A sluice gate discharges water into a horizontal rectangular channel with a velocity of 8 m/s and depth of flow is 0.5 m. The width of the channel is 6 m. Determine whether a hydraulic jump will occur, and if so, find its height and loss of energy per kg of water. Also determine the horse power lost in the hydraulic jump.

Hydraulic Jump Analysis in a Rectangular Channel

Problem Statement

Given Data & Constants

Initial velocity, $V_1 = 8 \, \text{m/s}$
Initial depth, $d_1 = 0.5 \, \text{m}$
Channel width, $B = 6 \, \text{m}$
Density of water, $\rho = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Determine if a Hydraulic Jump Will Occur

A hydraulic jump can only occur if the initial flow is supercritical, which is determined by the Froude number ($Fr_1$).

$$ Fr_1 = \frac{V_1}{\sqrt{g d_1}} = \frac{8}{\sqrt{9.81 \times 0.5}} = \frac{8}{2.2147} \approx 3.612 $$

Since $Fr_1 > 1$, the flow is supercritical and a hydraulic jump will occur.

2. Find the Height of the Jump

First, we calculate the depth of flow after the jump ($d_2$).

$$ d_2 = \frac{d_1}{2} \left[ \sqrt{1 + 8 Fr_1^2} - 1 \right] = \frac{0.5}{2} \left[ \sqrt{1 + 8 \times (3.612)^2} - 1 \right] $$ $$ d_2 = 0.25 \left[ \sqrt{1 + 104.34} - 1 \right] = 0.25 [\sqrt{105.34} - 1] $$ $$ d_2 = 0.25 [10.263 - 1] \approx 2.316 \, \text{m} $$

The height of the jump is the difference in depths.

$$ \text{Jump Height} = d_2 - d_1 = 2.316 - 0.5 = 1.816 \, \text{m} $$

3. Find the Loss of Energy Head

We calculate the specific energy before ($E_1$) and after ($E_2$) the jump.

$$ V_2 = \frac{Q}{A_2} = \frac{V_1 A_1}{A_2} = \frac{8 \times (6 \times 0.5)}{6 \times 2.316} \approx 1.727 \, \text{m/s} $$ $$ E_1 = d_1 + \frac{V_1^2}{2g} = 0.5 + \frac{8^2}{2 \times 9.81} \approx 3.762 \, \text{m} $$ $$ E_2 = d_2 + \frac{V_2^2}{2g} = 2.316 + \frac{(1.727)^2}{2 \times 9.81} \approx 2.468 \, \text{m} $$

The head loss ($h_L$) is the difference in specific energy.

$$ h_L = E_1 - E_2 = 3.762 - 2.468 = 1.294 \, \text{m} $$

4. Determine the Horsepower Lost

First, calculate the total power loss in Watts, then convert to metric horsepower (1 HP = 735.5 W).

$$ \text{Discharge, } Q = V_1 A_1 = 8 \times (6 \times 0.5) = 24 \, \text{m}^3/\text{s} $$ $$ \text{Power Loss (W)} = \rho g Q h_L = 1000 \times 9.81 \times 24 \times 1.294 \approx 304665 \, \text{W} $$ $$ \text{Horsepower Lost} = \frac{304665}{735.5} \approx 414.2 \, \text{HP} $$

Final Results:

A hydraulic jump will occur.

Height of the jump: $ \approx 1.816 \, \text{m} $

Loss of energy head: $ \approx 1.294 \, \text{m} $

Horsepower lost: $ \approx 414.2 \, \text{HP} $