Chapter 13, Problem 85P

During a hydraulic jump in a wide channel, the flow depth increases from 1.1 to 3.3 in. Determine the velocities and Froude numbers before and after the jump, and the energy dissipation ratio.

To determine

The velocity before the jump.

The velocity after the jump.

The Froude number before the jump.

The Froude number after the jump.

The energy dissipation ratio.

Answer to Problem 85P

The velocity before the jump is $8.402\text{m}/\text{s}$.

The velocity after the jump is $2.80\text{m}/\text{s}$.

The Froude number before the jump is $2.449$.

The Froude number after the jump is $0.492$.

The energy dissipation ratio is $0.212$.

Explanation of Solution

Given Information:

The flow depth after the jump is $3.3\text{m}$ and the flow depth before the jump is $1.1\text{m}$.

Write the expression for the flow depth after the hydraulic jump.

  $y_2=0.5y_1\left(-1+\sqrt{1+8F{r}_1{}^2}\right)$...... (I)

Here, the Froude number before jump is $F{r}_1$, the flow depth before the jump is $y_1$.

Write the expression for the Froude number before jump.

  $F{r}_1=\frac{V_1}{\sqrt{g{y}_1}}$...... (II)

Here, the velocity before the jump is $V_1$ and the acceleration due to gravity is $g$.

Write the expression for the average velocity after the hydraulic jump.

  $\begin{array}{l}{y}_1{V}_1=y_2{V}_2\\ V_2=\frac{y_1{V}_1}{y_2}\end{array}$...... (III)

Write the expression for the Froude number after jump.

  $F{r}_2=\frac{V_2}{\sqrt{g{y}_2}}$...... (IV)

Here, the velocity after the jump is $V_2$.

Write the expression for the head loss associated with the hydraulic jump.

  $h_L=y_1-y_2+\frac{V_1^2-V_2^2}{2g}$...... (V)

Here, the average velocity before hydraulic jump is $V_1$, the velocity after hydraulic jump is $V_2$, the flow depth before the jump is $y_1$, the flow depth after the hydraulic jump is $y_2$ and the acceleration due to gravity is $g$.

Write the expression for the specific energy of water before the jump.

  $E_{c1}=y_1+\frac{V_1^2}{2g}$....... (VI)

Write the expression for the dissipation ratio.

  $D_r=\frac{h_L}{E_{c1}}$...... (VII)

Calculation:

Substitute $3.3\text{m}$ for $y_2$ and $1.1\text{m}$ for $y_1$ in Equation (I).

  $\begin{array}{l}3.3\text{m}=0.5\left(1.1\text{m}\right)\left(-1+\sqrt{1+8{\left(F{r}_1\right)}^2}\right)\\ \sqrt{1+8{\left(F{r}_1\right)}^2}=\frac{3.3\text{m}}{\left(5.5\text{m}\right)}+1\\ {\left(F{r}_1\right)}^2=\frac{{\left(1.6\right)}^2-1}{2}\\ F{r}_1=2.449\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.449$ for $F{r}_1$ and $1.2\text{m}$ for $y_1$ in Equation (II).

  $\begin{array}{c}2.449=\frac{V_1}{\sqrt{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1.2\text{m}\right)}}\\ V_1=\left(2.449\right)\sqrt{\left(11.772{\text{m}}^2/{\text{s}}^{\text{2}}\right)}\\ V_1=\left(2.449\right)\left(3.43\text{m}/\text{s}\right)\\ V_1=8.402\text{m}/\text{s}\end{array}$

Substitute $8.402\text{m}/\text{s}$ for $V_1$, $3.3\text{m}$ for $y_2$ and $1.1\text{m}$ for $y_1$ in Equation (III).

  $\begin{array}{c}{V}_2=\frac{\left(1.1\text{m}\right)\left(8.402\text{m}/\text{s}\right)}{\left(3.3\text{m}\right)}\\ =\frac{\left(9.24\text{m}/\text{s}\right)}{\left(3.3\right)}\\ =2.80\text{m}/\text{s}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.80\text{m}/\text{s}$ for $V_2$ and $3.3\text{m}$ for $y_2$ in Equation (IV).

  $\begin{array}{c}F{r}_2=\frac{2.80\text{m}/\text{s}}{\sqrt{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(3.3\text{m}\right)}}\\ =\frac{2.80\text{m}/\text{s}}{\sqrt{\left(32.37{\text{m}}^2/{\text{s}}^{\text{2}}\right)}}\\ =\frac{2.80}{5.68}\\ =0.492\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.80\text{m}/\text{s}$ for $V_2$, $8.402\text{m}/\text{s}$ for $V_1$, $3.3\text{m}$ for $y_2$ and $1.1\text{m}$ for $y_1$ in Equation (V).

  $\begin{array}{c}{h}_L=\left(1.1\text{m}\right)-\left(3.3\text{m}\right)+\frac{{\left(8.402\text{m}/\text{s}\right)}^2-{\left(2.80\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =-\left(2.2\text{m}\right)+\frac{\left(62.75{\text{m}}^2/{\text{s}}^2\right)}{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)}\\ =-\left(2.2\text{m}\right)+3.198\text{m}\\ =0.998\text{m}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $8.402\text{m}/\text{s}$ for $V_1$ and $1.1\text{m}$ for $y_1$ in Equation (VI).

  $\begin{array}{c}{E}_{c1}=1.1\text{m}+\frac{{\left(8.402\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =1.1\text{m}+\frac{\left(70.5936{\text{m}}^2/{\text{s}}^2\right)}{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)}\\ =1.1\text{m}+3.59\text{m}\\ =4.69\text{m}\end{array}$

Substitute $4.69\text{m}$ for $E_{c1}$ and $0.998\text{m}$ for $h_L$ in Equation (VII).

  $\begin{array}{c}{D}_r=\frac{0.998\text{m}}{4.69\text{m}}\\ =0.212\end{array}$

Conclusion:

The velocity before the jump is $8.402\text{m}/\text{s}$.

The velocity after the jump is $2.80\text{m}/\text{s}$.

The Froude number before the jump is $2.449$.

The Froude number after the jump is $0.492$.

The energy dissipation ratio is $0.212$.