Question

The Peak rate of runoff expected from the catchment area of a farm pond is 4 m\(^3\)s\(^{-1}\). Assuming no temporary storage, find the length of the surplus weir, if the depth of flow over the weir is not to exceed 0.75 m.

• A. 2.5 m
• B. 3.0 m
• C. 3.5 m
• D. 3.7 m

Hint:

For broad-crested conservation weir designs, always use the standard discharge coefficient \(C = 1.71\).
Converting the exponent \(H^{1.5}\) to \(H \cdot \sqrt{H}\) simplifies manual calculations during exams.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A surplus weir acts as an emergency spillway for a farm pond to safely discharge excess runoff.
To find the required length of the weir, we use the principles of open-channel hydraulics for flow over a broad-crested or rectangular weir.
Key Formula or Approach:
The discharge capacity (\(Q\)) of a standard rectangular surplus weir is given by: \[ Q = C \cdot L \cdot H^{3/2} \] where:
- \(Q\) is the peak runoff rate (\(\text{m}^3/\text{s}\)).
- \(C\) is the discharge coefficient (typically \(1.71\) for broad-crested weirs used in soil conservation).
- \(L\) is the length of the weir (\(\text{m}\)).
- \(H\) is the depth of flow over the crest (\(\text{m}\)).

Step 2: Detailed Explanation:
Given values:
- Peak runoff rate (\(Q\)) = \(4 \text{ m}^3/\text{s}\)
- Maximum depth of flow (\(H\)) = \(0.75 \text{ m}\)
- Discharge coefficient (\(C\)) = \(1.71\)
Let's substitute these values into our discharge equation: \[ 4 = 1.71 \cdot L \cdot (0.75)^{1.5} \] First, calculate the term \((0.75)^{1.5}\): \[ (0.75)^{1.5} = 0.75 \times \sqrt{0.75} \approx 0.75 \times 0.866 = 0.6495 \] Substitute this back into the equation: \[ 4 = 1.71 \cdot L \cdot 0.6495 \] \[ 4 = 1.11 \cdot L \] Solve for the length \(L\): \[ L = \frac{4}{1.11} \approx 3.604 \text{ m} \] Rounding to the nearest standard design option gives \(3.7 \text{ m}\).
(Note: Using the Francis formula with \(C = 1.84\) and accounting for end contractions would also yield a design length in the range of \(3.5\text{ m}\) to \(3.7\text{ m}\)).

Step 3: Final Answer:
The required length of the surplus weir is 3.7 m.