Mass Transfer Part

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12. ADSORPTION 12.1 Introduction

 Adsorption operation involves contact of solids with either liquids or gases in which the mass transfer transfer is towards towards solids. The reverse of this this operation is called called  Desorption.. Adsorption operations exploit the ability of certain solids to concentrate  Desorption specific substances from fluid on to their surfaces. The adsorbed substance is called ‘adsorbate adsorbate’’ and the solid substance is called ‘adsorbent  ‘ adsorbent ’. ’. Typical applications of this solid-liquid operation are as follows:



removal of moisture dissolved in gasoline



de-colorization of petroleum products and sugar solutions



removal of objectionable taste and odour from water.

solid - gas operations include:



dehumidification of air and gases



removal of objectionable odours and impurities from gases

• •

recovery of valuable solvent vapours from dilute gas mixtures to fractionate mixtures of hydrocarbon gases such as methane, ethane and

 propane. 12.2

Types of adsorption:

The two types of adsorption are  physical adsorption or  or  physi-sorption  physi-sorption (van der Waals adsorption) and chemi-sorption (activated adsorption).  Physical adsorption is a readily reversible phenomenon, which results from the intermolecular forces of attraction between a solid and the substance adsorbed. Chemi-sorption is the result of chemical interaction, generally stronger than  physi-sorption between the solid and the adsorbed substance. This process is irreversible. It is of importance importance in catalysis. 12.3

Nature of adsorbents:

Adsorbentss are usually in granular form with their size ranging from 0.5 mm Adsorbent to 12 mm. They must must neither neither offer high pressure pressure drop drop nor get carried carried away by flowing stream. stream. They must must not loose loose their their shape shape and size while while handling. handling. They must have larger surface area per unit mass and also lot of pores. Some of the commonly used adsorbents, their sources and applications are given below:

157

 

Sl. No.

Adsorbent

1.

Fuller’s ea earth

2.

Activated Clay

3.

Bauxite

4.

Alumina

5.

Bone-char

6.

Activated carbon

7.

Silica gel

8.

Molecular   sieves

  12.4 12.4

Source

Application

Naturally occ occurring cla clay is he heated De-colorizing, drying of  and dried to get a porous structure. lubr lubricat icating ing oils, oils, kerosene kerosene and engine oils. for de-colorizing Bentonite or other activated clay Used products . which activated by treatment with  petroleum products. sulfuric acid and further washing, drying and crushing. A naturally occurring hydrated alumina, alum ina, activat activated ed by heating heating to o 230 to 815 C. A hard hydrated aluminium oxide, which is activated by heating to drive dri ve off the the moist moistur uree and then then crushed to desired size. Obtained by destructive disti distill llati ation on of cr crush ushed ed bones bones at o 600-900 C. i) Vegetable matter is mixed with calcium chloride, carbonized and finally fina lly the inorgani inorganicc compoun compounds ds are leached away. ii) Organic matter is mixed with  porous pumice stones and then heated and carbonized to deposit the carbonaceous matter   throughout the porous particle. iii) iii) Carboni Carbonizing zing substanc substances es like like wood, wo od, sawdu sawdust st,, co cocon conut ut shell shells, s, fruit pits, coal, lignite and subsequent activation with hot air  steam. stea m. It is availa available ble in granula granular  r  or pellated form. A ha hard gr granular an and po porous  product obtained from sodium sil silic icate ate solut solution ion af after ter trea treatm tment ent with wit h acid. acid. Normally Normally has has 4 to 7% water in the product. These are porous synthetic zeolite crystals, metal alumino - silicates.

Used for de-colorizing  petroleum products and for  drying gases. Used as desiccant.

Used for refining sugar and can  be reused after washing and  burning. De-colorizing of sugar   solu soluti tion ons, s, chem chemic ical als, s, drug drugs, s, water wat er purific purificatio ation, n, refinin refining g of  veg vegeta etable and anim nimal oil oils, reco recove very ry of gold gold and and silv silver  er  from cyanide ore-leach soluti sol ution, on, recov recovery ery of solven solventt vapo vapour ur from from gasgas-m mixtu ixture res, s, colle col lecti ction on of gasol gasoline ine hydro hydro-carb arbons ons from natu atural gas, gas, fract fra ction ionati ation on of hydro hydrocar carbon bon gases.

Used for de-hydration of air and othe otherr ga gase ses, s, frac fracti tion onat atio ion n of  hydro-carbons.

Dehy hyd dration of gases and liquids, and separation of gasliquid hydrocarbon mixture.

Adso Adsorp rpti tion on equi equili libr bria ia:: Different gases and vapors are adsorbed to different extent under comparable

conditions as shown in Fig. 12.1.

158

 

  g    H     m   m  ,   e    t   a    b   r   o   s    d   a    f   o   r   e   u   s   s   e   r   p    l   a    i    t   r   a   p     m   u    i   r    b    i    l    i   u   q    E (B) 100ºC

(B) 70ºC

(A) 100ºC 

100˚C ºC

Fig.12.1 Equilibrium adsorption on activated carbon

(A)

As a general  rule, vapors and gases with higher molecular weight and lower  critical temperature temperature are more readily readily adsorbed. To some extent, level level of saturation (A) 60ºC

also influences influences the degree of adsorption. adsorption. The adsorption adsorption isotherms isotherms are generally concave to pressure axis. However, other shapes are also exhibited as shown in Fig 12.2.

 

   f   o   e   r   u   s   s   e   r   p    l   a   g    i    t   r    H     p   a     m   m  m  ,   u   e    i   r    t   a    b   b    i    l    i   r   u   o   s   q   d    E   a

kg adsorbed /kg adsorbent

Fig.12.2:Adsorption Isotherms Fig.12.2:Adsorption Repeated adsorption and desorption studies on a particular adsorbent will

change the shape of isotherms isotherms due to gradual change change in pore-structure. pore-structure. Further, Further, adsorption is an exothermic process and hence the concentration of adsorbed gas decrease decr easess with increase increase in temper temperatur aturee at a constant constant pressur pressure. e.

Similar Similarly ly an

increase in pressure increases the concentration of adsorbed gas in the adsorbent kg adsorbed /kg adsorbent 159

30ºC

 

at a consta constant nt te temp mpera eratu ture. re.

Ther Theree are three three co comm mmonl only y used used math mathem emati atical cal

expressi expr essions ons to describ describee vapor vapor adsorpti adsorption on equilibr equilibria ia viz. Langmu Langmuir, ir, Brunaver Brunaver-Emmattt - Teller (BET) and Freundlich Emmat Freundlich isotherms. The first two are derived from theory whereas the last one is by a fit technique from the experimental data. 12.5 12.5

Adso Adsorp rpti tion on hyst hyster eres esis is::

The adsorpti adsorption on and desorpt desorption ion operatio operations ns exhibit exhibit differen differentt equilib equilibrium rium  phenomena as shown in Fig 12.3 and is called Adsorption Adsor ption hysteresis.    l   a    i    t   r   g   a   H     p     m   m  m   u  ,    i   r   e    b   r    i   u    l    i   s   s   u   e   q   r    E   p

Adsorption

Desorption kg adsorbed /kg adsorbent Fig.12.3: Adsorption Hysteresis This may be due to the shape of the openings to the capillaries and pores

of the solid or due to the complex phenomena of wetting of the solid by the adsorbate.. Whenever hysteresis adsorbate hysteresis is observed, observed, the desorption desorption curve curve is below the adsorption curve. 12.6 12.6

Hea Heat of adso sorrptio ption: n:

The differential heat of adsorption (-H) is defined as the heat liberated at constant temperature when unit quantity of vapor is adsorbed on a large quantity of solid already containing containing adsorbate. adsorbate. Solid so used is in such a large large quantity that the adsorbate concentration remains unchanged. The integral heat of adsorption, (ΔH) at any concentration X is defined as the enthal enthalpy py of the adsorb adsorbat ate–a e–adso dsorbe rbent nt combi combinat nation ion minus minus the sum sum of the enthalpies of unit weight of pure solid adsorbent and sufficient pure adsorbed substance (before adsorption) to provide the required concentration X, at the same temperature. The differenti differential al heat of adsorpti adsorption on and integra integrall heat of adsorpt adsorption ion are functions of temperature and adsorbate concentration. 12.7 12.7

Effe Effect ct of Temp Temper erat atur ure: e:

160

 

Increase Incr ease of tempera temperature ture at constant constant pressure pressure decrease decreasess the amount amount of  solute adsorbed adsorbed from a mixture. mixture. However, However, a generalization generalization of the result result is not easy. Fig 12.1 also indicates the effect of temperature. 12.8 12.8

Ef Effe fecct of Pre Press ssur ure: e:

Generally Generall y lowering of pressure reduces the amount of adsorbate adsorbed upon the adsorbent. However, However, the relative adsorption adsorption of paraffin hydrocarbon hydrocarbon on carbon decreases at increased pressures. 12.9

Liquids

The impurities are present both at low and high concentrations in liquids. These The se are normally normally removed removed by adsorpt adsorption ion techniq technique. ue. The charact characteris eristics tics of  adsorpti adso rption on of low and high concentrat concentration ion impurit impurities ies are differen different. t. They are discussed below. 12.9.1 12.9. 1 Adsorptio Adsorption n of solute solute from dilute dilute solutions solutions::

Whenever a mixture of solute and solvent is adsorbed using an adsorbent,  both the solvent and solute are adsorbed. Due to this, only relative or apparent adsorption of solute can alone be determined. Hence, it is a normal practice to treat a known volume of solution of  original concentration concentration C0, with a known known weight of adsorbent. adsorbent. Let C* be the final final equilibrium concentration of solute in the solution. If ‘v’ is the volume of solution per unit mass of adsorbent (cc/g) and C 0 and C* are the initial and equilibrium concentrations (g/cc) of the solute, then the apparent adsorption of solute per unit mass of adsorbent, neglecting any change in vo volu lume me is v( v(C C0-C*), -C*), (g/g). (g/g).

This This expressi expression on is mainly mainly appli applicab cable le to dilut dilutee

solutions. solutio ns. When the fractio fraction n of the original original solvent solvent which can be adsorbed adsorbed is small sm all,, the C* value value de depen pends ds on the the temp temper erat ature ure,, nature nature and proper properti ties es of  adsorbent. In the the case case of dilut dilutee solut solution ionss an and d over over a small small conce concentr ntrat ation ion range range,,  Freundich adsorption Isotherm describes the adsorption phenomena, C* = K [v (C0-C*)]n

(12.1)

Freundlich adsorption equation is also quite useful in cases where the actual identity of the solute is not known (eg.) removal of colouring substance from sugar sugar solutions, solutions, oils oils etc. The color color content in the solution solution can easily be measured using spectrophotome spectrophotometer ter or colorimeter. The interpretation interpretation of this data is illust illustra rated ted in worked worked example example 2. If the value value of ‘n’ ‘n’ is high, high, say 2 to 10, 10,

161

 

adsorption is good. If it lies between 1 and 2, moderately difficult and less than 1 indicates poor adsorption characteristics. A typical adsorption isothermal for the adsor ads orpti ption on of variou variouss adsorb adsorbat ates es A, B an and d C in dilut dilutee solut solution ion at the same temperature for the same adsorbent is shown in Fig 12.4.   n    i   n    i   o    t   a   r    t   n   e   c   n   o   c   e    t   u    l   o   s    m   u    i   r    b    i    l    d    i    i   u   u   q   q    l    E   i

(C)

(B)

(A)

kg Solute apparently adsorbed /kg adsorbent

Fig.12.4: Adsorption Isotherms for various adsorbates 12.9.2 Adsorption from from concentrated concentrated solution: solution:

When the apparent adsorption of solute is determined over the entire range of concentr concentrati ations ons from pure solvent solvent (0% solute concent concentrati ration) on) to pure solute (100% solute concentration concentration), ), curves as shown in Fig.12.5 Fig.12.5 will occur. Curve ‘1’ occurs when the solute is more strongly adsorbed in comparison to solvent at all solute concentration. concentration. Whenever Whenever both solute and solvent are adsorbed adsorbed to nearly samee extent, sam extent, the ‘S’ shaped shaped curve ‘2’occur ‘2’occurs. s. In the range PQ solute solute is more more strongly adsorbed adsorbed than than solvent. At point Q both are equally equally well well adsorbed. adsorbed. In the range QR solvent is more strongly adsorbed.

162

 

 

   d    i   u   q    i    l   n    i   e    t   u    l   o   s    f   o   n R  adsorbed /kg adsorbent   o kg Solute apparently    i    t (1)   a   r    t   n   e   c   n   o    C

Q (2)

P

Fig.12.5: Adsorption of Solute in concentrated concentrated solutions 12.9.3: Other adsorption Isotherms:  Langmuir adsorption adsorption Isotherm:

163

 

The theory proposed by  Langmuir  postulates that gases being adsorbed by a solid surface cannot form a layer more more than a simple simple molecule in depth. His theory visualizes adsorption as a process consisting two opposing actions, a condensation of molecules from the gas phase on to the surface and an evaporation of molecules from the surface  back into the body of the gas. When adsorption starts, every molecul moleculee colliding with the surface may condense condense on it. However, as as adsorption proceeds, proceeds, only those those molecules molecules which strike the the uncovered area surface surface can be adsorbed. adsorbed. Due to this the initial initial rate of  condensation condensati on of molecules molecules on the surface is very high and decreases as time progresses. progresses. The molecules attached to the surface also get detached by factors like thermal agitation. The rate at which desorption occurs depends on the amount of surface covered by moleculess and will increase as the surface becomes mo molecule more re fully saturated. When the rate of adsorption and desorption become equal, adsorption equilibrium is said to be reached. If ‘θ’ is the fraction of surface covered by adsorbed molecules at any instant, the fractional area available for adsorption is ‘(1- θ)’. The rate at which the molecules strike the unit area of surface is proportional to Pressure. Therefore the rate of condensation = k 1(1- θ)P Where k 1 is a constant Similarly, the rate of evaporation α k 2 θ Where, k 2 is a constant Under adsorption equilibrium, k 1(1- θ)P = k 2 θ (i.e.) θ = k 1(1- θ)P / k 2 + k 1P = bP/1+bP where, b = k 1/k 2.2.

(12.2)

 Now the gas adsorbed per unit area or unit mass of adsorbent, y, must obviously be  proportional to the fraction of surface covered. Hence, y = k θ = k{ bP/1+bP }= aP/1+bP

(12.3)

where, a and b are constants. This is Langmuir Langmuir adsorption Isotherm BET Adsorption Isotherm:

This Th is postu postula late tess that that the the adsorp adsorpti tion on phenom phenomeno enon n involv involves es the forma formati tion on of many many multilayers on the surface rather than a single one. Based on this Brunauer, Emmett and Teller Tel ler derived derived the followin following g adsorpti adsorption on isotherm isotherm popularl popularly y called called BET adsorpti adsorption on Isotherm.

164

 

+  0 [v( P  − P )] [ vmc]   P 

=

1

( c − 1)    P  

(12.4)

  P o 

vmc

where, v is the volume, reduced to standard conditions of gas adsorbed at pressure P and temperature T P° the saturated vapor pressure of the adsorbate at temperature T, vm

the volume of gas reduced to standard conditions, adsorbed when the surface is

covered with a unimolecular layer, c is a constant at any given temperature equal to approximately by exp[(E1-E2)/RT]

12 12.1 .10 0 Type Typess of Ope Opera rati tion on::

Adsorption operations are carried out either on batch or continuous basis. Batch process process is not very much used. However, However, a batch operation operation is quite useful in obtaining equilibrium. equilibrium. Much widely used continuous continuous operation operation can either be a single stage or a multistage multistage operation. operation. The multistage multistage operation could could once again  be either a cross-current cross- current operation or o r a counter current operation. oper ation.

12.10.1 Single Stage Operation:

A schematic arrangement for a single stage operation is shown in Fig 12.6. LS, X0

GS, Y0

GS, Y1

LS, X1

Fig.12.6: Single stage operation

The concentration concentration of solute increases in the adsorbent from X0 to X 1 (g/g) and the concentration of solute in the solution decreases from Y 0 to Y1 (g/g).

165

 

The mass balance for solute gives, Gs [Y0 – Y1] = LS [X1 – X0] (i.e)

LS −

=

GS

( Y0 ( X0





(12.5)

Y1)

(12.6)

X1)

where (LS/GS) indicates the slope of the operating operating line passing through the  points (X0,Y0) and (X1,Y1). If the leaving streams streams are in perfect equilibrium equilibrium,, then th thee po poin intt (X1,Y1) wi will ll lie lie on the equi equili libr briu ium m adso adsorp rpti tion on Isot Isothe herm rm.. If the equili equ ilibr brium ium is not re reach ached ed due to facto factors rs like like poor poor contac contacti ting ng then then point point P represents the conditions of leaving streams as shown in Fig.12.7.

Y

X

Y0

Y1

0

X0 ,

Fig.12.7: Adsorption Isotherm and operating line for a single stage operation

166

 

Assumin Assu ming g the validit validity y of Freundli Freundlich ch equatio equation, n, especia especially lly when a low concentration of solute is involved, the equation can be written as, Y* = m xn

(12.7)

and at the final equilibrium conditions, 1/n

 Y1   X1 =      m  

(12.8)

when the pure adsorbent is used, (i.e.) X0 = 0 The above equation (12.8) yields LS GS

=

( Y0 − Y1) 1/n

 Y1        m  

(12.9)

12.10.2 Multistage cross-current cross-current operation:

A schematic arrangement of multistage crosscurrent operation is shown in Fig.12.8. LS, X0

LS, X0

LS,X0 G,Y

GS, Y0

1

Feed

S

GS,Y1

LS,X1

2

2

LS,X2

3

GS,Y3

LS,X3

Fig.12.8:Multistage Fig.12.8:Mu ltistage cross current operation

Making a material balance of solute for stage 1 and use of Freundlich equation for  the entry of pure adsorbent gives GS (Y0-Y1) = LS1 (X1 – X0)

(12.10)

According to Eq.(12.9), LS1 GS

1 ) = ( Y0 − Y 1/ n  Y1        m  

(12.11)

A material balance of solute for stage 2 yields, GS (Y1 – Y2) = LS2 (X2-X0)

(12.12)

Use of Freundlich equation for the entry of pure adsorbent gives LS2 GS

=

( Y1 − Y2 ) 1/n

 Y2        m  

(12.13)

A similar material balance for stage ‘p’ yields GS( Y p − 1 Y p ) LSp( X p X0 )



=



(12.14)

Using Freundlich equation as before gives

167

 

LSp GS

=

( Y p − Y p ) - 1

1/n

 Y p        m  

(12.15)

This operation is represented graphically as shown in Fig.12.9.

Y0 Operating line slope = - LS1/GS

Y

Oper slop

Y1 Y2 0

X0

,

X

X1

X2

,

Fig.12.9:Adsorption Fig.12.9:Adsor ption Isotherm and operating line for a two stage cross curre current nt operation 12.10.2.1 Steps involved in the determination of number of stages needed for a cross-current adsorption process:

1) Draw Draw the the equil equilibri ibrium um curve curve (X Vs Vs Y) 2) Loca Locate te the the poi point nt (X0,Y0) and draw the operating line with a slope

 − LS1        GS  

168

 

3) The The inter intersec sectio tion n of opera operati ting ng line line an and d equil equilibr ibrium ium curve curve yield yields, s, (X1, Y1) –  conditions of stream leaving stage (I).

   

4) Locate (X (X0, Y1) and draw the operating line with a slope of   −

LS2  

  (since X0

GS  

remains constant for adsorbent for II stage). 5) Inter Intersec secti tion on of opera operati ting ng line line and equili equilibri brium um curve curve yield yieldss (X2,Y2) – the conditions of leaving stream from stage II. 6) Proc Proceed eed in in the sam samee way til tilll the X NP point is crossed and count the number of  stages for the specified amount of adsorbent for each stage. 12.10.2.2 Optimisation of a two stage cross-current operat operation: ion:

In a typical two stage operation, the concentrations of solute both in the inlet solution stream and outlet solution stream are fixed along with the feed rate of solution. The objective objective will be to use the minimum minimum amount of adsorbent adsorbent for  this. If the quantity of the adsorbent adsorbent is changed the exit conc concentratio entration n of solution from each stage will also vary. However, the terminal conditions are always fixed and only the intermediate intermediate concentration is a variable. Hence, with with one particular  interm int ermedi ediate ate value value,, if the amoun amounts ts of adsorb adsorbent ent used used in both both the the stages stages are are estimated, it will result in the minimum amount of adsorbent being used. For schemat schematic ic arrange arrangemen mentt shown shown in Fig.12.1 Fig.12.10, 0, the materi material al balance balance equations for stage I and II are from Eqs.(12.8) and (12.9) as LS, X0

Feed

GS, Y0

LS,X0

GS, Y1

1

LS,X1

2

GS, Y2

LS,X2

Fig. 12.10:Two stage cross current operation

LS1 GS

=

( Y0 − Y1) 1/n

 Y1        m  

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LS2 GS

(Y

=

−Y  Y       m   1

2

)

1 /  n

2

Adding equations (12.11) and (12.12) we get,

 LS1 + LS2  = ( Y0 − Y1) + ( Y1 − Y2 )    1/n 1/n 2 S 1 G Y Y                   m   m      

(12.16)

The total amount of adsorbent used can be optimised with respect to ‘Y 1’ (the intermediate concentration), the only variable on the R.H.S. of Eq. (12.16). The other parameters Y0, Y2, m and n are all fixed for a specified operation involving a specific adsorbent.

    d  [ LS1 + LS2]  d  ( Y0 − Y1) ( Y1 − Y2 )  (i.e.)  = dY1  Y1 1 / + Y2 1/n  dY1  GS                    m     m    d  ( Y0 − Y1) ( Y1 − Y2 )  = m1/n +   dY1  ( Y1) 1/n ( Y2 ) 1/n 

(12.17)

n

1/n

(i.e.) m

1/n

-1/n

m (Y2)

m1/n(Y2)-1/n

 [(Y0 − Y1)Y21/n + (Y1 − Y2)Y11/n ]    dY1  [(Y2)1/n (Y1)1/n ]  d

[(Y0Y21/n ) − (Y1Y21/n )] + [Y1(1+1/n) − Y2Y11/n ]    dY1  Y11/n  d

d dY1

[Y Y Y − − Y 0

2

1

1/n

2

1/n

.Y1 (1−1/n)

+ Y1 − Y2]

(12.18)

(12.19)

m1/n (Y2)-1/n [Y0.Y21/n (-1/n) Y1-1-1/n – Y21/n (1-1/n)Y 1-1/n + 1 – 0] (12.20) For minimum adsorbent R.H.S. of (12.20) should be zero. (i.e.) Y0Y21/n (-1/n) Y1-1-1/n – (1-1/n)Y21/n Y1-1/n + 1 – 0 (12.21) (since m, n and Y2 have definite values) 1/n

 Y2   Dividing by     Y1  

we get, 1/n

 Y0   − 1  −  1 − 1  +  Y1   = 0             Y1    n     n    Y2   1/n

(i.e.)

 Y1    1   Y0    1     Y2     −   n      Y1    = 1 −   n    

(12.22)

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Thee abov Th abovee Eq.( Eq.(12 12.2 .22) 2) ca can n be solv solved ed by tria triall and and erro errorr to ge gett the the interme inte rmediat diatee concentr concentrati ation on ‘Y1’ whic which h wi will ll op opti timi mise se the the tota totall qu quan anti tity ty of  adsorbent to be used. used. However, However, using the chart shown shown in Fig.12.11 Fig.12.11 below also, also, we can get the intermediate concentration.

Fig. 12.11 Minimum total adsorbent two stage crosscurrent Eq. (12.22) 12.10.3 Multistage Counter current adsorption:  GS,Y0 Solution to be treated Spent adsorbent LS, X1

GS,Y1

GS,Y2

G ,Y S

1

2 X2

3

Y  NP - 1

3 X3

X4

X Np

GS,Y Np Final solution  NP  Fresh LS, X Np+1 adsorbent

Fig. 12.12: Multistage counter-current counter-current operation

The schematic arrangement as shown in Fig. 12.12 represents the multistage counter current operation Solute balance for the system as a whole gives, GS (Y0 – Y NP) = LS (X1 – X NP+1)

(12.23)

 LS   ( Y0 − Y Np )   GS    = ( X1 − X Np + 1)

(12.24)

(i.e.)

171

 

Eq. (12.24) gives the slope of the operating line passing through the terminal conditi cond itions ons (X1,Y0) and (X Np+1,Y Np).

By conven convention tional al stepwis stepwisee constru constructi ction on

starting from the point (X 1,Y0) the number of theoretical stages are estimated. This operation is represented graphically as shown in Fig.12.14 B

Y0

Operating line slope = LS/GS

1

Y1

Y 2 Equilibrium curve Y2 Y Np -1

Y Np

A

X Np+1

 N p

X Np X3

X2

X1

X

Fig.12.13: Counter current multistage adsorption

In order to determine the minimum amount of adsorbent for the process, draw a line from the point ‘P’ (X Np+1, Y Np) which could be a tangent to the equilibrium equilibr ium curve. In such cases, the slope of the line gives the ratio of (L S/GS)min. However, Howev er, in the case case of eq equil uilib ibriu rium m curve curve being being a str strai aight ght line line or concav concavee upwards, draw a horizontal line from ‘Y0’ to intersect the equilibrium curve, (or  line) at a point by Q and then join PQ which gives the slope of the operating line (LS/GS)min. The above two cases have been shown graphically in Fig.12.13 (a) and Fig.12.13 (b).

172

 

 

Y   Operating line slope = LS/G S 

P

Y   Equilibrium curve

Y

 

Y

Equilibrium curve P Operating line slope = L S/G S 

T

Q Y Np 

Y N  Np p 

X Np+1 

X Fig.12.14 (a)

X Np+  Np+1 1 

X Fig.12.14 (b)

Fig. 12.14: Operating line and minimum adsorbent/solvent ratio for infinite stages 12.10. 12. 10.3.1 3.1 Steps Steps involv involved ed in deter determini mining ng the numbe numberr of stages stages in a multis multistag tagee Counter Current operation:

1) Draw Draw tthe he equ equil ilibr ibrium ium cu curve rve 2) Loca Locate te the point point ‘P’ ‘P’ ((X X Np+1, Y Np) 3) Draw Draw a llin inee with with a slo slope pe of of (LS/GS), where LS is the mass flow rate of solute free adsorbent and GS is the mass flow rate of solution on solute free basis. 4) St Star arti ting ng fro from m (X1,Y0) by stepwise construction estimate the number of stages till the point (X Np+1, Y Np) is crossed. The operation is graphically shown in Fig. 12.14. 5) If it is desired desired to determ determine ine the amount amount of adsorbe adsorbent nt needed needed for a specifie specified d level of solute removal from a solution stream with a fixed number of stages, draw the operating line of different slopes by trial and error and choose the one which gives exactly the same number of specified stages and the specified concentration concentr ation in the liquid stream. stream. From the slope slope of the operating line, line, thus chos chosen en,, de dete term rmin inee the the amou amount nt of adso adsorb rben entt to be us used ed and and the the solu solute te concentration in the adsorbent. 12.10.3.2 Optimization of two stage counter current adsorption

For a typical two stages counter-current counter-current operation as shown schematically schematically in Fig. 12.15,

173

 

Y1

Y0 1

2

X3 = 0

X2

X1

Y2

Fig. 12.15: Two stage counter current adsorption

Solute balance for the system as a whole with pure adsorbent yields, LS (X1 – 0) = GS [Y0 – Y2] Applying Eq. (12.8) yields

(12.25)

1/n

 Y1   LS   = GS( Y0 − Y2 )   m   LS ( Y0 − Y2 ) = 1/n GS  Y1        m  

(12.26)

(12.27)

Applying a similar balance for stage 2 yields, 1/n

GS (Y1-Y2) =LS(X2 – 0) =

 Y2   LS      m  

 LS  = ( Y1 − Y2 )     GS    Y2  1/n      m  

(12.28)

(12.29)

Equating (12.27) and (12.29) we get

( Y0 − Y2 ) 1/n

 Y1        m   ( Y0 − Y2 ) Y2

=

( Y1 − Y2 ) 1/n

 Y2        m   1/n -1/n (Y1 − Y2)  Y1   Y2      =    ×     Y2   m     m    1/n

     Y1    − 1 =    Y0    Y2    Y2    

   − 1  Y1   Y 2     

(12.27)

Since Y0, Y2, n are all specific values for a specified level of adsorption

and also a specific adsorbent, the only unknown ‘Y 1’, can be estimated by trial and error. Alternately, ‘Y1’, can be estimated by the following chart as shown in Fig.12.16.

174

 

Fig.12.16: Two – stage counter current adsorption Eq. (12.27) 12.11 12. 11 Contin Continuo uous us adsorp adsorptio tion: n:

In these these adsorber adsorbers, s, the fluid and adsorben adsorbentt are in continuo continuous us contact contact without any separation separation of the the phases. This is quite quite analogous to gas gas absorption with the solid adsorbent replacing the liquid solvent. The operation can be carried out in strictly continuous, continuous, steady state fashion with both fluid and solid moving at constant rate rate and the composition composition remains const constant ant at a particul particular ar point. It can also als o be operat operated ed on semi-c semi-cont ontin inuou uouss basis basis wi with th solid solid parti particl cles es remai remainin ning g stationary stationa ry and fluid in moving moving condition. condition. Such operations operations constitute constitute unsteady unsteady adsorption process. 12.11.1

Steady state adsorption:

Continuo Cont inuous us differe differentia ntiall contact contact tower tower is schemat schematical ically ly represen represented ted in Fig.12.17.

175

 

X2

Y2 2

Y dz

z

X

1

Y1

X1

Fig.12.17: Continuous differential contact tower

Solute balance for the entire tower is GS (Y1 – Y2) = LS (X1 – X2)

(12.31)

Solute balance for the upper part of the tower is GS (Y-Y2) = LS (X- X2)

(12.32 )

Using Eq. (12.31), one can draw the operating line and Eq. (12.32) gives us the concentration of the two phases at any point in the tower . Making a solute balance across the element of thickness ‘dz’, LS (dx) = GS dY = K y a (Y – Y*) dZ

(12.33 )

where K y a is mass transfer coefficient based on the outside surface area ‘a’ of particl particles, es, kg/m3.s .s.( .(∆Y ∆Y)) and Y* is equil equilibr ibrium ium co conce ncentr ntrat ation ion of the fluid fluid corresponding to its concentration X. Eq.(12.33) on integration yields, Y2 K Ya dY =  NtoG = GS Y2 (Y − Y*)

∫ 

Z

∫ 

dz

0

=

Z HtoG

(12.34 )

where HtoG = GS/K ya  NtoG can be determined graphically as usual. 12.11.2 Unsteady state adsorbers:

When a fluid mixture is passed through a stationary bed of adsorbent, the adsorbent adsorbs solute continuously and it results in an unsteady state operation. Ultimately Ultima tely,, the bed may get saturated saturated and no further further adsor adsorptio ption n results results.. The change chan ge in concentr concentratio ation n of effluent effluent stream is shown in Fig.12. Fig.12.18. 18. The system system

176

 

indicates an exit concentration concentration varying from C, to a final concentration concentration very close to inlet concentration. The point ‘A’ indicates break point. The portion from A to B is termed termed the break through through curve. Beyond Beyond this, very little little adsorpti adsorption on takes  place, indicating that the system has more or less reached equilibrium or  saturation.   o   n    i    t   a   r    t   n   e   c   n   o    C

B

A 0

Volume of effluent

Fig.12.18: The adsorption wave 12.12 Equipments for adsorption:

Equipments are available for adsorption of a solute from a gaseous or a liquid stream. When the solute (which could be colouring colouring matter, odorous substances, valuable solutes etc) to be adsorbed is relatively strongly adsorbed from a liquid stream one can use contact filtration equipments which can be operated as batch units, semi continuous or as continuous ones. Continuous ones can be realized by fluidized bed techniques. These are similar to mixer-settler mixer-settler units units used in extraction operations. operations. Generally Generally gases are treated with fluidized bed techniques. 12.12.1 Contact filtration equipments: The equipment consists of a mixing tank in which the liquid to be treated and the adsorbent are thoroughly mixed at the operating temperature and for a specified duration of time. In some cases like ion exchange sparging is done with air Subsequently the slurry is filtered filtered off to separate the solids solids from the solution.. solution.. The filtration filtration is done in a filter press or centrifuge or in a continuous rotary filter. Multi stage operations could easily be done by providing a number of tanks and filter combinations. The filter cake is usually washed to displace the solution. If the adsorbate is the desired product, then it can be desorbed by contact with a solvent other than the one which constituted the solution and the one in which the solute is more readily soluble. When the solute is more

177

 

volati vol atile, le, it ca can n re remo moved ved by the passa passage ge of steam steam or warm warm air air throu through gh the solid. solid. Whenever, the adsorbent is activated carbon care must be taken so that the adsorbent does not burn away at high temperatures of desorption operation operation.. Adsorbent can also be regenerated by burning away the adsorbate. 12.12.2 Fluidised beds:

When the mixtures of gases are to be treated on a continuous basis it is preferable to use fluidize flui dized d beds. beds. This This is done by passing passing the gases gases at high velociti velocities es through through a bed of  granular solids in which the adsorption occurs. The beds of solids remain in suspended condition throughout the operation. The bed can be regenerated by passing steam/air at high temperatures. To improve the effectiveness of operation, one can go in for the multistage counter operation with regeneration. In these operations one has to take care to minimize or prevent the carry over of solids. 12.12.3 Steady- state moving bed adsorbers:

In th this is ca categ tegory ory of adsorb adsorber erss both both the solids solids and flui fluid d move move contin continuou uously sly..

The The

composition at any particular point is independent of time. They are operated with the solids moving down wards and the liquid in upward direction. The flow of solids is plug flow in nature and not in fluidized state. The Higgins contactor developed for ion exchange is an excellent facility for adsorption. The figure 12.19 shown indicates the arrangement. This consists of two sections. In the top section to star with adsorption adsorption takes place. Simultaneously the bottom section of the  bed under goes regeneration. After some pre-calculated duration of operation, the flow of liquids is stopped and the positions positions of the valves are changed as indicated. The liquid  –filled piston pump is moved and this th is leads to the clockwise movement of solids. so lids. Once Onc e again the valves are moved to their original position and the movement of solid also stops. The adsorption cycle once again starts in the top section of the unit and desorption at the bottom section.

178

 

Fig 12.19 12.19 Higgins Contactor Worked Examples:

One litre flask is containing air and acetone at 1 atm.and 303 K with a relative

1.

humidity of 35% of acetone. 2 grams of fresh activate carbon is introduced and the flask  is scaled. Compute the final vapour composition and final pressure neglecting adsorption of air. Equilibrium data: gm adsorbed/gm.carbon

0

0. 0.2 1 Partial pressure of acetone, mm Hg 0 2 12 Vapour pressure of acetone at 30°C is 283 mm Hg.

0. 3 42

0.35 92

Solution:

Lett us co Le conve nvert rt the the data data from from parti partial al pressu pressure re to concen concentra tratio tion n in term termss of  gm.acetone/gm.of air. (i.e.)

2

( 760 − 2)

×

58 28.84

= 5.28 × 10 -3 g acetone/ g air 

Likewise the other values can also be converted. Hence, X, g adsorbed/g carbon 0

0.1

0.2

0.3

0.35

179

 

Y, g acetone/g air 0 5.28×10 -3 32.1×10 -3 Originally the feed contains 35% RH acetone

117×10 -3

276×10 -3

(i.e.) Partial pressure/Vapour pressure = 0.35

∴ Partial pressure of acetone = 283 × 0.35 = 99 mm Hg. ∴ Y0 =

99

( 760 − 99 )

×

58 28.84

= 0.301 g of acetone/g of air.

LS = 2gm The feed point is (Xo, Yo) = (0.0, 0.301) Volume fraction of air in original mixture =

( 760 − 99 ) 760

= 0.87

(i.e.) volume of air = 0.87 lit. (At 1 atm and 303 K) (i.e.) moles of air =

( 0.87 ×1) 303

×

273 1

×

1 22.414

= 0.03496 g moles

(i.e.) mass of air in original mixture = 1.008 gm

LS

2

∴ GS = 1.008 = 1.984 . Y1 (from graph) = 13 × 10

-3

g acetone/g air 

Grams of acetone left behind after adsorption, per gram of air = 0.013 (i.e.)

(i.e.)

Partial pressure of  acetone Partial pressure of  air 

×

Partial pressure of  acetone 661

58 28.84

×

58 28.84

=

g acetone g air 

= 0.013

g acetone g air 

 

∴ Partial pressure of acetone in flask after adsorption = 4.27 mm Hg ∴ Total pressure = 661 + 4.27 = 665.27 mm Hg.

180

 

Fig. 12.20 Example 1

2)

A solid adsorbent is used to remove colour impurity from an aqueous solution. solution. The original original value of colour on an arbitrary arbitrary scale is 48. It is required to reduce reduce this to 10% of its original value. Using the following data, find the quantity a fresh adsorbent used for 1000 kg of a solution for (a) a single stage and (b) a two stage cross current operation when the intermediate colour value is 24. Equilibrium data: kg adsorbent/kg of of solution

0

Equilibrium colour (Y)

4 8

0.00 1 43

0.00 4 31.5

0.00 8 21.5

0.0 2 8.5

0.04 3.5

Solution:

The given data will be converted to enable us to handle it more easily. The initial values are Xo = units of colour/kg adsorbent = 0 Yo = units of colour/kg solution = 48

181

 

When, 0.001 kg of adsorbent is added to1 kg of solution the colour reduces from 48 units to 43 units. These 5 units of colour are thus transferred to 0.001 kg adsorbent.

∴ X,

units of  colour  kg adsorbent

=

( 48 − 43) 0.001

= 5000

Similarly by adding 0.004 kg adsorbent colour drops by 16.5 units 0 500 412 3312. 197 X, colour adsorbent/kg adsorbent 0 5 5 5 Y, colour/kg solution 4 43 31.5 21.5 8 .5 8

(i.e.) X =

16.5 0.004

1112.5 3 .5

= 4125

The final solution has 4.8 units of colour  Slope

LS GS

= – 0.030 (from graph)

GS is 1000 kgs of solution

∴ Dosage of carbon = 0.03 × 1000 = 30 kgs. ii) A two stage cross current operation: LS   ( 48 − 24 ) 24 = = 6.76 ×10 −3 −     1 =  GS   ( 3550 + 0 ) 3550 ( 24 − 4.8) LS   = 0.01324 −     2 =  GS   ( - 1450 + 0) GS is 1000 kgs of solution

∴ LS1 +LS2 = 6.76 + 13.24 = 20.00 kgs.

182

 

Fig 12.21 Example 2

3)

The The equil equilib ibriu rium m decol decolour ourisa isati tion on dat dataa for for a cert certai ain n syste system m using using ac acti tivat vated ed car carbon bon is given by the equation, Y = 0.004 X2 where Y is g. colouring impurity/kg impurity free solution and X is g. colouring impurity/kg pure activated carbon Calculatee the amount of activated Calculat activated carbon required per 1000 kg of impurity free solution to reduce the impurity concentration from 1.2 to 0.2 g/kg of impurity free solution using (i) a single stage operation and (ii) a two stage cross current operation with intermediate composition of 0.5 g. of colouring impurity per kg of  impurity free solution. Y = 0.004 X2 Solution:

Feed, GS = 1000 kg of impurity free solution i) Y0 = 1.2 gm/kg of impurity free solution

Y1 = 0.2 gm/kg of impurity free solution X0 = 0

183

 

0.5

0.5

  Y1   =   0.2   = 7.07 X1 =        0.004    0.004   LS ( Y0 − Y1) (1.2 − 0.2 ) 1 = = −0.1414 = − = GS ( X0 − X1) ( 0 − 7.07 ) − 7.07 ∴ LS = 0.1414 × 1000 = 141.4 kgs of adsorbent (ii)

Intermediate colour concentration is 0.5 gm/kg of impurity free solution 0.5

0.5   ∴ X1 =      = 11.18  0.004   LS   (1.2 − 0.5) 0.7 = = −0.06261 ∴ −    1 =  GS   ( 0 − 11.18) 11.18 X2 = X final = 7.07 LS   ( 0.5 − 0.2 ) = −0.04243 −     2 = G ( 0 7.07 ) S −    

∴ LS total = 0.06261+ 0.04243 = 0.10504 S

G The adsorbent needed, LS = 105.04 kgs of adsorbent. 4)

A solution of washed raw cane sugar of 48% sucrose by weight is coloured by the presence of small quantities of impurities. impurities. It is to be decolourised at 80ºC by treatment with an adsorptive carbon in a contact filtration plant. The data for an equilibrium adsorption isotherm were obtained by adding various amounts of the carbon to separate batches of the original solution and observing the equilibrium colour reached in each case. The data with the quantity of carbon expressed on the basis of the sugar content of the solution are as follows: kg carbon 0 kg dry sugar  % colour removed 0

0.00 5

0.0 1

0.01 5

0.0 2

0.03

47

70

83

90

95

 

The original solution has a colour concentration of 20 measured on an arbitrary scale and it is desired to reduce the colour to 2.5% of its original value. (i) Convert the equilibrium data to Y and X (ii) Calculate the amount of carbon required for a single stage process for a feed of  1000 kg solution (iii) Estimate the amount of carbon needed for a feed of 1000 kg solution in a two stage counter current process.

184

 

Solution:

Feed solution contains 48% sucrose kg carbon

0

0.005

0.01

0.015

0.02

0.03

0

47

70

83

90

95

0

0.0024

0.0048

0.0072

0.0096

0.0144

53 × 0.2 = 10.6

30 × 0.2 =6

17 × 0.2 = 3.4

kg dry sugar  % color  removed kg carbon kg of  solution Y, X,

color  kg of  solution color  kg  carbon

 

 

100 ×0.2 = 20

(20 - 10.6) −

0.0024 = 3916.7

 

(20 - 6)

 

0.0048 = 2916.7

(20 - 3.4) 0.0072 = 2305.6

10 × 0.2 = 2.0  

( 20 − 2) 0.0096 = 1875

5 ×0 .2 = 1.0  

( 20 − 1) 0.0144

 

=1319.4

Feed is (X0, Y0) = (0, 20) Final product is to have 2.5% original colour (i.e.) 0.5 units = Y1 S ∴ −   = ( 20 − 0.5) = − 19.5  L    GS   ( 0 − 1000) 1000 ∴ LS = GS × 0.0195 = 19.5 kgs.

ii)

   

The operating line is fixed by trial and error for exactly two stages.

 LS  = 19.5 = 6.142 ×10 −3     GS   3175 ∴ LS = 6.142 kgs.

185

 

Fig 12.22 Example 4

(5)

NO2 produced by a thermal process pr ocess for fixation of nitrogen n itrogen is to be removed from f rom a dilute mixture with air by adsorption on silica gel in a continuous counter  current adsorber. The gas entering at the rate of 0.126 kg/s contains 1.5 % NO 2  by volume and 90% of NO2 is to be removed. Operation is isothermal at 25°C and 1 atm. pressure. The entering gel will be free of NO2  Partial Pressure of NO2 , mm Hg Kg NO2 / 100 kg gel

2 0 .4

4 0.9

6 1.65

8 2.6

10 12 3.65 4.85

(a) Calculate the minimum weight of gel required/hr  (b) For twice the minimum gel rate, calculate the number of stages required. Solution:

Entering gas rate  NO2 present

: 450 kg/hr = 0.126 kg/s :

1.5 % by volume

Temperature

: 25oC

Pressure

: 1 std. atm.

yin = 1.5 % Partial Pressure of   NO2 , mm Hg kgNO2 / 100 kg gel K g NO2/kg gel, X (pp NO2/pp air) × (46/28.84),Y

0

2

0 0.4 0 0.004 0 0.0042

4

6

8

10

12

0.9 0.009 0.00844

1.65 0.0165 0.01269

2.6 0.026 0.01697

3.65 0.0365 0.02127

4.85 0.0485 0.0256

186

 

Yin (kg/kg) =

1.5 98.5

×

46 28.84

= 0.0243

yin (kg/kg of mixture) = 0.0237 Gs = 450 (1– 0.0237) = 439.3 kg/hr  90% of NO2 is to be recovered = 1.5 × 0.1 × 46 98.5 28.84

Yout

= 0.00243

 LS  min = 0.025 = 0.667    GS   0.0375

 

Weight of absorbent required, Ls = 0.667 × 439.3

= 291.1 kg/hr 

Number of stages needed for twice the adsorbent rate

 LS  act = 582.2 = 1.334      GS   439.3

  ∴  

 NO. of Stages = 3

Fig 12.23 Example 5

(6)

500 kg/min of dry air at 20oC and carrying 5 kgs of water vapour/min. is to be dehumidified with silica gel to 0.001 kg of water vapour/kg of dry air. The operation has to be isothermally and counter currently with 25 kg/min. of dry silica gel. How many theoretical stages are required and what will be the water  content in the silica gel leaving the last stage

187

 

Kg.of water vapour/ kg.of dry silica gel, X

0

0.05

0.10

0.15

0.20

Kg. of water -vapour/ kg.of dry air, Y

0

0.0018

0.0036

0.0050

0.0062

Solution:

Quantity of dry air entering at 20 oC, GS = 500 kg/min Quantity of water vapour entering Y1 =

5 500

= 5 kg/min

= 0.01 kg water vapour/kg dry air 

Concentration of water vapour in leaving air, Y 2 = 0.001 Quantity of silica gel entering, LS = 25 kg/min X2 = 0 (ie)

 LS  = 25 = 0.05       GS   500

Making a material balance LS [X1 – X2] = GS [Y1 – Y2]

∴ X1 =

500 × 0.009 25

= 0.18

Total number of stages needed = 4

188

 

Fig 12.24 Example 6 Exercise:

1)

The The equi equili libr briu ium m rela relati tion onsh ship ip for for the the ads adsor orpt ptio ion n of colo colour ur for form m a carr carrie ierr gas gas is given by y = 0.57x

0.5

, where where y is the the gram of coloured coloured substance removed per 

gram of adsorbent and x is the gram of colour/100 grams of colour free carrier. If  100 kg of the carrier containing 1 part of colour per 3 parts of total carrier is contacted with 25 kgs of adsorbent, calculate the percent of colour removed by (i) single contact (ii) Two stage cross current contact dividing the adsorbent equally per contact. 2)

Experim Experiments ents on decolour decolourisat isation ion of oil yielded yielded the followi following ng relat relation ion ship y = 0.5x

0.5

, where where y is the the gram of colour colour removed removed per gram gram of of adsorbent adsorbent and and x is

the gram of colour/1000 grams of colour free oil. If 1000 kg of oil containing 1  part of colour per p er 3 parts of colour free oil is contacted with 250 kgs of adsorbent ad sorbent calculatee the percent of colour removed by (i) single stage process (ii) Two stage calculat cross current contact process using 125 kg adsorbent in each stage. 3)

A solid adsorbent is used to remove colour impurity from an aqueous solution. The original original value of colour on an arbitrary arbitrary scale is 48. It is required to reduce reduce this to 10% of its original value. Using the following data, find the number of  stages stag es needed needed for 1000 kg of a solutio solution n in a counter current current operat operation ion if 1.5 times the minimum adsorbent needed is used. Equilibrium data:

4)

kg adsorbent/kg of of solution

0

Equilibrium colour (Y)

4 8

0.00 1 43

0.00 4 31.5

0.00 8 21.5

0.0 2 8.5

0.04 3.5

An aqueous solution containing containing valuable solute is colored by the presence of  small amounts of impurity. It is decolourised using activated carbon adsorbent. The equilibrium relationship is Y = 0.00009X

1.7

where, X = Colour units/Kg

carbon and Y = colour colour units/kg solution. solution. If the original original solution has 9.8 9.8 colour  units/kg solution, calculate (i) the amount of colour removed by using 30 kgs of  adsorbent in a single stage operation and (ii) the amount of carbon needed for a two stage counter current operation operation if the final colour in the solution is to be 10%

189

 

of original value and the solution leaving the first stage has 4 times the final colour of the solution

5)

The adsorption of moisture using silica gel varies with moisture content as follows: Y = 0.035X 1.05 where, X = kg water adsorbed /kg dry gel and Y = humidity of air, kg moisture/kg moisture/kg dry air. 1 kg of silica gel containing containing 2% (Dry basis) of moisture is  placed in a vessel v essel of volume 5 m3

containing containi ng moist air. air. The partial partial pressure of 

water wat er is 15 mmHg. The total total pressure pressure and tempera temperature ture are 1 atm. and 298 K  respectively. What is the amount of water picked up by silica gel from the moist air in the vessel? Estimate the final partial pressure of moisture and final total  pressure in the vessel. ves sel.

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