Question 1

Raw-water intake design: The entry opening is screened to exclude debris, and its sill elevation is placed at what typical level relative to the source bed to minimise sediment intake while ensuring submergence?

Accepted Answer

Introduction / Context:Intake structures must limit debris and sediment ingress while maintaining reliable submergence. Correct vertical placement of the intake opening reduces sediment load and avoids drawing surface scum.Given Data / Assumptions:Screened intake in a river/lake/reservoir.Typical engineering rule-of-thumb clearances from bed and surface.Goal is to minimise sediment uptake and floating debris.Concept / Approach:Sediment concentrations are highest close to the bed; floating debris accumulates near the surface. Placing the intake several metres above the bed, commonly around 2.5 m (subject to local hydraulics), strikes a balance between submergence and sediment avoidance.Step-by-Step Solution:Identify the undesirable zones: near-bed (sediment) and near-surface (scum).Adopt a typical clearance: about 2.5 m above bed for many designs.Hence select option (c).Verification / Alternative check:Design guides present intake sill elevations based on depth, velocity, and sediment regimes, often converging on a few metres above the bed unless special conditions apply.Why Other Options Are Wrong:(a) At water surface would entrain floating matter and air.(b) At bed maximises sediment entry.(d) and (e) ignore hydraulics and sediment transport principles.Common Pitfalls:Ignoring seasonal depth variations and setting sill too close to bed.Underestimating the need for multiple-level offtakes in reservoirs.Final Answer:Approximately 2.5 m above the bed of the source.

Question 2

Plain chlorination in water treatment In public health engineering, “plain chlorination” refers to dosing disinfectant without prior coagulation–sedimentation or filtration. It is typically adopted for which types of supplies/situations?

Accepted Answer

Introduction / Context:Plain chlorination is a minimal-treatment strategy used in water supply engineering when raw water is already quite clear or when time and infrastructure are limited. It focuses solely on microbial safety by adding chlorine without the preceding clarification steps of coagulation, flocculation, and sedimentation.Given Data / Assumptions:Turbidity threshold for effective primary disinfection is relatively low.Clear natural sources and temporary/emergency arrangements often rely on rapid solutions.Goal is pathogen inactivation, not removal of colour/turbidity/organics.Concept / Approach:Chlorine is highly effective against bacteria and many viruses, provided adequate contact time and a reasonable chlorine demand. High turbidity shields microorganisms, increases chlorine demand, and risks by-product formation; thus plain chlorination is best for low-turbidity sources or urgent deployments where maintaining a residual is the priority.Step-by-Step Solution:Identify appropriate contexts: low turbidity sources and time-critical deployments.Match to options: clear lakes/springs, emergency supplies, and military field supplies match typical use.Therefore the most inclusive correct choice is “All the above.”Verification / Alternative check:Standard texts recommend turbidity generally below about 20 NTU (historical mg/L as silica convention) for reliable disinfection by chlorination alone; otherwise prior clarification is advised.Why Other Options Are Wrong:The individual options A–D are not wrong; they are each subsets of the correct comprehensive scenario, hence E is the correct selection.Common Pitfalls:Assuming chlorination alone will handle high turbidity or organic colour; neglecting contact time and residual monitoring; overlooking taste/odour issues in highly coloured waters.Final Answer:All the above

Question 3

Pressure capacity of PVC (UPVC) water mains Approximate maximum hydrostatic pressure head that standard PVC pressure pipes can withstand under normal water-supply service conditions is closest to:

Accepted Answer

Introduction / Context:PVC (UPVC) pressure pipes are widely used in distribution networks due to corrosion resistance, hydraulically smooth bore, and convenient jointing. Designers often convert pressure ratings expressed in bar to an equivalent water head for quick checks in field problems.Given Data / Assumptions:1 bar is approximately equal to 10 m of water head.Common PVC pressure classes include ratings up to about 10 bar in many standard municipal uses (with higher classes also available, but 10 bar is a typical design figure in textbook problems).Working conditions are steady with appropriate safety factors and temperature allowances.Concept / Approach:To translate pressure class to head: head (m) ≈ 10 * pressure (bar). Thus a 10-bar pipe corresponds to about 100 m water head. While higher-class pipes exist, the conventional “rule-of-thumb” ceiling often cited for standard mains in didactic questions is around 10 bar (≈ 100 m of head).Step-by-Step Solution:Relate bar to head: 1 bar ≈ 10 m water.Assume common class: 10 bar → ≈ 100 m.Select the nearest option: 100 m of water head.Verification / Alternative check:Manufacturer catalogues and codes list multiple classes; for general exam purposes, 10 bar (≈ 100 m) is a standard benchmark.Why Other Options Are Wrong:25–75 m are below typical higher service classes; 125 m exceeds the common 10-bar textbook figure unless specially specified.Common Pitfalls:Ignoring derating for temperature and surge; selecting pipe class without checking transient pressures (water hammer) and test pressures.Final Answer:100 m of water head

Question 4

Terminology for channel structures on trestles Open channels carried above ground on a series of trestles are generally referred to as:

Accepted Answer

Introduction / Context:Conveyance structures that carry water across depressions, valleys, or obstacles may be supported above ground. Different names are used depending on whether the section is open channel or closed conduit and on how it is structurally supported.Given Data / Assumptions:Structure is an open channel element.It is supported on trestles (bents or frames).It spans over ground rather than being buried.Concept / Approach:A flume is an elevated open channel, commonly of wood, metal, or concrete, supported by trestles or bents. An aqueduct is often a larger bridge-like structure conveying water (sometimes masonry arches), but in hydrotechnical parlance “flume” specifically signals an open channel on supports. A siphon is a closed conduit that flows under pressure.Step-by-Step Solution:Identify that the cross-section remains open to atmosphere.Recognize trestle support → typical of flume construction.Choose the term “Flumes.”Verification / Alternative check:Design manuals and hydraulic engineering texts consistently define flumes as open channels supported above ground, frequently on trestles.Why Other Options Are Wrong:Raised canal is generic and not standard terminology; aqueducts are broader bridge structures and not necessarily on trestles; siphons are pressurized closed conduits; “conduits” are enclosed and non-specific.Common Pitfalls:Using “aqueduct” for all elevated conveyances even when the element is a lightweight trestle-supported open channel.Final Answer:Flumes

Question 5

Allowable non-verticality of deep well bores For a tube well of 200 m depth, what is the typical maximum permissible horizontal deviation from vertical (non-verticality) at the bottom?

Accepted Answer

Introduction / Context:Verticality of well bores matters for proper screen placement, pump alignment, and longevity. Standards specify allowable deviation as a ratio (e.g., 1 in N) or as an equivalent lateral displacement at the bottom for a given depth.Given Data / Assumptions:Well depth = 200 m.Typical verticality criterion is of the order of 1 in a few hundred (illustrative for design questions).We must choose from given discrete lateral deviations.Concept / Approach:If a commonly cited tolerance is about 1 in 200, the maximum lateral displacement allowed at the bottom equals depth / 200. For 200 m depth: displacement = 200 / 200 = 1 m = 100 cm. This value aligns practical drilling tolerances while safeguarding pump operations and screen integrity.Step-by-Step Solution:Assume allowable deviation ratio ≈ 1:200.Compute displacement: 200 m / 200 = 1 m.Convert to centimetres: 1 m = 100 cm → choose 100 cm.Verification / Alternative check:Many specifications express verticality in percent or per-centimetre deviation per metre; for illustrative problems, 1:200 is a standard proxy.Why Other Options Are Wrong:25–75 cm are stricter than the typical illustrative limit; 150 cm would exceed the reasonable tolerance and risk equipment misalignment.Common Pitfalls:Confusing verticality (lateral offset) with bore drift angle; ignoring that small angular errors accumulate significantly over deep wells.Final Answer:100 cm

Question 6

Reservoir sedimentation concepts Which combination of statements about trap efficiency, capacity–inflow ratio, and siltation tendency is correct?

Accepted Answer

Introduction / Context:Understanding how reservoirs trap sediment is crucial for estimating useful life and planning desilting strategies. Two key measures are trap efficiency and the capacity–inflow ratio, which together indicate how effectively a reservoir captures incoming sediment.Given Data / Assumptions:Sediment inflow varies with watershed characteristics.Trap efficiency generally increases with capacity–inflow ratio.We compare relative tendencies, not precise percentages.Concept / Approach:Trap efficiency = (sediment retained) / (total sediment inflow). Capacity–inflow ratio = (reservoir capacity) / (annual inflow). A small reservoir on a large river has a small capacity–inflow ratio, hence lower trap efficiency and less percentage of inflow trapped (appearing to “silt less” per unit inflow). Conversely, a large reservoir on a small river has a higher capacity–inflow ratio and traps a larger fraction, appearing to “silt more” relatively.Step-by-Step Solution:Confirm definitions in A and B: both standard.Relate capacity–inflow ratio to trapping: higher ratio → higher trap efficiency.Evaluate C and D accordingly; both are consistent with the relationship.Verification / Alternative check:Empirical curves (e.g., Brune’s) show trap efficiency rising with capacity–inflow ratio, supporting statements C and D.Why Other Options Are Wrong:Since A–D are all correct, selecting any single option alone would be incomplete; “All the above” captures the full, correct set.Common Pitfalls:Confusing absolute silt volume with percentage trapped; a large river can still deliver enormous absolute silt volumes even if the fraction trapped is low.Final Answer:All the above

Question 7

Evaporation reduction from reservoirs Which chemical is used to form a monomolecular surface film to minimize evaporation losses from reservoirs?

Accepted Answer

Introduction / Context:In arid and semi-arid regions, open-water evaporation from reservoirs is a significant loss. One mitigation technique is spreading a monomolecular film of long-chain fatty alcohols on the surface to reduce evaporation by lowering surface tension and suppressing capillary waves.Given Data / Assumptions:Objective is evaporation reduction, not disinfection or pH modification.Candidates include chemicals with low volatility and ability to spread as a thin film.Concept / Approach:Cetyl alcohol (hexadecanol) and stearyl alcohol (octadecanol) are classic film-forming agents used to reduce evaporation. They are hydrophobic, spread spontaneously into a monolayer, and are effective under calm to moderate conditions. Acids, methane, or simple alcohol spirits are unsuitable for sustained monolayer film formation.Step-by-Step Solution:Recognize the monomolecular film method for evaporation control.Identify long-chain fatty alcohols as the appropriate chemicals.Select cetyl alcohol (hexadecanol) as the correct choice.Verification / Alternative check:Engineering literature documents successful trials using cetyl or stearyl alcohol, sometimes in blends, to reduce evaporation rates.Why Other Options Are Wrong:Industrial spirit evaporates quickly and does not form durable films; hydrochloric acid and methane are unsafe/irrelevant; “None of these” is incorrect because an effective agent (cetyl alcohol) exists.Common Pitfalls:Confusing “acetyl” with “cetyl”: the effective film former is cetyl alcohol (C16H34O), not acetyl compounds.Final Answer:Cetyl alcohol (hexadecanol)

Question 8

Sources of alkalinity in water chemistry Alkalinity in a water sample may arise from which of the following constituents?

Accepted Answer

Introduction / Context:Alkalinity is a measure of a water’s capacity to neutralize acid. It is a key parameter for coagulation control, corrosion tendency, and biological stability in distribution systems.Given Data / Assumptions:In natural and treated waters, alkalinity typically results from bicarbonates, carbonates, and hydroxides.Metal cations (Ca, Mg, Na, K) pair with these anions to form the salts present.Concept / Approach:Bicarbonate alkalinity (HCO3−) commonly arises from Ca and Mg bicarbonates. Carbonate alkalinity (CO3^2−) may be present as sodium or potassium carbonates. Hydroxide alkalinity (OH−) can appear when lime is added or pH is adjusted upward. The total alkalinity equals the sum of these contributions expressed as CaCO3 equivalent.Step-by-Step Solution:Link each listed compound class to alkalinity type: HCO3−, CO3^2−, OH−.Recognize their combined contribution to titratable alkalinity.Thus, the inclusive choice “All the above” is correct.Verification / Alternative check:Standard methods (acid titration to pH endpoints) apportion alkalinity among bicarbonate, carbonate, and hydroxide species.Why Other Options Are Wrong:Each individual option is only a subset; full characterization requires acknowledging all contributing species.Common Pitfalls:Assuming alkalinity equals hardness: magnesium/calcium bicarbonates contribute to both, but sodium/potassium carbonates add alkalinity without hardness.Final Answer:All the above

Question 9

Spacing of tube wells to limit regional drawdown For minimal cumulative impact on the groundwater table in an aquifer, a common planning thumb-rule is to space tube wells approximately one well per:

Accepted Answer

Introduction / Context:Well spacing affects interference (overlap of drawdown cones) and long-term aquifer response. Over-densification causes excessive regional drawdown, higher pumping lifts, and declining yields.Given Data / Assumptions:Homogeneous unconfined or semi-confined aquifer for a rule-of-thumb estimate.Moderate pumping rates typical of municipal/agricultural wells.Planning guidance rather than an exact analytical design.Concept / Approach:Interference theory shows that superposition of cones of depression increases drawdown when wells are too close. Many textbook rules adopt a nominal spacing near 1 well per square kilometre to reduce interference at community-scale abstractions, pending detailed aquifer testing and modeling.Step-by-Step Solution:Recognize that the problem asks for a practical planning density, not a computed value from aquifer parameters.Among the discrete choices, 1.0 km² per well is the widely cited “safe” preliminary figure.Select 1.0 km² as the answer, subject to refinement by pumping tests.Verification / Alternative check:Field designs refine spacing via transmissivity, storativity, and permissible drawdown; preliminary spacing near 1 km² is commonly used where data are limited.Why Other Options Are Wrong:Closer densities (0.5–0.75 km²) increase interference; very sparse networks (≥1.25 km²) may be conservative but not a “least effect” standard rule.Common Pitfalls:Ignoring well-field layout (line vs staggered grid) and boundary effects; neglecting seasonal recharge variability.Final Answer:1.0 km²

Question 10

Basic water-quality treatment facts Which set of statements about boiling, lime softening, pH effects, and bacterial survival is collectively correct?

Accepted Answer

Introduction / Context:Basic water-quality control employs physical, chemical, and biological mechanisms. Disinfection, softening, and pH adjustment are routine interventions that influence microbial survival and scaling/corrosion tendencies.Given Data / Assumptions:Boiling pertains to point-of-use emergency treatment.Lime softening refers to chemical precipitation that removes carbonate hardness and, with soda ash, part of the non-carbonate hardness.Excess lime raises pH and shifts carbonate equilibrium.Many waterborne bacteria are sensitive to high pH.Concept / Approach:Boiling achieves thermal inactivation of vegetative bacteria and many pathogens. Lime (CaO/Ca(OH)2) addition precipitates calcium as CaCO3 and magnesium as Mg(OH)2 under appropriate conditions, thus reducing hardness. Overdosing lime increases pH significantly. Elevated pH (above ~9.5) is hostile to many bacteria, complementing disinfection processes but not replacing them entirely.Step-by-Step Solution:Assess each statement against standard practice: all four are correct.Therefore, the collective correct choice is “All the above.”Verification / Alternative check:Jar tests and plant operations confirm pH rise with lime and the softening reactions; public health guidance recognizes boiling as effective emergency disinfection.Why Other Options Are Wrong:Choosing any single statement ignores the fact that all listed statements are accurate in their scope.Common Pitfalls:Assuming lime alone removes all permanent hardness (soda ash is often needed), and assuming high pH alone ensures complete pathogen safety without contact time or residual disinfectant.Final Answer:All the above

Question 11

Rapid sand filter design – under-drain laterals: For the lateral pipes in a rapid sand filter under-drain system, the ratio of lateral length to its diameter (L/D) should not be greater than which limiting value to avoid excessive headloss and maldi…

Accepted Answer

Introduction / Context:Under-drain systems in rapid sand filters distribute backwash water and collect filtered water. An appropriate geometry of laterals (length and diameter) ensures uniform velocities, acceptable headloss, and even cleaning of the bed. A practical limit on the length-to-diameter (L/D) ratio of laterals is used in design handbooks to avoid maldistribution.Given Data / Assumptions:Conventional rapid sand filter with a manifold and perforated laterals.Backwash and filtration both pass through the under-drain.Headloss uniformity and velocity limits control sizing.Concept / Approach:As L/D increases, friction losses along the lateral cause flow to decline toward the closed end, creating uneven distribution. Design uses a maximum L/D to keep headloss variation small. A commonly taught conservative limit for many exam problems is L/D about 20 for laterals, with additional checks on orifice loss versus pipe friction loss to maintain a desired ratio (often orifice loss several times pipe friction loss).Step-by-Step Solution:Set design objective: maintain near-uniform discharge along lateral.Recognize that excessive L/D raises friction loss, reducing uniformity.Adopt limiting L/D value used in standard practice problems.Select the closest upper bound from choices: 20.Verification / Alternative check:Checking orifice headloss to pipe loss ratio and computing head distribution along the lateral with L/D = 20 typically yields acceptable uniformity (e.g., within ±10%).Why Other Options Are Wrong:10, 15: Overly restrictive; not necessary for uniformity in typical designs.25: Higher L/D risks noticeable maldistribution and larger end loss.Common Pitfalls:Ignoring that under-drain design must satisfy both filtration and backwash modes.Letting lateral spacing and perforation layout substitute for proper L/D control.Final Answer:20

Question 12

Definition check in water-supply engineering: Per-capita water demand is calculated in litres on which time basis for each person?

Accepted Answer

Introduction / Context:Per-capita demand (often noted as LPCD) is a foundational planning parameter for source sizing, treatment capacity, and distribution design in municipal water-supply engineering.Given Data / Assumptions:Per-capita demand refers to average consumption attributed to one person.Used in feasibility studies, DPRs, and design standards.Concept / Approach:By definition, LPCD stands for litres per capita per day. It captures domestic and sometimes a proportion of non-domestic uses depending on the adopted standard, and it is averaged on a daily basis for planning and comparison across systems.Step-by-Step Solution:Recognize the unit acronym: LPCD.L = litres, P = per-capita, D = per day.Therefore, it is litres per person per day.Verification / Alternative check:Most national and international manuals, including urban design guidelines, define demand targets explicitly in LPCD terms (e.g., 135 LPCD for domestic consumption under typical Indian urban conditions with sewerage).Why Other Options Are Wrong:Per month / per year: not standard; lacks the daily operational granularity used for design.None of these: incorrect because a standard definition exists.Common Pitfalls:Confusing daily design demand with hourly peak factors; peak factors are applied to LPCD-based averages.Final Answer:per person per day

Question 13

Potable water quality – hardness limit: What is the maximum permissible total hardness (as CaCO3) commonly adopted for public water supplies?

Accepted Answer

Introduction / Context:Hardness in potable water arises mainly from calcium and magnesium salts. While not a direct health hazard, it affects soap consumption, scaling in pipes and heaters, and consumer acceptability. Engineers use guideline limits to judge the need for softening.Given Data / Assumptions:Total hardness is expressed as mg/litre as CaCO3.Question targets the upper permissible value commonly cited in exam standards.Concept / Approach:Typical textbook/exam practice accepts up to around 150 mg/litre as CaCO3 as a permissible level for public supplies without mandatory softening, although “desirable” limits may be lower. Above this, hardness-related issues become more evident and treatment may be considered depending on context.Step-by-Step Solution:Identify the parameter: total hardness as CaCO3.Recall the accepted public-supply threshold often taught: approximately 150 mg/litre.Select the matching option: 150 mg/litre.Verification / Alternative check:Field practice tolerates moderate hardness in many systems; customer complaints and appliance scaling increase for higher values, prompting softening or blending strategies.Why Other Options Are Wrong:95–125 mg/litre: represent lower thresholds; water at these levels is typically acceptable without softening.Common Pitfalls:Confusing “desirable” versus “permissible” limits; the latter allows a higher ceiling before mandatory action.Final Answer:150 mg/litre

Question 14

Process placement in conventional treatment: A rapid sand filter should be placed after which pretreatment steps in a conventional water-treatment train?

Accepted Answer

Introduction / Context:Conventional treatment for surface waters typically follows the sequence: coagulation, flocculation, sedimentation, filtration, and disinfection. Rapid sand filters are designed to remove the fine turbidity remaining after clarification.Given Data / Assumptions:Turbid surface water requiring full conventional treatment.Granular media filtration stage is rapid sand filtration.Concept / Approach:Coagulation and flocculation convert colloids into settleable flocs. Sedimentation removes most suspended solids. The rapid sand filter polishes the clarified water by removing remaining fine particles. Disinfection is a separate, final step; it is not “combined” inside the rapid sand filter unit.Step-by-Step Solution:Establish train: coagulation → flocculation → sedimentation → rapid sand filtration → disinfection.Therefore, rapid sand filtration should be preceded by coagulation and sedimentation.Verification / Alternative check:Design manuals specify maximum allowable turbidity to filters to achieve target effluent turbidities and acceptable filter runs, confirming the need for prior clarification.Why Other Options Are Wrong:“Uses rapid sand as filter media”: wording is imprecise; media is graded silica sand (often with anthracite/GS support), not “rapid sand.”“Used after slow sand filtering”: slow sand is an alternative technology, not a pretreatment to rapid sand.“Can combine disinfection also”: disinfection is a separate process.Common Pitfalls:Feeding high turbidity directly to filters causes short runs and breakthrough.Final Answer:should be preceeded by coagulation and sedimentation

Question 15

Metal pipe materials – properties and usage: Which set of statements about copper and wrought-iron (galvanised) pipes is correct in water-supply practice?

Accepted Answer

Introduction / Context:Choosing appropriate pipe material requires balancing mechanical strength, corrosion resistance, weight, and durability. Copper and wrought-iron (galvanised) pipes have long histories in building plumbing and distribution networks.Given Data / Assumptions:Statements address relative properties compared to cast iron and common nomenclature.General potable-water environments (not highly aggressive industrial fluids).Concept / Approach:Copper exhibits good general corrosion resistance in typical potable water ranges and is widely used in building plumbing. Wrought iron is lighter than cast iron but, if unprotected, corrodes faster and is less durable; galvanising (zinc coating) produces galvanised iron (G.I.) pipes with improved corrosion resistance.Step-by-Step Solution:Compare densities and wall thickness: wrought iron pipes are lighter than cast iron.Assess corrosion: unprotected wrought iron corrodes relatively quickly.Define G.I.: zinc-coated wrought-iron/steel pipes are termed galvanised iron pipes.Copper’s general resistance in potable service is good (within typical pH ranges).Verification / Alternative check:Building codes and material standards document widespread use of copper and G.I. with the above general properties and limitations (e.g., copper not recommended for highly acidic/alkaline waters without controls).Why Other Options Are Wrong:Each separate statement is correct in general potable contexts; the most complete answer is “All the above.”Common Pitfalls:Interpreting “highly resistant” as universally inert; extreme pH or specific ions can still attack copper.Final Answer:All the above

Question 16

Breakpoint chlorination – residual behaviour: At the breakpoint in a chlorination curve for treated water, which statement best describes the observed residual?

Accepted Answer

Introduction / Context:Breakpoint chlorination is the process of adding chlorine until the demand from reducing agents, ammonia, and chloramine formation is satisfied, after which a free chlorine residual appears and increases with further dosage.Given Data / Assumptions:Typical post-filtration chlorination of potable water.Presence of ammonia/organics initially consumes added chlorine.Concept / Approach:As dosage rises, chlorine first reacts with reducing substances and forms chloramines; residuals may remain low or non-free. Near the breakpoint, chloramines are destroyed and nitrogen is released. Beyond this point, additional chlorine produces a measurable free-chlorine residual that “reappears” and then increases with dose.Step-by-Step Solution:Add chlorine incrementally; observe residual pattern.Approaching breakpoint: combined residual drops as chloramines are oxidized.Past breakpoint: free residual reappears and increases with dose.Verification / Alternative check:Chlorine residual curves consistently show the characteristic dip at breakpoint followed by a rising free-chlorine residual region, confirming the “reappearance.”Why Other Options Are Wrong:“Residual chloride is zero/maximum”: chloride ion is not the measured disinfectant residual; the focus is free/combined chlorine.“Chlorine is used to oxidise” is true in general but does not describe the breakpoint residual behaviour.Common Pitfalls:Confusing chloride (Cl−) with chlorine residual (Cl2 as free/combined forms).Final Answer:residual chlorine reappears

Question 17

Measuring low turbidity in finished water: Which instrument is best suited for detecting low turbidity levels (nephelometric range) in potable water?

Accepted Answer

Introduction / Context:Finished water quality targets very low turbidity to ensure effective disinfection and aesthetic acceptability. Instruments differ in measurement principle and sensitivity range.Given Data / Assumptions:Low turbidity typical of post-filtration potable water.Need for high sensitivity and repeatability.Concept / Approach:The Jackson candle method is suitable for higher turbidities and is not sensitive at the low end. Turbidity tubes are crude field devices. Nephelometric (Hellige-type) turbidimeters measure light scattered at an angle and are appropriate for low turbidity monitoring in drinking water treatment.Step-by-Step Solution:Eliminate Jackson/tube for low ranges.Select nephelometric instrument designed for low-NTU accuracy: Hellige turbidimeter.Verification / Alternative check:Standards for drinking water turbidity compliance are typically based on nephelometric turbidity units (NTU), measured by nephelometers (e.g., Hellige/modern equivalents).Why Other Options Are Wrong:Turbidity tube and Jackson meter lack required sensitivity for very low turbidity.Baylis is not the common standard for low-level compliance monitoring.Common Pitfalls:Using candle method results for regulatory reporting where nephelometric methods are required.Final Answer:Hellige (nephelometric) turbidimeter

Question 18

Pump sizing – minimum horsepower: A well discharges 3.25 × 10^-2 m^3/s and must be lifted against a total head of 30 m. What is the minimum required pump horsepower (assume practical overall efficiency so the installed power meets duty)?

Accepted Answer

Introduction / Context:Selecting a pump–motor set requires estimating hydraulic power and adjusting for realistic efficiency. Oversights here can result in underpowered installations, poor service pressures, and shortened equipment life.Given Data / Assumptions:Discharge Q = 3.25 × 10^-2 m^3/s.Total dynamic head H = 30 m.Water density ρ ≈ 1000 kg/m^3; g ≈ 9.81 m/s^2.Overall (pump + motor + transmission) efficiency taken as a practical value to estimate minimum installed H.P.Concept / Approach:Hydraulic power Ph = ρ * g * Q * H. Brake power Pb = Ph / η_overall. Convert watts to horsepower by dividing by 746.Step-by-Step Solution:Compute Ph: Ph = 1000 * 9.81 * 0.0325 * 30.Numerical value: 9.81 * 0.0325 = 0.318825; 0.318825 * 1000 * 30 ≈ 9564.75 W.Hydraulic horsepower ≈ 9564.75 / 746 ≈ 12.8 H.P.Assume realistic η_overall ≈ 0.6–0.7 for minimum installed rating.At 0.6 efficiency: required ≈ 12.8 / 0.6 ≈ 21.3 H.P.; nearest standard option ≈ 20 H.P.Verification / Alternative check:If η_overall were 0.7, Pb ≈ 18.3 H.P.; selecting 20 H.P. still provides margin for wear, voltage variation, and minor head underestimation.Why Other Options Are Wrong:10 or 15 H.P.: insufficient once efficiency and safety margin are included.18 H.P.: borderline if η is lower or headlosses are slightly higher.25 H.P.: feasible but not the minimum required per given duty.Common Pitfalls:Forgetting to include overall efficiency; choosing by hydraulic power alone underestimates motor rating.Ignoring extra minor losses and future performance drift.Final Answer:20 H.P.

Question 19

Limits on suction lift: The achievable suction lift of a centrifugal pump depends primarily on which factors?

Accepted Answer

Introduction / Context:The maximum suction lift of pumps is limited by available Net Positive Suction Head (NPSH), which is a function of ambient conditions and hydraulic losses. Misjudging these factors leads to cavitation and performance loss.Given Data / Assumptions:Pump drawing water from a sump via a suction line.Standard atmospheric conditions unless otherwise stated.Concept / Approach:NPSH available = (atmospheric head − vapour pressure head) − static suction lift − entrance/friction losses − velocity head at pump eye. Higher atmospheric pressure increases available suction head; higher water temperature raises vapour pressure and reduces margin; higher velocities and friction losses reduce available head at the impeller eye.Step-by-Step Solution:Account for atmospheric pressure: sets absolute ceiling on suction head.Account for water temperature: vapour pressure increases with temperature, lowering allowable lift.Account for velocity and friction: losses in suction line reduce pressure at pump inlet.Conclusion: all listed factors govern suction lift.Verification / Alternative check:Applying the NPSH formula confirms changes in any of these variables alter cavitation margin and thus safe suction lift.Why Other Options Are Wrong:Considering only one factor leads to incorrect prediction of permissible suction conditions; the phenomenon is multi-factorial.Common Pitfalls:Placing pumps far above water level and ignoring summer temperature effects.Using undersized suction lines causing high entrance/pipe losses.Final Answer:all the above

Question 20

Biochemical stabilization timeline: Approximately how long does it take for complete stabilization (ultimate BOD exertion) of organic matter in polluted water at standard conditions used in design problems?

Accepted Answer

Introduction / Context:Biochemical Oxygen Demand (BOD) reflects the oxygen required by microorganisms to oxidize biodegradable organic matter. While the standard BOD test is 5 days at 20°C (BOD5), full biochemical stabilization (ultimate BOD) takes longer.Given Data / Assumptions:Standard wastewater engineering convention at 20°C.Ultimate BOD approaches asymptote over several weeks.Concept / Approach:BOD exertion follows first-order kinetics: BOD(t) = L0 * (1 − e^(−k*t)), where L0 is ultimate BOD and k is the deoxygenation rate constant. At 5 days, only a fraction of L0 is exerted; practical “complete stabilization” for design approximations is taken near 20 days for typical domestic wastewater at 20°C.Step-by-Step Solution:Recognize BOD5 is a standard test period, not full stabilization.Use the conventional design approximation: ultimate exertion within roughly 20 days at 20°C.Select the corresponding option.Verification / Alternative check:Plotting BOD kinetics with a representative k (say 0.23 day^-1 at 20°C) shows BOD exerted by 20 days is very close to L0, supporting the rule-of-thumb period.Why Other Options Are Wrong:5 or 10 days: too short for ultimate BOD; used for standard sampling and reporting.30 or 40 days: beyond the customary engineering reference; offers little additional design insight.Common Pitfalls:Equating BOD5 with complete stabilization—this underestimates oxygen demand for long retention processes.Final Answer:20 days

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