Question
An isolated storm of 3 hours occurred over a catchment area in the following manner Total rainfall excess (in cm, rounded off to one decimal place) over the catchment for the above storm is ________.% Catchment Area $\phi$-index (cm/hour) Rainfall (cm) 1st hour 2nd hour 3rd hour 15 0.5 0.4 2.5 1.6 35 1.0 0.8 3.0 2.1 50 0.8 0.6 2.6 1.9
Calculating Catchment Rainfall Excess Using $\phi$-index
The problem requires calculating the total rainfall excess over a catchment area during a 3-hour storm using the provided data on catchment area percentages, $\phi$-indices, and hourly rainfall. Rainfall excess is the amount of rainfall that becomes surface runoff, calculated as Rainfall - $\phi$-index, only when Rainfall exceeds the $\phi$-index.Storm Data Table
| Catchment Area (%) | $\phi$-index (cm/hour) | Rainfall Hour 1 (cm) | Rainfall Hour 2 (cm) | Rainfall Hour 3 (cm) |
| 15 | 0.50 | 2.5 | 1.6 | 1.9 |
| 35 | 1.00 | 3.0 | 2.1 | 2.1 |
| 50 | 0.80 | 2.6 | 1.9 | 1.9 |
Step 1: Calculate Rainfall Excess for Each Hour and Area Segment
We calculate the excess rainfall for each segment by subtracting the $\phi$-index from the rainfall, provided rainfall is greater than the $\phi$-index. The contribution of each segment to the total hourly excess is its percentage area multiplied by its excess rainfall.Hour 1 Excess Calculation
- Area 1 (15%, $\phi$=0.50): Rainfall = 2.5 cm. Excess = $2.5 - 0.50 = 2.0$ cm. Contribution = $0.15 \times 2.0 = 0.30$ cm.
- Area 2 (35%, $\phi$=1.00): Rainfall = 3.0 cm. Excess = $3.0 - 1.00 = 2.0$ cm. Contribution = $0.35 \times 2.0 = 0.70$ cm.
- Area 3 (50%, $\phi$=0.80): Rainfall = 2.6 cm. Excess = $2.6 - 0.80 = 1.8$ cm. Contribution = $0.50 \times 1.8 = 0.90$ cm.
- Total Weighted Excess Hour 1 = $0.30 + 0.70 + 0.90 = 1.90$ cm.
Hour 2 Excess Calculation
- Area 1 (15%, $\phi$=0.50): Rainfall = 1.6 cm. Excess = $1.6 - 0.50 = 1.1$ cm. Contribution = $0.15 \times 1.1 = 0.165$ cm.
- Area 2 (35%, $\phi$=1.00): Rainfall = 2.1 cm. Excess = $2.1 - 1.00 = 1.1$ cm. Contribution = $0.35 \times 1.1 = 0.385$ cm.
- Area 3 (50%, $\phi$=0.80): Rainfall = 1.9 cm. Excess = $1.9 - 0.80 = 1.1$ cm. Contribution = $0.50 \times 1.1 = 0.550$ cm.
- Total Weighted Excess Hour 2 = $0.165 + 0.385 + 0.550 = 1.100$ cm.
Hour 3 Excess Calculation
- Area 1 (15%, $\phi$=0.50): Rainfall = 1.9 cm. Excess = $1.9 - 0.50 = 1.4$ cm. Contribution = $0.15 \times 1.4 = 0.210$ cm.
- Area 2 (35%, $\phi$=1.00): Rainfall = 2.1 cm. Excess = $2.1 - 1.00 = 1.1$ cm. Contribution = $0.35 \times 1.1 = 0.385$ cm.
- Area 3 (50%, $\phi$=0.80): Rainfall = 1.9 cm. Excess = $1.9 - 0.80 = 1.1$ cm. Contribution = $0.50 \times 1.1 = 0.550$ cm.
- Total Weighted Excess Hour 3 = $0.210 + 0.385 + 0.550 = 1.145$ cm.
Step 2: Calculate Total Rainfall Excess
Sum the weighted excess rainfall from all three hours. Total Excess = Excess Hour 1 + Excess Hour 2 + Excess Hour 3 Total Excess = $1.90 + 1.100 + 1.145 = 4.145$ cm.Step 3: Round the Result
Rounding the total excess rainfall to one decimal place gives 4.1 cm. Final Answer: The total rainfall excess is calculated to be 4.1 cm.Was this answer helpful?
Important Questions from Runoff Estimation
- A $0.80\text{ km}^2$ agricultural watershed has a slope of $0.30\%$ and the maximum length of travel of water is $1\text{ km}$. The 10-year maximum depth of rainfall is tabulated below:
Duration (minutes) 5 10 20 30 40 60 Maximum depth of rainfall (mm) 16 24 39 50 60 65
Half of the watershed has row crops (runoff coefficient = $0.40$), whilst the other half of the watershed has pasture (runoff coefficient = $0.35$). The peak flow rate for the watershed (in $\text{m}^3\text{.s}^{-1}$) for the 10-year return period is ________. (Rounded off to two decimal places) - The mass curve of a rainfall event of 100 min duration over a catchment is given in the table.
If the initial loss is 0.6 cm and $\phi$-index is 0.6 cm/hour, the total surface runoff from the catchment is ______ cm (rounded off to one decimal place).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 - The Rational method is used to estimate
- A $100 \text{ ha}$ watershed received rainfall at a rate of $5 \text{ cm h}^{-1}$ for $2 \text{ hours}$. If the runoff generated by the storm was at the rate of $1 \text{ m}^3 \text{ s}^{-1}$ for $10 \text{ hours}$, the runoff coefficient for the watershed would be
The Curve Number (CN) of a watershed of $40$ ha area under given hydrologic soil group, land use and management practices, and Antecedent Moisture Condition (AMC)-II is $80$. The initial abstraction is $20$% of maximum retention. For the rainfall event of $40$ mm, the direct runoff in mm (round off to $2$ decimal places) is _____________
Need Expert Advice?
