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 — from the start of effective rainfall to the peak.
- Basin lag — from the centroid of effective rainfall to the peak (sometimes to the centroid of the hydrograph).
- Time of concentration — time for runoff from the hydraulically most remote point to reach the outlet.
- Time base — 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 days after the peak.
- Fixed base method — extend the pre-storm recession to below the peak, then join to the point days after the peak.
- Variable slope method — separate groundwater recession curves before and after the storm.
Recession curve: , with recession constant (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
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:
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 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
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:
T-hour UH from a D-hour S-curve:
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 ( = main stream length, = distance along the stream to the point nearest the catchment centroid, in km; in h); peak discharge (m³/s per cm).
- SCS dimensionless unit hydrograph — time to peak ; peak (m³/s per cm, A in km², in h); time base ≈ 5 for the curvilinear form (2.67 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 — 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
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. → (Check: equilibrium S-curve discharge m³/s = sum of the UH ordinates ✓)
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: m³ = 5 cm over 216 km² ✓
Find for a catchment of 1000 km².
Solution.
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
Frequently tested points
- Hydrograph: rising limb, crest, recession; recession depends only on catchment storage.
- Fan-shaped catchment and downstream-moving storm → higher peak.
- days; recession .
- UH: 1 cm effective rain, uniform, D hours; assumptions of linearity (proportionality, superposition) and time invariance.
- Area under UH = 1 cm × A: .
- S-curve equilibrium ; .
- Snyder ; IUH is the UH with zero duration.
- 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 when deriving a T-hour UH from an S-curve.
- 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.