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Water flows at a steady and uniform depth of 2m in an open channel of rectangular cross section having base width of 4m and laid at a slope of 1 in 10000. If it is desired to obtain critical flow in the channel by providing a hump in the bed. Take Manning’s coefficient = 0.01. Calculate the height of hump (in meters) required for the critical flow
  • a)
    0.5
  • b)
    2.5
  • c)
    2
  • d)
    0.94
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Water flows at a steady and uniform depth of 2m in an open channel of ...
Area A = 4 ×2 = 8, Perimeter P = 4 + 4 = 8
R = (A/P) = 1 m
Discharge per unit width q = 8/4 = 2
We know that the minimum specific energy corresponds to the point of critical depth and is given by 1.5 times the critical depth
Critical depth = (q2/g)(1/3) = 0.74 m
Minimum specific energy is 1.5 × 0.74 = 1.11 m
Specific energy should be same at all points (neglecting hydraulic losses)
So, E = Eminimum + h (this energy is at position of hump)
E = y + v2/(2*g) = 2.05
h = 2.05 – 1.11 = 0.94 m
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Most Upvoted Answer
Water flows at a steady and uniform depth of 2m in an open channel of ...
's roughness coefficient (n) as 0.03 and acceleration due to gravity (g) as 9.81 m/s^2.

To obtain critical flow in the channel, the Froude number (Fr) needs to be equal to 1.

The Froude number is given by the equation:

Fr = V / sqrt(g * d)

where V is the velocity of the water, g is the acceleration due to gravity, and d is the depth of the water.

Since the water depth is given as 2m, we can rearrange the equation to solve for V:

V = Fr * sqrt(g * d)

To determine the Froude number for critical flow, we can use the equation:

Fr = sqrt(g * d) / sqrt(R)

where R is the hydraulic radius of the channel.

The hydraulic radius (R) is given by the equation:

R = (A / P)

where A is the cross-sectional area of the channel and P is the wetted perimeter.

The cross-sectional area (A) is given by the equation:

A = b * d

where b is the base width of the channel.

The wetted perimeter (P) is given by the equation:

P = b + 2d

Substituting these values into the equations, we have:

A = 4m * 2m = 8m^2
P = 4m + 2(2m) = 8m

R = 8m^2 / 8m = 1m

Now, we can solve for the Froude number (Fr) using the hydraulic radius (R) and the depth (d):

Fr = sqrt(9.81 m/s^2 * 2m) / sqrt(1m) = sqrt(19.62) / 1 = 4.43

Since the desired Froude number for critical flow is 1, we need to reduce the velocity by a factor of 4.43.

Using the Manning's equation:

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

where V is the velocity, n is Manning's roughness coefficient, R is the hydraulic radius, and S is the slope of the channel.

Substituting the values:

4.43V = (1 / 0.03) * (1m^(2/3)) * (1/10000)^(1/2)

Simplifying:

4.43V = 33.33 * 0.1 * 0.01^(1/2)

4.43V = 0.03333

V = 0.03333 / 4.43

V ≈ 0.0075 m/s

Therefore, to obtain critical flow in the channel, a hump in the bed should be provided to reduce the velocity to approximately 0.0075 m/s.
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Water flows at a steady and uniform depth of 2m in an open channel of rectangular cross section having base width of 4m and laid at a slope of 1 in 10000. If it is desired to obtain critical flow in the channel by providing a hump in the bed. Take Manning’s coefficient = 0.01. Calculate the height of hump (in meters) required for the critical flowa)0.5b)2.5c)2d)0.94Correct answer is option 'D'. Can you explain this answer?
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