Question 1

Equivalent egg-shaped sewer: If D is the diameter of a circular sewer and D' is the top horizontal diameter of an hydraulically equivalent egg-shaped section, which relation is typically used?

Accepted Answer

Introduction / Context: Egg-shaped sections are sometimes designed to be hydraulically equivalent to a circular pipe—meaning they deliver similar discharge under specified conditions. The proportions relate key egg-shaped dimensions to an equivalent circular diameter for comparison and substitution in design problems. Given Data / Assumptions: Hydraulic equivalence considered at running-full condition under the same slope and roughness. D is equivalent circular diameter; D' is top horizontal diameter of the egg-shaped section. Standard proportion set for the “new egg-shaped” geometry. Concept / Approach: Design tables provide dimensional ratios for egg-shaped sections to match the carrying capacity of a reference circular pipe. Among these, the top horizontal diameter D' commonly relates to circular D by a fixed multiple. A frequently cited relation in exam-oriented references is D' ≈ 0.84 D for equivalence based on capacity at full flow. Step-by-Step Solution: Identify that equivalence is based on discharge running full.Use standard proportion relation linking top horizontal diameter to the equivalent circular diameter.Select D' = 0.84 D as the typical value. Verification / Alternative check: Cross-check with proportion tables for egg-shaped sewers used in civil engineering texts and past competitive examinations that employ the 0.84 D ratio. Why Other Options Are Wrong: 0.64 D and 0.74 D are too small and would understate capacity; 0.94 D is too large for the top horizontal dimension in the standard equivalence set; 1.04 D overstates the dimension relative to circular D. Common Pitfalls: Mixing “equivalent” for full-flow capacity with geometric similarity; confusing top horizontal diameter with overall depth or crown radius of the egg-shaped section. Final Answer: D' = 0.84 D

Question 2

Sewer gradient from sight-rail levels: The reduced levels (R.L.) of the string at consecutive sight rails A and B are 203.575 m and 203.475 m, respectively. If the difference in their R.D.s (chainage spacing) is 10 m along the alignment, determine t…

Accepted Answer

Introduction / Context: Setting sewer gradients accurately in the field is crucial to ensure adequate self-cleansing velocity while preventing excessive scouring. Sight rails (and a traveller/boning rod) are used to transfer design levels along the alignment. This problem checks your understanding of how level differences translate to gradient over a known chainage interval. Given Data / Assumptions: R.L. at sight rail A = 203.575 m. R.L. at sight rail B = 203.475 m. Horizontal distance (R.D. difference) A → B = 10 m. Rail strings are set to represent design line; fall or rise between rails equals design gradient over the spacing. Concept / Approach: Gradient is defined as fall (or rise) per unit horizontal distance. If downstream is lower, the line has a downward gradient in the direction of flow. Numerical gradient is reported as 1 in N where N = horizontal distance / vertical fall (for downward gradients). Step-by-Step Solution: Compute fall from A to B: fall = 203.575 - 203.475 = 0.100 m.Horizontal distance = 10 m.Gradient N = distance / fall = 10 / 0.100 = 100.Since B is lower than A, gradient is downward from A to B. Verification / Alternative check: Express as percentage: fall% = (0.100 / 10) * 100 = 1%. A 1% slope equals 1 in 100, confirming the calculation. Why Other Options Are Wrong: 1 in 100/500/503 upward: levels are lower at B, so the line is not upward.1 in 50 downward: would require a 0.20 m fall over 10 m, which is not the case. Common Pitfalls: Reversing sign when interpreting which point is higher. Confusing gradient notation (1 in N) with percentage slope. Using pipe invert instead of string level without accounting for traveller height; here we directly compare string R.L.s. Final Answer: 1 in 100 downward

Question 3

Design depth for sedimentation basins: In conventional plain sedimentation tanks used in water or wastewater treatment, what is the typical maximum liquid depth adopted for design?

Accepted Answer

Introduction / Context: Sedimentation tanks (clarifiers) remove settleable solids by gravity. Their geometry—especially depth—affects sludge storage, vertical velocity profile, and constructability. Many codes and handbooks recommend standard depth ranges to balance performance and cost. Given Data / Assumptions: Plain sedimentation (no tube modules or high-rate internals). Conventional design practice. Looking for the typical upper bound of depth range. Concept / Approach: Typical liquid depths for conventional rectangular or circular basins are about 3.0–4.5 m. A higher depth can increase cost and complicate sludge removal without proportional performance gains. Hence, designers cap depth around well-established maxima. Step-by-Step Solution: Identify standard range: approximately 3.0 to 4.5 m.Select the maximum commonly used value within this band.Hence choose 4.5 m as the typical maximum design depth. Verification / Alternative check: Worked examples for conventional clarifiers routinely adopt 3.0–4.5 m depth with side water depth not exceeding ~4.5 m unless special design features are provided. Why Other Options Are Wrong: 3.0–4.0 m: valid within range but not the maximum.5.5 m: exceeds common practice for plain sedimentation without special internals. Common Pitfalls: Confusing high-rate clarifiers (lamella/tube settlers) with conventional tanks. Assuming deeper is always better; sludge handling and mixing considerations impose practical limits. Final Answer: 4.5 m

Question 4

Detention time definition: In the context of a settling (sedimentation) tank, the detention time t is best defined as the time required for which of the following?

Accepted Answer

Introduction / Context: Detention time is a fundamental hydraulic design parameter for sedimentation basins. It relates plant flow to tank volume and serves as a first-cut indicator of whether a basin will provide sufficient quiescent time for settling before short-circuiting occurs. Given Data / Assumptions: Steady average flow rate through the tank. Completely mixed inflow distribution is not assumed; definition remains V/Q regardless. No reaction or flocculation kinetics included in the definition. Concept / Approach: By definition, detention time t = V / Q, where V is the tank volume and Q is the flow rate. Physically, it is the time the flow would take to fill the tank once, not the time a specific particle takes to settle or traverse the basin (which depends on velocity fields and particle properties). Step-by-Step Solution: Start from t = V / Q.Interpretation: time to displace a volume equal to the tank volume at the prevailing flow.Hence, the correct qualitative description is the time required for the flow to fill the tank once. Verification / Alternative check: Detention time is also called hydraulic retention time (HRT) in many unit processes and is universally defined as V/Q, confirming the selection. Why Other Options Are Wrong: Particle travel times (options a, b, e) refer to settling velocity and trajectory, not the hydraulic definition of detention time.None of these: incorrect because option (c) precisely states the definition. Common Pitfalls: Equating detention time with settling time of a specific particle class. Ignoring that short-circuiting reduces effective detention time compared to theoretical t = V/Q. Final Answer: The incoming sewage flow to fill the tank volume once

Question 5

Storm overflow design: In combined sewerage, which arrangement is most effective for diverting excess storm water into a natural drainage while allowing dry-weather flow to continue toward the treatment works?

Accepted Answer

Introduction / Context: Combined sewer overflows (CSOs) protect treatment plants and trunk sewers during heavy storms by spilling the excess to natural drains or watercourses. The hydraulic arrangement must reliably separate dry-weather flow (DWF) from storm surges with minimal maintenance and clogging risk. Given Data / Assumptions: Combined sewer carrying DWF and storm runoff. Need to pass DWF onward while diverting excess to a drain. Simplicity and self-cleansing preferred. Concept / Approach: The leaping weir uses hydraulic momentum of higher storm flows to leap into an overflow channel, whereas low DWF hugs the invert and continues to the outfall works. The configuration has proven effective for automatically apportioning flows without movable parts. Step-by-Step Solution: At low flow: stream does not leap; DWF continues to interceptor.At high flow: increased velocity and depth cause the jet to leap into the overflow, diverting excess.Therefore, among listed options, the leaping weir provides the intended selective diversion. Verification / Alternative check: Design manuals describe leaping weirs as standard CSO structures due to hydraulic reliability and low headloss for DWF. Why Other Options Are Wrong: Broad-crested/side weirs may work but are less selective and prone to setting issues for DWF control.Siphon spillways are for dams/reservoirs and not typical for urban CSOs.None of these: incorrect since a known effective arrangement exists. Common Pitfalls: Confusing dam spillway devices with urban sewer overflow structures. Ignoring DWF bypass requirement in favour of storm-only diversion. Final Answer: Leaping weir overflow

Question 6

Urban runoff estimation: The empirical rational formula Q = 0.0278 K P A (with units consistent to yield discharge in m^3/s) was proposed by which drainage engineer?

Accepted Answer

Introduction / Context: The rational method relates peak runoff to rainfall intensity, catchment area, and a runoff coefficient. Different empirical forms and unit conversions appear in literature; exam questions often probe authorship and form awareness to avoid confusion across textbooks. Given Data / Assumptions: Formula stated: Q = 0.0278 K P A. Q in m^3/s when P is in mm/h and A in hectares (via 0.0278 factor), K is runoff coefficient. Identify proponent of this particular presentation. Concept / Approach: Historically, Lloyd-Davies popularized a rational-type relation for urban drainage in the U.K., and the 0.0278 factor is a common SI unit converter (from mm·ha/h to m^3/s). Recognizing this pairing helps attribute authorship correctly. Step-by-Step Solution: Note empirical form with 0.0278.Recall that Lloyd-Davies is typically associated with this variant.Select Lloyd Davis as the correct answer. Verification / Alternative check: Numerous civil engineering references present the Lloyd-Davies form with 0.0278 for SI, confirming authorship. Why Other Options Are Wrong: Kinchling and Frubling are distractors not tied to this exact empirical presentation.All of the above: incorrect since a single author is credited.Mulvany: credited with the original rational method, but not with this specific 0.0278 K P A expression. Common Pitfalls: Mixing Mulvany’s original method with later unit-specific forms. Forgetting the role of unit conversion constant 0.0278 in SI formulations. Final Answer: Lloyd Davis

Question 7

Street inlets and spacing: Which statements about road-surface drainage inlets are correct for urban stormwater design?

Accepted Answer

Introduction / Context: Stormwater inlets collect runoff from streets and convey it to the underground drainage network. Correct siting and selection of inlet types (curb vs grate) reduce ponding, improve safety, and limit nuisance flooding. Given Data / Assumptions: Urban roadway with kerbs and underground storm sewer. Typical design practice for inlet placement and types. Moderate rainfall intensities and standard crossfall. Concept / Approach: Designers place inlets at local sag points and space them periodically along grades to prevent spread of water beyond acceptable widths. Curb inlets present vertical openings in the kerb face; grate inlets accept flow through horizontal grates in the gutter flowline. Combining both can improve capture efficiency. Step-by-Step Solution: Evaluate (a): Low points (sags) accumulate runoff; inlets are essential there.Evaluate (b): Spacing of roughly 30–60 m on grades is typical for many urban streets (adjusted for intensity and crossfall).Evaluate (c): Curb inlets indeed have vertical openings at the kerb line.Evaluate (d): Grate inlets provide horizontal intake via grating—statement is correct.Therefore (e) All of the above is correct. Verification / Alternative check: Municipal design guides illustrate similar spacing ranges and distinguish curb vs grate inlet configurations, often recommending combinations at sags. Why Other Options Are Wrong: Individual statements (a–d) are each correct; selecting a single one would be incomplete. Common Pitfalls: Ignoring sags, where inlet capacity must be highest. Relying only on grates where bicycle or debris concerns suggest using combination (curb + grate) inlets. Final Answer: All of the above

Question 8

Detritus tank sizing: A treatment plant detritus tank carries sewage discharge of 576 L/s at a flow velocity of 0.2 m/s. If the width-to-depth ratio (B:D) is 2, calculate the required depth of the tank.

Accepted Answer

Introduction / Context: Detritus tanks remove heavy inorganic particles (grit) by maintaining a controlled, relatively low velocity. Geometry is selected to achieve the target velocity and detention, often expressed via cross-sectional area and width-to-depth proportions. Given Data / Assumptions: Sewage discharge Q = 576 L/s = 0.576 m^3/s. Design flow velocity v = 0.2 m/s. Width-to-depth ratio B:D = 2 (i.e., B = 2D). Steady uniform section with area A = B * D. Concept / Approach: The required area A is determined by continuity: A = Q / v. With the ratio B = 2D, substitute into A = B * D to solve for D. This is a straightforward hydraulic sizing step in preliminary design. Step-by-Step Solution: Compute area: A = Q / v = 0.576 / 0.2 = 2.88 m^2.Express area via D: A = B * D = (2D) * D = 2D^2.Solve for D: 2D^2 = 2.88 → D^2 = 1.44 → D = 1.20 m.Convert to cm: 1.20 m = 120 cm. Verification / Alternative check: Width B = 2D = 2.40 m. Check velocity: v = Q / A = 0.576 / 2.88 = 0.2 m/s (matches design), confirming correctness. Why Other Options Are Wrong: 100 cm and 110 cm: yield areas too small; velocity would exceed 0.2 m/s.150 cm: produces area larger than required; velocity would drop below 0.2 m/s.90 cm: clearly undersized for the given flow and velocity. Common Pitfalls: Forgetting to convert litres per second to m^3/s. Confusing B:D = 2 with D:B = 2; always confirm which dimension is which. Final Answer: 120 cm

Question 9

Settling behavior: In sedimentation, discrete (granular) particles exhibit what change while settling under quiescent conditions?

Accepted Answer

Introduction / Context: Sedimentation theory distinguishes between discrete (Type I) settling and flocculent (Type II) settling. Understanding the difference is vital for selecting and designing clarification processes and for correctly interpreting jar tests and settling column data. Given Data / Assumptions: Discrete (granular) particles are non-flocculent. Chemistry and surface conditions remain constant during settling. No coalescence or breakage is assumed. Concept / Approach: Discrete settling implies each particle settles independently without changing its physical characteristics. The drag force and terminal velocity are functions of particle size, shape factor, and density difference, but these parameters are treated as constant during the observation period for Type I settling. Step-by-Step Explanation: Consider Type I (discrete) regime: particles do not coalesce or flocculate.Size, shape, and specific gravity remain constant for each particle.Therefore, there is no change in size, shape, or weight while settling. Verification / Alternative check: Textbook definitions of Type I settling explicitly state that particles settle as individual entities with invariant properties, in contrast to flocculent settling where agglomeration changes effective size and settling velocity. Why Other Options Are Wrong: Options a–c: contradict the definition of discrete settling.Option e: describes flocculent settling, not discrete settling. Common Pitfalls: Assuming all suspended solids flocculate; many mineral particles behave discretely in plain sedimentation. Confusing biological flocs with inert granular particles. Final Answer: None of these

Question 10

Sewer ventilation detail: In a ventilating column used on a sewer line, where is the 'cowl' (rain cap and airflow hood) provided?

Accepted Answer

Introduction / Context: Cowl fittings are used on vent pipes to prevent ingress of rain and debris while permitting passage of air. In sewerage, they help relieve gases and reduce odour accumulation by promoting safe venting to the atmosphere. Given Data / Assumptions: Standard ventilating column in a gravity sewer network. Objective is to protect the top opening from weather and enhance airflow. Concept / Approach: The cowl is a protective hood installed at the top end of the vent stack/column. It serves as a rain cap and may aid airflow by inducing slight draft, depending on design. Placement at the upper end ensures free exchange between sewer air and the atmosphere while preventing water ingress. Step-by-Step Explanation: Identify the open end exposed to weather: top of the column.Provide cowl at this location to cap the vent while allowing airflow.Hence the correct location is the upper end of the ventilating column. Verification / Alternative check: Standard details for sewer vent stacks depict cowls or mushroom caps at the top termination above roofline or grade, confirming the correct placement. Why Other Options Are Wrong: Lower end/manhole step/interceptor-only: do not describe the weather-exposed vent termination.Manhole opening may have grates, not a cowl on a vent column. Common Pitfalls: Confusing manhole ventilation gratings with dedicated vent columns. Omitting cowls in windy or rainy climates leading to water ingress. Final Answer: At the upper end of the ventilating column

Question 11

Reinforcement placement in large-diameter sewers: For large circular sewers subjected to external pressure only (no internal pressure), where should the primary reinforcement be placed?

Accepted Answer

Introduction / Context: Reinforced concrete pipes and large-diameter sewers experience different stress distributions under internal and external loads. Correct placement of steel ensures that tensile zones are reinforced, preventing cracking and extending service life. Given Data / Assumptions: Loading case: external pressure only (e.g., soil and surcharge), negligible internal pressure. Concrete is strong in compression, weak in tension; steel should be located where tensile stress develops. Ring action of a pipe section under uniform external pressure. Concept / Approach: Under internal pressure, hoop tension is maximum at the inner surface; reinforcement is therefore placed near the inner face. Conversely, under external pressure alone, tensile hoop stress develops near the outer surface, so primary reinforcement should be concentrated there. This aligns steel with the tensile zone, counteracting crack formation. Step-by-Step Reasoning: Identify loading: external pressure (soil, groundwater head, traffic surcharge) on the outside.Determine tensile zone under ring compression: tension occurs closer to the outer surface.Place main reinforcement accordingly near the outer face to resist tensile hoop stresses. Verification / Alternative check: Standard R.C.C. pipe design practices differentiate reinforcement cages for internal vs external pressure cases, with outer-face steel specified for external-load-only design. Why Other Options Are Wrong: Inner-face steel suits internal pressure, not external-only loading.Both faces or elliptical cages may be used for combined or special loads but are not required for external-only case here.None of these: incorrect as a clear correct placement exists. Common Pitfalls: Assuming symmetric steel regardless of load case, which wastes material or misplaces capacity. Ignoring construction tolerances that can shift steel cover—outer-face placement must still meet cover requirements. Final Answer: Near the outer surface of the pipe

Question 12

Circular sewer, partial flow: If the angle subtended at the sewer centre by the free water surface is 120°, what is the depth of flow (depth of sewerage) for a 1 m diameter circular sewer?

Accepted Answer

Introduction / Context: This question tests the geometric relationships used when a circular sewer runs partially full. Designers often express depth of flow as a function of the central angle of the wetted segment. Knowing how to convert angles to depths allows quick checks of capacity and hydraulic performance in partially full conditions. Given Data / Assumptions: Circular sewer, diameter D = 1 m (so radius R = 0.5 m). The angle subtended at the centre by the free water surface (the chord angle for the water surface) is 120°. Depth of flow y is measured vertically from invert (lowest inner point) to the water surface. Concept / Approach: For a circular segment defined by a central angle 2α (radians or degrees), the depth of the segment is y = R * (1 - cos α). If the chord (free surface) subtends 120° at the centre, then 2α = 120°, hence α = 60°. Substituting α into the depth formula gives the fraction of the radius occupied by water above the centreline to the surface. Express the final depth in metres (or centimetres) using R = D/2. Step-by-Step Solution: Let 2α = 120° ⇒ α = 60°.Depth formula: y = R * (1 - cos α).Use R = D/2 = 1/2 = 0.5 m.Compute cos 60° = 0.5.y = 0.5 * (1 - 0.5) = 0.5 * 0.5 = 0.25 m.Convert to centimetres: 0.25 m = 25 cm. Verification / Alternative check: Since α = 60°, the depth equals R/2. Because R = 0.5 m, R/2 = 0.25 m = 25 cm. This quick mental check matches the computed value. Why Other Options Are Wrong: 20 cm: Would require a smaller α (cos α closer to 0.6).40 cm or 50 cm: Exceed the R/2 result for α = 60°; 50 cm equals the full radius, which would imply α = 90°.60 cm: Greater than the radius; impossible for a 1 m diameter pipe in this configuration. Common Pitfalls: Confusing the 120° with α itself; remember 2α is the full wetted central angle. Using diameter instead of radius in y = R * (1 - cos α). Mixing degrees and radians when using calculators; here, simple angles allow exact values. Final Answer: 25 cm

Question 13

Sewage putrefaction gases: During the putrefaction (anaerobic decomposition) of sewage, which gases are typically given off?

Accepted Answer

Introduction / Context: In the absence of oxygen, sewage undergoes putrefaction (anaerobic decomposition). Recognizing the characteristic gases evolved is important for ventilation design, odor control, worker safety, and corrosion mitigation in sewers and treatment units. Given Data / Assumptions: Environment is anaerobic (limited or no free oxygen). Decomposition of organic matter by anaerobic bacteria. Typical byproducts include various reduced and oxidized gases. Concept / Approach: During putrefaction, organic compounds are broken down to produce a mixture of gases. Common constituents are methane (from methanogenesis), carbon dioxide (from fermentation steps), hydrogen (from certain fermenters), along with other malodorous gases such as hydrogen sulfide and traces of ammonia and volatile organics. The presence of these gases dictates the need for gas handling and monitoring systems. Step-by-Step Solution: Identify anaerobic pathways: fermentation and methanogenesis.Major gases: CH4 (methane), CO2 (carbon dioxide), H2 (hydrogen).Ancillary gases: H2S, NH3, and others may also be present depending on sewage composition.Since H2, CH4, and CO2 are all plausible, the inclusive option is correct. Verification / Alternative check: Operational data from anaerobic digesters and septic environments confirm methane and carbon dioxide as dominant gases with hydrogen and traces of other compounds also present, validating the comprehensive choice. Why Other Options Are Wrong: Single-gas options (Hydrogen, Methane, Carbon dioxide) omit other well-known byproducts.“Ammonia only” disregards the principal methane and carbon dioxide components in anaerobic environments. Common Pitfalls: Confusing aerobic oxidation (which mainly produces CO2) with anaerobic putrefaction (which also yields CH4 and H2). Ignoring safety; methane is combustible and hydrogen sulfide (often present) is toxic and corrosive. Final Answer: All of the above

Question 14

Nitrogen form in well-oxidized sewage: A well-oxidized sewage contains nitrogen predominantly in which form?

Accepted Answer

Introduction / Context: Nitrogen species in wastewater evolve as treatment progresses. Understanding whether nitrogen appears as ammoniacal, organic, nitrite, or nitrate forms helps diagnose treatment performance and oxygenation status in the system. Given Data / Assumptions: “Well-oxidized sewage” implies sufficient aeration and biological oxidation. Nitrification proceeds from ammonia → nitrite → nitrate under aerobic conditions. Concept / Approach: In aerobic environments, nitrifying bacteria convert ammonia first to nitrite (by ammonia-oxidizing bacteria) and then to nitrate (by nitrite-oxidizing bacteria). As oxidation becomes more complete, nitrate becomes the dominant inorganic nitrogen form, indicating advanced nitrification. Step-by-Step Solution: If oxygen present and biology active: NH3 → NO2− → NO3−.“Well-oxidized” means the terminal product of nitrification predominates.Therefore, nitrates are the main nitrogen form. Verification / Alternative check: Process control in activated sludge systems uses low ammonia and nitrite with higher nitrate as markers of complete nitrification in aerobic zones, matching the statement. Why Other Options Are Wrong: Nitrites: Intermediate; should be low in fully oxidized conditions.Free ammonia: Characteristic of reduced or early-stage oxidation.None/Organic nitrogen only: Inconsistent with “well-oxidized” status. Common Pitfalls: Assuming elevated nitrite persists; it should be further oxidized to nitrate under stable aerobic conditions. Confusing nitrification (aerobic) with denitrification (anoxic), where nitrate is reduced. Final Answer: Nitrates

Question 15

Hierarchy of sewerage network: In a conventional system, from which component does the sewerage network originate (i.e., where does collection begin)?

Accepted Answer

Introduction / Context: Sewer networks are hierarchical. Knowing the correct order—from building connections to final outfall—helps in layout, capacity checks, and maintenance planning. Given Data / Assumptions: Standard municipal sewerage hierarchy. Collection starts at individual properties. Concept / Approach: The system begins with house sewers (building laterals) that carry wastewater from fixtures to the public right-of-way. These discharge into small branch or lateral sewers, then into mains, then trunks/interceptors, and finally to the outfall sewer conveying to treatment. Step-by-Step Solution: 1) Identify smallest element collecting flow: house sewer (building lateral).2) Flow path: house sewer → branch/lateral → main → trunk/interceptor → outfall to treatment.3) Therefore, the origin of the network is the house sewer. Verification / Alternative check: Typical standards and textbooks illustrate this sequence; no exceptions in conventional systems. Why Other Options Are Wrong: Branch/Main/Outfall/Trunk: These are progressively downstream components and do not represent the system origin at the parcel level. Common Pitfalls: Confusing “lateral” (public) with “house sewer” (private); terminology varies, but the house connection is the source. Final Answer: House sewers (building laterals)

Question 16

Typical gases in sewers: Which of the following gases can be evolved within a sewer system due to biochemical activity?

Accepted Answer

Introduction / Context: Sewers create environments conducive to both aerobic and anaerobic microbial processes. Gas evolution affects corrosion, odor, safety, and ventilation design (e.g., risk of explosion with methane and toxicity of hydrogen sulfide). Given Data / Assumptions: Presence of organic matter, biofilms, slime layers, and variable oxygen conditions. Typical municipal sewage composition. Concept / Approach: Under anaerobic niches, methanogens produce methane; sulfate-reducing bacteria generate hydrogen sulfide; aerobic and anaerobic processes yield carbon dioxide. Identifying all likely gases helps assess hazards and corrosion potential (e.g., H2S oxidizing to sulfuric acid in moist headspace, attacking concrete). Step-by-Step Solution: List likely gases: CO2 (ubiquitous), H2S (from sulfate reduction), CH4 (from methanogenesis).Confirm that all three are plausible in sewer headspace.Select inclusive option recognizing multiple gas species. Verification / Alternative check: Empirical measurements in sewer headspaces commonly report variable CH4, H2S, and CO2 concentrations, confirming the selection. Why Other Options Are Wrong: Single-gas answers omit other well-known sewer gases.Oxygen alone is not “evolved” by sewage; it is consumed. Common Pitfalls: Underestimating H2S dangers; even low ppm can be hazardous. Ignoring explosion risks where CH4 accumulates. Final Answer: All of these

Question 17

Design assumption for sewage quantity: For sewer design, what percentage of water supply is commonly assumed to appear as sewage discharge?

Accepted Answer

Introduction / Context: In preliminary sewer design, the expected sewage quantity is often estimated as a percentage of the water supply. Accounting for losses (outdoor use, evaporation, leakage, consumptive use) determines the proportion that reaches sewers. Given Data / Assumptions: Municipal conditions with typical indoor plumbing and fixtures. Conventional design practice references. Concept / Approach: Historically, 75–80% of the supplied water is assumed to return as wastewater (sanitary sewage), allowing for non-sewered uses and system losses. This rule-of-thumb aids early sizing prior to detailed metering or modeling. Step-by-Step Solution: Recognize loss mechanisms: landscape irrigation, leakage, consumption.Apply the standard planning fraction for sanitary flow: approximately 0.75–0.80 of water supply.Select the choice consistent with practice: 75 to 80%. Verification / Alternative check: Many municipal manuals adopt 0.8 of supply as an upper bound for sanitary contribution, with I/I (infiltration/inflow) added separately for peak wet-weather analysis. Why Other Options Are Wrong: 65–75% ranges may understate flows in many urban contexts.85% or higher usually overstates returns unless special conditions exist. Common Pitfalls: Failing to add separate allowances for infiltration/inflow and industrial discharges when applicable. Using a single percentage for all land uses without refinement. Final Answer: 75 to 80%

Question 18

Sizing a sewer main from flow and velocity: For a peak discharge of 0.0157 m^3/s conveyed at 0.9 m/s, what is the approximate internal diameter of the sewer main (assume full circular section)?

Accepted Answer

Introduction / Context: Quick diameter estimates are routine in sewer design when target velocities are specified for self-cleansing and hydraulic efficiency. The continuity relationship Q = A * V directly links discharge, area, and velocity. Given Data / Assumptions: Peak discharge Q = 0.0157 m^3/s. Mean velocity V = 0.9 m/s. Full circular flow assumed for sizing (conservative first-pass estimate). Concept / Approach: Use Q = A * V to find area A, then back-calculate diameter from A = (π/4) * d^2. Convert to centimeters for selection among the discrete options. Step-by-Step Solution: Compute area: A = Q / V = 0.0157 / 0.9 ≈ 0.017444 m^2.Relate to diameter: A = (π/4) * d^2 ⇒ d = sqrt(4A / π).Calculate: 4A ≈ 0.069776; divide by π ≈ 0.02222.d ≈ sqrt(0.02222) ≈ 0.149 m ≈ 0.15 m.Therefore, diameter ≈ 15 cm. Verification / Alternative check: Reverse check: A = π*(0.15^2)/4 ≈ 0.01767 m^2; Q = A * V ≈ 0.01767 * 0.9 ≈ 0.0159 m^3/s, close to the given 0.0157 m^3/s. Why Other Options Are Wrong: 10–12 cm: Too small; would require higher velocity than 0.9 m/s for the same flow.18–20 cm: Oversized; would give lower velocity than desired for self-cleansing at this flow. Common Pitfalls: Using nominal diameters without checking velocity limits for self-cleansing. Confusing internal diameters with external (nominal) diameters; always use internal for hydraulics. Final Answer: 15 cm

Question 19

Why circular sections are preferred for sewers: Identify the most comprehensive reason set for selecting a circular cross-section in gravity sewers.

Accepted Answer

Introduction / Context: The choice of sewer cross-section affects hydraulic capacity, self-cleansing, constructability, and cost. Circular sections are most common for small to medium diameters in municipal systems. Given Data / Assumptions: Gravity sewer applications with standard construction methods. Comparison of geometric and economic attributes. Concept / Approach: Among shapes, a circle offers a high area-to-perimeter ratio, which implies a larger hydraulic radius R = A/P and, hence, lower resistance for a given perimeter. Circles are also structurally efficient and straightforward to manufacture and install using pipes, leading to cost effectiveness. Step-by-Step Solution: Geometric efficiency: For given perimeter, circle maximizes enclosed area.Hydraulic implication: Larger hydraulic mean depth (A/P) yields better flow capacity.Practicality: Circular pipes are widely available, easy to lay, and economical.Therefore, the inclusive choice captures the full rationale. Verification / Alternative check: Textbook derivations show the circle maximizes area for a given perimeter, which correlates with a favorable hydraulic radius, and industry costing demonstrates economies of scale for circular pipe products. Why Other Options Are Wrong: Single-attribute options do not reflect the totality of hydraulic and economic advantages.“Only because it is easy to clean” is not the primary or sole reason. Common Pitfalls: Assuming non-circular shapes are never used; egg-shaped sections can be advantageous for combined sewers. Ignoring partial flow hydraulics where circular conduits maintain reasonable performance. Final Answer: All of the above

Question 20

Definition check: What do we call the sewer that carries sewage from the collection system to the point of treatment or final discharge?

Accepted Answer

Introduction / Context: Terminology in sewerage systems distinguishes between local collection conduits and the final conveyance to treatment. Correct naming ensures clarity in drawings, specifications, and operation manuals. Given Data / Assumptions: Standard definitions in municipal engineering. Focus on conduit that conveys to treatment or final outfall. Concept / Approach: An out-fall sewer (also spelled outfall sewer) is the final trunk conduit that transports the collected wastewater to the treatment plant or discharge point. It is distinct from mains, branches, and laterals that make up the upstream collection network. Step-by-Step Solution: Identify network hierarchy: house → branch/lateral → main → trunk/interceptor → out-fall.The conduit delivering to treatment is the out-fall sewer. Verification / Alternative check: Textbook glossaries and standards consistently define “out-fall sewer” as the final conveyance to treatment/discharge. Why Other Options Are Wrong: Main/Branch/Lateral: Upstream network elements; not the terminal conveyance.House sewer: Private connection from premises, far upstream of the outfall. Common Pitfalls: Confusing “trunk sewer” with the final out-fall; in many systems the out-fall is the terminal segment of the trunk. Final Answer: Out-fall sewer

Waste Water Engineering Questions