Presentation on theme: "Absorption Bringing the dirty effluent gas into contact with the scrubbing liquid and subsequently separating the cleaned gas from the contaminated liquid."— Presentation transcript:
1 AbsorptionBringing the dirty effluent gas into contact with the scrubbing liquid and subsequently separating the cleaned gas from the contaminated liquidAbsorption is a basic chemical enginnering unit operation which in the APC field is reffered as scrubbingThey have wide use in controlling SO2, H2S, and light hydrocarbons
2 Absorption _Wet scrubbers can be categorized into 3 groups: Packed-bed counterflow scrubbersCross-flow scrubbersBubble plate and tray scrubbers
4 Concept of AbsorptionGas absorption is the removal of one or more pollutants from a contaminated gas stream by allowing the gas to come into intimate contact with a liquid that enables the pollutatants to become dissolved by the liquid. The principal factor dictating performance is the solubility of the pollutants in the absorbing liquid. The rate of transfer in the liquid is dictated by the diffusion processes occurring on each side of the gas liquid interface.
5 Liquid WastePollutants removed from the gas stream transferred into liquid phase whose disposal is another issue to deal with. Therefore scrubber needs other units such as storage vessels, additives to treat the scrubbing liquid according to required discharge standards.
6 AbsorptionAbsorption units must provide large surface area of liquid-gas interfaceTherefore the units are designed to provide large liquid surface area with a minimum of gas pressure drop
11 Absorption TheoryPhysically, the absorption of a pollutant gas from a moving gas stream into an appropriate liquid stream is quite complexBasically the transfer process into each fluid stream is accomplished by 2 mechanisms:The pollutant species is transferred from the bulk of the gas stream toward the gas-liquid interface by turbulent eddy motionsVery close to the interface laminant flow is valid and transfer is accomplished by molecular diffusionOn the liquid side of the interface process is reversed
13 Absorption TheoryOn the basis of Fick's Law, the diffusion of one gas (A) through a second stagnant gas B, NA, the molar rate of transfer of A per unit cross-sectional area is given by;NA = -DAB (dcA/dz)/(1-(cA/c)DAB: molecula diffusion coef. (m2/t)cA: molar concentration of species A (mol/L)c: molar concentration of the gas mixture (mol/L)z: the direction of mass transfer (m)DAB tables are available for a number of binary gas mixtures
14 Absorption TheoryMass transfer rate per unit area for molecular diffusion of A through a second liquid is given by:NA = -DL/z (cA2-cA1)DL: liquid phase molecular diffusion coef. (m2/t)cA2-cA1: concentration difference of A over the distance zTypical values of DL for binary mixtures are tabulated in the literature.
15 The Equilibrium Distribution Curve Before entering into details of mass transfer, let's summarize the method of presenting equilibrium data for a pollutant A distributed between liquid and gas phaseP=cInject solute AInert LiquidSolventInert carriergasMole fraction in gas, yAExp. Equilib. Distribution curveMole fraction in lqiuid, xA
16 The Equilibrium Distribution Curve After sufficient time, no further change in the concentration of A in two phases. These concentration can be measured and converted into mole fraction xA in the liquid phase and yA in the gas phaseP=cInject solute AInert LiquidSolventInert carriergasMole fraction in gas, yAExp. Equilib. Distribution curveMole fraction in lqiuid, xA
17 Mass transfer coefficients based on Interfacial Concentrations When mass transfer occurs in moving liquid and gaseous streams, it is difficult to evaluate the separate effects of molecular and turbulent diffusionAn alternative to this is to express NA for each phase in terms of mass transfer coefficient k and a driving force based on the bulk and interfacial concentrations for that phase
18 Mass transfer coefficients based on Interfacial Concentrations For the liquid phase:NA = kL(cAi-cAL) = kx(xAi-xAL)kL(is the liquid mass transfer coeff. Based on concentration, in length per unit of time, cAiis the concentration of A in the liquid phase at the interface, cALis the concentration of A in the bulk of the phase, in moles per unit volume.kx is the liquid mass transfer coefficient based on mole fractions, in moles per units of time and length squared, xA is the mole fraction of A in the liquid interface, and xAL is the mole fraction of A in the bulk of the liquid phase
19 Mass transfer coefficients based on Interfacial Concentrations For the gas phase:NA = kG(pAG-cAi) = ky(yAG-yAi)kGis the gas phase mass transfer coeff. based on partial pressures, in moles/length2 time, pAGis the partial pressure of A in the bulk of gas phase pAi is the partial pressure of A in the gas interfaceky is the gas phase mass transfer coefficient based on mole fractions, in moles per units of time and length squared, yAG is the mole fraction of A in the bulk of the gas phase, and yAi is the mole fraction of A in the gas phase interface
20 Mass transfer coefficients based on Interfacial Concentrations However this approach to determining NA is not practical since kx and ky are difficult to obtain and no way to measure the values of yAi and xAi experimentally since any attempt to do it will perturb the equilibrium between the two streams
21 Overall Mass Transfer Coefficients When mass transfer rates are reasonably low, NA can be expressed as:NA = KG(pAG-pA*) = Ky(yAG-yA*)KG and Ky are local overall mass transfer coefficientspA* : equilibrium partial pressure of solute A in a gas phase which is in contact with a liquid having the composition of cAL of the main body of the absorption liquidyA*: defined similarly in terms of a liquid with mole fraction xAL of the bulk liquid
22 Overall Mass Transfer Coefficients Point P represents the state of the bulk phase of the 2 fluid streams, yAG and xAL.The point M represents the state (yAi and xAi) associated with equilibrium at the interfaceThe distance between P and C is a measure of the driving force.slope=m'
23 Overall Mass Transfer Coefficients NA = KG(pAG-pA*) = Ky(yAG-yA*) This equation is usually restricted the resistance to mass transfer is primarily in the gas phase, which characterizes the majority of absorption problems in air pollution workThe solubility of the polutant gas normally determines the liquid that is chosenThe major physical problem is getting the pollutant to diffuse through the gas phase to the interface, consequently gas phase controls the process.If the liquid phase controls:NA = KL(cA*-cAL) = Kx(xA*-xAL)
24 Overall Mass Transfer Coefficients It is important to note that the quantities pA*,yA*,cA*,xA* do not represent any actual condition in the absorption process but are related in each case to a real concentration in one of the bulk fluids through the equilibrium data for the two-phase system. From the geometry of the previous figure:yAG-yA*= yAG-yAi+(yAi-yA*)yAi-yA*=m'(xAi-xAL)yAG-yA*= yAG-yAi+m'(xAi-xAL)1/Ky=1/ky+m'/kx
25 Mass Balances and the Operating Line for Packed Towers Gm molar total gas flow rate (carrier gas + pollutant)Gc molar inert carrier gas flow rateLm molar total solvent flow rate (solvent + absorbed pollutant)Ls molar solvent flow ratex is the liquid mole fraction of pollutant, y is the gas phase mole fraction of the pollutants, X is the liquid phase mole ratio and Y is the gas phase mole ratioLm,2Lsx2X2Gm,2Gcy2Y2T = constP = constCross-sectional area, AdzGm,1Gcy1Y1Lm,1Lsx1X1
26 Mass Balances and the Operating Line for Packed Towers Mole fraction and mole ratio:X = x/(1-x) Y = y/(1-y)Subscript m denotes that rates are in the units of mole basisThe conservation of mass principle applied to the pollutant species in terms of total mass flow rates at top and bottom yields:Gm,1y1+ Lm,2x2 = Gm,2y2+ Lm,1x1or Gm,1y1 -Gm,2y2 = Lm,1x1 -Lm,2x2
27 Mass Balances and the Operating Line for Packed Towers In Gm,1y1 -Gm,2y2 = Lm,1x1 -Lm,2x2 total gas and liquid flow rates are not equal at the top and the bottom of the column, therefore we cannot further simplify this equation.When we write the equation in terms of the carrier gas and liquid solvent rates then:GC,m(Y2-Y1) = LS,m(X2-X1)These two equations above gives a straight line on Y-X coordinates with a slope of Lsm/GCm and called operating lines.
28 The operating line lies above the equilibrium line for absorption For a stripping (removal of gas from liquid stream) the operating line must lie below the equilibrium line in order for the drving force to act from the liquid phase toward the gas phaseDirty airClean airClean waterDirty water
29 The Minimum and Design Liquid- Gas Ratio At the bottom and top of the absorber, parameters Gm,1, Gc, y1, Gm2, y2, and x2 are known.We need to determine Ls, and x1So we have one equation with 2 unknowns...However selection of one of these values, obviously fixes the other.How to select a value?
30 The Minimum and Design Liquid- Gas Ratio The minimum rate is highly undesirable. At this point driving force is almost 0. Hence it would take an infinetely tall absorber to accomplish the desired separationAs a general operating principle an absorber is typically designed to operate at liquid rates which are 30 to 70 % greater than minimum rate.
31 Tower Diameter and Pressure Drop per Unit Tower Height For a given packing and liquid flow rate in an absorption tower variation in the gas velocity has a significant effect on the pressure dropAs the gas velocity is increased, the liquid tends to be retarded in its downward flow, giving rise to term liquid holdup (LH)A LH increases, the free cross-sectional area for gas flow decreases and pressure drop per unit height increases.
32 Problems with High Gas Velocity Channeling: the gas or liquid flow is much greater at some points than at othersLoading: the liquid flow is reduced due to the increased gas flow; liquid is held in the void space between packing•Flooding: the liquid stops flowing altogether and collects in the top of the column due to very high gas flowTO AVOID this condition experience dictates operating at gas velocities which are 40 to 70 % of those which causing flooding
33 Flood PointThe relationship between DP/Z and other important tower variables-liquid and gas rates, liquid and gas stream densities and viscosities, and type of packing has been extensively studied on an experimental basis.A widely accepted correlation among these parameters can be seen in below figureWhere G' and L': superficial gas and liquid mass flow rate defined as actual flow rates divided by the empty cross-sectional area of the tower.
35 G’:gas mass flux (lb/s- ft2) F:packing factor (ft2/ft3) mL:liquid viscosity, cpgc: proportıonality constant, ft- lb/s2-lbfrL:liquid density, lb/ft3rG:gas density, lb/ft3L’/G’√(rG/rL-rG) L’: liquid mass flux (lb/s-ft2)In Cooper and Alley’s book, Figure 13.6 can be used. Note that in Figure 13.6 Gx and Gy are liquid and gas flux (lg/s-ft2), respectively. In our notation G’ and L’ correspond to Gx and Gy
36 Packing Factor FThe top line in the figure represents the general flooding condition for many packings. The flooding condition however has been found to vary as a function of the packing factor F (dimensionless packing factor tabulated below)Recent studies showed that when F is in the range of 10 to 60, the pressure drop can be expressed by:DPflood = 0.115F0.7
38 Determining Tower Diameter First abscissa value is calculated(L'/G')(pG/(pL-pG))0.5Where this value intercepts the flooding line on Figure A, move horizontally to the left and read the value of the ordinate:(G')2F(mL)0.1/gc(pL-pG)pGCalculate the G’ and take 30 to 70% of it to prevent floodingTower crossectional area: A = G/G‘Evaluate the tower diameter
39 Determining Expected Pressure Drop per Unit Height of Tower First calculate actual G’ and L’ and then calculate the abscissa and the ordinate for use in Figure 13.6From those values the intersection on the figure defines the pressure drop per foot of packed heightAnother emprical correlation found in the litrature for the DP in packing when operating below the load point isDP/Z = 10-8m[10nL’/rL](G’2/rG) m and n are packing constants see Table 6.2
40 Determining Tower Diameter and Expected Pressure Drop per Unit Height of Tower
41 ExampleA packed tower is to be designed to remove 95% of the ammonia from a gaseous mixture of 8 percent ammonia and 92% air, by volume. The flow rate of the gas mixture entering the tower at 68 F and 1 atm is 80 lb-moles/hr. Water containing no ammonia is to be the solvent, and 1-in. Ceramic raschig rings will be used as the packing. The tower is to operated at 60% of theflood point and the liquid water rate is to be 30% greater than the minimum rate. Determine1. The gas-phase flow rates, in lb-moles/hr, for the solute and carrier gas2. The mole ratios of the gas and liquid phases at inlet and outlet and the required water rate in lbmoles/hr.3.The gas and liquid rates (lb/hr) for carrier gas, solute gas, total gas, liquid solvent, solute in liquid, and total liquid4. The tower area and diameter5. The pressure drop based on the two methods given in the lecture notes.
42 Example Removal efficiency: 95% Effluent Stream Composition: 8% ammonia and 92% airGas T and P: 68F and 1 atmFlowrate: 80 lb-moles/hrLiquid phase: Containing no ammonia
43 ExampleDetermine composition of the liquid at the exit (X1)(Inlet liquid concentration since pure water is used is x2=X2=0)Use equilibrium data for ammonia-air-water mixtures which are given below for 68 F and 14,7 psia. :X0.02060.03100.04070.05020.07350.0962Y0.01580.0240.03290.04180.06600.0920In order to determine composition of liquid at the exit, we need to calculate the minimum solvent flow rate first.By plotting X vrs Y at the equilibrium, we can evaluate the minimum solvent and then operating solvent rate.In Cooper and Alley’s book, use Table B4 in the Appendix.
44 Example0,90Y2=0.092Since the liquid rate is to be 30% greater than the miniumu rate(Lm,S)/Gm,C)design = 1,30(0.90) = 1.17 mole/moleLm,S = Gm,C*1,17 = 1.17*73.6 = 86.1 lb moles/hr
45 Example Now, X1 can now be found. 0,020,040,060,080,1X, moles solute per mole solventy, moles solute per molecarrier gasX2,Y2(Lm,S/Gm,C)=1.17Y1=0.0870,90X1:Now, X1 can now be found.Graphically by drawing operating line with a slope of 1.17 with starting point of (0, ) and the point crosses Y1=0.087 can be read. ORFrom Lms/Gm,C = Y2-Y1/(X2-X1)= /(0-X1) = 1.17X1 = lm mole A/lm mole water or x1 = lb mole A /lb moles solution
47 Tower AreaTo determine the tower area, we need to use Figure flooding correlation plot. Therefore we need to calculate gas and liquid phase densities at the top and bottom of the tower. Since the ammonia content is very low in liquid phase, use the density of pure water, 62.3 lb/ft3 as the solution density through the tower. For the gas phase assume ideal gas behavior: r= P/RT = MwP/RT At the top: Mw= SyiMi = * *29 = r= 28.95*14.7/(10.73*528) = lb/ft3 At the bottom Mw= 0.08* *29 = r= 28.04*14.7/(10.73*528) = lb/ft3 Now calculate the abscissa of Flooding Figure
50 Pressure DropPressure drop can be determined from the flooding figure or from an emprical equation
51 Determination of an Absorption Tower Height Height of a packed tower = f(the overall resistance to mass transfer between the gas and liquid phases, the average driving force and interfacial area)Consider a differential height of the absorber dZ. In height dZ, the rate of mass transfer of species Aa: interfacial area available to mass transfer per unit volume of the packingA: cross-sectional area of the tower
52 Tower HeightThe equation can be also written for liquid resistance part.
53 Tower HeightTo solve the above equation we can determine the overall value of Kya (Kga) based on experimental “pilot plant” operated with a certain packing and gas/liquid rate. The right side of the equation can be integrated from the knowledge of the operating line and equilibrium line chracteristics. This method can be modified to deal with the “height of a transfer unit” and “the number of transfer units” by modifiying the equation somewhat
54 Tower HeightTo solve the above equation we can determine the overall value of Kya (Kga) based on experimental “pilot plant” operated with a certain packing and gas/liquid rate. The right side of the equation can be integrated from the knowledge of the operating line and equilibrium line chracteristics. This method can be modified to deal with the “height of a transfer unit” and “the number of transfer units” by modifiying the equation somewhat
55 Tower HeightThe equation can be expressed in terms of height of transfer unit (HTU ) and number of transfer units :HTU is reaonably constant through the absorber and has unit of length. NTU is dimensionless.NTU or NoyHTU or Hoy
56 For dilute gas streams, transfer unit equation can be simplified: x1, y12017/4/14x1, y1*xZ, yZAerosol & Particulate Research LabxZ, yZ*
57 Pure amine 0.04% CO2 Lm = 0.46 gmole/s 1.27% CO2 Gm = 2.31 gmole/s 2017/4/14Q: A Packed tower using organic amine at 14 oC to absorb CO2. The entering gas contains 1.27% CO2 and is in equilibrium with a solution of amine containing 7.3% mole CO2. The gas leaves containing 0.04% CO2. The amine, flowing counter-currently, enters pure. Gas flow rate is 2.31 gmole/s and liquid flow rate is 0.46 gmole/s. The tower’s cross-sectional area is 0.84 m2. KOGa = 9.34×10-6 s-1atm-1cm-3. The pressure is 1 atm. Determine the tower height that can achieve this goal.Aerosol & Particulate Research Lab1.27% CO2Gm = 2.31gmole/sC* = 7.3% CO2 in amine
58 Absorption of concentrated vapor Mole balance on the controlled volume2017/4/14x1, y1x1, y1*Gas fluxLiquid fluxAerosol & Particulate Research LabxZ, yZxZ, yZ*
59 ExampleA 1‐ft diameter packed column is used to scrub a soluble gas (MW = 22) from an air‐gas mixture. Pure water enters the top of the column at 1000 lbm/hr. The entering gas stream contains 5% soluble gas and 95% air. Ninety‐five percent of the soluble gas is removed. Both the operating line and equilibrium curve may be assumed to be straight. The equation for the equilibrium curve is y = 1.2x, where x, y = mole fractions. The entering gas mixture flow rate is 800 lbm/hr. The column operates at 30 °C and 1 atm, andKya = 4.29 lbmol/hr‐ft3‐Δy φ
60 Example Calculate or find: a) Concentration of the soluble gas in the effluent liquid if the column is operated at minimum liquid flow rateb) Concentration of soluble gas in the liquid at a point in column where y = 0.02c) Height of packed section, ZTd) Hoye) Whether columnis in danger of flooding if it is packed with 1⁄2‐in. ceramic Raschig rings