Question 1

Basic water chemistry: Water having pH = 6 at ambient conditions is classified as

Accepted Answer

Introduction / Context:The pH scale quantifies acidity/alkalinity of water on a logarithmic basis. Understanding pH classifications is essential for corrosion control, coagulation, disinfection efficacy, and consumer acceptability in water-supply engineering.Given Data / Assumptions:Neutral pH at 25°C is approximately 7.0.Values below 7 are acidic; values above 7 are alkaline (basic).Measured pH is 6.Concept / Approach:Classify pH numerically relative to neutrality. Consider implications for pipe materials and treatment chemistry (e.g., alkalinity adjustment, corrosion indices).Step-by-Step Solution:Compare measured pH (6) with neutral value (7).Since 6 Acidic water may require pH adjustment/alkalinity addition to reduce corrosion potential.Verification / Alternative check:Logarithmic nature: each pH unit represents a tenfold change in hydrogen-ion activity, so pH 6 is ten times more acidic than pH 7 (in terms of [H+]).Why Other Options Are Wrong:Alkaline and neutral categories contradict the numeric definition; “none of these” is unnecessary because “acidic” is exact.Common Pitfalls:Ignoring temperature effects on neutral pH (slight shift from 7.0), which does not change classification here.Final Answer:acidic

Question 2

Groundwater hydraulics: For discharge of a confined (artesian) tube well under steady conditions, which classical formula is used?

Accepted Answer

Introduction / Context:Estimating the discharge of tube wells requires groundwater flow theory. For steady, radial flow to a well in a confined aquifer, the classical Thiem equation relates drawdown and discharge. Recognizing which expression applies to which aquifer type is a core hydrogeology skill.Given Data / Assumptions:Confined aquifer of thickness b with transmissivity T = K * b.Steady-state pumping with two observation wells at radii r1 and r2 giving piezometric heads h1 and h2.Radial flow, Dupuit–Thiem assumptions (horizontal flow in aquifer, negligible vertical gradients).Concept / Approach:For confined conditions, Thiem’s formula for discharge is commonly written as Q = 2 * π * K * b * (h1 − h2) / ln(r2 / r1) = 2 * π * T * (Δh) / ln(r2 / r1). It stems from integrating Darcy’s law in cylindrical coordinates under steady conditions.Step-by-Step Solution:Apply radial Darcy flow: q_r = −K * (dh/dr).Conservation for steady flow: Q = 2 * π * r * b * q_r.Integrate between r1, r2 and h1, h2 to obtain Thiem’s expression for Q.Verification / Alternative check:Plot drawdown versus ln(r); the slope gives Q / (2 * π * T), a standard field method for confined aquifers.Why Other Options Are Wrong:Darcy’s law is the basis but not, by itself, the discharge formula; Tolman is not the standard reference here; Dupuit’s unconfined form uses heads squared (h^2) rather than linear heads.Common Pitfalls:Mixing unconfined and confined forms; using base-10 logarithms without converting constants; ignoring well losses and partial penetration effects.Final Answer:Thiem’s formula

Question 3

Pipe-flow formulae (SI units): Which expression correctly represents the Hazen–Williams velocity relation for full pipes?

Accepted Answer

Introduction / Context:The Hazen–Williams equation is an empirical relation widely used for turbulent water flow in full pipes. In SI form, it provides average velocity or discharge in terms of hydraulic radius and slope, using a material coefficient C.Given Data / Assumptions:Water at ordinary temperatures; full circular pipe flow.Hydraulic radius R and energy-grade-line slope S are known.Coefficient C reflects internal roughness (dimensionless in SI usage).Concept / Approach:In SI units, a common form is V = 0.85 * C * R^0.63 * S^0.54, with V in m/s, R in m, and S dimensionless. The constant 0.85 reconciles unit systems from the original U.S. customary form. The equation is empirical and valid for water (not other fluids) in typical municipal ranges.Step-by-Step Solution:Select the expression containing R^0.63 and S^0.54.Ensure no extraneous parameters (e.g., “H”) appear.Choose the SI constant 0.85 (not 1.85, which belongs to the headloss exponent in another form).Verification / Alternative check:Alternative discharge form: Q = 0.278 * C * A * R^0.63 * S^0.54, where A is area in m^2 and Q in m^3/s; both are consistent if V = Q/A.Why Other Options Are Wrong:Option b introduces an undefined “H”; option c changes the exponents; option e misplaces the constant as 1.85; “None” is invalid because a correct expression exists.Common Pitfalls:Mixing SI and US constants; applying Hazen–Williams to fluids other than water or outside recommended Reynolds number ranges.Final Answer:V = 0.85 C R^0.63 S^0.54

Question 4

Perched aquifers in groundwater hydrology are typically located:

Accepted Answer

Introduction / Context:A perched aquifer is a localized saturated zone separated from the regional (main) water table by an unsaturated interval. It forms when infiltrating water accumulates atop a relatively impermeable lens within the vadose zone.Given Data / Assumptions:Heterogeneous stratigraphy including discontinuous clay or silt lenses.Infiltration from precipitation or irrigation.Regional water table exists at greater depth.Concept / Approach:The defining characteristic is a perched saturated zone above the main phreatic surface due to a low-permeability layer that impedes vertical percolation locally. Thus, perched aquifers are “above” the main water table, not coincident with it.Step-by-Step Solution:Recognize the need for an impervious/intervening lens to hold water.Place perched saturation above the regional water table within the vadose zone.Conclude correct location as below ground surface but above the main water table.Verification / Alternative check:Observation wells may show shallow saturation in local zones while nearby deeper wells reflect a lower regional water table; perched water disappears laterally where the lens pinches out.Why Other Options Are Wrong:At the surface implies surface water; below the regional water table describes the phreatic zone; “everywhere” disregards the local nature; “none” is invalid given the correct description.Common Pitfalls:Confusing perched aquifers with seasonal water tables; assuming perched conditions extend basin-wide.Final Answer:Below the ground surface but above the regional water table, resting on a local impervious lens.

Question 5

Sources delivering filtered or naturally filtered water to consumers can include which of the following?

Accepted Answer

Introduction / Context:Riverbank filtration and riverbed infiltration are natural pre-treatment methods. Properly designed galleries and radial collector wells can provide relatively low-turbidity water with moderated temperature and reduced pathogen loads before additional treatment.Given Data / Assumptions:Alluvial river setting with permeable sand/gravel deposits.Hydraulic connectivity between river and adjacent aquifer materials.Proper setback to avoid direct surface intrusion during floods.Concept / Approach:Infiltration galleries (horizontal drains) and infiltration wells tap bank-side groundwater induced by river stage gradients, providing natural filtration through alluvium. Ranney wells extend multiple lateral collectors into the aquifer, greatly increasing intake area and capacity.Step-by-Step Solution:Recognize common design variants: galleries, vertical infiltration wells, and radial collectors.All rely on bank/bed filtration processes that improve raw-water quality.Therefore, each listed option is a valid consumer supply source (with appropriate treatment).Verification / Alternative check:Performance data show reductions in turbidity and microbes downstream of bank filtration; subsequent disinfection is still required for safety.Why Other Options Are Wrong:“None of these” contradicts established practice; “All of the above” correctly aggregates the viable methods.Common Pitfalls:Underestimating pathogen breakthrough during floods; neglecting clogging (schmutzdecke) management at the riverbed interface; ignoring wellhead protection.Final Answer:All of the above.

Question 6

A strainer-type well that penetrates three pervious aquifer layers separated by three impervious aquicludes will draw water from which strata?

Accepted Answer

Introduction / Context:Strainer (screened) wells are designed to admit groundwater selectively from pervious zones while excluding fine materials from surrounding formations. When multiple aquifer layers are present, screens can be placed to tap each layer separated by aquicludes.Given Data / Assumptions:Three distinct pervious layers separated by impervious aquicludes.Strainers (screens) are installed opposite each pervious layer.Pumping induces radial flow into the screened intervals.Concept / Approach:Each screened interval hydraulically communicates with its adjacent permeable stratum. The total discharge is the sum of contributions from all open intervals, governed by transmissivity and drawdown distribution. Aquicludes prevent vertical leakage between pervious beds except through the wellbore itself at screened sections.Step-by-Step Solution:Identify the screened zones → one per pervious layer.Under pumping, head near screens drops, inducing inflow from each layer.Total yield equals sum of inflows, adjusted for well losses and partial penetration effects.Verification / Alternative check:Multi-layer well analysis (Jacob, Hantush) allocates contributions per layer based on transmissivity and storativity; field logs confirm inflow with spinner or heat-pulse flowmeter surveys.Why Other Options Are Wrong:Limiting inflow to only one layer ignores purpose of multiple screens; claiming “none” contradicts the defined design.Common Pitfalls:Screening continuously across aquicludes (not permitted); neglecting sand control and proper gravel packing leading to fines intrusion.Final Answer:All of the pervious layers (through their respective strainers).

Question 7

Urban water demand planning: A city with a population of 50,000 has an average daily demand of 20 million litres per day (20 mld). Using the conventional peak-factor approach, what is the expected maximum daily demand for design purposes?

Accepted Answer

Introduction / Context:In water supply engineering, systems are sized not only for the average day but also for peak demands. A commonly used guideline is to estimate the maximum daily demand by applying a peak factor to the average daily demand so that storage, pumps, and mains can handle short-term surges without service deficits.Given Data / Assumptions:Population = 50,000 (used only to give context).Average daily demand = 20 mld (million litres per day).Typical design practice: maximum daily demand ≈ 1.8 * average daily demand.Water losses and diurnal variations are implicitly contained in the peak factor.Concept / Approach:Design handbooks widely adopt a peak factor of about 1.8 for the maximum day relative to the average day in municipal systems, unless local data justify another value. This captures diurnal peaks and seasonal effects while remaining conservative for sizing facilities.Step-by-Step Solution:Let Q_avg = 20 mld.Use maximum day factor k_md ≈ 1.8.Compute Q_max-day = k_md * Q_avg = 1.8 * 20 = 36 mld.Verification / Alternative check:Many utility design standards also apply a maximum hourly factor to the maximum day. Our result focuses only on the maximum day, consistent with the question. Sensitivity checks with k_md = 1.6–2.0 would give 32–40 mld; 36 mld sits near the center of common practice.Why Other Options Are Wrong:24 mld and 30 mld: Too low; factors of 1.2–1.5 underestimate typical peak-day conditions.54 mld: Implies a factor of 2.7, which is excessive for most municipal systems.18 mld: Below average demand; not a peak.Common Pitfalls:Confusing maximum day with maximum hour.Ignoring local demand data; use local peaking where available.Final Answer:36 mld.

Question 8

Steel pipes in water supply: Which statement is the most accurate and generally applicable for design selection and specification?

Accepted Answer

Introduction / Context:Choosing pipe materials involves pressure ratings, jointing, corrosion resistance, installation method, and life-cycle cost. Steel is a common choice for high-pressure mains and long river crossings because of its strength and weldability.Given Data / Assumptions:General municipal water transmission context.No special linings/coatings assumed unless stated.Buried versus above-ground lines impose different thermal and corrosion considerations.Concept / Approach:Steel’s high tensile strength and ductility allow economical design at higher internal pressures and for special conditions (e.g., surge, seismic restraints). However, jointing, thermal movements, and corrosion protection must be managed through welding/flanging, anchors/expansion loops, and linings/coatings/cathodic protection.Step-by-Step Solution:Identify universally correct property: high internal pressure capability.Check the other statements for over-generalization or obsolescence (riveted joints are outdated; thermal movement may matter above ground; unprotected steel will corrode before 100 years in ordinary soils).Select the option that remains correct across typical scenarios.Verification / Alternative check:Design manuals specify steel for high-pressure transmission and surge-prone systems, confirming its pressure suitability. Modern practice favors welded/flanged joints, not riveted joints.Why Other Options Are Wrong:Riveted-only joints: obsolete in modern practice.No expansion provisions: false for exposed lines; buried lines see less thermal swing.100-year life without protection: unrealistic; coatings/cathodic protection are needed.Unsuitable for buried service: incorrect; steel is widely used when protected.Common Pitfalls:Ignoring corrosion control in steel pipe design.Assuming thermal issues are identical for buried and above-ground lines.Final Answer:Steel pipes are suitable for withstanding high internal pressure.

Question 9

Coagulants in water treatment: “Alum” used for coagulation during sedimentation and filtration is chemically which compound?

Accepted Answer

Introduction / Context:Coagulation is a key step in conventional water treatment. The commonly used coagulant, known as “alum,” destabilizes colloids so they can aggregate into flocs and be removed by sedimentation and filtration.Given Data / Assumptions:Alum refers to the water treatment coagulant used widely in municipal plants.No modifiers (e.g., polyaluminium chloride) are implied.Concept / Approach:In water treatment, “alum” is the commercial name for aluminium sulphate, generally Al2(SO4)3·nH2O. When added to water with alkalinity, it produces aluminium hydroxide floc and reduces turbidity and natural organic matter through adsorption and sweep flocculation.Step-by-Step Solution:Identify alum’s chemical identity: aluminium sulphate.Recall its reaction with alkalinity to form Al(OH)3 floc.Confirm other sulphates listed are different salts used for other purposes (e.g., copper sulphate for algae control).Verification / Alternative check:Standard water treatment texts consistently define alum as aluminium sulphate, not ferric/ferrous salts or copper sulphate.Why Other Options Are Wrong:Copper sulphate: An algicide, not the primary coagulant “alum.”Ferrous/ferric sulphate: Iron-based coagulants (effective alternatives) but not alum.Magnesium sulphate: Not used as a primary coagulant.Common Pitfalls:Confusing “alum” with “alumina” or with iron coagulants.Ignoring alkalinity needs for effective coagulation with alum.Final Answer:Aluminium sulphate.

Question 10

Chlorination operations: Chlorine supply cylinders are commonly maintained around 38–40 °C in service. What operational problem does this prevent?

Accepted Answer

Introduction / Context:Chlorine is typically stored as a liquefied gas. During dosing, heat input is needed to sustain evaporation; otherwise, refrigeration from vaporization can chill surfaces so much that frost and solid phases impede flow. Operators often maintain evaporators or water-baths at about 38–40 °C to stabilize gas withdrawal.Given Data / Assumptions:Cylinders are used with vaporizers or water-bath heaters.Ambient conditions may be cool enough to cause icing and solidification phenomena at valves/evaporators.Concept / Approach:When chlorine vaporizes, latent heat is absorbed, lowering temperature. Without external heat, local cooling can cause icing and even formation of chlorine “hydrates” or crystals at valves/flow paths, restricting gas flow. Controlled heating counters this effect and ensures steady feed and accurate control.Step-by-Step Solution:Recognize latent heat effects during vaporization.Identify operational symptom: icing/crystal formation obstructing valves and regulators.Apply practical remedy: maintain cylinder/evaporator bath at ~38–40 °C to prevent solidification and maintain stable evaporation.Verification / Alternative check:Equipment manuals for chlorinators specify water-bath temperatures in this range to avoid icing and flow instability.Why Other Options Are Wrong:Combustion (“burning”) is not relevant; chlorine is not a fuel.Heating is not primarily for explosion prevention when systems are operated within rated pressures.Corrosion control is handled by materials selection, not by heating.Common Pitfalls:Overheating beyond recommended limits can elevate internal pressure; always follow manufacturer settings.Confusing heater use for cold climates only—icing can occur from vaporization even in mild weather.Final Answer:Conversion of chlorine into crystals that impede gas withdrawal.

Question 11

Settling tank design terminology: The ratio of the overflow (effluent) discharge to the plan surface area of a continuous-flow settling tank is commonly termed what?

Accepted Answer

Introduction / Context:Clarifier sizing uses the concept of surface overflow rate (SOR), which is the volumetric flow divided by the plan area. This parameter relates hydraulic loading to settling performance of particles and flocs in primary/secondary clarifiers and water treatment basins.Given Data / Assumptions:SOR is expressed typically as m³/m²·day or litres/hr/m².We refer to conventional continuous-flow settling basins.Concept / Approach:Different texts use “surface loading rate,” “overflow rate,” or simply “overflow” (as shorthand) to represent the same design ratio, Q/A. Because these are synonymous in the clarifier design context, an option including all terms is correct.Step-by-Step Solution:Define SOR = Q / A_plan.Recognize synonyms: surface loading = overflow rate.Select the option that encompasses the synonymous terms.Verification / Alternative check:Design standards, manuals, and textbooks interchange these terms, though some prefer “surface overflow rate” for clarity.Why Other Options Are Wrong:Each single term is correct, but the most complete answer is the inclusive choice.“Downflow index” is not a standard term for Q/A in settling tanks.Common Pitfalls:Confusing SOR with weir loading rate, which uses effluent weir length.Using inappropriate units—always check for consistent units.Final Answer:all the above.

Question 12

Hydrologic yield concept: In basin water resources assessment, the “yield of a drainage basin” generally refers to which quantity?

Accepted Answer

Introduction / Context:The term “yield” is used in water resources to indicate dependable water available from a catchment, commonly over an annual period. Correctly identifying yield is crucial for planning reservoirs, water supply allocations, and irrigation schemes.Given Data / Assumptions:Yield is related to runoff leaving a defined catchment boundary.We consider the hydrologic (not economic) definition.Concept / Approach:Annual yield is the accumulated discharge volume over a hydrologic year. It integrates baseflow and quickflow from all events, making it the relevant quantity for long-term resource planning rather than momentary or short-period rates.Step-by-Step Solution:Identify time scale: annual (entire year).Define yield = ∫Q(t) dt over the year.Select the option stating “total volume annually.”Verification / Alternative check:Reservoir safe yield and firm yield analyses begin from the annual runoff series, not instantaneous values.Why Other Options Are Wrong:Instantaneous rate: A flow rate, not a yearly volume.Short-period average: Inadequate for annual planning.Net infiltration volume: Not the same as basin yield at the outlet.Common Pitfalls:Confusing “yield” with “peak discharge.”Ignoring inter-annual variability and dependability analysis.Final Answer:Total volume of water flowing from the basin over an entire year.

Question 13

Drinking-water standards: The typical upper limit for nitrate concentration in domestic water supplies is closest to which value?

Accepted Answer

Introduction / Context:Nitrate in drinking water is regulated because high levels can cause health issues such as methemoglobinemia in infants. Standards are usually expressed as mg/L (ppm) as nitrate (NO3−) or as nitrate-nitrogen (N). Understanding the common limit is important for source protection and treatment design.Given Data / Assumptions:Concentration units in options are mg/L (ppm) as NO3−.Domestic supply typical guideline values are considered.Concept / Approach:Many national standards (e.g., IS 10500) set nitrate (as NO3−) at 45 mg/L as a permissible limit, roughly equivalent to about 10–11 mg/L as nitrate-nitrogen. This aligns with health-based guidance and provides a practical screening target for raw and treated sources.Step-by-Step Solution:Convert mental reference: 10 mg/L as N ≈ 45 mg/L as NO3−.Match the provided options to the common standard.Select 45 ppm (mg/L) as the closest typical upper limit.Verification / Alternative check:Utility compliance reports often list nitrate near or below 45 mg/L (as NO3−) where agricultural runoff is controlled.Why Other Options Are Wrong:10–30 ppm: Too low if expressed as NO3−; these may align with “as N” values when converted.Common Pitfalls:Confusing “as N” and “as NO3−” units.Ignoring seasonal spikes from fertilizer application.Final Answer:45 ppm.

Question 14

Aqueduct cross-sections in irrigation/water conveyance: Which of the following cross-sectional shapes are commonly designed for aqueducts?

Accepted Answer

Introduction / Context:Aqueducts convey water over obstacles such as valleys and roads. The selection of cross-section depends on hydraulics, structural economy, freeboard, and constructability. Several shapes are standard in practice.Given Data / Assumptions:Reinforced concrete aqueducts in open channel or pressure flow.Hydraulic performance and structural formwork considerations.Concept / Approach:Circular, rectangular, and horse-shoe sections are all used. Circular sections are efficient in pressure flow, rectangular sections suit open channels and easier formwork, and horse-shoe sections balance structural arching with hydraulic efficiency, often used in tunnels and large conduits.Step-by-Step Solution:Identify common shapes from design manuals: circular, rectangular, horse-shoe.Match the option that includes all of them.Select “all the above.”Verification / Alternative check:Case studies and standard drawings in irrigation departments show all three shapes in service depending on site conditions.Why Other Options Are Wrong:Single-shape options ignore the diversity of typical designs.“Triangular only” is not a standard aqueduct section.Common Pitfalls:Confusing open-channel aqueduct sections with closed pressure mains.Neglecting freeboard and air-entrainment in high-velocity open channels.Final Answer:all the above.

Question 15

Centrifugal pump performance at constant speed: Which set of statements best describes how head, discharge, efficiency, and brake horsepower vary with discharge?

Accepted Answer

Introduction / Context:Centrifugal pumps have characteristic head–discharge, efficiency–discharge, and power–discharge curves at constant speed. Understanding these relationships is essential for selection, control-valve sizing, and energy optimization.Given Data / Assumptions:Single-stage centrifugal pump operating at constant rotational speed.Clear liquid, negligible viscosity effects beyond water-like behavior.No variable-speed drives or system interactions altering basic trends.Concept / Approach:The pump head–Q curve typically slopes downward; at zero flow, the shut-off head is achieved. Efficiency peaks near the “best efficiency point” (BEP), often called normal discharge for selection. Input power (and thus brake horsepower) usually rises with discharge up to near-BEP and may increase further depending on specific-speed and impeller design.Step-by-Step Solution:Confirm H–Q: as Q increases, H decreases.Identify shut-off head at Q = 0.Define normal discharge ≈ BEP.Note BHP trend: generally increases with Q across the operating region.Verification / Alternative check:Manufacturer curves consistently display these trends. System curves intersect pump curves to determine operating point; throttling shifts the operating point along the pump curve accordingly.Why Other Options Are Wrong:Any single statement alone is incomplete; together they correctly describe the pump behavior, hence “all the above.”Common Pitfalls:Assuming head increases with discharge at constant speed.Confusing shut-off head with maximum efficiency point.Final Answer:all the above.

Question 16

Reservoir storage zoning: From bottom to top, what is the correct order of storage zones in a reservoir used for water resources projects?

Accepted Answer

Introduction / Context:Reservoirs are conceptually divided into storage zones to guide operation and design. Correct identification of these zones ensures reliable supply while providing flood cushion and sediment management.Given Data / Assumptions:Conventional reservoir with low-level outlet(s) and a designated full-reservoir level (FRL) and maximum water level (MWL).Sediment accumulation occurs primarily in lower zones over time.Concept / Approach:Dead storage lies at the bottom and cannot be released by gravity through normal outlets (often allocated for sediment deposition). Above it is useful or live storage, which can be utilized for supply/power/irrigation. At the top is surcharge storage, a temporary flood cushion above FRL up to MWL, not normally used for conservation storage.Step-by-Step Solution:Order zones by functionality: non-releasable → releasable → flood cushion.Map to vertical position: bottom (dead), middle (useful/live), top (surcharge).Select the option that states this sequence.Verification / Alternative check:Reservoir operation manuals and design textbooks consistently adopt this stratification for storage accounting and rule-curve development.Why Other Options Are Wrong:Any option placing surcharge or useful storage below dead storage is physically inconsistent with outlet control and gravity release.Common Pitfalls:Confusing “useful” with “active” or “live”—they refer to the same releasable volume.Assuming surcharge is part of dependable supply.Final Answer:dead storage → useful storage (live storage) → surcharge storage.

Question 17

Design of distribution main: compute the approximate internal diameter required to convey 7.2 MLD (million litres per day) at an average velocity of 1.2 m/s. Choose the nearest standard size.

Accepted Answer

Introduction / Context:In water-supply engineering, a quick way to size a distribution main is to convert the demand to a discharge in m^3/s, select a practical flow velocity (often 0.8–1.5 m/s), and then compute the pipe area and equivalent internal diameter. This checks hydraulic feasibility before detailed head-loss analysis.Given Data / Assumptions:Total flow = 7.2 MLD (million litres per day).Velocity V = 1.2 m/s.Steady flow; circular pipe flowing full.1 m^3 = 1000 litres; 1 day = 86,400 s.Concept / Approach:First convert 7.2 MLD to m^3/s, then compute required area A = Q / V. Finally, determine internal diameter D from A using D = sqrt(4A / pi). Round to the nearest standard size.Step-by-Step Solution:Convert MLD to m^3/s: 7.2 MLD = 7.2 × 10^6 L/day = 7200 m^3/day.Discharge Q = 7200 / 86400 = 0.08333 m^3/s.Required area A = Q / V = 0.08333 / 1.2 = 0.06944 m^2.Diameter D = sqrt(4A / pi) = sqrt(0.27778 / 3.1416) ≈ sqrt(0.0885) ≈ 0.298 m ≈ 30 cm.Verification / Alternative check:Back-check velocity using D = 0.30 m: A = pi * D^2 / 4 ≈ 0.0707 m^2; V = Q / A ≈ 0.08333 / 0.0707 ≈ 1.18 m/s, consistent with the assumed 1.2 m/s.Why Other Options Are Wrong:24–28 cm (smaller) would increase velocity above the target and raise head loss.32 cm (larger) would lower velocity and increase capital cost unnecessarily at this stage.Common Pitfalls:Forgetting to convert MLD to m^3/s; using pi incorrectly; not checking that the selected diameter gives a reasonable velocity band.Final Answer:30 cm

Question 18

Pump sizing concept: If w is the weight of water per cubic metre, Q the discharge (m^3/s), and H the total head (m), what is the required water horsepower (output at water) of the pump?

Accepted Answer

Introduction / Context:Water horsepower (WHP) is the hydraulic power imparted to the water by the pump. It excludes motor and pump efficiencies and is useful for preliminary power estimates before accounting for losses.Given Data / Assumptions:w = weight of water per m^3 in kilogram-force per cubic metre (kgf/m^3), commonly ≈ 1000 kgf/m^3.Q in m^3/s; H in metres.1 metric horsepower (HP) = 75 kgf·m/s.Concept / Approach:Hydraulic power P (kgf·m/s) equals w * Q * H. Converting to metric horsepower requires dividing by 75. If w were in newtons per m^3, division would be by 735.5 W per HP; but the variable here is in kgf units.Step-by-Step Solution:Hydraulic power (kgf·m/s) = w * Q * H.Water HP = (w * Q * H) / 75.Brake HP = Water HP / (pump efficiency * motor efficiency), if later required.Verification / Alternative check:Dimensional check: w (kgf/m^3) * Q (m^3/s) * H (m) gives kgf·m/s, then divide by 75 to get HP. Consistent.Why Other Options Are Wrong:(a) missing conversion to horsepower.(c) uses 735.5 appropriate only if w is in N/m^3, not kgf/m^3.(d) is dimensionally incorrect.(e) is false because (b) is correct.Common Pitfalls:Mixing kilogram-force and newton systems; always confirm which specific weight unit the formula uses.Final Answer:(w * Q * H) / 75

Question 19

Temporary hardness removal by boiling: hardness will be removed by simple boiling if it is primarily due to which dissolved salts of calcium?

Accepted Answer

Introduction / Context:Hardness in water is commonly classified as temporary (carbonate) and permanent (non-carbonate). Understanding which salts cause which type of hardness dictates the appropriate treatment method at household or plant scale.Given Data / Assumptions:Boiling precipitates bicarbonates as carbonate scale.Permanent hardness salts remain dissolved after boiling.Concept / Approach:Temporary hardness arises from bicarbonates of calcium and magnesium. On boiling, bicarbonate ions decompose, releasing carbon dioxide and precipitating carbonates, thus reducing hardness. Sulphates, chlorides, and nitrates of calcium or magnesium constitute permanent hardness and are not removed by boiling.Step-by-Step Solution:Identify species removable by boiling: Ca(HCO3)2 and Mg(HCO3)2.Boiling reaction (conceptual): bicarbonate → carbonate + CO2 + H2O; carbonate precipitates.Conclude that calcium bicarbonate is the correct choice.Verification / Alternative check:Household experience: kettles develop carbonate scale when boiling hard water, indicating precipitation of carbonates from bicarbonates.Why Other Options Are Wrong:Sulphates, chlorides, nitrates of calcium (b–d) persist on boiling; they require softening (lime-soda, ion exchange).(e) is incorrect because (a) is valid.Common Pitfalls:Confusing carbonate hardness (temporary) with non-carbonate hardness (permanent). Always identify the anion to choose the treatment.Final Answer:Calcium bicarbonates

Question 20

Control of algal growth in service reservoirs and open storage: which chemical is typically dosed to suppress algae (with due care)?

Accepted Answer

Introduction / Context:Algal blooms in open reservoirs and slow-moving basins can cause taste, odour, and filter-clogging problems. A common operational control is the careful application of algicides, notably copper sulphate, alongside preventive measures.Given Data / Assumptions:Dose rates are low and must respect aquatic toxicity limits.Applied in quiescent waters; mixing ensured for uniform contact.Concept / Approach:Copper ions act as an algicide by disrupting cellular processes of algae and certain cyanobacteria. While chlorine (bleaching powder) is an oxidant and disinfectant, it is not typically the primary algicide for reservoir-wide control; alum and lime are for coagulation and pH/alkalinity adjustment.Step-by-Step Solution:Identify the algicide: copper sulphate (CuSO4).Recognize limits: avoid over-dosing to protect fish and comply with environmental norms.Coordinate with treatment: remove dead biomass via coagulation-sedimentation-filtration to prevent downstream issues.Verification / Alternative check:Utility manuals commonly recommend controlled CuSO4 application, often seasonally, with jar tests to determine dose.Why Other Options Are Wrong:Bleaching powder mainly disinfects; persistent algae control is limited.Lime and alum modify pH/alkalinity and coagulate particles, not direct algicides.Hence copper sulphate is the appropriate choice.Common Pitfalls:Over-dosing copper leading to aquatic toxicity; failing to remove dead algae, causing taste and odour episodes.Final Answer:Copper sulphate

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