A rectangular channel of 5 m has a depth of 1.5 m. The slope of bed channel is 1 in 1500. It is desired to increase the discharge to maximum by changing the dimension of the section for constant area of cross section, slope of the bed and roughness of the channel. Find the new dimensions of the channel and increase in discharge. Assume other values if not given.?

Question

Answer

Problem Statement

A rectangular channel with a width of 5 m and a depth of 1.5 m has a slope of 1 in 1500. The objective is to increase the discharge to its maximum by changing the dimensions of the section while keeping the area of cross-section, slope of the bed, and roughness of the channel constant. Determine the new dimensions of the channel and the increase in discharge.

Solution

Given,

Width of the rectangular channel (b) = 5 m

Depth of the rectangular channel (d) = 1.5 m

Slope of the bed (S) = 1/1500

Area of cross-section (A) = constant

Roughness of the channel = constant

Step 1: Calculation of the existing discharge

The existing discharge (Q) can be calculated using the Manning's equation as follows:

Q = (1/n) x A x (R^(2/3)) x S^(1/2)

where,

n = Manning's roughness coefficient

A = b x d = 5 x 1.5 = 7.5 m^2

R = A/P = A/(b+2d) = 7.5/(5+2x1.5) = 1.071 m

Assuming a Manning's roughness coefficient of 0.03 for the channel, the existing discharge can be calculated as follows:

Q = (1/0.03) x 7.5 x (1.071^(2/3)) x (1/1500)^(1/2) = 1.093 m^3/s

Step 2: Calculation of the new dimensions of the channel

Assuming that the area of cross-section is constant after changing the dimensions of the channel, the new dimensions can be calculated as follows:

A = b x d = constant = 7.5 m^2

Let the new width of the channel be (b1) and the new depth be (d1).

Therefore, b1 x d1 = 7.5

Solving for d1, we get:

d1 = 7.5/b1

Substituting the value of d1 in the Manning's equation, we get:

Q = (1/n) x b1 x (7.5/b1) x ((b1+2(7.5/b1))/3)^(2/3) x (1/1500)^(1/2)

Simplifying the equation, we get:

Q = (7.5/2.303) x ((b1^2)/(b1+15))^(2/3)

To maximize the discharge, we need to differentiate the

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