Chapter 13, Problem 54P

To determine

The maximum flow rate corresponding to maximum channel height.

Answer to Problem 54P

The maximum flow rate corresponding to maximum channel height is $244{\text{m}}^{\text{3}}/\text{sec}$

Explanation of Solution

Given information:

The maximum height of the channel is $3.2\text{m}$.

Write the expression for the cross- sectional area of the channel.

   $A_C=\left(\frac{a+b}{2}\right)h$     ...... (I)

Here, the top length of the channel is $a$, bottom length is $b$ and the height of liquid in channel is $h$.

Write the expression for total volume flow rate through channel 1:

   ${\dot{V}}_1=\frac{a}{n}{A}_{C_1}{R}_{h_1}{}^{2/3}{S}_0{}^{1/2}$     ...... (II)

Here, the slope of the channel is $S_0$, the surface roughness is $n$, hydraulic radius though channel $1$ is $R_{h_1}$, cross- section area of flow through channel $1$

   $A_{C_1}$, and the constant is $a$.

Write the expression for total volume flow rate through channel 2

   ${\dot{V}}_2=\frac{a}{n}{A}_{C_2}{R}_{h_2}{}^{2/3}{S}_0{}^{1/2}$     ...... (III)

Here, the slope of the channel is $S_0$, the surface roughness is $n$, hydraulic radius though channel $2$ is $R_{h_2}$, cross -section area of flow through channel $2$

   $A_{C_2}$, and the constant is $a$.

Divide Equation (II) and Equation (III)

   $\begin{array}{l}\frac{{\dot{V}}_1=\frac{a}{n}{A}_{C_1}{R}_{h_1}{}^{2/3}{S}_0{}^{1/2}}{{\dot{V}}_2=\frac{a}{n}{A}_{C_2}{R}_{h_2}{}^{2/3}{S}_0{}^{1/2}}\\ \frac{{\dot{V}}_1}{{\dot{V}}_2}=\frac{A_{C_1}{R}_{h_1}{}^{2/3}}{A_{C_2}{R}_{h_2}{}^{2/3}}\end{array}$     ...... (IV)

Write the expression for hydraulic depth.

   $R_h=\frac{A_C}{P}$     ...... (III)

Here, the perimeter is $P$ and the cross-section area is $A_C$.

Write the expression for the perimeter of channel.

   $P=\frac{a}{2}+2\sqrt{{\left(\frac{b}{2}\right)}^2+h^2}$     ...... (IV)

Calculation:

Substitute $12\text{m}$ for $a$, $6\text{m}$ for $b$, $2.2\text{m}$ for $h$ and $A_{C_1}$ for $A_C$ in Equation (I).

   $$

   $\begin{array}{c}{A}_{C_1}=\left(\frac{6\text{m}+12\text{m}}{2}\right)2.2\text{m}\\ =\left(9\times 2.2\right){\text{m}}^2\\ =19.8{\text{m}}^{\text{2}}\end{array}$

Substitute $12\text{m}$ for $a$, $6\text{m}$ for $b$, $2.2\text{m}$ for $h$ and $P_1$ for $P$ in Equation (IV).

   $\begin{array}{c}{P}_1=\left(6\text{m}+2\left(\sqrt{{\left(3\text{m}\right)}^2+{\left(2.2\text{m}\right)}^2}\right)\right)\\ \text{=6}\text{m}+7.44\text{m}\\ \text{=13}\text{.44}\text{m}\end{array}$

Substitute $13.44\text{m}$ for $P$, $19.8{\text{m}}^{\text{2}}$ for $A_C$ and $R_{h_1}$ for $R_h$ in Equation (III).

   $\begin{array}{c}{R}_{h_1}=\frac{19.8{\text{m}}^{\text{2}}}{13.44\text{m}}\\ =1.47\text{m}\end{array}$

Substitute $14.73\text{m}$ for $a$, $6\text{m}$ for $b$, $3.2\text{m}$ for $h$ and $A_{C_2}$ for $A_C$ in Equation (I)

   $\begin{array}{c}{A}_{C_2}=\left(\frac{14.73\text{m+6}\text{m}}{2}\right)3.2\text{m}\\ \text{=10}\text{.365}\text{m}\times \text{3}\text{.2}\text{m}\\ \text{=33}\text{.168}{\text{m}}^2\end{array}$

Substitute $12\text{m}$ for $a$, $\left(14.73\text{m}-\text{6}\text{m}\right)$ for $b$, $3.2\text{m}$ for $h$ and $P_2$ for $P$ in Equation (IV)

   $\begin{array}{c}{P}_2=\left(6\text{m}+2\left(\sqrt{{\left(3.2\text{m}\right)}^2+{\left(\frac{\left(14.73\text{m}-6\text{m}\right)}{2}\right)}^2}\right)\right)\\ =6\text{m}+10.82\text{m}\\ \text{=16}\text{.82}\text{m}\end{array}$

Substitute $16.82\text{m}$ for $P$, $33.168{\text{m}}^{\text{2}}$ for $A_C$ and $R_{h_2}$ for $R_h$ in Equation (III).

   $\begin{array}{c}{R}_{h_2}=\frac{33.168{\text{m}}^{\text{2}}}{16.82\text{m}}\\ =1.97\text{m}\end{array}$

Substitute $19.8{\text{m}}^{\text{2}}$ for $A_{C_1}$, $33.168{\text{m}}^{\text{2}}$ for $A_{C_2}$, $1.47\text{m}$ for $R_{h_1}$, $1.97\text{m}$ for $R_{h_2}$ and $120{\text{m}}^{\text{3}}/\text{sec}$ for ${\dot{V}}_1$ in Equation (IV).

   $\begin{array}{l}\frac{120{\text{m}}^{\text{3}}/\text{sec}}{{\dot{V}}_2}=\frac{19.8{\text{m}}^{\text{2}}\times {\left(1.47\text{m}\right)}^{2/3}}{33.168{\text{m}}^{\text{2}}\times {\left(1.97\text{m}\right)}^{2/3}}\\ \frac{120{\text{m}}^{\text{3}}/\text{sec}}{{\dot{V}}_2}=0.5969\times 0.822\\ {\dot{V}}_2=244{\text{m}}^{\text{3}}/\text{sec}\end{array}$

Conclusion:

The maximum flow rate corresponding to maximum channel height is $244{\text{m}}^{\text{3}}/\text{sec}$.