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).
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).
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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.
Updated On: Jan 30, 2026
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Correct Answer: 1.08
Solution and Explanation
Given:
Area of watershed = $8$ ha = $80{,}000$ msuperscript{2}
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 msuperscript{3}
Step 1: Total rainfall depth (cm)
\[
{Total Rainfall} = 0.5 \times (2.5 + 3.6 + 5.4 + 3.3 + 2.6 + 1.2) = 0.5 \times 18.6 = 9.3 \, {cm}
\]
Step 2: Convert rainfall depth to volume
\[
{Total Rainfall Volume} = \frac{9.3}{100} \times 80{,}000 = 7440 \, {m}^3
\]
Step 3: Calculate W-index
\[
W = \frac{{Rainfall Volume} - {Runoff Volume}}{{Area}} \times \frac{100}{{Time}}
\]
\[
W = \frac{7440 - 4800}{80{,}000} \times \frac{100}{3} = \frac{2640}{80{,}000} \times 33.33 \approx 1.11 \, {cm/h}
\]
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