Chapter 13, Problem 62P

To determine

(a)

The dimension for the circular channel.

Answer to Problem 62P

The diameter of the circular channel is $3.446\text{m}$.

The depth of the circular channel is $1.732\text{m}$.

Explanation of Solution

Given Information:

The water flow rate is $10{\text{m}}^{\text{3}}/\text{s}$ and the bottom slope is $0.0015$

Write the expression for the volume flow rate of the water.

  $Q=\frac{a}{n}{A}_c{R}_h^{2/3}{S}_o^{1/2}$  ...... (I)

Here, the manning coefficient is $n$, the constant is $a$, the hydraulic radius is $R_h$, the bottom slope is $S_o$ and the cross sectional flow area is $A_c$.

Write the expression for the radius of the semicircle.

  $R_h=\frac{A_c}{P}$...... (II)

Here, the diameter of the semicircle is $D$.

Write the expression for the area of the circular channel.

  $A_c=\frac{\pi }{8}{D}^2$...... (III)

Here, the diameter of the circle is $D$.

Write the expression for the perimeter of the circular channel.

  $P=\frac{\pi }{2}D$...... (IV)

Substitute $\frac{\pi }{8}{D}^2$ for $A_c$ and $\frac{\pi }{2}D$ for $P$ in Equation (II).

  $\begin{array}{c}{R}_h=\frac{\frac{\pi }{8}{D}^2}{\frac{\pi }{2}D}\\ =\frac{D}{4}\end{array}$...... (V)

Write the expression for the flow depth of the circular channel.

  $y=\frac{D}{2}$...... (VI)

Calculation:

Refer to table 13-1 "Mean values of the manning coefficient $n$ for water flow in open channels" to obtain the manning coefficient as $0.016$.

Substitute $\frac{D}{4}$ for $R_h$, $1{\text{m}}^{\text{1}/\text{3}}/\text{s}$ for $a$, $0.016$ for $n$, $\frac{\pi }{8}{D}^2$ for $A_c$, $0.0015$ for $S_o$ and $10{\text{m}}^{\text{3}}/\text{s}$ for $Q$ in Equation (I).

  $\begin{array}{l}10{\text{m}}^{\text{3}}/\text{s}=\frac{\left(1{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)}{0.016}\left(\frac{\pi }{8}{D}^2\right){\left(\frac{D}{4}\right)}^{2/3}{\left(0.0015\right)}^{1/2}\\ 10{\text{m}}^{\text{3}}/\text{s}=62.5{\text{m}}^{\text{1}/\text{3}}/\text{s}\left(0.3926D^2\right)\left(0.3967D^{2/3}\right)\left(0.03872\right)\\ 10{\text{m}}^{\text{3}}/\text{s}=\left(0.369{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)D^{8/3}\\ D=3.446\text{m}\end{array}$

Substitute $3.446\text{m}$ for $D$ in Equation (VI).

  $\begin{array}{c}y=\frac{3.446\text{m}}{2}\\ =1.732\text{m}\end{array}$

Conclusion:

The diameter of the circular channel is $3.446\text{m}$.

The depth of the circular channel is $1.732\text{m}$.

To determine

(b)

The dimension for the rectangular channel.

Answer to Problem 62P

The dimension for the rectangular channel are-

The bottom breadth of the rectangular channel is $3.124\text{m}$.

The depth of the rectangular channel is $1.562\text{m}$.

Explanation of Solution

Given Information:

The water flow rate is $10{\text{m}}^{\text{3}}/\text{s}$ and the bottom slope is $0.0015$

Write the expression for the area of the rectangular channel.

  $A_c=yb$...... (VII)

Here, the depth of the rectangular is $y$ and the bottom breath of the rectangular channel is $b$.

Write the expression for the flow depth of the rectangular channel.

  $y=\frac{b}{2}$...... (VIII)

Substitute $\frac{b}{2}$ for $y$ in Equation (VII).

  $\begin{array}{c}{A}_c=b\left(\frac{b}{2}\right)\\ =\frac{b^2}{2}\end{array}$...... (IX)

Write the expression for the perimeter of the rectangular channel.

  $P=b+2y$...... (X)

Substitute $\frac{b}{2}$ for $y$ in Equation (X).

  $\begin{array}{c}P=b+2\left(\frac{b}{2}\right)\\ =2b\end{array}$...... (XI)

Substitute $\frac{b^2}{2}$ for $A_c$ and $2b$ for $P$ in Equation (II).

  $\begin{array}{c}{R}_h=\frac{\frac{b^2}{2}}{2b}\\ =\frac{b}{4}\end{array}$...... (XII)

Calculation:

Substitute $\frac{b}{4}$ for $R_h$, $1{\text{m}}^{\text{1}/\text{3}}/\text{s}$ for $a$, $0.016$ for $n$, $\frac{b^2}{2}$ for $A_c$, $0.0015$ for $S_o$ and $10{\text{m}}^{\text{3}}/\text{s}$ for $Q$ in Equation (I).

  $\begin{array}{l}10{\text{m}}^{\text{3}}/\text{s}=\frac{\left(1{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)}{0.016}\left(\frac{b^2}{2}\right){\left(\frac{b}{4}\right)}^{2/3}{\left(0.0015\right)}^{1/2}\\ 10{\text{m}}^{\text{3}}/\text{s}=62.5{\text{m}}^{\text{1}/\text{3}}/\text{s}\left(0.5b^2\right)\left(0.3967b^{2/3}\right)\left(0.03872\right)\\ 20.864=b^{8/3}\\ b=3.124\text{m}\end{array}$

Substitute $3.124\text{m}$ for $b$ in Equation (VIII).

  $\begin{array}{c}y=\frac{3.124\text{m}}{2}\\ =1.562\text{m}\end{array}$

Conclusion:

The bottom breath of the rectangular channel is $3.124\text{m}$.

The depth of the rectangular channel is $1.562\text{m}$.

To determine

(c)

The dimension for the trapezoidal channel.

Answer to Problem 62P

The bottom breath of the trapezoidal channel is $1.902\text{m}$.

The depth of the trapezoidal channel is $1.647\text{m}$.

Explanation of Solution

Given Information:

The water flow rate is $10{\text{m}}^{\text{3}}/\text{s}$ and the bottom slope is $0.0015$

Write the expression for the area of the trapezoidal channel.

  $A_c=y\left(b+\frac{y}{\tan \theta }\right)$...... (XIII)

Here, the trapezoidal angle is $\theta$, the depth of the trapezoidal is $y$ and the bottom breath of the trapezoidal channel is $b$.

Write the expression for the flow depth of the trapezoidal channel.

  $y=\frac{\sqrt{3}}{2}b$...... (XIV)

Write the expression for the inclined height of the trapezoidal section.

  $c=\frac{y}{\sin \theta }$...... (XV)

Write the expression for the perimeter of the trapezoidal channel.

  $P=b+2c$...... (XVI)

Calculation:

Substitute $60°$ for $\theta$ and $\frac{\sqrt{3}}{2}b$ for $y$ in Equation (XV).

  $\begin{array}{c}c=\frac{\frac{\sqrt{3}}{2}b}{\sin 60°}\\ =\frac{\frac{\sqrt{3}}{2}b}{\frac{\sqrt{3}}{2}}\\ =b\end{array}$

Substitute $60°$ for $\theta$ and $\frac{\sqrt{3}}{2}b$ for $y$ in Equation (XIII).

  $\begin{array}{c}{A}_c=\frac{\sqrt{3}}{2}b\left(b+\frac{\frac{\sqrt{3}}{2}b}{\tan 60°}\right)\\ =\frac{\sqrt{3}}{2}b\left(b+\frac{\frac{\sqrt{3}}{2}b}{\sqrt{3}}\right)\\ =\frac{3\sqrt{3}}{4}{b}^2\end{array}$

Substitute $b$ for $c$ in Equation (XVI).

  $\begin{array}{c}P=b+2b\\ =3b\end{array}$

Substitute $3b$ for $P$ and $\frac{3\sqrt{3}}{4}{b}^2$ for

  $A_c$ in Equation (II).

  $\begin{array}{c}{R}_h=\frac{\frac{3\sqrt{3}}{4}{b}^2}{3b}\\ =\frac{\sqrt{3}}{4}b\end{array}$

Substitute $\frac{\sqrt{3}}{4}b$ for $R_h$, $1{\text{m}}^{\text{1}/\text{3}}/\text{s}$ for $a$, $0.016$ for $n$, $\frac{3\sqrt{3}}{4}{b}^2$ for $A_c$, $0.0015$ for $S_o$ and $10{\text{m}}^{\text{3}}/\text{s}$ for $Q$ in Equation (I).

  $\begin{array}{l}10{\text{m}}^{\text{3}}/\text{s}=\frac{\left(1{\text{m}}^{\text{1}/\text{3}}/\text{s}\right)}{0.016}\left(\frac{3\sqrt{3}}{4}{b}^2\right){\left(\frac{\sqrt{3}}{4}b\right)}^{2/3}{\left(0.0015\right)}^{1/2}\\ 10{\text{m}}^{\text{3}}/\text{s}=62.5{\text{m}}^{\text{1}/\text{3}}/\text{s}\left(1.299b^2\right)\left(0.5723b^{2/3}\right)\left(0.03872\right)\\ 5.55=b^{8/3}\\ b=1.902\text{m}\end{array}$

Substitute $1.902\text{m}$ for $b$ in Equation (XIV).

  $\begin{array}{c}y=\frac{\sqrt{3}}{2}\left(1.902\text{m}\right)\\ =0.8660\left(1.902\text{m}\right)\\ =1.647\text{m}\end{array}$

Conclusion:

The bottom breath of the trapezoidal channel is $1.902\text{m}$.

The depth of the trapezoidal channel is $1.647\text{m}$.