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1

Wastewater and Bioremediation Wastewater and Bioremediation

CHNG 3804

Fariba Dehghani

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Particle Settling VelocitiesParticle Settling VelocitiesParticle Settling Velocities

gVF sG ρ=

gVF wB ρ=

2

swpDD vAC5.0F ρ=

DBG FFF =−

GFDF

DF

BF

Drag force

Gravity force

Buoyancy force

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Particle Settling VelocitiesParticle Settling VelocitiesParticle Settling Velocities

DBG FFF =−

GFDF

DF

BF

2/1

wD

wss

C3

d)(g4v

ρ

ρ−ρ=

ν== s

e

dvRR

If R < 0.5If R < 0.5

If 0.5 <R < 10If 0.5 <R < 1044

If R > 10If R > 1044

R/24CD ====

4.0CD ====

34.0R/3R/24C 5.0D ++++++++====

d3

2

)2/d(

)2/d(3/4A/V

2

3

p =π

π=

Reynolds No.Reynolds No.

4

Stokes’ LawStokesStokes’’ LawLaw

sv

µ

ρ−ρ=

18

d)(gv

2ws

s

ν= sdv

R

If R < 0.5If R < 0.5

R/24CD =

For Spherical ParticlesFor Spherical Particles

Settling Settling

VelocityVelocity

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ExampleExampleExample

Will a grit particle with a diameter of 0.2 mm and

specific gravity of 2.65 be collected in a horizontal

grit chamber?

The chamber length = 13.5 m

The chamber flow = 0.15 m3/s

Width of the chamber = 0.56 m

The horizontal velocity = 0.25 m/s

Temperature = 22°C (µ µ µ µ = 9.95××××10-4 Pa.s)

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SolutionSolutionSolution

µ

ρ−ρ=

18

d)(gv

2ws

s

s/m0361.0)000995.0(18

0002.0)10002650(8.9v

2

s ====−−−−

====

45.7dv

R s =ν

=Iterative Iterative

solutionsolutions/m028.0vs =

Cross sectional area:Cross sectional area:2

h

c m60.025.0

15.0

v

QA ===

Depth of flowDepth of flow m07.156.0

60.0

w

Ah c ===

2

7

SolutionSolutionSolution

It will take It will take h/vh/vss seconds to reach bottomseconds to reach bottom

s2.38028.0

07.1

v

ht

s

===

Since the chamber is 13.5 m in length and the horizontal velocity is 0.25 m/s the liquid remains in the chamber

s5425.0

5.13t ==

Thus the particle will be captured in the grit chamber.8

BIOREMIDITIONBIOREMIDITIONBIOREMIDITION

1. Using Bacteria To Clean Up Contaminated Soils

2. Using Bacteria To Clean Up Contaminated Groundwater

3. Using Plants To Clean Up Contaminated Sites

1.1. Using Using BacteriaBacteria To Clean Up Contaminated SoilsTo Clean Up Contaminated Soils

2.2. Using Using BacteriaBacteria To Clean Up Contaminated To Clean Up Contaminated

GroundwaterGroundwater

3.3. Using Using PlantsPlants To Clean Up Contaminated SitesTo Clean Up Contaminated Sites

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Bioremediation is the use of

microorganisms to destroy or im-mobilise

waste materials.

Microorganisms include:

� Bacteria (aerobic and anaerobic)

� Fungi

� Yeast

� Algae

Bioremediation Bioremediation

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1

3

2CO2 and H2O

Degradation of contaminant by microbes

(From USEPA 1991)

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1. The presence of the appropriate micro-

organisms in the correct amounts and

combinations

2. The appropriate environmental conditions.

Bioremediation Depends on: Bioremediation Depends on:

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Presence of AirPresence of Air

� Under Aerobic Bioremediation

Contaminants are converted to:

Carbon dioxide and water

� Under Anaerobic Bioremediation

Contaminants are converted to:

Methane, limited amounts of CO2 and traces of

hydrogen

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Types of

Contamination

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Types of ContaminationTypes of Contamination

� Bioremediation is commonly used for the treatment of

soils contaminated with Organic compounds e.g.

petroleum hydrocarbons (NAPLS) especially mono-

aromatic hydrocarbons (BTEX)

� But poly-nuclear aromatic hydrocarbons (PAHs) and

poly-chlorinated biphenyls (PCBs) are resistant to

degradation.

� Bioremediation cannot degrade inorganic contaminants

such as heavy metals but it can be used to change the

valence states of these metals thus converting them into

immobile form.

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Dealing with a Contaminated SiteDealing with a Contaminated Site

Site Characterisation

Site Characterisation

A site can be contaminated as a result of:

� Disposal of waste materials

� Spilling toxic materials during transportation

� Leakage from waste disposal or storage sites

� Leakage from industrial facilities, and so on

Selection of an Effective Remedial

Action

Selection of an Effective Remedial

Action

Impact (or Risk) Assessment

Impact (or Risk) Assessment

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Non-aqueous Phase Liquids (NAPL)

in Soils

Non-aqueous Phase Liquids (NAPL)

in Soils

NAPL is a term used to denote any liquid which is immiscible with water.

In site remediation NAPL is usually associated with petroleum hydrocarbons such as those leaking from oil storage tanks or accidental spills of organic chemicals.

In spite of their immiscible nature, NAPLs dissolve in water. Even slight amounts of dissolved NAPLs in potable watersupplies may result in detrimental health effects.

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Light and Dense NAPLsLight and Dense NAPLs

DNAPLDNAPLLNAPLLNAPL

Benzene ( Gs = 0.88)

Toluene (0.0.86)

Diesel fuels (0.83)

Gasoline (0.73)

m-Xylene (0.86)

o-Xylene (0.88)

Vinyl chloride (0.91)

Petroleum distillates (0.73)

Kerosene (0.82)

Chloroform (Gs = 1.48)

Carbon tetrachloride (1.56)

Phenol (1.07)

Naphthalene (1.03)

Trichloroethylene, TCE (1.46)

Coal tar (1.08)

Bromobenzene (1.49)

Methylene chloride (1.33)

Ethylene choloride (1.24) 18

Mono-aromatic hydrocarbons: BTEXMono-aromatic hydrocarbons: BTEX

0.70.867Ethyl benzene (106)

100.870Xylene (106)

10.866Toluene (92)

0.0050.874Benzene (78)

Max Conc. Level

Allowed (mg/L)ρ ρ ρ ρ (g/cm3)Compound (MW)

NOTE: Contamination of potential drinking water sources

by BTEX represent a serious threat to public health.

CH3

CH2CH3

C6H6

CH3

CH3*

4

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O-Xylene m-Xylene p-Xylene

ortho-Xylene meta-Xylene para-Xylene

CH3

CH3

CH3

CH3

CH3

CH3

*Xylenes*Xylenes

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Naphthalene

Pyrene

Antracene

Fluorene

Benzo(a)pyrene

Naphthalene

Pyrene

Antracene

Fluorene

Benzo(a)pyrene

Potential Sources

Oil production and storage

Landfills

Gas work sites

Power stations

Coke plants

Engine works

Tar production and storage

Boiler ash dump sites

Poly Aromatic Hydrocarbons (PAHs)Poly Aromatic Hydrocarbons (PAHs)

Examples:

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4-Chloro-biphenyl

4,4-Dichloro-biphenyl

4-Chloro-biphenyl

4,4-Dichloro-biphenyl

Potential Sources

Electrical manufacturing

Power station

Railway yards

Poly Chlorinated Biphenyls (PCBs)Poly Chlorinated Biphenyls (PCBs)

Examples:

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Advantages of BioremediationAdvantages of Bioremediation

� Minimum mechanical equipment are required

� Does not require excavation of contaminated

soils and does not disturb the natural

surrounding of the site (in-situ bioremediation)

� Cost is low compared to other remediation

techniques

� It may result in complete degradation of organic

compounds to nontoxic by-products

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Disadvantages of BioremediationDisadvantages of Bioremediation

� The process is highly sensitive to toxins and

environmental conditions.

� Extensive monitoring is required to determine

biodegradation rates.

� It generally requires a longer treatment time than of other

remediation technologies.

� There is a potential for degradation to equally toxic, more

highly mobile products.

� It may be difficult to control volatile organic compounds

during an ex-situ process. 24

Environmental Conditions for

Microorganisms to Survive and Grow

Environmental Conditions for

Microorganisms to Survive and Grow

1. pH (near 7 is very efficient but pH values between 5.5 and 8.5 are still OK)

2. Temperature (15 °°°° to 45°°°°)*

3. Oxygen (concentration > 2 mg/L for aerobic microorganisms)

4. Nutrients (C, N & P plus H & O are required, depends on BOD; Ratio: C:N:P by weight should be 120:10:1)

5. Toxicity (high concentration of contaminant can be toxic to microbes)

6. Saturation (between 40% and 80% of field capacity)

7. Clay content: Low clay and silt

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Source: USAEC, 2000. Biopiles of POL Contaminated Soils. U.S. Army Engineer

Environmental Center.

Requirements for Soil BioremediationRequirements for Soil Bioremediation

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Practical Ways to Improve the

Environmental Conditions

Practical Ways to Improve the

Environmental Conditions

If the soil has too much acid it is possible to rinse the pH by adding lime (CaO).

Plastic covering can be used to enhance solar warming in late spring, summer, and autumn.

Available water is essential for all the living organisms, and irrigation is needed to achieve the optimal moisture level.

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A site has been contaminated with 158 kg spill of gasoline. The volume of the contaminated soil is 375 yd3 (286.33 m3)

Assume soil porosity, n = 30%

The initial degree of saturation, S = 20%

Find the required amount of:

• water,

• nutrients, and

• air.

NOTE: 1 yard = 3 ft = 24 in = 0.914 m

Example Design CalculationsExample Design Calculations

n = Vv/Vt

S = Vw/Vv

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Example design calculations

The required water:

porosity, n = 30%

initial saturation, S = 20%

Desired degree of saturation = 25% to 85%, use 60%

S = Vw / Vv & n = Vv / Vt ⇒ Vw = Vt. n . S

Water needed = (375) (0.30) (0.6 – 0.2) = 45

yd3 = 34 m3 water

= 1215 ft3 = 9,090 gallons

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Example design calculations

Nutrient requirement:

158 kg spill of gasoline (≅ C7H16)

Nutrient sources:

Ammonium sulphate ((NH4)2SO4)

Tri-sodium phosphate (Na3PO4•12H2O)

MW of gasoline = 7 x 12 + 1 x 16 = 100 g/mol

Moles of gasoline = 158 x 1000 / 100 = 1580 mol

Moles of C = 7 x 1580 mol = 1.1 x 104 mol

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Example design calculations

Molar ratio C:N:P = 120:10:1

Moles of N needed = 10/120 x 1.1 x 104 = 920 mol

Moles of (NH4)2SO4 needed = 920 / 2 = 460 mol

MW of (NH4)2SO4 = (14 + 4) x 2 + 32 + 4 x 16 = 132

Mass of (NH4)2SO4 needed = 132 g/mol x 460 mol

= 6.1 x 104 g = 61 kg

By similar calculation:

Mass of (Na3PO4.12H2O) needed = 35 kg

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Example design calculations

Oxygen requirement:

C7H16 + 22O2 → 7CO2 + 8H2O

1 mole (100 g) gasoline requires 22 moles (704 g) O2

Oxygen content of air = 21% by volume

= 210,000 ppm (vol.)

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Example design calculations

Oxygen needed for 158 kg spill of gasoline (≅ C7H16)

100 g gasoline needs ~700 g oxygen (the ratio is 1:7)

158 kg gasoline x 7 = 1106 kg O2 = 1.1 x 106 g O2

Water in soil pile = (286.3) (0.30) (0.6) = 52 m3

= 52,000 L

At saturation at 20ºC and 1 atm, DO = 9.2 mg/L

Mass of oxygen in soil moisture =

= 52,000 L x 9.2 mg/L x 0.001 g/mg

= 480 g O2

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Example design calculations

480 g O2 in soil moisture is much less than 1.1 x 106 g

O2 required.

At 0.28 g/L air, air requirement is:

1.1 x 106 g / 0.28 g/L = 3.95 x 106 L air

= 3,950 m3 air

Air void volume in pile = (286.3) (0.30) (0.4) = 34 m3

Need to exchange 3950 / 34 = 116 void volumes to

fulfill oxygen requirement34

General Approach to BioremediationGeneral Approach to Bioremediation

Contaminated soilContaminated soilContaminated soil

Addition of nutrients,

Moisture, oxygen and bacteria

Addition of nutrients, Addition of nutrients,

Moisture, oxygen and bacteriaMoisture, oxygen and bacteria

Measure pH, temperature and Redox

potential

Measure pH, temperature and RedoxMeasure pH, temperature and Redox

potentialpotential

Monitor the concentrations of

the contaminants and biological growth

Monitor the concentrations of Monitor the concentrations of

the contaminants and biological growththe contaminants and biological growth

Is the remediation

complete?

Is the remediation

complete?

Abandon

remediation

AbandonAbandon

remediationremediation

No

Yes

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BIOREMEDIATION STRATEGIESBIOREMEDIATION STRATEGIES

In-situ Bioremediation Ex-situ Bioremediation

Bio-venting

In situ biodegradation

Bio-sparging

Bio-augmentation

Land-farming

Composting

Biopiles

Bioreactors

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Contaminant BiodegradabilityContaminant Biodegradability

01.0COD

BOD5 ≥

01.0COD

BOD5 <

Biodegradable

Nonbiodegradable

BOD5 = 5-day biochemical oxygen demand

COD = chemical oxygen demand

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BOD5/COD Ratios for various organic compound

BOD5/COD Ratios for various organic compound

0.55Acetone0.02Gasoline

0.38Benzene0.01O-Xylene

0.31Acetic acid0.01TCE

0.2Soybean oil0Propane

0.12Toluene0Butane

RatioCompoundRatioCompound

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BOD5 and BODUBOD5 and BODU

P

DODOBOD 21

5

−=

The most widely used parameter of organic pollution is 5-day biochemical oxygen demand (BOD5).

At 20°°°°C after 5 days:

)e1(BODBOD ktUt

−−=Ultimate BOD = BODU

DO1 = dissolved oxygen of diluted sample at t = 0

DO2 = dissolved oxygen of diluted sample at t = 5 days

P = decimal volumetric fraction of sample used

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Example: Calculation of BODExample: Calculation of BOD

200)e1(BODBOD 23.05U5 =−=

×−

Determine the ultimate BOD for a wastewater whose 5-day, 20°C BOD is 200 mg/L. (the reaction constant, k = 0.23 d-1)

L/mg293BOD

)316.01/(200BOD

U

U

=

−=

)e1(BODBOD ktUt

−−=

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Chemical Oxygen Demand (COD)Chemical Oxygen Demand (COD)

OHCOCr)heat()catalist(HCrOOHC 2232

7cba ++→++++ ++−

The oxygen equivalent of the organic matter that can be oxidize is measured by using a strong chemical oxidizing agent (e.g. potassium dichromate) in an acidic medium.

Usually COD can be determined in 3 hours.

The associated unbalanced equation is as follows:

Organic matter Silver sulphate

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If Refractory Index (RI) is greater than 0.5 the contaminated soil is generally biodegradable.

5.0UOD

BODRI u >=

)NNH6.4()BOD5.1(UOD 45 ×+×=

Contaminant Biodegradability:

Refractory Index (RI)

Contaminant Biodegradability:

Refractory Index (RI)

BODu = ultimate biochemical oxygen demand UOD = ultimate oxygen demand.