Water Supply: Demand, Sources, Reservoirs, Distribution, Treatment and Quality
2. Fire demand, design period and population forecast
Fire demand is a large rate for a short time. It decides pipe diameters and the fire reserve in the reservoir, not the yearly volume. Empirical formulas relate it to the population P in thousands. Always read the unit: four give a flow in litres per minute, the Indian formula gives kilolitres per day.
| Name | Formula | Unit of Q |
|---|---|---|
| Kuichling | Q = 3182 √P | L/min |
| Freeman | Q = 1136 (P/10 + 10) | L/min |
| Buston | Q = 5663 √P | L/min |
| National Board of Fire Underwriters | Q = 4637 √P (1 − 0.01 √P) | L/min |
| Indian (CPHEEO) formula | Q = 100 √P | kL/day |
Worked example: a town of 64,000 people has P = 64, √P = 8. Kuichling gives Q = 3182 × 8 = 25,456 L/min; Freeman gives 1136 × (6.4 + 10) = 18,630 L/min; the Indian formula gives 100 × 8 = 800 kL/day. A fire storage in the reservoir is then the fire rate × the duration the authority adopts. The mistake to avoid is putting P = 64,000 instead of 64, which inflates Kuichling’s answer about thirty-fold.
The design period is the future year up to which a component is sized. The rule is economic: works that are costly and disruptive to duplicate, large civil works and buried pipes, are sized for 30 years or more (dams and large storage longer, up to about 50 years); mechanical and electrical plant and treatment units, which wear out, can be added in modules and are cheaper to extend, are sized for shorter periods of about 15 years. Sizing for the future needs a population forecast.
- Arithmetic increase: P_n = P₀ + n·x̄, x̄ the mean decadal increase; fits old, large, near-saturated cities.
- Geometric increase: P_n = P₀ (1 + r)ⁿ, r the mean decadal growth rate as a fraction; fits young, growing towns, but overestimates when room runs out.
- Incremental increase: P_n = P₀ + n·x̄ + n(n+1)/2 · ȳ, ȳ the mean of the second differences; it lets the increment itself change.
- Also: decreasing rate of growth, the logistic (S-curve), graphical extension and comparison with similar cities, the ratio method (the town’s share of a larger region’s forecast), and employment-based forecasts for a planned new town.
Worked example: counts of 40,000, 48,000 and 56,000 at successive decades give x̄ = 8,000. Three decades after the last count, P = 56,000 + 3 × 8,000 = 80,000. At 135 lpcd the average demand is 10.8 ML/d, the maximum day is 1.8 × 10.8 = 19.44 ML/d, and the maximum hour runs at 2.7 × 10.8 = 29.16 ML/d, which is 337.5 L/s.
3. Sources of water and intakes
Sources are surface water (rivers, lakes, impounding reservoirs, canals), groundwater (open wells, tube wells, springs, infiltration galleries), harvested rain, and, on the coast or in scarcity, desalinated or reclaimed water. A source is chosen on four tests: it must give the design quantity in the driest year, of a quality that treatment can correct at acceptable cost, at an elevation and distance that keep pumping and conveyance affordable, with legal right to the water.
| Feature | Surface water | Groundwater |
|---|---|---|
| Turbidity and organic load | high and seasonal; full treatment needed | clear and steady; often disinfection alone |
| Hardness and dissolved minerals | usually lower | usually higher; local fluoride, arsenic, iron, nitrate |
| Pathogens and pollution risk | high; open to discharges | low if the aquifer is protected |
| Quantity and reliability | large; falls in a dry season | limited; falls with over-extraction |
An intake is the structure that draws raw water from the source. Its design aims at the best quality water, at all levels of the source, with little silt and debris.
- Reservoir or lake intake (intake tower): ports at several levels let the operator draw from the best layer, avoiding surface algae and bottom sediment.
- River intake: a submerged intake pipe or a well (jack-well) on the bank linked by inlet pipes; screens and low entry velocity keep out debris and fish. The inlet sits below the lowest water level but clear of the bed.
- Infiltration gallery or well: a perforated conduit in the river bed or bank takes water already filtered by the sand.
- Spring: a protected spring box over the eye of the spring, with a fenced catchment.
4. Groundwater yield and recharge
An aquifer is a saturated formation that stores and yields usable water (sand, gravel, fractured rock). An aquiclude (clay) holds water but does not transmit it; an aquitard transmits it poorly. In an unconfined aquifer the top of the saturated zone is the water table, open to the atmosphere. A confined (artesian) aquifer lies between impermeable layers, the water is under pressure, and its piezometric surface can stand above the ground, so a well flows by itself. Porosity n is the void fraction; the part that drains by gravity is the specific yield S_y, the part held back is the specific retention S_r, and n = S_y + S_r. Clay has high porosity but tiny specific yield, which is why it is a poor aquifer.
Darcy’s law governs flow through the aquifer: v = K·i, so Q = K·i·A, where K is the hydraulic conductivity, i = dh/dl the hydraulic gradient, and A the whole cross-section of flow. v is the Darcy velocity, discharge per unit gross area; the water in the pores really moves at the faster seepage velocity v/n. The product T = K·b (b the aquifer thickness) is the transmissivity. Example: a confined sand aquifer 10 m thick and 400 m wide, K = 30 m/day, gradient 0.002, carries Q = 30 × 0.002 × (10 × 400) = 240 m³/day.
| Aquifer | Formula | What it shows |
|---|---|---|
| Unconfined (Dupuit) | Q = π K (h₂² − h₁²) / ln(r₂/r₁) | yield varies with the square of the water-column heights; the saturated thickness itself falls |
| Confined (Thiem) | Q = 2π K b (h₂ − h₁) / ln(r₂/r₁) | yield is directly proportional to the drawdown; thickness b stays constant |
Specific capacity is yield per unit drawdown, Q/s, and measures how productive a well is. Yield cannot be taken for ever: the safe yield is the extraction that recharge can replace. Natural recharge comes from rain infiltrating through pervious ground and seepage from rivers, tanks and irrigated fields. Over-extraction lowers the water table, dries wells, raises pumping cost, causes land subsidence, and draws saltwater into coastal aquifers. Artificial recharge uses percolation tanks, check dams, recharge pits, trenches and shafts, spreading basins, and injection wells for confined aquifers. For a planner the implication is spatial: impervious cover over recharge zones cuts recharge, so such land is protected, and roof and road runoff is led into recharge structures.
5. Service reservoirs and pumps
A service (distribution) reservoir stores treated water close to the demand. It does three jobs: it balances the constant rate at which water is pumped or treated against the varying rate at which it is used; it holds pressure in the network, especially when elevated; and it keeps a fire and emergency reserve. Types are ground-level or underground tanks (cheap, but the network then needs pumping), and elevated reservoirs (overhead tanks, giving gravity head). The best site is high ground at the centre of the zone served, which shortens mains and evens out pressure.
| Part | What it covers | How it is found |
|---|---|---|
| Balancing (equalising) storage | the gap between inflow and draw-off over the day | mass curve: cumulative inflow minus cumulative demand; storage = largest surplus + largest deficit; it falls as pumping hours rise |
| Fire storage | water for fighting a fire while the network is under strain | fire demand rate × the duration adopted |
| Emergency (breakdown) storage | failure of pumps, power or a main, and repair time | a share of daily demand or of storage, fixed by the designer or the standard |
Worked example: daily demand 4,800 m³. In four six-hour blocks the draw is 10%, 40%, 30% and 20% of the day’s demand, while pumping gives 25% in each block. Cumulative (inflow − demand) is +15%, 0, −5%, 0. The largest surplus is 15% and the largest deficit is 5%, so balancing storage = 15% + 5% = 20% of 4,800 = 960 m³. Add a 400 m³ fire reserve and an emergency reserve of 10% of daily demand (480 m³): capacity = 960 + 400 + 480 = 1,840 m³.
Pumps lift water or add head. The centrifugal pump (impeller, smooth flow, wide range, compact) is the usual choice in water supply; a submersible pump, with its motor below water, serves tube wells; reciprocating pumps give high head at small, pulsating flow; jet pumps suit shallow wells; the hydraulic ram runs on falling water with no power. The power needed is P (kW) = ρ g Q H / (1000 η) = 9.81 Q H / η, with Q in m³/s, H the total head in m (static lift plus friction and minor losses) and η the overall efficiency. For Q = 0.04 m³/s, H = 25 m and η = 0.5, P = 9.81 × 0.04 × 25 / 0.5 = 19.62 kW. A pump works where its head–discharge curve cuts the system curve H = H_static + K Q²; two pumps in parallel add discharge at the same head, in series they add head at the same discharge.
6. Distribution layout and network design
The distribution system takes treated water to the consumer at adequate pressure. By source of energy it is a gravity system (source above the town), a pumping system, or a dual system, pumped to a service reservoir and fed by gravity from it, which is the commonest and gives the steadiest pressure. A continuous (24 × 7) supply keeps the pipes pressurised so that polluted water cannot be sucked in through a leak, which is why intermittent supply is a public-health risk as well as a convenience problem. Large towns are divided into zones or metered areas to control pressure and measure loss.
| Layout | Shape | Strength | Weakness | Fits |
|---|---|---|---|---|
| Dead-end (tree) | main with sub-mains and branches ending in dead ends | cheapest, fewest valves, flow in each pipe known directly | stagnation at the ends, a break cuts off everything downstream, weak fire flow | irregular, organically grown towns |
| Grid-iron | mains and sub-mains interconnected in a grid | no dead ends, supply from two or more directions, a break isolates a small area, good fire flow | more pipe and valves, design needs trial and error | regular, planned towns with a rectangular street pattern |
| Ring | main laid as a ring around the area, sub-mains inside | each point fed from two sides, valves isolate a section, good pressure | longer pipe, complex design | planned areas with ring or circular roads, campuses |
| Radial | zones, each with a reservoir at its centre, pipes radiating outward | high, quick and even pressure, smaller pipes, simple design | one reservoir per zone, so higher cost | planned, zoned towns with several elevated tanks |
Network design rests on a head-loss law. The Hazen–Williams equation is empirical, for water at ordinary temperature in turbulent flow: in SI, V = 0.849 C R^0.63 S^0.54 (V in m/s, R = D/4 in m for a full pipe, S = h_f/L), and h_f = 10.67 L Q^1.852 / (C^1.852 D^4.87) (L in m, Q in m³/s, D in m). C is the roughness coefficient, larger for smoother pipe, smaller for old tuberculated pipe. The Darcy–Weisbach equation, h_f = f L V² / (2 g D), is general for any fluid and regime, with f from the Moody chart or f = 64/Re in laminar flow. Because h_f falls as D^−4.87, doubling a diameter at the same flow cuts head loss about 2^4.87 ≈ 29 times: small diameters are expensive in head. Example: D = 0.3 m, C = 100, S = 0.002 gives R = 0.075 m and V = 0.849 × 100 × 0.075^0.63 × 0.002^0.54 = 0.58 m/s.
In a looped network the flows are not known directly, so the Hardy Cross method finds them by successive correction. The two laws are continuity at every junction (inflow equals outflow) and energy around every closed loop (the algebraic sum of head losses is zero). The steps: (1) assume flows in every pipe that satisfy continuity at each node, clockwise positive; (2) for each loop compute h = K Q |Q|^(n−1) with its sign, n = 1.852 for Hazen–Williams and 2 for Darcy–Weisbach; (3) compute ΔQ = − Σh / (n Σ|h/Q|); (4) add ΔQ to every pipe of the loop, and to a pipe shared by two loops add both corrections, the neighbour’s with opposite sign; (5) repeat until Σh is within tolerance.
7. Water treatment: the train, settling and the settling tank
Treatment makes raw water safe, clear and palatable. The train depends on the source: groundwater may need only aeration and disinfection, or a specific removal of iron, fluoride or arsenic, while turbid surface water needs the full sequence below.
| Unit | Job |
|---|---|
| Screening | coarse bars then fine screens take out floating debris, leaves, fish |
| Aeration (where needed) | drives off CO₂, H₂S, taste and odour; oxidises dissolved iron and manganese so they can be filtered |
| Plain sedimentation | heavy silt and sand settle under gravity in still water, relieving later units |
| Coagulation and flocculation | chemical dose and mixing turn fine colloids into settleable floc |
| Clarification (sedimentation) | floc settles in rectangular, circular or tube-settler tanks |
| Filtration | remaining fine floc, turbidity and some bacteria are strained out |
| Disinfection | pathogens are killed or inactivated, and a residual protects the network |
A grain settles by Stokes’ law: v_s = g (ρ_s − ρ_w) d² / (18 μ), valid for small spheres in laminar flow (Reynolds number below 1). Settling speed rises with the square of the diameter, so tiny colloids, which would take days to years to settle, must first be clumped by coagulation.
The ideal settling tank (Hazen, Camp) has four zones, inlet, settling, sludge and outlet, with horizontal flow and no mixing. Its key quantity is the overflow rate (surface loading) v₀ = Q / A_s, with A_s the plan area. A particle whose settling speed v_s ≥ v₀ is removed completely; one with v_s < v₀ is removed in the ratio v_s / v₀ (if it is spread evenly over the inlet depth). The depth does not appear: removal depends on plan area, not depth. Depth only fixes the detention time t = V/Q = H / v₀. Example: Q = 12,000 m³/d and v₀ = 30 m³/m²/d give A_s = 12,000/30 = 400 m². With t = 2.4 h = 0.1 day, V = 12,000 × 0.1 = 1,200 m³ and H = V/A_s = 3.0 m. A rectangular tank is typically long, with length to width of about 3:1 to 5:1.
8. Coagulation, filtration and disinfection
Colloids carry like charges and repel each other, so they stay suspended. Coagulation destabilises them. Alum, Al₂(SO₄)₃·18H₂O, is the commonest coagulant: it reacts with the alkalinity of the water (bicarbonate) to form the gelatinous Al(OH)₃ floc, so a water short of natural alkalinity needs lime or soda ash as well. Ferrous sulphate with lime, ferric chloride and polyaluminium chloride are alternatives, and polyelectrolytes serve as coagulant aids. The dose and pH are found by the jar test: a row of beakers is dosed differently, mixed fast, then slowly, allowed to settle, and the one with the least turbidity gives the optimum. Rapid mixing lasts seconds to disperse the chemical; flocculation is slow paddling for about half an hour, gentle enough that the growing floc is not sheared apart.
| Feature | Slow sand (SSF) | Rapid sand (RSF) |
|---|---|---|
| Filtration rate | about 0.1 to 0.2 m/h (100 to 200 L/h/m²) | about 5 m/h (3,000 to 6,000 L/h/m²) |
| Sand effective size | about 0.2 to 0.3 mm | about 0.45 to 0.7 mm |
| Pre-treatment | plain sedimentation only; low-turbidity raw water | coagulation, flocculation and sedimentation |
| How it works | biological film (schmutzdecke) on top plus straining | straining and adsorption through the bed depth |
| Cleaning | scrape off the top layer every few weeks | backwash with reversed flow, often with air scour, every day or two |
| Area and skill | large area, little skill, almost no chemicals; suits villages | compact, skilled operation, chemicals; suits towns and cities |
Disinfection kills or inactivates pathogens; it is not sterilisation. Chlorine in water forms hypochlorous acid, Cl₂ + H₂O → HOCl + HCl, which ionises, HOCl ⇌ H⁺ + OCl⁻. HOCl is the far stronger disinfectant, and low pH favours it, so chlorination works better below about pH 7.5 than at pH 9. The chlorine demand is the dose minus the residual. The residual is either free (HOCl and OCl⁻) or combined (chloramines, formed with ammonia, longer-lived but weaker). In break-point chlorination the residual first rises, then falls as ammonia is oxidised, and beyond the break-point every extra dose appears as free chlorine. Bleaching powder, CaOCl₂, carries a quarter to a third of its weight as available chlorine; the chlorine mass needed is dose (mg/L) × flow (m³/d) in g/d, divided by the available fraction to get the powder. Example: 10 ML/d at 2 mg/L needs 10,000 m³/d × 2 g/m³ = 20 kg/d of chlorine, or 20/0.25 = 80 kg/d of powder at 25% available chlorine.
9. Water quality and testing
Quality is judged by physical, chemical and biological parameters against the drinking-water specification, IS 10500, whose concept is a two-level standard: an acceptable limit, which consumers accept without objection, and a permissible limit in the absence of an alternate source, beyond which water is rejected. The headline figures (as in the 2012 edition) are below.
| Parameter | What it tells and how it is tested | Standard figure |
|---|---|---|
| Turbidity | cloudiness from suspended matter; nephelometer, in NTU; particles shelter microbes from the disinfectant | 1 NTU acceptable, 5 NTU permissible |
| pH | acidity or alkalinity; electrometric pH meter | 6.5 to 8.5 |
| Total hardness | calcium and magnesium as mg/L CaCO₃; EDTA titration | 200 mg/L acceptable, 600 permissible |
| Total dissolved solids | dissolved salts; evaporation or conductivity | 500 mg/L acceptable, 2,000 permissible |
| Fluoride | too little weakens teeth, too much causes fluorosis | 1.0 mg/L acceptable, 1.5 permissible |
| Residual chlorine | free chlorine at the tap; DPD colorimetric test | at least 0.2 mg/L free |
| Coliforms and E. coli | indicators of faecal contamination; multiple-tube MPN or membrane filter | E. coli not detectable in any 100 mL sample |
Hardness is the soap-consuming power of Ca²⁺ and Mg²⁺, expressed as mg/L of CaCO₃: each ion is scaled by 50 over its equivalent weight (Ca 20, Mg 12), so hardness = 2.5 Ca + 4.1 Mg approximately. For Ca = 60 mg/L and Mg = 12 mg/L: 60 × 50/20 + 12 × 50/12 = 150 + 50 = 200 mg/L as CaCO₃. Carbonate (temporary) hardness is the part matched by alkalinity and removed by boiling or lime; non-carbonate (permanent) hardness from sulphates and chlorides is removed by lime–soda or ion exchange. With total hardness 200 and alkalinity 120 mg/L as CaCO₃, carbonate hardness is 120 and non-carbonate is 80.
BOD (biochemical oxygen demand) is the oxygen micro-organisms use to oxidise the biodegradable organic matter in water in 5 days at 20 °C, written BOD₅; it follows a first-order law, BOD_t = L₀ (1 − e^(−kt)), with L₀ the ultimate demand. COD (chemical oxygen demand) is the oxygen equivalent of the matter oxidised by a strong chemical oxidant (dichromate), so COD is at least BOD, and a high COD/BOD ratio signals organic matter that bacteria cannot break down. Both measure organic pollution, mostly in rivers and sewage rather than in treated drinking water, where dissolved oxygen is the other key indicator. Water is tested for indicator organisms, coliforms and E. coli, because pathogens are too varied and scarce to test for one by one: their presence signals faecal contamination.
10. Water-sensitive urban planning
The conventional approach treats water as three separate pipe systems: bring supply in, take sewage out, and carry storm water away as fast as possible. Water-sensitive urban planning (water-sensitive urban design) treats supply, storm water, wastewater and groundwater as one urban water cycle and plans it together with land use, street layout and open space. The aim is a city that conserves water, recharges its aquifer, cleans its runoff and floods less, and that keeps its post-development runoff close to what the land gave before.
- Conserve potable water: metering, leakage control, efficient fixtures, demand management.
- Fit-for-purpose supply: use greywater, treated sewage or harvested rain for flushing, gardens and cooling, and keep potable water for drinking and cooking.
- Manage runoff at source: infiltrate, detain and retain on the site so post-development peak and volume stay near pre-development values.
- Protect the natural water system: lakes, wetlands, streams, floodplains and recharge zones form a blue–green network that master plans keep free of development.
- Clean runoff where it falls: swales, rain gardens and wetlands remove sediment and pollutants before the water reaches a drain.
Key takeaways
- Demand = population × per capita rate (135 lpcd for a town with sewerage, stated); maximum daily is 1.8 × average and maximum hourly is 2.7 × average; distribution is sized on the larger of maximum hourly, or maximum daily plus fire, never on fire plus the peak hour.
- Read the unit of every fire formula (P in thousands; L/min for Kuichling, kL/day for the Indian one); long design periods go to costly, hard-to-duplicate works, short ones to modular plant; match the forecast method to the town.
- Intakes sit upstream of outfalls, on the concave bank, below low water and clear of the bed; groundwater follows Darcy (Q = K i A, seepage velocity v/n), with Dupuit and Thiem for wells, and a planner protects recharge zones.
- Reservoir capacity = balancing (largest surplus + largest deficit from the mass curve) + fire + emergency; pump power = 9.81 Q H / η kW; the best network layout depends on the town (dead-end irregular, grid or ring regular and planned, radial zoned).
- Hazen–Williams and Darcy–Weisbach give head loss; Hardy Cross corrects loop flows by ΔQ = −Σh / (n Σ|h/Q|), applied to every pipe of the loop, shared pipes taking both loops’ corrections; head loss varies as D^−4.87.
- Settling removal depends on overflow rate Q/A, not on depth; coagulation (alum, jar test) then filtration and chlorination (HOCl strongest, low pH favours it, residual kept); quality is tested against IS 10500’s acceptable and permissible limits, and water-sensitive planning treats the whole urban water cycle as one.
Practice questions (20)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
For a town’s distribution mains, the design flow is taken as the larger of the maximum hourly demand and which other case?
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Answer: B — Maximum daily demand plus the fire demand
The mains are checked for two cases, the peak hour alone, and the maximum day coinciding with a fire, and the larger governs. Maximum hourly plus fire is the tempting answer, but it assumes that the daily peak hour and a major fire fall together, which design does not assume. Average daily plus fire understates the day on which the fire is supposed to occur.Using Kuichling’s formula Q = 3182 √P (Q in L/min, P in thousands), the fire demand of a town of 36,000 people is
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Answer: D — 19,092 L/min
P is in thousands, so P = 36 and √36 = 6, giving Q = 3182 × 6 = 19,092 L/min. The 114,552 figure forgets the square root; 603,742 puts P = 36,000 and so inflates the answer about thirty-fold.On a river bend, a town’s intake is best placed on
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Answer: A — the concave bank, upstream of the town’s own sewage and drain outfalls
The outer (concave) bank has the deeper water and a current that scours silt, while the convex bank builds a silt bar. The intake must also lie upstream of the town’s own outfalls, or it would draw the town’s sewage back. The calm convex bank is the tempting choice, but calm water is exactly where silt settles.A sand aquifer has porosity 0.30 and specific retention 0.08. Its specific yield is
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Answer: C — 0.22
Porosity n = S_y + S_r, so S_y = 0.30 − 0.08 = 0.22. The 0.38 answer wrongly adds the two, and 0.30 is the porosity itself, which includes the water the grains hold back.A sand has hydraulic conductivity 20 m/day and porosity 0.25. Under a hydraulic gradient of 0.005, a dissolved tracer moves through it at about
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Answer: C — 0.40 m/day
Darcy velocity v = K i = 20 × 0.005 = 0.10 m/day, but that is discharge per gross area. The tracer moves through the pores at the seepage velocity v/n = 0.10/0.25 = 0.40 m/day. The 0.10 answer is the Darcy velocity, the common slip; 0.025 multiplies by n instead of dividing.A new planned township is divided into zones; each zone has an elevated tank at its centre and the pipes run outward from it. This distribution layout is
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Answer: A — Radial
A reservoir at the centre of each zone with pipes radiating to the periphery is the radial layout; it gives quick, even pressure but needs one reservoir per zone. A grid-iron or ring layout interconnects mains rather than radiating them from a central tank, and a dead-end layout has branches ending blindly.A water main carries a fixed discharge. It is replaced by a pipe of twice the diameter, with the same length and the same Hazen–Williams C. The friction head loss falls by a factor of about
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Answer: B — about 29 times
In the Hazen–Williams equation h_f varies as Q^1.852 / D^4.87 at fixed length and C, so doubling D divides h_f by 2^4.87, about 29. Four times is only the area ratio. Sixteen squares the fall in velocity and forgets the extra 1/D. Thirty-two is the D⁵ of Darcy–Weisbach with constant f, close but not the Hazen–Williams exponent.In an ideal horizontal-flow settling tank the discharge and plan area are fixed and the depth is doubled. The removal of any given particle size
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Answer: D — stays the same, because it depends on the overflow rate Q/A
In Hazen’s theory a particle is removed if its settling speed is at least the overflow rate Q/A, independent of depth; a deeper tank gives more detention time, but the particle must also fall through more depth, so the two cancel. The detention-time argument is the usual trap: time and path length grow together.At the same free chlorine concentration, disinfection of water is more effective at
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Answer: B — pH 6.5, where hypochlorous acid predominates
HOCl is the far more powerful disinfectant and low pH keeps chlorine in that form, so effectiveness is better near pH 6.5 than at 9 or 11. The ‘fully ionised is most reactive’ idea fails because the charged OCl⁻ penetrates microbes poorly and is a weaker disinfectant.A hill village of 800 people has a spring-fed stream of low, steady turbidity, free land, no trained operator and no regular chemical supply. The most suitable filtration is
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Answer: A — a slow sand filter preceded by plain sedimentation
A slow sand filter needs no coagulant, works through a biological film and is run with little skill, and its large area is not a problem where land is free and turbidity is low. The rapid and pressure options all depend on chemicals, backwash machinery and a trained operator, which this village lacks.A water has Ca²⁺ 80 mg/L and Mg²⁺ 24 mg/L. Taking equivalent weights Ca 20, Mg 12 and CaCO₃ 50, its total hardness as CaCO₃ is
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Answer: C — 300 mg/L
Hardness = Σ (mg/L × 50 / equivalent weight) = 80 × 50/20 + 24 × 50/12 = 200 + 100 = 300 mg/L as CaCO₃. 104 simply adds the two concentrations, and 200 counts the calcium alone and forgets the magnesium.Which of the following statements about a service reservoir are correct?
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Answer: A — Balancing storage falls as the pumping hours per day rise, for the same daily demand; B — Balancing storage equals the largest cumulative surplus plus the largest cumulative deficit of inflow against demand; D — A high site near the centre of the zone served shortens mains and evens out pressure
Longer pumping brings inflow closer to the demand curve, so less balancing is needed; the mass-curve rule adds the largest surplus and largest deficit; a central, high site is the preferred location. Emergency storage covers pump, power and main failures and the repair time, and an elevated tank still empties if the inflow stops, so it is not made unnecessary by height.Which of the following are genuine advantages of a grid-iron layout over a dead-end layout?
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Answer: A — Water reaches each point from more than one direction, so a pipe break isolates only a small area; B — There are no dead ends, so there is less stagnation and the fire flow is better
Interconnection gives two-way supply and sectional isolation, and removes stagnant ends. The other two are advantages of the dead-end layout: its flows can be found directly, and it uses less pipe and fewer valves, while the grid needs a trial-and-error method such as Hardy Cross.Which of the following statements about coagulation and disinfection are correct?
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Answer: A — Alum needs alkalinity in the water to form its floc, so lime or soda ash is added when alkalinity is short; C — Chloramines last longer in the network but disinfect more weakly than free chlorine; D — The jar test is used to find the optimum coagulant dose and pH
Alum consumes bicarbonate to form Al(OH)₃, chloramines trade strength for persistence, and the jar test is the standard way to fix dose and pH. Hypochlorous acid, not the hypochlorite ion, is the stronger disinfectant, so that statement reverses the facts.Which of the following are measures of water-sensitive urban planning?
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Answer: B — Permeable paving and vegetated swales along streets, to infiltrate and filter runoff; C — Treated wastewater for flushing and gardens, keeping potable water for drinking; D — Reserving lakes, wetlands and floodplains as no-development areas in the master plan
Water-sensitive planning slows, infiltrates, cleans and reuses water, and protects the natural water system. Covering and straightening natural drains does the opposite: it speeds the flow, raises the downstream peak and removes the storage and infiltration that the natural channel gave.Populations at three successive decennial counts are 25,000, 31,000 and 37,000. By the arithmetic increase method, the population four decades after the last count is forecast, and the demand is taken as 135 lpcd. The average daily demand, in m³/day, is ____.
Numerical answer — type the value.
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Answer: 8235
Mean decadal increase = 6,000. Population = 37,000 + 4 × 6,000 = 61,000. Demand = 61,000 × 135 = 8,235,000 L/day = 8,235 m³/day. Check: 61,000 × 100 = 6.1 million and 61,000 × 35 = 2.135 million, which sum to 8.235 million litres.A service reservoir serves a daily demand of 4,800 m³. In four successive six-hour blocks the demand is 10%, 40%, 30% and 20% of the day’s total, and pumping delivers 25% of the day’s total in each block. A fire reserve of 400 m³ and an emergency reserve of 10% of the daily demand are added to the balancing storage. The reservoir capacity, in m³, is ____.
Numerical answer — type the value.
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Answer: 1840
Cumulative (inflow − demand) in % of the day: 25 − 10 = +15; then +15 + (25 − 40) = 0; then 0 + (25 − 30) = −5; then −5 + (25 − 20) = 0. Largest surplus 15, largest deficit 5, so balancing = 20% of 4,800 = 960 m³. Emergency = 0.10 × 4,800 = 480 m³. Total = 960 + 400 + 480 = 1,840 m³.In a Hardy Cross analysis with head loss h = r Q |Q| (Q positive clockwise), a loop has four pipes with (r, Q): (2, +15), (1, +5), (2, −10) and (1, −10), in consistent units. The magnitude of the first flow correction ΔQ for this loop, to two decimal places, is ____.
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Answer: 1.35
h = 2 × 15² = 450; 1 × 5² = 25; −2 × 10² = −200; −1 × 10² = −100. Σh = 175. Σ r|Q| = 2×15 + 1×5 + 2×10 + 1×10 = 65, so n Σ|h/Q| = 2 × 65 = 130. ΔQ = −175/130 = −1.346, magnitude 1.35. Re-check: 450 + 25 = 475, minus 300 gives 175.A plain settling tank treats 9,000 m³/day at an overflow rate of 28 m³/m²/day with a detention time of 3 hours. Its water depth, in m, is ____.
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Answer: 3.5
Plan area A = Q/v₀ = 9,000/28 = 321.4 m². Volume V = Q × t = 9,000 × (3/24) = 1,125 m³. Depth H = V/A = 1,125/321.4 = 3.5 m. Direct check: H = v₀ × t = 28 m/day × 0.125 day = 3.5 m.A plant treats 15 ML/d and applies chlorine at a dose of 2.4 mg/L, using bleaching powder that holds 25% available chlorine by weight. The bleaching powder needed, in kg/day, is ____.
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Answer: 144
Flow = 15 ML/d = 15,000 m³/d. Chlorine = 15,000 m³/d × 2.4 g/m³ = 36,000 g/d = 36 kg/d. Powder = 36/0.25 = 144 kg/d. Check: 144 × 0.25 = 36.