Chapter 6 Adsorption

Chapter 6: ADSORPTION

MISS. RAHIMAH BINTI OTHMAN (Email: [email protected])

COURSE OUTCOMES CO APPLY principles of batch adsorption and fixedbed adsorption. CALCULATE and EXAMINE adsorption isotherms. DEVELOP basic design of gas or liquid adsorber.

OUTLINES  √ Introduction to adsorption.  √ Adsorption equipments.  √ Adsorption Isotherms Analysis.  √ Principles of Adsorption.  Basic Equation for Adsorption.  Adsorber Design Calculation.

INTRODUCTION TO ADSORPTION ADSORPTION ≠ ABSORPTION ! 

Absorption – a fluid phase is transferred from one medium to another.



Adsorption – certain components of a fluid (liquid or gas) phase are transferred to and held at the surface of a solid (e.g. small particles binding to a carbon bed to improve water quality)



Adsorbent – the adsorbing phase (carbon, silica gel, zeolite)



Adsorbate – the material adsorbed at the surface of adsorbent .

INTRODUCTION TO ADSORPTION APPLICATION OF ADSORPTION: 

Used in many industrial processes:  Adsorbing the desired product from fermentation broths.  Isolation of proteins.  Dehumidification.  odour/colour/taste removal.  gas pollutant removal (H2S).  water softening and deionisation.  hydrocarbon fractionation.  pharmaceutical purification.

INTRODUCTION TO ADSORPTION * NATURE OF ADSORBENT 

Porous material - Large surface area per unit mass - internal surface area greater than the external surface area - often 500 to 1000 m 2/g.



Granular (50μm - 12 mm diameter), small pellets or beads.



Suitable for packed bed use.



Activated carbon, silica gel, alumina, zeolites, clay minerals, ion exchange resins.



Separation occurs because differences in molecular weight, shape or polarity of components.



Rate of mass transfer is dependent on the void fraction within the pores.

Silica structure

Zeolite structure

INTRODUCTION TO ADSORPTION TYPES OF ADSORPTION 1. Ion exchange 

Electrostatic attachment of ionic species to site of the opposite charge at the surface of an adsorbent

2. Physical Adsorption    

result of intermolecular forces causing preferential binding of certain substances to certain adsorbents Van der Waal forces, London dispersion force reversible by addition of heat (via steam, hot inert gas, oven) Attachment to the outer layer of adsorbent material

3. Chemisorption   

result of chemical interaction Irreversible, mainly found in catalysis change in the chemical form of adsorbate

ADSORPTIO N EQUIPMENT  

Fixed-bed adsorbers



Gas-drying equipment



Pressure-swing adsorption

Fixed-bed adsorbers

Gas-drying Equipment

ADSORPTION EQUIPMENT Adsorption From Liquids

Pressureswing Adsorption

FIXED-BED ADSORBERS 

Adsorbent particles: 0.3 – 1.2 m deep supported on a perforated plate



Feed gas passes down through the bed.



Downflow is preffered because upflow at high rates may fluidize the particles, causing attrition and loss of fines.



The feed gas is switched to the other bed when the conc. Of solute in exit gas reaches a certain value.



The bed is regenerate by steam / hot inert gas.

FIXED-BED ADSORBERS Regeneration 

To remove unwanted particles from the adsorbent surface after the adsorption process.



using steam/hot inert gas.



Steam condenses in the bed, raising the temp. of the solid, provide energy for desorption.



The solvent is condensed, separated from water.



Then the bed is cooled and dried with inert gas.

Fixed-bed adsorbers

Gas-drying Equipment

ADSORPTION EQUIPMENT Adsorption From Liquids

Pressureswing Adsorption

GAS-DRYING EQUIPMENT

Fig. 25.1 Vapor-phase Adsorption System



The equipment for drying is similar to the shown in Fig. 25.1, but hot gas is used for regeneration.



The moist gas from the bed being generated may be vented, or much of the water may be removed in a condenser and the gas recirculated through a heater to the bed.



For small dryers, electric heaters are sometimes installed inside the bed to provide the energy for regeneration.

Fixed-bed adsorbers

Gas-drying Equipment

ADSORPTION EQUIPMENT Adsorption From Liquids

Pressureswing Adsorption

PRESSURE-SWING ADSORPTION 

Most often, adsorption is used as a purification process to remove small amounts of material, but, there is a number of applications involve separations of gas mixtures with moderate to high concentration of adsorbates.



These are called bulk separations, and they often use different operating procedures than for gas purification.



Pressure-swing adsorption (PSA) is a bulk separation process that is used for small-scale air separation plants and for concentration of hydrogen in process streams.

Fixed-bed adsorbers

Gas-drying Equipment

ADSORPTION EQUIPMENT Adsorption From Liquids

Pressureswing Adsorption

ADSORPTION FROM LIQUIDS 

Use of activated carbon to remove pollutants from aqueous wastes.



Use carbon beds up to 10 m tall, several ft in diameter, several bed operating in parallel.



Tall beds are needed to ensure adequate treatment.

PACKED EXTRACTION TOWERS

Tower packings; (a) Raschig rings, (b) metal Pall ring, (c) plastic Pall ring, (d) Berl saddle, (e) ceramic Intalox saddle, (f) plastic Super Intalox saddle, (g) metal Intalox saddle

ADSORPTION ISOTHERMS ANALYSIS 

Adsorption isotherm – equilibrium relationship between the concentration in the fluid phase and the concentration in the adsorbent particles.



For gas – concentration in mole % or partial pressure



For liquid – concentration in mg/L (ppm) or μg/L (ppb)



Concentration of adsorbate on the solid = mass adsorbed (g) per unit mass of original adsorbent (g).

TYPES OF ISOTHERMS Amount of adsorbed is independent of concentration down to very low values. Amount of adsorbed is proportional to the concentration in the fluid.

Concave upward; low solid loadings are obtained and because it leads to quite long mass-transfer zones in the bed. (this shape are rare)

TYPES OF ISOTHERMS Nearly linear isotherm up to 50 percent humidity, and the ultimate capacity is about twice that for the other solids. Water is held most strongly by molecular sieves, and the adsorption is almost irreversible, but the pore volume not as great as for silica gel

Fig. 25.3 Adsorption isotherms for water in air at 20 to 50 oC

ADSORPTI ON DATA FOR VAPORS ON ACTIVATED CARBON Sometimes fitted to Freundlich isotherms, but data for wide range of pressures show isotherm slopes gradually decrease as the pressure is increased.

Amount of adsorbed depends on (T/V) log (fs/f), where: T: adsorption temperature (Kelvin). V: molar volume of the liquid at the boiling point fs: fugasity of the saturated liquid at adsorption temperature f: fugasity of the vapor

For adsorption at atmospheric pressure; * fugasity = partial pressure = vapor pressure

Volume adsorbed is converted to mass by assuming the adsorbed liquid has the same density as liquid at the boiling point.

QUESTION 1 EXAMPLE EXAMPLE 25.1. 25.1. Adsorption Adsorption on on BPL BPL carbon carbon is is used used to to treat treat an an airstream airstream containing containing 0.2 0.2 percent percent n-hexane n-hexane at at 20 20 ooC. C. (a) (a) Estimate Estimate the the equilibrium equilibrium capacity capacity for for aa bed bed operating operating to to 20 20 ooC. C. (b) (b) How How much much would would the the capacity capacity decrease decrease ifif the the heat heat of of adsorption adsorption raised raised the the bed bed temperature temperature to to 40 40 ooC. C.

ANSWER (a) (a)

Estimate Estimate the the equilibrium equilibrium capacity capacity for for aa bed bed operating operating to to 20 20 ooC. C.

The The MW MWn-hexane (C66H H14 )= 86.17, 86.17, At At 20 20oo C C (from (from Perry’s Perry’s Handbook, Handbook, 77ththed.) ed.) n-hexane (C 14)= Pʹ=120mm Pʹ=120mm Hg Hg ≈≈ ffss.. At At the the normal normal boiling boiling point point (68.7 (68.7 ooC), C), ρρLL=0.615 =0.615 g/cm g/cm33.. The The adsorption adsorption pressure pressure P P is is 760 760 mm mm Hg. Hg. p  0.002 x 760  1.52 mmHg  f

V

86.17  140.1 cm 3 /gmol 0.615

T f 293 120 log s  log  3.97 V f 140.1 1.52 from Fig 25.4, volume adsorbed is 31 cm 3 liquid per100 g carbon; W  0.31 x 0.615  0.19 g/g carbon.

(b) (b) At At 40 40 ooC, C, Pˊ= Pˊ= 276 276 mm mm Hg Hg T f 313 276 log s  log  5.05 V f 140.1 1.52 from Fig 25.4, volume adsorbed is 27 cm 3 liquid per100 g carbon; W  0.27 x 0.615  0.17 g/g carbon.

4 TYPES OF ADSORPTION ISOTHERMS 1. Linear Isotherms - Adsorption amount is proportional to the concentration in the fluid 2. Irreversible – independent of concentration 3. Langmuir Isotherm – favorable type 4. Freundlich Isotherm – strongly favorable type

LANGMUIR ISOTHERM • •

Often been used to correlate equilibrium adsorption data for protein. Isotherms that convex upward are called favorable.



Kc  W  Wmax    1  Kc 

•

Where: W = adsorbate loading (g absorbed/g solid) c = the concentration in the fluid (mg/L) K = the adsorption constant K >> 1 : the isotherm is strongly favorable.

•

Wmax and K are constants determined experimentally by plotting 1/W against 1/c

FREUNDLICH ISOTHERM  strongly favourable  Describe the adsorption of variety of antibiotics, steroids and hormones.  high adsorption at low fluid concentration

W  bc

m

where b and m are constant  Linearize the equation: Log W = log b + m log c  Constant determined from experimental data by plotting log W versus log c  Slope = m, intercept = b



PRINCIPLES OF ADSORPTION In fixed bed adsorption, the concentrations in the fluid phase and the solid phase change with: a) time b) as well as the position in the bed.



At first, most of the mass transfer takes place near the inlet of the bed, where the fluid contacts the adsorbent.



After a few minutes, the solid near the inlet is nearly saturated.



Most of the mass transfer takes place farther from the inlet.



The concentration gradient become S-shaped.



The region where most of the change in concentration occurs is called the mass-transfer zone (MTZ), and the limits are often taken as c/c0 values of 0.95 to 0.05.

MASS TRANSFER ZONE AND BREAKTHROUGH

Concentration Profile In Fixed Beds

Concentration Profile In Fixed Beds • t11: no part of the bed is saturated. • From t11 to t22: the wave had moved down the bed. • t22: the bed is almost saturated for a distance LSS, but is still clean at LFF. • Little adsorption occurs beyond LFF at time t22, and the adsorbent is still unused. • The MTZ where adsorption takes place is the region between LSS and LFF. • The concentration of the adsorbate on the adsorbent is related to the adsorbate concentration in the feed by the thermodynamic equilibrium. • Because it is difficult to determine where MTZ begins and ends, LFF can be taken where C/CFF = 0.05, with LSS at C/CFF = 0.95. • tBB: the wave has moved through the bed, with the leading point of the MTZ just reaches the end of the bed. This is known as the breakthrough point. • Rather than using C/CFF = 0.05, the breakthrough concentration can be taken as the minimum detectable or maximum allowable solute concentration in the effluent fluid, e.g. as dictated by downstream



Concentration profile in fixed beds

Is the ratio of solute concentration to inlet solute concentration in the fluid.

Figure 25.6(a)

BREAKTHROUGH CURVES

t1, t2, t3: the exit concentration is practically zero.

Time for fluid living the bed.

BREAKTHROUGH CURVES

•

• • • •

BREAKTHROUGH CURVES t – time when the concentration reaches break b

point The feed is switched to a fresh adsorbent bed Break point – relative concentration c/co of 0.05 or 0.10 Adsorption beyond the break point would rise rapidly to about 0.50 Then, slowly approach 1.0 (concentration liq in = liq out)

BREAKTHROUGH CURVES • t* is the ideal adsorption time for a vertical breakthrough curve • t* is also the time when c/co reaches 0.50 • Amount of adsorbed is proportional to the rectangular area to the left of the dashed line at t*



BREAKTHROUGH CURVES

Solute feed rate (FA) = superficial velocity (uo) X concentration (co)

Where: Wo = initial adsorbate loading Wsat = adsorbate at equilibrium with the fluid (saturation) L = length of the bed ρb = bulk density of the bed

LENGTH OF UNUSED BED (LUB) 

For systems with favorable isotherm, the concentration profile in the mass-transfer zone acquires a characteristic shape and width that do not change as the zone moves down the bed.



Test with different bed lengths have breakthrough curve of the same shape, but with longer beds, the MTZ is a smaller fraction of the bed length, and greater fraction of the bed is utilized.



The scale-up principles “The amount of unused solid or length of unused bed does not change with the total bed length.”

LENGTH OF UNUSED BED (LUB) To calculate

LUB, determine the total solute adsorbed up to the break point by integration;

c   1  0  co  dt t

The



break point time, tb is calculated from the ideal time and the fraction of bed utilized:

QUESTION 2

EXAMPLE EXAMPLE EXAMPLE25.2 25.2. 25.2. The The adsorption adsorption of of n-butanol n-butanol from from air air was was studied studied in in aa small small (10.16 (10.16 cm cm diameter) diameter) with with 300 300 and and 600 600 gg carbon, carbon, corresponding corresponding to to bed bed lengths lengths of of 88 and and 16 16 cm. cm.

(a) (a) From From the the following following data data for for effluent effluent concentration, concentration, estimate estimate the the saturation saturation capacity capacity of of the the carbon carbon and and the the fraction fraction of of the the bed bed used used at at c/c c/c00== 0.05. 0.05. (b) (b) Predict Predict the the break-point break-point time time for for aa bed bed length length of of 32 32 cm. cm. Data Data for for n-butanol n-butanol on on Columbia Columbia JXC JXC 4/6 4/6 carbon carbon are are as as follows follows:: 300 g

600 g

t,h

c/c0

t,h

c/c0

1

0.005

5

0.0019

u0  58 cm/s

D p  0.37 cm

1.5

0.01

5.5

0.003

c0  365 ppm

S  1,194 m 2 /g

2

0.027

6

0.0079

T  25 o C

 b  0.461 g/cm 3   0.457

2.4

0.050

6.5

0.018

2.8

0.10

7

0.039

3.3

0.20

7.5

0.077

4

0.29

8

0.15

P  737 mmHg

ANSWER

The concentration profiles are plotted in Fig 25.8, and extended to c/c0=1.0 assuming the curves are symmetric about c/c0=0.5. Per square centimeter of bed cross section, the solute feed rate is F u c M A

0 0

cm  365 x 10 6 273 737  mol    58 x x x 74.12 g/mol s  22,400 298 760  cm 3  6.22 x10 -5 g/cm 2 .s or 0.224 g/cm 2 .h

The total solute adsorbed is the area above the graph multiplied by FA. For the 8 cm bed, the area is;



8.5

0



c  1   dt  4.79 h c0  

ANSWER

This area corresponds to the ideal time that would be required to adsorb the same amount if the breakthrough curve were a vertical line. The mass of carbon per unit cross-sectional area of bed is; 8 x 0.461 = 3.69 g/cm 2 Thus, • Trapezoidal rule:

h  f ( x0 )  f ( x1 ) x2 2 where h  x1-x0 x1

At the break point, where c/c0 = 0.05, and t = 2.4 h

f ( x)dx 

The amount adsorbed up to the break point is then

Thus 50 percent of the bed capacity is unused, which can be represented by a length of 4 cm.

ANSWER (-cont’)

For the 16-cm bed the breakthrough curve has the same initial slope as the curve for the 8-cm bed, and although data were not taken beyond c/c0 = 0.25, the curves are assume to be parallel. For the entire bed,

At c/c0 = 0.05, t = 7.1 h, and

At the break point, 74 percent of the bed capacity is used, which corresponds to an unused section of length 0.26 x 16 = 4.2 cm. Within experimental error, the lengths of unused bed agree, and 4.1 cm is the expected value for a still longer bed.

ANSWER (-cont’)

(b) For L = 32 cm, the expected length of the fully used bed is; 32 - 4.1 = 27.9 cm. The fraction of the bed used is:

The break-point time is

QUESTION 3 A A waste waste stream stream of of n-butanol n-butanol vapor vapor in in air air from from aa process process was was adsorbed adsorbed by by activated activated carbon carbon particles particles in in aa packed packed bed bed having having aa diameter diameter of of 44 cm cm and and length length of of 14 14 cm cm containing containing 79.2 79.2 gg of of carbon. carbon. The The density density of of the the activated activated carbon carbon is is 0.461 0.461 g/cm g/cm33.. The The inlet inlet gas gas stream stream having having aa concentration, concentration, C C00 of of 600 600 ppm ppm and and aa density density of of 0.00115 0.00115 g/cm g/cm33 entered entered the the bed bed at at the the solute solute feed feed rate, rate, F FAA of of 0.063 0.063 g/cm g/cm22.s. .s. Data Data in in Table Table 3.1 3.1 give give the the concentrations concentrations of of the the fluid fluid in in the the bed, bed, C. C. The The break break point point concentration concentration is is set set at at C/C C/Coo == 0.05. 0.05. QUESTION; QUESTION; 1. 1. Plot Plot aa breakthrough breakthrough curve. curve. 2. 2. Determine Determine the the break-point break-point time. time. 3. 3. Calculate Calculate the the saturation saturation capacity capacity of of the . the carbon, carbon, W Wsat sat. 4. 4. Calculate Calculate the the length length of of unused unused bed bed (LUB). (LUB).

Time (hour) 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.2 6.5 6.8

Concentration of fluid, C (ppm) 0.1 1.2 18.0 93.0 237.6 394.8 541.8 559.8 585.0 595.8

ANSWER Given; Packed bed, D = 4 cm L = 14 cm adsorbent = 79.2 g ρ carbon = 0.461 g/cm3 Inlet gas stream, C0 = 600 ppm ρ = 0.00115 g/cm3 FA = 0.063 g/cm2.s C/C0 = 0.05 1. Plot a breakthrough curve. t (h)

C

C/C0

3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.2 6.5 6.8

0.1 1.2 18.0 93.0 237.6 394.8 541.8 559.8 585.0 595.8

1.667E-04 0.002 0.030 0.155 0.396 0.658 0.903 0.933 0.975 0.993

1.0 0.9 0.8 0.7 0.6

C/C0 0.5 0.4 0.3 0.2 0.1 0.0 0

1

2

3

4

t, h

5

6

7

ANSWER- cont’ 2.

Determine the break-point time. From the breakthrough curve, breakthrough time at C/C0 = 0.05 is tb = 4.1 h 1.0 0.9 0.8 0.7 0.6

C/C0

0.5 0.4 0.3 0.2 0.1 0.0 0

1

2

3

4

t, h

5

6

7

ANSWER- cont’ 3. Calculate the saturation capacity of the carbon, Wsat. The total solute adsorbed is the area above the graph multiplied by FA



C  0  1  C0  dt 7



Simpson’s Rule of integration. (pp. 872)

h  f 0  4 f1  2 f 2  4 f 3  f 4  3 x  x0 7  0 h 4   1.75 4 4

x4

f ( x)dx 

x0

From the graph plotted, the following data is obtained;





4

0

 1  

Wsat 

C C0

 

 dt 

t (h)

C/C0

f(x) = (1-C/C0)

0.00

0

1

1.75

0

1

3.50

0.002

0.998

5.25

0.5

0.5

7.00

1

0

1.75 1  41  2(0.998)  4(0.5)  0   5.248h 3

FA x total solute adsorbed mass of carbon per unit cross  sectional area

0.063 g/cm 2 .s (5.248 h) 3600 s  x 14 cm x 0.461 g/cm 3 1h g solute  184.42 g carbon

 W  LUB  L 1  b  Wsat   At break point, where C / C0  0.05 and tb  4.1 h

4.

Calculate the length of unused bed (LUB).

4 .1  C Area    1   0 C0   Trapezoida l rule



x1

x2

f ( x)dx 

h  f ( x0 )  f ( x1)  2

h  x1  x0

h  4.1  0  4.1 C  4.1   1  0.95  3.9975 h 1  dt  t0  C0  2 t1



The amount adsorbed up to the break point is, Wb 0.063 g / cm 2 .s (3.9975h) 3600 s Wb  x 14 cm x 0.461 g / cm3 1h g solute  140.48 g carbon Wb 140.48   0.7617 Wsat 184.42  W  LUB  L 1  b   14(1  0.7617)  3.33 cm Wsat   Thus, 23.8 percent of the bed capacity is unused.

t (h)

C/C0

f(x) = (1-C/C0)

0.00

0

1

4.1

0.05

0.95

OUTLINES  √ Introduction to adsorption.  √ Adsorption equipments.  √ Adsorption Isotherms Analysis.  √ Principles of Adsorption.  √ Basic Equation for Adsorption.  √ Adsorber Design Calculation.

BASIC EQUATION FOR ADSORPTION Rate of Mass Transfer Internal and External Mass-transfer Coefficients Solution to MassTransfer Equations Irreversible Adsorption Linear isotherm YIELD YIELD

RATE OF MASS TRANSFER Equation for mass transfer in fixed-bed adsorption are obtained by making a solute material balance for a section dL of the bed, as in Fig. 25.9. The rate of accumulation in the fluid and in the difference between input and output flows. The change in superficial velocity is neglected: Fig.25.9 Mass balance for a section of a fixed bed

(25.5) (25.6)

• ε is the external void fraction of the bed • solute dissolved in the pore fluid is included with the particle fraction 1ε.

Adsorption from a gas or a dilute solution Accumulation in the fluid is negligible compare to accumulation on the solid.

The transfer process is approximated using an overall volumetric coefficient and an driving force: The mass transfer area, a is taken as the external surface of the particles, which is 6(1-ε)/DP for spheres. The concentration c* is the value that would be in equilibrium with the average concentration W in the solid. (25.7) Kca= volumetric overall mass-transfer coefficient.

BASIC EQUATION FOR ADSORPTION Rate of Mass Transfer Internal and External Mass-transfer Coefficients Solution to MassTransfer Equations Irreversible Adsorption Linear isotherm YIELD YIELD

INTERNAL AND EXTERNAL MASS-TRANSFER COEFFICIENTS  The overall coefficient, Kc depends on the external coefficient kc, ext and on an effective internal coefficient, kc, int.  Diffusion within the particle is actually an unsteady-state process, and the value of kc, int decreases with time, as solute molecules must penetrate farther and farther into the particle to reach adsorption sites.  Average effective coefficient can be used to give an approximate fit to uptake data for spheres.

This leads to (25.8) De = effective diffusion coefficient, depends on particle porosity, the pore diameter, the tortuosity, and the nature of the diffusing species.  Dp = diameter of particle.

BASIC EQUATION FOR ADSORPTION Rate of Mass Transfer Internal and External Mass-transfer Coefficients Solution to MassTransfer Equations Irreversible Adsorption Linear isotherm YIELD YIELD

SOLUTION TO MASS-TRANSFER EQUATIONS  There are many solutions to Eq. (25.6) and (25.7) for different isotherm shapes and controlling steps, and all solution involve a dimensionless time, τ and a parameter N representing the overall number of transfer units: (25.9) (25.10)  The term Lε/u0 in Eq. (25.9) is the time to displace fluid from external voids in the bed, which is normally negligible.  ρp(1- ε) is the bed density, ρb  τ is the ratio of the time to the ideal time t* from Eq. (25.3). (25.3)  If there were no mass-transfer resistance, the adsorber could be operated with complete removal of solute up to τ = 1.0, and the the concentration would jump from 0 to c/c0 = 1.0.

BASIC EQUATION FOR ADSORPTION Rate of Mass Transfer Internal and External Mass-transfer Coefficients Solution to MassTransfer Equations Irreversible Adsorption Linear isotherm YIELD YIELD

IRREVERSIBLE ADSORPTION  Irreversible adsorption with a constant mass –transfer coefficient is the simplest case to consider, since the rate of mass transfer is then just proportional to the fluid concentration.  Strongly favorable adsorption gives almost the same results as irreversible film, because the equilibrium concentration in the fluid is practically zero until the solid concentration is over one-half the saturation value.  If the accumulation term for the fluid is neglected, Eq. (25.6) and (25.7) are combine to give; (25.6)

(25.12)

(25.7)

 The initial shape of the concentration profile is obtained by integration Eq. (25.12) (25.13)  Since the term KcaL/u0 is defined as N in Eq. (25.10), the concentration at the end of the bed is given by (25.14)

(25.10)

IRREVERSIBLE ADSORPTION  The rate of mass transfer to the first layer of particles is assumed to be constant until the particles reach equilibrium with the fluid, and until this happens, the concentration profile in the bed remains constant.  The time to saturate the first portion of the bed, t1 is the equilibrium capacity divided by the initial transfer rate (W0= 0 to simply the analysis): (25.15)  After this time, the concentration profile moves steadily down the bed, keeping the same shape.  The transfer zone moves at a velocity vz, which is equal to the amount of solute removed per unit time divided by the amount retained on the solid per unit length of bed: (25.16)  The concentration is constant at c0 for the saturated portion of the bed and then falls exponentially in the mass-transfer zone, as shown Fig. 25.10. Fig. 25.10

IRREVERSIBLE ADSORPTION  To predict the break point, Eq. (25.17) is applied for a bed of length L with c/c0 set at 0.05 or another selected value. (25.17)  The length of the saturated bed is the product of transfer zone velocity and the time since the zone started to move: (25.18) (25.19)  Substituting the equation for Lsat in Eq. (25.17) and using the dimensionless terms τ and N [Eq. (25.9) and (25.10) give: (25.20) (25.21)

IRREVERSIBLE ADSORPTION  The predicted breakthrough curve is shown as a solid line in Fig. 25.11.  The slope increases with time, and c/c0 becomes 1.0 at N(τ-1)=1.0.  In practice, the breakthrough curves are usually S-shaped, because the internal diffusion resistance is not negligible, and it increases somewhat when the solid becomes nearly saturated.

Fig 25.11

 When both internal and external resistances are significant, the breakthrough curve is Sshaped, as shown by the dashed line in Fig 25.11.  For this plot, the value of N is based on the overall mass-transfer coefficient given by Eq. (25.8), or it can be expressed in Hall’s terminology as;

(25.22)

EXAMPLE 25.3 (a) (a) Use Use the the breakthrough breakthrough data data in in Example Example 25.2 25.2 to to determine determine N N and and K Kccaa for for the the 8-cm 8-cm bed, bed, assuming assuming irreversible irreversible adsorption. adsorption. (b) (b) Compare Compare K Kccaa with with the the predicted predicted kkccaa for for the the external external film. film. SOLUTION: SOLUTION: (a) = 0.495, τ-1= -0.505. Assume equal (a) From From Example Example 25.2, 25.2, at at c/c c/c00 == 0.05, 0.05, W/W W/Wsat sat = 0.495, τ-1= -0.505. Assume equal internal internal and and external external resistances resistances to to determine determine N N from from Fig. Fig. 25.11: 25.11:

= 23.0 s-1

(b) (b) Prediction Prediction of of kkccaa from from Re, Re, Sc Sc (k (kcc is is the the external external coefficient): coefficient): D Dpp == 0.37 0.37 cm cm At At 25 25ooC, C, 11 atm, atm, μ/ρ μ/ρ == 0.152 0.152 cm cm22/s /s and and D Dvv == 0.0861 0.0861 cm cm22/s. /s. Then Then

EXAMPLE 25.3 (a) (a) Use Use the the breakthrough breakthrough data data in in Example Example 25.2 25.2 to to determine determine N N and and K Kccaa for for the the 8-cm 8-cm bed, bed, assuming assuming irreversible irreversible adsorption. adsorption. (b) (b) Compare Compare K Kccaa with with the the predicted predicted kkccaa for for the the external external film. film. SOLUTION: SOLUTION: (a) = 0.495, τ-1= -0.505. Assume equal (a) From From Example Example 25.2, 25.2, at at c/c c/c00 == 0.05, 0.05, W/W W/Wsat sat = 0.495, τ-1= -0.505. Assume equal internal internal and and external external resistances resistances to to determine determine N N from from Fig. Fig. 25.11: 25.11:

= 23.0 s-1

(b) (b) Prediction Prediction of of kkccaa from from Re, Re, Sc Sc (k (kcc is is the the external external coefficient): coefficient): D Dpp == 0.37 0.37 cm cm At At 25 25ooC, C, 11 atm, atm, μ/ρ μ/ρ == 0.152 0.152 cm cm22/s /s and and D Dvv == 0.0861 0.0861 cm cm22/s. /s. Then Then

SOLUTION: From Eq. (17.74)

Since Kca is slightly less than one-half the predicted value of kca, the external resistance is close to one-half the total resistance, and the calculated value of N need to be revised. The internal coefficient can be obtained from;

If diffusion into the particle occurred only in the gas phase, the maximum possible value of De would be about Dv/4, which leads to;

Since the measured value of kc, int is an order of magnitude greater than this value, surface diffusion must be the dominant transfer mechanism.

BASIC EQUATION FOR ADSORPTION Rate of Mass Transfer Internal and External Mass-transfer Coefficients Solution to MassTransfer Equations Irreversible Adsorption Linear isotherm YIELD YIELD

ADSORBER DESIGN CALCULATION  The The design design of of adsorber adsorber for for gas gas or or liquid liquid purification purification involves; involves; -- choosing choosing the the adsorbent adsorbent and and the the particle particle size, size, selecting selecting an an appropriate appropriate velocity velocity to to get get the the bed bed area, area, and and either either determining determining the the bed bed length length for for aa given given cycle cycle time time or or calculating calculating the the break-through break-through time time for for aa chosen chosen length. length.  For For gas gas purification/ purification/ adsorption: adsorption: -- 44 xx 66- or or 44 xx 10-mesh 10-mesh carbon carbon is is needed needed and and pressure pressure drop drop is is not not aa problem. problem. -- The The gas gas velocity velocity is is usually usually between between 15 15 and and 60 60 cm/s cm/s (0.5 (0.5 and and 22 ft/s) ft/s) -- Because and k c, int increase as D Because the the external external area area varies varies with with 1/D 1/Dpp and and both both kkc,ext c,ext and kc, int increase as Dpp decreases, decreases, kkccaa is is expected expected to to vary vary with with the the -1.5 -1.5 to to -2.0 -2.0 power power of of D Dp.p.  For For liquid liquid adsorption: adsorption: -- smaller smaller particle particle sizes sizes are are chosen, chosen, and and the the fluid fluid velocity velocity is is much much lower lower than than with with gases. gases. -- Typical Typical conditions conditions for for water water treatment treatment are are 20 20 xx 50-mesh 50-mesh carbon carbon (D (Dpp == 0.3 0.3 to to 0.8 0.8 mm) mm) and and aa superficial superficial velocity velocity of of 0.3 0.3 cm/s cm/s (0.01 (0.01 ft/s ft/s or or about about 44 gal/min. gal/min. ft ft22)) -- Even Even with with these these conditions conditions K Kcca/u a/u00 is is smaller smaller than than for for typical typical gas gas adsorption, adsorption, and and LUB LUB may may be be 10 10 to to 20 20 cm cm or or even even as as much much as as 11 m m if if internal internal diffusion diffusion controls. controls.

QUESTION 25.4 Adsorption on activated carbon is being considered to treat a process airstream that has 0.12 volume percent methyl ethyl ketone (MEK), C44H88O. The gas is at 25ooC and 1 atm, and the flow is 16,000 ft33/min. The pressure drop across the bed should not exceed 12 in. H22O. a. If BPL 4 x 10-mesh carbon is used, predict the saturation capacity and the working capacity if the average bed temperature is 35 ooC and the regeneration is stopped when W = 1/3 Wsat . sat b. What gas velocity and bed size could be used to give reasonable cycle time if the length of unused bed is 0.5 ft? How much carbon is needed?

ANSWER (a) From the handbooks, Pˊ = fs = 151 mmHg at 35 oC and ρL = 0.805 g/cm3 at 20 oC. The normal boiling point is 79.6 oC, and the estimated density at this temperature is ρL = 0.75 g/cm3. The molecular weight is 72.1. At 35 oC.

From Fig. 25.4, the volume adsorbed is 24 cm3 per 100 g carbon:

Working capacity = Wsat –W0 = 12g/ 100 g carbon = 0.12 lb/lb carbon. (b) What gas velocity and bed size could be used to give reasonable cycle time if the length of unused bed is 0.5 ft? How much carbon is needed? Try u0 = 1 ft/s:

Amount of adsorbed depends on (T/V) log (fs/f), where: T: adsorption temperature (Kelvin). V: molar volume of the liquid at the boiling point fs: fugasity of the saturated liquid at adsorption temperature f: fugasity of the vapor

For adsorption at atmospheric pressure; * fugasity = partial pressure = vapor pressure

Volume adsorbed is converted to mass by assuming the adsorbed liquid has the same density as liquid at the boiling point.

ANSWER For a circular cross section, D = 18.4 ft. a rectangular bed 10 ft x 27 ft might be more suitable if the bed depth is only 3 to 4 ft. Try L = 4 ft. From Eq. (25.3)

(25.3)

At 25 oC,

=18.1 h If the length of unused bed is 0.5 ft, 3.5 ft is used, and

If the bed length is 3 ft with 2.5 ft used,

ANSWER Allowing for uncertainties in the calculations, a bed length of 3 ft would be satisfactory with regeneration once per 8-h shift. Check ∆P using the Ergun equation, Eq. (7.22). Note that gc is needed when fps units are used. For granular carbon, assume Φs = 0.7 (see Table 7.1). Assume external void fraction ε = 0.35 (see Table 7.2). From handbooks, the properties of air at 25 oC are

From Perry, 7th ed., p. 19-20, for 4 x 10-mesh carbon,

ANSWER For L = 3 ft, ∆P = 9.9 in H2O, which is satisfactory. A velocity of 1.5 ft/s would give ∆P/L = 6.06 in. H 2O/ft and require L ≤ 2 ft to keep ∆P < 12 in. H2O. However, the breakthrough time would be reduced to 11.3/1.5 x (1.5/2.5) = 4.5 h, and the bed would have to be regenerated twice each shift. This design might be satisfactory but does not give as great a margin for error. The recommended design is for two beds 10 x 27 x 3 ft placed in horizontal cylinders. The total inventory of carbon is; mc = 2 (270 x 3)ft3 x 30 lb/ft3 = 48, 600 lb

QUESTION 25.5 Water contaminated with 1.2 ppm TCE is to be purified in a fixed bed of 20 x 50-mesh Ambersorb 563. (a) For a bed length of 2 ft and a flow rate of 4.5 gal/min.ft22, estimate the breakthrough time if the length of the unused bed is 0.6 ft. (b) What is the effective capacity in volume treated per unit bed volume? The adsorbent will be regenerated by steam to remove 85 percent of the TCE. The bulk density of the adsorbent is 0.53 g/cm33.

ANSWER (a) From Fig. 25.5

From Eq. (25.3),

Breakthrough time is; (b) Bed volumes treated,

THANK YOU Prepared by, MISS RAHIMAH OTHMAN