Question

An unlined canal under regime conditions along with a slit factor of 1 has a width of flow 71.25 m. Assuming the unlined anal as a wide channel, the corresponding average depth of flow (in m, round off to two decimal places) in the canal will be

Answer 1

Explanation 

According to Lacey, the design formulas for canal design is as follows:

1) silt factor ⇒ \(\rm{f = 1.76√ {{d_{mm}}}}\)

2) velocity of flow ⇒ \(V = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}\)

3) hydraulic mean depth ⇒ \(R = \frac{{5{V^2}}}{{2f}}\)

4) wetted perimeter ⇒ \(P = 4.75√ Q\)

Calculation

Given,

Silt factor (f) = 1, Width of flow = 71.25 m

Depth of channel =?

To get the Depth of the channel we need to determine the area of the channel

It is mentioned that channel as a wide channel. 

a) For a wide channel, we know

Perimeter (P) ≈ Width of the channel (B)

∴ Perimeter P = 71.25 m

b) We know Perimeter P = 4.75√Q

71.25 = 4.75 √Q

Q = 225 m³/sec

c) Also we know. 

\(V = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}\)

→  \(V = {\left[ {\frac{{225×{1^2}}}{{140}}} \right]^{\frac{1}{6}}}\) = 1.08 m/sec

d) We know Q = AV

Area = Q/V= 225/1.08 = 208.33 m³/sec

e) Area = Width × Depth 

 208 = 71.25 × Depth 

Depth = 208/71.25 = 2.92 m

∴ Depth of the channel is 2.92 m