Question

A most economical trapezoidal section is required to give a maximum discharge

of 2M mP/s of water. The slope of the channel bottom is 1 in 15MM. Taking

C = 7MI in Chezy’s equationI determine the dimension of the channel.
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Answer 1

Answer:

Conditions which should be satisfied by a trapezoidal section to be most efficient

1) Top width of channel (T) = 2 × Length of side slope

2) Hydraulic radius (R) = ½ × depth of flow (y)

3) Side slope = 60°

Formulas to be remembered

1) \(R = \frac{y}{2}\)

2) \(Width\;of\;channel\;\left( B \right) = \frac{{2y}}{{\sqrt 3 }}\)

3) \(Area\;of\;channel = \sqrt 3 \times {y^2}\)

Relation between manning coefficient and Chezy constant is given by

\(n = \frac{{{R^{\frac{1}{6}}}}}{C}\)

We know, from chezy equation

\(V = C \times \sqrt {R \times S} \)

\(Q = A \times C \times \sqrt {R \times S} \)

Calculation

Given,

Q = 21.5 m3/s, S = 1 in 2500, C = 70 m1/2/s

\(Q = A \times C \times \sqrt {R \times S} \)

\(21.5 = \sqrt 3 \times {y^2} \times C \times \sqrt {\frac{y}{2}} \times \sqrt S \)

\(21.5 = \sqrt 3 \times {y^2} \times 70 \times \sqrt {\frac{y}{2}} \times \sqrt {\frac{1}{{2500}}} \)

\(21.5 = \sqrt 3 \times {y^2} \times 70 \times \sqrt {\frac{y}{2}} \times \frac{1}{{50}}\)

\({y^{\frac{5}{2}}} = \frac{{21.5\; \times \;\sqrt 2\; \times \;50}}{{70\sqrt 3 }}\)

y = 2.75 m

∴ Depth of the flow is 2.75 m

Now, R = y/2 = 2.75/2 = 1.375 m

∴ Hydraulic Radius is 1.375 m

\(B = \frac{{2y}}{{\sqrt 3 }} = 2 \times \frac{{2.75}}{{\sqrt 3 }} = 3.175\;m\)

Value of manning’s coefficient

\(n = \frac{{{R^{\frac{1}{6}}}}}{C} = \frac{{{{1.375}^{\frac{1}{6}}}}}{{70}} = 0.015\)

∴ manning’s coefficient = 0.015