Chapter 13, Problem 59P

To determine

The diameter of the third channel.

Answer to Problem 59P

The diameter of the channel 3 is $2.33\text{m}$.

Explanation of Solution

Given Information:

The figure below represents the cross -section of three channels.

  

  Figure-(1)

The diameter of channel 1 is $1.8\text{m}$, the diameter of channel 2 is $\text{1}\text{.8}\text{m}$ and the slope of the draining system is $0.0025$.

Write the expression for flow rate from channel 1 and channel 2.

  $\dot{V}=\frac{a}{n_1}{A}_{c1}{R}_{h1}{}^{\frac{2}{3}}{S}_o{}^{\frac{1}{2}}+\frac{a}{n_2}{A}_{c2}{R}_{h2}{}^{\frac{2}{3}}{S}_o{}^{\frac{1}{2}}$...... (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 $R_{h1}$ 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 $R_{h2}$.

Write the expression for flow rate through the channel 3.

  ${\dot{V}}^{\prime }=\frac{a}{n_3}{A}_{c3}{\left(R_{c3}\right)}^{\frac{2}{3}}{\left(S_o\right)}^{\frac{1}{2}}$...... (II)

Here, the manning's coefficient for channel 3 is $n_3$, the hydraulic radius for channel 3 is $R_{c3}$, the area of the channel 3 is $A_{c3}$.

Write the expression for the cross sectional flow area of channel1.

  $A_{c1}=\frac{\pi D_1^2}{8}$...... (III)

Here, the radius of the channel 1 is $D_1$.

Write the expression for the cross sectional flow area of channel 2.

  $A_{c2}=\frac{\pi D_2^2}{8}$...... (IV)

Here, the radius of the channel 2 is $D_2$.

Write the expression for the cross sectional flow area of channel 3.

  $A_{c3}=\frac{\pi D_3^2}{8}$...... (V)

Here, the radius of the channel 3 is $D_3$.

Write the expression to calculate the wetted perimeter of the channel 1.

  $$

  $p_1=\frac{\pi D_1}{2}$...... (VI)

Write the expression to calculate the wetted perimeter of the channel 2.

  $p_2=\frac{\pi D_2}{2}$...... (VII)

Write the expression to calculate the wetted perimeter of the channel 3.

  $p_3=\frac{\pi D_3}{2}$...... (VIII)

Write the expression to calculate the hydraulic radius of channel 1.

  $R_{h1}=\frac{A_{c1}}{p_1}$...... (IX)

Write the expression to calculate the hydraulic radius of channel 2.

  $R_{h2}=\frac{A_{c2}}{p_2}$...... (X)

Write the expression to calculate the hydraulic radius of channel 3.

  $R_{h3}=\frac{A_{c3}}{p_3}$...... (XI)

Write the expression to calculate the diameter of the channel 3.

  $D_3=2R_3$...... (XII)

Calculation:

Substitute $\text{1}\text{.8}\text{m}$ for $D_1$ in Equation (III).

  $\begin{array}{c}{A}_{c1}=\frac{\pi {\left(1.8\text{m}\right)}^2}{8}\\ =\frac{10.17{\text{m}}^{\text{2}}}{8}\\ =1.272{\text{m}}^{\text{2}}\end{array}$

Substitute $\text{1}\text{.8}\text{m}$ for $D_2$ in Equation (IV).

  $\begin{array}{c}{A}_{c2}=\frac{\pi {\left(1.8\text{m}\right)}^2}{8}\\ =\frac{10.17{\text{m}}^{\text{2}}}{8}\\ =1.272{\text{m}}^{\text{2}}\end{array}$

Substitute $\text{1}\text{.8}\text{m}$ for $D_3$ in Equation (VI).

  $\begin{array}{c}{p}_1=\pi \left(\frac{1.8\text{m}}{2}\right)\\ =2.82\text{m}\end{array}$

Substitute $\text{0}\text{.9}\text{m}$ for $R_2$ in Equation (VII).

  $\begin{array}{c}{p}_1=\pi \left(\frac{1.8\text{m}}{2}\right)\\ =2.82\text{m}\end{array}$

Substitute $1.272{\text{m}}^{\text{2}}$ for $A_{c1}$ and $\text{2}\text{.82}\text{m}$ for $p_1$ in Equation (IX).

  $\begin{array}{c}{R}_{h1}=\frac{1.272{\text{m}}^{\text{2}}}{2.82\text{m}}\\ =0.45\text{m}\end{array}$

Substitute $1.272{\text{m}}^{\text{2}}$ for $A_{c2}$ and $\text{2}\text{.82}\text{m}$ for $p_2$ in Equation (X).

  $\begin{array}{c}{R}_{h2}=\frac{1.272{\text{m}}^{\text{2}}}{2.82\text{m}}\\ =0.45\text{m}\end{array}$

Substitute $\frac{\pi R_3^2}{2}$ for $A_{c3}$ and $\pi R_3$ for $p_3$ in Equation (XI).

  $\begin{array}{c}{R}_{h3}=\frac{\frac{\pi R_3^2}{2}}{\pi R_3}\\ =\frac{R_3}{2}\end{array}$

Refer to Table (13.1), "Mean values of the Manning Coefficient for water flow in open channels" to obtain the value of $n_1$ as $0.012$ and $n_2$ as $0.012$.

Substitute $\text{1}{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}$ for $a$, $1.272{\text{m}}^{\text{2}}$ for $A_{c1}$, $1.272{\text{m}}^{\text{2}}$ for $A_{c2}$, 0.012 for $n_1$, 0.012 for $n_2$, $0.45\text{m}$ for $R_{h1}$

  $0.45\text{m}$ for $R_{h2}$ and $0.0025$ for $S_o$ in Equation (I).

  $\begin{array}{c}\dot{V}=\left[\begin{array}{l}\frac{\text{1}{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{0.012}\times 1.272{\text{m}}^{\text{2}}\times {\left(0.45\text{m}\right)}^{\frac{2}{3}}{\left(0.0025\right)}^{\frac{1}{2}}\\ +\frac{\text{1}{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{0.012}\times 1.272{\text{m}}^{\text{2}}\times {\left(0.45\text{m}\right)}^{\frac{2}{3}}{\left(0.0025\right)}^{\frac{1}{2}}\end{array}\right]\\ =2\left(\frac{\text{1}{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{0.012}\times 1.272{\text{m}}^{\text{2}}\times {\left(0.45\text{m}\right)}^{\frac{2}{3}}{\left(0.0025\right)}^{\frac{1}{2}}\right)\\ =6.22{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $6.22{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}^{\prime }$, $\text{1}{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}$ for $a$, 0.012 for $n$, $\frac{R_3}{2}$ for $R_{h3}$, $\frac{\pi R_3^2}{2}$ for $A_{c3}$ and 0.0025 for $S_o$ in Equation(II).

  $\begin{array}{l}6.22{\text{m}}^{\text{3}}/\text{s}=\frac{\text{1}{\text{m}}^{\frac{\text{1}}{\text{3}}}/\text{s}}{0.012}\left(\frac{\pi R_3^2}{2}\right){\left(\frac{R_3}{2}\right)}^{\frac{2}{3}}{\left(0.0025\right)}^{\frac{1}{2}}\\ 6.22{\text{m}}^{\text{3}}/\text{s}=\left(83.33\times 0.5\pi \times {\left(0.5\right)}^{\frac{2}{3}}\times 0.005\right)R_3^{\frac{8}{3}}\\ R_3=1.166\text{m}\end{array}$

Substitute $1.166\text{m}$ for $R_3$ in Equation (XII).

  $\begin{array}{c}{D}_3=2\times 1.166\text{m}\\ =2.33\text{m}\end{array}$

Conclusion:

The diameter of the channel 3 is $2.33\text{m}$.