How are the rainfall depths in each 20-min interval obtained from the mass curve?

By differences between successive cumulative values

By adding successive cumulative values

By dividing cumulative rainfall by 100

How is the 0.6 cm initial loss handled?

Taken first: all 0.5 cm of interval 1 and 0.1 cm of interval 2, then phi loss is applied

Added to the phi loss in every interval

Subtracted from the final runoff only

2081 Bhadra · 6 marks

Surface runoff with initial loss and phi-index, 100-min storm

Short answer

After subtracting initial loss of 0.6 cm from early rain, φ-index losses of 0.2 cm per 20-min interval applied to remaining rainfall yielded excess depths summing to 2.1 cm, so total surface runoff = 2.1 cm.

The mass curve of the rainfall of 100 min duration is given below. If the catchment had an initial loss of 0.6 cm and $\phi$ -index of 0.6 cm/hr, calculate the total surface runoff from the catchment.

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

Rainfall in each 20-minute interval (difference of the mass curve):

Interval (min)	0-20	20-40	40-60	60-80	80-100
Rain (cm)	0.5	0.7	1.4	0.7	0.2

$\phi$ loss per 20 min $= 0.6\times\frac{20}{60} = 0.2$ cm.

The initial loss of 0.6 cm is satisfied first: all 0.5 cm of interval 1 and 0.1 cm of interval 2. After that the $\phi$ loss is applied.

Interval	Rain after initial loss (cm)	$\phi$ loss (cm)	Excess (cm)
1	0	0	0
2	0.6	0.2	0.4
3	1.4	0.2	1.2
4	0.7	0.2	0.5
5	0.2	0.2	0
Total			2.1

Check: total rain 3.5 - initial loss 0.6 - infiltration (0.2 + 0.2 + 0.2 + 0.2 = 0.8) = 2.1 cm.

Answer: Total surface runoff = 2.1 cm (depth over the catchment).

✓ Answer checked by Claude Sonnet 5.5

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