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A rectangular channel having a bed slope of 0.0001, width 3.0 m and Manning’s coefficient ‘n’ 0.015, carries a discharge of 1.0 m3/s. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses)
- a)0.64
- b)0.84
- c)1.04
- d)1.24
Correct answer is option 'B'. Can you explain this answer?
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A rectangular channel having a bed slope of 0.0001, width 3.0 m and Ma...
's roughness coefficient of 0.025 is designed to carry a flow rate of 10.0 m3/s. Determine the depth of flow in the channel.
To determine the depth of flow in the channel, we can use the Manning's equation:
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = Flow rate (m3/s)
n = Manning's roughness coefficient
A = Cross-sectional area of flow (m2)
R = Hydraulic radius (m)
S = Bed slope
First, we need to calculate the cross-sectional area of flow (A). Since the channel is rectangular, the cross-sectional area is equal to the product of the width (W) and the depth of flow (D):
A = W * D
A = 3.0 m * D
Next, we need to calculate the hydraulic radius (R). The hydraulic radius is defined as the ratio of the cross-sectional area to the wetted perimeter (P). For a rectangular channel, the wetted perimeter is equal to the sum of the width and depth:
P = W + D
P = 3.0 m + D
R = A / P
R = (3.0 m * D) / (3.0 m + D)
Now, we can substitute the values into the Manning's equation and solve for the depth of flow (D):
10.0 m3/s = (1/0.025) * (3.0 m * D) * ((3.0 m * D) / (3.0 m + D))^(2/3) * 0.0001^(1/2)
Simplifying the equation:
10.0 = (40) * D * ((9D) / (3 + D))^(2/3) * 0.01^(1/2)
Dividing both sides by 40:
0.25 = D * ((9D) / (3 + D))^(2/3) * 0.01^(1/2)
Simplifying the equation:
0.25 = D * ((9D) / (3 + D))^(2/3) * 0.1
Raising both sides to the power of 3/2:
(0.25)^(3/2) = D^3 * ((9D) / (3 + D))
Simplifying the equation:
0.125 = D^3 * ((9D) / (3 + D))
Dividing both sides by D:
0.125 / D = D^2 * (9 / (3 + D))
Simplifying the equation:
0.125 / D = 9D / (3 + D)
Cross-multiplying:
0.125(3 + D) = 9D^2
0.375 + 0.125D = 9D^2
Rearranging the equation:
9D^2 - 0.125D - 0.375 = 0
Using the quadratic formula:
D = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 9, b = -0.125, and c = -0.375
D = (-(-0.125) ± √((-0.125)^2 - 4(
To determine the depth of flow in the channel, we can use the Manning's equation:
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = Flow rate (m3/s)
n = Manning's roughness coefficient
A = Cross-sectional area of flow (m2)
R = Hydraulic radius (m)
S = Bed slope
First, we need to calculate the cross-sectional area of flow (A). Since the channel is rectangular, the cross-sectional area is equal to the product of the width (W) and the depth of flow (D):
A = W * D
A = 3.0 m * D
Next, we need to calculate the hydraulic radius (R). The hydraulic radius is defined as the ratio of the cross-sectional area to the wetted perimeter (P). For a rectangular channel, the wetted perimeter is equal to the sum of the width and depth:
P = W + D
P = 3.0 m + D
R = A / P
R = (3.0 m * D) / (3.0 m + D)
Now, we can substitute the values into the Manning's equation and solve for the depth of flow (D):
10.0 m3/s = (1/0.025) * (3.0 m * D) * ((3.0 m * D) / (3.0 m + D))^(2/3) * 0.0001^(1/2)
Simplifying the equation:
10.0 = (40) * D * ((9D) / (3 + D))^(2/3) * 0.01^(1/2)
Dividing both sides by 40:
0.25 = D * ((9D) / (3 + D))^(2/3) * 0.01^(1/2)
Simplifying the equation:
0.25 = D * ((9D) / (3 + D))^(2/3) * 0.1
Raising both sides to the power of 3/2:
(0.25)^(3/2) = D^3 * ((9D) / (3 + D))
Simplifying the equation:
0.125 = D^3 * ((9D) / (3 + D))
Dividing both sides by D:
0.125 / D = D^2 * (9 / (3 + D))
Simplifying the equation:
0.125 / D = 9D / (3 + D)
Cross-multiplying:
0.125(3 + D) = 9D^2
0.375 + 0.125D = 9D^2
Rearranging the equation:
9D^2 - 0.125D - 0.375 = 0
Using the quadratic formula:
D = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 9, b = -0.125, and c = -0.375
D = (-(-0.125) ± √((-0.125)^2 - 4(
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A rectangular channel having a bed slope of 0.0001, width 3.0 m and Manning’s coefficient ‘n’ 0.015, carries a discharge of 1.0 m3/s. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses)a)0.64b)0.84c)1.04d)1.24Correct answer is option 'B'. Can you explain this answer?
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A rectangular channel having a bed slope of 0.0001, width 3.0 m and Manning’s coefficient ‘n’ 0.015, carries a discharge of 1.0 m3/s. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses)a)0.64b)0.84c)1.04d)1.24Correct 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 rectangular channel having a bed slope of 0.0001, width 3.0 m and Manning’s coefficient ‘n’ 0.015, carries a discharge of 1.0 m3/s. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses)a)0.64b)0.84c)1.04d)1.24Correct 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 rectangular channel having a bed slope of 0.0001, width 3.0 m and Manning’s coefficient ‘n’ 0.015, carries a discharge of 1.0 m3/s. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses)a)0.64b)0.84c)1.04d)1.24Correct answer is option 'B'. Can you explain this answer?.
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