Question

A peak flow of a flood hydrograph due to a six-hour storm is \(470\) m\(^3\)/s. The corresponding average depth of rainfall is \(8\) cm. Assume infiltration of \(0.25\) cm/hour and a constant base flow of \(15\) m\(^3\)/s. What is the peak discharge of 6 hour unit hydrograph for this catchment?

• A. \(60\) m\(^3\)/s
• B. \(70\) m\(^3\)/s
• C. \(80\) m\(^3\)/s
• D. \(90\) m\(^3\)/s

Hint:

For unit hydrograph problems, \[ \boxed{ P_e=P-\text{Losses} } \] and \[ \boxed{ Q_{UH}=\frac{Q_{\text{Direct Runoff}}}{P_e} } \] where \[ Q_{\text{Direct Runoff}}=Q_{\text{Observed}}-Q_{\text{Base Flow}}. \]

Answer 1

The Correct Option is B

Concept: A unit hydrograph represents the direct runoff hydrograph produced by one unit depth of effective rainfall. The effective rainfall is \[ \boxed{ P_e=P-I } \] where \[ I=\text{Infiltration Loss}. \] The peak ordinate of the unit hydrograph is \[ \boxed{ Q_{UH}=\frac{Q_p-Q_b}{P_e} } \] where \[ Q_p=\text{Observed peak flow}, \] \[ Q_b=\text{Base flow}. \]

Step 1: Calculate effective rainfall. Rainfall \[ P=8\text{ cm} \] Infiltration loss \[ 0.25\times6=1.5\text{ cm} \] Therefore, \[ P_e=8-1.5=6.5\text{ cm} \]

Step 2: Determine the direct runoff peak. \[ Q_d = 470-15 = 455\text{ m}^3/\text{s} \]

Step 3: Calculate the unit hydrograph peak. \[ Q_{UH} = \frac{455}{6.5} = 70\text{ m}^3/\text{s} \] Hence, \[ \boxed{Q_{UH}=70\text{ m}^3/\text{s}} \] Therefore, the correct option is \[ \boxed{(B)\;70\text{ m}^3/\text{s}.} \]