Question 1

Laboratory practice in wastewater analysis: Which apparatus is used for evaporation and measurement of settlable solids in sewage?

Accepted Answer

Introduction / Context:Settlable solids testing helps assess grit and heavier particulate fractions in wastewater primary treatment. The Imhoff cone is a standardized device for measuring the volume of settleable solids after a defined settling period.Given Data / Assumptions:Objective: identify the correct apparatus for settlable solids measurement.Standard settling interval (typically 60 minutes) applies to Imhoff cone tests.Concept / Approach:The Imhoff cone is a clear, conical vessel graduated at the tip to read settled solids volume directly (e.g., mL/L). It is specifically designed for wastewater settleability tests.Step-by-Step Solution:Recognize the specific test—sewage settlable solids.Recall the conical, graduated apparatus used is the Imhoff cone.Select “an Imhoff cone.”Verification / Alternative check:Wastewater lab manuals list the Imhoff cone method for primary settling performance checks.Why Other Options Are Wrong:Jar / test tube / generic beaker (breaker): Not standardized for direct volume reading of settled solids in mL/L.Graduated cylinder only: Lacks the conical tip and standardized graduation for this test.Common Pitfalls:Confusing jar tests for coagulation with settlable solids measurement.Final Answer:an Imhoff cone

Question 2

Hydraulics of sewers: In a flowing sewer, where is the scouring action (abrasive wear by grit and high velocities) typically the most pronounced?

Accepted Answer

Introduction / Context:Scouring refers to abrasive wear caused by grit and suspended solids moving with the flow. Understanding where scouring is most severe guides material choice and maintenance of sewer lines.Given Data / Assumptions:Gravity sewers with bed shear stress acting along the invert (bottom).Presence of grit (sand, small stones) that travels along the bed.Concept / Approach:Under gravity flow, the highest contact between moving particulates and pipe occurs along the invert. Bed shear, rolling and sliding of particles, and occasional saltation cause maximum abrasion at the bottom. The crown (top) experiences corrosion risks (e.g., H2S to sulfuric acid) but not scouring.Step-by-Step Solution:Identify the direction of gravity-driven flow and sediment motion.Recognize that abrasive contact is concentrated along the invert.Conclude the scouring effect is greatest at the bottom side.Verification / Alternative check:Field inspections often show invert wear and need for hard inverts or improved self-cleansing velocities in grit-prone networks.Why Other Options Are Wrong:Top side / horizontal side / all sides: Do not match the mechanics of bed load abrasion.Only at manholes: Local turbulence exists, but scouring dominance is along the invert in pipe reaches.Common Pitfalls:Confusing chemical crown corrosion with mechanical scouring.Final Answer:bottom side

Question 3

Historical sewer sections: In older practice for trunk and outfall sewers, which cross-sectional shape was commonly adopted to manage variable flows?

Accepted Answer

Introduction / Context:Historically, egg-shaped (oviform) sewers were used for large trunk and outfall lines to perform well across very low to high flow conditions. The geometry helps maintain higher velocities at low flow, aiding self-cleansing.Given Data / Assumptions:Older construction methods using masonry or brickwork.Combined sewers with large diurnal and seasonal flow variations.Concept / Approach:The narrow invert of an egg-shaped sewer concentrates low flows to increase depth and velocity, improving transport of solids. As flow increases, the upper, wider portion provides capacity. This makes it superior to simple circular sections under highly variable flows in historical contexts.Step-by-Step Solution:Consider low-flow performance and self-cleansing needs.Recognize egg-shaped geometry concentrates low flows but offers upper capacity for peaks.Select “egg shaped” as the historically favored form.Verification / Alternative check:Masonry oviform sewers in classic European systems are well-documented for favorable hydraulics under variable loads.Why Other Options Are Wrong:Circular is common today (pipe economics), but not the hallmark of older trunk/outfall works for variable flows.Parabolic/horse shoe/rectangular: Used in some cases but egg-shaped had distinctive low-flow advantages.Common Pitfalls:Assuming modern circular pipe dominance applied historically to all major outfalls.Final Answer:egg shaped

Question 4

House drainage practice: What is the recommended minimum gradient (fall) for a house drain to maintain self-cleansing velocities under typical domestic flows?

Accepted Answer

Introduction / Context:Adequate gradient is essential in house drains to prevent deposition and blockages. The “minimum gradient” refers to the flattest slope normally allowed while still aiming for self-cleansing velocities in small-diameter domestic pipes.Given Data / Assumptions:Typical house drain diameters (around 100–150 mm).Domestic sanitary flows with diurnal variability and low overnight flows.Concept / Approach:Empirical practice suggests gradients between 1 in 40 and 1 in 80 for small house drains. The flattest commonly accepted minimum in many codes/guides is around 1 in 80 (i.e., 12.5 mm fall per metre), balancing constructability and self-cleansing criteria.Step-by-Step Solution:Identify code-based recommended ranges for house drainage gradients.Select the flattest value that is still commonly recommended for self-cleansing.Hence, choose 1 in 80 as the minimum gradient.Verification / Alternative check:Many plumbing and public works handbooks tabulate 1 in 80 as a minimum for 100–150 mm house drains, with steeper slopes preferred when feasible.Why Other Options Are Wrong:1 in 60: Acceptable but not the “minimum”; it is steeper than 1 in 80.1 in 10: Unrealistically steep for house drainage.1 in 400: Too flat for self-cleansing in small pipes.Common Pitfalls:Confusing recommended with absolute minimum; local codes may vary but 1 in 80 is a widely cited minimum.Final Answer:1 in 80

Question 5

River water quality for aquatic life: The minimum dissolved oxygen (DO) concentration typically prescribed for the survival of fish in a river is approximately:

Accepted Answer

Introduction / Context:Dissolved oxygen (DO) is critical for aquatic organisms. Rivers receiving treated effluents must maintain adequate DO to sustain fish and benthic life. Regulatory targets and ecological studies establish minimum thresholds.Given Data / Assumptions:Question asks for the minimum DO for fish survival, not the desirable or optimum DO.Units ppm and mg/L are equivalent for dilute aqueous solutions.Concept / Approach:While 5–6 mg/L is often cited as desirable for healthy fisheries, a minimum around 4 mg/L is commonly taken as a survival threshold for many freshwater species. Below this level, stress and mortality risks rise rapidly, especially for sensitive species and at higher temperatures.Step-by-Step Solution:Identify survival threshold vs. optimal conditions.Select the minimum commonly prescribed level: about 4 mg/L (4 ppm).Note that higher DO is preferable for spawning and for cooler waters.Verification / Alternative check:Standard environmental guidelines and fisheries literature frequently recommend maintaining river DO ≥ 4 mg/L at all times, with targets ≥ 5 mg/L preferred.Why Other Options Are Wrong:3 ppm or 2 ppm: Too low; severe stress and fish kills can occur.5 ppm or 10 ppm: Desirable/healthy levels, but the question asks the minimum for survival.Common Pitfalls:Confusing minimum survival with recommended design or regulatory targets.Final Answer:4 ppm

Question 6

BOD calculation with dilution: A 2% diluted sewage sample is incubated for 5 days at 20°C and shows an oxygen depletion of 5 ppm. What is the BOD of the original sewage?

Accepted Answer

Introduction / Context:BOD (Biochemical Oxygen Demand) quantifies the oxygen required to biologically oxidize biodegradable organic matter. High-strength sewage is often diluted for the 5-day, 20°C test (BOD5) to keep DO within measurement range.Given Data / Assumptions:Dilution fraction f = 2% = 0.02 of original sewage.Observed DO depletion in the bottle after 5 days = 5 mg/L (ppm).Nitrification inhibition is assumed or negligible; we report BOD5 of the original sample.Concept / Approach:The BOD of the original undiluted sample equals the measured depletion divided by the dilution fraction.Step-by-Step Solution:Dilution fraction f = 0.02.Measured depletion (diluted) = 5 mg/L.BOD_original = measured depletion / f = 5 / 0.02 = 250 mg/L (ppm).Verification / Alternative check:Sanity check: A 2% sample causing 5 mg/L depletion implies full-strength BOD ≈ 50 * 5 = 250 mg/L, a plausible municipal value.Why Other Options Are Wrong:200 or 225 ppm: Underestimates; does not reflect the 50× dilution factor.None of these: Incorrect because 250 ppm is directly obtained from the calculation.Common Pitfalls:Using 2 (not 0.02) as the dilution factor—remember it is a fraction, not percent value.Confusing ppm with % or mg/L; here 1 ppm ≈ 1 mg/L.Final Answer:250 ppm

Question 7

On-site sanitation design: What is the recommended rate of accumulation of sludge in septic tanks (storage allowance) per person per year for design purposes?

Accepted Answer

Introduction / Context:Septic tank design requires allowance for sludge and scum storage between desludging operations. Using standard per capita accumulation rates helps size tanks for reliable operation without frequent overflows or carry-over.Given Data / Assumptions:Domestic sewage conditions with routine desludging intervals (often 1–3 years).Per capita sludge accumulation values used for preliminary sizing.Concept / Approach:Empirical design guidelines recommend about 25–35 L/person/year for sludge accumulation under typical conditions. A widely used round figure is 30 L/person/year for conservative sizing.Step-by-Step Solution:Select a per capita accumulation that ensures adequate storage.Adopt the commonly recommended value of 30 L/person/year.Confirm units (per year) match design period assumptions.Verification / Alternative check:Public health engineering texts often present 30 L/person/year as a standard figure, with adjustments for user habits and desludging frequency.Why Other Options Are Wrong:25 L/person/year: Slightly on the low side; may be used in specific contexts but 30 is more conservative and widely taught.Monthly values: Unrealistically high; misinterpretation of units.10 L/person/year: Underestimates storage needs, risking frequent desludging.Common Pitfalls:Confusing sludge accumulation with total tank volume (which also includes working volume and scum space).Final Answer:30 litres/person/year

Question 8

Building drainage practice in civil engineering: House service connections to the street laterals are generally made using which pipe material for durability, corrosion resistance, and smooth hydraulics?

Accepted Answer

Introduction / Context:House sewer connections convey domestic wastewater from individual buildings to street laterals. Selecting an appropriate material is critical for resistance to sewage corrosion, abrasion by grit, watertight joints, and long service life under shallow cover and variable ground conditions.Given Data / Assumptions:Small-diameter house connections discharging by gravity into municipal laterals.Need for smooth internal surface, good corrosion resistance, and easy jointing.Typical historical/standard practice for traditional systems.Concept / Approach:Glazed stoneware pipes provide a dense ceramic body with a smooth, glazed interior that resists chemical attack (including sulfide-related corrosion) and root intrusion when properly jointed. They have long been a preferred choice for house connections, particularly before widespread PVC adoption, due to their hydraulics and durability.Step-by-Step Solution:Identify functional needs: corrosion resistance, smooth hydraulics, leak-tight joints.Match properties: glazed stoneware satisfies these requirements for house drains to laterals.Conclude material choice: glazed stoneware is the standard answer in classical design manuals.Verification / Alternative check:Historic public health engineering texts list glazed stoneware for house connections; RCC is used for larger gravity sewers; CI and AC are typically pressure or special-use lines, not the norm for small gravity house connections.Why Other Options Are Wrong:Cast iron: Excellent for pressure/vertical stacks, but costly and not standard for small gravity house laterals.R.C.C./P.C.C.: Used for larger trunk/branch sewers or encasement, not for small house connections.Asbestos–cement: Used historically for pressure lines; not the typical choice for small gravity house connections.Common Pitfalls:Assuming materials for large trunk sewers apply unchanged to house connections.Overlooking chemical corrosion in domestic sewage environments.Final Answer:Glazed stonewares

Question 9

Self-cleansing gradient for small sewers: For a gravity sewer of diameter 225 mm, what longitudinal gradient is typically required to achieve self-cleansing velocity under domestic flows?

Accepted Answer

Introduction / Context:Self-cleansing velocity is the minimum flow velocity that prevents deposition of solids in sewers. It depends on pipe size, roughness, and typical solids load. For small community sewers (e.g., 225 mm), design guides tabulate recommended minimum slopes (gradients) to achieve these velocities during low flows.Given Data / Assumptions:Diameter D = 225 mm (0.225 m).Domestic wastewater with typical solids and grit content.Standard roughness and Manning/Chézy considerations implicit in tabulated practice.Concept / Approach:Empirical tables in public health engineering recommend slopes steep enough to secure ~0.6–0.75 m/s at part-full conditions for small sizes. For 225 mm, a commonly cited self-cleansing gradient is about 1 in 120 (≈0.0083), balancing excavation cost and hydraulic performance.Step-by-Step Solution:Identify the standard tabulated minimum gradient for D = 225 mm.Select 1 in 120 as the accepted value yielding self-cleansing velocities.Cross-check: 1/120 is steeper than 1/200 and flatter than 1/60, aligning with typical design guidance.Verification / Alternative check:Back-of-envelope Manning checks with n ≈ 0.013 at part-full conditions show velocities in the expected range at 1:120 for 225 mm sewers.Why Other Options Are Wrong:1 in 60: Unnecessarily steep; increases excavation and may cause high velocities and wear.1 in 100: Sometimes acceptable, but 1 in 120 is the standard recommended minimum in many tables.1 in 200 or none: Too flat for reliable self-cleansing in small sewers.Common Pitfalls:Ignoring that recommended slopes are for minimum-velocity conditions, not peak flows.Final Answer:1 in 120

Question 10

Urban drainage design: drop manholes are typically provided in sewers in which type of locality?

Accepted Answer

Introduction / Context:Where ground slopes are steep, conventional sewer gradients may become too steep, causing high velocities, scouring, and difficult depth transitions. Drop manholes are a practical detail to accommodate large elevation differences while controlling internal flow conditions.Given Data / Assumptions:Gravity sewer system following natural topography.Need to connect higher incoming sewers to deeper downstream sewers.Objective: prevent excessive velocities and erosion inside manholes and pipes.Concept / Approach:A drop manhole allows the incoming sewer to fall vertically (or via a drop pipe) before joining the lower invert, dissipating energy and keeping the downstream grade manageable. Such configurations are common in hilly townships where street grades fall rapidly between successive manholes.Step-by-Step Solution:Identify steep terrain with large invert differences between adjacent manholes.Provide an inside or outside drop connection to bring the higher inflow down to the lower channel.Detail energy dissipation and splash protection to avoid turbulence-related damage.Maintain safe access and ventilation provisions in the manhole design.Verification / Alternative check:Hydraulic grade line analysis shows controlled velocities and acceptable head losses compared with a continuous steep gradient that might exceed permissible velocity limits.Why Other Options Are Wrong:(a), (b), and (d) do not specifically require drop arrangements; flat areas can use mild gradients without drops.Common Pitfalls:Improper drop alignment leading to splashing and odor; inadequate corrosion-resistant materials; ignoring access and safety clearances.Final Answer:Hilly townships (steep ground levels)

Question 11

In fully oxidized sewage (end of nitrification), the nitrogen is predominantly present in which form?

Accepted Answer

Introduction / Context:Understanding nitrogen transformations in wastewater is essential for process control and environmental compliance. As wastewater undergoes aerobic biological treatment, reduced nitrogen forms are progressively oxidized, culminating in nitrate under fully aerobic, fully oxidized conditions.Given Data / Assumptions:Aerobic conditions sufficient for complete nitrification.Microbial community includes nitrifying bacteria (e.g., Nitrosomonas, Nitrobacter).Organic matter has been largely oxidized.Concept / Approach:The nitrification pathway converts ammonia to nitrite and then to nitrate. In a fully oxidized (stable) effluent, nitrite is transient and typically low; nitrate is the dominant oxidized inorganic nitrogen species. Ammonia would be low due to oxidation, and “albuminoid nitrogen” refers to organic nitrogen that has largely been mineralized during treatment.Step-by-Step Solution:Ammonia → Nitrite via ammonia-oxidizing bacteria.Nitrite → Nitrate via nitrite-oxidizing bacteria.Fully oxidized stage: nitrate accumulates as the stable end product.Therefore the dominant nitrogen form is nitrate (NO3−).Verification / Alternative check:Typical secondary effluents post-nitrification show NO3− in the tens of mg/L as N, with negligible NO2− and low NH3-N unless anoxic conditions or short-circuiting occur.Why Other Options Are Wrong:(a) Nitrite is intermediate and unstable under strong aeration; (c) ammonia indicates incomplete nitrification; (d) albuminoid nitrogen is largely removed by oxidation/biodegradation.Common Pitfalls:Confusing “fully oxidized” with denitrification (which reduces nitrate to nitrogen gas under anoxic conditions); misreading grab samples where short-term nitrite spikes can occur.Final Answer:Nitrates (NO3−)

Question 12

For very large trunk and out-fall sewers carrying high discharges, which cross-sectional shape is generally preferred in practice?

Accepted Answer

Introduction / Context:Sewer cross-sections are selected based on hydraulic performance, structural feasibility, construction practicality, and maintenance. While egg-shaped sewers are excellent for combined systems at small to medium sizes, very large trunk and out-fall conduits often use alternative shapes.Given Data / Assumptions:Out-fall/trunk mains with very large discharge.Gravity flow with wide variation in flows.Constructability and structural stability are key considerations.Concept / Approach:The horse-shoe section offers a broad invert for structural stability and ease of construction at very large sizes, while still providing favorable hydraulic characteristics. Circular sections are common overall, but at very large spans, the horse-shoe provides better roof and sidewall behavior under loads. Egg-shaped is ideal for small to medium combined flows (wet-weather performance at low dry-weather flows) but is not the standard for the largest out-fall sizes.Step-by-Step Solution:Match section choice to discharge scale and structural considerations.For very large spans, select a shape accommodating construction formwork and lining.Horse-shoe gives adequate hydraulic radius and stable roof geometry.Thus, choose the horse-shoe section for large trunk/out-fall sewers.Verification / Alternative check:Historic large out-falls and interceptors (masonry or concrete) frequently adopt horse-shoe geometry due to formwork practicality and load paths.Why Other Options Are Wrong:(a) Best for combined small/medium conduits; (b) feasible but less common at very large spans compared to horse-shoe; (d) and (e) are uncommon choices for main out-falls.Common Pitfalls:Assuming one shape is optimal at all scales; ignoring maintenance access and ventilation needs in very large conduits.Final Answer:Horse-shoe section

Question 13

Wastewater quality: the oxygen required by aerobic bacteria to biologically oxidize biodegradable organics in sewage is called

Accepted Answer

Introduction / Context:BOD is a foundational parameter in wastewater engineering, indicating the amount of oxygen microorganisms will consume while stabilizing biodegradable organic matter. It is central to plant sizing, discharge permitting, and river oxygen-balance studies.Given Data / Assumptions:BOD test typically conducted over 5 days at 20°C (BOD5) unless otherwise noted.Measures biologically oxidizable organic load (carbonaceous stage primarily).Aerobic bacterial activity is assumed.Concept / Approach:BOD quantifies the oxygen demand exerted by biodegradable organic matter. It differs from COD, which measures oxygen equivalent of total oxidizable matter via chemical oxidant, and from DO, which is the concentration of oxygen present in the water at a given time. High BOD indicates strong sewage requiring more treatment capacity.Step-by-Step Solution:Collect sample and seed as needed.Measure initial DO and incubate at prescribed temperature/time.Measure final DO; BOD = DO_initial − DO_final (adjusted for dilution and seed).Interpret BOD as a proxy for biodegradable organic strength.Verification / Alternative check:Compare BOD with COD and TOC; BOD/COD ratios indicate biodegradability. Secondary effluents typically show significant BOD reduction relative to raw sewage.Why Other Options Are Wrong:(b) DO is oxygen present, not consumed; (c) COD is a chemical oxidation measure that includes non-biodegradable fractions; (d) is incorrect since BOD is the standard definition.Common Pitfalls:Assuming BOD equals COD; neglecting nitrification stage unless inhibited; mishandling incubation conditions.Final Answer:Bio-Chemical Oxygen Demand (B.O.D.)

Question 14

In a combined sewer system, “dry weather flow” (DWF) is understood to mean which of the following?

Accepted Answer

Introduction / Context:Combined systems convey sanitary (domestic + industrial) wastewater and storm runoff in the same conduit. However, design and operation often differentiate dry-weather conditions from wet-weather events for capacity, storage, and treatment decisions.Given Data / Assumptions:Combined sewer with separate hydraulic regimes for dry and wet weather.No storm precipitation during “dry weather.”Continuous domestic and industrial discharges exist regardless of rainfall.Concept / Approach:Dry Weather Flow (DWF) includes sanitary flows—domestic and industrial—and excludes storm water. During rainfall, additional inflow/infiltration and surface runoff create Wet Weather Flow (WWF), which can be many times higher than DWF.Step-by-Step Solution:Define DWF = Q_domestic + Q_industrial + base infiltration (if considered).Exclude Q_storm (surface runoff) for dry conditions.Use DWF for sizing minimum velocities and self-cleansing criteria.Use WWF/peaking factors for storage, CSO controls, and treatment bypass assessments.Verification / Alternative check:Flow monitoring during non-rain periods reflects DWF; hydrographs spike during storms, distinguishing WWF from DWF.Why Other Options Are Wrong:(a) and (b) are partial; (c) is specifically excluded by definition of DWF.Common Pitfalls:Ignoring groundwater infiltration as part of base sanitary flow estimate; confusing combined and separate systems terminology.Final Answer:Domestic and industrial sewage together, excluding storm water

Question 15

Recognizing sewage quality: which statements about appearance and odour correctly characterize strong, fresh, and septic sewage?

Accepted Answer

Introduction / Context:Visual and olfactory indicators provide quick field assessments of sewage condition when laboratory analysis is unavailable. These cues help operators identify onset of septic conditions, verify aeration performance, and anticipate odor complaints.Given Data / Assumptions:Typical municipal sewage without unusual industrial discharges.Warm-weather conditions can accelerate septic onset.Normal conveyance times from sources to treatment.Concept / Approach:“Strong” sewage has higher suspended and dissolved organic load, giving greater turbidity. Fresh sewage (recently generated) is generally grayish and nearly odourless. Without aeration, biological oxygen demand rapidly depletes dissolved oxygen; anaerobic processes then produce black color (sulfides) and offensive odors (e.g., hydrogen sulfide), often within a few hours depending on temperature and retention time.Step-by-Step Solution:Associate turbidity with strength: higher solids → higher turbidity.Note fresh state: minimal septic odours shortly after generation.Progression to septic: DO falls, sulfate-reducing bacteria form sulfides → black colour and odour.Timeframe: offensive smells can appear within roughly four hours under warm, stagnant conditions.Verification / Alternative check:Routine plant observations: increasing retention with poor ventilation correlates with blackening and odor complaints; laboratory measures (DO, ORP) corroborate field signs.Why Other Options Are Wrong:Each individual statement is correct; only “All the above” captures the full characterization.Common Pitfalls:Assuming clear sewage is low-strength; some dissolved organics do not add turbidity. Misattributing black colour to industrial dyes rather than sulfide formation without testing.Final Answer:All the above.

Question 16

In sanitary sewer systems, which gases are most commonly encountered during inspection and operation? Select the best comprehensive choice that reflects typical sewer atmospheres (toxicity and explosibility considerations apply).

Accepted Answer

Introduction / Context:Safe sewer maintenance requires a clear understanding of typical sewer gases. Common constituents like hydrogen sulphide, methane, and carbon dioxide affect worker safety (toxicity, asphyxiation, and explosion risks) and corrosion of infrastructure. This question tests practical environmental engineering knowledge used in confined-space entry procedures.Given Data / Assumptions:Sewer conveys domestic wastewater under gravity with limited ventilation.Biochemical activity is ongoing (anaerobic zones are common).No unusual industrial discharge is assumed.Concept / Approach:Sewer atmospheres typically contain: hydrogen sulphide (H2S) from anaerobic sulphate reduction; methane (CH4) from anaerobic digestion; and carbon dioxide (CO2) from biodegradation. H2S is highly toxic and corrosive, CH4 is flammable/explosive, and CO2 displaces oxygen causing asphyxiation. Confined-space protocols therefore require gas monitoring for O2, H2S, CO, and combustible gases before entry.Step-by-Step Solution:Recognize typical anaerobic by-products: H2S and CH4.Account for metabolic CO2 produced during biological oxidation.Identify the option that groups these typical gases together.Conclude the comprehensive correct option is the trio H2S, CH4, and CO2.Verification / Alternative check:Standard sewer safety training and manuals list these gases as primary hazards; many municipalities require meters to alarm at H2S thresholds and lower explosive limits (LEL) for CH4, with oxygen deficiency checks due to CO2.Why Other Options Are Wrong:(b), (c), and (d) each list a single gas and ignore the others that commonly co-exist. (e) Petrol vapours are not typically present in sanitary sewers and are not considered a defining sewer gas.Common Pitfalls:Assuming oxygen levels are always safe; overlooking the simultaneous risks of toxicity and explosion; neglecting corrosion effects of H2S on concrete and metals.Final Answer:Hydrogen sulphide (H2S), methane (CH4), and carbon dioxide (CO2)

Question 17

In open drains carrying stormwater, a maximum permissible mean velocity of about 1.5 m/s is generally adopted for which types of beds/linings to avoid erosion and structural damage?

Accepted Answer

Introduction / Context:Designing safe non-eroding velocities in open drains is central to stormwater engineering. Excess velocity can scour bed and sides; too low a velocity promotes deposition. The permissible velocity depends on material and lining type.Given Data / Assumptions:Drains intended for stormwater or combined flows.Materials considered: rock and gravel beds, stone lining.Permissible velocity target ≈ 1.5 m/s.Concept / Approach:Empirical permissible velocities correlate to material resistance: rock/gravel or stone lining tolerates higher velocities than bare earth. Around 1.0–1.5 m/s is often acceptable for stone-lined or rock/gravel sections, while unlined silty or sandy beds require much lower velocities to prevent scour.Step-by-Step Solution:Identify materials capable of withstanding 1.5 m/s without significant erosion.Rock/gravel beds and stone lining both provide high shear resistance.Therefore both (a) and (b) are suitable for the stated velocity.Verification / Alternative check:Design handbooks provide tables of safe mean velocities: stone-pitched channels allow higher velocities than bare cohesive soils. Many municipal standards accept ≈1.5 m/s for stone-lined sections depending on gradation and jointing.Why Other Options Are Wrong:(a) or (b) alone are incomplete. (d) is false because 1.5 m/s is not generally permissible for unlined erodible soils but is permissible for stone/rock lining. (e) Bare cohesive clay often needs lower velocities to avoid erosion.Common Pitfalls:Using peak (instantaneous) rather than mean velocity; ignoring side-slope protection and turbulence at bends, junctions, or culvert outlets.Final Answer:Both (a) and (b)

Question 18

In the classic stream self-purification profile below a wastewater outfall, in which zone do algae typically die out due to lack of light and dissolved oxygen?

Accepted Answer

Introduction / Context:After sewage is discharged to a river, the water quality changes longitudinally through characteristic zones: degradation, active decomposition, recovery, and clean water. Understanding biota responses (like algal presence) helps interpret DO sag curves and biological indices.Given Data / Assumptions:Typical warm-season conditions with significant organic load.Limited reaeration immediately downstream of outfall.Standard zone nomenclature from stream sanitation theory.Concept / Approach:In the active decomposition zone, conditions are often anaerobic or severely oxygen-depleted. Black color, odours (H2S), and high reducing conditions prevail. Photosynthetic organisms like algae need light and dissolved oxygen; they typically die out or are greatly suppressed in this zone. They reappear downstream as reaeration restores DO in the recovery zone.Step-by-Step Solution:Identify zones and their DO characteristics.Match biological requirements of algae to DO and light availability.Conclude that algae die out in the active decomposition zone.Verification / Alternative check:DO sag models (Streeter–Phelps) show minimum DO within or near active decomposition; field observations align with low algal biomass there, increasing toward recovery.Why Other Options Are Wrong:(a) Degradation zone still has some DO and turbulence; algae need not “die out.” (c) Recovery zone sees improving DO and algal resurgence. (d) Cleaner water zone supports normal aquatic life.Common Pitfalls:Confusing “degradation” with “decomposition,” and assuming the lowest DO is immediately at the outfall rather than further downstream where BOD exertion peaks.Final Answer:Zone of active decomposition

Question 19

Municipal solid waste composting methods: which statements are correct about Indore and Bangalore processes and the product formed?

Accepted Answer

Introduction / Context:Composting stabilizes organic fractions of municipal solid waste, reducing volume, odour, and pathogens while producing soil-like amendments. Two historically important Indian methods are Indore and Bangalore, each with distinct aeration and handling practices.Given Data / Assumptions:Aerobic conditions imply frequent turning and oxygen supply.Anaerobic trench composting limits air ingress.End product is expected to be stable, humus-like material.Concept / Approach:The Indore method uses aerobic decomposition with turning and layering of refuse and night soil, encouraging thermophilic bacteria. The Bangalore method uses sealed trenches promoting anaerobic digestion; after sufficient time, the stabilized residue resembles humus, suitable for soil conditioning.Step-by-Step Solution:Identify Indore → aerobic with turning and air supply.Identify Bangalore → anaerobic trench, slower but simpler.Recognize stabilized product → humus-like powdery mass.Verification / Alternative check:Classic sanitation texts describe temperature profiles and pathogen reduction under both methods, with different retention times but similar stabilized outcomes.Why Other Options Are Wrong:Since (a), (b), and (c) are correct, “All of these” is the most complete answer; “None of these” is false.Common Pitfalls:Confusing composting (stabilization) with mere drying; assuming both methods are aerobic; overlooking retention time differences.Final Answer:All of these

Question 20

A circular sewer of diameter 2.0 m runs partially full at depth y = 0.50 m. What is the wetted perimeter (in metres) of the flowing segment? (Use geometric relations for a circular segment.)

Accepted Answer

Introduction / Context:Hydraulic elements for partially full circular sewers (area, wetted perimeter, hydraulic radius) are needed to compute discharge and velocity using Manning or Chezy equations. This problem asks for the wetted perimeter from basic geometry.Given Data / Assumptions:Diameter D = 2.0 m → radius R = 1.0 m.Flow depth y = 0.50 m.Steady cross-sectional geometry; no deformation.Concept / Approach:For a circular segment of depth y in a circle of radius R, the half-central angle θ satisfies y = R(1 − cos θ). The wetted perimeter Pw equals arc length subtended by 2θ: Pw = R * 2θ (with θ in radians).Step-by-Step Solution:Compute cos θ = 1 − y/R = 1 − 0.50/1.0 = 0.5.Thus θ = arccos(0.5) = π/3 radians (60°).Wetted perimeter Pw = R * 2θ = 1.0 * 2 * (π/3) = 2π/3 m.Verification / Alternative check:At y = R (i.e., 1.0 m), Pw would be half the circumference = πR = π m; since 0.5 m is shallower than R, Pw should be less than π m; 2π/3 m satisfies this expectation.Why Other Options Are Wrong:(a) π/3 m is the half-angle, not the arc length. (c) π m corresponds to half-full, not y = 0.5 m. (d) 4π/3 m exceeds half-circumference for this shallow depth. (e) 2 m is not an arc length derived from geometry.Common Pitfalls:Mixing the central angle with arc length; using degrees instead of radians in arc computations; confusing depth y with rise from invert to water surface across the diameter.Final Answer:2π/3 m

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