Question

An isolated 3-hour storm occurred over a catchment.
Given the $\phi$–index and rainfall data, compute the total rainfall excess (rounded off to one decimal place).

Hint:

Rainfall excess = rainfall – $\phi$ index (but not below zero), weighted by area.

Answer 1

Correct Answer: 3

Catchment 1: 15% area
$\phi = 0.5$ cm/hr
Rainfall: 0.4, 2.5, 1.6 cm
Excess (negative → 0):
\[ (0.4-0.5)\rightarrow 0,\quad 2.5-0.5=2.0,\quad 1.6-0.5=1.1. \]
Weighted excess:
\[ 0.15(0 + 2.0 + 1.1) = 0.465\ \text{cm}. \]
Catchment 2: 35% area
$\phi = 1.0$ cm/hr
Rainfall: 0.8, 3.0, 2.1 cm
Excess:
\[ 0.8-1.0=0,\quad 3.0-1.0=2.0,\quad 2.1-1.0=1.1. \]
Weighted:
\[ 0.35(0 + 2.0 + 1.1) = 1.085. \]
Catchment 3: 50% area
$\phi = 0.8$ cm/hr
Rainfall: 0.6, 2.6, 1.9 cm
Excess:
\[ 0.6-0.8=0,\quad 2.6-0.8=1.8,\quad 1.9-0.8=1.1. \]
Weighted:
\[ 0.50(0 + 1.8 + 1.1) = 1.45. \]
Total rainfall excess:
\[ 0.465 + 1.085 + 1.45 = 3.00\ \text{cm}. \]
Thus:
\[ \boxed{3.0\ \text{cm}} \]