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An incompressible fluid (kinematic viscosity, 7.4 x 10-7 m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate is
- a)0.651 x 10-3
- b)0.651
- c)6.51
- d)0.651 x 103
Correct answer is option 'B'. Can you explain this answer?
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An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific...
Shear stress
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An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific...
To find the shear stress on the surface of the top plate, we can use the concept of viscosity and linear velocity profile.
Given data:
Kinematic viscosity (ν) = 7.4 x 10^(-7) m^2/s
Specific gravity (SG) = 0.88
Velocity of the top plate (V) = 0.5 m/s
Gap between the plates (h) = 0.5 mm = 0.5 x 10^(-3) m
1. Understanding the problem:
We have an incompressible fluid trapped between two parallel plates, with one plate moving and the other stationary. The fluid attains a linear velocity profile, meaning the velocity increases linearly from the bottom plate to the top plate.
2. Calculating the dynamic viscosity:
Dynamic viscosity (μ) can be calculated using the formula:
μ = ν x ρ
where ν is the kinematic viscosity and ρ is the density of the fluid.
Since specific gravity (SG) is given, we can calculate the density as:
ρ = SG x ρ_water
where ρ_water is the density of water (1000 kg/m^3).
Substituting the values, we get:
ρ = 0.88 x 1000 = 880 kg/m^3
Now, calculating the dynamic viscosity:
μ = 7.4 x 10^(-7) x 880 = 0.0006512 kg/(m.s)
3. Determining the shear rate:
Shear rate (γ) can be calculated using the formula:
γ = V / h
where V is the velocity of the top plate and h is the gap between the plates.
Substituting the values, we get:
γ = 0.5 / (0.5 x 10^(-3)) = 1000 s^(-1)
4. Calculating the shear stress:
Shear stress (τ) can be calculated using the formula:
τ = μ x γ
where μ is the dynamic viscosity and γ is the shear rate.
Substituting the values, we get:
τ = 0.0006512 x 1000 = 0.6512 Pa
5. Converting shear stress to Pascals:
Since 1 Pascal (Pa) is equal to 1 N/m^2, the shear stress can be expressed as:
0.6512 Pa = 0.6512 x 10^(-3) N/m^2
The shear stress on the surface of the top plate is approximately 0.651 x 10^(-3) N/m^2, which is option 'B'.
Therefore, the correct answer is option 'B' (0.651).
Given data:
Kinematic viscosity (ν) = 7.4 x 10^(-7) m^2/s
Specific gravity (SG) = 0.88
Velocity of the top plate (V) = 0.5 m/s
Gap between the plates (h) = 0.5 mm = 0.5 x 10^(-3) m
1. Understanding the problem:
We have an incompressible fluid trapped between two parallel plates, with one plate moving and the other stationary. The fluid attains a linear velocity profile, meaning the velocity increases linearly from the bottom plate to the top plate.
2. Calculating the dynamic viscosity:
Dynamic viscosity (μ) can be calculated using the formula:
μ = ν x ρ
where ν is the kinematic viscosity and ρ is the density of the fluid.
Since specific gravity (SG) is given, we can calculate the density as:
ρ = SG x ρ_water
where ρ_water is the density of water (1000 kg/m^3).
Substituting the values, we get:
ρ = 0.88 x 1000 = 880 kg/m^3
Now, calculating the dynamic viscosity:
μ = 7.4 x 10^(-7) x 880 = 0.0006512 kg/(m.s)
3. Determining the shear rate:
Shear rate (γ) can be calculated using the formula:
γ = V / h
where V is the velocity of the top plate and h is the gap between the plates.
Substituting the values, we get:
γ = 0.5 / (0.5 x 10^(-3)) = 1000 s^(-1)
4. Calculating the shear stress:
Shear stress (τ) can be calculated using the formula:
τ = μ x γ
where μ is the dynamic viscosity and γ is the shear rate.
Substituting the values, we get:
τ = 0.0006512 x 1000 = 0.6512 Pa
5. Converting shear stress to Pascals:
Since 1 Pascal (Pa) is equal to 1 N/m^2, the shear stress can be expressed as:
0.6512 Pa = 0.6512 x 10^(-3) N/m^2
The shear stress on the surface of the top plate is approximately 0.651 x 10^(-3) N/m^2, which is option 'B'.
Therefore, the correct answer is option 'B' (0.651).
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An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 x 10-3b)0.651c)6.51d)0.651 x 103Correct answer is option 'B'. Can you explain this answer?
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An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 x 10-3b)0.651c)6.51d)0.651 x 103Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 x 10-3b)0.651c)6.51d)0.651 x 103Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 x 10-3b)0.651c)6.51d)0.651 x 103Correct answer is option 'B'. Can you explain this answer?.
An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 x 10-3b)0.651c)6.51d)0.651 x 103Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 x 10-3b)0.651c)6.51d)0.651 x 103Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An incompressible fluid (kinematic viscosity, 7.4 x 10-7m2/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate isa)0.651 x 10-3b)0.651c)6.51d)0.651 x 103Correct answer is option 'B'. Can you explain this answer?.
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