A 1-m wide rectangular channel is carrying a flow of 5 m³ /s at a flow depth of 2 m. Determine the height of a surge wave and its velocity if the discharge is suddenly increased to 10 m³ /s at the upstream end.

Given:

$Q_{1}$ = 5 m³ /s;
$Q_{2}$ = 10 m³ /s;
$y_{1}$ = 2 m;
B = 1 m.

Determine:

$y_{2}$ = ?
$V_{w}$ = ?

Question

A 1-m wide rectangular channel is carrying a flow of 5 m³ /s at a flow depth of 2 m. Determine the height of a surge wave and its velocity if the discharge is suddenly increased to 10 m³ /s at the upstream end.

Given:

$Q_{1}$ = 5 m³ /s;
$Q_{2}$ = 10 m³ /s;
$y_{1}$ = 2 m;
B = 1 m.

Determine:

$y_{2}$ = ?
$V_{w}$ = ?

Given:

[latex]Q_{1}[/latex] = 5 m³ /s;
[latex]Q_{2}[/latex] = 10 m³ /s;
[latex]y_{1}[/latex] = 2 m;
B = 1 m.

Determine:

[latex]y_{2}[/latex] = ?
[latex]V_{w}[/latex] = ?

Accepted Answer

The flow velocity at section 1,
[latex]\quad\quad\quad\quad V_{1}=Q_{1}/(By_{1})=5/(1.x2.)=2.5 m/s[/latex]
Now, [latex]Q_{2} = By_{2}V_{2}[/latex] or [latex]V_{2}y_{2}[/latex] = 10/1 = 10. By substituting the values of [latex]y_{1}[/latex] and [latex]V_{1}[/latex] and [latex]V_{2}y_{2}[/latex] = 10 into Eq. 11-22, and simplifying, we obtain
[latex]\quad\quad\quad\quad V_{w}=\frac{V_{2}y_{2}-V_{1}y_{1}}{y_{2}-y_{1}}[/latex] (11-22)
[latex]\quad\quad\quad\quad (1-\frac{4}{y_{2}})^{2}=\frac{2.452(y_{2}-2)}{y_{2}}(y^{2}_{2}-4)[/latex]
Solution of this equation by trial and error gives [latex]y_{2}[/latex] = 2.334 m. Thus, the height of the surge = 2.334-2. = 0.334 m.
[latex]\quad\quad[/latex]Substituting values of [latex]y_{1}[/latex], [latex]y_{2}[/latex], [latex]V_{1}[/latex] and [latex]V_{2}y_{2}[/latex] = 4 into Eq. 11-21, we obtain
[latex]\quad\quad\quad\quad c=\sqrt{gy}[/latex] (11-21)
[latex]\quad\quad\quad\quad V_{w}=\frac{2-4}{2.0-2.334}\\quad\quad\quad\quad\quad=\frac{2}{0.334}\\quad\quad\quad\quad\quad=5.99 m/s[/latex]

Question 11.1: A 1-m wide rectangular channel is carrying a flow of 5 m³ /s......