Question

A \(0.80\) km\(^2\) agricultural watershed has a slope of \(0.30\%\) and the maximum length of travel of water is \(1\) 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 m\(^3\)/s) for the \(10\)-year return period is ________. (Rounded off to two decimal places)

Hint:

Find the time of concentration first, then read intensity from the depth-duration table before applying the Rational method.

Answer 1

Correct Answer: 7.69

Step 1: Find the time of concentration.
Use Kirpich's formula, \(T_c(\text{min}) = 0.01947\, L^{0.77}\, S^{-0.385}\), with \(L = 1000\) m (max travel length) and \(S = 0.30\% = 0.003\).
\(L^{0.77} = 1000^{0.77} \approx 204.2\), and \(S^{-0.385} = (0.003)^{-0.385} \approx 9.36\).
\[ T_c = 0.01947 \times 204.2 \times 9.36 \approx 37.2\ \text{min} \]
Step 2: Read the design rainfall depth for this duration from the table.
\(T_c = 37.2\) min lies between the \(30\) min (\(50\) mm) and \(40\) min (\(60\) mm) rows, so interpolate:
\[ \text{depth} = 50 + \frac{37.2-30}{40-30}\times(60-50) = 50+7.2=57.2\ \text{mm} \]
Step 3: Convert this depth to an intensity.
\[ i = \frac{57.2 \times 60}{37.2} \approx 92.3\ \text{mm/hr} \]
Step 4: Weight the runoff coefficient over the split land use.
Half the watershed is row crop (\(C=0.40\)) and half pasture (\(C=0.35\)), so the area-weighted value is \[ C_{avg} = \frac{0.40+0.35}{2} = 0.375 \]
Step 5: Apply the Rational method.
With area \(A = 0.80\) km\(^2\) = \(80\) ha, \[ Q_p = \frac{C\, i\, A}{360} = \frac{0.375 \times 92.3 \times 80}{360} \approx 7.69\ \text{m}^3/\text{s} \]
Final Answer:
The watershed's peak flow for the \(10\)-year storm comes out near \(7.69\) m\(^3\)/s. \[ \boxed{Q_p \approx 7.69\ \text{m}^3/\text{s}} \]