Question

An 8 ha watershed receives rainfall intensities of 2.5, 3.6, 5.4, 3.3, 2.6 and 1.2 cm.h\(^{-1}\) at the successive intervals of 30 minutes. The corresponding surface runoff volume is estimated to be 4800 m\(^3\). Neglecting initial abstraction losses, the W-index for this watershed is _________ cm.h\(^{-1}\) (Rounded off to 2 decimal places).

Hint:

When calculating the W-index, ensure that rainfall volumes and surface runoff are calculated in the same units, and then use the formula to find the average intensity.

Answer 1

Given:

• Area of watershed = 8 ha = 80,000 m²
• Rainfall intensities (cm/h) = 2.5, 3.6, 5.4, 3.3, 2.6, 1.2
• Duration of each interval = 0.5 h
• Total runoff volume = 4800 m³

Step 1: Total Rainfall Depth

\( \text{Total Rainfall} = 0.5 \times (2.5 + 3.6 + 5.4 + 3.3 + 2.6 + 1.2) = 0.5 \times 18.6 = 9.3\, \text{cm} \)

Step 2: Convert Rainfall Depth to Volume

\( \text{Total Rainfall Volume} = \frac{9.3}{100} \times 80{,}000 = 7440\, \text{m}^3 \)

Step 3: Calculate W-index

\( W = \frac{\text{Rainfall Volume} - \text{Runoff Volume}}{\text{Area}} \times \frac{100}{\text{Time}} \)

\( W = \frac{7440 - 4800}{80{,}000} \times \frac{100}{3} = \frac{2640}{80{,}000} \times 33.33 \approx \boxed{1.11\, \text{cm/h}} \)