Hydrology and Irrigation: Runoff, Hydrographs, Groundwater, Duty and Canals

Hydrology is the one part of civil engineering where the loading is measured rather than specified, and everything about the subject follows from that. A structural engineer is told the live load; a hydrologist has to infer the design flood from a record that is short, incomplete and non-stationary, and the honest answer always carries a return period rather than a number. The chain runs: rainfall falls on a catchment, some of it is lost to interception, depression storage, evaporation and above all infiltration, and what remains becomes the rainfall excess that appears at the outlet as direct run-off. Separating those two is the central practical problem, and the crudest useful tool — the φ-index, a constant loss rate that makes the excess equal the observed runoff — is worth understanding precisely because its crudeness shows where the difficulty lies. The unit hydrograph then converts any excess into a discharge hydrograph on the assumption that the catchment is a linear, time-invariant system, which is false in detail and serviceable in practice. For small catchments the rational method Q = CiA/3.6 short-circuits all of it into one coefficient, and the coefficient carries every assumption. Underground, Darcy’s law governs, and steady well hydraulics gives the drawdown a logarithm of distance — which is why a well’s influence extends far but weakly. Then irrigation, whose whole vocabulary is one relation: duty, the area a unit discharge can serve, and delta, the depth of water a crop receives, are inverses of each other through the base period, Δ = 8.64B/D. Canals are sized by regime theory, weirs on permeable ground by creep, and both are empirical in a way that is worth being frank about rather than dressing up.

The Hydrologic Cycle, the Watershed, and What Gets Measured

The hydrologic cycle is the closed accounting the whole subject sits inside: water evaporates from the oceans and the land, is transported and falls as precipitation, and returns by evaporation, transpiration, surface runoff and groundwater flow. Written as a balance over any bounded area and any period, P − R − G − E − T = ΔS: precipitation in, less runoff out, less groundwater outflow, less evaporation and transpiration, equals the change in storage. The area it is written over is the watershed or catchment — the region draining to a single outlet, bounded by the ridge line, and it is a topographic definition rather than an administrative one, which is why river basin management crosses district and state boundaries and administrative water management is so often at odds with it. The catchment is the natural unit because it is the only area over which that balance closes.

Precipitation is measured at points and needed over areas, and the three standard ways of getting from one to the other differ in what they assume. The arithmetic mean is adequate only when the gauges are evenly spread and the terrain flat. The Thiessen polygon method weights each gauge by the area closer to it than to any other, which respects an uneven network and still assumes rainfall is uniform within each polygon. The isohyetal method contours the rainfall itself and weights the area between contours, which is the only one of the three that can represent orographic effects — and the only one that requires judgement, since the contours are drawn rather than computed. Whichever is used, the evaporation and evapotranspiration terms are the hardest to measure and are usually estimated: a pan coefficient applied to a measured pan evaporation, or an energy-balance or empirical formula for crop evapotranspiration. It is worth being blunt that in most catchment balances the residual error is concentrated in these terms, which is why a water balance closes to within a few per cent at best.

🧠 Flood routing: what a reach does to a flood wave, in two constants
A flood wave entering a river reach or a reservoir does not leave it unchanged: it is delayed, and its peak is attenuated, because the reach stores water on the rising limb and gives it back on the falling one. Routing computes the outflow hydrograph from the inflow, and it is the continuity equation I − Q = dS/dt closed by some relation between storage and flow. For a reservoir the water surface is level, so storage depends on the outflow alone and the routing is a one-variable problem — level pool routing. For a river reach the surface is sloped, so storage depends on both, and the Muskingum method writes S = K[xI + (1 − x)Q] with two constants. K is a travel time: at steady state I = Q and S = KQ, so K is the storage the reach holds per unit of steady discharge — for K = 12 h at 60 m³/s that is 720 m³/s·h, about 2.6 million m³. x is a weighting between inflow and outflow, and it is the more interesting of the two because it sets how much attenuation happens. At x = 0 the storage depends only on outflow, the reach behaves as a level-pool reservoir, and the attenuation is maximum. At x = 0.5 inflow and outflow are weighted equally, storage is a pure wedge, and the wave is translated without attenuation at all. Real river reaches sit between, typically 0.1 to 0.3. So the two constants separate the two things a reach does — K says how long it delays the wave and x says how much it flattens it — and a candidate who reports a peak reduction without knowing x has guessed the answer to the question that was asked.

From Rainfall to Runoff: Losses, the φ-index and the Rational Method

Not all rain becomes runoff. Some is intercepted by vegetation, some fills surface depressions, some evaporates, and much of it infiltrates. What is left is the rainfall excess, and it is the excess — not the rainfall — that produces the direct runoff hydrograph at the catchment outlet. Infiltration itself decreases with time during a storm as the soil wets up, following something like Horton’s exponential decay, but for design work a much cruder device is used: the φ-index, a single constant loss rate in cm per hour chosen so that the resulting excess equals the observed direct runoff. It has no physical basis beyond that definition, and its value is that it can be extracted from a rainfall record and a stream gauge without knowing anything about the soil.

⚠️ The φ-index is not the average loss, and here they differ by 17 per cent
Take a storm of 0.5, 3.5, 2.5 and 1.5 cm in four successive hours — total 8 cm — producing 4 cm of direct runoff. The tempting move is to divide the losses evenly: (8 − 4)/4 = 1.0 cm/h. Test it before believing it. At φ = 1.0 the excess is 0 + 2.5 + 1.5 + 0.5 = 4.5 cm, not 4. The arithmetic average has failed, and the reason is the one that makes this a real technique rather than a division: an hour whose rainfall is below φ contributes zero excess, not negative excess. The first hour delivers only 0.5 cm, so at any φ above 0.5 it is entirely lost and contributes nothing — but the naive average has quietly credited it with absorbing a full 1.0 cm it never had. Solve it properly by assuming the first block is entirely lost and letting the other three carry the excess: (3.5 + 2.5 + 1.5) − 3φ = 4, so φ = 7.5/3 = 1.1667 cm/h. Then check the assumption — 0.5 < 1.1667 ✓ — and verify the total: 0 + 2.333 + 1.333 + 0.333 = 4.000 cm ✓. The two answers differ by 16.7 per cent, which on a design flood is not a rounding matter. The method generalises to a rule: guess which blocks are active, solve, then check that every block you excluded really is below φ and every block you included really is above it. If the check fails, move a block and solve again. That iteration is the whole technique, and a candidate who reaches for the average is skipping the only part of it that requires thought.

For a small catchment — a culvert, an urban drain, a storm sewer — the whole chain is collapsed into the rational method, Q = CiA/3.6 with i in mm/h, A in km² and Q in m³/s. The rainfall intensity i is taken at a duration equal to the time of concentration, the time for runoff from the remotest point to reach the outlet, on the argument that this is the duration at which the whole catchment contributes at once. Everything else — infiltration, storage, antecedent wetness, the shape of the catchment — is inside the runoff coefficient C. At C = 0.6, i = 50 mm/h and A = 2 km², Q = 16.7 m³/s. Now urbanise the same catchment so C rises to 0.9: the peak becomes 25.0 m³/s, half as much again, with no change whatever in the rainfall. That is the arithmetic behind urban flooding, and it is why drainage designed for a green catchment fails once the catchment is paved. The method’s honest limits are worth stating: it gives a peak only, not a hydrograph, assumes uniform rainfall over the whole area, and is reliable for catchments of a few square kilometres at most.

The Unit Hydrograph, and the Assumption It Rests On

A unit hydrograph is the direct runoff hydrograph produced by one unit depth of rainfall excess — conventionally 1 cm — falling uniformly over the catchment in a specified duration. The duration is part of its name, so a 2-hour unit hydrograph and a 6-hour one are different curves for the same catchment and are not interchangeable. Its power comes from two assumptions. Linearity: doubling the excess doubles every ordinate, so 2.5 cm of excess produces a hydrograph 2.5 times as tall with the same time base. Time invariance: the same excess always produces the same shape, whenever it falls. Together these let you convolve any storm — decomposed into unit-duration blocks — into a complete hydrograph by scaling and superposing shifted copies. Neither assumption is true in detail: a large storm routes through the catchment faster than a small one, and a wet catchment responds differently from a dry one. They are close enough to be useful and their failure is where the method’s error lives.

One arithmetic check makes unit hydrograph problems safe: the area under the direct runoff hydrograph must equal the volume of rainfall excess. For a unit hydrograph that volume is 1 cm × A, so on a 100 km² catchment it is 100 × 10⁶ × 0.01 = 1.0 × 10⁶ m³ exactly. Integrate the ordinates you have derived, in m³/s multiplied by the time interval in seconds, and if the total does not come back to that figure the derivation has gone wrong somewhere — usually in the base flow separation, which is the step that requires judgement and is therefore the step that fails. Base flow is the groundwater contribution that was in the stream before the storm and would have been there without it; it must be removed before the direct runoff hydrograph can be scaled, and added back after. The volume check catches an error in that separation immediately, which is why it is worth running before doing anything else with the result.

Groundwater: Darcy, Aquifers and the Logarithm of Distance

Groundwater moves under Darcy’s law, v = ki, and the relevant properties of an aquifer combine into two: the transmissivity T = kb, the product of permeability and saturated thickness, and the storativity, the volume released per unit area per unit head decline. A confined aquifer lies between two impermeable layers and is under pressure, so water rises above the top of the aquifer in a well and the water released comes from expansion of the water and compression of the matrix — hence a very small storativity. An unconfined aquifer has a free water table, and the water released is actual drainage of pores, so its storativity is a hundred to a thousand times larger. That difference decides how far and how fast a pumping well affects its neighbours.

ℹ️ A well reaches far because the drawdown is logarithmic
For steady radial flow to a fully penetrating well in a confined aquifer, the discharge across any cylinder of radius r is the same, so Q = 2πrb k (dh/dr). Integrating between two observation wells gives the Thiem equation, Q = 2πkb(h₂ − h₁)/ln(r₂/r₁). Put numbers on it: Q = 1000 m³/day from an aquifer with k = 25 m/day and b = 20 m, so T = 500 m²/day. Between r = 10 m and r = 100 m the head difference is 1000 × ln(10)/(2π × 500) = 0.733 m. Now take the outer well further out, to 200 m: the difference rises only to 0.954 m. Doubling the distance added 0.22 m, while the first factor of ten added 0.73. That is the logarithm at work, and it has two consequences worth holding. The cone of depression is deep and narrow near the well and very shallow and very wide far from it, so a pumping well affects neighbouring wells over a large radius but only slightly — which is exactly the pattern that makes aquifer over-abstraction hard to attribute to any one user and easy for everyone to blame on someone else. And because the relation is logarithmic, the radius of influence is a soft concept: there is no distance at which the drawdown becomes zero, only a distance beyond which it is too small to measure, which is why the term is defined by convention rather than by physics. For an unconfined aquifer the same derivation with h in place of b gives the Dupuit equation, Q = πk(h₂² − h₁²)/ln(r₂/r₁), in which the squares appear because the saturated thickness itself is what is being drawn down.

Irrigation: Duty and Delta, Canals, and Weirs on Sand

The vocabulary of irrigation looks like several ideas and is really one. Duty D is the area in hectares that a discharge of one cumec, running continuously through the base period, can irrigate. Delta Δ is the total depth of water the crop receives over that period. They are two views of the same water, and the relation between them is pure unit conversion: one cumec running for B days delivers 86 400 B m³, spread over D hectares, that is 10⁴ D m², giving Δ = 8.64 B/D metres. At B = 120 days and D = 1800 ha/cumec, Δ = 0.576 m = 57.6 cm. Halve the duty to 900 and delta doubles to 115.2 cm — duty and delta are inverses, so a high duty is an efficient system and a large delta is a thirsty crop. Rice runs at a delta of well over a metre; wheat at a fraction of it. The crop water requirement underneath both is the evapotranspiration the crop needs, and delta must supply that plus the losses in conveyance and application.

An unlined canal in alluvium cannot be designed like a rigid channel, because its own bed and banks are made of the material it carries. Regime theory treats it as a channel that has adjusted its width, depth and slope until it neither silts nor scours, and Lacey’s relations give those from the discharge and a silt factor f describing the sediment. At Q = 30 m³/s and f = 1.0: wetted perimeter P = 4.75√Q = 26.02 m, hydraulic radius R = 0.47(Q/f)^(1/3) = 1.460 m, and regime slope S = f^(5/3)/(3340 Q^(1/6)) = 1.70 × 10⁻⁴, that is 1 in 5888. The area follows as A = PR = 38.0 m² and the mean velocity as 0.79 m/s, which is the velocity at which this sediment stays in motion without scouring the bed. These relations are empirical, fitted to canals in northern India in the early twentieth century, and they should be described that way rather than as theory — they carry no derivation and their dimensional inconsistency is a sign of it.

Ahead of any of that sits the choice of irrigation system and method, and the syllabus names it first for a reason. Surface methods — border, check basin, furrow — deliver water by flooding and letting gravity spread it: cheap in equipment, expensive in water, with application efficiencies commonly around 40 to 60 per cent because so much is lost to deep percolation and evaporation from wetted soil. Sprinkler irrigation pressurises the water and distributes it as rain, which suits undulating ground and sandy soils that would not spread water sideways, and reaches perhaps 70 to 80 per cent. Drip or trickle irrigation delivers water slowly to the root zone alone, wetting only a fraction of the surface, and is the most efficient at around 90 per cent — and the most capital-intensive and the most vulnerable to clogging. The pattern to hold is that efficiency is bought with capital and with dependence: the more efficient the method, the more it costs to install and the more completely a failure of pumps, filters or emitters stops the irrigation altogether. Efficiency also feeds straight back into the arithmetic above, because delta must supply the crop requirement plus whatever the method wastes, so the same crop has a larger delta under furrow than under drip.

One structure type completes the list and exists for a purely geometric reason: a canal and a natural drainage line, both running across country, must eventually meet, and one of them has to give way. That is a cross drainage work, and which of the three forms is used depends only on the relative levels. An aqueduct carries the canal over the drain, and is chosen when the canal bed is comfortably above the drain’s high flood level. A super-passage does the reverse, carrying the drain over the canal, when the canal is the lower of the two. A level crossing admits both into a common pool and lets them separate downstream through regulators, and is used when the levels are too close for either to pass cleanly — the least satisfactory of the three, because it mixes silt-laden flood water into the canal. The decision is not a design preference but an arithmetic one taken from a levelling survey, and getting it wrong is not a matter of efficiency: an aqueduct built where the flood level is higher than assumed is drowned, and the canal it carries is lost with it.

🎯 A weir on sand fails by seepage under it, not by force against it
A weir founded on rock is a stability problem — sliding, overturning, bearing. A weir founded on permeable alluvium is something else: the water does not merely push against it, it travels underneath it, and the two ways that kills a structure are both invisible from above. At the downstream exit the upward seepage can lift the grains and start piping, a channel that erodes backwards under the floor until the structure loses its foundation; and along the base the residual pore pressure creates uplift that must be resisted by the weight of the floor. Bligh’s creep theory is the simplest treatment: it assumes the head is dissipated uniformly along the shortest seepage path, so the required creep length is L = C H with C a coefficient of the soil. For a 4 m head, gravel at C = 5 needs 20 m of creep, coarse sand at C = 12 needs 48 m, and fine sand at C = 18 needs 72 m — the finer the material, the longer the path required, because the finer material is the easier to pipe. Lane’s weighted creep improves on it by recognising that seepage does not treat all directions alike: vertical creep is counted at full value and horizontal creep at only one third, so a vertical cut-off is three times as effective per metre as an equal length of horizontal floor. That is the design consequence, and it is why sheet-pile cut-offs are driven at the upstream and downstream ends of a weir floor rather than the floor simply being made longer. Khosla’s later analysis replaced the creep assumption altogether with a proper potential-flow solution, showing that the head is not dissipated uniformly and that the exit gradient concentrates sharply at the downstream cut-off — which is where the piping check actually belongs.

Streamflow measurement

The GATE 2027 revision adds streamflow measurement to this section, and it belongs here for a reason the chapter has already made: hydrology is the one subject whose loading is measured rather than specified, so how the measurement is made is part of the subject and not a laboratory detail. The workhorse is the area–velocity method. Divide the cross-section into vertical strips, measure the depth and the mean velocity in each, and sum Σ (width × depth × velocity). Velocity comes from a current meter, and the sampling rule is fixed by depth: one point at 0.6 of the depth for a shallow vertical, and the mean of 0.2 and 0.8 for a deeper one — two readings averaging out the logarithmic velocity profile that a single reading cannot.

🧠 The mid-section method, worked once
Take a stream gauged at 2 m intervals across a 12 m width, with the water’s edge at 0 m and 12 m. Depths and mean velocities at the six interior verticals are (2 m, 0.50 m, 0.30 m/s), (4, 1.20, 0.65), (6, 1.50, 0.80), (8, 1.10, 0.60) and (10, 0.40, 0.25). In the mid-section method each vertical is credited with the strip reaching halfway to its neighbours on each side — here 2 m for every interior vertical — so the partial discharges are 2 × 0.50 × 0.30 = 0.300, then 1.560, 2.400, 1.320 and 0.200 m³/s, and the total is 5.780 m³/s. The area is 2 × (0.50 + 1.20 + 1.50 + 1.10 + 0.40) = 9.400 m², so the mean velocity of the whole section is 5.780/9.400 = 0.615 m/s. Three things this makes visible. The deepest, fastest vertical dominates: the strip at 6 m contributes 2.400 of the 5.780, over 40% of the discharge from a sixth of the width, which is why the gauging interval must be closer where the bed changes fastest and why a uniformly spaced gauging can miss a scour hole entirely. The edge verticals contribute nothing in this method, since depth and velocity are both zero there — the alternative mean-section method averages adjacent verticals instead and treats the edges differently, giving a slightly different answer on the same data, so a question must be read for which method it wants. And the whole exercise is far too slow to repeat daily, which is the point of a stage–discharge rating curve: gauge properly a few dozen times, fit Q = C(G − a)^β to the pairs, and thereafter read the stage off a gauge board and infer the discharge. The honest caveat is that a rating curve is only valid while the channel keeps the shape it had when the curve was fitted — a flood that scours or silts the section invalidates it, and extrapolating one above the highest gauged flow is a guess dressed as a measurement.

Earthen dams, river training and well irrigation

The 2027 revision adds three more items to this section, and each is a place where the water goes somewhere it was not intended to. Seepage through earthen dams is the first. An earth dam is permeable by construction, so the question is never whether water passes through it but where the seepage surface sits and whether the exit is controlled. The top streamline is the phreatic line, located by Casagrande’s construction, and the discharge follows from q = k a sin²α, where a is the length of the seepage face on the downstream slope and α its angle. For k = 10⁻⁶ m/s, α = 30° and a = 8 m, q = 2.0 × 10⁻⁶ m³/s per metre — 172.8 litres per day per metre of dam, which is a negligible loss of water and an entirely non-negligible risk, because the danger from seepage is never the volume. It is piping: seepage emerging on the downstream face with enough exit gradient to carry soil particles with it, which erodes a channel backwards into the dam and fails it. That is why every earth dam has a filter — a graded rock toe, a chimney drain, an inclined blanket — whose job is to let the water out while holding the soil in, and why the design criterion is the exit gradient rather than the discharge.

River training is the second, and it is the acknowledgement that an alluvial river will not stay where it is put. Its works are all attempts to fix in place one property of a channel that would otherwise migrate: guide banks hold the approach to a bridge so the river presents itself square to the waterway; groynes or spurs project into the flow to push the deep channel away from a bank being attacked; revetment and pitching armour a bank against the velocity it cannot resist unaided; and cut-offs shorten a meander that has grown too long. Each of them works by changing where the erosive velocity goes rather than by resisting it everywhere, and each has the same failure mode — the flow attacks the structure’s own downstream end, which is why every groyne field is designed as a series rather than as one structure.

🧠 Well irrigation: a recuperation test measures the aquifer, not the well
Well irrigation is the third addition, and it splits into two devices with two different limits. A tube well taps a confined or deep unconfined aquifer and is limited by the aquifer’s transmissivity — the Thiem analysis of the previous section. An open (dug) well taps a shallow water table and is limited by how fast water seeps in through its walls and floor, which is measured by a recuperation test: pump the well down, stop, and time the recovery. If the depression head falls from h₁ to h₂ in time T, then K/A = (1/T) ln(h₁/h₂). For h₁ = 2.5 m, h₂ = 1.0 m and T = 1.5 hours, K/A = (1/1.5) ln 2.5 = 0.6109 per hour, and the safe yield at a working depression head H is (K/A) × A × H: with a well area of 4 m² and H = 3 m, that is 7.330 m³/h, or 175.93 m³/day. The name of the quantity is the trap. K/A is called the specific yield, it has units of per unit time, and it is a property of the soil surrounding the well and not of the well — which is why the same test on a wider well in the same ground gives the same K/A and a proportionally larger yield, and why deepening a well in a poorly permeable stratum buys much less than widening it. Note also that the yield is proportional to the depression head, so a well pumped harder yields more only until the water table is drawn down to the point where the head cannot be maintained — which is the physical form of the over-abstraction that has emptied shallow aquifers across northern India.

Key takeaways

  • The φ-index is not the average loss. On the worked storm the average gives 1.0 cm/h and the correct answer is 1.167 — because a block below φ contributes zero excess, not negative. Guess the active blocks, solve, then check.
  • Q = CiA/3.6 puts every assumption inside C. Urbanising a catchment from C = 0.6 to 0.9 raises the peak by 50% with no change in rainfall — the arithmetic of urban flooding.
  • A unit hydrograph assumes the catchment is linear and time-invariant. Always check that the area under the direct runoff equals the excess volume — 1 cm over 100 km² is exactly 10⁶ m³, and a mismatch means the base flow separation is wrong.
  • Steady well drawdown goes as ln(r₂/r₁), so distance matters weakly: taking the outer well from 100 m to 200 m added only 0.22 m against 0.73 m for the first factor of ten. The cone of depression is deep and narrow, then shallow and very wide.
  • Δ = 8.64 B/D — duty and delta are inverses. 1800 ha/cumec over 120 days is a delta of 57.6 cm; halving the duty doubles the water the crop gets.
  • A weir on sand fails by seepage under it: piping at the exit and uplift on the floor. Bligh needs L = CH; Lane counts vertical creep at full value and horizontal at a third, which is why cut-offs beat a longer floor.

Practice questions (8)

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.

  1. A storm delivers 0.5, 3.5, 2.5 and 1.5 cm of rain in four successive hours and produces 4 cm of direct runoff. The φ-index, in cm/h, is ______ (round to three decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 1.167

    1.167 cm/h. The tempting answer is the average loss, (8 − 4)/4 = 1.0 cm/h, and it is wrong — test it and see. At φ = 1.0 the excess is 0 + 2.5 + 1.5 + 0.5 = 4.5 cm, not the 4 cm observed. The reason the average fails is the feature that makes this a real technique: an hour whose rainfall falls below φ contributes zero excess, not a negative one. The first hour delivers only 0.5 cm, so at any φ above 0.5 it is lost entirely and can absorb no more than the 0.5 cm it received — but the arithmetic average has credited it with soaking up a full 1.0 cm it never had, and that missing 0.5 cm is exactly the discrepancy. Solve it properly. Assume the first block is wholly lost and let the other three carry all the excess: (3.5 + 2.5 + 1.5) − 3φ = 4, so φ = 7.5/3 = 1.1667 cm/h. Then verify the assumption, which is the step candidates skip: is 0.5 < 1.1667? Yes, so the first block really is inactive, and recomputing gives 0 + 2.333 + 1.333 + 0.333 = 4.000 cm ✓. The general procedure is an iteration — assume which blocks are active, solve for φ, then check that every excluded block is genuinely below φ and every included one genuinely above; if not, move a block and repeat. The two answers here differ by 16.7 per cent, which on a design flood is not a rounding question, and it is why the check is the part of the method that matters.

  2. By the rational method, a catchment of 2 km² with a runoff coefficient of 0.6 receives rainfall of intensity 50 mm/h. The peak discharge, in m³/s, is ______ (round to three decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 16.667

    16.667 m³/s. Q = CiA/3.6 = 0.6 × 50 × 2/3.6 = 60/3.6 = 16.667 m³/s. The 3.6 is nothing but the unit conversion that lets you use the convenient units directly — mm/h for intensity and km² for area, giving m³/s. The substance of the method is elsewhere. The intensity i must be read at a storm duration equal to the time of concentration, the time runoff takes to travel from the hydraulically remotest point of the catchment to the outlet: a shorter storm never has the whole catchment contributing at once, and a longer one has a lower intensity for the same return period, so this duration maximises the peak. And C carries every simplification the method makes — infiltration, depression storage, antecedent wetness, catchment shape and slope are all inside one number. That is where its sensitivity lives. Urbanise the same catchment so C rises from 0.6 to 0.9 and the peak becomes 25.0 m³/s, 50 per cent larger with no change whatever in the rainfall — which is the arithmetic behind urban flooding and the reason a culvert sized for fields fails once the fields become roofs and roads. Three honest limits to state alongside the answer: the method gives a peak only and no hydrograph, so it cannot size a detention basin; it assumes rainfall uniform over the whole area, which fails for large catchments; and it is generally trusted only up to a few square kilometres. Beyond that a unit hydrograph or a full runoff model is needed.

  3. The unit hydrograph method rests on the assumption that a catchment behaves as a

    1. linear, time-invariant system
    2. non-linear system whose response depends on storm magnitude
    3. system whose response depends only on the total rainfall depth
    4. steady-state system with no storage
    Show answer

    Answer: A — linear, time-invariant system

    Linear and time-invariant. Linearity means the response scales with the input: 2.5 cm of rainfall excess produces a hydrograph with every ordinate 2.5 times the unit hydrograph’s, and with the same time base. Time invariance means the shape does not depend on when the storm occurs, so the same excess gives the same hydrograph in June and in October. Together they permit superposition: decompose any storm into blocks of the unit duration, scale a copy of the unit hydrograph for each, shift each in time, and add. That convolution is the whole method. Option B describes the truth about real catchments, which is why it is a tempting answer, but it describes what the unit hydrograph assumes away — a large storm does route through a catchment faster than a small one, since higher flows travel at higher velocities, and that is precisely the assumption’s known failure. Option C ignores the duration, which is part of the unit hydrograph’s identity: a 2-hour UH and a 6-hour UH for the same catchment are different curves and cannot be substituted for one another without conversion through the S-curve. Option D denies the storage that produces the hydrograph shape at all — a catchment with no storage would return the rainfall instantly and there would be nothing to plot. The useful arithmetic companion to all this is the volume check: the area under the direct runoff hydrograph must equal the excess volume, and for 1 cm over 100 km² that is exactly 10⁶ m³. When it does not, the error is almost always in the base flow separation.

  4. A crop has a base period of 120 days and a duty of 1800 hectares per cumec. The delta, in cm, is ______ (round to one decimal)

    Numerical answer — type the value.

    Show answer

    Answer: 57.6

    57.6 cm. Δ = 8.64 B/D = 8.64 × 120/1800 = 1036.8/1800 = 0.576 m = 57.6 cm. The formula is worth deriving once rather than memorising, because the derivation is only unit conversion and the constant then stops being arbitrary. One cumec running continuously for B days delivers 1 × 86 400 × B cubic metres. Spread over a duty of D hectares, that is D × 10⁴ square metres, so the depth is 86 400 B/(10⁴ D) = 8.64 B/D metres — and the 8.64 is simply 86 400/10⁴. The relationship to hold is that duty and delta are inverses: halve the duty to 900 ha/cumec and the delta doubles to 115.2 cm, because the same cumec is now watering half the area and each hectare therefore receives twice as much. A high duty describes an efficient system or an undemanding crop; a large delta describes a thirsty one. Rice, ponded through its growing season, runs at a delta well above a metre; wheat at a fraction of that. Two cautions. The base period is the period over which the water is actually supplied, and it is not the same as the crop period, the full time from sowing to harvest — they differ by the interval at the ends when the crop is living on stored soil moisture and rainfall. And duty quoted at the field is not duty quoted at the head of the canal: conveyance and application losses mean the head duty is lower, so a system’s efficiency is precisely the ratio between the two.

  5. A fully penetrating well in a confined aquifer of thickness 20 m and permeability 25 m/day is pumped at 1000 m³/day. The difference in piezometric head between observation wells at 10 m and 100 m from the well, in m, is ______ (round to three decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 0.733

    0.733 m. Thiem’s equation for steady radial flow to a well in a confined aquifer is Q = 2πkb(h₂ − h₁)/ln(r₂/r₁), so h₂ − h₁ = Q ln(r₂/r₁)/(2πkb) = 1000 × ln(10)/(2π × 25 × 20) = 2302.6/3141.6 = 0.733 m. The transmissivity T = kb = 500 m²/day is the combination that actually governs, which is why pumping tests report T rather than k and b separately — the two cannot be separated from a drawdown record alone. The derivation is worth seeing: the same discharge crosses every cylindrical surface around the well, so Q = 2πrb k (dh/dr), and integrating that from r₁ to r₂ produces the logarithm. That logarithm is the result to carry. Take the outer observation well from 100 m out to 200 m and the head difference rises only to 0.954 m — doubling the distance added 0.22 m, where the first factor of ten had added 0.73. So the cone of depression is deep and steep close to the well and extremely flat far from it, which has two practical consequences. A pumping well affects its neighbours over a very large radius but only slightly, which is why over-abstraction from a shared aquifer is easy to cause collectively and almost impossible to attribute individually. And the radius of influence is a convention rather than a physical boundary: the drawdown never actually reaches zero, it only becomes too small to measure. For an unconfined aquifer the same derivation with the saturated thickness itself varying gives the Dupuit equation Q = πk(h₂² − h₁²)/ln(r₂/r₁), where the squares appear because the flow area shrinks as the water table falls.

  6. In Lane’s weighted creep theory for a weir on a permeable foundation, vertical creep is weighted relative to horizontal creep as

    1. 1 : 1, both counted equally
    2. 3 : 1, vertical creep being three times as effective
    3. 1 : 3, horizontal creep being three times as effective
    4. in proportion to the permeability in each direction
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    Answer: B — 3 : 1, vertical creep being three times as effective

    3 : 1 — vertical creep counts three times as effective per metre. Bligh’s original creep theory assumed the head is dissipated uniformly along the shortest seepage path, so the required creep length is simply L = C H: a 4 m head needs 20 m of creep on gravel (C = 5), 48 m on coarse sand (C = 12) and 72 m on fine sand (C = 18), the finer material demanding more because it pipes more easily. Lane refined it by observing what Bligh’s equal treatment ignored — natural alluvium is stratified, with the horizontal permeability far greater than the vertical, so a metre of horizontal path resists seepage much less than a metre of vertical path. Lane therefore counts vertical creep at full value and horizontal at one third, which is the same statement as vertical being three times as effective. The design consequence is direct and visible on any barrage: instead of simply extending the floor, sheet-pile cut-offs are driven at the upstream and downstream ends, since a 3 m deep cut-off does the work of 9 m of floor. Option D sounds sophisticated and is not what the theory does; Lane’s weights are empirical constants, not computed from measured anisotropy. What is worth adding is that both theories were superseded: Khosla’s analysis solved the potential flow properly and showed that the head is not dissipated uniformly at all — the gradient concentrates sharply at the downstream cut-off, which is exactly where the exit gradient must be checked against the critical value to guard against piping. Creep theories survive as quick checks; the exit gradient is where the real design decision sits.

  7. By Lacey’s regime theory, the wetted perimeter of a channel carrying 30 m³/s, in m, is ______ (round to three decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 26.017

    26.017 m. P = 4.75√Q = 4.75 × √30 = 4.75 × 5.4772 = 26.017 m. Note what this relation does not contain: no silt factor, no slope, no roughness — the wetted perimeter of a regime channel depends on the discharge alone. The companion relations do bring in the sediment: R = 0.47(Q/f)^(1/3) = 0.47(30)^(1/3) = 1.460 m at f = 1.0, and the regime slope S = f^(5/3)/(3340 Q^(1/6)) = 1.70 × 10⁻⁴, that is 1 in 5888. From those, A = PR = 38.0 m² and the mean velocity is Q/A = 0.79 m/s, which is the velocity at which this material stays in motion without either settling out or scouring the bed. The idea behind the theory is worth stating plainly because it is unlike anything else in the paper: an unlined canal in alluvium is made of the same material it carries, so it is not a fixed boundary at all — it widens, deepens and adjusts its slope until it reaches an equilibrium in which silting and scouring balance. Regime theory describes that equilibrium state rather than deriving it. And that is the honest caveat: these are empirical relations fitted to canals in northern India in the early twentieth century, not results derived from mechanics. Their dimensional inconsistency — a constant of 4.75 with units buried in it, a 3340 that works only in SI — is the visible sign of it. They remain in use because they work well for the conditions they were fitted to, and they should be applied with care to sediments and discharges outside that range.

  8. Which of the following are correct about duty and delta in irrigation? (One or more)

    1. They are related by Δ = 8.64 B/D with Δ in metres and B in days
    2. For a given base period, a higher duty means a smaller delta
    3. Duty measured at the field is higher than duty measured at the canal head
    4. Delta is independent of the crop
    Show answer

    Answer: A — They are related by `Δ = 8.64 B/D` with Δ in metres and B in days; B — For a given base period, a higher duty means a smaller delta; C — Duty measured at the field is higher than duty measured at the canal head

    A, B and C. A is unit conversion, not a law to memorise: one cumec for B days delivers 86 400 B m³ over D hectares, that is 10⁴ D m², so the depth is 8.64 B/D metres, and the 8.64 is just 86 400/10⁴. B follows immediately — duty is in the denominator, so duty and delta are inverses: at B = 120 days, a duty of 1800 ha/cumec gives 57.6 cm and halving the duty to 900 doubles delta to 115.2 cm, since the same cumec now serves half the area. A high duty therefore describes an efficient system or an undemanding crop. C is the one that requires thinking through rather than recalling. Water is lost between the canal head and the field — seepage through the bed and banks, evaporation, spillage, and inefficiency in application — so a cumec measured at the head irrigates less area than a cumec measured at the field would suggest. Field duty is therefore the higher figure, and the ratio between the two is the conveyance efficiency of the system, which is why lining a canal raises the head duty without changing the crop’s requirement at all. D is false, and it is the trap: delta is a property of the crop above all. It is the depth of water the crop needs over its base period, driven by evapotranspiration, so rice standing in ponded water runs well over a metre while wheat needs a fraction of that. Delta, duty and base period are three views of the same quantity of water, and fixing any two fixes the third.