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A hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)
Correct answer is 'Range: 4.6 to 4.7'. Can you explain this answer?
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A hydraulic jump is formed in a 2.5 m wide rectangular channel which ...
Given data y2 = 0.7 m v2 = 0.9 m/s
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A hydraulic jump is formed in a 2.5 m wide rectangular channel which ...
Hydraulic Jump in a Rectangular Channel
To determine the pre-jump velocity in a hydraulic jump, we can use the principles of conservation of energy and momentum. The hydraulic jump occurs when a high-velocity flow transitions to a low-velocity flow, resulting in a sudden rise in water depth.
Given Data:
- Width of the rectangular channel (b): 2.5 m
- Post-jump depth (y2): 0.7 m
- Post-jump velocity (V2): 0.9 m/s
- Acceleration due to gravity (g): 10 m/s^2
Conservation of Energy:
In a hydraulic jump, the total energy before and after the jump remains constant. The total energy consists of kinetic energy and potential energy.
The energy equation can be written as:
E1 = E2
Kinetic Energy:
The kinetic energy per unit weight (K) can be defined as:
K = V^2 / (2g)
Potential Energy:
The potential energy per unit weight (P) can be defined as:
P = y
Conservation of Energy Equation:
K1 + P1 = K2 + P2
Substituting the values:
(V1^2) / (2g) + y1 = (V2^2) / (2g) + y2
Conservation of Momentum:
The momentum equation can be written as:
Q1 = Q2
Where Q is the discharge or flow rate, which is equal to the product of velocity and area.
Discharge Equation:
Q = V * A
For a rectangular channel, the area (A) can be defined as:
A = b * y
Substituting the values:
V1 * (b * y1) = V2 * (b * y2)
Calculating the Pre-jump Velocity:
We can rearrange the momentum equation to solve for V1:
V1 = (V2 * y2) / y1
Substituting the given values:
V1 = (0.9 * 0.7) / 2.5
Calculating the result:
V1 ≈ 0.252 m/s
Therefore, the pre-jump velocity is approximately 0.252 m/s.
Answer Explanation:
The correct answer range provided is 4.6 to 4.7 m/s. However, based on the calculations and given data, the pre-jump velocity is determined to be approximately 0.252 m/s. It is possible that there is an error in the answer or the question itself. It is recommended to double-check the data and calculations to ensure accuracy.
To determine the pre-jump velocity in a hydraulic jump, we can use the principles of conservation of energy and momentum. The hydraulic jump occurs when a high-velocity flow transitions to a low-velocity flow, resulting in a sudden rise in water depth.
Given Data:
- Width of the rectangular channel (b): 2.5 m
- Post-jump depth (y2): 0.7 m
- Post-jump velocity (V2): 0.9 m/s
- Acceleration due to gravity (g): 10 m/s^2
Conservation of Energy:
In a hydraulic jump, the total energy before and after the jump remains constant. The total energy consists of kinetic energy and potential energy.
The energy equation can be written as:
E1 = E2
Kinetic Energy:
The kinetic energy per unit weight (K) can be defined as:
K = V^2 / (2g)
Potential Energy:
The potential energy per unit weight (P) can be defined as:
P = y
Conservation of Energy Equation:
K1 + P1 = K2 + P2
Substituting the values:
(V1^2) / (2g) + y1 = (V2^2) / (2g) + y2
Conservation of Momentum:
The momentum equation can be written as:
Q1 = Q2
Where Q is the discharge or flow rate, which is equal to the product of velocity and area.
Discharge Equation:
Q = V * A
For a rectangular channel, the area (A) can be defined as:
A = b * y
Substituting the values:
V1 * (b * y1) = V2 * (b * y2)
Calculating the Pre-jump Velocity:
We can rearrange the momentum equation to solve for V1:
V1 = (V2 * y2) / y1
Substituting the given values:
V1 = (0.9 * 0.7) / 2.5
Calculating the result:
V1 ≈ 0.252 m/s
Therefore, the pre-jump velocity is approximately 0.252 m/s.
Answer Explanation:
The correct answer range provided is 4.6 to 4.7 m/s. However, based on the calculations and given data, the pre-jump velocity is determined to be approximately 0.252 m/s. It is possible that there is an error in the answer or the question itself. It is recommended to double-check the data and calculations to ensure accuracy.
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A hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)Correct answer is 'Range: 4.6 to 4.7'. Can you explain this answer?
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A hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)Correct answer is 'Range: 4.6 to 4.7'. 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 hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)Correct answer is 'Range: 4.6 to 4.7'. 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 hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)Correct answer is 'Range: 4.6 to 4.7'. Can you explain this answer?.
A hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)Correct answer is 'Range: 4.6 to 4.7'. 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 hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)Correct answer is 'Range: 4.6 to 4.7'. 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 hydraulic jump is formed in a 2.5 m wide rectangular channel which is horizontal and frictionless. The post jump depth and velocity are 0.7 m and 0.9 m/s respectively. The pre-jump velocity is _____________ m/s (g = 10 m/s2)Correct answer is 'Range: 4.6 to 4.7'. Can you explain this answer?.
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