Rapid adsorption

1. Batch adsorption

dtdqVyyH

dtdyV F )1()( εε −−−=

VrdtdqV)1( =ε−

Mass Balance

• On the solute in the liquid

• On the adsorbent

V : tank volumey : effluent concentrationyF : fed concentrationH : feed rateq : adsorbed concentration

r : adsorption rate per tank volume

(1)

(2)

What r ?

Mechanisms

1. Adsorption is controlled by diffusion from the solutionto the adsorbent.

2. Adsorption is controlled by diffusion and reaction withinthe adsorbent particles.

Diffusion controls

*)yy(kar −=

n*)y(Kq =

k : mass transfer coefficienta : surface area

(adsorbent per tank volume)y*: concentration(equilibrium)

Freundlich isotherm

(3)

(4)

Diffusion and reaction

*)yy)(aD(r −κ=

D : diffusion coefficientκ : first order irreversible reaction

• Adsorption rate-independent of stirring-strongfunction of temperature

(5)

Solve y(t), q(t)

• Combining and integrating

3or5

Eqs. 1 and 2 +

+ Equilibrium isotherm

t

21

1t

21

2

F

F 21 eey

K/qy σ−σ−σ−σ

σ+σ−σ

σ−=−

ε−−

ε−ε++

ε±

ε−ε++

ε=σ

V)1(KkaH4

K)1(1ka

VH

K)1(1ka

VH

21

2

i

t

21

1t

21

2

F

F 21 eV/HeV/Hy

yy σ−σ−σ−σ

ε−σ+σ−σ

σ−ε=−

*Kyq=

• Assume adsorption isotherm as

• Integrate equation analytically

+ : σ1- : σ2

(6)

(7)

(8)

Example 1 Novobiocin Adsorption

H : 2.7 liters/hr qo : 1.35g/liter, yF : 640μg/cm3,T : 35oC, pH : 6.8∼7.2Calculate the resin loading q(t).Then estimate the rate constant ka predicted from the simplified theory.

Time(hr) y(μg/cm3)0.5 1381.0 1811.5 2172.0 2463.0 3134.0 3485.0 3756.0 4057.0 418

Solution : To find the resin loading, we integrate Eq. (1) :

)t(y1

ydtV)1(

HV)1(tHyqq

t

0F

o

ε−ε−

ε−−

ε−+=

)t(yliter250.0liter876.0ydt

liter25.0hr/liters7.2t

litercm10

liter25.0cm/g640

hrliters7.2liter/g35.1

t

0

333

−−

μ+=

To find k, plot logarithm (yF-y) & time(Fig. 1,next slide)From Eq.6., slop of plot=σ2(at larger times)

hr13.0

2 =σ

From y and q, equilibrium constant K :

110mg

g10cm/g418cm/mg45

yqK

3

3

3

=μμ

==

Fig.1. Novobiocin adsorption.

Result

ε−−

ε−

ε++ε

−

ε−ε++

ε=σ

V)1(KkaH4

1(1ka

VH

11ka

VH

21

2

2

−

++−

++=

)250.0(110)7.2(ka4

)25.0(110876.01ka

hr76.87.2

)liter250.0(110liter876.01ka

liter876.0hr/liters7.2

21

hr13.0

2

Thus

hr3.1ka =

Eq. 8

2. FIXED BEDS

y(t) : effluent conc.YB : breakthrough conc.(10%)yE : exhaustion conc.(90%)

yF

0

l

Basic Equations

tq)1(

zyE

zyv

ty

2

2

∂∂ε−−

∂∂+

∂∂−=

∂∂ε

rtq)1( =

∂∂ε−

*)yy(kar −=

n*)y(Kq =

Mass balance in liquidε : void fraction in the bedv : superficial velocity(H/A)E : dispersion coefficient

Mass balance on the adsorbed solute

r : adsorption rate(controlled by mass transferfrom bulk solution to the surface of the adsorbent)

r : rate per bed volumeka : rate constanty* : concentration in solution at equilibrium

Isotherm

Equilibrium(adsorbent and solution concentrations)

(9)

(10)

(11)

(12)

• 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

2.1 An approximate analysis

q(equilibrium)=q(yF)

)y(q21)adsorption(q F=

BF

BFBF

t2t1

l)y(q

))t/t1(ll)(y(21)t/t1(l)y(q Δ−=

Δ−−+Δ−=Θ

• Equilibrium zone

• Adsorption zone contains half of q(yF) as a good approximation

• Fraction of the bed which is loaded

• Depend completely on experiments•1. Equilibrium zone saturated

2. Adsorption zone long

(A)

2.2 Two parameter model

σ−+=

o

o

F t2tterf1

21

yy

to : time at half feed concentrationσto : standard deviation(slope of curve)

TABLE 1. Typical characteristics of the standard deviation for breakthrough curves Controlling The Quantity σ2 is

Proportional to The Variance (toσ)2 is Proportional to

Equilibrium 1/l l/v2 Kinetics of adsorption v/l l/v Mass transfer v1/2/l l/v3/2 Dispersion v/l l/v Diffusion 1/lv l/v3

Two parameters : 1. Characteristic time2. Standard deviation

2.3 Linear adsorption model

*Kyq =

−−

∂∂−=

Kqyka

zyv0

−=

∂∂ε−

Kqyka

tq)1(

t=0, all z, q=0t>0, z=0, y=yF

Eqs. 9,10,11 and 13 combined

Linear isotherm

and

conditions

Neglects terms : E∂2/ ∂z2 and∂y/ ∂t in Eq. 9

(13)

• Five parameters :(y/yF), t, v, K, k

- four parameter found- then fifth found

2.4 Differential contacting model

)0q(W)0y(H −=−

*)yy(zkaA)yy(Av0zzz

−Δ−−=Δ+

*)yy(kadzdyv0 −−−=

Z=0, y=yF

accumulation=solute flow(in-out) - solute adsorbed

• Mass balance

• Steady state mass balance on the solute in solution

Initial condition

-Moving countercurrently-Infinitely long, steady state-equilibrium, yF and q(yF)-exit solution conc. zero

Between the exit and some arbitrary position

Negligible dispersion

Dividing by the volume AΔz and taking the limit as this volume goes to zero

−==Fy

y

l

0 ))q(*yy(dy

kavdzl Evaluate

- choosing y, integral numerically andreading q

- operating line- using q to find y* from equilibrium

Example 1. Lactate dehydrogenase adsorption

L=1.3m, diameter=7cm, void fraction=0.3, feed conc.=1.7mg/literlinear isotherm :q(in mg/cm3)=38y(in mg/cm3)breakthrough 6.4hr, bed exhausted 10hr.(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) 10-6.4=3.6hr, so (3.6hr/6.4hr)×1.3m=0.73mb) 1.3-0.73=0.57mc) By approximate analysis and from Eq. A

θ=1-(3.6hr/2(6.4hr))=72%

Example 2. Cephalosporin adsorption

q(g/liter resin)=32(y(g/liter solution))1/3bed length : 1.0m, diameter : 3.0, density : 0.67cm3 resin/cm3bedfeed conc. : 4.3g/liter, superficial bed velocity : 2m/hr

a) Calculate how much of the feed is lost if we stop the adsorption when y=0.4g/liter

b) Estimate what fraction of the bed’s capacity is used at this breakthrough.Assume the bed is exhausted when y=4.0g/liter

c) Estimate the rate constant in the bedd) If we double the flow rate, how much of the will have been lost when

the exit concentration is 0.4g/liter?

Solution

a) From next slide , breakthrough occurs at 6.3hrthus fraction lost by numerical integration =0.02 )3.6(

3.6

0

hry

ydt

F

hr

=

b) From the figure, breakthrough = 6.3hr and exhausted = 9.0hrfrom Eq. (A)

%79)3.6(23.60.91

t2t1B

=−−=Δ−=Θ

c) By linear adsorption model

)hr5.0t(ka124.0hr/m2

m1t)67.0(12

kavzt

Kka

hrka5.1

)hr/m2(33.0)ka(m1

vzka

−=

−=

−

ρ=τ

==ε

=ξ

d) By two parameter model

to=7.8hr(inversely proportional to the velocity)σ=0.15(σ2 will be directly proportional to the velocity(cf. Table 1))toσ : slope of curve

−+=15.02

1))2/8.7/t(erf121

yy

F

y/yF=(0.4/4.0), t=2.4hrlost : about 5%