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A short reach of a 3 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000 it carries a discharge of 5 m3 /s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance dy/dx at this section is __________ (use g = 10 m/s2)
  • a)
    3.959 × 10−3
  • b)
    2.959 × 10−3
  • c)
    3.449 × 10−3
  • d)
    3.786 × 10−4
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A short reach of a 3 m wide rectangular open channel has its bed leve...
Given data b = 3m,
So = −1 / 10000
Q = 5 m3⁄s , n = 0.01, y = 0.5 m
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A short reach of a 3 m wide rectangular open channel has its bed leve...
To find the rate of change of water depth with distance at a certain section in the open channel, we can use the Gradually Varied Flow (GVF) equation. The GVF equation is derived from the energy equation and assumes that the flow is gradually varying, meaning that the water depth changes gradually along the channel.

The GVF equation is given by:

(dA/dx) = (Q^2 / gA) * ((Sf - So) / (1 - (Q^2 / gA^3) * dA/dx))

Where:
- (dA/dx) is the rate of change of cross-sectional area with distance (unknown)
- Q is the discharge (5 m^3/s)
- g is the acceleration due to gravity (10 m/s^2)
- A is the cross-sectional area (unknown)
- Sf is the friction slope (unknown)
- So is the slope of the channel bed (1/10000)

To solve for (dA/dx), we need to determine the values of A and Sf at the given section.

Given:
- Width of the channel (B) = 3 m
- Depth of flow (y) = 0.5 m

We can calculate the cross-sectional area (A) using the equation:

A = B * y
A = 3 * 0.5
A = 1.5 m^2

Next, we need to calculate the friction slope (Sf) using the Manning's equation:

Sf = (1 / n) * (Q / A)^(2/3) * R^(2/3)

Where:
- n is the Manning's roughness coefficient (0.01)
- R is the hydraulic radius (unknown)

To find R, we can use the equation:

R = A / P
R = A / (B + 2y)
R = 1.5 / (3 + 2(0.5))
R = 1.5 / 4
R = 0.375 m

Substituting the values into the Manning's equation:

Sf = (1 / 0.01) * (5 / 1.5)^(2/3) * (0.375)^(2/3)
Sf = 100 * 1.974 * 0.717
Sf = 142.086

Now we can substitute the known values (Q = 5 m^3/s, g = 10 m/s^2, A = 1.5 m^2, Sf = 142.086, So = 1/10000) into the GVF equation:

(dA/dx) = (5^2 / 10 * 1.5) * ((142.086 - 1/10000) / (1 - (5^2 / 10 * 1.5^3) * (dA/dx)))

Simplifying the equation:

(dA/dx) = (25 / 15) * ((142.086 - 1/10000) / (1 - (25 / 15^3) * (dA/dx)))

(dA/dx) = 1.667 * (142.086 - 1/10000) / (1 - 25 / 675 * (dA/dx))

Now,
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A short reach of a 3 m wide rectangular open channel has its bed leve...
Given data b = 3m,
So = −1 / 10000
Q = 5 m3⁄s , n = 0.01, y = 0.5 m
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A short reach of a 3 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000 it carries a discharge of 5 m3 /s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance dy/dx at this section is __________ (use g = 10 m/s2)a) 3.959 × 10−3b) 2.959 × 10−3c) 3.449 × 10−3 d) 3.786 × 10−4Correct answer is option 'C'. Can you explain this answer?
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A short reach of a 3 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000 it carries a discharge of 5 m3 /s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance dy/dx at this section is __________ (use g = 10 m/s2)a) 3.959 × 10−3b) 2.959 × 10−3c) 3.449 × 10−3 d) 3.786 × 10−4Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A short reach of a 3 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000 it carries a discharge of 5 m3 /s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance dy/dx at this section is __________ (use g = 10 m/s2)a) 3.959 × 10−3b) 2.959 × 10−3c) 3.449 × 10−3 d) 3.786 × 10−4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A short reach of a 3 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000 it carries a discharge of 5 m3 /s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance dy/dx at this section is __________ (use g = 10 m/s2)a) 3.959 × 10−3b) 2.959 × 10−3c) 3.449 × 10−3 d) 3.786 × 10−4Correct answer is option 'C'. Can you explain this answer?.
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