Question

A rectangular weir is installed in a gully to measure runoff. The length of weir crest is \(3 \, \text{m}\) and the head of water over the weir is \(40 \, \text{cm}\). Find the discharge through weir. Take the value of coefficient of discharge (\(C_d\)) as \(0.60\).

• A. \(1.07 \, \text{m}^3/\text{sec}\)
• B. \(1.34 \, \text{m}^3/\text{sec}\)
• C. \(0.45 \, \text{m}^3/\text{sec}\)
• D. \(4.22 \, \text{m}^3/\text{sec}\)

Hint:

For a rectangular sharp crested weir: \[ \boxed{ Q = \frac{2}{3} C_d L \sqrt{2g} H^{3/2} } \] Discharge varies directly with:
• Crest length \(L\)
• Head raised to power \(3/2\)

Answer 1

The Correct Option is B

Concept: A rectangular weir is used to measure flow discharge in open channels. The discharge over a sharp crested rectangular weir is given by: \[ Q = \frac{2}{3} C_d L \sqrt{2g} H^{3/2} \] where:
• \(Q\) = discharge through weir
• \(C_d\) = coefficient of discharge
• \(L\) = length of crest
• \(H\) = head of water over crest
• \(g\) = acceleration due to gravity This formula is derived from integrating elementary discharge strips over the depth of flow.

Step 1: Writing the given data carefully. Length of crest: \[ L = 3 \, \text{m} \] Head over weir: \[ H = 40 \, \text{cm} = 0.40 \, \text{m} \] Coefficient of discharge: \[ C_d = 0.60 \] Acceleration due to gravity: \[ g = 9.81 \, \text{m/sec}^2 \]

Step 2: Writing the discharge formula. For rectangular weir: \[ Q = \frac{2}{3} C_d L \sqrt{2g} H^{3/2} \] Substituting values: \[ Q = \frac{2}{3} \times 0.60 \times 3 \times \sqrt{2 \times 9.81} \times (0.40)^{3/2} \]

Step 3: Calculating \(\sqrt{2g}\). \[ \sqrt{2g} = \sqrt{19.62} \] \[ \sqrt{19.62} \approx 4.43 \]

Step 4: Calculating \(H^{3/2}\). \[ H^{3/2} = (0.40)^{3/2} \] \[ = \sqrt{(0.40)^3} \] \[ = \sqrt{0.064} \] \[ = 0.253 \]

Step 5: Substituting numerical values. \[ Q = \frac{2}{3} \times 0.60 \times 3 \times 4.43 \times 0.253 \] First: \[ \frac{2}{3} \times 0.60 \times 3 = 1.2 \] Now: \[ Q = 1.2 \times 4.43 \times 0.253 \] \[ Q = 1.2 \times 1.12079 \] \[ Q \approx 1.34 \, \text{m}^3/\text{sec} \]

Step 6: Checking all options carefully. Option (A): Incorrect discharge. \[ \boxed{ \text{Option (A) is incorrect} } \] Option (B): Matches calculated value. \[ \boxed{ \text{Option (B) is correct} } \] Option (C): Too small. \[ \boxed{ \text{Option (C) is incorrect} } \] Option (D): Too large. \[ \boxed{ \text{Option (D) is incorrect} } \] Final Conclusion: The discharge through the rectangular weir is: \[ \boxed{ 1.34 \, \text{m}^3/\text{sec} } \] Hence the correct answer is: \[ \boxed{ (B) } \]