4. DIFFUSION AND MASS TRANSFER

Transcription

1 4. DIFFUSION AND MASS TRANSFER Design Problem 3 Dry spinning of PAN/DMF solution (Fig 21) l PAN-dry spinning, wet spinning, melt spinning DMF into air DMF into water No solvent l Three mass transfer mechanisms in dry spinning Flash evaporation Diffusion thru spinning line Controlling step Interface mass transfer Scan Fig 21 here Data: m = 2 10 g/s (Dope output) ω = 0.74 (Dope solvent mass fraction) T = 100 (Dope Temp) T = 200 (Air Temp) V = 2 m/s (Cross flow air velocity)

2 V = 50 cm/s (Parallel flow air velocity) x = 0 (Solvent mole fraction in air) D. = exp ( 2,360/T) (Diffusivity of DMF in PAN) Dope cross section= 0.001cm Final cross section= cm Mass transfer coefficients k = 0.26[D ( ) ]( ) (R(z)) V k = 0.53[D ( ) ]( ) (R(z)) V R(z) : Radius of spinning z=z. Two dimensionless numbers: Schmidt # Prandtl # Assume Sc= = ( Pr= Sc= 1.81 Pr= 0.69 ( Calculate axial distance to evaporate all solvent! (To be worked out at the end of this chapter) ) ) 4.1 MASS TRANSFER FUNDAMENTALS Definition Mass concentration ρ = Molar concentration C = Mass faction Mole fraction ω = = x = = Mass average velocity v = = Molar average velocity (1) (1) [=ω v + ω v (binary)] (2)

3 v = = Volume average velocity Note that v = ρ v (4) [=x v + x v (binary)] (3) molar volume ρv = local rate of mass transfer through a plane perpendicular to v cv = local rate of molar transfer v ρv [=] cv [=] = = mass flux molar flux Diffusion and bulk motion Diffusion of i th species can be written as v = v v + v (5) diffusion bulk motion velocity of species i with respect to a fixed coordinate See Table 4.2 for binary system Ex 4.1

4 x =, v = 12, v = 15, M = 5M Calculate v and v? Solution v = x v + x v v = = () = ( ) v = ω v + ω v = v + v (ρ = c M ) = v + v (c = cx ) = (cx )v + (cx )v (M = ) = x v + x v (M = M x + M x = M + M = + M = 3) = 3x v + x v = Fluxes

5 n = ρ v mass flux w.r.t fixed coordinate N = C v molar flux j = ρ (v v) mass flux w.r.t v J = C (v v) molar flux w.r.t v j = ρ (v v ) mass flux w.r.t v J = C (v v ) molar flux w.r.t v Fick s First law concentration gradient J = D C = CD x (C = constant)(= D for 1D in x direction) Diffusion coefficient, binary diffusivity D[=]? mole flux [D] = = [D] j = D ρ = ρd ω (Constant ρ) Remember τ = μ( ) = / velocity gradient Momentum : momentum flux => flux gradient (concentration or velocity gradient) Most frequently used form N = x N + N cd x (13) Table 4.3 here for various fluxes

6 Ex 4.2 Penetrant(A)-polymer membrane(p) system Find n in terms of ω Solution j = ρ (v v)= ρ v ρ v 0 (fixed)

7 = n ρ ( ) = n ω n = ρd ω (15) (= D ρ ) ρ=const n = Typically for gas (A) membrane (P) ω 0 So, 1 n ρd ω (16) Microscopic material balance For a control volume (dxdydz) fixed in space : dxdydz= V Rate of accumulation of A in (dxdydz) = rate of A into the (dxdydz) rate of A out of (dxdydz) + rate of generation in V rate of consumption in V By chemical reaction γ (mass/vol/ time) (dxdydz) = n n dydz + n n dxdz + n n dxdy + γ (dxdydz) => γ = 0

8 => + n γ = 0 (20) (continuity eq.) In terms of molar flux + N R = 0 (21) For all species (binary) in terms of ρ + n γ = n γ = 0 ( ) = n + n = n + n = n

9 γ + γ = 0 (for mass) For const ρ => + n = 0 or + ρv = 0 = 0, ρv = ρ v + ρ v => v = 0 (23) Same with the incompressible fluid In terms of C + N R = N R = 0 ( ) = N + N = N + N = N = Cv v Cv = N + N Distribution law = + Cv R + R = 0 (24) [ = 0 C=const R ] 0 Cv = C v + C v = v v = R (25) Useful form of continuity eq. From (20) From (10) (10) (20) + n γ = 0 (20) j = ρ v v = n ρ v (10) n = j + ρ v (10) + j + ρ v γ = 0 N

10 + j + (ρ v) γ = 0 (26) j = ρd ω = D ρ D ρ + (ρ v) γ = 0 (28 From (21) + N R = 0 (21) J = C v v = C v C v = N C v N = J + C v = D C + C v (also, J = CD x = D C ) + D C + C v R = 0 (29) (28) for no molar flow ρ v = 0 For no chemical reaction γ = 0 So, = D ρ (30) For D =const 0 D ρ = D ρ + D ρ = D ρ = D ρ (32) Fick s second law of diffusion Substitute for ρ = C A /M A, = D C (32) Example 4.5 Sorption of a dye (A) into a falling polymer film (P) Assumptions Fully developed Laminar Steady flow No chemical reaction

11 Fig 4.6 here Macroscopic balance in (dxdz) (no mass transfer in y-direction) (N N )Wdz + (N N )Wdx = 0 (40), lim (40) => + = 0 (41) Also from Table 4.6 B.C z=0, for all x, C = C (42) x=0, for all z, C = C (43) C : interface concentration x=, for all z, C = C (44) C : shallow penetration In x-direction (Fick s law) N = D + x (N + N ) (53) 0 0 Slightly soluble (x A 0) N D (46) Similarly in z-direction

12 N = D + C v (47) N = CD x + x (N + N ) diffusion bulk motion C v = C v, (48) v = = ( ) Since penetration depth δ is very small, v = v, (surface) (46), (48) (41) + = 0 (41) D + C v, = 0 Letting t =,, dz = v, dt B.C. becomes = D (50) PDF (Partial Differential Equation) 1) z = 0, C = C t = 0, C = C (C : almost 0) 2) x =, C = C for all z (Same as above) 3) x = 0, C = C Methods of solution for PDF 1) Method of image 2) Separation of variable 3) Similarity transform 4) Laplace transform Rewrite the equation in dimensionless form: Then = ϕ = t = x (dimensionless concentration) (T=a characteristic time) (X=a characteristic distance)

13 ( ) = ( ) = D = ( ) Dimensionless form of eq dimensionless So, this dimensionless group is important. The similarity variable (η) is a function of this, i.e., or. Let η = and try to find ϕ = f(η) = = = t as ϕ = f(η) only = ( ) = η = = = = = = = ( + = ( + = ( + + η ) + η + ) ) See η = So, = So, eq.(50) with ϕ = = D becomes = D ϕ t = D AP 2 ϕ x 2 (50)

14 + = 0 subset x = η 4D t + (η 4D t) = 0 + 2η = 0 (A) ODE (Ordinary Differential Equation) B.C (3) x = 0, C = η = 0, ϕ = (3) η =, t =, B.C (1) + (2) (t 0, and x η, C = C, ϕ = 1 (1) Let ψ = dϕ dη Then ψ = = (A) becomes So, ψ + 2ηψ = 0 d = 2ηdη lnψ = η + constant ψ = c e Integration gives ϕ = c 1 e dη 0 + c 2 c constant Apply B.C (3) c η = 0, ϕ =

15 Apply B.C η, ϕ = 1 1 = c e η2 dη + c = = = = 1 ϕ = = e dη + = ( ) e dη = e dη = e dη ( )( )( ) ( ) = e dη = 1 e dη = 1 erf η = erf cη (51) erfc= Complementary error function Error function (Also called an engineering function) i) erf q = e dx x : Dummy variable ii) Gaussian type curve, symmetric w.r.t y axis

16 Now e dx = π So e e i.e., erf = 1 dx = dx = 1 iii) Area = e dx = erf q iv) as the graph is symmetric erfq q v) erf q = (1 x + = x +! ()(!)! + )dx ()(!) + (Taylor series expansion) So, erf q = [q + + ]

17 vi) For small value of q erf q = q if q 0.2 vii) For large value of q erf q 1 In fact, if q 2, erf q = 1 viii) erf cq = 1 erf q ix) = = e dx = e dx = e dx = e 0 + e + 0 (Leibnitz formula) Molar flux or diffusion of dye A along the x-axis. N D (46) Try to evaluate N (Err function differentiation). J = D C x N = x (N + N ) D C x D C x Concentration distribution was obtained as: C C = 2 C C π e dx = erfη C = C + (C C ) C x = x (C + (C C ) erfη) = (C C ) erf η x = (C C ) e

18 Then = (C C ) e = 2(C C ) e < 0 4πDt N = D = ( ) e > 0 H.W. Prob. 4c-3 (a) (see Fig. 4.27,28) Fig. 4.27, 28 here (Make it small) Calculate the dopant concentration as a functioning time. Tip: Define a similarity variable as η =

19 0 Hint Table 4.6 or 4.7 = D = 0 C = C for < < 0 1 C = 0 for 0 < < + 2 B.C. C = 0 as x for all t C = C as x for all t 3 4 Define a dimensionless concentration C = IC 1 C = 0, 2 C = 1 BC 3 C = 1, 4 C = 0 Answer = erfη +

20 4.2 Diffusivity, Solubility, Permeability in Polymer Mass transport applied in polymer processing: Gas-polymer Liquid-polymer water swell, solvent swell (gel) Polymer-polymer interdiffusion, adhesion, healing Diffusivity and Solubility of Simple Gases Simple gas Noneasily condensable gas (following Fickian diffusion) with weak intermolecular interactions, low boiling, low critical point: N, O, He. (permanent gas) Also, CO, SO, CH considered simple gas at low pressure Intermolecular energy of two molecules (Lenard-Jones Eq.)(See LN 256) ϕ(r) = 4ϵ[( ) ( ) ] (52) r intermolecular distance ϵ potential energy constant Lennard-Jonse scaling factor σ potential length const (collision dia) : [-] Lennard-Jonse temperature ( K) See LN 296 for detail (@r=r m, ϕ(r)=- ϵ )

21 Simple gas noninterative Fickian diffusion D D(concentration) D = D(temperature only) Arrhenius type i.e. D = D e (53) E = Activation energy for diffusion= Energy for the gas molecule to jump into a new position (hole) Large molecules large energy to move low diffusivity Size of gas molecule (x) is determined by its collision diameter (σ ) : E D : Van Krevelen (1990) For elastomer 10 = ( ) T ± 0.6 (54) For glassy amorphous polymers 10 = ( ) T 298 ± 1.0 (55) R = J/mole K E [=]kj/mole So, the ration of E /R = = 1000 K σ = 37.98(nm), Reference material (See Table 4.8) Table 4.8 here

22 D o : Van Krevelen (1990) logd = 0.4 ± 0.4 (elastomer) (56) = 5.0 ± 0.8 (amorphous glassy polymer) (57) D = D (1 ϕ ) (semi-crystalline) (58) crystallinity Ex 4.6 Diffusivity of O in 298K σ = 34.67(nm) (Table 4.8) σ = T of PC = 150 So, PC is in its amorphous glassy 298K.

23 Solution Eq. (4.55) From (57) 10 = (.. ) [ ( ) ] ± 1.0 = 5.96 ± 1.0 = 5960 ± 1000 (A) R=8.3 J/molK=. KJ/ molk E = ± 1000 = 49.6 ± 8.3 kj/molk LogD = 5.0 ± 0.8 = (.±.) 5.0 ± 0.8. LogD = 0.96 ± 0.8 (B) So, subset (A),(B) (53) D(298) = 10. e ( D = D e (53)) = ( ) Table 4.9 here (Diffusivity)

24 Solubility, S = Amount of gas dissolved in polymer at equilibrium = S(P)P (61) Partial pressure of gas A V Volume of gas A (@STP; 298K, 1atm) dissolved in polymer matrix per unit volume of solution V Volume of polymer per unit volume of solution (polymer+gas) ρ = d V ρ = () d = density of species A = V = (). mass of gas A vol of gas A (STP) S[=]cm (STP)/[cm Pa] Solubility follows Arrhenius eq. S = S e (62) H = molar heat of sorption H R [=] K Van Krevelen (1990) Elastomers: 10 = () ± 0.5 (63) L-J temperature logs = ± 0.8 (64) Amorphous, glassy polymers: 10 = ( ) ± 1.2 (65) logs = ± 1.8 (66) For semi-crystalline polymers:

25 (, ) S = S (1 ϕ ) (67) semi-crystalline polymer Gas is soluble only in amorphous region only Particularily at 298K S = S(T) logs(298) = ± 0.25 (Elastomer) (68) logs(298) = ± 0.6 (Glassy amorphous polymers) (69) S[=]cm /[cm Pa] Ex. 4.7 Solubility of O in PC and 298K? Solution Glassy state of PC and PVAc logs(298) = ± 0.6 (69) Table 4.10 here = 106.7K (Table 4.8) logs(298) = (106.7) ± 0.25 = ~ (cm /cm Pa) Compare this with exp. data (Table 4.10)

26 4.2.2 Permeability and Permachor Permeability, P P DS (71) Determines the amount of diffusing species passing thru a polymer film of unit thickness per unit cross sectional area, per unit time, at a unit pressure difference. Applications Membrane separation - selective diffusion Packaging film - minimize mass transport barrier Fig. 4.7 here

27 Thickness t = 2b Diffusion of gas A Pressure P > P diffusion to the right j = ρ v v = ρ v = n ω n = n (1 ω ) = ρd ω n = ω (1 ω ) 1 (16) = ρd = ρd ω = (72) = 0 = D (v v ) ( = S(P)P (61)) = D S (P P ) (73) 1 i.e. n = D S (P P ) = D S P = P P (74) P Mass transfer thru polymer membrane depends on P

28 P is the difference in pressure of species A of the left and right region. v 1 (volume of the polymer per unit volume of solution) Unit of P: P=DS, D[=], S[=] () P[=] or 1 barrer = 10 cm (STP) = cm /s Pa (76cmHg = 10 Pa) Finally, P = P e / = DS = D e / S e / = D S e ( )/ (75) So, P = D S, E = (E + H ) Activation energy of permeability is sum of diffusion & solubility Approximation based on experiments: logp = = ± 0.25 (Elastomer) (76) ± 0.75 (Amorphous glassy polymers) (77) Experimental data of nitrogen permeability for various polymers showed great deviation depending on the type of polymers, viz. elastomer, amorphous glassy polymer, or semicrystalline polymer: Silicon rubber showed the highest, P = 10 cm /s Pa PVC the lowest, P = 4 10 cm /s Pa Table 4.11 is for rule of thumb Table 4.11 here (Relative Permeability Parameter)

29 A more accurate way to correlate polymer structure with permeability = Permachor, Permeability is defined by P(298) = P (298)e (78) s : A scaling factor P Pre-exponential permeability factor (Constant for specific gas (Table 4.12)) Table 4.12 here

30 π = π (79) π Permachor for amorphous polymer n Total # of individual groups per structural unit of the macromolecule π Individual group value (Table 4.13) For semi-crystalline polymers For oriented crystallites π = π 18ln (1 ϕ ) (80) P, = ( = P ). τ tortuosity of crystallite Group contribution to the molar permachor (Table 4.13) π (Table 4.14) (81) ()

31 (For elastomer, glassy amorphous, semicrystalline polymers) Note π(nr) = 0 π(oriented and crystalline polyvinylidene chloride) = 100 CH 2 CH 2 CH CH Example 4.8 H 3 C CH CH 2 n Cl Cl n CO 2 permeability of nylon 6 and nylon 6,6? (ϕ = 0.4)

32 Solution: Nylon 6 O (CH 2 ) 5 C NH n 2 Nylon 6,6 NHCO (CH 2 ) 4 CONH (CH 2 ) 6 n Group # Contribution CH 5 (10) 15 5 (10) CONH 1 (2) (2) π = π = (75(150) + 309(618)) = = 64 () π = π 18 ln(1 ϕ ) = ln(1 0.4) = 73(cal/cm ) ( ): For nylon 12 So, the permeability is obtained as P(298) = P (298)e =2.48x10-11 (exp-0.122x73)=3.4x10-15 cm 2 /s-pa Moisture Sorption & Diffusion Simple gas - no interaction with polymer Water - interacts with polymer hydrophilic hydrophobic - follow rule of simple gas diffusivity increases with its content So, for hydrophilic polymers logd = logd ω (83) Increases with wt% of water And pre-exponential factor: logd = 7 (84) (D = D e / ) Similar to simple gas For intermediate (hydrophilic- hydrophobic) ones

33 logd = logd 0.08ω (85) Diffusivity decreases with increasing water content Solubility of water in polymers Estimated by group contribution (Table 4.15) Unit : g/g or cm (STP)/g Heat of sorption = 25.4 kj/mole (non-polar) = 40 kj/mole (polar polymer) Example. 4.9 Estimation of saturation moisture content of polymer at 50% RH, 25.

34 CH 2 CH C O n O CH 3 For PMA Table 4.15 (Moles of water/100g of polymer) CH, CH, CH O C 0.05 O ( ) = 0.05mole Or (. ). =. (g/g) (Exp data: 0.99/100) For PS CH 2 CH n = 0.032% (Exp data: 0.032) Solution to the design problem Consider for one jet.

35 Coordinate origin: Maximum diameter position of the cylindrical jet. As z increases diameter decreases due to solvent evaporation and winding-up (stretching). At z=z w, solvent is completely evaporated and the fiber is solidifies. Solvent diffuses from the spinning line to the surface by diffusion (D) and is carried away by the flowing air by convection. Air is flowing cross to the fiber but entrained in parallel direction due to the high speed of fiber. Convective mass transfer occurs in r as well as in z direction (k cc >>k cp ). But the controlling mass transfer in parallel direction due to the small mass transfer coefficient. Estimate D air (Arhenius eq), get ρ and μ from Perry H/B Calculate k as Mass transfer coefficients k = 0.26[D ( ) ]( ) (R(z)) V k = 0.53[D ( ) ]( ) (R(z)) V R(z) : Radius of spinning z=z. Two dimensionless numbers: Schmidt # Prandtl # Sc= = ( Pr= Assume Sc= 1.81, Pr= 0.69 V = 2 m/s V = 50 cm/s ( ) ) k c,p =0.402R(z) -2/3 cm/s Given: R(z)= (at the dope), cm (at the wind up) The corresponding k values are k= 5.9 and k= 23.8cm/s. The resistance to mass transfer scales to 1/5.9=0.17 and 1/23.8=0.04 Note for interface mass transfer: Mass flow rate (mass/time)=ka Δρ A (A=Cross sectional area)

36 Mass flux=mass flow rate/a= k Δρ A (=n A ) Mole flux=mole flow rate/a= k ΔC A = k (C Ai - C A ) (=N A ) Diffusion coefficient of DMF in PAN at average temperature 90 is D=9.03x10-4 exp(-2360/(273+90))=1.36x10-6 cm 2 /s. The internal diffusion reistance >> mass transfer resistance in air So, solution to the diffusion eq (Apply from z=z o ) would give realistic solution. The continuity eq (Table 4.5) becomes t = 1 r r (4.119) with the IC and BC as I. C.: ω (r, 0) = ω B. C. 1: ω (R, t) = 0 B. C. 2: = 0. (4.120) The solution is available elsewhere (Crank: Mathematics of Diffusion) as: = 4 μ exp (4.121) where μ k are the root of Bessel function. When the series converges rapidly, only the first term is taken to be = exp (4.122) where μ 1 = Time t can be replaced by t = z - (4.123)

37 where V is the polymer velocity. Then the mass flow rate is obtained as m = πρr V(1 ω ) (4.124) Combining (4.122)-(4.124) gives dz = (1 ω )ω ( ) Subject to the following BC B. C.: ω A z 0 = ω A0. (4.126) This is nonlinear and can only be solved by numerical method (IMSL). See Fig 4.22 or the solutions. Fig 4.22 here

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