Chapter 13, Problem 140P

A sluice gate with free outflow is used to control the discharge rate of water through a channel Determine the flow rate per unit width when the gate is raised to yield a gap of 50 cm and the upstream flow depth is measured to be 2.8 m. Also determine the flow depth and the velocity downstream.

To determine

The flow rate per unit width.

The flow depth.

The velocity for downstream.

Answer to Problem 140P

The flow rate per unit width is $2.068{\text{m}}^3/\text{s}$.

The flow depth is $0.293\text{m}$.

The velocity for downstream is $7.06\text{m}/\text{s}$.

Explanation of Solution

Given Information:

The height for the sluice gate is $50\text{cm}$ and the flow depth for upstream is $2.8\text{m}$.

Write the expression for the flow depth.

  $D_r=\frac{y_1}{a}$...... (I)

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

Write the expression for the discharge rate.

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

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

Write the expression for the cross- sectional area at upstream.

  $A_{c1}=b{y}_1$...... (III)

Write the expression for the cross -sectional area at downstream.

  $A_{c2}=b{y}_2$...... (IV)

Write the expression for the upstream velocity of the fluid.

  $V_1=\frac{\dot{Q}}{A_{c1}}$...... (V)

Write the expression for the downstream velocity of the fluid.

  $V_2=\frac{\dot{Q}}{A_{c2}}$...... (VI)

Write the expression for the upstream specific energy.

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

Write the expression for the downstream specific energy.

  $E_{c2}=y_2+\frac{V_2^2}{2g}$...... (VIII)

Calculation:

Substitute $2.8\text{m}$ for $y_1$ and $50\text{cm}$ for $a$ in Equation (I).

  $\begin{array}{c}{D}_r=\frac{2.8\text{m}}{50\text{cm}}\\ =\frac{2.8\text{m}}{50\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =5.6\end{array}$

Refer to Figure 13.44 "Discharge coefficient for the drowned and free discharge from underflow gates" to obtain the coefficient of the discharge as $0.56$ at depth ratio $R=5.6$.

Substitute $0.56$ for $C_d$, $1\text{m}$ for $b$, $0.5\text{m}$ for $a$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $2.8\text{m}$ for $y_1$ in Equation (II).

  $\begin{array}{c}\dot{Q}=\left(0.56\right)\left(1\text{m}\right)\left(0.5\text{m}\right)\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(2.8\text{m}\right)}\\ =\left(0.28{\text{m}}^2\right)\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(2.8\text{m}\right)}\\ =\left(0.28{\text{m}}^2\right)\left(7.38\text{m}/\text{s}\right)\\ =2.068{\text{m}}^3/\text{s}\end{array}$

Substitute $2.8\text{m}$ for $y_1$ and $1\text{m}$ for $b$ in Equation (III).

  $\begin{array}{c}{A}_{c1}=\left(2.8\text{m}\right)\left(1\text{m}\right)\\ =2.8{\text{m}}^2\end{array}$

Substitute $2.8{\text{m}}^2$ for $A_{c1}$ and $2.068{\text{m}}^3/\text{s}$ for $\dot{Q}$ in Equation (V).

  $\begin{array}{c}{V}_1=\frac{2.068{\text{m}}^3/\text{s}}{2.8{\text{m}}^2}\\ =0.738\text{m}/\text{s}\end{array}$

Substitute $1\text{m}$ for $b$ in Equation (IV).

  $\begin{array}{c}{A}_{c2}=\left(y_2\right)\left(1\text{m}\right)\\ =y_2\text{m}\end{array}$...... (IX)

Substitute $y_2\text{m}$ for $A_{c2}$ and $2.068{\text{m}}^3/\text{s}$ for $\dot{Q}$ in Equation (VI).

  $\begin{array}{c}{V}_2=\frac{2.068{\text{m}}^3/\text{s}}{y_2\text{m}}\\ =\frac{2.068{\text{m}}^2/\text{s}}{y_2}\end{array}$

Substitute $2.8\text{m}$ for $y_1$, $0.738\text{m}/\text{s}$ for $V_1$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VII).

  $\begin{array}{c}{E}_{c1}=2.8\text{m}+\left(\frac{{\left(0.738\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)\\ =2.8\text{m}+0.0277\text{m}\\ =2.827\text{m}\end{array}$

Since, the specific energy at downstream is equal the specific stream at upward stream hence $E_{c2}=2.827\text{m}$.

Substitute $2.827\text{m}$ for $E_{c2}$, $\frac{2.068{\text{m}}^2/\text{s}}{y_2}$ for $V_2$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VIII).

  $\begin{array}{l}2.827\text{m}=y_2+\left(\frac{{\left(\frac{2.068\text{m}/\text{s}}{y_2}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)\\ 2.827\text{m}=\frac{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)y_2+\left(\left(\frac{4.276{\text{m}}^2/{\text{s}}^2}{y_2^2}\right)\right)}{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)}\\ y_2^3-2.827y_2^2+0.218=0\\ y_2=0.293\text{m}\end{array}$

Substitute $0.293\text{m}$ for $y_2$ in Equation (IX).

  $\begin{array}{c}{A}_{c2}=\left(0.293\text{m}\right)\text{m}\\ =0.293{\text{m}}^2\end{array}$

Substitute $0.293{\text{m}}^2$ for $A_{c2}$ and $2.068{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ in Equation (VI).

  $\begin{array}{c}{V}_2=\frac{2.068{\text{m}}^{\text{3}}/\text{s}}{0.293{\text{m}}^2}\\ \approx 7.06\text{m}/\text{s}\end{array}$

Conclusion:

The flow rate per unit width is $2.068{\text{m}}^3/\text{s}$.

The flow depth is $0.293\text{m}$.

The velocity for downstream is $7.06\text{m}/\text{s}$.