

28.41 0

(1 + R D ) dD θ

0.20

0

1,929

-

0.16

8.59

2.443

0.5(1.929+2.443)(8.59-0)

0.135

13.38

2.932

0.5(2.443+2.932)(13.38-8.59) = 12.87

0.1

19.45

3.665

0.5(2.932+3.665)(19.45-13.38) = 19.69

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

28.41

26.18

0.5(7.33+26.18)(28.41-23.84) = 76.57

= 18.78

GT = 152.37 k-mol Then, latent heat required, Total heat required,

qL = (152.37)(32 000)= 4 875 840 kJ 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 12 min. For the calculation of distillation time, equation (5-146) is used along with equation (5-147). θL =

(1.1)(4 875 840) = 26 163 sec ≈ 7 h 16 min (1.0)(10)(121 − 100.5)

(10 % excess energy was taken to account for the heating of the column itself).


226

Mole fraction of chloroform in vapor, y

xy-diagram of chloroform-toluene at 760 mmHg

D

Initial operating line

Final operating line

xD = 0.733

xF = 0.20 xW = 0.04

Mole fraction of chloroform in liquid, x

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 Operation mode

Example-5.14

Example-5.15

Simple distillation Batch distillation Batch distllation with RD= constant with xD= constant

Purity of top product (mole fraction)

0.34

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 less energy and batch time than the batch distillation under varying reflux ratio. Hence this mode of operation is preferred for this system.

Calculation of Number of Real Plates: After determining the number of the ideal plates needed for a given separation, the number of the real plates can be found by


227

Over-all column efficiency, Eo

either applying plate or over-all column efficiency defined in Chapter-4. These efficiencies are found experimentally in pilot size plate columns. The graphical correlation giving by O’Connell and shown in Fig.5.38 may be used to calculate overall column efficiency for rectification when experimentally obtained value is not available. Although this correlation was obtained for bubble-cap columns, it can also

α µFL Fig.5.38 Over-all column efficiency at rectification

be used for sieve and valve tray columns for approximate calculations. If the relative volatility of the system changes along the column, the relative volatility calculated at the column average temperature is used. The liquid viscosity is taken as the molar average viscosity of the feed liquid, which is calculated from the viscosities of pure components (kg/ms) at column average temperature. Example-5.16 Estimation of Number of Real Plates Estimate the numbers of the real plates needed in each section of the column given in Example 5-10 from O’Connell correlation. Viscosities of liquid methanol and n-propanol at 81 OC are 0.298 cP and 0.61 cP respectively Solution : From t-xy diagram of the system, temperature at the top of the column, for y1=0.95 is read as 67 oC, and at the bottom of the column, for x9 =0.04 is read as 95 oC. Then, average column temperature is 81 0C. At this temperature, from t-xy diagram : x = 0.342 and y =0.658 are read. Then relative volatility of methanol to n-propanol from equation (5-13), (0.658)(1 − 0.342) = 3.7 (0.342)(1 − 0.658) Molar average viscosity of the feed liquid at 81oC,

α=

µ FL = (0.35)(0.298 * 10 −3 ) + (1 − 0.35)(0.61 * 10 −3 ) = 0.50 * 10 −4 kg / ms Then, αµFL= (3.7)(0.50*10-4) = 1.85*10-4


228

Now enter this value into Fig.5.38 and read Eo=0.74. Number of the real plates needed in the enriching section =

3 =5 0.74

Number of the real plates needed in the stripping section =

6 =9 0.74

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 out in continuous contact type of equipment. Especially, when the flow rate of solution to be rectified is small, the packed column is preferred as in this case construction of plate column is rather difficult due to the small diameter, although the possibility of cartridge type construction exists. For rectification at fine vacuum, packed columns are in many cases preferred because of the fact that special structured packings, developed recently have very low pressure drop along the packing at even fairly high liquid and vapor loads. To give an idea, for Montz-Pak, type: A3-500 packing element, pressure drop 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

m3/m2h

∆pG mbar/m.

L

u G ρG

Pa0.5

Fig.5.39 Pressure drop at Montz-Pak, type: A3-500

much smaller than the plate column, the residence time of the solution in the packed column is low and as a result of this, decomposition of heat sensitive components of solution is rather low. Except these special cases, the choice of column is done by looking 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 of the diameter of a packed rectification column is accomplished by using the same correlation for flooding or pressure drop per unit height of packing as for packed absorption column (Fig.4.14), as gas and vapor hydrodynamically are the same. But, as the vapor and liquid loads are generally quite different in enriching and stripping sections of the column, the diameter calculations must be done for each section separately.


229

The calculation of the height of packing required for a giving rectification is done by using the same procedure as used in gas absorption. But here again the enriching and stripping section must be considered separately. If the constant molar over-flow case is applicable, then xy-diagram of the system is sufficient for the calculation, otherwise enthalpy-composition diagram of the system is also required. The total material, component and enthalpy balance equations written for rectification in a plate column are also valid for rectification in packed column. But, as in the packed column the contact of two phases is continuous, the subscripts in the equations denoting the plates 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 crosssectional area of the empty column in enriching section. By substituting this into equation (5-148) d(Gy) = -d(Lx) = NAavAcdhe (5-149) is obtained. G and L remain approximately constant even constant molar over-flow does not apply. Hence NA= - NB and then from Chapter-3 for NA, N A = k ′y (y i − y) = k ′x (x − x i ) = K ′y (y∗ − y) = K ′x (x − x ∗ ) (5-150) can be written. By substituting all these into equation (5-148) finally; Ze =

∫

Ze

0

( Gy ) 2

( Lx ) 2

( Gy ) 2

( Lx ) 2

( Gy ) q

q

( Gy ) q

q

d (Gy) d (Lx ) d (Gy) d (Lx ) ⌠ ⌠ ⌠ ⌠ dh e = ⎮ = ⎮ = ⎮ = ⎮ ∗ ∗ ⌡ k ′y a v A c ( y − y i ) ( Lx⌡) k ′x a v A c ( x − x i ) ⌡ K ′y a v A c ( y − y ) ( Lx⌡) K ′x a v A c ( x − x )

is obtained. (5-151) In the case of varying over-flow case, the integral terms are evaluated as follows: as example consider the evaluation of first integral on the right hand side. From point ∆e arbitrary lines (operating lines) are drawn on the right hand side of ∆eF∆s line (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 (- k ′x a v /k ′y a v ), 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/ k ′y a v A c (y i − y) values against the Gy, between the limits of (Gy)q and (Gy)2 gives the value of integral hence the ze.


230

qC G2,y2

P=cons.

∆e

H-y

he=ze

L2,x2

Enriching sec.

D,xD Top product dhe

Specific enthalpies, H, h

G

G G

W F

L L L

L

L

h-x

L

D

0

he=o

xD=x2

zF

xw=x1

Lq,xq

Gq,yq

F,zF

G

x,y

∆S

hs=zs

1.0

Stripping sec.

(b)

dhs

1.0

y2 yi y

hs=o

G1,y1

M

D

q-line

P y=x

y qS

yq

Q Enr.sec.op.line

Bottom product

F

L1,x1

W xw=x1

Strip.sec.op.line y1 0

(a)

W 0

xw=x1

xq

zF

x

xi

xD=x2

1.0

x (c)

Fig.5.40 Rectification in packed column

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; G L L L HL = HL = H OL = (5-152) HG = k ′y a v A c k ′x a v A c k ′x a v A c K ′x a v A c 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 units simplify to:


231

y2

⌠ dy NG = ⎮ ⌡ yi − y yq

x2

⌠ dx NL = ⎮ ⌡ x − xi xq

y2

N OG

⌠ dy =⎮ ∗ ⌡y − y

x2

⌠ dx N OL = ⎮ ⌡ x − x∗

yq

(5-153)

xq

The solutions of these integrals are again accomplished by using graphical integration technique by selecting arbitrary P points on the enriching section operating line, which is a straight line in this case, and drawing the lines with slopes (- k ′x a v /k ′y a v ) and locating the M points on the equilibrium curve. If the steps from equation (5-148) are repeated for the stripping section of the column, the height of packing of this section, (zs) is found. The integral limits in this section must be taken as (Gy)q - (Gy)1 and (Lx)q - (Lx)1. In the case of varying over-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 from point ∆s on the left hand side of ∆eF∆s line. For constant molar over-flow case, G and L must be used in the equations (5-152) along with the 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 Furnas and Taylor at atmospheric rectification of ethanol-water system are given in Table.5.2 for two different packing types.

Table 5-2. HOG values for ethanol-water system

Packings 50 mm Raschig rings 25 mm Raschig rings 9.5 mm Raschig rings 9.5 mm Raschig rings 25 mm Berl saddles 25 mm Berl saddles 12.5 mm Berl saddles 12.5 mm Berl saddles

Depth of Liquid flow flux packing(m) (kg/m2s) 3.0 1.06 3.0 1.02 2.44 0.416 2.44 0.780 2.75 0.195 2.75 1.25 3.0 0.25 3.0 1.196

HOG (m) 0.670 0.366 0.396 0.305 0.427 0.335 0.457 0.274

Another method, which is used at the calculation of height of packing, is the height 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 or Ponchon-Savarit method depending upon the conditions. If the height of selected packing, which makes a separation equivalent to one theoretical plate under the operating conditions, is known; by multiplying this height with the 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 in the vapor is above 17 mm liquid head per meter of packing.


232

Nideal/m. packing

Nideal/m. packing

A3-500

u G ρG

Pa 0.5

u G ρG

Pa 0.5

Fig.5.41 The separation powers of two different type of packing

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 function of flow factor, which is defined as uG√ρG. As an example, in Fig.5.41 the separation powers of two different types of packing produced by Montz GmbH are shown. 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 separation of the solution is not possible. Depending upon the compositions of the feed and azeotrope, one of the t-y tB components may be obtained in pure state. As an example consider the binary solution having tA t Az minimum boiling azeotrope at 65% MVC, as t-x shown in Fig.5.42. If a 100 k-mol/h feed, containing 30% MVC, is subjected to rectification, the top product can approach to zF 0.8 1.0 0 0.2 zF 0.4 0.6 the azeotropic composition (of course, if sufficient number of the plates or packing x,y height is provided in the enriching section) and 1.0 with the assumption that all the MVC appears in the top product, a top product of y Az D=(100x30/0.65)= 46.15 k-mol/h is withdrawn from the column, leaving behind 100D 46.15 = 53.85 k-mol/h bottom product which is y=x almost pure in the LVC. As it is seen, in this F case a pure product consisting of LVC can be obtained, the top product being an azeotrope. 0 zF 0.8 1.0 0.2 zF 0.4 0.6 Now, let us assume that the feed flow is again x 100 k-mol/h, but it contains this time 75% Fig.5.42 Rectification of a minimum MVC. In this case, the top product may again boiling azeotrope 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 the MVC.


233

As it is seen, in this case a product, which is almost pure in the MVC is produced, while the other product is not pure. The azeotropes having maximum boiling point behave similarly, but in this case bottom products are azeotropes and the top products are pure products. It follows from these that azeotropic solutions cannot be separated into pure products by ordinary rectification. As it will be considered later, only by adding a third component to the solution, complete separation 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 or their azeotropic azeotropic composition change with pressure. If an azeotropic vapor forms two liquid phases upon condensation, this type of azeotropism is called as heteroazeotropism. The reason for hetero-azeotropism is the limited solubilities of the 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)

5 10 15 20 25 30 35 40 50 60 70

Water layer Butanol layer (mass % of alcohol) (mass % of alcohol) 80.38 80.33 80.14 79.93 79.73 79.38 78.94 78.59 77.58 76.38 74.79

9.95 8.92 8.21 7.81 7.35 7.08 6.83 6.60 6.46 6.52 6.73

Water/n-butanol solutions hence can be completely separated in a distillation system containing two columns as shown in Fig.5.43. As it is seen, the feed is introduced to a decanter, which is located beneath the condenser. While the bottom layer (water layer) of the decanter is pumped to the top of a column, the upper layer (butanol layer) is introduced to the top of another column. As it is seen, both columns consist of only stripping sections and there is a common condenser for both columns. The vapors coming from top of the columns have almost azeotropic compositions (approximately 23 mol % butanol) and upon condensing in this condenser, they flow to the decanter, where they separate in two insoluble liquid phases. The water vapor rising from the reboiler of the water column stripes off the butanol from the liquid flowing down the column, giving it the water. By providing 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

234

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

GI yDI ≈0.77

azeotrope

GII yDII ≈0.77

qc

condenser

decanter

I

II

760 mmHg 1

GI

760 mmHg LI xoI

LI

LII xoII

n-Butanol column

1

GII LII

Water column Feed F, zF GI

N

GII

qsI

N

qsII reboilers

LII

LI

n-Butanol

Water

WI xwI

WII xwII

WII

1.0 slope=LII/GII

Mole fraction of water in vapor, y

P =760 mm Hg

Az

yDI

N 1

yDII

Water column

1

n-Butanol column y=x

slope=LI/GI

N

WI 0 0 xwI

0.2

xoI

0.4

0.6

0.8

xoII

xwII 1.0

Mole fraction of water in liquid, x

Fig.5.43 Rectification of n-butanol/water solution


235

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 whether they are plate or packed, can be made with the methods given above. In another group of azeotropic solutions, the composition of azeotrope changes with pressure. In this case, by using two columns, each operating at 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 aceto nitrile at 760 mmHg, this rises to 84 mole percent at 150 mmHg. Aqueous aceto nitrile solution, containing approximately 30 mole percent aceto nitrile, is mixed with the top product of the second column and is fed to the first column, which operates 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, which operates at 760 mmHg. 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 may be plate or packed can easily be designed by using the methods giving above.

Azeotrope-I

1

Azeotrope-II

qcI

I

II

P=150 mmHg

P=760 mmHg

1

xoI

qcII

xoII

DII xDII

FII =DII zFII = xDII

FEED F, zF

Aceto nitrile column

Water column FI , zFI N

N

qsI

qsII

Reboilers

WI ,xwI

WII , xwII

Water

Aceto nitrile

Fig.5.44 Rectification of aceto nitrile/water solution


236

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 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 here. The design is accomplished by compromising various opposing tendencies, which require large number of trial and error.

Bubble Cap Plate

Sieve Plate

Valves Tray

Valves Tray

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 vapor loads, 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 without losing the efficiency. Below, design principles of a sieve plate column are to be given as an example. Although the equations for other types of plates are 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 a liquid pool on the plate. 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, where Ac is the total cross-sectional area of the plate. No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

237

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 Down-comer apron Vapor

Projected down-comer area Down-comer area

Weir

hd

how hw

hdc

liquid

Plate

Ad

holes

LW

Ad

Column Aa

Down-comer

Net area (An)

(a)

Active area Weir

Plate (Ac)

(b)

Fig.5.46 (a) Vertical cut of a sieve plate column, (b)view of a sieve plate from the top

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. 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 then decreases 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 downcomer 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 and which reduces No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

238

Liquid entrainment Flooding

Conning

Flow rate of gas

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.

Satisfactory operating region

Limitation of down-comer

Weeping

Flow rate of liquid

Fig.5.47 Satisfactory operating region of sieve-plate column

5.4.1 Column Diameter: The diameter of column Dc (m) is computed from equation, & ⎡ ⎤ 4G Dc = ⎢ ⎥ ⎣ πρ G u G (1 − a) ⎦

0.5

(5-154)

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 & , L& , ρ , ρ , P.S., σ, A /A . Here, L& (kg/s) is the liquid flow rate, ρL (kg/m3) is the G G L h a 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& ρ G FLG = & G ρL

(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.


239

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 f 1.0

0.08 0.9

0.06 0.8

Then from equation below the gas velocity, which will cause the flooding, is calculated. ρL − ρG (5-156) u GF = K1c ρG 0.2 Capacity factor, K1

Plate spacing, m 0.1 0.08 0.06 0.04

0.02

0.01 0.01

0.02

0.04 0.06 0.08

0.1

Flow parameter, FLG

0.2

0.4 0.6 0.8

L& ⎛ ρ = ⎜⎜ G & ρ G ⎝ L

⎞ ⎟⎟ ⎠

1.0

2.0

5.0

0.5

Fig.5.48 Capacity factor

By taking some percent of this velocity (usually 60-80 %), normal operating gas velocity 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 commonly used range in the designs. 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 consecutive plates, is the most important parameter fixing the height of a column and hence it must be kept as small as possible, if there are no other reasons such as construction of manholes etc. In practice, plate spacing is selected 0.45m or smaller for columns 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.


240

Fractional entrainment, E

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 E= & (5-157) L+e must be kept below 0.1 for a satisfactory design. In the equation above, e is the liquid as 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 percent of flooding gas velocity and these relationships found experimentally are given in Fig.5.49. 5.4.4 The Holes: The diameters of the holes, dh (mm) punched in the active area change from 3 to 15 mm, the 5 mm 1.0 0.8 being the most widely used one. They 0.6 are drilled at the corners of either a 0.4 square or an equilateral triangle, the Percent of flooding pitch (distance between the hole 0.2 centres), PT (mm) being 2.5 to 5 hole diameters. While smaller pitch causes 0.1 0.08 the coalescence of the gas bubbles 0.06 issuing from the holes, greater pitch 0.04 makes the diameter of the column unduly large. Between down-comer and 0.02 the first row of holes and also between the last row of holes and the weir 100 0.01 0.008 mm’s un-drilled spaces must be left 0.006 (why?). 0.004 By observing these, the holes must be distributed evenly in the active area. 0.002 Total holes area accounts for generally between 5-15% of the active area. The 0.001 ratio of total holes area to the active 0.01 1.0 0.02 0.04 0.06 0.1 0.2 area is function of the ratio of hole Flow parameter, FLG diameter to the hole pitch, the constant Fig.5.49 Liquid entrainment being dependent on the holes configuration. For equilateral hole configuration the relationship is ; 2 ⎡ dh ⎤ Ah (5-158) = 0.907 ⎢ ⎥ Aa ⎣ PT ⎦ and for square hole configuration, 2 ⎡ dh ⎤ Ah (5-159) = 0.785 ⎢ ⎥ Aa ⎣ PT ⎦ 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, n π d 2h −6 Ah = 10 (5-160) 4


241

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 increases the pressure drop in the gas side. Hence, an optimum height must be found for the weir. This is as low as 20 mm 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 m3/sm.

Orifice constant, Co

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 experimentation was given as; 2/3 ⎡ L& ⎤ 0.95 (mm) (5-161) h ow = 750 ⎢ ⎥ ⎣ ρL Lw ⎦ Plate Thickness/Hole diameter= χ/dh For a reasonable plate efficiency, hw+how 0.90 should be no smaller than 50 mm. 5.4.6 Pressure Drop in the Gas along 0.85 the Plate: The pressure drop that the gas phase will encounter as it passes the plate, can be divided in two parts: dry pressure 0.80 drop and the pressure drop that occurs as the gas passes through aerated liquid. Dry pressure drop that occurs during the 0.75 passage of gas through the holes of dry plate can easily be found from very well and less known orifice equation, which for this 0.70 case can be written in terms of liquid head ho (mm) as; 0.65 2 ⎛ 1 ⎞ ρG 2 0 5 10 15 20 (5-162) h o = 51⎜ ⎟ uh (Ah/Aa)*100 ⎝ Co ⎠ ρ L Fig.5.50 Orifice constant where, uh (m/s) is the velocity of gas through the holes which can be calculated from, & G (5-163) uh = ρG A h The orifice coefficient, Co is a function 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 Fig.5.50. The pressure drop that is due to the passage of gas through aerated liquid, ha is given as mm liquid head by, ha = Qp(hw+how) (5-164) where, Qp is known as aeration factor and can be found from Fig.5.51.


242

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 =

K 2 − 0.90(25.4 − d h ) ρ 0.5 G

(5-166)

where, K2 is a constant dependent on (hw+how) and the relationship between them is given in Fig.5.52. Down-comer back-up: the liquid level in the down-comer, which is also known as 32

31 1.0

K2

0.8

Qp 0.6

30

29

0.4

28 0.0

0.5

1.0

1.5

2.0

& G FGa = A a ρG

Fig.5.51 Aeration factor

2.5

3.0 27 0

20

40

60

80

100

120

hw+how , mm

Fig.5.52 Constant 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 gas and liquid phases as they flow, to the liquid level on the plate, which is (hw+how). Total pressure drop in the gas side was obtained as hT. The pressure drop in the liquid side as it flows through the down-comer and 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 neglected as the velocity of the liquid and the distance it travels, are very small. But the pressure drop encountered during the flow through the restriction at plate entrance may be significant. The restriction at the plate entrance is created deliberately by immersing the down-comer apron into the liquid on the plate to prevent the escape of gas bubbles through the down-comer. Generally an opening corresponding to hdc (mm) = hw-10 is left on the plate between plate deck and tip of down-comer apron. Hence, the cross-sectional area through which liquid flows at the plate entrance


243

Down-comer hpe ha

hda hd

Weir

ho how

how hw

hw

liquid

Plate

Column Vapor

Fig.5.53 Down-comer back-up

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; ⎡ L& ⎤ hpe= 166 ⎢ ⎥ ⎢⎣ ρ L A pe ⎥⎦

2

(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 safety, finally liquid level in the down-comer is given as, h da =

h w + h ow + h T + h pe Qp

(5-168)

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 designed such that hda ≤ ½ (P.S)*103 (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; θ dr =

A d (P.S. ) L& /ρ L

(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.


244

& as kg/s, which are no longer changed. 5.4.7 Design Steps: 1o) First calculate L& and G 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, PT 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 was 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 assumptions. 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 than 5 s., then go back to step 4 and change the relevant ones in the assumed 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, internal design of a sieve plate column is accomplished by first making some assumptions and then checking them one by one. In the design of absorption and desorption columns, the design is realized at the & are the greatest. As for the rectification columns, diameter is section where L& and G calculated 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 sections is quite great and/or the construction material of the column is expensive, then the diameters and the plate lay-outs of each section maybe made different. Example-5.17) Internal Design of a Sieve-Plate Rectification Column For the separation of methanol/n-propanol solution given in Example-5.10 , a sieve-plate column with 4 mm plate thickness will be used. Design the stripping section of the column with the values at the bottom plate.

Solution: From Example-5.10, L = 493.84 k − mol / h , G = 296.84 k − mol / h , tN= 95 oC, xN=0.022 (mass frac.), xN = 0.04 (mol frac.), yN = 0.12 (mol frac.) Mass flow rate of liquid, L& = L M L = (493.84 / 3 600)[(0.04)(32) + (0.96)(60)] = 8.1 kg / s & = G M = (296.84 / 3 600)[(0.12)(32) + (0.88)(60)] = 4.7 kg / s Mass flow rate of vapor, G G

o

At 95 C

ρ meth L

= 716 kg / m , ρ prop = 735 kg / m 3 , σ meth = 0.016 N / m , σ prop = 0.018 N / m L 3

ρ L = (0.022)(716) + (0.978)(735) = 734.6 kg / m 3 , σ = (0.022)(0.016) + (0.978)(0.018) ≅ 0.018 N / m

Pbottom M G (1.07)(56.64) = = 2.0 kg / m 3 (Pressure at the bottom was assumed as 1.07 atm) RT (0.082)(273 + 95) Assumptions: P.S.=0.60 m, Lw/Dc=0.69, hw=50 mm, dh=5 mm, PT =2.5 dh(∆), Percent of flooding =70 ρG =


245

8.1 2.0 = 0.09 , Then, uncorrected capacity factor 4.7 734.6 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), Flow parameter, from equation(5-155)

FLG =

0.5

⎡ ⎤ (4)(4.7) Dc = ⎢ ⎥ = 1.573 m . Take Dc = 1.60 m. ⎣ (π )(2.0)(1.32)(1 − 0.0833) ⎦ 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, Aa = 1.843-0.167 = 1.676 m2 Total hole area, Ah = (0.145)(1.676) = 0.243 m2 Length of weir, Lw = (0.69)(1.60) = 1.10 m

-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 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 + how = 50 + 34.87 = 84.87 mm. From Fig.5.52 K2 = 30.8 is read. Minimum vapor velocity through the holes from equation (5-166), 30.8 − 0.90(25.4 − 5) u hm = = 8.8 m / s 2 .0 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, Ah will have new value. This is : Ah = (0.116)(1.676) = 0.194 m2. Now, vapor velocity through the holes rises to uh = 4.7/(2.0)(0.194) = 12.11 m/s. This velocity is much secure than the previous one to prevent the weeping. -Check for pressure drop in the vapor along the plate; Dry pressure drop at the vapor side along the plate, For (Ah/Aa)*100 = 11.6 and with χ/dh = 4/5 = 0.8, from Fig.5.50 Co = 0.80 is read. Then, from 2 ⎛ 1 ⎞ ⎛ 2 .0 ⎞ 2 equation (5-162); h o = 51⎜ ⎟ ⎜ ⎟(12.11) = 31.82 mm liquid head 0 . 80 734 . 6 ⎝ ⎠ ⎝ ⎠ FGa = 4.7/[(1.676)(2.0)0.5] = 1.98 ≈ 2. From Fig.5.51 aeration factor, Qp = 0.61 is read. Pressure drop in the vapor through the aerated liquid, from equation (5-164); ha = 0.61 (84.87) = 51.77 mm liquid head. Total pressure drop in the vapor through the plate, hT = 31.82 + 51.77 = 83.59 mm liquid head. This pressure drop corresponds to (83.59)(734.6)/13 600 = 4.52 mm Hg, which can be considered reasonable for atmospheric rectification column. Pressure drop in the liquid at plate entrance; Clearance at plate entrance, hdc = 50-10 = 40 mm. Then, flow area 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),


246

50 + 34.87 + 83.59 + 10.43 = 293.3 mm 0.61 As hda < ½ (P.S.)*103 , assumptions made are valid. -Check for down-comer residence time from equation (5-170); (0.167)(0.60) θ dr = = 9.1 s (8.1 / 734.6) 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 (1.07 atm) at the start of the design is quite close to this. h da =

Finalised Design: P.S. hw dh χ (thickness of plate) PT (∆) Dc LW hdc n (number of the holes/plate) NReal

= 0.60 m = 50 mm = 5 mm = 4 mm = 14 mm = 1.60 m = 1.10 m = 40 mm = 5 908 = 14

Percent of flooding E uh uhm θdr Pbottom hda

= 70 = 0.021 = 12.11 m/s = 8.8 m/s = 9.1 s = 1.083 atm = 293.3 mm

Problems 5.1 The variation of vapor pressure with temperature for n-hexane and n-heptane is as follows:

po(mbar) t ( oC)

13.3 hexane - 25.0 heptane - 2.1

26.7 - 14.1 9.5

53.3 - 2.3 22.3

80 5.4 30.6

133 15.8 41.8

267 31.6 58.7

533 49.6 78.0

1013 68.7 98.4

2026 93.0 124.8

Assuming ideal behaviour, calculate and plot the vapor-liquid equilibrium data for the system at 1 atm. 5.2 a) Calculate the bubble point of a binary liquid solution containing 0.30 mole fraction of the MVC at 1 bar pressure. b) What is the dew point of the vapor mixture containing 30 mol percent MVC at 1 bar. c) Find the compositions of 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 pressure of the less volatile component with temperature is given by the expression

log p oB = 7.620 − o

where p B is in mbar and T in Kelvin.

1750 T [Ans.: a) 90 0C, b) 98 oC, c) x = 0.206, y = 0.438]

5.3 From the vapor-liquid equilibrium measurements of three different binary solutions at 760 mmHg, the relative volatilities at two different liquid compositions were calculated as follows:

1 2 3

system αAB Ethylacetate(A)-Ethanol(B) Methanol(A)-Water(B) Etyhlenedichloride(A)-Benzene(B)

at x = 0.3 1.49 4.63 0.898

αAB at x = 0.8 0.66 2.69 0.902


247

Which of these three systems shows azeotropism between the given concentrations, why? 5.4 1 200 kg n-propyl benzoate (M.W.= 164 ; b.p.=231 oC) at 20 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 20 oC will be continuously supplied to the vessel to maintain a liquid water level. N-propyl benzoate is insoluble in water. 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.

Temperature ( oC) Water (A)

97.93

98.62

99.31

99.66

99.83

110

p oA (mmHg) 705.69 723.44 741.56

751.47 755.40 1060

λA (kJ/kg)

2267

n-Propyl p o (mmHg) B benzoate λB (kJ/kg) (B)

7.89

8.17

8.46

-

-

-

8.53 352.2

2238 8.68

-

-

-

Over the temperature range of operation specific heats of water and n-propyl benzoate may be taken as 4.18 kJ/kgK and 1.87 kJ/kgK. 5.5 An acetone-water solution containing 60 % acetone by weight is separated by flash distillation at atmospheric pressure in two equilibrium stages. Half of the solution (by weight) is vaporized in the first stage, the remaining liquid solution being passed to the second stage, where its temperature is raised to 80 oC. a) Determine the temperature in the first stage, and the fractional vaporisation in the second, together with the compositions of vapor and liquid in each stage. b) What is the percentage recovery of the acetone in the two vapor products together? Equilibrium data for acetone-water at atmospheric pressure is given below, where x and y show the mass fractions of acetone in liquid and vapor respectively. [Ans.: a) 67.5 oC, ε2 = 0.28, b) 94 %]

80 71.9 67.6 65.2 63.4 62.0 60.8 59.7 58.2 56.4 t(oC) 100 88.2 x 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 volatility of the MVC to the LVC is constant at 12.5 at the operating concentration range. Calculate: a) The compositions of top and bottom products, b) The percentage recovery of the MVC. [ Ans. a) yD = 0.63, xw = 0.12 ] 5.7 Cumen is an important precursor for the production of phenol and acetone. It is produced by the catalytic reaction between benzene and propylene. Composition of the mixture leaving the catalytic reactor is given below; Propane 40.0 mole percent Propylene 2.0 mole percent Benzene 27.0 mol percent Cumen 30.0 mole percent Less volatile organics 1.0 mole percent


248

In order to separate cumen from the others, first flash distillation is used to remove the propane and propylene, later benzene is recovered and recycled. Calculate 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: Temperature,oC Propan Propylene Benzene Cumen

37.8 12.8 15.3 0.217 0.013

65.6 23.3 27.6 0.62 0.05

93.3 39 46 1.45 0.16

5.8 A 60 k-mol/h liquid solution which is at 25 oC, contains 32 mol percent ethylene glycol (EG) and 68 mol percent water (W). This solution will be flash- distilled at 1.013 bar constant pressure. For a fractional vaporization of 0.50 calculate: a) The flash temperature, b) The compositions of top and bottom products, c) The percentage recovery of EG. (EG is recovered in bottom product) 5.9 A 40 k-mol liquid solution containing 45 mol percent acetaldehyde (AL) and 55 mol percent ethanol (ET) will be subjected to differential (simple) distillation at 1.013 bar pressure. Operation will be stopped when the mol fraction of AL in the still drops to 0.15. a) Find the temperature of the still at the end of the operation, b) Calculate the average composition of distillate, c) Find the percentage recovery of AL. 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

with Rayleigh Equation.

F ⎛ xF =⎜ W ⎜⎝ x w

1

⎞ m −1 ⎟⎟ where, F and W are the total moles of the feed ⎠

(charge) and bottom product, xF and xw are the mole fraction of the MVC in the feed and bottom product respectively. b) A 80 k-mol binary solution containing 23 mole percent MVC will be subjected to simple distillation at 760 mmHg. Distillation will be stopped when the mole fraction of the MVC in the batch-still drops to 0.03. Calculate the average composition of top product (distillate) and percentage recovery of the MVC. c) If the same solution is subjected to flash distillation under the same conditions with the same percent 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! Within the concentration range involved vapor-liquid equilibrium at 760 mmHg may be given as 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 distillation at the same pressure, what will be the purity and percentage recovery of acetone?


249

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 orthocresol 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 equilibrium relation at the operating pressure and over range of concentration is represented 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 solution at 15 oC containing 45 mole percent benzene and 55 mole percent heptane is to be fed to a rectification column which will operate at atmospheric pressure. It is required to produce a top product containing not more than 5 mole percent heptane. a) Calculate the minimum reflux ratio, b) What will be the composition of the bottom product, if the top product flow rate is 40 % of the feed rate? Latent heat of vaporization of the solutions may be considered constant at 23 000 kJ/k-mol. The average specific heat of the feed over the required temperature range is 185 kJ/k-mol oC. 5.14 A solution of benzene and methylisobutylketone (MIBK) containing 60 mole percent benzene is to be separated by rectification in a sieve-plate column operating at atmospheric pressure. The feed is 40 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 % of the recirculated liquid is to be vaporised. The feed rate is 400 k-mol/h and it is suggested to operate the column at a reflux ratio 1.3 times its minimum value. Determine: a) the composition of the bottom product, b) the composition of vapor entering the condenser and of the reflux liquid entering the column, c) 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 [Ans.: a) xw = 0.046, b) y1 = 0.905, xo = 0.85 c) yN+1 = 0.092, xN = 0.072] 5.15 An ethyl acetate-acetic acid solution, containing 35 mole percent ethyl acetate will be rectified in a plate column operating at 760 mmHg. The column will be equipped with a total condenser and a total reboiler. The feed will be introduced as liquid-vapor mixture in which mole fraction of liquid is 0.65. 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 as operating reflux ratio. 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 the number of the ideal plates by using McCabe-Thiele Method. d) Calculate the condenser and reboiler loads (duties) as kW. Molar latent heats of vaporization of both components are the same at 28 500 kJ/k-mol. 5.16 A plate rectification column is to be designed to separate a 12 500 kg/h ethyl acetate-acetic acid solution, 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 reflux ratio which is 1,50 times the minimum is to be selected for the operation. The column will be equipped with a total condenser and a kettle-type reboiler. Average latent heat of vaporization of the solution remains constant at 28 500 kJ/k-mol. Liquid heat capacities of ethyl acetate and acetic acid at 60oC are 2.06 kJ/kg oC and 2.21 kJ/kg oC respectively. a) Calculate the flow rates of top and bottom products as kg/h and k-mol/h and their compositions as mass and mole percent. b) Find the minimum reflux ratio from xy-diagram and Underwood equation and compare them.


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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 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 temperature of each ideal plate and write these on the schematic column above. h) Calculate the quantities of ethyl acetate and acetic acid transferred between liquid and vapor phases on 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 loads as kW. 5.17 A continuous rectification column is used to separate a binary solution. Enriching and stripping section operating lines for this operation are given as;

yn+1 = 0.739 xn + 0.226 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 condenser, it condenses only the part of the vapor required as reflux liquid. [Ans. : a) xD = 0.866, xw = 0.051, D = 34.05 k-mol/h, W = 40.95 k-mol/h] 5.18 An existing column providing 7 equilibrium plates and equipped with a kettle-type reboiler and a total 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-flow, a) Calculate the necessary rate of vapor production in the reboiler as k-mol/h, b) What would be the percentage recovery of methanol at the same purity if reboiler capacity were limited to 90 k-mol vapor/h. 5.19 Fruit juice concentrates are prepared commercially by evaporation. But during the evaporation flavoring volatile components are lost with the vapor. In order to recover these flavoring components, saturated vapor coming from the evaporator is sent to a rectification column where essence is withdrawn 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 per k-mol of feed to the column. b) 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 only of methyl antranilate and water vapor, and relative volatility of methyl antranilate to water at high dilution is 3.5. [Ans.: a) G / F = 12 ]


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5.20 An ethanol- propanol solution is to be separated in a plate rectification column at atmospheric pressure. The feed is 40 mole percent vapor and contains 45 mole percent ethanol. 95 % of the ethanol should be recovered in the top product, which should be 92 % pure by mole. The column is provided with a partial condenser and a kettle-type reboiler, both of which have Murphree vapor phase efficiencies 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 minimum value, calculate: a) 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 constant at 47 000 kJ/k-mol. 5.21 It is required to design a rectification column to separate methanol from water at atmospheric pressure. The following table gives the design requirements for the feeds and products;

Stream Feed-1 Feed-2 Top product Bottom product Side product

Quality Saturated vapor Saturated liquid Saturated liquid Saturated liquid Saturated liquid

Flow rate, k-mol/h 400 200 150

Methanol mol fraction 0.50 0.30 0.96 0.04 0.70

The column has a total condenser and a reboiler. Assuming constant molar over-flow; a) Calculate the minimum reflux ratio, 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 from which the side stream 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 mmHg, the mole fractions of methanol in the liquid samples taken from the exit of the 1st, 2nd and 3rd plates 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 in a batch column at 760 mmHg, to produce a top product of 90 mole percent SCl2. The operation will be carried out keeping the top product composition constant and increasing the reflux ratio steadily until the mole fraction of SCl2 in the batch-still drops to 0.15. The reflux ratio at the last moment of the distillation will be 20 % greater than the minimum reflux ratio corresponding to the 0.15 mole fraction of SCl2 in the batch-still. Calculate: a) 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-moloC and 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


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liquid. For the conditions stated HG= 0.5 m and HL=1.0 m. For a reflux ratio 25 % greater than the minimum 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 assumed as follows: P.S. = 0.45 m, dh = 6 mm, □ 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 occur or not. If yes, what would you change in the assumed plate lay-out? c) Calculate the total pressure drop in the gas per plate. Densities of liquid and gas and surface tension of the liquid are: 850 kg/m3, 2.2 kg/m3 and 0.40 N/m respectively. 5.26 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. At the temperature of the top plate the densities and surface tensions of the ethyl acetate and acetic acid are 825 kg/m3, 984 kg/m3, 0.017 N/m and 0.022 N/m respectively.


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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.


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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 phase x xB xC -

E x t r a c t y yB -

p h a s e yC -

Weight ratio in extract phase, Y'=y /1-y

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 t = cons.

X′ = x / x values against Y' = y/yB values c

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 0.0 which special millimetric papers are 0.0 Weight ratio in raffinate phase, X'=x /1-x required and enlargement of part of it is rather difficult. As shown in Fig.6.2, the Fig.6.1 Liquid-liquid equilibrium when B apexes of the triangle represent pure and C are insoluble within each other 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


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Mass fraction of A in extract phase,y

A Mass fractions of A in raffinate and extract phases, x,y

1.0

G

0.5

M

L

L1 ∆

0.0

C

G1

0.0

B 0.5

Mass fractions of B in raffinate and extract phases, xB,yB

1.0

1.0

0.5

0.0 0.0

0.5

1.0

Mass fraction of A in raffinate phase, x

Fig.6.2 (a) Right angle triangular diagram, (b) distribution diagram

L * LM = G * GM , 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 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

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y

A 1.0

x,y

1.0

x=y

Distribution curve

P

xP=yP

yP

G y M Raffinate phase curve

G,L

Extract phase curve Tie line

L

x

x

xB

0.0

C

P

y

0.0

yB

R

T

J

1.0

0.0

B

0.0

x x

1.0

xp

xB,yB (b)

(a)

Fig.6.3 (a) Solubility, (b) Distribution diagram for one pair partially soluble system

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 y

A 1.0

x,y

1.0

U x=y Raffinate phase curve

x

L

x

V

M

Extract phase curve

Tie line

G,L

0.0

C

Distribution curve

y

G

y 0.0

R

xB

J

B 1.0

T yB

xB,yB (a)

0.0 0.0

x

x

1.0

(b)

Fig.6.4 (a) Solubility, (b) Distribution diagram of two pairs partially soluble system


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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), x ′B = xB/(x+xC) and x ′C = xC/(x+xC) in the raffinate phase and as y'= y/(y+yC), y′B = yB/(y+yC) and y′C = yC/(y+yC) in the extract phase and then by carrying x ′B values against x' values and y′B 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 t =cons. between extract and raffinate curves, y′B − y′ such as point M represents Extract phase curve G ′ y heterogeneous ternary mixture, which B T upon settling gives equilibrium y′B V Tie line extract and raffinate phases, which x ′B are located on the diagram by M drawing the tie line passing through R point M by trial and error with the x ′B help of diagram below. While their L U x ′B − x ′ compositions are directly read from Raffinate phase curve the diagram, their amounts are ′ y calculated by applying inverse leverx′ 1.0 0 x ′, y′ arm rule to the line LMG. 6.3 Selection of Solvent: In many 1.0 cases, there is more than one solvent that can be used for a specific y′ extraction operation. The following y =x L,G properties of the solvents are looked y′ 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 0 separation power of solvent B of x′ 1.0 0 solute A from raffinate solvent C. x′ Whatever is the α in rectification is Fig.6.5 Liquid-liquid equilibrium on solvent-free coordinates No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

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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. y/yC β= (6-1) x/x C 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 mixerFeed Solvent,S,ys settler unit shown in Fig.6.6 is a F, xF typical stage-wise contact type of equipment used in extraction operations. The equipment, which consists of two main sections, one Extract Mixer being mixing and the other settling phase,G,y section, can be constructed as single piece of equipment as shown in the Raffinate Figure or, as two separate units. In phase,L,x the mixing section, the two phases Settler are mixed thoroughly with the help of a stirrer to disperse one of the phases as small droplets in the other Fig.6.6 Mixer-settler unit No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

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Exit of light phase

Main interface

Exit of light phase

Heavy phase

Dispersed light phase Perforated plate

Perforated plate

Down-comer

Coalesed light phase

Light phase

Heavy phase

Exit of heavy phase

Coalesed heavy phase

Up-comer

Dispersed heavy phase Light phase

Main interface

Exit of heavy 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


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relationship and the total and components balance equations which can be written along the equilibrium mixer-settler unit shown in Fig.6.8 as; (6-2) Total material balance: F+S = M1 = L1+G1 (6-3) Solute A balance: FxF + Sys = M1xM1 = L1x1 +G1y1 (6-4) Solvent B: FxFB+ SysB = M1xM1B = L1x1B + G1y1B 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 Extract product, G1,y1 continuously. When the solvent does not Solvent, S,ys contain solute A then ys=0. Points F and S can easily be located on the diagram as their 1 compositions are known. It follows from Fig.6.9 that the feed is a binary solution of A Feed, F,xF Raffinate and C and the solvent contains small amount product, L1,x1 of solute A. Then either by applying inverse Fig.6.8 An Equilibrium Stage 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 y

A 1.0

x,y

1.0

x=y

Gm

P

G1

y1 xF F

G1,L1 D

xM1

x1 L1

M1

x1 E

yS Lm 0.0

C

P

y1

0.0 x1B

xM1B

y1B

S

B

xB,yB

ySB 1.0

0.0 0.0

x x1

1.0

(b)

(a)

Fig.6.9 Solution of single stage extraction

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. 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


261

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; Fx − L1x1 P.R. = F .100 (6-5) Fx F 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;

F,xF

L1,x1

S1

L2,x2

S2

S,ys Solvent

S3

Feed

S1 Feed 1

3

2

G1 y1

G3 y3

G, y Extract product

(a)

2

1

G3 y3

L3,x3 Raffinate product

3

G1 y1

G2 y2

S,ys Solvent L3,x3

L2,x2

L1,x1

F,xF G2 y2

S3

S2

G3 y3

Raffinate product G, y Extract product

(b)

Fig.6.10 Flow diagram of cross-current multi-stage extraction

(6-6) Total material balance: Ln-1 + Sn = Mn = Ln + Gn (6-7) Solute A balance: Ln-1xn-1 + Snys = MnxMn = Lnxn + Gnyn By substituting Lo = F and xo = xF, these equations for the first stage (n=1) becomes, F + S 1 = M1 = L 1 + G 1 FxF + 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;


262

L1 + S2 = M2 = L2 + G2 L1x1 + 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. y

A 1.0

x,y

1.0

x=y P y1

xF

F

M1

G1

y1

x1 L1 y2

x1

2

y3 x2 x3 yS 0.0 0.0

P

1 G2

M2

3 L2 L3

y2 y3

G3

x2

M3

x3

S

C

xB,yB (a)

1.0

B

0.0 0.0

x3

x2

x1

x 1.0

(b)

Fig.6.11 Solution of three-stage cross-current extraction on right angle triangular diagram

The percentage recovery of solute A is calculated from, Fx F − L N x N .100 (6-8) P.R. = Fx F If the solution is done on solvent-free coordinated diagram; for any stage n; L′n −1 + S′n = M′n = L′n + G ′n (6-9) Total material balance: Solute A balance: L′n−1x ′n−1 + S′n y′s = M′n x ′Mn = L′n x ′n + G ′n y′n (6-10) Solvent B balance: L′n−1x′(n−1)B + S′n y′sB = M′n x ′MnB = L′n x′nB + G′n y′nB (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 L′o = F′ and x ′o = x ′F can be written as; F′ + S1′ = M1′ = L1′ + G1′ F′x ′F + S1′ y′s = M1′ x ′M1 = L1′ x1′ + G1′ y1′ F′x ′FB + S1′ y′sB = M1′ x ′M1B = L1′ x1B ′ + G1′ y1B ′


263

Points F and S are first located on the diagram with the help of their known compositions as shown in Fig.6.12. Then point M1 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 L1′ and G1′ 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; S

y'SB

t =cons.

P.R.=

y'B-y' G3

y′B x ′B

G2

M3

1 M2

x'3B

0

x'3

y'1B

2

3

L3

G1

L2

M1

x'B-x'

L1

' y'S x2' y 3 x'1

F

x'F y'2

y'1

1.0

F′x ′F − L′N x ′N .100 F′x ′F

(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; C X ′n −1 + B n Ys′ = C X ′n + B n Yn′ (6-13)

x',y'

From which,

1.0 y' y'

−

y'2 y =x

Ys′ − Yn′ C = Bn X′n−1 − X′n

(6-14)

is obtained. This equation represents a straight line on X'Y'diagram passing through points ( X′n−1 ; Ys′) and (X′n ; Yn′ ) with a 0 ' ' ' x1 1.0 x2 slope of (-C/Bn), which is known F x3 x' as operating line for stage n. Fig.6.12 Solution of cross-current three-stage Remember that the equilibrium extraction on solvent-free coordinated diagram relationship in this case is also given on X'Y'-diagram. Hence, solution is done as follows: first, equilibrium relationship is drawn on X'Y'-diagram, then equation (6-14) is written for the first stage by taking X′o = X′F as; y'3

−

Ys′ − Y1′ C = B1 X′F − X1′

with the known values, point (X′F ; Ys′) is located on the diagram and the line with slope of (–C/B1) passing through this point is drawn. The coordinates of intersection


264

point of this line with equilibrium curve give X1′ and Y1′ . Then by writing equation (614) 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; CX′F − CX′N X′ − X′N .100 = F .100 (6-15) P.R. = C X′F X′F Y'

Y'1

Equilibrium curve

Y'2 slope= -C/B1 1 Y'3

slope = -C/B3

2 slope = -C/B2

3 Y'S 0.0 0.0

X'2

X'3

X'1

X'

X'F

Fig.6.13 Determination of equilibrium stages at cross-current multi-stage extraction when the solvents are insoluble

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 required solvent rate is less. n shows any stage, N shows the last stage of the cascade. Extract product G1, y1

Solvent

1 F, xF Feed

G3, y3

G2, y2

L2, x2

Ln-1, x n-1

S, yS

GN, yN

n

2 L1, x1

Gn+1, yn+1

Gn, yn

N Ln, x n

LN-1, xN-1

LN, xN Raffinate product

∆

(+)

Fig.6.14 Flow diagram of counter-current multi-stage extraction

Total material and solute A balances along the cascade are written as; Total material balance: F + S = M = LN + G1 Solute A balance: FxF+Sys= MxM = LNxN+ G1y1 No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

(6-16) (6-17) 265

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; (6-18) F - G1 = LN - S = ∆s where, ∆s is the total net flow which remains constant along the cascade, hence equation (6-18) can be generalized to; (6-19) F - G1 = Ln - Gn+1 = LN - S = ∆s If F > G1 or LN > S, the total net flow is from left to right as shown in Fig.6.14, 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. y

A 1.0

x,y

1.0

P

y1 G1

1

xF F

2 M

L1

L3

x1

0.0

y3

G3

x2 y4

G4

4

x3

S

LN=L4

0.0

C

x=y

G2

3

L2

∆S

y2

xB,yB

B 1.0

x4 0.0 0.0

x4

x3

x2

x x1

Fig.6.15 Solution of four-stage counter-current extraction on right angle triangular diagram

With the help of equations given above the number of the equilibrium stages required for 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


266

(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; P.R. =

Fx F − LN x N .100 Fx F

(6-20)

For the solution of the problem on the solvent-free coordinated diagram; total material balance on solvent-free basis and the solute A balance along the cascade can be written as; (6-21) F' + S' = M' = G1′ + L′N F′x ′F + S′y′s = M′x ′M = G1′ y1′ + L′N x ′N (6-22) Equation (6-21) can also be written as, F′ − G1′ = L′n − G ′n +1 = L′N − S′ = ∆′s (6-23) This means that total net flow on solvent-free basis remains constant along the cascade. If the feed does not contain any solvent then F' = F and if the solvent is pure, S' = 0, S' y′s = 0, F' = M', X′F = X′M . The solution on solvent-free S y'SB coordinated diagram of the system is done as follows: first, points F, S G3 and LN are located on the diagram y′B G2 G1 y'1B with the help of known x ′B compositions (in Fig.6.16). Points 3 y'B-y' F and S are joined and point M is located on this line in the known 2 1 M way. By joining points LN and M and extending this line, point G1 is L3 LN L2 located. According to equation (6L1 ' ' 23), point ∆s is obtained by x B-x x'N drawing FG1 and SLN lines. Then, ' ' ' y 1.0 y 1 0 y'S ' drawing of equilibrium stages is 2 y 3 F x3 ' started. By drawing the tie line xF x',y' passing through point G1, point L1 1.0 is obtained. Equation (6-23) x'∆SB indicates that points ∆s, L1 and G2 ∆S should be on the same straight line; y' y' y =x 2 hence by joining ∆s with L1 and extending the line, point G2 is ' y3 found. The tie line passing through point G2 gives point L2. The solution is continued by drawing 0 ' ' ' once an operating line (the lines x2 x1 1.0 0 x3 ' drawn from point ∆s are known as x operating lines) once a tie line until Fig.6.16 Solution of counter-current extraction on solvent-free coordinated diagram No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

267

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; P.R. =

F′x′F − LN′ x′N .100 F′ x′F

(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; C X′N + B Yn′+1 = C X′n + B Ys′ (6.25)

If this equation is solved for Yn′ +1 ,

C X′N C X′n + Ys′ − B B (6-26) ' ' is obtained. This equation represents a straight line on X Y -diagram passing through points ( X′N ; Ys′) and (X′F ; Y1′) 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'-diagram, then the operating line given by equation (6-26) is drawn on the same diagram as the values of C, B, Ys′ , X ′N and X ′F 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 Y' extract phase is given by; Yn′+1 =

Y'

Equilibrium curve

1

P.R. =

X′F − X N′ .100 X′F

(6-27)

Minimum solvent rate: In counter-current multi-stage extraction if the solvent rate S is Operating line reduced, point M shifts toward F and point ∆s slope= C/B comes closer to the diagram as seen from 3 Fig.6.15. The closer ∆s means more equilibrium stages for the same extraction. 4 Hence, for the same extraction, less number of Y's ' equilibrium stages is required, if the solvent X X'F X'N rate is increased; and more number of Fig.6.17 Determination of number of the equilibrium stages is needed, if the solvent equilibrium stages in counter-current rate is decreased. If one of the operating lines extraction, when solvents are insoluble 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 2


268

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 shown by point (G1)m. Intersection point of lines (G1)mLN and FS locates point Mm. By y

A 1.0

x,y

1.0

P (y1)m

(G1)m

x=y F Mm

∆S

∆Sm

0.0

C

S

LN 0.0

1.0

xB,yB

B

0.0

x

0.0

Fig.6.18 Determination of minimum solvent rate

applying inverse lever-arm rule to line FMmS, 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. 6.4.1.4 Counter-Current Multi-Stage Extraction under Reflux: It is obvious that, the 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. Stri pping section

Enriching section

Se,ye

Gn,yn

L1,x1 Lo xo reflux

Gn+1,yn+1

Gf,yf

n

2

1

G,y

G3,y3

G2,y2

G1,y1 solvent separat

L2,x2

Ln-1,xn-1

Gf+1,yf+1

Ln,xn

Lm-1,xm-1

S,yS N

m

Lf,xf

Lf-1,xf-1

Gm+1,ym+1 GN,yN

Gm,ym

f

solvent

Lm,xm

LN,xN

LN-1,xN-1

Raffinate product Extract D, xD product

Feed F, xF

Fig.6.19 Flow diagram of multi-stage extraction under reflux No part of this CD-book may be multiplied for commercial purposes. E.Alpay & M.Demircioğlu

269

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; G ′n +1 = L′n + D′ (6-28) This equation can also be written as, G ′n +1 − L′n = D′ = ∆ ′e (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: G ′n +1 y′n +1 = L′n x ′n + D′x ′D (6-30) which can also be written as; G ′n +1 y′n +1 − L′n x ′n = D′x ′D = ∆ ′e y′∆e (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, G ′n +1 y′(n +1)B − L′n x ′nB = D′x ′DB = ∆′e y′∆eB (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; L′m + S′ = G ′m+1 + L′N (6-33) which can be written also as, L′m − G ′m+1 = L′N − S′ = ∆′s (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; L′m x ′m + S′y′s = G ′m+1 y′m+1 + L′N x ′N (6-35) which can also be written as, L′m x ′m − G ′m+1 y′m+1 = L′N x ′N − S′y′s = ∆′s x ′∆s (6-36) This means that in the stripping section there is a net flow of solute A whose value equals ∆′s x ′∆s . From the solvent B balance in this section, L′m x ′mB − G′m+1 y′(m+1)B = L′N x′NB − S′y′sB = ∆′s x ′∆sB (6-37) can also be written. Total material balance for the whole cascade;


270

F' + S' = D' + L′N (6-38) With the help of equations (6-29) and (6-34) this equation can also be written as; F′ = ∆′e + ∆′s (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 x′D ( = x′o = y′∆e ) and x′N , points, D(Lo) and LN are located. S ∆e y′SB y′∆eB Point LN must lie on the y′B − y′ raffinate phase curve but not G5 G4 y′B the point Lo. Points S and LN G3 G2 G1 are joined and extended x ′B y′1B beyond LN. A vertical from 5 4 x ′D (= x ′o = y′∆e ) fixes the 2 3 1 point G1, from which y1B ′ is L5 read. The values of y1B ′ , x ′oB LN L4 L3 L2 are and selected RD L1 substituted into equation Lo,D x ′B − x ′ (6-40) and y′∆eB is solved. 1.0 0 x′N y′S x′F F With the help of this, point ∆e x ′D = y′∆e x ′, y′ is located. 1.0 L′ y′ − y1B ′ y′1 y′ R D = o = ∆eB (6-40) D′ y1B ′ − x ′oB y′2 By joining point ∆e with point y′3 x′∆SB F and extending the line until y =x y′4 ∆S cutting the extension of SLN line, point ∆s is found. Then y′5 drawing of equilibrium stages is started. By drawing the tie 0 line passing through point G1, 1.0 0 ′ x ′3 x′2 x′1 x 5 x′ x′4 point L1 is located. Equation x′ ∆S (6-29) indicates that points G2, L1 and ∆e must be on the Fig.6.20 Solution of multi-stage extraction under reflux same straight line (operating on solvent-free coordinated diagram 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 ∆eF∆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,


271

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 x ′D (= x ′o = y′∆e ) . The tie line, which when extended cuts the vertical at the farthest point from the diagram, gives point ∆em from which value of y′∆emB is read and finally from, y′ − y1B ′ R Dm = ∆emB (6-41) y1B ′ − x ′oB 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, ρ M = ρs φs + ρ d φ d µ 6 µ oφo µ M = w (1 + ) , φ w > 0.4 (6-42) φw µw + µo µ 1.5 µ w φ w µ M = o (1 + ) , φ w < 0.4 φo µw + µo 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 φ d , is usually smaller than its value in the feed ( φ dF ) which is giving as; φ dF = q dF /(q dF + q sF )

(6-43)

The ratio of φ d to φ dF is given as,

⎞ ⎛ ρ s ⎞0.430 ⎛ σ 3 ρ s ⎞ ⎛ µ d ⎞ ⎛ P q dF µ s2 ⎞ ⎛ µ 3s φd ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ = 3.39 (6-44) 3 2 4 φ dF ∆ρ V σ q ρ σ 9.81 µ ⎠ ⎝ ⎝ µs ⎠ ⎝ L ⎠ ⎝ dF s ⎠ ⎝ s ⎠ 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 D 2v z ], 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). 0.247

0.427

0401


0.0987

272

100

1

0.1 10000

10

100000

Re

1000000

P d n 3ρ M 5 i

1

0.1 1

10

100

Re =

1000

10000

100000

d nρ M µM 2 i

Fig.6.21 Power requirment in un-baffled vessels equipped with turbine type impeller 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; 0.0473 0.274 −0.204 ⎛σ⎞ ⎛ ⎞ ⎛ P ⎞ − 2.066 + 0.732 φd µ s ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (6-45) d p = 10 ⎝ VL ρ M ⎠ ⎝ ρs ⎠ ⎝ ρs ⎠ 6 φd (6-46) av = dp 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, 2/3 ⎤0.62 0.17 1/3 ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ k Ls d p ρ d P s i ⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ Sc s0.36 (6-47) = 2 + 0.47 ⎢ d 4/3 Sh s = p D As ⎝ VL ⎠ ⎝ µ s ⎠ ⎦ ⎝ D v ⎠ ⎣ 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; ⎡3 ∞ ⎛ λ 64 D θ ⎞⎤ d (6-48) k Ld = p ln ⎢ ∑ Tn2 exp⎜⎜ − n 2 Ad ⎟⎟⎥ d 6 θ ⎣⎢ 8 1 p ⎝ ⎠⎦⎥ 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


273

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; 1 1 1 = + (6-49) K Ld k Ld m sd k Ls 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; c Ad1 dc Ad ⌠ N od = ⎮ (6-50) ∗ ⌡c Ad 2 c Ad − c Ad

where it is assumed that molar concentration of solute A in dispersed phase changes from cAd1 to cAd2 and c ∗Ad 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; c Ad1 1 c −c 2 N od = dc Ad = Ad1 Ad (6-51) ∗ ∫ c Ad2 − c Ad2 cAd 2 c Ad 2 − c∗Ad 2 Table 6-1. The values of Tn and λn in equation (6-48) kLsdp/DAs 3.20 5.33 8.00 10.7 16.0 21.3 26.7 53.3 107 213 320 ∞

λ1 0.262 0.386 0.534 0.680 0.860 0.982 1.082 1.324 1.484 1.560 1.600 1.656

λ2 0.424

4.92 5.26 5.63 5.90 7.04 7.88 8.50 8.62 9.08

λ3

15.7 17.5 19.5 20.8 21.3 22.2

T1 1.49

T2 0.107

1.49 1.48 1.47 1.49 1.43 1.39 1.31 1.31 1.29

0.300 0.382 0.428 0.495 0.603 0.603 0.588 0.583 0.596

T3

0.205 0.298 0.384 0.396 0.391 0.386

On the other hand from definition equation, N od =

z z = H od u d / K Ld a v

(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


274

[ ud= qdF /0.785 D 2v ]. As the Murphree stage efficiency for the dispersed phase is, c −c E Md = Ad1 ∗Ad2 c Ad1 − c Ad2 Then, E Md =

(6-53)

c Ad1 − c Ad 2 (c Ad1 − c Ad 2 ) /(c Ad 2 − c ∗Ad 2 ) = (c Ad1 − c Ad 2 ) + (c Ad 2 − c ∗Ad 2 ) (c Ad1 − c Ad 2 ) /(c Ad 2 − c ∗Ad 2 ) +1

and finally with the help of equation (6-51); E Md =

N od N od + 1

(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; (6-55) Dd = 8.4 (qs+ qd)0.5 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.


275

APPENDICES

276

Table.App. 2.1 Laplace Transforms No .

Transform Function f(θ)

∞

₤(s) = ∫ e −sθ f (θ)dθ 0

1

1

2

θ

1 s 1 s2

θ n −1 (n − 1)! 1 (πθ)

3

4

1 s 1

s3 / 2

1 s−a

6

eaθ

7

⎛ k2 ⎞ exp⎜ − ⎟ 3 ⎝ 4θ ⎠ 2 (πθ )

k

8

9

10

⎛ k ⎞ ⎟ ⎝2 θ⎠ ⎛ k2 ⎞ 1 exp⎜⎜ − ⎟⎟ (πθ) ⎝ 4θ ⎠ θ ⎡ ⎛ k 2 ⎞⎤ ⎛ k ⎞ 2 ⎢exp⎜⎜ − ⎟⎟⎥ − k erfc⎜ ⎟ π ⎣⎢ ⎝ 4θ ⎠⎦⎥ ⎝2 θ⎠

erfc⎜

⎛

11

12

13

n = 1, 2, 3,...

θ π

2

5

1 sn

− exp(ak ) exp(a 2θ)erfc⎜ a θ + ⎝

k ⎞ ⎛ k ⎞ ⎟ + erfc⎜ ⎟ 2 θ⎠ ⎝2 θ⎠

k ⎞ ⎛ exp(ak ) exp(a θ)erfc⎜ a θ + ⎟ 2 θ⎠ ⎝ 2

⎛ θ ⎞ ⎟ ⎝ 2k ⎠

erf ⎜

e −k

k >0

s

1 −k e s 1 −k e s

k ≥0

s

s

k ≥0

− 3 −k s s 2e

k ≥0

ae − k s k ≥0 s (a + s ) e −k s k ≥0 s (a + s ) 1 exp(k 2 s 2 )erfc(ks) k > 0 s

277

Table. App. 2.2 Error Function

erf u =

2 π

u

∫ e − z dz ;

o

2

d 2 − u du (erf u ) = − e . ; erf(-u) =-erf u; erf(o)=o; dz dz π 2

erf(∞)=1

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

erf u 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

u 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

erf u 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

kg NH3 ----------100 kg H2O 100 90 80 70 60 50 40 30 25 20 15 10 7.5 5 4 3 2 1

Partial pressure of ammonia, mmHg 0oC 947 785 636 500 380 275 190 119 89.5 64 42.7 25.1 17.7 11.2

10oC

987 780 600 439 301 190 144 103.5 70.1 41.8 29.9 19.1 16.1 11.3

20oC

30oC

945 686 470 298 227 166 114 69.6 50.0 31.7 24.9 18.2 12.0

719 454 352 260 179 110 79.7 51.0 40.1 29.6 19.3

40oC

50oC

60oC

692 534 395 273 167 120 76.5 60.8 45.0 30.0 15.4

825 596 405 247 179 115 91.1 67.1 44.5 22.2

834 583 361 261 165 129.2 94.3 61.0 30.2

Table. App. 4.2 Solubility of sulphur dioxide in water

kg SO2 ---------100 kg H2O 20 15 10 7.5 5.0 2.5 1.5 1.0 0.7 0.5 0.3 0.1 0.05 0.02

Partial pressure of sulphur dioxide, mmHg 0oC 646 474 308 228 148 69 38 23.3 15.2 9.9 5.1 1.2 0.6 0.25

7oC 657 637 417 307 198 92 51 31 20.6 13.5 6.9 1.5 0.7 0.3

10oC

15oC

20oC

30oC

40oC

50oC

726 474 349 226 105 59 37 23.6 15.6 7.9 1.75 0.75 0.3

567 419 270 127 71 44 28.0 19.3 10.0 2.2 0.8 0.3

698 517 336 161 92 59 39.0 26.0 14.1 3.2 1.2 0.5

688 452 216 125 79 52 36 19.7 4.7 1.7 0.6

665 322 186 121 87 57 7.5 2.8 0.8

458 266 172 116 82 12 4.7 1.3

279

Table. App. 4.3 Solubility of carbondioxide in 15.3% Monoethanolamine solution

Partial pressure of CO2,mmHg 1 5 10 30 50 70 100 200 300 400 500 600 760 1000 2000 3000 5000 7000

k-mol CO2/ k-mol amine 40oC 0.383 0.438 0.471 0.518 0.542 0.558 0.576 0.614 0.639 0.657 0.672 0.686 0.705 0.727

60oC

0.412 0.459 0.482 0.498 0.516 0.552 0.574 0.591 0.605 0.615 0.631 0.650 0.702

80oC

0.379 0.405 0.422 0.442 0.481 0.505 0.523 0.538 0.550 0.566 0.584 0.637 0.669 0.712 0.742

100oC 0.096 0.152 0.194 0.265 0.299 0.322 0.347 0.393 0.423 0.442 0.458 0.472 0.489 0.509 0.562 0.596 0.641 0.672

120oC

140oC

0.200 0.227 0.281 0.314 0.336 0.355 0.370 0.390 0.413 0.476 0.513 0.562 0.597

0.109 0.162 0.194 0.219 0.237 0.254 0.275 0.300 0.366 0.408 0.464 0.500

Table. App. 4.4 Solubility of H2S in 15.3% Monoethanolamine solution

Partial pressure of H2S, mmHg 1 3 5 10 30 50 70 100 200 300 400 500 600 700 800

k-mol H2S / k-mol amine 40oC 0.128 0.212 0.271 0.374 0.579 0.683 0.750 0.802 0.890 0.931 0.949 0.959 0.970 0.980 -

60oC 0.137 0.171 0.240 0.386 0.472 0.534 0.600 0.722 0.790 0.836 0.871 0.900 0.921 0.942

80oC

0.141 0.243 0.314 0.364 0.422 0.545 0.617 0.666 0.706 0.738 -

100oC 0.029 0.050 0.065 0.091 0.160 0.203 0.238 0.279 0.374 0.439 0.490 0.536 0.575 0.607 0.636

120oC

140oC

0.025 0.036 0.056 0.101 0.139 0.153 0.182 0.256 0.312 0.357 0.393 0.426 0.453 -

0.016 0.025 0.040 0.072 0.091 0.106 0.124 0.167 0.200 0.226 -

280

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

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.500

0.632

0.735

0.816

0.878

0.925 0.959

0.983

1.0

x

0

0.05

y

0

0.187 0.317

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

x

0.0

0.05

0.1

0.2

0.4

0.5

0.6

y

0.0 0.148

0.795

0.856 0.902

0.3

0.272 0.467 0.61 0.716

70.2

68.0

66.1

65.3

64.5

0.7

0.8

0.9

0.95

1.0

0.939

0.970

0.984

1.0

Table.App.5.4 Vapor- liquid equilibrium for methanol-water system at 760 mmHg t (oC) x y

100 93.7 0 0.05 0 0.243

89.2 0.1 0.390

83.2 0.2 0.562

79.1 76.1 0.3 0.4 0.664 0.735

73.6 0.5 0.791

71.5 69.6 67.8 66.1 0.6 0.7 0.8 0.9 0.839 0.882 0.923 0.962

64.5 1.0 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 77,2 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 0,18 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 80.1 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 0.267 0.455 0.598 0.69 0.778 0.838 0.89 0.933 0.968 1.0

281

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 21.1 x 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 0 0.0951 0.20 0.30 0.40 0.5 0.6 0.7 0.8 0.9 1.0

Specific enthalpy of y, mass fraction of Specific enthalpy of saturated liquid, h (kJ/kg) ammonia in vapor saturated vapor,H (kJ/kg) 790 0 2 810 600 0.622 2 280 420 0.824 1 960 310 0.903 1 775 200 0.94 1 650 140 0.958 1 580 100 0.97 140 0.975 160 0.98 1 530 240 0.99 310 1.0 1 500

282

Table.App.5 Periodic Table Group Period

1

1

2

3

4

5

6

7

Ia

IIa

IIIb

IVb

Vb

VIb

VIIb

8

9

10

VIIIb

11

12

13

14

15

16

17

18

Ib

IIb

IIIa

IVa

Va

VIa

VIIa

VIIIa

1 H 1.01

2 He 4.00

2

3 Li 6.94

4 Be 9.01

3

11 12 Na Mg 22.99 24.31

4

19 20 21 K Ca Sc 39.10 40.08 44.96

22 23 24 25 26 27 28 Ti V Cr Mn Fe Co Ni 47.88 50.94 52.00 54.94 55.85 58.93 58.70

5

37 38 39 Rb Sr Y 85.47 87.62 88.91

40 41 42 43 44 45 46 47 48 49 50 51 52 53 Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I 91.22 92.91 95.94 98.91 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.75 127.60 126.90

6

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 * Hf Cs Ba La Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po 132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.20 192.22 195.08 196.97 200.59 204.38 207.20 208.98 (209)

85 At (210)

86 Rn (222)

7

87 88 89 104 105 106 107 108 109 110 ** Rf Fr Ra Ac Ha Sg Ns Hs Mt Uun (223) 226.03 (227) (261) (262) (263) (262) (265) (266)

117

118

Lanthanoide

Actinoide

29 Cu 63.55

111 Uuu

30 Zn 65.39

112 Uub

5 B 10.81

6 C 12.01

7 N 14.01

8 O 15.99

9 10 F Ne 19.00 20.18

13 Al 26.98

14 Si 28.09

15 P 30.97

16 S 32.07

17 18 Cl Ar 35.45 39.95

31 Ga 69.72

32 Ge 72.61

33 As 74.92

34 Se 78.96

113

114

115

116

35 Br 79.90

58 59 60 61 62 63 64 65 66 67 68 69 70 * Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb 140.12 140.91 144.24 (145) 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.04

71 Lu 174.97

90 91 92 93 94 95 96 ** Th Pa U Np Pu Am Cm 232.04 231.04 238.03 237.05 (244) (243) (247)

103 Lr (260)

97 Bk (247)

98 Cf (251)

99 Es (254)

100 Fm (257)

101 Md (258)

102 No (259)

36 Kr 83.80 54 Xe 131.29

283

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.

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

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