Engineering Hydrology quantifies the movement and availability of water through the hydrological cycle — from precipitation measurement and losses (evapotranspiration, infiltration), through runoff generation and streamflow measurement, unit hydrograph theory, flood frequency analysis and flood routing, to groundwater hydraulics. Every formula, coefficient, diagram, solved example and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: Hydraulics & Fluid Mechanics (open-channel flow, Manning's formula and flow measurement introduced there underpin stream-flow gauging and flood routing here). Leads to: Irrigation Engineering, which applies runoff and groundwater estimation to canal and reservoir water demand.
The hydrological cycle describes the continuous circulation of water on, above, and below the surface of the Earth. Solar energy drives evaporation; gravity drives precipitation and runoff. The cycle has no beginning or end — it is a closed system.
| Component | Description | Key Parameter |
|---|---|---|
| Precipitation | All forms of water falling from atmosphere — rain, snow, sleet, hail | Rainfall depth (mm), intensity (mm/hr) |
| Interception | Precipitation caught by vegetation canopy, evaporated before reaching ground | Interception loss (mm) |
| Infiltration | Water entering soil surface | Infiltration rate f (mm/hr) |
| Evapotranspiration | Evaporation from water bodies + transpiration by plants | PET, AET (mm/day) |
| Runoff | Water flowing over land surface into streams | Direct runoff, base flow |
| Groundwater flow | Sub-surface flow through aquifers | Hydraulic head, Darcy flux |
The water balance equation is the hydrological analogue of conservation of mass applied to a catchment: precipitation input equals runoff output plus evapotranspiration loss plus any change in storage (soil moisture, groundwater, surface water bodies). Over a long averaging period (a year or more), the change in storage term becomes negligible compared to the other terms, simplifying the equation considerably — the fraction of precipitation that becomes runoff (the runoff coefficient) is a useful single-number summary of a catchment's response.
A catchment's shape and drainage network strongly influence its flood response. The form factor and compactness coefficient quantify shape (elongated catchments attenuate floods more than compact/circular ones, which concentrate flow arrival times). Drainage density measures how finely a basin is dissected by streams — high drainage density indicates impermeable soil and rapid runoff response, while low drainage density suggests permeable soil and more infiltration. The time of concentration (time for water from the most hydraulically remote point to reach the outlet) governs the critical storm duration for peak flow estimation.
\( P = R + ET + \Delta S \)
Long-term (\(\Delta S\to0\)): \( P \approx R + ET \)
Runoff coefficient: \( C = \dfrac{R}{P} \) (typically 0.1–0.6 for Indian catchments)
Form factor: \( F = A/L^2 \) (\(A\)=area, \(L\)=axial length)
Compactness coefficient: \( K_c = \dfrac{P}{2\sqrt{\pi A}} \) (\(P\)=perimeter)
Drainage density: \( D_d = \sum L/A \) (\(\sum L\)=total stream length)
Stream frequency: \( F_s = N/A \) (\(N\)=number of streams)
\( t_c = 0.0195\times \dfrac{L^{0.77}}{S^{0.385}} \)
\(L\) in m, \(S\)=slope, \(t_c\) in minutes.
Given: A catchment receives 1200 mm annual rainfall and generates 420 mm of annual runoff. Find the runoff coefficient.
Solution:
\( C = \dfrac{R}{P} = \dfrac{420}{1200} = 0.35 \)
Answer: \(C = 0.35\) — within the typical Indian catchment range (0.1–0.6).
Given: A 250 km² catchment has a total stream network length of 625 km. Find the drainage density and comment on the likely soil permeability.
Solution:
\( D_d = \dfrac{625}{250} = 2.5\ \text{km/km}^2 \)
Answer: \(D_d = 2.5\ \text{km/km}^2\) — a moderately high value suggesting relatively impermeable soil with rapid runoff response.
Given: The longest flow path in a catchment is 800 m with an average slope of 0.015. Find the time of concentration.
Solution:
\( t_c = 0.0195\dfrac{\times800^{0.77}}{0.015^{0.385}} \)
\( 800^{0.77} \approx 179.8 \); \( 0.015^{0.385} \approx 0.1935 \)
\( t_c = 0.0195\dfrac{\times179.8}{0.1935} \approx 18.1\ \text{min} \)
Answer: \(t_c\approx18.1\) minutes.
Fig. 1.1 — The hydrological cycle: evaporation from oceans and land, condensation into clouds, precipitation, runoff to streams, and infiltration to groundwater — a closed loop driven by solar energy and gravity.
| Type | Mechanism | Characteristics |
|---|---|---|
| Frontal | Warm moist air lifted over cold air mass at a front | Large area, moderate intensity, long duration |
| Orographic | Moist air forced upward over mountains | Windward slope heavy rain; leeward (rain shadow) dry |
| Convective | Local surface heating causes air mass to rise | Small area, very high intensity, short duration; thunderstorms |
| Cyclonic | Convergence of air into low-pressure system | Very large area; tropical cyclones / depressions |
| Type | Working | Key Feature |
|---|---|---|
| Symon's gauge | Cylindrical collector, manual daily reading | Standard gauge in India; 127 mm dia orifice |
| Tipping bucket | 0.25 mm tips recorded electronically | Continuous automatic record; suited for intensity |
| Weighing type | Spring balance / float records weight of collected water | Continuous record; measures snow also |
| Float type | Rising float records depth on chart drum | Self-recording; 24-hour chart |
The optimum number of rain gauges depends on the spatial variability of rainfall across existing stations (captured by the coefficient of variation) and the acceptable error in the estimated catchment-mean rainfall. A catchment with widely varying rainfall (high \(C_v\)) needs many more gauges than a uniform one for the same target accuracy.
When a station's data is missing for some period, it is estimated from nearby stations. If the normal annual rainfall of any surrounding station differs from the missing station's normal by more than 10%, the Normal Ratio Method is used (weighting each surrounding station's observed value by the ratio of normals); if all normals are within 10% of each other, the simpler arithmetic mean of surrounding stations suffices.
Point rainfall measurements must be converted to a single representative catchment-average value. The simple arithmetic mean assumes uniformly distributed gauges. The Thiessen polygon method weights each gauge by its "zone of influence" (found by constructing perpendicular bisectors between neighbouring gauges), correctly handling non-uniform gauge networks. The isohyetal method (drawing contours of equal rainfall and weighting by the area between them) is the most accurate, since it can directly account for orographic and other spatial rainfall patterns that gauge-based weighting alone cannot capture.
IDF curves relate rainfall intensity, storm duration, and return period through an empirical relationship (Sherman's formula) fitted to historical data at a given location — essential input for design storm selection in urban drainage and small-catchment flood estimation. The return period is the reciprocal of the annual exceedance probability, and the risk of at least one occurrence within a design life of \(n\) years follows directly from the probability of non-occurrence in each of those \(n\) independent years.
\( N = (\dfrac{C_v}{\varepsilon})^2 \)
\( C_v = (\dfrac{\sigma}{\bar x})\times100\% \)
\(\varepsilon\)=allowable % error (typically 10%); additional gauges needed = \(N-n\)
Normal Ratio (normals differ >10%): \( P_x = \dfrac{N_x}{3}\left[\dfrac{P_A}{N_A}+\dfrac{P_B}{N_B}+\dfrac{P_C}{N_C}\right] \)
Arithmetic mean (normals differ <10%): \( P_x = \dfrac{P_A+P_B+P_C}{3} \)
Arithmetic mean: \( \bar P = \tfrac1n\sum P_i \)
Thiessen: \( \bar P = \dfrac{\sum(A_iP_i)}{A_{total}} \), weightage \( w_i=A_i/A_{total} \)
Isohyetal (most accurate): \( \bar P = \dfrac{\sum(a_i\bar P_i)}{A_{total}} \)
Sherman formula: \( i = \dfrac{kT^x}{(t+a)^n} \)
Return period: \( T=\dfrac{1}{P} \)
Non-occurrence probability: \( q=(1-\dfrac{1}{T})^n \)
Risk of ≥1 occurrence in \(n\) years: \( R=1-(1-\dfrac{1}{T})^n \)
Given: A catchment has 8 existing rain gauges with \(C_v=32\%\). Find the optimum number of gauges for a 10% error, and how many additional gauges are needed.
Solution:
\( N = (\dfrac{32}{10})^2 = 3.2^2 = 10.24 \approx 10 \)
Additional gauges \(= N-n = 10-8=2\)
Answer: 10 gauges total needed; 2 additional gauges required.
Given: Station X (missing data) has normal annual rainfall \(N_x=750\ \text{mm}\). Three surrounding stations: A (\(P_A=800, N_A=780\)), B (\(P_B=650, N_B=720\)), C (\(P_C=900, N_C=850\)) mm. Find \(P_x\).
Solution:
\( P_x = \dfrac{750}{3}\left[\dfrac{800}{780}+\dfrac{650}{720}+\dfrac{900}{850}\right] = 250\times[1.026+0.903+1.059] \)
\( = 250\times2.988 \approx 747\ \text{mm} \)
Answer: \(P_x\approx747\ \text{mm}\).
Given: A structure has a design life of 50 years and is designed for the 100-year flood. Find the risk of the design flood being exceeded at least once during the structure's life.
Solution:
\( R = 1-(1-\dfrac{1}{100})^{50} = 1-(0.99)^{50} \)
\( (0.99)^{50} \approx 0.605 \)
\( R \approx 1-0.605 = 0.395 \)
Answer: \(R\approx39.5\%\) — a substantial risk despite designing for a "100-year" event.
Fig. 2.1 — Thiessen polygon construction: perpendicular bisectors between neighbouring gauges define each gauge's zone of influence (polygon area), used to weight its contribution to the catchment-average rainfall.
| Term | Definition |
|---|---|
| Evaporation | Water loss from open water/soil surface to atmosphere (liquid → vapour) |
| Transpiration | Water loss through plant stomata |
| Evapotranspiration (ET) | Combined evaporation + transpiration from a catchment |
| Potential ET (PET) | ET from an extensive green surface with unlimited water supply |
| Actual ET (AET) | ET under actual soil moisture conditions; \(AET \le PET\) |
Penman's combination method computes PET as a weighted average of an energy-balance term (net radiation available for evaporation) and an aerodynamic term (driven by wind speed and the vapour-pressure deficit between the surface and the air), with weights set by the slope of the saturation vapour-pressure curve and the psychrometric constant. This physically-based combination makes it more broadly applicable across climates than purely empirical methods.
The Class A evaporation pan is a simple, widely-used field instrument: evaporation measured from the pan is scaled by an empirical pan coefficient to estimate evaporation from a real lake or reservoir, since a small pan loses heat and gains/loses energy differently than a large water body (a pan is more exposed to advected heat from its rim and surroundings, generally over-estimating evaporation relative to a lake).
The Blaney-Criddle method estimates crop consumptive use (water used by a crop through evapotranspiration) from mean monthly temperature and daytime-hour fraction alone, scaled by an empirical crop coefficient that varies through the growing season — a simple method requiring only temperature data, making it useful where radiation, humidity and wind data (needed for Penman) are unavailable.
\( E_0 = \dfrac{\Delta}{\Delta+\gamma}H_n + \dfrac{\gamma}{\Delta+\gamma}E_a \)
\(\Delta\)=slope of sat. vapour pressure vs. T curve; \(\gamma\)=psychrometric constant (≈0.49 mm Hg/°C)
Aerodynamic term: \( E_a = 0.35(e_s-e_a)(1+\dfrac{u_2}{16}) \)
Class A pan: 1.22 m dia, 25.5 cm deep
\( E_{lake} = C_p\times E_{pan} \), \(C_p\approx0.6\text{–}0.8\) (typically 0.7 for India)
\( CU = k\times f \)
\( f = \dfrac{p(0.46T+8.13)}{40.6} \)
\(p\)=% daytime hours of the year for the month; \(T\)=mean monthly temperature (°C); \(k\)=crop coefficient
Given: A Class A pan records 8.5 mm/day evaporation. Using \(C_p=0.7\), find the estimated lake evaporation.
Solution:
\( E_{lake} = 0.7\times8.5 = 5.95\ \text{mm/day} \)
Answer: \(E_{lake}\approx5.95\ \text{mm/day}\).
Given: For a given month, \(p=8.5\%\), \(T=28°\text{C}\), crop coefficient \(k=0.75\). Find the monthly consumptive use factor \(f\) and CU.
Solution:
\( f = 8.5\dfrac{\times(0.46\times28+8.13)}{40.6} = 8.5\dfrac{\times(12.88+8.13)}{40.6} = 8.5\dfrac{\times21.01}{40.6} \)
\( f \approx 8.5\times0.5175 \approx 4.40\ \text{cm} \)
\( CU = 0.75\times4.40 = 3.30\ \text{cm} \)
Answer: \(f\approx4.40\ \text{cm}\), \(CU\approx3.30\ \text{cm}\) for the month.
Given: A field has PET = 6 mm/day, but soil moisture is well below field capacity due to a dry spell. Comment on AET.
Solution:
Since soil moisture is limited, plants and soil cannot supply water at the potential (unlimited-supply) rate, so \(AET < PET\).
Answer: \(AET < 6\ \text{mm/day}\) — the exact value depends on the soil moisture deficit, but it is strictly less than PET under water-stressed conditions.
Fig. 3.1 — Penman's combination method: PET is the weighted sum of an energy-balance term (net radiation) and an aerodynamic term (wind speed, vapour pressure deficit).
Infiltration capacity \(f\) is the maximum rate at which soil can absorb water (mm/hr). The actual infiltration rate equals \(f\) when rainfall intensity meets or exceeds it; otherwise the actual rate simply equals the rainfall intensity (all rainfall infiltrates, since supply — not soil capacity — limits it). Infiltration capacity decreases with time during a storm as the soil becomes progressively saturated, approaching a constant final (ultimate) rate \(f_c\).
Horton's equation is a purely empirical exponential decay model, fitted to observed infiltration curves, and is the most widely used in practice for its simplicity. Integrating it over time gives the cumulative infiltrated depth for a given storm duration, useful for computing effective rainfall in unit hydrograph analysis.
The Green-Ampt model is a physically-based alternative to Horton's purely empirical decay, derived from Darcy's law applied to a simplified "piston" wetting front advancing into the soil. Because it is grounded in physical soil parameters (hydraulic conductivity, suction head, moisture deficit) rather than curve-fitted constants, it generalises better to variable-intensity rainfall than Horton's equation.
The φ-index is a simplified, single-value approximation: the uniform infiltration rate such that the total rainfall in excess of this rate over the storm exactly equals the observed direct runoff volume. It is found by trial and error or graphically, and is widely used because of its simplicity, despite not capturing the time-varying nature of real infiltration. The W-index is a related but distinct concept — it additionally accounts for the initial abstraction (interception, depression storage) separately from ongoing infiltration.
\( f(t) = f_c+(f_0-f_c)e^{-kt} \)
Cumulative: \( F(t) = f_ct+\dfrac{(f_0-f_c)(1-e^{-kt})}{k} \)
\( f = K_s\left[1+\dfrac{\psi\,\Delta\theta}{F}\right] \)
\(K_s\)=saturated hydraulic conductivity; \(\psi\)=suction head; \(\Delta\theta=\theta_s-\theta_i\); \(F\)=cumulative infiltration
φ-index: uniform rate such that \(\sum(P_i-\phi)\) over the storm = observed direct runoff
W-index: \( W = \dfrac{P-Q-I_a}{t_r} \)
\(P\)=total rainfall; \(Q\)=direct runoff; \(I_a\)=initial abstraction; \(t_r\)=storm duration
Given: \(f_0=10\ \text{cm/hr}\), \(f_c=1\ \text{cm/hr}\), \(k=2\ \text{hr}^{-1}\). Find the infiltration rate at \(t=1\ \text{hr}\).
Solution:
\( f(1) = 1+(10-1)e^{-2\times1} = 1+9\times0.1353 = 1+1.218 = 2.218\ \text{cm/hr} \)
Answer: \(f(1)\approx2.22\ \text{cm/hr}\).
Given: Using the same parameters as Example 1, find the cumulative infiltration depth over the first 2 hours.
Solution:
\( F(2) = 1\times2+\dfrac{(10-1)(1-e^{-4})}{2} = 2+\dfrac{9\times(1-0.0183)}{2} = 2+\dfrac{9\times0.9817}{2} \)
\( F(2) = 2+4.417 = 6.417\ \text{cm} \)
Answer: \(F(2)\approx6.42\ \text{cm}\).
Given: A storm has hourly rainfall of 3, 5, 4, 2 cm over 4 hours (total 14 cm), producing 7 cm of direct runoff. Find the φ-index (assume φ is exceeded in all 4 hours).
Solution:
\( \sum(P_i-\phi) = (3-\phi)+(5-\phi)+(4-\phi)+(2-\phi) = 14-4\phi = 7 \)
\( 4\phi = 7 \Rightarrow \phi = 1.75\ \text{cm/hr} \)
Answer: \(\phi\approx1.75\ \text{cm/hr}\) (valid since all hourly values exceed 1.75 cm/hr, consistent with the initial assumption).
Fig. 4.1 — Horton's infiltration capacity curve: an exponential decay from the initial rate f₀ toward the final constant rate f_c, with decay constant k.
| Component | Description |
|---|---|
| Surface runoff (overland flow) | Precipitation excess flowing over land surface; quickest response |
| Interflow (subsurface storm flow) | Lateral flow through upper soil layers; intermediate response time |
| Base flow (groundwater runoff) | Groundwater discharge to streams; slow, sustained flow |
| Direct runoff | Surface runoff + interflow; the rapidly responding portion |
| Total runoff / streamflow | Direct runoff + base flow |
Runoff generation depends on rainfall characteristics (intensity, duration, spatial distribution), catchment characteristics (area, shape, slope, soil type, land use, vegetation cover), and climatic/antecedent conditions (temperature, humidity, and — critically — how wet the soil already was before the storm).
The SCS (Soil Conservation Service) Curve Number method estimates direct runoff from total rainfall using a single empirical parameter, the Curve Number (CN), which encodes soil group, land use/cover, and antecedent moisture condition (AMC) into a single 0–100 scale. A CN of 100 represents a fully impervious surface (all rainfall becomes runoff); a CN of 0 would represent infinite storage capacity (no runoff ever). This method's simplicity and reliance on widely available soil/land-use maps makes it the standard for practical runoff estimation in the US and India alike.
The area-velocity method (current meter gauging) is the standard field technique: mean velocity is measured at one or two depths per vertical (chosen to average out the natural parabolic velocity profile), and the product of area and mean velocity gives discharge. Once a stage-discharge (rating curve) relationship is established at a gauging site, ongoing discharge can be estimated from a simple water-level (stage) reading alone, without repeating the labour-intensive current-meter survey each time. Dilution gauging (tracer/salt dilution) is an alternative particularly suited to turbulent mountain streams where current-meter access is difficult.
\( Q = \dfrac{(P-I_a)^2}{P-I_a+S} \)
\( S = \dfrac{25400}{CN}-254 \) (mm); \( I_a=0.2S \) (standard assumption)
So: \( Q = \dfrac{(P-0.2S)^2}{P+0.8S} \)
AMC I (dry), AMC II (normal/standard), AMC III (wet)
Area-velocity: \( Q = A\times\bar V \)
One-point method: velocity at \(0.6d\); two-point: \(\dfrac{0.2d+0.8d}{2}\)
Rating curve: \( Q = a(H-H_0)^b \)
\( Q = q\times\dfrac{C_1-C_2}{C_2-C_0} \)
\(q\)=injection rate; \(C_1\)=injected concentration; \(C_2\)=plateau concentration; \(C_0\)=background
Given: A storm with \(P=100\ \text{mm}\) falls on a catchment with \(CN=75\). Find the direct runoff.
Solution:
\( S = \dfrac{25400}{75}-254 = 338.7-254 = 84.7\ \text{mm} \)
\( Q = \dfrac{(100-0.2\times84.7)^2}{100+0.8\times84.7} = \dfrac{(100-16.94)^2}{100+67.76} = \dfrac{83.06^2}{167.76} \)
\( Q = \dfrac{6898.96}{167.76} \approx 41.13\ \text{mm} \)
Answer: \(Q\approx41.1\ \text{mm}\) of direct runoff.
Given: A stream vertical has depth 3 m; velocities measured at 0.2d and 0.8d are 0.85 m/s and 0.55 m/s respectively. Find the mean velocity at that vertical.
Solution:
\( \bar V = \dfrac{0.85+0.55}{2} = 0.70\ \text{m/s} \)
Answer: \(\bar V = 0.70\ \text{m/s}\).
Given: A tracer is injected at \(q=15\ \text{L/s}\) with concentration \(C_1=200\ \text{g/L}\); background concentration \(C_0=0\); the plateau concentration downstream is \(C_2=0.5\ \text{g/L}\). Find the stream discharge.
Solution:
\( Q = 15\times\dfrac{200-0.5}{0.5-0} = 15\times\dfrac{199.5}{0.5} = 15\times399 = 5985\ \text{L/s} \)
Answer: \(Q\approx5985\ \text{L/s}\approx5.99\ \dfrac{\text{m}^3}{\text{s}}\).
Fig. 5.1 — Current meter velocity measurement points: 0.6d depth for the one-point method, or the average of 0.2d and 0.8d depths for the more accurate two-point method.
A Unit Hydrograph (UH) is the direct runoff hydrograph resulting from 1 cm (or 1 unit) of effective rainfall distributed uniformly over the entire catchment at a uniform rate for a specified duration D. Three key assumptions underpin the method: time-invariance (the same UH applies for a storm of duration D regardless of when it occurs), linearity (runoff scales directly with effective rainfall depth), and superposition (hydrographs from successive rainfall increments can simply be added). These assumptions let a single derived UH be reused to predict runoff from any storm of the same duration, of any magnitude, and even from complex multi-period storms.
Deriving a UH from an actual observed flood hydrograph requires first separating base flow (the slow groundwater contribution) from the direct runoff hydrograph (DRH), then determining the effective rainfall depth that produced that DRH (typically via the φ-index). Dividing every DRH ordinate by the effective rainfall depth (in cm) gives the unit hydrograph ordinates — a consistency check is that the total volume under the UH, summed as (ordinate × time interval), must equal the catchment area (in consistent units), since a UH represents exactly 1 cm of runoff.
Sometimes a UH of one duration is available, but a different duration is needed for a specific analysis. The S-curve (summation hydrograph) technique — formed by lagging and adding an infinite series of the known-duration UH — provides a bridge: lagging the S-curve by the new target duration and taking the difference (then rescaling by the ratio of durations) yields the UH for that new duration. This is more general than simple lagging, which only works cleanly when the new duration is an exact multiple of the original.
When no observed streamflow records exist for a catchment (ungauged basin), a synthetic unit hydrograph must be constructed from catchment geometry alone. Snyder's method (widely calibrated for Indian rivers) relates the basin lag time to stream lengths through an empirical coefficient, then estimates peak discharge and various time parameters (standard rainfall duration, base time) from the lag time and basin area — providing a complete synthetic hydrograph shape without requiring any observed flood record.
The SCS dimensionless unit hydrograph is an alternative synthetic method, expressing the hydrograph shape as a universal dimensionless curve (applicable to any catchment) scaled by the catchment's own peak discharge and time-to-peak — the time to peak itself is estimated from half the rainfall duration plus the basin lag, which is in turn correlated with the time of concentration.
UH ordinate = DRH ordinate / rainfall excess (cm)
Check: \( Q_{total} = P_{eff}\times A \) (volume of DRH = volume of effective rainfall)
To get UH of duration \(D'\) from an S-curve derived from \(D\)-hr UH:
1. Lag S-curve by \(D'\) hours to get \(S'\)
2. Offset: \(S-S'\)
3. Scale: UH(\(D'\)) ordinate \(=(S-S')\times(\dfrac{D}{D'})\)
Basin lag: \( t_p = C_t(LL_c)^{0.3} \) (\(C_t=1.35\text{–}1.65\) for Indian rivers)
Peak discharge: \( Q_p = \dfrac{C_pA}{t_p} \) (\(C_p=2.75\text{–}4.0\))
Standard duration: \( t_r=\dfrac{t_p}{5.5} \); Base time: \( t_b=72+3t_p \) (hours)
\( Q_p = \dfrac{2.08A}{T_p} \)
\( T_p = \dfrac{D}{2}+t_{lag} \), \( t_{lag}\approx0.6\,t_c \)
Given: A DRH has ordinates 0, 20, 60, 40, 15, 0 m³/s at 3-hr intervals, produced by 3 cm of effective rainfall in a 3-hr storm. Find the 3-hr UH ordinates.
Solution:
UH ordinate = DRH ordinate / 3: \(0, 6.67, 20, 13.33, 5, 0\ \dfrac{\text{m}^3}{\text{s}}\)
Answer: 3-hr UH ordinates: 0, 6.67, 20, 13.33, 5, 0 m³/s.
Given: A catchment has \(L=40\ \text{km}\), \(L_c=18\ \text{km}\), \(A=350\ \text{km}^2\), \(C_t=1.5\), \(C_p=3.2\). Find the basin lag and peak discharge.
Solution:
\( t_p = 1.5\times(40\times18)^{0.3} = 1.5\times(720)^{0.3} \)
\( 720^{0.3} \approx 6.79 \); \( t_p \approx 1.5\times6.79 \approx 10.2\ \text{hr} \)
\( Q_p = 3.2\dfrac{\times350}{10.2} \approx 109.8\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(t_p\approx10.2\ \text{hr}\), \(Q_p\approx109.8\ \dfrac{\text{m}^3}{\text{s}}\).
Given: Using \(t_p=10.2\ \text{hr}\) from Example 2, find the base time of the synthetic hydrograph.
Solution:
\( t_b = 72+3\times10.2 = 72+30.6 = 102.6\ \text{hr} \)
Answer: \(t_b\approx102.6\ \text{hr}\) (\(\approx4.3\) days).
Fig. 6.1 — Separating base flow from the observed hydrograph gives the direct runoff hydrograph (DRH); dividing DRH ordinates by the rainfall excess gives the unit hydrograph.
The return period (recurrence interval) of a flood magnitude is the average number of years between events of that magnitude or greater — a statistical, not deterministic, concept (a "100-year flood" can occur in consecutive years; it simply has a 1% chance of occurring in any given year). Plotting positions (Weibull's and Hazen's formulas being the two most common) assign an empirical return period to each ranked historical observation, forming the basis for fitting a frequency distribution.
Gumbel's method is the most widely used flood frequency distribution, assuming annual maximum floods follow the Extreme Value Type I distribution. The flood magnitude for a chosen return period is estimated from the sample mean and standard deviation, scaled by a frequency factor that itself depends on a "reduced variate" (a function purely of the return period) and tabulated reduction statistics that depend on the sample size — a larger historical record gives more reliable reduction statistics and hence a more reliable design flood estimate.
The Log-Pearson Type III distribution — recommended by the US Water Resources Council and widely adopted internationally — first transforms the flood data to logarithms, then fits a three-parameter (mean, standard deviation, skewness) distribution to the logarithms. Because it explicitly accounts for the skewness of the data (Gumbel's method implicitly assumes a fixed skew), it can better represent flood series that depart from the standard Gumbel shape.
The Rational Formula estimates peak discharge directly from catchment area, a runoff coefficient, and design rainfall intensity — a much simpler alternative to full frequency analysis, appropriate only for small (typically under 50 km²) urban or semi-urban catchments where the storm duration used equals the time of concentration (ensuring the entire catchment is contributing simultaneously at the moment of peak flow, the critical/steady-state assumption behind the method).
Return period: \( T=\dfrac{1}{P} \)
Probability of occurrence in \(n\) years: \( P_n=1-(1-\dfrac{1}{T})^n \)
Weibull: \( T=\dfrac{N+1}{m} \); Hazen: \( T=\dfrac{2N}{2m-1} \)
\( x_T = \bar x+K\sigma \)
\( K = \dfrac{y_T-\bar y_n}{S_n} \)
\( y_T = -\ln[-\ln(1-\dfrac{1}{T})] \)
\(\bar y_n, S_n\)=tabulated mean/SD of reduced variate for sample size \(N\)
\( y=\log x \) (or \(\ln x\))
Compute \(\bar y\), \(\sigma_y\), skewness \(G_s\)
\( x_T = \text{antilog}(\bar y+K\sigma_y) \), \(K\) from Pearson III table (\(G_s\), \(T\))
\( Q_p = \dfrac{CiA}{3.6} \)
\(C\)=runoff coefficient (0.1–0.95); \(i\)=intensity at \(t_c\) and design \(T\) (mm/hr); \(A\)=area (km²)
Applicable for catchments < 50 km²; assumes steady state (storm duration ≥ \(t_c\)).
Given: An annual flood series has \(\bar x=450\ \dfrac{\text{m}^3}{\text{s}}\), \(\sigma=180\ \dfrac{\text{m}^3}{\text{s}}\). For \(N=30\) years, \(\bar y_n=0.5362\), \(S_n=1.1124\). Find the 50-year flood.
Solution:
\( y_{50} = -\ln[-\ln(1-\dfrac{1}{50})] = -\ln[-\ln(0.98)] = -\ln[0.0202] \approx 3.902 \)
\( K = \dfrac{3.902-0.5362}{1.1124} = \dfrac{3.366}{1.1124} \approx 3.026 \)
\( x_{50} = 450+3.026\times180 = 450+544.7 = 994.7\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(x_{50}\approx995\ \dfrac{\text{m}^3}{\text{s}}\).
Given: A 40-year flood record has its largest value ranked \(m=1\). Find the Weibull return period for this observation.
Solution:
\( T = \dfrac{N+1}{m} = \dfrac{40+1}{1} = 41\ \text{years} \)
Answer: \(T=41\) years for the largest recorded flood.
Given: A small urban catchment, \(A=8\ \text{km}^2\), \(C=0.65\), design rainfall intensity \(i=60\ \text{mm/hr}\) (for the 25-year storm, at the catchment's \(t_c\)). Find the peak discharge.
Solution:
\( Q_p = \dfrac{0.65\times60\times8}{3.6} = \dfrac{312}{3.6} \approx 86.7\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(Q_p\approx86.7\ \dfrac{\text{m}^3}{\text{s}}\).
Fig. 7.1 — Flood frequency curve: observed annual maxima (plotted at their empirical return period) are fitted with a theoretical distribution (Gumbel or Log-Pearson III) to extrapolate design floods for return periods beyond the observed record.
Flood routing determines the outflow hydrograph at a downstream point given the inflow hydrograph upstream, accounting for the storage effects of the intervening reach (or reservoir), which both delay and flatten (attenuate) the flood wave. Reservoir (level pool) routing applies to lakes and reservoirs, where the water surface is essentially horizontal at any instant, giving storage as a unique function of outflow alone. Channel routing applies to river reaches, where storage depends on both inflow and outflow (the channel develops a "wedge" of extra storage during the rising limb that is not present during recession) — a fundamentally more complex situation requiring methods like Muskingum's.
All flood routing methods rest on the same continuity principle: the difference between inflow and outflow equals the rate of change of storage in the reach or reservoir. Expressed in finite-difference form over a routing time step, this single equation — combined with a storage-outflow (or storage-inflow-outflow) relationship specific to the method — allows the outflow at the end of each time step to be solved sequentially from known conditions at the start.
The Muskingum method models channel storage as a weighted combination of inflow and outflow, governed by a travel-time parameter \(K\) and a weighting factor \(x\) — \(x=0\) corresponds to pure reservoir-type (level-pool) behaviour, while \(x=0.5\) corresponds to pure translation (no attenuation, just delay), with natural rivers typically falling in between (\(x\approx0.2\text{–}0.3\)). The resulting routing equation expresses the new outflow as a weighted combination of the current and previous inflows and the previous outflow, with three coefficients that must always sum to exactly 1 (a useful arithmetic check on any computation).
Because reservoir storage depends only on outflow (via a known stage-storage-discharge relationship for the reservoir), the routing procedure reduces to constructing a single auxiliary curve — the storage-indication curve — relating outflow to a combined storage/time-step term. Once built, this curve lets each time step's outflow be read off directly, without needing to iterate or solve simultaneous equations, making level-pool routing considerably more straightforward than channel routing.
\( I-O = \dfrac{dS}{dt} \)
Finite difference: \( \dfrac{I_1+I_2}{2}-\dfrac{O_1+O_2}{2} = \dfrac{S_2-S_1}{\Delta t} \)
Rearranged: \( \dfrac{2S_2}{\Delta t}+O_2 = (I_1+I_2)+\left(\dfrac{2S_1}{\Delta t}-O_1\right) \)
Storage: \( S = K[xI+(1-x)O] \)
\(x=0\)→pure reservoir; \(x=0.5\)→pure translation; natural rivers: \(x\approx0.2\text{–}0.3\)
Routing: \( O_2 = C_0I_2+C_1I_1+C_2O_1 \)
\( C_0 = \dfrac{-Kx+0.5\Delta t}{K-Kx+0.5\Delta t} \)
\( C_1 = \dfrac{Kx+0.5\Delta t}{K-Kx+0.5\Delta t} \)
\( C_2 = \dfrac{K-Kx-0.5\Delta t}{K-Kx+0.5\Delta t} \)
Check: \(C_0+C_1+C_2=1\).
Storage-indication curve: plot \(O\) vs. \((\dfrac{2S}{\Delta} t+O)\)
Each step: \(\text{RHS} = (I_1+I_2)+(\dfrac{2S_1}{\Delta} t-O_1)\)
\( \dfrac{2S_1}{\Delta} t-O_1 = (\dfrac{2S_1}{\Delta} t+O_1)-2O_1 \)
Given: A river reach has \(K=12\ \text{hr}\), \(x=0.25\), routing interval \(\Delta t=6\ \text{hr}\). Find \(C_0\), \(C_1\), \(C_2\).
Solution:
\( Kx = 12\times0.25=3 \); \( K-Kx=9 \); denominator \(=9+0.5\times6=9+3=12\)
\( C_0 = \dfrac{-3+3}{12} = \dfrac{0}{12} = 0 \)
\( C_1 = \dfrac{3+3}{12} = \dfrac{6}{12} = 0.5 \)
\( C_2 = \dfrac{9-3}{12} = \dfrac{6}{12} = 0.5 \)
Answer: \(C_0=0\), \(C_1=0.5\), \(C_2=0.5\); check: \(0+0.5+0.5=1\) ✓.
Given: Using \(C_0=0\), \(C_1=0.5\), \(C_2=0.5\) from Example 1, with \(I_1=50\), \(I_2=90\), \(O_1=55\ \dfrac{\text{m}^3}{\text{s}}\). Find \(O_2\).
Solution:
\( O_2 = 0\times90+0.5\times50+0.5\times55 = 25+27.5 = 52.5\ \dfrac{\text{m}^3}{\text{s}} \)
Answer: \(O_2\approx52.5\ \dfrac{\text{m}^3}{\text{s}}\).
Given: A channel reach's Muskingum analysis gives \(x=0.5\). Comment on the routing behaviour.
Solution:
At \(x=0.5\), storage \(S=K[0.5I+0.5O]=K\times\)(average of \(I\) and \(O\)) — the reach behaves as pure translation, meaning the flood wave is delayed by time \(K\) but not attenuated (its peak shape is preserved).
Answer: \(x=0.5\) represents pure translation (delay without attenuation) — the theoretical upper limit of \(x\).
Fig. 8.1 — Flood routing: the outflow hydrograph is delayed and attenuated (lower peak, broader shape) relative to the inflow, due to channel or reservoir storage.
| Zone | Description |
|---|---|
| Vadose zone (aeration zone) | Above water table; pores partly filled with air and water |
| Capillary zone | Just above water table; water held by capillarity; pressure < atmospheric |
| Saturated zone (phreatic zone) | Below water table; all pores saturated; pressure > atmospheric |
| Aquifer | Characteristics | Piezometric surface |
|---|---|---|
| Unconfined (phreatic) | Water table is the upper boundary; free surface | Water table itself |
| Confined (artesian) | Bounded above by an impermeable layer (aquitard); water under pressure | Above the top of the aquifer; may rise above ground → flowing artesian well |
| Leaky (semi-confined) | Semi-permeable layer above; leakage from adjacent layer | Between confined and unconfined behaviour |
| Perched | Local saturated zone above the main water table, on an impermeable lens | — |
Darcy's law states that groundwater flow rate is proportional to the hydraulic gradient, with hydraulic conductivity as the proportionality constant — remarkably, this simple linear law (originally derived for sand filters) holds accurately for almost all natural groundwater flow, since flow velocities are so low that Reynolds numbers stay well within the laminar range. Two related but distinct quantities matter: the Darcy velocity (specific discharge, based on the full cross-sectional area including solids) and the seepage velocity (the actual, faster velocity of water through the pore spaces alone) — the two differ by a factor equal to the porosity.
For steady-state radial flow to a pumped well, the Thiem equation gives discharge directly from the drawdown observed at two different radii — a different form applies for confined aquifers (where transmissivity is constant with depth) versus unconfined aquifers (where the Dupuit-Thiem approximation accounts for the varying saturated thickness). The radius of influence (Sichardt's empirical formula) estimates how far the cone of depression extends from the pumped well, useful for well-interference and well-spacing design.
Unlike the Thiem equation (which requires the system to have reached steady state), the Theis equation describes the time-varying drawdown around a pumped well from the very start of pumping, using the well function \(W(u)\) — an infinite series that must generally be evaluated from tables or curve-matching, though the Cooper-Jacob approximation gives a much simpler closed-form expression valid once pumping has continued long enough (small \(u\)), making late-time drawdown analysis considerably more tractable.
\( Q = K\,i\,A \)
Darcy velocity: \( v = \dfrac{Q}{A} = Ki \)
Seepage velocity: \( v_s = \dfrac{v}{n} \) (\(n\)=porosity)
Validity: \( Re = \dfrac{vd_{50}}{\nu} < 1\text{–}10 \) (laminar)
Transmissivity: \( T=Kb \); Storativity: \(S\)=specific storage×\(b\) (confined) or specific yield (unconfined)
Confined: \( Q = \dfrac{2\pi T(h_2-h_1)}{\ln(r_2/r_1)} \)
Unconfined (Dupuit-Thiem): \( Q = \dfrac{\pi K(h_2^2-h_1^2)}{\ln(r_2/r_1)} \)
Radius of influence (Sichardt): \( R = 3000\,s_w\sqrt K \) (\(s_w\)=drawdown at well, m; \(K\) in m/s; \(R\) in m)
\( s = \dfrac{Q}{4\pi T}W(u) \)
\( u = \dfrac{r^2S}{4Tt} \)
\( W(u) = -0.5772-\ln u+u-\dfrac{u^2}{2\cdot2!}+\ldots \)
Cooper-Jacob (small \(u<0.05\)): \( W(u)\approx-0.5772-\ln u \)
Given: A confined aquifer has \(K=25\ \text{m/day}\), hydraulic gradient \(i=0.002\), cross-sectional area \(A=500\ \text{m}^2\), porosity \(n=0.3\). Find the discharge and seepage velocity.
Solution:
\( Q = 25\times0.002\times500 = 25\ \dfrac{\text{m}^3}{\text{day}} \)
\( v = Ki = 25\times0.002 = 0.05\ \text{m/day} \)
\( v_s = \dfrac{v}{n} = \dfrac{0.05}{0.3} \approx 0.167\ \text{m/day} \)
Answer: \(Q=25\ \dfrac{\text{m}^3}{\text{day}}\), \(v_s\approx0.167\ \text{m/day}\).
Given: A confined aquifer, \(T=300\ \dfrac{\text{m}^2}{\text{day}}\). Observation wells at \(r_1=15\ \text{m}\) and \(r_2=90\ \text{m}\) show heads \(h_1=20\ \text{m}\) and \(h_2=23\ \text{m}\). Find the well discharge.
Solution:
\( Q = \dfrac{2\pi\times300\times(23-20)}{\ln(90/15)} = \dfrac{2\pi\times300\times3}{\ln6} = \dfrac{5654.9}{1.792} \approx 3155\ \dfrac{\text{m}^3}{\text{day}} \)
Answer: \(Q\approx3155\ \dfrac{\text{m}^3}{\text{day}}\).
Given: A well has drawdown \(s_w=4\ \text{m}\), aquifer \(K=2\times10^{-4}\ \text{m/s}\). Find the radius of influence.
Solution:
\( R = 3000\times4\times\sqrt{2\times10^{-4}} = 12000\times0.01414 \approx 169.7\ \text{m} \)
Answer: \(R\approx170\ \text{m}\).
Fig. 9.1 — Steady-state radial flow to a pumped well in a confined aquifer: the Thiem equation relates discharge to the observed heads at two radii, forming a cone of depression in the piezometric surface.
| Topic | Key Formula / Value |
|---|---|
| Water balance | \(P = R + ET + \Delta S\) |
| Optimum rain gauges | \(N = (\dfrac{C_v}{\varepsilon})^2\) |
| Missing rainfall | \(P_x=(\dfrac{N_x}{3})[\dfrac{P_A}{N_A}+\dfrac{P_B}{N_B}+\dfrac{P_C}{N_C}]\) |
| Pan evaporation | \(E_{lake}=0.7\times E_{pan}\) |
| Horton's equation | \(f=f_c+(f_0-f_c)e^{-kt}\) |
| SCS runoff | \(Q=(P-0.2S)\dfrac{^2}{P+0.8S}\); \(S=\dfrac{25400}{CN}-254\) |
| Gumbel distribution | \(x_T=\bar x+K\sigma\); \(y_T=-\ln[-\ln(1-\dfrac{1}{T})]\) |
| Muskingum routing | \(O_2=C_0I_2+C_1I_1+C_2O_1\); \(C_0+C_1+C_2=1\) |
| Darcy's law | \(Q=KiA\); \(T=Kb\); \(v_s=\dfrac{v}{n}\) |
| Thiem (confined) | \(Q=2\pi \dfrac{T(h_2-h_1)}{\ln(r_2/r_1)}\) |
| Thiem (unconfined) | \(Q=\pi \dfrac{K(h_2^2-h_1^2)}{\ln(r_2/r_1)}\) |
| Rational formula | \(Q_p=\dfrac{CiA}{3.6}\) (m³/s; i in mm/hr; A in km²) |
| Topic | GATE Focus | ESE Focus | SSC JE Focus |
|---|---|---|---|
| Hydrological Cycle | Water balance numericals; catchment characteristics | Full water balance with storage terms | Water cycle components; basic definitions |
| Precipitation | Thiessen/isohyetal method; Normal Ratio numericals | IDF curves; return period/risk derivation | Rain gauge types; Thiessen concept |
| Evapotranspiration | Penman formula application; pan coefficient | Full Penman derivation; Blaney-Criddle | PET vs AET; pan coefficient ≈0.7 |
| Infiltration | Horton's equation numericals; φ-index | Green-Ampt model; W-index | Horton's equation concept |
| Runoff | SCS-CN numericals; current meter methods | Full SCS-CN with AMC conditions; dilution gauging | SCS-CN basic formula; rating curve concept |
| Unit Hydrograph | UH derivation from DRH; superposition problems | S-curve method; Snyder's synthetic UH | UH definition and assumptions |
| Flood Frequency | Gumbel's method numericals; plotting positions | Log-Pearson III; full frequency analysis | Return period concept; Rational formula |
| Flood Routing | Muskingum C₀,C₁,C₂ numericals | Reservoir routing; storage-indication curve | Muskingum concept; continuity equation |
| Groundwater | Darcy's law; Thiem equation numericals | Theis equation; Cooper-Jacob method | Aquifer types; Darcy's law basic |
1. A catchment has 6 existing gauges with Cv=28%. Find the optimum number of gauges for 10% error.
2. A storm of P=80 mm falls on a catchment with CN=80. Find the direct runoff (SCS-CN method).
3. Find the Weibull return period for the 5th ranked flood in a 45-year record.
4. A reach has K=8 hr, x=0.2, Δt=4 hr. Find C₀, C₁, C₂.
5. An unconfined aquifer well has K=15 m/day, h1=18 m at r1=20 m, h2=22 m at r2=100 m. Find discharge (Dupuit-Thiem).