Question

A 12-h unit hydrograph of a catchment is triangular in shape with a base width of 144h and a peak discharge of 23 cumec. This unit hydrograph refers to a catchment area of

• A. 786 km\(^2\)
• B. 596 km\(^2\)
• C. 900 km\(^2\)
• D. 1200 km\(^2\)

Hint:

Remember that the area under a unit hydrograph must equal 1 cm of runoff depth spread over the whole catchment.

Answer 1

The Correct Option is B

Step 1: For any unit hydrograph, the total volume of direct runoff under the hydrograph equals the volume produced by 1 cm of rainfall excess spread uniformly over the whole catchment area. This is the defining property used to size a catchment from its UH shape.
Step 2: The hydrograph is triangular, so its area (the runoff volume) is simply \(\tfrac{1}{2} \times \text{base} \times \text{peak discharge}\). Convert the base width to seconds first: \(144\ h = 144 \times 3600\ s = 518400\ s\).
Step 3: Compute the volume: \[V = \tfrac{1}{2} \times 518400\ s \times 23\ m^3/s = \tfrac{1}{2} \times 11923200 = 5961600\ m^3\]
Step 4: This volume corresponds to 1 cm (0.01 m) of rainfall excess depth over the catchment area \(A\), so \(V = A \times 0.01\). Solve for \(A\): \[A = \frac{5961600}{0.01} = 596160000\ m^2\]
Step 5: Convert to square kilometres by dividing by \(10^6\): \[A = \frac{596160000}{1000000} = 596.16\ km^2 \approx 596\ km^2\]
Step 6: This matches option 2 exactly, and the other listed values (786, 900, 1200 km\(^2\)) do not satisfy the volume balance, so they are ruled out.