Question 1

Acceptable hardness range for potable water (as CaCO3):\r Within which range (in parts per million, ppm) is total hardness typically prescribed for safe drinking water quality?

Accepted Answer

Introduction / Context:Total hardness in potable water mainly arises from calcium and magnesium salts expressed as mg/L (ppm) of calcium carbonate equivalent. While hardness does not pose a direct health risk within moderate limits, it affects aesthetic acceptability, scaling, and soap consumption. This question tests awareness of a commonly prescribed acceptable band for hardness in drinking water.Given Data / Assumptions:Hardness is expressed as CaCO3 equivalent in ppm.The ranges shown mirror typical textbook guidance for acceptable palatability and scale control.Regulatory limits can vary; we focus on commonly cited teaching ranges.Concept / Approach:Very soft water ( 300 ppm) causes significant scale and soap wastage. Many standard references present 75 to 115 ppm as a comfortable range, offering acceptable taste with limited scaling in household use. Therefore, this middle range is often singled out in exam-style questions as “prescribed”.Step-by-Step Solution:Compare each option range to a commonly acceptable zone.Recognize that 75 to 115 ppm aligns with standard guidance in teaching texts.Select 75 to 115 ppm as the correct range.Verification / Alternative check:Utilities commonly target hardness near 100 ppm for balanced operation where softening is used, supporting the educational range provided here for general acceptability.Why Other Options Are Wrong:50 to 75 ppm is on the softer side and may increase corrosion potential.100 to 150 ppm and 150 to 200 ppm are acceptable to many consumers but are usually above the “prescribed” teaching band noted in many syllabi.“None of these” is incorrect as a reasonable range exists.Common Pitfalls:Confusing aesthetic hardness guidance with strict legal limits, which vary by jurisdiction.Not distinguishing between carbonate (temporary) and non-carbonate (permanent) hardness when diagnosing scaling problems.Final Answer:75 to 115 ppm

Question 2

Effects of high pH on water distribution systems:\r A high pH value in water does not typically cause which of the following issues?

Accepted Answer

Introduction / Context:Water pH significantly influences corrosion, scaling, and microbial stability in distribution networks. Higher pH values tend to make water less corrosive and more scale forming, while lower pH increases corrosivity. The question asks which problems are not produced by high pH water.Given Data / Assumptions:“High pH” refers to water on the alkaline side (pH > 8).Metallic distribution mains may include cast iron, ductile iron, or steel.Typical dissolved mineral content supports carbonate equilibria.Concept / Approach:At higher pH, water is less aggressive; calcium carbonate tends to precipitate, forming protective scales (incrustation). This reduces corrosion rate and tuberculation (localized nodular corrosion). Conversely, low pH enhances corrosion and iron tuberculation. Sediment deposits can occur under a variety of pH conditions due to solids entering the system or formed by precipitation at high pH.Step-by-Step Solution:Identify phenomena promoted by high pH: scale formation and possible sedimentation due to precipitates.Identify phenomena inhibited by high pH: corrosion and iron tuberculation.Therefore, the effects that high pH does not produce are tuberculation and corrosion.Verification / Alternative check:Stability indices (e.g., Langelier Saturation Index) become positive at higher pH, indicating scale forming tendencies and reduced corrosivity, which aligns with the selected choice.Why Other Options Are Wrong:Incrustation and sediment deposits are more likely in high pH water due to precipitation of carbonates and hydroxides.Corrosion and tuberculation are characteristic of more acidic, aggressive water, not high pH.Common Pitfalls:Assuming any deposit equals corrosion. Scale deposits may actually reduce corrosion by forming a protective film.Ignoring alkalinity and calcium hardness, which interact with pH to determine scaling versus corrosivity.Final Answer:Both tuberculation and corrosion

Question 3

Detention period in sedimentation (settling) tanks:\r Which description best represents the detention period used in design and analysis of settling basins?

Accepted Answer

Introduction / Context:Detention period (also called theoretical residence time) is a fundamental parameter in the design of sedimentation tanks. It directly relates tank volume to flow rate and indicates the time available for particles to settle under quiescent conditions.Given Data / Assumptions:Tank volume V is known; influent flow Q is approximately steady.Plug flow is assumed for theoretical definition, knowing that real flow deviates.Multiple equivalent descriptions are presented.Concept / Approach:The detention period t_d is defined by t_d = V / Q. This equals the theoretical time to fill the tank at flow Q, the average retention time, and the ratio of volume to discharge. These statements are equivalent ways to express the same definition used in preliminary design.Step-by-Step Solution:Start with t_d = V / Q.Interpretation 1: theoretical passage time for a parcel of water through the tank.Interpretation 2: time to fill from empty at constant flow Q.Interpretation 3: average time water is retained in the basin.All are consistent, so “All of the above” is correct.Verification / Alternative check:Tracer tests in real tanks show mean residence time near V / Q but effective flow through period may be less due to short-circuiting; however, the definition of detention period remains V / Q.Why Other Options Are Wrong:Each individual statement is true but incomplete; the comprehensive option captures the equivalence explicitly.Common Pitfalls:Confusing theoretical detention with effective flow through period; the latter is lower because of non-ideal hydraulics.Assuming detention period alone guarantees settling performance without considering particle characteristics and overflow rates.Final Answer:All of the above

Question 4

How turbidity is reported in water quality tests:\r Turbidity values are expressed as what type of measurement?

Accepted Answer

Introduction / Context:Turbidity indicates the scattering of light by suspended and colloidal particles in water. It does not directly equal the mass concentration of particles; rather, it is measured optically using standardized devices and scales such as NTU (nephelometric turbidity units) or historical JTU (Jackson Turbidity Units).Given Data / Assumptions:Turbidity is to be reported using standard water quality conventions.Options include mass concentration, pH, and color code, which are not turbidity units.Recognition that the scale is instrument dependent.Concept / Approach:Nephelometric instruments measure light scattered at 90 degrees and report NTU. The Jackson candle turbidimeter historically reported JTU. Both are numeric values tied to instrument calibration standards, not direct mg/L of solids and not pH or color alone.Step-by-Step Solution:Match definition: turbidity is an optical property, not a direct gravimetric concentration.Identify correct reporting: numeric units on an arbitrary scale (NTU/JTU).Therefore select the option specifying an instrument scale.Verification / Alternative check:Standard Methods for the Examination of Water and Wastewater specify NTU as the standard reporting unit for turbidity with nephelometry, confirming the chosen option.Why Other Options Are Wrong:ppm (mg/L) relates to concentration, not optical scattering magnitude.pH measures acidity/alkalinity.Color code alone is not a turbidity measurement unit.Common Pitfalls:Assuming a simple conversion exists between NTU and mg/L of solids; particle size, shape, and refractive index affect scattering.Using color as a surrogate for turbidity; colored but particle free water can still have low turbidity.Final Answer:As numbers on an arbitrary instrument scale (e.g., NTU or JTU)

Question 5

Well testing in groundwater engineering:\r The recuperation (recovery) test is carried out primarily to determine which parameter?

Accepted Answer

Introduction / Context:Groundwater development requires estimating how much water a well can sustainably supply. When pumping equipment is limited or steady pumping tests are impractical, the recuperation (recovery) test provides a simple method to gauge well yield by observing the rate at which water level recovers after partial dewatering.Given Data / Assumptions:Open well or large diameter well is typical for this method.Water level is depressed by bailing or pumping, then allowed to recover.Time series of water level recovery is recorded.Concept / Approach:The governing idea is that the recovery rate reflects aquifer recharge to the well and well hydraulics. By applying recovery equations, one estimates the yield corresponding to permissible drawdown. The test thus aims at determining well yield rather than water quality metrics like pH or turbidity.Step-by-Step Solution:Lower the well water level to a measurable drawdown.Stop withdrawal and measure rise in water level at regular intervals.Use recovery data to compute inflow characteristics and estimate sustainable yield at chosen drawdown.Report the yield value for design of pump capacity and command area planning.Verification / Alternative check:Where possible, compare yield from recuperation with that from a constant rate pumping test. Consistency validates the estimate; discrepancies indicate issues like boundary effects or storage anomalies.Why Other Options Are Wrong:Turbidity and pH are laboratory water quality parameters, unrelated to the recovery method’s purpose.“Discharge from a well at a fixed drawdown” is related but the recuperation test specifically targets overall yield capacity rather than single operating point discharge measurement.Common Pitfalls:Not allowing sufficient observation time for recovery, leading to underestimated yields.Ignoring interference from nearby pumping wells which alters recovery curves.Final Answer:Yield of a well

Question 6

Nitrite limit in potable water:\r What is the permissible amount of nitrites (NO2–) present in safe drinking water supplied to consumers?

Accepted Answer

Introduction / Context:Nitrite (NO2–) is an intermediate in the nitrogen cycle and a potential indicator of incomplete nitrification or contamination. Elevated nitrite poses health risks, especially methemoglobinemia in infants. Water supply standards therefore set very stringent limits for nitrite in finished water.Given Data / Assumptions:Finished water after treatment is being considered.Public health protection is the priority.Conservative exam-oriented guidance often treats nitrite as undesirable and to be absent.Concept / Approach:While specific numerical limits vary by jurisdiction, many education syllabi emphasize that nitrite should be absent in distributed water. Hence, the safest and commonly accepted response in basic water quality questions is “Nil”. This reflects the operational target rather than a nuanced regulatory table that may list very small allowable concentrations.Step-by-Step Solution:Recognize nitrite as a contaminant indicating biological instability or pollution.Recall that potable supply management aims for nondetectable nitrite.Select “Nil” as the permissible amount for exam purposes.Verification / Alternative check:Utilities closely monitor nitrification in chloraminated systems to keep nitrite undetectable; early detection triggers corrective actions (e.g., boosting disinfectant, flushing, converting to free chlorine temporarily).Why Other Options Are Wrong:10–45 ppm values are far above health-based guideline levels and would be unacceptable in drinking water.Common Pitfalls:Confusing nitrate (NO3–) limits with nitrite limits; nitrate limits are much higher than nitrite.Assuming a small allowable concentration is acceptable in practice; operations typically target non-detect for safety.Final Answer:Nil

Question 7

Sedimentation control for reservoirs and dams Before constructing a dam and planning the reservoir, which measures are considered for controlling siltation and sediment entry into the pool?

Accepted Answer

Introduction / Context:Sedimentation is a primary life-cycle risk for reservoirs. If unmanaged, deposited silt reduces live storage, affects intakes, and shortens project life. Designers therefore integrate catchment and hydraulic measures to control the sediment load that reaches and remains in the reservoir.Given Data / Assumptions:New storage project in a sediment-bearing catchment.Typical riverine sediments (sands, silts, clays).Conventional gravity or embankment dam with bottom outlets.Concept / Approach:Controls act at three levels: (1) catchment treatment to reduce erosion and sediment yield, (2) in-stream structures (check dams) to trap coarse fractions before they reach the pool, and (3) hydraulic provisions at the dam (under-sluices) to flush deposited material from near the intake and thalweg when flows are available.Step-by-Step Solution:Identify catchment-level actions → vegetation screens, afforestation, contour bunds reduce soil loss.Add in-channel traps → check dams slow flow, promote deposition of sand/gravel upstream.Provide dam-level features → under-sluices at low level allow density current and bottom flushing during high flows.Site selection → choose alignment and site with lower specific sediment yield (geology/land use), and with feasible flushing hydraulics.Verification / Alternative check:Reservoir sediment management guidelines consistently recommend a portfolio approach: watershed treatment + structural traps + sluicing / flushing through low-level outlets when hydrology permits.Why Other Options Are Wrong:Each of (a)–(d) is indeed a recognized measure; none alone is sufficient. Hence the comprehensive option is correct.Common Pitfalls:Relying solely on under-sluices without catchment treatment; ignoring density currents; providing bottom outlets but without adequate discharge capacity during floods.Final Answer:All the above

Question 8

Slow sand filter operation — terminal head loss for cleaning A slow sand filter should be taken out for cleaning (scraping) when the filter head (loss of head) rises beyond which range?

Accepted Answer

Introduction / Context:Slow sand filters remove turbidity and pathogens primarily through the biologically active schmutzdecke and fine sand matrix. Over time, clogging increases head loss. A maximum permissible head loss (terminal head) is specified to decide when to stop and clean the unit.Given Data / Assumptions:Conventional slow sand filter (not rapid sand filter).Standard bed depth and effective size sand in municipal practice.Cleaning by scraping 1–2 cm of top sand.Concept / Approach:The filter runs until head loss approaches the terminal value. Typical terminal head for slow sand filters lies around 0.7 m to about 1.0–1.2 m of water column, beyond which operation is inefficient and risk of breakthrough increases during rate upsets.Step-by-Step Solution:Monitor inlet and outlet heads to compute loss of head across the bed.When head loss approaches 70–120 cm, schedule scraping and resting.After cleaning, restore bed depth with make-up sand at intervals.Verification / Alternative check:Operating manuals and exam standards cite ≈0.75–1.0 m as the trigger, aligning with the 70–120 cm range.Why Other Options Are Wrong:10–40 cm and 40–70 cm are too low; frequent cleaning would be uneconomic.120–150 cm is above typical recommended terminal head for slow filters.Common Pitfalls:Confusing slow sand filters (biological) with rapid sand filters (higher rates, different head-loss criteria). Running past terminal head increases risk of media disturbance and short-circuiting.Final Answer:70 cm to 120 cm

Question 9

Surface overflow rate (SOR) for sedimentation with coagulation What is the normal range of surface overflow rate for a plain sedimentation tank when coagulants are used?

Accepted Answer

Introduction / Context:In clarifier design, the surface overflow rate (SOR) is a key sizing parameter. With coagulation, fine particles form settleable floc, allowing higher SOR than plain sedimentation without coagulant.Given Data / Assumptions:Conventional coagulation–flocculation followed by sedimentation.Municipal water quality of moderate turbidity.Design for average conditions (not peak shock loads).Concept / Approach:Typical exam and handbook values place SOR for coagulated water around 1000–1250 L/hr/m^2 (which equals about 24–30 m^3/m^2/day). This range balances tank area versus settling performance and floc quality.Step-by-Step Solution:Recall standard SOR ranges for coagulated waters.Select the range 1000–1250 L/hr/m^2 as the normal design band.Verification / Alternative check:Converting to m/day: 1000–1250 L/hr/m^2 ≈ 24–30 m/day, matching many design manuals for conventional clarifiers.Why Other Options Are Wrong:750–1000 L/hr/m^2 is on the conservative/low side for coagulated water.1250–1500 or a single figure of 1500 L/hr/m^2 are relatively high and not “normal” baseline values.Common Pitfalls:Confusing SOR for primary sedimentation in wastewater with potable-water clarifiers; also mixing up hr-based and day-based units.Final Answer:1000 to 1250 litres/hr/m^2

Question 10

Yield from a well using specific capacity A well has specific capacity 0.3183 × 10^-3 per second. For a 4 m diameter well under a depression head (drawdown) of 4 m, estimate the discharge.

Accepted Answer

Introduction / Context:Specific capacity is a convenient index in well hydraulics. For some formulations, it is expressed per unit area per unit drawdown, allowing quick discharge estimates when geometry and drawdown are known.Given Data / Assumptions:Specific capacity, Cs = 0.3183 × 10^-3 s^-1.Well diameter, d = 4 m → area A = π d^2 / 4.Drawdown (depression head), h = 4 m.Concept / Approach:Assuming the given Cs represents discharge per unit area per unit drawdown, use Q = Cs * A * h. Compute area and multiply by drawdown to obtain discharge in m^3/s; then convert to L/s by multiplying by 1000.Step-by-Step Solution:A = π d^2 / 4 = π * 4^2 / 4 = π * 4 = 12.566 m^2 (approx).Q = Cs * A * h = 0.3183 × 10^-3 * 12.566 * 4.Compute product of area and head: 12.566 * 4 = 50.264.Q ≈ 0.3183 × 10^-3 * 50.264 ≈ 0.015999 m^3/s ≈ 16 L/s.Verification / Alternative check:Back-check: 16 L/s corresponds to 0.016 m^3/s. Dividing by A * h (≈ 50.264) gives about 0.318 × 10^-3 s^-1, matching the given specific capacity.Why Other Options Are Wrong:8–14 L/s underestimate the computed discharge for the stated area and drawdown.Common Pitfalls:Misinterpreting the meaning of specific capacity; forgetting to include the well cross-sectional area; unit conversion errors between m^3/s and L/s.Final Answer:16 litres/sec

Question 11

Pressure terminology in water mains The maximum pressure that a pipe experiences during normal operation (including expected surges) is called:

Accepted Answer

Introduction / Context:Pipeline design uses distinct pressure terms. Confusing them can lead to under- or over-design. Exam questions often ask which term corresponds to the pressure actually seen during service.Given Data / Assumptions:Normal operation of a distribution or transmission main.Transient surges are expected within limits but catastrophic water hammer is excluded.Concept / Approach:Working pressure: the maximum service pressure during operation.Design pressure: a chosen limit including allowances for surges, used to specify pipe class; sometimes equal to or greater than the working pressure.Test pressure: higher hydrostatic pressure applied during commissioning to verify integrity.Thus, the term corresponding to the maximum during normal operation is “working pressure”.Step-by-Step Solution:Identify operational context → not a factory/field test (exclude test pressure).Select the term used for in-service maxima → working pressure.Verification / Alternative check:Standards distinguish test pressure (often 1.5 × working) from service conditions; design/class selections are tied to working plus surge criteria.Why Other Options Are Wrong:Design pressure is a specification limit, not necessarily the realized service maximum.Test pressure is non-operational and temporary.“Pipe pressure” is not a precise technical term; “All the above” cannot be correct.Common Pitfalls:Using test-pressure values to size pumps or surge equipment; ignoring surge allowances in setting design class.Final Answer:Working pressure

Question 12

Selecting a source for urban water supply Which of the following factors must be evaluated while selecting a city’s water source and its conveyance route?

Accepted Answer

Introduction / Context:Source selection drives capital and operating costs for decades. The hydrology, topography, and raw-water characteristics govern treatment needs, pumping energy, and resilience.Given Data / Assumptions:City-scale potable supply.Multiple potential sources (river, lake, groundwater, regional scheme).Concept / Approach:Quantity determines reliability during drought; quality dictates treatment train and operating cost; elevation affects gravity advantage versus pumping; terrain affects alignment feasibility, tunneling, and environmental impacts.Step-by-Step Solution:Assess safe yield and seasonal variability (quantity).Analyze raw-water contaminants and treatability (quality).Compare available hydraulic head for gravity conveyance (elevation).Survey terrain for constructability, risk, and maintenance access (route).Verification / Alternative check:Feasibility studies explicitly balance these criteria in multi-objective analysis.Why Other Options Are Wrong:Any single factor alone is insufficient; a complete evaluation requires all listed factors.Common Pitfalls:Underestimating quality variability (algal blooms, turbidity spikes); neglecting headloss and future expansion corridors.Final Answer:All the above

Question 13

Typical breakdown of urban water demand Choose the most appropriate distribution of total system consumption among end-uses and losses in a typical planning study.

Accepted Answer

Introduction / Context:While actual water-use fractions vary by city and year, planning guidelines often adopt representative percentages for preliminary sizing and tariff studies.Given Data / Assumptions:No unusual industrial base dominating demand.Reasonable non-revenue water levels around 15% in planning figures.Concept / Approach:Domestic demand typically dominates. Commercial/institutional, public purposes, and unavoidable losses (non-revenue water) comprise the remainder. The provided split (50% + 25% + 10% + 15%) sums to 100% and matches a commonly cited planning template.Step-by-Step Solution:Check each fraction and confirm they form a plausible total.Recognize this is a representative planning breakdown, not a universal rule.Verification / Alternative check:Many bylaws and manuals use similar splits for initial estimates before calibrating with metered data.Why Other Options Are Wrong:Choosing any single item ignores that all listed proportions are used together as a standard planning set.Common Pitfalls:Assuming losses are negligible; underestimating firefighting allowance and seasonal public uses.Final Answer:All the above

Question 14

Dead storage allowance in reservoir capacity Dead storage (reserved for silt deposition over project life) is generally taken as what fraction of the total capacity Q?

Accepted Answer

Introduction / Context:Reservoir capacity is divided into dead storage (below outlets) and live/active storage. Dead storage often includes allowance for silt deposition during the design life so that active storage is preserved as long as possible.Given Data / Assumptions:General planning guidance; actual projects calibrate with sediment yield studies.Design life where significant siltation is expected.Concept / Approach:Textbook and exam practice frequently take a dead-storage allowance on the order of about one-third of total for preliminary sizing in silt-laden rivers, subject to later refinement. This ensures intake function and maintains live storage for much of the project life.Step-by-Step Solution:Relate dead storage to projected sediment volume over life.Select the commonly adopted preliminary fraction ≈ 1/3 Q when no site-specific study is given.Verification / Alternative check:More detailed designs use trap efficiency curves and catchment sediment yield to refine the fraction, but 1/3 Q remains a standard exam value.Why Other Options Are Wrong:Fractions 1/8 to 1/4 may under-allow silt in high-sediment basins for preliminary design.Common Pitfalls:Using a rule-of-thumb instead of a sedimentation study at DPR stage; omitting dead-storage elevation relative to intakes.Final Answer:1/3 of Q

Question 15

Hydrogeology terminology — ability of a formation to transmit water The capability of a saturated soil or aquifer of full width and depth to transmit water across its thickness is termed:

Accepted Answer

Introduction / Context:Groundwater hydraulics distinguishes between properties of the medium per unit thickness and properties of the entire saturated thickness. Correct terminology is essential for well design and yield prediction.Given Data / Assumptions:Saturated, laterally extensive aquifer.“Full width and depth” implies considering the whole saturated thickness.Concept / Approach:Permeability (hydraulic conductivity, K) is a material property per unit thickness.Transmissibility (often transmissivity, T) = K * b, where b is saturated thickness; it represents the ability of the entire thickness to transmit water.Porosity is volumetric void fraction and does not directly measure ease of flow.Step-by-Step Solution:Interpret “full width and depth” → use property integrated over thickness.Recall T = K * b → measure of total capacity to transmit flow through the aquifer section.Therefore, the correct term is “Transmissibility”.Verification / Alternative check:Pumping-test analyses (Theis, Cooper-Jacob) use transmissivity directly to relate drawdown to discharge.Why Other Options Are Wrong:Permeability refers to unit thickness; porosity is not a flow parameter; “None” is inapplicable; specific yield is a storage parameter.Common Pitfalls:Using K and T interchangeably; forgetting to multiply by saturated thickness when estimating well fields.Final Answer:Transmissibility

Question 16

Sluice valves in water distribution mains — purpose and siting Which statements correctly describe the use and placement of sluice valves on main water pipelines?

Accepted Answer

Introduction / Context:Sluice valves (gate valves) are isolation devices on water mains. They allow sections to be shut for repair, tie-ins, or operational control. Their correct spacing and location differ from that of air valves and washouts.Given Data / Assumptions:Trunk main and distribution network for potable water.Standard practice for long pipelines and urban grids.Concept / Approach:Sluice valves control or isolate flow. Spacing depends on network topology, consumer nodes, and the need to minimize service disruption; typical guidance may be a few kilometers on trunk mains, but exact spacing is not always fixed at 5 km. Placement at summits is characteristic of air valves, not sluice valves; washouts (blow-offs) go at low points.Step-by-Step Solution:Identify primary function → isolation/regulation → (a) true.Evaluate spacing → (b) is not a universal rule; spacing is context-specific.Check siting → (c) is incorrect; summits require air valves, not sluice valves by default.Therefore, only statement (a) stands universally correct.Verification / Alternative check:Design manuals show valve clusters at junctions, branches, and boundary points to sectionalize networks efficiently rather than strictly by a fixed kilometer spacing.Why Other Options Are Wrong:(b) Over-specific; not a standard invariant distance.(c) Misattributes summit placement (that is for air valves).(d) Cannot be correct since (b) and (c) are incorrect.Common Pitfalls:Confusing roles of sluice valves, air valves (summits), and washouts (low points). Over-spacing valves can lead to large outage areas during maintenance.Final Answer:They are used to regulate or isolate the flow of water in pipes

Question 17

Pumps and appurtenances in water-supply engineering: Which one of the following statements is INCORRECT about common pump types and their principles of operation?

Accepted Answer

Introduction / Context:Pumps used in water-supply systems include air-lift pumps, jet (ejector) pumps, hydraulic rams, centrifugal pumps, and rotary positive-displacement pumps. Each type has a distinct principle, typical application, and installation constraint. This question checks recognition of correct operational principles and a common distractor about “cams” in rotary pumps.Given Data / Assumptions:Deep and small-diameter wells often require special pump arrangements.Hydraulic ram devices use pressure surges (water hammer) to lift water.Rotary pumps generally use gears, lobes, vanes, or screws rather than “cams”.Concept / Approach:Identify the statement that mischaracterizes a pump’s working mechanism. Air-lift and jet pumps move water by entrained air or jet-induced suction respectively; hydraulic rams exploit water hammer. Rotary pumps are positive-displacement units employing gears/lobes/vanes/screws, not opposing “cams”.Step-by-Step Solution:Check air-lift: compressed air reduces density in the riser, lifting water → statement is correct.Check jet pump: nozzle–venturi ejector down the well permits small-bore use → correct.Check hydraulic ram: uses water hammer and waste valve oscillation → correct.Check rotary pump: uses gears/lobes/vanes/screws; “cams rotating opposite” is not the accepted description → incorrect.Verification / Alternative check:Manufacturer catalogs and standard texts depict internal gear, lobe, vane, and screw pumps; none rely on counter-rotating “cams” as the primary displacement element in waterworks service.Why Other Options Are Wrong:Option a: Describes air-lift principle accurately.Option b: Correctly notes small-well suitability of jet pumps.Option c: Properly cites water-hammer basis for hydraulic rams.Option e: Not applicable because one statement (d) is wrong.Common Pitfalls:Confusing rotary positive-displacement internals with cam-and-follower mechanisms; mixing centrifugal and rotary PD concepts; assuming all deep-well pumping uses submersible centrifugal pumps.Final Answer:Rotary pumps use cams that rotate in opposite directions to move water.

Question 18

Drinking-water microbiology: Which of the following water-borne diseases are caused by non-pathogenic bacteria?

Accepted Answer

Introduction / Context:Water-borne diseases result from ingestion of water contaminated with pathogenic organisms (disease-causing microbes). Understanding which agents are pathogenic is essential for public-health protection, disinfection design, and monitoring programs.Given Data / Assumptions:“Non-pathogenic bacteria” by definition do not cause disease.Cholera is caused by Vibrio cholerae (pathogenic bacteria).Typhoid fever is caused by Salmonella Typhi (pathogenic bacteria).Infectious hepatitis (commonly Hepatitis A or E) is caused by viruses, not non-pathogenic bacteria.Concept / Approach:Clarify terminology: pathogenic vs non-pathogenic organisms. Relate each listed disease to its causative agent. If any disease is caused by pathogens (bacteria or viruses), it cannot be attributed to non-pathogenic bacteria.Step-by-Step Solution:Identify cholera → bacterial and pathogenic → excludes non-pathogenic cause.Identify typhoid → bacterial and pathogenic → excludes non-pathogenic cause.Identify infectious hepatitis → viral pathogen, not bacterial.Therefore, none of the listed diseases is caused by non-pathogenic bacteria.Verification / Alternative check:Public-health references list Vibrio cholerae and Salmonella Typhi among notifiable pathogens; Hepatitis A/E are enteric viruses controlled by source protection and disinfection.Why Other Options Are Wrong:Cholera and typhoid are explicitly caused by pathogenic bacteria.Infectious hepatitis is viral, not bacterial.“All of these” contradicts definitions; “None of these” is consistent with causation.Common Pitfalls:Confusing “indicator bacteria” (e.g., coliforms) with pathogens; assuming all water-borne illnesses are bacterial; overlooking viruses and protozoa in control strategies.Final Answer:None of these.

Question 19

Intake siting for surface-water supply: Which combination of criteria should be satisfied when selecting the intake location?

Accepted Answer

Introduction / Context:Correct intake location substantially reduces raw-water variability, sediment load, and pollution risk, improving treatment reliability and cost. This question tests standard siting criteria for surface-water intakes.Given Data / Assumptions:An existing river/lake source with potential pollution and sediment transport.A downstream water-treatment plant (WTP) that benefits from shorter conveyance.Meandering river morphology yielding deeper concave banks.Concept / Approach:Good siting minimizes contamination and sedimentation. Proximity reduces head losses and pipeline costs. Upstream placement with respect to discharges avoids direct pollution plumes. On meandering rivers, concave banks experience deeper flow and more consistent depths, reducing clogging from shoals/bars common on convex banks.Step-by-Step Solution:Evaluate proximity: shorter raw-water mains → lower losses and O&M costs.Evaluate water quality: purer zone reduces treatment challenges.Evaluate hydraulics: upstream of effluent outfalls prevents intake of diluted wastewater.Evaluate morphology: concave bank is deeper and less prone to silting of bell mouth.Verification / Alternative check:Planning manuals recommend data collection on bathymetry, sediment transport, and outfall mapping prior to intake finalization; pilot screens and model studies may be used.Why Other Options Are Wrong:Each single criterion is necessary but not sufficient; the comprehensive, correct choice is to satisfy all listed criteria simultaneously where practicable.Common Pitfalls:Ignoring future upstream developments; underestimating sediment dynamics; placing intakes in stagnant coves leading to algae and taste-odor issues.Final Answer:All of the above.

Question 20

Pressure conduits and conveyance works: Which statement below is INCORRECT in general water-supply practice?

Accepted Answer

Introduction / Context:Water-conveyance systems include pressure pipes, tunnels, and aqueducts. Section shapes and hydraulic controls differ depending on whether the flow is pressurized or open-channel. This question distinguishes typical practice from an incorrect generalization.Given Data / Assumptions:Pressure pipes convey under head; alignment is not constrained by HGL grade.Tunnels may carry either open-channel or pressurized flow; circular shapes are common for pressure and structural reasons.Aqueducts in water-supply/irrigation typically operate as open channels (rectangular/trapezoidal), not circular.Concept / Approach:Separate pressure systems from open-channel carriers. Circular is excellent for pressure (isotropic hoop strength), but open-channel aqueducts use noncircular prismatic sections that optimize free-surface flow, construction, and maintenance.Step-by-Step Solution:Evaluate (a): In pressure pipelines, invert can rise or fall; HGL governs energy, not necessarily pipe slope → essentially correct.Evaluate (b): Pipes may crest above HGL (siphon routing), provided vapor pressure constraints are respected → acceptable in principle.Evaluate (c): Tunnels may be circular, but aqueducts (open channel) are commonly trapezoidal/rectangular; calling both “generally circular” is incorrect.Verification / Alternative check:Design manuals illustrate aqueduct flumes as rectangular/trapezoidal concrete channels; pressure tunnels and steel liners are often circular for structural efficiency.Why Other Options Are Wrong:Options (a) and (b) reflect accepted practice; “None/All” are distractors because (c) is demonstrably inaccurate as a blanket statement.Common Pitfalls:Confusing pressure conduits with open channels; assuming circular sections are universal regardless of flow regime.Final Answer:Aqueducts (open-channel carriers) and tunnels are generally kept circular in section.

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