Chapter 13, Problem 54P

To determine

The flow rate through the channel and the equivalent manning's coefficient.

Answer to Problem 54P

The flow rate is $116.42{\text{m}}^{\text{3}}/\text{s}$ and the equivalent manning's coefficient is $0.0217$.

Explanation of Solution

Given Information:

The figure below shows the cross -section of the channel.

  

  Figure-(1)

The bottom slope of the channel is $0.002$.

Write the expression for flow rate.

  $\dot{V}=\frac{a}{n_1}{A}_{c1}{R}_1{}^{\frac{2}{3}}{S}_o{}^{\frac{1}{2}}+\frac{a}{n_2}{A}_{c2}{R}_2{}^{\frac{2}{3}}{S}_o{}^{\frac{1}{2}}$...... (I)

Here, the manning's coefficient for channel 1 is $n_1$, the manning's coefficient for channel 2 is $n_2$, the area of cross -section of channel 1 is $A_{c1}$, the hydraulic radius of channel 1 is $R_1$ and the bottom slope is $S_o$, the area of cross section of channel 2 is $A_{c2}$, the hydraulic radius of channel 1 is $R_2$.

Write the expression for area of channel 1.

  $A_{c1}=A_a+A_b+A_c+A_d$...... (II)

Here, the area of section a is $A_a$, the area of section b is $A_b$, the area of section c is $A_c$ and the area of section d is $A_d$.

Write the expression to calculate the area of channel a.

  $A_a=b_a\times h_a$...... (III)

Here, the breadth of section a is $b_a$ and the height of the section a is $h_a$.

Write the expression to calculate the area of channel b.

  $A_b=\frac{1}{2}\left(b_b\times h_b\right)$...... (IV)

Here, the breadth of section b is $b_b$ and the height of the section b is $h_b$.

Write the expression to calculate the area of channel c.

  $A_c=b_c\times h_c$...... (V)

Here, the breadth of section c is $b_c$ and the height of the section c is $h_c$.

Write the expression to calculate the area of channel d.

  $A_d=\frac{1}{2}\left(b_d\times h_d\right)$...... (VI)

Here, the breadth of section d is $b_d$ and the height of the section d is $h_d$.

Write the expression to calculate the perimeter of the concrete channel.

  $p_1=2\left(h_a+\sqrt{{\left(h_b\right)}^2+{\left(h_a\right)}^2}\right)$...... (VII)

Write the expression to calculate the perimeter of the light brush channel.

  $p_2=b+h_a$...... (VIII)

Here, the breadth of the light brush channel is $b$.

Write the expression to calculate the total perimeter.

  $p=p_1+p_2$...... (IX)

Write the expression to calculate the equivalent hydraulic radius.

  $R=\frac{A}{p}$...... (X)

Write the expression to calculate the area of light brush channel.

  $A_{c2}=b\times h_a$...... (XI)

Write the expression for total area.

  $A=A_{c1}+A_{c2}$...... (XII)

Write the expression for hydraulic radius for channel 1.

  $R_1=\frac{A_{c1}}{p_1}$...... (XIII)

Write the expression for the hydraulic radius for channel 2.

  $R_2=\frac{A_{c2}}{p_2}$...... (XIV)

Write the expression for equivalent flow rate.

  $\dot{V}=\frac{a}{n^{\prime }}A{\left(R\right)}^{\frac{2}{3}}{\left(S_0\right)}^{\frac{1}{2}}$...... (XV)

Here, the equivalent manning's coefficient is $n^{\prime }$.

Calculation:

Substitute $6\text{m}$ for $b_a$ and $1.5\text{m}$ for $h_b$ in Equation (III)

  $\begin{array}{c}{A}_a=6\text{m}\times 2\text{m}\\ \text{=12}{\text{m}}^{\text{2}}\end{array}$

Substitute $2\text{m}$ for $b_b$ and $1.5\text{m}$ for $h_a$ in Equation (IV)

  $\begin{array}{c}{A}_b=\frac{1}{2}\left(2\text{m}\times \text{1}\text{.5}\text{m}\right)\\ =1.5{\text{m}}^2\end{array}$

Substitute $2\text{m}$ for $b_c$ and $1.5\text{m}$ for $h_c$ in Equation (V)

  $\begin{array}{c}{A}_c=2\text{m}\times \text{1}\text{.5}\text{m}\\ =\text{3}{\text{m}}^2\end{array}$

Substitute $2\text{m}$ for $b_d$ and $1.5\text{m}$ for $h_d$ in Equation (VI)

  $\begin{array}{c}{A}_d=\frac{1}{2}\left(2\text{m}\times \text{1}\text{.5}\text{m}\right)\\ =1.5{\text{m}}^{\text{2}}\end{array}$

Substitute $12{\text{m}}^{\text{2}}$ for $A_a$, $1.5{\text{m}}^{\text{2}}$ for $A_b$, $3{\text{m}}^{\text{2}}$ for $A_c$, $1.5{\text{m}}^{\text{2}}$ for $A_d$ in Equation (II).

  $\begin{array}{c}{A}_{c1}=12{\text{m}}^{\text{2}}+1.5{\text{m}}^{\text{2}}+3{\text{m}}^{\text{2}}+1.5{\text{m}}^{\text{2}}\\ =18{\text{m}}^{\text{2}}\end{array}$

Substitute $10\text{m}$ for b and $2\text{m}$ for $h_a$ in Equation (XI)

  $\begin{array}{c}{A}_{c2}=10\text{m}\times \text{2}\text{m}\\ =20{\text{m}}^{\text{2}}\end{array}$

Substitute $18{\text{m}}^{\text{2}}$ for $A_{c1}$ and $20{\text{m}}^{\text{2}}$ for $A_{c2}$ in Equation (XII)

  $\begin{array}{c}A=18{\text{m}}^{\text{2}}+20{\text{m}}^{\text{2}}\\ =38{\text{m}}^{\text{2}}\end{array}$

Substitute $\text{2}\text{m}$ for $h_a$ and $1.5\text{m}$ for $h_b$ in Equation (VII)

  $\begin{array}{c}{p}_1=2\left(2\text{m}+\sqrt{{\left(2\text{m}\right)}^2+{\left(1.5\text{m}\right)}^2}\right)\\ =2\left(2\text{m}+2.5\text{m}\right)\\ =9\text{m}\end{array}$

Substitute $10\text{m}$ for $b$ and $2\text{m}$ for $h_a$ in Equation (VIII)

  $\begin{array}{c}{p}_2=10\text{m}+2\text{m}\\ =\text{12}\text{m}\end{array}$

Substitute $\text{9}\text{m}$ for $p_1$ and $12\text{m}$ for $p_2$ in Equation (IX).

  $\begin{array}{c}p=9\text{m}+\text{12}\text{m}\\ =\text{21}\text{m}\end{array}$

Substitute $21\text{m}$ for $p$ and $38{\text{m}}^{\text{2}}$ for $A$ in Equation (X).

  $\begin{array}{c}R=\frac{38{\text{m}}^{\text{2}}}{21\text{m}}\\ =1.81\text{m}\end{array}$

Substitute $\text{18}{\text{m}}^{\text{2}}$ for $A_{c1}$ and $\text{9}\text{m}$ for $p_1$ in Equation (XIII).

  $\begin{array}{c}{R}_1=\frac{18{\text{m}}^{\text{2}}}{9{\text{m}}^{\text{2}}}\\ =2\text{m}\end{array}$

Substitute $\text{20}{\text{m}}^{\text{2}}$ for $A_{c2}$ and $\text{12}\text{m}$ for $p_2$ in Equation (XIV).

  $\begin{array}{c}{R}_2=\frac{20{\text{m}}^{\text{2}}}{12\text{m}}\\ =1.667\text{m}\end{array}$

Refer to the Table (13.1), "Mean values of Manning coefficient for water flow open channels" to obtain the value of $n_1$

  $0.014$ and $n_2$ as $0.050$.

Substitute $0.014$ for $n_1$, $0.050$ for $n_2$, $1{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}$ for $a$

  $18{\text{m}}^{\text{2}}$ for $A_{c1}$, $20{\text{m}}^{\text{2}}$ for $A_{c2}$, 0.002 for $S_o$, $2\text{m}$ for $R_1$, $1.667\text{m}$ in Equation (III)

  $\begin{array}{c}\dot{V}=\frac{\text{1}{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{\text{0}\text{.014}}\times \left(18{\text{m}}^{\text{2}}\right){\left(2\text{m}\right)}^{\frac{2}{3}}{\left(0.002\right)}^{\frac{1}{2}}+\frac{1{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{0.050}\times \left(20{\text{m}}^{\text{2}}\right){\left(1.667\text{m}\right)}^{\frac{2}{3}}{\left(0.002\right)}^{\frac{1}{2}}\\ =91.27{\text{m}}^{\text{3}}/\text{s}+25.15{\text{m}}^{\text{3}}/\text{s}\\ =116.42{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $1{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}$ for $a$, $38{\text{m}}^{\text{2}}$ for $A$, $1.81\text{m}$ for $R$ and 0.002 $S_o$.and $116.42{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (XV).

  $\begin{array}{l}116.42{\text{m}}^{\text{3}}/\text{s}=\frac{1{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{n^{\prime }}\left(38{\text{m}}^{\text{2}}\right){\left(1.81\text{m}\right)}^{\frac{2}{3}}\left(0.002\right)\\ n^{\prime }=\frac{1{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{116.42{\text{m}}^{\text{3}}/\text{s}}\left(38{\text{m}}^{\text{2}}\right){\left(1.81\text{m}\right)}^{\frac{2}{3}}\left(0.002\right)\\ n^{\prime }=0.0217\end{array}$

Conclusion:

The flow rate is $116.42{\text{m}}^{\text{3}}/\text{s}$ and the equivalent manning's coefficient is $0.0217$.