Gas Absorption in Packed Tower (S1 2015) (Note)

CHE3165Separation Processes S1 2015 GAS ABSORPTION IN PACKED TOWERS 1 Leaning Outcomes By the end of the lecture you should be able to: 1. Define rate based method for packed columns 2. Define Height Equivalent to a Theoretical Plate (HETP) and explain how it and the number of equilibrium stages differ with the height of a transfer unit (HTU) and number of transfer units (NTU) 3. Explain the differences between loading point and flooding point in a packed column 4. Estimate the packed height 2 Contents 1 Introduction 2 Mole balance 3 Absorption of very dilute mixtures 4 Transfer unit 5 Absorption of concentrated mixtures 6 Absorption of semi-dilute mixtures 7 Controlling resistance 3 1. Introduction  Absorption: Solute A absorbed from the gas phase into the liquid phase OR a process involves molecular/mass transfer of solute A through a stagnant, non-diffusing gas B into a stagnant liquid C Gas-Liquid System: Solute transfer from Gas  Liquid  Adsorption: Components of a liquid or gas stream adsorbed on the surface or in the pores of a solid adsorbent. Gas/Liquid – Solid System: Solute transfer from Gas/Liquid Solid 4  Stripping: Reverse of absorption Liquid – Gas System: Solute transfer from Liquid Gas 5 Tray and packed bed columns 6 . Vo xi. Lo 7 . V’.yo. V’. Li L = Liquid stream total flow rate V = Gas stream total flow rate L’ = Inert (carrier) liquid flow rate V’ = Inert gas flow rate yA = Mass/mole fraction of A in gas stream xA = Mass/mole fraction of A in liquid stream yi. Vi xo. L’. L’. 9) = 0.27/90.73 = 0. Li yi.27 kgmol/h V0 = 29.01) (1-0.Example 1 (Geankoplis: Example 10. L’. Identify all the unknowns: xi.9) = 0. The total inlet gas flow to the tower is 30.3-2) It is desired to absorb 90% of the acetone in a gas containing 1.01) (0.003 yo.7 + 0.7 kgmol/h L’ = 90 kgmol/h xi = 0 Li = 90 kgmol/h Ac in V0 = 30 (0.0 kg mol/h and the total inlet pure water flow to be used to absorb the acetone is 90kg mol H2O/h. Vo.00101 x0 = 0.03 kgmol/h Ac in L0 = 30 (0. xo. V’.0 mol% acetone in air in a countercurrent stage tower. Vi xo.27 = 0. yo. L’. L’. V’. yi.01) = 29. V’.01 V’ = 30 (1-0.03 kgmol/h = 29. Li.73 kgmol/h L0 = 90 + 0.27 kgmol/h = 90. Vi. Lo 8 .27 kgmol/h y0 = 0.03/29. Lo Solution: Vi = 30 kgmol/h yi = 0. Vo xi. 0 atm pressure abs containing air and CO2 is contacted in a single-stage mixer continuously with pure water at 293 K. Calculate the amounts and compositions of the two outlet phases.Example 2 (Geankoplis: Example 10.3-1) A gas mixture at 1. The inlet gas flow rate is 100 kgmol/h. with a mole fraction of CO2 of yA2 = 0. Assume that water does not vaporize to the gas phase. Solution: L=300 xA2=0 V1? yA1? V=100 yA2=0. The two exit gas and liquid streams reach equilibrium.20.2 V’ =? L1? XA1? 9 . The liquid flow rate entering is 300 kgmol water/h. of packed tower = a x (S x Z) Function of packing is to generate largest possible interfacial area for the smallest possible gas pressure drop. 10 . x1 A = a x vol. x2 Gas Liquid (2) z2=Z V and L = total molar flow rates Packing A – total interfacial area (m2) a – interfacial area per unit volume of packed tower (m2/m3) S – cross sectional area of tower (m) Z – Bed height (m) z1=0 (1) Gas Liquid V1.V2. y1 L1. y2 L2. Liquid can no longer flow Optimum economic: V ≥ 0.5 flooding velocity = f(equipment cost.Packed Tower Operation Flooding velocity: Upper limit to the rate of gas flow. processing variables) Loading point: Gas flow rate where liquid down flow starts to be hindered by gas 11 . ∆P. Packing Materials 12 . Packing Materials 13 . Packing Materials http://finepacstructures.com 14 .tradeindia. Packing Materials Random Structured Raschig rings and saddles “Through flow” Relative cost Low Moderate High Pressure drop Moderate Low Very low Efficiency Moderate High Very high Vapor capacity Fairly high High High 15 . 6-2) A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 293K.m2.3 kPa. The inert air flow rate is 150 kg air/h.m2 and the entering water flow rate is 6000 kg water/h.Example 3 (Geankoplis: Example 10. how many theoretical trays and actual trays are needed? Assume that the tower operates at 293 K. y1 Solution L’ xo V’ yn+1 xn ?? . The entering gas contains 20 mol% SO2 and that leaving 2 mol% at a total pressure of 101. Assuming an overall tray efficiency of 25%. We can choose either gas or liquid phase.2. V’Y = V’(Y+dY) (in at z) + dń (out at z+dz) (transfer to liquid) V’dY = .dń 17 . Mole Balance Mole balance for solute over differential volume of tower: . Write ń in terms of flux and area.dA V d  1.dn '  y    .y) 18 .y) 2  .N.NaS dz (1 .y  V ' dy (1.NaS dz Volume dy V  . and Y in terms of y: V ' dY  . 19 .y) Integrating from (1) to (2) y2 z2  z1 dz   y1  V dy  Z NaS (1.y) This is the most general equation relating total packed height (z) to gasphase variables.Separating variables V dy dz   NaS (1 . If we balance the liquid phase alone. N(y) or L(x).x) To evaluate the integrals. we need V(y). N(x) 20 . an analogous equation is obtained: x2 Z  x1 L dx  NaS (1 . y y  y  * K 'y * *LM Across gas film 21 .For gas phase analysis:   N  K y y  y * or Solute flux N  k y y  y i  or   N  K G PA .PAi  Choose overall Ky as typical case N K 'y y BLM y  y   1.PA* or N  k G PA . y   ln  1.1.y *    1.1.y *LM  1.y *LM varies up the tower 22 .y *   f(y) 1.y  . need to make simplifying assumptions!!! 23 .y2 z  y1 y2 z  V dy  NaS (1 . with y) K’y is a function of flowrate V V’ solvent-only flow and S tower cross section area are the only true constant To further progress.e.y *LM * V' V 1 y  a may vary with V (i.y)2 (y  y * )  y1 y y2    K 'y y  y  N  K y y  y* 1.y *LM (1.y *LM dy   K ' aS(1.y) N  dy (y  y * )   1.y) V  K aS ' y y1 V z S  '  1. y less than 0.e.3.y *LM (1. Linear equilibrium (i.1 or 10%) 2. Absorption of very dilute mixtures Assumptions: 1.y) V varies only slightly from (1) to (2) Use V1  V2 VAVE  2  dy (y  y * ) V = f(y) V' V 1 y  K’ya will be nearly constant because V ~ constant 24 . Very dilute solutions (x. Henry) y2 Start with z V  K aS  y1 ' y 1. y) y2 z 1 V  K aS  y1 z 1 1.y *LM (1 .y *LM ' y VAVE y2  (1.y) .dy K 'y aS y1 (y y ) *  dy (y  y * ) Approximate 25 .(1 – y) ~ 1 (1 – y)*LM ~ 1 1. y *1 ) y 2 .This integral can be evaluated analytically if Henry’s law holds Mass balance (operating line): straight line if V & L are constant from (1) to (2) (y  y*)  f(y) is stright line Say (y . y*  y *1 ) & (y  y 2 .y*)  αy  β d(y .y *2 )  (y 1 . y*  y *2 ) α (y 2 .y1  Δ(y  y*) Δy 26 .y*) α put (y  y1.y*)  α dy dy  d(y . y *1 ) y 2 .y * )  y  y *  2 2 1 1  1 1  y    These terms make up a log mean V  (-1)(y  y ) 2 1 z   'AVE    K aS  (y .d(y .y * )  (y .y *2 y1  y1* α     (y 2 .y*) α(y  y*) V   1   z   'AVE     ln  K aS   α     y  y 2 .z VAVE y2  K 'y aS y1 .y *2 )  (y 1 .y*)LM (1)-(2)  y  27 .y1 *  V   (y  y )(  1) y y 2 1  2 2 z   'AVE   ln  K aS  (y . x) LM (1) -(2)  x  We usually choose the phase which is controlling the rate (i.y i )LM (1)-(2)  y   L AVE  (-1)(x 2  x1 )  z '  k aS  (x .x) LM (1) -(2)  x  i  L AVE  (-1)(x 2  x1 )  z '  K aS  (x * .Similar expressions hold for the various forms of N V  (-1)(y  y ) AVE 2 1  z '  k aS  (y .e. the phase with the greatest mass transfer resistance) 28 . x 2   k 'x az(xi  x)M S V y1 .y 2   K 'y az(y  y*)M S L x1 .Re-arrange V y1 .x 2   K 'x az(x * -x)M S 29 .y 2   k 'y az(y  y i )M S L x1 . x2 Dilute solution yi2 y*2 xi1 xi2 x 30 .How to find mole fraction at interface and equilibrium ??? y y1.x1 yi1 y*1 y2. 16 x 10-2 kgmol/s.186 m2 at 293K and 101. The inlet air contains 2. (c) Calculate K’ya and the tower height.mol frac.x 2   K 'x az(x * -x)M S 31 .78 x 10-2 kgmol/s.m3.32 kPa. Solution V y1 .6-4) Acetone is being absorbed by water in a packed tower having a cross-sectional area of 0.5%.m3. (a) Calculate the tower height using k’ya. (b) Repeat using k’xa.6 mol% acetone and outlet 0.y 2   k 'y az(y  y i )M S L x1 .y 2   K 'y az(y  y*)M S L x1 .x 2   k 'x az(xi  x)M S V y1 . The gas flow is 13. Pure water inlet flow is 45.mol frac and k’xa = 6. Film coefficients for the given flows in the tower at k’ya = 3.36 kg mol water/h.Example 4 (Geankoplis: Example 10.65 kgmol inert air/h. Transfer Units Z = Hy Ny z (m) VAVE k 'y aS y2  y1 . V. packing type and size) Z = Hox Nox Depending on concentration units and coefficients used Z = HG NG = H L NL Z = HOG NOG = HOL NOL HG ↔ Hy.e.dy (y  y i ) Z = Hx Nx If OVERALL coefficients are used to describe the flux: Z = Hoy Noy (m) HEIGHT of a gas film transfer unit Dimensionless Number of a gas film transfer unit Heights of transfer units have been correlated with operating variables (i. HOG ↔ Hoy. L.4. NG ↔ Ny. NOG ↔ Noy 32 . x)iM S HOG HOL V V  '  K y aS K y a(1.x)M L HL  '  k x aS k x a(1.y i )M L x1 .y)iM S y1 .V V HG  '  k y aS k y a(1.y)*M S L L  '  K x aS K x a(1.x 2 NL  (x i .x 2  (x * -y)M 33 .y*)M NOL x1 .y 2 NG  (y .x)*M S NOG y1 .y 2  (y . 34 . Example 5 (Geankoplis: Example 10.6-5) Repeat Example 5 using transfer units and height of a transfer unit as follows: (a) Use HG and NG to calculate tower height (b) Use HOG and NOG to calculate tower height Solution 35 . 5 S S  L HL  θ  SμL η   Sc 0.8b) 36 .Correlation form: β γ V L HG  α    Sc 0. η for different packings and sizes (Geankoplis 10.5  Constants α. θ. γ. β. correlation for ½ inch Raschig rings with NH3/H2O 37 .43 on log-log graph eg.V Hy α  S β Slope β ~ 0. gas pressure drop . ~1/2 flooding 38 .Absorption Column Size Height – Mass transfer (Hy Ny) Diameter – Hydraulics – Two phase flow over packing.Design (Flooding and loading) Difficult to locate precisely (deviation from standard line) Loading – curve nearly vertical Operate close to loading. 39 . y1/m   N ln (1/A) A1  L1/m1V1.mx2   (1  1/A)  1/A ln   y 2 .Analytical equation to calculate theoretical number of trays For transfer of the solute from phase V to phase L (absorption)  y1 . A 2  L 2 /m2 V2 and A  A1A 2 40 .y1/m   (1  A)  A  ln   x1 .mx2   N ln A For transfer of the solute from phase L to phase V (stripping)  x 2 . mx 2  1   1/A  ln (1  1/A) 1.y1/m  1   A   ln (1  A) 1.mx2  Stripping NOL    x 2 .1/A    y 2 .y1/m  41 .A     x1 .Analytical equation to calculate packed bed height Operating and/or equilibrium lines are slightly curved Absorption NOG    y1 . 6-5 42 .A /A z = HOG x NOG z = N x HETP See Example 10. in m is related to HOG by HETP  HOG ln(1/A) 1. HETP.Analytical equation to calculate packed bed height When the operating and equilibrium lines are straight NOG N lnA  1.1/A The height of a theoretical tray or stage. Example 6: Experimental data have been obtained for air containing 1.5m2 cross-sectional area and 3. Entering gas and liquid flow rates are 0.6% of SO2 being scrubbed by pure water in a packed bed column of 1.004. m = 40. Solutions 43 . respectively.2 kmol/s. Calculate NOG.5 m in packed height. If the outlet mole fraction of SO2 in the gas is 0.062 and 2. Then.y) (y  y i ) y1 y  Approximate z VAVE y2    V  1.y) (y  y i )  y1 y y2  The only constant .y  dy iLM   z   '  k aS  (1 . 44 .5. Absorption of concentrated mixtures Dilute mixtures y2 Concentrated mixtures 1 1.y iLM V dy z   ' k aS (1 . evaluate the integral numerically (or graphically) in order to find yi between y1 and y2.dy K 'y aS y1 (y  y * ) Each term (or bracketed terms) needs to be calculated for a range of values of y between y1 and y2. We need the whole operating curve from (1) to (2). 45 . This is obtained from a series of mass balances between (1) and a range of points up to (2). The * values are much easier to get since ky need not be known.Sherwood suggests that (1-y)*LM and (1-y*) can often replace the true interface values if the liquid resistance is small. 46 . 47 . The inert air flow is 6.0594Gy0.53 x 10-4 kg mol air/s and the inert water flow is 4. The entering gas contains 20 mol% SO2 and that leaving 2 mol%.7Gx0.013 x 105Pa abs pressure. the film mass-transfer coefficients at 293 K are.4 mm rings: k’ya = 0. for 25.Example 6 (Geankoplis: Example 10. For dilute SO2.7-1) A tower packed with 25. Calculate the tower height.82 Gx and Gy are kg total liquid/gas per sec per m2 tower cross section.152Gx0.20 x 10-2 kg mol water/s.4 mm ceramic rings is to be designed to absorb SO2 from air by using pure water at 293K and 1. Solution 48 .0929 m2. The tower cross-sectional area is 0.25 k’xa = 0. y  (-dy) AVE  *LM  z    K ' aS  (1. If yes. Absorption of semi-dilute mixtures Dilute enough to use VAVE or LAVE and constant K’ya or K’xa   V  1.0.y  dy *LM  z  '   K aS  (1. need only evaluate Which is relatively simple (numerical or graphical) y2  y1 dy yy* 49 .y) (y  y*)  y1 y y2  V  y2 1.y) (y  y*)  y  y1  Always check this for ~ 1.6. 7. Controlling resistance i) Liquid film controlling (eg. gas almost insoluble – O2/H2O) 50 . 7. gas very soluble or react with liquid – NH3/H2O) 51 . Controlling resistance ii) Gas film controlling (eg.