CIVE 350Water Quality Management in Rivers & Lakes Pascal E. Saikaly Department of Civil and Environmental Engineering American University of Beirut Outline • Water Supply & Usage • Water Pollutant and Sources • Water Quality Management in Rivers • Effect of oxygen-demanding waste on rivers • Measures of oxygen demand • Biochemical oxygen demand – Carbonaceous BOD – Nitrogenous BOD • Laboratory test for carbonaceous BOD • DO sag curve • Effects of nutrients on water quality in rivers Water Supply Subsystem • Two major sources to supply community and industrial needs • Surface water (e.g. Streams, lakes, and rivers) • Groundwater (pumped from wells) Water supply resource system Water Supply Subsystem • Municipal water demand: • Domestic: water furnished to houses, hotels, etc. • Commercial and industrial: e.g. factories, offices, and stores. • Public use: water furnished to public buildings (e.g. city buildings, schools, flushing streets, fire protection. DEMAND Unit of Measure Lpcd = Liters Per Capita Per Day The U.S. total water withdrawal (fresh and saline) for all uses (agricultural, commercial, domestic, mining, and thermoelectric power) for the year 2000 was 5,400 Lpcd (Hutson et al., 2001). The U.S public supply (domestic, industrial, agricultural) was estimated to be 580 Lpcd in 2000 (Hutson et al., 2001). Example • Estimate the per capital daily water withdrawal for public supply in the United States in 2005 (in Lpcd). Use the following population data (McGeveran, 2003) and water supply data (Houston, 2001): Year Population Public supply withdrawal, m 3 /d 1950 1.51E+08 5.30E+07 1960 1.79E+08 7.95E+07 1970 2.03E+08 1.02E+08 1980 2.27E+08 1.29E+08 1990 2.49E+08 1.46E+08 2000 2.81E+08 1.64E+08 Example Estimated = 575 Lpcd The average per-capita household water use in the U. S. is about 400 liters per day. Water demand (Lpcd) by sector in Lebanon 1990 1994 2015 Lpcd % Lpcd % Lpcd % Agriculture 599 72 651 74 1164 60 domestic 186 22 140 16 616 32 Industry 45 5 89 10 164 8 Total 829 100 880 100 1945 100 Source: Lebanon State of the Environment Report. Lebanese Ministry of Environment, http://www.moe.gov.lb/Reports/SOER2001.htm, 2001. Fresh water use in the United States, 1990 Water Usage Water quality management is concerned with the control of pollution from human activity so that the water is not degraded to the point that is not suitable for intended uses (e.g. drinking, recreation, agriculture). Water that has been withdrawn, used for some purpose, and then returned will be polluted in one way or another: •Agricultural return water: pesticides, fertilizers, and salts •Municipal return water: human sewage, pharmaceuticals, and surfactants. •Power plant: discharges water that is elevated in temperature. •Industry: chemical pollutants and organic wastes. Pollutants also enter water from: •Natural sources: e.g. Arsenic from natural mineral deposits. •Human sources via nonaqueous routes: e.g. mercury in water is deposited from the air (coal combustion). Water Pollutants & Sources To know how much waste can be tolerated by a water body (lakes, rivers, ponds, and streams), you must know the type of pollutants discharged Point sources: collected by a network of pipes or channels and conveyed to a single point of discharge into the receiving water. Non-point sources: multiple discharge points. The polluted water flows over the surface of the land or along drainage channels to the nearest water body. Oxygen-Demanding Wastes One of the most important measures of the quality of a water is the amount of dissolved oxygen (DO) present. Oxygen-demanding wastes: are substances that oxidize in the receiving water body with the consumption of DO. The waste material is normally biodegradable organic matter (e.g. municipal wastewater) , certain inorganic compounds, and naturally occurring organic matter (e.g. leaves, animal droppings) Bacteria + waste + DO New bacteria + oxidized waste + drop in DO Importance of DO: Higher forms of aquatic life must have DO to live. The critical level of DO varies with species. Trout and salmon: 8 mg/L Bluegill and bass: 5 mg/L Nutrients Nutrients are chemicals, such as nitrogen, phosphorous, carbon, sulfur, calcium, iron, manganese, boron, and cobalt, that are essential to the growth of living things. They are considered as pollutants when their concentrations are sufficient to allow excessive growth of some organisms particularly algae. Nitrogen: sources include municipal wastewater discharge, runoff from animal feedlots, chemical fertilizers, and nitrogen deposition from the atmosphere, especially in the vicinity of coal-fired power plants. Seawater is often limited by nitrogen. NO 3 can pose a serious threat when found in drinking water. Bacteria convert nitrate to nitrite in the alkaline digestive tract of infants. Hemoglobin is oxidized by nitrite to methemoglobin, which cannot carry oxygen. This causes a bluish discoloration of the infant: “blue baby syndrome” Nutrients Phosphorous: sources include agricultural runoff in fertilized areas, discharge from animal feedlots, and domestic sewage (human feces + detergents). Most of the phosphorous is from non-point sources. Freshwater is often limited by phosphorous. Nutrient enrichment can lead to blooms of algae, which eventually die and decompose. The process of nutrient enrichment, called Eutrophication, is especially important in lakes. Pathogen = An organism which causes disease Potable = Safe to drink (not necessarily tastes good) Palatable = Pleasing to drink (not necessarily safe) Basic Definitions Pathogens Typical pathogens excreted in human feces: Virus: e.g. Adenoviruses, Eneteroviruses, Hepatitis A virus. Bacteria: e.g. Salmonella typhi (Typhoid fever), Salmonella paratyphi (Paratyphoid fever), Shigella species (dysentery), Vibrio cholera (cholera). Protozoa: e.g. Giardia lamblia, Cryptosporidiumspecies. Helminths (parasitic worms): e.g. Ascaris lumbricoides (Ascariasis). Waterborne diseases: spread by ingestion of contaminated water. Water-based diseases: involve water contact but don’t require ingestion. Water-related diseases (e.g. malaria) : involve a host that depends on water for its habitat (e.g. mosquitoes). Human contact with water is not required Pathogens • 80% of US population turn their tap on every day to drink from a publicly supplied water. They assume the water they drink is safe. • In developing countries, clean water is the exception rather than the rule. • 2005, every 15 sec a child under the age of 5 dies from water-related illness • 17% of the earth’s population don’t have reliable drinking water • 40% of the population do not have access to adequate sanitation • The fact that the US and developed countries have an outstanding water supply record is not an accident Typhoid fever cases per 100,000 population from 1890 to 1935, Philadelphia Slow sand filters Disinfection Untreated Water from rivers Many organisms associated with deaths or serious illness (e.g. typhoid, polio virus) have been eliminated Organisms with the potential to cause sickness and occasionally death in the US are still being found (see figure) Many are Gastrointestinal symptoms (diarrhea, fatigue, cramps) Waterborne disease outbreaks in the US, 1980-1996 Milwaukee, WI Microbiological Characteristics • Water for drinking and cooking must be free of pathogens (bacteria, virus, protozoa, and worms) • Protozoa of concern: Giardia cysyts and Cryptosporidium oocysts (gastrointestinal illness) • Some organisms which cause disease in people originate with the fecal discharge of infected individuals and animals • Techniques for comprehensive bacteriological examination are complex and time consuming • Indicator organisms Why Coliforms Were Selected As An indicator Organism? • Common inhabitants of intestinal tract •Their presence is an indication of fecal contamination • Excreted in large quantities •Easy to culture • Can survive in water for long periods of time, however, they don’t grow effectively • Are relatively easy to culture • Some agencies use E. coli as a better indicator of bacteriological contamination than total coliforms Salts Water accumulates a variety of dissolved solids, or salts, as it passes through soils and rocks on its way to the sea. These salts include cations (sodium, calcium, magnesium, and potassium) and anions (chloride, sulfate, and bicarbonate). Salinity is the list of the concentrations of the primary cations and anions. Total dissolved solids (TDS) is a measure of salinity. Fresh water: TDS < 1,500 mg/L Saline water : TDS > 5,000 mg/L Seawater: TDS ≈ 30,000-34,000 mg/L The concentrations of dissolved solids is an important indicator of the usefulness of water for various applications Drinking water: recommended maximum TDS ≈ 500 mg/L. Suspended Solids Organic and inorganic particles that are carried by the wastewater into a receiving water are termed suspended solids (SS) Colloidal particles that do not settle readily, cause the turbidity found in many surface waters. Toxic Metals Toxic metals: Most metals are toxic. The most important heavy metals in terms of their environmental impacts are mercury, lead, cadmium, and arsenic. The most important route for the elimination of metals after they are inside a person is via the kidney. Metals are totally nondegradable. Mercury vapor is more toxic than liquid mercury (damage to central nervous system). Lead dissolved in blood is transferred to vital organs, including the kidneys and brains, and it readily passes from a pregnant women to her fetus. Children and fetuses are the most at risk since their brains are growing rapidly (can cause permanent brain damage). Toxic Organic Compounds Toxic Organic Compounds: Pesticides. (insecticides, herbicides, rodenticides, and fungicides) Pesticides is used to cover a range of chemicals that kill organisms that human consider undesirable. Toxic Volatile Organic Chemicals (VOC): are among the most commonly found contaminants in groundwater. They are often used as solvents in industrial processes and a number of them are either known or suspected carcinogens or mutagens. Five VOCs are especially toxic, and their presence in drinking water is cause for special concern: vinyl chloride, trichloroethene (suspected carcinogen), tetrachloroethylene, carbon tetrachloride (very toxic if ingested; few milliliters can produce death), 1,2-dichloroethene. The most toxic of the five is vinyl chloride (known human carcinogen). Toxic Organic Compounds H H H H H H H H H H H H PCE TCE cis-DCE trans-DCE Vinyl chloride Ethene Dehalococcoides (PCE to ethene) Emerging contaminants How is it possible to identify and prioritize pollutants of most environmental concern? 1. Environmental persistence 2. Relative toxicity (to humans and other biota) 3. Occurrence frequency and concentration 4. Immediacy of impact Since the 1990s , many previously little-recognized pollutants have become classified as emerging contaminants: contaminants that by meeting some combination of the foregoing attributes warrant particular interest and concern. Much of this trend in identifying and characterizing emerging contaminants impacts depend on improvement in instrumentation, sampling, and analytical techniques. Emerging contaminants Foremost among the emerging contaminants are endocrine disrupting chemicals (EDCs) . New pathogenic organisms have been identified (human adenoviruses). Rapidly increasing incidences of antibiotic-resistant pathogens. Nanoparticles: materials with functionalities at the near-atomic scale. Endocrine-Disrupting Chemicals They alter the normal physiological function of the endocrine system in humans and wildlife either by being or acting like a natural hormone, blocking the action of a natural hormone, or increasing or reducing the production of natural hormones (i.e. interferes with the regulation of reproductive and developmental process in mammals, birds, fish) Pharmaceutical and EDCs Endocrine-Disrupting Chemicals They alter the normal physiological function of the endocrine system in humans and wildlife either by being or acting like a natural hormone, blocking the action of a natural hormone, or increasing or reducing the production of natural hormones (i.e. interferes with the regulation of reproductive and developmental process in mammals, birds, fish) Pharmaceutical and EDCs Thermal Pollution • Large steam-electric power • Nuclear plant If the heat is released into local river or lake, the resulting rise in temperature can adversely affect life in the vicinity of the thermal plume. As water temperature increases: 1. Metabolic rates increases 2. Amount of DO that the water can hold decreases with temperature As T o Demand for O 2 DO Water Quality Management in Rivers Objective: To control the discharge of pollutants so that the quality of water is not degraded to an unacceptable extent below the natural background level. Oxygen-demanding wastes and nutrients have a profound impact on almost all types of rivers. Bacteria + organic waste + DO New bacteria + CO 2 + H 2 O + drop in DO If DO falls below a critical point threat to higher forms of aquatic life Oxygen-Demanding Wastes 1. What are the factors that affect oxygen consumption during the degradation of organic matter? 2. Inorganic nitrogen oxidation 3. How to predict the DO concentration in rivers from degradation of organic matter? Therefore, it is important to determine the amount of O 2 required to oxidize a substance. Oxygen-Demanding Wastes The are several measures of oxygen demand. 1. Theoretical oxygen demand (ThOD) 2. Chemical oxygen demand (COD) 3. Biochemical oxygen demand (BOD) ThOD If the chemical composition of the substance is known then the amount of O 2 required to completely oxidize a particular organic substance may be calculated from stoichiometry. This amount of oxygen is known as the theoretical oxygen demand (ThOD) ThOD has limitations because: – Some of the carbon is incorporated into new cells – When the amount of remaining wastes diminishes, bacteria begin to draw on their own tissue for energy in order to survive (endogenous respiration) – Dead bacteria becomes food for other bacteria Example 1 Calculate the ThOD of 108.75 mg/L of glucose. C 6 H 12 O 6 + 6O2 6CO 2 + 6H 2 O 180 g 192g 264g 108 g It takes 192 g of oxygen to oxidize 180 g of glucose The ThOD of 108.75 mg/L of glucose is (108.75 mg/L)(192 g O 2 /180 g Glucose) = 116 mg/L O 2 . COD Amount of dichromate consumed in the oxidation of inorganic and organic matter. COD is a chemical oxidation (no microorganisms). Does not depend either on the ability of microorganisms to degrade the organic matter or on knowledge of the particular substances in question. Organic matter = slowly biodegradable + nonbiodegradable + easily biodegradable + inert organic matter BOD Amount of oxygen consumed by microorganisms to oxidize the waste aerobically: Carbonaceous oxygen demand (CBOD) and nitrogenous oxygen demand (NBOD). The BOD test is an indirect measure of organic measure because we actually measure only the change in dissolved oxygen concentration caused by the microorganisms as they degrade the organic matter. BOD test is the most widely used method for measuring organic matter because of the direct conceptual relationship between BOD and oxygen depletion in receiving waters. BOD requires microorganisms that consume oxygen in the process of degrading organic matter. Laboratory Test for CBOD Standard BOD test: 1. Special 300 mL bottles that is completely filled with a waste sample that has been appropriately diluted with a certain dilution water that contains an inoculumof microorganisms. The bottle is stoppered to exclude air bobbles. The sample requires dilution because the oxygen demand of a waste is several hundreds mg/L and the maximum DO that can dissolve in water at 20 0 C is about 9 mg/L. Samples are diluted with a special dilution water that contains all the necessary trace elements required for bacterial metabolism. The appropriate sample size to use is determined by dividing 4 mg/L (midpoint of desired range of diluted BOD) by the estimated BOD concentration in the sample being tested size(%) sample 100 sample undiluted of vol. sample diluted of vol. factor Dilution = = Laboratory Test for CBOD Standard BOD test: 2. Blank samples containing only the inoculated dilution water are also placed in BOD bottles and stoppered. Blanks are required to estimate the amount of oxygen consumed by the added inoculum of microorganisms (called seed) in the absence of the sample. 3. The two BOD bottles are incubated in the dark at 20 o C for the desired number of days (normally 5 days, BOD 5 ). It is incubated in the dark to prevent photosynthesis from adding oxygen to the water. Blank: seeded dilution water Seeded dilution water + sample waste Laboratory Test for CBOD Standard BOD test: 4. After the desired number of days has elapsed, the samples and blanks are removed and dissolved oxygen concentration in each bottle is measured. The BOD of the undiluted sample is then calculated using the following equation: where DO b,t = dissolved oxygen concentration in blank after t days of incubation, mg/L DO s,t = dissolved oxygen concentration in sample after t days of incubation, mg/L factor dilution ) DO (DO BOD s,t b,t t × − = Laboratory Test for CBOD Example 2 The BOD of a wastewater sample is estimated to be 180 mg/L. What volume of undiluted sample should be added to a 300 mL bottle? Also, what are the sample size and dilution factor using this volume? Assume that 4 mg/L BOD can be consumed in the BOD bottle. The volume of diluted sample is 300 mL Vol. of undiluted sample = 0.0222 X 300 mL = 6.66 mL. Therefore a convenient sample volume would be 7.00 mL. The actual sample size and dilution factor: 2.22% 100 180 4 size Sample = × = 2.33% 100 300mL 7.00mL size Sample = × = 42.9 7.00mL 300mL factor Dilution = = Example 3 What is the BOD 5 of the wastewater sample of Example 4 if the DO values for the blank and diluted sample after five days are 8.7 and 4.2 mg/L, respectively?. factor dilution ) DO (DO BOD s,t b,t t × − = 193mg/L 42.9 4.2) (8.7 BOD 5 = × − = NBOD Oxygen consumption due to nitrogen oxidation is called nitrogenous BOD (NBOD). Many organic compounds contain proteins, which contains nitrogen, and the nitrogen is released to the surrounding water as ammonia (NH 3 ). At normal pH values, this ammonia is in the form of ammonium cation (NH 4 + ). The ammonia released by organic compounds, plus that from other sources such as industrial wastewater and agricultural runoff (fertilizers), is oxidized to nitrate (NO 3 - ) by a special group of microorganisms called nitrifying bacteria. The process is called nitrification. NH 4 + + 2O 2 + nitrifying bacteria NO 3- + H 2 O +2H + The theoretical NBOD can be calculated as follows: The actual NBOD is slightly less than the theoretical value due to incorporation of some of the nitrogen into new bacterial cells. N /g O g 4.57 14 16 4 oxidized nitrogen of grams used oxygen of grams NBOD 2 = × = = Example 4 Compute the theoretical NBOD of a wastewater containing 30 mg/L of ammonia as nitrogen (we often say “ammonia nitrogen” and write the expression as NH 3 -N). If the wastewater analysis reported as 30 mg/L of ammonia (NH 3 ), what would the theoretical NBOD be? In the first part of the problem, the amount of ammonia was reported as NH 3 -N. Therefore Theo. NBOD = (30 mg N/L)(4.57 mg O 2 /mg N) = 137 mg O 2 /L In the second part, we must convert mg/L of ammonia to NH3-N by multiplying by the ratio of gram molecular weight of N to NH3. N/L mg 24.7 NH 17g gN 14 /L NH mg 30 3 3 = × Theo. NBOD = (24.7 mg N/L)(4.57 mg O 2 /mg N) = 133 mg O 2 /L NBOD vs CBOD The rate at which NBOD is exerted depends heavily on the number of nitrifying organisms present. In untreated sewage, there are few of these organisms, while in a well-treated effluent, the concentration is high. Lag: time it takes for nitrifying bacteria to reach a sufficient population. No Lag: higher population of nitrifying organisms reduces the lag time Lag No Lag CBOD When a water sample containing degradable organic matter is placed in a closed container and inoculated with bacteria, the oxygen consumption typically follows the pattern shown in the figure below. Organic matter remaining Organic matter oxidized CBOD The amount of organic matter remaining in the container will decrease with time. Another way to describe the organic matter in the container is to say as time goes on, the amount of organic matter already oxidized goes up until finally all of the original organic matter has been oxidized Organic matter remaining Organic matter oxidized CBOD The remaining demand for oxygen to decompose wastes decreases with time until there is no more demand, or we could say the amount of oxygen demand already exerted, or utilized, starts at zero until all of the original oxygen demand has been satisfied Organic matter remaining Organic matter oxidized CBOD The translation of the figure into a mathematical form is straightforward. To do so, it is assumed that the rate of decomposition of organic waste is directly proportional to the concentration of degradable organic matter remaining at any time t, (L t ). Assuming a first-order reaction, we can write where k = BOD rate constant, d -1 The solution to the above equation after rearranging and integrating where L 0 = oxygen demand of organic compound at time t = 0 L o is often referred to as the ultimate BOD, that is, the total amount of oxygen required by microorganisms to oxidize the waste completely to carbon dioxide and water. t A t kL r dt dL − = − = kt 0 t e L L − = CBOD Rather than L t , we are interested in the amount of oxygen utilized in the consumption of the organics (BOD t ) . From the figure, BOD t is the difference between L 0 and L t The above equation is called the BOD rate equation and is often written in base 10: ) e (1 L e L L L L BOD kt 0 kt 0 0 t 0 t − − − = − = − = ) (1 L BOD Kt 0 t − − = 10 2.303K k = Example 5 If the BOD 3 of a waste is 75 mg/L and the K is 0.150 d -1 , what is the ultimate BOD? For base 10 For base e ) (1 L BOD Kt 0 t − − = 10 ) (1 L 7 0 ) 3 )( 150 . 0 ( 10 5 − − = 116mg/L L o = 0.345 2.303K k = = ) (1 L BOD kt 0 t − − = e ) (1 L 7 0 ) 3 )( 345 . 0 ( 5 − − = e 116mg/L L o = CBOD The reaction rate constant k indicates how rapidly oxygen will be depleted in a receiving water. As k increases, the rate at which dissolved oxygen is used increases. The numerical value of the rate constant is dependent on the following: 1. The nature of the waste: simple sugars and starches degrade easily while cellulose does not. 2. Ability of the available microorganisms to degrade the waste 3. Temperature: rate of biodegradation increases with increasing T where T = temperature of interest, o C k T = BOD rate constant at the temperature of interest, d -1 k 20 = BOD rate constant determined at 20 o C, d -1 20 20 ) ( − = T θ k k T CBOD Ө = temperature coefficient. This has a value of 1.135 for temperature between 4 and 20 o C and 1.056 for temperatures between 20 and 30 o C (Schroepfer, et al., 1964). Example 6 A waste is being discharged into a river that has a temperature of 10 o C. What fraction of the maximum oxygen consumption has occurred in four days if the BOD rate constant determined in the laboratory under standard conditions (20 o C) is 0.115 d -1 (base e)? 20 20 ) ( − = T θ k k T 1 20 10 C 10 d 0.032 5) 0.115(1.13 k o − − = = ) (1 L BOD kt 0 t − − = e 0.12 ] e [1 L BOD (0.032)(4) 0 4 = − − = NBOD vs CBOD Once nitrification begins, NBOD can be described by the equations used for CBOD with a BOD rate constant comparable to that for CBOD with well-treated effluent (K = 0.04 to 0.10 d -1 ). When measurements of only CBOD is required, chemical inhibitors are added to stop the nitrification process. DO Sag Curve • Introduction • Mass-balance approach • Oxygen deficit • DO sag equation • Deoxygenation rate constant • Reaeration rate constant • Management strategy • Nitrogenous BOD Introduction One of the major tools of water quality management in rivers is the ability to assess the capability of a stream to absorb a waste load. This is done by determining the profile of DO concentration downstream from a waste discharge. This profile is called the DO sag curve. Introduction The biota of the stream are often a reflection of the DO conditions in the stream The DO concentration decreases near the point of discharge of waste as oxygen-demanding material are oxidized. As we move further downstream, less and less organic material remains and the oxygen is replenished from the atmosphere Introduction To develop a mathematical expression for the DO sag curve the following factors need to be considered: 1. Sources of oxygen : reaeration from the atmosphere and photosynthesis of aquatic plants 2. Factors affecting oxygen depletion • CBOD and NBOD • BOD already in the river upstream of the waste discharge 3. DO in the waste discharge is usually less than that in the river • Thus DO in the river is lowered as soon as the waste is discharged (Initial DO reduction) 4. Non-point source pollution 5. Respiration of organisms living in the sediments 6. Respiration of aquatic plants Mass-Balance Approach Simplified mass balances help us understand and solve the DO sag curve problem. A. Two conservative (no chemical rxn) mass balances may be used to account for the initial mixing of the waste stream and the river. 1. Mass balance for DO 2. Mass balance for CBOD DO and CBOD change as the result of mixing of the waste stream and the river. B. Once these are accounted for, the DO sag curve may be viewed as a nonconservative mass balance. Conservative Mass-Balance for DO and CBOD Conservative, no accumulation, and complete and instantaneous mixing: Mass balance for DO: where Q w = volumetric flow rate of wastewater, m 3 /s Q r = volumetric flow rate of the river, m 3 /s DO w = dissolved oxygen concentration in the wastewater, g/m 3 DO r = dissolved oxygen concentration in the river, g/m 3 DO = dissolved oxygen concentration after mixing, g/m 3 Mass of DO and BOD in Wastewater Mass of DO and BOD in River Mass of DO and BOD in River after Mixing out Mass in Mass = )DO Q (Q DO Q DO Q r w r r w w + = + Conservative Mass-Balance for DO and CBOD The concentration of DO after mixing equals: Mass balance for BOD: where L w = ultimate BOD of the wastewater, mg/L L r = ultimate BOD of the river, mg/L L a = initial ultimate BOD after mixing, mg/L The concentration of ultimate BOD after mixing equals: Qr Q DO Q DO Q DO w r r w w + + = a r w r r w w )L Q (Q L Q L Q + = + Qr Q L Q L Q L w r r w w + + = a Example 7 The town of State College discharges 17,360 m 3 /d of treated wastewater into the Bald Eagle Creek. The treated wastewater has a BOD 5 of 12 mg/L and a k of 0.12d -1 at 20 o C. Bald Eagle Creek has a flow rate of 0.43 m 3 /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. Compute the DO and initial ultimate BOD after mixing. Example 7 Solution: Assume complete and instantaneous mixing. The DO after mixing is given by the equation below: Q w is given in m 3 /d, to use the above equation the units of Q w should be in m 3 /s The DO after mixing is then Qr Q DO Q DO Q DO w r r w w + + = /s 0.20m ) (86,400s/d /d) (17,360m Q 3 3 w = = 4.75mg/L /s 0.43m /s 0.20m L) /s)(6.5mg/ (0.43m L) /s)(1.0mg/ (0.20m DO 3 3 3 3 = + + = Example 7 The initial ultimate BOD after mixing is given by the equation below: The ultimate BOD of the wastewater is determined using the following equation: where The initial ultimate BOD after mixing is 11.86mg/L /s 0.43m /s 0.20m L) /s)(5.0mg/ (0.43m /L) /s)(26.6mg (0.20m L 3 3 3 3 a = + + = Qr Q L Q L Q L w r r w w + + = a ) (1 L BOD kt 0 t − − = e 26.6mg/L 0.55) (1 12 ) e (1 12mg/L ) e (1 BOD L (0.12)(5) kt 5 0 = − = − = − = − − Oxygen Deficit The DO sag equation has been developed using oxygen deficit rather than dissolved oxygen concentration, to make it easier to solve the integral equation that results from the mathematical description of the mass balance. The oxygen deficit is given by the following equation: where D = oxygen deficit, mg/L DO s = saturation concentration of dissolved oxygen at the temperature of the river after mixing, mg/L DO = actual concentration of DO, mg/L Initial deficit: The initial deficit is calculated as the difference between saturated DO and the concentration of the DO after mixing where D a = initial deficit after river and waste mixed, mg/L DO DO D s − = r w r r w w s a Q Q DO Q DO Q DO D + + − = DO Sag Curve DO s Values For Fresh Water Example 8 Calculate the initial deficit of the Bald Eagle Creek after mixing with the wastewater from the town of State College (see Example 7). The stream temperature is 10 o C and the wastewater is 10 o C. Solution: with the stream temperature, the saturation value of dissolved oxygen (DO s ) can be determined from the table in the previous slide. At 10 o C, DO s = 11.33 mg/L. Since we calculated the concentration of DO after mixing as 4.75 mg/L from Example 7, the initial deficit after mixing is DO DO D s − = 6.58mg/L 4.75mg/L 11.33mg/L D = − = DO Sag Equation Figure a is a comprehensive mass balance diagram of DO in a small reach (stretch) of river that accounts for all the inputs and outputs. Figure b is a simplified mass balance diagram developed by Streeter-Phelps. The DO sag equation is also known as the Streeter-Phelps equation. DO Sag Equation The mass balance equation for figure b is: where RDO in = mass of DO in river flowing into reach W = mass of DO in wastewater flowing into reach A = mass of DO added from atmosphere M = mass of DO removed by microbial degradation of CBOD RDO out = mass of DO in river flowing out of reach The rate at which DO disappears from the stream as a result of microbial action (M) is exactly equal to the rate of increase in the deficit. With the assumption that the saturation value for DO remains constant [d(DO s )/dt =0], differentiation of the following equation yields: 0 RDO - M - A W RDO out in = + + DO DO D s − = 0 dt dD dt d(DO) = + DO Sag Equation DO Sag Equation and The rate at which DO disappears coincides with the rate that BOD is degraded, so Remember (L 0 is constant) 0 dt dD dt d(DO) = + dt dD dt d(DO) − = dt d(BOD) dt dD dt d(DO) − = − = t 0 t L L BOD − = dt dL dt d(BOD) t − = DO Sag Equation Thus, the rate of change in deficit at time t due to BOD (rate of deoxygenation) is a first order reaction proportional to the BOD remaining after the waste enters the rivers (k d is the deoxygenation rate constant) (k is the BOD rate constant determined in laboratory) dt dL dt d(BOD) t − = t t kL dt dL − = t d L k dt dD = DO Sag Equation The rate oxygen mass transfer into solution from the air (A) (Rate of reaeration) is a first order reaction proportional to the difference between the saturation value and the actual concentration remember and then k r = reaeration rate constant DO) (DO k dt d(DO) s r − = DO) (DO D s − = dt dD dt d(DO) − = D k dt dD r − = DO Sag Equation The deoxygenation caused by microbial decomposition of waste (M) and oxygenation caused by reaeration (A) are competing processes that are simultaneously removing and adding oxygen to a stream Rate of increase of the deficit = rate of deoxygenation – Rate of reaeration where dD/dt = the change in oxygen deficit (D) per unit time, mg/L . d k d = deoxygenation rate constant, d -1 L t = BOD remaining t (days) after waste enters the river, mg/L k r = reaeration rate constant, d -1 D = oxygen deficit in river water after utilization of BOD for time, t, mg/L D k L k dt dD r t d − = DO Sag Equation By integrating and using the initial condition (at t = 0, D = D a ) we obtain the DO sag equation: where L a = initial ultimate BOD after river and wastewater have mixed, mg/L k d = deoxygenation rate constant, d -1 k r = reaeration rate constant, d -1 D a = initial deficit after river and wastewater have mixed, mg/L t = time of travel of wastewater discharge downstream, d When k r = k d D k L k dt dD r d − = ) (e D ) e (e k k L k D t k a t -k t k d r a d r r d − − + − − = ) )(e D tL (k D t k a a d d − + = DO Sag Equation Once D is found at any point downstream, the DO can be found using the equation: Physically it is impossible for the DO to be less than zero. If the deficit (D) is greater than DO s , then all the oxygen was depleted at some earlier time and the DO is zero. ( ¸ ( ¸ + − − − = − − ) (e D ) e (e k k L k DO DO t k a t k - t k d r a d s r r d DO DO D s − = DO Sag Equation Deoxygenation rate constant: (Bosko, 1996) where k d = deoxygenation rate constant at 20 o C, d -1 v = average speed of stream, m/s k = BOD rate constant determined in laboratory at 20 o C, d -1 H = average depth of stream, m η = bed-activity coefficient The bed-activity coefficient may vary from 0.1 for stagnant or deep water to 0.6 or more for rapidly flowing streams. A deep, slowly moving rivers have a much lower k d than a shallow, rapidly flowing river. In general, BOD is utilized more rapidly in a river compared to a BOD bottle because of turbulent mixing and larger numbers of “seed” organisms. k d is affected by temperature and can be adjusted to the river temperature using the equation: η H ν k k d + = 20 20 ) ( − = T θ k k T Example 9 Determine the deoxygenation rate constant for the Bald Eagle Creek (Example 7 and 8) below the wastewater outfall (discharge pipe). The average speed of the stream in the creek is 0.03 m/s. The depth is 5.0 m and the bed-activity coefficient is 0.35. Solution: from Example 7, the value of k is 0.12d -1 Using the equation: The deaoxygenation rate constant at 20 o C is Note that the units are not consistent and the bed-activity coefficient includes a conversion factor to make the second term dimensionally correct. η H ν k k d + = 1 1 d 0.1221d (0.35) 5.0m 0.03m/s 0.12d k − − = + = Example 9 The stream temperature in Example 8 was 10 o C. Thus we must correct the estimated k d using the equation 20 20 ) ( − = T θ k k T 1 20 10 1 o d 0.034d )(1.135) (0.1221d C 10 at k − − − = = DO Sag Equation Reaeration rate constant: (O’Connor and Dobbins, 1958) where k r = reaeration rate constant at 20 o C, d -1 v = average speed of stream, m/s H = average depth of stream, m k r depends on the degree of turbulent mixing and amount of water surface exposed to the atmosphere compared to the volume of water in the river. A narrow, deep river will have a much lower k r than a wide, shallow river. k r is affected by temperature and can be adjusted to the river temperature using the equation: where ө = 1.024 1.5 0.5 r H 3.9ν k = 20 20 ) ( − = T θ k k T DO Sag Equation Critical point: is the lowest point on the sag curve and it is of major interest since it indicates the worst conditions in the river. Therefore, the location of the critical point is of obvious importance. The time to the critical point (t c ) can be found by setting the derivative of the oxygen deficit equal to zero, and solving for t using base e values for k r and k d : When k r = k d ( ( ¸ ( ¸ | | . | \ | − − − = a d d r a d r d r c L k k k D 1 k k In k k 1 t | | . | \ | − = a a d c L D 1 k 1 t DO Sag Curve 1. Near the outfall: the rate of removal of oxygen by bacteria from the water is higher than the rate of reaeration. 2. At the critical point: the rate of removal of oxygen is equal to the rate of reaeration. 3. Beyond the critical point: reaeration begins to dominate and the rate of reaeration is faster than the rate of removal of oxygen by bacteria. DO Sag Equation Example 10 Determine the DO concentration at a point 5 Km downstream from the State College discharge into the Bald Creek (Example 7, 8, and 9). Also determine the critical DO and the distance downstream at which it occurs. Solution: L a , k d , and D a were calculated in previous examples. K r and travel time, t will be calculated as follows: The stream temperature is 10 o C: ( ¸ ( ¸ + − − − = − − ) (e D ) e (e k k L k DO DO t k a t k t k d r a d s r r d 1 1.5 0.5 1.5 0.5 o r 0.0604d (5.0m) m/s) (3.9)(0.03 H 3.9ν C 20 at k − = = = 1 20 10 1 o r 0.04766d )(1.024) (0.0604d C 10 at k − − − = = Example 10 The travel time t is computed from the distance downstream and the speed of the stream : Where x = distance downstream v = average stream velocity t = elapsed time between discharge point and distance x downstream t × =ν x 1.929d 86,400s/d) (0.03m/s)( 0m/km) (5km)(1,00 t = = ( ¸ ( ¸ + − − − = − − ) (e D ) e (e k k L k DO DO t k a t k - t k d r a d s r r d 4.60mg/L ) 6.58(e ) e (e 0.03442 0.04766 11.86) (0.03442)( 11.33 DO 1.929) (0.04766)( 1.929) (0.04766)( - 1.929) (0.03442)( = ( ¸ ( ¸ + − − − = − − Example 10 The critical time is computed using the equation: The critical deficit is calculated using the equation 6.85mg/L ) 6.58(e ) e (e 0.03442 0.04766 11.86) (0.03442)( D 6.45) (0.04766)( 6.45) (0.04766)( - 6.45) (0.03442)( c = ( ¸ ( ¸ + − − = − − ( ( ¸ ( ¸ | | . | \ | − − − = a d d r a d r d r c L k k k D 1 k k In k k 1 t 6.45d 11.86) (0.03422)( 0.03422 0.04766 6.58 1 0.03442 0.04766 In 0.03442 0.04766 1 t c = ( ( ¸ ( ¸ | | . | \ | − − − = ) (e D ) e (e k k L k D c r c r c d t k a t -k t k d r a d c − − + − − = Example 10 The critical DO is then calculated using the equation DO c = 11.33-6.85 = 4.48 mg/L The critical DO occurs downstream at a distance of c s c DO DO D − = t × =ν x 16.7km ) 1,000m/km 1 .03m/s)( ,400s/d)(0 (6.45d)(86 x = = Management Strategy 1. A DO standard is set to protect the most sensitive species that exist or could exist in a particular river. 2. For a known waste discharge and a known set of river characteristics, the DO sag equation can be solved to find the DO at the critical point. 3. If the DO at the critical point is higher than the standard, the stream can adequately assimilate the waste. 4. If the DO at the critical point is less than the standard DO, then additional waste treatment is needed. 5. Environmental engineers and scientists have control over just two parameters, L a and D a . • Improving treatment efficiency or adding additional treatment steps will reduce L a . • Adding oxygen to wastewater to bring it to saturation level before discharge. This will reduce D a . Management Strategy 6. The new values of L a and D a are used to determine if DO standard will be violated at the critical point. 7. When using the DO sag curve to determine the adequacy of wastewater treatment, it is important to use river conditions that will cause the lowest DO concentration. These conditions are: • Late summer when the river flows are low and temperatures are high. • Low rivers flows results in higher values for L a and D a . • k r is reduced more than k d by low river flow because of reduced velocities. • Higher temperatures increase k d more than k r . • Higher temperature reduces DO saturation. Example 11 The Pitts Canning Company is considering opening a new plant at one of two possible locations: the Green River and its twin, the White River. Among the decisions to be made are what effect the plant discharge will have on each river and which river would be impacted less. The effluent parameter and river parameters at summer low flow conditions are available. Example 11 Effluent Parameter Plant A Plant B Flow (m 3 /s) 0.05 Flow (m 3 /s) 0.05 Ultimate BOD at 25 0 C, mg/L 30.00 Ultimate BOD at 25 0 C, mg/L 30.00 DO, mg/L 0.90 DO, mg/L 0.90 Temperature, 0 C 25.00 Temperature, 0 C 25.00 k at 20 o C,d -1 0.12 k at 20 o C,d -1 0.07 River Parameter Green River White River Flow,m 3 /s 0.50 Flow,m 3 /s 0.50 Ultimated BOD at 25 o C, mg/L 19.00 Ultimated BOD at 25 o C, mg/L 19.00 DO, mg/L 5.85 DO, mg/L 5.85 Temperature, o C 25.00 Temperature, o C 25.00 Speed, m/s 0.10 Speed, m/s 0.20 Average depth, m 4.00 Average depth, m 4.00 Bed-Activity coefficient 0.20 Bed-Activity coefficient 0.20 ө (k d ) 1.056 ө (k d ) 1.056 ө (k r ) 1.024 ө (k r ) 1.024 Example 11 Solution (See attached Excel file “Do Sag Curve”) : we have four combinations (A-Green, A-White, B-Green, B-White) 1. Calculate L a (ultimate BOD after mixing), g/L 2. Calculate Initial DO after mixing, mg/L 3. Calculate Initial deficit (D a ), mg/L 4. Calculate deoxygenation rate constant at 20 o C, k d 5. Calculate deoxygenation rate constant at 25 o C, k d 6. Calculate reaeration rate constant at 20 o C, k r 7. Calculate reaeration rate constant at 25 o C, k r 8. Calculate critical time, d 9. Calculate critical deficit, mg/L 10. Calculate critical DO, mg/L Example 11 A-Green A-White B-Green B-White k d 0.1578 0.1643 0.0973 0.1039 k r 0.1736 0.2455 0.1736 0.2455 t c 5.09 4.00 5.96 4.47 D c 8.14 6.93 6.28 5.32 DO c 0.24 1.45 2.10 3.06 Nitrogenous BOD The NBOD can be incorporated into the DO sag equation by adding an additional term to the equation Where k n = nitrogenous deoxygenation coefficient, d -1 L n = ultimate nitrogenous BOD after waste and river have mixed, mg/L ) e (e k - k L k ) (e D ) e (e k k L k D t k t k n r n n t k a t -k t k d r a d r n r r d − − − − − + + − − = Effect of Nutrients on Water Quality in Rivers Effects of nitrogen: 1. In high concentration, NH 3 -N is toxic to fish 2. NH 3 , in low concentrations, and NO 3 - serve as nutrients for excessive growth of algae. 3. The conversion of NH 4 + to NO 3 - consumes large quantities of DO. Effect of phosphorous: Serves as a vital nutrient for the growth of algae. When algae dies, they become an oxygen-demanding organic material for bacteria. This will cause further reduction in the DO supply of water body. Outline • Water Supply & Usage • Water Pollutant and Sources • Water Quality Management in Rivers • Effect of oxygen-demanding waste on rivers • Measures of oxygen demand • Biochemical oxygen demand – Carbonaceous BOD – Nitrogenous BOD • Laboratory test for carbonaceous BOD • DO sag curve • Effects of nutrients on water quality in rivers Water Supply Subsystem • Two major sources to supply community and industrial needs • • Surface water (e.g. Streams, lakes, and rivers) Groundwater (pumped from wells) Water supply resource system . • Commercial and industrial: e. flushing streets. fire protection. city buildings. • Public use: water furnished to public buildings (e.g. hotels.Water Supply Subsystem • Municipal water demand: • Domestic: water furnished to houses. factories. offices. etc. schools. and stores. .g. DEMAND Unit of Measure Lpcd = Liters Per Capita Per Day . agricultural) was estimated to be 580 Lpcd in 2000 (Hutson et al.S public supply (domestic. mining. domestic. 2001). total water withdrawal (fresh and saline) for all uses (agricultural. commercial.. The U. 2001). and thermoelectric power) for the year 2000 was 5..The U. .S. industrial.400 Lpcd (Hutson et al. 2003) and water supply data (Houston.81E+08 Public supply withdrawal. 2001): Year 1950 1960 1970 1980 1990 2000 Population 1.Example • Estimate the per capital daily water withdrawal for public supply in the United States in 2005 (in Lpcd).95E+07 1. m3/d 5.49E+08 2.46E+08 1.79E+08 2.30E+07 7.02E+08 1.51E+08 1.64E+08 . Use the following population data (McGeveran.03E+08 2.27E+08 2.29E+08 1. Example Estimated = 575 Lpcd . S.The average per-capita household water use in the U. . is about 400 liters per day. Lebanese Ministry of Environment.moe. .gov. http://www.htm. 2001.lb/Reports/SOER2001.Water demand (Lpcd) by sector in Lebanon 1990 Lpcd 599 186 45 829 % 72 22 5 100 1994 Lpcd 651 140 89 880 % 74 16 10 100 2015 Lpcd 1164 616 164 1945 % 60 32 8 100 Agriculture domestic Industry Total Source: Lebanon State of the Environment Report. 1990 .Fresh water use in the United States. recreation.g. •Power plant: discharges water that is elevated in temperature.g. Arsenic from natural mineral deposits. mercury in water is deposited from the air (coal combustion). •Industry: chemical pollutants and organic wastes. Water that has been withdrawn.g. . Pollutants also enter water from: •Natural sources: e. pharmaceuticals. and salts •Municipal return water: human sewage. and surfactants.Water Usage Water quality management is concerned with the control of pollution from human activity so that the water is not degraded to the point that is not suitable for intended uses (e. fertilizers. drinking. agriculture). and then returned will be polluted in one way or another: •Agricultural return water: pesticides. used for some purpose. •Human sources via nonaqueous routes: e. The polluted water flows over the surface of the land or along drainage channels to the nearest water body. and streams). ponds. you must know the type of pollutants discharged Point sources: collected by a network of pipes or channels and conveyed to a single point of discharge into the receiving water. rivers. .Water Pollutants & Sources To know how much waste can be tolerated by a water body (lakes. Non-point sources: multiple discharge points. Oxygen-Demanding Wastes One of the most important measures of the quality of a water is the amount of dissolved oxygen (DO) present. The critical level of DO varies with species.g. municipal wastewater) . and naturally occurring organic matter (e. leaves. The waste material is normally biodegradable organic matter (e. Trout and salmon: 8 mg/L Bluegill and bass: 5 mg/L . Oxygen-demanding wastes: are substances that oxidize in the receiving water body with the consumption of DO. animal droppings) Bacteria + waste + DO New bacteria + oxidized waste + drop in DO Importance of DO: Higher forms of aquatic life must have DO to live.g. certain inorganic compounds. Bacteria convert nitrate to nitrite in the alkaline digestive tract of infants. boron. carbon. calcium. and nitrogen deposition from the atmosphere. that are essential to the growth of living things. This causes a bluish discoloration of the infant: “blue baby syndrome” . NO3 can pose a serious threat when found in drinking water. which cannot carry oxygen. iron. runoff from animal feedlots. especially in the vicinity of coal-fired power plants. such as nitrogen. Nitrogen: sources include municipal wastewater discharge. manganese. Seawater is often limited by nitrogen. and cobalt.Nutrients Nutrients are chemicals. chemical fertilizers. They are considered as pollutants when their concentrations are sufficient to allow excessive growth of some organisms particularly algae. Hemoglobin is oxidized by nitrite to methemoglobin. phosphorous. sulfur. . Freshwater is often limited by phosphorous. Nutrient enrichment can lead to blooms of algae. and domestic sewage (human feces + detergents). called Eutrophication. Most of the phosphorous is from non-point sources. is especially important in lakes. The process of nutrient enrichment. which eventually die and decompose.Nutrients Phosphorous: sources include agricultural runoff in fertilized areas. discharge from animal feedlots. Basic Definitions Pathogen = An organism which causes disease Potable = Safe to drink (not necessarily tastes good) Palatable = Pleasing to drink (not necessarily safe) . Salmonella typhi (Typhoid fever). Salmonella paratyphi (Paratyphoid fever). Eneteroviruses. Bacteria: e. malaria) : involve a host that depends on water for its habitat (e. Vibrio cholera (cholera).g.g.g. Giardia lamblia.g. Waterborne diseases: spread by ingestion of contaminated water. Protozoa: e. Water-related diseases (e.g. Adenoviruses. Hepatitis A virus. Ascaris lumbricoides (Ascariasis). Helminths (parasitic worms): e.Pathogens Typical pathogens excreted in human feces: Virus: e. Shigella species (dysentery). Water-based diseases: involve water contact but don’t require ingestion. Human contact with water is not required .g. Cryptosporidium species. mosquitoes). • In developing countries. • • • 2005. They assume the water they drink is safe. every 15 sec a child under the age of 5 dies from water-related illness 17% of the earth’s population don’t have reliable drinking water 40% of the population do not have access to adequate sanitation • The fact that the US and developed countries have an outstanding water supply record is not an accident . clean water is the exception rather than the rule.Pathogens • 80% of US population turn their tap on every day to drink from a publicly supplied water. 000 population from 1890 to 1935.Slow sand filters Disinfection Untreated Water from rivers Typhoid fever cases per 100. Philadelphia . cramps) Many organisms associated with deaths or serious illness (e. polio virus) have been eliminated Organisms with the potential to cause sickness and occasionally death in the US are still being found (see figure) . typhoid.g. fatigue.Many are Gastrointestinal symptoms (diarrhea. WI .Waterborne disease outbreaks in the US. 1980-1996 Milwaukee. Microbiological Characteristics • Water for drinking and cooking must be free of pathogens (bacteria. and worms) • Protozoa of concern: Giardia cysyts and Cryptosporidium oocysts (gastrointestinal illness) • Some organisms which cause disease in people originate with the fecal discharge of infected individuals and animals • Techniques for comprehensive bacteriological examination are complex and time consuming • Indicator organisms . virus. protozoa. Why Coliforms Were Selected As An indicator Organism? • Common inhabitants of intestinal tract •Their presence is an indication of fecal contamination • Excreted in large quantities •Easy to culture • Can survive in water for long periods of time. coli as a better indicator of bacteriological contamination than total coliforms . they don’t grow effectively • Are relatively easy to culture • Some agencies use E. however. . Fresh water: TDS < 1. These salts include cations (sodium. and bicarbonate). Salinity is the list of the concentrations of the primary cations and anions. magnesium. Total dissolved solids (TDS) is a measure of salinity.000-34. as it passes through soils and rocks on its way to the sea.000 mg/L Seawater: TDS ≈ 30. sulfate. and potassium) and anions (chloride.000 mg/L The concentrations of dissolved solids is an important indicator of the usefulness of water for various applications Drinking water: recommended maximum TDS ≈ 500 mg/L. calcium.Salts Water accumulates a variety of dissolved solids. or salts.500 mg/L Saline water : TDS > 5. Suspended Solids Organic and inorganic particles that are carried by the wastewater into a receiving water are termed suspended solids (SS) Colloidal particles that do not settle readily. cause the turbidity found in many surface waters. . Mercury vapor is more toxic than liquid mercury (damage to central nervous system). lead. and it readily passes from a pregnant women to her fetus. and arsenic. The most important heavy metals in terms of their environmental impacts are mercury. cadmium.Toxic Metals Toxic metals: Most metals are toxic. The most important route for the elimination of metals after they are inside a person is via the kidney. Lead dissolved in blood is transferred to vital organs. Children and fetuses are the most at risk since their brains are growing rapidly (can cause permanent brain damage). including the kidneys and brains. Metals are totally nondegradable. . herbicides.2-dichloroethene. Five VOCs are especially toxic. Toxic Volatile Organic Chemicals (VOC): are among the most commonly found contaminants in groundwater. few milliliters can produce death). The most toxic of the five is vinyl chloride (known human carcinogen). . They are often used as solvents in industrial processes and a number of them are either known or suspected carcinogens or mutagens. carbon tetrachloride (very toxic if ingested. tetrachloroethylene. (insecticides.Toxic Organic Compounds Toxic Organic Compounds: Pesticides. rodenticides. trichloroethene (suspected carcinogen). and their presence in drinking water is cause for special concern: vinyl chloride. and fungicides) Pesticides is used to cover a range of chemicals that kill organisms that human consider undesirable. 1. Toxic Organic Compounds H H cis-DCE H PCE TCE H H trans-DCE Ethene H H H Vinyl chloride H H H Dehalococcoides (PCE to ethene) H . Environmental persistence 2. many previously little-recognized pollutants have become classified as emerging contaminants: contaminants that by meeting some combination of the foregoing attributes warrant particular interest and concern. and analytical techniques. Immediacy of impact Since the 1990s . Much of this trend in identifying and characterizing emerging contaminants impacts depend on improvement in instrumentation.Emerging contaminants How is it possible to identify and prioritize pollutants of most environmental concern? 1. Occurrence frequency and concentration 4. Relative toxicity (to humans and other biota) 3. . sampling. Emerging contaminants Foremost among the emerging contaminants are endocrine disrupting chemicals (EDCs) . New pathogenic organisms have been identified (human adenoviruses). . Rapidly increasing incidences of antibiotic-resistant pathogens. Nanoparticles: materials with functionalities at the near-atomic scale. or increasing or reducing the production of natural hormones (i. interferes with the regulation of reproductive and developmental process in mammals. birds.Endocrine-Disrupting Chemicals They alter the normal physiological function of the endocrine system in humans and wildlife either by being or acting like a natural hormone. blocking the action of a natural hormone. fish) Pharmaceutical and EDCs .e. birds. or increasing or reducing the production of natural hormones (i.e. blocking the action of a natural hormone. interferes with the regulation of reproductive and developmental process in mammals. fish) Pharmaceutical and EDCs .Endocrine-Disrupting Chemicals They alter the normal physiological function of the endocrine system in humans and wildlife either by being or acting like a natural hormone. Amount of DO that the water can hold decreases with temperature As To Demand for O2 DO . the resulting rise in temperature can adversely affect life in the vicinity of the thermal plume. As water temperature increases: 1. Metabolic rates increases 2.Thermal Pollution • Large steam-electric power • Nuclear plant If the heat is released into local river or lake. Oxygen-demanding wastes and nutrients have a profound impact on almost all types of rivers.Water Quality Management in Rivers Objective: To control the discharge of pollutants so that the quality of water is not degraded to an unacceptable extent below the natural background level. Bacteria + organic waste + DO If DO falls below a critical point New bacteria + CO2 + H2O + drop in DO threat to higher forms of aquatic life . . it is important to determine the amount of O2 required to oxidize a substance. Inorganic nitrogen oxidation 3. What are the factors that affect oxygen consumption during the degradation of organic matter? 2. How to predict the DO concentration in rivers from degradation of organic matter? Therefore.Oxygen-Demanding Wastes 1. 1. Theoretical oxygen demand (ThOD) 2.Oxygen-Demanding Wastes The are several measures of oxygen demand. Biochemical oxygen demand (BOD) . Chemical oxygen demand (COD) 3. ThOD If the chemical composition of the substance is known then the amount of O2 required to completely oxidize a particular organic substance may be calculated from stoichiometry. bacteria begin to draw on their own tissue for energy in order to survive (endogenous respiration) – Dead bacteria becomes food for other bacteria . This amount of oxygen is known as the theoretical oxygen demand (ThOD) ThOD has limitations because: – Some of the carbon is incorporated into new cells – When the amount of remaining wastes diminishes. C6H12O6 + 6O2 180 g 192g 6CO2 + 6H2O 264g 108 g It takes 192 g of oxygen to oxidize 180 g of glucose The ThOD of 108.75 mg/L of glucose.Example 1 Calculate the ThOD of 108.75 mg/L)(192 g O2/180 g Glucose) = 116 mg/L O2.75 mg/L of glucose is (108. . Does not depend either on the ability of microorganisms to degrade the organic matter or on knowledge of the particular substances in question.COD Amount of dichromate consumed in the oxidation of inorganic and organic matter. Organic matter = slowly biodegradable + nonbiodegradable + easily biodegradable + inert organic matter . COD is a chemical oxidation (no microorganisms). The BOD test is an indirect measure of organic measure because we actually measure only the change in dissolved oxygen concentration caused by the microorganisms as they degrade the organic matter. BOD test is the most widely used method for measuring organic matter because of the direct conceptual relationship between BOD and oxygen depletion in receiving waters.BOD Amount of oxygen consumed by microorganisms to oxidize the waste aerobically: Carbonaceous oxygen demand (CBOD) and nitrogenous oxygen demand (NBOD). . BOD requires microorganisms that consume oxygen in the process of degrading organic matter. . The sample requires dilution because the oxygen demand of a waste is several hundreds mg/L and the maximum DO that can dissolve in water at 200C is about 9 mg/L. of diluted sample 100 = vol. The bottle is stoppered to exclude air bobbles. of undiluted sample sample size(%) The appropriate sample size to use is determined by dividing 4 mg/L (midpoint of desired range of diluted BOD) by the estimated BOD concentration in the sample being tested . Dilution factor = vol. Samples are diluted with a special dilution water that contains all the necessary trace elements required for bacterial metabolism.Laboratory Test for CBOD Standard BOD test: 1. Special 300 mL bottles that is completely filled with a waste sample that has been appropriately diluted with a certain dilution water that contains an inoculum of microorganisms. Blank samples containing only the inoculated dilution water are also placed in BOD bottles and stoppered. Blanks are required to estimate the amount of oxygen consumed by the added inoculum of microorganisms (called seed) in the absence of the sample. BOD5). It is incubated in the dark to prevent photosynthesis from adding oxygen to the water. . The two BOD bottles are incubated in the dark at 20oC for the desired number of days (normally 5 days. Blank: seeded dilution water Seeded dilution water + sample waste 3.Laboratory Test for CBOD Standard BOD test: 2. t = dissolved oxygen concentration in blank after t days of incubation. After the desired number of days has elapsed. mg/L .t − DOs. The BOD of the undiluted sample is then calculated using the following equation: BODt = (DOb.Laboratory Test for CBOD Standard BOD test: 4.t ) × dilution factor where DOb.t = dissolved oxygen concentration in sample after t days of incubation. mg/L DOs. the samples and blanks are removed and dissolved oxygen concentration in each bottle is measured. Laboratory Test for CBOD . what are the sample size and dilution factor using this volume? Assume that 4 mg/L BOD can be consumed in the BOD bottle. What volume of undiluted sample should be added to a 300 mL bottle? Also. Therefore a convenient sample volume would be 7.00mL × 100 = 2.33% 300mL 300mL = 42. Sample size = 4 × 100 = 2. of undiluted sample = 0.66 mL.00mL Dilution factor = .22% 180 The volume of diluted sample is 300 mL Vol.0222 X 300 mL = 6.00 mL. The actual sample size and dilution factor: Sample size = 7.9 7.Example 2 The BOD of a wastewater sample is estimated to be 180 mg/L. respectively?.7 − 4.9 = 193mg/L .7 and 4.2) × 42.t ) × dilution factor BOD5 = (8.2 mg/L.Example 3 What is the BOD5 of the wastewater sample of Example 4 if the DO values for the blank and diluted sample after five days are 8. BODt = (DOb.t − DOs. . and the nitrogen is released to the surrounding water as ammonia (NH3). this ammonia is in the form of ammonium cation (NH4+).57 g O2 /g N grams of nitrogen oxidized 14 The actual NBOD is slightly less than the theoretical value due to incorporation of some of the nitrogen into new bacterial cells. NH4+ + 2O2 + nitrifying bacteria NO3. which contains nitrogen. The ammonia released by organic compounds.NBOD Oxygen consumption due to nitrogen oxidation is called nitrogenous BOD (NBOD). Many organic compounds contain proteins. The process is called nitrification. plus that from other sources such as industrial wastewater and agricultural runoff (fertilizers). At normal pH values. is oxidized to nitrate (NO3-) by a special group of microorganisms called nitrifying bacteria.+ H2O +2H+ The theoretical NBOD can be calculated as follows: NBOD = grams of oxygen used 4 × 16 = = 4. 7 mg N/L 17g NH 3 Theo.Example 4 Compute the theoretical NBOD of a wastewater containing 30 mg/L of ammonia as nitrogen (we often say “ammonia nitrogen” and write the expression as NH3-N).57 mg O2/mg N) = 133 mg O2/L . the amount of ammonia was reported as NH3-N. NBOD = (24. what would the theoretical NBOD be? In the first part of the problem. 30 mg NH 3 /L × 14 gN = 24.57 mg O2/mg N) = 137 mg O2/L In the second part. Therefore Theo.7 mg N/L)(4. NBOD = (30 mg N/L)(4. If the wastewater analysis reported as 30 mg/L of ammonia (NH3). we must convert mg/L of ammonia to NH3-N by multiplying by the ratio of gram molecular weight of N to NH3. NBOD vs CBOD The rate at which NBOD is exerted depends heavily on the number of nitrifying organisms present. No Lag: higher population of nitrifying organisms reduces the lag time . the concentration is high. In untreated sewage. while in a well-treated effluent. there are few of these organisms. Lag No Lag Lag: time it takes for nitrifying bacteria to reach a sufficient population. CBOD When a water sample containing degradable organic matter is placed in a closed container and inoculated with bacteria. Organic matter oxidized Organic matter remaining . the oxygen consumption typically follows the pattern shown in the figure below. the amount of organic matter already oxidized goes up until finally all of the original organic matter has been oxidized Organic matter oxidized Organic matter remaining . Another way to describe the organic matter in the container is to say as time goes on.CBOD The amount of organic matter remaining in the container will decrease with time. or utilized. or we could say the amount of oxygen demand already exerted.CBOD The remaining demand for oxygen to decompose wastes decreases with time until there is no more demand. starts at zero until all of the original oxygen demand has been satisfied Organic matter oxidized Organic matter remaining . it is assumed that the rate of decomposition of organic waste is directly proportional to the concentration of degradable organic matter remaining at any time t.CBOD The translation of the figure into a mathematical form is straightforward. (Lt). the total amount of oxygen required by microorganisms to oxidize the waste completely to carbon dioxide and water. To do so. d-1 The solution to the above equation after rearranging and integrating dLt = −rA = −kLt dt Lt = L0 e − kt where L0 = oxygen demand of organic compound at time t = 0 Lo is often referred to as the ultimate BOD. . Assuming a first-order reaction. that is. we can write where k = BOD rate constant. 303K . we are interested in the amount of oxygen utilized in the consumption of the organics (BODt) . From the figure.CBOD Rather than Lt. BODt is the difference between L0 and Lt BODt = L0 − Lt = L0 − L0 e − kt = L0 (1 − e − kt ) The above equation is called the BOD rate equation and is often written in base 10: BODt = L0 (1 − 10 − Kt ) k = 2. Example 5 If the BOD3 of a waste is 75 mg/L and the K is 0.150)(3) ) Lo = 116mg/L For base e k = 2.150 d-1.345)(3) ) Lo = 116mg/L .345 BODt = L0 (1 − e − kt ) 75 = L0 (1 − e −(0. what is the ultimate BOD? For base 10 BODt = L0 (1 − 10 − Kt ) 75 = L0 (1 − 10 −(0.303K = 0. 2. Ability of the available microorganisms to degrade the waste Temperature: rate of biodegradation increases with increasing T kT = k 20 (θ )T − 20 where T = temperature of interest. d-1 . The nature of the waste: simple sugars and starches degrade easily while cellulose does not. As k increases. oC kT = BOD rate constant at the temperature of interest.CBOD The reaction rate constant k indicates how rapidly oxygen will be depleted in a receiving water. d-1 k20 = BOD rate constant determined at 20oC. The numerical value of the rate constant is dependent on the following: 1. 3. the rate at which dissolved oxygen is used increases. CBOD Ө = temperature coefficient. 1964).056 for temperatures between 20 and 30oC (Schroepfer. et al.135 for temperature between 4 and 20oC and 1. . This has a value of 1.. 135)10 − 20 = 0.032)(4) ] = 0.Example 6 A waste is being discharged into a river that has a temperature of 10oC. What fraction of the maximum oxygen consumption has occurred in four days if the BOD rate constant determined in the laboratory under standard conditions (20oC) is 0.12 L0 .032 d −1 BODt = L0 (1 − e − kt ) BOD4 = [1 − e −(0.115 d-1 (base e)? kT = k 20 (θ )T − 20 k10 o C = 0.115(1. NBOD vs CBOD Once nitrification begins.04 to 0.10 d-1). When measurements of only CBOD is required. NBOD can be described by the equations used for CBOD with a BOD rate constant comparable to that for CBOD with well-treated effluent (K = 0. chemical inhibitors are added to stop the nitrification process. . DO Sag Curve • • • • • • • • Introduction Mass-balance approach Oxygen deficit DO sag equation Deoxygenation rate constant Reaeration rate constant Management strategy Nitrogenous BOD . This is done by determining the profile of DO concentration downstream from a waste discharge. . This profile is called the DO sag curve.Introduction One of the major tools of water quality management in rivers is the ability to assess the capability of a stream to absorb a waste load. less and less organic material remains and the oxygen is replenished from the atmosphere .Introduction The biota of the stream are often a reflection of the DO conditions in the stream The DO concentration decreases near the point of discharge of waste as oxygen-demanding material are oxidized. As we move further downstream. DO in the waste discharge is usually less than that in the river • Thus DO in the river is lowered as soon as the waste is discharged (Initial DO reduction) 4. Sources of oxygen : reaeration from the atmosphere and photosynthesis of aquatic plants 2. Non-point source pollution 5. Respiration of organisms living in the sediments 6.Introduction To develop a mathematical expression for the DO sag curve the following factors need to be considered: 1. Factors affecting oxygen depletion • CBOD and NBOD • BOD already in the river upstream of the waste discharge 3. Respiration of aquatic plants . Mass balance for CBOD DO and CBOD change as the result of mixing of the waste stream and the river. . Two conservative (no chemical rxn) mass balances may be used to account for the initial mixing of the waste stream and the river. 1. Mass balance for DO 2. A. Once these are accounted for. the DO sag curve may be viewed as a nonconservative mass balance. B.Mass-Balance Approach Simplified mass balances help us understand and solve the DO sag curve problem. no accumulation. m3/s Qr = volumetric flow rate of the river. and complete and instantaneous mixing: Mass in = Mass out Mass of DO and BOD in Wastewater Mass of DO and BOD in River Mass of DO and BOD in River after Mixing Mass balance for DO: Qw DOw + Qr DOr = (Qw + Qr )DO where Qw = volumetric flow rate of wastewater.Conservative Mass-Balance for DO and CBOD Conservative. g/m3 DO = dissolved oxygen concentration after mixing. g/m3 . g/m3 DOr = dissolved oxygen concentration in the river. m3/s DOw = dissolved oxygen concentration in the wastewater. Conservative Mass-Balance for DO and CBOD The concentration of DO after mixing equals: Qw DOw + Qr DOr DO = Qw + Qr Mass balance for BOD: Qw Lw + Qr Lr = (Qw + Qr )La where Lw = ultimate BOD of the wastewater, mg/L Lr = ultimate BOD of the river, mg/L La = initial ultimate BOD after mixing, mg/L The concentration of ultimate BOD after mixing equals: Qw Lw + Qr Lr La = Qw + Qr Example 7 The town of State College discharges 17,360 m3/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 20oC. 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. Compute the DO and initial ultimate BOD after mixing. Example 7 Solution: Assume complete and instantaneous mixing. The DO after mixing is given by the equation below: DO = Qw DOw + Qr DOr Qw + Qr Qw is given in m3/d, to use the above equation the units of Qw should be in m3/s 3 Qw = (17,360m /d) = 0.20m 3 /s (86,400s/d) The DO after mixing is then DO = (0.20m 3 /s)(1.0mg/ L) + (0.43m 3 /s)(6.5mg/ L) 0.20m /s + 0.43m /s 3 3 = 4.75mg/L 20m 3 /s)(26.43m 3 /s)(5.55) The initial ultimate BOD after mixing is La = (0.43m /s 3 3 = 11.6mg/L (1 − 0.Example 7 The initial ultimate BOD after mixing is given by the equation below: Qw Lw + Qr Lr La = Qw + Qr The ultimate BOD of the wastewater is determined using the following equation: BOD = L (1 − e − kt ) t 0 where L0 = (1 − e − kt ) BOD5 = 12mg/L (1 − e −(0.0mg/ L) 0.86mg/L .12)(5) ) = 12 = 26.20m /s + 0.6mg /L) + (0. Oxygen Deficit The DO sag equation has been developed using oxygen deficit rather than dissolved oxygen concentration, to make it easier to solve the integral equation that results from the mathematical description of the mass balance. The oxygen deficit is given by the following equation: D = DOs − DO where D = oxygen deficit, mg/L DOs = saturation concentration of dissolved oxygen at the temperature of the river after mixing, mg/L DO = actual concentration of DO, mg/L Initial deficit: The initial deficit is calculated as the difference between saturated DO and the concentration of the DO after mixing where Qw DOw + Qr DOr Q w + Qr Da = initial deficit after river and waste mixed, mg/L Da = DOs − DO Sag Curve DOs Values For Fresh Water 75 mg/L from Example 7.33mg/L − 4.Example 8 Calculate the initial deficit of the Bald Eagle Creek after mixing with the wastewater from the town of State College (see Example 7). Solution: with the stream temperature. The stream temperature is 10oC and the wastewater is 10oC. DOs = 11. Since we calculated the concentration of DO after mixing as 4.33 mg/L. the initial deficit after mixing is D = DOs − DO D = 11. the saturation value of dissolved oxygen (DOs) can be determined from the table in the previous slide.75mg/L = 6. At 10oC.58mg/L . Figure b is a simplified mass balance diagram developed by Streeter-Phelps.DO Sag Equation Figure a is a comprehensive mass balance diagram of DO in a small reach (stretch) of river that accounts for all the inputs and outputs. . The DO sag equation is also known as the Streeter-Phelps equation. differentiation of the following equation D = DOs − DO yields: d(DO) dD + =0 dt dt . With the assumption that the saturation value for DO remains constant [d(DOs)/dt =0].RDOout = 0 where RDOin = mass of DO in river flowing into reach W = mass of DO in wastewater flowing into reach A = mass of DO added from atmosphere M = mass of DO removed by microbial degradation of CBOD RDOout = mass of DO in river flowing out of reach The rate at which DO disappears from the stream as a result of microbial action (M) is exactly equal to the rate of increase in the deficit.M .DO Sag Equation The mass balance equation for figure b is: RDOin + W + A . DO Sag Equation . so d(DO) dD d(BOD) =− =− dt dt dt Remember BODt = L0 − Lt (L0 is constant) dLt d(BOD) =− dt dt .DO Sag Equation d(DO) dD + =0 dt dt and d(DO) dD =− dt dt The rate at which DO disappears coincides with the rate that BOD is degraded. DO Sag Equation Thus. the rate of change in deficit at time t due to BOD (rate of deoxygenation) is a first order reaction proportional to the BOD remaining after the waste enters the rivers dD = kd Lt (kd is the deoxygenation rate constant) dt dLt d(BOD) =− dt dt dLt = −kLt dt (k is the BOD rate constant determined in laboratory) . DO Sag Equation The rate oxygen mass transfer into solution from the air (A) (Rate of reaeration) is a first order reaction proportional to the difference between the saturation value and the actual concentration d(DO) = k r (DOs − DO) dt remember and D = (DOs − DO) d(DO) dD =− dt dt then dD = −kr D dt kr = reaeration rate constant . DO Sag Equation The deoxygenation caused by microbial decomposition of waste (M) and oxygenation caused by reaeration (A) are competing processes that are simultaneously removing and adding oxygen to a stream Rate of increase of the deficit = rate of deoxygenation – Rate of reaeration dD = k d Lt − k r D dt where dD/dt = the change in oxygen deficit (D) per unit time. t. d-1 Lt = BOD remaining t (days) after waste enters the river. d-1 D = oxygen deficit in river water after utilization of BOD for time. mg/L . mg/L kr = reaeration rate constant. mg/L.d kd = deoxygenation rate constant. d-1 kr = reaeration rate constant. mg/L kd = deoxygenation rate constant.DO Sag Equation dD = kd L − k r D dt By integrating and using the initial condition (at t = 0. D = Da) we obtain the DO sag equation: kd La D= (e − k d t − e -k r t ) + Da (e − k r t ) k r − kd where La = initial ultimate BOD after river and wastewater have mixed. d-1 Da = initial deficit after river and wastewater have mixed. d When kr = kd D = (kd tLa + Da )(e − k d t ) . mg/L t = time of travel of wastewater discharge downstream. the DO can be found using the equation: D = DOs − DO kd La DO = DOs − (e − k d t − e .k r t ) + Da (e − k r t k r − kd ) Physically it is impossible for the DO to be less than zero. then all the oxygen was depleted at some earlier time and the DO is zero. .DO Sag Equation Once D is found at any point downstream. If the deficit (D) is greater than DOs. DO Sag Equation Deoxygenation rate constant: ν η kd = k + H (Bosko. d-1 H = average depth of stream. m/s k = BOD rate constant determined in laboratory at 20oC. m η = bed-activity coefficient The bed-activity coefficient may vary from 0. d-1 v = average speed of stream. A deep. BOD is utilized more rapidly in a river compared to a BOD bottle because of turbulent mixing and larger numbers of “seed” organisms.1 for stagnant or deep water to 0. In general. kd is affected by temperature and can be adjusted to the river temperature using the equation: kT = k20 (θ )T − 20 . 1996) where kd = deoxygenation rate constant at 20oC. slowly moving rivers have a much lower kd than a shallow.6 or more for rapidly flowing streams. rapidly flowing river. 03 m/s.35) = 0. Solution: from Example 7. the value of k is 0.12d-1 ν Using the equation: kd = k + H η The deaoxygenation rate constant at 20oC is kd = 0.0 m and the bed-activity coefficient is 0.35.12d −1 + 0.0m Note that the units are not consistent and the bed-activity coefficient includes a conversion factor to make the second term dimensionally correct. The average speed of the stream in the creek is 0. The depth is 5.1221d −1 5.03m/s (0.Example 9 Determine the deoxygenation rate constant for the Bald Eagle Creek (Example 7 and 8) below the wastewater outfall (discharge pipe). . Thus we must correct the estimated kd using the equation kT = k20 (θ )T − 20 kd at 10 o C = (0.Example 9 The stream temperature in Example 8 was 10oC.034d −1 .135)10 − 20 = 0.1221d −1 )(1. 5 H 1. kr is affected by temperature and can be adjusted to the river temperature using the equation: kT = k20 (θ )T − 20 where ө = 1.024 . m/s H = average depth of stream. m kr depends on the degree of turbulent mixing and amount of water surface exposed to the atmosphere compared to the volume of water in the river. deep river will have a much lower kr than a wide. d-1 v = average speed of stream. shallow river.DO Sag Equation Reaeration rate constant: kr = 3.5 (O’Connor and Dobbins. 1958) where kr = reaeration rate constant at 20oC. A narrow.9ν 0. Therefore. The time to the critical point (tc) can be found by setting the derivative of the oxygen deficit equal to zero. and solving for t using base e values for kr and kd : kr 1 In tc = k r − kd kd k − kd 1 − Da r kd La When kr = kd 1 tc = kd Da 1 − La . the location of the critical point is of obvious importance.DO Sag Equation Critical point: is the lowest point on the sag curve and it is of major interest since it indicates the worst conditions in the river. 2.DO Sag Curve 1. At the critical point: the rate of removal of oxygen is equal to the rate of reaeration. Near the outfall: the rate of removal of oxygen by bacteria from the water is higher than the rate of reaeration. . Beyond the critical point: reaeration begins to dominate and the rate of reaeration is faster than the rate of removal of oxygen by bacteria. 3. DO Sag Equation . 024)10 − 20 = 0. Also determine the critical DO and the distance downstream at which it occurs. Kr and travel time. kd La DO = DOs − (e − k d t − e k r t ) + Da (e − k r t k r − kd ) Solution: La. and 9).04766d −1 .0604d − 1 The stream temperature is 10oC: k r at 10 o C = (0.9ν 0. 8.9)(0. t will be calculated as follows: k r at 20 C = o 3.5 = (3.Example 10 Determine the DO concentration at a point 5 Km downstream from the State College discharge into the Bald Creek (Example 7. and Da were calculated in previous examples.0604d −1 )(1.5 H 1.5 = 0.0m)1.03m/s)0.5 (5. kd. 929) ) = 4.929) .Example 10 The travel time t is computed from the distance downstream and the speed of the stream : x =ν × t Where x = distance downstream v = average stream velocity t = elapsed time between discharge point and distance x downstream t= (5km)(1.929d (0.33 − 0.k r t ) + Da (e − k r t k r − kd ) (0.03m/s)(86.03442 .04766)(1.03442)(11.04766 − 0.03442)(1.60mg/L −e DO = 11.400s/d) kd La DO = DOs − (e − k d t − e .04766)(1.(0.58(e −(0.86) −(0.929) (e ) + 6.000m/km) = 1. 03422 1 = 6.58(e −(0.86) 0.04766)(6.03442)(11.04766)(6.Example 10 The critical time is computed using the equation: kr k − kd 1 1 − Da r In tc = kd La k r − kd kd 0.04766 0.03442 0.04766 − 0.03442 The critical deficit is calculated using the equation Dc = kd La (e − k d t c − e -k r t c ) + Da (e − k r t c ) k r − kd (0.86) −(0.45) .58 In tc = (0.04766 − 0.03422)(11.85mg/L −e Dc = 0.03442 .03442)(6.45) ) = 6.45d 1 − 6.04766 − 0.45) (e ) + 6.(0. 03m/s)( 1 ) = 16.400s/d)(0.33-6.Example 10 The critical DO is then calculated using the equation Dc = DOs − DOc DOc = 11.000m/km .7km 1.85 = 4.48 mg/L The critical DO occurs downstream at a distance of x =ν × t x = (6.45d)(86. 5. the stream can adequately assimilate the waste. A DO standard is set to protect the most sensitive species that exist or could exist in a particular river. 2. If the DO at the critical point is higher than the standard. Environmental engineers and scientists have control over just two parameters. This will reduce Da.Management Strategy 1. For a known waste discharge and a known set of river characteristics. • Improving treatment efficiency or adding additional treatment steps will reduce La. 3. . • Adding oxygen to wastewater to bring it to saturation level before discharge. 4. If the DO at the critical point is less than the standard DO. then additional waste treatment is needed. the DO sag equation can be solved to find the DO at the critical point. La and Da. The new values of La and Da are used to determine if DO standard will be violated at the critical point. . Higher temperatures increase kd more than kr. • kr is reduced more than kd by low river flow because of reduced velocities. 7. it is important to use river conditions that will cause the lowest DO concentration. • Low rivers flows results in higher values for La and Da. When using the DO sag curve to determine the adequacy of wastewater treatment.Management Strategy 6. • • Higher temperature reduces DO saturation. These conditions are: • Late summer when the river flows are low and temperatures are high. The effluent parameter and river parameters at summer low flow conditions are available. Among the decisions to be made are what effect the plant discharge will have on each river and which river would be impacted less. the White River. .Example 11 The Pitts Canning Company is considering opening a new plant at one of two possible locations: the Green River and its twin. 00 5. m Bed-Activity coefficient ө (kd) ө (kr) 0. mg/L Temperature.05 30.00 0. oC Speed.056 1. mg/L DO.00 0.90 25.50 19.00 0.50 19. m Bed-Activity coefficient ө (kd) ө (kr) 3 White River 0.05 30.d-1 0.Example 11 Effluent Parameter Plant A Flow (m /s) Ultimate BOD at 250C.m3/s Ultimated BOD at 25oC.056 1. m/s Average depth.d-1 3 Plant B 0. 0C k at 20oC.00 0. 0C k at 20oC. oC Speed.10 4. mg/L DO.20 1.85 25.m /s Ultimated BOD at 25oC.00 0. mg/L Temperature.00 0.024 . m/s Average depth. mg/L Temperature.12 Flow (m3/s) Ultimate BOD at 250C. mg/L DO.024 Flow.20 4.00 0.00 5. mg/L DO.85 25.90 25. mg/L Temperature.20 1.00 0.07 River Parameter Green River Flow. B-Green. kr 8. mg/L 3. mg/L . Calculate deoxygenation rate constant at 20oC. mg/L 10. A-White. Calculate Initial deficit (Da). mg/L 4. Calculate reaeration rate constant at 25oC. d 9. kd 6. Calculate critical DO. Calculate reaeration rate constant at 20oC. Calculate Initial DO after mixing. g/L 2. B-White) 1. Calculate deoxygenation rate constant at 25oC.Example 11 Solution (See attached Excel file “Do Sag Curve”) : we have four combinations (A-Green. Calculate La (ultimate BOD after mixing). kd 5. Calculate critical time. Calculate critical deficit. kr 7. 2455 4.47 5.00 6.1736 5.1578 0.0973 0.1643 0.2455 4.10 B-White 0.Example 11 A-Green 0.96 6.1736 5.14 0.93 1.45 B-Green 0.32 3.06 kd kr tc Dc DOc .09 8.28 2.24 A-White 0.1039 0. d-1 Ln = ultimate nitrogenous BOD after waste and river have mixed. mg/L .kn Where kn = nitrogenous deoxygenation coefficient.Nitrogenous BOD The NBOD can be incorporated into the DO sag equation by adding an additional term to the equation D= kd La k L (e − k d t − e -k r t ) + Da (e − k r t ) + n n (e − k n t − e − k r t ) k r − kd kr . serve as nutrients for excessive growth of algae.Effect of Nutrients on Water Quality in Rivers Effects of nitrogen: 1. This will cause further reduction in the DO supply of water body. in low concentrations. they become an oxygen-demanding organic material for bacteria.consumes large quantities of DO. Effect of phosphorous: Serves as a vital nutrient for the growth of algae. The conversion of NH4+ to NO3. In high concentration. NH3-N is toxic to fish 2. and NO3. . NH3. When algae dies. 3.
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