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A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?
- a)1.44 m3/s
- b)1.62 m3/s
- c)1.92 m3/s
- d)2.24 m3/s
Correct answer is option 'B'. Can you explain this answer?
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A hydraulically efficient trapezoidal section of open channel flow car...
To calculate the discharge through the channel, we can use the Manning's equation, which relates the discharge (Q) to the hydraulic radius (R), bed slope (S), and the Chezy coefficient (C). The Manning's equation is given as:
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = Discharge
n = Manning's roughness coefficient
A = Cross-sectional area of flow
R = Hydraulic radius
S = Bed slope
Given:
Chezy coefficient (C) = 75
Optimal depth (y) = 0.6 m
Bed slope (S) = 1 in 250
We first need to calculate the hydraulic radius (R) using the given data. The hydraulic radius is given by:
R = A / P
Where:
P = Wetted perimeter
To calculate the cross-sectional area (A) and wetted perimeter (P), we need to know the dimensions of the trapezoidal section. Since the dimensions are not given, we cannot directly calculate A and P. However, we know that the trapezoidal section is hydraulically efficient, which implies that the side slopes are close to the critical slope. In such cases, we can assume the side slopes to be at the critical slope of 1:1.
Assuming the side slopes to be 1:1, we can calculate the cross-sectional area (A) and wetted perimeter (P) using the formula for a trapezoid:
A = (y + B) * y = (0.6 + B) * 0.6
P = y + 2 * √((0.6^2) + 1) + B = 0.6 + 2 * √((0.6^2) + 1) + B
Where B is the bottom width of the trapezoidal section.
Now, we can substitute the values of A, R, and S into the Manning's equation and solve for Q:
Q = (1/n) * A * R^(2/3) * S^(1/2)
= (1/n) * (0.6 + B) * 0.6 * (A / P)^(2/3) * (1/250)^(1/2)
Since the discharge is dependent on the bottom width (B), we can differentiate the equation with respect to B and set it equal to zero to find the optimal value of B. However, since we don't have the derivative equation, we can solve this problem by trial and error.
By trying different values of B, we can find that the discharge is maximum when B is approximately 0.3 m.
Substituting this value of B into the equation, we get:
Q = (1/75) * (0.6 + 0.3) * 0.6 * (A / P)^(2/3) * (1/250)^(1/2)
= (1/75) * 0.9 * 0.6 * (A / P)^(2/3) * (1/250)^(1/2)
Now, we need to find the ratio of A/P, which is called the hydraulic radius (R). We can use the formula for the hydraulic radius of a trape
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = Discharge
n = Manning's roughness coefficient
A = Cross-sectional area of flow
R = Hydraulic radius
S = Bed slope
Given:
Chezy coefficient (C) = 75
Optimal depth (y) = 0.6 m
Bed slope (S) = 1 in 250
We first need to calculate the hydraulic radius (R) using the given data. The hydraulic radius is given by:
R = A / P
Where:
P = Wetted perimeter
To calculate the cross-sectional area (A) and wetted perimeter (P), we need to know the dimensions of the trapezoidal section. Since the dimensions are not given, we cannot directly calculate A and P. However, we know that the trapezoidal section is hydraulically efficient, which implies that the side slopes are close to the critical slope. In such cases, we can assume the side slopes to be at the critical slope of 1:1.
Assuming the side slopes to be 1:1, we can calculate the cross-sectional area (A) and wetted perimeter (P) using the formula for a trapezoid:
A = (y + B) * y = (0.6 + B) * 0.6
P = y + 2 * √((0.6^2) + 1) + B = 0.6 + 2 * √((0.6^2) + 1) + B
Where B is the bottom width of the trapezoidal section.
Now, we can substitute the values of A, R, and S into the Manning's equation and solve for Q:
Q = (1/n) * A * R^(2/3) * S^(1/2)
= (1/n) * (0.6 + B) * 0.6 * (A / P)^(2/3) * (1/250)^(1/2)
Since the discharge is dependent on the bottom width (B), we can differentiate the equation with respect to B and set it equal to zero to find the optimal value of B. However, since we don't have the derivative equation, we can solve this problem by trial and error.
By trying different values of B, we can find that the discharge is maximum when B is approximately 0.3 m.
Substituting this value of B into the equation, we get:
Q = (1/75) * (0.6 + 0.3) * 0.6 * (A / P)^(2/3) * (1/250)^(1/2)
= (1/75) * 0.9 * 0.6 * (A / P)^(2/3) * (1/250)^(1/2)
Now, we need to find the ratio of A/P, which is called the hydraulic radius (R). We can use the formula for the hydraulic radius of a trape
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A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?a)1.44 m3/s b)1.62 m3/sc)1.92 m3/s d)2.24 m3/sCorrect answer is option 'B'. Can you explain this answer?
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A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?a)1.44 m3/s b)1.62 m3/sc)1.92 m3/s d)2.24 m3/sCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?a)1.44 m3/s b)1.62 m3/sc)1.92 m3/s d)2.24 m3/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?a)1.44 m3/s b)1.62 m3/sc)1.92 m3/s d)2.24 m3/sCorrect answer is option 'B'. Can you explain this answer?.
A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?a)1.44 m3/s b)1.62 m3/sc)1.92 m3/s d)2.24 m3/sCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?a)1.44 m3/s b)1.62 m3/sc)1.92 m3/s d)2.24 m3/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydraulically efficient trapezoidal section of open channel flow carries water at the optimal depth of 0.6m.Chezy coefficient is 75 and bed slope is 1 in 250. What is the discharge through the channel?a)1.44 m3/s b)1.62 m3/sc)1.92 m3/s d)2.24 m3/sCorrect answer is option 'B'. Can you explain this answer?.
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