Chapter 13, Problem 61P

Repeat Prob. 13-60 for a weedy excavated earth channel with n = 0.030.

To determine

The depth.

Answer to Problem 61P

The depth is $1.073\text{m}$.

Explanation of Solution

Given information:

The channel is made of unfinished concrete, the bottom angle of trapezoidal channel is $1°$, the base width is $5\text{m}$, the side surface slop is $1:1$, the flow rate is $25{\text{m}}^{\text{3}}/\text{s}$, the angle is $45°$ and the manning coefficient is $0.030$.

Below figure represent the cross- section of the trapezoidal channel.

Figure-1

Write the expression of height of water in the channel as shown in Figure-1.

  $h=x\tan \theta$........ (I)

Here, the parallel distance of water from below corner is $x$ and the angle is $\theta$.

Substitute $45°$ for $\theta$ in Equation (I).

  $\begin{array}{c}h=x\tan 45°\\ =x\times 1\\ =x\end{array}$

Write the expression of surface width of the channel as shown in Figure-1.

  $c=\left(\text{5}\text{m}\right)+2x$........ (II)

Here, the parallel distance of water from below corner is $x$.

Substitute $x$ for $h$ in Equation (II).

  $c=\left(5\text{m}\right)+2h$

Write the expression of bottom slop.

  $S_0=\tan \alpha$........ (III)

Here, the slop angle or bottom angle is $\alpha$.

Substitute $1°$ for $\alpha$ in Equation (III).

  $\begin{array}{c}{S}_0=\tan 1°\\ =0.0175\end{array}$

Write the expression of area of cross -section of flow.

  $A_c=\left(\frac{b+c}{2}\right)h$........ (IV)

Here, the width is $b$, the surface width of channel is $c$ and height of the channel is $h$.

Substitute $5\text{m}$ for $b$ and $\left(5\text{m}\right)+2h$ for $c$ in Equation (IV).

  $\begin{array}{c}{A}_c=\left(\frac{\left(5\text{m}\right)+\left(5\text{m}\right)+2h}{2}\right)h\\ =\left(\frac{10\text{m+2h}}{2}\right)h\\ =\left\{\left(5\text{m}\right)\text{+h}\right\}h\\ =\left(5\text{m}\right)h+h^2\end{array}$

Write the expression of wetted perimeter of the channel.

  $p=\frac{h}{\sin \theta }+b+\frac{h}{\sin \theta }$........ (V)

Here, the height of the channel is $h$, the width of the channel is $b$ and the angle is $\theta$.

Substitute $5\text{m}$ for $b$ and $45°$ for $\theta$ in Equation (V).

  $\begin{array}{c}p=\frac{h}{\sin 45°}+\left(5\text{m}\right)+\frac{h}{\sin 45°}\\ =\frac{h}{0.707}+\left(5\text{m}\right)+\frac{h}{0.707}\\ =\frac{2h}{0.707}+\left(5\text{m}\right)\\ =2.828h+\left(5\text{m}\right)\end{array}$

Write the expression of hydraulic radius of the channel.

  $R_h=\frac{A_c}{p}$........ (VI)

Here, the cross-sectional area of flow is $A_c$ and the perimeter of the channel is $p$.

Substitute $\left(5\text{m}\right)h+h^2$ for $A_c$ and $2.828h+\left(5\text{m}\right)$ for $p$ in Equation (VI).

  $R_h=\frac{\left(5\text{m}\right)h+h^2}{2.828h+\left(5\text{m}\right)}$

Write the expression of flow rate.

  $\dot{V}=\frac{a}{n}{A}_c{R}_h{}^{2/3}{S}_0{}^{1/2}$........ (VII)

Here, the constant is $a$, the manning coefficient is $n$, the area of cross -section of flow is $A_c$, the hydraulic radius is $R_h$ and the slop is $S_0$.

Substitute $25{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $1{\text{m}}^{\text{1}/\text{3}}/\text{s}$ for $a$, $0.030$ for $n$, $\left(5\text{m}\right)h+h^2$ for $A_c$, $\frac{\left(5\text{m}\right)h+h^2}{2.828h+\left(5\text{m}\right)}$ for $R_h$ and $0.0175$ for $S_0$ in Equation (VII).

  $\begin{array}{l}25{\text{m}}^{\text{3}}/\text{s}=\frac{\left(1{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)}{0.030}\times \left\{\left(5\text{m}\right)h+h^2\right\}{\left\{\frac{\left(5\text{m}\right)h+h^2}{2.828h+\left(5\text{m}\right)}\right\}}^{2/3}{\left(0.0175\right)}^{1/2}\\ 25{\text{m}}^{\text{3}}/\text{s}=\left(33.33{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)\times \left(0.1322\right)\times \frac{\left\{\left(5\text{m}\right)h+h^2\right\}{\left\{\left(5\text{m}\right)h+h^2\right\}}^{2/3}}{{\left\{2.828h+\left(5\text{m}\right)\right\}}^{2/3}}\\ 25{\text{m}}^{\text{3}}/\text{s}=\left(4.4{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)\times \frac{{\left\{\left(5\text{m}\right)h+h^2\right\}}^{5/3}}{{\left\{2.828h+\left(5\text{m}\right)\right\}}^{2/3}}\\ \frac{25{\text{m}}^{\text{3}}/\text{s}}{\left(4.4{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)}=\frac{{\left\{\left(5\text{m}\right)h+h^2\right\}}^{5/3}}{{\left\{2.828h+\left(5\text{m}\right)\right\}}^{2/3}}\end{array}$

  $\begin{array}{l}\left(5.68{\text{m}}^{\text{2}\text{.66}}\right)=\frac{{\left\{\left(5\text{m}\right)h+h^2\right\}}^{5/3}}{{\left\{2.828h+\left(5\text{m}\right)\right\}}^{2/3}}\\ \left(5.68{\text{m}}^{\text{2}\text{.66}}\right)\times {\left\{2.828h+\left(5\text{m}\right)\right\}}^{2/3}={\left\{\left(5\text{m}\right)h+h^2\right\}}^{5/3}\\ {\left\{\left(5\text{m}\right)h+h^2\right\}}^{5/3}-\left(5.68{\text{m}}^{\text{2}\text{.66}}\right)\times {\left\{2.828h+\left(5\text{m}\right)\right\}}^{2/3}=0\end{array}$........ (VIII)

Apply hit and trial method in Equation (VIII).

Substitute $1\text{m}$ for $h$ in Equation (VIII).

  $\begin{array}{l}{\left\{\left(5\text{m}\right)\times \left(1\text{m}\right)+{\left(1\text{m}\right)}^2\right\}}^{5/3}-\left(5.68{\text{m}}^{\text{2}\text{.66}}\right)\times {\left\{2.828\left(1\text{m}\right)+\left(5\text{m}\right)\right\}}^{2/3}=0\\ {\left(6{\text{m}}^{\text{2}}\right)}^{5/3}-\left(5.68{\text{m}}^{\text{2}\text{.66}}\right)\times {\left(7.828\text{m}\right)}^{2/3}=0\\ \left(36{\text{m}}^{3.32}\right)-\left(44.46{\text{m}}^{3.32}\right)=0\\ -8.46{\text{m}}^{3.32}\ne 0\end{array}$

Hence $1\text{m}$ for $h$ does not satisfy the Equation.

Substitute $1.073\text{m}$ for $h$ in Equation (VIII).

  $\begin{array}{l}{\left\{\left(5\text{m}\right)\times \left(1.073\text{m}\right)+{\left(1.073\text{m}\right)}^2\right\}}^{5/3}-\left(5.68{\text{m}}^{\text{2}\text{.66}}\right)\times {\left\{2.828\times \left(1.073\text{m}\right)+\left(5\text{m}\right)\right\}}^{2/3}=0\\ {\left(6.52{\text{m}}^{\text{2}}\right)}^{5/3}-\left(5.68{\text{m}}^{\text{2}\text{.66}}\right)\times {\left(4\text{m}\right)}^{2/3}=0\\ \left(22.74{\text{m}}^{3.32}\right)-\left(22.74{\text{m}}^{3.32}\right)=0\\ 0=0\end{array}$

Hence $1.073\text{m}$ for $h$ satisfies the equation.

Conclusion:

The depth is $1.073\text{m}$.