Question

Water flows through a 90° V-notch weir having a discharge coefficient of 0.6. If the depth of water above the notch is 49 cm and the acceleration due to gravity is 9.81 m.s\(^{-2}\), the discharge over the notch is _________ m\(^3\).s\(^{-1}\) (Rounded off to 2 decimal places).

Hint:

When calculating discharge over a V-notch weir, ensure that the depth is in meters and use the correct discharge coefficient for accurate results.

Answer 1

Given:

• Angle of the V-notch, \( \theta = 90^\circ \)
• Discharge coefficient, \( C_d = 0.6 \)
• Head over the notch, \( H = 0.49 \, \text{m} \)
• Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)

Formula:

The discharge \( Q \) for a V-notch weir is given by:

\( Q = \frac{8}{15} C_d \sqrt{2g} \tan\left(\frac{\theta}{2}\right) H^{5/2} \)

For a 90° V-notch: \( \tan\left( \frac{90^\circ}{2} \right) = \tan(45^\circ) = 1 \), so:

\( Q = \frac{8}{15} C_d \sqrt{2g} H^{5/2} \)

Calculations:

1. Compute \( \sqrt{2g} \):

   \( \sqrt{2 \times 9.81} = \sqrt{19.62} = 4.429 \, \text{m}^{0.5}/\text{s} \)

2. Compute \( H^{5/2} \):

   \( (0.49)^{5/2} = 0.49^2 \times \sqrt{0.49} = 0.2401 \times 0.7 = 0.16807 \)

3. Multiply all terms:

   \( Q = \frac{8}{15} \times 0.6 \times 4.429 \times 0.16807 \)

   \( Q = 0.5333 \times 0.6 \times 4.429 \times 0.16807 \)

   \( Q = 0.32 \times 4.429 \times 0.16807 = 1.41728 \times 0.16807 = 0.238 \, \text{m}^3/\text{s} \)

Final Answer:

\( \boxed{0.24 \, \text{m}^3/\text{s}} \)