Chapter 13, Problem 146P

To determine

The velocity over the bump.

The flow depth over the bump.

The Froude number over the bump.

Answer to Problem 146P

The velocity over the bump is $1.42794\text{m}/\text{s}$.

The flow depth over the bump is $1.5757\text{m}$.

The Froude number over the bump is $0.3632$.

Explanation of Solution

Given information:

The height of bump is $20\text{cm}$, initial velocity of flow is $1.25\text{m}/\text{s}$, the flow depth is $1.8\text{m}$.

Write the expression for Froude number.

  $\text{Fr}=\frac{V}{\sqrt{gy}}$...... (I)

Here, velocity is $V$, acceleration due to gravity is $g$, flow depth is $y$.

Write the expression for critical flow of depth.

  $y_c={\left[\frac{y_1^2{V}_1^2}{g}\right]}^{1/3}$...... (II)

Here, velocity before bump is $V_1$, depth before bump is $y_1$, acceleration due to gravity is $g$.

Write the expression for specific energy before bump.

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

Write the expression for specific energy after the bump.

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

Here, height of bump is $\Delta z_b$.

Write the expression for critical specific energy.

  $E_c=\frac{3}{2}{y}_c$...... (V)

Write the relation for flow depth after the bump.

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

Write the expression for velocity after the bump.

  $V_2=\frac{y_1}{y_2}{V}_1$...... (VII)

Calculation:

Substitute ${\text{Fr}}_1$ for $\text{Fr}$, $1.25\text{m}/\text{s}$ for $V_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.8\text{m}$ for $y_1$ in Equation (I)

  $\begin{array}{c}{\text{Fr}}_1=\frac{1.25\text{m}/\text{s}}{\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\times 1.8\text{m}}}\\ =\frac{1.25\text{m}/\text{s}}{\sqrt{17.658{\text{m}}^2/{\text{s}}^{\text{2}}}}\\ =0.2975\end{array}$

The Froude number is less than $1$, so flow is sub-critical and surface level over bump drops.

Substitute $1.25\text{m}/\text{s}$ for $V_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.8\text{m}$ for $y_1$ in Equation (II).

  $\begin{array}{c}{y}_c={\left[\frac{{\left(1.8\text{m}\right)}^2{\left(1.25\text{m}/\text{s}\right)}^2}{9.81\text{m}/{\text{s}}^{\text{2}}}\right]}^{1/3}\\ ={\left[\frac{5.0625{\text{m}}^{\text{4}}/{\text{s}}^{\text{2}}}{9.81\text{m}/{\text{s}}^{\text{2}}}\right]}^{1/3}\\ =0.8023\text{m}\end{array}$

Substitute $1.25\text{m}/\text{s}$ for $V_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.8\text{m}$ for $y_1$ in Equation (III).

  $\begin{array}{c}{E}_{s1}=1.8\text{m}+\frac{{\left(1.25\text{m}/\text{s}\right)}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =1.8\text{m}+0.07963\text{m}\\ =\text{1}\text{.8796}\text{m}\end{array}$

Substitute $\text{1}\text{.8796}\text{m}$ for $E_{s1}$ and $20\text{cm}$ for $\Delta z_b$ in Equation (IV).

  $\begin{array}{c}{E}_{s2}=\text{1}\text{.8796}\text{m}-20\text{cm}\\ \text{=1}\text{.8796}\text{m}-20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\\ =\text{1}\text{.8796}\text{m}-0.2\text{m}\\ =1.6796\text{m}\end{array}$

Substitute $0.8023\text{m}$ for $y_c$ in Equation (V).

  $\begin{array}{c}{E}_c=\frac{3}{2}\times 0.8023\text{m}\\ =\text{1}\text{.20345}\text{m}\end{array}$

Substitute $\text{1}\text{.8796}\text{m}$ for $E_{s1}$, $20\text{cm}$ for $\Delta z_b$, $1.25\text{m}/\text{s}$ for $V_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.8\text{m}$ for $y_1$ in Equation (VI).

  $\begin{array}{l}{y}_2^3-\left(\text{1}\text{.8796}\text{m}-0.20\text{m}\right)y_2^2+\frac{{\left(1.25\text{m}/\text{s}\right)}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}{\left(1.8\text{m}\right)}^2=0\\ y_2^3-\left(1.6796\right)y_2^2+0.258=0\end{array}$

Solve for $y_2$.

  $y_2=1.5757\text{m}$

  $y_2=0.4599\text{m}$

  $y_2=-0.356\text{m}$

Here, depth of flow should be greater than critical depth.

  $y_2=1.5757\text{m}$

Substitute $1.25\text{m}/\text{s}$ for $V_1$, $1.5757\text{m}$ for $y_2$ and $1.8\text{m}$ for $y_1$ in Equation (VII).

  $\begin{array}{c}{V}_2=\frac{1.8\text{m}}{1.5757\text{m}}\times 1.25\text{m}/\text{s}\\ =2.83626\times 1.25\text{m}/\text{s}\\ =1.42794\text{m}/\text{s}\end{array}$

Substitute ${\text{Fr}}_2$ for $\text{Fr}$, $1.42794\text{m}/\text{s}$ for $V_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.5757\text{m}$ for $y_2$ in Equation (I)

  $\begin{array}{c}{\text{Fr}}_2=\frac{1.42794\text{m}/\text{s}}{\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\times 1.5757\text{m}}}\\ =\frac{1.42794\text{m}/\text{s}}{\sqrt{15.4576{\text{m}}^2/{\text{s}}^{\text{2}}}}\\ =0.3632\end{array}$

Conclusion:

The velocity over the bump is $1.42794\text{m}/\text{s}$.

The flow depth over the bump is $1.5757\text{m}$.

The Froude number over the bump is $0.3632$.