1. Batch adsorption. Rapid adsorption

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2 . Batch adsorption Rapid adsorption

3 Mass Balance On the solute in the liquid dy εv H ( yf y) ( ε ) V dt On the adsorbent dq dt () V : tank volume y : effluent concentration y F : fed concentration H : feed rate q : adsorbed concentration dq ( ε)v dt Vr () r : adsorption rate per tank volume

4 What r? Mechanisms. Adsorption is controlled by diffusion from the solution to the adsorbent.. Adsorption is controlled by diffusion and reaction within the adsorbent particles.

5 Diffusion controls r ka ( y y*) (3) k : mass transfer coefficient a : surface area (adsorbent per tank volume) y*: concentration(equilibrium) n q K(y*) (4) Freundlich isotherm

6 Diffusion and reaction r ( D κ a )( y y*) (5) D : diffusion coefficient κ : first order irreversible reaction Adsorption rate-independent of stirring-strong function of temperature

7 Solve y(t), q(t) Combining and integrating Eqs. and + 3 or 5 + Equilibrium isotherm

8 Assume adsorption isotherm as qky* Integrate equation analytically y y y F σt σt e e F H / εv σ σ σ σ H / εv + σ σ (6) σ i y q / K y F σ t σt F ε H V σ σ σ e + e σ ε + ka + ± ( ε )K σ σ ε H V (7) ε + ka + ( ε )K + : σ - : σ 4kaH K( ε)v (8)

9 Example Novobiocin Adsorption H :.7 liters/hr q o :.35g/liter, y F : 640μg/cm 3, T : 35 o C, ph : Calculate the resin loading q(t). Then estimate the rate constant ka predicted from the simplified theory. Time(hr) y(μg/cm3)

10 Solution : To find the resin loading, we integrate Eq. () : HyFt H q qo + ( ε)v ( ε)v t 0 ydt ε ε y(t) liters 640μg / cm 0 cm.7liters / hr.35g / liter + t hr 0.5liter liter 0.5liter To find k, plot logarithm (y F -y) & time(fig.,next slide) From Eq.6., slop of plotσ (at larger times) σ 0.3 hr From y and q, equilibrium constant K : t 0 ydt 0.876liter 0.50liter y(t) K q y 45mg / cm 48μg / cm μg mg 0

11 Fig.. Novobiocin adsorption.

12 Result ε ε ε + + ε ε ε + + ε σ )V K( 4kaH ( ka V H ka V H (0.50) 4ka(.7) 0(0.5) ka 8.76hr.7 0(0.50liter) 0.876liter ka 0.876liter hr.7liters / hr 0.3 Thus hr.3 ka Eq. 8

13 . FIXED BEDS y(t) : effluent conc. Y B : breakthrough conc.(0%) y E : exhaustion conc.(90%) y F l 0

14 Basic Equations Mass balance in liquid y ε t y y q v + E ( ε) z z t (9) ε : void fraction in the bed v : superficial velocity(h/a) E : dispersion coefficient Mass balance on the adsorbed solute q ( ε ) t r (0) r : adsorption rate(controlled by mass transfer from bulk solution to the surface of the adsorbent) r ka(y Isotherm y*) () r : rate per bed volume ka : rate constant y* : concentration in solution at equilibrium q K ( y*) n () Equilibrium(adsorbent and solution concentrations)

15 Nonlinear Coupled Equations. Numericallyn solved For aproximate analysis - first : breakthrough curves as a ramp - second : two parameters ( characteristic time, standard deviation ) - third : adsorption equilibrium is linear - fourth : mimic graphical analysis

16 . An approximate analysis Equilibrium zone Depend completely on experiments. Equilibrium zone saturated. Adsorption zone long q(equilibrium)q(y F ) Adsorption zone contains half of q(yf) as a good approximation q(adsorption) q(y Fraction of the bed which is loaded F ) q(y Θ F )l( Δt / t B ) + (yf)(l l( Δt / t q(y )l F B )) Δt t B (A)

17 . Two parameter model y y F + erf t t o σt o Two parameters :. Characteristic time. Standard deviation t o : time at half feed concentration σt o : standard deviation(slope of curve) TABLE. Typical characteristics of the standard deviation for breakthrough curves Controlling The Quantity σ is Proportional to The Variance (t o σ) is Proportional to Equilibrium /l l/v Kinetics of adsorption v/l l/v Mass transfer v / /l l/v 3/ Dispersion v/l l/v Diffusion /lv l/v 3

18 .3 Linear adsorption model Linear isotherm q Ky* (3) Eqs. 9,0, and 3 combined y 0 v z ka y and q ( ε) ka y t conditions t0, all z, q0 t>0, z0, yy F q K q K Neglects terms : E / z and y/ t in Eq. 9

19 Five parameters : (y/y F ), t, v, K, k - four parameter found - then fifth found

20 .4 Differential contacting model Mass balance H(y 0) W(q 0) Steady state mass balance on the solute in solution accumulationsolute flow(in-out) - solute adsorbed -Moving countercurrently -Infinitely long, steady state -equilibrium, y F and q(y F ) -exit solution conc. zero Between the exit and some arbitrary position 0 Av(y y z z+ Δz ) kaaδz(y y*) Negligible dispersion dy 0 v ka(y dz Initial condition Z0, yy F y*) Dividing by the volume AΔz and taking the limit as this volume goes to zero

21 l l v y F dy dz 0 ka y (y y*(q)) Evaluate - choosing y, integral numerically and reading q - operating line - using q to find y* from equilibrium

22 Example. Lactate dehydrogenase adsorption L.3m, diameter7cm, void fraction0.3, feed conc..7mg/liter linear isotherm : q(in mg/cm 3 )38y(in mg/cm 3 ) breakthrough 6.4hr, bed exhausted 0hr. (a) the length of the adsorption zone at breakthrough (b) the length of the equilibrium zone at breakthrough (c) the fraction of the bed s capacity which is being used. Solution a) hr, so (3.6hr/6.4hr).3m0.73m b) m c) By approximate analysis and from Eq. A θ-(3.6hr/(6.4hr))7%

23 Example. Cephalosporin adsorption q(g/liter resin)3(y(g/liter solution)) /3 bed length :.0m, diameter : 3.0, density : 0.67cm 3 resin/cm 3 bed feed conc. : 4.3g/liter, superficial bed velocity : m/hr a) Calculate how much of the feed is lost if we stop the adsorption when y0.4g/liter b) Estimate what fraction of the bed s capacity is used at this breakthrough. Assume the bed is exhausted when y4.0g/liter c) Estimate the rate constant in the bed d) If we double the flow rate, how much of the will have been lost when the exit concentration is 0.4g/liter? Solution 6.3hr a) From next slide, breakthrough occurs at 6.3hr ydt 0 thus fraction lost by numerical integration 0.0 y (6.3hr) F

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25 b) From the figure, breakthrough 6.3hr and exhausted 9.0hr from Eq. (A) Δt Θ t B % (6.3) c) By linear adsorption model zka ξ εv τ ka K t ρ m(ka) 0.33(m / hr) z v ka (0.67).5ka hr t m m / hr 0.4ka(t 0.5hr)

26 d) By two parameter model t o 7.8hr(inversely proportional to the velocity) σ0.5(σ will be directly proportional to the velocity(cf. Table )) t o σ : slope of curve y yf (t / 7.8 / )) + erf 0.5 y/y F (0.4/4.0), t.4hr lost : about 5%