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click here for freelancing tutoring sitesYANGON TECHNOLOGICAL UNIVERSITYDEPARTMENT OF CIVIL ENGINEERINGCE 5018ENVIRONMENTAL ENGINEERING (II)SAMPLE QUESTIONS AND SOLUTION Question 1Define the following terms(a)Aeration period(b)BOD loading(c)F/M ration(d)Return sludge rateSolution(a) Aeration periodAeration period in an activated sludge process is equal to volume of the aeration basindivided by flow of raw water. Since biological filters do not contain a liquid volume, hydraulicloading is presented as the amount of waste water applied per unit of surface area, for example, cubicmeters per square meter per day.(b) BOD loadingOrganic loading on biological treatment units are stated in terms of kilograms of applied 5day BOD. Loading on an aeration basin is commonly expressed as grams BOD applied per cubicmeter of tank volume per day. A biological filter loading uses the same units except the volumerefers to the quantity of media rather than liquid volume.(c) F/M rationThe food to micro-organism ration (F/M) is a way of expression BOD loading with regard tothe microbial mass in the system. The F/M value as grams of BOD applied per day per gram ofMLSS in the aeration tank. F/M ration as an expression of BOD loading relates to the metabolic stateof the biological system rather than to the volume of aeration basin.(d) Return sludge rateThe rate of return studge from the final clarifier to the aeration basin is expressed inpercentage of the raw waste water influent. If the return activated sludge rate is 30 percent and theraw waste water flow into the plant is 1.0 m/s, the recirculated flow equal 0.3 m/s, BOD removed33in aeration and subsequent setting by the raw BOD entering.*************************** Question 2Explain briefly the effects of organic pollution on stream.SolutionMany waste discharges from municipalities and industries contain organic compound thatdecompose using dissolved oxygen. The rate of biological stabilization is a time-temperaturefunction with deoxygenation increasing as the temperature rises. Oxygen is replenished in thestream water primarily by reaeration from the atmosphere. Thus, quantity of flow, time of passagedown the river, water temperature and reaeration and reaeration are the four major factors thatgovern self-purification from organic wastes.A stream polluted from a substantial point source of organic matter exhibits four fairly welldefined zones. The zone of degradation, immediately following the sewer outfall, has a progressivereduction of dissolved oxygen used up in satisfying BOD. The zone of active decomposition exhibitsthe characteristics of significant pollution. Dissolved oxygen is at a minimum level and oftenanaerobic decomposition of bottom mud results in offensive odors. Higher forms of life, particularlyfishes, find the environment of these polluted ones undesirable. Bacteria and fungi thrive on thedecomposition of organic decreasing the BOD and increasing ammonia nitrogen. In the zone ofrecovery, reaeration exceeds the rate of deoxygenation and the level of dissolved oxygen increasesslowly. Ammonia nitrogen is converted biologically to nitrate. Rotifer crustaceans and tolerant fishspecies reappear. Algae thrive on the increase in inorganic nutrients that result from the stabilizationof the matter. The zone of clean water supports a wide variety of aquatic plants and animals andmore sensitive fishes. Dissolved oxygen returns to its original value, and the BOD has been nearlyeliminated. The permanent changes in water quality prior to the waste discharge and the clear waterzone include an increase in inorganic compounds, such as nitrate, phosphates, and dissolved salts.These nutrient produce and support higher algal population other environmental conditions ofsunlight, pH and temperature are adequate.*********************** Question 3What is waste stabilization pond? Explain briefly about facultative ponds.SolutionStabilization ponds, also called lagoons or oxidation ponds are generally employed assecondary treatment in rural areas. Ponds are classified as facultative, tertiary, aerated, and anaerobicaccording to the type of biological activity that takes place in them.Facultative Ponds are the most common lagoons employed for stabilizing municipal wastewater. The bacterial reactions include both aerobic and anaerobic decomposition and, hence, the termfacultative pond. Waste organics in suspension are broken down by bacteria releasing nitrogen andphosphorus nutrients, and carbon dioxide. Algae use these inorganic compounds for growth, alongwith energy from sunlight, releasing oxygen to solution. Dissolved oxygen is in turn taken up by thebacteria, thus closing the symbiotic cycle. Oxygen is also introduced by reaeration through windaction. Settleable solids decomposed under anaerobic conditions on the bottom yield inorganicnutrients and odorous compounds, for instance, hydrogen sulfide and organic acids. The latter aregenerally oxidized in the aerobic surface water thus preventing their emission to the atmosphere.Bacterial decomposition and algal growth are both severely retarded by cold temperature.During winter when pond water is only a few degrees above freezing, the entering waste organicsaccumulate in the frigid water. Microbial activity is further reduced by ice and snow cover thatprevents sunlight penetration and wind reaeration. Under this environment, the water can becomeanaerobic causing odorous conditions during the spring thaw, until algae become reestablished. Thismay take several weeks depending on climatic conditions and the amount of waste organicsaccumulated during the cold weather.Operating water depths range from 0.6 to 1.5 m, with 0.9m of dike freeboard above the highwater level. The minimum 0.6 m depth is needed to prevent growth of rooted aquatic weeds, butexceeding a depth of 1.5 m may create excessive odors because of anaerobiosis on the bottom.Facultative ponds treating only domestic waste water normally operate odor free except for ashort period of time in the spring of the year. On the other hand, lagoons treating municipal wastewater that include industrial wastes can produce persistent obnoxious odors. Often this is the result oforganic overload from food-processing industries or a result of the odorous nature of the industrialwastes itself or both. The best solution is to require pretreatment of the offending waste waters priorto discharge to the sewer system.Facultative ponds are best suited for small towns that do not anticipate industrial expansion,and where extensive land area is available for construction and effluent disposal. The advantages oflow initial cost and ease of operation, as compared to a mechanical plant, can be offset byoperational difficulties. The key problems are poor assimilative capacity for industrial wastes,odorous emission, and meeting the minimum effluent standards for disposal in surface waters. Question 4.Explain briefly about extended aeration.SolutionThe most popular application of this process is in treating small flows from schools,subdivisions, trailer parks, and villages. Aeration basins may be cast-in-place concrete or steel tanksfabricated in a factory. Continuous complete mixing is either by diffused air or mechanical aerators,and aeration periods are 24 to 36 h. Because of these conditions, as well as low BOD loading, thebiological process is very stable and can accept intermittent loads without upset. For example, a unitserving a school may receive waste water during a 10 h period each day, for only five days a week.Clarifiers for small plants are conservatively sized with low overflow rates, ranging from 8 to24m3/m2.d, and long detention times. Sludge may be returned to the aeration chamber through a slotopening by employing an air-lift pump. A slot return requires periodic cleaning to prevent pluggingby settled solids. Although satisfactory performance can be achieved, returning settled solids bypumping provides a more positive process control. Sludge that floats to the surface of thesedimentation chamber is returned to the aeration tank by either hydraulic action or through askimming device attached to the air-lift pump return.There is usually no provision for wasting of excess activated sludge from small extendedaeration plants. Instead, the mixed liquor is allowed to increase in solids concentration over a periodof several months and then is discharged directly from the aeration basin. This is performed byallowing the suspended solids to settle in the tank with the aerators off, and then pumping theconcentrated sludge from the bottom into a vehicle for hauling away. The MLSS operating rangevaries from a minimum of 1,000 mg/l to a maximum of about 10,000 mg/l. In treating domesticwaste water under normal loading, the mixed liquor concentration increases at the rate ofapproximately 50 mg/l suspended solids per day.Larger extended aeration plants consisting of an aeration basin, clarifier and aerobic digesterare used by small municipalities. The basin may be a concrete tank with diffused aeration, a linedearth basin with mechanical aerators, or a race-track-shaped oxidation ditch. Final clarifiers aregenerally separate circular concrete tanks with mechanical sludge collectors. In most cases, the BODloading of these systems is at the upper limit of 500 g/m-d of BOD for extended aeration, and the3aeration period is as low as 12h. The treatment plants, in addition to lighter loading, are distinguishedfrom conventional and step aeration processes by applying unsettled waste water directly to aerationwithout primary settling.********************************* Question 5What is eutrophication? Explain briefly the process of eutrophication.SolutionEutrophication is the process whereby lakes become enriched with nutrients that result inwater quality characteristics undesirable for mans useof the water, both for water supplies andrecreation. Limnologists categorize lakes according to their boilogical productivity. Oligotropic lakesare nutrient poor. Typical examples are a cold-water mountain lake and a sand-bottomed, spring-fedlake characterize by transparent water, very limited plant growth, and low fish production. A slightincrease in fertility results in a mesotrophic, lake with some aquatic plant growth, greenish water,and moderate production of game fish. Eutrophic lakes are nutrient rich. Plant growth, in the formsof microscopic algae and rooted aquatic weeds, produces a water quality undesirable for body-contact reaeration.The process of eutrophication is directly related to the aquatic food chain. Algae use carbondioxide, inorganic nitrogen, orthophosphate, and trace nutrients for growth and reproduction. Theseplants serve as food for microscopic animals (zooplankton). Small fishes feed on zooplankton, an dlarge fishes consume small ones. Productivity of the aquatic food chain is keyed to the availability ofnitrogen and phosphorus, often in short supply in natural waters. The amount of plant growth andnormal balance of the food chain are controlled by the limitation of plant nutrient. Abundantnutrients unbalance the normal succession and promote blooms of blue-green algae that are noteasily utilized as food by zooplankton. Thus the water becomes turbid and under extreme conditionstakes on the appearance of pea soup. Floating masses of algae are windblown to the shore wherethey decompose producing malodors. Decaying algae also settle to the bottom reducing dissolvedoxygen. Shorelines and shallow bays become weed-choked with the prolific growth of rootedaquatics. Preferred food-fishes cannot survive in these unfavorable increasingly tolerant fishes. Troutare succeeded by warm-water fishes such as perch, walleye, and bass, and these in turn aresucceeded by coarse fishes like bullheads and carp.Relatively mild algal bloom can result in accumulation of substantial decaying scum alongthe windward shoreline because of the lakes vast surface area. Gentle winds passing over the lakecan pick up fishy odors from algae blooms. Algal growth developed in the sunlit epilimnion cansettle into the hypolimnion, which is dark and stagnant during lake stratification. Bacterialdecomposition of these cells and organic bottom muds deplete dissolved oxygen in the bottom zone.Reduced dissolved oxygen has been associated with the reduction of commercial fishing in someeutrophic lakes, and treatment of water supplies is complicated by removal of tastes and odorscreated by algal blooms in the epilimnion and anaerobic conditions in the hypolimnion. Perhaps themost devastating aspect of eutrophication is that process appears to be difficult to retard, expect inunusual cases. Once a lake has become eutrophic it remains so, at any rate for a very long time, even if nutrients from point sources are reduced. In part, this results from the long turnover time (detentiontime) which reduces the rate of flushing.Macronutrients for plant growth are carbon dioxide, inorganic nitrogen, and phosphate; avariety of trace elements, such as iron, are also needed for growth. The key to controlling rate of lakeeutrophication lies in limiting plantnutrients. Natural waters contain sufficient carbon in thebicarbonate alkalinity system to provide carbon dioxide in excess of growth needs. At present,emphasis is being placed on phosphorus reduction to control the extent of plant growth in lakes.Removal of algae has been suggested as a means for reducing nuisance species andwithdrawing nutrients. Copper sulfate is commonly used for control of algae in water supplyreservoirs. This algicide has several shortcomings as a eutrophication control device. Copper sulfatepoisons fish when used in excessive concentrations and has been demonstrated to accumulate inbottom muds of lakes following application over a period of several years. In fertile lakes, the coppersulfate must be applied at intervals throughout the growing season to ensure effective algalcontrol;blooms must be anticipated and treated before they occur. This process is very expensive inboth man-power and chemical costs. A large number of herbicides are available to provide relieffrom aquatic weeds, both preemergent and emergent.***************************** Question 6How to determine the performance evaluation of treatment plants?SolutionComprehensive studies are needed to determine treatment efficiency and economicaloperation in waste-water processing. As-built drawings of all treatment units provide dimensions oftanks and interconnecting piping. From these, a process flow diagram can be sketched showingnormal plant operation. Changes in plant operation to meet unusual flow and load conditions shouldbe noted. Physical facilities for flow-measuring and sampling at various points are essential. Theinfluent par shall flume should be checked accuracy, since significant error in raw waste-water flowsprecludes satisfactory results. Flow measuring at various points within the plant may beaccomplished by calibrating pump discharges, or by installing temporary weirs in flow channels.Sampling points must be carefully selected to insure collection of representative portions forcomposting. Often, a lack of adequate flow-measuring facilities and accessibility for sampling in-plant jeopardizes or prevents the study of individual unit operations. Physical modifications may beneeded in some plants to permit evaluation. Finally, a laboratory facility is needed to perform at leastroutine tests, such as total and suspended solids, BOD, pH, fecal coliforms, and chlorine residual.Metropolitan plants require additional equipment for analyses of COD, grease, alkalinity, phosphates,various forms of nitrogen, sulfides, volatile acids, gas analysis, sludge filterability, and biologicaloxygen uptake. In addition it may be desirable to have testing facilities for heavy metals and totalorganic carbon.The person supervising a plant evaluation must be thoroughly familiar with each unitoperation and how it fits into the overall plant process. Characteristics of the raw waste water mustbe completely defined by flow patterns, waste strength parameters, types of industrial waste, andinfiltration-inflow quantities. Gathering these data relies on a year-round testing and samplingprogram of influent waste water and monitoring of industrial wastes discharged to the sewer system.To insure that the degree of treatment is satisfactory, the superintendent should be familiar with local,state, and federal water quality and effluent standards. A complete study also includes a review ofmaintenance procedures and accurate records of operating costs.The process diagram for a typical treatment plant in figure indicates the minimum testingprogram for evaluation. Effluent standards require daily monitoring of average BOD and suspendedsolids concentrations, pH, and fecal coilform counts. In some locations, tests for chlorine residual,presence of heavy metals, phosphates, and ammonium nitrogen content may also be specified byregulatory agencies. Percentage of organic matter removal is traditionally calculated by comparinginfluent and effluent BOD and suspended solids. Examination of individual unit operations within a treatment plant requires testing of allinfluent and effluent flow streams. Also, loading parameters should be calculated for each unitprocess to determine whether the system is being stressed beyond its intended design capacity.Complete and accurate records of all phases of plant operation and maintenance are essential.************************** Question 7Define the terms of solid waste, municipal waste, hazardous waste and its source.SolutionSolid wastes are all the wastes arising from human and animal activities that are normallysolid and that are discarded as useless or unwanted.Industrial wastes are those wastes arising fromindustrial activities and typically include rubbish, ashes, demolition and construction wastes, specialwastes, and hazardous wastes. Wastes that pose a substantial danger immediately of over a period oftime to human, plant, or animal life are classified as hazardous wastes. A waste is classified ashazardous it if exhibits any of the following characteristics: (1) ignitability, (2) corrosivity, (3)reactivity, or (4) toxicity.In the past, hazardous wastes were often grouped into the following categories :(1)radioactive substances, (2) chemicals, (3) biological wastes. (4) Flammable wastes, and (5)explosives. The chemical category includes wastes that are corrosive reactive, or toxic. One lieprincipal sources of hazardous biological wastes are hospitals and biological research facilities.Hazardous wastes are generated in limited amounts throughout most industrial activities. Interms of generation, the concern is with the identification of the amounts and types of hazardouswastes developed at each source, with emphasis on those sources where significant waste quantitiesare generated. Unfortunately, very little information is available on the quantities of hazardouswastes generated in various industries.The spreading of hazardous wastes by spillage must also be considered. The quantities ofhazardous wastes that are involved in spillages usually are not known. After a spill, the wastesrequiring collection and disposal are often significantly greater than the amount of spilled wastes,especially where an absorbing material, such as straw, is used to soak up liquid hazardous wastes orwhere the soil into which a hazardous liquid waste has percolated must be excavated. Both the strawand (lie liquid and the soil and the liquid are classified as hazardous wastes).**************************** Question 8Write down the sampling procedure for analyzing solid waste.SolutionPerhaps the most difficult task facing anyone concerned with the design and operation ofsolid-waste management systems is to predict the composition of solid wastes that will be collectednow and in the future. The problem is complicated because of the heterogeneous nature of wastematerials and the fact that unpredictable externalities such as world oil prices can affect the long-term abundance of the individual waste components.The load-count and the mass-volume methods of analysis are recommended. The followingtechnique is recommended to assess the individual components within a given waste category (e.g.,domestic wastes).1.Unload a truckload of wastes in a controlled area away from other operations.2.Quarter the waste load.3.Select one of (lie quarters and quarter that quarter.)4.Select one of the quartered quarters and separate all of the individual components ofthe waste into pre-selected components.5.Place the separated components in a container of known volume and tarce mass andmeasure the volume and mass of each component. The separated components shouldbe compacted tightly to simulate the conditions in the storage containers from whichthey were collected.6.Determine the percentage distribution of each component by mass and the discardeddensity. Typically, from 100 to 200 kg (200 to 400 lb) of waste should be sorted toobtain a representative sample. To obtain a more representative distribution ofcomponents, samples should be collected during each season of the year. Clearly, nomatter how many samples are analyzed, common sense is needed in selecting theloads to be sorted, in analyzing the data, and in preparing projections.************************** Question 9Write down the sources of air pollutant.SolutionAll air contains natural contaminants such as pollen, fungi spores, salt spray, and smoke anddust particles from forest fires and volcanic eruptions. It contains also naturally occurring carbonmonoxide (CO) from the breakdown of methane (CH); hydrocarbons in the form of trepans from4pine trees; and hydrogen sulfide (HS) and methane (CH) from the anaerobic decomposition of24organic matter.In contrast to the natural sources of air pollution there are contaminants of anthropogenicorigin. The use of fossil fuels for heating and cooling, for transportation, for industry, and for energyconversion, and the incineration of the various forms of industrial, municipal, and private waste allcontribute to the pollution of the atmosphere. So do the handling and processing operations ofvarious and sundry industries. The sources of these pollutants are so numerous and varied that theyhave been categorized into four main groups-mobile transportation. (i.e., motor vehicles, aircraft,railroads, ships, and the handling and/or evaporation of gasoline) stationary combustion (i.e.,residential, commercial, and industrial power and heating, including steam-powered electric powerplants), industrial processes (i.e., chemical, metallurgical, and pulp-paper industries and petroleumrefineries), and solid-waste disposal (i.e., household and commercial refuse, coal refuse, andagricultural burning).It will be noted that while transportation was the single largest source of air pollution, fuelcombustion in stationary sources (for power and heating) was the second major contributor. Powergeneration and heating accounted for about 80 percent of the oxides of sulfur and 51 percent of theoxides of nitrogen emitted to the ambient air, while industrial processes contributed 50 percent of thehydrocarbons.********************* Question 10Explain the effect of acid rain.SolutionThe effects of acid deposition vary according to the sensitivity of the ecosystems upon whichthe deposits fall. In some highly buffered areas acidic compounds could be deposited for yearswithout causing any appreciable increase insoil or surface-water acidity, but the same depositioncould cause sharp increases in acidity in poorly buffered areas. Acid rain has cause considerabledamaged to buildings and monuments in highly industrialized areas, but damage is not limited to theimmediate area of industrialization. Tall stacks disperse pollutants into the upper reaches of thetroposphere where they may remain for days, often being carried long distances. Pollutants that aregenerated in one country and deposited in another have become a matter of international concern andof international negotiation.Other changes in the atmosphere as a whole may not be quite so obvious. For example, thetwentieth century has seen side spread use of radioactive materials, and concern over the long-rangeeffects of release of these substances into the atmosphere has led to investigation of possible methodsof safe disposal by deep burial in the earth or ocean.The ozone (O) layer in the stratosphere is being depleted as ozone reacts with chlorine3released from the fluorocarbons used as aerosol spray propellants. Since the Oin the atmosphere3reduces the ultraviolet radiation that reaches the earths surface, and since ultraviolet radiation at highlevels can damage plants and animals, loss of Orepresents a potentially serious problem. In light of3this danger, some industrialized nations have banned the use of fluorocarbons.The amount of tropospheric carbon dioxide (CO) is reported to be increasing at a rate of 1.82mg/mper year, a process that may not be reversible. Furthermore, this increase has been3accompanied by an equivalent decrease in atmospheric oxygen (O). Currently, there is more than2700 billion tons of carbon in the form of COin the atmosphere. Each year this figure increases by22.3 billion tons, the equivalent of a 3 percent increase every decade.Fossil fuel consumption and agricultural, forestry and land use practices of various typescontribute to the CO build up. CO strongly absorbs long wave (infrared) terrestrial radiation, andcontinued CObuildup could lead to a significant enough rises in earths surface temperatures to melt2the Arctic ice pack. If the warming trend can be confirmed and positively linked to CObuildup,2then global action such as reforestation may eventually have to be pursued to remove COfrom the2atmosphere.*************************** Question 11Write down the carbon monoxide in air and discuss the sources of carbon monoxide.SolutionColorless, tasteless, and odorless, carbon monoxide gas is chemically inert under normalconditions and has an estimated atmospheric mean life of about 2months. The total emission ofCO on a mass basis in 1977 accounted for slightly, over all (53 percent) of all the anthropogenic airpollutants.Carbon monoxide at present ambient levels has little if any effect on property, vegetation, ormaterials. At higher concentrations, it can seriously affect human aerobic metabolism, owing to itshigh affinity for hemoglobin, the component of the blood responsible for the transport of oxygen.Carbon monoxide reacts with the hemoglobin (Hb) of blood to give carboxyhemoglobin(COHb).Thus reducing the capability of the blood to carry oxygen. Since the affinity of hemoglobin forcarbon monoxide is more than 200 times as great as its affinity for oxygen, CO can seriously impairthe transport of O, even when present at low concentrations. As COHb levels increase, effects2become more and more severe.The absorption of CO by the body increases with CO concentration, exposure duration, andthe activity being performed. Carbon monoxide concentrations are especially high in congestedurban areas where traffic is heavy and slow-moving.Carbon monoxide sources are both natural and anthropogenic. 3billion tones of CO areproduced in nature yearly by the oxidation of methane gas from decaying vegetation. Still anothersource is human metabolism. The exhalations of a resting person contain approximately 1 ppm CO.Applied to the entire nation, this would total about 2.9 tones of CO produced each day.Yet this production is still a great deal less than the estimated 69.1 million tones produced in1980 by transportation sources primarily gasoline-powered internal combustion engines. Thequantities of CO emission from the four major groups transportation, fuel combustion in stationarysources (power, heating), industrial purposes and solid waste disposal. On a weight basis, the totalestimated emission of CO from transportation in 1968, 1970, 1975, 1977, and 1980 was about 78.3percent of the total CO emitted by all sources combined. In 1980, the next largest source ofanthropogenic carbon monoxide was solid-waste disposal and miscellaneous causes, which includedforest fires, structural fires, coal refuse, and agricultural burning.On a mass basis, the emissions of carbon monoxide from anthropogenic sources havedropped from 137 million tones in 1968 to 85 million tones in 1980. This reduction has taken placemainly in the automotive area, owing to the initiation in 1968 of pollution control devices. Even atpresent levels of emission, were it not for the natural processes of removal, the CO content of theatmosphere would be increasing at the rate of about 0.5 ppm yearly.************************* Question 12Discuss conventional and step aeration briefly.SolutionThese processes are similar to the activated sludge systems that were constructed forsecondary treatment of municipal waste water. The aeration basin is a long rectangular tank with airdiffusers along one side for oxygenation and spiral flow mixing. In a conventional basin, the airsupply is tapered along the length of the tank to provide greatest aeration at the head end where rawwastewater and return activated sludge are introduced. Air is provided uniformly in step aerationwhile wastewater is introduced at intervals, or steps, along the first portion of the tank. Fine-bubbleair diffusers are set at a depth of 2.5m or more to provide deep mixing and adequate oxygen transfer.The air header is connected to a jointed arm so that diffusers can be swung out of the tank forcleaning and maintenance. A variety of diffusers are marketed; two popular types are synthetic clothtubes, easily removed for laundering, and diffuser nozzles that can be detached from the air headerpipe.Plug-flow pattern of long rectangular tanks produces an oscillating biological growth pattern.The relatively high food -to-microorganism ratio at the head of the tank decreases as mixed liquorflows through the aeration basin. Since the aeration period, is 6 to 8 h, and can be considerablygreater during low flow, the microorganisms move into the endogenous growth phase before theirreturn to the head of the aeration basin. This weak starving microbial population must quickly adaptto a renewed supply of waste organics. The process has few problems of instability where waste-water flows are greater than 2,000 m3/d; however, because of wide hourly variations in waste loadsfrom small cities, the conventional plug-flow system can experience serious problems of biologicalinstability. This phenomenon was a major factor contributing to the development of complete mixingaeration for handling small flows.**************************** Question 13(a)Two rapid sludge return final clarifiers following high rate aeration are 18m in diameter witha 2.7 m side water depth. The effluent weir is an inboard channel set on a diameter of 16.5 m. For atotal flow of 13000 m3/d, calculate the overflow rate, detention time, and weir loading.SolutionQQ13000V====25.54m/m-d32Overflow rateApdp180222244pdp1822H22.72V44Detention time t =24=24=24=2.54hQQ13000QQ13000Weir loading ====57.47m/m-d3weirlength2pd22p182Trickling FilterLow rate T/F ? no Q, BOD loading = 250 g/ m-d3RSingle stage T/FHigh rate T/F ? Allow QRTwo stage T/FHigh BOD loading750 g/ m3-dQuestion 13(b)What is Hydrocarbons? Write down the types of Hydrocarbons.SolutionOrganic compounds containing only carbon and hydrogen are classified as hydrocarbons.Most of the major chemicals in gasoline and other petroleum products are hydrocarbons, which aredivided into two major classifications- aliphatic and aromatic.******************** Question 14A trickling filter plant has the following: a primary clarifier with a 16.8 m diameter, 2.1 mside water depth, and single peripheral weir; a 26.0 m diameter trickling filter with a 2.1 m deeprock-fill setting tank with a 15.2 m diameter, 2.1 m side water depth, and single peripheral weir. Thenormal operating recirculation ratio is 0.5 with return to the wet well from the bottom of the finalduring periods of low influent flow. The daily waste-water flow is 5220 m/d with average BOD of3180 mg/l, essentially all domestic waste. Calculate the loadings on all of the units, and theanticipated effluent BOD at 20 C and 16 C.SolutionPrimary Settling TankQQ5220Overflow rateV====23.55m/m-d(Without Q)32RApdp16.802244Q+Q5220+52200.5V===35.32m/m-d(With Q)32RRAp16.8024pdp16.822H2.1V44Detention time t =24=24=24=2.14hQQ5220QQ5220Weir loading ====98.9m/m-d3weirlengthpdp16.8BOD removal efficiency = 35%Remaining BOD = 65% of 180 mg/l = 0.65 x 180 = 117 mg/lTrickling FilterQSettledBOD5220117BOD loading ===547.77g/m-d3pVol262.124Q+Q5220+52200.5Hydraulic loading ===14.75m/m-dR32Ap2624BOD removal efficiency at 20 C = 78 % (Table 2-17)BOD removal efficiency at 16 C = 68 % (Table 2-19)Remaining BOD at 20 C = 22% of 117 mg/l = 0.22 x 117 = 25.74 mg/lRemaining BOD at 16 C = 32% of 117 mg/l = 0.32 x 117 = 37.44 mg/l Final ClarifierQQ5220Overflow rateV====28.44m/m-d(Without Q)32RApdp15.202244Q+Q5220+52200.5V===43.15m/m-d(With Q)32RRAp15.2024pdp15.222H2.1V44Detention time t =24=24=24=1.75hQQ5220QQ5220Weir loading ====109.31m/m-d3weirlengthpdp15.2Overall Treatment Plant efficiency at 20 C,180-25.74(OR)E=100=85.7%18035E3578.0E=100-1001-1-=100-1001-1-=85.7%1100100100100Overall Treatment Plant efficiency at 16 C,180-37.44(OR)E=100=79.2%18035E3568.0E=100-1001-1-=100-1001-1-=79.2%1100100100100************************* Question 15The design flow for a two-stage trickling filter process is 4400 m3/d with an average BODconcentration of 400 mg/l. Calculate the unit loadings and treatment plant efficiency of waste-watertemperature of 17 C.Primary tank surface area = 220 m2First-stage filter area = 450 m2Volume = 830 m3Intermediate clarifier area = 140 m2Second-stage filter is identical to first stage.Final clarifier area = 140 m2Recirculation pattern return to wet well is 1100 m3/d under flow of each clarifier for total'Q=2200m3/d direct recirculation around each filterQis 2930 m3/d.RRSolutionPrimary ClarifierQ4400Overflow rateV===20m/m-d32A2200BOD removal = 35%Remaining BOD = 65% of 400 mg/l = 0.65 x 400 = 260 mg/lFirst-stage T/FQSettledBOD4400260BOD loading ===1378.3g/m-d3Vol830'Q+2Q+Q4400+2200+2930Hydraulic loading ===21.18m/m-d32RRA4502Q+Q'2200+2930Recirculation Ratio = 1.17R===RRQ4400BOD removal efficiency at 20 C = 72 %BOD removal efficiency at 17 C = 65 %Remaining BOD = 35% of 260 mg/l = 0.35 x 260 = 91 mg/lIntermediate ClarifierQ+Q4400+1100Overflow rateV===39m/m-d32RA1400Second-stage T/FQSettledBOD440091BOD loading ===482.41g/m-d3Vol830 Q+Q+Q'4400+1100+2930Hydraulic loading ===18.73m/m-dRR32A450Q+Q'1100+2930Recirculation Ratio = 0.92R===RRQ4400ActualBODloading482.41Adjusted BOD loading ===3938.04100-E100-65221100100BOD removal efficiency at 20 C = 58 %BOD removal efficiency at 17 C = 53 %Remaining BOD = 47% of 91 mg/l = 0.47 x 91 = 42.77 mg/lFinal ClarifierQ4400Overflow rateV===31.43m/m-d32A1400Overall Treatment Plant efficiency at 17 C,400-42.77(OR)E=100=89.3%18035EE3565.053E=100-1001-1-1-=100-1001-1-1-=89.3%12100100100100100100**************************** Question 16The design flow for a two-stage trickling filter process is 4400 m3/d with an average BODconcentration of 280 mg/l. Calculate the unit loadings and treatment plant efficiency of waste-watertemperature of 17 C.Primary tank surface area = 220 m2First-stage filter area = 450 m2Volume = 830 m3Intermediate clarifier area = 140 m2Second-stage filter is identical to first stage.Final clarifier area = 140 m2Recirculation pattern return to wet well is 1100 m3/d under flow of each clarifier for total'Q=2200m3/d direct recirculation around each filterQis 2930 m3/d.RRSolutionPrimary ClarifierQ4400Overflow rateV===20m/m-d32A2200BOD removal = 35%Remaining BOD = 65% of 280 mg/l = 0.65 x 280 = 182 mg/lFirst-stage T/FQSettledBOD4400182BOD loading ===904.82g/m-d3Vol830'Q+2Q+Q4400+2200+2930Hydraulic loading ===21.18m/m-d32RRA4502Q+Q'2200+2930Recirculation Ratio = 1.17R===RRQ4400BOD removal efficiency at 20 C = 75 %BOD removal efficiency at 17 C = 68 %Remaining BOD = 32% of 182 mg/l = 0.35 x 182 = 58.24 mg/lIntermediate ClarifierQ+Q4400+1100Overflow rateV===39m/m-d32RA1400Second-stage T/FQSettledBOD440058.24BOD loading ===308.74g/m-d3Vol830 Q+Q+Q'4400+1100+2930Hydraulic loading ===18.73m/m-dRR32A450Q+Q'1100+2930Recirculation Ratio = 0.92R===RRQ4400ActualBODloading482.41Adjusted BOD loading ===3015.04100-E100-68221100100BOD removal efficiency at 20 C = 62 %BOD removal efficiency at 17 C = 56 %Remaining BOD = 44% of 58.24 mg/l = 0.47 x 58.24 = 25.63 mg/lFinal ClarifierQ4400Overflow rateV===31.43m/m-d32A1400Overall Treatment Plant efficiency at 17 C,400-25.63E=100=89.3%180******************************** Question 17(a)A high-rate trickling filter is a 21 m diameter and 2.1 m depth. The raw waste-water flow is3000 m/d with 130 mg/l BOD. Indirect recirculation is during low flow period is 1500 m/d and33direct recirculation is 38 l/s. Calculate the BOD and hydraulic loadings, recirculation ratio, BODremoval efficiency and effluent BOD at a temperature of 16 C.SolutionQSettledBOD3300130==536.19g/m-d3BOD loading =Volp2122.14'Q+Q+Q3000+1500+3283.2Hydraulic loading ===22.47m/m-d32RRAp2124'Q+Q1500+3283.2Recirculation Ratio = 1.59R===RRQ3000BOD removal efficiency at 20 C,1+R1+1.59F===1.93(1+0.1R)(1+0.11.59)22100100E===81%BODload0.5360.50.51+0.4331+0.433F1.93BOD removal efficiency at 16 C = Ex 1.035T-2020 C= Ex 1.035= 81.07 x 1.035= 70.64%16-2016-2016 CRemaining BOD effluent = 29 % of 130 mg/l = 0.29 x 130 = 37.7 mg/lQuestion 17(b)Name the four major layers of the atmosphere.SolutionFour major layers of the atmosphere are:(i)troposphere(ii) stratosphere(iii) mesosphere(iv) thermosphere*************************** Question 18(a)Two rapid sludge return final clarifier following high rate T/F are 18 m diameter and 2.7 mside water depth. The effluent weir is inboard channel set on a diameter of 16.5 m for raw w/w flowof 13000 m3/d, with 210 mg/l BOD. Calculate the loading on both T/F and F/C.(a)Recirculation = 0.5(b)BOD Removal effluent in primary clarifier = 35%(c)BOD Removal efficiency in T/F = 85%(d)Area of T/F = 650 m2and depth is 2.5 mSolutionTrickling FilterSettled BOD = 65% of 210 mg/l = 0.65 x 210 = 136.5QSettledBOD13000136.5BOD loading ===1092g/m-d3Vol6502.5Q+Q13000+130000.5Hydraulic loading ===30m/m-d32RA650Remaining BOD = 15% of 136.5 mg/l = 0.15 x 136.5 = 20.48 mg/lFinal ClarifierQ13000Overflow rate dV===25.54m/m-32Ap18024pdp18222H22.7V44Detention time t =24=24=24=2.54hQQ13000QQ13000Weir loading ====62.7m/m-d3weirlength2pd22p15.22Overall Treatment Plant efficiency at 20 C,210-20.48E=100=90.25%210Question 18(b)What are the major components of the troposphere?SolutionIn troposphere the air which we breathe, consists by volume of about 78% of nitrogen, 21%of oxygen, 1% of argon and 0.3% of carbon dioxide.************************* Question 19(a)Estimate the design flow and BOD load for a high-rate filled trickling filter plant with thefollowing sized units.Diameter (m) Area (m) Depth (m)2Primary Clarifier 18 254 2.4Trickling Filter 27 573 2.1Final Clarifier 14 154 2.1SolutionFinal ClarifierAssume V= 32 m3/m2-d0V= Q/A0Q = Vx A = 32 x 154 = 4928 m/d30Checking in Primary Sed: tank (without Q)RQ4928Overflow rate(16 ~ 32 m/m-d)32V===29.1m/m-d32A2540Q+Q4928+49280.5(16 ~ 32 m/m-d)32V===29.1m/m-d32RA2540V2542.4Detention time t = h24=24=2.97(>2.0 hrs)Q4928Take Q = 4928 m/d3BOD loading on Trickling Filter = 750 g/m3-dQSettledBODBOD loading =Vol750Vol7505732.1BOD (mg/l) = mg/l==183.13Q4928183.13BOD in Raw w/w ==281.74mg/l0.65Question 19(b)What is acid deposition?SolutionAcid rain or acid deposition results when gaseous emissions of sulfur oxide, nitrogen oxideinteract with water vapor and sunlight are chemically converted to strong compounds such as sulfuricacid and nitric acid. Those compounds along with other organic chemicals are deposited on the earthas aerosols and particulates (dry deposition) or are carried to the earth by raindrops, snowflakes, fogor dew (wet deposition).************************* Question 20The following are average operating data from a conventional activated sludge secondary.Waster-water flow = 2900 m/dRecirculated sludge flow = 10000 m/d33V = 8500 m3Waste sludge quantity = 20 m3/dInfluent total solid = 599 mg/lSuspended solids in waste sludge = 9800 mg/lInfluent suspended solids = 100 mg/lt = ?Influent BOD = 173 mg/lBOD loading = ?Effluent total solids = 497 mg/lF=?MEffluent S.S = 22 mg/lS.S, BOD removal efficiency = ?Effluent BOD = 20 mg/lMLSS = 2500 mg/lSolutionV8500Aeration Period t = h24=24=7.03Q29000QSettledBOD29000173BOD loading ===590.24g/m-d3Vol8500599-497T/S removal efficiency =100=17.03%599100-22S/S removal efficiency =100=78%100173-20BOD removal efficiency =100=88.44%173FQBOD29000173g/dofBOD===0.236MVMLSS85002500gofMLSSF1BOD sludge age ===4.24daysM0.236gMLSSS.S sludge age = sludge)g/d(SSineffluent+SSin wasteg MLSS = 2500 x 8500 = 21.25 x 10g6S.S in effluent = 22 x 29000 = 63.8 x 104g/dS.S in waste sludge = 9800 x 200 = 1.96 x 106g/d21.25106( )S.S sludge age ==8.18days63.8101.961046Q10000Return sludge ratio R = 0.35==RQ29000*********************** Question 21Estimate the design flow and BOD loading for a step aeration activated-sludge plant based onthe following data.Primary clarifier A = 2800 m, Depth = 2.4 m2Aeration basins, V = 17000 m3Aeration capacity, V = 12 mof air per second + stand by compressor3Final Clarifiers A = 2090 m, Depth = 2.4 m2SolutionFinal ClarifierTry V= 24 m/m-d ( 6000 m/d)330Check detention timeV20902.4t = hr24=24=1.8hr2OkQ66880In aeration tankCheckV17000Aeration period t = commmon)24=24=6.1hr(6-7.5OkQ66880Primary SedimentationQ66880Overflow rateV===23.89m/m-d32A28000V28002.4Detention time t = hr)24=24=2.41hr(>2Q66880Design flow Q = 66880 m3/dBOD loading in Aeration Tank = ?95 m3of air ? 1kg BOD12m/s12312 m3/s of air ? BOD1=kg/s95m953121BOD loading (g/m3-d) =10243600=641.98g/m-d339517000121BOD concentration (mg/l) =10243600=163.18g/m=mg/l339566880i.e. 65% of raw BOD (35% removal in primary sedimentation)raw BOD = 163.18/0.65 = 251.05 mg/l Question 22(a)A conventional activated-sludge basin is 7.3 m wide, 30 m long and has 4 m liquid depth.The influent flow is 344o m/d containing 450 kg BOD. Compute the BOD loading and aeration3period. The operating MLSS is 2200 mg/l and the settled sludge volume in the SVI test is 230 ml/l.Compute the F/M ratio, SVI, recommended sludge recirculation rate and solids concentration in thereturn sludge.Solution450103BOD loading ==513.7g/m-d3( )307.34V7.3304Aeration period t = hr24=24=6.11Q3440FQBOD45010g/dofBOD3===0.233( )MVMLSS22007.3304gofMLSSV10002301000SVI ===104.55ml/gmMLSS2200VQ230QRecommended sludge recirculation rateQ===1027.53m/d31000-V1000-230RQ1027.53Ratio 0.299R===RQ3440101066Solid concentration return sludge = 9564.8==SVI104.55Question 22(b)What are primary and secondary pollutants?SolutionAccording to their origin, pollutants are considered as either primary or secondarycontaminants. Primary pollutants such as sulfur oxide, nitrogen oxide and hydrocarbons are thoseemitted directly to the atmosphere and found there in the form in which they were emitted.Secondary pollutants such as ozone and peroxyacetyl (PAN) nitrate are those formed in theatmosphere by a photochemical reaction of hydrolysis or oxidation.****************************** Question 23(a)A step-aeration activated-sludge secondary is operating under the following conditions:influent waste-water flow = 20800 m/d, average influent BOD = 126 mg/l, volume of aeration basin3is 4250 m3and MLSS concentration 2500mg/l. Determine the following. BOD loading, aerationperiod, F/M ratio, and estimated effluent BOD assuming good operation.SolutionQSettledBOD20800126BOD loading ===616.66g/m-d3Vol4250V4250Aeration period t = hr24=24=4.9Q20800FQBOD20800126g/dofBOD===0.25MVMLSS42502500gofMLSSFor good operationBOD removal efficiency = 85% to 95%Effluent BOD = 5% to 15% of raw BOD= 0.05 x 126 = 6.3 mg/l to 0.15 x 126 = 18.9 mg/lQuestion 23(b)What are the important factors that to be considered in the solid waste management?SolutionTo access the management possibilities it is important to consider:(i)materials flow in society(ii) reduction in raw materials usage(iii) reduction in solid waste quantities(iv) reuse of materials(v) materials recovery(vi) energy recovery(vii) day-to-day solid waste management************************** Question 24(a)A 170 m3complete mixing aeration basin treats 1330 m3/d with an average BOD of 200 mg/l.The operating MLSS is 4000 mg/l, the effluent suspended solids concentration equals 40 mg/l, andwaste sludge solids average 80 kg/day. Compute the BOD loading, aeration period, F/M ratio, BODand SS sludge age.SolutionQSettledBOD1330200BOD loading ===1564.71g/m-d3Vol170V170Aeration period t = hr24=24=3.07Q1330FQBOD1330200g/dofBOD===0.391MVMLSS1704000gofMLSSM1BOD sludge age ===2.56daysF0.391gMLSSS.S sludge age = sludge)g/d(SSineffluent+SSin wasteg MLSS = 4000 x 170 = 680 x 10g3S.S in effluent = 40 x 1330 = 53200 g/dS.S in waste sludge = 80 x 1000 = 80 x 103g/d680103( )S.S sludge age ==5.1days5320080103Question 24(b)Describe the effect of sulfur oxide on human health.SolutionSulfuric acid, sulfur oxide and sulfate salts tend to irritate the mucous membranes of therespiratory tract and foster the development of chronic respiratory diseases particularly bronchitisand pulmonary emphysema. In a dusty atmosphere, SOis particularly harmful because both sulfur2dioxide and sulfuric acid molecules paralyze the hairlike cilia which line the respiratory tract.Without the regular sweeping action of the cilia, particulates are able to penetrate to the lungs andsettle there. These particulates usually carry with them concentrated amounts of SOthus bringing2this irritant into direct, prolonged contact with delicate lung tissues. The SOparticulate combination2has been cited as cause of death in several air pollution tragedies.************************ Question 25(a)Estimate the design flow and BOD loading for an extended aeration system with floatingmechanical aerators. The units sizes are volume of aeration basins = 9500 m3, Oxygen transfercapacity of aerators = 150 kg/h at 2.0 mg/l DO and 3.0 m deep final clarifiers with a surface area of310 m.2SolutionFinal ClarifierTry V= 12 m3/m2-d V= Q/A00Q = Vx A = 12 x 310 = 3726 m3/d (to 200 m3/d) (200 ~ 600)0Try V= 24 m/m-d320Q = Vx A = 24 x 310 = 7440 m/d (> 600 m/d) Ok330Check detention timeV3103t = hr24=24=3hr>2OkQ7440Check aeration periodV9500Aeration period t = hr24=24=30.65(24 ~36hr) OkQ7440Design flow = 7440 m3/dBOD loading = 150 kg/hr1 kg O= 1kg BOD2150 kg O= 150 kg BOD21BOD loading =150kg/hr1024=378.95g/m-d339500( )1BOD concentration =150kg/hr1024=483.87mg/l37440Question 25(b)What are the major components of the troposphere?SolutionIn troposphere the air which we breathe, consists by volume of about 78% of nitrogen, 21%of oxygen, 1% of argon and 0.3% of carbon dioxide.************************* Question 26A waste-water effluent of 560 l/s with a BOD = 50 mg/l, DO = 3mg/l and temperature of23C enters a river where the flow is 2.8 m/s with BOD of 4.0 mg/l, DO = 8.2 mg/l and temperature3of 17C. From laboratory BOD testing, kof the waste is 0.1 per day 20C. The river downstream has1an average velocity of 0.18m and depth of 1.2 m. Calculate the minimum dissolved oxygen level andits distance downstream by using the oxygen sag equation.SolutionAt discharging pt,CQ+CQ42.8+500.56BOD,C===11.67mg/l1122Q+Q2.8+0.56128.22.8+30.56DO,C==7.337mg/l2.8+0.56172.8+230.56T,C==18C2.8+0.561kk-kt=log1-D221k-kkkLc21110DO = DO sat DO mix = 9.5 7.33 = 2.17 mg/lBOD= L(1- 10)-k1t5011.67 = L(1- 10)-0.1 x 50L= 17.1 mg/l0k, T = k, 20C x 1.047T-2011k, 18C = 0.1 x 1.047= 0.091 /day18-201k, T = k, 20C x 1.015T-2022k, 18C = 0.311 x 1.01518-20= 0.302 /day2V0.18k=2.2=2.2=0.311/dayH1.221 .3 31. 3 31kk-k10.3020.302-0.091t=log1-D=log1-2.17=1.75days221k-kkkL0.302-0.0910.0910.09117.1c21110~1.8( ) ( )kLD=10-10+DO1010-kt-kt-kt1c2c2ck-kc21( ) ( )0.09117.1=10-10+2.1710=3.57mg/l-0.0911.8-0.3021.8-0.3021.80.302-0.091Min DO conc: = DO sat D= 9.5 3.57 = 5.93 mg/lc?= V x t= 0.18 x 1.8 x 24 x 3600 x 10= 27.99 km-3cc********************************** Question 27(a)The following treated effluent is discharged to a stream; Q = 30 l/s, DO = 2.0 mg/l, 5 dayBOD = 40.0 mg/l, k= 0.1/day and T = 20C. Upstream from the outfall the water course has the1following characteristics; Q = 0.27 m3/s, DO = 8.0 mg/l, 5 day BOD = 2.0 mg/l and T = 20C. Thestream channel has a k= 0.30/day. Calculate the critical dissolved oxygen concentration2downstream and the distance from the outfall to this point assuming a mean velocity of 0.60 m/s inthe river. Saturation DO at 20C is 9.2 mg/l.SolutionAt mixing pt,CQ+CQ80.27+20.03DO,C===7.4mg/l1122Q+Q0.27+0.031220.27+400.03BOD,C==5.8mg/l0.27+0.03T, C = 20CDO = DO sat DO mixing = 9.2 7.4 = 1.8 mg/lBOD= L(1- 10-kt)5015.8 = L(1- 10-0.1 x 5)0L= 8.48 mg/l01kk-k10.30.3-0.1t=log1-D=log1-1.8=1.19days221k-kkkL0.3-0.10.10.18.48c21110( ) ( )kLD=10-10+DO10-kt-kt-kt101c2c2ck-kc21( ) ( )0.18.48=10-10+1.810=2.15mg/l-0.11.19-0.31.19-0.31.190.3-0.1Min DO conc: = DO sat D= 9.2 2.15 = 7.05 mg/lc?= V x t= 1.19 x 0.6 x 24 x 3600 x 10-3= 61.69 kmccQuestion 27(b)What are primary and secondary pollutants?SolutionAccording to their origin, pollutants are considered as either primary or secondarycontaminants. Primary pollutants such as sulfur oxide, nitrogen oxide and hydrocarbons are thoseemitted directly to the atmosphere and found there in the form in which they were emitted.Secondary pollutants such as ozone and peroxyacetyl (PAN) nitrate are those formed in theatmosphere by a photochemical reaction of hydrolysis or oxidation. Question 28The following treated effluent is discharged to a stream; Q = 45 l/s, DO = 2.0 mg/l, 5 dayBOD = 40.0 mg/l and T = 20C. Upstream from the outfall the water course has the followingcharacteristics with saturated DO; Q = 3 m3/s, 5 day BOD = 3.0 mg/l and T = 19C. From laboratoryBOD testing kof the waste is 0.1/day at 20C. The river downstream has an average velocity of10.6m/s and depth of 1.5 m. Calculate the minimum dissolved oxygen level and its distancedownstream by using the oxygen sag equation. Sat DO at 19C = 9.3 mg/l.SolutionCQ+CQ33+400.045BOD,C===3.55mg/l1122Q+Q3+0.045129.33+20.045DO,C==9.19mg/l3+0.045193+200.045T,C==19C3+0.045DO = DO sat DO mix = 9.3 9.19 = 0.11 mg/lBOD= L(1- 10-kt)5013.55 = L(1- 10-0.1 x 5)0L= 5.19 mg/l0k, T = k, 20C x 1.047T-2011k, 19C = 0.1 x 1.047= 0.096 /day19-201k, T = k, 20C x 1.015T-2022k, 19C = 0.76 x 1.01519-20= 0.758 /day2V0.6k=2.2=2.2=0.76/dayH1.521 . 331. 331kk-k10.760.76-0.096t=log1-D=log1-0.11=1.25days221k-kkkL0.76-0.0960.0960.0965.19c21110( ) ( )kLD=10-10+DO1010-kt-kt-kt1c2c2ck-kc21( ) ( )0.0965.19=610-10+0.1110=0.49mg/l-0.0961.25-0.761.5-0.761.250.76-0.091Min DO conc: = DO sat D= 9.3 0.49 = 8.81 mg/lc?= V x t= 0.6 x 1.25 x 24 x 3600 x 10= 64.8 km-3cc Question 29A stream with BOD 3 mg/l and saturated with DO has a normal flow of 2.8 m3/s and receivea sewage effluent also saturated with DO of 0.9 m/s with BOD 35 mg/l. Determine the DO deficit3over the next 5 days and hence plot the DO sag curve. Calculate the critical DO deficit throughoutsaturation DO at 24C 8.4 mg/l. kat 20C is 0.1 per day, kat 20C is 0.3 per day.12SolutionCQ+CQ32.8+350.9BOD,C===10.78mg/l1122Q+Q2.8+0.912DO, C = sat, 8.4 mg/lT, C = 24CDO = 0 mg/l= L(1- 10)-k1tBOD200510.78 = L(1- 10)-0.1 x 50L= 15.77 mg/l0( ) ( )kLD=10-10+D10-kt-kt-kt10122k-kt21k, T = k, 20C x 1.047T-2011k, 24C = 0.1 x 1.04724-20= 0.12 /day1k, T = k, 20C x 1.015T-2022k, 24C = 0.3 x 1.01524-20= 0.32 /day2( ) ( )kLD=10-10+DO10-kt-kt-kt10122k-kt21( )0.1215.77=610-10+0=2.14mg/l-0.125-0.3250.32-0.121kk-k10.32( )t=log1-D=log1-0=2.13days221k-kkkL0.32-0.120.12c21110( ) ( )kLD=10-10+DO10-kt-kt-kt101c2c2ck-kc21( )0.1215.77=610-10+0=3.28mg/l-0.122.13-0.322.130.32-0.12 ( ) ( )kLD=10-10+DO10(for table)10-kt-kt-kt122k-kt21( )0.1215.77=610-10-0.12t-0.32t0.32-0.12DO conc: = DO sat D= 8.4 - DttTime (days) DO deficit DO conc0 0 8.41 2.65 5.752 3.28 5.123 3.09 5.314 2.64 5.765 2.14 6.266 1.69 6.717 1.31 7.098 1.01 7.399 0.77 7.6310 0.59 7.8111 0.45 7.9512 0.34 8.0613 0.26 8.1414 0.2 8.215 0.15 8.25DO conc9876Oxygen sag curve5DO conc432101 2 3 4 5 6 7 8 9 10111213141516 Question 30Why sewer maintenance and program is necessary.SolutionSewer maintenance requires a through knowledge of the layout and appurtenances used incollection systems, waste water flows, infiltration, and inflow. Prerequisite information regarding thefunction of various unit operations and how they relate to each other is required.Serious and expensive sewer problems can result from improper design or poor construction.Adequate slopes to maintain self cleaning velocity is essential to minimizing maintenance. Selectionof a suitable pipe joint is vital to prevent penetration of roots and excessive infiltration. Cutting oftree roots from sewer lines can be an expensive and reoccurring cleaning process. Ground waterentering joints carries with it soil from around the pipe which ultimately causes structural failure. Inaddition to review of new design and supervision of construction, building permits should requirecareful inspection of all service connections before backfilling. It is important to make certain thatunused service lines are properly capped when buildings are demolished.A successfully maintenance program operates on a planned schedule and requires keepingeffective records. Maps are used to show location of manholes, flushing inlets, service connectionsand the appurtenances. Records should be kept on maintenance performed with particular stress ontroublesome lines that are known to require more frequent inspection or cleaning. While large seweron adequate slopes may never require flushing or cleaning, others must be placed on regular schedulethat may range every month to once a year. The number of emergency sewer blockages can bematerially reduced by such preventive maintenance.Sewer stoppages are caused chiefly by sand, grease materials, sticks, stones and tree roots.The latter are most troublesome. Common cleaning techniques are flushing with water, scraping withmechanical tools, hydraulic scoring with high pressure jets and addition of chemicals. Periodicflushing helps to keep lines clear and is often performed in association with inspection. The usualprocedure for developing scouring velocity is to inert a fire hose into the sewer through manhole.This is most advantageous in cleaning lines in residual sections that do not have sufficientconnections to provide cleaning flow of waste water. Flushing has limitations, since it merely movesdebris from one section of a sewer into another; it is assumed that flows in downstream pipes aresufficient to suspend the solids and keep them moving.*********************************