双语课后题答案合集.doc

水与废水物化处理课后题答案合集by franky&lhySolutions for Homework Assignment of Topic 21. List the most common water quality indicators in surface water and wastewater as possible as you can.Key point:Surface water: pH, Turbidity, Color, TasteWastewater: COD, BOD, SS, TSS (usually these two, not TDS, TFS,etc),Total bacteria, total coliform, fecal coliform, not Bacteria, Protozoa, virusesthey are not indicators.Radium 226/228 (5 pCi/L); beta emitters (4 mrems); gross alpha standard (15 pCi/L); and uranium (30 g/L).2. Calculate the hardness of a water sample containing 20 mg/L K+, 30 mg/L Ca2+, 50 mg/L Mg2+, 100 mg/L Cl-, and 40 mg/L SO42-.Key point:Hardness: A total concentration of all divalent metallic ions (M2+), such as Ca2+, Mg2+, Fe2+, Cu2+, and etc., expressed in mg/L as CaCO3.Solution: Ions concerned are Ca2+ and Mg2+ Hardness mg/L as CaCO3 3. Calculate the TN of a solution containing 30 mg/L glycine (C2H5O2N), 100 mg/L NaNO3 and 50 mg/L NH4Cl.Key point:Total nitrogen (TN): A total concentration of ammonia nitrogen, organic nitrogen, nitrate nitrogen and nitrite nitrogen, expressed in mg/L as N. Solution: Compounds concerned are C2H5O2N, NaNO3 and NH4Cl4. A water with a pH 10.3 contains HCO3- and CO32- at 6.1 and 6.0 mg/L, respectively. What is its total alkalinity in mg/L as CaCO3?Key point:Alkalinity: The measure of the capacity of a water to neutralize acids, in other words, to absorb hydrogen ions without significant pH change, expressed in mg/L as CaCO3.Alk = HCO3- + 2CO32- + OH- 50 mg/L as CaCO3Solution: OH- = 10-3.7mol = 10-0.7 mmol Alk = (0.1 + 20.1 + 10-0.7) 50 = 24.976 mg/L as CaCO3 Solutions for Homework Assignment of Topic 3Question 1:(Just example, not a standard answer)Raw water-Mixing-coagulation-flocculation-sedimentation-filtration-disinfection-usersMixing: to disperse coagulants quickly;Coagulation: to destabilize the suspended particles;Flocculation: to agglomerate small floc;Sedimentation: to settle settleable floc by gravity;Filtration: to remove small particles and floc;Disinfection: to make the water biologically safe;Question 2: The government is planning to build a new airport with a airport hotel and a small residential area on an isolated island. No industry is allowed to be developed in the planned site. Please calculate the daily water consumption rate and sizing the capacity of an on-site wastewater treatment plant based on the following criteria and the tables of water usage in your class note.Criteria:Population of 20,000; designed airport passenger of 100,000/day; total employee of 5,000; parking space of 1,000; 20 lavatories with a total using frequency of 80/min; one airport hotel with a designed capacity of 2,000 guests/day; 6 restaurants with a designed customer size of 50/hr in each; 80% of the consumed water becomes wastewater; and the unaccounted system losses should not be counted in wastewaterResidential water demand:Domestic:230L/capita-dayPublic Service:40L/capita-daySystem Losses and Leakages:100L/capita-dayTotal:370L/capita-dayCalculation of water demand:Residential:20000People370L/capita-day 10-3 m3/L=7400m3/dayAirportPassengers:100000Persons/day15L/person 10-3 m3/L=1500m3/dayEmployee:5000Persons40L/person-day 10-3 m3/L=200m3/dayParking:1000Numbers8L/number-day 10-3 m3/L=8m3/dayLavatories:806024 People/day20L/person 10-3 m3/L=2304m3/dayHotel:2000Guests/day190L/guest 10-3 m3/L=380m3/dayRestaurant:65024 Customer/day25L/customer 10-3 m3/L=180m3/dayDaily Water Demand=11972m3/dayDaily water consumption = 11972 m3/day.System losses and leakages= 20000 people 100 L/capita-day 10-3 m3/L = 2000 m3/dayWater actually consumed = (11972 2000) m3/day = 9972 m3/dayDesign wastewater flow rate = 0.80 9972 m3/day = 7978 m3/dayCapacity of the on-site wastewater treatment plant = 7978 m3/dayQuestion 3: There are two raw water supply sources available for City T: river A and reservoir B. It is know that the former has a 7-day minimum flow rate of 2 cu.m./sec and the latters max storage capacity and a static (dead) capacity are 9 million cu.m. and 1.5 million cu.m., respectively. The population of City T is anticipated to grow up to 120,000 in the next 20 years. The average water demand per capita per day (including industrial and other uses and losses) is currently 300 L/person/day and may increase to 120% in the next 20 years. The maximum daily water demand is 220% of the average daily demand while the max hourly demand is 400% of the average daily demand. You are assigned to design a water treatment plant for this city. Therefore, you need to solve the following problems:(1) Does the river A have enough water to supply?(2) Can enough water be taken from the reservoir in a continuous 3-month drought period without any rainfall ?(3) What is the design capacity of water treatment plant (expressed in cu.m./day)?Population = 120000Average water demand= 1.20 120000 persons 300 L/capita-day 10-3 m3/L = 43200 m3/dayMaximum daily water demand= 2.20 43200 m3/day = 95040 m3/day(1) Required river flow rate = 95040 m3/day (1 day/86000 s) = 1.1 m3/s 3.89 106 m3Hence enough water can be taken from the reservoir during 3 months drought period.(3)Design capacity of the water treatment plant = Maximum daily water demand= 95040 m3/dayAssignment Solution for topic 4Question 1 a) Primary TreatmentPrimary Treatment mainly includes three major processes such as bar screen, grit chamber, and primary sedimentation.Bar Screen the first unit operation in wastewater treatment plant. A screen is a device with openings, generally of uniform size, that is used to retain the coarse solids found in wastewater.Grit Chamber are designed to remove grit, consisting of sand, gravel, cinders, or other heavy solid materials that have subsiding velocities or specific gravities substantially greater than those of the organic putrescible solids in wastewater. Primary Sedimentation Tanks The objective of treatment by sedimentation is to remove readily settleable solids and floating material and thus reduce the suspended-solids content.b) Secondary TreatmentSecondary Treatment mainly includes three major processes such as Biological Treatment Processes, Secondary Clarifier, and Disinfection.Biological Treatment Processes The major biological processes used for wastewater treatment have five major groups: aerobic process, anoxic processes, anaerobic processes, combined aerobic, anoxic, and anaerobic processes, and pond processes. Table 1 shows the list of these processes. TypeCommon NameUseAerobic processes: (suspended-growth)Activated-sludge processCarbonaceous BOD removal Suspended-growth nitrificationNitrificationAerated lagoonsStabilization, carbonaceous BOD RemovalAerobic DigestionStabilization, carbonaceous BOD removal (attached-growth)Trickling filtersCarbonaceous BOD removal, nitrificationRough filtersCarbonaceous BOD removalRotating Biological contactorsCarbonaceous BOD removal, nitrificationPacked-bed reactorsCarbonaceous BOD removal, nitrification (combined suspended-growth and attached growth)Activated biofilter processCarbonaceous BOD removal, nitrificationAnoxic Processes: (Suspended-growth)Suspended growth denitrificationDenitrification (Attached-growth)Fixed film denitricationDenitrificationAnaerobic processes: (Suspended-growth)Anaerobic digestionStabilization, carbonaceous BOD removalAnaerobic contact processCarbonaceous BOD removalUpflow anaerobic sludge-blanketCarbonaceous BOD removal (attached growth)Anaerobic filter processCarbonaceous BOD removal, waste stabilization (denitrification)Expanded bedCarbonaceous BOD removal, waste stabiliziationCombined aerobic, anoxic, and anaerobic process (suspended growth)Single- or multi-stage processes, Various proprietary processesCarbonaceous BOD removal, nitrification, denitrification, phosphorus removal (Combined suspended- and attached growth)Single- or multi- stage processesCarbonaceous BOD removal, nitrification, denitrification, and phosphorus removalPond processesAerobic pondsMaturation (tertiary) pondsFacultative pondsAnanerobic pondsCarbonaceous BOD removalCarbonaceous BOD removal, (nitrification)Carbonaceous BOD removalCarbonaceous BOD removal, (waste stabilization)The most common used in biological treatment is aerobic sludge treatment process. Secondary Clarifier The function of secondary clarifiers is to separate the activated-sludge solids from the mixed liquor. Solids separation is the final step in the production of a well-clarified, stable effluent low in BOD and suspended solids and, as such, represents a critical link in the operation of an activated-sludge treatment process. Also, Secondary clarifier could be used as a sludge thickening process before the sludge discharges to digestion or belt-press process.Disinfection - Disinfection refers to the selective destruction of disease-causing organism. In the field of wastewater treatment, the three categories of human enteric organisms of the greatest consequence in producing disease are bacteria, viruses and amoebic cysts. Disfection in wastewater is most commonly accomplished by the use of (1) chemical agents, and (2) physical agents. (Reading materials) Chemical Agents that have been used as disinfectants include (1) chloride and its compounds, (2) bromine, (3) iodine, (4) ozone, (5) phenol and phenolic compounds, (6) alcohols, (7) heavy metals and related compounds, (8) dyes, (9) soaps and synthetic detergents, (10) quaternary ammonium compounds, (11) hydrogen peroxide, and (12) various alkalies and acids. Of these, the most common disinfectants are the oxidizing chemicals, and chlorine is the one most universally used. Bromine and iodine have also been used for wastewater disinfection. Ozone is a highly effective disinfectant, and its use is increasing even though it leaves no residual. Highly acidic or alkaline water can also be used to destroy pathogenic bacteria because water with a pH greater than 11 or less than 3 is relatively toxic to most bacteria.Physical Agents Physical disinfectants that can be used are heat and light. Heating water to the boiling point, for example, will destroy the major disease-producing monspore-forming bacteria. Heat is commonly used in the beverage and dairy industry, but it is not a feasible means of disinfecting large quantities of wastewater because of the high cost. Sunlight is also a good disinfectant. In particular, ultraviolet radiation can be used. Special lamps that emit ultraviolet rays have been used successfully to sterilize small quantities of water. The efficiency of the process depends on the penetration of the rays into water. The contact geometry between the ultraviolet-light source and the water is extremely important because suspended matter, dissolved organic molecules, and water itself, as well as the microorganisms, will absorb the radiation. It is therefore difficult to use UV radiation in aqueous systems, especially when large amounts of particulate matter are present.Question 2The three major parameters are BOD, COD and TSS. In sewage, usually the TSS: 200mg/L; BOD5: 250mg/L; COD: 450mg/L, and BOD5/COD =0.40.7.BOD5/COD is used to see the biodegradability of wastewater. And TSS is used to assess the performance of conventional treatment process and the need for effluent filtration.Question 3Flow rate, Q = 20,000 m3/dayConcentration = C mg/LLoading (kg/day) = Q (m3/day) C (mg/L) (1 kg/106 mg) (103 L/1 m3)= Q (m3/day) C 10-3 (kg/m3)Waste strengthBODSuspended SolidsConcentrationLoading (kg/day)ConcentrationLoading (kg/day)Strong450 mg/L=20000 450 10-3 = 9,000350 mg/L=20000 350 10-3 = 7,000Medium220 mg/L= 4,400220 mg/L= 4,400Weak100 mg/L= 2,000100 mg/L= 2,000Solutions for Assignment of topic 6Question 1:Aeration is the process by which a gaseous phase, usually air, and water are brought into intimate contact with each other for the purpose of transferring substance to or from the water. Gravity aerator (cascade tower, Spray aerator, Tray Tower) packed tower Surface (mechanical) aerationAir-in-Water: Air is dispersed into water Water-in-Air: Water is dispersed into air Diffused (bubble) aeration Surface (mechanical) aeration Tube mixer Jet aerationSolutions for Assignment of topic 7A flocculator paddle of the design and dimensions shown below is rotated through water at 20oC with an angular speed of 4.0 r/min.(a) How much power is dissipated into the water?(b) If the tank in which this paddle is rotating has the dimensions of 4 4 4 m and the flow through the tank is 5000 m3/d, determine the Gt value for the flocculator.(1) Total PowerPaddle velocity for Inner board = p x 2 x 4 x 1/60 = 0.419 m/sPaddle velocity for outer board = p x 3 x 4 x 1/60 = 0.628 m/sPower for each paddle Paddle tip speed (Vp)= 0.75 x paddle velocity Co-efficient of drag (Cd)=1.8 for flat bladeDensity of water (r)= 998.2 kg/m3 Viscosity, m= 1.002 x10-3 N.s/m2P = (Cd Ap r Vp3)/2Power, P1 for inner board = 1.8 x (0.1 x3 x 4) x 998.2 x (0.75 x 0.419)3/2= 33.46 wattPower, P2 for outer board = 1.8 x (0.1 x3 x 4) x 998.2 x (0.75 x 0.628)3/2= 112.64 wattTotal power P = P1 + P2 = 33.46 + 112.64 = 146.1 watt(2) GtRetention Time, t = V/Q= 4x4x4/5000 = 0.128 days = 1106 secG = P/(Vxm)1/2= 146.1/(4x4x4x1.002x10-3) 1/2= 47.75 s-1Therefore, Gt = 1106 x 47.75 = 52811.5 Solutions for Assignment of topic 111. Monochloramine is found to be able to achieve 4-log kill of E. coli with a dose of 35 mg/L*min. Please calculate the amount of NaOCl (mg/L) to achieve 99.9% E. coli inactivation with a contact time of 60 minutes and the breakpoint curve in the class note.Solution:Log (N/N0) = -kCt, wherek : coefficient of specific lethalityC : concentration of chemical disinfectantt : contact time99.9% inactivation means 3-log kill. From the graph below, we can find the dose for achieving 3-log kill is 26.25mg/L*min, thus:C = 26.25mg/L*min /60 min = 0.4375 mg/L, say 0.45 mg/LThe dose of chlorine required to obtain a mon