Respuesta :
Answer:
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, [tex]M_{unit}[/tex] is 6.01 m²
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
[tex]y_1+\frac{V_1^2}{2\cdot g} = y_2+\frac{V_2^2}{2\cdot g} +h_L[/tex]
[tex]-h_L = y_2+\frac{V_2^2}{2\cdot g} -(y_1+\frac{V_1^2}{2\cdot g})[/tex]
[tex]-h_L = 3+\frac{\frac{20}{9} ^2}{2\cdot 9.81} -(1+\frac{\frac{20}{3} ^2}{2\cdot 9.81 })[/tex] = -0.1357
y₂ - y₁ = 2.00 m
We have energy loss per unit width of the channel give by
[tex]E = \frac{q^{2} }{2gy^2} +y[/tex]
Where:
q = Unit discharge = Q/b = 20/3
b = Widtjh of the chammel
Therefore
[tex]E = \frac{\frac{20}{3} ^{2} }{2\times 9.81 \times 3^2} +3[/tex] = 3.252 m
The specific force per unit width of the channel is given by
[tex]M_{unit} = \frac{y^2}{2} +\frac{q^2}{g\cdot y}[/tex]
Where:
[tex]M_{unit}[/tex] = Specific force per unit width of the channel
[tex]M_{unit} = \frac{3^2}{2} +\frac{(20/3)^2}{9.81\cdot 3}[/tex] = 6.01 m²
