Water Quality Control PDF

Water Quality Control Wastewater Characteristics What is Wastewater? Wastewater is a dilute mixture of various wastes from residential, commercial , industrial and other public places. It is also defined as the used water from an...

Water Quality Control Wastewater Characteristics What is Wastewater? Wastewater is a dilute mixture of various wastes from residential, commercial , industrial and other public places. It is also defined as the used water from any combination of domestic, agricultural activities, surface runoff or stormwater, and any sewer inflow or sewer infiltration. Wastewater contains organic matter and inorganic matters which may be in dissolved, suspended and colloidal state.  Urban City Sewer Wastewater Sewage Wastewater Treatment Plant Characteristics of WW and their sources Characteristic Sources 1. Physical Characteristics i) Color Domestic and industrial wastes; natural decay of organic materials ii) Odor Decomposing wastewater; industrial wastes iii) Solids Domestic water supply; domestic and industrial wastes; soil erosions; iv) Temperature inflow-infiltration Domestic and industrial wastes Characteristics of WW and their sources Characteristic Sources 2. Chemical Characteristics a) Organic Domestic, commercial and industrial i) Carbohydrates (~50%) ii) wastes; Decomposing wastewater; Fats, oils and greases (~10%) iii) industrial wastes Pesticides Agricultural wastes iv) Phenols Industrial wastes v) Protein (~40%) Domestic and commercial wastes vi) Surfactants Domestic and industrial wastes vii) Others Natural decay of organic materials Characteristic Sources 2. Chemical Characteristics b) Inorganic Domestic wastes, domestic water supply, ground i) Alkalinity water infiltration Domestic wastes, domestic water supply, ground ii) Chlorides water infiltration, water softners Industrial wastes iii) Heavy metals Domestic and agricultural wastes iv) Nitrogen Industrial wastes v) pH Domestic and industrial wastes, natural runoff vi) Phosphorus Domestic water supply, domestic and industrial vii) Sulfur wastes Industrial wastes viii) Toxic compounds c) Gases Decomposition of domestic wastes i) Hydrogen sulfide Decomposition of domestic wastes ii) Methane Domestic water supply, surface water infiltration iii) Oxygen Characteristics of WW and their sources Characteristic Sources 3. Biological Characteristics i) Animals Open water courses and treatment plants ii) Plants Open water courses and treatment plants iii) Protista Domestic wastes; treatment plants iv) Viruses Domestic wastes Oxygen Demand Oxygen is essential fro the livelihood of organisms Required for oxidation of both inorganic as well as organic matter The demand of oxygen may be expressed in the following ways: Biochemical Oxygen demand (BOD) Chemical Oxygen demand (COD) Total Oxygen demand (TOD) Theoretical Oxygen demand (Th.OD) The amount of organic matter present may also be determined by the total organic carbon (TOC) test. Biochemical Oxygen Demand (BOD)  Biochemical Oxygen Demand (BOD): oxygen required for the micro-organisms to carry out biological decomposition of dissolved solids or organic matter in the wastewater under aerobic conditions at standard temperature. Widely used parameter of organic pollution applied to both wastewater as well as surface water. BOD test results are used for the following purposes: Determination of approximate quantity of oxygen required for the biological stabilization of organic matter present in the wastewater. Determination of size of wastewater treatment facilities. Measurement of efficiency of some treatment process. Determination of strength of sewage. Determination of amount of clear water required for the efficient disposal of wastewater by dilution. Diagram showing the combined BOD(Carbonaceous and Nitrification) demand curve  BOD The organic matter present in wastewater may belong to two groups: i) Carbonaceous matter ii) Nitrogenous matter The ultimate carbonaceous BOD of a liquid waste is the amount of oxygen necessary for the microorganisms in the sample to decompose the carbonaceous matter that are subject of microbial decomposition. This is the first stage oxidation and the corresponding BOD is also sometimes called the first stage demand. Biochemical oxidation is a slow process and theoretically takes an infinite time to go to completion, though the ultimate first stage BOD of a given wastewater is equal to the initial oxygen equivalent of the organic matter present. Generally, a 5 day period (at 200C) is chosen for standard BOD test, during which oxidation of 60- 70% is complete, while within 20 days period, the oxidation is about 95-99% complete. First stage BOD formulation At given temp. the rate of BOD at any time given to be directly proportional to the amount of organic matter present in sewage. In other words, the exertion of BOD is considered to be first order reaction defined by: ′ = − = amount of first stage BOD remaining in the sample at any time t (or oxygen equivalent of carbonaceous oxidisable organic matter present at any time t), mg/L. ′ = rate constant signifying the rate of oxidation of organic matter, day-1; depends on temperature and nature of organic matter present t = time, days. First stage BOD formulation Integrating between time t=0 (as Lt=L0, say) to t = t, we get: or, 0 − ′ or, = 0 loge  0= − ′ First stage BOD formulation or, 0= − ′ = 10− Where, rate constant K= ′ 2.303 In the above equation, L0is the oxygen equivalent of organic matter present in sewage at the beginning. Also, K is known as base 10 rate constant (or deoxygenation constant) while K’is base e rate constant. First stage BOD formulation The amount of BOD remaining at any time t is: = 0 10− Hence , the amount of BOD that has been exerted at any time t is given by: = = 0 − = 0 1 − 10− The ultimate first stage BOD (i.e. yu = BODult) will be obtained by substituting t = ∞; ∞ yu= 0 1 − 10− = 0  First stage BOD formulation yu = BODult = initial oxygen equivalent of organic matter present in the sample of WW. Fixed for the given specimen, and does not depend on the temperature during the reaction. K determines the speed of BOD reaction and depends upon: Type of WW Temperature during the reaction K at 20°C may vary between 0.05 to 0.3 day-1, depending upon type of waste. First stage BOD formulation For the same ultimate BOD (yu = BODult = L0)  First stage BOD formulation Typical value of K at 20°C for various type of water and wastewater.  First stage BOD formulation BOD test is usually conducted at 20°C, the reaction constant KT, at any temperature T°C, is related to the reaction constant K20(at 20°C) by the following approximate relationship, known as van’t Hoft-Arrhenious relationship: = 20 ( −20°) = 1.056 ( for 20°C - 30°C) = 1.135 ( for 4°C - 20°C) Sometimes a constant value of = 1.047 is adopted,  though this value does not apply in cold temperatures. Second stage BOD formulation Carbonaceous matter is oxidized in the first stage of biochemical reaction, while nitrogenous matter is oxidized in the second stage. Some of the autotrophic bacteria are capable of using oxygen to oxidized non carbonaceous matter, such as ammonia to nitrites and nitrates. This second stage reaction is called nitrification. Thus, the nitrogenous oxygen demand caused by the autotrophic bacteria is called the second stage BOD. At 20°C, the reproduction rate of nitrifying bacteria is very slow and it takes about 6-10 days to reach the significant numbers and to exert a measurable oxygen demand. The interference caused by their presence can be eliminated by pretreatment of sample, such as pasteurization, chlorination or acid treatment. Alternatively, certain inhibitory agents (such as methylene blue, thiourea, etc.) may be used.  Problem 1. Determine ultimate BOD for a sewage having 5-day BOD at 20°C as 160ppm. Assume the deoxygenation constant as 0.12 day-1. 5 = = 0 1 − 10− 160 = 0 1 − 10−0.12 ×5 0 = 213.7 ppm (or mg/L) Problem Determine the 3-day BOD of the above sample at 27°C Solution: First: or, =  20 ( −20°) 10− or, 3 = 213.7 (1 − 10−0.18 ×3) = 152 mg/L o , = 0 1 − or, 27 = 0.12 × 1.056(27−20°) or, 27 = 0.18 Homework The 4 day 15℃ BOD of a sample of sewage is 200 mg/L. Draw a graph of 5 day BOD as a function of temperature in the range of 10 ℃ t0 30 ℃ in steps of 5 ℃. (Take K at 20 ℃ = 0.1 day-1) Chemical Oxygen Demand (COD) Chemical Oxygen Demand (COD) is the measure of organic matter. It is oxygen equivalent of the organic matter when it is oxidized by  using a strong chemical oxidizing agent (potassium dichromate). The major advantage of this test over BOD is its short determination times i.e. 3 hours instead BOD requires 5 days for determination. In many cases this method is used as a substitute for BOD test where there is requirement for quick determination of the water quality. Catalyst (CaHbOc) + Cr2O7-- +H+ Cr3+ + CO2 + H2O Organic matter Heat Chemical Oxygen Demand (COD) There is less interference that can disturb during measuring compared with measuring BOD. However, the data for both have their individual significance and sometimes correlations between COD and BOD values are used to interpret the quality of the water sample.  The major drawback of COD test is its inability to differentiate between the biologically oxidizable and biologically inert organic matter. Similarly, this test doesn’t give any information on the rate at which the biologically active material would oxidized (stabilized) under conditions that existed in nature. COD test is specifically more suitable to measure organic matter present in industrial wastes having compounds that are toxic to biological life. COD > BOD, COD test will oxidize materials such as fats and lignins which are only slowly biodegradable. Chemical Oxygen Demand (COD) Inorganic substance that are oxidized by the dichromate increase the apparent organic content of the sample. Typical values for the ratio of BOD/COD for untreated municipal wastewater are in the range from 0.3 to 0.8. If the BOD/COD ratio for untreated wastewater is 0.5 or greater, the waste is considered to be easily treatable by biological means.  If the ratio is below about 0.3, either the waste may have some toxic components or acclimated microorganisms may be required in its stabilization. Total Organic Carbon (TOC) TOC test is applicable to small concentrations of organic matter. The TOC test methods utilize heat and oxygen, UV radiation, chemical oxidants, or some combination of these methods to convert organic carbon to carbon dioxide which is measured with an IR analyzer or by other means.  Theoretical Oxygen Demand (ThoD) Organic matter of animal or vegetable origin in wastewater is generally a combination of carbon, hydrogen, oxygen and nitrogen. Principal groups of these present in WW are: carbohydrates, proteins, oils and grease and products of their decomposition.  If chemical formula of the organic matter is known, the ThOD can be easily computed. Theoretical Oxygen Demand (ThoD) Methods of WW disposal  After conveying the WW through sewers, the next step is its disposal, either after treatment or even before its treatment.. Natural Methods: By Dilution By land treatment Artificial methods: Primary Treatment Secondary treatment Treated WW or effluent is either discharged in large water bodies such as rivers or streams or lake or sea. The discharged WW is purified in due course of time, by the so called self purification process of natural waters. Disposal by dilution Conditions favoring dilution without treatment: Where the WW is quite fresh, i.e. discharged within 2-3h of its collection. Where the floating matter and settleable solids have been removed Where water  body has large volume in comparison to the volume of wastewater Where the diluting water has high content of DO, so that not only the BOD is satisfied, but sufficient DO remains available for the aquatic life Where it is possible to thoroughly mix or diffuse the WW through the water body Where swift forward currents are available, so that there is no deposition of sewage at site Where the WW does not contain industrial WW having toxic substances, and Where the receiving water is not a source of drinking water collection immediately to the downstream side Self purification of Natural Streams  Self purification of Natural Streams Organic matter is broken down by bacteria to ammonia, nitrates, sulphates, carbon dioxides, etc.  In this process of oxidation, the DO content in natural water is utilized; defiency is created. As excess of organic matter is stabilized, the normal cycle will be restabilized in a process known as self purification wherein the oxygen is replenished by its reaeration of wind. Also, stable biproducts of oxidation mentioned above are utilized by plants and algae to produce carbohydrates and oxygen. Actions involved in self purification Dilution Dispersion due to currents Sedimentation Oxidation Reduction  Temperature Sunlight Problem (Water) A stream, (shown in the following figure) flowing at 150 L/s and 20 mg/L suspended solids, receives wastewater from three separate sources: What are the flow rate and suspended solids concentration downstream at the sampling point? [2+2]  In the beginning, let’s draw a MBD C0 = 20 mg/L QA = 100 lps CA = 200 mg/L QC= 50 lps QS = ? lps CC = 200 mg/L CS = ? mg/L Q0 = 150 lps QB= 300 lps CB = 50 Firstly, = mg/L MBE is: − − + Balancing the volumetric rate:(i.e. Discharge): =  − 39 + − Assuming the steady state condition and conservative pollutant: 0 = or, = +0−0 − or, Q0 + QA + QB+ QC= QS or, QS = 150 +100 + 300 + ∴ QS= 600 lps 50 Balancing the Mass flow rate (i.e. Rate of Suspended Solid (SS)) in the system: Assuming same as above:  0= − +0-0 or, Q0C0 + QACA + QBCB + QCCC = QSCS (150×20+100×200+300×50+50×200) QS= × Q or, CS = 0C0+QACA+QBCB+ QCCC ∴ 600 = / Unit Operations and Processes for Wastewater Treatment Methods of treatment in which the application of physical forces predominates are known as unit operations; chemical/biological activities are involved are known as unit processes.  Combination of unit operation and processes can reduce the objectionable properties of water-carried waste and render it less dangerous and repulsive to man. There are three types of unit operation and processes: Physical unit operations Chemical unit processes Biological unit processes  For industrial WW, the objective is to remove/reduce the concentration of organic and inorganic compounds which may be otherwise toxic. Kinetics of Bacterial Growth  The general term that describes all of the chemical activities performed by a cell is metabolism. This in turn is divided into two parts: catabolism and anabolism. Catabolism includes all the biochemical processes by which a substrate is degraded to end products with the release of energy. Anabolism includes all the biochemical processes by which the microorganisms synthesizes new chemical compounds needed by the cells to live and reproduce. The synthesis process is driven by the energy obtained from catabolism. The performance of biological waste treatment can be measured by the rate at which microorganisms metabolize the waste which, in turn, is directly related to their rate of growth. Kinetics of Bacterial Growth Bacteria reproduce by binary fission (i.e. by dividing, the original cell becomes two new organisms) and the generation time is 15-20 minutes.  The relationship of cell growth and food (substrate) utilization can be illustrated by a simple batch reactor such as stoppered bottle. A given quantity of food containing all the necessary nutrients is placed in the bottle and inoculated with a mixed culture of microorganisms. If S represents the quantity of soluble food (in mg/L) and X represents the quantity of biomass (in mg/L), the rate of utilization of food dS/dt and the rate of biomass growth dX/dt can be represented by growth curves. 54 Kinetics of Bacterial Growth Biomass growth  and food utilization 55 Kinetics of Bacterial Growth Rate of growth varies depending on the growth phase.  During the lag phase immediately after inoculation, rate of growth is essentially zero. Cells use the lag phase to adapt to their new environment; new enzymes or structural components may be synthesized. Following the lag period, growth starts in the acceleration phase and continues through the growth and decline phases. 56 Kinetics of Bacterial Growth There are several distinct segments in the biomass curve as discussed below. i. Lag phase :  The lag phase represents the time required for the microorganisms to acclimate to their new environment. If the organisms have been accustomed to a similar environment and similar food, the lag phase will be very brief. 57 Kinetics of Bacterial Growth ii. Log-growth phase: During this period the cells divide at a rate determined by their generation time and their ability to process food. There is always excess amount of food surrounding the microorganisms, and the rate of metabolism and growth is only a function of the ability of the microorganisms to process the substrate. During this phase maximum growth occurs at a logarithmic rate. Anish Ghimire, Eng., Ph.D. 58 Kinetics of Bacterial Growth iii. Stationary phase : Maximum growth cannot continue indefinitely. The food supply may become limiting, environmental conditions may change and a population of grazers may develop. Cells that are unable to obtain food from external sources begin endogenous catabolism. The stationary phase represents the time during which the production of new cellular material is roughly offset by death and endogenous respiration.  59 Kinetics of Bacterial Growth iv. Endogenous phase: During this phase, the death rate of organisms exceeds the production of new cells. During this phase, a phenomenon known as 'lysis' can occur in which the nutrients remaining in the dead cells diffuse out to furnish the remaining cells with food. 60  Kinetics of Bacterial Growth In both batch and continuous culture, the rate of growth of bacterial cell can be represented as: = = where rg = rate of bacterial growth (mass/unit volume. time) (mg/L. day-1 ) (cells / L. day) X = bacterial concentration (mg/L) (cell/L) = growth rate constant, day-1 61 Monod Equation  Direct evaluation of the growth rate constant is impossible for mixed cultures of microorganisms metabolizing mixed organics. Several models have been developed to establish a value of. The most widely accepted is the Monod equation. This equation assumes that the rate of food utilization, and therefore the rate of biomass production, is limited by the enzyme reactions involving the food compound. 62 Monod Equation The Monod equation is = m + Where m = maximum growth rate constant, day-1 S = concentration of growth limiting substrate, mg/L Ks = Half velocity constant (saturation constant), mg/L or concentration of limiting substrate when = ½ m 63