GATE AG 2012 • Q51
MCQ • 2 Marks
Section 4: Soil and Water Conservation Engineering
Hydrology
Subtopic: Hydrograph analysis, unit hydrograph theory and application
Question Statement:
The peak of the 1-h unit hydrograph, in m³/s, for the watershed of Q.50, and its time of occurrence in h, respectively are:
Options:
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(A) 6, 1
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(B) 7, 2
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(C) 8, 2
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(D) 9, 1
Step-by-Step Solution & Explanation
• **Concept (Unit Hydrograph Deconvolution):**
- The storm has excess rainfall depths $R_1 = 2\text{ cm}$ in hour 1 ($0-1\text{ h}$) and $R_2 = 3\text{ cm}$ in hour 2 ($1-2\text{ h}$).
- Base flow is constant at $1\text{ m}^3/\text{s}$. Direct Runoff Hydrograph (DRH) ordinates $Q_t = \text{Total} - 1\text{ m}^3/\text{s}$ are: $Q_0 = 0, Q_1 = 6, Q_2 = 25, Q_3 = 36, Q_4 = 26, Q_5 = 12, Q_6 = 0$.
• **System of Equations for 1-h Unit Hydrograph Ordinates $U_t$:**
- $t = 1\text{ h}: Q_1 = R_1 U_1 \implies 6 = 2 U_1 \implies U_1 = 3\text{ m}^3/\text{s}$.
- $t = 2\text{ h}: Q_2 = R_1 U_2 + R_2 U_1 \implies 25 = 2 U_2 + 3(3) \implies 2 U_2 = 16 \implies U_2 = 8\text{ m}^3/\text{s}$.
- $t = 3\text{ h}: Q_3 = R_1 U_3 + R_2 U_2 \implies 36 = 2 U_3 + 3(8) \implies 2 U_3 = 12 \implies U_3 = 6\text{ m}^3/\text{s}$.
- $t = 4\text{ h}: Q_4 = R_1 U_4 + R_2 U_3 \implies 26 = 2 U_4 + 3(6) \implies 2 U_4 = 8 \implies U_4 = 4\text{ m}^3/\text{s}$.
- $t = 5\text{ h}: Q_5 = R_1 U_5 + R_2 U_4 \implies 12 = 2 U_5 + 3(4) \implies 2 U_5 = 0 \implies U_5 = 0$.
• **Unit Hydrograph Peak:** The maximum ordinate is $U_2 = 8\text{ m}^3/\text{s}$ occurring at $t = 2\text{ h}$. **Correct Answer: C**.
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