Separationsteknik / Separation technology

Transcription

1 Separationsteknik / Separation technology Fyllkroppskolonner / Packed columns Ron evenhoven Åbo Akademi University Thermal and Flow Engineering aboratory / Värme- och strömningsteknik tel ; ron.zevenhoven@abo.fi 5.1 Principle of operation, Packings 2/56

2 Packed tower columns /1 Random packing materials Pictures: T68, SH06 3/56 Packed tower columns /2 Mass transfer from gas to liquid or vice versa where the liquid forms a (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 may be considered constant. Picture: WK92 4/56

3 Packed tower columns /3 Mellapak Structured packing materials Support plate Pictures: T68 5/56 Packed tower columns /4 Structured packings Pictures: SH06 6/56

4 Packed tower columns /5 Structured packings or a random packed material can be used. Typical size of filling material mm. Mostly applied for absorption / desorption or distillation, with low pressure drop as an advantage. Compared to tray columns, more liquid with respect to gas flow. Typically countercurrent operation (gas upwards, liquid flows downwards). Also co-current application as a gas-liquid reactor, and socalled trickle-beds where the packing acts as catalyst. The use of packed tower columns is prefered over tray columns if 1) small units are needed (H < 10 m, D < 1 m); 2) corrosive fluids must be handled (using a ceramic or plastic packing); 3) pressure drop must be low; or 4) foaming can be a problem Operation is limitated by flooding, where the amount of gas flow prevents downflow of liquid. Typical operation at ~ 70% of flooding limit. Important factors are liquid and gas flow per unit cross-section, and liquid viscosity. 7 Example: Raschig ring packing For Raschig rings with a 1 (inch) diameter and height, calculate the contact surface a, m 2, per unit column volume, m 3. Assume a cubic packing (see Figure). (1 = 2.54 cm) Picture: WK92 Answer: The surface of inside + outside of the cylinders equals 2 π d l m 2 ; in a cubic packing one cylinder occupies a volume equal to d² l m³. Thus, a = 2 π d l / (d² l ) = 2 π / d with d = m gives a ~ 250 m 2 /m 3 with respect to column volume. The packing occupies fraction (1-ε) of the column volume (with voidage ε), thus the contact surface with respect to packing volume equals a p = a (1-ε). For Raschig rings: length l d 8

5 Column packing characteristics NOTE: instead of a p /ε 3 = a (1-ε)/ε 3 nowadays PACKIN FACTOR F p is used F p = a p / ε 3 unit. m -1 enerally, the column diameter to packing size ratio D/d should be > 30 for Raschig rings, > 15 for ceramic saddles, and > 10 for plastic rings or saddles. Source: 9/56 Packed columns: flooding limits (random-dumped packings) Coulson & Richardson vol. 2 (1983) See also expression in Öhman (1996) Fig. 6.2 (extended to X = 100): Y = ( 10 log X)/1.14 (( 10 log X)/2.04) 2 (( 10 log X)/2.88) 3 symbols: and : mass rate of liquid and gas (kg/s); densities ρ, ρ (kg/m 3 ); S B is surface of packing per unit volume of bed (m 2 /m 3 ) = S (1-e) with packing voidage e = ε (m 3 /m 3 ) and S is surface of packing per unit volume of column (m 2 /m 3 ); g is gravity; u = gas velocity calculated over whole cross-section of the column; µ is liquid viscosity (Pa.s), µ W is viscosity of water at 20 C (~ Pa.s) 10/56

6 Packed columns: pressure drop Often, column diameter is calculated from the gas velocity for a certain pressure drop! F μ Y ρ ρ ρ X g ρ ρ ρ Error: unit: 1/m (1/ft) Coulson & Richardson vol. 2 (1983) 11/56 Packed columns: wetting rate There is a minimum liquid rate for effective use of the packing surface area. A wetting rate can be defined as Volumetric liquid rate per unit cross - sectional area of the column Packing surface area per unit volume of the column 3 2 ( m / s) / A column ( m ) u ( m / s) 2 ( m W a p ( m / m ) ap ( m / m ) / s) Recommended minimum liquid wetting rates are W = m 2 /s for mm rings and grids with less than 50 mm spacings, and W = m 2 /s for larger packings. 12/56

7 Packed tower columns /6 In a packed column no discrete, identifiable stages are seen Similar to a plate column the goal is, however, not that equilbrium is reached at each stage More importantly, mass transfer is accomplished that brings the phases closer to equilibrium Pictures: BS60 K71 13/ Mass balance, Mass transfer 14/56

8 Task ö6.1 : the question (swedish) vatten y 1 = 0.2 % 70 mol/s luft y N+1 = 1.8% Picture: after SH06 15 Task ö6.1 /1 a) Assume that air = aceton-free air, A = aceton Mass flow gas in: ṅ, in = ṅ air, in + ṅ A, in = mol/s % ṅ, in ṅ, in = 70 / ( ) = mol/s and ṅ A, in = mol/s Mass flow gas out: ṅ, out = ṅ air, out + ṅ A, out = mol/s % ṅ,out ṅ, out = 70 / ( ) = mol/s and ṅ A, out = mol/s Transferred: ṅ A = ṅ A,in ṅ A,out = = mol/s 16/56

9 Task ö6.1 /2 b) With incoming gas stream (1.8 % %) = mol/s A With outgoing gas stream (1.8 % %) = mol/s A Transferred (1.8 % % ) = mol/s Exact: ṅ, in y A,in ṅ,out y A,ut = ṅ A ṅ, in y A,in (ṅ,in ṅ A ) y A,out = ṅ A gives ṅ A = [ ṅ,in (y A,in -y A,out )] / (1 y A,out ) ṅ,in (y A,in -y A,out ) 17/56 Mass transfer in a packed column /1 Transport equation: transport across surface A ṅ = k oy A (y - y*) = k ox A (x*- x) = k oy A (c -c *) = k ox A (c *- c ) (mol/s) (mol/s) defined like this: ṅ > 0 for, ṅ < 0 for Equilibrium constant K: y* = K x and y = K x* at equilibrium: y = y*= K x*= K x = end of mass transfer process K is related to k x and k y 18/56

10 Mass transfer in a packed column /2 1. With K = β = H c /p total (Henry), or K = β = p /p total (Raoult), or K = β = γ p /p total (Modified Raoult) 1 k oy 1 k y β k unit k ox, k oy, k x, k y = (m/s) (mol/m 3 ) 2. With volumetric distribution coefficient m: m c = c x β k ox k k' m k' m oy y x k ox Distribution coefficient β linked by with ρ mol,x k ' x molar density k ρ x mol,x 3 3 ρx(kg/m ) (mol/m ) M(kg/mol) mass density molar mass unit k ox, k oy,k x, k y = m/s 19/56 Volumetric distribution coefficient m The volumetric distribution coefficient, m, is usually defined as c liquid = m c gas This is related to equilibruim constant K = y/x using c liq. = x ρ mol, liq. = x ρ liquid /M liq. c gas = y ρ mol, gas = y (p/r T) with molar densities ρ mol, (mass) densities ρ, molar mass M, pressure p, temperature T ρliquid m M K p RT This gives: Not always, is sometimes defined as c gas = m.c liquid (see e.g. WK92) Volumetric distribution coefficient data for several gases in water at 20 C c eq water = m c gas See also #4 slides 6-10 Table: BMH99 20/56

11 Mass balance for a packed column Mass balance for a height section dl : ṅ = V dy = k (c -c i ) a A dl = k (c i -c ) a A dl = dx with mass transfer coefficients k and k for gas and liquid side, cross-section area A and packing surface per volume a. With overall mass transfer coefficients k o or k o (eliminating the phase interface concentrations), for the gas side, using k o : V dy = k o (y* - y) (p/r T) a A dl yin yout dy/(y* - y) = 0 (1/V) k o ( p/r T) a A dl and similar for the liquid phase Picture: SH06 21/56 eneral approach height calculation general : flux N k (c c for height section d this gives mass flow dn N dsurface N A a d and also dn dc for volume flow k (c c with c c c c out in in int 0 and dc c c interface interface 2 ) mol/(m s) for each phase ) A a d dc 0 c c k A a d This is based on the mass balance for one phase only; concentration c interface usually unknown and should be eliminated overall mass transfer coefficient k oy or k ox.

12 5.3 Transfer units: HTU, NTU (Height of a transfer unit, Number of transfer units) Method 1, based on mass transfer resistances Transfer units HTU, NTU /1 contact surface a m 2 /m 3 dz Φ Φ cross-section area A m 2 Absorption of a component from a gas into a liquid (or similarly, for desorption / stripping from a liquid to a gas) Mass balance for transfer in a column height section with volume A dz, with concentrations c =c (z) and c =c (z), volume streams Φ, Φ c dc koy a (c ) A dz m with distribution constant m, c / m = c *at equilbrium, and mass transfer surface a m 2 /m 3 apparatus, and total height k oy (m/s) (or k o ) is the overall mass transfer coefficient for the gas phase side A similar mass balance can be set up for the liquid phase, using k ox (or k o ) 24

13 Transfer units HTU, NTU /2 Integration defines the overall Height of a Transfer Unit HTU, here: HTU o, subscript o = overall, from gas phase c ( z ) z z oy dz c z z ( 0) c 0 z0 ( c ) dc a A Integrating the left-hand part, called the Number of Transfer Units NTU o gives m k dz HTU o c Δc Δc c Δc c Δc ln Δc c m and HTU Δc o c where Column height = HTU o NTU o. c m z Operation line & equilibrium line must be straight lines! 25/56 Transfer units HTU, NTU /3 Column height = HTU o NTU o. In a column section with volume HTU o A (m 3 ), the mass transfer capacity k oy a A HTU o (m 3 /s) is equal to Φ (m 3 /s), i,e. HTU o is the height of a section for which k oy a A HTU o = Φ NTU equals the ratio between horizontal mass transfer and vertical transport by convection note: 1) HTU o = HTU o (z), if k oy varies with position; 2) axial dispersion is neglected contact surface a m 2 /m 3 dz Φ Φ cross-section area A m 2 z 26/56

14 Transfer units HTU, NTU /4 The total mass balance gives (for counter-current plug flow) c c mass c c c c 0 0 (using 0 ) c ln 0 HTUo c c0 c c0 c koy a A koy a A c c0 HTU c ln o ln 0 c c with a logarithmic mean concentration driving force which can be used if the equilibrium line and operating line are straight (constant value for m), (see also #9: Φ = k A ΔY lm ) otherwise: plot 1/(c -c /m) versus c and integrate: see next Without thermodynamic limitations in the other phase (m=0 or = ) the fraction transferred within one HTU is 1 - e -1 63%. Similarly, = HTU o NTU o = HTU o NTU o, but HTU o HTU o lm z 27/56 Transfer units HTU, NTU /5 y A y AE Illustration for procedure of determining NTU o by integrating the reciprocal of the driving force curve. Here: NTU o = 6.2. y AE = equilibrium value for y A Source: King 1971, /56

15 Transfer units HTU, NTU /6 Alternatively, HTU o can be calculated from the individual values for the gas phase HTU and for liquid phase HTU. Combining these gives HTU o or HTU o, using the equilibrium data (K or m). (m) u (m/s) u (m/s) Φ The mass transfer capacity for the (m 3 /s) cross gas phase equals k a A (m 3 /s), sectional with gas-side mass transfer area A (m 2 ) coefficient k (m/s). Φ (m 3 /s) packing with contact surface a (m 2 /m 3 ) The ratio Φ / (k a A) = u / (k a) is the height of a transfer unit HTU, (or H ) for the gas phase. Similarly, the dimensionless ratio Φ /(k a A) = u /(k a) is the height of a transfer unit HTU (or H ) for the liquid phase. 29 Transfer units HTU, NTU /7 Combining the gas-side and liquid-side HTUs: HTU o 1 a A k o 1 a A k a A k m k a A k a A m k 1 HTU HTU S a A k m with separation factor S. Similary, for the overall liquidphase: u Φ HTU note : o HTU HTU HTU o o HTU S S (m) (m/s) (m 3 /s) cross sectional area A (m 2 ) u Φ (m/s) (m 3 /s) packing with contact surface a (m 2 /m 3 ) 30/56

16 Transfer units HTU, NTU /8 For the NTUs, for gas-side and liquid-side: as film : NTU iquid film : NTU Usually the mass transfer is mainly limited by the gas-side transfer, k o is more constant than k o more convenient to use NTU o and HTU o than NTU o and HTU o Column height = HTU o NTU o = HTU o NTU o dy y y dx x x i i as overall : NTU o iquid overall : NTU (m) u u (m/s) Φ Φ (m 3 /s) o dy y y * dx x * x (m/s) (m 3 /s) cross sectional area A (m 2 ) packing with contact surface a (m 2 /m 3 ) 31 Transfer units HTU, NTU /9 With straight operation lines and equibrium lines and liquid and gas molar streams V and ~ constant: y = x (/V) + y out -x in (/V) for an absorber y = x (/V) + y in -x out (/V) for a stripper y = K x for the equilibrium For an absorber, this gives the following for NTU (Colburn, 1939): NTU o yin yout dy y y * S 1 yin K x ln S yout K x S 1 in yin yout in (1 1 S KV ) y y K x for an absorber, with S A S For a stripper (with S = KV/) a similar expression can be derived for NTU o or stick to NTU o... dy out KV in x = (y-y out ) V/ - x in y* = K x KV 32/56

17 Example HTU, NTU Air containing 1.6 %-vol SO 2 is scrubbed with clean water in an absorber, which is a packed bed column with cross-sectional area A = 1.5 m 2 and packing height = 3.5 m. Incoming gas and liquid flow rates are = 62 mol/s and V = 2200 mol/s. For a concentration of 0.4 mol-% in the outgoing air and K = y/x = 40: calculate NTU o, HTU o and k oy a (as mol/m 3 s) for the SO 2. Answer: Separation factor S = A = /KV = 0.89; with y in = 0.016, y out = 0.004, and x in = 0 Colburn s expression (see previous slide) gives NTU o = 3.75 = NTU o HTU o = 3.5 m HTU o = 0.93 m k oy a A HTU o = Φ, with unit for k oy (m/s), V = Φ (p/rt) k oy a A HTU o = V, with unit for k oy (m/s) (mol/m 3 ) k oy a = 44 mol/m 3.s 33/56 Transfer units HTU, NTU /10 Alternatively (notation Öhman 1996, ) using molar fractions y and x, describing equilibrium with y = β x notation for molar streams ṅ (mol/s), contact surface a (m 2 /m 3 ) transport coefficients k x and k y for each phase (mol/m 3 m/s) column height and cross sectional area A tv notation H = HTU ; H = HTU ; = H o N o = H o N o ṅ = ṅ (y 0 -y 1 ) = K y a A tv (y-y*) lm = [ṅ /(K y a A tv )] [(y 0 -y 1 )/(y-y*) lm ] H A H H tv n k y n n n H n and a H x H y y A0 A A0 y* x A y A1 tv A n k A1 lm x * A x A lm x a = H o N o. But note: using logarithmic mean concentrations so that N = Δy/Δy lm and N = Δx/Δx lm requires that the equilibrium line and operating lines are straight! 34/56

18 5.4 Theoretical stages ( plates ): HETP Method 2, based on equilibrium stages Height equivalent to a theoretical plate, HETP Important for mass transfer equipment is the Height Equivalent to a Theoretical Plate (or equilibrium stage), HETP. This is the height needed for equilibrium for outgoing flows. For liquid 1 and gas 2 this implies c 1 (x) = m c 2 (x + HETP) HETP and HTU are related via HETP HTU o xhetp x dc2 c1 c m 2 lns... S 1 Φ v c mass balance Φ v c V Picture: BMH99 constant S = 1: HETP = HTU o S > 1: HETP < HTU o S < 1: HETP > HTU o 36/56

19 HETP, HTU, NTU, N theoretical HETP HTU o NTU N theor. o xhetp x dc2 c1 c m since HETP N 2 ln S... S 1 theor. HTU o NTU S = 1: HETP = HTU o S > 1: HETP < HTU o S < 1: HETP > HTU o o Picture: SH06 (a) NTU = N theor. (b) NTU > N theor. (c) NTU < N theor. Absorber S = A = 1 S = A < 1 S = A > 1 37/56 HETP values for common column packings For Raschig rings 25 < d < 50 mm HETP 18 d + 12 K ((V/ ) - 1) CRBH83, dhb07 Note: smaller packing elements give lower value for HETP but increase pressure drop... 38/56

20 Definitions revisited... HTU: height of mass transfer apparatus over which a concentration is reduced by a factor 1/e, for c 1 >> m c 2 or for c 1 << m c 2 (i.e. small thermodynamic limitation in other phase) (= also height needed to reach equilibrium between opposite streams at equilibrium if separation factor S =1) NTU: how many HTUs in apparatus height, is equal to residence time / time needed for transfer HETP: height in transfer apparatus between opposite streams at equilibrium Column height = HTU NTU; or = HETP N theoretical stages from counting stages in x,y diagram, or Kremser equation. 39/56 Example: exam question 341 (7+3 p.) I en motströmsprocess absorberas SO 3 från en upgående luftström med 7 %-mol SO 3 i koncentrerad H 2 SO 4 vätske som rinner ner i en fyllkroppskolonn med Rashig-ring element som ger kontaktytan a = 100 m 2 /m 3. asens hastighet (ytlig, dvs för tornet utan packning) är v = 2 m/s. Partialtrycket p* för SO 3 i jämvikt med koncentrerad H 2 SO 4 är p* 0 bar. Massöverföringskoefficienten för gassidan är k = 0.02 m/s medan massöverförings-resistansen för vätskesidan kan försummas. Också värmeeffekter kan försummas. a. Om 98% av SO 3 ska absorberas från luftströmmen, beräkna kolonnhöjden (i m). b. Beräkna totala överföringshöjden HTU o (= H o ) (i m), och antalet överföringsenheter NTU o (= N o ) hänförd till gassidan. Beräkna även antalet teoretiska steg, N theory 40/56

21 a. Mass K SO3 Example: exam question 341 answer dc 1 transfer coefficient k p * 1 0 m ~ p K mass balance k : SO c in k a ln ln(0.02) m. cin v 2 v v b. HTU HTUo 1m. NTUo NTU a k a k HTU NTU HTU N a c theory dc k A dz HETP. 1 k k a (c 0.02c m : k ln(1/e) -1for h HETP, or -1 o o o c 1 m k o in k ; 0 SO3, out SO3, in no thermodyn. limitation: a HETP v c c c ) A dz gives m in dc k a A dz c 0.02 (for 98% removal) - HETP 0 k a dz v c(h) c in HETP 1m N /e for h HETP. theory / Design calculations

22 Dimensioning packed columns /1 Example A column for stripping aceton from water using an air stream at T = 20 C, 1 bar. x 0 = 10-3 mol/mol in incoming water stream, x n = 10-4, y n+1 = 0 (aceton-free air). K = y*/x = 3 for the equilibrium k x = m/s, k y = m/s. iven v = 1m/s. ρ mol, = mol/m 3, ρ mol, from ideal gas law. Use a = 200 m 2 /m 3. How many (theoretical) equilibrium stages N th if /V = (/V) max /1.2, using Kremser s equation. Calculate HTU o and HETP and column height. 43/56 Dimensioning packed columns /2 answer: y eq.line 1 = K x 0 = (/V) max = (y 1 -y n+1 )/(x 0 -x n ) = ( )/( ) = 3.3. /V = 3.3/1.2 = 2.75, y new 1 = ; S = K/(/V)= 3 / 2.75 = 1.08; not transferred f = 0.1 Kremser N th = 6.64 y 1 0 V See also #10 p. 20 y = Kx (/V) max (/V) ρ mol, = mol/m 3, ρ mol, = p/rt = 41 mol/m 3 equilibrium constant K = (ρ mol, / ρ mol, )/m m = 452 1/k oy = 1/(mk x )+ 1/k y k oy = 0.01 m/s Mass transfer resistance: 1/100 = 1/ /0.04. gas 25%, liq. 75% 44/56

23 Dimensioning packed columns /3 answer: HTU o = Q /(k oy a k oy A) = v /(k oy a) = 1/( ) = 0.5 m y = Kx (/V) HETP =(ln S)/(S-1) HTU o =(ln1.08)/(0.08) 0.5 = 0.48 m = N theor HETP = m = 3.2 m Picture: 45/56 Dimensioning packed columns /4 Similar to tray column design, the most important features are column diameter and pressure drop. Also here, a load diagram can be used. Raschig rings Packing factor Pictures: WK92 ceramic metal load diagram for a packed column for Raschig rings 46/56

24 Dimensioning packed columns /5 The load diagram follows from a pressure drop analysis: Assume the packing to be a set of vertical plates at a distance d, on which the liquid flows down as a layer with thickness δ (m), and velocity υ (m/s) υ is related to the superficial (= with respect to empty column) liquid velocity v (m/s) as v = 2 υ δ/d. iquid volume fraction λ = 2 δ/d (m 3 /m 3 ). The gas moves upwards with superficial velocity v (m/s). Picture: WK92 47/56 Dimensioning packed columns /6 Forces as a result of flow in a packed column (per m 2 surface): 1. Turbulent shear at the wall : f 1 = c 1 ρ υ ² 2. ravity: f 2 = - ρ δ g 3. Turbulent shear gas/liquid: f 3 = c 2 ρ (υ +v ) 2 c 2 ρ v 2 With total force = 0 for stationary flow this gives: c 1 ρ (½ v d/δ)² - ρ δ g + c 2 ρ v 2 = 0 with λ = 2 δ/d c 3 (v 2 /(g.d)) ⅓ gives final result c 1 ρ v ² - c 4 ρ d g + c 2 ρ v 2 = 0 (packing factor F p is included in c 4!) Picture: WK92 48/56

25 Dimensioning packed columns /7 Design procedure: Determine Q, Q, ρ, ρ, column height and diameter d and packing factor F p for the packing Calculate parameter (Q /Q ) (ρ /ρ ) ½ and read v (ρ F p / ρ g) ½ from load diagram Calculate the maximum gas velocity v and to allow for some margin take v 0.70 Column cross-sectional area A = Q /v = (π/4) D 2 Calculate liquid superficial velocity v = Q /A and volume fraction liquid λ =1.2 (v 2 /(g d) ⅓ Dry pressure drop Δp dry 0.8 ρ v 2 F p Wet pressure drop Δp Δp dry (1-3.5 λ) -3 49/56 Dimensioning packed columns /8 Example (almost the same as above) A column for stripping aceton from water using an air stream at T = 20 C, 1 bar. = 0.4 mol/s, V = 0.25 mol/s, M = 18 kg/kmol, M = 29 kg/kmol, ρ = 1000 kg/m 3, ρ = 1.2 kg/m 3. Column height = 2 m, packing material 10 mm ceramic Raschig rings. Calculate column diameter and pressure drop Picture: 50/56

26 Dimensioning packed columns /9 answer: Q = M /ρ = m 3 /s Q = V M /ρ = m 3 /s. F p = /m and (Q /Q ) (ρ /ρ ) ½ = oad diagram: v (ρ F p /ρ g) ½ = 0.37 v = 0.90 m/s, 0.7 v = 0.63 m/s. A = Q /V = m 2 D = 0.11 m v = Q /A = m/s, λ = 1.2 (v 2 /(g d)) ⅓ = 0.021, Δp dry 0.8 ρ v 2 F p = 1075 Pa, Δp Δp dry (1-3.5 λ) -3 = 1360 Pa Picture: 51/56 Dimensioning packed columns /10 Typical design procedure for packed columns Calculate the number of theoretical stages N th. Select the type and size of the packing material Calculate the flooding limit from the two flows Calculate the column diameter for a gas velocity that corresponds to 70% of the flooding limit. Calculate the pressure drop. Determine the values for HTU o and HETP. Calculate the height of packing as HETP N th. Add m above and below packing. for further details (inlet/outlet distributors etc.): see the literature 52/56

27 5.6 Concentrated solutions, Distillation Concentrated solutions; Distillation When concentrations are high and significant amounts of mass is transferred, the operation line and/or the equilibrium line will be curved. Taking the inert parts of the gas and liquid streams = (1-x) and V = V (1-y) and use these in equations for HTU and NTU etc. In distillation applications, the equilibrium constant K changes strongly from place to place HTU and HETP etc. depend on position and change dramatically at the feed point. In calculations, instead of K use the local slope ς = dy/dx of the equilbrium curve 1/NTU o = 1/NTU + ς/ntu etc. More detail: see SH & /56

28 Tray vs. packed columns for distillation Source: King 1980, p. 605 Sources BMH99 Beek, W.J., Muttzall, K.M.K., van Heuven, J.W. Transport phenomena Wiley, 2nd edition (1999) BS60 Bird., R.B., Stewart, W.E., ightfoot, E.N, Transport phenomena Wiley (1960) CRBH83 J.M. Coulson, J.F. Richardson, J.R. Blackhurst, J.H. Harker Chemical engineering, vol. 2, 3rd ed. Pergamon Press (1983) Ch. 11,12 dhb07 A.B. De Haan, H. Bosch Fundamentals of industrial separations 2nd Ed., TU Eindhoven / U Twente, the Netherlands (2007) Ch. 4 K71, K80 King, C.J. Separation processes Mcraw-Hill (2nd ed. 1980, reprinted 2013) MSH93 W.. McCabe, J.C. Smith. P. Harriott Unit operations of chemical engineering 5th ed. Mcraw-Hill (1993) Ch. 22 SH06 J.D. Seader, E.J Henley Separation process principles John Wiley, 2nd edition (2006) 6.1, , 7.5 ** T68 R.E. Treybal Mass transfer operations Mcraw-Hill 2nd edition (1968) WH92 J.A. Wesselingh, H.H. Kleizen Separation processes (in Dutch: Scheidingsprocessen) Delft University Press (1992) 97 F. uiderweg Physical separation methods (in Dutch: Fysische Scheidingsmethoden) TU Delft 1987 (vol. 1, vol. 2) Ö96. Öhman Massöverföring Åbo Akademi Värmeteknik (1996) Ch. 6 Available on - line (28 Mb): * * See ÅA course library 56/56