*INTRODUCTION TO GAS ABSORPTIONThe common in all mass transfer operations in our curriculum areAll mass transfer operations.Two phases are involved.Equilibrium process (attempt is made to establish an equilibrium between two phases).Definition: Absorption is a separation process where a mixture of gases is separated into its component by absorbing one component into its particular solvent.A (solute)Liquid In (C)Liquid Out (C + A)Gas In (Feed) (A + B)Gas Out (B + lower Concentration of A)Component A in Gas feed is soluble in liquid solvent CComponent B is insoluble in CConcentrations of Solute A are high at bottom of column in both liquid and Gas streams which is opposite to the distillation

*DIFFERENCES BETWEENDISTILLATIONDepends on difference in volatilities.

Starts with single phase and second phase is generated by application of energy.Heat energy is required (Simultaneous heat and mass transfer process).There is a diffusion of molecules in both directions, so that for an ideal system equimolecular counter diffusion exists.GAS ABSORPTIONDepends on difference in solubility of gases in selective solvent.We always have two phases (liquid is well below its boiling point).No heat energy is required (pure mass transfer process).Gas molecules are diffusing into the liquid, and the movement in the reverse direction is negligible.Note: The ratio of liquid to gas flow rate is considerably greater in absorption than in distillation with the result that layout of trays are different for the two operations. Furthermore with higher liquid rates in gas absorption packed type columns are commonly used.

*MECHANISM OF GAS ABSORPTIONTwo Film Theory: Material is transferred in the bulk of fluid by convection and concentration differences are regarded as negligible except in the vicinity of the interface between two phases. On either side of interface the currents will die out and a thin film exists. Through the thinfilm the transfer is effected solely by molecular diffusion.The direction of transfer of material across the interface is not dependent solely on the concentration difference but also on the equilibrium relationship. The controlling factor will be the rate of diffusion through the two films where all the resistanceis considered to be present.Penetration Theory: In this theory it is assumed that the eddies in the fluid bring an element of fluid to the interface where it is exposed to the second phase for a definite interval of time after which the surface element is mixed with the bulk again. It is assumed that equilibrium is immediately attained by the surface layers, that a process of unsteady state molecular diffusion than occurs and that the element is remixed after a fixed interval of time.PAGPAiCAiCALABDEBulk of GasBulk of LiquidGas film boundaryLiquid film boundaryLiquid filmGas filmInterfacePartial Pressure PA of Solute Gas AMolar Concentration CA of A in Liquid

*OPERATING LINELiquid is solvent and x1 and x2 are the mol fraction of soluble component in liquid.Feed is gas mixture in which y1 and y2 are the mol fractions of soluble component in liquid.Let Gm1 and Gm2 are the molar flow rates of gas at points 1 and 2 and Lm1 and Lm2 are the molar flow rates of liquid at points 1 and 2.Now by soluble component balance:Gm2y2 - Gm1y1 = Lm2x2 - Lm1x1If we calculate the molar ratios:X1,2 = Moles of Solute/ Moles of SolventY1,2 = Moles of Solute/ Moles of inert gasGm = Molar flow rate of Inert gasLm = Molar flow rate of SolventThen by solute balance:Gm(Y2 Y1) = Lm(X2 X1)Slope of line = (Y2 Y1)/(X2 X1) = Lm/GmFor most of the cases, solvent is pure e.g., X2 = 0 (Y1 Y2) = (Lm/Gm) X1 Mol fr. y = Moles of Solute (A)/{Moles of Solute (A) + Inert (B)}Dividing Numerator & Denominator by moles of inert (B)y = {moles of A/moles of B}/{(moles of A/moles of B) + 1}y = Y/(Y+1)For dilute solutions; Y

*GAS ABSORPTIONDiffusion through a stagnant gas: According to the two film theory, solute first has to diffuse through a stagnant gas film and then through liquid film before entering bulkof liquid. Let A is soluble in liquid absorbent C and B is insoluble.Then according to the Stefans Law: N/A = -Dv(CT/CB)(dCA/dz) whereN/A =Overall molar flux, Dv =Gas phase diffusivity, Z=Distance in the direction of MTCA, CB & CT are the molar concentration of A, B & total gas.Integrating over the whole thickness zG of the film. N/A = (DvCT/zG)ln(CB2/CB1)=(DvP/RTzG)ln(PB2/PB1) since CT = P/RT (for ideal gas)R is gas constant, T is absolute temperature & P is total pressure.If PBM = logmean difference of partial pressure.PBM = (PB2-PB1)/{ln(PB2/PB1)}N/A = (DvP/RTzG){(PB2-PB1)/PBM}Hence the rate of absorption of A per unit time over unit area.N/A = k/GP{(PA1-PA2)/PBM} N/A = kG(PA1-PA2) = kG(PAG PAi)where kG = DvP/{RTzGPBM} kG = k/G(P/PBM)Gas film transfer coefficient for absorption. It is a direct measure of rate of absorption per unit area of interface with a driving force of unit partial pressure.Diffusion in the liquid phase: The rate of diffusion in liquids is much slower than in gases, and mixtures of liquids may take a long time to reach equilibrium unless agitated. This is due to much closer spacing of molecules as a result of which the molecular attractions are more important. For dilute solutions,N/A = -DLdCA/dzOn integration N/A = -DL{(CA2 CA1)/zL} where CA, CB are the molar concentrations of A & B, ZL is the thickness of liquid film & DL is the liquid phase diffusivity. Since film thickness is rarely known, N/A = kL(CA1- CA2) = kL(CAi-CAL)where kL = DL/ZL (m/s) is the liquid film transfer coefficient.PAGPAiCAiCALABDEBulk of GasBulk of LiquidGas film boundaryLiquid film boundaryLiquid filmGas filmInterfacePartial Pressure PA of Solute Gas AMolar Concentration CA of A in Liquid12zG

*Rate of AbsorptionIn a steady state process; Rate of material transfer through gas film=Rate of material transfer through liquid film General equation of mass transfer of a component A may be written as:N/A = kG(PAG PAi) = kL(CAi CAL)PAG = Partial pressure of A in the bulk of gas.PAi = Partial pressure of A at the interface.CAi = Concentration of A at the interface.CAL = Concentration of A in the bulk of liquid.Note: Change in liquid composition is positiveand in gas composition is negative.Therefore, kG/kL = (CAi CAL)/(PAG PAi)These conditions are shown in figure:ABF is the equilibrium curve for component APoint D (CAL, PAG ) represents conditions in thebulk of the gas and liquid phases.Point B (CAi, PAi) represents the equilibrium concentrations present at the interface.Point A (CAe,PAG) represents the concentration in liquid phase in equilibrium with PAG.Point F (CAL, PAe) represents partial pressure of gas in equilibrium with CAL.The driving force causing transfer in the gas phase is = (PAG - PAi) =DEand the driving force causing transfer in liquid phase is = (CAi - CAL) =BEThen (PAG - PAi) / (CAI - CAL) = kL/kG = -Slope of the line DBand the concentration at the interface (point B) are found by drawing a line through D of slope - kL/kG to cut the equilibrium curve in B.Overall Coefficients: To obtain a direct measurement of kL and kG requires the measurement of the concentration at the interface. These values can only be obtain in very special circumstances and it has been found of considerable value to choose to overall coefficients KG and KL defined by the following equations. N/A = KG(PAG PAe) = KL(CAe CAL) where KG and KL are known as the overall gas and liquid phase coefficients respectively.PAePAiPAGCALCAiCAePartial Pressure (PA)Concentration in liquid phase (CA)ABFDEPAG-PAiCAi-CALDriving forces in gas and liquid phases

*Rate of AbsorptionRelationship between Film and Overall Coefficients: The net rate of transfer of A is N/A = kG(PAG PAi) = kL(CAi CAL) = KG(PAG PAe) = KL(CAe CAL) 1/KG = (1/kG)[(PAG PAe)/(PAG PAi)] 1/KG = (1/kG)[(PAG PAi)/(PAG PAi)] + (1/kG)[(PAi PAe)/(PAG PAi)] 1/KG = (1/kG) + (1/kG)[(PAi PAe)/(PAG PAi)] But we know that: (1/kG) = 1/(kL)[(PAG PAi)/(CAi CAL)]1/KG = (1/kG) + (1/kL)[(PAG PAi)/(CAi CAL)]X[(PAi PAe)/(PAG PAi)] 1/KG = (1/kG) + (1/kL)[(PAi PAe)/(CAi CAL)][(PAi PAe)/(CAi CAL)] = Average slope of equilibrium curve.When the solution obeys Henrys Law: H = dPA/dCA = [(PAi PAe)/(CAi CAL)] Therefore: 1/KG = (1/kG) + (H /kL) Similarly: 1/KL = (1/kL) + (1/H kG)From equations A and B: 1/KG = (H /KL) The validity of using equations A and B in order to obtain an overall transfer coefficient has been examined and the equilibrium constant H must not vary over the equipment, there must be no significant interfacial resistance and there must be no independence over the values of the two film coefficients.Rates of Absorption in terms of mol fractions: The mass transfer equations can be written in terms of mol fractions as:N/A = k//G(yA yAi) = k//L(xAi xA) = K//G(yA yAe) = K//L(xAe xA)where xA& yA are the mol fractions of soluble component A in the liquid and gas phases. k//G, k//L, K//G and K//L are transfer coefficients defined in terms of mol fractions.If m is the slope of equilibrium curve [approximately (yAi yAe)/(xAi xA)], it can be shown that1/ K//G = (1/k//G) + (m/k//L)ABC

*Rate of AbsorptionFactors influencing the transfer coefficient: The general influence of the solubility of the gas on the shape of equilibrium curve and the resulting influence on the film and overall coefficients, will be seen by considering very soluble, almost insoluble and moderately soluble gases.Very soluble gas: Here the equilibrium curve lies close to the concentration axis and the point E (in figure) approaches very close topoint F. The driving force over the gas film (DE)is then approximately equal to the overall drivingforce (DF), so that kG is approximately equal to KG.Almost insoluble gas: Here the equilibrium curverises very steeply so that the driving force (EB) inthe liquid film becomes approximately equal to the Overall driving force (AD). In this case kL will be approximately equal to KL.Moderately soluble gas: Here both films offer an appreciable resistance, and the point B at the interface must be located by drawing a line through D of slope (kL/kG) = -(PAG PAi)/(CAi CAL)In most experimental work, the concentration at the interface cannot be measured directly, and only overall coefficients will therefore be found. To obtain values for the film coefficients, the relations between kG, kL and KG are utilized.PAePAiPAGCALCAiCAePartial Pressure (PA)Concentration in liquid phase (CA)ABFDEPAG-PAiCAi-CALDriving forces in gas and liquid phases

*HEIGHT OF PACKED COLUMNRate of moles transfer = N/A x Interfacial areaIn a packed column it is very difficult to measure the area of contact. This interfacial area is characteristics of each packing (area/unit volume)The interfacial area for transfer = adV = aAdZ wherea = Surface area of interface per unit volume of columnA = Cross-Sectional area of column, and Z = Height of packed section.Rate of moles transfer = N/A x aAdZ = kG(PAG PAi)aAdZ = kL(CAi CAL)aAdZHeight of Column based on conditions in Gas Film: Let Gm = Rate of moles of inert gas/(unit cross sectional area)Lm = Rate of moles of soluble-free liquor/(unit cross sectional area)Y = Moles of soluble gas A/ moles of inert gas B in gas phase.X = Moles of solute A/ moles of solvent in liquid phase.Consider any plane at which the molar ratios of the diffusing material in the gas and liquid phases are Y and X. Then over a Small height dZ, the moles of gas leaving the gas phase will equal to the moles taken up by the liquid. Then ,AGmdY = ALmdXBut AGmdY = N/A x adV = kG(PAi PAG)aAdZ It should be noted that in a gas absorption process, gas and liquid concentrations will decrease in the upward direction and both dX and dY will be negative.SincePAG = {Y/(1+Y)}PGmdY = kGaP[{Yi/(1+Yi)}-{Y/(1+Y)}]dZ = kGaP[(Yi-Y)/{(1+Y)(1+Yi)}]dZHence height of column Z required to achieve a change in Y from Y1 at the bottom to Y2 at the top of the column is given by:0ZdZ = Z = (Gm/kGaP)Y1Y2{(1+Y)(1+Yi)/(Yi-Y)}dYWhich for weak solutions:Z = {Gm/(kGaP)}Y1Y2 {dY/(Yi-Y)}Note: In this analysis it is assumed that kG is constant throughout the column and provided the concentration changes are not too large dZXX+dXYY+dYGm Y2Gm Y1Lm X2Lm X1

*HEIGHT OF PACKED COLUMNHeight of column based on conditions in liquid film: Similarly for liquid film case:ALmdX = -kLa(CAi - CAL)AdZWhere molar concentrations C are in terms of moles of solute per unit volume of liquor. If CT = (moles of solute + moles of solvent)/(volume of liquid)Then CAL/(CT CAL) = Moles of solute/ Moles of solvent = XTherefore: CAL = {X/(1+X)}CT The transfer equation may be written as: LmdX = kLaCT[{X/(1+X)}-{Xi/(1+Xi)}]dZLmdX = kLaCT[(X Xi)/{(1+X)(1+Xi)}]dZTherefore:0ZdZ = Z = {Lm/ (kLaCT)}X1X2{(1+X)(1+Xi)/(X Xi)}dXand for dilute conditions this gives:Z = {Lm/ (kLaCT)}X1X2{dX/(X Xi)} where CT and kL have been taken constant over the column.Height based on Overall Coefficients: If the driving force based on the gas concentration is written as (Y-Ye) and the overall gas transfer coefficient as KG, then the height of the tower for dilute concentrations becomes:Z = {Gm/(KGaP)}Y1Y2 {dY/(Ye-Y)}Or in terms of liquid concentration as: Z = {Lm/ (KLaCT)}X1X2{dX/(X Xe)} Equations for dilute concentrations: As the mole fraction is approximately equal to the molar ratio at dilute concentrations than considering the gas film.Z = {Gm/(kGaP)}y1y2 {dy/(yi - y)} = {Gm/(KGaP)}y1y2 {dy/(ye-y)}and considering the liquid film:Z = {Lm/(kLaCT)}x1x2{dx/(x xi)} = {Lm/(KLaCT)}x1x2{dx/(xxe)}

*HEIGHT OF PACKED COLUMNTHE OPERATING LINE AND GRAPHICAL INTEGRATION:Taking a material balance on the solute from the bottom of the column to any plane where the mole fractions are Y and X gives for unit area of cross-section:Gm(Y1 Y) = Lm(X1 X)Or (Y1 Y) = (Lm/Gm)(X1 X)This is the equation of straight line of slope Lm/Gm, whichpasses through the point (X1,Y1).If we apply balance over the whole column it will pass throughthe points (X2,Y2). This is known as the equation of operating line.Following is the figure for the case of moist air and sulfuric acid orcaustic soda solution, where the main resistance lies in gas phase.The equilibrium curve is represented by the line FR, and theoperating line is given by ADB, A corresponding to the concentrations at the bottom of the column and B to thoseat the top of the column. D represents the condition of thebulk of the liquid and gas at any point in the column, andhas coordinates X and Y. Then if the gas film iscontrolling the process, Yi equals Ye, and is given by a point F on the equilibrium curve, with coordinates X andYi. The driving force causing transfer is then given by thedistance DF. It is therefore possible to evaluate the expression Y1Y2 {dY/(Yi - Y)}by selecting values of Y, reading of from the diagram the correspondingvalues of Yi, and thus calculating 1/(Yi - Y). Then plot Y vs 1/(Yi - Y) and find the area under the curve to find thevalue of the integral term. Note that for gas absorption Y>Yi and Yi-Y and dY in the integral are both negative.For liquid film controlling the process, Xi = Xe and thedriving force X-Xi=DR. Then the evaluation of integral X1X2{dx/(XXi)} may be effected in the same way as for the gas film.dZXX+dXYY+dYGm Y2Gm Y1Lm X2Lm X1Moles solute/mole inert gasMoles solute/moles solventY1YY2Yi=YeX2=0XXi = XeX1Operating LineADBRFEquilibrium curve

*HEIGHT OF PACKED COLUMNTHE OPERATING LINE AND GRAPHICAL INTEGRATION:Special Case: When equilibrium curve is straight lineOver the range of concentration considered if the equilibriumcurve is a straight line, it is possible to use an average value ofdriving force over the column. For dilute concentrations, overa small height dZ of column, dZ = {Gm/(KGaP)}{dY/(Ye-Y)}If Ye = mX + CThenYe1 = mX1 + Cand Ye2 = mX2 + CSo that m = (Ye1 Ye2)/(X1 X2)Further, applying balance over the lower portion of column:Lm(X1 - X) = Gm(Y1 Y)andX = X1 + (Gm/Lm)(Y Y1)Therefore from the equation 1:(KGaP/Gm)0ZdZ = Y1Y2 {dY/(Ye-Y)}= Y1Y2 dY/[m{X1 +(Gm/Lm)(Y-Y1)} + C Y]

Therefore the height of column, Z = {Gm/(KGaP)}{(Y1 Y2)/(Y Ye)lm}dZXX+dXYY+dYGm Y2Gm Y1Lm X2Lm X1Moles solute/mole inert gasMoles solute/moles solventY1YY2Yi=YeX2=0XXi = XeX1Operating LineADBRFEquilibrium curve12345

* PROBLEM 12.1Some experiments are made on the absorption of Carbon dioxide from carbon dioxide-air mixture in 2.5 normal caustic soda, using a 250 mm diameter tower packed to a height of 3m with 19 mm Raschig rings. In one experiment at atmosphere pressure, the results obtained were:Gas rate G/ = 0.34 kg/m2s;Liquid rate L/= 3.94 kg/m2s.The carbon dioxide in the inlet gas is 315 parts per million and in the exit gas 31 parts per million. What is the value of the overall gas transfer coefficient KGa?Solution: At the top of the tower: y2=31x10-6, x2=0,L/ = 3.94 kg/m2s;2.5 N NaOH contains = 2.5x40g/l = 100g/l = 100kg/m3 NaOHMean molecular weight of liquid = (100x40 + 900x18)/1000MW = 20.2 kg/kgmolLm = L//MW = 3.94/20.2 = 0.195 kmol/m2s.At the bottom of the tower: y1=315x10-6 G/=0.34 kg/m2s, Gm = G//MW = 0.34/29 = 0.0117 kmol/m2sNow from operating line relation:x1=x2+(Gm/Lm)(y1-y2)=0+(0.0117/0.195)(315-31)x10-6= 0(approximately)It may be assumed that as the solution of NaOH is fairly concentrated, there will be negligible vapor pressure over the solution, therefore all resistance to transfer lies in the gas phase. Also it is a dilute mixture, height of the tower will beZ = (Gm/KGaP){(y1-y2)/(y-ye)lmKGa = (Gm/ZP){(y1-y2)/(y-ye)lmDriving force at the top of column, (y-ye)2 = 31x10-6 0 = 31x10-6Driving force at the bottom of column, (y-ye)1 = 315x10-6 0 = 315x10-6Log-mean driving force difference, (y-ye)lm = {(y-ye)1-(y-ye)2}/ln{(y-ye)1/(y-y3)2}= (315-31)x10-6/ln(315/31) = 122.5x10-6Thus,KGa = (0.0117/3x101.3)(315-31)x10-6/(122.5x10-6) = 8.93x10-5D=250 mmH=3m19 mm RaschigRingsx2=0L/=3.94 kg/m2sx1y2=31x10-6y1=315x10-6G/=0.34kg/m2sKGa = 8.93x10-5

*THE TRANSFER UNITConsidering the height of packed column consist of stages of definite height such as for a plate column. These hypothetical stages are called transfer unit. Therefore for a packed columnHeight of packed column = (Height of a transfer unit) x (Number of transfer units)For gas phase based on overall driving force: Chilton and Colburn defined the number of overall gas phase transfer units as NOG = y1y2 {dy/(ye - y)}Therefore NOG is the integral of change in concentration per unit driving force.The driving force (ye-y) is very small at the top of column which result in a large number of transfer units at the top of the column.Height of overall gas phase transfer units: HOG = Z/ NOG= Gm/KGaPFor conditions based of on gas film: Z = HG x NG Where number of gas film transfer unit, NG = y1y2 {dy/(yi - y)} and height of gas film transfer unit HG = Gm/kGaPFor condition based on overall driving force in liquid phase: Z = HOL x NOL Where height of overall liquid phase transfer unit, HOL = Lm/KLaCT and number of overall liquid phase transfer unit, NOL = x1x2{dx/(xxe)} For conditions based on Liquid film: Z = HL x NL Where height of liquid film transfer unit, HL = Lm/kLaCT and number of liquid film transfer unit, NL = x1x2{dx/(xxi)} The relationships between the height of overall coefficients and film coefficients can be defined as HOG = HG + (mGm/Lm)HLHOL = HL + (Lm/mGm)HGWhere m is the slope of equilibrium curve

*THE TRANSFER UNITThe Importance of Liquid and Gas Flowrates and Slope of Equilibrium CurveFor a packed tower operating with dilute concentrations, since x X1 and y Y1, then:Gm(y1 y2) = Lm(x1 x2) where, as before, x and y are the mole fractions of solute in the liquid and gas phases, and Gm and Lm are the gas and liquid molar flowrates per unit area on a solute free basis.A material balance between the top and some plane where the mole fractions are x, y gives:Gm(y y2) = Lm(x x2) If the entering solvent is free from solute, then x2 = 0 and:x = (Gm/Lm)(y y2) But the number of overall transfer units is given by:NOG = y1y2 {dy/(ye - y)}For dilute concentrations, Henrys law holds and ye = mx. Thus: NOG = y1y2 [dy/{m(Gm/Lm)(y y2)- y}] =y1y2 [dy/{m(Gm/Lm-1) y m(Gm/Lm) y2}]and: NOG = 1/ {1-m(Gm/Lm)} x ln [{1-m(Gm/Lm)} (y1/ y2)+m(Gm/Lm)]COLBURN has shown that this equation may usefully be plotted as shown in figure above which is taken from his paper. In this plot the number of transfer units NOG is shown for values of y1/y2 using mGm/Lm as a parameter and it may be seen that the greater mGm/Lm, the greater is the value of NOG for a given ratio of y1/y2. From the above equation:Lm/Gm = (y1 y2)/x1 = (y1 y2)/(ye1/m)Thus:mGm/Lm = ye1/(y1 y2) where ye1 is the value of y in equilibrium with x1.On this basis, the lower the value of mGm/Lm, the lower will be ye1, and hence the weaker the exit liquid. Colburn has suggested that the economic range for mGm/Lm is 0.7-0.8. If the value of HOG is known, the quickest way of obtaining a good indication of the required height of the column is by using figure above.

* PROBLEM 12.2An acetone-air mixture containing 0.015 mol fraction of acetone has the mol fraction reduced to 1 percent of this value by countercurrent absorption with water in a packed tower. The gas flow rate G/ is 1 kg/m2s of air and the water entering is 1.6 kg/m2s. For this system, Henrys Law holds and ye=1.75x, where ye is the mol fraction of acetone in the vapor in equilibrium with a mol fraction x in the liquid. How many overall transfer units are required?Solution: At the top of the tower: y2=0.00015, x2=0,L/ = 1.6 kg/m2s, Lm = L//MW = 1.6/18 = 0.0889 kmol/m2s.At the bottom of the tower: y1=0.015,G/= 1.0 kg/m2s, Gm = G//MW = 1.0/29 = 0.0345 kmol/m2sNow from operating line relation:x1=x2+(Gm/Lm)(y1-y2)=0+(0.0345/0.0889)(0.015-0.00015=0.00576For dilute system, height of column:Z = (Gm/KGaP){(y1-y2)/(y-ye)lmHOG = Gm/KGaPNOG = Z/HOG= (y1-y2)/(y-ye)lmDriving force at the top of column, (y-ye)2 = y2 mx2 = 0.00015 - 1.75x0 = 0.00015Driving force at the bottom of column, (y-ye)1 = y1mx1 = 0.0151.75x0.00576= 0.0049Log-mean driving force difference, (y-ye)lm = {(y-ye)1-(y-ye)2}/ln{(y-ye)1/(y-ye)2}= (0.0049-0.00015)/ln(0.0049/0.00015) = 0.00136Therefore NOG = (0.015-0.00015)/0.00136 = 10.92 = 11(approximately)Also, NOL = NOG(mGm/Lm) = 10.92(1.75x0.0345/0.0889) = 7.42 = 7 (approximately)x2=0L/=1.6 kg/m2sx1y2=0.00015y1=0.015G/=1 kg/m2sNOG = 11 & NOL = 712

*PLATE COLUMNS FOR GAS ABSORPTIONWhen the load is more than can be handle in a packed column of about 1 m diameter and when there is any likelihood of deposition of solids which would quickly choke a packing, plate columns are preferred. Plate columns are particularly useful when the liquid rate is sufficientto flood a packed tower. Since the ratio of liquid rate to gas rate isgreater than with distillation, the slot area will be rather less and thedowncomers rather larger.On the whole, plate efficiencies have been found to be less than withthe distillation equipment, and to range from 20 to 80 percent.;The plate column is a common type of equipment for large installations,but when the diameter of the column is less than 2m, packed columns aremore often used. For the handling of very corrosive fluids, packedcolumns are frequently preferred for larger units.The essential arrangement of the unit is indicated in figure: LetLm = molar flux of solute free liquidGm = molar flux of inert gas.n = refers to the plate numbered from the bottom upwards.X = molar ratio of absorbed component in liquid.Y = molar ratio of absorbed component in gas.s = total number of plates in the column.Assumption: Each plate will be taken as an ideal unit.Material balance for the absorbed component from the bottom plate to a plane above plate n:Gm Yn + LmX1 = GmY0 + LmXn+1yn = (Lm/Gm)Xn+1 + {Y0 (Lm/Gm)X1}This equation is a straight line of slope (Lm/Gm), relating thecompositions of passing streams, known as the equation of operating line.Point A represents the conditions at the bottom of the column. The gasrising from the bottom plate is in equilibrium with a liquid ofconcentration X1 and is shown as point B on operating line. The point 4indicates the concentration of the liquid on the second plate from bottom. In this way the steps may be drawn to point B, giving the gas Ys risingfrom the top plate and the liquid Xs+1 entering the top of absorber.

ABLiquid Xs+1Liquid X1Gas YsGas Y0Ys-1YnYn-1Y1X21n-1nn+1sXnXn+1XsYXY0X1Xs+1YsA234BEquilibrium curveOperating line

*PROBLEM 12.3An oil containing 2.55 mol percent of a hydrocarbon is stripped by running the oil down a column up which live steam is passed, so that 4 kmol of steam are used per 100 kmol of oil stripped. Determine the number of theoretical plates required to reduce the hydrocarbon content to 0.05 mol percent, assuming that the oil is non-volatile. The vapor-liquid relation of the hydrocarbon in the oil is given by ye=33x, where ye is the mol-fraction in the vapor and x the mol-fraction in the liquid. The temperature is maintained constant by internal heating, so that the steam does not condense in tower.Solution: Data: Steam to oil ratio = 4kmol/100kmolVapor-liquid equilibrium relationship for hydrocarbon, ye = 33xSteam does not condense due to internal heating systemFrom steam to oil ratio, Lm/Gm = 100/4 = 25 kmol/kmolSince, xs+1 = 0.0255, therefore Xs+1 = xs+1/(1-xs+1)Xs+1 = 0.0255/(1 - 0.0255) = 0.0262And x1 = 0.0005, then X1 = x1/(1x1) =0.0005/(10.0005)= 0.0005Operating line equation is: Yn = (Lm/Gm)Xn+1 + {Y0 - (Lm/Gm)X1}Yn = 25Xn+1 + 0 (25)0.0005 = 25Xn+1 0.0125From operating line equation, Ys = 25Xs+1 0.0125 = 25(0.0262) - 0.0125 = 0.6423Plot the operating line on X-Y diagram by points A and B.From the equilibrium equation, ye = 33xYe/(1+Ye) = 33{X/(1+X)}Ye = 33X/(1-32X)From the above equation, make the table and draw the curve.Stripperxs+1= 0.0255x1= 0.0005Steam y0=0ys=0XY81Equilibrium curveOperating lineNumber of Stages = 08

X0.00.0020.0040.0060.0080.0100.0120.014Y0.00.0710.1510.2450.3550.4850.6430.837

**