Chapter 13, Problem 129P

To determine

The flow rate through the channel.

The effective manning coefficient.

Answer to Problem 129P

The flow rate through the channel is $37.2{\text{m}}^{\text{3}}/\text{s}$.

The effective manning coefficient is $0.0386$.

Explanation of Solution

Given information:

The channel angle is $0.5°$.

Write the expression for flow rate using manning's equation.

  $\dot{V}=\frac{a}{n}{A}_{\text{c}}{R}_{\text{h}}^{2/3}\tan \alpha$...... (I)

Here, dimensional constant is $a$, area is $A_{\text{c}}$, manning's coefficient is $n$, hydraulic radius is $R_{\text{h}}$,channel angle is $\alpha$.

Write the expression for net volume flow rate.

  $\dot{V}=\frac{a}{n}{A}_{\text{c1}}{R}_{\text{h1}}^{2/3}\tan \alpha +\frac{a}{n}{A}_{\text{c2}}{R}_{\text{h2}}^{2/3}\tan \alpha$...... (II)

Here, hydraulic radius for first channel is $R_{\text{h1}}$, hydraulic radius for second channel is $R_{\text{h2}}$, area of first channel is $A_{\text{c1}}$ and area of first channel is $A_{\text{c2}}$.

Write the expression for area of rectangular section.

  $A_c=l\times b$...... (III)

Here, length is $l$ and width is $b$.

Write the expression for total area.

  ${A^{\prime }}_c=A_{\text{c1}}+A_{\text{c2}}$...... (IV)

Write the expression for wetted perimeter.

  $p=l+b+d$...... (V)

Here, depth is $d$.

Write the expression for total wetted perimeter.

  $p^{\prime }=p_{\text{1}}+p_{\text{2}}$...... (VI)

Write the expression for hydraulic radius.

  $R_h=\frac{A_{\text{c}}}{p}$...... (VII)

Write the expression for equivalent hydraulic radius.

  ${R^{\prime }}_h=\frac{{A^{\prime }}_{\text{c}}}{p^{\prime }}$

Calculation:

Substitute $3\text{m}$ for $l$, $2\text{m}$ for $b$ and $A_{c\text{1}}$ for $A$ in Equation (III)

  $\begin{array}{c}{A}_{\text{c1}}=3\text{m}\times 2\text{m}\\ =\text{6}{\text{m}}^2\end{array}$

Substitute $10\text{m}$ for $l$, $1\text{m}$ for $b$ and $A_{c2}$ for $A$ in Equation (III)

  $\begin{array}{c}{A}_{\text{c2}}=10\text{m}\times 1\text{m}\\ =\text{10}{\text{m}}^2\end{array}$

Substitute $\text{6}{\text{m}}^2$ for $A_{\text{c1}}$ and $\text{10}{\text{m}}^2$ for $A_{\text{c2}}$ in Equation (IV)

  $\begin{array}{c}{A^{\prime }}_{\text{c}}=6{\text{m}}^{\text{2}}+10{\text{m}}^{\text{2}}\\ =16{\text{m}}^{\text{2}}\end{array}$

Substitute $3\text{m}$ for $l$, $2\text{m}$ for $b$, $p_1$ for $p$ and $1\text{m}$ for $d$ in Equation (V).

  $\begin{array}{c}{p}_1=3\text{m}+2\text{m}+1\text{m}\\ =6\text{m}\end{array}$

Substitute $10\text{m}$ for $l$, $1\text{m}$ for $b$ and $p_2$ for $p$ in Equation (V).

  $\begin{array}{c}{p}_2=10\text{m}+\text{1}\text{m}\\ =11\text{m}\end{array}$

Substitute $6\text{m}$ for $p_1$ and $11\text{m}$ for $p_2$ in Equation (VI).

  $\begin{array}{c}{p}^{\prime }=6\text{m}+11\text{m}\\ =\text{17}\text{m}\end{array}$

Substitute $\text{10}{\text{m}}^2$ for $A_{\text{c2}}$, $11\text{m}$ for $p_2$ and $R_{\text{h2}}$ for $R_{\text{h}}$ in Equation (VII).

  $\begin{array}{c}{R}_{h2}=\frac{10{\text{m}}^2}{11\text{m}}\\ =0.909\text{m}\end{array}$

Substitute $\text{6}{\text{m}}^2$ for $A_{\text{c1}}$, $6\text{m}$ for $p_1$ and $R_{\text{h1}}$ for $R_{\text{h}}$ in Equation (VII).

  $\begin{array}{c}{R}_{h1}=\frac{\text{6}{\text{m}}^2}{6\text{m}}\\ =1\text{m}\end{array}$

Substitute $\text{16}{\text{m}}^2$ for ${A^{\prime }}_{\text{c}}$, $17\text{m}$ for $p^{\prime }$ and ${R^{\prime }}_h$ for $R_{\text{h}}$ in Equation (VII).

  $\begin{array}{c}{R}_{h1}=\frac{16{\text{m}}^2}{17\text{m}}\\ =0.94\text{m}\end{array}$

Substitute $\text{6}{\text{m}}^2$ for $A_{\text{c1}}$, $\text{10}{\text{m}}^2$ for $A_{\text{c2}}$, $1\text{m}$ for $R_{h1}$, $0.909\text{m}$ for $R_{h2}$ $1{\text{m}}^{\text{1}/\text{3}}/\text{s}$ for $a$, $0.022$ for $n$ and $0.5°$ for $\alpha$ in Equation (II).

  $\begin{array}{c}\dot{V}=\frac{a}{n}\tan \alpha \left(A_{\text{c1}}{R}_{\text{h1}}^{2/3}+A_{\text{c2}}{R}_{\text{h2}}^{2/3}\right)\\ =\frac{1{\text{m}}^{\text{1}/\text{3}}/\text{s}}{0.022}\tan 0.5°\left(\left(\text{6}{\text{m}}^2\right){\left(1\text{m}\right)}^{2/3}+\left(\text{10}{\text{m}}^2\right)\left(0.909\text{m}\right)\right)\\ =0.396675{\text{m}}^{\text{1}/\text{3}}/\text{s}\left(\left(\text{6}{\text{m}}^2\right){\left(1\text{m}\right)}^{2/3}+\left(\text{10}{\text{m}}^2\right)\left(0.909\text{m}\right)\right)\\ =37.2{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $\text{16}{\text{m}}^2$ for ${A^{\prime }}_{\text{c}}$, $11\text{m}$ for $p_2$, $0.941\text{m}$ for ${R^{\prime }}_h$$1{\text{m}}^{\text{1}/\text{3}}/\text{s}$ for $a$, $n^{\prime }$ for $n$, $37.2{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ and $0.5°$ for $\alpha$ in Equation (I).

  $\begin{array}{l}37.2{\text{m}}^{\text{3}}/\text{s}=\frac{1{\text{m}}^{\text{1}/\text{3}}/\text{s}}{n^{\prime }}\left(\text{16}{\text{m}}^2\right){\left(0.941\text{m}\right)}^{2/3}\tan 0.5°\\ n^{\prime }=\frac{\left(1{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)\left(\text{16}{\text{m}}^2\right){\left(0.941\text{m}\right)}^{2/3}\tan 0.5°}{37.2{\text{m}}^{\text{3}}/\text{s}}\\ n^{\prime }=0.0386\end{array}$

Conclusion:

The flow rate through the channel is $37.2{\text{m}}^{\text{3}}/\text{s}$.

The effective manning coefficient is $0.0386$.