ABSORPTİON Bringing the dirty effluent gas into contact with the scrubbing liquid and subsequently separating the cleaned gas from the contaminated liquid Absorption is a basic chemical enginnering unit operation which in the APC field is reffered as scrubbing They have wide use in controlling SO 2, H 2 S, and light hydrocarbons
ABSORPTİON _Wet scrubbers can be categorized into 3 groups: 1. Packed-bed counterflow scrubbers 2. Cross-flow scrubbers 3. Bubble plate and tray scrubbers
SCRUBBER TYPES Packed Tower Spray tower Venturi Absorber
CONCEPT OF ABSORPTİON Gas 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.
LİQUİD WASTE Pollutants 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.
ABSORPTİON Absorption units must provide large surface area of liquid-gas interface Therefore the units are designed to provide large liquid surface area with a minimum of gas pressure drop
PACKED TOWER
PACKİNG MATERİAL AND SHAPES Packing material (must be inert) is designed to increase the liquid-film surface Many geometric shapes are available : Raschig ring, pall ring, berl saddle, tellerete etc.
PACKİNG MATERİAL AND SHAPES
PACKİNG MATERİAL PROPERTİES
ABSORPTİON THEORY Physically, the absorption of a pollutant gas from a moving gas stream into an appropriate liquid stream is quite complex Basically 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 motions Very close to the interface laminant flow is valid and transfer is accomplished by molecular diffusion On the liquid side of the interface process is reversed
ABSORPTİON THEORY
ABSORPTİON THEORY On 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 (dc. A/dz)/(1 -(c. A/c) DAB: molecula diffusion coef. (m 2/t) c. A: 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
ABSORPTİON THEORY Mass transfer rate per unit area for molecular diffusion of A through a second liquid is given by: NA = -DL/z (c. A 2 -c. A 1) DL: liquid phase molecular diffusion coef. (m 2/t) c. A 2 -c. A 1: concentration difference of A over the distance z Typical values of DL for binary mixtures are tabulated in the literature.
THE EQUİLİBRİUM DİSTRİBUTİON 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 phase P=c Inject solute A Inert carrier gas Inert Liquid Solvent Mole fraction in gas, y. A Exp. Equilib. Distribution curve Mole fraction in lqiuid, x. A
THE EQUİLİBRİUM DİSTRİBUTİON 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 x. A in the liquid phase and y. A in the gas phase P=c Inject solute A Inert carrier gas Inert Liquid Solvent Mole fraction in gas, y. A Exp. Equilib. Distribution curve Mole fraction in lqiuid, x. A
MASS TRANSFER COEFFİCİENTS BASED ON INTERFACİAL CONCENTRATİONS When mass transfer occurs in moving liquid and gaseous streams, it is difficult to evaluate the separate effects of molecular and turbulent diffusion An 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
MASS TRANSFER COEFFİCİENTS BASED ON INTERFACİAL CONCENTRATİONS For the liquid phase: NA = k. L(c. Ai-c. AL) = kx(x. Ai-x. AL) k. L(is the liquid mass transfer coeff. Based on concentration, in length per unit of time, c. Aiis the concentration of A in the liquid phase at the interface, c. ALis 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, x. A is the mole fraction of A in the liquid interface, and x. AL is the mole fraction of A in the bulk of the liquid phase
MASS TRANSFER COEFFİCİENTS BASED ON INTERFACİAL CONCENTRATİONS For the gas phase: NA = k. G(p. AG-c. Ai) = ky(y. AG-y. Ai) k. Gis the gas phase mass transfer coeff. based on partial pressures, in moles/length 2 time, p. AGis the partial pressure of A in the bulk of gas phase p. Ai is the partial pressure of A in the gas interface ky is the gas phase mass transfer coefficient based on mole fractions, in moles per units of time and length squared, y. AG is the mole fraction of A in the bulk of the gas phase, and y. Ai is the mole fraction of A in the gas phase interface
MASS TRANSFER COEFFİCİENTS BASED ON INTERFACİAL CONCENTRATİONS 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 y. Ai and x. Ai experimentally since any attempt to do it will perturb the equilibrium between the two streams
OVERALL MASS TRANSFER COEFFİCİENTS When mass transfer rates are reasonably low, NA can be expressed as: NA = KG(p. AG-p. A*) = Ky(y. AG-y. A*) KG and Ky are local overall mass transfer coefficients p. A* : equilibrium partial pressure of solute A in a gas phase which is in contact with a liquid having the composition of c. AL of the main body of the absorption liquid y. A*: defined similarly in terms of a liquid with mole fraction x. AL of the bulk liquid
OVERALL MASS TRANSFER COEFFİCİENTS Point P represents the state of the bulk phase of the 2 fluid streams, y. AG and x. AL. The point M represents the state (y. Ai and x. Ai) associated with equilibrium at the interface slope=m' The distance between P and C is a measure of the driving force.
OVERALL MASS TRANSFER COEFFİCİENTS NA = KG(p. AG-p. A*) = Ky(y. AG-y. A*) 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 work The solubility of the polutant gas normally determines the liquid that is chosen The 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(c. A*-c. AL) = Kx(x. A*-x. AL)
OVERALL MASS TRANSFER COEFFİCİENTS to note that the quantities It is important p. A*, y. A*, c. A*, x. A* 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 twophase system. From the geometry of the previous figure: y. AG-y. A*= y. AG-y. Ai+(y. Ai-y. A*) y. Ai-y. A*=m'(x. Ai-x. AL) y. AG-y. A*= y. AG-y. Ai+m'(x. Ai-x. AL) 1/Ky=1/ky+m'/kx
MASS BALANCES AND THE OPERATİNG LİNE FOR PACKED TOWERS Lm, 2 Ls x 2 X 2 Gm, 2 Gc y 2 Y 2 T = const P = const Cross-sectional area, A dz Lm, 1 Ls x 1 X 1 Gm, 1 Gc y 1 Y 1 Gm molar total gas flow rate (carrier gas + pollutant) Gc molar inert carrier gas flow rate Lm molar total solvent flow rate (solvent + absorbed pollutant) Ls molar solvent flow rate x 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 ratio
MASS BALANCES AND THE OPERATİNG LİNE 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 basis The conservation of mass principle applied to the pollutant species in terms of total mass flow rates at top and bottom yields: Gm, 1 y 1+ Lm, 2 x 2 = Gm, 2 y 2+ Lm, 1 x 1 or Gm, 1 y 1 -Gm, 2 y 2 = Lm, 1 x 1 -Lm, 2 x 2
MASS BALANCES AND THE OPERATİNG LİNE FOR PACKED TOWERS In Gm, 1 y 1 -Gm, 2 y 2 = Lm, 1 x 1 -Lm, 2 x 2 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(Y 2 -Y 1) = LS, m(X 2 -X 1) These two equations above gives a straight line on Y -X coordinates with a slope of Lsm/GCm and called operating lines.
Dirty air Clean water Dirty water 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 phase
THE MİNİMUM AND DESİGN LİQUİDGAS RATİO At the bottom and top of the absorber, parameters Gm, 1, Gc, y 1, Gm 2, y 2, and x 2 are known. We need to determine Ls, and x 1 So we have one equation with 2 unknowns. . . However selection of one of these values, obviously fixes the other. How to select a value?
THE MİNİMUM AND DESİGN LİQUİDThe minimum rate GAS RATİO is highly undesirable. At this point driving force is almost 0. Hence it would take an infinetely tall absorber to accomplish the desired separation As a general operating principle an absorber is typically designed to operate at liquid rates which are 30 to 70 % greater than minimum rate.
TOWER DİAMETER AND PRESSURE DROP PER UNİT TOWER HEİGHT For a given packing and liquid flow rate in an absorption tower variation in the gas velocity has a significant effect on the pressure drop As 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.
PROBLEMS WİTH HİGH GAS VELOCİTY Channeling: the gas or liquid flow is much greater at some points than at others Loading: 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 flow TO AVOID this condition experience dictates operating at gas velocities which are 40 to 70 % of those which causing flooding
FLOOD POİNT The 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 figure Where 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.
G’: gas mass flux (lb/sft 2) F: packing factor (ft 2/ft 3) m. L: liquid viscosity, cp gc: proportıonality constant, 32. 17 ftlb/s 2 -lbf r. L: liquid density, lb/ft 3 r. G: gas density, lb/ft 3 L’/G’√(r. G/r. L-r. G) L’: liquid mass flux (lb/s-ft 2) 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-ft 2), respectively. In our notation G’ and L’ correspond to Gx and Gy
PACKİNG FACTOR F The 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. 115 F 0. 7
PACKİNG DATA
DETERMİNİNG TOWER DİAMETER First abscissa value is calculated (L'/G')(p. G/(p. L-p. G))0. 5 Where this value intercepts the flooding line on Figure A, move horizontally to the left and read the value of the ordinate: (G')2 F(m. L)0. 1/gc(p. L-p. G)p. G Calculate the G’ and take 30 to 70% of it to prevent flooding Tower crossectional area: A = G/G‘ Evaluate the tower diameter
DETERMİNİNG EXPECTED PRESSURE DROP PER UNİT HEİGHT OF TOWER First calculate actual G’ and L’ and then calculate the abscissa and the ordinate for use in Figure 13. 6 From those values the intersection on the figure defines the pressure drop per foot of packed height Another emprical correlation found in the litrature for the DP in packing when operating below the load point is DP/Z = 10 -8 m[10 n. L’/r. L](G’ 2/r. G) m and n are packing constants see Table 6. 2
DETERMİNİNG TOWER DİAMETER AND EXPECTED PRESSURE DROP PER UNİT HEİGHT OF TOWER
EXAMPLE A 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. Determine 1. The gas-phase flow rates, in lb-moles/hr, for the solute and carrier gas 2. 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 liquid 4. The tower area and diameter 5. The pressure drop based on the two methods given in the lecture notes.
EXAMPLE Removal efficiency: 95% Effluent Stream Composition: 8% ammonia and 92% air Gas T and P: 68 F and 1 atm Flowrate: 80 lb-moles/hr Liquid phase: Containing no ammonia
EXAMPLE Determine composition of the liquid at the exit (X 1) (Inlet liquid concentration since pure water is used is x 2=X 2=0) Use equilibrium data for ammonia-air-water mixtures which are given below for 68 F and 14, 7 psia. : X Y 0. 0206 0. 0158 0. 0310 0. 0407 0. 0502 0. 0735 0. 0962 0. 024 0. 0329 0. 0418 0. 0660 0. 0920 In 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 B 4 in the Appendix.
EXAMPLE 0, 90 Y 2=0. 00435 0. 092 Since the liquid rate is to be 30% greater than the miniumu rate (Lm, S)/Gm, C)design = 1, 30(0. 90) = 1. 17 mole/mole Lm, S = Gm, C*1, 17 = 1. 17*73. 6 = 86. 1 lb moles/hr
y, moles solute per mole carrier gas EXAMPLE 0, 1 Y 1=0. 087 0, 08 0, 06 (Lm, S/Gm, C)=1. 17 0, 04 0, 90 0, 02 X 2, Y 2 0. 00435 0 X 1: 0. 0707 0, 02 0, 04 0, 06 0, 08 0, 1 X, moles solute per mole solvent Now, X 1 can now be found. 1. Graphically by drawing operating line with a slope of 1. 17 with starting point of (0, 0. 00435) and the point crosses Y 1=0. 087 can be read. OR 2. From Lms/Gm, C = Y 2 -Y 1/(X 2 -X 1)=0. 00435 -0. 087/(0 -X 1) = 1. 17 X 1 = 0. 0707 lm mole A/lm mole water or x 1 = 0. 066 lb mole A /lb moles solution
FLOW RATES The gas and liquid rates: Lm, 2 GC = 73. 6*29 = 2134 lb/hr Ls GA, 1 = 6. 4*17 = 109 lb/hr x 2 X 2 GA, 2 = 0. 32*(17) = 5. 4 lb/hr LS = 86. 1*18=1550 lb/hr LA, 1 = DGA= 109*0. 95=104 lb/hr Therefore: G 1 = 2134 +109=2243 lb/hr bottom L 1 = 1550 +104 = 1654 lb/hr G 2 = 21345+5=2139 lb/hr top Lm, 1 L 2=1550 + 0 = 1550 lb/hr top Ls x 1 X 1 Gm, 2 Gc y 2 Y 2 T = const P = const Cross-sectional area, A dz Gm, 1 Gc y 1 Y 1
TOWER AREA To 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/ft 3 as the solution density through the tower. For the gas phase assume ideal gas behavior: r= P/RT = Mw. P/RT At the top: Mw= Syi. Mi = 0. 00435*17 + 0. 9957*29 = 28. 95 r= 28. 95*14. 7/(10. 73*528) = 0. 075 lb/ft 3 At the bottom Mw= 0. 08*17 + 0. 92*29 = 28. 04 r= 28. 04*14. 7/(10. 73*528) = 0. 0728 lb/ft 3 Now calculate the abscissa of Flooding Figure
TOWER AREA
TOWER AREA
PRESSURE DROP Pressure drop can be determined from the flooding figure or from an emprical equation
DETERMİNATİON OF AN ABSORPTİON TOWER HEİGHT 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 d. Z. In height d. Z, the rate of mass transfer of species A a: interfacial area available to mass transfer per unit volume of the packing A: cross-sectional area of the tower
TOWER HEİGHT The equation can be also written for liquid resistance part.
TOWER HEİGHT To 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
TOWER HEİGHT To 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
Tower Height The equation can be expressed in terms of height of transfer unit (HTU ) and number of transfer units : HTU or Hoy NTU or Noy HTU is reaonably constant through the absorber and has 55 unit of length. NTU is dimensionless.
x 1, y 1 x. Z, y. Z* Aerosol & Particulate Research Lab x 1, y 1* 2018/3/19 For dilute gas streams, transfer unit equation can be simplified: 56
0. 04% CO 2 Pure amine Lm = 0. 46 gmole/s 2018/3/19 Aerosol & Particulate Research Lab Q: A Packed tower using organic amine at 14 o. C to absorb CO 2. The entering gas contains 1. 27% CO 2 and is in equilibrium with a solution of amine containing 7. 3% mole CO 2. The gas leaves containing 0. 04% CO 2. 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 m 2. KOGa = 9. 34× 10 -6 s-1 atm-1 cm-3. The pressure is 1 atm. Determine the tower height that can achieve this goal. 1. 27% CO 2 Gm = 2. 31 gmole/s C* = 7. 3% CO 2 in amine 57
Absorption of concentrated vapor Mole balance on the controlled volume x 1, y 1 2018/3/19 Liquid flux x. Z, y. Z* Aerosol & Particulate Research Lab Gas flux x 1, y 1* 58
EXAMPLE A 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. 2 x, where x, y = mole fractions. The entering gas mixture flow rate is 800 lbm/hr. The column operates at 30 °C and 1 atm, and Kya = 4. 29 lbmol/hr‐ft 3‐Δy φ
EXAMPLE Calculate or find: a) Concentration of the soluble gas in the effluent liquid if the column is operated at minimum liquid flow rate b) Concentration of soluble gas in the liquid at a point in column where y = 0. 02 c) Height of packed section, Z d) H T oy e) Whether columnis in danger of flooding if it is packed with 1⁄2‐in. ceramic Raschig rings
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