Question 1

Minimum internal diameter for sanitary sewers: What minimum diameter is typically adopted for small gravity sanitary sewers to avoid blockage and to facilitate maintenance?

Accepted Answer

Introduction / Context: Minimum sizes are standardized so that sewers are less likely to clog, are maintainable with standard equipment, and can pass typical base flows without excessive deposition. Knowledge of the typical minimum diameter is a core design guideline. Given Data / Assumptions: Gravity sanitary sewer serving residential/commercial premises. Conventional design standards and self-cleansing velocity criteria. Concept / Approach: Industry practice commonly adopts 150 mm (15 cm) as the minimum for lateral and small branch gravity sewers. Smaller diameters increase risk of clogging and complicate maintenance. The 150 mm size allows reasonable velocities under normal flows and compatibility with cleaning tools. Step-by-Step Solution: Identify the code/practice minimum: 150 mm.Compare against options and choose 15 cm. Verification / Alternative check: Public works standards frequently state 150 mm as the minimum internal diameter for new gravity sewers; larger sizes are chosen as flows warrant. Why Other Options Are Wrong: 10–12.5 cm: Too small for maintenance and self-cleansing velocities.20–30 cm: Possible but not the minimum; used when capacity demands it. Common Pitfalls: Confusing house plumbing pipe sizes with public sewer minimums. Ignoring that storm or combined sewers may have different practical minima. Final Answer: 15 cm

Question 2

Adopted minimum diameter for gravity sewers: In municipal practice, what is the minimum sewer diameter generally adopted for new gravity sanitary sewers?

Accepted Answer

Introduction / Context: Minimum diameter rules help ensure operability, maintenance access, and reduced blockage risk. This question reaffirms the standard adopted minimum size used at the start of system planning and design. Given Data / Assumptions: Gravity sanitary sewer context. Typical municipal standards and best practices. Concept / Approach: A 150 mm minimum is widely accepted for public sanitary sewers. It balances hydraulic capacity, cleaning access, and constructability, while smaller sizes can hamper maintenance and increase clog risks. Step-by-Step Solution: Recall practice minimum: 150 mm.Select 15 cm among options. Verification / Alternative check: Design manuals and local authority specifications commonly list 150 mm as the minimum diameter for gravity sewers; larger diameters are used for higher contributing populations or combined flows. Why Other Options Are Wrong: 10–12.5 cm: Below accepted minimum; escalates blockage risk.20–25 cm: Valid sizes but not the minimum. Common Pitfalls: Assuming house drain sizes apply to public sewers. Overlooking that pressure sewers or special systems may differ. Final Answer: 15 cm

Question 3

Activated sludge process (ASP): Which statement best describes the essence of the activated sludge process in biological wastewater treatment?

Accepted Answer

Introduction / Context: The activated sludge process is the workhorse of secondary (biological) wastewater treatment. Understanding its core feature—mixing influent with return activated sludge under aeration—helps differentiate ASP from other processes like trickling filters or primary sedimentation. Given Data / Assumptions: Conventional ASP with aeration tank and secondary clarifier. Use of return activated sludge (RAS) to maintain biomass concentration. Air (or oxygen) supply supports aerobic oxidation of organics. Concept / Approach: In ASP, mixed liquor (influent + RAS) is aerated to oxidize organics and synthesize biomass. After clarification, part of the settled sludge is returned (RAS) to the aeration tank; the remainder is wasted (WAS). This loop is central to maintaining the right mixed liquor suspended solids (MLSS) and sludge age. Step-by-Step Solution: Identify the hallmark: RAS + aeration.Select the option explicitly stating aeration with previously aerated sludge. Verification / Alternative check: Standard ASP schematics show influent + RAS entering aeration, followed by secondary clarification and RAS return—confirming the correct statement. Why Other Options Are Wrong: Pure stirring without aeration is insufficient; oxygen is needed for aerobic metabolism.Centrifugation is a solids-liquid separation, not the ASP core mechanism.“Aeration until full chemical stability” is vague and omits the critical RAS loop. Common Pitfalls: Confusing primary treatment (settling) with secondary biological oxidation. Overlooking the importance of sludge age and MLSS control via RAS/WAS. Final Answer: Aeration is carried out with an admixture of previously aerated sludge (return activated sludge)

Question 4

Drop manholes: In sewer networks, “drop manholes” are provided primarily when…

Accepted Answer

Introduction / Context: Where a branch sewer joins a deeper main at a substantial level difference, direct plunging inflow can cause turbulence, odor, and structural issues. “Drop manholes” use a vertical drop connection to dissipate energy and improve hydraulics and safety. Given Data / Assumptions: Two or more sewers with different invert elevations. Need to connect higher-level flow to a lower-level main safely. Concept / Approach: A drop manhole routes the higher sewer through a vertical or near-vertical drop pipe into the manhole, discharging near the benching level. This reduces free-fall splashing inside the shaft, controls scouring, and protects the structure. Step-by-Step Solution: Identify the condition: different levels (inverts) at junctions.Engineering response: provide a drop (inside or outside) to handle elevation difference.Select the option describing drains at different levels. Verification / Alternative check: Design manuals illustrate internal or external drop pipes where level differences warrant, confirming the purpose of drop manholes. Why Other Options Are Wrong: Ladders/covers: Safety and access features do not define “drop” manholes.Depth > 3.5 m alone is not the defining criterion; it is the level difference at junction.Diameter changes do not necessitate a drop unless levels differ. Common Pitfalls: Confusing deep manholes with drop manholes; the key is level difference and drop connections. Overlooking the need for energy dissipation and ventilation in drop configurations. Final Answer: Drains meet at significantly different invert levels

Question 5

Combined sewers cross-section: Which cross-section is most suitable for carrying combined flow (sanitary + storm) over a wide range of discharges?

Accepted Answer

Introduction / Context: Combined sewers must perform satisfactorily at both low dry-weather flows and high wet-weather flows. Cross-section choice affects velocity distribution, self-cleansing, and surcharge behavior across this wide operating range. Given Data / Assumptions: Combined sewer (sanitary + stormwater). Need stable velocities at low flows and adequate capacity at peaks. Concept / Approach: The egg-shaped section narrows at the invert, producing higher velocities for low flows to limit deposition, yet provides large capacity higher up for storm events. This dual benefit makes it a classic choice for combined systems compared with simple circular or rectangular sections. Step-by-Step Solution: Assess low-flow behavior: egg-shaped has smaller wetted width → higher velocity.Assess peak behavior: taller crown provides storage/conveyance for storms.Select egg-shaped as the balanced solution. Verification / Alternative check: Historical sewer design literature consistently recommends egg-shaped sections for combined sewers to maintain self-cleansing across flow regimes. Why Other Options Are Wrong: Circular: Good general choice but less effective than egg-shaped at very low flows.Rectangular/horse-shoe/triangular: Have disadvantages in sediment control or structural efficiency for buried conduits. Common Pitfalls: Assuming circular always suffices; combined flow needs special attention at low flows. Ignoring maintenance issues tied to sediment deposition. Final Answer: Egg-shaped

Question 6

Components of a sewerage system: Which list correctly represents the hierarchical elements that make up a municipal sewerage system?

Accepted Answer

Introduction / Context: Understanding terminology and hierarchy in sewerage systems is essential for layout, capacity planning, O&M, and documentation. The system aggregates flow from premises progressively to outfall and treatment. Given Data / Assumptions: Standard municipal sewer network components. House → lateral → branch → main → trunk/interceptor → outfall. Concept / Approach: The upstream-most element is the house sewer, which discharges into a lateral or branch. These collect into mains, then trunks/interceptors, finally to the outfall. A complete list should include multiple upstream elements, not just a subset. Step-by-Step Solution: Identify all major hierarchical levels.Choose the option that includes house, lateral, branch, and main sewers. Verification / Alternative check: Typical standard drawings and textbooks map the exact chain from building laterals to the outfall, confirming the comprehensive list. Why Other Options Are Wrong: Partial lists omit key elements and thus do not represent the full sewerage system. Common Pitfalls: Confusing “lateral” and “branch” terms; usage varies by jurisdiction but both are upstream collectors. Equating “sewerage” (system) with “sewage” (wastewater) in terminology. Final Answer: House sewer, lateral sewer, branch sewer, main sewer (all of these)

Question 7

Sedimentation tank design parameter: The discharge per unit plan area of a sedimentation (settling) tank is generally called what?

Accepted Answer

Introduction / Context: Settling basin performance is governed by the surface overflow parameter that compares inflow to plan area. Different terms are used interchangeably in practice. Recognizing these synonyms avoids confusion when consulting varied standards and texts. Given Data / Assumptions: Parameter is discharge divided by plan area (Q/A). Used as a sizing criterion for sedimentation tanks and clarifiers. Concept / Approach: Surface overflow rate, surface loading rate, and overflow velocity all denote the same conceptual parameter: the upward superficial velocity against which particle settling must compete. If the particle settling velocity exceeds this value (with safety factors), it will be removed in an ideal clarifier. Step-by-Step Solution: Define the parameter: v_s = Q / A.Recognize synonymous terminology in literature: overflow rate = surface loading = overflow velocity (superficial).Choose the option that includes all synonymous terms. Verification / Alternative check: Design manuals routinely use these terms interchangeably and provide typical design ranges by process type (primary, secondary, tertiary clarification). Why Other Options Are Wrong: Individual terms (a, b, or c) are correct but incomplete; the question asks what it is “generally called,” acknowledging synonyms.Basin turnover number is unrelated. Common Pitfalls: Confusing overflow rate with weir loading rate (Q per unit weir length) or hydraulic retention time. Ignoring that real clarifier performance also depends on flocculation, inlet/outlet design, and sludge removal. Final Answer: All of the above

Question 8

In open-channel hydraulics, the numerical value of Chezy's constant C is specifically used in which empirical resistance–velocity relation?

Accepted Answer

Introduction / Context: Chezy's constant C is a classic resistance parameter used in uniform open-channel flow calculations. Many examination questions test whether you can distinguish where C appears directly and where it is predicted indirectly by other relations. Given Data / Assumptions: Topic: uniform flow in open channels and pipes flowing partly full. Chezy's constant C relates mean velocity to hydraulic radius and slope. Other formulae (Bazin, Kutter, Manning) are also used in resistance computations. Concept / Approach: The Chezy equation expresses mean velocity V as V = C * sqrt(R * S), where R is hydraulic radius and S is energy slope. In contrast, Bazin and Kutter give empirical expressions to estimate C, and Manning gives V directly without mentioning C, i.e., V = (1/n) * R^(2/3) * S^(1/2). Step-by-Step Solution: Identify the equation that explicitly contains C.Chezy: V = C * sqrt(R * S).Bazin/Kutter: provide C as a function of roughness and R, used to substitute into Chezy, but not the velocity formula by themselves.Manning: V = (1/n) * R^(2/3) * S^(1/2), no C. Verification / Alternative check: If you see C in the velocity formula itself, it is Chezy's relation. Other formulae either compute C (Bazin, Kutter) or bypass C (Manning). Why Other Options Are Wrong: Bazin's formula: estimates C; it does not present V unless combined with Chezy. Kutter's formula: also yields C empirically; again, V requires Chezy's relation. Manning's formula: gives V directly using n, not C. All of the above: incorrect because only Chezy's formula uses C directly in V. Common Pitfalls: Confusing the formulas that define C with the formula that uses C. Remember: Bazin/Kutter → compute C; Chezy → uses C; Manning → no C. Final Answer: Chezy's formula.

Question 9

In regions where the rainy (monsoon) season is short and limited to only a few months, which sewerage system is generally recommended for urban drainage and wastewater collection?

Accepted Answer

Introduction / Context: Choosing an appropriate sewerage system depends on rainfall pattern, cost, operation, and public health. Where rainfall is brief or infrequent, designers prefer arrangements that keep foul sewage flows in smaller, more economical conduits. Given Data / Assumptions: Rainy season lasts only a few months. Objective: recommend a suitable sewerage system. Systems considered: combined, partially separate, and separate. Concept / Approach: A combined system carries sanitary sewage and storm water together, requiring large pipelines sized for storm peaks all year. A separate system carries sanitary sewage and storm runoff in separate conduits, allowing the sanitary sewer to be sized only for wastewater (plus infiltration), reducing cost and avoiding storm-induced overflows. This is ideal when stormwater loads are seasonal or small. Step-by-Step Solution: Assess rainfall duration → short season implies most of the year only sanitary flows occur.Economics and hydraulics → separate system keeps sanitary sewers small and efficient.Storm drainage can be handled by open channels or separate storm sewers used mainly during rainy months. Verification / Alternative check: Compare life-cycle costs and overflow risks: combined systems risk combined sewer overflows; separate systems minimize pollution during storms and allow targeted stormwater controls. Why Other Options Are Wrong: Combined: oversizes pipes year-round and risks overflows. Partially separate: reduces but does not eliminate combined issues. None of these: there is a standard solution—separate system. Combined with detention only: still conveys foul flow with stormwater. Common Pitfalls: Assuming combined systems are always cheaper. In short rainy seasons, they often are not due to large diameters and treatment burdens during storms. Final Answer: Separate system (sanitary sewage and storm water in different sewers).

Question 10

Which set correctly describes the environmental conditions preferred by different microbial groups in wastewater treatment: anaerobic, aerobic, and facultative bacteria?

Accepted Answer

Introduction / Context: Microbiology underpins sewage treatment. Understanding which organisms thrive with or without oxygen is essential when selecting unit processes (e.g., anaerobic digesters vs. activated sludge). Given Data / Assumptions: Three microbial groups: anaerobic, aerobic, facultative. We are matching each group to its oxygen preference. Concept / Approach: Anaerobic bacteria function in the absence of free dissolved oxygen (e.g., digesters, septic tanks). Aerobic bacteria require oxygen (e.g., aeration tanks, trickling filters). Facultative bacteria can live in both conditions, shifting metabolism as oxygen availability changes. Step-by-Step Solution: Map each group to oxygen requirement.Anaerobic → no oxygen; Aerobic → oxygen needed; Facultative → either condition.Select the option that states all three correctly. Verification / Alternative check: Check typical locations: anaerobic digesters (no aeration), activated sludge aeration basins (bubbled air), facultative ponds (upper aerobic layer, lower anaerobic layer). Why Other Options Are Wrong: Reversing oxygen needs contradicts microbiology. Claiming all require oxygen ignores anaerobes. Sunlight is not the defining requirement for these bacteria. Common Pitfalls: Confusing anoxic (no dissolved oxygen, nitrate present) with anaerobic (no dissolved oxygen and no nitrate). The question refers broadly to absence/presence of oxygen. Final Answer: Anaerobic bacteria flourish without oxygen; aerobic bacteria flourish with oxygen; facultative bacteria can adapt to either condition.

Question 11

Which statement about sanitary sewers is most universally correct for design and public health protection?

Accepted Answer

Introduction / Context: Sewer design balances hydraulics, public health, and economics. This question focuses on which guidance is broadly applicable across most projects. Given Data / Assumptions: We consider normal gravity sewers (not force mains). Statements include size adequacy, flow condition, and burial depth. Concept / Approach: Adequate sizing to pass peak sanitary flow (plus infiltration/inflow) prevents surcharge and overflows. Gravity sanitary sewers are typically designed to run part full at design peak, not full. Burial depth depends on local topography, frost depth, and service connections—not a universal fixed depth. Step-by-Step Solution: Assess (a): Adequate sizing is a universal requirement to avoid overflows and backups.Assess (b): Normal operation aims for part-full flow with free surface; full flow indicates potential surcharge.Assess (c): Depth varies project-to-project; 2–3 m is not a universal rule.Hence only (a) is consistently correct. Verification / Alternative check: Standard manuals specify minimum slopes and capacities to maintain self-cleansing velocities while avoiding surcharge—implying part-full, not full, operation. Why Other Options Are Wrong: (b) Incorrect: design for full flow is atypical for gravity sewers. (c) Incorrect: burial depth is site-specific; sometimes shallower or deeper. (d) Incorrect: because (b) and (c) are not universally true. (e) Incorrect: pressurized operation is for force mains, not gravity sewers. Common Pitfalls: Assuming that “full” signals good capacity. In sewers, steady part-full flow with a free surface is desired to allow ventilation and minimize surcharge risk. Final Answer: Sewers must be of adequate size to avoid overflow under peak conditions.

Question 12

In standard colorimetric testing of sewage, which reagent combination is used to develop a matching color specifically for detecting nitrites (NO₂⁻)?

Accepted Answer

Introduction / Context: Colorimetric tests for nitrogen species in wastewater are routine in environmental engineering labs. Nitrite detection relies on a diazotization–coupling reaction commonly known as the Griess reaction. Given Data / Assumptions: Analyte: nitrite (NO₂⁻) in sewage. We are choosing the reagent system that develops the diagnostic pink–red azo dye. Concept / Approach: The Griess reaction converts nitrite to a diazonium salt (often via sulfanilic acid in acidic medium), which then couples with an aromatic amine such as α-naphthylamine to produce a colored azo dye. The question’s phrasing points to the acid plus naphthylamine system used for nitrite color development. Step-by-Step Solution: Identify the species: nitrite, not nitrate.Associate the correct reaction: Griess reaction reagents (acid medium + aromatic amine like naphthylamine).Select the option corresponding to this chemistry. Verification / Alternative check: Laboratory practice distinguishes nitrite (Griess) versus nitrate (phenol disulphonic acid after reduction) to avoid cross-interference. Why Other Options Are Wrong: Potassium permanganate is an oxidant, not the standard color reagent for nitrite. Phenol disulphonic acid with KOH is used for nitrate determination. None of these is wrong because a correct reagent system is listed. Orthotolidine relates to chlorine residual testing. Common Pitfalls: Confusing nitrite and nitrate methods. Ensure the correct analyte–reagent pairing to avoid erroneous results. Final Answer: Sulphuric acid with naphthylamine (Griess reaction reagents).

Question 13

In which location is aerobic biological activity typically at its maximum in municipal wastewater systems?

Accepted Answer

Introduction / Context: Aerobic microorganisms require oxygen to oxidize organic matter efficiently. Understanding where their activity peaks helps in designing treatment systems. Given Data / Assumptions: Locations: household lines, gravity sewers, aeration tanks, septic tanks, rising mains. Aerobic activity correlates with available dissolved oxygen and biomass. Concept / Approach: In treatment plants, aeration basins intentionally supply air (or oxygen) and maintain high biomass concentrations, creating ideal conditions for maximum aerobic activity. In contrast, sewers often lose dissolved oxygen as travel time increases, tending toward septic conditions. Step-by-Step Solution: Identify the unit with forced aeration and high microbial mass → aeration tanks.Compare to other locations (house connections, long sewers, septic tanks), where DO is lower or absent.Therefore, maximum aerobic activity occurs in aerated treatment units. Verification / Alternative check: Process control measurements (DO profiles, oxygen uptake rates) confirm highest aerobic activity within engineered aeration systems. Why Other Options Are Wrong: Fresh sewage may have some DO, but not engineered aeration. Long sewers lose DO and may become anoxic or anaerobic. Septic tanks are predominantly anaerobic. Rising mains are enclosed and typically not aerated. Common Pitfalls: Assuming “fresh” always means “aerobic.” Without aeration, DO can be depleted quickly by biochemical demand. Final Answer: Sewage treatment process units with aeration (e.g., activated sludge tanks).

Question 14

Hazen’s settling relation in water and wastewater treatment is typically applicable to particles of which size range?

Accepted Answer

Introduction / Context: Different empirical/analytical laws govern settling velocity depending on particle size and flow regime. Recognizing which law applies is essential for sedimentation basin design. Given Data / Assumptions: Hazen’s relation is empirical for larger, non-colloidal particles. Stokes’ law applies to very small, laminar-settling particles. Concept / Approach: Stokes’ law is valid for small particles (Reynolds number for particle settling in laminar regime), commonly for diameters below about 0.1 mm in water. For larger grains (sand-sized), inertial effects increase and Stokes’ assumptions break; empirical relations like Hazen's are used. Step-by-Step Solution: Classify size → boundary near 0.1 mm for water treatment practice.Hazen’s relation → applicable for particles larger than approximately 0.1 mm.Choose the option reflecting this threshold. Verification / Alternative check: Design manuals differentiate Stokes (fine) vs. Hazen or Newton regimes (coarser). Why Other Options Are Wrong: Options (a)–(c) refer to smaller sizes where Stokes is more appropriate. (e) Colloidal particles do not follow simple gravity settling; they require coagulation-flocculation. Common Pitfalls: Using Stokes for coarse sand leads to significant errors because laminar assumptions fail. Final Answer: Diameters greater than 0.1 mm.

Question 15

Given n as the roughness (rugosity) coefficient, r as hydraulic radius (or hydraulic depth in wide channels), and s as bed slope, which resistance formula gives the mean velocity of flow in m/s?

Accepted Answer

Introduction / Context: Engineers often use empirical resistance equations to compute mean velocity in open-channel flow. Recognizing which variables belong to which formula is fundamental. Given Data / Assumptions: Variables provided: n (roughness), r (hydraulic radius/depth), s (slope). We must identify the formula that directly gives V in terms of n, r, and s. Concept / Approach: Manning’s formula provides velocity directly as V = (1/n) * r^(2/3) * s^(1/2). Chezy's formula uses V = C * sqrt(r * s), where C is separate; Bazin and Kutter predict C, not V directly without Chezy. Froude relations address critical conditions, not resistance. Step-by-Step Solution: Match variables (n, r, s) → Manning's equation is the only one with all three directly.Therefore, select Manning’s formula. Verification / Alternative check: Check dimensions: r^(2/3) * s^(1/2) gives velocity units when multiplied by 1/n (n is dimensionally adjusted empirically). Why Other Options Are Wrong: Chezy: requires C, not n. Bazin/Kutter: give C for Chezy, not V directly in the form shown. Froude: relates velocity to depth at critical flow, not roughness. Common Pitfalls: Mixing up formulas that estimate C (Bazin/Kutter) with those that yield V directly (Manning). Final Answer: Manning's formula: V = (1/n) * r^(2/3) * s^(1/2).

Question 16

The length of overland flow from the critical point to a drain mouth is 13.58 km, and the elevation drop is 10 m. Estimate the inlet time (time of concentration) to the drain using the Kirpich relation. Choose the nearest value.

Accepted Answer

Introduction / Context: Time of concentration (or inlet time) estimates how long runoff from the hydraulically most distant point takes to reach a drain. Empirical formulas such as Kirpich’s are commonly used when only length and slope are known. Given Data / Assumptions: Length of overland/flow path L = 13.58 km = 13,580 m. Elevation drop H = 10 m → bed slope S = H / L. Use Kirpich formula in minutes: t_c = 0.01947 * L^0.77 * S^(-0.385), where L in m and S is dimensionless slope. Concept / Approach: Kirpich's empirical relation links time of concentration to length and slope. Lower slopes and longer lengths increase travel time, raising peak arrival time at the drain. Step-by-Step Solution: Compute slope S = H / L = 10 / 13,580.Calculate t_c (min) = 0.01947 * (13,580)^0.77 * (S)^(-0.385).Numerically, t_c ≈ 476 minutes.Convert to hours: 476 / 60 ≈ 7.94 hours ≈ 8 hours. Verification / Alternative check: The very shallow slope (10 m drop over 13.58 km) implies slow runoff—an 8-hour inlet time is plausible. Any steeper slope would reduce the computed time markedly. Why Other Options Are Wrong: 2 h and 4 h: too small for such a long, shallow path. 6 h: still underestimates travel time for the given gentle slope. 10 h: larger than the Kirpich estimate; not the nearest value. Common Pitfalls: Mixing units (km vs. m) or confusing H/L with L/H. Always use L in metres and dimensionless slope S = H/L in the formula given. Final Answer: 8 hours.

Question 17

Which factors significantly influence the gravitational settling (sedimentation) of a particle in a sedimentation tank handling sewage or water?

Accepted Answer

Introduction / Context: Settling behavior in sedimentation tanks depends on properties of both the fluid and the particles. Designers must account for hydrodynamics and particle characteristics to predict removal efficiency. Given Data / Assumptions: Parameters considered: tank velocity field, water viscosity, particle size/shape, particle specific gravity. Concept / Approach: Settling velocity depends on the net gravitational force minus buoyancy and drag. Drag depends on fluid viscosity and particle size/shape (which determines drag coefficient). The velocity field affects residence time and potential resuspension, thereby affecting overall removal. Step-by-Step Solution: Fluid property: viscosity influences drag and Reynolds number.Particle properties: size and shape affect projected area and drag coefficient; specific gravity affects net driving force.Hydraulics: velocity distribution controls detention time and short-circuiting.Therefore, all listed factors matter. Verification / Alternative check: Compare Stokes’ regime vs. transitional/newtonian regimes: each still depends on viscosity, size, and density contrast; tank hydraulics governs whether a particle can traverse the basin depth in the available time. Why Other Options Are Wrong: Single-factor answers ignore the coupled nature of settling physics. Common Pitfalls: Assuming particle size alone dictates removal. In cold weather, higher viscosity can reduce settling velocity noticeably; poor hydraulics can also limit removal even for relatively heavy particles. Final Answer: All of the above (velocity field, viscosity, size/shape, and specific gravity).

Question 18

Sewer manholes: At which locations along a sewer line are manholes typically provided for access, inspection, and maintenance?

Accepted Answer

Introduction / Context: Manholes are essential appurtenances in urban sewerage. They provide access for inspection, cleaning, maintenance, flow monitoring, and ventilation. Proper placement reduces blockages, shortens maintenance time, and improves safety for workers. Given Data / Assumptions: Sewer lines include horizontal and vertical geometric changes. Flow may carry grit and debris that can settle at hydraulic transitions. Periodic access is required for cleaning and repairs. Concept / Approach: Hydraulic transitions and connections are the locations most prone to energy losses, turbulence, and deposition. Therefore, design practice places manholes at all points where grade, direction, or pipe size changes, and at junctions. Spacing is also limited (e.g., 30–100 m depending on size) to ensure routine access even on straight, uniform reaches. Step-by-Step Solution: Identify geometry change points: changes in slope (gradient) and direction (horizontal bends).Identify network nodes: junctions where laterals connect to mains or two mains combine.Provide manholes at each of these points to allow entry and cleaning equipment passage.Confirm additional spacing-based manholes on long uniform runs for operational access. Verification / Alternative check: Operations records show that blockages cluster near bends, junctions, and slope changes. Providing manholes at these locations reduces time to clear blockages and improves safety. Why Other Options Are Wrong: Each of options (a), (b), and (c) is individually correct; the comprehensive correct choice is (d) because all listed locations require manholes. Common Pitfalls: Placing manholes too far apart on long straight runs; omitting manholes at diameter transitions; poor alignment causing difficult entry or unsafe benching inside the structure. Final Answer: All of these

Question 19

Survey control for sewer construction: Which primary instrument is used to set out the alignment and grade while laying sewers on site?

Accepted Answer

Introduction / Context: Laying sewers requires accurate alignment (direction) and grade (invert slope). Survey instruments provide the control needed to ensure gravity flow, avoid backfalls, and meet design levels. Two instruments are commonly involved: theodolite for alignment and level for elevations. Given Data / Assumptions: Gravity sewer with designed longitudinal slope. Need for accurate direction between consecutive manholes. Site control points available. Concept / Approach: Theodolite precisely sets out straight lines and angles, establishing the sewer centerline between manholes. Elevations are then checked with a level (or total station). The question asks for the primary instrument used for laying out sewers—interpreted in practice as establishing alignment with a theodolite, followed by level checks for grade using boning rods or laser levels. Step-by-Step Solution: Set up theodolite over control point; sight to backsight and turn required angle.Fix pegs along the centerline for trench excavation.Use a level to transfer elevations and set invert grade with boning rods or lasers.Check as-built alignment and reduce levels to confirm compliance. Verification / Alternative check: Total stations and lasers integrate angle and distance measurement; however, conceptually theodolite remains the alignment instrument of reference, with a level for elevations. Why Other Options Are Wrong: Magnetic compass and plane table lack required precision; clinometer gives slopes but not alignment angles; a level alone cannot set horizontal direction. Common Pitfalls: Poor instrument centering; failure to protect reference pegs; neglecting temperature and refraction on long sights. Final Answer: Theodolite

Question 20

Empirical sectioning for small drains (Q ≤ 15 m³/s): What relation between depth (d) and top width (B) is commonly adopted for preliminary proportioning?

Accepted Answer

Introduction / Context: For roadside, storm-water, or small irrigation drains up to about 15 m³/s, designers often begin with empirical proportions before iterating with Manning or Chézy equations. A typical starting proportion is a depth roughly half the top width, providing a stable, maintainable section. Given Data / Assumptions: Design discharge up to approximately 15 m³/s. Trapezoidal or prismatic open-channel section. Need for an initial proportion prior to hydraulic checking. Concept / Approach: Using a rule-of-thumb like d = 0.5 B (equivalently B ≈ 2 d) provides a reasonable hydraulic radius and side-slope economy for first sizing. The section is then refined to satisfy permissible velocities, freeboard, side slopes, and material constraints using Manning’s formula and stability checks. Step-by-Step Solution: Start with d = 0.5 B as a preliminary proportion.Select side slopes based on soil stability (e.g., 1V:1.5H).Apply Manning: Q = (1/n) * A * R^(2/3) * S^(1/2) and iterate B and d.Check velocity limits and provide freeboard; adjust as needed. Verification / Alternative check: Sensitivity analysis against B/d ratios near 2 shows minor differences in area and hydraulic radius; final sizing is governed by Manning and stability criteria. Why Other Options Are Wrong: d = 0.2 B produces too shallow a section; B = 0.2 d and B = 0.5 d invert the proportion unrealistically, giving very narrow or very deep sections. Common Pitfalls: Keeping empirical proportions without hydraulic verification; ignoring freeboard; neglecting sediment transport and maintenance access. Final Answer: d = 0.5 B

Waste Water Engineering Questions