pascal saikaly lecture water quality management 2010
TRANSCRIPT
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, m3/d1950 1.51E+08 5.30E+071960 1.79E+08 7.95E+071970 2.03E+08 1.02E+081980 2.27E+08 1.29E+081990 2.49E+08 1.46E+082000 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 2015Lpcd % Lpcd % Lpcd %
Agriculture 599 72 651 74 1164 60domestic 186 22 140 16 616 32Industry 45 5 89 10 164 8Total 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 UsageWater 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 WastesOne 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/LBluegill and bass: 5 mg/L
NutrientsNutrients 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.
NO3 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”
NutrientsPhosphorous: 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
PathogensTypical 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, Cryptosporidium species.
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
SaltsWater 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/LSaline water : TDS > 5,000 mg/LSeawater: TDS ≈ 30,000-34,000 mg/L
The concentrations of dissolved solids is an important indicator of the usefulness of water for various applicationsDrinking water: recommended maximum TDS ≈ 500 mg/L.
Suspended SolidsOrganic 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 CompoundsToxic 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
HH
H
HH
HH
HH
PCE TCE
cis-DCE
trans-DCE
Vinyl chloride
Ethene
Dehalococcoides (PCE to ethene)
Emerging contaminantsHow is it possible to identify and prioritize pollutants of most environmental concern?1. Environmental persistence2. Relative toxicity (to humans and other biota)3. Occurrence frequency and concentration4. 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 contaminantsForemost 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 ChemicalsThey 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 ChemicalsThey 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 To Demand for O2 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 + CO2 + H2O + drop in DO
If DO falls below a critical point threat to higher forms of aquatic life
Oxygen-Demanding Wastes1. What are the factors that affect oxygen
consumption during the degradation of organic matter?
2. Inorganic nitrogen oxidation3. How to predict the DO concentration in rivers
from degradation of organic matter?
Therefore, it is important to determine the amount of O2 required to oxidize a substance.
Oxygen-Demanding WastesThe are several measures of oxygen demand.
1. Theoretical oxygen demand (ThOD)2. Chemical oxygen demand (COD)3. Biochemical oxygen demand (BOD)
ThODIf 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. 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 1Calculate the ThOD of 108.75 mg/L of glucose.
C6H12O6 + 6O2 6CO2 + 6H2O180 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 O2/180 g Glucose) = 116 mg/L O2.
CODAmount 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
BODAmount 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 CBODStandard 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. 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 200C 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(%)sample100
sampleundilutedofvol.sampledilutedofvol.factorDilution ==
Laboratory Test for CBODStandard 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 20oC for the desired number of days (normally 5 days, BOD5). 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 CBODStandard 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 DOb,t = dissolved oxygen concentration in blank after t days of incubation, mg/LDOs,t = dissolved oxygen concentration in sample after t days of incubation, mg/L
factordilution)DO(DOBOD s,tb,tt ×−=
Laboratory Test for CBOD
Example 2The 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 mLVol. 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%100180
4sizeSample =×=
2.33%100300mL7.00mLsizeSample =×=
42.97.00mL300mLfactorDilution ==
Example 3What is the BOD5 of the wastewater sample of Example 4 if the DO values for the blank and diluted sample after five days are 8.7 and 4.2 mg/L, respectively?.
factordilution)DO(DOBOD s,tb,tt ×−=
193mg/L42.94.2)(8.7BOD5 =×−=
NBODOxygen 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 (NH3). At normal pH values, this ammonia is in the form of ammonium cation (NH4
+).
The ammonia released by organic compounds, plus that from other sources such as industrial wastewater and agricultural runoff (fertilizers), is oxidized to nitrate (NO3
-) by a special group of microorganisms called nitrifying bacteria. The process is called nitrification.
NH4+ + 2O2 + nitrifying bacteria NO3- + H2O +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/gOg4.5714
164oxidizednitrogenofgramsusedoxygenofgramsNBOD 2=
×==
Example 4Compute 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). If the wastewater analysis reported as 30 mg/L of ammonia (NH3), what would the theoretical NBOD be?
In the first part of the problem, the amount of ammonia was reported as NH3-N. ThereforeTheo. NBOD = (30 mg N/L)(4.57 mg O2/mg N) = 137 mg O2/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/Lmg24.7NH17ggN 14/LNHmg30
33 =×
Theo. NBOD = (24.7 mg N/L)(4.57 mg O2/mg N) = 133 mg O2/L
NBOD vs CBODThe 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
CBODWhen 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
CBODThe 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
CBODThe 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
CBODThe 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, (Lt). 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 L0 = oxygen demand of organic compound at time t = 0
Lo 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.
tAt kLr
dtdL
−=−=
kt0t eLL −=
CBODRather than Lt, we are interested in the amount of oxygen utilized in the consumption of the organics (BODt) . From the figure, BODt is the difference between L0 and Lt
The above equation is called the BOD rate equation and is often written in base 10:
)e(1LeLLLLBOD kt0
kt00t0t
−− −=−=−=
)(1LBOD Kt0t
−−= 10
2.303Kk =
Example 5If the BOD3 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
)(1LBOD Kt0t
−−= 10)(1L7 0
)3)(150.0(105 −−=
116mg/LLo =
0.3452.303Kk ==
)(1LBOD kt0t
−−= e
)(1L7 0)3)(345.0(5 −−= e
116mg/LLo =
CBODThe 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 waste3. Temperature: rate of biodegradation increases with increasing T
whereT = temperature of interest, oCkT = BOD rate constant at the temperature of interest, d-1
k20 = BOD rate constant determined at 20oC, d-1
2020 )( −= TθkkT
CBODӨ = temperature coefficient. This has a value of 1.135 for temperature between 4 and 20oC and 1.056 for temperatures between 20 and 30oC (Schroepfer, et al., 1964).
Example 6A 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 0.115 d-1
(base e)?20
20 )( −= TθkkT
12010C10 d0.0325)0.115(1.13k o
−− ==
)(1LBOD kt0t
−−= e
0.12]e[1L
BOD (0.032)(4)
04 =−−=
NBOD vs CBODOnce 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
IntroductionOne 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.
IntroductionThe 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
IntroductionTo 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 sediments6. Respiration of aquatic plants
Mass-Balance ApproachSimplified 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 DO2. 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 Qw = volumetric flow rate of wastewater, m3/sQr = volumetric flow rate of the river, m3/s
DOw = dissolved oxygen concentration in the wastewater, g/m3
DOr = dissolved oxygen concentration in the river, g/m3
DO = dissolved oxygen concentration after mixing, g/m3
Mass of DO and BOD in Wastewater
Mass of DO and BOD in River
Mass of DO and BOD in River after Mixing
outMassinMass =
)DOQ(QDOQDOQ rwrrww +=+
Conservative Mass-Balance for DO and CBOD
The concentration of DO after mixing equals:
Mass balance for BOD:
where Lw = ultimate BOD of the wastewater, mg/LLr = ultimate BOD of the river, mg/LLa = initial ultimate BOD after mixing, mg/L
The concentration of ultimate BOD after mixing equals:
QrQDOQDOQDO
w
rrww++
=
arwrrww )LQ(QLQLQ +=+
QrQLQLQL
w
rrww++
=a
Example 7The 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 7Solution: Assume complete and instantaneous mixing. The DO after mixing is given by the equation below:
Qw is given in m3/d, to use the above equation the units of Qwshould be in m3/s
The DO after mixing is then
QrQDOQDOQDO
w
rrww++
=
/s0.20m)(86,400s/d/d)(17,360mQ 3
3w ==
4.75mg/L/s0.43m/s0.20m
L)/s)(6.5mg/(0.43mL)/s)(1.0mg/(0.20mDO 33
33=
+
+=
Example 7The 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/s0.43m/s0.20m
L)/s)(5.0mg/(0.43m/L)/s)(26.6mg(0.20mL 33
33a =
+
+=
QrQLQLQL
w
rrww++
=a
)(1LBOD kt0t
−−= e
26.6mg/L0.55)(1
12)e(1
12mg/L)e(1
BODL (0.12)(5)kt5
0 =−
=−
=−
=−−
Oxygen DeficitThe 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/LDOs = saturation concentration of dissolved oxygen at the
temperature of the river after mixing, mg/LDO = 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 Da = initial deficit after river and waste mixed, mg/L
DODOD s −=
rw
rrwwsa QQ
DOQDOQDOD++
−=
DO Sag Curve
DOs Values For Fresh Water
Example 8Calculate 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 10oC and the wastewater is 10oC.
Solution: with the stream temperature, the saturation value of dissolved oxygen (DOs) can be determined from the table in the previous slide. At 10oC, DOs = 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
DODOD s −=
6.58mg/L4.75mg/L11.33mg/LD =−=
DO Sag EquationFigure 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 EquationThe mass balance equation for figure b is:
where RDOin = mass of DO in river flowing into reachW = mass of DO in wastewater flowing into reachA = mass of DO added from atmosphereM = 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. With the assumption that the saturation value for DO remains constant [d(DOs)/dt =0], differentiation of the following equation
yields:
0RDO-M-AWRDO outin =++
DODOD s −=
0dtdD
dtd(DO)
=+
DO Sag Equation
DO Sag Equation
and
The rate at which DO disappears coincides with the rate that BOD is degraded, so
Remember (L0 is constant)
0dtdD
dtd(DO)
=+
dtdD
dtd(DO)
−=
dtd(BOD)
dtdD
dtd(DO)
−=−=
t0t LLBOD −=
dtdL
dtd(BOD) t−=
DO Sag EquationThus, 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
(kd is the deoxygenation rate constant)
(k is the BOD rate constant determined in laboratory)
dtdL
dtd(BOD) t−=
tt kL
dtdL
−=
td LkdtdD
=
DO Sag EquationThe 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 kr = reaeration rate constant
DO)(DOkdt
d(DO)sr −=
DO)(DOD s −=
dtdD
dtd(DO)
−=
DkdtdD
r−=
DO Sag EquationThe 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.dkd = deoxygenation rate constant, d-1
Lt = BOD remaining t (days) after waste enters the river, mg/Lkr = reaeration rate constant, d-1
D = oxygen deficit in river water after utilization of BOD for time, t, mg/L
DkLkdtdD
rtd −=
DO Sag Equation
By integrating and using the initial condition (at t = 0, D = Da) we obtain the DO sag equation:
where La = initial ultimate BOD after river and wastewater have mixed, mg/Lkd = deoxygenation rate constant, d-1
kr = reaeration rate constant, d-1
Da = initial deficit after river and wastewater have mixed, mg/Lt = time of travel of wastewater discharge downstream, d
When kr = kd
DkLkdtdD
rd −=
)(eD)e(ekk
LkD tka
t-ktk
dr
ad rrd −− +−−
=
))(eDtL(kD tkaad d−+=
DO Sag EquationOnce 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 DOs, then all the oxygen was depleted at some earlier time and the DO is zero.
+−
−−= −− )(eD)e(e
kkLkDODO tk
atk-tk
dr
ads rrd
DODOD s −=
DO Sag EquationDeoxygenation rate constant: (Bosko, 1996)
where kd = deoxygenation rate constant at 20oC, d-1
v = average speed of stream, m/sk = BOD rate constant determined in laboratory at 20oC, d-1
H = average depth of stream, mη = 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 kd 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.
kd is affected by temperature and can be adjusted to the river temperature using the equation:
ηHνkkd +=
2020 )( −= TθkkT
Example 9Determine 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 20oC is
Note that the units are not consistent and the bed-activity coefficient includes a conversion factor to make the second term dimensionally correct.
ηHνkkd +=
11d 0.1221d(0.35)
5.0m0.03m/s0.12dk −− =+=
Example 9The stream temperature in Example 8 was 10oC. Thus we must correct the estimated kd using the equation
2020 )( −= TθkkT
120101od 0.034d)(1.135)(0.1221dC10atk −−− ==
DO Sag EquationReaeration rate constant:
(O’Connor and Dobbins, 1958)
where kr = reaeration rate constant at 20oC, d-1
v = average speed of stream, m/sH = 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.
A narrow, deep river will have a much lower kr than a wide, shallow river.
kr is affected by temperature and can be adjusted to the river temperature using the equation:
where ө = 1.024
1.5
0.5r
H3.9νk =
2020 )( −= TθkkT
DO Sag EquationCritical 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 (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:
When kr = kd
−−
−=
ad
dra
dr
drc Lk
kkD1kkIn
kk1t
−=
a
a
dc L
D1k1t
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 10Determine 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: La, kd, and Da were calculated in previous examples. Kr and travel time, t will be calculated as follows:
The stream temperature is 10oC:
+−
−−= −− )(eD)e(e
kkLkDODO tk
atktk
dr
ads rrd
11.5
0.5
1.5
0.5o
r 0.0604d(5.0m)
m/s)(3.9)(0.03H
3.9νC20atk −===
120101or 0.04766d)(1.024)(0.0604dC10atk −−− ==
Example 10The travel time t is computed from the distance downstream and the speed of the stream :
Where x = distance downstreamv = average stream velocityt = elapsed time between discharge point and distance x downstream
t×=νx
1.929d86,400s/d)(0.03m/s)(
0m/km)(5km)(1,00t ==
+−
−−= −− )(eD)e(e
kkLkDODO tk
atk-tk
dr
ads rrd
4.60mg/L)6.58(e)e(e0.034420.0476611.86)(0.03442)(11.33DO 1.929)(0.04766)(1.929)(0.04766)(-1.929)(0.03442)( =
+−
−−= −−
Example 10The critical time is computed using the equation:
The critical deficit is calculated using the equation
6.85mg/L)6.58(e)e(e0.034420.0476611.86)(0.03442)(D 6.45)(0.04766)(6.45)(0.04766)(-6.45)(0.03442)(
c =
+−
−= −−
−−
−=
ad
dra
dr
drc Lk
kkD1kkIn
kk1t
6.45d11.86)(0.03422)(
0.034220.047666.5810.034420.04766In
0.034420.047661tc =
−−
−=
)(eD)e(ekk
LkD crcrcd tka
t-ktk
dr
adc
−− +−−
=
Example 10The critical DO is then calculated using the equation
DOc = 11.33-6.85 = 4.48 mg/L
The critical DO occurs downstream at a distance of
csc DODOD −=
t×=νx
16.7km)1,000m/km
1.03m/s)(,400s/d)(0(6.45d)(86x ==
Management Strategy1. 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, La and Da.
• Improving treatment efficiency or adding additional treatment steps will reduce La.
• Adding oxygen to wastewater to bring it to saturation level before discharge. This will reduce Da.
Management Strategy6. The new values of La and Da 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 La and Da.• kr is reduced more than kd by low river flow because of reduced
velocities.• Higher temperatures increase kd more than kr.• Higher temperature reduces DO saturation.
Example 11The 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 11Effluent Parameter
Plant A Plant BFlow (m3/s) 0.05 Flow (m3/s) 0.05
Ultimate BOD at 250C, mg/L 30.00 Ultimate BOD at 250C, mg/L 30.00DO, mg/L 0.90 DO, mg/L 0.90
Temperature, 0C 25.00 Temperature, 0C 25.00k at 20oC,d-1 0.12 k at 20oC,d-1 0.07
River Parameter
Green River White RiverFlow,m3/s 0.50 Flow,m3/s 0.50
Ultimated BOD at 25oC, mg/L 19.00 Ultimated BOD at 25oC, mg/L 19.00DO, mg/L 5.85 DO, mg/L 5.85
Temperature, oC 25.00 Temperature, oC 25.00Speed, m/s 0.10 Speed, m/s 0.20
Average depth, m 4.00 Average depth, m 4.00Bed-Activity coefficient 0.20 Bed-Activity coefficient 0.20
ө (kd) 1.056 ө (kd) 1.056ө (kr) 1.024 ө (kr) 1.024
Example 11Solution (See attached Excel file “Do Sag Curve”) : we have four combinations (A-Green, A-White, B-Green, B-White) 1. Calculate La (ultimate BOD after mixing), g/L2. Calculate Initial DO after mixing, mg/L3. Calculate Initial deficit (Da), mg/L4. Calculate deoxygenation rate constant at 20oC, kd
5. Calculate deoxygenation rate constant at 25oC, kd
6. Calculate reaeration rate constant at 20oC, kr
7. Calculate reaeration rate constant at 25oC, kr
8. Calculate critical time, d9. Calculate critical deficit, mg/L10.Calculate critical DO, mg/L
Example 11
A-Green A-White B-Green B-Whitekd 0.1578 0.1643 0.0973 0.1039kr 0.1736 0.2455 0.1736 0.2455tc 5.09 4.00 5.96 4.47Dc 8.14 6.93 6.28 5.32DOc 0.24 1.45 2.10 3.06
Nitrogenous BODThe NBOD can be incorporated into the DO sag equation by adding an additional term to the equation
Where kn = nitrogenous deoxygenation coefficient, d-1
Ln = ultimate nitrogenous BOD after waste and river have mixed, mg/L
)e(ek-k
Lk)(eD)e(ekk
LkD tktk
nr
nntka
t-ktk
dr
ad rnrrd −−−− −++−−
=
Effect of Nutrients on Water Quality in Rivers
Effects of nitrogen:1. In high concentration, NH3-N is toxic to fish2. NH3, in low concentrations, and NO3
- serve as nutrients for excessive growth of algae.
3. The conversion of NH4+ to NO3
- 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.