A trapezoidal channel has side slope 1 in 1. It is desired to convey discharge of 13.75 $m^3/sec$ of water with bed gradient 1 in 1000. If unlined Chezy’s constant is taken as 44 and if lined, Chezy’s constant is 60. The cost per m3 of excavation is 4 times cost of per $m^2$ of lining.

The channel has to be designed efficiently, find whether lined canal or unlined canal will be cheaper, Also find the dimensions of most efficient section.

Solution :

Given :

Side slope, $\quad n=\frac{1}{1}=1$

Discharge, $\quad Q=13.75 \mathrm{~m}^{3} / \mathrm{s}$

Slope of bed, $\quad i=\frac{1}{1000}$

For unlined, $\quad C=44 $

For lined, $\quad C=60$

Let the cost per $\mathrm{m}^{2}$ of lining $=x$

Then cost per $\mathrm{m}^{3}$ of excavation $=4 x$

As the channel is most efficient,

Hydraulic mean depth,

$$m=\frac{d}{2}$$

where, $d=$ depth of channel

Let $\quad b=$ width of channel

Also for the most efficient trapezoidal channel,we have

Half of top width = length of sloping side

$$or$$

$\begin {aligned}\frac{b+2 n d}{2} &=d \sqrt{n^{2}+1}\\ or \\ \quad \frac{b+2 \times 1 \times d}{2} &=d \sqrt{1^{2}+1}=\sqrt{2 d} \\ b &=2 \times \sqrt{2 d}-2 d\\ &=0.828 d ....(i)\\ \text { Area, } \\ \quad A &=(b+n d) \times d\\ &=(0.828 d+1 \times d) \times d \\ &=1.828 d^{2} .......(ii) \end {aligned}$

1) For unlined channel

Value of C=44

The discharge, $Q$ is given by, or $ \begin{aligned} \space\space\space\space\space\space Q =A \times V =A \times C \sqrt{mi} \end{aligned}$ $\begin{aligned}13.75 =1.828 d^{2} \times 44 \times \sqrt{\frac{d}{2} \times \frac{1}{1000}} \end{aligned}$ $\left(\because A =1.828 d^{2}, m=\frac{d}{2}\right) $ $\space\space\space\space\space\space\space\space\space\space \begin{aligned}=\frac{1.828 \times 44}{\sqrt{2000}} \times d^{5 / 2} \end{aligned}$ $\begin{aligned} \space d^{5 / 2} =\frac{13.75 \times \sqrt{2000}}{1.828 \times 44} =7.6452 \end{aligned} $ $\begin{aligned}\space\space\space\space\space\space d =(7.6452)^{2 / 5}\ =2.256 \mathrm{~m} \end{aligned} $ Substituting this value in equation $(i)$, we get $$ b=0.828 \times 2.256=1.868 \mathrm{~m} $$

Now cost of excavation per running metre length of unlined channel

= Volume of channel $\times$ cost per $\mathrm{m}^{3}$ of excavation

$=( Area \space of \space channel \times 1) \times 4 x=[(b+n d) \times d \times 1] \times 4 x$

$=(1.868+1 \times 2.256) \times 2.256 \times 1 \times 4 x=37.215 x ..........(iii)$

2) For lined channels

Value of C=60

The discharge is given by the equation,

$Q=A \times C \times \sqrt{m i}$

Substituting the value of $A$ from equation (ii) and value of $m=\frac{d}{2}$, we get $$\begin{aligned} 13.75 =1.828 d^{2} × 60 × \sqrt{\frac{d}{2} × \frac{1}{1000}} \end{aligned}$$ $$ \space\space\space\space\space\space
\space\space \begin{aligned} =1.828 × 60 × \frac{1}{\sqrt{2000}} × d^{5 / 2} \end{aligned}$$ $\space\space\space\space \begin{aligned} d^{5 / 2} =\frac{13.75 × \sqrt{2000}}{1.828 × 60}=5.606 \end{aligned}$ $\space\space\space\space\space\space \space\space\space\begin{aligned} d =(5.606)^{2 / 5}=1.992 \mathrm{~m} \end{aligned}$

( $\because Q=13.75)$

Substituting this value in equation (i), we get

$$b=0.828 \times 1.992=1.649 \mathrm{~m} $$

In case of lined channel, the cost of lining as well as cost of excavation is to be consid Now cost of excavation $= $(Volume of channel $) \times$ cost per $\mathrm{m}^{3}$ of excavation

$ \begin{aligned} &=(b+n d) \times d \times 1 \times 4 x \\ &=(1.649+1 \times 1.992) \times 1.992 \times 4 x\\ &=29.01 x \end{aligned} $

Cost of lining

$=$ Area of lining $\times $ cost per ${m}^{2} $ of lining $ \begin{aligned} =(\text { Perimeter of lining } \times 1) \times x \ =\left(b+2 d \sqrt{1+n^{2}}\right) \times 1 \times x\ =\left(1.649+2 \times 1.992 \sqrt{1+1^{2}}\right) \times x \end{aligned} $ $ \begin{aligned} \text{Total cost } =29.01 x+7.283 x=36.293 x \end{aligned} $ The total cost of lined channel is $36.293 x$ whereas the total cost of unlined channel is $37.215 x$. Hence lined channel will be cheaper. The dimensions are $$b=1.649 \mathrm{~m} $$

$$d=1.992 \mathrm{~m}$$