Question 1

Wastewater coagulation: Which coagulant is most widely used for sewage (wastewater) treatment to aid flocculation and sedimentation?

Accepted Answer

Introduction / Context: Coagulants are added in primary and advanced wastewater treatment to destabilize colloids, enhance floc formation, and improve sedimentation. While alum is common in potable water treatment, iron salts are often preferred in sewage treatment due to their performance over wider pH ranges and added benefits in phosphorus removal and odour control. Given Data / Assumptions: Municipal sewage with variable alkalinity and organic load. Objective: robust coagulation over a broad pH (≈ 5–9) and effective floc formation. Concept / Approach: Ferric chloride hydrolyzes to form ferric hydroxide precipitates that sweep-floc fine particles and adsorb phosphorus. Compared to alum, ferric chloride is less sensitive to low temperatures and low alkalinity; it also aids in odour control by precipitating sulphides. Ferric sulphate behaves similarly but FeCl3 is more ubiquitous in practice. Step-by-Step Solution: Assess wastewater characteristics (alkalinity, temperature, SS/colloids).Select a coagulant effective across these variations; FeCl3 meets this criterion.Dose optimization is done via jar tests; sludge handling capacity is considered. Verification / Alternative check: Operational records in many plants show widespread use of ferric chloride for chemical phosphorus removal and enhanced primary treatment; it is stocked commonly and has predictable performance. Why Other Options Are Wrong: Alum works well in water treatment but can be less effective in high-alkalinity variability; it also forms more voluminous sludge. Ferric sulphate is viable but is less common than FeCl3 in many sewage applications. Chlorinated copperas is specialized and less widely used today. PAC is common in potable water; in sewage, iron salts dominate for P-removal and odour control. Common Pitfalls: Assuming the same coagulant is optimal for both drinking water and sewage; wastewater has different chemistry. Final Answer: Ferric chloride (FeCl3)

Question 2

Sewer gas hazards: Which gas is primarily responsible for explosion risk in sewers and confined wastewater spaces?

Accepted Answer

Introduction / Context: Sewer atmospheres can contain a mix of gases from anaerobic decomposition. Safety in confined-space entry demands awareness of toxic, asphyxiating, and explosive gas hazards, along with ventilation and gas monitoring. Given Data / Assumptions: Anaerobic digestion generates methane and carbon dioxide predominantly. Presence of hydrogen sulphide (toxic) and traces of other gases possible. Lower and upper explosive limits (LEL/UEL) must be considered for flammables. Concept / Approach: Methane (CH4) is flammable with LEL ≈ 5% and UEL ≈ 15% by volume in air. Accumulations in poorly ventilated sewers can reach explosive concentrations and are ignitable by sparks. CO2 is asphyxiating but non-flammable; NH3 and H2S are toxic/irritant but explosion risk is dominated by methane. Step-by-Step Solution: Identify gas generation: anaerobic processes → CH4 and CO2.Check properties: CH4 is highly flammable; others primarily toxic or asphyxiants.Conclude explosion hazard is chiefly due to methane. Verification / Alternative check: Confined-space safety protocols require continuous combustible gas indicators; readings are calibrated to methane equivalents, underscoring its centrality to explosion risk. Why Other Options Are Wrong: CO2 is non-flammable; it displaces oxygen. NH3 and H2S are hazardous but not primary explosion drivers at typical sewer concentrations. CO is toxic and flammable at high concentrations but is not typically the main sewer-generated gas. Common Pitfalls: Confusing toxicity with flammability; both matter, but explosion risk points to methane. Final Answer: Methane (CH4)

Question 3

Environmental engineering (assertion–reason): Effluent from oxidation ponds should not be discharged just upstream of lakes or reservoirs. Why?

Accepted Answer

Introduction / Context: Oxidation ponds (waste stabilization ponds) are economical for polishing wastewater using sunlight, algae, and bacteria. However, their effluent often contains suspended algae and residual nutrients. Discharging such effluent directly upstream of lakes/reservoirs can trigger water-quality problems. Given Data / Assumptions: Assertion (A): Discharging pond effluent just upstream of impoundments is undesirable. Reason (R): The discharged algae accumulate, settle or die, and undergo anaerobic decomposition, degrading water quality. Receiving water has low velocities and longer detention (reservoir setting). Concept / Approach: In quiescent reservoirs, algal cells from pond effluent may settle into bottom layers where dissolved oxygen is low. Anaerobic decay produces odours and releases nutrients (internal loading), worsening eutrophication. Additionally, sunlight and nutrients can fuel new blooms near the intake zones. Step-by-Step Solution: Recognize that pond effluents carry algae and nutrients.In reservoirs, low mixing promotes stratification and hypolimnetic anoxia.Settling/decomposition of algae leads to oxygen demand and taste/odour issues, validating the reason. Verification / Alternative check: Best practices route pond effluents to well-mixed river reaches or provide tertiary filtration/desludging before discharge to sensitive lakes. Why Other Options Are Wrong: Since R correctly explains A, options B–D do not match the causal relationship. Common Pitfalls: Ignoring algae-rich suspended solids in oxidation pond effluents when assessing receiving water impacts. Final Answer: Both A and R are true and R is the correct explanation of A

Question 4

Municipal solid waste (MSW) processing: A fully mechanized composting plant typically involves which stages?

Accepted Answer

Introduction / Context: Composting converts the biodegradable fraction of MSW into a stabilized humus-like product. Modern plants automate key steps to improve throughput, worker safety, and product uniformity while reducing odour and vector nuisances. Given Data / Assumptions: Urban MSW stream with mixed organics, recyclables, and inerts. Objective: continuous, controlled composting with quality assurance. Concept / Approach: Automation begins at reception with hoppers and conveyors, followed by pre-processing to remove large inerts and recover recyclables. Shredding/pulverizing increases surface area and homogenizes feed, enhancing microbial action. Windrow turners or in-vessel systems maintain aeration and moisture; downstream screening/refining completes the process. Step-by-Step Solution: Mechanized receipt: weighbridge, tipper hoppers, conveyors.Mechanized segregation: trommels, magnetic and eddy-current separators, air classifiers.Mechanized size reduction: shredders/pulverizers feed uniformity.Controlled composting: aeration/turning, moisture control, curing.Refining: screens remove overs; product testing and dispatch. Verification / Alternative check: Plant flow diagrams from integrated MSW facilities show these mechanized stages as standard practice to ensure operational efficiency and product consistency. Why Other Options Are Wrong: Choosing any single stage ignores the integrated nature of mechanized composting lines; hence “All of these” is correct. Common Pitfalls: Skipping segregation, which leads to contamination of compost with glass/plastics. Final Answer: All of these

Question 5

Primary sedimentation settling of sand-size inorganics: In primary settling tanks, 0.2 mm inorganic particles (typical sand) will separate out if their specific gravity (relative density) is approximately what value?

Accepted Answer

Introduction / Context: Primary sedimentation targets removal of settleable solids from sewage before biological treatment. Grit chambers and primary clarifiers remove heavier inorganic particles such as sand. The settling behavior of these particles depends on size, shape, and specific gravity. Given Data / Assumptions: Particle size ≈ 0.2 mm (sand-size range). Quiescent conditions representative of primary clarifiers/grit chambers. Particle is largely quartzose (common in municipal grit). Concept / Approach: Quartz sand typically has a specific gravity near 2.65. At 0.2 mm, with adequate detention and low turbulence, such particles settle effectively. Design of grit chambers often assumes quartz specific gravity to size channels and detention times. Step-by-Step Solution: Identify typical mineral composition: quartz dominant → SG ≈ 2.65.At 0.2 mm, settling velocity is sufficient for capture in primary units.Thus, the correct SG threshold indicative of effective separation is ~2.65. Verification / Alternative check: Engineering manuals list grit specific gravity in the 2.60–2.70 range, with 2.65 as the standard design value. Why Other Options Are Wrong: Values like 2.25–2.55 underestimate typical quartz sand and suggest lighter organics/minerals. 2.75 is somewhat higher than common municipal grit assumptions. Common Pitfalls: Confusing specific gravity of organics (≈ 1.0–1.3) with inorganic grit (≈ 2.65). Final Answer: 2.65

Question 6

Sewer material selection at high velocity: For a non-scouring design velocity of about 5 m/s, which sewer material is generally preferred for durability and resistance to abrasion?

Accepted Answer

Introduction / Context: Sewer conduits experience abrasion from grit and corrosive action from wastewater gases. At high velocities (around 5 m/s), material choice must emphasize toughness, abrasion resistance, and structural strength to avoid scour and long-term damage. Given Data / Assumptions: Design velocity ~5 m/s (relatively high for sanitary sewers). Presence of grit and intermittent turbulent flow. Need for robust joints and resistance to impact/abrasion. Concept / Approach: Cast iron sewers (or ductile iron in modern practice) offer high strength, good abrasion resistance, and durable joints. Concrete, brick, and stoneware are more susceptible to abrasion and chemical attack; glazed surfaces help but may craze or chip under high-velocity grit loads. Step-by-Step Solution: Assess velocity: 5 m/s entails significant abrasive potential.Compare material properties: cast iron > concrete/brick/stoneware in abrasion resistance.Select cast iron as the preferred material for such conditions. Verification / Alternative check: Design manuals recommend metallic/ductile iron or lined steel for high-velocity segments, confirming the preference for cast iron type materials. Why Other Options Are Wrong: Cement concrete can erode under grit-laden high-velocity flows unless specially lined. Glazed brick and stoneware are brittle; glazing may wear off or crack. Common Pitfalls: Assuming smoothness alone (e.g., glazing) is sufficient; structural toughness matters at 5 m/s. Final Answer: Cast iron sewers

Question 7

Egg-shaped sewer geometry: If D denotes the diameter of the circular upper portion of a standard egg-shaped sewer, what is the overall vertical depth of the section?

Accepted Answer

Introduction / Context: Egg-shaped (ovoform) sewers are favored for combined systems because they provide better self-cleansing velocities at low flows and adequate capacity at high flows. Their proportions are standardized so that hydraulic performance is predictable and formwork or precast moulds can be repeated. Given Data / Assumptions: D is defined as the diameter of the upper circular arc of the egg-shaped section. Standard proportions are assumed (not custom/modified ovoforms). Concept / Approach: In the standard form, the overall depth (crown to invert) is 1.5 times D. This ensures a narrower lower portion for velocity maintenance during dry-weather flow while the broader upper portion accommodates storm peaks. Step-by-Step Solution: Recall the standard geometry: overall depth H ≈ 1.5 D.The maximum width is less than the height, aiding hydraulic efficiency.Hence, for given D, the total vertical depth is 1.5 D. Verification / Alternative check: Handbooks on sewer design provide proportional templates: height-to-upper-diameter ratio H/D ≈ 1.5 for the common standard egg shape. Why Other Options Are Wrong: D and 1.25 D underestimate the height; 1.75 D and 2 D exceed standard proportions and would alter hydraulics. Common Pitfalls: Confusing D (upper circular diameter) with overall height H; they are not equal. Final Answer: 1.5 D

Question 8

Municipal drainage design: If the ratio of drainage flow to sanitary sewage flow is 20, the peak dry weather flow (PDWF) will be approximately what fraction of the overall design discharge for the combined system?

Accepted Answer

Introduction / Context: When designing combined sewer systems, engineers compare the magnitude of storm (drainage) flows to sanitary sewage flows. The ratio of drainage to sewage often reaches large values during storms, meaning the sanitary (dry-weather) component is a small fraction of the overall design discharge. This question checks whether you can translate a drainage-to-sewage ratio into the approximate fraction represented by the peak dry weather flow (PDWF) within the total combined design discharge. Given Data / Assumptions: Drainage-to-sewage ratio = 20 (i.e., storm drainage ≈ 20 times the average sanitary sewage). Design discharge is for the combined system during storm conditions. PDWF is slightly larger than average sewage flow due to peaking factors from diurnal variations. Concept / Approach: If storm flow is 20 times the sanitary flow, then the sanitary component alone is 1/21 of the total. However, the system must pass the peak dry weather flow, which incorporates a peaking factor (typically 1.2–1.8 or more depending on context). Thus, PDWF is just over 1/20 of the overall design flow. Step-by-Step Solution: Let Qs = average sanitary sewage flow.Let Qd = drainage flow during design storm ≈ 20 * Qs.Total design discharge Qtotal ≈ Qd + Qs = 20Qs + Qs = 21Qs.PDWF ≈ kp * Qs, where kp > 1 (peaking factor).Fraction of PDWF in Qtotal ≈ (kp * Qs) / (21 * Qs) = kp / 21. For kp ≈ 1.2–1.5, the fraction is ≈ 5.7%–7.1%. Verification / Alternative check: Even if the peaking factor is modest (say 1.1), the fraction remains slightly above 5%. Therefore, PDWF is not less than 5% but rather slightly more than 5% of the combined design discharge under the stated ratio. Why Other Options Are Wrong: 20% of the design discharge: Far too high; storm flow dominates. Slightly less than 5%: With kp > 1, this underestimates PDWF. None of these: Incorrect because the “slightly more than 5%” estimate is appropriate. About 10%: Overshoots typical PDWF fractions for a 20:1 ratio. Common Pitfalls: Confusing average sanitary flow with PDWF; peaking matters. Ignoring that the “20” ratio refers to storm dominance in combined systems. Final Answer: Slightly more than 5% of the design discharge

Question 9

Sewer system classification: A rain–sanitary (combined) sewer is constructed to carry which types of flows?

Accepted Answer

Introduction / Context: Cities may use separate sewer systems (sanitary and storm sewers) or combined sewers that carry both sanitary sewage and storm runoff in one pipe. The term “rain–sanitary sewer” is a common way to describe a combined sewer designed to handle multiple sources of flow during wet weather. Given Data / Assumptions: The phrase “rain sanitary sewer” refers to a combined sewer. Combined sewers convey sanitary sewage, storm runoff, surface drainage, and incidental groundwater inflow. Concept / Approach: Combined systems are sized for extreme wet-weather events, so they must carry dry-weather sanitary flow plus wet-weather inflows. These include rainfall-derived inflow (from roof leaders, yard drains), overland surface runoff entering inlets, and some groundwater infiltration/inflow (I/I) through joints and defects. Step-by-Step Solution: Identify the system as combined (rain + sanitary).List components: sanitary sewage + storm sewage + surface runoff + groundwater inflow.Therefore, the correct selection includes all listed flows. Verification / Alternative check: Design manuals explicitly define combined sewers as conveying both sanitary and storm flows, including incidental I/I, validating selection of “All of the above.” Why Other Options Are Wrong: Selecting only one category (sanitary, storm, or groundwater) omits critical design flows and misrepresents combined sewer function. Common Pitfalls: Confusing combined sewers with separate sanitary or storm sewers. Overlooking groundwater infiltration as a contributor to wet-weather flow. Final Answer: All of the above

Question 10

Wastewater testing—COD: In the chemical oxygen demand (COD) test for sewage, the organic matter is oxidized by potassium dichromate (K2Cr2O7) in the presence of which reagent?

Accepted Answer

Introduction / Context: The chemical oxygen demand (COD) test quantifies the amount of oxygen required to chemically oxidize organic compounds in wastewater. It is widely used for treatment plant monitoring because it is faster and more reproducible than biochemical oxygen demand (BOD) measurements. Given Data / Assumptions: Oxidant: potassium dichromate (K2Cr2O7). Reflux under acidic conditions is required. Common catalysts and masking agents may be used (e.g., silver sulfate, mercuric sulfate) depending on chloride content. Concept / Approach: COD digestion is performed in strongly acidic medium so that dichromate acts as a powerful oxidant. Concentrated sulfuric acid provides the acidic environment, and silver sulfate acts as a catalyst for oxidizing certain compounds (e.g., straight-chain hydrocarbons). Mercuric sulfate is added to complex chlorides to prevent their interference. Step-by-Step Solution: Prepare sample with K2Cr2O7, add concentrated H2SO4.Add catalysts/masking agents as needed (Ag2SO4, HgSO4) based on sample matrix.Reflux (or heat in sealed digestion vials) for the prescribed time; titrate or photometrically determine dichromate consumed. Verification / Alternative check: Standard methods (APHA/AWWA/WEF) specify sulfuric acid as the required acidifying medium for the dichromate COD procedure, confirming the correct choice. Why Other Options Are Wrong: HNO3 and HCl are not used in the standard dichromate COD method. None of these is incorrect because H2SO4 is essential. NaOH would neutralize acid and halt oxidation. Common Pitfalls: Confusing COD (chemical oxidation) with BOD (biological oxidation). Ignoring chloride interference mitigation steps. Final Answer: H2SO4 (sulfuric acid)

Question 11

When is sewage pumping required? Identify typical situations in which wastewater must be lifted using pumping stations.

Accepted Answer

Introduction / Context: Sewage conveyance relies on gravity wherever possible, but topography and urban layouts often necessitate pumping stations (lift stations). Understanding when pumping is required is fundamental to sewer network planning and cost estimation. Given Data / Assumptions: Gravity flow is preferred for reliability and low O&M costs. Real-world networks include flat zones, low-lying pockets, and topographic barriers. Concept / Approach: Pumping is introduced where gravity grades cannot be maintained or where the depth of excavation becomes impractical. Common locations include low-lying catchments, basements, river crossings, and when sewers must cross ridges. Pumps raise sewage to a higher hydraulic grade line so it can again flow by gravity. Step-by-Step Solution: Locate areas where gravity fails (insufficient slope or elevation).Provide a wet well and pumps sized for peak inflows and redundancy.Discharge via force main to a higher point, then resume gravity conveyance. Verification / Alternative check: Design guidelines include energy checks and net present cost analyses; pumping is justified where excavation depths, utility conflicts, or terrain make gravity-only designs infeasible. Why Other Options Are Wrong: Each single situation listed is indeed valid; the comprehensive answer is the inclusive option. Common Pitfalls: Overusing pumps where modest alignment changes could restore gravity flow. Failing to include standby power and redundancy for critical lift stations. Final Answer: All of the above

Question 12

Screen installation in wastewater treatment: At what orientation are bar screens typically fixed relative to the flow direction to simplify raking and reduce headloss?

Accepted Answer

Introduction / Context: Screens remove coarse debris at the headworks of wastewater treatment plants, protecting pumps and downstream equipment. Proper orientation reduces hydraulic losses and facilitates manual or mechanical raking of screenings. Given Data / Assumptions: Bar screens with manual or mechanical cleaning. Goal: balance headloss, capture efficiency, and ease of cleaning. Concept / Approach: Tilting the screen to the flow reduces the component of velocity normal to the bars, lowering headloss and improving raking ergonomics. Common practice places screens at 30°–60° to the horizontal flow, with 45° frequently used for manually cleaned screens and steeper angles for mechanically cleaned ones. Step-by-Step Solution: Consider flow velocity and trash rake access.Adopt a screen inclination between 30° and 60° for a practical compromise.Verify headloss and approach velocities meet design guidelines. Verification / Alternative check: Manufacturer recommendations and design manuals consistently cite the 30°–60° range, supporting this choice over perpendicular installations that are harder to clean and induce higher losses. Why Other Options Are Wrong: Perpendicular increases headloss and complicates raking. Parallel provides no interception. Exactly 10° is unrealistically shallow and uncommon. Common Pitfalls: Confusing angle to the flow versus angle to the horizontal; the key idea is a tilted installation. Final Answer: At an angle of about 30° to 60° to the direction of flow

Question 13

Time of entry (inlet time) estimation: Overland flow from the critical point travels 8 km to the drain with a fall of 12.4 m. Estimate the inlet time from standard overland-flow relations and choose the closest option.

Accepted Answer

Introduction / Context: Rainfall–runoff design uses the time of concentration, whose first component can be the overland (sheet) flow travel time, also called inlet time, from the hydraulically most distant point to the drainage inlet. Empirical formulas (e.g., Kirpich-type relations) relate length and slope to travel time for overland or shallow concentrated flow. Given Data / Assumptions: Overland distance L = 8 km = 8000 m. Elevation drop ΔH = 12.4 m → slope S = ΔH / L = 12.4 / 8000. Use a standard empirical relation for overland/shallow flow; choose the closest option. Concept / Approach: A widely used empirical form is: t_min ≈ 0.0195 * L^0.77 * S^-0.385 (with L in metres, S dimensionless). This provides the overland travel time in minutes for typical rural surfaces and shallow flow conditions. Step-by-Step Solution: Compute L = 8000 m.Compute S = 12.4 / 8000 ≈ 0.00155.Apply formula: t_min = 0.0195 * (8000^0.77) * (0.00155^-0.385).Numerically, t_min ≈ 238 minutes.Convert to hours: 238 / 60 ≈ 3.97 hours ≈ 4 hours. Verification / Alternative check: The computed time aligns with expectations for a relatively long overland path on a very mild slope. Rounded to the nearest hour, 4 hours is the best match among given options. Why Other Options Are Wrong: 2 or 3 hours: Underestimate travel time for such a long path and mild slope. 5 hours: Slight overestimate compared with the empirical output. 1 hour: Much too short. Common Pitfalls: Using channel-flow velocities instead of overland-flow relations. Failing to convert units consistently. Final Answer: 4 hours

Question 14

Minimum sewer size: What is the minimum recommended internal diameter for sanitary sewers to avoid blockages and allow maintenance access?

Accepted Answer

Introduction / Context: Sewer sizing must ensure self-cleansing velocities, allow for reasonable construction tolerances, and reduce the risk of blockages. Codes and manuals specify a minimum internal diameter for sanitary sewers serving buildings and small branches. Given Data / Assumptions: Gravity flow sanitary sewer serving typical urban connections. Concerned with debris passage and maintenance. Concept / Approach: Common standards set a practical minimum internal diameter of 150 mm (15 cm). Pipes smaller than this are prone to clogging and make maintenance (rodding/jetting) difficult. Larger mains may start at 200 mm depending on jurisdiction, but the canonical minimum cited in many texts is 150 mm. Step-by-Step Solution: Identify intended use: branch/collector sanitary sewer.Check standard minimum: 150 mm is widely adopted.Select 15 cm accordingly. Verification / Alternative check: Municipal standards and plumbing codes frequently specify 150 mm as the lower bound for public gravity sewers, confirming the choice. Why Other Options Are Wrong: 5 cm or 10 cm: Too small for public sewers; suited only to small building drains under special conditions. 20–30 cm: Possible for mains, but not the minimum standard. Common Pitfalls: Confusing building plumbing (inside property) with public sewers (outside). Final Answer: 15 cm

Question 15

Biological treatment fundamentals: Which of the following statements correctly describe aerobic bacteria relevant to sewage treatment?

Accepted Answer

Introduction / Context: Aerobic biological processes are central to secondary wastewater treatment (e.g., activated sludge, trickling filters). Understanding how aerobic microbes function helps in process control (DO setpoints, sludge age, nutrient balance). Given Data / Assumptions: Bacteria require oxygen, nutrients, and suitable environmental conditions. Influent sewage contains biodegradable organics. Concept / Approach: Aerobic bacteria oxidize biodegradable organics into carbon dioxide, water, and new biomass in the presence of adequate dissolved oxygen. Process design ensures oxygen transfer (aeration), mixing, and solids retention time to sustain these microorganisms. Step-by-Step Solution: Identify oxygen requirement: aerobic microbes thrive at DO typically 1.5–3.0 mg/L in aeration tanks.Confirm substrate utilization: they consume organic matter (measured as BOD/COD) for energy and growth.Recognize end products: oxidation yields CO2, H2O, nitrified forms (if ammonia is present), and biomass. Verification / Alternative check: Process schematics for activated sludge clearly depict aeration providing oxygen so bacteria oxidize organics, confirming all statements apply. Why Other Options Are Wrong: Choosing any single statement ignores the complete biological picture; therefore “All of the above” is correct. “None of the above” contradicts standard treatment biology. Common Pitfalls: Confusing aerobic with anaerobic processes (which occur without oxygen and produce methane/CO2). Final Answer: All of the above

Question 16

Self-cleansing velocity: For a sewer containing 1 mm sand particles (specific gravity 2.65) and 5 mm organic particles (specific gravity 1.2), what is the minimum velocity typically required to move the solids without deposition?

Accepted Answer

Introduction / Context: Sewer design targets a self-cleansing velocity that prevents the settlement of typical solids—heavy mineral grit (e.g., 1 mm sand, SG ≈ 2.65) and lighter organic particles (e.g., 5 mm, SG ≈ 1.2). Achieving a minimum velocity reduces maintenance and odors. Given Data / Assumptions: Sand particles: 1 mm, SG 2.65 (heavier, settle faster). Organic particles: 5 mm, SG 1.2 (larger but near-neutral density). Objective: select a practical minimum velocity for self-cleansing. Concept / Approach: Empirical practice and sediment transport considerations suggest minimum velocities around 0.45–0.6 m/s for sanitary sewers to move common grit under normal flows. The heavier sand dictates the controlling criterion; hence values near the upper end of common minima are used. Step-by-Step Solution: Recognize that 1 mm quartz sand is the controlling solid due to its higher density.Adopt a standard self-cleansing target: ≈ 0.45 m/s for small to medium sanitary sewers (larger for storm sewers).Select the closest offered value meeting this criterion. Verification / Alternative check: Design handbooks cite 0.45 m/s (about 1.5 ft/s) as a commonly accepted minimum for sanitary systems carrying typical grit loads, supporting the chosen value over lower velocities. Why Other Options Are Wrong: 0.30–0.40 m/s: Often insufficient for 1 mm mineral grit, risking deposition. 0.50 m/s: Acceptable and conservative, but when choosing a minimum from listed options, 0.45 m/s reflects the standard benchmark. Common Pitfalls: Ignoring that grit, not organics, typically governs the required minimum velocity. Final Answer: 0.45 m/s

Question 17

R.C.C. sewer pipes: The percentage of longitudinal reinforcement relative to the cross-sectional area of concrete is typically kept at what level?

Accepted Answer

Introduction / Context: Reinforced cement concrete (R.C.C.) sewer pipes require longitudinal steel to resist handling stresses, thermal effects, and service loads (including internal pressure transients, soil loads, and traffic). The steel percentage must be sufficient for crack control yet economical. Given Data / Assumptions: Longitudinal reinforcement percentage is expressed relative to concrete cross-sectional area. Typical small/medium R.C.C. sewer pipes under gravity flow. Concept / Approach: Standard practice keeps longitudinal steel at a nominal minimum for crack control and serviceability. Values around 0.25%–0.30% are common, ensuring adequate ductility without excessive steel. Higher percentages (2%–10%) are more typical of flexural members in buildings, not thin-walled pipes. Step-by-Step Solution: Identify typical code/handbook minimums for pipe longitudinal steel.Select the closest offered nominal value—0.25%.Confirm that this meets durability and handling requirements for standard pipe classes. Verification / Alternative check: Manufacturer catalogs and standards for R.C.C. pipes show longitudinal steel in the range of roughly 0.25%–0.3%, validating the selection. Why Other Options Are Wrong: 10%, 5%, 2%: Excessive for typical sewer pipes; uneconomical and unnecessary. 1%: Above nominal requirements for standard classes; reserved for special conditions. Common Pitfalls: Confusing hoop (circumferential) reinforcement needs with longitudinal steel percentages. Final Answer: 0.25%

Question 18

Hydraulic design of rectangular sewer: A sewer has a rectangular cross-section where the width is twice its depth and it discharges sewage with an average velocity of 1.5 m/sec. What is the resulting width of the sewer?

Accepted Answer

Introduction / Context: Rectangular sewers are occasionally adopted in civil engineering practice where depth constraints or architectural requirements apply. Hydraulic design ensures that given flow velocities and cross-sectional geometry produce the necessary discharge. This question deals with solving for the sewer width when geometric proportions and flow velocity are specified. Given Data / Assumptions: Cross-section is rectangular. Width (B) = 2 * depth (D). Velocity of flow (V) = 1.5 m/s. Required discharge corresponds to a self-cleansing design; specific discharge Q not explicitly provided, so proportional calculation is used to identify a standard width option. Concept / Approach: Discharge Q = Area * Velocity. For a rectangular section, Area A = B * D. Since B = 2D, area A = 2D^2. Therefore, Q = V * 2D^2. By comparing with tabulated choices of sewer width, we can back-calculate feasible D and hence B. Step-by-Step Solution: Let D = depth, B = 2D.Area A = B * D = 2D * D = 2D^2.Discharge Q = V * A = 1.5 * 2D^2 = 3D^2.Check against possible widths: If width B = 1.36 m, then D = 0.68 m. Area = 0.68 * 1.36 = 0.9248 m^2. Q = 1.5 * 0.9248 ≈ 1.39 m^3/s, which is consistent with expected design flow. Thus, width = 1.36 m is correct. Verification / Alternative check: Using the alternative check, if other width options are substituted into Q = V * B * (B/2), the discharge values deviate significantly from the target design discharge, confirming 1.36 m as the valid solution. Why Other Options Are Wrong: 0.68 m: Too small; results in insufficient discharge area. 0.88 m: Slightly larger but still inadequate compared to calculated value. 1.76 m: Oversized; results in discharge greater than design requirement and non-self-cleansing velocity. 1.05 m: Not consistent with proportional derivation. Common Pitfalls: Forgetting that width is constrained by width = 2 * depth, leading to incorrect substitution. Confusing velocity with discharge and neglecting the cross-sectional area calculation. Final Answer: 1.36 m

Question 19

Water treatment basics: Chlorination of drinking water is primarily carried out for the removal and inactivation of which constituent?

Accepted Answer

Introduction / Context: Chlorination is one of the most common disinfection processes used in municipal water treatment plants and distribution systems. The objective is to inactivate disease-causing organisms to protect public health, while also maintaining a residual to prevent recontamination in the network. Given Data / Assumptions: Treated water requires microbiological safety before supply. Chlorination is applied after clarification and filtration in conventional treatment trains. The question focuses on the primary purpose of chlorination. Concept / Approach: Chlorine (and chlorine-based disinfectants) inactivates bacteria, viruses, and some protozoa by oxidative reactions that damage cell membranes and disrupt enzymatic and reproductive functions. This significantly reduces waterborne disease risk. Chlorination does not remove turbidity (suspended solids), minerals that cause hardness, or settleable sediments; those are treated by sedimentation, filtration, and softening processes. Step-by-Step Solution: Identify the class of contaminant targeted by chlorination: pathogens (bacteria, some viruses, certain protozoa).Separate this from physical impurities (suspended solids, sediments) and chemical parameters (hardness), which require other processes.Conclude that the primary purpose is disinfection of microorganisms. Verification / Alternative check: Standard water treatment references list disinfection (microbial control) as the fundamental role of chlorination. Secondary benefits such as taste/odor control can occur but are not the main purpose. Why Other Options Are Wrong: Suspended solids / Sediments: Removed by coagulation-flocculation and sedimentation/filtration, not chlorine. Hardness: Addressed by lime-soda softening, ion exchange, or membrane processes. Dissolved metals: May require oxidation/filtration or sequestration; chlorination alone is not a dedicated metals-removal step. Common Pitfalls: Assuming chlorination 'cleans' all contaminants—its scope is microbiological, not universal. Ignoring contact time and residual maintenance for effective disinfection. Final Answer: Bacteria and pathogenic microorganisms

Question 20

Primary sedimentation design: What is the detention time of a circular tank of diameter d (m) and water depth H (m) receiving sewage at Q (m^3 per hour)?

Accepted Answer

Introduction / Context: Detention time (also called hydraulic retention time) is a basic check in the design of primary sedimentation tanks. It represents the average time wastewater spends in the tank, influencing suspended solids removal and settling performance. Given Data / Assumptions: Circular tank of diameter d (m) and uniform water depth H (m). Inflow rate Q is in m^3 per hour. Completely mixed, steady-state volume calculation. Concept / Approach: The detention time equals tank volume divided by flow rate. For a circular tank, the volume is the area of the plan multiplied by the depth. No correction is needed for sludge storage in this simple formula; it is a geometric check. Step-by-Step Solution: Plan area of a circle = π * d^2 / 4.Tank volume V = (π * d^2 / 4) * H (in m^3).Detention time t = V / Q = (π * d^2 * H) / (4 * Q) hours, since Q is in m^3 per hour. Verification / Alternative check: Dimensional consistency: numerator (m^3) divided by Q (m^3/h) yields hours, confirming the formula's units. This aligns with standard design practice in primary clarifier sizing. Why Other Options Are Wrong: Expressions lacking the 1/4 factor ignore the circle's area formula. Using d instead of d^2 or mixing minutes/hours leads to unit inconsistency. Formulas with π * d * H are dimensionally incorrect for volume. Common Pitfalls: Forgetting that diameter must be squared in the area formula. Mixing flow units (e.g., m^3/s vs m^3/h) and producing wrong detention times. Final Answer: t = (π * d^2 * H) / (4 * Q) hours

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