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Question

A 10 ha watershed received a 4-hour rainfall storm with rainfall intensity of 2, 1, 1.5, and 1 cm.h$^{-1}$ in each subsequent hour, respectively. Assuming phi-index ($\phi$) of 1 cm.h$^{-1}$, the volume of direct runoff, in m$^3$, is _________. (Rounded off to 2 decimal places)

The correct answer is
1500

Calculate Direct Runoff Volume using Phi-Index

This solution calculates the direct runoff volume for a given watershed using the phi-index method based on the provided rainfall data.

Watershed Rainfall Analysis

  • Watershed Area: 10 ha
  • Rainfall Duration: 4 hours
  • Rainfall Intensities: $I_1 = 2$ cm/h, $I_2 = 1$ cm/h, $I_3 = 1.5$ cm/h, $I_4 = 1$ cm/h
  • Phi-Index ($\phi$): 1 cm/h

Step 1: Determine Effective Rainfall

Calculate the excess rainfall for each hour where intensity ($I$) exceeds the phi-index ($\phi$). The excess rainfall depth is $I - \phi$.

  • Hour 1: $I_1 = 2$ cm/h. Since $2 > 1$, Excess = $2 - 1 = 1$ cm.
  • Hour 2: $I_2 = 1$ cm/h. Since $1 \le 1$, Excess = 0 cm.
  • Hour 3: $I_3 = 1.5$ cm/h. Since $1.5 > 1$, Excess = $1.5 - 1 = 0.5$ cm.
  • Hour 4: $I_4 = 1$ cm/h. Since $1 \le 1$, Excess = 0 cm.

The total depth of direct runoff ($R$) is the sum of the hourly excesses: $R = 1 + 0 + 0.5 + 0 = 1.5$ cm.

Step 2: Calculate Runoff Volume

Convert the watershed area and runoff depth to compatible units (meters).

  • Area ($A$) = 10 ha = $10 \times 10,000$ m$^2$ = 100,000 m$^2$.
  • Runoff Depth ($R$) = 1.5 cm = 0.015 m.

Calculate the volume ($V$) using the formula $V = A \times R$.

$V = 100,000 \text{ m}^2 \times 0.015 \text{ m}$

$V = 1500 \text{ m}^3$.

Result

The volume of direct runoff is 1500 m$^3$.

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Important Questions from Runoff Estimation

  1. 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)51020304060
    Maximum depth of rainfall (mm)162439506065

    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)
  2. 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)020406080100
    Cumulative rainfall (cm)00.51.22.63.33.5
  3. The Rational method is used to estimate
  4. 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
  5. 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 _____________

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