Chapter 3. Water Pollution - KOCWcontents.kocw.net/KOCW/document/2015/hankyong/yun... · Chapter 5. Water Pollution This dilution will be satisfactory for 60210 mg/L of BOD5 values

Chapter 5. Water Pollution

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<Unusual Properties of Water>

Figure 2 A water molecule is diploar

Strong H bonding → Dipole moment

→ Unusual properties of water

▶ Unusual Properties of Water

- High bp

- High surface tension → occur capillary action

- Maximum density at 4℃ ※ unique common liquid that expands when it freezes.

- High difference between mp and bp

※ Only substance that appears in all three states within room temperatures

on earth.(ex, H2S : mp –82.9 / bp –59.6)

- High specific heat(4,184 J/(kg℃))

※ higher heat capacity than any other known liquid except NH3. (5 times

higher than rock or concrete → protecting life from rapid thermal

fluctuations)

- High heat of vaporization(2,258 kJ/kg) : stores big energy

※ one of the highest of all liquid

- Very effective solvent : dissolving many substances

- Green house effect by H-O-H bending vibration


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<Hydrology Cycle>

Figure 3 The hydrology cycle.(unit : 103 km3/yr)

TABLE 1 Stocks of water on earth.

※ Main source for human use : 93,000 m3(0.0072 %)


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<Water Usage>

FIGURE 4 Per capita water availability for North America, Africa, and Asia,

showing the implications of growing population.

(Source: Shiklomanov, 1993. Reprinted by permission of Oxford University

Press.)

TABLE 2 Dependence on imported surface water for selected countries


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FIGURE 5 Fresh water use in the United States, 1990.

The proportion of withdrawals by category of use was unchanged in 2000.

Annual 2000 freshwater withdrawals are about 476 km3.

(Sources : From Gleick, 1993; U.S. Water Resources Council, 1978; and Hutson

et al., 2004.)

『Notes』

Industry includes mining, agriculture includes irrigation and livestock, and

municipal includes public, domestic, and commercial uses.


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TABLE 3 Examples of water withdrawals to supply various end uses


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3. Water Pollutants

TABLE 4 Typical pathogens excreted in human faces

<Oxygen-Demanding Wastes>

▶ Measurement of O2 demand(OD)

- COD : chemical oxygen demand

- BOD : biochemical oxygen demand


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▶ DO(Dissolved Oxygen)

- Saturated DO : 8∼15mg/L(depending on temperature & salinity)

- Minimum recommended DO for heathy fish : 5 mg/L

<Nutrients>

▶ Nutrients : essential to the growth(N, P, C, S, Ca, K, Fe, Mn, B, Co)

▶ Sufficient concentration of nutrients

→ excessive growth of aquatic plants(particularly Algae)

▶ Nutrient enrichment(Eutrophication) → blooms of algae → algae

decomposition → DO↓, turbidity↓, odor, objectionable tastes

▶ Controlling nutrients : N or P

※ C is usually available from natural source like alkalinity, CO2.

- Sea water is most often limited by N

- Freshwater is most often limited by P

▶ Source of N

- Municipal wastewater discharge

- Runoff from animal feedlots

- N deposition from NOx from atmosphere(coal fire power plants)

- Direct from atmosphere N2 through bacteria or blue-green algae.

▶ P source : most coming from human activity

- Fertilizer

- Detergent

<Salts>

▶ TDS : Total Dissolved Solids

- Fresh water : < 1,500 mg/L

- Brackish water : ∼5,000 mg/L

- Saline water : > 5,000 mg/L

- Sea water : 30,000 ∼ 34,000 mg/L


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▶ TDS limitation for living things

- Drinking water for Human : 500 mg/L

- Livestock

∙Poultry : 2,860 mg/L

∙Pig : 4,290 mg/L

∙Beef cattle : 10,100 mg/L

- Corps

∙Irrigation without impact : 500 mg/L

∙Irrigation with small lose of yield : 500 mg/L

※ At above 2,100 mg/L, water is generally unsuitable for irrigation

except for the most salt tolerant of crops.

<Heavy Metals>

▶ Heavy metals : sp. gr.> 4 or 5(Hg, Pb, Cd, As)

▶ Toxic metals : Al, As, Be, Bs, Cd, Cr, Co, Cu, Pb, Mn, Hg, Ni, Ce, Sr,

Ta, Sn, Ti, Zn

<Pesticides>

▶ 3 main group of synthetic organic pesticides

- Organo-chlorines(chlorinated hydrocarbon) : DDT

- Organo-phosphates

- Carbamates

▶ Organo-chlorines : Methoxychlor, Chlordane, Heptachlor, Aldrin, Dieldrin,

Endrin, Endosulfan, Kepone, Dioxin

- Used to control insects and herbicides

- Disruptive to food chain

- Persistent

- Soluble in hydrocarbon solvent(ex, fatty tissue)


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▶ Organo-phosphates : Parathion, Malathion, Diazinon, TEPP

- effective against wide range of insects

- not persistent

- more acutely toxic to humans than organo-chlorines

▶ Carbamates : Propoxur, Carbaryl, Aldicarb

- similar with organo-phosphates

- not persistent

- more acutely toxic to humans than organo-chlorines


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<Volatile Organic Chemicals(VOCs)>

▶ Most commonly found in ground water

∙in surface water : a few µg/L

∙in ground water : a few 100∼1000 µg/L

▶ Most of VOCs are suspected carcinogens or mutagens

▶ 5 toxic VOCs


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4. Status of Surface Water Quality

TABLE 7 Beneficial Uses of Surface Water


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FIGURE 7 Fraction of impaired river and stream miles broken down.


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FIGURE 8 Extent of impaired lake acres broken down.


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5. Biochemical Oxygen Demand(BOD)

▶ Aerobic Decomposition

OrganicMatter + O2

Microorganisms→ CO2+H2O+New cells + Stable products

(NO3, PO4, SO4,...)

Carbonaceous Oxygen Demand(CBOD)BOD

Nitrogenous Oxygen Demand(NBOD)

▶ Anaerobic Decomposition

OrganicMatter

Microorganisms→ CO2+H2O+New cells + Unstable products

(NH3, H2S, CH4,...)

<5 day BOD Test without Seeding>

▶ 5 day BOD(BOD5)

Total amount of O2 consumed by microorganisms during the first five days

of biodegradation.

▶ Standard Method to Measure BOD5

Putting a sample of waste into a stoppered bottle↓

Measure DO in the sample at the beginning of the test↓

Keep the bottle at 20℃ and keep out it from light for 5 days↓

Measure DO in the sample↓

Difference in DO divided by volume of the waste is BOD5

BOD5 = DOi–DOf

P (2)

where DOi = the initial DO of the diluted wastewater

DOf = the DO of the diluted wastewater, 5 days later


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P = dilution fraction = volume of wastewatervolume of wastewater + dilution water

『Note』

1. Oxygen demand of typical wastewater : several hundred mg/L

2. Saturated value of DO for water at 20℃ = 9.1 mg/L

3. It is necessary to dilute the sample to keep the final DO above 0

4. If during 5 days, the DO drops to 0, the test is invalid.

5. Vol. of std. BOD bottle : 300 mL→P=vol. of wastewater/300mL

<EXAMPLE 1> Unseeded 5 day BOD Test

<Given Data>

1. volume of sewage = 10 ml

2. DO after diluted to 300 ml = 9.0 mg/L

<Question> What range of BOD5 would this dilution produce the desired

results?

1. To help accurate test, it is desirable to have at least 2 mg/L drop in DO

during 5 day run

2. the final DO should be at least 2 mg/L.

<Solution> Dilution fraction(P) = 10/300

To get at least 2 mg/L drop in DO, the minimum BOD needs to be

BOD5 ≥DOi–DOf = 2.0 mg/L = 60 mg/LP 10/300

To assure at least 2 mg/L of DO remains after 5 days requires that

BOD5 ≤DOi–DOf = (9.0–2.0) mg/L = 210 mg/LP 10/300


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This dilution will be satisfactory for 60∼210 mg/L of BOD5 values.

<5 day BOD Test with Seeding>

In some case it is necessary to seed the dilution water with microorganism to

assure that there is a adequate bacterial population to carry out the

biodegradation.

BOD of waste itself = OD in mixture – OD caused by the seed.

BOD bottle containing the mixture

of both the wastewater and seeded

dilution water

verse

BOD bottle containing just the

seeded dilution water

(called the “blank”)

FIGURE 9 Laboratory test for BOD using seeded dilution water.

BODmVm = BODwVw + BODdVd (3)

where BODm = BOD of (wastewater + seeded dilution water)

BODw = BOD of the wastewater alone

BODd = BOD of the seeded dilution water alone(the blank)


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Vm = the volume of mixture = Vw + Vd

Vw = the volume of wastewater in the mixture

Vd = the volume of seeded dilution water in the mixture

Let P = Vw/Vm = the fraction of the wastewater

(1 – P) = Vd/Vm = the fraction of mixture

from Eq(3)

BODw = BODmVm – BODd

Vd × Vm

Vw Vw Vm

= BODm – BODd(Vd/Vm)

(Vw/Vm) (Vw/Vm)

= BODm – BODd(1 – P)P

(6)

Because BODm = DOi – DOf and BODd = Bi – Bf

where Bi = initial DO in the seeded dilution water(blank)

Bf = final DO in the seeded dilution water(blank)

BODw = (DOi – DOf) – (Bi – Bf)(1 – P)P (7)

<EXAMPLE 2> A Seeded BOD Test

<Given Data>

1. DO drop in Blank = 1.0 mg/L

2. Volume of wastewater = 15 mL → diluted 20 times


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3. DO drop in mixture = 7.2 mg/L

<Question> BOD5 = ?

<Solution> Dilution factor(P) = 15/200 = 0.05

From Eq(7),

BODw = (DOi – DOf) – (Bi – Bf)(1 – P)P

= 7.2 – 1.0 × (1 – 0.05) = 125 mg/L0.05

<Modeling BOD as a 1st Order Reaction>

FIGURE 10 Two equivalent ways to describe the time dependence of organic

matter in a flask.

▶ Assume : bio-decomposition = the 1st order

dLt = - kLt Decomposition rate ∝ Remaining wastedt (8)

where Lt = amount of O2 demand left after time t

k = BOD reaction constant(time–1)

Solution of Eq(8) → Lt = L0e–kt (9)

where L0 = ultimate carbonaceous oxygen demand


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※ Oxygen demand can be described by the BOD remaining.

L0 = BODt + Lt → BODt = L0–Lt (10)

Combining Eq(9) and (10) → BODt = L0(1–e–kt) (11)

BODt = L0(1–10–Kt) (12)

where k = K ln10 = 2.303K (13)

(a) BOD remaining (b) BOD utilizedFIGURE 11 Idealized carbonaceous O2 demand.

<EXAMPLE 3> Estimating L0 from BOD5

<Given Data> Unseeded BOD

1. dilution factor = 0.03 2. DOi = 9.0 mg/L

3. DOf = 3.0 mg/L after 5 days 4. k = 0.22 d-1

<Question> 1. BOD5 = ? / 2. L0 = ? / 3. Lt = ?


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<Solution>

1. From Eq(2), BOD5 = DOi–DOf = 9.0–3.0 = 200 mg/LP 0.03

2. From Eq(11),

BODt = L0(1–e–kt) → L0 = BODt = 200 = 300 mg/L(1–e–kt) (1–e–0.22×5)

3. Remaining oxygen demand = 300 – 200 = 100 mg/L

<BOD Reaction Rate Constant, k>

TABLE 8 Typical value for the BOD rate constant at 20℃

Sample k(d-1)a K(d-1)b

Raw sewage 0.35 ∼ 0.70 0.15 ∼ 0.30Well-treated sewage 0.12 ∼ 0.23 0.05 ∼ 0.10Polluted river water 0.12 ∼ 0.23 0.05 ∼ 0.10

a : based on exponent

b : based on 10

▶ Temperature dependency : kT = k20 θ (T–20) (14)

where k20 = reaction rate constant(k) at 20℃

kT = reaction rate constant(k) at a different temperature T(℃)

※ θ = 1.047 as most commonly used value

Example 4 Temperature Dependence of BOD5

<Given Data> In Example 3,

1. Ultimate BOD = 300 mg/L 2. BOD5 = 200 mg/L at 20℃

3. k = 0.22 d-1

<Question> BOD5 at 25℃ = ?

<Solution>


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From Eq(25) → k25 = k20 θ(25–20) = 0.22 × (1.047)5 = 0.277 d-1

From Eq(11) → BODt = L0(1–e–kt) = 300(1–e–0.277×5) = 225 mg/L

<Nitrification>

Dead body of living things or excrete waste product

Bacteria or Fungi↓

Nitrogen complex with organic molecules

↓

2NH3 + 3O2

Nitrosomonas↓

2NO2–(Nitrite)+ 2H+ + 2 H2O (15)

Nitrobacter↓+ O2

2NO3–(Nitrate) (16)

▶ BOD

Carbonaceous Oxygen Demand(CBOD) Organics→CO2

Nitrogenous OxygenDemand(NBOD) NH3→NO3–


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FIGURE 12 Changes in nitrogen forms in polluted water under aerobic

conditions

N2Nitrogen Fixation

→ NH3Nitrosomonas

→ NO2–

↑ Nitrobacter ↓

→NO3

–

Denitrifying bacteria


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FIGURE 13 Illustrating the carbonaceous and nitrogenous biochemical oxygen

demand. Total BOD is the sum of the two.

※ NBOD does not normally begin to exert itself for at least 5 to 8 days →

most 5 day tests are not affected by nitrification.

<EXAMPLE 5> Nitrogenous Oxygen Demand

<Given Data> N : 30 mg/L in the form of either organic-N or NH3.

<Assumption> No cell growth during the nitrification of the waste

<Question>

1. Ultimate nitrogenous oxygen demand = ?

2. Ratio of ultimate NBOD to N in the waste = ?


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<Solution>

1. Combining Eq (15) and (16) → NH3+2O2 → NO3–+H++H2O (17)

MW of NH3 = 17(MW of N = 14)

MW of O2 = 32

NBOD = 30 mg N 17 g NH3 64 g O2 = 137 mg O2

L 14 g N 17 g NH3 L

2. 137 mg O2 L = 4.57 mg O2

L 30 mg N mg L

▶ Total Kjeldahl Nitrogen(TKN) = Organic-N + NH3-N

▶ Ultimate NBOD ≈ 4.57 × TKN (18)

※ TKN of raw sewage : 15∼50 mg/L

→ O2 demand for nitrification : 70∼230 mg/L

※ L0 of raw sewage : 250∼350 mg/L

<Other Measurement of Oxygen Demand>

▶ ThOD : Theoretical Oxygen Demand

- calculated value by stoichiometry

- normally over-estimate O2 demand

- limited usefulness in practice because it pre-supposes a particular, single

pollutant with a known formular.

▶ COD : Chemical Oxygen Demand

- Chemical oxidation of the waste using strong oxidant like sodium

dichromate

- Very quick test(1 hour)

- COD > BOD

※ Non-biodegradable organics(toxic to micro-organism)

Pesticides, Various Industrial chemicals

※ Biodegradable resist organics


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Cellulose, Phenol, Benzene, Tannic acid

- sometimes used as a way to estimate the ultimate BOD


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6. The Effect of Oxygen Demanding Waste on River

▶ DO : One of the major indicator of a river’s health

- DO < 4 or 5 mg/L → aquatic life begin to die

▶ Simplest modeling of DO

- Removal of DO by micro-organism

- Replenishment of DO through re-aeration at the interface between the river

and the atmosphere.

▶ Assumption : Plug flow(valid for deep and slowly moving river)

- Uniform concentration at any given cross section of river.

- No dispersion of waste in the direction of flow.

FIGURE 14 Point source, plug flow model for DO calculation.

<De-oxygenation>

Rate of de-oxygenation = kd∙Lt (19)

where kd = de-oxygenation rate constant(d-1) ≈ BOD rate constant k

Lt = BOD at time t(days) after the waste enter the river(mg/L)

By substituting Eq(9) to Eq(19), Lt = L0e–kt,


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Rate of de-oxygenation = kd∙L0∙exp(–kd t) (20)

where L0 = BOD of the mixture at the point of discharge.

By weight average concept , L0 = QwLw + QrLr

Qw + Qr (21)

where L0 = ultimate BOD of mixture(mg/L)

Lw = ultimate BOD of wastewater(mg/L)

Lr = ultimate BOD of river before mixing(mg/L)

Qw = Flow rate of wastewater(m3/s)

Qr = Flow rate of river before mixing(m3/s)

<EXAMPLE 6> Downstream BOD

<Given Data> for 20,000 residents

Qw = 1.1 m3/s, Qr = 8.7 m3/s

L0 = 50 mg/L, Lr = 6 mg/L, kd = 0.2 d-1

<Question>

1. L0 = ? just after mixing

2. BOD at d = 30,000m down stream when v = 0.3 m/s.

<Solution>

1. From Eq(21),

L0 = QwLw + QrLr = 1.1m3/s × 50mg/L + 8.7m3/s × 6.0mg/L =10.9 mg/LQw + Qr (1.1 + 8.7)m3/s


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2. t = 30,000 m s hr d = 1.16 days0.3 m 3600 s 24 hr

From Eq(9), Lt = L0e–kt = 10.9 × e–(0.2 ×1.16) = 8.7 mg/L

<Re-aeration>

▶ rate of re-aeration ∝ DO difference between actual and saturation

※ difference = oxygen deficit

Rate of Re-aeration = kr∙D (22)

where kr = re-aeration constant(time-1)

D = dissolved oxygen deficit = (DOs – DO) (23)

DOs = saturated dissolved oxygen

DO = actual dissolved oxygen at given location downstream

▶ kr = very dependent on the particular conditions in the river

Empirical Equation → kr = 3.9 u 1/2

H 3/2 (24)

where kr = re-aeration constant at 20℃(day-1)

u = average stream velocity(m/s)

H = average stream depth(m)

TABLE 9 Typical re-aeration constant(kr) for various water

a : Based on exponent, Source : Tchobanoglous and Schroeder, 1985.

▶ For other temperature, using Eq(14) with θ = 1.024


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※ Temperature dependency : kT = k20 θ (T–20) (14)

TABLE 10 Solubility of oxygen in water(mg/L) at 1 atm pressure

▶ Initial deficit of the river by weight average

D0 = DOs –QwDOw + QrDOr

Qw + Qr (25)

where D0 = initial oxygen deficit of mixture

DOs = saturated DO in water

DOw = DO in wastewater

DOr = DO in river just after mixing

<EXAMPLE 7> Initial Oxygen Deficit

<Given Data> Wastewater : Example 6

1. DOw = 2.0 mg/L, Qw = 1.1 m3/s

2. DOr = 8.3 mg/L, Qr = 8.7 m3/s at 20℃

<Question> D0(initial oxygen deficit of mixture) = ?


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<Solution>

From definition of the initial DO in mixture by weight average

DO= QwDOw+QrDOr = 1.1m3/s × 2.0mg/L + 8.7m3/s × 8.3mg/L =7.6 mg/LQw + Qr 1.1 m3/s + 8.7 m3/s

From EQ(25) and saturated DO from Table 10,

D0 = DOs–DO = 9.09 – 7.6 = 1.5 mg/L

<The Oxygen Sag Curve>

▶ De-oxygenation verse Oxygenation → Competitive Process

▶ By combining Eq(20 and (22),

Rate of increase of deficit = Rate of de-oxygenation–Rate of re-aeration

dD/dt = kd∙L0∙exp(–kdt) – kr∙D (26)

∴ D = kd∙L0 [exp(–kdt)–exp(–krt)] + D0exp(–krt)kr–kd (27)

From the definition of deficit, D = DOs–DO → DO = DOs–D,

DO = DOs–{ kd∙L0 [exp(–kdt)–exp(–krt)] + D0exp(–krt) }kr–kd (28)

FIGURE 15 Streeter-Phelps oxygen sag curve.


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▶ For special case(kr = kd),

denominator of Eq(27)=0→Not acceptable

From Eq(26), D = (kd∙L0∙t + D0)exp(–kd t) (29)

▶ Maximum oxygen deficit

For constant cross-sectional area river, x = u × t (30)

where x = distance downstream at time t

u = stream speed

t = elapsed time after mixing

From Eq(27),

D = kd∙L0 [exp(–kdx/u)–exp(–krx/u)] + D0exp(–krx/u)kr–kd (31)

To solve critical time(tc) at minimum D, derivative of D set 0,

then,

ln

(32)

Substituting Eq(32) to (28) → Minimum DO

EXAMPLE 8 Streeter-Phelps Oxygen Sag Curve

<Given Data>

Temperature = 20℃

At x = 0, BOD(=L0) = 10.9 mg/L and DO = 7.6 mg/L


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kd = 0.2 d-1

u = 3.0 m with average depth = 3.0 m

<Question>

1. tc and xc = ?

2. DOmin = ?

<Solution>

From Table 10, DOs = 9.1 mg/L at 20℃.

Initial deficit(D0) = 9.1 – 7.6 = 1.5 mg/L

To calculate re-aeration constant(kr),

From Eq(24) → kr = 3.9 u 1/2

= 3.9×(3.0)1/2

= 0.41 d-1

H 3/2 (3.0)3/2

1. To calculate critical time(tc), from Eq(32)

ln

=1

ln{0.41

[1–1.5(0.41–0.2)

]} = 2.67 days(0.41–0.20) 0.20 0.41×10.9

2. To maximum deficit(Dmax), from Eq(27)

Dmax = kd∙L0 [exp(–kdt)–exp(–krt)] + D0exp(–krt)kr–kd

= 0.2 × 10.9 [e(–0.2×2.67) – e(–0.41×2.67)] + 1.5e(–0.41×2.67) = 3.1mg/L(0.41–0.2)

So, DOmin = 9.1 – 3.1 = 6.0 mg/L


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At the critical point,

those rates are equal.

Beyond the critical point,

reaeration exceeds

decomposition, the DO

curve climbs toward

saturation, and the river

recovers.

FIGURE 16 When the rate of deoxygenation exceeds the rate of reaeration, the

DO in the river drops.


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<Change in Oxygen Sag Curve>

FIGURE 17 As a river gets more polluted, the oxygen sag curve below an

acceptable level. In extreme case, anaerobic conditions can occur

FIGURE 18 Changes in the oxygen sag curve as temperature increases.

Temperature↑ → DO↓


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7. Water Quality in Lake and Reservoirs

<Controlling Factors in Eutrophication>

▶ Euphotic Zone : Top layer of water, O2 generation > respiration

▶ Profundal Zone : Below euphotic zone

▶ Light compensation level : transition between the two zone.

※ Light compensation level ≈ depth at which light intensity is about 1% of

full sunlight

▶ Liebig’s Law of the Minimum

“Growth of a plant is dependent on the amount of foodstuff that is presented

to it in minimum quantity.”

→ quickest way to control eutrophication = reduce limiting nutrient

▶ Most fresh water systems are phosphorus limited

Nitrogen could be easily supplied from atmosphere through blue-green

Ctanophyta and NOx.

▶ Relative amount of nitrogen and phosphorous for algae growth

106CO2 + 16NO3–+ HPO4

2–+ 122H2O + 18H+→C106H263O110N16P + 138O2 (33)

N = 16 × 14 = 7.2P 1 × 31

▶ Limiting substance

∙N : P > 10 → phosphorous limited(Fresh water)

∙N : P < 10 → nitrogen limited(Marine water : < 5)

▶ Index for eutrophication


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StatusItems

Oligotrophic(few food) Mid Eutrophic

(well food) Reference

TP(mg/l)

< 0.01 0.01∼0.03 0.03∼0.1 Vollenweider(1967)< 0.01 0.01∼0.02 > 0.02 USEPA(1976)

< 0.012 0.012∼0.024 > 0.024 Carlson(1977)< 0.01 0.01∼0.035 0.035∼0.1 OECD

< 0.015 0.015∼0.025 0.025∼0.1 Fordberg & Ryding(1980)TN(mg/l) < 0.4 0.4∼0.6 0.6∼1.5 Fordberg & Ryding(1980)

Chl-a(mg/l) < 0.004 0.004∼0.01 > 0.01 USEP(1974)< 0.003 0.003∼0.007 0.007∼0.04 Fordberg & Ryding(1980)

Transparency(m)

> 3.7 2.0∼3.7 < 2.0 USEPA(1974)> 4.0 2.0∼4.0 < 2.0 USEPA(1977)> 4.0 2.5∼4.0 < 2.5 Fordberg & Ryding(1980)

※ Sawyer(1947) : P>0.015 and N>0.3 mg/L → eutrophic

※ Vollenweider(1975) : P=0.01 mg/L→“acceptable”, P=0.02 mg/L→“excessive”

<Simple Phosphorous Model>

FIGURE 19 Well-mixed lake phosphorus mass balance.

<Assumption>


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1. Qin = Qout

2. Qs = Negligible(No flow input from the point-source)

3. Cp = uniform(well-mixed) lake

4. Steady state

- No accumulation

- Ignore the dynamic behavior of lakes as weather and seasons change

Simplest Modeling

Rate of addition of P = Rate of removal of P

QCin + S = QC + νSAC (34)

∴ C = QCin + SQ + νSA (35)

where S = rate of addition of phosphorus from the point-source (g/s)

C = concentration of phosphorus in the lake(g/m3)

Cin = concentration of P in the incoming stream flow(g/m3)

Q = stream inflow and outflow rate(Qin = Qout)

νS = phosphorus settling rate, 10 ∼16 m/s

A = the surface area of the lake (m2)

EXAMPLE 9 Phosphorus Loading in a Lake

A phosphorus limited lake with surface area equal to 80×106 m2 is fed by a

15.0 m3/s stream that has a phosphorus concentration of 0.010 mg/L. In addition,

effluent from a point-source adds 1 g/s of phosphorus. The phosphorus settling

rate is estimated at 10 m/yr.

a. Estimate the average total phosphorus concentration.

b. What rate of phosphorus removal at the wastewater treatment plant would be

required to keep the concentration of phosphorus in the lake at an acceptable

level of 0.010 mg/L?


- 38 -

<Given Data>

S = 1 g/s

Cin = 0.01 mg/L = 0.01 g/m3

Q = 15 m3/s

νs = 10 m yr d hr = 3.17 × 10–7 m/syr 365 d 24 hr 3600 s

νs = 10(m/s)

A = 80 × 106 m2

<Question>

1. C(g/m3) = ?

2. P removal rate(g/s) at Waste Water Treatment Plant(WWTP) to keep 0.01

mg/L ?

<Solution>

1. C = 15 m3/s × 0.01 g/m3 + 1 g/s = 0.028 g/m3

15 m3/s + 3.17 × 10–7 m/s × 80 × 106 m2

2. S = C(Q + νS A)╶ QCin = 0.01 g/m3(15 m3/s + 3.17×10╶7 m/s × 80×106

m2)╶ 15 m3/s × 0.01 g/m3 = 0.25 g/s


- 39 -

<Thermal Stratification>

FIGURE 21 Thermal stratification of a lake showing winter and summer

stratification temperature profiles.

▶ Thermal Stratification

Very stable layering effect caused by density difference between surface and

bottom, which inhibit vertical mixing in the lake.

▶ Epilimnion : Upper layer

Completely mixed by the action of wind and waves, causing an almost

uniform temperature profile

▶ Thermocline, or Metalimnion : Middle layer below the epilimnion

Transition layer in which the temperature drops rather quickly

▶ Hypolimnion : Bottom layer below the thermocline

<Stratification and Dissolved Oxygen>

FIGURE 22 Dissolved oxygen curves for eutrophic and oligotrophic lakes during

summer thermal stratification.

<Bicarbonate Buffering>


- 40 -

▶ Buffer : Substances capable of neutralizing added H+.

▶ Source of buffer of an aquatic ecosystem

- CO2 from air(※ buffered by HCO3╶)

- Nearby soils through which water percolates

▶ General behavior of lake by adding acid

- Bicarbonate lake(pH > 6.3) : little decrease in pH

- Transitional lake(5 < pH < 6) : very sensitive in pH

- Acidic lake(pH < 5) : unbuffering

FIGURE 23 Bicarbonate buffering strongly resists acidification until pH drops

below 6.3.


- 41 -

FIGURE 24 Frequency histograms of fish status for 684 Norwegian lakes

categorized as bicarbonate, transition, or acid lakes.

<Importance of the Local Watershed>

▶ Limestone is an effective buffer to acidification.

▶ Calcareous lakes

Lake having abundance of CaCO3 ⇨ invulnerable to acidification.

▶ Other watershed

- Thin, relatively impermeable soils

provide little contact between soil and runoff

⇨ reduce the ability of natural soil buffers to affect acidity.

- Steep slopes

create conditions in which the runoff has little time to interact with soil

buffers, increasing the likelihood of lake acidification

- Deciduous foliage : decrease acidity

- Evergreen foliage : more acidic than the rain itself.

▶ The most vulnerable lakes


- 42 -

- shallow soil of low permeability

- granite bedrock

- a steep watershed, and a predominance of conifers.


- 43 -

9. Aquifers

▶ Unsaturated Zone

Presence of both air and water in the spaces between soil particles.

▶ Vadose Water : Water in the unsaturated zone

Essentially unavailable for human use.(cannot be pumped)

▶ Saturated Zone : All spaces between soil particles are filled with water.

▶ Ground Water : Water in the saturated zone

▶ Water Table : the upper boundary of the saturated zone

▶ Capillary Fringe : Transition region between these two zones

Water rises into small cracks as a result of the attraction between water and

rock surfaces.

▶ Aquifer : Saturated geologic layer

- Permeable enough to allow water to flow fairly and easily

- An aquifer sits on top of a confining bed(aquitard or aquiclude) which is a

relatively impermeable layer that greatly restricts the movement of ground

water.


- 44 -

FIGURE 26 Identification of subsurface regions.(unconfined aquifer)

FIGURE 27 Confined and unconfined aquifers, a perched water table, water table

well, and an artesian well.

▶ Confined Aquifer


- 45 -

- Aquifer sandwiched between two confining layers

- Water in a confined aquifer can be under pressure

▶ Artesian Well : Well drilled into water in a confined aquifer

▶ Piezometric Surface or Potentiometic Surface.

Surface level to which water would rise in an artesian well

▶ Flowing Artesian Well

Artesian Well of which ground water raised above the ground level and flow

without pumping

▶ Porosity(η) = Volume of voidsTotal volume of solids and voids (39)

▶ Specific Yield(Effective Porosity)

The volume of water that can be drained from an unconfined aquifer per

unit of area per unit decline in the water table

▶ Storage Coefficient : Specific yield for confined aquifers

TABLE 11 Representative values of porosity and specific yield

EXAMPLE 10 Specific yield

For an aquifer of sand, having characteristics given in Table 11, what volume of

water would be stored in a saturated column with a cross-sectional area equal to

1.0 m2 and a depth of 2.0 m? How much water could be extracted from that

volume?


- 46 -

<Given Data>

1. Material : Sand(η = 34%, specific yield = 25%)

2. A = 1 m2

3. D = 2 m

<Question>

1. Volume of water stored = ?

2. Volume of water extracted = ?

<Solution>

Total volume of material = 2 m3

Volume of water stored = volume of material × porosity

= 2 m3 × 34% = 0.68 m3

Volume of water extracted = volume of material × specific yield

= 2 m3 × 25% = 0.5 m3


- 47 -

10. Hydraulic Gradient

▶ Hydraulic Gradient : Slope of the water table in an unconfined aquifer

Hydraulic gradient = Change in head = h2–h1 = dhHorizontal distance L dL (41)

FIGURE 28 Hydraulic head and gradient in an unconfined aquifer.

FIGURE 29 A two-dimensional flow net consisting of streamlines and

equipotential lines.

▶ Calculation of Gradient Using 3 Wells


- 48 -

Step 1 Draw a line between the two wells(highest head and lowest head)↓

Identify equipotential head spot on the line which is same with the 3rd(intermediate head) well’s head.

↓

Step 2Draw equipotential line between 3rd well and the equipotential head spot

↓

Step 3Draw ground water flow line from the lowest head well ⊥ equipotential line

↓

Step 4Determine the gradient(Δh / ΔL)Δh = head of equipotential–head at the lowest wellΔL = distance from the equipotential line to lowest well.

FIGURE 30 Using three wells to determine the gradient

EXAMPLE 11 Estimating the Hydraulic Gradient from Three Wells

Two wells are drilled 200 m apart along an east-west axis. The west well has a

total head of 30.2 meters, and the east well has a 30.0 m head. A third well

located 100 m due south of the east well has a total head of 30.1 m. Find the

magnitude and direction of the hydraulic gradient.

<Given Data>

1. Distance between high head well(west) and low head well(east) = 200 m

2. The 3rd well location = 100 m south of low head well(east)

3. Head of highest well = 30.2 m


- 49 -

4. Head of highest well = 30.0 m

5. Head of the 3rd well = 30.1 m

<Question> magnitude and direction of the hydraulic gradient = ?

<Solution> Hydraulic gradient = (30.1 – 30.0) m = 0.00141100 / √2 m

FIGURE 31 Finding the gradient for the well field in Example 11.


- 50 -

11. Darcy's Law

Flow rate ∝ (cross-sectional area) × (hydraulic gradient)

Darcy’s Law : Q = KA ( dh )dL (42)

where Q = flow rate(m3/d)

K = hydraulic conductivity(or coefficient of permeability)(m/day)

A = cross-sectional area(m2)

( dh )dL = hydraulic gradient

『Note in using Darcy’s Law』

① Invalid for turbulent flow

② Invalid for extremely fine-grained medium materials(ex, colloidal clays)

③ Invalid when the medium is not fully saturated with water

④ It also depends on temperature

『About K』

① very rough

② easily vary over several orders of magnitude according to

- particle orientation

- particle shape

- relative amounts of silt and clay

- vary from place to place within the aquifer

- direction of flow

※ It is common, for example, to have higher hydraulic conductivities in

the horizontal direction than in the vertical.

▶ Homogeneous aquifers : the same hydraulic conductivity throughout

▶ Heterogeneous aquifers : hydraulic conductivity differs from place to place

▶ Isotropic Aquifers : the same hydraulic conductivity in any flow direction


- 51 -

▶ Anisotropic Aquifiers : conductivity depends on direction

TABLE 12 Approximate Values of Hydraulic Conductivity

EXAMPLE 12 Flow Through an Aquifer

A confined aquifer 20.0 m thick has two monitoring wells spaced 500 m apart

along the direction of ground water flow. The difference in water level in the

wells is 2.0 m (the difference in piezometric head). The hydraulic conductivity is

50 m/day.

Estimate the rate of flow per meter of distance perpendicular to the flow.

<Given Data>

Δh = 2 m

ΔL = 500 m

K = 50 m/d

Thickness of confined aquifer = 20 m

<Question> Q/W = ?

<Solution>

Cross-sectional Area = 20W m2

Darcy’s Law : Q = KA ( dh ) = 50 m 20W m2 2 m = 4W m3/ddL d 500 m

∴ Q/W = 4 m3/d


- 52 -

FIGURE 32 Example 12, flow through a confined aquifer.

<Flow Velocity>

FIGURE 33 The cross-sectional area available for flow, A' ≪ A.

▶ Darcy’s velocity(v)

From Darcy’s law and Q = Av,

Darcy’s velocity(v) = Q = KAdh = K dhA AdL dL (44)

▶ Average Linear Velocity(Seepage Velocity), v´

Q = Av = A´v’ (45)


- 53 -

v´ = Av = ALv = Total Volume × v = vA´ A´L Void Volume η (46)

where Porosity(η) = Void VolumeTotal Volume

From Eq(44) and (46),

Average Linear Velocity(v´) = K ( dh )η dL (48)

EXAMPLE 13 A Groundwater Plume

Suppose the aquifer in Example 12 has become contaminated upgradient of the

two wells. Consider the upgradient well as a monitoring well whose purpose is

to provide early detection of the approaching plume to help protect the second,

drinking water well. How long after the monitoring well is contaminated would

you expect the drinking water well to be contaminated? Make the following

three assumptions (each of which will be challenged later):

<Assumption>

1. Ignore dispersion or diffusion of the plume (that is, it does not spread out).

2. Assume the plume moves at the same speed as the ground water.

3. Ignore the "pulling" effect of the drinking water well.

The aquifer has a porosity of 35 percent.

<Question> Time to travel from monitoring well to drinking water well = ?

<Given Data>

Δh = 2 m

ΔL = 500 m

K = 50 m/d

η = 35%

<Solution>

From Eq(44), Darcy’s velocity(v) = Q = K dh = 50 m 2m = 0.2 m/dA dL d 500 m


- 54 -

Average Linear Velocity(v´) = v = 0.2 m/d = 0.57 m/dη 0.35

Time to Travel(t) = 500 m = 877 day = 2.4 yr0.57 m/d

※ Ground water moves so slowly.


- 55 -

12. Contaminant Transport

<Dispersion and Diffusion>

▶ Diffusion : Solutes are moved by concentration difference between regions

▶ Dispersion : Solutes are moved by solvent bulk force.

FIGURE 34 Dispersion and retardation as a continuous feed of tracer passes

through a column.(Step input)


- 56 -

FIGURE 35 An instantaneous (pulse) source in a flow field creates a plume that

spreads as it moves downgradient: (a) in one dimension

(longitudinal), (b) in two dimensions (longitudinal + orthogonal).

<Retardation>

▶ In real process, contaminants are absorbed and are adsorbed by solids.

▶ Retardation Factor : Ratio of to

▶ Retardation Factor(R) = Total contaminant in unit volume of aquiferContaminant dissolved in ground water

= Average ground water velocity(ν´)≥ 1Velocity of the sorbed material(νs) (49)

▶ Linear Sorption : Contaminant sorbs at a rate ∝ its concentration


- 57 -

FIGURE 36 Influence of sorption on contaminant transport.


- 58 -

FIGURE 37 Plume separation for chloride(CL), carbon tetrachloride(CTET), and

tetrachloroethylene(PCE) 21 months after injection.


- 59 -

FIGURE 38 Retardation factors for carbon tetrachloride(CTET), etrachloroethylene

(PCE), bromoform(BROM), dichlorobenzene(DCB), and

hexachloroethane(HCE).

※ Retardation factors(from FIGURE 38)

- are not constant according to time or concentration

- increase over time

- eventually reach at steady-state value


- 60 -

FIGURE 39 Retardation increases the operating cost of a pump-and-treat system.

(Based on $650,000 initial capital cost, $180,000 annual O&M, 3.5

percent discount factor, 25-year equipment life, and 1 pore-volume

pumping per year.)

(Source: National Research Council, 1994.)


- 61 -

13. Cone of Depression

Pumping water from the aquifer through well↓

Generate a gradient toward the well to provide flow to the well↓

Affect the hydraulic gradient(※ Faster pumping → Steeper gradient)

FIGURE 40 Cone of depression in an unconfined aquifer

<Assumption>

1. Pumping has been steady for a long enough time

→ Shape of the cone is no longer changing(equilibrium conditions)

2. Original water table is horizontal

3. Drawdown is small relative to the depth of the aquifer

→ Well draws from the entire depth of the aquifer

→ Flow to the well is horizontal and radial.


- 62 -

FIGURE 41 Flow to the well passes through a cylinder of area A = 2πrhc

From Darcy’s Law,

Q = KA dhc = K(2πrhc)dhc

dr dr (50)

where r = cylinder of radius

hc = depth of cylinder

A = 2πrhc = outside area of an imaginary cylinder around well

dhc/dr = hydraulic gradient

= slope of the water table at radius r

K = hydraulic conductivity

▶ Q through the cylinder toward the well = Q of pumping from the well

▶ Hydraulic head(h)

measured from any suitable horizontal datum such as sea level.

▶ Depth of cylinder(hc)

height of the water table above the confining layer at the bottom of the

aquifer.(the depth of an unconfined aquifer at a distance r from the well.)


- 63 -

Integration of Eq(50) after rearranging,

(51)

ln

(52)

∴ Uncomfined aquifer : ln

(53)

▶ Application of EQ(53) for an unconfined aquifer

- Estimate the shape of the cone of depression for a given pumping rate

- Determine the aquifer hydraulic conductivity K

<To estimate K for an unconfined aquifer>

- Set up two observation wells at distances r1 and r2 apart from a pumping

well

- Measure heads(hc,2 and hc,1)

- Measure constant pumping rate(Q)

- Put the data into Eq(53)

EXAMPLE 14 Determining K from the Cone of Depression

Suppose a well 0.30 m in diameter has been pumped at a rate of 6,000 m3/day

for a long enough time that steady-state conditions apply. An observation well

located 30 m from the pumped well has been drawn down by 1.0 m, and

another well at 100 m is drawn down by 0.50 m. The well extends completely

through an unconfined aquifer 30.0 m thick.

<Given Data>

Steady-state

Total Depth of the aquifer = 30 m


- 64 -

Radius of pumping well(rw) = 0.3 m/2 = 0.15 m

Q = 6,000 m3/day r1 = 100 m r2 = 30 m

hr1 = 29.5 m ⇐ (30m – 0.5m) hr2 = 29.0 m ⇐ (30m – 1.0m)

<Question>

1. K = ?

2. The drawdown at the well(ht – hw) = ?

<Solution>

FIGURE 42 Data from Example 14 used to determine hydraulic conductivity(K).

It helps to put the data onto a drawing, as has been done in

Figure 42.

1. From EQ(53), ln

,

ln


- 65 -

=(6000 m3/d) ln(100m/30m)

= 78.6 m/dπ(29.52 – 29.02) m2

2. To estimate the drawdown, let r1 = 30 m and hr1 = 29.0 m

ln

6,000 m3/d = π(78.6 m/d)(292 – hw2) = 72.71(841 – hw

2)ln(30 – 0.15)

So, hw = 27.5 m

∴ Drawdown(ht – hw) = 30.0m – 27.5m = 2.5 m


- 66 -

<To estimate K for an confined aquifer>

FIGURE 43 Cone of depression for a confined aquifer.

▶ Height of cylinder at radius r (hc) = thickness of the aquifer(B) = constant

Also, as shown in Figure 43, the cone of depression now appears in the

piezometric surface

From Darcy’s law for the unconfined aquifer,

Q = KA dhc = K(2πrB) dhc

dr dr (54)

∴ Comfined aquifer : ln

(55)


- 67 -

15. Control of Ground Water Plumes

FIGURE 50 Manipulating the hydraulic gradient with multiple wells

16. Contaminants in Ground Water

TABLE 13 The 25 most frequently detected groundwater contaminants.


- 68 -

<Non Aqueous-Phase Liquids(NAPLs)>

Dense NAPLs (DNAPLs)

NAPLs

Light NAPLs (LNAPLs)

TABLE 14 Examples of nonaqueous-Phase liquids(NAPLs)

FIGURE 51 Nonaqueous-phase liquids do not dissolve very well in groundwater:

(a) LNAPLs float on top of groundwater; (b) DNAPLs form pools.

EXAMPLE 17 Estimated Time to Remove a TCE Spill1


- 69 -

Suppose 1 m3 of aquifer is contaminated with 30 L of trichloroethylene (TCE).

The aquifer has porosity of 0.3, groundwater moves through it with an actual

speed of 0.03 m/day, and the TCE has a dissolved concentration equal to 10

percent of its aqueous solubility.

<Given Data>

Volume aquifer = 1 m3

Volume of pollutant : 30 L of trichloroethylene (TCE)

Porosity aquifer : 0.3

vp = 0.03 m/day

Dissolved concentration of TCE = 10 % of water solubility

<Question>

1. Mass of dissolved TCE = ? and Mass of undissolved DNAPL = ?

2. Time for TCE to be removed = ?

<Solution>

1. From Table 14, the aqueous solubility of TCE = 1,100 mg/L

→ actual dissolved TCE = 10% of 1,100 mg/L = 110 mg/L

Porosity of the aquifer is 0.3

→ volume of fluid in aquifer = 0.3 × 1 m3 = 0.3 m3

∴ Total mass of dissolved TCE = 110 g/m3 × 0.3 m3 = 33 g

It means that 33 g of TCE was dissolved in 1 m3 aquifer.

From Table 14, specific gravity of TCE = 1.47

→ total mass of TCE in the aquifer = 1.47 kg/L × 30 L = 44.1 kg

∴ Total mass of DNAPL remained = 44,100 g – 33 g = 44,097 g

It means 99.92% of the TCE has not dissolved yet.

2.


- 70 -

Fluid leaving = A m2 × 0.03 m/d

Q of TCE = 0.03A m3/d × 33 g/m3 = 0.99A g/day

Time to remove TCE = t = 44,100 g / 0.99A = 44,545/A day

If A = 1 m2, t = 44,545 day = 122 yr


- 71 -

17. Ground Water Remediation Technology

<Conventional Pump-and-Treatment Systems>

FIGURE 52 Impact of cleanup goal on (present value) cost and duration of a

conventional pump-and-treat system.

Upper data points correspond to 99 percent and 99.99 percent removal of TCE.

Drawn for a plume volume of 190 × 106 L, a pumping rate of 380 L/min, a

retardation factor of 4.8 (TCE), and a 4 percent discount factor.

(Source: Based on data in NR Council, 1994.)


- 72 -

TABLE 15 Relative ease of cleaning up contaminated aquifers as a function of

contaminant chemistry and hydrogeology


- 73 -

<Soil Vapor Extraction(SVE)>

FIGURE 53 A soil vapor extraction system.

▶ SVE system

- effective at VOCs such as leaky gasoline storage tanks at service stations.

- remove some NAPLs in the unsaturated zone.

▶ Combination of SVE and pump-and-treat

- VOC remove some NAPLs in the saturated zone.


- 74 -

FIGURE 54 Combining air sparging with soil vapor extraction.

FIGURE 55 Air sparging with horizontal wells.


- 75 -

<In Situ Bio-remediation>

FIGURE 56 In situ bio-remediation.


- 76 -

<Permeable Reactive Barriers(PRBs)>

FIGURE 57 A permeable reactive barrier for remediation.

▶ PRB is a permeable wall to remove specially chosen pollutant

▶ Removing process

abiotic redox degradation, biotransformation, adsorption, and precipitation.

▶ Hydraulic conductivity of the PRB = greater than that of the aquifer

▶ The most common material = granular zero valent iron (ZVI) wall(2′∼6′)

▶ Application for redox-sensitive inorganic(Cr(VI) to Cr(III) state).

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