Water Quality

5/20/2018 Water Quality

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CHAPTER 2WATER QUALITY

Water Quality Standards

and

Parameters

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Content

PART 1: INTRODUCTION

i) Beneficial Water Useii) Water Resources

PART 2:WATER QUALITY

i) Definitionii) Objectivesiii) Water Quality Parameters

PART 3: WATER QUALITY PARAMETERS

i) Physicalii) Chemicaliii) Microbiological

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PART 1

INTRODUCTION

(i) Beneficial Water Use

(ii) Water Resources

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Beneficial Water Uses

Municipal Uses

A ricultural Uses

Industrial Uses

Rural Uses

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Water Resources

1. Snow / Rain

2. Surface Water

(i) Watershed Management

(ii) Lake /River /Reservoir(iii) Intake Structure

(iv) Pump(v) Treatment Facilities

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Cont.

3. Imported water

(i) Transmission Pipeline

(ii) Treatment Facilities

4.Groundwater(i) Basin Management

Natural and artificial recharge

Quality Control

(ii) Wells

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PART 2

Water Qualityi) Definition

ii) Objective

iii) Water quality parameters

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Water Quality Definition

Is the technical term that is basedupon the characteristics of water in

suitable for human consumption and

for all usual domestic purpose

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Objective of Water Quality

To control the discharge of

pollutants so that water quality is

not degraded to an unacceptableextent below the natural

background level

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Water Quality Parameters

Are the natural and man-made

chemical, biological and

m cro o og ca c arac er s cs orivers, lakes and groundwater.

It provides important information

about the health of a water body.

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Are used to find out if the qualitywater is good enough for drinking

water, recreation, irri ation and


aquatic life.

These include chemical, physical

and biological parameters

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PART 3WATER QUALITY

PARAMETERS

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1. Physical 2. Chemical

arameters

3. Biological

parameters

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Physical Parameters

This parameters respond to the sense of

sight, touch, taste orsmell

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Total Solids (TS)

TSS (Total Suspended Solids - TSS)

Dissolved Solids (DS)

Volatile Solids (VS)

Volatile Disolved Solids (VDS)

Unit: mg/l

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Classification of Solids in Water

and Wastewater

Solids

Suspended (> 1mm)

Colloid (1mm 0.01mm)

Dissolved (


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SolidsFilterable

Non-filterable

Settleable

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Non-settleable

SolidsVolatile

Non-Volatile


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Suspended solids (SS) SS contain

Organic matters

Biological Solids

Inorganic mattersClay (size 2 m)

Silt (size 2-60 m)

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Analysis of Solids The following test were obtained for a wastewater taken from a headwork

to a WTP. All the test were performed using sample size of 50 mL.

Determine the concentration of total solids (TS), total volatile solids (TVS),suspende solids (SS), volatile suspended solids (VSS), total dissolved

solids (TDS) and volatile dissolved solids.

Data:

Tare mass of evaporating dish = 53.5433 gMass of evaporating dish + residue after evaporation at 105oC =53.5794 g

Mass of evaporating dish + residue after ignition at 550oC = 53.5625 g

Tare mass of Whatman GF/C filter after drying at 105oC = 1.5433 g

Mass of Whatman GF/C filter + residue after drying at 105oC = 1.5554 gMass of Whatman GF/C filter + residue after ignition at 550oC = 1.5476 g

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Caused by foreign matters such as organics compounds,inorganic salts, bacteria, algae and dissolved gases

Measurement: Threshold Odor Number (TON)

Examples:

(i) addition of ammonia in the pipes(ii)excessive manganese & iron present in the finishedwater.

** manganese & iron often found in groundwatersupplies where the overall quality of the water is good butthere is a high amount of soluble salt. These metals thenreact with O2 in the distribution system to produced thereduced and insoluble form of the metal**

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Taste problems relating to watercould be indicators of changes in

water sources or treatment rocess

Inorganic compound such asmagnesium, calcium, sodium, copper,

iron and zinc are generally detected

by taste of water.

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It is due to the presence of dissolved andsuspended matter(metallic ions, chemical

pollutants, plankton and plant pigments from

humus and peat). These substance do not threaten stream

water quality, but indicate INCREASED

DEVELOPMENT in watershed.

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Dissolved organic material fromhumic substances generally lend a

brown or tea color to water

Dissolved organic material fromvegetation and certain inorganic

matter may cause color in water

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Is a measure of the amount of particulate

matter that is suspended in water. Unit-NTU (Nephlometric Turbidity Unit)

Water that has HIGH turbidity appears

. HIGH turbidity can cause INCREASED ofwater TEMPERATURE and DECREASEDDO

WHY???

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More suspended particles will absorbmore heat which in turn lowers dissolved

O2 levels.

It is because

Such particles (SS clay, silt, finelydivided organic material, plankton) can

also prevent sunlight from reaching plants

below surface hence DECREASE the rateof PHOTOSYNTHESIS.

So, LESS O2 is produced by plant

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It is a major factor in determining which

species are present in the stream Temperature will impacts:

(i) the rates of metabolism and growth of

aquat c organ sm(ii) rate of plant photosynthesis

(iii) solubility of O2 in water[0C,

DO = 14.6 mg/l; 20C, DO = 9.1 mg/l](iv) organisms sensitivity to disease,

parasites and toxic materials

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Cool water tastes better

Temperature affects rate of chemicaland microbiological reactions

The most suitable drinking waters are

consistently cool and do not havetemperature fluctuations of more thana few degrees

Groundwater and surface water frommountain area generally meet thesecriteria

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CHEMICAL PARAMETERS

()

pH

Alkalinit

Hardness

Biochemical Oxygen

Demand (BOD)

Nitrites and Nitrates

Chemical Oxygen

Demand (COD)31


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Chemical Parameters pH

It is a measure of the concentration ofhydrogen ions

The term pH was derived from the

pH

manner in which the hydrogen ionconcentration is calculated

pH scale ranges from 0 to 14. A pH of 7is considered to be neutral.

Substances with pH of less than 7 areacidic; substances with pH greater than7 are basic

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pH Continues..

Affects chemical and

biolo ical reactions

pH

Low pH is corrosive

High pH cause deposits

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Example 1

Calculate the concentration of hydrogen ion (H+) for a water sample with pHof 10.

pH = -log [H+]

10 = -log [H+]

pHContinues

..

, + = ant og -

= 10-10mol/liter

Example 2

Calculate the pH value of a water sample which has hydrogen ion

concentration of 1 x 10-6.4 mol/liter.

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Work Example Find the Hydrogen ion

concentration and thehydroxide ion concentration in

. .

Concentration unit mol/L then

mg/L

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4.1 = +

, + = 4.1

= 104.1 /

+ = 14

= 144.1 = 9.9

=

9.9 =

, = 9.9

= 109.9 /

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Conversionof Unit:

Mol/L to mg/L

H+ == atomic weight = 1 g/mol

OH- == atomic wei ht = 17

g/mol

Mol/L (conc. ) x g/mol (atomicwt) x 1000mg/1g = mg/L

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Hardness- Stream water hardness is the total

concentration of cations, specificallycalcium (Ca2+ ),magnesium (Mg2+), iron

Fe2+ man anese Mn2+ in the water.

- Water rich in these cations is said to behard. Stream water hardness reflects

the geology of the catchment area.

- Sometimes it also provides a measure

of the influence of human activity

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Hardness For instance, acid mine drainage

often results in the release of iron intoa stream. The iron producesextraordinarily high hardness is ause u wa er qua y n ca or.

Hardness is a reflection of the amountof calcium and magnesium enteringthe stream through the weathering ofrock such as limestone (CaCO3).

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Carbonate hardness

Source: Combination of Ca and Mg

ions with ions of CO32-, or HCO3-. These carbonate components can be

as boiling, or by adding lime When the carbonate components

settled then the water have become

soft water.

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Ca 2+ Mg 2+ HCO3- CO3

2-+

Carbonate hardness

Mg(HCO3)2Ca(HCO3)2 CaCO3 MgCO3

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Noncarbonate Hardness Source: Combinations of Ca and Mg ions with ions of Cl-, SO4

2-,

or NO3-.

The non-carbonate ions cannot be eliminated by ordinary softening

methods as done on carbonate hardness.

Usuall elimination of non carbonate hardness is done chemicall b

adding softening soda (soda ash or sodium carbonate)

Total hardness is measured in mg/L CaCO3:

Mg/L material X = Conc. of X (mg/L) (50 mg CaCO3/meq)

as CaCO3 (Equivalent wt of X (mg/meq))

Total Hardness as CaCO3 :

Total Hardness = Ca2+ + Mg2+

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+Ca 2+ Mg 2+ Cl- SO42- NO3

-

Noncarbonate Hardness

MgCl2Ca(NO3)2 MgSO4 Mg(NO3)2

CaCl2 CaSO4

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KeliatanKeliatan BukanBukan KarbonatKarbonat

((NoncarbonateNoncarbonate hardness)hardness)

Keliatan air dinyatakan dalam unit CaCO3 setara.

Pengkelasan keliatan: Lembut (soft) 50mg/l CaCO3 setara

Sederhana keras

(moderately hard) 50-150mg/l CaCO3

setara

Keras (hard) 150-300 mg/l CaCO3 setara

Sangat keras (very hard) >300 mg/l CaCO3 setara

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Calculation of Water Hardness Find the equivalent weight (EW) of each of the following:

Ca2+, CO3

2-, CaCO3

.

Solution:

EW = atomic or molecular weight / n (valence)units: grams/equivalent (g/eq) or milligrams/milliequivalent (mg/meq)

For calcium, n=2 (valence or oxidation state in water).Atomic weight = 40.08, therefore the EW is then

EW = 40.08/2 = 20.04 g/eq or mg/meq

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..For carbonate ion (CO3

2-), the oxidation state of 2- isused for n since the base CO3

2- can potentially

accept 2 H+. The molecular weight is 60.01.Therefore,

EW = 60.01/2 = 30 g/eq or mg/meq

In CaCO3, n=2 since it would take 2H+ to replace the

cation (Ca2+) to form carbonic acid, H2CO3. the MWis 100.

Therefore,EW = 100/2 = 50 g/eq or mg/meq

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A sample of groundwater has 100 mg/L of Ca2+ and 10

mg/L of Mg2+. Express it hardness in unit of mg/L as

CaCO3.

Solution:

recalled: Mg/L of X = concentration of X (mg/L) (50 mg CaCO3/meq)

as CaCO3 (equivalent weight of X (mg/meq))

1. Convert Ca2+ and Mg2+ to mg/L as CaCO3

Ca2+

: MW = 40, n=2, EW=40/2 = 20 g/eq or mg/meqMg2+ : MW = 24.3, n=2, EW=24.3/2 = 12.2 g/eq or mg/meq

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Cont..

Now, find the mg/L as CaCO3 of cations

Ca2+ = 100 (50/20) = 250

Mg2+

= 10 (50/12.2) = 41Total Carbonate Hardness = Ca2+ + Mg2+

= 250 + 41

= 291 mg/L as CaCO3

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Alkalinity Is measured to determine the ability of a

stream to resist changes in pH.

Alkalinity results from the dissolution ofcalcium carbonate (CaCO3) from limestonebedrock which is eroded during the naturalprocesses of weathering

Alkalinity values of 20 -200 ppm are commonin freshwater ecosystems. Alkalinity levelsbelow 10 ppm indicate poorly bufferedstreams.

These stream are the least capable of

resisting changes in pH, therefore they aremost susceptible to problems which occur asa result of acidic pollutants

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Kealkalian (Alkalinity) Mengukur kebolehan air menyerap (absorbs) ion

Hidrogen tanpa melibatkan perubahan pH yang

ketara.

Jadi, kealkalian adalah ukuran kapasiti penimbal

buffer ca acit ba i air.

spesies Karbonat yang menyumbang kepadaalkaliniti adalah:

Hydroksil (OH-), Ion Karbonat (CO32-), Ion

Bikarbonat (HCO3-),

Alkalinity (mol/L) = [HCO3-] + 2[CO3

2-] + [OH-] [H+]

Alkalinity (mg/L as CaCO3) = (HCO3-) + (CO3

2-) +

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,

Nitrogen is an essential nutrient that is

required by all plants and animals for the

formation of amino acids. In its molecular form, nitrogen cannot be usedby most aquatic plants, therefore it must beconverted to another form.

One such form is ammonia (NH3). Ammoniamay be taken up by plants or oxidized bybacteria into nitrate (NO3) or nitrite (NO2). Ofthese two forms, nitrate is usually by the mostimportant.

Nitrosomonas Nitrobacter

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Nutrients Nitrogen is often the limiting nutrient in

ocean waters and some streams Nitrogen can exist in numerous forms, but

- - 3 , 2 ,

(NH3) are most commonly measured

Sources are primarily from fertilizers and

acid deposition

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Biochemical Oxygen Demand


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(BOD)

It is a measure of the quantity of oxygen

used by microorganisms (eg.aerobicbacteria) in the oxidation of organic matter.

In other words: BOD measures the change

in dissolved oxygen concentration causedby the microorganisms as they degrade the

organic matter.

High BOD is an indication of poor waterquality

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(BOD)

BOD bottles

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Bi h i l O


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Biochemical Oxygen

Demand Measurement Take sample of waste; dilute with

oxygen saturated water; add nutrientsand microorganisms (seed)

days Temperature 20 C

In dark (prevents algae from growing)

Final DO concentration must be > 2 mg/L Need at least 2 mg/L change in DO over 5 days

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Example 1 A BOD test was conducted in the

laboratory using wastewater being

dumped into Lake Spartan. The

samples are prepared by adding 3.00

mL of wastewater to the 300.0 mL BODbottles. The bottles are filled to capacity

with seeded dilution water.

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E l 1 R D t


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Example 1: Raw Data

Time(days)

Dilutedsample

DO (mg/L)

Blank SeededSample DO

(mg/L)

0 7.95 8.15

1 3.75 8.102 3.45 8.05

3 2.75 8.00

4 2.15 7.955 1.80 7.90

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Example 1: Calculations What is the BOD5 of the sample?

volumetotalsample/volumefactordilutionP

PP)])(1B(B-)DO[(DOBOD fifim

==

=

Plot the BOD with respect to time.samplediluted

theofionsconcentratDOfinalandinitialDO,DO

(blank)waterdilutedseededtheofionsconcentratDOfinalandinitialB,B

fi

fi

=

=

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Example 1: Time Concentration Plot

400

500

600

700

g/L)

0

100

200

300

0 1 2 3 4 5 6

time (days)

BOD

(

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Modeling BOD as a First-order Reaction


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Modeling BOD as a First-order Reaction

45

6

mg/L)

Organic matter oxidized

0

1

2

0 10 20 30

time (days)

Conc

.

Organic matter remaining

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.

tt - kL

dL=

demandoxygenuscarbonaceoultimatetheiswhere

:yieldsequationthisSolving

(timeconstantrateBODthe

ttimeafterleftdemandoxygenofamountwhere1-

o

kt

ot

t

L

eLL

k

L

=

==

)

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Ultimate BOD


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Ultimate BOD

34

56

nc.(mg/L)

LoLo- Lt BOD exerted

BODt

0

1

0 10 20 30time (days)

C

Lt L remaining

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Ultimate Biochemical Oxygen


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yg

Demand

Lt = amount of O2 demand left in sample at

time, tLo = amount of O2 demand left initially (at time 0,

=,

At any time, Lo = BODt + Lt (that is the amountof DO demand used up and the amount

of DO that could be used up eventually)

Assuming that DO depletion is first orderBODt = Lo(1 - e

-kt)

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Example 2


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p If the BOD5 of a waste is 102 mg/L and the

BOD20 (corresponds to the ultimate BOD) is

158 mg/L, what is k (base e)?

kteL = 10tBOD

kteL

=0

1 tBOD

ktL

=

0

1ln tBOD

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Example 2 (cont)

tLk

= 0

1ln tBOD

day

mg/Lmg/L

5

1581021ln

=k

-1day21.0=k65

Biological Oxygen Demand:


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Biological Oxygen Demand:

Temperature Dependence

Temperature dependence of biochemical

oxygen demand

As temperature increases, metabolism

increases, utilization of DO also increases

kt = k20T-20

= 1.135 if T is between 4 - 20 oC

= 1.056 if T is between 20 - 30 oC

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Example 3The BOD rate constant, k, was determined

empirically to be 0.20 days-1

at 20o

C.What is k if the temperature of the water

increases to 25 oC?

What is k if the temperature of the waterdecreases to 10 oC?

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Example 3 : Solution


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Example 3 : Solution

2025

25 )056.1(20.0 = -1dayk

-1da26.0=k

2010

10 )135.1(20.0 = -1dayk

-1day056.010 =k

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Nitrogenous Oxygen Demand

So far we have dealt only withcarbonaceous demand (demand to

Many other compounds, such asproteins, consume oxygen

Mechanism of reactions are different

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Nitrogenous Oxygen Demand Nitrification (2 step process)

2 NH3 + 3O2 2 NO2- + 2H+ + 2H2O2 NO2

- + O2 2 NO3-

NH3 + 2O2 NO3- + H+ + H2O

Theoretical NBOD =

N/gOg4.5714

16x4

oxidizednitrogenofgrams

usedoxygenofgrams2

==

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Nitrogenous Oxygen Demand

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Nitrogenous oxygen demand Untreated domestic wastewater

ultimate-CBOD = 250 - 350 mg/Lultimate-NBOD = 70 - 230 mg/L

Total Kjeldahl Nitrogen (TKN) = totalconcentration of organic and ammonia

nitrogen in wastewater: 15 - 50 mg/L as N

Ultimate NBOD 4.57 x TKN

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Other Measures of Oxygen

eman

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Example 5 Theoretical oxygen demand (ThOD)

C6H

12O

6(glucose) + 6O

26 CO

2+ 6H

2O

If concentration of glucose is 10 mg/L what is the

theoretical oxygen demand (amount of DO

required?)

LOmg

glucosemmol

glucosemg

Ommol

Omg

glucosemmol

Ommol

Lglucosemg 22

22

7.10180

326

10 =

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Chemical Oxygen Demand Chemical oxygen demand - similar to BOD

but is determined by using a strongoxidizing agent to break down chemical

Still determines the equivalent amount ofoxygen that would be consumed

Value usually about 1.25 times BOD

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Thomas method: Graphical determination of


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BOD rate ConstantsAccording to Thomas,

(1-e-kt) = kt(1+kt/6)-3

Therefore

BODt=Lo(1-e-kt)BODt=Lo(kt)[1+(1/6)kt]

-3 .....(1)

By rearranging terms & taking the cube root of both sides,equa on can rans orme o:

(t/BODt)1/3=(kLo)

-1/3 + (k)2/3/6(Lo)1/3 (t) ....(2)

A plot of (t/BODt)1/3 versus tis linear. The intercept is

defined as:

A = (kLo)-1/3

.(3)

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..


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..

a slope is defined by:B = (k)2/3/6(Lo)

1/3 .(4)y=(t/BOD) 1/3

Recalled , y= c + mx

=

m=a/b = (k2/3/6Lo1/3)

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..


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Solving Lo1/3 in Eq.(3) substituting into Eq. (4) and solving for k

yields:

k= 6(B/A) (5) Likewise, substituting Eq. (5) into Eq.(3) and solving for Lo yields:

Lo = 1/6(A)2

(B) (6)

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,


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,

:

(/) 125 200 220 230 237

79

,


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,

(/)

1/3

.

(/) 125 200 220 230 237

(/)1/3

.

(/)1/3 0.252 0.271 0.301 0.326 0.348

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0.3

0.4

0.5

= 2/3 1/3

)

= 0.224

0

2 4 6 8 10

0.1

0.2

=0.0125

81

..


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() ()

.

= 0.224

= 0.3480.224 100

= 0.0124

k= 6 (B/A) Lo = 1/ 6(A)2(B)

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..


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= 6(/)= 6(0.0125/0.224)

= 0.335 /

= 1/6()2()

= 266 /

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WATER QUALITY


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MEASUREMENTThOD theoretical oxygen demand

(i) It is the amount of O2 required tooxidize a substance to CO2 and H2O

ii Calculated b stoichiometr if the

chemical composition of the substanceis known

(iii) The ThOD of X in mg/L

= (amount of X in mg/L )( MW of O2 in g/MW of X in g)

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Example:


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Compute the ThOD of 108.75 mg/L

of glucose (C6H12O6) STEPS:

wr e a ance equa on or e

reaction

(ii) Compute the grams molecular

weights of the reactants(iii) Determine ThOD

BFC 3103

ENVIRONMENTAL ENGINEERING85

Example:


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The balanced equation for the reaction

C6H12O6 + O2 CO2 + H2O

The molecular weights (grams) of thereactants

Oxygen =Thus, it takes ? of O2 to oxidize ? g of

glucose to CO2 and H2O.

The ThOD of 108.75 mg/L of glucose is(108.75 mg/L glucose)( ? g O2/ ? g

glucose) =116 mg/L O286

Example:

Th b l d ti f th ti


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The balanced equation for the reaction

C6H12O6 + 6O2 6CO2 + 6H2O

The molecular weights (grams) of the reactants

Glucose = 6C=72, 12 H=12, 6O=96, = 180

Oxygen =6(2)O=192

Thus, it takes 192 of O2 to oxidize 180 g of

glucose to CO2 and H2O.

The ThOD of 108.75 mg/L of glucose is

(108.75 mg/L glucose)(192 g O2/180 g glucose)

=116 mg/L O2

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Exercise

(5 1 5 2 5 3 i th t t b k)


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(5.1, 5.2, 5.3-in the text book)

1. Glutamic acid (C5H904N) is used as one of

the reagents for a standard to check the BOD

test. Determine the ThOD of mg/L of glutamic

acid. Assume the following reaction apply:

C5H904N + 4.5O2 5CO2 + 3H2O + NH3

NH3 + 2O2 NO3+ H+ + H2O

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Pathogenic Organisms


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Many organims that cause human oranimal diseases colonize the instinal tract

but can live for a period of time outside thebody

disease symptoms) excrete theseintestinal tract organims in very largenumbers

When water is contaminated by excretia,the organisms can be transmitted to thosewho contact the water

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Pathogenic Organisms


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Classification of Water

A i t d Di


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Associated Diseases

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Disease Rates and Risk


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Biological Parameters


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It is biomonitor: defined as an organism

that provides quantitative information on

the quality of the environmental around

it.

It can be deduced through the study ofthe content of certain elements or

compounds, morphological or cellular

structure, metabolic-biochemicalprocess behavior or population structure

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.. There are several types of bioindicators:

(i) plant indicators

- the presence or absence of certain plant or othervegetative life in an ecosystem can provide important

- lichens are organism comprising both fungi and algae.

Lichens are found on rocks and tree trunks, and theyrespond to environmental changes in forest, includingchanges in forest structure conservation biology, airquality and climate

- The disappearance of lichens in a forest may indicateenvironmental stresses, such as high level of sulfurdioxide, sulfur-based pollutants and nitrogen oxides

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Lichens

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(ii)Animal indicator and toxins- an increase or decrease in an animal

population may indicate damage to theecosystem caused by pollutant. For eg; ifpopulation causes the depletion ofim ortant food sources animal s eciesdependent upon these food sources will

also be reduced in number: populationdecline- Submerged aquatic vegetation (SAV)

provides invaluable benefits to aquaticecosystems. It not only provides food andshelter to fish and invertebrates but alsoproduces oxygen, trap sediment andabsorbs nutrients such as nitrogen andphosphorus

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Cont


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Cont(iii) Microbial indicators and chemical

pollutants

- Microorganisms can be used asindicators of aquatic or terrestrialecosystem health

- Found in large quantities, microorganism

will produce new proteins, called stressproteins when exposed to contaminantslike cadmium and benzene

- These stress proteins can be used as anearly warning system to detect highlevels of pollution

BFC 3103


Cont


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Cont(iv) Macroinvertebrate bioindicators- Macroinvertebrate are useful and

convenient indicators of the ecologicalhealth of a waterbody or river. They arealmost alwa s resent and are eas tosample and identify

- Benthic refers to the bottom of awaterway. Example of benthicmacroinvertebrates include insects intheir larval or nymph form, crayfish,

claims, snails and worms. Most live partor most of their life cycle attached tosubmerged rocks, logs and vegetation.

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BFC 3103

ENVIRONMENTAL ENGINEERING

Macroinvertebrata

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101

Cont


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Cont- The basic principle behind the

study of macroinvertebrates is that

some are more sensitive topollution than others

- Therefore, if a stream site isinhabited by organism that cantolerate pollution and the morepollution-sensitive organisms are

missing a pollution is likely

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Microbiological


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Microbiological

Bacteria( coliform test)

Virus

Algae

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Bacteria


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Bacteria Pathogenic bacteria causing cholera,

typhoid fever etc

Indicator bacteria

o orm

Fecal Coliform( E. Coli)

104

Virus


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Virus One virus can cause

illness

Hard to detect

pec y rea men

process ( disinfectiondose and contact time)

instead of measuring

virus concentration

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Algae


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Algae Taste and odor

Some algae could be harmful to animals fish

r s

BFC 3103



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EFFECTS ON WATER

QUALITY

2) Nontoxic organic3) Toxic organic

107

Toxic inorganic elements and

radicals


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radicals Arsenic, Mercury, Cadmium,

Chromium, Lead-- accumulates in

body

Nitrate--Blue baby Perchlorate ( ClO4

-) --Thyroid

disorder, cancer

108

Nontoxic organics


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Nontoxic organics NOC ( Natural organic matter)

Decayed vegetation etc

Form toxic disinfection by-products

Lower concentrations up to 4 mg/lmay be removed by Enhanced

Coagulation.

109

Toxic organics


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Toxic organicsCausing cancer, mutation or

miscarriage

chlorinated hydrocarbons

orop enoxy er c es

Trihalomethanes

VOCs and SOCs.

110

Other Contaminants


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Other Contaminants Asbestos

Radionuclides Alpha and Beta radioactivity

ran um, a um, a on

111

Dissolved oxygen (DO)It i ti l f th i l f


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yg ( ) It is an essential for the survival of

nearly all aquatic life and measured

in mg/L If oxygen levels are high, it was

presume that pollution levels in the

water are low. Conversely, if oxygen levels are low,

one can presume there is a high

oxygen demand and that the body ofwater is not of optimal health

112

Cont


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Levels of DO vary depending on factorsincluding water temperature, time of day,

season, depth, altitude and rate of flow.(i) water at higher temp and altitudes will haveLESS DO. so, demand O2 will increased

ecause a g er emp, e ra e o

metabolisme is increased.(ii) at night, DO decreased as photosynthesishas stopped while oxygen consumingprocess such as respiration, oxidation

(iii) DO reaches its peak (HIGH) during theday

113

DO sag curve


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The concentration of DO in a river is an

indicatorof the general health of the river.

All rivers have CAPACITY for self purification.(i) As long as the discharge of oxygen

purification capacity, the DO level remain HIGH

and a diverse population of plants and animals

(ii) As the amount of waste increase, the self

purification capacity can be exceeded, causing

detrimental changes in plant and animal life

114

.(iii) then the stream losses its ability to clean


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(iii) then, the stream losses its ability to cleanitself and the DO level DECREASES.

(iv) when the DO drops below 4 to 5 mg/L,most game fish will have been driven out.

,other higher animals are killed or driven

(v) The water become blackish and foulsmelling as the sewage and dead animal lifedecompose under anaerobic condition

(without O2)

BFC 3103


Dissolved Oxygen Depletion


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(From: Environmental Science: A Global Concern, 3rd ed. by W.PCunningham and B.W. Saigo, WC Brown Publishers, 1995)

116


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117

Mass Balance Approach

O i i ll d l d b H W St t d


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Originally developed by H.W. Streeter and

E.B. Phelps in 1925

River described as plug-flow reactor

ass a ance s s mp e y se ec on o

system boundaries Oxygen is depleted by BOD exertion

Oxygen is gained through reaeration

118

Steps in Developing the DO

Sag Curve


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1. Determine the initial conditions

2. Determine the reaeration rate from streamgeometry

3. Determine the deoxygenation rate from

BOD test and stream geometry

4. Calculate the DO deficit as a function of

time

5. Calculate the time and deficit at the critical

point

119


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120


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Qw = waste flow (m3/s)DOw = DO in waste (mg/L)

Qr= river flow (m3/s)

DOr= DO in river (mg/L)

Lr= BOD in river (mg/L)

Qmix = combined flow (m3/s)

DO = mixed DO (mg/L)

La = mixed BOD (mg/L)

w

121

1.

a. Initial dissolved oxygen concentrationDOQDOQ +


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b. Initial dissolved oxygen deficit

rw

rrww

QQ

DOQDOQDO

+

+=

where D = DO deficit (mg/L)

DOs = saturation DO conc. (mg/L)

mix

rrwwsa

Q

DOQDOQDOD +=

s

122

1. Determine Initial Conditions

DO t is a function of temperature Values


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DOsat is a function of temperature. Values

can be found in Table A-2.

c. Initial ultimate BOD concentration

rw

rrwwa

QQ

L

+

+=

123

2. Determine Reaeration Rate

a. OConnor-Dobbins correlation2/19.3 u

k


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where kr= reaeration coefficient @ 20C (day-1)

u = avera e stream velocit m/s

2/3hkr=

h = average stream depth (m)

b. Correct rate coefficient for stream temperature

where = 1.024

20

20,

= Trr kk

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. = kdLt

kd


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kd

(1)

Lt

. kd

() = k( )

tk

tdeLL

= 0

tk

ddeLk

= 0tiondeoxygentaofrate125

3. Determine the Deoxygenation Rate

c. However, k= kd only for deep, slow moving


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, d y p, gstreams. For others,

hukkd +=

where = bed activity coefficient (0.1 0.6)

d. Correct for temperature

where = 1.135 (4-20C) or 1.056 (20-30C)

20

20,

= Trr kk

126

4. DO as function of time

Mass balance on moving element


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g

DkLkdt

dDrtd

=

( ) ( )tkatktkdr

adt

rrd eDeekk

LkD +=

127

5. Calculate Critical time and DO

drr kkD

k1l

1


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=

ad

dra

d

r

dr

cLk

Dkkk

t 1ln

( ) crcrcd tk

a

tktk

ar

ad

c eDeekk

Lk

D

+=

128

Example

A city of 200,000 people discharges 37.0cfs of treated sewage having an ultimate


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BOD of 28.0 mg/L and 1.8 mg/L DO into a

river with a flow of 250 cfs and velocity of1.2 ft/sec. U stream of the dischar e

point, the river has a BOD of 3.6 mg/L and

a DO of 7.6 mg/L. The saturation DO is8.5 mg/L, kd = 0.61 day

-1, and kr= 0.76

day-1. Determine a) the critical DO and

critical distance, and b) the DO at 10 miles

downstream.129


a. Initial dissolved oxygen concentration


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rw

rrww

QQ

DOQDOQ

DO +

+

=

b. Initial dissolved oxygen deficit

DODOD s=

L

85.6

37250

...=

+

+=DO

L

mg6.185.65.8 ==aD

130


c. Initial ultimate BOD concentration


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rw

rrwwa

QQLQLQL

+

+=

( )( ) ( )( )Lmg75.6

37250

2506.30.3728 =++=aL

131

Step 1. Variations

Flow given in cfs, not m3/s does not


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g

matter for mixing calculations

Saturation DO given no need to look up

ma e g ven no nee o ca cu a e

from BOD5

132

2. Determine Reaeration Rate

kr = 0.76 day-1 given


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r y g

no need to calculate from stream

geometry

assume g ven va ue s a e s ream

temperature (since not otherwisespecified), so no need to correct

133

3. Determine the Deoxygenation Rate

kd = 0.61 day-1 given


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d

no need to calculate corrections from

stream geometry

assume g ven va ue s a e s ream

temperature (since not otherwisespecified), so no need to correct

134


= drrkk

Dk

t 1ln1


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=

ad

a

ddr

cLk

Dkkk

t 1ln

( )( )

=

75.661.0

..6.11

61.0

.ln

61.076.0

ct

day07.1=ct

135



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( )

= ft

midhr

hrs

sftd

528012436002.107.1cx

mi9.20=c

x

136


( ) tktktkad DLkD


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( ) crcrcd tkatktkar

adc eDee

kk

D +

=

( )( ) ( )( )

( )( )( )07.176.0

07.176.007.161.0

6.1

61.076.0

..

+

=

e

eeDc

L

mg58.2=cD137


csc DDODO =


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mg7.558.25.8 ==cDO

138

4. DO as function of time (at 10 miles)

( )ft

mi 528010


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( )

dhrsft

mimi

miles10 51.0

528010

=

=t

dhrs.

( )( ) ( )( ) ( )( )( )( )( )07.176.0

07.176.007.161.0

6.1

61.076.0

75.661.0

+

=

e

eeDc

139

4. DO as function of time (at 10 miles)

mg582D


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L

gmiles10 58.2=D

Lmiles10 9.558.25.8 ==DO

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Water Quality

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