Diffusion, convection; Mass transfer coefficient

Transcription

1 Processteknikens grunder ( PTG ) Introduction to Process Engineering v /56 7. Short introductions to: Mass transfer; Separation processes; Particulate technology & multi-phase flow Ron Zevenhoven Åbo Akademi University Heat Engineering Laboratory / Värmeteknik tel ; ron.zevenhoven@abo.fi 2/ Mass transfer processes; Diffusion, convection; Mass transfer coefficient

2 Mass transfer process examples Drying of a solid using heat or dry gas Adsorption of gas or liquid using a solid Distillation using heat to create a vapour phase Gas absorption using a liquid absorbent Extraction of liquid or solid using a solvent that creates immiscible phases Crystallisation using cooling to create a solid phase Note the support phase : solvent, sorbent,, heat,... 3/56 Picture: Mass transfer mechanisms /1 Diffusion: Fick s Law* (for binary systems) Spreading of a substance from a region with high concentration to regions with lower concentration; more correct for concentration concentration is chemical potential Molecular diffusion coefficient Đ; for species A in medium B Đ = Đ AB c diffusion x m& = dm A dt surface dc = D dx time * Note analogy with Fourier s Law for heat conduction c x 4/56

3 Mass transfer mechanisms /2 Diffusion + (forced( or free) convection Flow as a result of a pressure difference, gravity,,... If needed, Đ = Đ mol + Đ turb can be used to include turbulent eddy diffusion c flow, v diffusion x dc m& D + v c dx v = dx / dt time c flow, v x 5/56 Transfer of matter as a result of a concentration (or density) difference, more accurately chemical potential difference ( gradient ) Main cause: Brownian motion of molecules Diffusion 6/56 Mass flux Φ ma,x (kg / s per m 2 ) through a surface (x 0 ) perpen-dicular to the transport direction (x) as a result of a gradient in mass concentration (ρ( A ). More conventional for gases: use concentration c A (mol/m 3 )=ρ A /M A =p A /RT Φ " mola, x = - D A dc A dx Picture: SSJ84

4 Diffusion coefficients (some values) A B D AB m 2 /s Temp C water CO water N water O air NH air CO air H 2 O vapour Ambient conditions: Đ ~ 10-5 m 2 /s in gases Đ ~ m 2 /s in liquids Đ ~ m 2 /s in solids 7/56 (Ö96 p. 40) Diffusion and convection An example: Flow of a fluid from a tube into a region where c = c 0 for a certain species convection, v Steady state, mass balance gives dc 0 = D A v A c ; c = x = 0 dx vx vx c = c 0. exp - with = Pe (Péclet number) D D vx ρvx η η Pe = = = Re Sc with Sc = (Schmidt D η ρ D ρ D x c 0 A m 2 diffusion x c 0 8/56 number) c

5 Mass transfer and boundary For mass transfer between two phases A and B, concentration gradients usually only exist near the physical boundary that separates A and B Thus,, the driving forces are active only in boundary layers at the separating surface boundary layers c A,bulkA A boundary layer medium A boundary layer medium B Equilibrium between c A,interfaceA and c A,interfaceB Φ mola phase boundary 9/56 c A,interfaceA B c A,interfaceB c A,bulkB T 1 Mass heat transfer analogy /1 Phase Phase 1 2 T i T 2 A temperature profile,, heat transfer µ 1 c 1 µ i µ 2 A chemical potential profile, mass transfer c 1,i c 2,i c 2 c 1 c 1,i c 2,i Concentration profiles, mass transfer /56 c 2

6 x C 1 Mass transfer coefficient interface a 1 (L) 2 (G) x i C 1.i Φ A y i C 2.i L,G example y C 2 11/56 Mass flow species A: ṅ A = Φ A mol/s Mass transfer rate per area: Ṅ A = ṅ A /a = Φ A mol/(m2 s) Mass transfer coefficients,, k, (unit:: m/s) for both sides of the interface: Ṅ A = k x (c 1,i -c 1 ) = k y (c 2 -c 2,i ) Interface concentrations can be eliminated using equilibrium constant K = c 1,i /c 2,i = c 1 */c 2 = c 1 /c 2 * c 1 * = c 1 at equilibrium with c 2, etc. The film model A mass transfer coefficient can be linked to the film model: Φ A,mol = k (ck A0 -c A1 ) = - Đ A dc A /dy = Đ A (c which gives k = Đ A /δ c (c A0 -c A1 )/ )/δ c Thus,, the boundary layer thickness can be estimated if k and Đ A are known. The mass transfer limitations are concentrated in a well-defined region. 12/56 Picture: SSJ84 See heat transfer Nusselt number Nu = hd/λ = D/δ T mass transfer Sherwood number Sh = kd/đ = D/δ c

7 Mass heat transfer analogy /2 Heat transfer Nu = f(re,, Pr, L/D, Gr,..) Convection around a sphere: Nu = Re ½ Pr ⅓ Transfer from a wall and a turbulent flow: : 2000 < Re < 10 5 and Pr > 0.7 Nu = 0.027Re 0.8 Pr 0.33 (η/η wall ) 1/7 General: Nu = CRe m Pr n, where m = , n /56 Mass transfer Sh = f(re, Sc,, L/D, Gr,..) Convection around a sphere: Sh = Re ½ Sc ⅓ Transfer from a wall wall and a turbulent flow: : 2000 < Re < 10 5 and Sc > 0.7 Sh = 0.027Re 0.8 Sc 0.33 General: Sh = CRe m Sc n, where m = , n 0.33 Chilton-Colburn Colburn analogies,, heat and mass transfer values j H, j D : j H = NuRe - 1 Pr -⅓ j D = ShRe - 1 Sc -⅓ j H = j D = CRe m-1 = ½ƒ ƒ = Fanning friction factor for pipe flow 14/ Phase equilibrium (gas-gas, gas-liquid liquid, liquid-liquid liquid) Henry s Law, Raoult s Law

8 Mass transfer and equilibrium wet gas Drying of wet gas in an glycol absorber c H2O dry gas c H2O,eq c H2O in liq glycol Equilibrium determined by thermodynamics Rate determined by transport processes and equipment design time 15/56 Air above a lake 16/56 z at equilibrium : K and: some air dissolves in the lake! x, y, T H2 O = y H2O << 1 y x H2O H2O Concentration jump Equilibrium at surface x H2O 1 x = fraction in liquid,, y = fraction in gas, z = position coordinate,, T = temperature y O y N T x O2 << 0.01 x N2 << 0.01

9 Gas-liquid phase equilibrium: Raoult s law Assume a liquid mixture of components A, B, C,... at a temperature T. A occupies a large fraction, say x A > 5 % At temperature T, saturation pressure of pure substance A (i.e. A vapour above liquid A) is p 0 A For the mixture,, the vapour pressure of A equals p A = y A.p tot = x A.p 0 A For a two-component mixture of A and B: p B =x B.p 0 B = (1-x A ).p 0 B, using x A +x B =1 17/56 Picture: Gas-liquid phase equilibrium: Henry s law Assume a liquid mixture of components A, B, C,... at a temperature T. A occupies a small fraction, say x A < 5 % For the mixture,, the vapour pressure of A equals p A = y A.p tot = H ca.x A with Henry constant H c (unit:: Pa, bar,...) y A /x A = H ca /p tot = β, y A = β.x A distribution coefficient β (mol/mol) H c is a function of temperature, but independent of pressure at p tot < 5 bar. 18/56 Picture:

10 Example: Water T = 40 C P tot =? GAS LIQUID x H2O = 0.3 mol/mol x NH3 = 0.7 mol/mol Water-ammoniaammonia vapour /1 19/56 A mixture of ammonia and water 40 C, total pressure unknown Liquid composition: 70 %-mol NH %-mol H 2 O What is the composition of the gas, y NH3, y H2O? At equilibrium,, no driving forces or temperature gradients, x NH3 + x H2O = 1, y NH3 + y H2O = 1 but x NH3 y NH3, x H2O y H2O!!! Relative volatility of NH 3 with respect to water: α = (y NH3 / x NH3 ) / (y H2O / x H2O ) = K NH3 / K H2O at equilibrium Example: Water Water-ammoniaammonia vapour /2 20/56 Gas above an NH 3 /H 2 O liquid mixture with x NH3 = 0.7 and x H2O = 0.3, T = 40 C. Questions: pressure & equilibrium composition of gas? From tabelised data for saturation pressures, at 40 C : p H2O = kpa; ; p NH3 = kpa x-values for liquid >> 5 % : use Raoult s Law : p H2O = x H2O p H2O = kpa = 2.22 kpa p NH3 = x NH3 p NH3 = kpa = kpa P total = p H2O + p NH3 = kpa = bar y H2O = 2.22 kpa / kpa = = 0.2 %-v y NH3 = kpa / kpa = = 99.8 %-v Source: ÇB98

11 Example: Water in air above a lake Assume a lake with T=17 C, p tot = 92 kpa at water surface level. At the surface,, the water will be saturated with water, which means p H2O = p H2O = 1920 Pa. Fraction of water in the air at the water surface at equilbrium with the air above is then y H2O = p H2O /p tot = 1.92 kpa / 92 kpa = 2.09 % note: % = %-v (volume %) 21/56 Source: ÇB98 Example: Air Air dissolved in lake water 22/56 Assume the same lake, T=17 C, p tot = 92 kpa at water surface level.. At the surface p H2O = p H2O = 1920 Pa. A small amount (<< 5 %-vol)) of air will be dissolved in the water: use Henry s Law to calculate the equilibrium: At T = 290 K, H c AIR,water = 6200 MPa p AIR = p tot p H2O = kpa x AIR,, water side = p AIR,air side / H c AIR, water = MPa / 6200 MPa = = %-v This means 1.45 moles air (molar mass ~29 kg/kmol kmol) ) in moles water (molar mass 18 kg/kmol kmol), which means 23.4 mg air / kg water Source: ÇB98

12 Two-component phase diagram (G/L) P, T, x diagram for a binary gas-liquid system critical points pure A, B x A = 1-x1 B y A = 1-y1 B 23/56 Picture: T68 Binary vapour-liquid equilibrium Constant pressure; ; the x-y diagram y x Temperature 24/56 Picture: T68

13 α = Picture: WK92 Relative volatility, α 25/56 Raoult (for not-small x i ): y A =x A p A, y B =x B p B x B = 1-x1 A, y B = 1-y1 A y B /y A = α x B /x A α = relative volatility α = p B /p A = α (T) result: (1-y A )/y A = α (1-x A )/x A y A = α x A / (1+ (α-1)( 1) x A ) 26/ Separation processes (gas-gas, gas-liquid liquid, liquid-liquid liquid): equilibrium stages

14 Separation of mixtures: 1 stage For example: separating phenol from water (L) by adding benzene (V) in a separation funnel. 27/56 x = phenol conc.. in L, y = phenol conc.. in V Equilibrium constant: K = y1 / x1 Separation factor: S = K V/LK Fraction of phenol separated = S / (S+1) for 1 stage Cu - ore 2,0 2,2 release from ore Diluted H 2 SO 4 extraction Organic solvent recovery 50 Concentrated H 2 SO 4 40 electrolysis 0,1 0,2 Cu Numbers are concentrations kg Cu/m 3 Separation and equilibrium stages Example Four-stage processing for Cu recovery from Cu- containing ore (liquid / liquid) 28/56 Source: WK92

15 Phase equilibrium stages /1 feed loaded solvent extraction refine solvent The extraction and the recovery process are combinations of mixing tanks and settling tanks 29/56 Picture: WK92 Phase equilibrium stages /2 feed loaded solvent Cu is transferred from feed stream L to solvent ( support phase ) V; the extraction unit can be described as a series of equilibrium stages. Equilibrium constant for Cu in feed stream andsolvent stream is K =y Cu Thus, for an equilibrium stage n: y n = Kx n Streams L and V are often roughly constant L 1 L 2.. L n = L; V 1 V 2... V n = V separation factor S = K V/LK equilibrium stage n+1 L n+1 x n+1 extraction equilibrium stage n refine solvent L n x n equilibrium stage n-1 30/56 Cu /x Cu V n y n V n-1 y n-1

16 Example: phase equilibrium stage For example: phase equilibrium constant K=5; L = 1000 kg/s, V = 800 kg/s S = KV/L= 4. Cu in feed x 0 = 0.2 %-wt = kg/kg For clean solvent, y 2 =0, then Vy 1 / Lx 1 = S Mass balance gives Lx 0 = (S+1)w-r Lx 0 +Vy 2 = Lx 0 = Lx 1 +Vy 1 = Lx 1 (1+KV/L) = Lx 1 (1+S) 31/56 Picture: WK92 Lx 0 = 2 kg Cu/s w Lx 1 = Lx 0 / (1+S) = 0.4 kg Cu/s 32/ Separation processes (gas-gas, gas-liquid liquid, liquid-liquid liquid): continuous distillation; packed tower columns

17 Continuous distillation 33/56 Pictures: T68 Liquid with composition a 1 boils at temperature T 2, giving vapour with composition a 2 which is enriched in component A Taking out and cooling vapour a 2 gives liquid a 4 at temperature T 4 In equilibrium with liquid a 4 is vapour a 4, again further enriched in component A Distillation, principle 34/56 Picture: A83

18 Continuous distillation (binary) Separating a mixture of 2 components A and B, with A more volatile Liquid for absorption section produced by condensing some top product Gas for stripping section produced by boiling some bottom product Roughly equimolar exchange: : 1 mol A liquid gas ~ 1mol B gas liquid Top product: more volatile component 35/56 Top section: rectifying section absorbing the less volatile component Bottom section: stripping section desorbing the more volatile component Bottom product: less volatile component Packed tower columns /1 Pictures: T68 Mellapak 36/56

19 Packed tower columns /2 Mass transfer from gas to liquid or vice versa where the liquid forms a (thin( thin) ) film on the surface of packing material elements, creating a large contact surface a (m 2 / m 3 apparatus) For relatively small amounts of material transferred (say,, < 2% of the streams) ) the process may be considered isotherm (vaporisation and condensation have a heat effect!)!) and streams V and L may be considered constant. 37/56 Picture: W92 38/ Particle technology; Multi-phase flows Consider a particle Prof. Brian Scarlett

20 Particle (or droplet) Particle 39/56 (or droplet) size distribution Frequency distribition X n dn/dx n=0,1,2,3 dn/dx XdN/dX=dL/dX X 2 dn/dx=ds/dx Particle size, X N = number L = length S = surface V = volume X 3 dn/dx=dv/dx Different distributions for number, length, surface and volume! Different particle size analysers give different distributions: some measure length, others measure surface, etc. Particle shape Shape factor, sphericity φ φ = (volume) 2/3 surface = surface of sphere with same volume surface of particle 40/56

21 Particles in fluid flows A smooth (a) and roughened (b) ball entering water at 25 C Picture: CR93 Turbulence as seen by Da Vinci Particles and turbulent eddies 41/56 Eulerian vs. Lagrangian particle representation time t 1 time t 2 time t 1 time t 2 time t 3 Focussing on a control volume (Euler), left, or focussing on particle trajectories (Lagrange), right Euler-Euler (for fluid and particulate phase) and Euler-Lagrange methods are both widely used time t 4 time t 4 42/56 Source: ZH00

22 Aerosols Aerosol: A suspension of solid or liquid particles in a gas. Aerosols are stable for at least a few seconds and in some cases may last a year or more.. The term aerosol includes both the particles and the gas, which is usually air. Particle size ranges from to over 100 µm. For example smoke is a dispersion of solid particles or droplets in air. Sol: particles dispersed in a liquid, for example ink Source: ZH00 43/56 Picture: Picture: Flow of particle swarms Drag coefficient for sphere in swarm, C D *, corrected for effect of neighbour particles Small particles, low Re: ƒ(ε) ) = ε -4.7 C D * = C f( ε) = D C ε = voidage,, porosity 44/56 Richardson - Zaki hindrance factor: D Re n < ~ Re 1 ~ Re > Source: ZH00 ε n Re Re -0.01

23 L Flow in packed beds Darcy s Law: with permeability K 45/56 Kozeny - Carman equation: u = u = K ε 3 5 ( 1 ε) S Δp L η fluid ( Δp) 2 vη fluid S v = specific surface = surface/volume L Source: ZH00 S v = 6 /d p for a sphere with diameter d p Fluidised beds 46/56 Source: ZH00 gas bubble in a gas/solid fluidised bed

24 1 Conveying systems: mechanic, pneumatic, hydraulic Pneumatic conveyor / drier Conveyor belt 2 4 pneumatic conveying regimes : - Solid Dense Phase - Discontinuous Dense Phase - Continuous Dense Phase - Dilute Phase 3 Picture 1: 47/56 Picture 2: Picture 3: Flow of powders in/from silos a. Mass flow b. Funnel flow c. Expanded flow d. Pipe e. Rathole f. Arching 48/56 Source: ZH00

25 A gas cyclone Advantages Simple, cheap and compact Large capacity Disadvantages Large pressure drop Low efficiency Catch removal problems No removal below ~5 μm Problems above ~ 400 C See also hydro-cyclones and other cyclones for liquid-solid, liquid-liquid and liquid-gas separations Source: ZH00 49/56 Electrostatic precipitators (ESPs 4 process steps: 1. Particle charging 2. Particle movement relative to the gas flow 3. Particle collection at deposition surface 4. Particle removal from deposition surface (often discontinuous) Note: : the electric properties of the particles to be removed should be suitable, otherwise use a filter system Typically quite large, mainly used at power plants for fly-ash removal from flue gas ESPs) 50/56 Pictures:

26 Baghouse filters Inside out / outside in operation Source: ZH00 51/56 Picture: Picture: Horizontal belt filter Liquid filtration Source: IGH91, ZH00 Δϕ Δϕ Rotary drum filter 52/56 ~ Constant pressure filtration N rotation /min (rpm), drum radius R (m),length L (m), submerged angle Δϕ= (0... π) volume element ΔA= R L Δϕ is submerged for a time Δt = Δϕ/ (2π N) (min)

27 Sedimentation of suspensions Batch sedimentation test Source: IGH91, ZH00 Clear zone Feed zone Feed Thickening zone Effluent Sludge discharge Continuous thickener Dry solids concentration 53/56 Picture: Solid-solid separations Used for separating unwanted materials or size fractions Equipment examples: Sieves, screens Cyclones, centrifuges Hydraulic separators Sink-float float, froth flotation separators Magnetic, electrostatic separators Screens Froth flotation 54/56 Picture: Picture

28 Crystallisers Solid product crystals can be produced from gases, liquid melts or solutions Advantages are: high product purity (crystallisation can be seen as a separation process!), and (except for liquid melts) low energy demand and low temperatures Import issues are crystal product morphology, crystal growth kinetics, inclusion of impurities (and crystal water), and process control (temperature product size distribution and quality) A continuous crystalliser 55/56 Picture Picture: Sources #7 A83: P.W. Atkins Physical chemistry 2nd Ed. Oxford Univ. Press (1983) BMH99: Beek,, W.J., Muttzall,, K.M.K., van Heuven,, J.W. Transport phenomena Wiley, 2nd ed. (1999) ÇB98: Y. A. Çengel,, M.A Boles Thermodynamics 3rd Ed. McGraw-Hill (1998) CR83: Coulson,, J.M., Richardson, J.F., Backhurst,, J.R., Harker,, J.H. Chemical Engineering, Vol. 2 : Unit Operations Pergamon Press, Oxford (1983) IGH91: Iinoya,, K., Gotoh,, K., Higashitani,, K. Powder technology handbook, Marcel Dekker, New York (1991) SH06: J.D. Seader,, E.J. Henley Separation process principles, Wiley, 2nd ed. (2006) SSJ84: J.M. Smith, E. Stammers,, L.P.B.M Janssen Fysische Transportverschijnselen I TU Delft,, D.U.M. (1984) T68: Mass transfer operations McGraw-Hill 2nd ed. (1968) WK92; J.A. Wesselingh,, H.H. Kleizen Scheidingsprocessen Delft Univ. Press (1992) ZH00: : R. Zevenhoven,, K. Heiskanen Particle technology for thermal power engineers part 1 & part 2, post-graduate course ene , TKK, Espoo,, Sept./Oct Oct Z06: : R. Zevenhoven,, Massöverföring & separationsteknik kursmaterial Åbo Akademi Univ. (2006) Ö96: : G. Öhman Massöverföring, Åbo Akademi Univ.. (1996) 56/56