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ADSORPTION

This unit operation involves the removal of a component(s) from a gas or liquid stream by a solid. The component is called adsorbate and the solid is the adsorbent. The fluid stream leaving the unit ( depleted in adsorbate) is the effleuent. There are many solids used as adsorbent e.g., activated carbon, silica gel, zeolites ……etc. When the adsorption involves chemical reaction it is called "chemisorptions". The process is used to separate fluid component or to remove impurities from one component such as in the case of removal of benzene from hexane by a zeolite to produce pure hexane. In many ways adsorption by a solid resembles gas absorption by a liquid . this, also, applies to their equilibrium isotherms. There are three modes of operations, vis, (i) Batch, (ii) Stage wise and (iii) continuous fixed bed. In this chapter we shall, only , deal with the latter.

8.1- Terminology

(i) Adsorption: The adhesion of an extremely thin layer of molecules (of gas or liquid) to the surfaces of granular porous solids.

(ii) Adsorbent: This is the solid material that collects the adsorbate (fluid molecules).

(iii) Activity: The maximum amount of adsorbate that can be adsorbed per unit mass of adsorbent at a certain temperature, pressure and concentration.

(iv) Breakthrough Curve: When a fluid is passed through a fixed adsorption bed, only the first part of the bed work at adsorbing the solute until this part is saturated. Then the following section adsorb the solute until it is, also, saturated. This trend continues until the whole bed is, nearly, saturated(usually about 90% of saturation). Plotting the concentration in the effluent vs volume of effluent gives the breakthrough curve.

(v) Breakthrough Point: When the adsorption zone (mass transfer zone) reaches the last (lowest) part of the bed, the system is said to have reached the breakthrough point.

……. Y0 ……………………………………………………….

YE

Concentration of

Solute in Effluent Breakthrough Curve

YB

Break Point

Volume of Effluent ω� ω�

Figure 8.1 Breakthrough Curve

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Beyond the break through point the column approaches saturation. This is shown in figure (8.2).

Cf Cf Cf

(a) (b) (c) Figure 8.2, (a) represent the start of operation where absorption is just starting at the top of column. (b) most of the bed is saturated at the break through point.(c) the bed is exhausted (completely saturated) .

8.2- Equilibrium Isotherms

In many respects the equilibrium adsorption isotherms resembles the equilibrium isotherms of a gas in a liquid. There are three commonly used mathematical expressions to describe adsorption equilibria. These are: the Langmuir, the Freundlich and the Brunauer-Emmett-Teller (BET) isotherms. The Langmuir and BET were derived with theory in mind, but none is applicable universally, nor can it be predicted to apply to a particular case.

(i) The Langmuir Equation

The Langmuir equation is the first equation of adsorption isotherms developed theoretically,

V = Vm

�.����.� (8.1)

Where V is the volume of fluid removed by adsorption, P is equilibrium pressure (or concentration) of adsorbate fluid, K is a constant that depends, mainly, on temperature, specific surface area of adsorbate and Vm is the volume needed for a monomolecular coverage of the whole surface.

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(ii) The Freundlich Equation

For many cases of physical adsorption Freundlich equation is applicable, V = K P1/n (8.2) Where, n is a constant that depends on temperature (n 1). P is the pressure (or concentration) of adsorbate.

(iii) The BET Equation

This equation was developed by applying Langmuir's ideas to multi-molecular adsorption,

��

= �.�

�����.������.�� (8.3)

Where, X is the relative pressure (or concentration) and C is a constant related to the heat of adsorption.

8.3- Factors Affecting Dynamic Adsorption

(i) Type of Adsorbent: Generally a better adsorbent is one with the highest surface area.

(ii) Granular Size: Smaller granules gives higher surface area and higher mass transfer, albeit, on the expense of higher pressure drop. Therefore a balance should be made between mass transfer area and pressure drop.

(iii) Depth of Bed: It is advantageous to size the adsorption bed to the maximum length allowed by pressure drop consideration.

(iv) Gas Velocity: Increasing the velocity lowers the capacity at the breakthrough point. Very high velocities may damage the adsorbent particles.

(v) Temperature Effect: increasing the temperature increases the volatility which means less material adsorbed or shorter residence time on the surfaces. Thus, lower temperatures give better adsorption.

(vi) Concentration of Adsorbent: Similar to other separation processes higher concentration of the solute results in higher rate of adsorption. Also low concentrations require deeper beds.

(vii) Pressure Effects: Generally, the adsorption capacity increases with increasing pressure, if the partial pressure of the solute. However, at high pressures (above 500 psig) the capacity decreases due to condensation and adsorption of the carrier phase.

(viii) Decomposition, polymerization and contamination of the adsorbate: The process fluids may react with the adsorbate causing decomposition or polymerization, or they contain contaminants. In such cases the correct

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choice of the adsorbent should be made. Or the bed should be changed periodically.

8.3.1- Steps in the Adsorption Operation (i) Bulk (External Diffusion): This is the molecular diffusion of the solute in the fluid surrounding the adsorption granules. (ii) Surface diffusion: This is the spreading diffusion over the exterior surface of the granules.

(iii)Pore (Knudsen) Diffusion: This is the transport of molecules within the small pores and to come within the force field of an active site.

(iv)Physical Adsorption: The molecules is held by and occupy the active site.

NOTE: If there is a chemical reaction, the reaction takes place on the active site and the product species moves according to reverse order to that above until it reaches the external bulk of the fluid.

8.4- Analysis of Adsorption Columns

We shall review the most common methods for adsorption bed calculations. These are,

(i) Length of Unused Bed (LUB) Equations:

If u is the velocity of advancement of the adsorption zone (mass transfer zone), then at saturation time ts, Z = u.ts, and at breakthrough point time, tb, Zs = u.tb, therefore,

LUB = Z – Zs = u.(ts – tb) = Z - u.tb = ���

(ts – tb) (8.4)

Where Z is the total bed length and Zs, is the saturated part.

A solute balance gives,

Gs (Y0 – Y0*).t = Zs ρ�(Xt – X0) (8.5)

Where ρ� is adsorbent bulk density, Gs is mass flux of unadsorbed fluid, X0 and Xt are initial and saturation concentrations in adsorbate, Y0 and Y0

*are initial and equilibrium (with X0) solute concentration in the carrier fluid.

But, at any time t, Zs = u.t, substitute this in equation (8.5) and rearrange,

u = ����

. ��� ������ ��

(8.6)

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Substitute u from (8.6) in (8.4) to get,

LUB = Z – ��.�

�� .

��� ������ ��

(8.7)

(ii) The Wheeler's Equation:

tb = !"

#.�� [ W –

#.$%&'�'().��

�* ] (8.8)

where ω+ is kinetic adsorption capacity (saturation concentration), Q is volumetric flow

rate of adsorbate, W is adsorbent bed weight, C0 is initial concentration and Kv is adsorption rate constant. And tb is the breakthrough time.

(iii) Scale-up of Resistance Time: This method depends on laboratory measurements from pilot data (p) to scale-up size (s),

tb (s) = tb (p) ,�-�,�.� .

#�.�.���.�#�-�.���-� * 2 (8.9)

Example 1

For assessment of the activated carbon filters in public shelters and for a time duration of 100 minutes ( this is the time recommended for dissipation of harmful or poisonous gases from the surrounding environment). From a laboratory test on the adsorption of carbon tetrachloride (CCl4) on granulated activated carbon, the following data were obtained,

Bed depth, Z = 1.67 cm = 0.0167 m

Initial CCl4 concentration in the carbon bed, X0 = 0

Bulk density of bed, ρ�= 640 kg/m3

Density of nitrogen = 1.09 kg/m3

Carrier (Nitrogen) gas flow rate, Gs = 309 kg/m2hr

Initial CCl4 concentration in the gas, C0 = 98 kg/m3

From the breakthrough curve, ts = 1430 s and tb = 170 s.

Saturation concentration, Xt = 0.2387 kg CCl4/kg carbon

Find the depth of bed for a breakthrough time, tb = 100 minutes

Solution:

From equation (8.4),

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LUB = ,��

(ts – tb) = �./0�123 (1430 – 170) = 1.47 cm

Using equation (8.5) and putting t = tb to obtain Zs,

Y0 = C0 / ρ45= 98 / 1.09 = 0.09 kgCCl4/kg N2

Zs = ��.�

�� .

��� ������ ��

= 236��33�3.36 � 3�/13�/3 �3.7280�3�

∴ Zs = 0.303 m

And Z = LUB + Zs

Z = 0.0147 + 0.3030 = 0.3147 m.

8.5- Height of Column Based on the Height of The Mass Transfer Zone (MTZ)

The analysis to obtain the height of the mass transfer zone is similar to that for absorption,

Z = �/;�<.α =

>�����

�?�5

And, therefore, the height of MTZ for adsorption in fixed packed bed we have,

Za = �/;�@.α =

>�����

�A�� (8.10)

The total height of the bed (Z) is given by,

,B, =

ωBωA� ���C�.ωB

(8.11)

Where, f = D

��.ωB =

= ���� ��.>ωωAω�

��.ωB (8.12)

Y0 : Initial solute concentration.

ωE: Inert gas effluent during the adsorption zone = FG H FI

ω�: Inert gas effluent until exhaust concentration, YE

ω�: Inert gas effluent until break point concentration, YB

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8.5.1- Degree of Saturation at Break Point, J

η = ,�C.,B

, (8.13)

Time taken to break point, tb = ,.;.ρ�.η.�M

�.�� (8.14)

Where, A: Cross section area of bed.

ρ�: Density of solid.

XT: Value in equilibrium with Y0 (from equilibrium relationship). Note that the units of Y and X are (kg solute / kg inert fluid) and (kg solute/kg solid).

8.5.2- Steps to Determine the Bed Height, Z

1- From the break through curve and the equilibrium relation, Y*= fn.(X) or equilibrium curve, construct the following table,

(1) (2) (3) (4) (5) (6) Y Y* 1

Y H Y� O dYY H Y�

�

��

ω H ω�ωE

YB

- - - YE

YB*

- - - YE

*

1Y� H Y��

- - - 1

Y� H Y��

0 - - -

O dYY H Y�

�A

��

0 - - - 1.0

-

-

-

NOTES

(i) In the above table column (2) is obtained from the equilibrium relation,

Y=fn.(X) and the operating line, Y = ���M

X. And, hence, column (3) is,

also, obtained. XT Is found from the equilibrium at Y0.

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(ii) Column (4) is obtained by numerical integration (e.g. Simpson Rule) for various values of Y from YB to YE.

(iii) Column (5) is obtained by dividing each entry in column (4) by the maximum (last) value in this column.

2- A plot of Y/Y0 vs (F H FI� / FQ to obtain (f) as the area under the curve or by numerical integration.

3- Find Za from equation (8.10) using G/A and KY.a from data given. And the integral from the table constructed above.

4- Find the bed height (Z) from equation (8.11) or from equation (8.13) if η .

5- Find degree of saturation from equation (8.13).

6- Find , t, time to break point from equation (8.14).

Example 2

Air with a flow rate of 0.1295 kg/m2.s and humidity of 0.00267 kg H2O/kg dry air is passed through affixed bed of silica gel (ρ�= 671 kg/m3). The break point is at

humidity of 0.0001 kg H2O/kg dry air. And the bed is considered exhausted when the effluent humidity is 0.0024 kg H2O/kg dry air. If the mass transfer coefficient, KY.a = 8.67 kg/m3.s and the saturation at the break point, η = 90%, find;

(i) The bed height,Z. (ii) The time to reach break point(t).

The equilibrium relation is Y = 0.158 X1.6367

Solution:

(i) Y0 = 0.00267, YB = 0.0001, YE = 0.0024, G/A = 0.1295

From the equilibrium relation at Y0 we have,

Y0 = 0.158 (XT)1.6367 = 0.00267

∴ XT = 0.0826 kg H2O/kg silica gel

The operating line, then, is,

Y = 0.032 X from (Y = ���M

X)

From the equilibrium and the operating lines we find Y* for values between YB and YE and, thus, construct the required table below,

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(1) (2) (3) (4) (5) (6) Y Y* 1

Y H Y� O dYY H Y�

�

��

ω H ω�ωE

YB=0.0001 0.0004 0.0008 0.0012 0.0016 0.0020 YE= 0.0024

0.00003 0.00016 0.00041 0.00076 0.00123 0.00175 0.0023

14300 4160 2560 2300 2700 4000 10000

0 2.22 3.49 4.44 5.43 6.73 9.3

0 0.2365 0.375 0.477 0.584 0.723 1.000

0.038 0.15 0.3 0.45 0.6 0.75 0.9

The last value of the integral, = >�

�� ���A

�� = 9.3

From step (2) above we find, f = 0.53.

From equation (8.10),

Za = (0.1295/8.67)* 9.3 = 0.14 m

From equation (8.13) we have,

0.9 = (Z – 0.53*0.14)/Z,

∴ Z = 0.742 m

(ii) From equation (8.14),

t = (0.742*1*671*0.9*0.0826)/(0.1295*0.00267) = 31.4 hours.

8.6- Volumetric Mass Transfer Coefficient in Fixed Bed Adsorption To determine the volumetric mass transfer coefficient in fixed bed adsorption column the procedure is the same as that given above in section (8.5.2) for steps (1) and (2), but the remaining steps are as follows,

3- Find Za from equation (8.11). Z is given or measured experimentally.

4- Find KY.a from equation (8.10). G/A is given and the integral is found in step (1) above.

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8.7- Ion Exchange

This is a case of adsorption with chemical reaction (chemisorptions). For example the removal of hardness from water,

Ca++(HCO3)2-- + 2 Na+(zeolite) ⟶ Na2CO3 + Ca++(zeolite)

For this case The height of MTZ (Za) from equation (8.10) is given by,

Za = �/;�@.α =

>�����

�A�� ; with Y* = 0

∴ Za = �/;�@.α ln(YE/YB) (8.15)