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A 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.
Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.
Correct answer is 'Range : 108 to 109'. Can you explain this answer?
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A 2 cm wide gap between two vertical plane surfaces is filled with an...
The shear stresses on two sides of the plate are
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τ1 = μ du / dy = μ V / t1 and τ2 = μ V / t2
Drag force or viscous resistance against the motion of plate,
F = (μ V / t1+ μ V / t2)A
= μAV [ 1 / t2 +1 / t 2]
Since the plate is placed midway in the gap,
t1 = t2 and therefore,
F = 2μAV / t
Where t = t1 = t2 = 2 − 0.2 / 2 = 0.9 cm = 0.009 m
∴ F = 2 × 2.5(1.25 × 1.25) × 0.12 / 0.009
= 104.17 N
Upthrust or buoyant force on the plate
= specific weight of oil × volume of oil displaced
= (0.85 × 9810) × (1.25 × 1.25 × 0.002)
= 26.06 N
Effective weight of the plate = 30 − 26.06 = 3.94 N
∴ Total force required to lift the plate at the given velocity,
= 104.17 + 3.94 = 108.11 N
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A 2 cm wide gap between two vertical plane surfaces is filled with an...
Problem Statement:
A 2 cm wide gap between two vertical plane surfaces is filled with oil of specific gravity 0.85 and dynamic viscosity 2.5 Ns/m^2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap. Find the force required to lift the plate up with a constant velocity of 0.12 m/s.
Given Data:
- Width of the gap = 2 cm
- Specific gravity of the oil = 0.85
- Dynamic viscosity of the oil = 2.5 Ns/m^2
- Dimensions of the metal plate:
- Length = 1.25 m
- Width = 1.25 m
- Thickness = 0.2 cm
- Weight of the metal plate = 30 N
- Velocity of the plate = 0.12 m/s
Assumptions:
- The flow between the surfaces is assumed to be laminar.
- The plate is assumed to be fully submerged in the oil.
- The plate is assumed to move parallel to the vertical plane surfaces.
Calculating the Area:
The area of the metal plate in contact with the oil can be calculated as:
Area = Length × Width = 1.25 m × 1.25 m = 1.5625 m^2
Calculating the Volume:
The volume of the oil displaced by the metal plate can be calculated as:
Volume = Area × Thickness = 1.5625 m^2 × 0.002 m = 0.003125 m^3
Calculating the Buoyancy Force:
The buoyancy force acting on the metal plate can be calculated as:
Buoyancy Force = Weight of the displaced oil = Density × Volume × g
Density = Specific Gravity × Density of water = 0.85 × 1000 kg/m^3 = 850 kg/m^3
Buoyancy Force = 850 kg/m^3 × 0.003125 m^3 × 9.81 m/s^2 = 25.91 N
Calculating the Drag Force:
The drag force acting on the metal plate can be calculated using the drag equation:
Drag Force = 0.5 × Drag Coefficient × Density of the fluid × Velocity^2 × Area
Calculating the Drag Coefficient:
The drag coefficient depends on the shape of the object and the Reynolds number. Since the plate is assumed to have a rectangular shape, the drag coefficient can be estimated as 1.2.
Calculating the Drag Force:
Drag Force = 0.5 × 1.2 × Dynamic Viscosity × (Velocity / Gap Width) × Area
Drag Force = 0.5 × 1.2 × 2.5 Ns/m^2 × (0.12 m/s / 0.02 m) × 1.5625 m^2 = 2.8125 N
Calculating the Required Force:
The force required to lift the plate up with a constant velocity can be calculated by adding the buoyancy force and the drag force:
Required Force = Buoyancy Force + Drag
A 2 cm wide gap between two vertical plane surfaces is filled with oil of specific gravity 0.85 and dynamic viscosity 2.5 Ns/m^2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap. Find the force required to lift the plate up with a constant velocity of 0.12 m/s.
Given Data:
- Width of the gap = 2 cm
- Specific gravity of the oil = 0.85
- Dynamic viscosity of the oil = 2.5 Ns/m^2
- Dimensions of the metal plate:
- Length = 1.25 m
- Width = 1.25 m
- Thickness = 0.2 cm
- Weight of the metal plate = 30 N
- Velocity of the plate = 0.12 m/s
Assumptions:
- The flow between the surfaces is assumed to be laminar.
- The plate is assumed to be fully submerged in the oil.
- The plate is assumed to move parallel to the vertical plane surfaces.
Calculating the Area:
The area of the metal plate in contact with the oil can be calculated as:
Area = Length × Width = 1.25 m × 1.25 m = 1.5625 m^2
Calculating the Volume:
The volume of the oil displaced by the metal plate can be calculated as:
Volume = Area × Thickness = 1.5625 m^2 × 0.002 m = 0.003125 m^3
Calculating the Buoyancy Force:
The buoyancy force acting on the metal plate can be calculated as:
Buoyancy Force = Weight of the displaced oil = Density × Volume × g
Density = Specific Gravity × Density of water = 0.85 × 1000 kg/m^3 = 850 kg/m^3
Buoyancy Force = 850 kg/m^3 × 0.003125 m^3 × 9.81 m/s^2 = 25.91 N
Calculating the Drag Force:
The drag force acting on the metal plate can be calculated using the drag equation:
Drag Force = 0.5 × Drag Coefficient × Density of the fluid × Velocity^2 × Area
Calculating the Drag Coefficient:
The drag coefficient depends on the shape of the object and the Reynolds number. Since the plate is assumed to have a rectangular shape, the drag coefficient can be estimated as 1.2.
Calculating the Drag Force:
Drag Force = 0.5 × 1.2 × Dynamic Viscosity × (Velocity / Gap Width) × Area
Drag Force = 0.5 × 1.2 × 2.5 Ns/m^2 × (0.12 m/s / 0.02 m) × 1.5625 m^2 = 2.8125 N
Calculating the Required Force:
The force required to lift the plate up with a constant velocity can be calculated by adding the buoyancy force and the drag force:
Required Force = Buoyancy Force + Drag
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A 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.Correct answer is 'Range : 108 to 109'. Can you explain this answer?
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A 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.Correct answer is 'Range : 108 to 109'. 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 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.Correct answer is 'Range : 108 to 109'. 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 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.Correct answer is 'Range : 108 to 109'. Can you explain this answer?.
A 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.Correct answer is 'Range : 108 to 109'. 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 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.Correct answer is 'Range : 108 to 109'. 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 2 cm wide gap between two vertical plane surfaces is filled with an oil specific gravity 0.85 and dynamic viscosity 2.5 Ns/m2. A metal plate 1.25 cm × 1.25 m × 0.2 cm thick and weighing 30 N is placed midway in the gap.Find the force required if the plate is to be lifted up with a constant velocity of 0.12 m/s.Correct answer is 'Range : 108 to 109'. Can you explain this answer?.
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