Chapter 13, Problem 106EP

To determine

The flow rate per unit width.

The Froude number of downstream gate.

Answer to Problem 106EP

The flow rate per unit width is $10.9{\text{ft}}^{\text{3}}/\text{s}$.

The Froude number of downstream gate is $3.7140$.

Explanation of Solution

Given Information:

The flow depth of upstream weir is $5\text{ft}$ and the height of the gate opening is $1.1\text{ft}$.

Write the expression for the depth ratio.

  $R=\frac{y_1}{a}$...... (I)

Here, the flow depth of upstream weir is $y_1$ and the height of the gate opening is $a$.

Write the expression for the volume flow rate.

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

Here, the breath of the channel is $b$ and the acceleration due to gravity is $g$.

Write the expression for the upstream specific energy.

  $E_{c1}=y_1+\frac{{\dot{Q}}^2}{2g{\left(b{y}_1\right)}^2}$....... (III)

Write the expression for the downstream specific energy.

  $E_{c2}=y_2+\frac{{\dot{Q}}^2}{2g{\left(b{y}_2\right)}^2}$....... (IV)

Write the expression for the downstream velocity.

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

Write the expression for the Froude number of downstream.

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

Calculation:

Substitute $5\text{ft}$ for $y_1$ and $1.\text{1}\text{ft}$ for $a$ in Equation (I).

  $\begin{array}{c}R=\frac{5\text{ft}}{1.1\text{ft}}\\ =4.55\end{array}$

Refer to Figure 13.44 "Discharge coefficient for the drowned and free discharge from underflow gates" to obtain the coefficient of the discharge as $0.55$ at depth ratio $R=4.55$.

Substitute $1\text{ft}$ for $b$, $1.1\text{ft}$ for $a$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $0.55$ for $C_d$ and $\text{5}\text{ft}$ for $y_1$ in equation (II).

  $\begin{array}{c}\dot{Q}=\left(0.55\right)\left(1.1\text{ft}\right)\left(1\text{ft}\right)\sqrt{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(5\text{ft}\right)}\\ =\left(0.55\right)\left(1.1\text{ft}\right)\left(1\text{ft}\right)\sqrt{\left(322{\text{ft}}^2/{\text{s}}^{\text{2}}\right)}\\ =0.605{\text{ft}}^2\left(17.94\text{ft}/\text{s}\right)\\ \approx 10.9{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Substitute $1\text{ft}$ for $b$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $10.9{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{Q}$ and $5\text{ft}$ for $y_1$ in Equation (III).

  $\begin{array}{c}{E}_{c1}=5\text{ft}+\frac{{\left(10.9{\text{ft}}^{\text{3}}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right){\left(1\text{ft}\times 5\text{ft}\right)}^2}\\ =5\text{ft}+\frac{\left(118.81{\text{ft}}^6/{\text{s}}^2\right)}{\left(1610{\text{ft}}^5/{\text{s}}^{\text{2}}\right)}\\ =5\text{ft}+0.07379\text{ft}\\ =5.074\text{ft}\end{array}$

Since, the specific energy of the fluid remains constant hence $E_{c2}=5.074\text{ft}$.

Substitute $1\text{ft}$ for $b$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $10.9{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{Q}$ and $5.074\text{ft}$ for $E_{c2}$ in Equation (IV).

  $\begin{array}{l}5.074\text{ft}=y_2+\frac{{\left(10.9{\text{ft}}^{\text{3}}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right){\left(1\text{ft}\times y_2\right)}^2}\\ 5.074\text{ft}=y_2+\frac{\left(118.81{\text{ft}}^6/{\text{s}}^2\right)}{\left(64.4{\text{ft}}^5/{\text{s}}^{\text{2}}\right)y_2^2}\\ \left(\left(326.76{\text{ft}}^5/{\text{s}}^{\text{2}}\right)y_2^2\right)=\left(\left(64.4{\text{ft}}^5/{\text{s}}^{\text{2}}\right)y_2^3\right)+118.81{\text{ft}}^6/{\text{s}}^2\\ y_2=0.643\text{ft}\end{array}$

Substitute $0.643\text{ft}$ for $y_2$, $1\text{ft}$ for $b$ and $10.9{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{Q}$ in Equation (V).

  $\begin{array}{c}{V}_2=\frac{10.9{\text{ft}}^{\text{3}}/\text{s}}{\left(0.643\text{ft}\times 1\text{ft}\right)}\\ =\frac{10.9{\text{ft}}^{\text{3}}/\text{s}}{\left(0.643{\text{ft}}^2\right)}\\ =16.9\text{ft}/\text{s}\end{array}$

Substitute $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $16.9\text{ft}/\text{s}$ for $V_2$ and $0.643\text{ft}$ for $y_2$ in Equation (VI).

  $\begin{array}{c}F{r}_2=\frac{16.9\text{ft}/\text{s}}{\sqrt{\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.643\right)}}\\ =\frac{16.9\text{ft}/\text{s}}{\sqrt{\left(20.7046{\text{ft}}^2/{\text{s}}^{\text{2}}\right)}}\\ =\frac{16.9}{4.5502}\\ =3.7140\end{array}$

Conclusion:

The flow rate per unit width is $10.9{\text{ft}}^{\text{3}}/\text{s}$.

The Froude number of downstream gate is $3.7140$.