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

Correct Answer: 1

Given: Angle of the V-notch, \(\theta = 90^\circ\) Discharge coefficient, \(C_d = 0.6\) Head over the notch, \(H = 0.49 \, {m}\) Acceleration due to gravity, \(g = 9.81 \, {m/s}^2\) \subsection{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 the formula simplifies to: \[ Q = \frac{8}{15} C_d \sqrt{2g} H^{5/2} \] Calculations: 1. Compute \(\sqrt{2g}\): \[ \sqrt{2 \times 9.81} = \sqrt{19.62} = 4.429 \, {m}^{0.5}/{s} \] 2. Compute \(H^{5/2}\): \[ (0.49)^{5/2} = (0.49)^{2.5} = 0.49^2 \times \sqrt{0.49} \] \[ 0.49^2 = 0.2401 \] \[ \sqrt{0.49} = 0.7 \] \[ 0.2401 \times 0.7 = 0.16807 \] 3. Multiply all terms: \[ Q = \frac{8}{15} \times 0.6 \times 4.429 \times 0.16807 \] \[ Q = 0.5333 \times 0.6 \times 4.429 \times 0.16807 \] \[ Q = 0.32 \times 4.429 \times 0.16807 \] \[ Q = 1.41728 \times 0.16807 \] \[ Q = 0.238 \, {m}^3/{s} \] Final Answer: The discharge over the notch (rounded to 2 decimal places) is: \[ \boxed{0.24 \, {m}^3/{s}} \]