A power canal of trapezoidal section has to be excavated through hard clay at the least cost. Determine the dimensions of the channel given, discharge equal to 15 m³/s, bed slope 1 : 2000 and Manning’s, N = 0.020.

Solution

1. Conditions for a Most Economical Trapezoidal Section

For a trapezoidal channel to be most economical (least cost for excavation and lining), two conditions must be met:

1. Half of the top width is equal to the length of one sloping side: \( \frac{B + 2nd}{2} = d\sqrt{1+n^2} \)
2. The hydraulic mean depth is half the depth of flow: \(m = d/2\)

2. Express Area and Hydraulic Mean Depth in Terms of Depth (d)

First, use the first condition with \(n = 1/\sqrt{3}\) to find a relationship between the bed width (B) and the depth (d).

$$ \frac{B + 2(1/\sqrt{3})d}{2} = d\sqrt{1+(1/\sqrt{3})^2} $$ $$ \frac{B + 1.1547d}{2} = d\sqrt{1+1/3} = d\sqrt{4/3} = \frac{2d}{\sqrt{3}} \approx 1.1547d $$ $$ B + 1.1547d = 2.3094d \implies B = 1.1547d = \frac{2d}{\sqrt{3}} $$

Now, express the area in terms of d.

$$ A = (B + nd)d = \left(\frac{2d}{\sqrt{3}} + \frac{1}{\sqrt{3}}d\right)d = \left(\frac{3d}{\sqrt{3}}\right)d = \sqrt{3}d^2 \approx 1.732d^2 $$

3. Use Manning's Formula to Solve for Depth (d)

We start with the discharge equation and substitute the expressions for A and m.

$$ Q = A \times \frac{1}{N} m^{2/3} i^{1/2} $$ $$ 15 = (\sqrt{3}d^2) \times \frac{1}{0.020} \times \left(\frac{d}{2}\right)^{2/3} \times \left(\frac{1}{2000}\right)^{1/2} $$ $$ 15 = \frac{\sqrt{3}}{0.020 \times 2^{2/3} \times \sqrt{2000}} d^{2 + 2/3} $$ $$ 15 = \frac{1.732}{0.020 \times 1.587 \times 44.72} d^{8/3} $$ $$ 15 \approx 1.218 d^{8/3} $$ $$ d^{8/3} = \frac{15}{1.218} \approx 12.315 $$ $$ d = (12.315)^{3/8} \approx 2.5 \, \text{m} $$

4. Calculate the Bed Width (B)

Using the relationship found in step 2:

$$ B = \frac{2d}{\sqrt{3}} = \frac{2 \times 2.5}{\sqrt{3}} \approx 2.887 \, \text{m} $$

Explanation of "Least Cost" Design

The term "least cost" for excavating a channel implies designing the **most economical or most efficient section**. This is the cross-sectional shape that carries the maximum possible discharge for a given area.

• Minimized Wetted Perimeter: The efficiency is achieved by minimizing the wetted perimeter (the length of the channel bed and sides in contact with the water). A smaller wetted perimeter means less frictional resistance to flow.
• Optimal Side Slope: While the material (hard clay) can support various side slopes, the most hydraulically efficient of all trapezoidal shapes is a half-hexagon. This shape has side slopes that make a 60° angle with the horizontal. This is the condition used in the calculation because the problem asks for the "least cost" design without providing a specific, pre-determined side slope.