I. Noun Explanation Questions:
1. Biochemical oxygen demand: Indicates the amount of oxygen required for the stabilization of degradable organic matter due to the activity of microorganisms in the case of aerobics
2. Chemical Oxygen Demand: Indicates the amount of oxygen required to oxidize organic matter with chemical oxidants.
3. Filter speed regulator: it is a device that maintains the same filter speed during the filtration cycle.
4. Precipitation: It is a process of solid-liquid separation or liquid-liquid separation, under the action of gravity, it is separated by the density difference between suspended particles or droplets and water.
5. Sedimentation ratio: Take 100mL of water sample from the contact agglomeration area with a measuring cylinder, let it stand for 5min, and the number of mL of the sunken alum flowers is expressed as a percentage, which is called the sedimentation ratio.
6. Social cycle of water: Human society takes a large amount of water from various natural water bodies, and after use, it becomes domestic sewage and industrial wastewater, which eventually flows into natural water bodies, so that water constitutes a circulatory system in human society, called ~.
7. Contact agglomeration area: In the clarification tank, the sludge that sinks to the bottom of the pool is lifted up, and this is in a uniformly distributed suspension state, forming a stable sludge suspension layer in the pool, and the concentration of suspended solids contained in this layer is about 3 to 10g/L, called ~.
8. Total hardness: The sum of Ca2+ and Mg2+ content in water is called total hardness.
9. Graded precipitation: If there are several ions in the solution that can form a precipitate with the same ion, the order of precipitation can be judged by the solubility product principle, which is called graded precipitation.
10. Chemical precipitation method: it is to add a certain chemical agent to the water, so that the exchange reaction with the dissolved substances in the water is generated, and the salts that are difficult to dissolve in the water are generated, forming sediment, thereby reducing the content of dissolved substances in the water.
11. Electrolysis method: it is a method of applying the basic principle of electrolysis to transform harmful substances in wastewater into harmless substances through the electrolysis process, oxidation and reduction reactions on the yang and cathodes respectively.
12. Electrodialysis: It is under the action of a direct current electric field that uses yin. The selective permeability of cations in solution by cation exchange membranes, and the separation of solutes from water in solution is a physicochemical process.
13. Sliding surface: When the particles are in motion, the counterions in the diffusion layer will disengage the particles, and this disengaged interface is called the sliding surface, which generally refers to the boundary of the adsorption layer.
14. Adsorption: It is the process by which a substance attaches to the surface of another substance, and it can occur between the two phases of gas-liquid, gas-solid, and liquid-solid.
15. Physical adsorption: It is an adsorption produced by the molecular gravitational force between the adsorbent and the adsorbent.
16. Chemical adsorption: it is the chemical action between the adsorbent and the adsorbent due to the chemical bond force, which changes the chemical properties.
17. Equilibrium concentration: When the adsorbent reaches a dynamic equilibrium on the surface of the adsorbent, that is, the adsorption rate is the same as the desorption rate, and the concentration of the adsorbent in the adsorbent and the solution is no longer changed, and the concentration of the adsorbent in the solution is called ~.
18. Semi-permeable film: In the solution, all one or more components can not penetrate the membrane, and other components can penetrate the membrane, is called semi-permeable film.
19. Membrane separation method: it is to separate a special semi-permeable membrane from the solution, so that a certain solute or solvent in the solution can penetrate out, so as to achieve the purpose of separating the solute.
20. Redox capacity: refers to the degree of difficulty of a substance to lose or acquire electrons, and can be uniformly used as an indicator of redox potential.
21. Biological treatment: it is a wastewater treatment method that mainly uses microorganisms to decompose and oxidize organic matter, and takes certain artificial measures to create a controllable environment, so that microorganisms can grow and multiply in large quantities to improve the efficiency of their decomposition of organic matter.
22. Biorespiration line: Represents a curve for oxygen consumption to accumulate over time.
23. Sludge age: refers to the average number of days that the new sludge remains in the aeration tank every day, that is, the time required for the average renewal of all activated sludge in the aeration tank, or the ratio of the total amount of activated sludge in work to the amount of remaining sludge discharged per day.
24. Oxidation ditch: is an activated sludge aeration tank with a closed ditch.
25. Total oxygen recharge: the total oxygen volume transferred to the aeration tank per unit time under stable conditions.
26. Activated sludge: Flocculent mud particles filled with microorganisms.
27. Biofilm reactor: A device that uses biofilm to purify wastewater.
28. Area load rate method: that is, the unit area can remove the same amount of organic matter in the wastewater per day.
29. Natural biological treatment method: it is a method of purifying wastewater by using natural water bodies and microorganisms in the soil.
30. Activated sludge method: A biological treatment method that purifies wastewater with activated sludge.
31. Suspension growth: In the activated sludge method, microorganisms form a flocculent shape, suspended in the mixture and constantly mixed and contacted with wastewater.
32. Sludge load rate: refers to the amount of organic matter that can be tolerated per unit of activated sludge (microorganism) in a unit of time.
33. Sludge concentration: refers to the weight of suspended solids contained in a unit volume of mixture in an aeration tank, commonly expressed in MLSS.
34. Sludge sedimentation ratio: refers to the sedimentation volume of the precipitated sludge in the aeration tank for 30 min, and the volume of the precipitated sludge accounts for the percentage of the total volume of the mixture.
35. Sludge volume index: referred to as sludge index, is the volume of wet sludge (in mL) occupied by 1g of dry sludge after 30 min precipitation of the aeration tank mixture.
36. Enthalpy of moist air: refers to the heat contained in dry air and 0 °C water contained in 1Kg dry air together with X Kg water vapor contained in it.
37. Two-stage biological filter: When the degree of wastewater treatment is high, when the first-level high-load biological filter cannot meet the requirements, two high-load filters can be connected in series, called ~.
38. Biological contact oxidation method: it is a treatment method between the activated sludge method and the biological filter, which has the advantages of both methods.
39. Anaerobic fluidized bed: When the expansion rate of the carrier in the bed reaches 40 to 50%, the carrier is in a fluidized state.
40. Anaerobic biological method: under the condition of no molecular oxygen, the organic pollutants in sludge and wastewater are degraded by the metabolic action of facultative bacteria and anaerobic bacteria, and the final product of decomposition is mainly biogas, which can be used as energy.
41. Gravity enrichment method: a method that uses gravity to separate the solids in the sludge from the water and reduce the moisture content of the sludge.
42. Land treatment system: is an engineering facility that uses the comprehensive purification capacity of soil and its microorganisms and plants to treat urban and certain industrial wastewaters, while using the sum of water in the wastewater to promote the growth of crops, pastures or trees and increase their yields.
43. Diffusion: Pollutants are transported from high concentrations to low concentrations, which is called diffusion.
44. Atmospheric reoxygenation: Oxygen in the air dissolves into water, which is called atmospheric reoxygenation.
45. Specific traffic: assume q1, q2... It is evenly distributed over the entire pipe segment, and the amount of water that should flow out of each meter of pipe segment length is calculated, and this flow rate is called the specific flow.
46. Rainfall curve: According to different rainfall periods, the highest value of the rainfall intensity selected each year is connected into a curve, such a set of curves, called the rainfall curve.
47. Combined drainage system: a system that collects domestic sewage, industrial wastewater and rainwater in the same drainage system.
48. Diversion system drainage system: a system in which domestic sewage, industrial wastewater and rainwater are removed in two or more separate canals.
49. Fullness: The ratio of the water depth h to the diameter D in the pipe canal, that is, h/D is called the fullness.
50. Drainage system: an engineering facility that systematically removes and treats wastewater and precipitation.
Second, water pollution control engineering multiple choice questions:
1. The following statement is incorrect ( C )
A. The external causes of water's natural circulation are solar radiation and earth's gravitational pull
B. Water is used in the social cycle of runoff water sources and seepage water sources
C. Domestic pollution sources are the most serious environmental pollution
D. Industrial pollution sources are point sources rather than non-point sources
2. The following statement is incorrect (C)
A. The regulation tank can adjust the flow, concentration, pH and temperature of the wastewater
The B diagonal outlet water regulation pool has the effect of automatic equalization
C. The weir overflow water regulation pool can regulate both water quality and water quantity
D. The external pump suction well regulation pool can regulate both water quality and water quantity
3, TOD refers to (A)
A. Total oxygen demand B Biochemical oxygen demand C Chemical oxygen demand D Total organic carbon content
4. The following statement is incorrect (D)
A. Degradable organic matter is partly oxidized by microorganisms, and part is synthesized by microorganisms
B. BOD is the sum of the oxygen consumed by microorganisms to oxidize organic matter and the oxygen consumed by microbial endogenous respiration
C. The decomposition process of degradable organic matter is divided into carbonization stage and nitrification stage
D. BOD is the sum of the oxygen required for carbonization and the oxygen required for nitrification
5. The following statement is incorrect (C)
A. COD determination usually uses K2Cr2O7 and KMnO7 as oxidants
B. COD determination not only oxidizes organic matter, but also inorganic reducing substances
The C. COD determination includes the amount of oxygen required for carbonization and nitrification
D. COD determination can be used in water where toxic substances are present
6. The following impurities that do not belong to the presence state in the water are ( D )
A. Suspension B colloidal C solute D precipitate
7. The following statement is incorrect ( B )
A. The grille is used to block coarse floating and suspended solids in the water
B. The loss of the head of the grille is mainly due to its own resistance
C. The bottom of the canal after the grid should be 10-15 cm lower than the bottom of the canal in front of the grid
D. The grille is tilted 50-60o, which can increase the grille area
8. The precipitation of particles in the sedimentation tank belongs to (A)
A free precipitation B flocculation precipitation C crowded precipitation D compression precipitation
9. Particles in the early stage of the initial sedimentation (A), in the late stage of the initial sedimentation (B)
10. The particles are in the early stage of the second sedimentation pool (B), in the late stage of the second sedimentation pool (C), in the mud bucket (D)
11, particles in the sludge thickening tank ( D )
12. The following statement is incorrect ( C )
A. The effective water depth of the sedimentation column for the free sedimentation test should be as much as possible the water depth of the actual sedimentation tank
B. The effective water depth of the sedimentation column for flocculation precipitation test should be as much as possible the water depth of the actual sedimentation tank
C. For the free precipitation E~t curve and the E~u curve are related to the test water depth
D. For the flocculation precipitation E~t curve and E~u curve are related to the test water depth
13, the contact adsorption of the material is one of the mechanisms of filtration, and what does not belong to this mechanism is (D)
A. Gravity precipitation B. Van der Waals force adsorption C. Brownian motion D. interception
14. According to the principle of the inclined plate (tube) sedimentation tank, if the pool depth H is divided into three layers, then ( B )
A. Without changing the flow rate v and the minimum sinking rate u0, the pool length L is shortened by 1/3
B. Without changing the flow rate v and the minimum sinking rate u0, the pool length L is shortened by 2/3
C. Without changing the flow rate v and the minimum sinking rate u0, the pool length L increases by 1/3
D. Without changing the flow rate v and the minimum sinking rate u0, the pool length L increases by 2/3
15. The water inlet device that does not belong to the advection sedimentation tank is ( D )
A. Transverse DTH B. Vertical DTH C. Perforated Wall D. Triangle Weir Filtration
16, the reason that is not stable in colloid is ( A )
A. Diffuse B. Brownian motion C. Charged D. Hydrated membrane
17. The following statement is correct ( C )
A. For inorganic salt coagulants, adsorption bridges play a decisive role
B. For inorganic salt coagulants, the mechanism of action of the double layer plays a decisive role
C. For organic polymers, adsorption bridges play a decisive role
D. For organic polymers, the electric double layer plays a decisive role
18. The following statement is incorrect (D)
A. In aerobic biological treatment wastewater systems, heterotrophic bacteria use organic compounds as carbon sources
B. In aerobic biological treatment wastewater systems, autotrophic bacteria use inorganic carbon as the carbon source
C. In the aerobic biological treatment wastewater system, endogenous respiration exists in the metabolic process of heterotrophic bacteria
D. In the aerobic biological treatment wastewater system, there is no endogenous respiration in the metabolic process of autotrophic bacteria
19, does not belong to the adsorbent surface adsorption force is ( B )
A. Van der Waals Force B. Surface Tension C. Chemical Bond Force D. Electrostatic Gravity
20, the statement about activated carbon is incorrect ( B )
A activated carbon is made of carbon-containing substances carbonized and activated at high temperatures
The larger the specific surface area of activated carbon, the better
C. Activated carbon itself is non-polar
D. Activated carbon also has catalytic oxidation and catalytic reduction
21. The following statement is incorrect (D)
A. Disinfection in water treatment is essential
B. The drinking water qualification standard is 3 coliforms < 3/L
C. HOCl has a stronger oxidation capacity than OCl-
The chlorine content in D. HOCl and OCl- is weighed as synthetic chlorine
22. The following statement is incorrect (C)
A. The net growth of microorganisms is equal to the amount of microbial growth synthesized by degrading organic matter - the amount of microorganisms reduced by the oxidation of microorganisms themselves
B. The unit of the oxidation rate of microorganisms themselves is 1/d
C. The oxygen consumption of microorganisms is equal to the amount of oxygen required by microorganisms to degrade organic matter - the amount of oxygen required by microorganisms to oxidize themselves
D. The unit of oxidation oxygen demand rate of microorganisms is gO2/g (microorganism.d)
23. The following factors affecting aerobic biological treatment are incorrect (D)
A. For every 10 to 15 °C increase in temperature, the activity capacity of microorganisms doubles
B. When the pH < 6.5 or pH >9, microbial growth is inhibited
C. Dissolved oxygen in water should be maintained at more than 2mg/l
D. The microbial requirement for nitrogen and phosphorus is BOD5:N:P=200:5:1
24. The figure on the right shows the relative oxygen consumption rate curve of microorganisms of different organic matter at different concentrations, and which types of organic matter they correspond to
A. Not biochemical non-toxic B. Not biochemically toxic
C. Biochemical D. Difficult to biochemical
25, about the process of activated sludge treatment of organic matter, the incorrect one is (D)
A. Activated sludge removal of organic matter is divided into two stages: adsorption and oxidation and synthesis
B. The quantity of organic matter in the previous stage changes, and the organic matter in the latter stage changes
C. The sludge lost its activity in the previous stage
D. The sludge in the latter stage loses its activity
26. The sludge sedimentation ratio of an aeration tank is 25%, the MLSS concentration is 2000mg/L, and the sludge volume index is (C)
A25 B.100 C.125 D.150
27, about the sludge volume index, the correct is (B)
A.SVI is high, and the activated sludge sedimentation performance is good
B.SVI is low, and the sedimentation performance of activated sludge is good
The C.SVI is too high and the sludge is small and tight
If the D.SVI is too low, the sludge will expand
28, the statement about the age of the sludge, incorrect is ( B )
A. Equivalent to the time required for the average renewal of all activated sludge in the aeration tank
B. The ratio of the total amount of sludge in operation to the amount of reflux sludge per day
C. Sludge age is not as old as possible
D. The age of sludge shall not be shorter than the generation period of microorganisms
29. BoD5/COD=50 of an industrial wastewater, and its biochemical properties are preliminarily judged to be (A)
A. Better B. Can C. Harder D. Not
30, the use of activated sludge growth curve can guide the design and operation of the treatment system, the following guidance is incorrect ( C )
A. The high-load activated sludge system is in the logarithmic growth period of the curve
B. The general load activated sludge system is in the deceleration growth period of the curve
C. The completely mixed activated sludge system is in the deceleration growth period of the curve
The D-delay aeration activated sludge system is in the endogenous metabolic phase of the curve
31. Between the activated sludge method and the natural water self-purification method is (D)
A. Aerobic pond B. Facultative pond C. Anaerobic pond D aeration pond
32, does not belong to the push flow type of activated sludge treatment system is ( D )
A. Gradual aeration method B. Multi-point water inlet method C. Adsorption regeneration method D. Delayed aeration method
33, the operation mode of the oxidation ditch is ( C )
A. Flat flow type B. Push flow type C. Circulating mixed type D. Complete mixed type
34, not belonging to the mechanical aeration device is ( A )
A. Riser B. Impeller C. Brush D. Rotary Disc
35. Large and medium-sized sewage plants mostly use (B) type (C) aeration
Small sewage plants mostly use (A) type (D) aeration
A. Complete mixing B. Push flow C. Blower D. Mechanical
36, between the activated sludge method and the biofilm method is ( B )
A. Biological filter B. Biological contact oxidation tank C. Biological turntable D. Biological fluidized bed
37. The following about the incorrect operation mode of various activated sludge is (B)
A. The gradual reduction aeration method overcomes the contradiction between oxygen supply and oxygen demand in ordinary aeration tanks
B. The multi-point inlet method is higher than the effluent quality of ordinary aeration tanks
C. The sludge reflux ratio used in the adsorption regeneration method is larger than that of ordinary aeration tanks
D. The complete mixing method overcomes the shortcomings of the ordinary aeration tank that is not resistant to impact load
38. Methanogenic bacteria in anaerobic digestion are (A)
A. Anaerobic bacteria B. Aerobic bacteria C. facultative bacteria D. neutral bacteria
39 The control step in the anaerobic fermentation process is (B)
A. Acidic fermentation stage B. Alkaline fermentation stage C. Neutral fermentation stage D. Facultative fermentation stage
40. The temperature of anaerobic fermentation increases, digestion time (B)
A. Increase B. Decrease C. Unchanged D. Cannot be judged
41. The temperature of anaerobic fermentation rises, and the amount of biogas produced (A)
42. The following devices that are not suitable for treating high concentrations of organic wastewater are (D)
A. Anaerobic contact cell B. Anaerobic filter C. Anaerobic fluidization tank D. Anaerobic high-speed digester
43. The main difference between a conventional digester and a high-speed digester is (B)
A. The former does not heat, the latter heats B. The former does not stir, the latter stirs C. The former has a layer, the latter does not layer D. The former effluent needs further treatment, and the latter does not need effluent
44. The device that should be in the secondary digester in the high-speed digester is ( C )
A. Steam pipe B. Water injector C. Pressure measuring tube D. Propeller
45. The role of the primary digester in the high-speed digester is (A)
A. Stirring and heating of sludge B. Storage of sludge C. Digestion sludge D. Thickening sludge
46. Which of the following is not a feature of modern anaerobic biological treatment wastewater process (B)
A. Longer sludge residence time B. Longer hydraulic residence time C. Less sludge yield D. Less N and P elements dosed
47. The following anaerobic biological treatment device for wastewater is not from the bottom of the pool, and the water from the pool is ( D )
A. Anaerobic fluidized bed B. UASB reactor C. Anaerobic filter D. Anaerobic contact cell
48. Which of the following is unsafe to operate an anaerobic digestive system (C)
A. Digester, gas storage cabinet, biogas pipeline, etc. are not allowed to leak air B. Digester, gas storage cabinet, biogas pipeline, etc. maintain positive pressure C. Gas storage cabinet connected branch pipe to provide energy use D. There is ventilation equipment near the biogas pipeline
49. Which of the following devices is not considered a source of sludge (B)
A. Primary sinking tank B. Aeration tank C. Second sedimentation tank D. Coagulated pool
50. The sludge should be transported in the pipeline in the (D) state
A. Laminar flow B. Intermediate C. Transition D. Turbulence
A.1/3 B.2/3 C.1/4 D.3/4
51.VS refers to (C).
A. Burning residue B. Containing inorganic substances C. Burning reduction D. Non-volatile solids
52. The hydraulic characteristics of the sludge flowing in the pipe are (A)
A. The resistance of sludge flow during laminar flow is greater than that of water flow, and it is smaller during turbulence B. The resistance of sludge flow is smaller than that of water flow during laminar flow C. The resistance of sludge flow is greater than that of water flow during laminar flow, and it is larger during turbulence D. The resistance of sludge flow is smaller than that of water flow during laminar flow, and it is smaller during turbulence
53. The moisture content of the sludge is reduced from 99% to 96%, and the volume of sludge is reduced (D)
54. Which of the following is not the treatment method of sludge (B)
A. High-speed digestion method B. Anaerobic contact method C. Aerobic digestion method D. Wet oxidation method
55. Which of the following is not the final treatment method of sludge (D)
A. Used as agricultural fertilizer B. Land filling C. Throwing into the sea D. Wet firing method to burn off
56. The following description of cooling water is incorrect (B)
A. Direct cooling water is polluted and heated, indirect cooling water only heats up without pollution B. Direct cooling water needs to be cooled and treated, indirect cooling water only needs to be cooled C. Direct cooling water and indirect cooling water will produce dirt D. Direct cooling water and indirect cooling water are cooled with wet air as the medium
57. The cooling of water is mainly through (A)
A. Contact heat transfer and evaporative heat dissipation B. Contact heat transfer and radiative heat dissipation C. Evaporative heat dissipation and radiative heat dissipation D. Radiative heat dissipation and convective heat transfer
58. The ideal gas equation of state P=rRT*10-3 (hPa) has the correct meaning of (C)
A.r is the specific gravity of the ideal gas at its pressure B.R is the gas constant, the unit is kg m / kg.°C C.T is the absolute temperature, T = 273 + t, t is the air dry bulb temperature D.R dry = 47.1 R gas = 29.3 kg.m / kg °C
59. Absolute humidity is (B), moisture content is (D)
A.1m3 Mass of water vapor in wet air B. Value is the bulk density of water vapor C.1kg Mass of water vapor in wet air D. Value is the ratio of water vapor bulk weight to dry air volumetric weight
60 Which of the following is not the driving force for water evaporation (C)
A. Difference between the water vapor content at the water and air interface and the water vapor content in the air B. The difference between the water vapor pressure and the water vapor pressure in the air at the water and air interface C. The difference between the temperature and air temperature at the water and air interface D. The difference between the moisture content at the water and air interface and the moisture content in the air
61. The specific heat of moist air is defined as (A)
A. Heat required to increase the temperature of 1 kg of moist air with a mass of 1 kg B. The heat required to increase the temperature of the humid air with a weight of 1 N 1 K C. The heat required to increase the temperature of the humid air with a weight of 1 kg D. The heat required to increase the temperature of the humid air with a mass of 1 N by 1 °C
62. The relative humidity is incorrect (B).
A. Ratio of absolute humidity of air to absolute humidity of saturated air at the same temperature B. Ratio of moisture content of air to the moisture content of saturated air at the same temperature C. Ratio of vapor pressure of air to vapor pressure of saturated air at the same temperature D. The ratio of the bulk density of water vapor to the absolute humidity of saturated air at the same temperature
63 . The following statement is incorrect (B)
A. When the water temperature T > temperature q, the evaporative heat dissipation of the water reduces the water temperature B. When the water temperature T < the temperature q, the evaporative heat dissipation of the water increases the water temperature C. When the water temperature T > the temperature q, the contact heat transfer of the water reduces the water temperature D. When the water temperature T < the temperature q, the contact heat transfer of the water increases the water temperature
64. What cannot speed up the evaporation and heat dissipation of water is (D)
A. Make water form water droplets B. Make water form a water film C. Blast D. Heating
65. The following statement about natural cooling tanks is incorrect (B)
A. Built with natural ponds or reservoirs B. It is not advisable to place hot water inlet and cooling outlet in the same lot C. Cooling of water on the ground D. The deeper the cooling pool, the better
66. The following statement about the water spray cooling pool is incorrect (C)
A. Water spraying equipment is installed on natural or artificial ponds B. It should be in the downwind direction of the dominant wind direction in summer C. The spray range of each nozzle should overlap each other D. The graph of the water-spray cooling tank is used for verification
67. The following statement about cooling towers is incorrect (C)
A. The effect of the cooling tower is higher than that of the cooling pool B. The water shower device is also called the packing C. If the air flow direction of the tower is perpendicular to the water flow direction, it is called the counter-flow type D. The natural ventilation cooling tower can not use the water collector
68. The cooling process of the cooling tower mainly occurs in (B)
A. Ventilation cartridge B. Water shower device C. Water distribution device D. Water collector
69. When the circulating water quality is unstable and prone to CaCO3 precipitation, (A) in the thin film sprinkling device can be used.
A. Waveform asbestos cement board B. Plastic spot wave type C. Plastic oblique wave type D. Plastic honeycomb type
70. The following conditions (D) are not suitable for setting up natural cooling tanks
A. Large amount of cooling water B. There are available depressions or reservoirs C. Not too far from the factory D. The water temperature requirements are more stringent in summer
71. The following conditions (A) are not suitable for setting up a water jet cooling tank
A. Large amount of cooling water B. There are available depressions or reservoirs C. There is a large enough open land D. The wind speed is small and stable
72. The following conditions (D) are not suitable for setting up natural ventilation cooling towers
A. Large amount of cooling water B. The water temperature requirements are not strict C. There is a large enough open land D. The wind speed is small and stable
73. The following effects produce both dirt and corrosion (C)
A. Deco2 effect B. Concentration effect C. Water temperature change effect D. Iron bacterial oxidation
74. What can be used as both a scale inhibitor and a corrosion inhibitor is (A)
A. Polyphosphate B. Sodium polyacrylate C. Sodium polycarboxylate D. Zinc salt
75. Which of the following methods does not stop limescale (A)
A. Reducing the rate of sewage loss B. Reducing the concentration of suspended solids in the supplementary water by clarifying C Adding acid to convert carbonates into carbonate D. into CO2
76. Of the following four water losses, the loss of water without the loss of salt is (A)
A. Evaporation B. Wind blowing C. Leakage D. Discharge
77. Select the processing object for the corresponding processing level
Primary Processing (D) Secondary Processing (C) Depth or Tertiary Processing (A)
Special Handling(B)
A. Residual suspensions and colloids, BOD and COD, ammonia nitrogen, nitrate. phosphate
B. Fine suspended particles and emulsions that cannot be separated by natural precipitation, organic matter that are difficult to biodegrade and toxic and harmful inorganic substances
C. Organic pollutants in a colloidal and dissolved state of water
D. Solid pollutants in suspended state in water
78. Biological methods are mainly used for (B)
A. Primary processing B. Secondary processing C. Deep processing D. Special processing
79. Secondary processing is mainly used (D)
A.Physical law B.Chemical law C. Physical chemistry D. Biological law
80. Lime by nitrogen removal method is generally (A)
A.CaO B.Ca(OH)2 C.CaCO3 D.Ca(HCO3)2
A. Activated carbon adsorption B. Bleaching powder C. ultrafiltration D. filtration
81. The following statement can be judged from the figure on the right that it is incorrect (D)
A. At the same temperature, as the pH increases, the amount of NH3 gradually increases to the amount of NH4+
B. In the summer, when pH = 7, only NH4 + ;p H = 11, only NH3
C. At the same pH, as the temperature increases, the amount of NH3 gradually increases to the amount of NH4+
D. In winter, when pH = 7, only NH4 + ;p H = 11, only NH3
82. Select the processing method for the corresponding processing object
Suspended Solids(D) Bacteria(B) Pigments(A) Inorganic Salts(C)
83. Lime added by phosphorus removal method is generally (B)
84. The statement about the biological nitrogen removal process is incorrect (B)
A. Nitrification and then denitrification B. Nitrifying bacteria are aerobic autotrophic bacteria, denitrifying bacteria are anaerobic heterotrophic bacteria C. Nitrification should have sufficient alkalinity, denitrification alkalinity can not be too high D. Nitrification should maintain enough DO, denitrification can be low oxygen or anaerobic
85. Conditions about denitrifying bacteria that are not required are (B)
A. Organic carbon as the C source B. There is sufficient alkalinity C. Hypoxia D. Temperature 0 ~ 50 °C
86. The following judgment is correct (D)
A. The pollution cloud emitted at the center of the river is half as wide as the discharge from the shore
B. Pollution clouds discharged at the center of the river are twice as wide as those discharged from the shore
C. The concentration of pollutants discharged from the river centre is half that discharged from the shore
D. The concentration of pollutants discharged from the center of the river is twice that of the shore discharge
87. The statement about the process of dilution of pollutants in water bodies is incorrect (B)
A. Contaminants are transported in still water caused by diffusion
B. Pollutants are transported in the laminar flow water body caused by diffusion and current, which cancel each other out
C. Pollutants should be calculated using eddy current diffusion coefficient in turbulent water
D. The accuracy of the two-dimensional diffusion equation for pollutants in the ocean is sufficient
89.Ex is (C).
A. Molecular diffusion coefficient B. Transverse diffusion coefficient C. Vertical diffusion coefficient D. Vertical diffusion coefficient
90. Regarding the biological denitrification process, the incorrect one is (C)
A. The nitrification tank in the primary nitrification and denitrification simultaneously oxidizes the carbon B. The first and second levels must be added to the organic carbon source before the denitrification C. The aeration tanks in the first and second stages mainly play an oxygenation role D. The sedimentation tanks in the first and second stages have reflux sludge discharge
91. Regarding the dilution of pollutants in the oceans, the incorrect statement is (C)
A. Conserved substance, the total dilution of the mass in the ocean is D1×D2
B. Initial dilution D1 is a function of the exhaust outlet depth, diameter, and Frauder number
C. The transfer diffusion dilution D2 is the ratio of the pollutant concentration at time t to the beginning
D. The total dilution of bacteria in the ocean is D1× D2×D3
92. Seabed porous diffusers should be associated with the seawater mainstream (B)
A. Parallel B. Vertical C. Tilt D. Casual
93. Regarding the oxoag curve, the incorrect statement is (A)
A. The stained point is the point where the oxygen loss is greatest
B. The oxygen consumption rate in the descending stage of the curve > the reoxygenation rate
C. The oxygen consumption rate in the ascending stage of the curve < the reoxygenation rate
D. Dissolved oxygen at the end of the curve returns to its initial state
94. With regard to natural degradation, the following statement is incorrect (B)
A. The rate of oxygen consumption during the degradation of organic matter is directly proportional to the content of organic matter at this time
B. The rate of reoxygenation during atmospheric reoxygenation is proportional to the dissolved oxygen content at this time
C. The constant of oxygen consumption rate and the constant of compound oxygen rate can be calculated by kt=k20× 1.047T-20
D. The rate of change of deficit oxygen is the algebraic sum of the oxygen consumption rate and the rate of reoxygenation
95. The following statement is correct (B)
A. Big city KH big, small city KH small B. big city KH small, small city KH big C. small city KH small, big city KH smaller D. small city KH large, big city KH bigger
96. Regarding the regulation of water quantity, the incorrect one is (C)
A. The difference between water theory and the amount of water coming from a pumping station is the total storage stock
B. The amount of water stored between the first pumping station and the second pumping station is called the storage capacity in the factory
C. The amount of water stored outside the second pumping station is the storage capacity of the clear water pool
D. Whether the amount of water stored or insufficient is called regulating the amount of water
97. The medium head required for the pump at the maximum transfer flow rate is compared with the head of the pump at the highest time (A)
A. > B.= C.< D. Is not easy to judge
98. Regarding the characteristics of the ring network, the statement is wrong (D)
A. There are connecting tubes between the main pipes B. Nodes meet THE SQ=0
C. Each ring satisfies Sh = 0 D. Number of nodes = number of pipe segments + number of rings - 1
99. Regarding the conversion of node traffic, the correct one is (B)
A. It is half of the concentrated traffic B. It is half of the traffic along the line C. It is half of the transfer traffic D. It is half of the pipeline traffic
100. The following judgment is correct (D)
A. When the flow rate is certain, the flow rate is reduced, the pipe diameter is reduced, and the head loss is also reduced
B. When the flow rate is certain and the flow rate decreases, the pipe diameter increases, and the water head loss also increases
C. When the flow rate is certain and the flow rate decreases, the pipe diameter decreases and the head loss increases
D. When the flow rate is certain, the flow rate decreases, the diameter of the pipe increases, and the loss of the head decreases
101. The following are not the causes of local resistance losses (D)
A. Turning B. Branching C. Pipe diameter enlargement D. Rough pipe wall
102. The most commonly used material in the water supply network is (A)
A. Cast iron pipe B. Steel pipe C. Reinforced concrete pipe D. Plastic pipe
103. The most commonly used circular pipe in sewage treatment systems, which has special hydraulic characteristics (C)
When A.h/D=1, Q max B. h/D=1, U max C. h/D=0.95, Q max D.h/D=0.95, U max
104. Taking the rainfall intensity i as the ordinate coordinate and the rainfall duration t as the abscissa coordinate, the third highest value of different rainfall durations in the rainfall data recorded in 20 years is marked, and the curve is linked, and the reproduction period given is P (C)
A. 20a B.10a C.5a D.6a
105. The most commonly used material in drainage pipes is (A)
A. Reinforced concrete pipe B. Cast iron pipe C. Steel pipe D. Clay pipe
3. Fill-in-the-blank questions
1. The internal causes that form the natural circulation of water are the physical properties of water, and the external causes are the radiation of the sun and gravity.
2. The continent's total water resources rank sixth in the world, but the per capita water resources are only 1/4 of the world's per capita water resources.
3. The most important source of surface and groundwater pollution is industrial pollution
4. Mercury pollution causes people to produce Minamata disease, and cadmium pollution produces bone pain disease.
5. Domestic sewage contains organic matter and pathogenic bacteria. Insect eggs, etc., discharged into the water body or infiltrated into the groundwater will cause pollution, when the dissolved oxygen is less than 3 ~ 4mg / L, it will affect the life of fish.
6. According to the World Health Organization, 80% of diseases worldwide are water-related.
7. Nitrogen, phosphorus and other nutrients in sewage are discharged into the water body, which will cause eutrophication of the water body.
8. Z1N2F2 When designing and operating a water pollution control facility, it is necessary to understand the composition of water and wastewater, as well as the flow of water and its variations.
9. All kinds of impurities can be divided into three categories according to their presence in water: suspended solids, colloids and dissolved substances.
10. The physical properties of wastewater include color, odor, solids, temperature.
11. The chemical composition of wastewater includes organics, inorganic substances, gases, etc.
12. Total solids in sewage include floating matter, suspended solids, colloids and dissolved matter.
13. The amount of residue dried by the sewage at 103~105 °C is the total solid.
14. The two indicators of biochemical oxygen demand BOD and chemical oxygen demand COD are generally used to express the content of organic matter.
15. Most countries now specify 20 °C as the temperature for the BOD measurement and 5 days as the standard time for the measurement.
16. TOD represents the oxygen consumed by burning a compound at high temperatures.
17. Municipal sewage includes domestic sewage and industrial wastewater, and also includes rain and snow water in the combined pipeline system.
18. Facilities to prevent pollution and public hazards must be designed, constructed and put into operation at the same time as the main project, referred to as the "three simultaneous" principle.
19. The methods of water and wastewater treatment can be summarized as physical methods, chemical methods, physical chemical methods, biological methods.
20. The regulating pool regulates water quality and quantity.
21. When calculating the regulation pool, it is calculated according to the most unfavorable situation, that is, the concentration and flow rate in the peak time interval.
22. The Z2N2F3 grille is a frame made of a set of parallel metal bars that is placed diagonally in the channel of the wastewater flow path, or at the inlet of the pumping station catchment tank, or at the inlet end of the water intake.
23. The water flow resistance of the grille itself is very small, generally only a few centimeters, and the resistance is mainly caused by the interception blocking the fence.
24. When the loss of the water head of the grille reaches 10 to 15 cm, it should be cleared.
25. There are many types of screen filters, including rotary, vibratory, rotary drum and so on.
26. Precipitation is a process of solid-liquid separation or liquid-liquid separation.
27. The precipitation of suspended solids in water can be divided into four basic types: free precipitation and flocculation precipitation. Crowded sedimentation and compression pellet.
28. According to the different directions of water flow in the pond, the form of the sedimentation tank can be divided into four types: advection type, vertical flow type, radial flow type and oblique flow type.
29. The inlet device of the sedimentation tank distributes the inlet water evenly across the entire pool as much as possible to avoid causing a flow.
30. The effluent device of the sedimentation tank is commonly used for the outlet of the weir.
31. The advection grit pond is also a sedimentation tank to remove sand particles, cinder, etc. from the sewage.
32. According to the relative direction of water flow and mud flow, the inclined plate (tube) sedimentation tank can be divided into three types: heterotropical flow, co-flow and transverse flow.
33. Z2N2F2 take a 100mL water sample from the contact agglomeration area, let it stand for 5 minutes, and the number of mL of the sunken alum flowers is expressed as a percentage, that is, the settlement ratio.
34. The most commonly used filter media are sand, anthracite, magnetite, etc.
35. The removal mechanism of suspended particles in water in the granular filter media is mechanical screening and contact adsorption with the filter material.
36. The filter material should have sufficient mechanical strength, sufficient chemical stability, a certain particle gradation and appropriate porosity.
37. d80 and d10 represent the diameter of the sieve holes that pass through 80% and 10% of the weight of the filter media, respectively.
38. The support layer is generally made of natural pebbles or gravel, and its role is to prevent the loss of filter material in the water distribution system. Distribute the water evenly during backwashing.
39. The expansion rate of the filter layer is the ratio of the increased thickness after expansion to the thickness before expansion.
40. The role of the filter water distribution system is to distribute the flush water evenly throughout the filter plane.
41. The filter water distribution system has a large resistance water distribution system and a small resistance water distribution system.
42. Oils exist in wastewater in three states: suspended, emulsified and dissolved.
43. According to the classification of the method of generating bubbles, there are pressurized dissolved air flotation, jet air flotation, pump suction pipe suction air flotation and dissolved air floating, etc., of which pressurized dissolved air flotation has the best effect and makes the most.
44. The general magnetic separation method is to use a magnet to suck out the strong magnetic material in the water, and stainless steel magnetic wire hairs are often used in the industry to generate a high magnetic field gradient.
45. Particles with a size of less than 10um cannot be separated by precipitation in water.
46. Adsorption bridging mainly refers to the adsorption and bridging of polymers and granules.
47. Coagulants can be divided into two categories: inorganic salts and organic polymer compounds.
48. Factors affecting the coagulation effect are: water temperature, pH value, alkalinity, hydraulic conditions and impurities in water.
49. The coagulant is dosed by gravity, water injector and pump.
50. The main design parameters of the mixing reaction equipment are the stirring strength and stirring time.
51. Agitation intensity is often expressed as the velocity gradient G of the adjacent two water layers, which essentially reflects the chance or number of particle collisions.
52. Disinfection is mainly to kill pathogenic microorganisms that are harmful to the human body.
53. The most commonly used disinfection methods in water treatment work are chlorine disinfection, ozone disinfection and ultraviolet disinfection.
54. Chlorinated disinfection can be used liquid chlorine, can also be used bleach powder.
55. After chlorine is added to the water, a part is consumed by impurities that can be chlorinated, and the remaining part is called residual chlorine.
56. The amount of chlorine added in the water during disinfection can be divided into two parts, namely the amount of chlorine required and the amount of residual chlorine.
57. For some industrial wastewater, there is a high probability of carcinogens produced after chlorine disinfection.
58. Lime neutralization methods include dry and wet doses.
59. There are many methods of ozone generation in the ozone oxidation method, which are generally taken by industrial and silent discharge methods.
60. Adsorption is the process by which a substance attaches to the surface of another substance, and the substance being adsorbed is called adsorbent substance.
61. The adsorption force on the surface of the adsorbent can be divided into three types, namely molecular gravity, chemical bond force and electrostatic gravity, so adsorption can be divided into three types: physical adsorption, chemical adsorption and ion exchange adsorption.
62. The factors affecting adsorption are the nature of the adsorbent, the nature of the adsorbent and the operating conditions of the adsorption process.
63. The main factors affecting detachment are: temperature, gas-liquid ratio. pH and oily substances.
64. Membrane separation methods include electrodialysis, reverse osmosis and ultrafiltration.
65. The energy required for electrodialysis is proportional to the salinity of the treated water.
66. Ultrafiltration is generally used to separate substances with molecular weight greater than 500.
67. Municipal sewage treatment with sedimentation method can generally only remove about 25 to 30% of BOD5
68. Microorganisms in aerobic biological treatment systems are mainly bacteria, in addition to algae, protozoa and higher microfauna. Bacteria are the main microorganisms that stabilize organic wastewater and are the most important in aerobic biological treatment.
69. In the process of aerobic biological treatment of wastewater, the removal of carbonaceous organic matter plays a major role in the removal of isooxygenic bacteria, and the largest number is also heterooxygenic bacteria.
70. While microorganisms degrade organic matter, the number of microorganisms themselves increases.
71. The factors affecting aerobic biological treatment are mainly temperature, pH, nutrients, oxygen supply, toxic and organic properties.
72. Aerobic biological treatment methods can be divided into two categories: natural and artificial.
73. Industrial wastewater is customarily divided into four categories: biochemical, difficult to biochemical, non-biochemical and non-toxic, and non-biochemically toxic.
74. The BOD5/COD ratio is generally used to initially evaluate the biochemical properties of wastewater.
75. The Activated Sludge Act was initiated in 1914 (UK).
76. Sludge concentration is often expressed in MLSS, sludge sedimentation ratio is expressed in SV, and sludge volume index is expressed in SVI or SI.
77. Under normal operating conditions of urban sewage, SVI is between 50 and 150.
78. The process of activated sludge removing organic matter from wastewater includes the adsorption stage and the oxidation and synthesis stage
79. Oxidation ditch is an activated sludge aeration tank with a closed ditch
80. Aeration methods can be divided into two categories: blower aeration and mechanical aeration.
81. The factors that affect the transfer of oxygen to water are temperature, composition of the solution, stirring strength, atmospheric pressure, concentration of suspended solids, etc.
82. Blower aeration is also known as compressed air aeration, and the blast air aeration system includes fans, ducts and aeration devices.
83. There are many forms of mechanical aeration, which can be summarized into two categories: vertical shaft aerator and horizontal shaft surface aerator.
84. The sludge load rate calculation method is commonly used in the volume calculation of aeration tanks.
85. The design of secondary sedimentation tanks must take into account both clarification and concentration.
86. The total depth of the secondary sedimentation tank should be at least 3 to 4 meters.
87. The operation and control of the activated sludge treatment plant is carried out by adjusting the oxygen supply, the sludge return flow and the removal of the remaining sludge.
88. Biofilm reactors can be divided into biological filters, biological turntables and biological contact oxidation tanks.
89. Biological filters can generally be divided into three types: ordinary biological filters, high-load biological filters and tower biological filters.
90. The main components of a biological filter include: filter media, pool wall, water distribution system and drainage system.
91. The number of stages of the biological turntable generally does not exceed 4 levels
92. The biological contact oxidation tank is composed of several parts, such as the pool body, filler, water distribution device and aeration system.
93. Oxidation ponds are also known as biological ponds or stabilization ponds.
94. The types of oxidation ponds are: aerobic ponds, facultative ponds, anaerobic ponds and aeration ponds.
95. The objects of anaerobic biological treatment are: organic sludge, organic wastewater, biomass.
96. Anaerobic biogenic degradation process: acidic fermentation stage and alkaline fermentation stage.
97. Anaerobic biological treatment devices include: septic tank, hidden pool, traditional anaerobic digester, high-speed anaerobic digestion tank, anaerobic contact method, anaerobic filter, ascending anaerobic sludge layer reactor, anaerobic expansion bed.
98. What are the measures taken by high-speed digesters compared to conventional digesters: mechanical agitation and heating.
99. Sludge is divided according to source: primary sedimentation pond sludge, residual sludge, humus sludge, anaerobic sludge, chemical sludge.
100. Sludge enrichment methods include: gravity concentration method, air flotation concentration method.
101. The United Kingdom often uses haifa to finally dispose of sludge.
102. Sludge drying methods include: natural drying method, drying method.
103. The calculation of sludge pipe hydraulics still adopts the calculation formula of general water pipes.
104. Sludge treatment methods are: anaerobic digestion method, aerobic digestion method, heat treatment method.
105. Sludge dewatering methods include filtration method (vacuum filter, plate and frame filter press, belt filter press), centrifugal method (centrifuge).
106. Circulating cooling water often has dirt and corrosion problems.
107. The structural composition of the cooling tower is: ventilation cylinder, water distribution system, water shower system, ventilation equipment, water collector, catchment pool.
108. The signature of water quality stability is commonly used for saturation index TL and stability index IR characterization.
109. Composition of urban water supply engineering system: water intake engineering, water purification engineering, water transmission and distribution engineering.
110. The water distribution network is: tree-like, network-like.
111. If the closure difference is >0, the flow rate allocated on the clockwise direction pipeline is large and should be appropriately reduced.
112. Drainage system: shunt system (complete shunting system, incomplete shunting system), combined flow system (direct discharge type, interception type).
113. The minimum cover thickness under the roadway is 0.7m.
114. Feedwater pumping stations are classified according to their functions: primary pumping stations, secondary pumping stations, pressurized pumping stations, circulating pumping stations.
115. Drainage pumping stations are classified according to the nature of water: sewage pumping stations, rainwater pumping stations, combined pumping stations, sludge pumping stations.
4. Short answer questions
1. What are the types of sedimentation of particles in water and their characteristics?
Answer: The precipitation of particles in water can be divided into four basic types according to their concentration and characteristics: (1) free precipitation: particles are in a discrete state during the precipitation process, their shape, size, and quality are not changed, the sinking speed is not disturbed, and the precipitation process is completed independently. (2) Flocculation precipitation: In the process of precipitation, suspended particles may collide with each other to produce flocculation, and their size and mass will increase with the increase of depth, and their sinking speed will increase with depth. (3) Crowded precipitation (layered precipitation): When the concentration of suspended particles is relatively large, they interfere with each other during the sinking process, and the particle groups combine into an overall downward precipitation to form a clear slurry interface, and the sedimentation is displayed as the interface sinking. (4) Compression precipitation: the concentration of particles in water increases to the point where the particles contact each other and support each other, and the upper particles squeeze out the gap water of the lower particles under the action of gravity, so that the sludge is concentrated.
2. Under what conditions does the individual's free sedimentation rate refer to the sinking rate, and how to analyze its influencing factors?
A: The rate of sinking of individual free settling refers to the rate of sinking of particles in the equilibrium state of force, which can be expressed by the Stokes equation. u=g (rs-ρ)d2/(18m) indicates: the factors affecting the free settling rate visible by the equation are: (1) rs-r: rs-r>0 particles sink, rs-r<0 particles float, rs-r = 0 particles suspended, increase the rs-r value, can strengthen the precipitation process. (2) d: free sedimentation rate uμd2, measures to increase particle size d can strengthen the precipitation process. (3) μ: free sedimentation rate uμ1/m, reducing the dynamic viscosity of water can improve the sedimentation rate, which is conducive to precipitation.
3When the water flowing out of the weir roof overflow in the regulation pool, can it play a role in water volume regulation? In what way can the purpose of water volume regulation be achieved?
A: No. In order to achieve the purpose of water volume regulation: one way is to take measures to stabilize the water level; the other is to set up a water pump outside the regulation pool.
4. Try to explain the design basis of the sinking velocity u when calculating the area of the sedimentation tank. A: Because u0t0 = H, V pool = HA surface = Qt0, so u0 = H / t0 = Qt0 / (A surface t0) = Q / A = q that is, the sinking velocity u is numerically equal to q (area load rate), you can determine the area of the sedimentation tank A according to u (= q) and the amount of treated water.
5. How to derive the basis for the generation of inclined plate (tube) sedimentation tank from the theoretical analysis of the ideal sedimentation tank?
A: Because there is H/u=L/v, that is, u/v=H/L, through the theoretical analysis of the ideal sedimentation tank. If the sedimentation tank with a water depth of H is divided into several sedimentation tanks with a water depth of H/n, when the length of the sedimentation tank is 1/n of the length of the original sedimentation area, the same amount of water can be treated as the original sedimentation tank, and the exact same treatment effect can be achieved. This shows that reducing the depth of the sedimentation tank can shorten the sedimentation time, thereby reducing the volume of the sedimentation tank, and the sedimentation efficiency can be improved. To facilitate the discharge of mud, tilt the spacer 60o.
6What are the similarities and differences between the working principle of the clarification tank and the sedimentation tank? What problems should be paid attention to in operation?
A: In the sedimentation tank, the suspended particles sink to the bottom of the pool, and the sedimentation process is completed. In the clarification tank, the sludge sinking to the bottom of the pool is lifted up, and this is in a uniformly analyzed suspension state, forming a stable sludge suspension layer (that is, contact condensation area) in the pool, when the raw water passes, the principle of contact condensation is used to adsorb the alum flowers in the water to achieve the purpose of removing suspended solids.
7What is the main role of the particles in the water being intercepted by the filter equipment (including the filter cell and other filtration equipment)?
A: The particles in the water are intercepted by the filter equipment (including the filter cell and other filtration equipment) mainly by the precipitation and contact adsorption on the surface of the filter layer, once a layer of filter film is formed on the surface of the filter layer, it will produce the interception effect of screening, but this interception effect will make the filter quickly blocked.
8How can the absoluteness of the backwash water distribution be uneven, how can the water distribution be relatively uniform?
A: According to the principle analysis of the large resistance water distribution system, the flow rate Qb = {(S1 +S2') Qa2/(S1+S2")+(u12+U2 2) / [ (S1+S2"))2g ]}0.5, which shows that the outlet flow of the first orifice a and the end orifice b cannot be equal. However, by reducing the total area of the orifice to increase S1, the effects of S2', S2" and the uneven pressure of the water distribution system can be weakened, so that Qa can be as close to Qb as possible. The small resistance water distribution system is by reducing the u1, u2 in the above formula to a certain extent, the second term on the uniformity of the right root number of the equation will be greatly weakened, in order to achieve the purpose of uniform water distribution, so whether it is a large resistance water distribution system, or a small resistance water distribution system can not achieve the absolute uniformity of the backwash water distribution system.
9Can any substance adsorb with air bubbles? What factors do you depend on?
Answer: Whether the substance can be adsorbed with bubbles mainly depends on the hydrophobic properties of the particles, the hydrophobic properties of the particles depend on the size of the gas-solid interface contact angle q, q>90o is a hydrophobic substance, q<90o is a hydrophilic substance, when q = 0, the solid particles are completely wet, and can not be adsorbed, when 0<q<90o, the adsorption of particles and bubbles is not firm, easy to desorb under the action of water flow, when q > 90o, particles are easy to be adsorbed by bubbles. For hydrophilic substances, they can be adsorbed with bubbles by adding a flotation agent and changing their contact angle so that their surface is converted into a hydrophobic substance.
10 From the hydrocyclone and various centrifuges to generate the magnitude of the centrifugal force to analyze what kind of particles they are suitable for separating?
A: From the principle of centrifugal separation, the particles are driven by the net movement force in the centrifugal force field, and move to the wall of the device, the distance of the particles moving in the direction is Dg = us t0 If Dr is greater than the starting distance of the particles and centrifugation, the equipment wall, the particles can reach the equipment wall and be removed. Because the net moving force c= (m-m0) u2/R (u=2 pRn/60 m/s); C and the rotation radius of the particles are inversely proportional, the hydrocyclone should not be too large, generally within f500mm, the rotational angular velocity generated in the device is also limited, so to achieve the purpose of separation, the particle size should be large, the density should be large, such as mineral sand, in order to achieve the purpose of being removed. For centrifuges, because it is a machine spin, there are high-end and low-speed centrifuges, because u s = kw2 (rs-ρ) d2/ (18m), in the case of W is very large, the density difference between particles and water r-ρ0, and particle size d has little effect on u s, so the high-speed centrifuge can separate emulsified oil and lanolin, and the medium-speed centrifuge can be used for sludge dewatering and wastewater treatment containing fibers or slurries. Low-speed centrifuges can use sludge and chemical sludge and dewatering.
11. In water treatment practice, what factors should be considered to choose the appropriate oxidant or reducing agent?
A: (1) It has a good redox effect on specific impurities in water. (2) The product after the reaction should be harmless and do not need to be treated twice. (3) Reasonable price and easy to obtain. (4) The reaction is rapid at room temperature and does not require heating. (5) The pH required for the reaction is not too high or too low.
12. What reactions occur on the cathodes in the electrolyzer? Write the ion equations for electrolytic reactions. (Take M+ as a metal cation as an example) Answer: Cathode: reduction reaction M++eM®2H++2e ®H2; Anode: oxidation reaction 4OH--4e2H2O®+O2 Cl--e-1 ®/2Cl2
13What are the advantages and disadvantages of electrolysis treatment of chromium-containing wastewater?
A: Advantages: the effect is stable and reliable, the operation and management is simple, the equipment footprint is small, and the wastewater water and heavy metal ions can also be reduced by electrolysis. Disadvantages: the need to consume electrical energy, the consumption of steel, the operating cost is high, and the problem of comprehensive utilization of sediment needs to be further studied and solved.
14What are the treatment methods for chromium-containing wastewater? (Write at least three)
A: Electrolysis method, chemical redox method, ion exchange method, etc.
15What are the factors that affect the coagulation effect?
Answer: (1) Water temperature: Water temperature affects the hydrolysis of inorganic salts. When the water temperature is low, the hydrolysis reaction is slow, the water viscosity is large, and the flocculate is not easy to form. (2) The PH value and alkalinity of water. Different pH values, the morphology of the hydrolysis products of the aluminum salt and iron salt coagulants is different, and the coagulation effect is also different. (3) The nature, composition and concentration of impurities in the water. The presence of positive ions above divalent in water is beneficial to the natural water compression double layer; the particle size of impurities is different, which will be conducive to coagulation; the concentration of impurities is too low (the number of particles is too small) will not be conducive to the collision between particles and affect the condensation. (4) Hydraulic conditions. The mixing stage of the coagulation process requires the coagulant to be mixed quickly and uniformly with muddy water; in the reaction stage, the stirring intensity is gradually reduced with the increase of the alum flower.
16Why is chlorine widely used as a disinfectant in disinfection? What problems have been identified in recent years?
A: The price of chlorine is low, the disinfection effect is good and the use is more convenient. In recent years, compounds such as chloroform, which may produce carcinogenicity in the process of chlorination and disinfection, have been discovered, so more ideal disinfectants are being explored at home and abroad.
17What are the deficiencies in adding chlorine gas to water that does not contain ammonia for disinfection? How to solve it?
A: Chlorinated gas disinfection, chlorine and water react to generate hypochlorous acid (HOCl), HOCl can have a strong oxidation capacity, kill bacteria, the more HOCl, the better the disinfection effect. Disadvantages: (1) Hypochlorous acid is easy to dissipate after staying in water for too long, and when the pipeline is very long, the end of the pipe network is not easy to meet the residual chlorine standard. (2) Free residual chlorine HOCl is prone to produce a heavier chlorine odor, especially when the water contains phenols, it will produce chlorophenols with a foul odor. Solution: When the water does not contain ammonia or has little content, artificial ammonia can be added to form chloramine. The disinfection effect of chloramines also depends on HOCL, but chloramine is gradually released HOCl, which is slower than HOCl disinfection, can be maintained for a long time, and it is easy to ensure the residual chlorine requirements at the end of the pipe network, and the chlorine odor is also lighter.
18What are the methods of neutralizing acidic wastewater?
Answer: (1) use the alkaline wastewater and alkaline waste residue of the plant to neutralize, (2) add alkaline reagents, and (3) filter through the filter media with neutralizing properties.
19What are the advantages and disadvantages of the drug neutralization method and the filtration neutralization method?
Answer: Drug neutralization method: Advantages: can treat any substance, any concentration of acidic wastewater, when adding lime, calcium hydroxide also has a coagulation effect on impurities in the wastewater. Disadvantages: poor labor and health conditions, complex operation and management, preparation concentration, dispensing agents require more mechanical equipment; lime quality is often not guaranteed, more ash residue, large sediment volume, dehydration trouble. Filtration neutralization method: Advantages: simple operation and management, stable PH value of effluent, does not affect environmental hygiene, less sediment. Disadvantages: The sulfuric acid concentration in the inlet water is limited.
20. What is the chemical precipitation method, and what substances are mainly removed in water treatment?
Answer: Add a certain chemical agent to the water to exchange reactions with the dissolved substances in the water, generate salts that are difficult to dissolve in water, form sediment, and reduce the content of dissolved substances in water. This method is called chemical precipitation. Remove calcium and magnesium hardness in the feedwater treatment. Removal of heavy metal ions such as mercury, cadmium, lead, zinc, etc. in wastewater treatment.
21 In the redox reaction, how to judge the redox reaction of organic matter?
A: The reaction with oxygenation or dehydrogenation is called oxidation, or the reaction of organic matter and strong oxidant to form CO2, H2O, etc. is judged to be an oxidation reaction, and the reaction of hydrogenation or deoxygenation is called a reduction reaction.
22What are the operation modes of the activated sludge method? Try to describe the characteristics of the traditional activated sludge method, and briefly describe the basis for the production of other activated sludge methods.
Answer: The main operation modes of the activated sludge method are: (1) ordinary activated sludge method (2) gradual aeration method (3) multi-point inlet activated sludge method (stage aeration method) (4) adsorption regeneration activated sludge method (5) complete mixed sludge method (6) delay aeration activated sludge method (7) oxidation ditch. For the traditional activated sludge method: sewage and biological sludge enter from the first end at the same time, and the tail end comes out; the growth of microorganisms shows a line segment on the growth curve; the method has the advantages of high treatment efficiency, good effluent quality, and small amount of residual sludge. But its shortcomings are that (1) the impact load difference and (2) the oxygen supply and the oxygen demand are unbalanced. ...
24. What is the relationship between sludge sedimentation ratio, sludge index and sludge concentration? To describe its significance in the operation of activated sludge.
A: SVI (Sludge Index) = SV% ×10/MLSS g/L. Sludge index SVI normal operation of the case of 50 ~ 150, sludge index MLSS certain case, can calculate the 30-minute sedimentation ratio SV, so by measuring the 30-minute settlement ratio, it is very easy to determine the sludge condition in the aeration tank, and treat accordingly.
25. Try to derive the relationship between X, Xr and r from the sludge balance in the aeration tank.
A: The sludge balance of the aeration tank is: QXr=(Q+rQ)X (inlet water X0=0) So there are: QXr=QX+rXQ, rQr=Q(Xr-X) So: r=Q(Xr-X)/QX=(Xr-X)/X
26. What is the role of the secondary sedimentation tank in the activated sludge method system? What are the requirements in the design?
Answer: In the activated sludge method system, the role of the secondary sedimentation tank is (1) the separation of sludge water; (2) the sludge has a certain thickening effect to ensure the amount of sludge required by the aeration tank. Therefore, after determining the sedimentation area according to the sedimentation rate and determining the area to meet a certain concentration effect, the larger of which is selected as the design sedimentation area to ensure the need for both.
27. Why does the activated sludge method need to return the sludge, and how to determine the reflux ratio?
Answer: The activated sludge method relies on the removal of biological sludge from the BOD, and the biomass carried by the inlet water is limited, so it should be refluxed from the sludge separated from the secondary sedimentation tank, and part of it should be returned to the aeration tank. The size of the reflux ratio depends on the sludge concentration X and the reflux sludge concentration Xr in the aeration tank, and the sludge balance r = (Xr-X)/X in and out of the aeration tank
28. How to calculate the oxygen demand of the activated sludge method, and what two aspects of the needs does it reflect? Answer: The oxygen demand of the activated sludge method is the decomposition and oxygen demand of boD by organisms, and the second is the endogenous respiratory demand of microorganisms. The calculation formula for daily oxygen demand is O2=aQ(S0-Se)+bVX0, and the amount of water treated in Q per day is the amount of water treated.
29. Explain the composition of boD in the treatment of water by activated sludge method, and explain the functional role of what kind of treatment structure they represent? A: The ACTIVATED Sludge Treatment of Water BOD consists of two components: (1) BOD5 of solubility (2) and BOD5 of non-solubility. Solubility BOD5 comes from the residual dissolved organic matter after biological treatment, which indicates the operation effect of the aeration tank, and the non-soluble BOD5 is derived from the solids brought out of the secondary sedimentation tank and the microbial suspension, which represents the effect of solid-liquid separation in the secondary sedimentation tank.
30. Try to analyze the main characteristics of biofilms. A: The biofilm method is a method of purifying the sewage by fixing the biofilm fixed on the surface of the carrier, and the sewage is purified through the biofilm reactor. The main characteristics of the biofilm method are as follows: (1) the diversification of microorganisms involved in the purification reaction has a certain nitrogen removal effect; (2) the sludge yield is low and the sedimentation performance is good; (3) the adaptability to changes in inlet water quality and water quantity is strong, and (4) it is easy to maintain and operate, and energy saving.
31. Try to briefly discuss the factors that affect anaerobic biological treatment.
Answer: The main factors affecting anaerobic digestion are: (1) temperature: there is medium temperature fermentation and high temperature fermentation, high temperature fermentation digestion time is short, high gas production, sterilization and egg sterilization effect is good; (2) PH and alkalinity: the best pH value of anaerobic fermentation should be controlled at 6.8 ~ 7.2, lower than 6 or higher, the anaerobic effect is significantly worse, the amount of volatile acid directly affects the PH value, so the reactor should maintain sufficient alkalinity; (3) carbon-nitrogen ratio, the appropriate carbon-nitrogen ratio is C/N = 10 ~ 20: 1 ;(4) organic load rate, The control rate of anaerobic digestion is the methane fermentation stage, so excessive organic load may lead to the accumulation of organic acids and is not conducive to methane fermentation; (5) stirring: proper stirring is necessary to prevent the formation of slag shells and ensure that the digestive biological environmental conditions are consistent; (6) toxic substances should be controlled within the maximum allowable concentration.
32. What is the object of anaerobic biological treatment? What can be achieved?
A: The objects of anaerobic biological treatment are: (1) organic sludge; (2) high concentration of organic wastewater; (3) biomass. After anaerobic treatment, (1) it can achieve the effect of sterilization and egg sterilization, anti-fly deodorization, and (2) it can remove a large amount of organic matter from sludge and wastewater to prevent pollution of water bodies. (3) Considerable bioenergy can be obtained at the same time as anaerobic fermentation - biogas (4) Through anaerobic fermentation, the amount of solids can be reduced by about 1/2, and the dewatering performance of the sludge can be improved.
33. Try to compare the advantages and disadvantages of anaerobic treatment with aerobic method.
A: In terms of high concentration of organic wastewater and sludge and stable treatment. The main advantages of anaerobic treatment are: (1) organic wastewater, sludge is recycled and (biogas production) ;(2) low power consumption (no oxygen supply). The shortcomings are: (1) the treatment is not complete, the concentration of organic matter in the effluent is still very high, can not be directly discharged, relatively speaking, aerobic activated sludge method, there are: (1) the treatment of organic matter is more thorough, the effluent can meet the standard discharge, (2) the power consumption is large.
34. What is the general process of sludge treatment? What is the role of sludge thickening?
A: The general process of sludge treatment is as follows: sludge ® thickening ® sludge digestion ® and dewatering and dry ® sludge utilization
The role of sludge thickening is to remove a large amount of water from the sludge, reduce the volume of sludge, facilitate subsequent treatment, reduce the volume of the anaerobic digestion tank, and reduce the amount of digestion and calorie consumption.
35. What are the methods of sludge thickening and compare them?
Answer: Concentrated sludge can be used gravity concentration method, can also use air flotation concentration method and centrifugal enrichment method, gravity enrichment method The main structure is gravity concentration tank, the equipment structure is simple, the management is convenient, and the operating cost is low; the main facilities for air flotation concentration are air flotation pool and compressed air system, with more equipment, more complex operation, and higher operating costs, but the moisture content of air flotation sludge is generally lower than that of gravity concentrated sludge. Centrifugal concentration can reduce the sludge moisture content to 80% to 85%, greatly reducing the volume of sludge, but compared with the larger power consumption.
36. What is the significance of the sludge ratio being hindered in the sludge dewatering? How to reduce the sludge specific resistance?
Answer: The sludge specific resistance is large, the sludge dewatering is not easy, the dewatering efficiency is low, and the power consumption is large; the treatment of reducing the sludge specific resistance is sludge pretreatment, which has (1) chemical coagulation method; (2) elution - chemical coagulation method, which can greatly reduce the amount of coagulant.
37.How to improve the cooling tower heat dissipation speed?
Answer: In order to improve the heat dissipation speed of the cooling tower, the water should be formed as much as possible into small water droplets or water films to expand the contact area between water and air; increase the flow rate of water surface air, so that the escaped water vapor molecules can quickly diffuse in order to improve the heat dissipation speed of the cooling tower.
39.What are the forms of water distribution network layout? What are their characteristics?
A: There are two types of water distribution pipe network layout, namely branch pipe network and ring pipe network. The length of the branch pipe network pipeline is the shortest, the structure is simple, the water supply is direct, and the investment is more economical. However, the water supply is less reliable. Water can come from two directions in each pipe in the ring pipe network, so the water supply is safe and reliable. The ring pipe network can also reduce the loss of water heads in the pipe network, save power, and greatly reduce the threat of water hammer in the pipe, but its pipeline is long and the investment is large.
40. What is the head of a water pump? How are I sure?
A: The head of the pump is also known as the total head or total head. The head of the pump is equal to the energy added by the unit weight of the liquid after passing through the pump. Denoted by H. The head of the pump is used in the actual process for two aspects: one is to lift the water from the suction well to the water tower or treatment structure (i.e. static head Hst); the other is to consume the loss of water head in the overcoming pipeline (Sh). Therefore, the total head of the pump should be H=Hst+Sh. For safety reasons, the calculation of the total head H is often followed by a rich head.
41. What is the task of elevation arrangement of sewage treatment plants? How can hydraulic calculations be carried out to ensure the self-flow of gravity of sewage?
A: The task of the elevation arrangement of the sewage treatment plant is: (1) to determine the elevation of each treatment structure and the pump room; (2) to determine the size and elevation of the connecting pipe canal between the treatment structures; (3) to calculate and determine the water surface elevation of each part, so that the sewage can flow by gravity between the treatment structures according to the treatment process. When calculating hydraulic power, choose a process with the longest distance and the largest loss of water head, and calculate the reverse wastewater treatment process backwards, and leave room appropriately to ensure that the treatment system can operate normally in any case.
V. Calculation questions for water pollution control projects:
1. The total hardness of raw water is 1.6mmol/L, alkalinity HCO3-=2.58 mmol/L, Mg2+ 1.0mmol/L, Na+ 0.03mmol/L, SO42-0.125mmol/L, Cl-0.4mmol/L, CO225.8mg/L, experimental calculation, carbonate hardness and non-carbonate hardness in water. If the lime-soda method is used for softening, try to find the amount of lime and soda (mmol/L). (Excess amount is taken at 0.1 mmol /L)
解：Ht=Ca2++Mg2+ =1.6 mmol /L
Mg2+ =1.0 mmol /L 故Ca2+ =0.6 mmol /L
Carbonate hardness Hc=HCO3- /2=2.58/2=1.29mmol/L
其中Ca(HCO3)20.6 mmol /L,Mg(HCO3)2=1.29-0.6=0.69 mmol /L
Non-carbonate hardness Hn=Ht-Hc=1.6-1.29=0.31mmol/L
Where MgSO4 + MgCl2 = 0.31 mmol / L
故[ CaO]=[CO2 ]+ [Ca(HCO3)2 ]+2 [ Mg(HCO3)2 ] +Hn+α
25.8/44+0.6+2×0.69+0.31+0.1=2.98 mmol/L
[ Na2 CO3] = Hn +β=0.31+0.1=0.41mmol /L
2. Hard water volume is 1000m3/d, the water quality is as follows: Ca2++Mg2 + 2.15 mmol /L, HCO3- 3.7mmol/L, Na + +K+ 0.35mmol/L, SO42-0.30mmol/L, Cl-0.35mmol/L, free CO2 22mg/L. The strong acid H-Na parallel debase softening was used to find the amount of water inlet of RH and RNa tanks, and the CO2 content (mg/L) of the CO2 remover (0.5 mmol/L of remaining alkalinity).
解：（1）QH（SO42－＋Cl－）＝QNa×[HCO3－]-QAr
<当量浓度>
QH（0.6+0.35）＝QNa×3.7-1000×0.5
0.95QH＝QNa×3.7－500
QH＝3.89QNa-526.32
又Q＝QH＋QNa
QH＝1000－QNa
QNa＝312m3/d
QH＝Q－QNa＝1000－312=688m3/d
(2) The amount of CO2 remover = the amount of CO2 in raw water + HCO3 - the amount of reduction = 22 + (3.7-0.5) ×44 = 22 + 140.8 = 162.8 (mg / L)
3. An advection sedimentation tank, the area of the clarification area is 20×4m2, and the flow rate is Q=120m3/h. If it is transformed into a slanted plate sedimentation tank, the flow rate is increased to 6.5 times the original flow rate, and other conditions remain unchanged. How many tilting wrenches do you need to install? (Tilt Wrench length L=1.2m, width B=0.8m, plate spacing d=0.1m, plate and horizontal angle q=60o, plate thickness negligible)
い:いいいA=20×4=80m2 , q=Q/A=120/80=1.5(m3/m2.h)
The slanted plate pool q=6.5Q/At is 1.5=6.5×120/At
At=520m2 (total area of inclined plate pool)
Set the total number of units of the inclined plate pool to n
则n（LBcosq）＋dB）＝At
n(1.2×0.8cos60o+0.1×0.8)=520
n=929 [number of units]
Therefore, the number of plates = n + 1 = 929 + 1 = 930 (blocks)
4. The maximum amount of sewage in a city is 1800m3/h, the original sewage suspension concentration C1=250mg/L, and the allowable concentration of discharged sewage suspension C2=80mg/L. It is proposed to be treated with a radiated sedimentation tank. Try to calculate the removal rate and the basic size of the sedimentation tank. (The sedimentation time is 1.5h, and the effective water depth of the sedimentation tank is 3.3m)
untie:
(1)E=(C1-C2)/C1=[(250-80/250]×100%=68%
(2) Sedimentation tank volume V=Qt=1800×1.5=2700m3
It is set as two pools, each with a volume of 1350m3
Surface area per pool F=V/2/h=1350/3.3=409(m2)
直径D=(4F/p)1/2=(4×409/3.14)1/2=23(m)
5. The concentration of the mixture liquid of an activated sludge aeration tank is MLSS=2500mg/L. The mixture was taken 100mL in a measuring cylinder, and the sludge volume measured was 30mL when standing for 30min. The SVI and moisture content of the activated sludge are obtained. (The density of activated sludge is 1g/mL)
(1) The sludge volume corresponding to the 100mL mixture is 30mL
The sludge volume corresponding to the 1L mixture is 300mL
Another 1 L of mixture contains 2500mg = 2.5g of mud
Therefore, SVI=300/2.5=120mL/g dry mud
(2) 1mL of the activated sludge contains dry sludge 1/SVI=1/120=0.008g
Because the density of activated sludge is 1g/mL, the quality of 1mL activated sludge is 1g
Then the moisture content is [(1-0.008)/1]×100% = 99.2%
6. The MLSS of the activated sludge aeration tank = 3g/L, the sludge volume of the mixture precipitated by 30min in the 1000mL measuring cylinder is 200mL, and the sludge sedimentation ratio, sludge index, required reflux ratio and reflux sludge concentration are calculated.
（1）SV＝200/1000×100%＝20%
(2)SVI=(SV%×10)/MLSS=(20×10)/3=66.7mL/g 干泥
（3）Xr＝106/SVI＝106/66.7＝15000（mg/L）
(4)因X(1+r)=Xr×r
即3(1+r)=15×r
r＝0.25＝25%
7. An urban domestic sewage is treated by activated sludge method, the amount of wastewater is 25000m3/d, the volume of the aeration tank V= 8000m3, the inlet water BOD5 is 300mg/L, the removal rate of BOD5 is 90%, and the solid concentration of the aeration tank mixture is 3000mg/L, of which volatile suspended solids account for 75%. Seek: Sludge load rate Fw, sludge age, daily residual sludge, daily oxygen demand. The relevant parameters of domestic sewage are as follows: a=0.60 b=0.07 A=0.45 B=0.14
(1)X=3000mg/l=3g/l Xv=0.75X=2.25g/l
Fw＝L0Q/VXv＝（300×10－3×2500）/（8000×2.25）＝0.42（KgBOD5/MLSS.d）
(2)e=(L0-Le)/L0 90%=(300-Le)/300 Le=30mg/L
u=QLr/XvV=25000×(300-30)×10-3/2.25×8000=0.375(KgBOD5/KgMLVSS.d)
1/θc=au-b , 1/θc=0.6×0.375-0.07, 1/θc=0.155
θc=6.45(d)
(3) Residual sludge DX=1/θc×(Xvv)=0.155×2.25×8000=2790(KgVsSS/d)
The corresponding MLSS amount is 2790/0.75=3720Kg/d
（4）O2/QLr＝A＋B/u O2/25000（300－30）×10－3＝0.45＋0.14/0.375
O2/6750=0.823, O2=5557(Kg/d)
8. What is the volume ratio of 99.5% sludge with a moisture content of 1% of the water before and after dewatering?
Solution: Sludge moisture content before dewatering 99.5%, sludge moisture content after dewatering 99.5%-1%=98.5%
The amount of mud remains constant according to the amount of mud before and after dehydration, ie
V1（1－99.5%）＝V2（1－98.5%）
V1×0.5%＝V2×1.5%
V1/V2＝3
9 In a certain place, the ordinary activated sludge method is used to treat urban sewage, the water volume is 20000m3/d, the raw water BOD5 is 300mg/L, the removal rate of the initial sedimentation tank BOD5 is 30%, and the BOD5 of the treated water is required to be 20mg/L. a=0.5, b=0.06 θc=10d, MLVSS=3500mg/L, the volume of the aeration tank and the amount of remaining sludge were determined
(1) E = (L1-L2) / L1 30% = (300-L2) / 300 L2 = 210mg / L of the primary sedimentation tank
That is, the BOD5 concentration of the effluent of the primary sedimentation tank is 210mg/L
Therefore, the aeration tank inlet water BOD5 is 270mg/L
Aeration tank volume V=aQ(L0-Le)/Xv(1/θc+b)=0.5×(210-20)×20000/[3500(0.1+0.06)]=3393(m3)
(2) DX=XvV/θc=3500×10-3/10×3393=1187.55(KgVSS/d)
10. High-load biological filter method is selected to treat sewage, the water volume Q=5200m3/d, the raw water BOD5 is 300mg/L, the removal rate of the primary sedimentation tank BOD5 is 30%, the inlet water BOD5 of the biological filter is required to be 150mg/L, the effluent BOD5 is 30mg/L, and the load rate Fw=1.2KgBOD5/m3 filter material d Try to determine the volume, return ratio and hydraulic load rate of the filter.
(1) E= (L1-L2)/L1, 30%=(300-L2)/300, L2=210mg/L, that is, the BOD5 concentration of the primary sedimentation tank
Due to the BOD5 concentration L0 =270 mg/L entering the filter> 150 mg/L
So it has to be reflowed
150=(210+30r)/[(1+r) r=0.5
（2） V＝Q（1＋r）×150/Fw＝5200（1＋0.5）×150/1.2×103＝975m3
(3) Take the effective water depth H = 2.0m, then the area A = 487.5m2
Therefore q=5200 (1+0.5)/487.5=16 (m3/m2.d)
11. It is known that the inlet water volume of the activated sludge aeration tank is Q=2400m3/d, the inlet water BOD5 is 180mg/L, the effluent BOD5 is 40mg/L, the mixture concentration MLVSS=1800mg/L, the aeration tank volume V=500m3, a=0.65, b =0.096d-1, the remaining sludge amount DX is found and the total oxygen recharge R is calculated accordingly.
(1)u=QLr/XvV=2400×(180-40)/(1800×500)=0.37 (KgBOD5/KgMLVSS.d)
1/θc=au-b =0.65×0.37-0.096 1/θc=0.1445
θc=6.92 (d)
Residual sludge DX=Xvv/θc=0.1445×1800×10-3×500=130(KgVSS/d)
（2）R＝QLr/0.68－1.42, DX＝2400（180－40）10－3/0.68－1.42×130
R=309.5 (KgO2/d)
12. The turntable diameter of the laboratory biological turntable device is 0.5 m, and the disc is 10 pieces. The amount of wastewater Q = 40L/h, the inlet water BOD5 is 200mg/L, and the outlet BOD5 is 50mg/L, and the load rate of the turntable is obtained.
Solution: The total platter area F=n×1/4×3.14 × D2×2
F＝10×0.25×3.14×0.52×2＝3.925（m2）
Q=40L/h=0.96m3/d
u=Q(L0-Le)/F=0.96(200-50)/3.925=36.7(gBOD5/m2.d)
13. The daily fresh sludge volume of a sewage treatment plant is 2268Kg, the moisture content is 96.5%, and the sludge dispensing rate is 5%. Calculate the anaerobic digester volume.
Solution: Sludge volume V'=2268/[(1-0.965)1000]=64.8(m3/d)
V=V'/(P×100)=64.8×100/5=1296 m3
14. Water quality data are as follows: CO2 30mg/L, HCO3-3.6mmol/L
Ca2+ 1.4mmol/L , SO42-0.55mmol/L
Mg2＋0.9mmol/L，Cl－0.3mmol/L，Na＋＋K＋ 0.4mmol/L
How many mg/L of CO2 and H2SO4+HCl produced by RH column softening are calculated?
Solution: Ca(HCO3)2 1.4mmol/L, Mg(HCO3)2 0.4mmol/L, Mg(SO42-+Cl-) 0.5mmol/L in water, and the CO2 produced by the reaction was 3.6mmol/L=158.4mg/L
总CO2=(30+158.4)mg/L=188.4mg/L
（H2SO4＋HCl）＝Mg（SO42－＋Cl－） ＝0.5 mmol/L
15. Treat the sludge with pressurized dissolved air flotation method, the amount of mud Q = 3000m3/d, the moisture content is 99.3%, the air flotation pressure is 0.3MPa (gauge pressure), the concentration of air flotation scum is required to be 3%, and the return flow of pressure water and air volume are obtained.
(A/S=0.025, Cs=29mg/L, air saturation coefficient f=0.5)
Solution: Sludge with a moisture content of 99.3%.
Sludge content 1-99.3% = 0.7%
The corresponding sludge concentration X0 = 7000mg/L
The absolute pressure P = 4.1Kgf/cm2 corresponding to the gauge pressure of 0.3MPa
A/S＝RCs（fP－1）/QX0 0.025＝R×29（0.5×4.1－1）/(3000×7000)
R=17241 m3/d
A=0.025S=0.025×3000×7=525Kg/d
16. How does the salt content change after NaR resin softening in Hard Water: Ca2+=1.4 mmol/L; Mg2+=0.9 mmol/L, water Q=100 m3/d? What is the daily variation?
(1) Because 1 Ca2+ and Mg2+ are replaced by 2 Na+ after RNa, the salt amount will increase
(2) Ca2+ replaced na+=2Ca2+=2× 1.4=2.8mmol/L
Therefore, the net increment is 2.8× 23-1.4×40 = 8.4 mg/L
Mg2+ replaced by Na+=2Mg2+=2×0.9=1.8mmol/L
Therefore, the net increment is 1.8× 23-0.9×24 =19.8 mg/L
Total 19.8+8.4=28.2 mg/L=28.2 g/m3
The daily net increment was 100× 28.2=2820 g/d=2.82 Kg/d
17. The amount of wastewater containing hydrochloric acid Q=1000m3/d, the concentration of hydrochloric acid is 7g/L, neutralized with limestone, the active ingredient of limestone is 40%, and the amount of limestone is used (per day).
解:[HCl]=7g/L=0.19mol/L
The corresponding [CaCO3] = 1/2 [HCl] = 0.085mol/L = 8.5 g/L
Actual limestone 8.5/40% = 21.25 g/L = 21.25Kg/m3
Daily Use 1000×21.25 Kg=21.25 (t)
18. What is the pH value required to reduce Cd2+ to 0.1 mg/L by treating Wastewater containing Cd2+ with Ca(OH)2? What is the Ca2+ concentration at this point? ( Atomic weight of Cd 112; KCd（OH）2＝2.2×10－14）
解:Cd2+=0.1mg/L=8.9×10-4mol/L
lg(8.9×10-4)=14×2-2×pH+lg(2.2×10-14)
pH＝8.7
(2) pH=8.7 [H+]=10-8.7 mol/L
因[OH－][H＋]＝10－14
故[OH－]＝10－5.3mol/L
Ca2＋＋2OH－®Ca（OH）2
[Ca2＋]＝1/2[OH－]＝1/2×10－5.3＝2.5×10－6 mol/L
19. A water plant uses refined aluminum sulfate as a coagulant, and its maximum amount is 35mg/L. The design water quantity of the water plant is 105m3/d, the coagulant is adjusted 3 times a day, the solution concentration is 10%, and the volume of the dissolved drug cell is 1/5 of the solution volume.
（1）W＝aQ/（1000×100×b×n）＝35×105/（1000×100×10%×3）＝11.7（m3）
(2)W1=W/5=11.7/5=2.34(m3)
20. The design flow rate of the isolated mixing pool is 75000 m3/d, and the effective volume of the mixing pool is 1100m3. Total head loss 0.26m. Water flow velocity gradient g and GT (kinematic viscosity of water at 20 °C r=10-6m2/s)
（1）T＝VQ＝1100/75000＝0.01467（d）＝1267（s）
G＝（gh/rT）1/2＝[9.8×0.26/(10－6×1267)]1/2＝45（s－1）
（2） GT＝45×1267＝57015
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