Sanitation, Sewerage, Wastewater Treatment and Storm-Water Drainage
2. Water-carriage systems: separate, combined and partially separate
A water-carriage system carries sewage by flowing water through underground conduits. What goes into the sewer decides its type.
| System | What the sewer carries | Strength | Weakness |
|---|---|---|---|
| Separate | sanitary sewage only; storm water goes in its own drains | small sewers, steady plant load, storm water needs no treatment, no sewage overflow | two networks to build and keep apart; storm drains may be misused for sewage |
| Combined | sewage and storm water together in one sewer | one network, cheaper in narrow streets, storms flush the pipe | large pipes, deposits in dry weather, huge swings at the plant, diluted sewage spilling to rivers in storms |
| Partially separate | sewage plus some storm water, from roofs and paved yards | larger flow flushes the sewer better, the storm drains are smaller | the plant and sewers must allow for the extra storm flow |
Indian practice favours the separate system. Monsoon storms are intense, so the storm flow can be many times the dry-weather flow, and a combined sewer sized for the storm would run nearly empty, and deposit solids, on all other days. The treatment plant, whose biology needs a steady feed, would be flooded or bypassed. Separate systems also let sewers and storm drains be built in either order as funds allow.
3. Sewage quantity and sewerage layout
The average dry-weather flow is taken as about 80% of the water supplied, the rest being lost to gardens, evaporation, leaks and retention. To it are added infiltration of groundwater through joints and manholes (larger with a high water table and poor joints) and any industrial wastewater. Sewers are sized for the peak flow = average × peak factor, and the factor falls as the population rises, from about 3 for small towns toward 2 for large ones. Harmon’s formula is M = 1 + 14 / (4 + √P), with P the population in thousands. Example: 100,000 people at 150 lpcd get 15 ML/d; sewage is 0.8 × 15 = 12 ML/d; P = 100 gives M = 1 + 14/14 = 2.0; the peak is 24 ML/d, which is 277.8 L/s. The minimum, night-time flow is only a fraction of the average, which is why self-cleansing must be checked at low flow.
The layout follows the ground slope and the outfall site, aiming at short, shallow, gravity routes with little pumping.
- Perpendicular: sewers run by the shortest line to a river or stream, with little or no treatment; suits storm drains.
- Interceptor: perpendicular sewers plus an intercepting sewer along the river that collects them for one treatment plant.
- Zone: the town is divided by elevation into zones, each draining to its own plant or to a main at a lower level, with pumping from the low zones.
- Fan: sewers converge like the ribs of a fan on a single point, for a town lying in a valley or on a slope toward one outfall.
- Radial: sewers diverge outward from a high central point to several outlets at the edge, for a town on a ridge or plateau.
4. Sewer section and design: Manning, velocity and hydraulic elements
A gravity sewer is designed as an open channel flowing part full, with air above, never as a pressure pipe (rising mains and siphons are the exceptions). Manning’s equation is V = (1/n) R^(2/3) S^(1/2), with V in m/s, the hydraulic radius R = A/P in m, S the bed slope in m/m and n the roughness (about 0.013 is the usual design value for concrete and stoneware); Q = A V. A circular pipe of diameter D flowing full or half full has R = D/4. Example: D = 0.6 m, n = 0.013, S = 0.005 gives R = 0.15 m, V = (1/0.013) × 0.15^(2/3) × 0.005^(1/2) = 1.54 m/s, Q_full = 1.54 × (π × 0.6²/4) = 0.434 m³/s, and half full the discharge is 0.217 m³/s at the same velocity.
- Hydraulic elements of a circular sewer: at half depth V equals V_full and Q is Q_full/2. The velocity peaks near 0.8 D at about 1.14 V_full, and the discharge peaks near 0.95 D at about 1.07 Q_full, so a full pipe is not the one carrying most. Design depth at peak flow is held to about two-thirds to four-fifths of D, leaving air space for ventilation.
- Self-cleansing velocity: the velocity that scours solids deposited at low flow, at least about once a day. It depends on the size and density of the particles, and a design minimum of the order of 0.6 to 0.8 m/s at design flow is commonly aimed at. The upper limit, of the order of 2.4 to 3 m/s, protects the pipe wall from abrasion.
- Minimum gradient follows from the velocity: S = (n V / R^(2/3))². For V = 0.8 m/s, n = 0.013 and D = 0.3 m (R = 0.075 m), S = 0.0034, about 1 in 290. A larger pipe has a larger R, so it needs a flatter slope; small sewers need the steeper minimum gradients.
- Shapes and materials: the circular section is commonest (strong, easy to make, best hydraulic radius when full); the egg-shaped section keeps up the velocity at low flow, which suits combined sewers; large sewers may be horseshoe, U or box sections. Materials are RCC pipe, vitrified clay or stoneware, cast or ductile iron for siphons and rising mains, PVC or HDPE for small lines, and brick or concrete for large ones.
5. Sewer appurtenances
Appurtenances are the structures that let a sewer be inspected, cleaned, ventilated, relieved of overload and carried past obstacles. A sewer is a buried pipe that cannot be seen, so without them it cannot be maintained. Each connection from a building ends in an inspection chamber with a trap, and the sewer itself is ventilated through the manholes and house vent pipes, so that sewer gases (hydrogen sulphide, methane) do not build up or enter buildings.
| Appurtenance | Purpose | Where and how |
|---|---|---|
| Manhole | access for inspection, cleaning and flushing | at changes of direction, gradient and size, at junctions and at intervals on straight runs; a shaped channel (benching) in the floor guides the flow |
| Drop manhole | lets a branch sewer entering far above the main drop down without scouring the manhole or splashing workers | a vertical drop pipe, used when the invert is higher than the main by roughly 0.6 m or more |
| Inverted siphon (depressed sewer) | carries a gravity sewer under a river, railway or valley | dips and rises again, flowing full under pressure; several barrels keep the velocity above self-cleansing at low flow; chambers at both ends allow cleaning |
| Street inlet and catch basin | takes street runoff into the drain and holds back grit | kerb or gutter inlet; the basin pit has a water seal against odour |
| Flushing tank, grease and oil trap | flushes dead-end sewers with too little flow; stops fat and oil from kitchens, hotels and garages from clogging the sewer | at the head of small sewers; at the source of the discharge |
| Storm overflow, pumping station | a combined-sewer overflow spills flow beyond several times the dry-weather flow; a pumping station lifts sewage where gravity cannot serve | a weir or leaping weir; a wet well with duty and standby pumps |
6. Wastewater treatment
Treatment cuts the BOD, suspended solids, pathogens and nutrients so the effluent meets its disposal or reuse standard. Preliminary treatment (screens, a grit chamber that holds a horizontal velocity of about 0.3 m/s so that heavy grit settles but organics stay suspended, and oil and grease skimming) protects pumps and later units. Primary treatment is plain sedimentation, removing roughly 60% of suspended solids and about a third of the BOD, the settleable part. Secondary treatment is biological, with micro-organisms oxidising the dissolved and colloidal organic matter, followed by settling, and it removes most of the remaining BOD. Tertiary (advanced) treatment polishes the effluent by filtration, nutrient removal and disinfection, for reuse or a sensitive water body. The sludge is thickened, digested anaerobically (producing biogas), dried and used or disposed of.
| Process | How it works | Strong and weak points |
|---|---|---|
| Activated sludge (ASP) | suspended growth: sewage mixes with flocculent micro-organisms (MLSS) in an aeration tank with air supplied; a clarifier separates them, and part of the sludge is returned | compact, high-quality effluent; needs power, skill, and care against bulking sludge |
| Trickling filter | fixed film: a rotating distributor sprays sewage over a bed of stone or plastic media coated by a biofilm; air enters by natural draught | no aeration power, simple; filter flies and odour; needs a settling tank after it |
| Waste stabilisation ponds (WSP) | anaerobic, then facultative, then maturation ponds; algae make oxygen by photosynthesis and bacteria use it, so no mechanical aeration is needed | cheap to run, little skill, good pathogen removal; needs large land and sunshine; odour from anaerobic ponds |
Two ASP numbers are used often. The food-to-micro-organism ratio is F/M = Q S₀ / (V X), the BOD applied per day per unit mass of MLSS in the tank. For Q = 5,000 m³/d, S₀ = 200 mg/L, V = 1,000 m³ and X = 2,500 mg/L: F/M = 5,000 × 200 / (1,000 × 2,500) = 0.4 per day. The sludge volume index is SVI = (30-minute settled volume in mL/L × 1000) / MLSS in mg/L, in mL/g: 250 mL/L at 2,500 mg/L gives 100 mL/g. Values near 100 mean good settling; a high SVI signals bulking.
7. Sewage disposal: dilution, self-purification and land
Treated effluent goes to a water body by dilution (river, lake, or the sea through an outfall pipe with a diffuser beyond the surf zone), to land (irrigation, subject to crop restrictions, salinity and pathogens, and the soil’s capacity), or into reuse (cooling, flushing, gardens, recharge). The choice rests on the assimilative capacity of the receiving water and on the effluent standards that the pollution-control authorities set under the Water (Prevention and Control of Pollution) Act, 1974 and the Environment (Protection) Act, 1986. Mixing is a weighted average: BOD of the mix = (Q_r BOD_r + Q_s BOD_s) / (Q_r + Q_s). A river of 5 m³/s at 2 mg/L receiving 1 m³/s of effluent at 30 mg/L reaches (5 × 2 + 1 × 30)/6 = 6.67 mg/L.
A stream purifies itself by dilution, sedimentation, oxidation, reaeration from the air, and sunlight. Below an outfall it passes through zones of degradation, active decomposition, recovery and clean water. The oxygen sag curve (Streeter–Phelps) shows the dissolved-oxygen deficit D rising as bacteria use oxygen, then falling as the air restores it: dD/dt = k₁ L − k₂ D, where k₁ is the deoxygenation constant, k₂ the reaeration constant and L the remaining BOD. The critical point, where the oxygen is lowest, is where dD/dt = 0, that is, where the rate of deoxygenation equals the rate of reaeration and the deficit is greatest. Treatment must be enough that this lowest dissolved oxygen stays above what aquatic life needs.
8. Surface-water runoff and discharge of streams
Runoff is the part of rain that flows over the surface into drains. Urbanising a catchment raises the runoff coefficient and shortens the time of concentration, so the peak is higher and arrives sooner. For a small catchment the rational method gives the peak: Q = C i A / 360, with Q in m³/s, i the rainfall intensity in mm/h, A the area in hectares and C the runoff coefficient (peak runoff over rainfall, 0 to 1). The same formula is Q = 0.278 C i A when A is in km². The units must be consistent: the divisor 360 belongs to mm/h and hectares only. Assumptions: rain falls uniformly over the catchment at a steady intensity; the design storm lasts as long as the time of concentration t_c, so that the whole catchment is contributing; C is constant; there is no storage. It is therefore for small areas such as urban drainage.
The time of concentration is the inlet time (overland flow to the first inlet) plus the time of flow in the drain, L/V. The intensity i is read from the intensity–duration–frequency (IDF) curve for the duration t_c and the chosen return period. For a mixed catchment the coefficient is the area-weighted mean, C = Σ C_j A_j / Σ A_j, with values from about 0.05 to 0.2 for lawns and parks to 0.8 to 0.95 for roofs and pavements. Example: 4 ha paved (C = 0.9), 6 ha housing (C = 0.5) and 2 ha lawn (C = 0.2): Σ C A = 3.6 + 3.0 + 0.4 = 7.0 ha; at i = 72 mm/h, Q = 7.0 × 72/360 = 1.4 m³/s.
Discharge of a stream is measured or estimated several ways. The velocity–area method divides the cross-section into strips and sums velocity (current meter, at 0.6 depth or the mean of 0.2 and 0.8 depth) times area. A stage–discharge (rating) curve, Q = K (h − h₀)ⁿ, then gives discharge from the water level alone. The slope–area method applies Manning’s equation to flood marks, and dilution gauging uses a tracer. For ungauged catchments, empirical formulas such as Dickens’ Q = C A^(3/4) (A in km², Q in m³/s, C a regional coefficient) are used, with Ryves’ A^(2/3) form for other regions. For a design flood, the return period T = 1/p, and the chance of at least one flood of that return period in n years is 1 − (1 − 1/T)ⁿ. For T = 50 and n = 25 this is 1 − 0.98²⁵ = 0.40, which is why a ‘50-year’ flood is no safe bet over a 25-year project life.
9. Flood protection and storm-water drains
Flood protection has two families. Structural measures change the flood: storage reservoirs and detention basins that cut the peak; embankments (levees) and flood walls that confine the river; channel improvement (widening, deepening, cut-offs); and diversion channels or floodways. Non-structural measures change exposure: flood-plain zoning, which bars permanent building in the floodway and allows only parks and farming there; flood forecasting and warning with evacuation plans; flood-proofing of buildings, such as raised plinths; flood insurance; and watershed management and preservation of lakes and wetlands that store flood water. For the planner the non-structural measures are central, since zoning and development control cost little and avoid the damage.
Storm drains are laid out on the ground slope to reach the nearest stream, lake or low point by short routes. Kerb-and-gutter carry road runoff to inlets (with catch basins that hold back grit); drains run along roads, open (rectangular, trapezoidal or U) where there is room, covered or piped where there is not; manholes sit at junctions, culverts carry them under roads, and the outfall to a stream has a flap gate and an energy dissipator against backflow and scour. Sewage and solid waste must be kept out of them.
The design flow is the rational-method peak for the chosen return period, longer for important or flood-prone areas. The section is designed with Manning’s equation, so that velocity stays above a non-silting minimum and below a non-scouring maximum, with a freeboard above the water. The most efficient rectangular section has the least wetted perimeter for its area, which gives b = 2y and R = y/2. Example: b = 2 m, y = 1 m, n = 0.015, S = 0.001 gives A = 2 m², P = 4 m, R = 0.5 m, V = (1/0.015) × 0.5^(2/3) × 0.001^(1/2) = 1.33 m/s and Q = 2.66 m³/s.
10. Rainwater harvesting
Rainwater harvesting (RWH) captures rain where it falls, to use it or to recharge the aquifer. Roof-top RWH takes the roof as the catchment: gutters and a down-pipe lead the water through a first-flush diverter, which discards the first, dirty runoff that washes the roof, then a filter of mesh, gravel and sand, and finally into either a storage tank (for use) or a recharge structure (pit, trench, shaft or recharge well) for the groundwater. Surface-runoff harvesting collects from larger open areas into ponds, check dams and percolation tanks.
The harvestable volume is V = R × A × C, with R the rainfall depth in m, A the catchment area in m² and C a runoff coefficient for losses (roofs are high, typically 0.8 or more). Example: R = 800 mm, A = 150 m², C = 0.8 gives V = 0.8 × 150 × 0.8 = 96 m³ a year. A storage tank is sized by the dry-spell method: daily demand × the number of dry days to be bridged, so 400 L/day over 120 dry days needs 48,000 L = 48 m³; but the tank is useful only up to the volume the roof can supply, so the smaller of the two limits governs.
Key takeaways
- On-site sanitation (septic tank with soak pit, twin-pit latrine) needs safe sludge management; decentralised sewerage treats near the source and reuses locally; the septic-tank effluent is not safe to discharge untreated.
- Separate systems suit intense monsoon storms; sewage is about 80% of supply, peak = average × a peak factor (Harmon M = 1 + 14/(4 + √P)); layouts are perpendicular, interceptor, zone, fan and radial.
- Sewers are open-channel designs by Manning, V = (1/n) R^(2/3) S^(1/2); half-full has the full-pipe velocity and half its discharge; maximum velocity and discharge come at 0.8 D and 0.95 D; the minimum gradient is set by the self-cleansing velocity and is steeper for small pipes.
- Appurtenances serve access (manholes), steep entries (drop manholes), obstacles (inverted siphons) and overflow; treatment runs preliminary, primary, secondary (ASP, trickling filter, WSP) and tertiary, with F/M = Q S₀/(V X) and SVI as the ASP control numbers.
- Disposal is by dilution, land or reuse within effluent standards; the oxygen sag’s critical point is where deoxygenation equals reaeration; the rational method is Q = C i A / 360 (m³/s, mm/h, ha) with i taken at the time of concentration; flood risk over n years is 1 − (1 − 1/T)ⁿ.
- Non-structural flood measures (zoning, warning, flood-proofing) are the planner’s main tools since levees only shift the flood; storm drains follow the slope, are designed by Manning for the return-period peak, and the efficient rectangle has b = 2y; RWH volume is R × A × C, and recharge pits take only clean water.
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.
At a junction of two sewers, or where a sewer changes direction, gradient or size, which appurtenance is normally provided so that the sewer can be inspected and cleaned?
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Answer: C — A manhole, which gives workers access and lets rods and cameras into the line
Manholes are placed at junctions, at changes of direction, gradient or size, and at regular intervals along a straight run, so that the sewer can be inspected, rodded and cleaned. A catch basin belongs to storm-water inlets, an inverted siphon crosses an obstruction, and an overflow weir relieves a combined sewer in a storm; none of them is there to give access to the line.A town has intense but short monsoon storms and means to build its sewers first, with storm drains following over the years. The water-carriage system that suits best is
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Answer: A — separate, because the plant sees only sewage and storm water needs no treatment
With intense storms the storm flow is many times the dry-weather flow, so a combined sewer would be oversized and run almost empty on dry days, and the plant, which needs a steady feed, would be swamped. Separate systems also allow sewers to be built before drains. Sizing a plant for the storm flow is the tempting idea, but it makes the plant uneconomic on all other days.The sewage of a town on a plateau is led outward by sewers in several directions to treatment works at the edge of the town. This layout is called
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Answer: D — radial
Sewers that diverge from a high central point to several outlets at the periphery form the radial layout. The fan layout is its reverse, with sewers converging on one point, as for a town in a valley. The perpendicular layout runs to a river by the shortest line, and the interceptor layout adds a collecting sewer along it.To give the same self-cleansing velocity, a small sewer must be laid at a steeper gradient than a large one, because
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Answer: C — its hydraulic radius is smaller, so Manning needs a larger slope for the same velocity
From Manning, S = (n V / R^(2/3))². For a circular pipe R = D/4, so a smaller D means a smaller R and a larger required S at the same V. The roughness n depends on the pipe material and finish, not on its diameter, so blaming n is the tempting error.On the oxygen sag curve below a sewage outfall, the critical point, where the dissolved oxygen is lowest, occurs where
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Answer: B — the rate of deoxygenation equals the rate of reaeration, so the deficit is greatest
The deficit grows while deoxygenation outpaces reaeration and shrinks when the reverse holds, so it peaks where the two rates are equal, dD/dt = 0. At that point BOD is still present, so the stream is not yet free of load; the zero-BOD idea confuses the end of the degradation with the point of greatest deficit.An inverted siphon (depressed sewer) is used when a gravity sewer
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Answer: A — dips beneath a river or railway and rises on the far side, running full under pressure between
An inverted siphon passes under an obstruction and runs full under pressure, with several barrels so that low flows still keep a self-cleansing velocity. Lifting sewage uphill needs a pump and rising main, not a siphon; a vertical fall into a main is a drop manhole; and spilling excess flow over a weir is a storm overflow.An aeration tank of 1,500 m³ receives 6,000 m³/day of settled sewage with a BOD of 250 mg/L, and the MLSS is 3,000 mg/L. The F/M ratio, in kg BOD per kg MLSS per day, is
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Answer: C — 0.33
F/M = Q S₀ / (V X) = (6,000 × 250) / (1,500 × 3,000) = 1,500,000 / 4,500,000 = 0.33 per day; in kg, the BOD load is 1,500 kg/day on 4,500 kg of MLSS. A value of 0.50 results from using a tank volume of 1,000 m³ instead of 1,500 m³.A small peri-urban town in a warm climate has cheap, plentiful land, an unreliable power supply and few skilled operators, and it needs moderate effluent quality with good pathogen removal. The most suitable secondary treatment is
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Answer: D — a series of waste stabilisation ponds: anaerobic, facultative and maturation
Waste stabilisation ponds need land, sunshine and warmth, which the town has, and almost no power or skill; the maturation pond also removes pathogens well. The aerated and pumped options all depend on reliable electricity and trained operators, which the town lacks, so they are the tempting but unsuitable choices.In the rational method, the rainfall intensity used for the design peak flow is that for a duration equal to
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Answer: B — the time of concentration of the catchment
The peak occurs when the whole catchment contributes, which first happens after the time of concentration, so the design storm lasts t_c and i is read from the IDF curve at that duration. A longer storm has a lower intensity and a shorter one does not yet engage the whole area; the return period is a different axis of the IDF curve altogether.Embankments (levees) are built along both banks of a river through a town. A consequence the planner must allow for is
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Answer: A — higher flood levels upstream, a faster, higher flood downstream, and sluices or pumps for interior drainage
Confining the river removes the storage on its flood plain, so water rises upstream and the flood passes downstream faster and higher. The town’s own rain cannot enter a high river, so the interior needs sluices with flap gates and pumps. A levee never ends the risk: overtopping and breach remain, which is why the ‘cannot be overtopped’ claim is wrong.For a given cross-sectional area, the most efficient rectangular drain section has
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Answer: C — width equal to twice the depth
The wetted perimeter b + 2y is least, for a fixed area b y, when b = 2y. Then R = A/P = 2y²/4y = y/2. A square section is less efficient because its perimeter for the same area is larger.Which of the following statements about a circular sewer flowing partly full are correct?
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Answer: A — At half depth the velocity equals the full-pipe velocity and the discharge is half the full-pipe discharge; B — The velocity is greatest at a depth of about 0.8 of the diameter; C — The discharge is greatest at a depth of about 0.95 of the diameter, slightly above the full-pipe discharge
At half depth R is the same as at full depth, D/4, so V is the same, and the area is half. Near the crown the wetted perimeter keeps growing while the added area becomes small, so R, and with it V, peaks a little below full depth, at about 0.8 D (about 1.14 V_full) and Q at about 0.95 D (about 1.07 Q_full). So a full pipe is not the one carrying most, which makes the claim that discharge is greatest at full depth false.Which of the following are disadvantages of a combined system compared with a separate system?
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Answer: A — The treatment plant must handle large and rapidly varying flows; B — In heavy storms diluted sewage may overflow into water bodies
The combined system pushes storm swings onto the plant and spills diluted sewage to rivers when the overflows operate. A combined sewer is larger, not smaller, because it carries storm flow, so the claim of smaller, cheaper sewers is false; and the need for two networks is a feature of the separate system, not the combined one.Which of the following are non-structural flood-management measures?
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Answer: B — Flood-plain zoning that bars permanent building in the floodway; C — Flood forecasting and warning with evacuation plans; D — Raising the plinths of buildings in flood-prone areas
Non-structural measures reduce exposure and damage without changing the flood: zoning, warning and evacuation, and flood-proofing such as raised plinths. Embankments are the classic structural measure, since they alter how the flood flows.Which of the following statements about roof-top rainwater harvesting are correct?
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Answer: A — A first-flush diverter discards the initial dirty runoff that washes the roof; B — The harvestable volume is rainfall depth × roof area × a runoff coefficient
The first flush carries dust and droppings, so it is diverted, and the volume is R × A × C. Sewage or septic overflow in a recharge pit pollutes the aquifer, and storage beyond what the roof can supply simply stays empty.A town of 36,000 people is supplied at 135 lpcd and 80% of the supply reaches the sewers. The peak factor is given by Harmon’s formula M = 1 + 14/(4 + √P), with P in thousands. The peak sewage flow, in L/s, is ____.
Numerical answer — type the value.
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Answer: 108
Water supplied = 36,000 × 135 = 4,860,000 L/d; sewage = 0.8 × 4,860,000 = 3,888,000 L/d. P = 36, √36 = 6, so M = 1 + 14/10 = 2.4. Peak = 3,888,000 × 2.4 = 9,331,200 L/d, and dividing by 86,400 s/d gives 108 L/s. Check: 3,888,000/86,400 = 45 L/s average, and 45 × 2.4 = 108.A circular sewer of diameter 0.3 m, with Manning’s n = 0.013, is laid at a slope of 0.004 and flows half full. Its velocity, in m/s, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.87
At half depth the hydraulic radius equals that of the full pipe, R = D/4 = 0.075 m. V = (1/0.013) × 0.075^(2/3) × 0.004^(1/2) = 76.92 × 0.1778 × 0.06325 = 0.865, so 0.87 m/s. The velocity is the same as for the full pipe, so the half-full velocity can be got from the full-pipe formula.A catchment has 5 ha paved (C = 0.85), 3 ha housing (C = 0.4) and 2 ha lawn (C = 0.15). The inlet time is 12 minutes, and the water then flows 600 m in a drain at 1.0 m/s. The design intensity is i = 2400/(t + 20) mm/h with t in minutes. By the rational method Q = C i A / 360 (Q in m³/s, i in mm/h, A in ha), the peak flow, in m³/s to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.91
Σ C A = 5×0.85 + 3×0.4 + 2×0.15 = 4.25 + 1.2 + 0.3 = 5.75 ha. Flow time = 600/1.0 = 600 s = 10 min, so t_c = 12 + 10 = 22 min. i = 2400/(22 + 20) = 57.14 mm/h. Q = 5.75 × 57.14/360 = 0.913, so 0.91 m³/s. Check: 5.75 × 57.14 = 328.6, and 328.6/360 = 0.913.A roof of 120 m² receives an annual rainfall of 900 mm. With a runoff coefficient of 0.85, the volume that can be harvested in a year, in m³ to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 91.8
V = R × A × C = 0.9 m × 120 m² × 0.85 = 108 × 0.85 = 91.8 m³. Convert 900 mm to 0.9 m first; using 900 directly would give 91,800, a thousand times too large.A flood of return period 50 years is adopted for design. The probability that at least one flood equal to or greater than it occurs during a 25-year project life, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.40
The yearly chance of exceedance is p = 1/50 = 0.02. The chance of none in 25 years is 0.98^25 = 0.6035. The chance of at least one is 1 − 0.6035 = 0.3965, so 0.40. The tempting answer 25/50 = 0.5 adds yearly chances, which overcounts.