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Calculate the velocity of flow in a triangular channel having depth 7m and the side slope of the channel is 1H:4V if the bed slope of the channel is 1 in 1200 and the slope of the energy line is 0.00010. Given:( dy)/dx=7.55m.
  • a)
    1 m/s
  • b)
    2 m/s
  • c)
    3 m/s
  • d)
    4 m/s
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Calculate the velocity of flow in a triangular channel having depth 7m...
To calculate the velocity of flow in a triangular channel, we can use the Manning's equation, which relates the flow velocity to the channel slope, cross-sectional area, and hydraulic radius.

Given:
- Depth of the channel (y) = 7 m
- Side slope of the channel = 1H:4V
- Bed slope of the channel = 1 in 1200
- Slope of the energy line = 0.00010

First, let's calculate the channel slope (S) using the bed slope (So) and the side slope (Sf):

S = (So^2 + Sf^2)^0.5
S = (1/1200^2 + 1/4^2)^0.5
S = 0.000833

Next, we can calculate the cross-sectional area (A) of the channel using the depth (y) and the side slope (Sf):

A = y * (y * Sf + y / Sf)
A = 7 * (7 * 1/4 + 7 / 1/4)
A = 7 * (7/4 + 28/4)
A = 7 * (35/4)
A = 61.25 m^2

Now, let's calculate the hydraulic radius (R) using the cross-sectional area (A) and the wetted perimeter (P):

P = y * (1 + Sf^2)^0.5
P = 7 * (1 + (1/4)^2)^0.5
P = 7 * (1 + 1/16)^0.5
P = 7 * (17/16)^0.5
P = 7 * 1.03078
P = 7.21546 m

R = A / P
R = 61.25 / 7.21546
R = 8.498 m

Finally, we can calculate the velocity (V) using Manning's equation:

V = (1.486 / n) * R^(2/3) * S^(1/2)
V = (1.486 / n) * 8.498^(2/3) * 0.000833^(1/2)

Here, 'n' represents the Manning's roughness coefficient, which is not given in the problem statement. However, we can assume a typical value of 0.03 for a natural channel.

V = (1.486 / 0.03) * 8.498^(2/3) * 0.000833^(1/2)
V ≈ 1.00 m/s

Therefore, the velocity of flow in the triangular channel is approximately 1 m/s. Hence, the correct answer is option 'A'.
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Calculate the velocity of flow in a triangular channel having depth 7m...
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