Chapter 4 Adsorption Operation

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Transcript Chapter 4 Adsorption Operation

Chapter 3 Adsorption Operation

By: In. Nurul Hasyimah Mohd Amin

Objectives

At the end of this lessons, students should be able to: • Define adsorption process and its mechanism.

• Explain and differentiate physical and chemical adsorption.

• Discuss the adsorption isotherms.

• Perform calculation related to this topic.

Introductions Adsorption: The accumulation of molecular species at the surface of a solid or liquid rather than in the bulk is called adsorption.

Bulk Surface

Adsorbed material Free materials

How it is happen?

Surface of solids and liquids has the tendency to attract and retain other molecules with which it is brought in to contact.

How it is happen? This is due to unbalanced residual inward forces of attraction at the surface of solids and liquids.

How it is happen?

Due to these residual forces, surface of solid or liquid has a higher concentration of other molecular species than the bulk.

Adsorbate: The molecular species that accumulate at the surface

.

Adsorbate

Adsorbent: The material on the surface of which adsorption takes place.

t Adsorben

Desorption: The removal adsorbate from the surface of

Examples for adsorption ………… 1. If a gas like H 2 , O 2 , Cl 2 etc is taken in a vessel containing powdered charcoal, pressure of the gas slowly decreases as the gas is adsorbed on the surface of charcoal.

2. Air becomes dry in presence of silica gel because adsorption of water takes place on the surface of the gel.

3. Aqueous solution of raw sugar becomes colourless when passed over a bed of charcoal. The coloring matter is adsorbed by the charcoal.

4. Litmus solution or a solution of a dye like methylene blue when shaken with charcoal turns colourless due to adsorption of coloring material.

Absorption: Sorption: If both adsorption and absorption takes place simultaneously is called sorption .

Distinction between Adsorption & Absorption

Adsorption Absorption

It is a bulk phenomenon It is a surface phenomenon Adsorbed species is accumulated in the surface It is a fast process It is uniformly distributed throughout the bulk It is a slow process Rate of adsorption decreases gradually Absorption takes place at steady rate

Physisorption Chemisorption

If accumulation of gas molecules on the surface of solids occurs due to weak van der Waals’ forces of attraction, the adsorption is called physisorption.

Characteristics of Physisorption

1. Non-specific nature: An adsorbent does not show any preference for a gas as the van der Waals’ forces are universal.

2. Easily liquefiable gases like CO 2 , SO 2 , NH 3 adsorbed.

etc, are readily 3. Reversible nature: Physisortion of a gas by a solid is reversible.

4. Increases by increase of pressure.

5. Surface area of adsorbant: When the surface area of the adsorbent increases, more gas is adsorbed, ie extent of adsorption increases.

6. Enthalpy of adsorption: Enthalpy of adsorption of physisorption is very low (20-40 KJ mol -1 )

Since adsorption is exothermic, physisorption takes place readily at low temperature and desorption takes place at higher temperature.

When atoms or molecules of gases are held by solids on its surface by chemical bonds, the adsorption is called chemisorption.

Characteristics of Chemisorption

1. High specificity: It is highly specific and will occur only if chemical bond formation takes between adsorbate and adsorbent.

2. Irreversibility: Chemisorption is irreversible because the chemical bond formed is difficult to break.

3. Chemisorption increases with temperature.

4. Increases by increase of pressure.

5. Surface area of adsorbent: When the surface area of the adsorbent increases, more gas is adsorbed, ie extent of adsorption increases.

6. Enthalpy of adsorption: Enthalpy of adsorption of chemisorption is high (80-240 KJ mol -1 )

1 2 3 4 5

Occurs due to van der Waals’ force Reversible

Physisorption

Not specific Irreversible Specific

Chemisorption

Chemical Bond Enthalpy of adsorption is low Enthalpy of adsorption is high More liquefiable gases are adsorbed readily Decreases with increase of temperature Gases which form compounds adsorbent alone undergo chemisorption with Increases with increase of temperature

6 7 8 9 10

No activation energy is needed

11

Low temperature is favorable.

High pressure favors physisorption and decrease of pressure causes desorption Results in multimolecular layers It is instantaneous High temperature is favorable High pressure is favorable but decreases of pressure does not cause desorption Only unimolecular layer are formed High activation energy is needed It is a slow process

Factors Influencing Adsorption

1. Surface Area of the Adsorbent

Same gas is adsorbed to different extent by different solids at identical conditions.

Greater the surface area of the adsorbent greater the volume of gas adsorbed.

2. Temperature

Adsorption of a gas generally decreases with rise in temperature. This is because adsorption is exothermic and increases of temperature favors the desorption.

backward process which is

Heat

3. Pressure

Adsorption of a gas by an adsorbent at constant temperature increases with increase of pressure.

Applications Adsorption

1. In Gas Masks

2. Production of High Vacuum

3. Softening of Hard water

Ca 2+ Ca 2+ Ca 2+ Ca 2+ Exchange resin

4. Heterogeneous Catalysis

5. Refining of Petroleum

6. Chromatographic Separation

Mechanisms of ad…

1 • diffuse from the bulk of phase to the exterior surface of the adsorbent 2 • diffuse inside the pore and to the surface of the pore.

3 • adsorbed on the surface

Adsorbents

Adsorbents

PAC Silica gel alumina EAC

Characteristics of Adsorbents

Adsorbents

Nature Pore structure Porosity

Characteristics of Adsorbents

Pore structure zeolite Silica gel Activated Carbon

Characteristics of Adsorbents

Porosity Mesopores D = 2-50 nm Macropores D > 50 nm Micropores D < 2 nm

Characteristics of Adsorbents

Nature Hydrophobic interaction Hydrophilic repulsion

Adsorption equilibrium

• If the adsorbent an adsorbate are contacted long enough, an equilibrium will be established • Eq  between amount of the adsorbate in solution and being adsorbed on the adsorbent • “isotherms”

Adsorption equilibrium

• q e = mass of material adsorbed per mass of adsorbent (at eq  ) • C e = concentration in solution when amount adsorbed equal to q e (at eq  ) (mg/L) • q e / C e depend on the types of adsorption (physical or chemical adsorption)

Adsorption equilibrium

Adsorption equilibrium

Isotherms Linear Henry’s Law Favorable Langmuir isotherm Strongly favorable Freundlich isotherm

Henry’s Law

• Assume that the concentration in one phase is proportional to the concentration in the other.

• Can be expressed by equation: • K is constant determined experimentally, m 3 /kg • Apply for a very low concentration.

• At very low concentration the molecules adsorbed are widely spaced over the adsorbent surface so that one molecule has no influence on another.

Langmuir isotherm

• Assume monolayer coverage and constant binding energy between surface and adsorbate. • Maximum adsorption occurs when the surface is covered by a monolayer of adsorbate.

Langmuir isotherm

• Can be expressed by:

q e

K Q o L

C e C e

• The linear form of equation: L/mg

q

1

e

K L Q o C e

 1

Q o

Max adsorption capacity (monolayer) (g solute/g adsorbent) Mg/L

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Freundlich isotherm

• For special case: heterogeneous surface energy. • Particularly for mixed wastes.

• Can be expressed by:

q e

K F C e

1 /

n

• The linearization will give

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Freundlich isotherm

• For freundlich isotherm, we use log-log version

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Freundlich isotherm

• Where q e C e K F = = = the amount of solute per unit weight of adsorbent the concentration at equilibrium Freundlich constant related to adsorption capacity n = Freundlich constant related to adsorption intensity

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Freundlich isotherm

• Adsorption capacity at equilibrium, q e • Adsorption capacity at time t, q t

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Example 1

Batch tests were performed in the laboratory using solutions of phenol in water and particles of granular activated carbon. The equilibrium data at room temperature are shown in Table 4.1.

Determine the isotherm that fits the data.

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Example 1

Table 4.1: Equilibrium data for example 1

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Answer

Langmuir Isotherm

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Answer

Freundlich Isotherm

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Batch adsoprtion

• Batch adsorption is often used to adsorb solutes from liquid solutions when the quantities treated are small in amount.

• Solid will not remove the entire contaminant unless it is infinitely good mixing.

• Total solute removed is the function of amount of contaminant to mass of solid.

• The material balance on the adsorbate is:

q F M

C F S

qM

CS

Where M S : the amount of adsorbent (kg) : the volume of feed solution (m 3)

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Batch adsoprtion

• Batch adsorption is often used to adsorb solutes from liquid solutions when the quantities treated are small in amount.

• Solid will not remove the entire contaminant unless it is infinitely good mixing.

• Total solute removed is the function of amount of contaminant to mass of solid.

• The material balance on the adsorbate is:

q F M

C F S

qM

CS

Where M S : the amount of adsorbent (kg) : the volume of feed solution (m 3)

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Batch adsoprtion

• An equilibrium relation and material balance are needed.

C F C : initial feed concentration : final equilibrium concentration q F q : initial feed adsorbate loading : final equilibrium adsorbate loading

q F M

C F S

qM

CS

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Example 2

A waste water (volume 1 m 3 ) contains 0.21

kg phenol/m 3 . A total of 1.40 kg of fresh granular activated carbon is added to the solution, mixed thoroughly to reach equilibrium. Using isotherm in Example 1, calculate the final equilibrium values and percentage of phenol adsorbed by the activated carbon.

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Answer

0.16

0.14

0.12

0.1

0.08

0.06

0.04

0.02

0 0 0.1

0.2

c (kg solute/m 3 )

0.3

0.4

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Answer

At the intersection; q = 0.106 and c = 0.062

% extracted = [(0.210 – 0.062)/(0.21)] x 100% = 70.5%

q e C e C

e q

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e e

Adsorption through fixed bed adsorber • The concentrations in the fluid phase and the solid phase change with time and position in the bed.

• As fluid enters the bed, it comes in contact with the first few layers of absorbent • Solute adsorbs, filling up some of the available sites until the sites become saturated • Fluid penetrates further into the bed before all solute is removed. • As the fluid passes though the bed, the concentration in this fluid drops very rapidly to almost zero • The active region shifts down through the bed as time goes on.

• As the solution continues to flow, this mass transfer zone, which is S-shaped, moves down the column.

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Breakthrough curve

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Breakthrough curve • The

break point

occurs when the concentration of the fluid leaving the bed as unadsorbed solute begins to emerge. The bed has become ineffective (very small but detectable).

• a breakpoint composition is set to be the maximum amount of solute that can be acceptably lost

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Breakthrough curve After the

t b

is reached, the rises very rapidly up to

c d

, (end of the breakthrough curve where the bed is judged ineffective)

c

c b t 1 t 2 t 3 t 4 t 5 c d

The outlet concentration (

point)

starts to rise to c , the exit bed starts rise to c b

break

at t 5 when the concentration of the fluid leaving the b

Breakthrough curve

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Breakthrough curve • The time required for a bed to become totally saturated is obtained by integrating as time goes to infinity: • In operation, you want to stop the process before solute breaks through, so integration to the breakpoint time gives the "usable" capacity:

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Breakthrough curve • The length of bed used up to break point is: • The length of unused bed is: • The total design height of a bed is determined by adding the required usable capacity to the unused height.

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Example 3

A waste stream of alcohol 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 bed of 14 cm containing 79.2 g of carbon. The inlet stream having a concentration, C o of 600 ppm and a density of 0.00115 g/cm 3 entered the bed at a flow rate of 754 cm 3 /s. Data in the table give the concentrations of the breakthrough curve. The break point concentration is set at C/C o = 0.01. Do as follows.

C e q e C

e q e

.

1

Q o a K

.

1

Q

o a C

o a e C Q o a e

Example 3

Time, h

0 3 3.5

4 4.5

5

C/C o

0 0 0.002

0.03

0.155

0.396

Time, h

5.5

6 6.2

6.5

6.8

C/C o

0.658

0.903

0.933

0.975

0.993

Example 3 a) Determine the breakpoint time, the fraction of total capacity used up to the break point, and the length of the unused bed. Also determine the saturation loading capacity of the carbon.

b) If the break point time required for a new column is 6.0 h, what is the new total length of the column required?

Example 3

Example 3

a) The break point as c/c o t d = 6.95

Area A1 = 3.65

Area A2 = 1.51

= 0.01, t = 3.65hr

Graphically numerical methods Simpson rule

t T

 0     1 

c o c

 

dt

 3 .

65  1 .

51  5 .

16

hr

Example 3

t u

t

0 

b

   1 

c c o

  

dt

 3 .

65

hr

The fraction capacity used to break up to break point;

t u t T

 3 .

65 5 .

16  0 .

707 The unused bed = (1.0 – 0.707) x 14 cm = 4.1 m