Chapter 6 ADSORPTION MISS RAHIMAH BINTI OTHMAN Email

Chapter 6: ADSORPTION MISS. RAHIMAH BINTI OTHMAN (Email: rahimah@unimap. edu. my)

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

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

INTRODUCTION TO ADSORPTION A DSORPTION ≠ A BSORPTION ! 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 A PPLICATIONOF A DSORPTION: Used in many industrial processes: Adsorbing the desired product from fermentation broths. Isolation of proteins. Dehumidification. odour/colour/taste removal. gas pollutant removal (H 2 S). 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 T YPES OF A DSORPTION 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

A DSORPTION E QUIPMENT 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

A DSORPTION I SOTHERM S 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 o. C

A DSORPTION 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 25. 1. Adsorption on BPL carbon is used to treat an airstream containing 0. 2 percent n-hexane at 20 o. C. (a) Estimate the equilibrium capacity for a bed operating to 20 o. C. (b) How much would the capacity decrease if the heat of adsorption raised the bed temperature to 40 o. C.

ANSWER (a) Estimate the equilibrium capacity for a bed operating to 20 o. C. The MWn-hexane (C 6 H 14)= 86. 17, At 20 o C (from Perry’s Handbook, 7 thed. ) Pʹ=120 mm Hg ≈ fs. At the normal boiling point (68. 7 o. C), ρL=0. 615 g/cm 3. The adsorption pressure P is 760 mm Hg. (b) At 40 o. C, Pˊ= 276 mm Hg

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. • 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 ü strongly favourable ü Describe the adsorption of variety of antibiotics, steroids and hormones. ü high adsorption at low fluid concentration 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

P RINCIPLESOF A DSORPTION 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/c 0 values of 0. 95 to 0. 05.

M ASS T RANSFER Z ONE AND B REAKTHROUGH

Concentration Profile In Fixed Beds

Concentration Profile In Fixed Beds • t 1: no part of the bed is saturated. • From t 1 to t 2: the wave had moved down the bed. • t 2: the bed is almost saturated for a distance LS, but is still clean at LF. • Little adsorption occurs beyond LF at time t 2, and the adsorbent is still unused. • The MTZ where adsorption takes place is the region between LS and LF. • The concentration of the adsorbate on the adsorbent is related to the adsorbate concentration in the feed by thermodynamic equilibrium. • Because it is difficult to determine where MTZ begins and ends, LF can be taken where C/CF = 0. 05, with LS at C/CF = 0. 95. • t. B: 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/CF = 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 processing unit.

Concentration Is the ratio of solute concentration to inlet solute concentration in the fluid. profile in fixed beds Figure 25. 6(a)

B REAKTHROUGH C URVES t 1, t 2, t 3: the exit concentration is practically zero. Time for fluid living the bed.

B REAKTHROUGH C URVES

B REAKTHROUGH C URVES • tb – time when the concentration reaches break • • 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)

B REAKTHROUGH C URVES • 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*

B REAKTHROUGH C URVES 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; The break point time, tb is calculated from the ideal time and the fraction of bed utilized:

EXAMPLE 25. 2. QUESTION 2 The adsorption of n-butanol from air was studied in a small (10. 16 cm diameter) with 300 and 600 g carbon, corresponding to bed lengths of 8 and 16 cm. (a) From the following data for effluent concentration, estimate the saturation capacity of the carbon and the fraction of the bed used at c/c 0 = 0. 05. (b) Predict the break-point time for a bed length of 32 cm. Data for n-butanol on Columbia JXC 4/6 carbon are as follows: 300 g t, h c/c 0 1 0. 005 1. 5 0. 01 2 0. 027 2. 4 0. 050 2. 8 0. 10 3. 3 0. 20 4 0. 29 5 0. 56 t, h 5 5. 5 6 6. 5 7 7. 5 8 8. 5 600 g c/c 0 0. 0019 0. 003 0. 0079 0. 018 0. 039 0. 077 0. 15 0. 24

ANSWER The concentration profiles are plotted in Fig 25. 8, and extended to c/c 0=1. 0 assuming the curves are symmetric about c/c 0=0. 5. Per square centimeter of bed cross section, the solute feed rate is The total solute adsorbed is the area above the graph multiplied by FA. For the 8 cm bed, the area is;

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: At the break point, where c/c 0 = 0. 05, and t = 2. 4 h 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/c 0 = 0. 25, the curves are assume to be parallel. For the entire bed, At c/c 0 = 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 waste stream of n-butanol vapor in air from a process was adsorbed by activated carbon particles in a packed bed having a diameter of 4 cm and length of 14 cm containing 79. 2 g of carbon. The density of the activated carbon is 0. 461 g/cm 3. The inlet gas stream having a concentration, C 0 of 600 ppm and a density of 0. 00115 g/cm 3 entered the bed at the solute feed rate, FA of 0. 063 g/cm 2. s. Data in Table 3. 1 give the concentrations of the fluid in the bed, C. The break point concentration is set at C/Co = 0. 05. QUESTION; 1. 2. 3. 4. Plot a breakthrough curve. Determine the break-point time. Calculate the saturation capacity of the carbon, Wsat. Calculate the length of unused bed (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/cm 3 1. Plot a breakthrough curve. t (h) C C/C 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. 667 E-04 0. 002 0. 030 0. 155 0. 396 0. 658 0. 903 0. 933 0. 975 0. 993 0. 9 0. 8 0. 7 C/C 0 Inlet gas stream, C 0 = 600 ppm ρ = 0. 00115 g/cm 3 FA = 0. 063 g/cm 2. s C/C 0 = 0. 05 1. 0 0. 6 0. 5 0. 4 0. 3 0. 2 0. 1 0. 0 0 1 2 3 4 t, h 5 6 7

ANSWER- cont’ Determine the break-point time. From the breakthrough curve, breakthrough time at C/C 0 = 0. 05 is tb = 4. 1 h 1. 0 0. 9 0. 8 0. 7 C/C 0 2. 0. 6 0. 5 0. 4 0. 3 0. 2 0. 1 0. 0 0 1 2 3 t, h 4 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 Simpson’s Rule of integration. (pp. 872) From the graph plotted, the following data is obtained; t (h) C/C 0 f(x) = (1 -C/C 0) 0. 00 0 1 1. 75 0 1 3. 50 0. 002 0. 998 5. 25 0. 5 7. 00 1 0

4. Calculate the length of unused bed (LUB). t (h) C/C 0 f(x) = (1 -C/C 0) 0. 00 0 1 4. 1 0. 05 0. 95

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

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

RATE OF MASS TRANSFER Equation for mass transfer in fixed-bed adsorption are obtained by making a solute material balance for a section d. L 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 1 -ε. 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 Mass. Transfer Equations Irreversible Adsorption Linear isotherm YIELD

INTERNAL AND EXTERNAL MASS-TRANSFER COEFFICIENTS Ø The overall coefficient, Kc depends on the external coefficient kc, ext and on an effective internal coefficient, kc, int. Ø 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. Ø Average effective coefficient can be used to give an approximate fit to uptake data for spheres. This leads to (25. 8) v. De = effective diffusion coefficient, depends on particle porosity, the pore diameter, the tortuosity, and the nature of the diffusing species. v Dp = diameter of particle.

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

SOLUTION TO MASS-TRANSFER EQUATIONS Ø 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) Ø The term Lε/u 0 in Eq. (25. 9) is the time to displace fluid from external voids in the bed, which is normally negligible. Ø ρp(1 - ε) is the bed density, ρb Ø τ is the ratio of the time to the ideal time t* from Eq. (25. 3) Ø If there were no mass-transfer resistance, the adsorber could be operated with complete removal of solute up to τ = 1. 0, and the concentration would jump from 0 to c/c 0 = 1. 0.

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

IRREVERSIBLE ADSORPTION Ø 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. Ø 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. Ø 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) Ø The initial shape of the concentration profile is obtained by integration Eq. (25. 12) (25. 13) Ø Since the term Kca. L/u 0 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 Ø 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. Ø The time to saturate the first portion of the bed, t 1 is the equilibrium capacity divided by the initial transfer rate (W 0= 0 to simply the analysis): (25. 15) Ø After this time, the concentration profile moves steadily down the bed, keeping the same shape. Ø 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) Ø The concentration is constant at c 0 for the saturated portion of the bed and then falls exponentially in the mass-transfer zone, as shown Fig. 25. 10

IRREVERSIBLE ADSORPTION Ø To predict the break point, Eq. (25. 17) is applied for a bed of length L with c/c 0 set at 0. 05 or another selected value. (25. 17) Ø 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) Ø 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 Ø The predicted breakthrough curve is shown as a solid line in Fig. 25. 11. Ø The slope increases with time, and c/c 0 becomes 1. 0 at N(τ-1)=1. 0. Ø 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 Ø When both internal and external resistances are significant, the breakthrough curve is Sshaped, as shown by the dashed line in Fig 25. 11. Ø 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) Use the breakthrough data in Example 25. 2 to determine N and Kca for the 8 -cm bed, assuming irreversible adsorption. (b) Compare Kca with the predicted kca for the external film. SOLUTION: (a) From Example 25. 2, at c/c 0 = 0. 05, W/Wsat = 0. 495, τ-1= -0. 505. Assume equal internal and external resistances to determine N from Fig. 25. 11: = 23. 0 s-1 (b) Prediction of kca from Re, Sc (kc is the external coefficient): Dp = 0. 37 cm At 25 o. C, 1 atm, μ/ρ = 0. 152 cm 2/s and Dv = 0. 0861 cm 2/s. 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 Mass. Transfer Equations Irreversible Adsorption Linear isotherm YIELD

ADSORBER DESIGN CALCULATION v. The design of adsorber for gas or liquid purification involves; - choosing the adsorbent and the particle size, selecting an appropriate velocity to get the bed area, and either determining the bed length for a given cycle time or calculating the break-through time for a chosen length. v. For gas purification/ adsorption: - 4 x 6 - or 4 x 10 -mesh carbon is needed and pressure drop is not a problem. - The gas velocity is usually between 15 and 60 cm/s (0. 5 and 2 ft/s) - Because the external area varies with 1/Dp and both kc, ext and kc, int increase as Dp decreases, kca is expected to vary with the -1. 5 to -2. 0 power of Dp. v. For liquid adsorption: - smaller particle sizes are chosen, and the fluid velocity is much lower than with gases. - Typical conditions for water treatment are 20 x 50 -mesh carbon (Dp = 0. 3 to 0. 8 mm) and a superficial velocity of 0. 3 cm/s (0. 01 ft/s or about 4 gal/min. ft 2) - Even with these conditions Kca/u 0 is smaller than for typical gas adsorption, and LUB may be 10 to 20 cm or even as much as 1 m if internal diffusion controls.

QUESTION 25. 4 Adsorption on activated carbon is being considered to treat a process airstream that has 0. 12 volume percent methyl ketone (MEK), C 4 H 8 O. The gas is at 25 o. C and 1 atm, and the flow is 16, 000 ft 3/min. The pressure drop across the bed should not exceed 12 in. H 2 O. 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 o. C and the regeneration is stopped when W = 1/3 Wsat. a. 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 mm. Hg at 35 o. C and ρL = 0. 805 g/cm 3 at 20 o. C. The normal boiling point is 79. 6 o. C, and the estimated density at this temperature is ρL = 0. 75 g/cm 3. The molecular weight is 72. 1. At 35 o. C. From Fig. 25. 4, the volume adsorbed is 24 cm 3 per 100 g carbon: Working capacity = Wsat –W 0 = 12 g/ 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 u 0 = 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) At 25 o. C, =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 o. C are From Perry, 7 th ed. , p. 19 -20, for 4 x 10 -mesh carbon,

ANSWER For L = 3 ft, ∆P = 9. 9 in H 2 O, which is satisfactory. A velocity of 1. 5 ft/s would give ∆P/L = 6. 06 in. H 2 O/ft and require L ≤ 2 ft to keep ∆P < 12 in. H 2 O. 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)ft 3 x 30 lb/ft 3 = 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. ft 2, 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/cm 3.

ANSWER (a) From Fig. 25. 5 From Eq. (25. 3), Breakthrough time is; (b) Bed volumes treated,

T HANK YOU Prepared by, MISS RAHIMAH OTHMAN