Question

The following figure shows a 2-hour unit hydrograph (1 cm rainfall excess) for a catchment area of 540 hectare. Find the peak discharge (in m$^3$/s, rounded to one decimal place).

Hint:

For triangular unit hydrographs, equate total runoff volume to triangular area: \[ V = \frac{1}{2}(\text{base})(Q_p). \]

Answer 1

Correct Answer: 4

A unit hydrograph of 1 cm rainfall excess must satisfy: \[ \text{Total runoff volume} = \text{Catchment area} \times 1\ \text{cm} \] Convert the catchment area: \[ 540\ \text{hectare} = 540 \times 10^4\ \text{m}^2 = 5.4 \times 10^6\ \text{m}^2 \] Rainfall excess depth: \[ 1\ \text{cm} = 0.01\ \text{m} \] Thus total runoff volume: \[ V = (5.4 \times 10^6)(0.01) = 5.4 \times 10^4\ \text{m}^3 \] Hydrograph shape: It is a triangle:
- Rising limb: 1 hour
- Falling limb: 2 hours
- Total base = 3 hours
Let peak discharge = \(Q_p\). Area of triangular hydrograph: \[ V = \frac{1}{2} \times \text{base} \times Q_p \] Convert base time to seconds: \[ 3\ \text{hr} = 3 \times 3600 = 10800\ \text{s} \] Thus: \[ 5.4 \times 10^4 = \frac{1}{2}(10800) Q_p \] \[ Q_p = \frac{5.4 \times 10^4 \times 2}{10800} \] \[ Q_p = 10\ \text{m}^3/\text{s} \] Rounded to one decimal: \[ \boxed{10.0\ \text{m}^3/\text{s}} \]