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Lecture No. 9

Physico-Chemical Processes

Chapter 9, p.447

Includes technologies that can be used for:

- hazardous waste treatment

- soil remediation

Each section includes:

- description of technology

- theory

- design

1. Air Stripping

A. Description

Air stripping is a mass transfer process that enhances the volatilization of compounds from

water by passing air through water to improve the transfer between the air and water phases.

One of the most common remediation methods for VOCs.

Suited for low concentrations < 200 mg/l.

Packed towers, tray towers, spray systems, diffused aeration, mechanical aeration. Packed

towers are generally used for remediating ground water.

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Lecture No. 1, Hazardous Waste, Page No. 2

B. Process Description

Process consists of counter-current flow of water and air through a packing material. The

packing material provides a high surface area for VOC transfer from the liquid to the gaseous

phase. F9-1, p.448.

Typical packing material consists of plastic shapes with high surface to volume ratios, specificvolume.

R = H'(Qa/Qw) units p.449

R = stripping factor

C. Theory

Two film theory. F9-2, p.449.

- Bulk film to liquid film

- Liquid film to air film

- Air film to bulk air

Sherwood Holloway equation

KLa = x DL x (305L/)1-n(/DL)

0.5 eq.9-4 units p.450

Also:

Z (depth of column) = HTU x NTU

HTU =L

MwKLa

NTU =R

R-1ln

(Cin/Cout)(R-1) + 1

Req.9-20 units p.455

Different packing shapes are available. See T9-2, p.451

Example: ** Problem 9-2, p.544

D. Design

Stripping towers have diameters of .5-3m and heights of 1-15m.

The air-to-water ratio is 5-several hundred and is controlled by pressure drop and flooding

considerations.

Distribution plates should be placed every 5-diameters to avoid channeling around the wall as

opposed to the packing media

It may be necessary to clean up the off-gas with activated carbon.

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The pressure drop in the tower should be between .25-.5 inches H2O/ft of tower to avoid

flooding.

Use F9-5, p.458 to simplify calculations.

2. Soil Vapor Extraction

A. Description

Soil Vapor Extraction (SVE) is a remedial technique to remove VOCs from soil in the vadose

zone or from stockpiled, excavated soil. The vadose zone is the zone above the GWT.

SVE consists of passing an air stream through the soil, thereby transferring the contaminants

fro the soil matrix to the air.

SVE systems can be enhanced:

- Install ground water extraction pumps to increase the vadose zone and perhapssimultaneously treat ground water.

- Impermeable barrier over the surface to minimize short-circuiting

- Install air recharge wells.

- Install wells into the ground water.

B. Theory

The removal of VOCs from the vadose zone can be modeled as a 5-step process:

- Gases desorb from the soil particles.

- Transfer to the soil water.- Volatize to the soil gas

- Gas migrates to surface

- Released to atmosphere

The movement of contaminants in the soil gas through the soil media can be described by two

processes

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- Advection. Movement with bulk airflow through the soil media and best describes

the flow through permeable soils with the unsaturated zone.

- Diffusion. Movement through the soil media via concentration gradients. Diffusion

tends to control in soils of low permeability.

Provided the leak is of sufficient quantity, the VOC contamination will tend to migrate

downward through the unsaturated zone, leaving globules, films and small droplets of thereleased material. Low density contaminants will tend to collect in the capillary fringe or float

on the ground water surface. Dense contaminants will tend to pass through the ground water

until encountering a impermeable layer.

A release of contaminants will result in residual contamination the soil pores. This residual

material in the unsaturated zone is the target contamination for cleanup via SVE.

Diffusion may be the rate limiting step for mass transfer.

Current practice is to utilize empirical models to select the most appropriate mechanical

system and then to use field data to refine system design.

Movement of VOCs through the soil is controlled in part by diffusion and Fick's Law:

J = -Dv

dC

dz eq.9-22 units p.464 The partition coefficient refers to the preference of contaminant for soil or water. A higher Kp

indicates that a contaminant is more likely to remain on the soil and not be transmitted

through soil moisture movement and is:

Kp =X

Ceq.9-26 units p. 465

Example:Given: Ethylene Dibromide and hexane.

Find: Based on H, which is a more likely candidate for SVE.

Assume T=20C, K=C+273.2=20+273.2

K=293.2

Ethylene Dibromide

From app. A, p.1046

A=5.70

B=3.24 x 103

p.1038 eq.A-2

H = exp {A-B/T}

H = exp{5.70-3.24x103/293.2}= exp{5.70-11.05}=exp{-5.35}= e-5.35

H = 4.748 x 10-3 atm.m3/mol

Hexane

From app. A, p.1046

A=25.3

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B=7.53 x 103

p.1038 eq.A-2

H = exp {A-B/T}

H = exp{25.3-7.53x103/293.2}= exp{25.3-25.68}=exp{-0.382}= e-0.382

H = .682 atm.m3/mol

Since hexane, .682 atm.m3/mol > Ethylene Dibromide, 4.748 x 10-3 atm.m3/mol,hexane would be best suited to vadose treatment by SVE.

Hartley equation estimates the volatilization of chemicals from soil:

J =

Asat(1-h)

1

D+2AsatM

kRT2

eq.9-32 units p.467

The second term in the denominator,AsatM

kRT2, indicates the resistance to volatilzation due to

thermal elements and may be neglected for compounds significantly less volatile than water.

Example: problem 9-11, 9-14

C. Design

SVEs primary benefit is that it is an in situ method and as such does not require the removal

and transportation of the hazardous waste. Also SVE can remediate soil beneath structures;

does not require reagents and employs conventional equipment, labor and materials.

SVE is not appropriate for:

- low-permeability soil

- low vapor pressure contaminants

- high ground water table

D. Process Description

A typical SVE system:

Infrastructure:

- vapor extraction wells, 6-11' deep

- piping

- monitoring wells

- gauges and valves

- impermeable cover

- vent wells

Equipment

- vacuum/blower unit, .5-30" Hg.

- moisture knockout drum

- off-gas treatment

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Three variables control performance, pilot testing is essential:

- well spacing, critical

- air flow rate

- subsurface pressure

Example:Given: A radius of 25'

Find: The vacuum requirement

From F9-12, p.474

.05 monitoring point vacuum/well vacuum

From F9-11, p.473

.05 inches water @ 75cfm



Given: The above data

Find: How much vacuum is required.

From F9-10, p.472.05 inches water @ 75cfm requires 9 inches water

.45 inches water @ 150cfm requires 20 inches water

.8 inches water @ 240cfm requires 34 inches water

3. Carbon Adsorption

Adsorption is a process in which a soluble contaminant is removed from water by contact with

a solid surface typically activated carbon usually in granulated form (GAC).

A. Process Description F9-14, p.477

The activated carbon is placed in cylindrical vessel, contaminated water enters the top,

contacts the carbon and exits the bottom.

Ancillary considerations include: a way to regenerate the carbon which is done thermally.

Extensively used in water and wastewater systems for the removal of non-biodegradable

organics and as a polishing step.

B. Theory

Sorption is he process by which a component moves from one phase to another across someboundary. In adsorption the process takes place at a surface.

Movement of an organic molecule to a surface involves 4 transport phenomena:

- bulk fluid transport

- film transport

- pore diffusion

- actual physical attachment

Driving forces that control adsorption:

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- chemical affinity between the pollutant and the activated carbon.

- electrical attraction

- van der Waal's forces

- hydrophobic nature of the organic

A plot of the amount of contaminant adsorbed per unit mass of carbon, X/M, against the

concentration of contaminant in the bulk fluid, C, is an adsorption isotherm. The Freundlichisotherm is an empirical model mathematically expressed:

X/M = KCf1/n eq.9-33 units p. 479

Example: **problem 9-16, 9-19

C. Design

The design of adsorption units requires column tests that simulate the actual operation of full

scale units.

In the lab, 2-inch diameter columns are filled with carbon and the contaminated ground water

is run through the columns. The effluent is monitored for the contaminants of interest.

The adsorption zone is where adsorption takes place. Breakthrough is the point where a

specified amount of the influent is detected in the effluent usually 5-10%.

Hutchins provides a method for analyzing the results of multiple columns in series F9-19,

p.483, by evaluating the bed-depth service time (BDST). A horizontal line is drawn where

breakthrough =90% of the influent concentration, Cout/Cin = .9. The Bohart-Adams equation

represents 90% removal.

t = aX + b eq.9-35 units p.483

4. Steam Stripping

A. Process Description

Steam stripping is the purging of contaminants from ground water by the use of steam.

Capable of reducing VOCs to very low concentrations.

T9-6. p.487. EPA established Best Available Technology Economically Achievable (BATEA)

for Organic Chemical, Plastics and Synthetic Fibers (OCPSF). In some cases, VOCs may be

below detectable limits.

The differences between steam stripping and air stripping:

- steam not air is the stripping gas

- the stripping gas, steam, is infinitely soluble in water

- much higher temperatures are used

- the organics in the water are recovered as a separate liquid phase

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Schematics F9-21,22, p.488. Based on distillation. The heated feed water is fed to the tank

and flows down where it encounters the steam which is flowing upward, counter-current to

the organics. The organics volatize and are carried upward; the mixture is condensed and

since the organics are supersaturated they separate and are disposed of.

B. Design

Too complicated based on:

- Thermodynamics

- Material Balance

- Mass transfer

- Process design selection

Generalized material balance:

F + S = B eq.9-49 units p.496

A steam stripper is designed as an equilibrium stage process; mass transfer resistances are notdirectly considered in the design approach.

Design considerations:

The strippalitity of the organics

- Simple for a single organic, but for a mixture of organics, a computerized process

simulator to assess the thermodynamics of the interactions between the various organics is

required.

- Rule of thumb. Any priority pollutant that is analyzed by direct injection on a gas

chromatograph can be considered a good candidate for high-efficiency stream stripping.

- Rule of thumb. Any compound with a boiling point

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Well established technology capable of treating a wide range of organics and inorganics:

chlorinated VOCs, mercaptans, phenols, cyanide.

Principle oxidants: ozone, hydrogen peroxide, chlorine. The chemicals that are reduced are the

contaminants.

Oxidation-reduction reactions occur in pairs to comprise an overall REDOX reaction.

Oxidizing agents are non-specific and will react with any reducing agents present in the wastestream.


Completely mixed or plug flow.

Mixing can be provided by :

- mechanical agitation

- pressure drop

- bubbling air

Hydrogen peroxide and UV, ultraviolet light are generally used together.

Common redox reactions T9-8, p.508, for example the reduction of hexavalent chromium to

tri-valent chromium using sulfur dioxide and ferrous sulfate.

Example:Given: A waste stream of 5000 gal/day containing 86 mg/l of hexavalent chromium.

Find: The stoichiometric amount of ferrous sulfate required to reduce it to trivalent chromium.

From T9-8, p.509

2CrO3 + 6FeSO4 + 6H2SO4 3Fe2(SO4)3 + Cr2(SO4)3 + 6H2O

From cover of book, atomic weights:

Cr = 51.996

O = 15.994

Fe = 55.847

S = 32.06

From the equation 6 moles of ferrous sulfate are required to reduce 2 moles of hexavalent chromium or a

ratio of 6/2=3.

Mass of hexavalent chromium reduced daily is:

lb/day = 86 mg/l x 8.34lb/MG

mg/lx 5000/106 MGD

lb/day = .359 hexavalent chromium

Lb moles of CrO3 discharged daily:

CrO3

Cr = 51.996x1=51.996

O = 15.994x3=47.982

CrO3 = 99.978 lb/lb mole

.359 lb/day / 99.978 lb/lb mole = 3.59 x 10-3 lb mole/day

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FeSO4 required daily:

Fe = 55.847x1=55.847

S = 32.06x1=32.06

O= 15.994x4=63.976

FeSO4 = 151.883

3 (molar ratio from equation) x 3.59 x 10-3 lb mole/day = .0108 lb mole FeSO4

= .0108 lb mole FeSO4 x 151.883 lb/lb mole= 1.64 lb/day of FeSO4 required, theoretical or stoichiometric amount

The power of an oxidizing or reducing agents is measure by it electrode potential. An

indication of how a reaction will proceed can be determined from the free energy

considerations:

G = -nFE = -RTlnK eq.9-52 units p. 510

E may be determined from T9-9, p.511.

It is possible to measure ORP, Oxidation Reduction Potential, directly by means of a galvanic

cell made up of a gold or platinum anode and a reference electrode, the cathode.

Nernst equation:

E = E - (RT/nf)ln Q eq.9-53 units p.510

B. Design

Ozone. Ozone is a blue gas with a pungent order and the most powerful oxidant available.

Ozone has a very high free energy which indicates that the oxidation reaction may proceed to

completion. Ozone dissociates to oxygen very rapidly and must be generated on site.

Example:Given: Hexachlorobiphenyl

Find: The time required for 60% removal using ozone with and without UV.

From F9-29, p51460% removal means 40% remaining or C/Co=.4

For no UV, Time=not doable

With UV, Time 50 min, UV make a significant difference

Hydrogen peroxide. Hydrogen peroxide is effective in oxidizing organic in soil through in situ

treatment. As with ozone, hydrogen peroxide is greatly enhance when used with UV.

Example:Given: It is desired to use hydrogen peroxide + UV to reduce chloroform to 90%.

Find: Time required.

From F-9-34, p.518

90% removal means 10% or .1 remaining

@ .1 on the graph for chloroform, time 150 minutes

Chlorine. Chlorine and its various compounds are used extensively in water and waste water

treatment and is the principal chemical involved in disinfection. However, when combined

with organic material, chlorine forms THM, TriHaloMethanes, which are carcinogenic.

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Chlorine is evaporated to a gas and mixed with water to provide a hypochlourous acid (HOCl)

solution.

6. Supercritical Fluids (SCF)

Supercritical fluids (SCF) is an emerging technology and are materials at elevated temperature

and pressure that have properties between those of a gas and liquid.

With SCF, organics in soils, sediments or water are dissolved in the fluid at elevated

temperatures and pressure conditions and released from the SCF at lower temperatures and

pressures.


F9-36, p.522. The contaminated stream is introduced into the extraction vessel, heated and

pressurized. The contaminant dissolves in the SCF which is then expanded which lowers the

solubility of the organic contaminant resulting in separation of the organic contaminant from

the extracting fluid.

B. Theory

The critical point is a temperature/pressure where the material exhibits properties between a

liquid and gas; densities approaching the liquid phase, dffusivities and viscosities approaching

the gas phase. As a result of these properties, organic compounds are highly soluble in SCFs

and can easily transfer to the SCF from their original medium.

C. Design

Selection of the SCF, solvent. Re-use is important. CO2 is a good choice because of lack of

toxicity, also T9-13, p,529.

7. Membrane Processes

Well established technology.

Membrane which is usually a solid matrix or swollen gel refers to a barrier to flow which will

allow the passage of water, ions or small molecules. Not a conventional, gravity, filtration

process as the driving force may be electrostatic or high pressure.

The membranes are subject to fouling and in hazardous waste management, membranes are

limited to extremely toxic materials that can not be removed by cost-effective technologies.

Processes include: electrodialysis, reverse osmosis and ultrafiltration.

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Electrodialysis. F.9-42, p.530. Electrodialysis consists of the separation of ionic species fro

water by applying a direct-current electric field. By alternating cation and anion exchange

membrane between two electrodes, alternate dilute and concentrated cells are created. The

membranes are about .5mm thick and the spacers are 1mm thick. Operated at 40-60psi and90% of the feed is turned into product water, the remainder is concentrate.

Reverse Osmosis. F9-46, p.534. In reverse osmosis, a solvent is separated fro a solution by

applying a pressure greater than the osmotic pressure, thus forcing the solvent through a

semipermeable membrane. The membrane will allow the water but not the salt, solvent, to

pass. In reverse osmosis, a pressure is applied to force the salt, solvent through the membrane,

thus leaving product water.

Ultrafiltration. F9-49. p.536. Ultrafiltration separates solutes from a solvent on the basis of

molecular size and shape by passing the solution through a membrane module where a

pressure difference is maintained across the membrane. Water molecules pass through, heavy

molecules are retained on the filter. Fouling is avoided by high velocities which in turn yields a

low efficacy requiring multiple passes. Molecules of molecular weight greater than 500 andless than 500,000 can be separated. Heavier molecules can be separated by conventional

filtration, the lower size reflects the opening size in commercially available membrane.

B. Theory

Electrodialysis. Faraday's Law yield required current:

I = (FQN/n) x (E1/E2) eq.9-69 units p.537

Voltage, Ohm's Law:

E = IR eq. 9-70 units p.537

Power:P = I2R eq. 9-71 units p.537

Current density, CD, is the current passing through a unit areas of membrane, amp/m2.

Reverse Osmosis. Osmotic pressure, solute rejections and flows are of interest. Osmotic

pressure determined by the van't Hoff equation:

= cNCsRT eq.9-72 units p.539

The flow is:

Jw = Wp x (P - ) eq.9-75 units p.540

Example:Given: Manganous nitrate, Mn(NO3)2, salt solution at 8400 mg/l at 30C. =.87, The wastewater has a

flow rate of 100gpm. A vendor gives the following data:

Wp=2.0 x 10-6 gmol/cm2.sec.atm

Area of a bundle = 500ft2

65% recovery rate

Optimal pressure across the membrane 625 psi

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Find:

1.) The Osmotic Pressure.

2.) Flow through the membrane, Jw

3.) Number of bundles required.

1.) Osmotic pressure.eq 9-72, p.539

= cNCsRT units p. 9-72

N= 3 (Mn(NO3)2, Mn=1, NO3=2)

Molecular weight

Mn=24.305x1=23.305

N= 14x 2=28

O=16x6=96

Mn(NO3)2=147.3 g/gmol

Cs = 8400 mg/l = 8.4 g/l / 147.3 g/gmol

Cs = .0570 gmol/l

K = C + 273.2 = 30 + 273.2

K = 303.2

= cNCsRT = .87 x 3 x .0570 gmol/l x .082 atm.l/gmol.K (p.1080) x 303.2 K

= 3.698 atmospheres x 14.7psi/atm. = 54.35psi

2.) Flow, Jw

Jw = Wp x (P - ) units p.540

Jw = 2.0 x 10-6 gmol/cm2.sec.atm x (625-54.35) (1atm/14.7psi)

Jw = 7.76 x 10-5 gmol/cm2.sec

3.) Number of bundles for 100gpm

Q = 100gpm x 3.79l/gal x 1min/60sec x 1000g/l x 1gmol/18g (AWof water) x cm2.sec/ 7.76 x 10-5 gmol

Q = 4.52 x 106 cm2 of membrane

Each bundle contains 500 ft2.

No. of bundles = 4.52 x 106 cm2 xBundle

500 ft2x

(3.28ft)2

m2x

1 m2

(100)2cm

No. of bundles = 9.73 bundles use 10

Ultrafiltration:

Jw =(P - )

(Rg + Rm)eq. 9-78 units p.542

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C. Design

Concentration Polarization results in an increased power requirement without an appreciable

increase in performance and occurs in all membranes.

HOMEWORKRead Chapter 9, Physico-Chemical Processes, pp. 447-554

Problems, p.544, 9-5, 9-7, 9-10, 9-14, 9-17, 9-18, 9-27, 9-29, 9-31, 9-32, 9-33, 9-34

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