Chapter 13, Problem 144P

Water is discharged from a 5-rn-deep lake into a finished concrete channel with a bottom slope of 0.004 through a sluice gate with a 0.7-m-high opening at the bottom. Shortly after supercritical uniform-flow conditions are established, the water undergoes a hydraulic jump. Determine the flow depth, velocity, and Froude number after the jump. Disregard the bottom slope when analyzing the hydraulic jump.

To determine

The flow depth after hydraulic jump.

The velocity after hydraulic jump.

The Froude number after hydraulic jump.

Answer to Problem 144P

The flow depth after hydraulic jump is $1.99\text{m}$.

The velocity after hydraulic jump is $2.02\text{m}/\text{s}$.

The Froude number after hydraulic jump is $0.457$.

Explanation of Solution

Given information:

The depth of the lack is $5\text{m}$, the bottom slop is $0.004$, high opening depth is $0.7\text{m}$.

The below figure represent the depths of flow.

Figure-(1)

Write the expression of depth ratio.

  $d_r=\frac{y_1}{a}$...... (I)

Here, the depth of lack is $y_1$ and the high opening depth is $a$.

Write the expression of discharge.

  $\dot{V}=C_{d}ab\sqrt{2g{y}_1}$...... (II)

Here, the coefficient of discharge is $C_d$, the width is $b$ and the acceleration due to gravity is $g$.

Write the expression of flow rate in the form of manning equation.

  $\dot{V}=\frac{a_1}{n}{A}_c{R}_h{}^{2/3}{S}_0{}^{1/2}$....... (III)

Here, the cross section are is $A_c$, the hydraulic radius is $R_h$, the slop is $S_0$ and the constant is $a$.

Write the expression of hydraulic radius.

  $R_h=y_2$

Here, the flow depth is $y_2$.

Substitute $y_2$ for $R_h$ in Equation (III).

  $\dot{V}=\frac{a}{n}{A}_c{\left(y_2\right)}^{2/3}{S}_0{}^{1/2}$...... (IV)

Write the expression of flow velocity when undergo a hydraulic jump.

  $V_2=\frac{\dot{V}}{b{y}_2}$...... (V)

Write the expression of Froude number when undergo a hydraulic jump.

  ${\text{Fr}}_2=\frac{V_2}{\sqrt{g{y}_2}}$...... (VI)

Write the expression of flow depth after hydraulic jump.

  $y_3=0.52y_2\left(-1+\sqrt{1+8{\text{Fr}}_2{}^2}\right)$...... (VII)

Write the expression of velocity after hydraulic jump.

  $V_3=\frac{y_2}{y_3}{V}_2$....... (VIII)

Write the expression of Froude number after hydraulic jump.

  ${\text{Fr}}_3=\frac{V_3}{\sqrt{g{y}_3}}$...... (IX)

Calculation:

Substitute $5\text{m}$ for $y_1$ and $0.7\text{m}$ for $a$ in Equation (I).

  $\begin{array}{c}{d}_r=\frac{\left(5\text{m}\right)}{\left(0.7\text{m}\right)}\\ =\frac{50}{7}\\ =7.143\end{array}$

Refer Figure 13.46 "Discharge coefficient for drowned and free discharge from underflow gates" to find the value of $C_d$ as $0.58$ corresponding to depth ration is $7.143$.

Consider width is $1\text{m}$ for per meter width.

Substitute $1\text{m}$ for $b$, $0.58$ for $C_d$, $0.7\text{m}$ for $a$, $5\text{m}$ for $y_1$ and $9.81{\text{m}/\text{s}}^2$ for $g$ in Equation (II).

  $\begin{array}{c}\dot{V}=\left(0.58\right)\times \left(1\text{m}\right)\times \left(0.7\text{m}\right)\times \sqrt{2\times \left(9.81{\text{m}/\text{s}}^2\right)\times \left(5\text{m}\right)}\\ =\left(0.406{\text{m}}^2\right)\times \left(9.905\text{m}/\text{s}\right)\\ =4.02{\text{m}}^{\text{3}}/\text{s}\end{array}$

Refer table 13.1 "Mean values of the manning coefficient $n$ for water flow in open channels" to find the value of $n$ as $0.012$ corresponding to steel smooth wall.

Consider the cross section area of the wall is $1{\text{m}}^{\text{2}}$ for unit area of the wall.

Substitute $0.012$ for $n$, $4.02{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $0.004$ for $S_0$, $1{\text{m}}^{\text{2}}$ for $A_c$ and $1{\text{m}}^{1/3}/s$ for $a$ in Equation (IV).

  $\begin{array}{l}\left(4.02{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\left(1{\text{m}}^{1/3}/s\right)}{\left(0.012\right)}\times \left(1{\text{m}}^{\text{2}}\right)\times {\left(y_2\right)}^{2/3}{\left(0.004\right)}^{1/2}\\ {\left(y_2\right)}^{2/3}=\frac{\left(4.02{\text{m}}^{\text{3}}/\text{s}\right)\times \left(0.012\right)}{\left(1{\text{m}}^{1/3}/s\right)\times \left(1{\text{m}}^{\text{2}}\right)\times {\left(0.004\right)}^{1/2}}\\ y_2={\left(\frac{\left(4.02{\text{m}}^{\text{3}}/\text{s}\right)\times \left(0.012\right)}{\left(1{\text{m}}^{1/3}/s\right)\times \left(1{\text{m}}^{\text{2}}\right)\times {\left(0.004\right)}^{1/2}}\right)}^{3/2}\\ y_2=0.67\text{m}\end{array}$

Substitute $4.02{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $0.67\text{m}$ for $y_2$ and $1\text{m}$ for $b$ in Equation (V).

  $\begin{array}{c}{V}_2=\frac{\left(4.02{\text{m}}^{\text{3}}/\text{s}\right)}{\left(1\text{m}\right)\left(0.67\text{m}\right)}\\ =\frac{\left(4.02{\text{m}}^{\text{3}}/\text{s}\right)}{\left(0.67{\text{m}}^2\right)}\\ =6\text{m}/\text{s}\end{array}$

Substitute $6\text{m}/\text{s}$ for $V_2$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $0.67\text{m}$ for $y_2$ in Equation (VI).

  $\begin{array}{c}{\text{Fr}}_2=\frac{\left(6\text{m}/\text{s}\right)}{\sqrt{\left(9.81{\text{m}/\text{s}}^2\right)\times \left(0.67\text{m}\right)}}\\ =\frac{\left(6\text{m}/\text{s}\right)}{\left(2.564\text{m}/\text{s}\right)}\\ =2.34\end{array}$

Substitute $0.67\text{m}$ for $y_2$ and $2.34$ for ${\text{Fr}}_2$ and in Equation (VII).

  $\begin{array}{c}{y}_3=0.52\left(0.67\text{m}\right)\left(-1+\sqrt{1+8{\left(2.34\right)}^2}\right)\\ =\left(0.35\text{m}\right)\times \left(5.69\right)\\ =1.99\text{m}\end{array}$

Substitute $1.99\text{m}$ for $y_3$, $0.67\text{m}$ for $y_2$ and $6\text{m}/\text{s}$ for $V_2$ in Equation (VIII).

  $\begin{array}{c}{V}_3=\frac{\left(0.67\text{m}\right)}{\left(1.99\text{m}\right)}\times \left(6\text{m}/\text{s}\right)\\ =\left(0.336\right)\times \left(6\text{m}/\text{s}\right)\\ =2.02\text{m}/\text{s}\end{array}$

Substitute $2.02\text{m}/\text{s}$ for $V_3$, $1.99\text{m}$ for $y_3$ and $9.81{\text{m}/\text{s}}^2$ for $g$ in Equation (IX).

  $\begin{array}{c}{\text{Fr}}_3=\frac{\left(2.02\text{m}/\text{s}\right)}{\sqrt{\left(1.99\text{m}\right)\left(9.81{\text{m}/\text{s}}^2\right)}}\\ =\frac{\left(2.02\text{m}/\text{s}\right)}{\sqrt{\left(19.52{{\text{m}}^2/\text{s}}^2\right)}}\\ =\frac{\left(2.02\text{m}/\text{s}\right)}{\left(4.42\text{m}/\text{s}\right)}\\ =0.457\end{array}$

Conclusion:

The flow depth after hydraulic jump is $1.99\text{m}$.

The velocity after hydraulic jump is $2.02\text{m}/\text{s}$.

The Froude number after hydraulic jump is $0.457$.