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The height of a hydraulic jump in the stilling pool of 1 : 25 scale model was observed to be 10 cm. The corresponding prototype height of the jump is
- a)not determinable from the data given
- b)2.5 m
- c)0.5 m
- d)0.1 m
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The height of a hydraulic jump in the stilling pool of 1 : 25 scale m...
Given Data:
- Height of hydraulic jump in the model = 10 cm
- Scale of the model = 1 : 25
To Find:
- Prototype height of the jump
Explanation:
A hydraulic jump occurs when there is a sudden change in flow velocity in an open channel flow. It is characterized by a sudden rise in water surface and energy dissipation. The height of the hydraulic jump can be determined using the Froude number, which is the ratio of flow velocity to the square root of gravity times the flow depth.
The Froude number can be expressed as:
Fr = V / √(gD)
Where:
- Fr = Froude number
- V = flow velocity
- g = acceleration due to gravity
- D = flow depth
1. Scale Conversion:
Since the height of the hydraulic jump is given in the model, we need to convert it to the corresponding prototype value. Since the scale of the model is 1 : 25, we need to multiply the model height by the scale factor to obtain the prototype height.
Prototype height = Model height × Scale factor
Prototype height = 10 cm × 25 = 250 cm
2. Froude Number:
Since the Froude number is a dimensionless quantity, it remains the same for both the model and the prototype. Therefore, we can write:
Fr_model = Fr_prototype
3. Flow Depth:
The flow depth in the model and prototype can be related using the scale factor.
Flow depth_prototype = Flow depth_model × Scale factor
Flow depth_prototype = D_model × 25
4. Hydraulic Jump Height:
Substituting the flow depth relation into the Froude number equation, we get:
Fr_model = V_model / √(gD_model)
Fr_prototype = V_prototype / √(gD_prototype)
Since the Froude number remains the same for both the model and the prototype, we can equate the two equations:
V_model / √(gD_model) = V_prototype / √(gD_prototype)
5. Conclusion:
From the above equation, we can see that the flow velocities cancel out, and we are left with the relation:
D_model = D_prototype
Therefore, the hydraulic jump height in the prototype is equal to the hydraulic jump height in the model, which is 10 cm. Converting this to meters, we get:
Prototype height = 10 cm = 0.1 m
Hence, the correct answer is option (B) 0.5 m.
- Height of hydraulic jump in the model = 10 cm
- Scale of the model = 1 : 25
To Find:
- Prototype height of the jump
Explanation:
A hydraulic jump occurs when there is a sudden change in flow velocity in an open channel flow. It is characterized by a sudden rise in water surface and energy dissipation. The height of the hydraulic jump can be determined using the Froude number, which is the ratio of flow velocity to the square root of gravity times the flow depth.
The Froude number can be expressed as:
Fr = V / √(gD)
Where:
- Fr = Froude number
- V = flow velocity
- g = acceleration due to gravity
- D = flow depth
1. Scale Conversion:
Since the height of the hydraulic jump is given in the model, we need to convert it to the corresponding prototype value. Since the scale of the model is 1 : 25, we need to multiply the model height by the scale factor to obtain the prototype height.
Prototype height = Model height × Scale factor
Prototype height = 10 cm × 25 = 250 cm
2. Froude Number:
Since the Froude number is a dimensionless quantity, it remains the same for both the model and the prototype. Therefore, we can write:
Fr_model = Fr_prototype
3. Flow Depth:
The flow depth in the model and prototype can be related using the scale factor.
Flow depth_prototype = Flow depth_model × Scale factor
Flow depth_prototype = D_model × 25
4. Hydraulic Jump Height:
Substituting the flow depth relation into the Froude number equation, we get:
Fr_model = V_model / √(gD_model)
Fr_prototype = V_prototype / √(gD_prototype)
Since the Froude number remains the same for both the model and the prototype, we can equate the two equations:
V_model / √(gD_model) = V_prototype / √(gD_prototype)
5. Conclusion:
From the above equation, we can see that the flow velocities cancel out, and we are left with the relation:
D_model = D_prototype
Therefore, the hydraulic jump height in the prototype is equal to the hydraulic jump height in the model, which is 10 cm. Converting this to meters, we get:
Prototype height = 10 cm = 0.1 m
Hence, the correct answer is option (B) 0.5 m.
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Community Answer
The height of a hydraulic jump in the stilling pool of 1 : 25 scale m...
The height of a hydraulic jump in the stilling pool of 1 : 25
hm = 10cm and hp = ?
hm/hp = Lr = 1/25
hp = 25 x hm = 25 x 10 = 250cm = 2.5m
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The height of a hydraulic jump in the stilling pool of 1 : 25 scale model was observed to be 10 cm. The corresponding prototype height of the jump isa)not determinable from the data givenb)2.5 mc)0.5 md)0.1 mCorrect answer is option 'B'. Can you explain this answer?
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