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Water flows in a rectangular, concrete, open channel that is 12 m wide. The channel slope is 0.0028. If the velocity of the flow is 6 m/s, find the depth of the flow. (n = 0.013)
- a)2.45 m
- b)2.35 m
- c)2.65 m
- d)2.55 m
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Water flows in a rectangular, concrete, open channel that is 12 m wid...
Given data b = 12 m, S = 0.0028, V = 6 m/s, n = 0.013
By Manning equation
P = b + 2y ⇒ 12 + 2y
y = 2.551 m
Most Upvoted Answer
Water flows in a rectangular, concrete, open channel that is 12 m wid...
Given data:
Width of the channel (b) = 12 m
Slope of the channel (S) = 0.0028
Velocity of flow (V) = 6 m/s
Manning's roughness coefficient (n) = 0.013
Formula used:
The depth of flow in an open channel can be calculated using Manning's formula:
Q = (1/n) * A * R^(2/3) * S^(1/2)
where,
Q = discharge
A = cross-sectional area of flow
R = hydraulic radius of flow
S = slope of the channel
To find the depth of flow (y), we need to first calculate the cross-sectional area of flow (A) and the hydraulic radius of flow (R).
Calculation:
Cross-sectional area of flow (A) can be calculated as:
A = b * y
where,
b = width of the channel
y = depth of flow
So, A = 12 * y
Hydraulic radius of flow (R) can be calculated as:
R = A / P
where,
P = wetted perimeter of flow
For a rectangular channel, wetted perimeter (P) can be calculated as:
P = b + 2y
So, P = 12 + 2y
Substituting the values of A and R in Manning's formula, we get:
Q = (1/n) * (12y) * [(12y)/(12+2y)]^(2/3) * (0.0028)^(1/2)
Q = (1/n) * 12^(5/3) * y^(5/3) * [(12+2y)^(-2/3)] * (0.0028)^(1/2)
As per continuity equation, discharge (Q) can be calculated as:
Q = A * V
where,
V = velocity of flow
Substituting the value of Q in the above equation, we get:
A * V = (1/n) * 12^(5/3) * y^(5/3) * [(12+2y)^(-2/3)] * (0.0028)^(1/2)
Substituting the given values of b, S, V, and n in the above equation, we get:
12y * 6 = (1/0.013) * 12^(5/3) * y^(5/3) * [(12+2y)^(-2/3)] * (0.0028)^(1/2)
Simplifying the above equation, we get:
y^(5/3) = 0.00488 * [(12+2y)^(-2/3)]
To solve for y, we need to use an iterative approach. Let's assume the value of y to be 2 m.
Substituting y = 2 m in the above equation, we get:
2^(5/3) = 0.00488 * [(12+2*2)^(-2/3)]
2.3783 = 0.00488 * 0.344
2.3783 = 0.00168
The above value is not equal, so we need to use another value of y. Let's assume the value of y to be 2.5 m.
Substituting y = 2
Width of the channel (b) = 12 m
Slope of the channel (S) = 0.0028
Velocity of flow (V) = 6 m/s
Manning's roughness coefficient (n) = 0.013
Formula used:
The depth of flow in an open channel can be calculated using Manning's formula:
Q = (1/n) * A * R^(2/3) * S^(1/2)
where,
Q = discharge
A = cross-sectional area of flow
R = hydraulic radius of flow
S = slope of the channel
To find the depth of flow (y), we need to first calculate the cross-sectional area of flow (A) and the hydraulic radius of flow (R).
Calculation:
Cross-sectional area of flow (A) can be calculated as:
A = b * y
where,
b = width of the channel
y = depth of flow
So, A = 12 * y
Hydraulic radius of flow (R) can be calculated as:
R = A / P
where,
P = wetted perimeter of flow
For a rectangular channel, wetted perimeter (P) can be calculated as:
P = b + 2y
So, P = 12 + 2y
Substituting the values of A and R in Manning's formula, we get:
Q = (1/n) * (12y) * [(12y)/(12+2y)]^(2/3) * (0.0028)^(1/2)
Q = (1/n) * 12^(5/3) * y^(5/3) * [(12+2y)^(-2/3)] * (0.0028)^(1/2)
As per continuity equation, discharge (Q) can be calculated as:
Q = A * V
where,
V = velocity of flow
Substituting the value of Q in the above equation, we get:
A * V = (1/n) * 12^(5/3) * y^(5/3) * [(12+2y)^(-2/3)] * (0.0028)^(1/2)
Substituting the given values of b, S, V, and n in the above equation, we get:
12y * 6 = (1/0.013) * 12^(5/3) * y^(5/3) * [(12+2y)^(-2/3)] * (0.0028)^(1/2)
Simplifying the above equation, we get:
y^(5/3) = 0.00488 * [(12+2y)^(-2/3)]
To solve for y, we need to use an iterative approach. Let's assume the value of y to be 2 m.
Substituting y = 2 m in the above equation, we get:
2^(5/3) = 0.00488 * [(12+2*2)^(-2/3)]
2.3783 = 0.00488 * 0.344
2.3783 = 0.00168
The above value is not equal, so we need to use another value of y. Let's assume the value of y to be 2.5 m.
Substituting y = 2
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Water flows in a rectangular, concrete, open channel that is 12 m wide. The channel slope is 0.0028. If the velocity of the flow is 6 m/s, find the depth of the flow. (n = 0.013)a)2.45 mb) 2.35 mc) 2.65 md) 2.55 mCorrect answer is option 'D'. Can you explain this answer?
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Water flows in a rectangular, concrete, open channel that is 12 m wide. The channel slope is 0.0028. If the velocity of the flow is 6 m/s, find the depth of the flow. (n = 0.013)a)2.45 mb) 2.35 mc) 2.65 md) 2.55 mCorrect answer is option 'D'. 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 Water flows in a rectangular, concrete, open channel that is 12 m wide. The channel slope is 0.0028. If the velocity of the flow is 6 m/s, find the depth of the flow. (n = 0.013)a)2.45 mb) 2.35 mc) 2.65 md) 2.55 mCorrect answer is option 'D'. 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 Water flows in a rectangular, concrete, open channel that is 12 m wide. The channel slope is 0.0028. If the velocity of the flow is 6 m/s, find the depth of the flow. (n = 0.013)a)2.45 mb) 2.35 mc) 2.65 md) 2.55 mCorrect answer is option 'D'. Can you explain this answer?.
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