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Question 1Integrated subject foundations, assessment and evidence

In EIA engineering analysis, what is the fundamental governing equation of the Law of Conservation of Mass in Material Balance calculations for a steady-state industrial process?

  1. $\sum G_{in} = \sum G_{out} \times 2.5$
  2. $\sum G_{loss} = \sum G_{in}^2 - \sum G_{out}^2$
  3. $\sum G_{in} = \sum G_{out} + \sum G_{accumulation} + \sum G_{loss}$ (Inputs = Outputs + Accumulation + Losses) Correct answer.
  4. $\sum G_{out} = \sum G_{in} / \pi$

Explanation

The Law of Conservation of Mass dictates that for any chemical process or unit operation, total material input equals total material output plus internal system accumulation and fugitive/uncontrolled losses: $\sum G_{in} = \sum G_{product} + \sum G_{byproduct} + \sum G_{waste} + \sum G_{accumulation} + \sum G_{loss}$. In continuous steady state, accumulation is zero ($G_{accumulation} = 0$).

Question 2Integrated subject foundations, assessment and evidence

A thermal power plant boiler consumes $100\text{ t/h}$ of raw coal containing $1.0\%\text{ total sulfur}$ ($S = 1.0\%$). The in-furnace ash sulfur retention rate is $15\%$ ($\eta_s = 0.15$), and the wet limestone-gypsum Flue Gas Desulfurization (FGD) system achieves a removal efficiency of $95\%$ ($\eta_{FGD} = 0.95$). What is the final atmospheric emission rate of $\text{SO}_2$?

  1. $1,700\text{ kg/h}$
  2. $2,000\text{ kg/h}$
  3. $17\text{ kg/h}$
  4. $85\text{ kg/h}$ ($0.085\text{ t/h}$) Correct answer.

Explanation

Calculated via stoichiometric sulfur balance: Total sulfur input $S_{in} = 100\text{ t/h} \times 1.0\% = 1.0\text{ t/h} = 1,000\text{ kg/h}$. Sulfur entering flue gas after furnace retention $= 1,000 \times (1 - 0.15) = 850\text{ kg/h}$. Since $1\text{ kg S}$ yields $2\text{ kg SO}_2$ (molar mass $32 \to 64$), $\text{SO}_2$ generated before FGD $= 850 \times 2 = 1,700\text{ kg/h}$. After $95\%$ FGD removal, atmospheric $\text{SO}_2$ emission $= 1,700 \times (1 - 0.95) = 1,700 \times 0.05 = 85\text{ kg/h}$.

Question 3Integrated subject foundations, assessment and evidence

In an industrial water balance flowsheet, an industrial plant uses $200\text{ m}^3/\text{h}$ of fresh makeup water ($Q_{fresh}$) and maintains an internal cooling and process recycled water circulation rate of $800\text{ m}^3/\text{h}$ ($Q_{recycled}$). What is the Industrial Water Recycling Rate ($R$)?

  1. $80.0\%$ ($R = \frac{800}{200 + 800} \times 100\%$) Correct answer.
  2. $25.0\%$
  3. $400.0\%$
  4. $20.0\%$

Explanation

Under standard industrial water balance accounting formulas: Water Recycling Rate $R = \frac{Q_{recycled}}{Q_{total\_water\_use}} \times 100\% = \frac{Q_{recycled}}{Q_{fresh} + Q_{recycled}} \times 100\% = \frac{800}{200 + 800} \times 100\% = \frac{800}{1000} \times 100\% = 80.0\%$.

Question 4Integrated subject foundations, assessment and evidence

In chemical engineering EIA analysis, what is the primary technical principle of a Leak Detection and Repair (LDAR) program for controlling fugitive Volatile Organic Compound (VOCs) emissions?

  1. Installing open discharge hoses on chemical storage tanks to vent fumes directly to ambient air
  2. Systematically detecting fugitive leaks from pipeline valves, flanges, pumps, and sampling connections using portable FID/PID instruments, and repairing components exceeding statutory leak thresholds within prescribed time limits Correct answer.
  3. Painting all process pipes with white reflective latex enamel without checking valves
  4. Increasing plant operating temperature to evaporate VOCs before they can be detected

Explanation

Under GB 37822-2019 (Standard for Fugitive Emission of Volatile Organic Compounds), LDAR (泄漏检测与修复) is an engineering compliance program requiring periodic instrumental screening (e.g. Flame Ionization Detectors FID) of thousands of potential piping leak points (valves, pump seals, flanges). Components showing VOC concentrations above statutory leak thresholds (typically 500-2,000 ppm) must be tagged, repaired, and re-tested within statutory timeframes (first attempt within 5 days, final repair within 15 days).

Question 5Integrated subject foundations, assessment and evidence

In atmospheric dispersion modeling, how is the Effective Stack Height ($H_e$) defined?

  1. $H_e = H_s - \Delta H$, subtracting the height of surrounding trees
  2. $H_e = H_s \times \Delta H$, multiplying height by stack diameter
  3. $H_e = \sqrt{H_s^2 + \Delta H^2}$
  4. $H_e = H_s + \Delta H$, where $H_s$ is the physical stack height and $\Delta H$ is the plume rise caused by thermal buoyancy and vertical exit momentum Correct answer.

Explanation

Effective stack height $H_e = H_s + \Delta H$ represents the actual centerline height of the bent-over smoke plume above ground level, calculated as the sum of the physical stack height ($H_s$) and the plume rise ($\Delta H$) generated by the flue gas thermal buoyancy flux ($F$) and initial discharge momentum.

Question 6Integrated subject foundations, assessment and evidence

Which atmospheric temperature inversion condition creates Fumigation Pollution (熏烟型污染), posing the most severe ground-level air pollution risks from elevated stacks?

  1. When a ground-based radiation inversion breaks up from the surface upward after sunrise while the upper layer remains inverted, causing rapid downward turbulent mixing into the newly formed shallow convective boundary layer Correct answer.
  2. When a strong dry adiabatic lapse rate extends uniformly to 10,000 meters altitude
  3. When heavy cyclonic rainfall creates intense downdrafts washing all pollutants out
  4. When high-altitude jet streams blow at velocities exceeding $150\text{ m/s}$

Explanation

Fumigation (熏烟型) occurs in morning hours after sunrise: solar heating creates an unstable, turbulent mixed layer near the ground that erodes the nocturnal radiation inversion from below. When this growing convective boundary layer reaches the elevated stack plume, the strong downward vertical turbulence rapidly drives high concentrations of pollutants to the ground while the remaining inversion aloft prevents upward dispersion.

Question 7Integrated subject foundations, assessment and evidence

A wastewater treatment plant discharges an effluent flow of $Q_p = 0.5\text{ m}^3/\text{s}$ with a Chemical Oxygen Demand of $C_p = 200\text{ mg/L}$ into a river with an upstream background flow of $Q_h = 4.5\text{ m}^3/\text{s}$ and background $\text{COD}$ of $C_h = 20\text{ mg/L}$. Assuming complete mixing, what is the resulting mixed river concentration ($C$)?

  1. $110.0\text{ mg/L}$
  2. $38.0\text{ mg/L}$ Correct answer.
  3. $220.0\text{ mg/L}$
  4. $22.2\text{ mg/L}$

Explanation

Applying the complete mixing mass balance equation: $C = \frac{Q_p C_p + Q_h C_h}{Q_p + Q_h} = \frac{(0.5 \times 200) + (4.5 \times 20)}{0.5 + 4.5} = \frac{100 + 90}{5.0} = \frac{190}{5.0} = 38.0\text{ mg/L}$.

Question 8Integrated subject foundations, assessment and evidence

Two identical uncorrelated industrial sound sources each produce a sound level of $60.0\text{ dB(A)}$ at a sensitive receiver. When both machines operate simultaneously, what is the combined total sound pressure level?

  1. $120.0\text{ dB(A)}$
  2. $60.0\text{ dB(A)}$
  3. $66.0\text{ dB(A)}$
  4. $63.0\text{ dB(A)}$ ($60.0 + 10 \log_{10}(2) = 63.01\text{ dB(A)}$) Correct answer.

Explanation

Noise superposition follows logarithmic acoustic energy addition: $L_{total} = 10 \log_{10}\left(10^{L_1/10} + 10^{L_2/10}\right) = 10 \log_{10}\left(10^{6.0} + 10^{6.0}\right) = 10 \log_{10}(2 \times 10^6) = 60 + 10 \log_{10}(2) = 60 + 3.01 = 63.01 \approx 63.0\text{ dB(A)}$. Superimposing two equal sound levels always increases the total level by $3.0\text{ dB(A)}$.

Question 9Integrated subject foundations, assessment and evidence

In acoustic noise barrier design, what dimensionless parameter is calculated from the path length difference ($\delta = A + B - d$) and acoustic wavelength ($\lambda$) to estimate barrier diffraction attenuation?

  1. Fresnel Number ($N = \frac{2\delta}{\lambda}$) Correct answer.
  2. Reynolds Number ($Re = \frac{\rho v d}{\mu}$)
  3. Froude Number ($Fr = \frac{v}{\sqrt{g h}}$)
  4. Mach Number ($Ma = \frac{v}{c}$)

Explanation

Under HJ 2.4-2021 Appendix B and standard acoustic physics (Maekawa barrier formula), the acoustic diffraction attenuation of a noise barrier is calculated from the Fresnel Number $N = \frac{2\delta}{\lambda} = \frac{2(A + B - d) f}{c}$, where $\delta$ is the acoustic path difference between the diffracted path over the barrier top ($A + B$) and direct line of sight ($d$).

Question 10Integrated subject foundations, assessment and evidence

In wastewater treatment engineering, which physical process is primarily utilized in a Primary Sedimentation Tank (初沉池) to remove settleable suspended solids (SS)?

  1. High-pressure reverse osmosis membrane filtration
  2. Gravity settling based on particle density difference and Stokes' Law Correct answer.
  3. Thermal incineration at $1,200^\circ\text{C}$
  4. Ultraviolet photocatalytic oxidation

Explanation

Primary sedimentation tanks rely on physical gravity sedimentation (重力沉降) governed by Stokes' Law: particulate matter and suspended solids denser than water settle to the tank bottom under quiescent hydraulic conditions and are removed by mechanical sludge scrapers.

Question 11Integrated subject foundations, assessment and evidence

In biological wastewater nutrient removal, what is the functional role of the Anaerobic (A), Anoxic (A), and Oxic (O) zones in an $A^2/O$ process?

  1. Anaerobic Zone: Ozone disinfection; Anoxic Zone: Sand filtration; Oxic Zone: Sludge drying
  2. Anaerobic Zone: Heavy metal evaporation; Anoxic Zone: Acid neutralization; Oxic Zone: Plastic shredding
  3. Anaerobic Zone: Phosphorus release by PAOs; Anoxic Zone: Denitrification of nitrate to nitrogen gas; Oxic Zone: Organic oxidation, nitrification of ammonia to nitrate, and luxury phosphorus uptake Correct answer.
  4. All three zones perform identical chemical chlorination

Explanation

The $A^2/O$ (Anaerobic-Anoxic-Oxic) process is designed for simultaneous nitrogen and phosphorus removal: In the Anaerobic zone ($DO < 0.2\text{ mg/L}$, no $\text{NO}_3^-$), phosphorus-accumulating organisms (PAOs) release phosphorus and store PHAs; In the Anoxic zone ($DO < 0.5\text{ mg/L}$, $\text{NO}_3^-$ present from nitrified internal recycle), heterotrophic denitrifiers reduce nitrate to $\text{N}_2$ gas; In the Oxic zone ($DO > 2\text{ mg/L}$), autotrophic nitrifiers oxidize ammonia to nitrate, organic matter is oxidized, and PAOs perform luxury phosphorus uptake.

Question 12Integrated subject foundations, assessment and evidence

What is the primary operational advantage of a Membrane Bioreactor (MBR) over a conventional activated sludge process with secondary clarifiers?

  1. Elimination of all electrical power and blower aeration requirements
  2. Complete conversion of wastewater directly into edible table sugar
  3. Ability to operate without producing any biological sludge forever
  4. Membrane microfiltration/ultrafiltration completely replaces secondary clarifiers, decoupling hydraulic retention time (HRT) from solids retention time (SRT) and maintaining high biomass concentrations (MLSS 8,000-12,000 mg/L) with superior effluent clarity Correct answer.

Explanation

MBR combines biological degradation with membrane filtration (pore size 0.04-0.4 $\mu\text{m}$), replacing bulky secondary sedimentation tanks. This allows independent control of HRT and SRT, operates at high mixed liquor suspended solids (MLSS 8,000-12,000 mg/L), reduces plant footprint by $>50\%$, and delivers ultra-clean effluent free of suspended solids and bacteria.

Question 13Integrated subject foundations, assessment and evidence

In advanced chemical oxidation for refractory organic industrial wastewater, what is the chemical reaction mechanism of the Fenton Reagent (芬顿试剂)?

  1. Reaction between ferrous iron ($\text{Fe}^{2+}$) and hydrogen peroxide ($\text{H}_2\text{O}_2$) under acidic conditions (pH 3-4) to generate highly reactive hydroxyl free radicals ($\cdot\text{OH}$): $\text{Fe}^{2+} + \text{H}_2\text{O}_2 \to \text{Fe}^{3+} + \cdot\text{OH} + \text{OH}^-$ Correct answer.
  2. Direct precipitation of heavy metals using limestone slurry at pH 12
  3. Biological fermentation by anaerobic methanogens at $55^\circ\text{C}$
  4. Absorption of volatile gases onto activated alumina pellets

Explanation

The Fenton process uses catalytic decomposition of hydrogen peroxide by ferrous ions under acidic conditions (optimal pH 3.0-4.0) to generate hydroxyl radicals ($\cdot\text{OH}$). With a standard oxidation potential of $2.80\text{ V}$, hydroxyl radicals non-selectively oxidize refractory, toxic aromatic rings and heterocyclic compounds into biodegradable organic acids, $\text{CO}_2$, and $\text{H}_2\text{O}$.

Question 14Integrated subject foundations, assessment and evidence

In thermal power plant flue gas desulfurization, what is the overall chemical reaction in the absorption tower of a Wet Limestone-Gypsum Flue Gas Desulfurization (FGD) system?

  1. $\text{SO}_2 + 2\text{NaOH} \to \text{Na}_2\text{SO}_4 + \text{H}_2$
  2. $\text{SO}_2 + \text{CaCO}_3 + \frac{1}{2}\text{O}_2 + 2\text{H}_2\text{O} \to \text{CaSO}_4\cdot 2\text{H}_2\text{O}\downarrow + \text{CO}_2\uparrow$ Correct answer.
  3. $\text{SO}_2 + \text{C} \to \text{CS}_2 + \text{O}_2$
  4. $\text{SO}_2 + \text{N}_2 \to 2\text{NO} + \text{S}$

Explanation

In wet limestone-gypsum FGD, flue gas $\text{SO}_2$ contacts a recirculating limestone ($\text{CaCO}_3$) slurry: $\text{SO}_2$ dissolves to form bisulfite ($\text{HSO}_3^-$), reacts with calcium ions to form calcium sulfite ($\text{CaSO}_3$), and is oxidized by injected forced oxidation air to form commercial-grade gypsum dihydrate ($\text{CaSO}_4\cdot 2\text{H}_2\text{O}$), releasing carbon dioxide ($\text{CO}_2$).

Question 15Integrated subject foundations, assessment and evidence

In Selective Catalytic Reduction (SCR) flue gas denitrification, what is the typical operating temperature window and main catalyst composition for achieving $>85\%$ $\text{NO}_x$ reduction?

  1. Operating temperature: $20^\circ\text{C}$ to $50^\circ\text{C}$; Catalyst: Crushed limestone gravel
  2. Operating temperature: $900^\circ\text{C}$ to $1,200^\circ\text{C}$; Catalyst: No catalyst (pure thermal injection)
  3. Operating temperature: $300^\circ\text{C}$ to $400^\circ\text{C}$; Catalyst: Vanadium-tungsten-titanium oxide ($\text{V}_2\text{O}_5\text{-WO}_3/\text{TiO}_2$) Correct answer.
  4. Operating temperature: $-40^\circ\text{C}$; Catalyst: Liquid nitrogen ice

Explanation

SCR injects ammonia ($\text{NH}_3$) into boiler flue gas upstream of a honeycomb or plate catalyst bed composed of vanadium pentoxide ($\text{V}_2\text{O}_5$) and tungsten trioxide ($\text{WO}_3$) on a titanium dioxide ($\text{TiO}_2$) carrier. The optimal catalytic reaction window is $300-400^\circ\text{C}$ ($4\text{NO} + 4\text{NH}_3 + \text{O}_2 \to 4\text{N}_2 + 6\text{H}_2\text{O}$), keeping ammonia slip below $2.5-3.0\text{ mg/m}^3$.

Question 16Integrated subject foundations, assessment and evidence

Why are Baghouse Fabric Filters (袋式除尘器) often preferred over Electrostatic Precipitators (ESP) for capturing high-resistivity fly ash and fine particles ($\text{PM}_{2.5}$) in modern industrial flue gas cleaning?

  1. Baghouse filters consume zero filter media and last for 100 years without maintenance
  2. Baghouse filters generate high-voltage electrostatic arcs that convert ash into helium gas
  3. Baghouse filters operate effectively only when flue gas temperature exceeds $1,000^\circ\text{C}$
  4. Baghouse filters rely on physical sieve barrier filtration through woven/felt fabrics and dust cake, achieving stable $\ge 99.9\%$ particulate removal efficiency independent of dust electrical resistivity Correct answer.

Explanation

Baghouses capture dust via interception, impaction, and diffusion across fabric media and dust cake. Because filtration is physical, its capture efficiency ($\ge 99.9\%$, outlet dust $\le 5-10\text{ mg/m}^3$) is not affected by dust specific electrical resistivity (which causes back-corona in ESPs on high-resistivity low-sulfur coal ash) and effectively captures fine $\text{PM}_{2.5}$ particles.

Question 17Integrated subject foundations, assessment and evidence

In industrial organic waste gas treatment, what is the operational principle and thermal efficiency characteristic of a Regenerative Thermal Oxidizer (RTO / 蓄热式热力燃烧装置)?

  1. VOCs are oxidized into $\text{CO}_2$ and $\text{H}_2\text{O}$ at high temperature ($\ge 760\text{--}800^\circ\text{C}$), while alternating ceramic honeycomb regenerative media beds preheat incoming gas and recover $\ge 95\%$ of thermal energy Correct answer.
  2. VOCs are absorbed into chilled mineral oil at $-10^\circ\text{C}$ without combustion
  3. VOCs are neutralized with liquid hydrochloric acid spray at room temperature
  4. Waste gas is vented through open cooling towers to dilute odors

Explanation

An RTO oxidizes organic compounds at $\ge 760-800^\circ\text{C}$ with destruction removal efficiencies (DRE) of $\ge 98-99\%$. By cyclically reversing the flow direction across multi-bed ceramic honeycomb regenerative matrices, outgoing hot clean gas preheats the incoming cold VOC-laden waste stream, achieving $>95\%$ thermal heat recovery and self-sustaining autothermal operation at moderate inlet concentrations ($>1.5-2.0\text{ g/m}^3$).

Question 18Integrated subject foundations, assessment and evidence

In groundwater hydrogeology, how is the Actual Linear Groundwater Seepage Velocity ($u$) related to Darcy Flux ($v$), Hydraulic Conductivity ($K$), Hydraulic Gradient ($I$), and Effective Porosity ($n_e$)?

  1. $u = v \cdot n_e = K \cdot I \cdot n_e$
  2. $u = \frac{n_e}{K \cdot I}$
  3. $u = \frac{v}{n_e} = \frac{K \cdot I}{n_e}$ Correct answer.
  4. $u = \sqrt{K \cdot I \cdot n_e}$

Explanation

Darcy's Law defines the apparent Darcy flux (discharge per total cross-sectional area of soil/rock) as $v = K \cdot I$. Because water flows only through interconnected pore spaces, the actual linear seepage velocity of groundwater particles through the pores is obtained by dividing the Darcy flux by the effective porosity ($n_e$): $u = \frac{v}{n_e} = \frac{K \cdot I}{n_e}$.

Question 19Integrated subject foundations, assessment and evidence

An unconfined sandy aquifer has a hydraulic conductivity of $K = 10.0\text{ m/day}$, a hydraulic gradient of $I = 0.005$ ($0.5\%\text{ slope}$), and an effective porosity of $n_e = 0.25$ ($25\%$). What is the actual linear groundwater seepage velocity ($u$)?

  1. $0.05\text{ m/day}$
  2. $0.0125\text{ m/day}$
  3. $1.25\text{ m/day}$
  4. $0.20\text{ m/day}$ ($u = \frac{10.0 \times 0.005}{0.25} = \frac{0.05}{0.25} = 0.20\text{ m/day}$) Correct answer.

Explanation

Darcy flux $v = K \cdot I = 10.0\text{ m/day} \times 0.005 = 0.05\text{ m/day}$. Actual linear pore seepage velocity $u = \frac{v}{n_e} = \frac{0.05\text{ m/day}}{0.25} = 0.20\text{ m/day}$ (equivalent to $73.0\text{ m/year}$).

Question 20Integrated subject foundations, assessment and evidence

In petrochemical and chemical project EIA case reviews, what are the three sequential defense barriers in the Three-Tier Emergency Environmental Risk Prevention System (三级防控体系)?

  1. Tier 1: Smoke detectors; Tier 2: Water sprinklers; Tier 3: Evacuation megaphones
  2. Tier 1: Equipment/tank dike containment bund (围堰/防火堤); Tier 2: Plant accident emergency retention basin (厂区事故应急池); Tier 3: Industrial park / terminal storm outfall shutoff weir and interceptor gate (园区/终端雨水排口切断闸门与防控池) Correct answer.
  3. Tier 1: Security guards; Tier 2: Perimeter fencing; Tier 3: CCTV cameras
  4. Tier 1: Export tax rebate; Tier 2: Corporate loan; Tier 3: Bankruptcy filing

Explanation

China's mandatory chemical environmental risk prevention guidelines establish a physical three-tier containment system: Tier 1 (Unit level) contains local leaks within storage tank bunds or workshop coffer dikes; Tier 2 (Plant level) routes contaminated firefighting water and major spill overflows to a dedicated plant accident retention basin ($V_{accident}$); Tier 3 (Park/Terminal level) intercepts contaminated flows at final stormwater/sewage boundary outfalls via emergency gate valves before reaching natural surface water bodies.

Question 21Integrated subject foundations, assessment and evidence

In a coal chemical project EIA case, how is high-salinity, refractory organic wastewater (e.g. coal gasification wastewater with high phenols, ammonia, and TDS $>30,000\text{ mg/L}$) typically processed to achieve Zero Liquid Discharge (ZLD)?

  1. Pretreatment (Phenol extraction + Ammonia stripping) $\to$ Biochemical degradation ($A/O$ or $A^2/O$) $\to$ Advanced oxidation $\to$ Dual-membrane concentration (UF + High-pressure RO) $\to$ Mechanical Vapor Recompression (MVR) / Multi-Effect Evaporation crystallization Correct answer.
  2. Direct discharge into local agricultural irrigation canals without treatment
  3. Pumping the raw wastewater untreated into deep unconfined drinking water aquifers
  4. Spraying raw toxic gasification wastewater onto public highways for road dust control

Explanation

Coal chemical ZLD systems use a rigorous multi-stage train: (1) solvent extraction to recover crude phenols and steam stripping for ammonia; (2) biological treatment for bulk organic removal; (3) Fenton/ozone oxidation to break recalcitrant organics; (4) ultrafiltration (UF) and high-pressure reverse osmosis (DTRO/STRO) to concentrate brine and recycle fresh water; and (5) MVR/MED thermal evaporative crystallization to produce solid sodium sulfate/chloride salts.

Question 22Integrated subject foundations, assessment and evidence

In thermal power plant EIA cases, which comprehensive utilization pathways are prioritized for coal combustion solid byproducts (fly ash, bottom slag, and FGD gypsum)?

  1. Dumping raw ash directly into scenic national lakes
  2. Spreading desulfurization gypsum across municipal drinking water reservoirs
  3. Packaging fly ash as dry baking flour for bakeries
  4. Production of cement, concrete admixtures, gypsum wallboards, road base materials, and sintered building bricks Correct answer.

Explanation

Under the Clean Production and Circular Economy Promotion Laws, coal ash and desulfurization gypsum from thermal power plants are classified as general industrial solid wastes (Class II). They are comprehensively utilized off-site in cement manufacturing, high-performance concrete supplementary cementitious materials, gypsum plasterboards, and road base aggregates.

Question 23Integrated subject foundations, assessment and evidence

In non-ferrous metal mining EIA cases, what chemical reaction mechanism causes Acid Mine Drainage (AMD / 酸性矿山废水) when sulfide minerals (such as pyrite $\text{FeS}_2$) are exposed to air and water?

  1. Oxidation of pyrite: $4\text{FeS}_2 + 15\text{O}_2 + 14\text{H}_2\text{O} \to 4\text{Fe(OH)}_3\downarrow + 8\text{SO}_4^{2-} + 16\text{H}^+$, generating low-pH sulfuric acid and dissolving heavy metals Correct answer.
  2. Dissolution of limestone generating pure alkaline drinking water
  3. Combustion of coal forming gaseous sulfur hexafluoride
  4. Condensation of ambient nitrogen into liquid ammonia fertilizer

Explanation

Acid Mine Drainage (AMD) occurs when pyrite ($\text{FeS}_2$) and other sulfide ores are exposed to oxygen, water, and iron-oxidizing bacteria (*Acidithiobacillus ferrooxidans*). Pyrite oxidizes to produce ferric hydroxide precipitate ('yellow boy') and sulfuric acid ($16\text{ H}^+$ per 4 moles $\text{FeS}_2$), dropping wastewater pH down to $2.0-3.5$ and leaching toxic heavy metals (copper, lead, zinc, cadmium, arsenic) into surrounding watersheds.

Question 24Integrated subject foundations, assessment and evidence

In mining EIA case analyses, which engineering measures are critical for preventing environmental risks and groundwater contamination from a Tailing Pond (尾矿库)?

  1. Discharging raw liquid tailing slurry directly down natural mountain streams during heavy rainstorms
  2. Constructing a perimeter flood interception ditch (截洪沟), anti-seepage geomembrane lining, downstream groundwater monitoring wells, seepage return pump station, and progressive ecological bench revegetation Correct answer.
  3. Using tailing sand as children's playground sandboxes
  4. Constructing tailings dams exclusively out of uncompacted dry straw bales

Explanation

Tailing pond environmental risk management requires: (1) diversion and flood interception channels (截洪沟/排洪系统) designed for 100-200 year return storm events; (2) bottom composite anti-seepage barrier systems to block heavy metal seepage into aquifers; (3) seepage collection and recycle return pumping stations; (4) down-gradient groundwater monitoring wells; and (5) progressive slope stabilization and revegetation.

Question 25Integrated subject foundations, assessment and evidence

In deep reservoir hydropower project EIA cases, what adverse downstream ecological impact is caused by Reservoir Thermal Stratification (水库水温分层), and which engineering measure mitigates it?

  1. Surface water overheats to boiling point; mitigated by adding ice cubes to the reservoir
  2. Thermal stratification creates underwater hurricanes; mitigated by installing submarine propellers
  3. Deep bottom-water release delivers low-temperature water ($4-8^\circ\text{C}$ colder in spring/summer), delaying downstream fish gonad development and spawning; mitigated by constructing a Multi-Level Intake Facility (分层取水设施 / 叠梁门) to draw warm surface water Correct answer.
  4. Thermal stratification permanently removes all hydrogen atoms from water molecules

Explanation

Deep reservoirs stratify in warm seasons into an epilimnion (warm surface), metalimnion/thermocline, and hypolimnion (cold, dense bottom layer). Conventional bottom turbine intakes draw cold hypolimnetic water ($4-8^\circ\text{C}$ below natural river temperature), suppressing downstream water temperature in spring and summer, which delays native fish gonad maturation and halts spawning. Multi-level intake structures (分层取水 / stoplog gates / selective withdrawal towers) draw warmer surface water to match natural downstream thermal regimes.

Question 26Integrated subject foundations, assessment and evidence

In hydropower dam EIA reviews, which suite of fish conservation measures is standardly implemented to mitigate dam barrier blockage on migratory fish?

  1. Installing high-speed shredding blades in turbine penstocks to pulverize fish
  2. Constructing continuous concrete dams across all tributary streams to prevent fish from entering any water
  3. Adding chemical fish poisons to clear river ecosystems before dam construction
  4. Fish passage facilities (pool-and-weir fish ladders 鱼道, fish locks 鱼闸, nature-like bypass channels), artificial fish breeding and restocking stations (增殖放流站), and establishment of downstream fish habitat protection reserves Correct answer.

Explanation

Under national ecological guidelines for water conservancy projects, dam mitigation combines: (1) fish passage engineering (vertical slot fishways, pool-and-weir fish ladders, nature-like bypass channels); (2) artificial breeding, spawning induction, and annual restocking stations (增殖放流); and (3) tributary habitat protection and seasonal operation coordination during spawning runs.

Question 27Integrated subject foundations, assessment and evidence

In MSW incineration EIA case analysis, what is the critical regulatory distinction between Incineration Fly Ash (焚烧飞灰) and Incineration Bottom Ash / Slag (炉渣)?

  1. Bottom Ash is classified as radioactive nuclear waste while Fly Ash is organic agricultural fertilizer
  2. Fly Ash is classified as Hazardous Waste (HW18) requiring chelation/solidification before landfilling in dedicated cells (GB 16889/GB 18598), whereas Bottom Ash is General Industrial Solid Waste that can be recycled into building materials Correct answer.
  3. Both Fly Ash and Bottom Ash are exempt from all environmental regulations and may be dumped into rivers
  4. Fly Ash must be discharged through the chimney into the air while Bottom Ash must be dissolved in drinking water

Explanation

Under the National Hazardous Waste Catalogue (HW18), Incineration Fly Ash (飞灰) collected from flue gas treatment contains concentrated heavy metals (lead, cadmium, mercury) and dioxins, legally classifying it as hazardous waste. It must be stabilized/chelated with cement/chelating agents before disposal in dedicated landfill cells under GB 16889. Bottom Ash (炉渣) from the furnace grate is general industrial solid waste (Class I/II) usable as brick aggregate or road base.

Question 28Integrated subject foundations, assessment and evidence

In groundwater remediation of a legacy volatile organic chemical (VOCs) and chlorinated solvent contaminated site, which in-situ remediation technology is standardly selected?

  1. Paving the site with ceramic bathroom tiles without subsurface remediation
  2. Flooding the subsurface aquifer with raw crude oil to coat the sand
  3. In-situ Air Sparging (空气曝气) coupled with Soil Vapor Extraction (SVE / 土壤气相抽提) and In-Situ Chemical Oxidation (ISCO) Correct answer.
  4. Planting plastic artificial flowers across the contaminated parcel

Explanation

For volatile chlorinated solvents (TCE, PCE) and petroleum hydrocarbons in groundwater, Air Sparging (AS) injects compressed air into the saturated zone to strip volatile contaminants into the vapor phase, where they are captured by Soil Vapor Extraction (SVE) wells in the vadose zone and treated via activated carbon/RTO off-gas systems. In-Situ Chemical Oxidation (ISCO using persulfate or permanganate) injects chemical oxidants to mineralize remaining dissolved plumes.

Question 29Integrated subject foundations, assessment and evidence

In contaminated site redevelopment EIA cases, which remediation technology is standardly chosen for heavy metal (lead, cadmium, arsenic) contaminated soils prior to commercial or public park reuse?

  1. Solidification / Stabilization (S/S / 固化/稳定化) using cementitious binders and chemical stabilizing agents to immobilize heavy metals, or Soil Washing (异位淋洗) Correct answer.
  2. High-temperature thermal desorption at $1,000^\circ\text{C}$ to evaporate non-volatile lead oxides
  3. Adding concentrated hydrochloric acid to dissolve heavy metals into open drinking fountains
  4. Covering the contaminated soil with $1\text{ mm}$ of clear polyethylene cling wrap

Explanation

For heavy metal contaminated soils, Solidification/Stabilization (S/S) mixes cement, fly ash, lime, and stabilizing agents (e.g. iron/phosphate compounds) to convert soluble heavy metal ions into insoluble, chemically stable mineral phases (e.g. lead phosphates, metal silicates), reducing leaching toxicity (HJ/T 299 extraction test) below regulatory standards. Ex-situ soil washing (淋洗) separates heavy metal fractions from coarse soil grains.

Question 30Integrated subject foundations, assessment and evidence

In national territorial spatial ecological environmental zoning, what are the core components of the 'Three Lines One Permit' (三线一单) ecological environmental zoning and control system in China?

  1. Railway Line, Highway Line, Airline, and Construction Permit
  2. Ecological Conservation Redlines (生态保护红线), Environmental Quality Bottom Lines (环境质量底线), Resource Utilization Upper Limits (资源利用上线), and Ecological Environmental Access Negative List (生态环境准入清单) Correct answer.
  3. Red Line, Yellow Line, Green Line, and Mining Concession Permit
  4. Tax Line, Budget Line, Profit Line, and Trade Permit

Explanation

China's 'Three Lines One Permit' (三线一单) system is the cornerstone of source-level environmental zoning and spatial planning governance: (1) Ecological Conservation Redlines (生态保护红线) define strictly protected ecological space; (2) Environmental Quality Bottom Lines (环境质量底线) establish ambient air, water, and soil quality targets; (3) Resource Utilization Upper Limits (资源利用上线) cap total water, energy, and land consumption; and (4) Ecological Environmental Access List (生态环境准入清单) establishes spatial negative lists for project approvals and industrial development.

Question 31Integrated subject foundations, assessment and evidence

For the activated sludge system in the previous problem (Q = 10,000 m^3/d, S_0 = 200 mg/L BOD_5, V = 3,040 m^3, MLVSS X = 2,500 mg/L), what is the Food-to-Microorganism ratio (F/M, organic loading rate on sludge) in kg BOD_5/(kg MLVSS·d)?

  1. 0.263 kg BOD_5/(kg MLVSS·d) Correct answer.
  2. 0.395 kg BOD_5/(kg MLVSS·d)
  3. 0.150 kg BOD_5/(kg MLVSS·d)
  4. 0.526 kg BOD_5/(kg MLVSS·d)

Explanation

Food-to-Microorganism ratio: F/M = (Q * S_0) / (V * X). Daily BOD_5 applied: Q * S_0 = 10,000 m^3/d * 200 mg/L * 10^-3 = 2,000 kg BOD_5/d. Total biomass in aeration tank: V * X = 3,040 m^3 * 2,500 mg/L * 10^-3 = 7,600 kg MLVSS. Therefore, F/M = 2,000 / 7,600 = 0.2632 kg BOD_5/(kg MLVSS·d). This is within the standard design range of 0.2–0.4 kg BOD_5/(kg MLVSS·d) for conventional activated sludge.

Question 32Integrated subject foundations, assessment and evidence

In an Anaerobic-Anoxic-Oxic (A^2/O) biological nutrient removal process, what is the primary biochemical role of the initial Anaerobic (厌氧) zone?

  1. To release intracellular polyphosphate and assimilate volatile fatty acids (VFAs) by Polyphosphate Accumulating Organisms (PAOs) Correct answer.
  2. To perform autotrophic nitrification of ammonia to nitrate under dissolved oxygen > 2.0 mg/L
  3. To achieve anoxic denitrification of recycled nitrate using incoming organic carbon
  4. To oxidize refractory COD through anaerobic digestion to generate methane gas

Explanation

In the A2O process, the first zone is strictly anaerobic (DO < 0.2 mg/L, NO_3-N ≈ 0). Polyphosphate-Accumulating Organisms (PAOs) break down intracellular polyphosphate chains to release orthophosphate into solution, generating energy to assimilate readily biodegradable volatile fatty acids (VFAs) and store them as intracellular poly-β-hydroxybutyrate (PHB). In the subsequent aerobic zone, PAOs oxidize PHB and perform 'luxury uptake' of phosphate far in excess of the released amount, which is removed via waste activated sludge.

Question 33Integrated subject foundations, assessment and evidence

What is the theoretical COD requirement (stoichiometric carbon-to-nitrogen ratio) for the complete heterotrophic denitrification of 1.0 g of nitrate-nitrogen (NO_3^--N) to nitrogen gas (N_2) using methanol (CH_3OH) as the external carbon source (neglecting biomass yield)?

  1. 4.57 g COD / g NO_3^--N
  2. 1.42 g COD / g NO_3^--N
  3. 2.86 g COD / g NO_3^--N Correct answer.
  4. 3.50 g COD / g NO_3^--N

Explanation

The stoichiometric reduction half-reaction is: 2 NO_3^- + 10 e^- + 12 H^+ -> N_2 + 6 H_2O. Each mole of N (14.0 g) accepts 5 moles of electrons. One mole of O_2 (32.0 g COD) accepts 4 moles of electrons (8 g COD per mole of electrons). Thus, electron equivalence = 5 mol e^- * 8 g COD/mol e^- = 40 g COD per 14 g N = 40 / 14 = 2.857 g COD / g NO_3^--N ≈ 2.86 g COD / g N. (When including cell synthesis, practical dosing ratios are typically 4.0–5.0 g COD/g N).

Question 34Integrated subject foundations, assessment and evidence

In a Membrane Bioreactor (MBR) wastewater treatment installation, what does the Critical Flux (临界通量) represent?

  1. The maximum permeate flux below which transmembrane pressure (TMP) remains stable without irreversible foulant cake layer formation Correct answer.
  2. The minimum flux required to maintain hydraulic circulation through the hollow-fiber membrane bundles
  3. The flux at which the membrane physical pores undergo permanent structural compaction under high hydraulic pressure
  4. The theoretical maximum water flux determined under pure distilled water conditions at 20°C

Explanation

Critical flux in MBR systems (per HJ 2010 Membrane Bioreactor Technical Specifications) is defined as the threshold permeate flux below which fouling rate is negligible and TMP remains steady over extended operational cycles. Operating above the critical flux causes rapid, irreversible deposition of extracellular polymeric substances (EPS) and colloidal cake fouling onto the membrane surface.

Question 35Integrated subject foundations, assessment and evidence

In Fenton advanced oxidation (Fenton高级氧化技术), what is the optimal operating pH range and the primary reactive radical species responsible for non-selective destruction of refractory organic pollutants?

  1. pH = 7.0–8.0; Superoxide radical (·O_2^-)
  2. pH = 10.0–11.0; Hydroperoxyl radical (·HO_2)
  3. pH = 3.0–3.5; Hydroxyl radical (·OH) Correct answer.
  4. pH = 1.0–2.0; Singlet oxygen (^1O_2)

Explanation

The classical Fenton reaction (Fe^2+ + H_2O_2 -> Fe^3+ + ·OH + OH^-) has an optimal pH window of 3.0 to 3.5. At pH > 4.0, Fe^3+ precipitates rapidly as Fe(OH)_3 sludge, terminating catalytic cycling. At pH < 2.5, H^+ scavenges hydroxyl radicals (·OH) and forms stable [Fe(H_2O)_6]^2+ complexes. The generated hydroxyl radical (·OH) has a very high standard redox potential (+2.80 V), enabling rapid, non-selective mineralization of complex aromatic rings.

Question 36Integrated subject foundations, assessment and evidence

During the operational monitoring of a high-rate mesophilic anaerobic sludge digester, which indicator ratio is universally used as the primary early warning parameter for digester acidification and failure?

  1. Chemical Oxygen Demand to Total Nitrogen ratio (COD / TN ≥ 20)
  2. Mixed Liquor Suspended Solids to Volatile Solids ratio (MLSS / MLVSS)
  3. Volatile Fatty Acids to Total Alkalinity ratio (VFA / ALK ≤ 0.30–0.40) Correct answer.
  4. Oxidation-Reduction Potential to pH ratio (ORP / pH)

Explanation

The Volatile Fatty Acids to Total Alkalinity ratio (VFA/ALK, 挥发酸/总碱度比值) is the most sensitive operational parameter in anaerobic digestion. In a healthy digester, VFA/ALK is maintained between 0.10 and 0.30. When VFA/ALK rises above 0.40, methanogen activity is being inhibited by accumulating acetic/propionic acids, signaling impending pH drop and digester souring before pH changes become evident in buffered systems.

Question 37Integrated subject foundations, assessment and evidence

A high-pressure chamber filter press (高压隔膜板框压滤机) dewaters digested sludge at feed flow Q = 100 m^3/d with moisture content P_1 = 97.0%. After mechanical pressure filtration, the dewatered sludge cake achieves a moisture content P_2 = 60.0%. Assuming solid recovery efficiency is 100% and sludge density is approximately 1000 kg/m^3 throughout, what is the daily mass of dewatered sludge cake produced?

  1. 3,000 kg/d (3.0 t/d)
  2. 10,000 kg/d (10.0 t/d)
  3. 5,000 kg/d (5.0 t/d)
  4. 7,500 kg/d (7.5 t/d) Correct answer.

Explanation

Mass of wet sludge feed: M_1 = 100 m^3/d * 1000 kg/m^3 = 100,000 kg/d. Dry solids mass in feed: M_dry = M_1 * (1 - P_1) = 100,000 kg/d * (1 - 0.97) = 100,000 * 0.03 = 3,000 kg dry solids/d. Since dry solids are conserved: Cake mass M_2 = M_dry / (1 - P_2) = 3,000 kg/d / (1 - 0.60) = 3,000 / 0.40 = 7,500 kg/d (7.5 t/d).

Question 38Integrated subject foundations, assessment and evidence

A Sludge Volume Index (SVI, 污泥体积指数) test is performed on mixed liquor from an activated sludge aeration tank. A 1,000 mL graduated cylinder filled with mixed liquor settles to a 30-minute settled sludge volume of SV_30 = 360 mL. If the MLSS concentration of the mixed liquor is determined to be 3.0 g/L (3,000 mg/L), what is the SVI, and what does this indicate regarding settling characteristics?

  1. SVI = 120 mL/g; normal, good sludge settleability Correct answer.
  2. SVI = 60 mL/g; pin-point floc settleability
  3. SVI = 240 mL/g; severe filamentous sludge bulking
  4. SVI = 180 mL/g; moderate sludge bulking

Explanation

Sludge Volume Index formula: SVI = (SV_30 (mL/L) * 1,000) / MLSS (mg/L) = (360 mL * 1,000) / 3,000 mg/L = 120 mL/g. An SVI between 70 and 150 mL/g indicates healthy activated sludge with excellent settling and compaction properties. SVI > 200 mL/g indicates filamentous sludge bulking (污泥膨胀), while SVI < 50 mL/g indicates small, dense pin-point flocs.

Question 39Integrated subject foundations, assessment and evidence

What is the primary operational mechanism of the Sequencing Batch Reactor (SBR) process that allows both biological nitrification and denitrification to occur in a single physical tank?

  1. Continuous counter-current recirculation between separate physical aeration and anoxic chambers
  2. High-pressure oxygen injection creating stratified aerobic surface and anoxic deep zones
  3. Time-sequenced alternation of aerobic aeration, anoxic mixing, and quiescent settling phases within a single reactor Correct answer.
  4. Fixed bio-carrier carriers that maintain anaerobic cores under continuous high DO aeration

Explanation

The SBR (序批式活性污泥法) is a fill-and-draw periodic batch system that operates through distinct time-phased cycles: Fill (进水), React (反应 - alternating aeration for nitrification and non-aerated mixing for denitrification), Settle (沉淀 - quiescent clarifier function), Decant (排水), and Idle (闲置). Space-oriented operations of continuous systems are thus replaced by time-oriented operations in a single tank.

Question 40Integrated subject foundations, assessment and evidence

In the Modified Bardenpho 5-stage biological nutrient removal process, what is the specific design purpose of the second anoxic (Zone 4) and second aerobic (Zone 5) tanks following the primary aeration tank?

  1. Zone 4 oxidizes residual ammonia; Zone 5 provides anaerobic phosphorus release
  2. Zone 4 performs chemical coagulation; Zone 5 serves as a gravity sludge thickener
  3. Zone 4 strips volatile organic compounds; Zone 5 provides enzymatic hydrolysis
  4. Zone 4 provides endogenous denitrification to remove residual nitrate; Zone 5 strips nitrogen gas and re-aerates mixed liquor before the clarifier Correct answer.

Explanation

In the 5-Stage Bardenpho configuration (Anaerobic -> 1st Anoxic -> 1st Aerobic -> 2nd Anoxic -> 2nd Aerobic): Zone 4 (2nd Anoxic) utilizes endogenous respiration of biomass to denitrify the remaining NO_3-N entering from the 1st aerobic zone without external recycle. Zone 5 (2nd Aerobic, re-aeration zone, HRT 10–30 min) strips entrained N_2 gas bubbles to prevent sludge floating in the clarifier, raises DO to 2–3 mg/L to prevent anaerobic phosphorus release in the secondary clarifier, and oxidizes any ammonia released during endogenous decay.

Question 41Integrated subject foundations, assessment and evidence

What is the primary biological mechanism of the Anammox (厌氧氨氧化) process in energy-neutral wastewater treatment?

  1. Autotrophic conversion of ammonium (NH_4^+) and nitrite (NO_2^-) directly into nitrogen gas (N_2) under strictly anaerobic conditions without organic carbon Correct answer.
  2. Heterotrophic reduction of nitrate (NO_3^-) to ammonia (NH_4^+) using volatile fatty acids
  3. Aerobic oxidation of ammonia directly to nitrate by Nitrospira bacteria at DO > 4.0 mg/L
  4. Enzymatic fixation of atmospheric nitrogen gas into microbial biomass proteins

Explanation

Anammox (Anaerobic Ammonium Oxidation) is carried out by specialized planctomycete bacteria (e.g., *Candidatus Brocadia*). Under anaerobic conditions, autotrophic anammox bacteria oxidize NH_4^+ using NO_2^- as the electron acceptor to form N_2 gas: NH_4^+ + 1.32 NO_2^- + 0.066 HCO_3^- + 0.13 H^+ -> 1.02 N_2 + 0.26 NO_3^- + 0.066 CH_2O_0.5N_0.15 + 2.03 H_2O. It eliminates the need for organic carbon (COD) and reduces aeration energy by 60% when paired with partial nitritation.

Question 42Integrated subject foundations, assessment and evidence

A tangential cyclone separator has inlet width W = 0.20 m, effective turns N_e = 5.0, and treats flue gas at 20°C (gas dynamic viscosity μ = 1.81 × 10^-5 Pa·s, gas density negligible) at an inlet velocity v_i = 20.0 m/s. The dust particles have density ρ_p = 2,000 kg/m^3. Using Lapple's formula, what is the cut diameter (d_pc) of this cyclone?

  1. 10.18 μm
  2. 2.55 μm
  3. 5.09 μm Correct answer.
  4. 15.27 μm

Explanation

Applying Lapple's cut diameter formula: d_pc = sqrt((9 * μ * W) / (2 * π * N_e * v_i * (ρ_p - ρ_g))). Numerator = 9 * (1.81 × 10^-5 Pa·s) * 0.20 m = 3.258 × 10^-5 kg/s. Denominator = 2 * π * 5.0 * 20.0 m/s * 2000 kg/m^3 = 2 * 3.14159 * 5.0 * 20.0 * 2000 = 1,256,637 kg/(m^2·s). Fraction = (3.258 × 10^-5) / (1.256637 × 10^6) = 2.5926 × 10^-11 m^2. Taking square root: d_pc = sqrt(2.5926 × 10^-11) = 5.0918 × 10^-6 m = 5.09 μm.

Question 43Integrated subject foundations, assessment and evidence

In coal-fired boiler flue gas dust collection (gas temperature 140–170°C, containing SO_2 and trace acid moisture), which filter bag material is widely selected for its superior acid corrosion resistance, hydrolytic stability, and high continuous operating temperature (190°C)?

  1. Polyphenylene Sulfide (PPS / 聚苯硫醚) Correct answer.
  2. Polyester (PET / 涤纶)
  3. Polypropylene (PP / 丙纶)
  4. Polyacrylonitrile (PAN / 亚克力)

Explanation

PPS (Polyphenylene Sulfide / 聚苯硫醚) needle felt is the industry standard filter media for coal-fired boiler flue gas filtration. It exhibits excellent continuous thermal resistance up to 160–190°C, outstanding acid resistance against SO_2/SO_3 attack, and high hydrolytic resistance under humid flue gas environments.

Question 44Integrated subject foundations, assessment and evidence

What dust electrical resistivity (粉尘比电阻) range is considered optimal for high-efficiency collection in standard Electrostatic Precipitators without experiencing back-corona (反电晕) or particle re-entrainment?

  1. 10^1 to 10^3 Ω·cm
  2. 10^11 to 10^14 Ω·cm
  3. 10^15 to 10^18 Ω·cm
  4. 10^4 to 10^10 Ω·cm Correct answer.

Explanation

The optimal dust resistivity operating window for dry ESPs is 10^4 to 10^10 Ω·cm. When dust resistivity is too low (< 10^4 Ω·cm), particles lose their charge instantaneously upon contacting the grounded collection plate, acquire a charge of the same polarity as the plate, and are repelled back into the gas stream (re-entrainment). When resistivity is too high (> 10^10 Ω·cm, typically > 10^11 Ω·cm), the accumulated dust layer retains charges, creating a massive electric field that triggers breakdown ionization and reverse positive ion discharge (back-corona / 反电晕), severely reducing collection efficiency.

Question 45Integrated subject foundations, assessment and evidence

A wet limestone FGD unit removes SO_2 at a mass rate of M_SO2 = 1,280 kg/h (20.0 kmol/h). If the Calcium-to-Sulfur molar stoichiometry is set at Ca/S = 1.03, and the raw limestone reagent contains 90.0% CaCO_3 by weight (molecular weight CaCO_3 = 100.1 g/mol, SO_2 = 64.1 g/mol), what is the hourly limestone consumption rate in kg/h?

  1. 2,062 kg/h
  2. 2,291 kg/h Correct answer.
  3. 1,856 kg/h
  4. 2,546 kg/h

Explanation

Molar rate of SO_2 removed: n_SO2 = (1,280 kg/h) / (64.066 kg/kmol) = 19.979 kmol/h. Required pure CaCO_3 molar feed: n_CaCO3 = n_SO2 * (Ca/S) = 19.979 * 1.03 = 20.578 kmol/h. Mass of pure CaCO_3: M_pure = 20.578 kmol/h * 100.09 kg/kmol = 2,059.7 kg/h. Accounting for 90.0% limestone purity: Raw limestone consumption = 2,059.7 kg/h / 0.90 = 2,288.6 kg/h ≈ 2,291 kg/h.

Question 46Integrated subject foundations, assessment and evidence

Why is the absorber slurry pH in a wet limestone-gypsum FGD system strictly controlled within the narrow optimal operating window of 5.2 to 5.8?

  1. To completely prevent CO_2 dissolution while maximizing NO_x oxidation
  2. To prevent bacterial biofouling and maintain anaerobic digestion of sulfur
  3. To maintain acidic conditions below pH 4.0 for catalytic mercury reduction
  4. To balance fast SO_2 gas absorption kinetics with adequate limestone dissolution and prevent scaling/blinding Correct answer.

Explanation

In wet limestone FGD chemistry, SO_2 absorption is favored at higher pH (higher liquid alkalinity), but limestone (CaCO_3) dissolution and sulfite oxidation to sulfate (CaSO_3 -> CaSO_4) require lower pH (< 5.5). Operating at pH > 6.0 causes unreacted limestone blinding and calcium sulfite scaling (hemihydrate blinding). Operating at pH < 5.0 severely drops SO_2 absorption efficiency. The optimum compromises both processes at pH 5.2–5.8.

Question 47Integrated subject foundations, assessment and evidence

A Selective Catalytic Reduction (SCR, 选择性催化还原) DeNO_x system installed at a thermal power plant operates with an active V_2O_5-WO_3/TiO_2 honeycomb catalyst. What is the standard optimal operating flue gas temperature window for this catalyst?

  1. 300°C to 400°C Correct answer.
  2. 150°C to 220°C
  3. 500°C to 650°C
  4. 850°C to 1100°C

Explanation

Standard commercial high-dust V_2O_5-WO_3/TiO_2 catalysts operate optimally within 300°C to 400°C (typically 320–380°C). Below 300°C (especially below 280°C), unreacted ammonia reacts with SO_3 in the flue gas to form sticky ammonium bisulfate (NH_4HSO_4 / ABS), which deposits on and blinds catalyst pores. Above 420°C, catalyst sintering occurs and ammonia begins to undergo direct thermal oxidation into NO (4 NH_3 + 5 O_2 -> 4 NO + 6 H_2O), reducing DeNO_x efficiency.

Question 48Integrated subject foundations, assessment and evidence

A fixed-bed granular activated carbon (GAC) adsorber treats VOC-laden exhaust gas at Q = 10,000 m^3/h with inlet toluene concentration C_in = 400 mg/m^3. The carbon bed contains M_carbon = 2,000 kg of activated carbon with an effective dynamic adsorption working capacity of q_w = 12.0% (0.12 kg toluene/kg carbon). Assuming 100% capture efficiency until breakthrough, what is the continuous operating breakthrough time (t_b) of the carbon bed?

  1. 30 hours
  2. 120 hours
  3. 48 hours
  4. 60 hours Correct answer.

Explanation

Total adsorption capacity of the bed: Total Capacity = M_carbon * q_w = 2,000 kg * 0.12 kg/kg = 240 kg of toluene. Hourly toluene mass loading: Loading rate = Q * C_in = 10,000 m^3/h * 400 mg/m^3 * 10^-6 = 4.0 kg/h. Breakthrough time: t_b = Total Capacity / Loading rate = 240 kg / 4.0 kg/h = 60 hours.

Question 49Integrated subject foundations, assessment and evidence

In a 3-bed Regenerative Thermal Oxidizer (RTO, 蓄热式热氧化炉), what is the function of the third ceramic media bed while Bed 1 is in inlet preheating mode and Bed 2 is in outlet heat recovery mode?

  1. Purge mode (吹扫), using clean air to flush residual untreated VOCs from the canister into the combustion chamber before it switches to outlet mode Correct answer.
  2. Standby cooling mode, isolated completely from the gas flow circuit
  3. Catalytic regeneration mode, burning off soot using direct fuel injection
  4. Exhaust reheating mode, blending hot flue gas directly with ambient bypass air

Explanation

A 2-bed RTO suffers from VOC 'puffing' (emission spikes) during valve switching because raw VOCs trapped in the inlet bed plenum are vented directly to the stack when switched to outlet mode. A 3-bed RTO eliminates this: while Bed 1 preheats inlet gas and Bed 2 recovers heat from clean exhaust, Bed 3 undergoes a Purge cycle (吹扫), where clean air or flue gas flushes trapped raw VOCs back into the combustion chamber, achieving VOC destruction efficiencies > 99%.

Question 50Integrated subject foundations, assessment and evidence

What is the key technical distinction between a Regenerative Thermal Oxidizer (RTO) and a Regenerative Catalytic Oxidizer (RCO)?

  1. RCO operates at higher combustion temperatures (1100–1300°C) to destroy chlorinated dioxins
  2. RCO incorporates a precious metal or transition metal catalyst bed, lowering required operating temperature to 300–400°C compared to 760–850°C for RTO Correct answer.
  3. RTO uses ceramic packing while RCO uses liquid chemical absorption scrubbing packing
  4. RCO requires continuous hydrogen gas injection as a reducing agent for flameless combustion

Explanation

RTO achieves direct thermal combustion of VOCs at 760–850°C in a refractory chamber. RCO (Regenerative Catalytic Oxidizer) places a catalytic layer (e.g., Pt/Pd on honeycomb ceramic) above the regenerative media, lowering the activation energy so VOC catalytic oxidation occurs flamelessly at 300–400°C. This drastically cuts auxiliary fuel consumption for low-concentration VOC streams.