Chapter 13, Problem 119P

Consider uniform water flow in a wide channel made of unfinished concrete laid on a slope of 0.0022. Now water flows over a 15-cm-high bump. If the flow over the bump is exactly critical (Fr = 1), determine the flow rate and the flow depth over the bump per m width.

To determine

The flow rate.

The flow depth over the bump per $\text{m}$ width.

Answer to Problem 119P

The flow rate is $20.3{\text{m}}^{\text{3}}/\text{s}$.

The flow depth is $3.48\text{m}$.

Explanation of Solution

Given information:

The slope of the wide channel made of unfinished concrete is $0.0022$, the height of the bump is $15\text{cm}$ and the Froude number is $1$.

The flow is steady and uniform, the channel is sufficiently wide so that the end effects are negligible, the frictional effects during the flow over the bump are negligible bottom slope is constant, roughness coefficient is constant along the channel, the width of the water channel is assumed to be $1\text{m}$, the manning coefficient for an open channel of unfinished concrete is $0.014$, the conversion factor is $1{\text{m}}^{1/3}/\text{s},$ and the acceleration due to gravity is $9.81{\text{m}/\text{s}}^2$.

Write the expression for the flow rate for which the hydraulic radius is equal to the flow depth from the manning equation.

   $\begin{array}{c}\dot{\vee }=\frac{a}{n}{A}_C{R}_h^{2/3}{S}_0^{1/2}\\ =\frac{a}{n}{A}_C\left(y_1^{2/3}\right)S_0^{1/2}\end{array}$    ...... (I)

Here, the manning coefficient for an open channel of unfinished concrete is $n$, the hydraulic radius of a wide channel is $R_h$, the bottom slope is $S_0$, the area of the wide channel is $A_C$ the flow depth is $y_1,$ and the conversion factor is $a$.

Write the expression for the critical depth corresponding to the flow rate.

   $\begin{array}{c}{y}_2=y_c\\ y_c={\left(\frac{{\dot{\vee }}^2}{g{b}^2}\right)}^{1/3}\end{array}$    ...... (II)

Here, the water flow rate through the channel per meter width is ${\dot{\vee }}_{}$, the width of the water channel is $b,$ and the acceleration due to gravity is $g$.

Write the expression for the average flow velocity.

   $V_1=\frac{\dot{\vee }}{A_c}$    ...... (III)

Write the expression for the specific energy before the bump.

   $E_{s1}=y_1+\frac{V_1^2}{2g}$    ...... (IV)

Here, the average flow velocity is $V_1$.

Write the expression for the critical specific energy.

   $E_{s2}=E_c$

As, the critical specific energy is equal to the specific energy after the bump because the critical depth corresponding to the flow rate is equal to the depth of the flow after the bump.

   $E_c=\frac{3}{2}{y}_c$    ...... (V)

Here, the critical depth of the flow is $y_c$ and the specific energy on the bump is $E_{s2}$.

Write the expression for the specific energy on the bump.

   $E_{s2}=E_{s1}-\Delta z_b$    ...... (VI)

Here, the height of the bump is $\Delta z_b$ and the specific energy before the bump is $E_{s1}$.

Calculation:

Substitute $0.014$ for $n$, $1{\text{m}}^{1/3}/\text{s}$ for $a$, $y_1$ for $A_c$ and $0.0022$ for $S_0$ in the Equation (I).

   $\begin{array}{c}\dot{\vee }=\frac{1{\text{m}}^{1/3}/\text{s}}{0.014}{y}_1\text{m}{\left(y_1\text{m}\right)}^{2/3}{\left(0.0022\right)}^{1/2}\\ =71.4285{\text{m}}^{1/3}/\text{s}{\left(y_1\text{m}\right)}^{5/3}{\left(0.0022\right)}^{1/2}\\ =3.350y_1^{5/3}{\text{m}}^3/\text{s}\end{array}$    ...... (VII)

Substitute $3.350y_1^{5/3}{\text{m}}^3/\text{s}$ for $\dot{\vee }$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $1\text{m}$ for $b$ in the Equation (II).

   $\begin{array}{c}{y}_c={\left(\frac{{\left(3.350y_1^{5/3}{\text{m}}^3/\text{s}\right)}^2}{9.81{\text{m}/\text{s}}^2\times {\left(1\text{m}\right)}^2}\right)}^{1/3}\\ ={\left(\frac{11.224y_1^{10/3}{\left({\text{m}}^{\text{3}}/\text{s}\right)}^2}{9.81{{\text{m}}^3/\text{s}}^2}\right)}^{1/3}\\ =1.046y_1^{10/9}\text{m}\end{array}$    ...... (VIII)

Substitute $3.350y_1^{5/3}{\text{m}}^3/\text{s}$ for $\dot{\vee }$ and $y_1{\text{m}}^{\text{2}}$ for $A_C$ in the Equation (III).

   $\begin{array}{c}{V}_1=\frac{3.350y_1^{5/3}{\text{m}}^3/\text{s}}{y_1{\text{m}}^{\text{2}}}\\ =3.350y_1^{2/3}\text{m}/\text{s}\end{array}$

Substitute $3.350y_1^{2/3}\text{m}/\text{s}$ for $V_1$ and $9.81{\text{m}/\text{s}}^2$ for $g$ in the Equation (IV).

   $\begin{array}{c}{E}_{s1}=y_1+\frac{{\left(3.350y_1^{2/3}\text{m}/\text{s}\right)}^2}{2\times 9.81{\text{m}/\text{s}}^2}\\ =y_1+\frac{{\left(3.350y_1^{2/3}\text{m}/\text{s}\right)}^2}{19.62{\text{m}/\text{s}}^2}\\ =y_1+0.5720y_1^{4/3}\text{m}\end{array}$

Substitute $1.046y_1^{10/9}\text{m}$ for $y_c$ in the Equation (V).

   $\begin{array}{c}{E}_c=\frac{3}{2}\left(1.046y_1^{10/9}\text{m}\right)\\ =\left(1.5\right)\left(1.046y_1^{10/9}\text{m}\right)\\ =1.569y_1^{10/9}\text{m}\end{array}$

Substitute $1.569y_1^{10/9}\text{m}$ for $E_{s2}$, $y_1+0.5720y_1^{4/3}\text{m}$ for $E_{s1}$ and $15\text{cm}$ for $\Delta z_b$ in the Equation (VI).

   $\begin{array}{l}1.569y_1^{10/9}\text{m}=\left(y_1+0.5720y_1^{4/3}\text{m}\right)-15\text{cm}\\ 1.569y_1^{10/9}\text{m}=\left(y_1+0.5720y_1^{4/3}\text{m}\right)-15\text{cm}\left[\frac{1\text{m}}{100\text{cm}}\right]\\ 1.569y_1^{10/9}\text{m}=\left(y_1+0.5720y_1^{4/3}\text{m}\right)-0.15\text{m}\end{array}$

By hit and trial method.

   $y_1=2.947\text{m}$

Substitute $2.947$ for $y_1$ in the Equation (VII).

   $\begin{array}{c}\dot{\vee }=3.350{\left(2.947\right)}^{5/3}{\text{m}}^3/\text{s}\\ =20.3{\text{m}}^3/\text{s}\end{array}$

Substitute $2.947\text{m}$ for $y_1$ in the Equation (VIII).

   $\begin{array}{c}{y}_c=1.046{\left(2.947\right)}^{10/9}\text{m}\\ \text{=3}\text{.48}\text{m}\end{array}$

Conclusion:

The flow rate is $20.3{\text{m}}^{\text{3}}/\text{s}$ and the flow depth is $3.48\text{m}$.