The mass curve of a rainfall event of 100 min duration over a catchment is given in the table.
Time from start of rainfall (min) 0 20 40 60 80 100
Cumulative rainfall (cm) 0 0.5 1.2 2.6 3.3 3.5
If the initial loss is 0.6 cm and φ-index is 0.6 cm/hour, the total surface runoff from the catchment is ______ cm (rounded off to one decimal place).

Question

Which one of the following gases is primarily responsible for the formation of photochemical smog?

Answer 1

A quicker cross-check uses the overall storm water balance instead of working interval by interval.

The total rainfall over the whole event is read directly off the mass curve as $P = 3.5$ cm (the cumulative value at $t=100$ min).

The total infiltration loss controlled by the $\phi$-index over the complete storm duration ($td = 100$ min $= 1.667$ hr) is $$F = \phi \times td = 0.6\ cm/hr \times 1.667\ hr = 1.0\ cm$$

Since this ongoing infiltration loss already exceeds the stated initial loss of 0.6 cm well before the storm ends (it reaches that depth by the 60-minute mark), the initial loss is already folded into this figure and is not an extra separate deduction. Applying the storm water balance $P = F + R$: $$R = P - F = 3.5 - 1.0 = 2.5\ cm$$

This total-balance shortcut gives the same figure as summing the interval-by-interval excess rainfall, which is a useful confirmation that no arithmetic slip crept into either method. \[\boxed{R = 2.5\ cm}\]

Answer 2

A quicker cross-check uses the overall storm water balance instead of working interval by interval.

The total rainfall over the whole event is read directly off the mass curve as $P = 3.5$ cm (the cumulative value at $t=100$ min).

The total infiltration loss controlled by the $\phi$-index over the complete storm duration ($td = 100$ min $= 1.667$ hr) is $$F = \phi \times td = 0.6\ cm/hr \times 1.667\ hr = 1.0\ cm$$

Since this ongoing infiltration loss already exceeds the stated initial loss of 0.6 cm well before the storm ends (it reaches that depth by the 60-minute mark), the initial loss is already folded into this figure and is not an extra separate deduction. Applying the storm water balance $P = F + R$: $$R = P - F = 3.5 - 1.0 = 2.5\ cm$$

This total-balance shortcut gives the same figure as summing the interval-by-interval excess rainfall, which is a useful confirmation that no arithmetic slip crept into either method. \[\boxed{R = 2.5\ cm}\]

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Correct Answer: 2.5

Solution and Explanation

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