Question

A watershed has an area of 600 ha. Due to a 10 cm rainfall event over the watershed a streamflow is generated and at the outlet of the watershed it lasts for 10 hours. Assuming a runoff/rainfall ratio of 0.2 for this event, the average stream flow rate at the outlet in this period of 10 hours is

• A. 1,20,000 m$^3$/h
• B. 200 m$^3$/minute
• C. 33.33 m$^3$/s
• D. 2.66 m$^3$/s

Hint:

Remember standard conversions:
- \(1 \text{ hectare (ha)} = 10,000 \text{ m}^2 = 10^4 \text{ m}^2\)
- Always write down the final value in multiple units (\(\text{m}^3/\text{h}\), \(\text{m}^3/\text{min}\), \(\text{m}^3/\text{s}\)) to match the options provided in the question.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The runoff volume generated from a watershed during a rainfall event is calculated using the watershed area, rainfall depth, and the runoff coefficient (runoff/rainfall ratio).
The average streamflow rate is then found by dividing the total runoff volume by the duration of the flow.
Key Formula or Approach:
1. Total Runoff Volume (\(V\)):
\[ V = A \times P \times C \] Where \(A\) is the watershed area, \(P\) is the rainfall depth, and \(C\) is the runoff coefficient.
2. Average Streamflow Rate (\(Q\)):
\[ Q = \frac{V}{T} \] Where \(T\) is the duration of flow.

Step 2: Detailed Explanation:
Given data:
- Watershed Area, \(A = 600 \text{ ha} = 600 \times 10^4 \text{ m}^2 = 6 \times 10^6 \text{ m}^2\)
- Rainfall Depth, \(P = 10 \text{ cm} = 0.1 \text{ m}\)
- Runoff/rainfall ratio (coefficient), \(C = 0.2\)
- Duration of flow, \(T = 10 \text{ hours} = 600 \text{ minutes}\)
First, calculate the total volume of runoff (\(V\)):
\[ V = (6 \times 10^6 \text{ m}^2) \times (0.1 \text{ m}) \times 0.2 \] \[ V = 120,000 \text{ m}^3 \] Next, calculate the average streamflow rate (\(Q\)) in different units to match the options:
- In \(\text{m}^3/\text{hour}\):
\[ Q = \frac{120,000 \text{ m}^3}{10 \text{ hours}} = 12,000 \text{ m}^3/\text{h} \] (Option A is \(1,20,000 \text{ m}^3/\text{h}\), which is incorrect).
- In \(\text{m}^3/\text{minute}\):
\[ Q = \frac{120,000 \text{ m}^3}{600 \text{ minutes}} = 200 \text{ m}^3/\text{min} \] (This matches Option B exactly).
- In \(\text{m}^3/\text{second}\):
\[ Q = \frac{120,000 \text{ m}^3}{36,000 \text{ seconds}} \approx 3.33 \text{ m}^3/\text{s} \] (Options C and D are \(33.33\) and \(2.66 \text{ m}^3/\text{s}\), which are incorrect).

Step 3: Final Answer:
The average streamflow rate is \(200 \text{ m}^3/\text{minute}\).