Unit Hydrographs, Flood Routing and Groundwater
Unit hydrograph theory and S-curve, peak flow from unit hydrographs, reservoir and channel routing (Muskingum), aquifers, Dupuit's well equations, and yield of wells.
The unit hydrograph turns any rainfall into a flood hydrograph. Flood routing follows the flood through reservoirs and rivers. Groundwater hydraulics sizes wells.
Unit hydrograph
A D-hour unit hydrograph (UH) is the direct runoff hydrograph from 1 cm of effective rainfall falling uniformly over the catchment in D hours.
- Linearity: ordinates scale with rainfall excess (3 cm in D hours → 3 × UH ordinates).
- Time invariance: later storms give the same shape, shifted in time.
- Area under a UH = catchment area × 1 cm. So for a catchment A (km²): Σ ordinates (m³/s) × Δt (h) × 3600 = A × 10⁶ × 0.01.
- To change the duration: superpose lagged UHs (for multiples of D), or use the S-curve (sum of an infinite series of D-hour UHs) for any duration.
- Synthetic UH (Snyder's method) for ungauged catchments.
Flood routing
- Reservoir (level-pool) routing: continuity I − O = dS/dt; solved step by step (Puls method). The outflow peak is lower and later than the inflow peak, and it occurs where the outflow curve crosses the falling inflow curve.
- Channel routing (Muskingum): S = K[xI + (1 − x)O]; O₂ = C₀I₂ + C₁I₁ + C₂O₁ with C₀ + C₁ + C₂ = 1. K is roughly the travel time; x = 0 to 0.5 (0 for reservoirs).
Groundwater
- Aquifer: a formation that stores and transmits water. Confined (between impermeable layers, may be artesian) or unconfined (water table aquifer).
- Specific yield (water that drains by gravity) + specific retention = porosity.
- Transmissivity T = k b (b = aquifer thickness). Storativity S.
Steady flow to a well (Dupuit–Thiem):
- Confined: Q = 2π k b (h₂ − h₁)/ln(r₂/r₁)
- Unconfined: Q = π k (h₂² − h₁²)/ln(r₂/r₁)
h are water levels at observation radii r. Using the well radius and radius of influence R gives the well's yield.
Worked example
A 6-hour unit hydrograph has a peak of 50 m³/s. A storm gives 4 cm of effective rainfall in 6 hours, with base flow 10 m³/s.
- Peak flood = 4 × 50 + 10 = 210 m³/s.
Practice
A 4-hour UH has ordinates at 4-hour intervals summing to 250 m³/s. What is the catchment area in km²?
Show answer and explanation
Answer: 360Volume = 250 × 4 × 3600 = 3.6 × 10⁶ m³ = A × 10⁶ × 0.01, so A = 360 km².
A confined aquifer 20 m thick has k = 10⁻⁴ m/s. What is its transmissivity in m²/day?
Show answer and explanation
Answer: 172.8T = kb = 10⁻⁴ × 20 = 0.002 m²/s = 172.8 m²/day.
For the Muskingum method, C₀ = 0.2 and C₁ = 0.5. What is C₂?
Show answer and explanation
Answer: 0.3C₀ + C₁ + C₂ = 1, so C₂ = 0.3.
A storm of 3 cm effective rainfall in the UH duration falls on a catchment whose UH peak is 40 m³/s; base flow is 5 m³/s. What is the flood peak in m³/s?
Show answer and explanation
Answer: 125Peak = 3 × 40 + 5 = 125 m³/s (by linearity).