B. Tech. Semester VCHE-307 : MASS TRANSFER-I (Chemical Engg.)

L T P CreditsClass Work : 50 Marks3 1 -4Exam : 100 MarksTotal : 150 MarksDuration of exam: 3 hrs.

Unit-I:Diffusion in gases and liquids, Equation of continuity.

Unit-II:Theories of mass transfer, Individual and overall mass transfer coefficients, Mass, heat and momentum transfer analogies. simultaneous heat and mass transfer.

Unit-III:Mass balance in co-current and counter-current continuous contact equipments. Concept of operating line. Multistage countercurrent operations. Concept of ideal stage. Stage efficiencies. Continuous conctact equipments. HTU and NTU concepts.

Unit-IV:Gas absorption: Design of plate and packed absorption columns. Non isothermal absorption.

Unit-V:Humidification, Design of cooling towers.

Unit-VI:Drying of solids. Rate of drying curves, Through circulation drying, Continuous drying, Types of driers.

TEXT BOOKS

1. Mass Transfer Operations: R.E. Treybal-Mcgraw-Hill Book Company New Delhi.2. Introduction to Chemical Engineering: W.L. Badger and J.T. Banchero-Mcgraw-Hill Book Company, New Delhi.

REFERENCE BOOKS

1. Unit Operations of Chemical Engineering: W.L. Mc Cabe & J.C. Smith- Mcgraw Hill, New Delhi. 2. Chemical Engineering: J.M.Coulson and J.F. Richardson Vol-I- Perganon, New York.

Note : Eight questions are to be set by taking at least one question from each unit but not more than two from any unit. Students have to attempt five questions in all.

Question: What is Ficks first law of diffusion?Answer: In the case of molecular diffusion, since mass transfer occurs from a region of high concentration to one of lower concentration, the flux is proportional to the concentration gradient.

or

Question: What is ?Answer: is the molar flux of component A in Z-directionQuestion: What is the unit of molar flux ?Answer: kmol/m2 sQuestion: What is ?Answer: is the proportionality constant, called the molecular diffusivity or diffusion coefficient of the molecule A in B.Question: What is the unit of diffusivity ?Answer: m2 /s.

Question: What is the effect of pressure and temperature on diffusivity?Answer: Since diffusive mobility is a function of the number of collisions, diffusivity increases with decrease in pressure because number of collisions is less at lower pressure. However in the case of liquid the effect of pressure is negligible. Again random thermal movement of molecules increases with increase in temperature, so diffusivity increases with increase in temperature.Question: What is the range of diffusivity for solids?Answer: Diffusivity for solids varies from about Question: What is the range of diffusivity for liquids?Answer: Diffusivity for liquids varies from about Question: What is the range of diffusivity for gases?Answer: Diffusivity for gases varies from about

Diffusion in GasesFrom, Ficks first law of diffusion

Consider the box of Fig. which is separated into two parts by the partition P. Into section I, 1 kg water (A) is placed and into section II, 1 kg of ethanol (B) ( the densities of the liquids are different, and the partition is so located that the depths of the liquids in each section are the same). Imagine the partition to be carefully removed, thus allowing diffusion of both liquids to occur. When diffusion stops, the concentration will be uniform throughout at 50 mass percent of each constituent, and the masses and moles of each constituent in the two regions will be as indicated in the figure. It is clear that while the water has diffused to the right and the ethanol to the left, there has been a net mass movement to the right, so that if the box had originally been balanced on a knife-edge, at the end of the process it would have tipped downward to the right. For a condition of steady state, the net flux is

P I II

The movement of A is made up of two parts, namely, that resulting from the bulk motion and the fraction of which is A and that resulting from diffusion :

Putting values from Eq. (1) and Eq. (2) in Eq. (3) we get

The counterpart of Eq. (4) for B is

Adding this gives

or If constant, it follows that at the prevailing concentration and temperature.From Eq. (4), we can readily separate the variables, and if is constant, it can be integrated

Letting , we get

or

Molecular Diffusion in GasesWhen the ideal-gas law can be applied Eq. (8) can be written in a form more convenient for use with gases. Thus,

where partial pressure of component A total pressure mole fraction concentrationFurther,

So that Eq. (9) becomes

or

Note: In order to use these equations the relation between and must be known. This is usually fixed by other considerations. For example, if methane is being cracked on a catalyst,CH4 C + 2H2Under circumstances such that CH4(A) diffuses to the cracking surface and H2 (B) diffuses back, the reaction stoichiometry fixes the relationship In the absence of chemical reaction, the ratio can be fixed by enthalpy considerations.

Steady-state diffusion of A through non-diffusing BThis might occur, for example, if ammonia (A) were being absorbed from air (B) into water.

And Eq. (12) becomes

Since then

If we let

then

Steady-state equimolal counterdiffusion This is a situation which frequently pertains in distillation operations. const. Eq. (12) becomes indeterminate, but we can go back to Eq. (4), which, for gases becomes

or for this case,

Steady-state diffusion in multicomponent mixturesThe expression for diffusion in multicomponent systems become very complicated, but they can frequently be handled by using an effective diffusivity in Eq. (12).

A common situation is when all the s except are zero. Eq. (22) then becomes

where is the mole fraction of component on an A-free basis.

Molecular Diffusion in LiquidsThe integration of Eq. (4) to put it in the form of Eq. (9) requires the assumption that and are constant. This is satisfactory for binary gas mixtures but not in the case of liquids, where both may vary considerably with concentration. Equation (9) is also conveniently written

Where and M are the solution density and molecular weight, respectively.1. Steady-state diffusion of A through nondiffusing B.

Where

2. Steady-state equimolal counterdiffusion

Equation of ContinuityOne should ask why is another form of the 'continuity equation' is needed. The previous form, the integral equation, work well with real defined volumes and known inflow and outflow.On the other hand, if information is needed throughout the flow field, then a differential form is more useful. However, boundary conditions and initial conditions are needed to solve the differential form, which can be difficult.

Consider the volume element of fluid of following Fig., where a fluid is flowing through the element. We shall need a material balance for a component of the fluid applicable to a differential fluid volume of this type.

z y x G z (x+x, y+y, z+z) E (x, y, z) x

yThe mass rate of flow of component A into the three faces with a common corner at E is

where signifies the -directed flux and its value at location Similarly the mass rate of flow out of the three faces with a common corner at G is

The total component A in the element is , and its rate of accumulation is therefore If, in addition, A is generated by chemical reaction at the rate mol/(volume)(time), its production rate is mass/time. Since in general

then

Dividing by and taking the limit as the three distances become zero gives

Similarly, for component B

The total material balance is obtained by adding those for A and B

where solution density, since the mass rate of generation of A and B must equal zero. Now we know that

In terms of masses and in the direction this equation becomes

where is the mass-average velocity such that

Therefore,

Eq.(6) therefore becomes

which is the equation of continuity, or a mass balance, for total substance. If the solution density is constant, it becomes

Returning to the balance for component A, we see from Eq. (8)

Eq. (4) then becomes

which is the equation of continuity for substance A. For a solution of constant density, we can apply Eq. (11) to the terms multiplying . Dividing by , we then have

In the special case where the velocity equals zero and there is no chemical reaction, it reduces to Ficks second law

Note: This is frequently applicable to diffusion in solids and to limited situations in fluids.

knowing and the interface values, can be calculatedOverall mass transfer coefficientsThe entire two phase mass transfer effect may be measured in terms of overall mass transfer coefficient. These are based on bulk compositions. Since the interface values are impossible to measure because of small distances, so equilibrium values are used for overall mass transfer coefficient to account for two phase mass transfer effect. based on gas stream

is the slope of chord CM

Based on liquid stream

Where is the slope of MD chord

The overall coefficients vary from place to place along the length of the tower. In practice average coefficients for the entire column is used.

Penetration theoryThis is developed by Higbie in 1935. He emphasized that in many situations the time of exposure of a fluid to mass transfer is short so the concentration gradient of film theory(which is steady state) would not develop.A bubble gas rises through liquid which absorbs the gas. A particle of liquid b is initially at the top of the gas bubble. The liquid particle is in contact with the gas for the time . is the time required for the bubble to rise a distance equal to its diameter while the liquid particles slip along the surface of gas bubble. If liquid is in turbulent motion, an eddy b rises from turbulent depths of the liquid and get exposed to the gas for a time . In this theory the time of exposure is taken as constant for all eddies of particles of liquid.Initially the concentration of dissolved gas in the eddy is uniformly . The eddy is considered a stagnant, internally. When the eddy is exposed to gas at the surface, the concentration in the liquid at the gas-liquid surface is . is taken as the equilibrium solubility of gas in liquid. During the time , the liquid particle is subjected to unsteady state diffusion of penetration of solute in z-direction.

For short exposure times and with slow diffusion in the liquid the molecules of dissolving solute are never able to reach the depth zb corresponding to the thickness of eddy. So from solute point of view is zb infinite.

Gas absorption (also known as scrubbing) is an operation in which a gas mixture is contacted with a liquid for the purpose of preferentially dissolving one or more components of the gas mixture and to provide a solution of them in the liquid.Therefore we can see that there is a mass transfer of the component of the gas from the gas phase to the liquid phase. The solute so transferred is said to be absorbed by the liquid.In gas desorption (or stripping), the mass transfer is in the opposite direction, i.e. from the liquid phase to the gas phase. The principles for both systems are the same.We will focus on the analysis for gas absorption, for the simple case whereby only one component of the gas solute is being absorbed. The other components of the gas are assumed to be non-soluble in the liquid (i.e. the other gas components are inert components), and the liquid is non-volatile, which means that there is no transfer of molecules from the liquid to the gas phase. In addition, we assume there is no chemical reaction in the system and that it is operating at isothermal condition.The process of gas absorption thus involves the diffusion of solute from the gas phase through a stagnant or non-diffusing liquid. [ Back on Top ]

The Table below showed representative commercial absorption applications.

From: Table 6.1 "Separation Process Principles", J.D. Seader and E.J. Henley, p.272Click here for more information on gas absorption operation.Gas absorption and desorption (stripping) can often integrated. Click here for an example.

Physical vs. Chemical AbsorptionThere are 2 types of absorption processes: physical absorption and chemical absorption, depending on whether there is any chemical reaction between the solute and the solvent (absorbent).When water and hydrocarbon oils are used as absorbents, no significant chemical reactions occur between the absorbent and the solute, and the process is commonly referred to as physical absorption. When aqueous sodium hydroxide (a strong base) is used as the absorbent to dissolve an acid gas, absorption is accompanied by a rapid and irreversible neutralization reaction in the liquid phase and the process is referred to as chemical absorption or reactive absorption. More complex examples of chemical absorption are processes for absorbing CO2 and H2S with aqueous solution of monoethanolamine (MEA), diethanolamine (DEA), diethyleneglycol (DEG) or triethyleneglycol (TEG), where a reversible chemical reaction takes place in the liquid phase. Chemical reactions can increase the rate of absorption, increase the absorption capacity of the solvent, increase selectivity to preferentially dissolve only certain components of the gas, and convert a hazardous chemical to a safe compound.HENRY'S LAW FOR GAS ABSORPTIONWhen the gas mixture in equilibrium with an ideal liquid solution follows the ideal gas behavior, we have - as seen previously - the Raoult's Law:p* = Pvp x[at this point, students should revise this concept previously covered in the topic on distillation]When the solution is non-ideal, Raoult's Law cannot be applied. For non-ideal solution, we must use Henry's Law which states that:p* = Hx or y* = p* / PT = m xwhere y* = equilibrium mole fraction in gas phasePT = total system pressurem = Henry's Law ConstantIn many cases, the superscript '*' is dropped for convenience:

Henry's Law is often used to represent equilibrium solubility curves.As we can see, Henry's Law predicts a linear equilibrium relationship. Still, most equilibrium relationships are actually non-linear. Henry's Law is only applicable over a modest liquid concentration range, especially when the solution is dilute. Counter-Current Gas Absorption

G2, y2 L2, x2

2

G, y L, x

1

G1, y1 L1, x1Gy and Lx are the molar flowrates of A in the gas and liquid respectively (kg-moles A/m2.s) at any point inside the column.Inside the column, mass transfer takes place as the solute (component A) is absorbed by the liquid. The quantities of L and x (for the liquid side) and G and y (for the gas side) varies continuously: as we gradually move up the column, component A is continuously being transferred from the gas phase to the liquid phase. Thus, in going up the column, there is a decrease in the total gas flowrate, and a decrease in the concentration of A in the gas phase. At the same time, in going down the column, there is an increase in the total liquid flowrate, and an increase in the concentration of A in the liquid phase. Thus,

For dilute systems, the solute content is small relative to the non-soluble inerts and non-volatile liquid. Thus, we can assume:G1 = G2 = G = constantL1 = L2 = L = constant

The relationship between these variables L, x, G and y is the operating line equation. The operating line equation is obtained by material balance around the column (as shown in Envelope 1 of the Figure above).At steady-state: IN = OUTThus, G y + L 1 x1 = L x + G1 y1Using the dilute system assumptions, we simply the equation and obtain:G y = L x + G y1 - L x1 Re-arranging:

Since L and G are assumed to be approximately constant, the operating line is a straight line of the form y = mx + c, with the gradient of L / G, the liquid-to-gas ratio.The operating line connects the 2 end points - point 1 (x1 , y1) that represents conditions at the bottom of the column, and point 2 (x2 , y2) that represents conditions at the top of the column.[ Back on Top ]For dilute solution, the equilibrium solubility line is also straight, as represented by Henry's Law, y = mx, where m is the Henry's Law constant which is also the gradient of the line.When these 2 lines are plotted on mole fraction coordinates, we have the following Figure

Operating Line

Mole fraction of solute A in Vapor, y y1(bottom)Equilibrium Line(Henrys Law)Any point y P(x, y) -kx/ky

y2(top)

x2 x x1 Mole fraction of solute A in Liquid, x

Any point P (x, y) on the operating line represents gas-liquid contact for which the analysis can be carried out using the 2-film theory covered in earlier section. The larger the distance between the operating line and equilibrium line, the larger the concentration difference for mass transfer, and thus, the easier the separation.Note: Operating line for gas absorption lies above the equilibrium line.Also, in the analysis of gas absorption, we will need to know the minimum liquid rate that can be used for a given separation, i.e. to remove a specified amount of solute from the gas. This is known as the minimum liquid-to-gas ratio. The analysis is applicable to both tray and packed column.

Co-Current Gas AbsorptionThis mode of operation is seldom used in practice. See the Figure below. G1, y1 L1, x1

G, y L, x

G2, y2 L2, x2

Mole Fractionof solute in vapor,yOperating Line(slope = -L/G ) y1 Equilibrium Liney = mx y2

(xe, ye)

Mole Fractionof solute in liquid, x x1 x2

The main points to note about this operation are as follow: The operating line has negative slope. The is no minimum liquid-to-gas ratio. To produce an exit liquid and gas streams at equilibrium (xe, ye) on the equilibrium curve, an infinitely tall column must be used. It is less efficient than counter-current operation.

Choice Of Solvent For Gas AbsorptionIf the principal purpose of the absorption operation is to produce a specific solution, as in the manufacture of hydrochloric acid, for example, the solvent is specified by the nature of the product, i.e. water is to be the solvent. If the principal purpose is to remove some components (e.g. impurities) from the gas, some choice is frequently possible. The factors to be considered are:Gas SolubilityThe gas solubility should be high, thus increasing the rate of absorption and decreasing the quantity of solvent required. Generally solvent with a chemical nature similar to the solute to be absorbed will provide good solubility. A chemical reaction of the solvent with the solute will frequently result in very high gas solubility, but if the solvent is to be recovered for re-use, the reaction must be reversible. For example, H2S can be removed from gas mixtures using amine solutions since the gas is readily absorbed at low temperatures and easily stripped at high temperatures. Caustic soda absorbs H2S excellently but will not release it in a stripping operation.VolatilityThe solvent should have a low vapour pressure to reduce loss of solvent in the gas leaving an absorption column.CorrosivenessThe materials of construction required for the equipment should not be unusual or expensive.CostThe solvent should be inexpensive, so that losses are not costly, and should be readily available.ViscosityLow viscosity is preferred for reasons of rapid absorption rates, improved flooding characteristics in packed column, low pressure drops on pumping, and good heat transfer characteristics.OthersThe solvent should be non-toxic, non-flammable and chemically stable.

Minimum Liquid/Gas Ratio For AbsorptionThe inlet gas has a solute mole fraction of y1. The solute mole fraction is reduced to y2 at the outlet. By material balance for the solute in the gas, the amount to be removed is G ( y1 - y2 ). The least amount of liquid Lmin that can remove this amount of solute is the minimum liquid rate, often expressed in terms of a liquid-to-gas ratio, Lmin/G. Understanding the effect of reducing liquid rate requires an analysis of the operating line equation. This is shown in the Figure below.

Mole Fractionof solute in vapor,y Operating Line at Min. Liq. Rate Equilibrium Line y1 (bottom) E F M

y2 D(top)

0

Mole Fractionof solute in liquid, x x2 x1 x1,max

The condition at the top of the column (point D) is known: x2 the mole fraction of entering liquid is known, and the mole fraction of gas leaving y2 is known. Hence point D is fixed.The mole fraction of gas entering y1 is known. The mole fraction of liquid leaving x1 obviously depends on the liquid rate used. For the same amount of solute to be removed, using a larger quantity of liquid will result in smaller value of x1, and vice versa. Hence, when the liquid rate is changed, the condition at the bottom of the column varies along the horizontal line through y1.Recall that the operating line has a gradient of L/G. By reducing the liquid rate, we are decreasing the slope of the operating line and increasing the exit concentration x1 . Therefore the operating line rotates around point D as L is decreased, e.g. from line DE to DF. Notice that the operating line has moved closer to the equilibrium curve. When this happens, the driving force for mass transfer is smaller, i.e. the absorption process becomes more difficult. [ Back on Top ]At point M, the operating line intersects the equilibrium line, and we have a condition of operation at zero driving force. At this point, we cannot reduce the liquid rate anymore. Hence, the liquid rate at this point of equilibrium is known as the minimum liquid rate, Lmin . At minimum liquid, the outlet liquid concentration is a maximum, x1(max) .

The minimum liquid rate results in infinite column height: infinite number of trays or packed height required for separation (at zero driving force).The minimum liquid rate, Lmin can be calculated from the gradient of the operating line:

Note: x1(max) can also be calculated using Henry's Law.

The actual liquid rate to be used is specified as multiples of the minimum liquid rate. If the liquid rate for absorption is initially unknown, then one must calculates the minimum liquid rate first.

PACKED COLUMN FOR GAS ABSORPTION (DILUTE SYSTEMS)Gas absorption can be carried out in a packed column. We will look at both column diameter and packed height in this Section.The Figure below showed a typical gas-liquid flow in a packed column. Click here for more information of packings.

Determination of column diameter involves the analysis of pressure drop across the packed bed.As for packed height, the design used in the early days was based on the HETP method. This was largely replaced by the Method of Transfer Units.Comparison between number of theoretical trays, HETP and Method of Transfer UnitsThe Number of Transfer Units (NTU) and Height of Transfer Units (HTU) such as NOG, HOG should not be confused with the number of theoretical trays (N), and the height equivalent to theoretical plate (HETP) respectively.When the operating line and equilibrium line are straight and parallel:NTU = N ; and HTU = HETPOtherwise, the NTU can be greater than or less than N as shown in the Figure below:

operating lineoperating lineoperating line yEqu. Line Equ. Line Equ. Line NTU = N NTU > N NTU < N

When the operating line is straight but not parallel, we have the following relationships:

where

is the Absorption Factor.

Column Diameter and Pressure DropIn determining the column diameter, we need to know what is the limiting (maximum) gas velocity that can be used. This is because the higher the gas velocity, the greater the resistance that will be encountered by the down-flowing liquid and the higher the pressure drop across the packings. Too high a gas velocity will lead to a condition known as flooding whereby the liquid filled the entire column and the operation became difficult to carry out. High pressure will crush and damage the packings in the column. We will begin our analysis by examining the relationship between the gas pressure drop and gas velocity. Refer to the Figure below that shows a typical gas pressure drop in a packed column.

L=20000L=15000L=10000L=5000L=0(Dry Packing)Log(P/Z), gas pressure drop per unit heightBe aAd

Log (G) , Superficial Gas Mass velocity c

b a

The horizontal axis is the logarithmic value of the gas velocity G, and the vertical axis is the logarithmic value of pressure drop per height of packing [ pressure drop in a packed bed is the result of fluid friction that is created by the flow of gas and liquid around the individual solid packing materials ]. Note: Each packing has its own characteristics pressure drop chart as reported by the manufacturer - for example, see the Figure above (right).[ Back on Top ]Analysis of Gas Pressure Drop in PackingWith a dry packing (i.e. no liquid flow, L = 0), pressure drop increases as gas velocity increases according to the linear relationship as shown by line a-a. This is a straight line on a log-log plot.With liquid flowing in the column, the packings now become wetted (irrigated). Part of void volume in the packings now filled with liquid, thereby reducing the cross-sectional area available for gas flow.At the same gas velocity, the pressure drop is higher for wetted packings compared to dry packings. For example, compare the case for L = 0 vs. L = 5. The line for DP/L under wetted condition lies to the left of line a-a.For a constant liquid flow (say L = 5), at low to moderate gas velocity G; the pressure drop characteristics is similar to that of dry packings, i.e. section b-c of the plot is still straight on log-log plot. Up to this point, there is an orderly trickling of the liquid down the packings. There is no observable liquid being trapped among the packings (no liquid hold-up).As the gas velocity is increased further, the pressure drop increased. Some liquid started to be retained in the packings. When point c is reached, the quantity of liquid retained in the packed bed increases significantly. There is a change in slope of the line at point c as pressure drop increases more rapidly with G. Point c is known as the loading point, as liquid starts to accumulate (load) in the packings.From point c to d to e, there is a sharp increase in pressure drop at higher G: there is a greater amount of liquid hold-up, a gradual filling of the packing voids with liquid (starting at the bottom of the column), and the column is slowly "drowned" in the liquid.At point e, there is another sharp change in the slope. At this point the entire column is filled liquid and the gas now has to bubble through the liquid in the packing voids. The gas pressure drop is now very high. Point e is known as the flooding point. The gas velocity at this point is known as the flooding velocity (limiting velocity). Points to note : - at constant liquid rate, gas pressure drop increases with gas velocity. - at constant gas velocity, the gas pressure drop is higher at larger liquid rate.- each liquid rate has its own loading and flooding points. - at higher liquid rate, the loading and flooding points occur at lower gas pressure drop. Operation of a gas absorption column is not practical above the loading point. For optimum design, the recommended gas velocity is 1/2 of the flooding velocity. Alternatively, some design can be based on a specified pressure drop condition, usually well below the pressure drop at which flooding would occur.

HETP (Height Equivalent to a Theoretical Plate)As we have noted, instead of a tray (plate) column, a packed column can be used for various unit operations such as continuous or batch distillation, or gas absorption. With a tray column, the vapours leaving an ideal plate will be richer in the more volatile component than the vapour entering the plate by one equilibrium "step". When packings are used instead of trays, the same enrichment of the vapour will occur over a certain height of packings, and this height is termed the height equivalent to a theoretical plate (HETP). As all sections of the packings are physically the same, it is assumed that one equilibrium (theoretical) plate is represented by a given height of packings. Thus the required height of packings for any desired separation is given by (HETP x No. of ideal trays required). HETP values are complex functions of temperature, pressure, composition, density, viscosity, diffusivity, pressure drop, vapour and/or liquid flowrates, packing characteristics, etc. Empirical correlations, though available to calculate the values of HETP, are restricted to limited applications. The main difficulty lies in the failure to account for the fundamentally different action of tray and packed columns.["Chemical Engineering Vol.2", 4th Ed., Coulson & Richardson, p.508-509]

In industrial practice, the HETP concept is used to convert empirically the number of theoretical trays to packing height. Most data have been derived from small-scale operations and they do not provide a good guide to the values which will be obtained on full-scale plant.[ For more information on HETP prediction, see pp.1-355 of "Handbook of Separation Techniques for Chemical Engineers", 3rd Ed., P.A. Schweitzer , or pp.335, "Separation Process Principles", J.D. Seader & E.J. Henley ]This method had been largely replaced by the Method of Transfer Units.

Packing Height : The Method of Transfer UnitsA newer concept in the analysis of packed column centred on the method of transfer units. This method is more appropriate because the changes in compositions of the liquid and vapour phases occur differentially in a packed column rather than in stepwise fashion as in trayed column. In this method, height of packings required can be evaluated either based on the gas-phase or the liquid-phase. The packed height (z) is calculated using the following formula:z = NTU x HTUwhere NTU = number of transfer units (NTU) - dimensionlessHTU = height of transfer units (HTU) - dimension of length

The number of transfer units (NTU) required is a measure of the difficulty of the separation. A single transfer unit gives the change of composition of one of the phases equal to the average driving force producing the change. The NTU is similar to the number of theoretical trays required for trayed column. Hence, a larger number of transfer units will be required for a very high purity product. The height of a transfer unit (HTU) is a measure of the separation effectiveness of the particular packings for a particular separation process. As such, it incorporates the mass transfer coefficient that we have seen earlier. The more efficient the mass transfer (i.e. larger mass transfer coefficient), the smaller the value of HTU. The values of HTU can be estimated from empirical correlations or pilot plant tests, but the applications are rather restricted.["Principles of Unit Operations" 2nd Ed., Foust et al, p.391][ Back on Top ]

The calculation of packing height follows the same nomenclature as before and this is shown in the Figure below.

G2,y2 L2,x2 21 = Bottom of column2 = Top of columnG,y L,x

1 G1, y1L1, x1

In this Section, we will focus on the applications of the equations rather than any derivation of them. Determination of the packed height can be based on either the gas-phase or the liquid-phase.For the gas-phase, we have: z = NOG x HOG

and KY is the overall gas-phase mass transfer coefficient. "a" is the packing parameter that we had seen earlier (recall the topic on column pressure drop, e.g. Table 6.3) that characterize the wetting characteristics of the packing material (area/volume).

Normally, packing manufacturers report their data with both KY and "a" combined as a single parameter. Since KY has a unit of mole/(area.time.driving force), and "a" has a unit of (area/volume), the combined parameter KY a will have the unit of mole/(volume.time.driving force), such as kg-mole/(m3.s.mole fraction). As seen earlier, other than mole fraction, driving force can be expressed in partial pressure (kPa, psi, mm-Hg), wt%, etc.[ Back on Top ]y1* is the mole fraction of solute in vapour that is in equilibrium with the liquid of mole fraction x1 and y2* is mole fraction of solute in vapour that is in equilibrium with the liquid of mole fraction x2 .The values of y1* and y2* can be obtained from the equilibrium line as previously covered (see section on Two-Film Theory). See the Figure below.

Operating Line y1Equilibrium Line

y1* y2 y2* x2 x2* x1 x1*(y1 - y1*) is the concentration difference driving force for mass transfer in the gas phase at point 1 (bottom of column) and (y2 - y2*) is the concentration difference driving force for mass transfer in the gas phase at point 2 (top of column). [ Point P (x, y) as shown is any point in the column. The concentration difference driving force for mass transfer in the gas phase at point P is (y - y*) as shown previously, this time no subscripts are shown. ]NOTE: Both equilibrium line and operating line are straight lines under dilute conditions.

Alternatively, equilibrium values y1* and y2* can also be calculated using Henry's Law ( y = m x, where m is the gradient) which is used to represents the equilibrium relationship at dilute conditions.Thus, we have: y1* = m x1 ; y2* = m x2[ Back on Top ]

Similarly for the liquid-phase we have: z = NOL x HOL

and KX is the overall liquid-phase mass transfer coefficient, and "a" is the packing parameter seen earlier. Again, normally both KX and "a" combined as a single parameter.Likewise, x1* is the mole fraction of solute in liquid that is in equilibrium with the vapour of mole fraction y1 and x2* is mole fraction of solute in liquid that is in equilibrium with the vapour of mole fraction y2 . Refer to Figure 134 for finding values of x1* and x2* from the equilibrium line. Alternatively, x1* = y1 /m and x2* = y2 /m.

(x1* - x1) is the concentration difference driving force for mass transfer in the liquid phase at point 1 (bottom of column) and (x2* - x2) is the concentration difference driving force for mass transfer in the liquid phase at point 2 (top of column). [ Back on Top ]Using either gas-phase or liquid-phase formula should yield the same required packing height :

whereN= number of transfer units(NTU)H=height of transfer units(HTU)

[ For more info on packed column design, see Chp. 4, "Process Plant Design", J.R. Backhurst & J.H. Harker, or for applications of various packed columns, refer to "Random Packings and Racked Towers", R.F. Strigle Jr. ]

Example 12.4Gas from a petroleum distillation column has its concentration of H2S reduced from 0.03 kmol H2S/kmol of inert hydrocarbon gas to 1% of this value by scrubbing with a triethanolamine-water solvent in a countercurrent tower, operating at 300K and at atmospheric pressure.H2S is soluble in such a solution and the equilibrium relation may be taken as Y = 2X, where Y is kmol of H2S/kmol of inert gas and X is kmol of H2S/kmol of solvent.The solvent enters the tower free of H2S and leaves containing 0.013 kmol of H2S/kmol of solvent. If the flow of inert hydrocarbon gas is 0.015 kmol/m2s of tower cross-section and the gas-phase resistance controls the process, calculate:(a) The height of the absorber necessary; and(b) The number of transfer units required.The overall coefficient for absorption may be taken as 0.04 kmol/s.m3 of tower volume.SolutionDriving force at the top of the column = (Y2-Y2e) = 0.0003.Driving force at the bottom of the column = (Y1-Y1e) = (0.03-0.026) = 0.004.Logarithmic mean driving force = (0.004-0.0003)/ln(0.004/0.0003)=0.00143.

7.8 mHeight of Transfer unit 0.375Number of Transfer unit (7.79/0.375) = 20.8 = 21

Example 5 Packed Column - Number of Transfer UnitsA gas mixture containing 0.015 mole fraction of solute S at the inlet (and the rest inerts) is subjected to counter-current absorption with water in a packed tower. The outlet concentration of the solute is to be 1% of the inlet value. The total gas inlet flow rate is 1.0 kg/m2.s (MW = 29) and the pure water entering is 1.6 kg/m2.s. The system can be considered as dilute. The equilibrium condition can be described by Henry's Law and is given as y = 1.75 x, where y and x are the mole fraction of solute S in the vapour and liquid respectively. The column uses a certain type of packings which provides an overall gas-phase mass transfer coefficient (KYa) of 0.06 kg-mole/(m3.s.mole fraction).Determine:(a) the height of packings required for the separation(b) the minimum liquid rate required for the separation.What would your answer for part (a) be if another type of packing is used that only provides a KYa value of 0.04 kg-mole/(m3.s.mole fraction)?What would your answers be if the solute concentration specification in the outlet gas is "relaxed" to 5% of the inlet value? Briefly discuss your results.Given:

Solution to Example 5Given the following informationInlet Gas:Total gas flow = 1.0 kg/m2.s (i.e. inerts + solute)

Solute concentration = 0.015 mole fraction

Outlet Gas:Solute concentration = 1% of inlet value

= 0.01 x 0.015 = 0.00015 mole fraction

Inlet Liquid:pure water rate = 1.6 kg/m2.s (i.e. assume no solute)

Thus, we have: y1 = 0.015 ; y2 = 0.00015 ; x2 = 0.00G1 = 1.0 / 29 = 0.03448 kg-mole/m2.sL2 = 1.6 / 18 = 0.08889 kg-mole/m2.sSince the solution is dilute, we can use the following assumptions:L1 = L2 = L = 0.03448 kg-mole/m2.s = constantG1 = G2 = G = 0.08889 kg-mole/m2.s = constantWe can obtain x1 from material balance around the column:L2 x2 + G1 y1 = G2 y2 + L1 x1Using the dilute solution simplification and with x2 = 0, the equation simplifies to:G y1 = G y2 + L x1

Thus; [ Back on Top ]This problem can be solved mathematically without plotting graphs by using Henry's Law: y = 1.75 xThus: y1* = 1.75 x 0.00576 = 0.01008y2* = 1.75 x 0.0 = 0.0Calculate HOG and NOG :

Using G1 to calculate the height of transfer unit: (this is a more conservative design, as using G1 gave a higher HOG value than G2 )

[ Back on Top ]If KYa = 0.04 kg-mole/(m3.s.mole fraction), the new HOG and z are:

Minimum liquid rate, Lmin can be determined by identifying the condition of zero driving force - when the operating line crosses the equilibrium line, as shown below:

[ Back on Top ]Use Henry's Law to obtain x1* :With y1 = 0.015, x1* = 0.015 / 1.75 = 0.008571Calculate Lmin from the gradient of the operating line:

Solving: Note: Actual water rate used is 1.60 kg/m2.s, which is approximately 1.5 times the minimum. Lmin is not affected by difference in KYa.[ Back on Top ]We can also check if the above assumptions on dilute solution is valid, as shown in the steps below:With y1 = 0.015, Amount of inerts entering = (1 - 0.015) (0.03448)= 0.03396 kg-mole/m2.sThe amount of inerts leaving the column would remain the same. The total gas leaving G2 would contain 0.03396 kg-mole/m2.s inerts + the unabsorbed solute. And we know that the mole fraction of solute in the outlet gas ( y2 ) is 0.00015.If n = kg-mole/m2.s of solute in outlet gas, then using the definition of mole fraction, we have:

Thus, G2 = 0.03396 + 5.095 x 10-6 kg-mole/m2.s = 0.033965 kg-mole/m2.sPercentage difference in G = ( G1 - G2 ) / G1 x 100% = 1.5%Changes in solute flow:Amount of solute entering =0.015 x 0.03448 kg-mole/m2.s = 0.000517 kg-mole/m2.s

Amount of solute leaving =5.095 x 10-6 kg-mole/m2.s

Amount of solute absorbed into water =0.000512 kg-mole/m2.s

Total amount of liquid leaving, L1 = L2 + solute absorbedL1 = 0.08889 + 0.000512 kg-mole/m2.sL1 = 0.089402 kg-mole/m2.sPercentage difference in L = ( L1 - L2 ) / L1 x 100% = 0.58%Note: For more conservative design, use L1 to calculate HOL .[ Back on Top ]Scenario Analysis:If the outlet gas mole fraction is 5% of the inlet value, theny2 = 0.05 x 0.015 = 0.00075This means that less solute was removed from the gas.The other parameters remain the same:y1 = 0.015 ; x2 = 0.00G1 = 0.03448 kg-mole/m2.sL2 = 0.08889 kg-mole/m2.sMole fraction solute in outlet liquid now becomes:

From Henry's Law, y1* = 1.75 x 0.00553 = 0.00967y2* = 1.75 x 0.0 = 0.0

The corresponding minimum liquid rate:With y1 = 0.015, x1* = 0.015 / 1.75 = 0.008571

[ Back on Top ]

Comparison of Results - Significance of NOG and HOG

CasesLminNOGHOGz

y2=0.00015; Kya=0.061.07510.840.5826.31

y2=0.00015; Kya=0.041.07510.840.8739.46

y2=0.00075; Kya=0.061.0326.100.5823.55

Number of transfer unit is a measure of difficulty of separation. A higher value of y2 (solute concentration in outlet gas) implied a less stringent separation requirement, thus a smaller NOG is needed. A lower minimum liquid rate is required for removing lesser amount of solute. Height of transfer unit is a measure of separation effectiveness of a particular packing. A more efficient packing (larger KYa value) will result in a smaller HOG.

HUMIDIFICATION OPERATIONS

Humidification and dehumidification involve the transfer of material between a pure liquid face and a fixed gas which is insoluble in liquid. These operations are somewhat simpler than those for absorption and stripping, for when the liquid contains only one component, there are no concentration gradients and no resistance to transfer in the liquid face. On the other hand, both heat and mass transfer are important and influence one another.Humidity H is the mass of vapor carried by a unit mass of vapor-free gas. If partial pressure of vapor is pA atm, the molal ratio of vapor to gas at 1 atm is . The humidity is therefore H Note : Vapor means the gaseous form of the component which is also present as a liquid and gas is the component which is present only in gaseous form.The humidity is related to the mole fraction in the gas phase by the equationy = (H /MA) / (1/ MB + H /MA)Since H /MA is usually small compared with 1/ MB , y may often be considered to be directly proportional to H .Saturated gas is gas in which the vapor is in equilibrium with the liquid at the gas temperature.

H s H s is saturation humidity and is the vapor pressure of the liquid .Relative humidity H R is defined as the ratio of the partial pressure of the vapor to the vapor pressure of the liquid at the gas temperature. It is usually expressed on a percentage basis. H R Percentage humidity H A is the ratio of the actual humidity H to the saturation humidity H s at the gas temperature.H A = 100(H / H s ) H RAt all humidity other than 0 or 100 percent, the percentage humidity is less than the relative humidity.Humid heat cs is the heat energy necessary to increase the temperature of 1 lb or 1 g of gas , plus whatever vapor it may contain, by 1F or 1C. Thuscs = cpB + cpA H where cpB and cpA are the specific heats of gas and vapor, respectively.Humid volume vH is the total volume of a unit mass of vapor-free gas plus whatever vapor it may contain, at 1 atm pressure and gas temperature.In fps unitsvH = 359 T / 492 [ (1/ MB) + (H / MA )]In SI unitsvH = 0.00224 T / 273 [ (1/ MB) + (H / MA )]Dew point is the temperature to which a vapor-gas mixture must be cooled ( at constant humidity) to become saturated.Total Enthalpy Hy is the enthalpy of a unit mass of gas plus whatever vapor it may contain. The total enthalpy is the sum of three items; the sensible heat of the vapor, the latent heat of the liquid at To, and sensible heat of vapor free gas. ThenHy = cpB ( T- To ) + H o + cpA H ( T - To )Hy = cs( T- To ) + H oPhase Equilibria Since the liquid is pure, xe is always unity. Equilibrium data are often presented as plots of ye vs. temperature at a given total pressure.

MOLE FRACTION H2O0.200.180.160.140.120.100.080.060.040.02 0

50 70 90 110 130 TEMPERATURE, F

Adiabatic-saturation temperature

AGas Inlet C Gas InletHumidity H Humidity H sTemperature T Temperature Ts B Make-up Liuid Temp Ts Temperature, Ts

Fig. Adiabatic saturator: A, spray chamber; B, circulating pump; C,sprays.

Consider the process in Fig. Gas, with an initial humidity H and temperature T, flows continuously through the spray chamber A. The chamber is lagged, so the process is adiabatic. Liquid is circulated by pump B from the reservoir in the bottom of the spray chamber through sprays C and back to the reservoir. The gas passing through the chamber is cooled and humidified. The temperature of the liquid reaches a definite steady-state temperature Ts called the adiabatic-saturation temperature. Unless the entering gas is saturated, the adiabatic-saturation temperature is lower than the temperature of the entering gas. If contact between liquid and gas is sufficient to bring the liquid and the exit gas into equilibrium, the gas leaving the chamber is saturated at temperature Ts. Since the liquid evaporated into the gas is lost from the chamber, make-up liquid is needed. To simplify the analysis, this liquid is assumed to be supplied to the reservoir at temperature Ts.An enthalpy balance can be written over this process. Pump work is neglected, and the enthalpy balance is based on temperature Ts as a datum.. Then the enthalpy of the make-up liquid is zero, and the total enthalpy of the entering gas equals that of the leaving gas.cs( T- Ts ) + H s = H s s(H - H s ) / ( T- Ts ) = - cs / s = - (cpB + cpA H ) / s

Wet- bulb temperature

Tw Make-up liquid, temperature Tw

Gas GasTemperature T Temperature THumidity H Humidity H

The wet-bulb temperature is the steady-state, nonequilibrium temperature reached by a small mass of liquid immersed under adiabatic conditions in a continuous stream of gas. The mass of the liquid is so small in comparison with the gas phase that there is only a negligible change in the properties of the gas, and the effect of the process is confined to the liquid. A thermometer is covered by a wick, which is saturated with pure liquid and immersed in a stream of gas having a definite temperature T and humidity H . Assume that initially the temperature of the liquid is about that of the gas. Since the gas is not saturated, liquid evaporates, and because the process is adiabatic, the latent heat is supplied first by cooling the liquid. As the temperature of the liquid decreases below that of the gas, sensible heat is transferred to the liquid. Ultimate a steady state is reached at such a liquid temperature that the heat needed to evaporate the liquid and heat the vapor to gas temperature is exactly balanced by the sensible heat flowing from the gas to the liquid. It is this steady-state temperature, denoted by Tw, that is called the wet-bulb temperature. It is a function of both T and H .Theory of wet-bulb temperatureAt the wet-bulb temperature the rate of heat transfer from the gas to the liquid may be equated to the product of the rate of vaporization and the sum of the latent heat of evaporation and the sensible heat of the vapor. Since radiation may be neglected, this balance may be written

where = rate of sensible heat transfer to liquid = molal rate of vaporization = latent heat of liquid at wet-bulb temperature The rate of heat transfer may be expressed in terms of the area, the temperature drop, and the heat-transfer coefficient in the usual way, or

where heat-transfer coefficient between gas and surface of liquid = temperature at interfaceA = surface area of liquidThe rate of mass transfer

where molal rate of transfer of vapor mole fraction of vapor at inter face mole fraction of vapor in air stream one way diffusion factor

The above equation may be simplified without serious error in the usual range of temperatures and humidities as follows:(1). The factor is nearly unity and can be omitted.(2). The sensible heat-items is small in comparison with w and can be neglected.(3). The terms H w /MA and H /MA are small in comparison with 1 /MB and may be dropped from the denominators of the humidity terms. With this simplifications above Eq. becomeshy(T-Tw)=MB ky w(H w - H )or(H w - H ) / (T-Tw) = - hy / MB ky wFor a given wet-bulb temperature, both w and H w are fixed. The relation between H and T then depends on the ratio hy / ky. The close analogy between mass transfer and heat transfer provides considerable information on the magnitude of this ratio and the factors that affect it.It has been shown that heat transfer by conduction and convection between a stream of fluid and a solid or liquid boundary depends on the Reynolds number DG/ and the Prandtl number cp /k. Also mass transfer coefficient depends on the Reynolds number and the Schmidt number /D.The rates of heat and mass transfer, when these processes are under the control of the same boundary layer, are given by equations which are identical in form. For turbulent flow of the gas stream these equations are

and

where b, n, m = constants = average molecular weight of gas streamassuming , gives

(H w - H ) / (T-Tw) = - hy / MB ky w and-hy / MB ky

Psychrometric line and Lewis relationFor a given wet-bulb temperature, Eq. (19) can be plotted on the humidity chart as a straight as a straight line having a slope of - hy / MB ky w and intersecting the 100% line at Tw. This line is called psychrometric line. When both a psychrometric line, from Eq. (19), and an adiabatic-saturation line, from Eq. (11), are plotted for the same point on the 100 percent curve, the relation between the lines depends on the relative magnitudes of cs and hy / MB ky.For the system air-water at ordinary conditions the humid heat cs is almost equal to the specific heat , and the following equation is very nearly correct:

Eq. (21) is known as the Lewis relation. When this relation holds, the psychrometric line and the adiabatic-saturation line become essentially the same.

2.01.81.61.41.21.00.80.60.40.2 0 Absolute humidity lb benzene vapor per lb of dry airHumid heat vs. abs humidityAdiabatic Saturat. LinePsychro. LineSaturated vol.Sp. Vol. of dry airLat. heat of vap. vs. temp.Saturated humidity line.

0 40 60 80 100 120 140 160 180 200 220 240

Dry Bulb TemperatureF

Measurement of Humidity(1) Dew-point methods If a cooled, polished disk is inserted into gas of unknown humidity and the temperature of the disk gradually lowered, the disk reaches a temperature at which mist condenses on the polished surface. The temperature at which this mist just forms is the temperature of equilibrium between the vapor in the gas and the liquid face. It is therefore the dew point.(2) Psychrometric methods- A vary common method of measuring the humidity is to determine simultaneously the wet-bulb and dry-bulb temperatures. From these readings the humidity is found by locating the psychrometric line intersecting the saturation line at the observed wet-bulb temperature and following the psychrometric line to its intersection with the ordinate of the observed dry-bulb temperature.(3) Direct methods- The vapor content of a gas can be determined by direct analysis, in which a known volume of gas is drawn through an appropriate analytical device.Equipment for Humidification OperationsWhen warm liquid is brought into contact with unsaturated gas, part of the liquid is vaporized and the liquid temperature drops. This cooling of the liquid is the purpose behind many gas-liquid contact operations, especially air-water contacts. Water is cooled in large quantities in spray ponds or more commonly in tall towers through which air passes by natural draft or by the action of a fan.The purpose of a cooling tower is to conserve cooling water by allowing the cooled water to be reused many times. Warm water, usually from a condenser or other heat-transfer unit, is admitted to the top of the tower and distributed by troughs and overflows to cascade down over slat gratings, which provide large areas of contact between air and water.Flow of air up through the tower is induced by the wind and by the buoyancy of the warm air in the tower. A cooling tower is, in principle, a special type of packed tower. The usual packing is, in principle, a special type of packed tower.Mechanism of interaction of gas and liquid

Liquid Gas

Constant temperature, TxTemperature, TyH i

TiHumidity, H Latent heatWater vaporSensible heat

Fig.Conditions in adiabatic humidifier

Liquid Gas

Humidity, H H i Temperature, Ty

TiTemp, TxVaporLatent heat(a)Sensible Sensible heat(b)heat(a+b) Fig. Conditions in dehumidifier

Liquid Gas

H i Humidity, H Temp, TxVaporTi Temperature, TyLatent heat(a)Sensible Sensible heat(b)heat(a+b) Fig. Conditions in top of cooling tower

Liquid Gas

H i Humidity, H Temp, TxVapor Sensible heat(a)SensibleLatent heat(b)heat(b-a) Temperature, TyTemp, Tx

Fig. Conditions in bottom of cooling tower

DRYING

Equilibrium moisture Curve

1.0 Relative humidity of gas Unbound Moisture

Bound Moisture

A

Free MoistureEqui. Moisture

0 0 X* X

Moisture content, kg moisture/kg dry solid

Moisture content, wet basis The moisture content of a solid or solution is usually described in terms of weight percent moisture, and unless otherwise qualified this is ordinarily understood to be expressed on the wet basis, i.e., as (kg moisture/ kg wet solid)100

Moisture content, dry basis This is expressed as kg moisture/ kg dry solid = XEquilibrium moisture X * - This is the moisture content of a substance when at equilibrium with a given partial pressure of the vapor.Bound Moisture This refers to the moisture contained by a substance which exert an equilibrium vapor pressure less than that of the pure liquid at the same temperature.Unbound Moisture This refers to the moisture contained by a substance which exert an equilibrium vapor pressure equal to that of the pure liquid at the same temperature.Free Moisture Free moisture is that moisture contained by a substance in excess of the equilibrium moisture: X - X* . Only free moisture can be evaporated.

Drying Characteristics of Wet SolidsThe Drying Characteristics of Wet Solids are usually described by the drying rate curve

Kg moisture/kg dry solidAA B

C D x* E

= time, hr Fig. (A) Batch drying, constant drying conditions0.40.30.20.1 0 N= rate of drying , 103 kg evaporation/m2.sFalling rateConstant rate

A

C B

Unsaturated surface drying A

D

Internal movement of moisture controls

E

X= kg moisture/kg dry solid Fig. (B) : Rate of Drying Curve 0 0.1 0.2 0.3 0.4

Time of drying- If one wishes to determine the time of drying a solid under the same conditions for which a drying curve such as Fig. (A) has been completely determined, one need merely read the difference in the times corresponding to the initial and final moisture contents from the curve.Within limits, it is sometimes possible to estimate the appearance of a rate of drying curve such as Fig. (B) for conditions different from those used in the experiments. In order to determine the time for drying for such a curve, we proceed as follows. The rate of drying is, by definition,

Rearranging and integrating over the time interval while the moisture content changes from its initial value to its final value gives

1. The constant- rate period If the drying takes place entirely within the constant-rate period, so that and > and

2. The falling-rate period If and are both less than , so that drying occurs under conditions of changing , we proceed as follows:a. General case. For any shape of falling-rate curve whatsoever, Eq.(2) can be integrated graphically by determining the area under a curve of as ordinate, as abscissa, the data for which can be obtained from the rate-of-drying curve.b. Special case. is linear in , as in the region of CD of Fig.(B). In this case,

where is the slope of the linear portion of the curve and is a constant.Substitution in Eq.(2) provides

But since, and and , Eq.(5) becomes

logarithmic average of the rate and .

Nusselt number is a dimensionless number. The characteristic length (L) is determined by the direction of the growth (thickness) of the boundary layer.

Then, what kind of physical meaning does this Nusselt number imply? If Nu=1, this means that the convection and conduction terms have relatively similar magnitude and thus is characterized by the laminar flow. On the other hand, largeNuimplies that the convective term is dominant, which typically characterized by turbulent flows (usuallyNuvalue in the range of 100-1000).

Therefore, by understanding the Nusselt number of a flow system, we can infer the dominance between convection and conduction heat transfer terms, and thus enabling us to design better and more efficient thermal engineering systems, especially in the convective heat transfer field.