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Water flows in a triangular channel of side slope 1 horizontal : 1 vertical and longitudinal slope of 0.001. Determine the type of channel slope when a discharge of 0.2 m3 /s flows through it. Assume Manning’s n = 0.015.
  • a)
    Steep
  • b)
    Mild
  • c)
    Critical
  • d)
    Cannot be determine
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Water flows in a triangular channel of side slope 1 horizontal : 1 ve...
Given data For a depth of flow of y in the channel
Area = y2
T = 2y
yn = 0.536 m
Since yn > yc the channel is a mild slope channel for this discharge.
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Most Upvoted Answer
Water flows in a triangular channel of side slope 1 horizontal : 1 ve...
Given data:
Discharge, Q = 0.2 m³/s
Side slope, z = 1 horizontal : 1 vertical
Longitudinal slope, S = 0.001
Manning's n = 0.015

To determine the type of channel slope, we need to calculate the hydraulic radius of the channel and then use it to calculate the critical slope.

1. Hydraulic radius (R):
The hydraulic radius is given by the formula:

R = A/P
where,
A = cross-sectional area of flow
P = wetted perimeter

For a triangular channel, the cross-sectional area and wetted perimeter can be calculated as follows:

A = (z/2)*y^2
P = z*y + 2*(y^2 + z^2)^0.5

where,
y = depth of flow

Substituting the given values, we get:

A = (1/2)*y^2
P = y + 2*(y^2 + 1)^0.5

Therefore, the hydraulic radius is given by:

R = A/P = (1/2)*y^2 / [y + 2*(y^2 + 1)^0.5]

2. Critical slope (Sc):
The critical slope is given by the formula:

Sc = (1/n)*[(Q/R)^2]^(1/3)

Substituting the given values, we get:

Sc = (1/0.015)*[(0.2/R)^2]^(1/3)

3. Type of channel slope:
If the longitudinal slope of the channel is greater than the critical slope, then the channel is said to be steep. If the longitudinal slope is less than the critical slope, then the channel is said to be mild. If the longitudinal slope is equal to the critical slope, then the channel is said to be critical.

Comparing the longitudinal slope and critical slope values, we get:

S < />
=> 0.001 < />

Simplifying, we get:

R < 1.376="" />

From the hydraulic radius formula, we know that the hydraulic radius is directly proportional to the depth of flow (y). Therefore, for a given discharge, a smaller hydraulic radius corresponds to a deeper flow. Since the maximum depth of flow cannot exceed the height of the channel (which is 1 in this case), it follows that the hydraulic radius cannot be less than 1. Therefore, we have:

R >= 1 m

Combining the above two inequalities, we get:

1 m <= r="">< 1.376="" />

This indicates that the channel slope is mild. Therefore, the correct option is (B) Mild.
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Water flows in a triangular channel of side slope 1 horizontal : 1 vertical and longitudinal slope of 0.001. Determine the type of channel slope when a discharge of 0.2 m3 /s flows through it. Assume Manning’s n = 0.015.a) Steepb) Mildc) Criticald) Cannot be determineCorrect answer is option 'B'. Can you explain this answer?
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