Chapter 13, Problem 142P

To determine

The flow rate of water through the gate per meter width.

The flow depth $y_1$ and $y_2$.

The energy dissipation ratio of the jump.

Answer to Problem 142P

The flow rate of water through the gate per meter width is $12{\text{m}}^{\text{3}}/\text{s}$.

The flow depth $y_1$ and $y_2$ are $9.6\text{m}$ and $1.969\text{m}$.

The energy dissipation ratio of the jump is $0.012$.

Explanation of Solution

Given information:

Water is flowing through a sluice gate, the velocity of water before reaching the gate is $1.25\text{m}/\text{s}$, the velocity of water after the jump is $4\text{m}/\text{s}$, the flow depth at $\left(3\right)$ is $3\text{m}$.

The flow is considered per meter width so $b=1\text{m}$.

The figure below shows the flow.

  

  Figure-(1)

Write the expression for the cross-sectional area at $\left(3\right)$.

  $A_{c,3}=b{y}_3$  ...... (I)

Here, the width is $b$, the flow depth at $\left(3\right)$ is $y_3$.

Write the expression for the cross-sectional area at $\left(3\right)$.

  $\dot{V}=V_3{A}_{c,3}$  ...... (II)

Here, the velocity at section $\left(3\right)$ is $V_3$ and area at section (3) is $A_{c,3}$.

Write the expression for the conservation of mass.

  $\begin{array}{l}{V}_1{y}_1=V_3{y}_3\\ y_1=\frac{V_3}{V_1}{y}_3\end{array}$  ...... (III)

Here, the velocity at section (3) is $V_3$, the velocity at section

Write the expression for the Froude number at section $\left(3\right)$.

  $F_{r,3}=\frac{V_3}{\sqrt{g{y}_3}}$  ...... (IV)

Here, the acceleration due to gravity is $g$.

Write the expression for the flow depth.

  $y_2=0.5y_3\left(-1+\sqrt{1+8F_r{}_3^2}\right)$  ...... (V)

Write the expression for the conservation of mass.

  $\begin{array}{l}{V}_2{y}_2=V_3{y}_3\\ V_2=V_3\frac{y_3}{y_2}\end{array}$ (VI)

The flow depth at section (2) is $y_2$.

Write the expression for the Froude number at section $\left(2\right)$.

  $F_{r,2}=\frac{V_2}{\sqrt{g{y}_2}}$  ...... (VII)

Write the expression for the head loss during jumping.

  $h_L=y_2-y_3+\frac{V_2^2-V_3^2}{2g}$  ...... (VIII)

Here, the velocity at section (2) is $V_2$ and the velocity at section (3) is $V_3$.

Write the expression for the specific energy before the jump.

  $E_{s2}=y_2+\frac{V_2^2}{2g}$  ...... (IX)

Write the expression for the dissipation ratio.

  $D_s=\frac{h_L}{E_{s2}}$....... (X)

Calculation:

Substitute $1\text{m}$ for $b$ and $3\text{m}$ for $y_3$ in Equation (I).

  $\begin{array}{c}{A}_{c3}=1\text{m}\times \text{3}\text{m}\\ ={\text{3m}}^2\end{array}$

Substitute $4\text{m}/\text{s}$ for $V_3$, $3{\text{m}}^{\text{2}}$ for $A_{c3}$ in Equation (II).

  $\begin{array}{c}\dot{V}=4\text{m}/\text{s}\times 3{\text{m}}^{\text{2}}\\ =12{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $4\text{m}/\text{s}$ for $V_3$, $1.25\text{m}/\text{s}$ for $V_1$ in Equation (III).

  $\begin{array}{c}{y}_1=\frac{4\text{m}/\text{s}}{1.25\text{m}/\text{s}}\left(3\text{m}\right)\\ =3.2\left(3\text{m}\right)\\ =9.6\text{m}\end{array}$

Substitute $4\text{m}/\text{s}$ for $V_3$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $3\text{m}$ for $y_3$ in Equation (IV).

  $\begin{array}{c}{F}_{r3}=\frac{4\text{m}/\text{s}}{\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\times 3\text{m}}}\\ =\frac{4\text{m}/\text{s}}{5.425\text{m}/\text{s}}\\ =0.7373\end{array}$

Substitute $3\text{m}$ for $y_3$

  $0.7373$ for $F_{r,3}$ in Equation (V).

  $\begin{array}{c}{y}_2=0.5\left(3\text{m}\right)\left(-1+\sqrt{1+8{\left(0.7373\right)}^2}\right)\\ =1.5\text{m}\left(1.3127\right)\\ =1.969\text{m}\end{array}$

Substitute $3\text{m}$ for $y_3$, $1.969\text{m}$ for $y_2$ and $4\text{m}/\text{s}$ for $V_3$ in Equation (VI).

  $\begin{array}{c}{V}_2=\frac{3\text{m}}{1.969\text{m}}\left(4\text{m}/\text{s}\right)\\ =1.5236\left(4\text{m}/\text{s}\right)\\ =6.094\text{m}/\text{s}\end{array}$

Substitute $6.094\text{m}/\text{s}$ for $V_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.969\text{m}$ for $y_2$ in Equation (VII).

  $\begin{array}{c}{F}_{r,2}=\frac{6.094\text{m}/\text{s}}{\sqrt{\left(9.81{\text{m}/\text{s}}^2\right)\left(1.969\text{m}\right)}}\\ =\frac{6.094\text{m}/\text{s}}{4.39498\text{m}/\text{s}}\\ =1.387\end{array}$

Here Froude number is greater than $1$ thus the flow before jump is super critical.

Substitute $1.969\text{m}$ for $y_2$, $3\text{m}$ for $y_3$

  $6.094\text{m}/\text{s}$ for $V_2$, $4\text{m}/\text{s}$ for $V_3$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VIII).

  $\begin{array}{c}{h}_L=1.969\text{m}-3\text{m}+\frac{{\left(6.9094\text{m}/\text{s}\right)}^2-{\left(4\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =1.969\text{m}-3\text{m}+1.6177\text{m}\\ =\text{0}\text{.0463}\text{m}\end{array}$

Substitute $1.9639\text{m}$ for $y_2$, $6.094\text{m}/\text{s}$ for $V_2$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (IX).

  $\begin{array}{c}{E}_{s2}=1.9639\text{m}+\frac{{\left(6.094\text{m}/\text{s}\right)}^2}{2\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =1.9639\text{m}+1.892\text{m}\\ =\text{3}\text{.862}\text{m}\end{array}$

Substitute $0.0463\text{m}$ for $h_L$ and $3.862\text{m}$ for $E_{s,2}$ in Equation (X).

  $\begin{array}{c}{D}_s=\frac{0.0463\text{m}}{3.862\text{m}}\\ =0.012\text{m}\end{array}$

Conclusion:

The flow rate of water through the gate per meter width is $12{\text{m}}^{\text{3}}/\text{s}$.

The flow depth $y_1$ and $y_2$ are $9.6\text{m}$ and $1.969\text{m}$.

The energy dissipation ratio of the jump is $0.012$.