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A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.
The longitudinal slope of the channel is
- a)0.00083
- b)0.00073
- c)0.00053
- d)0.00063
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under...
Width of rectangular channel
B = 6.5 m
Discharge Q = 18 m3/s
Normal depth = yn = 1.8 m
Manning’s n = 0.018
From Manning’s equation
S = 6.3033 × 10−4
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A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under...
Given data:
Width of rectangular channel (B) = 6.5 m
Discharge (Q) = 18 m3/s
Normal depth (Y) = 1.8 m
Manning’s n = 0.018
To find: Longitudinal slope of the channel
We can use Manning’s equation to find the slope of the channel.
Manning’s equation is given by:
Q = (1/n) * A * R2/3 * S1/2
Where,
Q = Discharge
n = Manning’s roughness coefficient
A = Cross-sectional area of flow
R = Hydraulic radius
S = Slope of the energy line
Cross-sectional area of flow (A) can be calculated as:
A = B * Y
Where,
B = Width of channel
Y = Normal depth
Hydraulic radius (R) can be calculated as:
R = A/P
Where,
P = Wetted perimeter
Wetted perimeter (P) can be calculated as:
P = B + 2Y
Substituting the above equations in Manning’s equation, we get:
Q = (1/n) * B * Y * [B + 2Y]2/3 * S1/2
Rearranging the equation to find slope (S), we get:
S = {Q / [(1/n) * B * Y * [B + 2Y]2/3]}2
Substituting the given values in the above equation, we get:
S = {18 / [(1/0.018) * 6.5 * 1.8 * [6.5 + 2(1.8)]2/3]}2
S = 0.00063
Therefore, the longitudinal slope of the channel is 0.00063. Hence, option (d) is the correct answer.
Width of rectangular channel (B) = 6.5 m
Discharge (Q) = 18 m3/s
Normal depth (Y) = 1.8 m
Manning’s n = 0.018
To find: Longitudinal slope of the channel
We can use Manning’s equation to find the slope of the channel.
Manning’s equation is given by:
Q = (1/n) * A * R2/3 * S1/2
Where,
Q = Discharge
n = Manning’s roughness coefficient
A = Cross-sectional area of flow
R = Hydraulic radius
S = Slope of the energy line
Cross-sectional area of flow (A) can be calculated as:
A = B * Y
Where,
B = Width of channel
Y = Normal depth
Hydraulic radius (R) can be calculated as:
R = A/P
Where,
P = Wetted perimeter
Wetted perimeter (P) can be calculated as:
P = B + 2Y
Substituting the above equations in Manning’s equation, we get:
Q = (1/n) * B * Y * [B + 2Y]2/3 * S1/2
Rearranging the equation to find slope (S), we get:
S = {Q / [(1/n) * B * Y * [B + 2Y]2/3]}2
Substituting the given values in the above equation, we get:
S = {18 / [(1/0.018) * 6.5 * 1.8 * [6.5 + 2(1.8)]2/3]}2
S = 0.00063
Therefore, the longitudinal slope of the channel is 0.00063. Hence, option (d) is the correct answer.
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A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.The longitudinal slope of the channel isa) 0.00083b) 0.00073c) 0.00053d) 0.00063Correct answer is option 'D'. Can you explain this answer?
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A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.The longitudinal slope of the channel isa) 0.00083b) 0.00073c) 0.00053d) 0.00063Correct answer is option 'D'. Can you explain this answer? for Civil Engineering (CE) 2025 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 rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.The longitudinal slope of the channel isa) 0.00083b) 0.00073c) 0.00053d) 0.00063Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.The longitudinal slope of the channel isa) 0.00083b) 0.00073c) 0.00053d) 0.00063Correct answer is option 'D'. Can you explain this answer?.
A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.The longitudinal slope of the channel isa) 0.00083b) 0.00073c) 0.00053d) 0.00063Correct answer is option 'D'. Can you explain this answer? for Civil Engineering (CE) 2025 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 rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.The longitudinal slope of the channel isa) 0.00083b) 0.00073c) 0.00053d) 0.00063Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular channel 6.5 m wide carries a discharge of 18 m3/s under uniform flow condition with normal depth of 1.8 m Manning’s n is 0.018.The longitudinal slope of the channel isa) 0.00083b) 0.00073c) 0.00053d) 0.00063Correct answer is option 'D'. Can you explain this answer?.
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