Hydrology: Precipitation, Runoff Estimation, Hydrographs, Stream Flow, Flood Routing, Infiltration and Drought
2. Runoff estimation: the rational method and the SCS curve number
For the small agricultural watersheds that terraces, waterways and culverts drain, the peak runoff is estimated by the rational method: Q = CIA/360, with Q in m³/s, C the dimensionless runoff coefficient (the fraction of rainfall that runs off, higher for bare, compacted and steep land), I the rainfall intensity in mm/h for a duration equal to the time of concentration (the time for water to travel from the most remote point to the outlet, estimated for instance by the Kirpich formula), and A the area in hectares. For C = 0.4, I = 50 mm/h and A = 36 ha, Q = 0.4 × 50 × 36/360 = 2.0 m³/s. The method assumes the rainfall is uniform in time and space and lasts at least the time of concentration, which is why it is limited to small areas.
The SCS (NRCS) curve-number method estimates the runoff depth from a storm. The potential maximum retention is S = 25400/CN − 254 (mm), where the curve number CN (0 to 100) is read for the hydrologic soil group (A, sandy and well drained, to D, clayey and slow), the land use and treatment, and the antecedent moisture condition (AMC I dry, II average, III wet). With initial abstraction I_a = 0.2S, the runoff is Q = (P − I_a)²/(P − I_a + S) for P > I_a, and zero otherwise. For CN = 80, S = 63.5 mm and I_a = 12.7 mm; a 100 mm storm gives Q = 87.3²/150.8 = 50.5 mm.
3. Hydrograph analysis and unit hydrograph theory
A storm hydrograph — discharge against time at the outlet — has a rising limb, a crest and a recession limb. It is split into base flow (groundwater contribution, separated by a straight line or other conventional method) and direct runoff, whose volume equals the effective rainfall depth times the catchment area. The unit hydrograph (UH) of duration D is the direct-runoff hydrograph from 1 cm of effective rainfall falling uniformly over the catchment in D hours. It rests on two assumptions: linearity (ordinates scale with the depth of effective rain, so 3 cm gives three times the ordinates) and time invariance (the same rain gives the same response whenever it falls), which permits superposition of successive blocks of rain lagged by D.
Two computations recur. The flood from a storm: a 4-h UH with a peak of 50 m³/s, under 3 cm of effective rain in 4 h and a base flow of 10 m³/s, gives a peak of 3 × 50 + 10 = 160 m³/s. The catchment area from a UH: its volume is 1 cm over the area, so if the ordinates of a 6-h UH at 6-h intervals sum to 250 m³/s, the volume is 250 × 6 × 3600 = 5.4 × 10⁶ m³, and A = 5.4 × 10⁶/0.01 = 5.4 × 10⁸ m² = 540 km². To change the duration of a UH, the S-curve (summation) method lags an S-curve by the new duration and scales the difference by the ratio of durations. Where no gauging exists, synthetic unit hydrographs (Snyder’s, and the SCS dimensionless UH) estimate one from catchment characteristics.
4. Stream flow measurement and flood routing
Stream flow is measured directly by the area–velocity method: the cross-section is divided into verticals, the velocity in each is measured by a current meter — at 0.6 of the depth from the surface in shallow water, or as the mean of readings at 0.2 and 0.8 of the depth in deeper water — and Σ(area × velocity) gives the discharge. Continuous records come from a stage–discharge (rating) curve, Q = C(G − a)^b, fitted to gaugings so that a stage recorder gives flow; small channels use weirs and flumes (the Parshall flume, which needs little head); turbulent mountain streams use the dilution (tracer) method; and floats give rough surface velocities.
Flood routing follows a flood wave through a reservoir or a channel reach, whose storage delays the peak and attenuates it. Both methods rest on the continuity equation over a time step Δt: (I₁ + I₂)/2 − (O₁ + O₂)/2 = (S₂ − S₁)/Δt. In hydrological reservoir (level-pool) routing — the Modified Puls or storage-indication method — storage and outflow are both functions of the water level alone, so the outflow peak falls on the recession limb of the inflow, at the moment the two hydrographs cross. In channel routing by the Muskingum method, storage is prism plus wedge, S = K[xI + (1 − x)O], with K the travel time of the reach and x a weighting factor between 0 and 0.5 (x = 0 is reservoir-type storage depending on outflow only; natural streams are typically 0.1 to 0.3). The routing equation is O₂ = C₀I₂ + C₁I₁ + C₂O₁, with C₀ = (0.5Δt − Kx)/D, C₁ = (0.5Δt + Kx)/D, C₂ = (K − Kx − 0.5Δt)/D and D = K − Kx + 0.5Δt, so that C₀ + C₁ + C₂ = 1.
5. Infiltration indices and equations; drought and its classification
Infiltration indices lump all losses into one rate. The φ-index is the constant loss rate above which the rainfall volume equals the runoff volume; hours whose rainfall is below φ contribute nothing and are excluded. For a storm of 2, 4, 3 and 1 cm in four successive hours producing 4 cm of runoff, trying all four hours gives φ = 1.5 cm/h, which exceeds the fourth hour’s 1 cm; excluding it, (2 − φ) + (4 − φ) + (3 − φ) = 4 gives φ = 1.67 cm/h, consistent with all three remaining hours. The W-index is the average infiltration rate excluding depression storage and interception as well.
| Equation | Form | Meaning of the constants |
|---|---|---|
| Horton | f = f_c + (f₀ − f_c) e^(−kt) | f₀ initial and f_c final (steady) rate; k a decay constant |
| Kostiakov | F = a t^b (cumulative), so f = ab t^(b−1) | a > 0 and 0 < b < 1, fitted to ring-infiltrometer data |
| Philip | F = S t^(1/2) + A t | S the sorptivity, A related to the saturated conductivity |
With Horton constants f₀ = 6 cm/h, f_c = 1 cm/h and k = 2 h⁻¹, the rate after 1 h is 1 + 5e⁻² = 1.68 cm/h. With Kostiakov constants a = 2.5 and b = 0.5 (F in cm, t in h), the cumulative infiltration after 4 h is 5 cm and the rate is ab t^(b−1) = 1.25 × 4^(−0.5) = 0.625 cm/h — differentiate F to get f; do not divide F by t, which gives the average rate. Drought is classified by what it deprives: meteorological drought is a rainfall deficiency (IMD calls a seasonal deficiency of 26 to 50 per cent of normal moderate and more than 50 per cent severe); hydrological drought is depleted streams, reservoirs and groundwater; agricultural drought is soil moisture too low for crops at critical stages; and socio-economic drought is the resulting shortfall of supply against demand.
Key takeaways
- Thiessen weights each gauge by its polygon area; the normal-ratio method fills a gap when normals differ by more than about 10 per cent.
- Rational method Q = CIA/360 (m³/s, mm/h, ha) with I for a duration equal to the time of concentration.
- SCS: S = 25400/CN − 254, I_a = 0.2S, Q = (P − I_a)²/(P − I_a + S) — a depth in mm.
- A unit hydrograph is 1 cm of effective rain in D hours; its volume fixes the catchment area, and linearity lets you scale and add.
- Muskingum: S = K[xI + (1 − x)O], 0 ≤ x ≤ 0.5, and the three routing coefficients sum to 1.
- Exclude hours below φ when computing the φ-index; Kostiakov’s rate is the derivative ab t^(b−1).
Practice questions (12)
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.
A 36 ha watershed has a runoff coefficient of 0.4. For a design rainfall intensity of 50 mm/h, what is the peak runoff by the rational method, in m³/s?
Numerical answer — type the value.
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Answer: 2
Q = CIA/360 = 0.4 × 50 × 36/360 = 2.0 m³/s. Using the km² form CIA/3.6 with A left in hectares gives 200 m³/s — a hundredfold error from mixing the unit systems.A watershed has a curve number of 80. Using the SCS method with I_a = 0.2S, what is the runoff depth from a storm of 100 mm, in mm (to one decimal place)?
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Answer: 50.5
S = 25400/80 − 254 = 317.5 − 254 = 63.5 mm; I_a = 12.7 mm. Q = (100 − 12.7)²/(100 − 12.7 + 63.5) = 7621.3/150.8 = 50.5 mm. Omitting the initial abstraction, Q = P²/(P + S) = 61.2 mm, overstates it.Three rain gauges in a catchment record 40 mm, 60 mm and 80 mm, and their Thiessen polygons cover 30, 50 and 20 per cent of the catchment respectively. What is the mean areal rainfall, in mm?
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Answer: 58
P = Σ(P_i × a_i) = 40 × 0.3 + 60 × 0.5 + 80 × 0.2 = 12 + 30 + 16 = 58 mm. The arithmetic mean, 60 mm, ignores the unequal areas the polygons represent.Station X, whose normal annual rainfall is 1000 mm, missed a storm. Three neighbouring stations recorded 80, 90 and 100 mm in that storm, and their normal annual rainfalls are 800, 1200 and 1000 mm. What is the estimated storm rainfall at X by the normal-ratio method, in mm (to one decimal place)?
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Answer: 91.7
P_x = (N_x/n)Σ(P_i/N_i) = (1000/3)(80/800 + 90/1200 + 100/1000) = 333.3 × (0.100 + 0.075 + 0.100) = 333.3 × 0.275 = 91.7 mm. The simple average, 90 mm, is used only when the normals are within about 10 per cent of each other, which here they are not.A storm produces rainfall of 2, 4, 3 and 1 cm in four successive hours, and the resulting direct runoff is 4 cm. What is the φ-index, in cm/h (to two decimal places)?
Numerical answer — type the value.
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Answer: 1.67
First trial over all four hours: φ = (10 − 4)/4 = 1.5 cm/h, but the last hour’s 1 cm is below 1.5, so it contributes no runoff and is excluded. Over three hours: 9 − 3φ = 4, φ = 1.67 cm/h, and each of 2, 4 and 3 exceeds it, so the answer is consistent. Stopping at 1.5 cm/h is the trap.Horton’s equation for a soil has f₀ = 6 cm/h, f_c = 1 cm/h and k = 2 h⁻¹. What is the infiltration rate 1 h after the start of rain, in cm/h (to two decimal places)?
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Answer: 1.68
f = f_c + (f₀ − f_c)e^(−kt) = 1 + 5e^(−2) = 1 + 5 × 0.1353 = 1.68 cm/h. Writing f₀e^(−kt) without the final rate gives 0.81 cm/h, below f_c, which Horton’s curve never reaches.Cumulative infiltration in a soil follows Kostiakov’s equation F = 2.5 t^0.5, with F in cm and t in h. What is the infiltration rate at t = 4 h, in cm/h (to three decimal places)?
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Answer: 0.625
f = dF/dt = 2.5 × 0.5 × t^(−0.5) = 1.25/√4 = 0.625 cm/h. Dividing F(4) = 5 cm by 4 h gives 1.25 cm/h, which is the average rate over the first 4 h, not the rate at 4 h.The 4-h unit hydrograph of a catchment has a peak of 50 m³/s. What is the peak of the flood hydrograph from an effective rainfall of 3 cm in 4 h, if the base flow is 10 m³/s, in m³/s?
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Answer: 160
By linearity the direct-runoff peak is 3 × 50 = 150 m³/s, and adding the base flow gives 160 m³/s. Scaling the base flow along with the direct runoff (3 × 60 = 180) is wrong: base flow is groundwater, not a response to the storm.The ordinates of a 6-h unit hydrograph (for 1 cm of effective rain), read at 6-h intervals, add up to 250 m³/s. What is the catchment area, in km²?
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Answer: 540
Runoff volume = ΣQ × Δt = 250 × 6 × 3600 = 5.4 × 10⁶ m³, which is 1 cm (0.01 m) over the area: A = 5.4 × 10⁶/0.01 = 5.4 × 10⁸ m² = 540 km². Using Δt in hours instead of seconds gives an area 3600 times too small.What is the probability that a 25-year storm occurs at least once in the next 2 years? Give the answer to four decimal places.
Numerical answer — type the value.
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Answer: 0.0784
p = 1/T = 0.04 per year; the probability of no occurrence in 2 years is 0.96² = 0.9216, so at least one occurrence is 1 − 0.9216 = 0.0784. Simply doubling p (0.08) overcounts the case of two occurrences.In the Muskingum method of channel routing, a weighting factor x = 0 means that:
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Answer: A — storage depends on outflow alone, as in a reservoir
With x = 0, S = K[0·I + O] = KO: storage is a function of outflow only — the level-pool reservoir case, giving maximum attenuation. The other limit, x = 0.5, weights inflow and outflow equally and gives a pure translation of the wave with no attenuation.Which of the following statements about the measurement and analysis of precipitation are correct?
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Answer: A — A tipping-bucket gauge is a recording rain gauge; B — The isohyetal method uses lines of equal rainfall; D — A double-mass curve is used to test the consistency of a rainfall record
The tipping bucket records each tip against time, so it gives intensity; isohyets are contours of equal rainfall; a break in the double-mass curve’s slope reveals inconsistency. Thiessen weights are purely geometric — distances between gauges — so they cannot represent orographic rainfall, which only the isohyetal method can follow.