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Chapter 3 of 12

Runoff & Hydrographs

In the IOCL Graduate Engineer Civil syllabus under Hydrology & Irrigation · 2 parts

Runoff & Rainfall-Runoff Relationships · Hydrographs & Unit Hydrograph

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📑 Contents (18 sections)

Part 1 of 2

Runoff & Rainfall-Runoff Relationships

Last reviewed 16 Sept 2026 · 7 min read

Runoff

Runoff is the part of precipitation that flows towards streams, rivers and the sea as surface or subsurface flow.

Component	Path
Surface runoff (overland flow)	Flows over the land surface into channels once rainfall exceeds infiltration and depression storage
Interflow (subsurface storm flow)	Infiltrated water moving laterally through upper soil layers and re-emerging in streams
Baseflow (groundwater flow)	Delayed contribution of groundwater to the stream; sustains flow in dry periods

Direct runoff = surface runoff + prompt interflow; it forms the flood hydrograph.
Natural flow (virgin flow) — runoff unaffected by human diversions and storage; observed flows must be corrected for upstream abstractions and return flows.

Types of streams

Perennial — flow throughout the year; the water table stays above the stream bed (effluent streams fed by baseflow).
Intermittent — flow during wet seasons only; water table above the bed in wet seasons.
Ephemeral — flow only during and immediately after rain; the bed is always above the water table (common in arid zones).

Factors affecting runoff

Climatic: type, intensity, duration and areal distribution of precipitation; direction of storm movement; antecedent rainfall; evaporation and transpiration.

Physiographic (catchment):

Area — total runoff volume increases, runoff per unit area and peak per unit area decrease.
Shape — fan-shaped (compact) catchments give higher, earlier peaks than fern-leaf (elongated) catchments.
Slope — steeper slopes give quicker, higher runoff.
Soil and geology, land use and vegetation (forests reduce and delay runoff; urbanisation increases it), drainage density (stream length per unit area — higher density gives quicker response), storage in lakes and swamps.

Yield of a catchment

The yield is the total quantity of water that can be expected from a stream in a given period (usually a year). The dependable yield is the yield that is equalled or exceeded in a stated percentage of years — commonly 75% dependable yield for irrigation planning and higher dependability (about 90% or more) for hydropower and drinking water.

Rainfall–runoff relationships

Runoff coefficient and correlation

Runoff coefficient $C = R / P$ .
A linear regression $R = a P + b$ (or exponential forms) is fitted to observed annual rainfall and runoff.

Empirical formulas and tables (Indian practice)

FormulaEmpirical runoff relations (R and P in cm)

Inglis and DeSouza (Western India):

Ghat regions: $R = 0.85 P - 30.5$
Deccan plateau: $R = \frac{( P - 17.8 ) P}{254}$

Khosla's formula (monthly):

R_{m} = P_{m} - L_{m}, L_{m} = 0.48 T_{m} (for T_{m} > 4.5 ° C)

$R_{m}$ , $P_{m}$ , $L_{m}$ in cm; $T_{m}$ = mean monthly temperature (°C). Annual runoff $= \sum R_{m}$ .

Tables: Binnie's percentages (Madhya Pradesh), Barlow's tables (Uttar Pradesh catchments classified by type and season), Strange's tables (Maharashtra, Karnataka).

SCS curve number method

FormulaSCS-CN method (SI units, mm)

Q = \frac{( P - I _{a} ) ^{2}}{P - I _{a} + S} for P > I_{a}, I_{a} = 0.2 S

\Rightarrow Q = \frac{( P - 0.2 S ) ^{2}}{P + 0.8 S}, S = \frac{25 400}{C N} - 254

$C N$ (0–100) depends on the hydrologic soil group (A–D), land use and antecedent moisture condition (AMC I dry, II average, III wet).

Part 2 of 2

Hydrographs & Unit Hydrograph

Last reviewed 16 Sept 2026 · 8 min read

The hydrograph

A hydrograph is a graph of discharge against time at a stream section. A storm hydrograph has:

Part	Description
Rising limb (concentration curve)	Discharge increases as runoff from progressively larger parts of the catchment arrives
Crest segment (peak)	Maximum discharge; occurs when runoff from all parts contributes most
Recession limb (falling limb)	Withdrawal of water from storage in the catchment and channels; independent of storm characteristics
Point of inflection on recession	Marks the end of direct runoff (approximately)

Time parameters:

Time to peak $t_{p}$ — from the start of effective rainfall to the peak.
Basin lag $t_{L}$ — from the centroid of effective rainfall to the peak (sometimes to the centroid of the hydrograph).
Time of concentration $t_{c}$ — time for runoff from the hydraulically most remote point to reach the outlet.
Time base $T_{B}$ — duration of direct runoff.

Factors affecting the hydrograph shape

Catchment shape — fan-shaped catchments: high, sharp peaks; fern-leaf (elongated): flatter, delayed hydrographs.
Size — large catchments have longer time bases and lower peaks per unit area.
Slope — steep channels and land slopes give steep rising limbs.
Drainage density — high density gives quicker, peakier response.
Land use — forests and vegetation flatten the hydrograph; urban areas sharpen it.
Storm characteristics — intensity, duration, areal distribution and direction of movement: a storm moving downstream gives a higher peak than one moving upstream.

Baseflow separation

To obtain the direct runoff hydrograph (DRH), baseflow is subtracted:

Straight-line method — join the start of the rising limb to a point on the recession limb $N$ days after the peak.
Fixed base method — extend the pre-storm recession to below the peak, then join to the point $N$ days after the peak.
Variable slope method — separate groundwater recession curves before and after the storm.

N = 0.83 A^{0.2} (N in days, A in km^{2})

Recession curve: $Q_{t} = Q_{0} K_{r}^{t}$ , with recession constant $K_{r} < 1$ (separately for surface, interflow and baseflow storage).

Effective rainfall hyetograph

Effective rainfall (rainfall excess) = rainfall − losses (using the φ-index or other loss models). Its volume equals the volume of direct runoff. The ERH plotted against time is the input to unit hydrograph computations.

Unit hydrograph

DefinitionUnit hydrograph (Sherman, 1932)

The D-hour unit hydrograph is the direct runoff hydrograph resulting from one unit (1 cm) depth of effective rainfall occurring uniformly over the catchment at a constant rate for D hours.

Assumptions

Time invariance — the DRH for a given effective rainfall is always the same regardless of when it occurs.
Linear response — ordinates are proportional to the effective rainfall depth (proportionality), and hydrographs from successive storms can be added (superposition).
Effective rainfall uniformly distributed over the catchment and within the duration.

Limitations

Precipitation must be nearly uniform — so the method is best for catchments of moderate size (large catchments are subdivided; the upper limit is often quoted as about 5000 km²).
Not suitable when snowmelt or large channel storage dominate, or for very small plots.
The effective rainfall must have a duration close to the unit duration used.

Area under the unit hydrograph

The volume of direct runoff equals 1 cm over the catchment:

A (km^{2}) = 0.36 \sum Q_{i} Δ t (Q in m^{3} / s, Δ t in h)

Derivation from an isolated storm

Select an isolated, uniform storm of duration about D hours.
Separate baseflow → DRH.
Compute the effective rainfall depth = DRH volume ÷ catchment area.
UH ordinate = DRH ordinate ÷ effective rainfall depth (cm).
Average the unit hydrographs from several storms (average peak and time to peak; adjust shape to keep unit volume).

Using the unit hydrograph

For effective rainfalls of $R_{1}, R_{2}, \dots$ cm in successive D-hour blocks, the DRH is obtained by multiplying the UH by each depth, lagging each by D hours, and adding. Baseflow is then added to get the flood hydrograph.

Changing the unit duration

Method of superposition

A nD-hour UH (n integer) = sum of n D-hour UHs lagged successively by D hours, divided by n.

S-curve method

FormulaS-curve (S-hydrograph)

The S-curve is the hydrograph from a continuous effective rainfall of 1 cm per D hours — the sum of an infinite series of D-hour UHs lagged by D hours. It rises to an equilibrium discharge:

Q_{s} = \frac{2.778 A}{D} (A in km^{2}, D in h, Q_{s} in m^{3} / s)

T-hour UH from a D-hour S-curve:

U_{T} (t) = \frac{D}{T} [S (t) - S (t - T)]

The S-curve method works for any T (larger or smaller than D, not necessarily a multiple).

Synthetic unit hydrographs

For ungauged catchments, UH parameters are related to catchment characteristics:

Snyder's method: basin lag $t_{p} = C_{t} (L L_{c a})^{0.3}$ ( $L$ = main stream length, $L_{c a}$ = distance along the stream to the point nearest the catchment centroid, in km; $t_{p}$ in h); peak discharge $Q_{p} = \frac{2.78 C _{p} A}{t _{p}}$ (m³/s per cm).
SCS dimensionless unit hydrograph — time to peak $T_{p} = D /2 + t_{L}$ ; peak $Q_{p} = \frac{2.08 A}{T _{p}}$ (m³/s per cm, A in km², $T_{p}$ in h); time base ≈ 5 $T_{p}$ for the curvilinear form (2.67 $T_{p}$ for the triangular form).
Regional methods — in India, the Central Water Commission's flood estimation reports give regional synthetic UH relations for hydro-meteorological sub-zones.

Instantaneous unit hydrograph (IUH)

The UH as $D \to 0$ — the response to 1 cm of effective rainfall applied instantaneously. It depends only on catchment characteristics. Conceptual models: Nash's cascade of linear reservoirs, Clark's model (time–area diagram routed through a linear reservoir). A D-hour UH is obtained by routing/averaging the IUH (or from its S-curve).

Worked examples

Worked ExampleExample 1 — catchment area from a UH

The ordinates of a 4-hour UH at 4-hour intervals are 0, 20, 60, 40, 20, 10, 0 m³/s. Find the catchment area.

Solution. $\sum Q Δ t = 150 \times 4 = 600$ → $A = 0.36 \times 600 = 216 k m^{2}$ (Check: equilibrium S-curve discharge $2.778 \times 216/4 = 150$ m³/s = sum of the UH ordinates ✓)

Worked ExampleExample 2 — flood hydrograph from the UH

For the same catchment, a storm gives 3 cm effective rainfall in the first 4 hours and 2 cm in the next 4 hours. Baseflow is 10 m³/s. Find the flood hydrograph and its peak.

Solution.

Time (h)	UH	3 × UH	2 × UH (lagged 4 h)	DRH	Flood (+10)
0	0	0	–	0	10
4	20	60	0	60	70
8	60	180	40	220	230
12	40	120	120	240	250
16	20	60	80	140	150
20	10	30	40	70	80
24	0	0	20	20	30
28	–	–	0	0	10

Peak flood = 250 m³/s at 12 h. Volume check: $\sum D R H Δ t = 750 \times 4 \times 3600 = 1.08 \times 1 0^{7}$ m³ = 5 cm over 216 km² ✓

Worked ExampleExample 3 — baseflow separation time

Find $N$ for a catchment of 1000 km².

Solution. $N = 0.83 \times 100 0^{0.2} = 0.83 \times 3.98 = 3.3 days$

Worked ExampleExample 4 — deriving a UH

A 6-hour storm producing 2.5 cm of effective rainfall gives a DRH with a peak ordinate of 150 m³/s. What is the peak of the 6-hour UH?

Solution. UH peak $= 150/2.5 = 60 m^{3} /s$

Frequently tested points

Hydrograph: rising limb, crest, recession; recession depends only on catchment storage.
Fan-shaped catchment and downstream-moving storm → higher peak.
$N = 0.83 A^{0.2}$ days; recession $Q_{t} = Q_{0} K_{r}^{t}$ .
UH: 1 cm effective rain, uniform, D hours; assumptions of linearity (proportionality, superposition) and time invariance.
Area under UH = 1 cm × A: $A = 0.36 \sum Q Δ t$ .
S-curve equilibrium $Q_{s} = 2.778 A / D$ ; $U_{T} = (D / T) [S (t) - S (t - T)]$ .
Snyder $t_{p} = C_{t} (L L_{c a})^{0.3}$ ; IUH is the UH with zero duration.

Common MistakeCommon mistakes

Adding baseflow before multiplying by rainfall depth (convert only the DRH).
Lagging subsequent rainfall blocks by the wrong interval (lag equals the UH duration).
Forgetting the factor $D / T$ when deriving a T-hour UH from an S-curve.

Revision SummaryChapter summary

Hydrograph shape reflects both storm and catchment characteristics.
Baseflow separation gives the direct runoff hydrograph; its volume equals effective rainfall.
The unit hydrograph converts effective rainfall to direct runoff by proportionality and superposition.
Superposition and S-curves change the unit duration; synthetic UHs and the IUH serve ungauged and conceptual analyses.

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