Chapter 13, Problem 104P

Consider the uniform flow of water in a wide channel with a velocity of 8 'n/s and now depth of 0.8 m. Now water flows over a 30-cm-high twmp_ Determine the change (increase or decrease) in the water surface level over the bump. Also deterrnine if the now over the twmp is subs upercritical_

To determine

The change in the water surface level over the bump.

If the flow over the bump is subcritical or supercritical.

Answer to Problem 104P

The raise over the bump is $0.346\text{m}$.

The flow is supercritical because Froude number is greater than $1$.

Explanation of Solution

Given information:

The flow is uniform, the velocity of the flow is $8\text{m}/\text{s}$, the depth of the flow is $0.8\text{m,}$ and the height of the bump is $30\text{cm}$.

The flow is steady, and the channel is sufficiently wide so that the end effects are negligible, the frictional effects are negligible so that there is no dissipation of mechanical energy, and the acceleration due to gravity is $9.81{\text{m}/\text{s}}^2$.

Write the expression for upstream Froude number.

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

Here the velocity of the flow is $V_1$, the acceleration due to gravity is $g,$ and the depth of the flow is $y_1$.

Write the expression for critical depth.

   $y_c={\left(\frac{{\dot{\vee }}^2}{g{b}^2}\right)}^{1/3}$    ...... (II)

Here, the thickness of the channel is $b$ and the rate of flow is $\dot{\vee }$.

Write the expression for the rate flow.

   $\dot{\vee }=b{y}_1{V}_1$

Substitute $b{y}_1{V}_1$ for $\dot{\vee }$ in the Equation (II).

   $\begin{array}{c}{y}_c={\left(\frac{{\left(b{y}_1{V}_1\right)}^2}{g{b}^2}\right)}^{1/3}\\ ={\left(\frac{{\left(y_1{V}_1\right)}^2}{g}\right)}^{1/3}\end{array}$    ...... (III)

Write the expression for the upstream specific energy.

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

Write the expression for over the bump specific energy.

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

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

Write the expression for the critical specific energy.

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

Here, the critical depth is $y_c$.

Write the expression for the flow depth over the bump.

   $y_2^3-\left(E_{s1}-\Delta z_b\right)y_2^2+\frac{V_1^2}{2g}{y}_1^2=0$    ...... (VII)

Here, the flow depth over the bump is $y_2$.

Write the expression for the rise over bump.

   $\text{Rise}\text{over}\text{bump}=y_2-y_1+\Delta z_b$    ...... (VIII)

Calculation:

Substitute $8\text{m}/\text{s}$ for $V_1$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $0.8\text{m}$ for $y_1$ in the Equation (I).

   $\begin{array}{c}F{r}_1=\frac{8\text{m}/\text{s}}{\sqrt{\left(9.81\text{m}/{\text{s}}^2\right)\left(0.8\text{m}\right)}}\\ =\frac{8\text{m}/\text{s}}{\sqrt{\left(70848{\text{m}}^2/{\text{s}}^2\right)}}\\ =2.856\end{array}$

Substitute $8\text{m}/\text{s}$ for $V_1$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $0.8\text{m}$ for $y_1$ in the Equation (III).

   $\begin{array}{c}{y}_c={\left(\frac{{\left(0.8\text{m}\right)}^2{\left(8\text{m}/\text{s}\right)}^2}{9.81{\text{m}/\text{s}}^2}\right)}^{1/3}\\ ={\left(\frac{\left(0.64{\text{m}}^{\text{2}}\right)\left(64{{\text{m}}^2/\text{s}}^2\right)}{9.81{\text{m}/\text{s}}^2}\right)}^{1/3}\\ =1.61\text{m}\end{array}$

Substitute $8\text{m}/\text{s}$ for $V_1$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $0.8\text{m}$ for $y_1$ in the Equation (IV).

   $\begin{array}{c}{E}_{s1}=\left(0.8\text{m}\right)+\frac{\left(8\text{m}/\text{s}\right)}{2\times 9.81{\text{m}/\text{s}}^2}\\ =\left(0.8\text{m}\right)+\frac{{\left(8\text{m}/\text{s}\right)}^2}{2\times 9.81{\text{m}/\text{s}}^2}\\ =\left(0.8\text{m}\right)+\left(3.26197\text{m}\right)\\ =4.06\text{m}\end{array}$

Substitute $4.06\text{m}$ for $E_{s1}$ and $30\text{cm}$ for $\Delta z_b$ in the Equation (V).

   $\begin{array}{c}{E}_{s2}=4.06\text{m}-30\text{cm}\\ \text{=}4.06\text{m}-30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =3.76\text{m}\end{array}$

Substitute $1.61\text{m}$ for $y_c$ in the Equation (VI).

   $\begin{array}{c}{E}_c=\frac{3}{2}\left(1.61\text{m}\right)\\ =2.42\text{m}\end{array}$

Substitute $4.06\text{m}$ for $E_{s1}$, $30\text{cm}$ for $\Delta z_b$, $8\text{m}/\text{s}$ for $V_1$, $9.81{\text{m}/\text{s}}^2$ for $g$ and $0.8\text{m}$ for $y_1$ in the Equation (VII).

   $\begin{array}{l}{y}_2^3-\left(4.06\text{m}-30\text{cm}\right)y_2^2+\frac{{\left(8\text{m}/\text{s}\right)}^2}{2\times 9.81{\text{m}/\text{s}}^2}{\left(0.8\text{m}\right)}^2=0\\ y_2^3-\left(4.06\text{m}-30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)y_2^2+\left(3.2619\text{m}\right){\left(0.8\text{m}\right)}^2=0\\ y_2=0.846\text{m}\end{array}$

Substitute $0.846\text{m}$ for $y_2$, $30\text{cm}$ for $\Delta z_b$ and $0.8\text{m}$ for $y_1$ in the Equation (VIII).

   $\begin{array}{c}\text{Rise}\text{over}\text{bump}=0.846\text{m}-0.8\text{m}+30\text{cm}\\ \text{=0}\text{.046}\text{m+}30\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.346\text{m}\end{array}$

Conclusion:

The raise over the bump is $0.346\text{m}$ and the flow is supercritical because Froude number is greater than $1$.