Mass transfer and separation technology Massöverföring och separationsteknik ( MÖF-ST ) , 7 sp

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1 Mass transfer and separation technology Massöverföring och separationsteknik ( MÖF-ST ) , 7 sp 9. Separation processes - intro Ron Zevenhoven Åbo Akademi University Thermal and Flow Engineering Laboratory tel ; ron.zevenhoven@abo.fi 9.1 Continuous heat and mass transfer processes 2

2 Transfer processes Process Transfer Driving force Apparatus Heat exchange Energy ΔT Heat exchanger Gas absorption Mass G L Δc, y-y* Packed tower, or tray column Gas desorption Mass L G Δc, y*-y Packed tower, or tray column Distillation Vaporisation cooling Extraction Mass, two components A and B from a mix Energy, water Mass from liquid I to liquid II y* A -y A, y B -y B * Δh (enthalpy) x I x II * Packed tower, or tray column Spray tower Extractor Drying Water, energy y H2O -y H2O *, ΔT Dryer 3/36 Describing processes inside mass transfer equipment Four important factors: 1. In- and outflow of phases batchwise continuously 2. The throughput and mixing 3. The interface 4. The transport process 4/36

3 Phase contacting and interfaces Plug flow Perfectly mixed By-passing A real (left) and modelled (right) interface 5/36 General transport equation /1 In general, a transport process involves (at least) two streams a and b, with different composition, temperature and/or velocity. A stream Φ of material or energy is transferred across an interface A as a result of a driving force (Y a -Y b ). In general: transport = transport coefficient interface driving force or: Φ = k A (Y a -Y b ) 6/36

4 General transport equation /2 It is often not possible to know or determine the interface area A (m 2 per m 3 volume), although it may be possible to determine the product k A (1/s) experimentally. Inside process equipment the driving force (Y a -Y b ) will vary with position; the transfer Φ is then found by integrating over the contact area A: dφ k Y Y A Φ total k a dφ Y Y a b da b 7/36 Models for continuous mass transfer Parallel flow Counter-flow Internal recirculation These allow for simple relations between (Y a -Y b ) and Φ 8/36

5 General transport equation /3 The relation between (Y a -Y b ) and Φ is not necessarily a straight line, and also the mass transfer coefficient k may vary with position inside transfer process equipment. If it can be assumed that k constant, a linearisation may be used, with ΔY 0 = (Y a -Y b ) at the inlet and ΔY 1 = (Y a -Y b ) at the outlet : With this gives Y a dφ k Y Y b a Y Y k A b 0 da Y ; total Y Y A 1 Y Φ total 1 0 total Y lm Y 1 ln Y 0 k A 0 k dφ Y Y a b with lm short for logarithmic mean 9/36 Task ö4.4 The question in Swedish 10/36

6 Task ö 4.4 A = aceton, transfer air (G) water (L) Index: 0 = column bottom, 1 = column top Input data: Air: ṅ G = 20 mol/s, water: ṅ L = 70 mol/s Fractions y A0 = 1.8 %, y A1 = 0.2 %, x A1 = 0.05% Transferred A, total: ṅ A,total = ṅ G (y A0 y A1 ) = 20 mol/s (1.8%-0.2%) = 0.32 mol/s (note that ṅ G decreases with an amount ṅ A,total!) Mole fraction x A0 : ṅ A,total = ṅ L (x A0 x A1 ) x A0 = x A1 + ṅ A,total / ṅ L = 0.05 % /70 = 0.05 % % = % 11/36 Task ö 4.4 (Fig. ö 4.4) Balance for aceton ṅ A ṅ G (y A0 -y A ) = ṅ L (x A0 -x A ) gives y A = y A0 - ṅ A /ṅ G = 1.8% - (ṅ A /20) 100% x A = x A0 - ṅ A /ṅ L = % - (ṅ A /70) 100% and y* A = 2.53 x A, x A *= y A /2.53 Note: more exact mass balance ṅ G y A0 + ṅ L x A1 = (ṅ G -ṅ A ) y A1 + (ṅ L + ṅ A ) x A0 gives ṅ A = [ṅ G (y A0 -y A1 ) - ṅ L (x A0 -x A1 )] / (y A1 x A0 ) 12/36

7 Task ö4.4 (Fig ö4.6) x,y - diagram 13/36 Task ö4.5 The question in Swedish 14/36

8 Task ö 4.5 Mass stream ṅ A = k y A.Δy A,ln = k x A. Δx A,ln = mol/s y A0 = 1.8 % x* A0 = % Δx A0 = % y A1 = 0.20 % x* A1 = % Δx A1 = % Δx A,ln = % x A1 = 0.05 % y* A1 = % Δy A1 = % x A0 = % y* A0 = % Δy A0 = % Δy A,ln = % k y A= ṅ A / Δy A,ln = mol/s k x A= ṅ A / Δx A,ln = mol/s Note: k x A/ k y A = / Separation of mixtures; equilibrium stages 16

9 Number of equilibrium stages: graphical method 1 transfer 1 transfer unit unit 1 transfer unit 17/36 Separation of mixtures: stages /1 In a system of two or more phases, relative amounts of different species are different. For example, water A and air B: for the gas phase y B /y A»1, for the liquid phase x B /x A «1. The equilibrium between two (or more) phases (gas/liquid, liquid/liquid,...) can be changed by changing temperature, total pressure or total volume (i.e. changing the total energy) adding more of one of the species, or introducing a new species (i.e. changing the total mass) 18/36

10 Separation of mixtures: stages /2 For example: separating phenol from water (L) by adding benzene (V) in a separation funnel. x = phenol conc. in L, y = phenol conc. in V 19/36 Separation of mixtures: stages /3 The ratio phenol transferred / phenol not-transferred = Vy/Lx = KV/L; can be referred to as S = KV/L separation factor, for mass w : Vy/Lx = Sw / w = S The fraction f not transferred from L to V, or not removed from L: S = (1-f) / f f = 1/(1+S) S can be changed by varying V (or L), or changing K At ambient conditions (1 bar, 20 C) the distribution coefficient for phenol across the phases equals K = y/x = 2 if L=V then S = 2 Phenol not removed = 1/3 = 33.33% 20/36

11 Separation of mixtures: stages /4 Adding the benzene support phase to the water/phenol in two steps S = 1 Phenol not removed = w/4w = 25% 21/36 Separation of mixtures: stages /5 Mixing the water/phenol and the benzene support phase in two steps S = 2 Phenol not removed = 3.33w / 18w = 19% 22/36

12 Cascade configurations for contacting stages Picture: SH06 23/36 Cu - ore 2,0 2,2 release from ore Diluted H 2 SO 4 extraction Organic solvent recovery electrolysis 0,1 0,2 50 Concentrated H 2 SO 4 40 Cu Numbers are concentrations kg Cu/m 3 Separation and equilibrium stages Example Four-stage processing for Cu recovery from Cucontaining ore (liquid / liquid) 24/36

13 Phase equilibrium stages /1 sv: raffinat feed refine extraction loaded solvent solvent The extraction and the recovery process are combinations of mixing tanks and settling tanks 25/36 Phase equilibrium stages /2 feed loaded solvent Cu is transferred from feed stream L to solvent phase ( support phase ) V; the process can be described as a series of equilibrium stages. L n+1 x n+1 extraction refine solvent L n x n equilibrium stage n+1 equilibrium stage n equilibrium stage n-1 V n y n V n-1 y n-1 26/36

14 Phase equilibrium stages /3 L n+1 x n+1 equilibrium stage n L n x n V n y n V n-1 y n-1 The equilibrium constant for Cu in the feed stream and in the solvent stream is K =y Cu /x Cu Thus, for an equilibrium stage n: y n = Kx n Streams L and V are often practically constant L 1 L 2.. L n = L; V 1 V 2... V n = V 27/36 Phase equilibrium stages /4 Lx n+1 equilibrium stage n Lx n Vy n Vy n-1 more accurately K V yn 1 S (1 ) L y while the objective is that The ratio Cu transferred / Cu not-transferred = Vy n /Lx n = KV/L = S = KV/L separation factor y n1 y n n or stripping factor A = L/(KV) 28/36

15 Phase equilibrium stages /5 Lx n+1 equilibrium stage n Lx n Vy n Vy n-1 more accurately K V yn 1 S (1 ) L y while the objective is that The fraction f not transferred from L to V, or not removed from L: S = (1/f)-1 f = 1/(1+S) for this 1-stage separation S can be changed by varying V (or L), or K = f(t) y n1 y n n or stripping factor A = L / (KV) 29/36 Phase equilibrium stages /6 For example: phase equilibrium constant K=5; L = 1000 kg/s, V = 800 kg/s S = K V/L= 4. Cu in feed x 0 = 0.2 %-wt = kg/kg For clean solvent, y 2 =0, then V y 1 / L x 1 = S Mass balance gives L x 0 = (S+1) w-r 30/36

16 Phase equilibrium stages /7 Calculation procedure Number of equilibrium stages is determined by the result w in the refine stream, and K, V and L i.e. the separation factor, stripping factor S Note: S= KV/L if V is support phase, stripping factor A = L/KV if L is support phase, absorption factor 31/ Kremser method 32

17 Phase equilibrium stages /8 Kremser For N-stage counter-flow separation with constant separation factor S, (S or A) Kremser showed that the not-separated fraction f equals S f N S N and S -1 ln f N ln S S large : f ~ N S S small : f ~ 1 - S 1 S 1 : f 1 N Fraction of not separated material, f, versus separation factor S and number of equilibrium stages N See also ÖS97 33/36 Phase equilibrium stages /9 Kremser Kremser s method can be applied to absorption (or adsorption) and desorption, using separation factor A or S depending on the phase that is added (L or V). Fraction of solute absorbed (gas liquid): N 1 yin yout A A fraction absorbed 1 fa N 1 y Kx A 1 Fraction of solute stripped (liquid gas) N 1 xin xout S S fracti on stripped 1 fs y N 1 in x S 1 in K in in 34

18 Phase equilibrium stages /10 Kremser For separation factor S = L/(KV) ~ 1, (or S = A = KV/L ~ 1), which can be expressed as S = 1 + α, with α <<1 : Kremsers equation gives S 1 2 N 1 f 1 ( N 1) for given N and f. 35 Sources #9 SH06 J.D. Seader, E.J Henley Separation process principles John Wiley, 2nd edition (2006) T68 R.E. Treybal Mass transfer operations McGraw-Hill 2nd edition (1968) WK92 J.A. Wesselingh, H.H. Kleizen Separation processes (in Dutch: Scheidingsprocessen) Delft University Press (1992) ÖS97 G. Öhman, H. Saxén Transportprocesser Åbo Akademi Värmeteknik (1997) 3.6 Ö96 G. Öhman Massöverföring Åbo Akademi Värmeteknik (1996) /36

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