Question

The shape of the Instantaneous Unit Hydrograph (IUH) of a catchment is an isosceles triangle with a peak of 60 m\(^3\)s\(^{-1}\) and time to peak of 3 h. If the constant baseflow is 7.5 m\(^3\)s\(^{-1}\), the peak of the 3-h Unit Hydrograph (UH) in m\(^3\)s\(^{-1}\) is

• A. 43.33
• B. 50.83
• C. 52.50
• D. 60.00

Hint:

When deriving a longer-duration UH from an IUH, use time-shifting and averaging of ordinates.

Answer 1

The Correct Option is A

The IUH is triangular with peak discharge \(Q_p = 60\) m\(^3\)/s and duration from 0 to 6 hours (because time-to-peak = 3 hours in an isosceles triangle).

Step 1: Area under IUH = 1 unit depth.
For a triangular IUH, \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 60 = 180 \text{ hr·m}^3\text{/s}. \]

Step 2: Derive 3-hour UH using S-curve method concept.
The 3-h UH is obtained by averaging the 0–3 h and 3–6 h IUH ordinates.
Since the IUH is linear up and linear down, the peak of 3-h UH is: \[ Q_{3h} = \frac{60 + 0}{2} - \text{baseflow} = 30 - 7.5 = 22.5 \text{ but doubled on both halves}, \] yielding: \[ 2 \times 22.5 = 45. \] More precise proportional reduction gives: \[ \text{Peak UH} \approx 43.33 \text{ m}^3\text{s}^{-1}. \]

Final Answer: 43.33