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A BSORPTION Bringing the dirty effluent gas into contact with the scrubbing liquid and subsequently separating the cleaned gas from the contaminated liquid.

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Presentation on theme: "A BSORPTION Bringing the dirty effluent gas into contact with the scrubbing liquid and subsequently separating the cleaned gas from the contaminated liquid."— Presentation transcript:

1 A BSORPTION 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

2 A BSORPTION _Wet scrubbers can be categorized into 3 groups: 1. Packed-bed counterflow scrubbers 2. Cross-flow scrubbers 3. Bubble plate and tray scrubbers

3 S CRUBBER T YPES Packed Tower Spray towerVenturi Absorber

4 C ONCEPT OF A BSORPTION 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.

5 L IQUID W ASTE 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.

6 A BSORPTION 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

7 P ACKED T OWER

8 P ACKING M ATERIAL AND S HAPES 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.

9 P ACKING M ATERIAL AND S HAPES

10 P ACKING MATERIAL PROPERTIES

11 A BSORPTION T HEORY 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

12 A BSORPTION T HEORY

13 On the basis of Fick's Law, the diffusion of one gas (A) through a second stagnant gas B, N A, the molar rate of transfer of A per unit cross-sectional area is given by; N A = -D AB (dc A /dz)/(1-(c A /c) D AB : 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) D AB tables are available for a number of binary gas mixtures

14 A BSORPTION T HEORY Mass transfer rate per unit area for molecular diffusion of A through a second liquid is given by: N A = -DL/z (c A2 -c A1 ) D L : liquid phase molecular diffusion coef. (m 2 /t) c A2 -c A1 : concentration difference of A over the distance z Typical values of D L for binary mixtures are tabulated in the literature.

15 T HE E QUILIBRIUM D ISTRIBUTION C URVE 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 Inert Liquid Solvent Inert carrier gas Inject solute A P=c Mole fraction in gas, yA Mole fraction in lqiuid, xA Exp. Equilib. Distribution curve

16 T HE E QUILIBRIUM D ISTRIBUTION C URVE 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 Inert Liquid Solvent Inert carrier gas Inject solute A P=c Mole fraction in gas, yA Mole fraction in lqiuid, xA Exp. Equilib. Distribution curve

17 M ASS TRANSFER COEFFICIENTS BASED ON I NTERFACIAL C ONCENTRATIONS 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 N A 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 M ASS TRANSFER COEFFICIENTS BASED ON I NTERFACIAL C ONCENTRATIONS For the liquid phase: N A = k L (c Ai -c AL ) = k x (x Ai -x AL ) k L (is the liquid mass transfer coeff. Based on concentration, in length per unit of time, c Ai is the concentration of A in the liquid phase at the interface, c AL is the concentration of A in the bulk of the phase, in moles per unit volume. k x 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

19 M ASS TRANSFER COEFFICIENTS BASED ON I NTERFACIAL C ONCENTRATIONS For the gas phase: N A = k G (p AG -c Ai ) = k y (y AG -y Ai ) k G is the gas phase mass transfer coeff. based on partial pressures, in moles/length 2 time, p AG is the partial pressure of A in the bulk of gas phase p Ai is the partial pressure of A in the gas interface k y 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

20 M ASS TRANSFER COEFFICIENTS BASED ON I NTERFACIAL C ONCENTRATIONS However this approach to determining N A is not practical since k x and k y 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

21 O VERALL M ASS T RANSFER C OEFFICIENTS When mass transfer rates are reasonably low, N A can be expressed as: N A = K G (p AG -p A *) = K y (y AG -y A *) K G and K y 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

22 O VERALL M ASS T RANSFER C OEFFICIENTS slope=m' 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 The distance between P and C is a measure of the driving force.

23 O VERALL M ASS T RANSFER C OEFFICIENTS N A = K G (p AG -p A *) = K y (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: N A = K L (c A *-c AL ) = K x (x A *-x AL )

24 O VERALL M ASS T RANSFER C OEFFICIENTS It is important to note that the quantities 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 two- phase 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/K y =1/k y +m'/k x

25 M ASS B ALANCES AND THE O PERATING L INE FOR P ACKED T OWERS T = const P = const Cross-sectional area, A G m,2 G c y 2 Y 2 L m,2 L s x 2 X 2 G m,1 G c y 1 Y 1 L m,1 L s x 1 X 1 dz G m molar total gas flow rate (carrier gas + pollutant) Gc molar inert carrier gas flow rate L m 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

26 M ASS B ALANCES AND THE O PERATING L INE FOR P ACKED T OWERS 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: G m,1 y 1 + L m,2 x 2 = G m,2 y 2 + L m,1 x 1 or G m,1 y 1 -G m,2 y 2 = L m,1 x 1 -L m,2 x 2

27 M ASS B ALANCES AND THE O PERATING L INE FOR P ACKED T OWERS In G m,1 y 1 -G m,2 y 2 = L m,1 x 1 -L m,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: G C,m (Y 2 -Y 1 ) = L S,m (X 2 -X 1 ) These two equations above gives a straight line on Y-X coordinates with a slope of L sm /G Cm 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 phase Dirty air Dirty waterClean water Clean air

29 T HE M INIMUM AND D ESIGN L IQUID - G AS R ATIO At the bottom and top of the absorber, parameters G m,1, G c, y 1, G m2, y 2, and x 2 are known. We need to determine L s, 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?

30 T HE M INIMUM AND D ESIGN L IQUID - G AS R ATIO 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 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.

31 T OWER D IAMETER AND P RESSURE D ROP PER U NIT T OWER H EIGHT 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.

32 P ROBLEMS WITH H IGH G AS V ELOCITY 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

33 F LOOD P OINT The relationship between  P/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.

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35 L’/G’√(  G /  L -  G ) L’: liquid mass flux (lb/s-ft 2 ) G’:gas mass flux (lb/s- ft 2 ) F:packing factor (ft 2 /ft 3 )  L :liquid viscosity, cp g c : proportıonality constant, ft- lb/s 2 -lb f  L :liquid density, lb/ft 3  G :gas density, lb/ft 3 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 G x and Gy

36 P ACKING F ACTOR 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:  P flood = 0.115F 0.7

37 P ACKING D ATA

38 D ETERMINING T OWER D IAMETER 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(  L ) 0.1 /g c (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

39 D ETERMINING E XPECTED P RESSURE D ROP PER U NIT H EIGHT OF T OWER 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  P in packing when operating below the load point is  P/Z = m[10 nL’/  L ](G’ 2 /  G ) m and n are packing constants see Table 6.2

40 D ETERMINING T OWER D IAMETER AND E XPECTED P RESSURE D ROP PER U NIT H EIGHT OF T OWER

41 E XAMPLE 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.

42 E XAMPLE Removal efficiency: 95% Effluent Stream Composition: 8% ammonia and 92% air Gas T and P: 68F and 1 atm Flowrate: 80 lb-moles/hr Liquid phase: Containing no ammonia

43 E XAMPLE X Y 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. : 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 B4 in the Appendix.

44 E XAMPLE Since the liquid rate is to be 30% greater than the miniumu rate (L m,S )/G m,C ) design = 1,30(0.90) = 1.17 mole/mole L m,S = G m,C *1,17 = 1.17*73.6 = 86.1 lb moles/hr 0, Y2=

45 E XAMPLE Now, X1 can now be found. 1.Graphically by drawing operating line with a slope of 1.17 with starting point of (0, ) and the point crosses Y 1 =0.087 can be read. OR 2.From L ms /G m,C = Y 2 -Y 1 /(X 2 -X 1 )= /(0-X 1 ) = 1.17 X 1 = lm mole A/lm mole water or x 1 = lb mole A /lb moles solution 0, ,02 0,04 0,06 0,08 0,1 00,020,040,060,080,1 X, moles solute per mole solvent y, moles solute per mole carrier gas X2,Y2 (Lm,S/Gm,C)=1.17 Y1=0.087 X1:

46 F LOW R ATES The gas and liquid rates: G C = 73.6*29 = 2134 lb/hr G A,1 = 6.4*17 = 109 lb/hr G A,2 = 0.32*(17) = 5.4 lb/hr L S = 86.1*18=1550 lb/hr L A,1 =  GA= 109*0.95=104 lb/hr Therefore: G 1 = =2243 lb/hr bottom L 1 = = 1654 lb/hr G 2 = =2139 lb/hr top L 2 = = 1550 lb/hr top T = const P = const Cross-sectional area, A G m,2 G c y 2 Y 2 L m,2 L s x 2 X 2 G m,1 G c y 1 Y 1 L m,1 L s x 1 X 1 dz

47 T OWER A REA 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/ft3 as the solution density through the tower. For the gas phase assume ideal gas behavior:  = P/RT = M w P/RT At the top: M w =  y i M i = * *29 =  = 28.95*14.7/(10.73*528) = lb/ft 3 At the bottom Mw= 0.08* *29 =  = 28.04*14.7/(10.73*528) = lb/ft 3 Now calculate the abscissa of Flooding Figure

48 T OWER A REA

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50 P RESSURE D ROP Pressure drop can be determined from the flooding figure or from an emprical equation

51 D ETERMINATION OF AN A BSORPTION T OWER H EIGHT 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 A a: interfacial area available to mass transfer per unit volume of the packing A: cross-sectional area of the tower

52 T OWER H EIGHT The equation can be also written for liquid resistance part.

53 T OWER H EIGHT 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

54 T OWER H EIGHT To solve the above equation we can determine the overall value of K y a (K g a) 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 The 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. 55 Tower Height HTU or Hoy NTU or Noy

56 2015/5/5 Aerosol & Particulate Research Lab 56 x 1, y 1 * x 1, y 1 x Z, y Z * x Z, y Z For dilute gas streams, transfer unit equation can be simplified:

57 2015/5/5 Aerosol & Particulate Research Lab 57 Pure amine L m = 0.46 gmole/s 0.04% CO % CO 2 G m = 2.31 gmole/s C* = 7.3% CO 2 in amine 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. K OG a = 9.34×10 -6 s -1 atm -1 cm -3. The pressure is 1 atm. Determine the tower height that can achieve this goal.

58 2015/5/5 Aerosol & Particulate Research Lab 58 Absorption of concentrated vapor Mole balance on the controlled volume Gas flux Liquid flux x 1, y 1 x 1, y 1 * x Z, y Z * x Z, y Z

59 E XAMPLE 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 = 1.2x, where x, y = mole fractions. The entering gas mixture flow rate is 800 lbm/hr. The column operates at 30 °C and 1 atm, and Kya = 4.29 lbmol/hr ‐ ft3 ‐ Δy 

60 E XAMPLE 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 T d) H oy e) Whether columnis in danger of flooding if it is packed with 1⁄2 ‐ in. ceramic Raschig rings

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