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Question

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 correct answer is
$0.36$

Solution: Runoff Coefficient Calculation

The runoff coefficient ($C$) represents the ratio of direct runoff volume to the total rainfall volume over a watershed. It is calculated using the formula:

$C = \frac{\text{Volume of Runoff}}{\text{Volume of Rainfall}}$

Step 1: Calculate Rainfall Volume

First, convert the watershed area and rainfall rate to consistent units.

  • Watershed Area ($A$): $100 \text{ ha} = 100 \times 10^4 \text{ m}^2 = 1 \times 10^6 \text{ m}^2$.
  • Rainfall Rate: $5 \text{ cm h}^{-1} = 0.05 \text{ m h}^{-1}$.
  • Rainfall Duration ($t_{rain}$): $2 \text{ hours}$.

Calculate the total rainfall depth:

$ \text{Rainfall Depth} = \text{Rainfall Rate} \times t_{rain} = 0.05 \text{ m h}^{-1} \times 2 \text{ h} = 0.1 \text{ m} $

Calculate the total volume of rainfall:

$ V_{\text{rainfall}} = A \times \text{Rainfall Depth} = (1 \times 10^6 \text{ m}^2) \times (0.1 \text{ m}) = 1 \times 10^5 \text{ m}^3 $

Step 2: Calculate Runoff Volume

Convert the runoff duration to seconds for consistency with the runoff rate.

  • Runoff Rate ($Q_{out}$): $1 \text{ m}^3 \text{ s}^{-1}$.
  • Runoff Duration ($t_{run}$): $10 \text{ hours} = 10 \times 3600 \text{ s} = 36000 \text{ s}$.

Calculate the total volume of runoff:

$ V_{\text{runoff}} = Q_{out} \times t_{run} = (1 \text{ m}^3 \text{ s}^{-1}) \times (36000 \text{ s}) = 36000 \text{ m}^3 $

Step 3: Calculate Runoff Coefficient

Now, use the formula for the runoff coefficient:

$ C = \frac{V_{\text{runoff}}}{V_{\text{rainfall}}} = \frac{36000 \text{ m}^3}{1 \times 10^5 \text{ m}^3} $

$ C = \frac{36000}{100000} = 0.36 $

The runoff coefficient for the watershed is $0.36$. This corresponds to Option 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. 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 _____________

  5. 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)
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