Question

A spillway has unit discharge of \(7.5 \text{ m}^3/\text{s/m}\). The flow depth at the downstream horizontal apron is \(0.5 \text{ m}\). The tail water depth (in meters) required to form a hydraulic jump is ______ (rounded {off to 2 decimal places).}

Hint:

For hydraulic jump calculations, ensure accurate determination of the Froude number and proper substitution into the depth relationship formula. Always verify units for consistency.

Answer 1

Step 1: Define the terms and equations. For a hydraulic jump, the relationship between the upstream depth (\( y_1 \)) and the downstream depth (\( y_2 \)) is given by the equation: \[ y_2 = \frac{y_1}{2} \left( \sqrt{1 + 8 \text{Fr}_1^2} - 1 \right), \] where: - \( y_1 = 0.5 \, \text{m} \) is the initial flow depth, - \( \text{Fr}_1 \) is the Froude number, calculated as: \[ \text{Fr}_1 = \frac{q}{y_1 \sqrt{g y_1}}, \] where: - \( q = 7.5 \, \text{m}^3/\text{s}/\text{m} \) (unit discharge), - \( g = 9.81 \, \text{m}/\text{s}^2 \) (acceleration due to gravity). Step 2: Calculate the Froude number (\( \text{Fr}_1 \)). Substitute the known values into the equation for \( \text{Fr}_1 \): \[ \text{Fr}_1 = \frac{7.5}{0.5 \cdot \sqrt{9.81 \cdot 0.5}}. \] Simplify: \[ \text{Fr}_1 = \frac{7.5}{0.5 \cdot \sqrt{4.905}} = \frac{7.5}{0.5 \cdot 2.214} = \frac{7.5}{1.107} \approx 6.78. \] Step 3: Calculate the downstream depth (\( y_2 \)). Using the equation for \( y_2 \): \[ y_2 = \frac{0.5}{2} \left( \sqrt{1 + 8 \cdot (6.78)^2} - 1 \right). \] Simplify: \[ y_2 = 0.25 \left( \sqrt{1 + 8 \cdot 45.97} - 1 \right), \] \[ y_2 = 0.25 \left( \sqrt{1 + 367.76} - 1 \right), \] \[ y_2 = 0.25 \left( \sqrt{368.76} - 1 \right), \] \[ y_2 = 0.25 \left( 19.21 - 1 \right), \] \[ y_2 = 0.25 \cdot 18.21 \approx 4.55 \, \text{m}. \] Conclusion: The tailwater depth required to form a hydraulic jump is approximately \( y_2 = 4.55 \, \text{m} \).