Question

A hydraulic jump occurs in a 15-m wide rectangular, concrete channel. The initial (normal) depth comes from a spillway discharge of 100 m^3/s with a slope of 10%. The sequent depth is 3.00 m.

Determine the energy loss and the specific force per unit width of the channel.

Answer 1

The head loss = 0.1357 m

The energy loss per unit width of the channel is 3.252 m

The Specific force per unit width of the channel,
`M_(unit)` is 6.01 m²

Step-by-step explanation:

Here, the slope is 10 %, therefore for every one unit traveled vertically we have 10 units traveled horizontally

Therefore when the depth is 3.00 m, we can select the initial depth as the reference = 1.00 m

We have V₁ = 100 m³/s ÷ (15×1) = 20/3 m/s

V₂ = 100 m³/s ÷ (15×3) = 20/9 m/s

`y_1+(V_1^2)/(2\cdot g) = y_2+(V_2^2)/(2\cdot g) +h_L`

`-h_L = y_2+(V_2^2)/(2\cdot g) -(y_1+(V_1^2)/(2\cdot g))`

`-h_L = 3+((20)/(9) ^2)/(2\cdot 9.81) -(1+((20)/(3) ^2)/(2\cdot 9.81 ))` = -0.1357

y₂ - y₁ = 2.00 m

We have energy loss per unit width of the channel give by

`E = (q^(2) )/(2gy^2) +y`

Where:

q = Unit discharge = Q/b = 20/3

b = Widtjh of the chammel

Therefore

`E = ((20)/(3) ^(2) )/(2* 9.81 * 3^2) +3` = 3.252 m

The specific force per unit width of the channel is given by

`M_(unit) = (y^2)/(2) +(q^2)/(g\cdot y)`

Where:

`M_(unit)` = Specific force per unit width of the channel

`M_(unit) = (3^2)/(2) +((20/3)^2)/(9.81\cdot 3)` = 6.01 m²