Chapter 1 Water Quality Sem 1 1819.pdf

CHAPTER 1 WATER QUALITY Dr Nor Adila Ab Aziz Prof Madya Dr Rafidah Hamdan

Content PART 1: INTRODUCTION TO WATER QUALITY i) Beneficial Water Use ii) Water Resources iii)Definition iv) Objectives v) Water Quality Parameters

PART 2:INTERIM NATIONAL WATER QUALITY STANDARDS PART 3: WATER QUALITY PARAMETERS i) Physical (Suspended Solids, Turbidity and Temperature) ii) Chemical (pH, Alkalinity, Hardness, Dissolved Oxygen (DO), Biochemical Oxygen Demand (BOD), Chemical Oxygen Demand (COD), Nitrogen and Phosphorus)

iii) Microbiological (Total Coliforms, Fecal Coliforms, E.Coli, Macroinvertebrate Bioindicator)

PART 4: SAMPLING (RIVER MONITORING) PART 5: DO SAG CURVE 2

PART 1 INTRODUCTION TO WATER QUALITY

3

Beneficial Water Uses • • • •

Municipal Uses Agricultural Uses Industrial Uses Rural Uses

4

Water Resources 1. Snow / Rain 2. Surface Water i) Watershed ii) Lake /River /Reservoir

3.Groundwater -

Sub-surface water, or groundwater, is fresh water located in the pore space of soil and rocks. Flowing within aquifers below the water table.

4. Desalination -

saline water is converted to fresh water.

5

Water Quality Definition Is the technical term that is based upon the characteristics of water in relation to guideline values of what is suitable for human consumption and for all usual domestic purpose

6

Objective of Water Quality To control the discharge of pollutants so that water quality is not degraded to an unacceptable extent below the natural background level

7

Water Quality Parameters • Are the natural and man-made chemical, biological and microbiological characteristics of rivers, lakes and groundwater. • It provides important information about the health of a water body. 8

Water Quality Parameters - Are the natural and man-made chemical, biological and microbiological characteristics of rivers, lakes and groundwater. - It provides important information about the health of a water body. 

Are used to find out if the quality water is good enough for drinking water, recreation, irrigation and aquatic life.

9

TYPES OF WATER POLLUTION POINT SOURCE  pollution flowing from a single and identifiable source such as discharge pipe from a factory, roadway, or leaking underground storage tank

NON-POINT SOURCE  pollution collected by rain falling over a larger watershed which is then carried by runoff to a nearby lake or stream, or by infiltration into the groundwater

POINT SOURCE POLLUTION

POINT SOURCE POLLUTION • Hazardous and toxic materials from manufacturing and industry discharged directly into the water - usually through a pipe or a leaky underground tank • Oil and gasoline • Solvents (toxic liquids) • Toxins and poisons • Heavy metals (arsenic, lead, mercury, etc.) • THERMAL POLLUTION heated water causes the dissolved oxygen (DO) content in a body of water to decrease - can result in fish kills

NON-POINT SOURCE POLLUTION

A HARDER PROBLEM TO SOLVE

THE CHANGING URBAN LANDSCAPE • •

•

Changing the landscape changes the amount of runoff in a watershed NON-POINT SOURCE POLLUTION is pollutants being collected by rainwater falling over a large watershed and carried directly to a river, lake or stream Gas, oil, chemicals, detergents containing phosphorus, trash and other pollutants collected off driveways, roads and city streets flow directly down drains and storm sewers to a nearby body of water untreated

THE CHANGING RURAL LANDSCAPE • MODERN FARMING IS A MAJOR SOURCE OF NON-POINT SOURCE POLLUTION • Pesticides (bug killer) and herbicides (weed killer) can wash into nearby lakes and rivers • Crop fields, especially after harvest, can wash large amounts of dirt and sediment into nearby lakes and rivers • Animal waste and manure can be a source of nutrients and harmful bacteria • Fertilizer can be a source of nutrients, such as nitrogen and phosphorus, entering nearby lakes and rivers leading to the serious

CONSTRUCTION & MINING • Clear-cutting trees and plowing a field to create a mining or construction site can be a major source of non-point source pollutants • Without the trees and the plants in the field to hold the soil in place, large amounts of dirt and sediment can be discharged into a nearby lake or stream • Can be a source of toxic chemicals, acids, or heavy metals used in the construction or mining process

PART 2 Interim National Water Quality Standards (INWQS)

18

Interim National Water Quality Standards • INWQS used for classification of rivers or river segments based on 5 classes of water quality. • INWQS helps Department of Environment (DOE) to identify problem areas & develop appropriate strategies for water quality management. • It is important to maintain high quality level for natural water. • Therefore, the DOE has set up the minimum quality standard that reflects its beneficial uses. 19

Malaysia : Interim National Water Quality Standard (INWQS) Parameter Ammoniacal Nitrogen Biochemical Oxygen Demand Chemical Oxygen Demand Dissolved Oxygen pH Color Electrical Conductivity* Floatables Odor Salinity Taste Total Dissolve Solid Total Suspended Solid Temperature Turbidity Faecel Coliform** Total Coliform

Unit

Classes I

IIA

IIB

III

IV

V

mg/L

0.1

0.3

0.3

0.9

2.7

>2.7

mg/L

1

3

3

6

12

>12

mg/L

10

25

35

50

100

>100

mg/L TCU

7 6.5-8.5 15

5-7 6-9 150

5-7 6-9 150

5-9 5-9 -

5-9 5-9 -

-

μS/cm

1000

1000

-

-

6000

-

% -

N N 0.5 N

N N 1 N

N N N

-

2 -

-

mg/L

500

1000

-

-

4000

-

mg/L

25

50

50

150

300

300

oC

-

NTU Count/ 100ml Count/ 100ml

5

Normal + 2 oC 50

50

10

100

400

100

5000

5000

Source: Environmental Quality Report 2010

-

Normal + 2 oC 5000 (20000)a

-

-

5000 (20000)a

-

50000

50000

>50000

-

20

Classification of Water Based on INWQS

mg/L

I <0.1

II 0.1-0.3

Class III 0.3-0.9

mg/L

<1

1-3

3-6

6-12

>12

mg/L mg/L mg/L

<10 >7 >7 <25

10-25 5-7 6-7 25-50

25-50 3-5 5-6 50-150

50-100 1-3 <5 150-300

>100 <1 >5 >300

Parameter

Unit

Ammoniacal Nitrogen Biochemical Oxygen Demand Chemical Oxygen Demand Dissolved oxygen pH Total Suspended Solid

IV 0.9-2.7

V >2.7

Water Classes and Uses Class I

IIA IIB III

IV V

Uses Conservation of natural environment. Water Supply I – Practically no treatment necessary. Fishery I – Very sensitive aquatic species. Water Supply II – Conventional treatment required. Fishery II – Sensitive aquatic species. Recreational use with body contact. Water Supply III – Extensive treatment required. Fishery III – Common, of economic value and tolerant species; livestock drinking. Irrigation None of the above

Source: Environmental Quality Report 2010

21

PART 3 WATER QUALITY PARAMETERS

22

Water Quality Parameters

1. Physical parameters

2. Chemical parameters

3. Biological parameters

23

Physical Parameters • This parameters respond to the sense of sight, touch, taste or smell

24

Solids

25

Solids Total Solid (TS) are the total of all solids in a water sample. They include the total suspended solids, total dissolved solids, and volatile suspended solids. Dissolved solids (DS) = calcium, chlorides, nitrate, phosphorus, iron, sulfur, and other ions and particles that will pass through a 2 micron filter Suspended solids (SS) = silt, clay, plankton, algae, fine organic debris, and other particulate matter that will not pass through a 2-micron filter

Interrelationships of solids

27

28

Determination of Suspended Solids

Example 1 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 g Mass 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 g Mass of Whatman GF/C filter + residue after ignition at 550oC = 1.5476 g

29

Solution:

1. Determine total solids (TS) TS = (mass of dish + residue, g)-(mass of dish, g) sample size, Liter TS = ((53.5794-53.5433) g)(103 mg/g) = 722 mg/L 0.050 L

2. Determine total volatile solids (TVS) TVS = (mass of dish + residue, g)-(mass of dish + residue after ignition, g) sample size, Liter TVS = ((53.5794-53.5625) g)(103 mg/g) = 338 mg/L 0.050 L

30

3. Determine total suspended solids (TSS) TS = (residue on filter after drying, g)-(tare mass of filter after drying, g) sample size, Liter TS = ((1.5554-1.5433) g)(103 mg/g) = 242 mg/L 0.050 L

4. Determine total volatile solids (TVS) TVS = (residue on filter after drying, g)-(residue on filter after ignition, g) sample size, Liter TVS = ((1.5554-1.5476) g)(103 mg/g) = 156 mg/L 0.050 L 31

• Is a measure of the amount of particulate matter that is suspended in water or the measurement of water clarity. • Measured by Turbidity meter. • Unit-NTU (Nephlometric Turbidity Unit) • Water that has high turbidity appears cloudy / opaque . • High turbidity can cause increased of water temperature and decreased DO WHY??? 32

It is because… • More suspended particles will absorb more heat from solar radiation than water molecules will. This heat is then transferred to the surrounding water by conduction. • Such particles (SS – clay, silt, finely divided organic material, plankton) can also prevent sunlight from reaching plants below surface hence decrease the rate of photosynthesis. So, less O2 is produced by plant 33

• 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 aquatic organism (ii) rate of plant photosynthesis (iii) solubility of O2 in water [0C, DO = 14.6 mg/l; 20C, DO = 9.1 mg/l] (iv) organism’s sensitivity to disease, parasites and toxic materials 34

• Temperature affects rate of chemical and microbiological reactions • The most suitable drinking waters are consistently cool and do not have temperature fluctuations of more than a few degrees • Groundwater and surface water from mountain area generally meet these criteria 35

CHEMICAL PARAMETERS pH

Hardness

Alkalinity Biochemical Oxygen Demand (BOD)

Nitrogen & Phosphorus Chemical Oxygen Demand (COD)

36

pH • affects chemical and biological reactions • It is a measure of the concentration of hydrogen ions [H+]

• The term pH was derived from the manner in which the hydrogen ion concentration is calculated pH= -log[H+] [H+] = [H3O+] pH=1 : [H+]=1 x 10-1 moles/liter (acidic) 37

pH

38

pH Acid – Base Concentrations concentration (moles/L)

10-1

pH = 11

pH = 3

OH-

H3O+

pH = 7 10-7 H3O+

OH-

OH-

H3O+

10-14

Timberlake, Chemistry 7th Edition, page 332

[H3O+] > [OH-] [H3O+] = [OH-] acidic neutral solution solution

[H3O+] < [OH-] basic solution

pH Example 1 Calculate the concentration of hydrogen ion (H+) for a water sample with pH of 10. pH = -log [H+] 10 = -log [H+] Therefore, [H+]= antilog -10 = 10-10 mol/liter Example 2 Calculate the pH value of a water sample which has hydrogen ion concentration of 1 x 10-6.4 mol/liter. pH = -log [H+] = -log (1 x 10-6.4) = -[log 1 + log 10-6.4] = -[0 + (-6.4)log 10] = 6.4 40

pH Example 3 Find the hydrogen ion ( H+) concentration and the hydroxide ion (OH) concentration in tomato juice having a pH of 4.1. Prepare answer in unit mol/L and mg/L [H+]

pH = -log 4.1 = -log [H+] Therefore, [H+] = antilog -4.1 = 10-4.1 mol/L

pH + pOH = 14 pOH = 14-4.1 = 9.9 pOH = -log [OH-] 9.9 = -log [OH-]

Therefore, [OH-] = antilog -9.9 = 10-9.9 mol/L 41

42

pH mol/L to mg/L – H+ == atomic weight = 1 g/mol ( refer Periodic Table) – OH- == atomic weight = 17 g/mol (refer Periodic Table) – mol/L (conc. ) x g/mol (atomic weight) x 1000mg/1g = mg/L --------------------------------------------------------------------------------------------------------------------------

Therefore, [H+]

= 10-4.1 mol/L = 10-4.1 mol/L x g/mol (1) x 1000mg/1g = 0.794 mg/L

[OH-]

= 10-9.9 mol/L = 10-9.9 mol/L x g/mol (17) x 1000mg/1g = 21.4 x 10-6 mg/L 43

Hardness - Stream water hardness is the total concentration of cations, specifically calcium (Ca2+ ),magnesium (Mg2+), iron (Fe2+), manganese (Mn2+) in the water. - Water rich in these cations is said to be ‘hard’. Stream water hardness reflects the geology of the catchment area. - Sometimes it also provides a measure of the influence of human activity 44

Hardness • For instance, acid mine drainage often results in the release of iron into a stream. The iron produces extraordinarily high hardness which is a useful water quality indicator. • Hardness is a reflection of the amount of calcium and magnesium entering the stream through the weathering of rock such as limestone (CaCO3). 45

Hardness: Carbonate hardness • Source: Combination of Ca and Mg ions with ions of CO32-, or HCO3-. • These carbonate components can be eliminated by softening methods such as boiling, or by adding lime • When the carbonate components settled then the water have become soft water.

46

Hardness: Carbonate hardness

Ca 2+

Ca(HCO3)2

Mg

2+

CaCO3

+

HCO3-

Mg(HCO3)2

CO32-

MgCO3

47

Hardness: Carbonate hardness • Source: Combination of Ca and Mg ions with ions of CO32-, or HCO3-. • These carbonate components can be eliminated by softening methods such as boiling, or by adding lime • When the carbonate components settled then the water have become soft water.

48

Hardness: Noncarbonate hardness • • •

Source: Combinations of Ca and Mg ions with ions of Cl-, SO42-, or NO3-. The non-carbonate ions cannot be eliminated by ordinary softening methods as done on carbonate hardness. Usually elimination of non carbonate hardness is done chemically by adding softening soda (soda ash or sodium carbonate)

49

Hardness: Carbonate hardness

Ca 2+

Mg 2+

+

CaCl2

CaSO4

Ca(NO3)2 MgCl2

Cl-

SO42-

MgSO4

NO3-

Mg(NO3)2

50

Hardness: Total Hardness • Because calcium and magnesium predominate, it is often convenient to define total hardness as the sum of Calcium and Magnesium elements.

• Therefore Total Hardness as CaCO3 : Total Hardness = Ca2+ + Mg2+

• 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))

51

Hardness • Water hardness is stated in equivalent unit of CaCO3 • Hardness classification: Hardness Soft Medium Hard Very Hard

Concentration Range (mg/L eq CaCO3) <50 50-150 150-300 >300 52

Hardness Example 1 Find the equivalent weight (EW) of each of the following: i) Ca2+, ii) CO32-, and iii) CaCO3.

Solution: Equivalent weight was defined as

EW = [atomic or molecular weight] / [n] (valence) units: grams/equivalent (g/eq) or milligrams/milliequivalent (mg/meq) i) For calcium, n=2 (valence or oxidation state in water). Atomic weight = 40.08 ( refer Periodic table) Therefore the EW is then EW of Ca2+ = 40.08/2 = 20.04 g/eq or mg/meq 53

Hardness ii) For carbonate ion (CO32-), the oxidation state of 2- is used for n since the base CO32- can potentially accept 2H+. The molecular weight (MW) is 60.01. Therefore, EW of CO32- = 60.01/2 = 30 g/eq or mg/meq iii) In CaCO3, n=2 since it would take 2H+ to replace the cation (Ca2+) to form carbonic acid, H2CO3. The MW of CaCO3 is 100. Therefore, EW of CaCO3 = 100/2 = 50 g/eq or mg/meq

54

Hardness Example 2 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: Since; mg/L of X as CaCO3 1.

=

concentration of X (mg/L)  (50 mg CaCO3/meq) (equivalent weight of X (mg/meq))

Convert Ca2+ and Mg2+ to mg/L as CaCO3 Ca2+ : MW = 40, n=2, EW=40/2 = 20 g/eq or mg/meq Mg2+ : MW = 24.3, n=2, EW=24.3/2 = 12.2 g/eq or mg/meq

55

Hardness 2.

Now, find the mg/L as CaCO3 of cations

Use formula; mg/L of X as CaCO3

=

concentration of X (mg/L)  (50 mg CaCO3/meq) (equivalent weight of X (mg/meq))

=>Ca2+ = 100 x 50 = 250 mg/L as CaCO3 20 =>Mg2+ = 10 x 50 = 41 mg/L as CaCO3 12.2 Total Carbonate Hardness = Ca2+ + Mg2+ = 250 + 41 = 291 mg/L as CaCO3 56

Hardness Example 3 From the water analysis below determine the total hardness of the water sample given the following water composition; Ca2+ = 95.2 mg/L HCO3 = 241.49 mg/L (50/(40/2)) 2+ 2Mg = 13.44 mg/L SO4 = 53.77 mg/L Na+ = 25.76 mg/L Cl- = 67.81 mg/L Solution:

Ion

mg/L as ion

EW CaCO3/ EW ion

mg/L as CaCO3

Ca2+

95.20

2.50

238.00

Mg2+

13.44

4.12

55.37

Na+

25.76

2.18

56.16

HCO3

241.46

0.82

198.00

53.77

1.04

55.92

67.81

1.41

95.61

SO4 Cl-

2-

Total Hardness (TH) = 293.37 mg/L as CaCO3 Carbonate Hardness (CH) = 198.00 as CaCO3 Non-carbonate Hardness (NCN) = TH-CH = 95.37 mg/L as CaCO3

EW CaCO3 = 50 mg/meq EW ion = Atomic wt valence

(50/(24.3/2))

(50/61)

(50/(96/2))

57

Alkalinity • Alkalinity is defined as the quantity of ions in water that will react to neutralise hydrogen ions (H+). Alkilinity is thus the measure of the ability of water to neutralise acids. • Therefore alkalinity is the buffer capacity of the water to remain neutral. • The carbonate species that contribute to alkalinity are: – Hydroxyl ions (OH-), Carbonate ions (CO32-), Bicarbonate ions (HCO3-), Alkalinity (mol/L) = [HCO3-] + 2[CO32-] + [OH-] – [H+] Alkalinity (mg/L as CaCO3) = (HCO3-) + (CO32-) + (OH-) – (H+) 58

Alkalinity • Is measured to determine the ability of a stream to resist changes in pH. • Alkalinity results from the dissolution of calcium carbonate (CaCO3) from limestone bedrock which is eroded during the natural processes of weathering • Alkalinity values of 20-200 ppm are common in freshwater ecosystems. Alkalinity levels below 10 ppm indicate poorly buffered streams. • In large quantities, alkalinity imparts bitter taste to water. • Reactions that can occur between alkaline water and certain ions resulting in precipitation of solids which can foul pipes and other water system appurtenances. 59

Alkalinity Example 1 A sample of water having a pH of 7.2 has the following concentrations of ions: Ca2+ 40 mg/L Mg2+ 10 mg/L Na+ 11.8 mg/L K+ 7.0 mg/L HCO3110 mg/L SO4267.2 mg/L Cl11 mg/L Calculate the TH, CH, NCH, Alkalinity, and construct a bar chart of the constituents. 60

Alkalinity Solution Ion

Conc. M.W. mg/L

n

mg/mole

Eq. Wt.

Conc.

Conc.

mg/meq

meq/L

mg/L as CaCO3

Ca2+

40.0

40.1

2

20.05

1.995

99.8*

Mg2+

10.0

24.3

2

12.15

0.823

41.2

Na+

11.8

23.0

1

23.0

0.510

25.7

K+

7.0

39.1

1

39.1

0.179

8.95

HCO3-

110.0

61.0

1

61.0

1.800

90.2

SO42-

67.2

96.1

2

48.05

1.400

69.9

Cl-

11.0

35.5

1

35.5

0.031

15.5

61

Alkalinity Sample Calculation: Conc in meq/L ( refer to Ca2+) meg/L = mg/L x meq/mg = 40.0 /20.05 = 1.995 Concentration of Ca2+ in mg/L as CaCO3 = (Concentration in meq/L)*(Equivalent Weight of CaCO3) = (1.995 meq/L) X (50 mg/meq) = 99.8 mg/L as CaCO3 Check Solution (Cation) s = (Anion)s 175.6 = 175.6 Note: to within  10% mg/L as CaCO3 Total Hardness =  of multivalent cations = (Ca2+) + (Mg2+) = 99.8 + 41.2 = 141 mg/L as CaCO3

62

Alkalinity •

Alkalinity = (HCO32-) + (CO32-) + (OH-) - (H+) Alkalinity  (HCO32-) = (1.80 x 10-3) eq/L Alkalinity = (1.80 x 10-3 eq/L)(50 g/eq)(1000 mg/g) = 90.1 mg/L as CaCO3

•

Carbonate Hardness (the portion of the hardness associated with carbonate or bicarbonate) Alkalinity = 90.1 mg/L as CaCO3 TH = 141 mg/L as CaCO3 CH = 90.1 mg/L as CaCO3 (Note: if TH > Alk then CH = Alk; and NCH = 0)

•

Non-carbonate Hardness NCH = TH - CH = 141 - 90.1 = 50.9 mg/L as CaCO3

63

Alkalinity Bar chart

64

ThOD (Theoretical oxygen demand )

Oxygen Demand

Measurements in form of i) Biochemical oxygen demand ii) Chemical oxygen demand iii) Nitrogenous oxygen demand 65

Oxygen Demand ThOD – theoretical oxygen demand (i) It is the amount of O2 required to oxidize a substance to CO2 and H2O (ii) Calculated by stoichiometry if the chemical composition of the substance is known (iii) The ThOD of X in mg/L  Molecular weight of O 2 in grams     Amount of X in mg/L  MW of X in grams   66

Oxygen Demand Example 1 Compute the ThOD of 108.75 mg/L of glucose (C6H12O6) • STEPS: (i) write balanced equation for the reaction (ii) Compute the grams molecular weights of the reactants (iii) Determine ThOD 67

Oxygen Demand The balanced equation for the reaction C6H12O6 + 6O2  6CO2 + 6H2O The molecular weights (grams , refer Periodic Table) of the reactants Glucose: 6C=72, 12H=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

 192 g of O 2    108.75 mg/L of glucose  180 g of glucose   116 mg/L O 2 68

Oxygen Demand Example 2 If concentration of glucose is 10 mg/L what is the theoretical oxygen demand (amount of DO required?)    mmol O mg O 2 2   6    32 mg O 2  mg glucose   mmol glucose  mmol O 2    10.7 10   mg glucose L L   180    mmol glucose  

69

Biochemical Oxygen Demand (BOD) • It is a measure of the quantity of oxygen used by microorganisms (eg.aerobic bacteria) in the oxidation of organic matter. • In other words: BOD measures the change in dissolved oxygen concentration caused by the microorganisms as they degrade the organic matter. • High BOD is an indication of poor water quality 70

Biochemical Oxygen Demand (BOD) • The term “decomposable” may be interpreted as meaning that the organic matter can serve as food for bacteria, and energy is derived from its oxidation. • The standard oxidation (or incubation) test period for BOD is 5 days at 20C or stated as a BOD5 and can be referred by using the Standard Methods for the Examination for Water and Wastewater (5210 B. 5- Day BOD Test). 71

Biochemical Oxygen Demand (BOD) Apply Eqn C to measure BOD5 for the sample (unseeded BOD test). BOD5 mg/L = Di – Df P

( Eqn C)

Where Di = initial DO ( mg/L) Df = final DO (mg/L) P = Volume of sample used / Total volume

Note: 1)DO in blanks should not deplete more than 0.2 mg/L 2)DO in sample should not less than 2 mg/L 3)Final DO in the samples should not less than 1 mg/L 72 –

Biochemical Oxygen Demand (BOD) A BOD measurement is to carried out for wastewater sample. The wastewater sample inserted to BOD bottle is 10 mL. The 300 mL BOD bottle is filled up with dilution water. Calculate the BOD5 for the wastewater by using the given data: Concentration of DO (mg/L) Day 1 Day 5 7.5 5.0 Solution: For each test bottle meeting the 2.0 mg/L minimum DO depletion and the 1.0 mg/L residual DO, then the; –

BOD5 , mg / L  

D1  D2 P 7.5  5.0 (10 / 300)

 75mg / L

73

Biochemical Oxygen Demand (BOD): BOD Rate Reaction Example 1 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)?



BODt  L0 1  e kt 1



BODt  e  kt L0

 BODt    kt ln 1  L0  

 BODt    ln 1  L0   k t

 102 mg/L    ln 1  158 mg/L   k 5 day

k  0.21 day -1  50 mg/L    ln 1   90 mg/L  k 8 day 74

Biochemical Oxygen Demand (BOD): BOD Rate Reaction • Temperature dependence of BOD rate reaction ! As temperature increases, metabolism increases, utilization of DO also increases kt = k20T-20  = 1.135 if T is between 4 - 20 oC  = 1.056 if T is between 20 - 30 oC 75

Biochemical Oxygen Demand (BOD): BOD Rate Reaction Example 1

The BOD rate constant, k, was determined empirically to be 0.20 days-1 at 20 oC. What is k if the temperature of the water increases to 25 oC? What is k if the temperature of the water decreases to 10 oC?





k25  0.20 day -1 (1.056) 2520

k25  0.26 day -1





1020

k10  0.20 day (1.135) -1

k10  0.056 day

-1 76

Biochemical Oxygen Demand (BOD): BOD Rate Reaction Graphical determination of BOD rate Constants According 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, equation (1) can be transformed to: (t/BODt)1/3=(kLo)-1/3 + (k)2/3/6(Lo)1/3 (t)

•

....(2)

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

A = (kLo)-1/3

….(3)

77

Biochemical Oxygen Demand (BOD): BOD Rate Reaction a slope is defined by:

B = (k)2/3/6(Lo)1/3

y=(t/BOD) 1/3

a

….(4)

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

b

C =(kLo)-1/3 x=t

• Recalled , y = c + mx 78

Biochemical Oxygen Demand (BOD): BOD Rate Reaction • 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/(6A2B) …(6)

79

Biochemical Oxygen Demand (BOD) Example 1 The following data were obtained from an experiment to determine the BOD rate constant and ultimate BOD for an untreated wastewater: Time (day)

2

4

6

8

BOD 125 200 220 230 (mg/L) Solution: 1. Calculate values of (t/BODt)1/3 for each day.

10 237

Time (day)

2

4

6

8

10

BOD (mg/L)

125

200

220

230

237

0.252

0.271

0.301

0.326

0.348

(t/BODt)1/3

80

Biochemical Oxygen Demand (BOD)

Y = (t/ BODt)1/3

2. Plot (t/BODt)1/3 versus t on graph paper and draw the line of best fit by eye. 3. Determine the intercept (A) and slope (B) from the plot.

0.5 0.4

Slope, B= (0.348-0.224)/(10-0) =0.0125

0.3 0.2

A= 0.224

0.1 0

2

4

6

8

10

Time (t)

81

Biochemical Oxygen Demand (BOD) 4. Calculate k and Lo using the following formula k = 6 (B/A)

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

k = 6(B/A) = 6(0.0125/0.224) = 0.335 / day Lo = 1/(6A2B) = 266 mg/L 82

Chemical Oxygen Demand (COD) • It is based on the fact that all organic compounds, with a few exception, can be oxidized by the action of strong oxidizing agents under acid condition (=Value usually about 1.25 times BOD) • Chemical oxygen demand is to measure the oxygen requirement of a sample that is likely to oxidation by strong chemical oxidant ( potassium dichromate). 83

Chemical Oxygen Demand (COD) • Organic matter in aqueous samples could be determined by oxidation with dichromate. • The amount of O2 that is chemically equivalent to the dichromate consumed is defined as the COD of the sample. • During the oxidation in which sample is heated with the known excess of dichromate, organic matter is converted to carbon dioxide and water while dichromate is reduced to Cr 3+. • The excess dichromate is determined by means of an oxidation-reduction titration with ferrous ammonium sulfate (FAS). 84

Nitrogenous Oxygen Demand • The total BOD of a wastewater is composed of two components – a carbonaceous oxygen demand and a nitrogenous oxygen demand. • Traditionally, because of the slow growth rates of those organisms (nitrosomonas and nitrobacter) that exert the nitrogenous demand, it has been assumed that no nitrogenous demand is exerted during the 5-day BOD5 test. 85

Nitrogenous Oxygen Demand

86

Nitrogenous Oxygen Demand • Nitrification (2 step process) 2 NH3 + 3O2  2 NO2- + 2H+ + 2H2O (nitrosomonas) 2 NO2- + O2  2 NO3- (nitrobacter)

– Overall reaction: NH3 + 2O2  NO3- + H+ + H2O

• Theoretical NBOD = grams of oxygen used 4 x 16   4.57 g O 2 /g N grams of nitrogen oxidized 14

87

Nitrogenous Oxygen Demand Untreated domestic wastewater ultimate-CBOD = 250 - 350 mg/L ultimate-NBOD = 70 - 230 mg/L Ultimate NBOD  4.57 x TKN TKN = is the sum of ammonia-nitrogen (NH3-N) plus organic nitrogen ( protein and urea) = 15 - 50 mg/L as N (Total Kjeldahl Nitrogen = TKN) 88

Nitrogen • Nitrogen is often the limiting nutrient in ocean waters and some streams • Nitrogen can exist in numerous forms, but nitrate (NO3-), nitrite (NO2-), ammonia (NH3) are most commonly measured • Sources are primarily from fertilizers and acid deposition

89

Nitrogen • Human intervention in the nitrogen cycle 1. Additional NO and N2O in atmosphere from burning fossil fuels; also causes acid rain 2. N2O to atmosphere from bacteria acting on fertilizers and manure 3. Destruction of forest, grasslands, and wetlands 4. Add excess nitrates to bodies of water 5. Remove nitrogen from topsoil 90

Nitrogen Cycle in a Terrestrial Ecosystem with Major Harmful Human Impacts

Phosphorus • Cycles through water, the earth’s crust, and living organisms • Limiting factor for plant growth • Impact of human activities 1. Clearing forests 2. Removing large amounts of phosphate from the earth to make fertilizers 3. Erosion leaches phosphates into streams

92

Phosphorus Cycle with Major Harmful Human Impacts

• Nitrogen & Phosphorus global concern Eutrophication

BIOLOGICAL PARAMETERS

Microbial Indicator

Macro-invertebrate indicators

95

Microbial indicators - Microorganisms can be used as indicators of the presence of pathogens or infectious agent that cause illness to aquatic or terrestrial ecosystem health. - The most used indicator: Microbial indicator

Origin

Total coliforms

Soil, plant

Fecal coliforms

fecal

E. coli

Fecal of human and

Indicator for Finished drinking water Recreational water Class 1: Primary contact recreational 200 cf/ 100 mL Class 1: Secondary contact recreational 2000 cf/100 mL 96 Freshwater recreational

Microbial indicators - Bacteria counting is a simple method to examine the microbial status of a sample Microbial indicator

Origin

Total coliforms

Soil, plant

Fecal coliforms

fecal

E. coli

Fecal of human and warm-blood animal

Indicator for Finished drinking water Recreational water Class 1: Primary contact recreational 200 cf/ 100 mL Class 1: Secondary contact recreational 2000 cf/100 mL Freshwater recreational

97

Microbial indicators Pathogenic Organisms • Many organims that cause human or animal diseases colonize the instinal tract but can live for a period of time outside the body • Carriers (who may or may not exhibit disease symptoms) excrete these intestinal tract organisms in very large numbers • When water is contaminated by excreta, the organisms can be transmitted to those who contact the water 98

99

100

101

Macro-invertebrate indicators - Macroinvertebrate are useful and convenient indicators of the ecological health of a waterbody or river. They are almost always present, and are easy to sample and identify - Benthic refers to the bottom of a waterway. Example of benthic macro-invertebrates include insects in their larval or nymph form, crayfish, claims, snails and worms. - Most live part or most of their life cycle attached to submerged rocks, logs and vegetation. 102

Macroinvertebrata

104

PART 4 WATER SAMPLING (RIVER MONITORING)

105

Sampling Preparations • Sample bottles, preservatives, labels and marker pens • Sample storage/transit containers and ice packs • Filtering apparatus (if required) • Samplers/sampling equipment • Rubber boots, waders, etc. • Standard operating procedures for sampling • Spares of all above items if possible and when appropriate 106

For documentation • Pens/wax crayons • Sample labels • Field notebook For on-site testing • List of analyses to be performed on site

107

COLLECTION &PRESERVATION OF WATER SAMPLES Standard Methods for the Examinations of Water and Wastewater

108

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques

Sequence of Sampling Matrices and Analytes • Project deals with multimedia and/or multiple parameters use following sequence: – Collect from least to most contaminated sampling locations – If sediment and water is being collected, collect water first to minimize effects from suspended bed materials – For shallow streams, start downstream and work upstream to minimize sediment effects due to sampling disturbances – If sampling at different depths, collect surface samples first and then proceed deeper – Always collect VOCs first, followed by SVOCs (e.g. pesticides, PCBs, oil, etc.), then total metals, dissolved metals, microbiological samples, and inorganic nonmetals

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques

Sample Amount • Minimum sample required depends on the concentration of the analytes present • Should take enough for all analyses and additional for any QA/QC work required • Heterogeneous samples generally require larger amounts to be representative of sample variations • Taking too much sample can lead to problems with storage and transportation

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques

Sample Amount - Water • 5 mL for total petroleum hydrocarbons (TPHs), 100 mL for metals, 1 L for trace organics (pesticides) • As a general rule the minimum volume collected should be 3-4 times the amount required for analysis (EPA, 1995)

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques

Sample Preservation and Storage • Purpose – minimize physical, chemical and biological changes • 3 approaches: – Refrigeration – Use of proper sample container – Addition of preserving chemicals

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques Sample Preservation and Storage •

Refrigeration is a universally accepted method to slow down loss processes

•

Container choice (material type and headspace) is critical to reduce – Volatilization – Adsorption – Absorption – Diffusion – Photodegradation

•

Addition of preservatives is critical to reduce losses due to chemical reactions and bacterial degradation

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques

Sample Preservation and Storage

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques

Sample Preservation and Storage • Maximum Holding Time (MHT) is the length of time a sample can be stored after collection and prior to analysis • Immediate: pH, temperature, salinity, DO • Within 1-2 days: careful pre-planning is required to avoid sampling on Friday, Saturday or near holidays

Environmental Sampling Techniques General Guidelines of Environmental Sampling Techniques Selection of Sample Containers Water • Glass vs. Plastics: – Glass may leach boron and silica, metals may stick to walls – Glass is generally used for organics and plastic for metals, inorganics and physical properties – For trace organics cap and liner should be made of inert materials (teflon) •

Headspace vs. no Headspace: – No headspace is allowed for VOC samples/DO – 40 mL vial with a teflon-lined septum – Oil and grease should only be half-filled in wide mouthed glass bottles

•

Special containers: – e.g. BOD/DO bottles and VOC vials

117

118

PART 5 DO SAG CURVE

120

Water purification system and DO sag curve • The concentration of DO in river is an indicator of the general health of the river. • All river have CAPACITY for water purification (self purification) 1.

2.

As long as the discharge of oxygen demanding wastes is well within the self purification capacity, the DO level remain HIGH and a diverse population of plants and animals As the amount of waste increase, the self purification capacity can be exceeded, causing detrimental changes in plant and animal life

121

Water purification system and DO sag curve 3. 4. 5. 6.

then, the stream losses its ability to clean itself and the DO level DECREASES. when the DO drops below 4 to 5 mg/L, most game fish will have been driven out. If the DO is completely removed, fish and other higher animals are killed or driven The water become blackish and foul smelling as the sewage and dead animal life decompose under anaerobic condition (without O2)

122

Dissolved Oxygen Depletion

(From: Environmental Science: A Global Concern, 3rd ed. by W.P Cunningham and B.W. Saigo, WC Brown Publishers, © 1995) 123

Dissolved Oxygen Sag Curve

124

Mass Balance Approach • Originally developed by H.W. Streeter and E.B. Phelps in 1925 • River described as “plug-flow reactor” • Mass balance is simplified by selection of system boundaries • Oxygen is depleted by BOD exertion • Oxygen is gained through reaeration 125

Steps in Developing the DO Sag Curve 1. Determine the initial conditions 2. Determine the reaeration rate from stream geometry 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 126

Selecting System Boundaries

127

Initial Mixing

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

128

1. Determine Initial Conditions a. Initial dissolved oxygen concentration Qw DOw  Qr DOr DO  Qw  Qr

b. Initial dissolved oxygen deficit Da  DOs  DO

where D = DO deficit (mg/L) DOs = saturation DO conc. (mg/L)

Qw DOw  Qr DOr D  DOs  Qmix

129

1. Determine Initial Conditions DOsat is a function of temperature. Values can be found in Table A-3. c. Initial ultimate BOD concentration

Qw Lw  Qr Lr La  Qw  Qr

130

Table A-3: Saturation values of DO in freshwater exposed to saturated atmosphere containing 20.9% O2 under a pressure of 101.325kPa (Davis & Cornwell, 2013)

131

2. Determine Re-aeration Rate a. O’Connor-Dobbins correlation 1/ 2

3.9u kr  3 / 2 h where kr = reaeration coefficient @ 20ºC (day-1) u = average stream velocity (m/s) h = average stream depth (m) b. Correct rate coefficient for stream temperature

k r  k r , 20

T  20

where Θ = 1.024 132

Determine the Deoxygenation Rate a. rate of deoxygenation = kdLt where kd = deoxygenation rate coefficient (day-1) Lt = ultimate BOD remaining at time (of travel downdstream) t b. If kd (stream) = k (BOD test)

Lt  L0e

 kd t

and

rate of deoxygentation  kd L0e kd t 133

3. Determine the Deoxygenation Rate c. However, k = kd only for deep, slow moving streams. For others,

u kd  k   h

where η = bed activity coefficient (0.1 – 0.6) d. Correct for temperature

k r  k r , 20

T  20

where Θ = 1.135 (4-20ºC) or 1.056 (20-30ºC) 134

4. DO as function of time • Mass balance on moving element dD  kd Lt  kr D dt • Solution is





 

k d La  kd t Dt  e  e kr t  Da e  kr t kr  kd

135

5. Calculate Critical time and DO  kr  k r  k d  1  tc  ln  1  Da kr  kd  kd  kd La 





k d La  kd tc  k r tc  k r tc Dc  e e  Da e kr  kd

136

Example • A discharges 37.0 cfs of treated sewage having an ultimate BOD of 28.0 mg/L and 1.8 mg/L DO into a river with a flow of 250 cfs and velocity of 1.2 ft/s. Upstream of the discharge point, the river has a BOD of 3.6 mg/L and a DO of 7.6 mg/L. Determine the critical DO and critical distance, and Given : The saturation DO (DOs) = 8.5 mg/L, kd = 0.61 day-1, kr = 0.76 day-1. 137

1. Determine Initial Conditions a.

Initial dissolved oxygen concentration

Qw DOw  Qr DOr DO  Qw  Qr

 37 1.8  250 7.6 mg DO   6.85 37  250

b.

L

Initial dissolved oxygen deficit

Da  DOs  DO

mg Da  8.50  6.85  1.65 L

138

1. Determine Initial Conditions c. Initial ultimate BOD concentration

Qw Lw  Qr Lr La  Qw  Qr

La

 37 (28)  250 3.6 mg   6.75 37  250

L

139

2. Determine Re-aeration Rate • kr = 0.76 day-1 given • no need to calculate from stream geometry • assume given value is at the stream temperature (since not otherwise specified), so no need to correct

140

3. Determine the Deoxygenation Rate • kd = 0.61 day-1 given • no need to calculate corrections from stream geometry • assume given value is at the stream temperature (since not otherwise specified), so no need to correct

141

5. Calculate Critical time (tc)  kr  k r  k d  1  tc  ln  1  Da kr  kd  kd  kd La   0.76  1 0.76  0.61  1  1.65  tc  ln  0.616.75  0.76  0.61  0.61 

tc  1.07 day 142

5. Calculate critical distance (as asked) ft  s  hr  1 mi   xc  1.07 d1.2  3600  24   s  hr  d  5280 ft  

xc  20.9 mi

143

5. Calculate DO critical (Doc) (as asked)





k d La  kd tc DOc  e  e  kr tc  De kr tc kr  kd DOc

 0.616.75 0.611.07 0.761.07 e  1.65e0.761.07  e 0.76  0.61

mg DOc  2.85 L

144

6. Calculate DO deficit (Da) during critical

Da  DOs  DOc

mg D  8.5  2.83  5.67 L

145

Try this!! 1. The town of State College discharges 17,630m3/d of treated wastewater into the Bald Eagle Creek. The treated wastewater has a BOD5 of 12 mg/L and a k of 0.12d-1 at 20C. Bald Eagle creek has a flow rate of 0.43 m3/s and an ultimate BOD of 5.0 mg/L. the DO of the river is 6.5 mg/L and the DO of the wastewater is 1.0 mg/L. the stream temperature is 10C and the wastewater temperature is 10C. Compute the DO and initial BOD after mixing. 2. Calculate the initial deficit of the Bald Eagle Creek after mixing with the wastewater from the town of State College. The stream temperature is 10C and the wastewater temperature is 10C. 3. Determine the deoxygenation rate constant for the reach of Bald Eagle Creek below the wastewater outfall (discharge pipe). The average speed of the stream flow in the creek is 0.03 m/s. the depth is 5.0 m and the bed-activity coefficient is 0.35. 4. Determine the DO concentration at a point 5km downstream from the State College discharge into the Bald Eagle Creek. Also determine the critical DO and the distance downstream at which it occurs. 146

END 147