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In a fluid the velocity measured at a distance of 75mm from the boundary is1.125m/s. the fluid has absolute viscosity 0.05 Pa s and relative density 0.93. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution? Calculate its kinematic viscosity,

Question

In a fluid the velocity measured at a distance of 75mm from the boundary is1.125m/s. the fluid has absolute viscosity 0.05 Pa s and relative density 0.93. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution? Calculate its kinematic viscosity,

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Solution 1

To solve this problem, we need to use the formulas for velocity gradient, shear stress, and kinematic viscosity.

  1. Velocity Gradient: The velocity gradient is the change in velocity with respect to the change in distance. Since we're assuming a linear velocity distribution, we can use the formula:

    Velocity Gradient = Δv/Δy

    Where Δv is the change in velocity and Δy is the change in distance. In this case, the velocity is 1.125 m/s and the distance is 75 mm or 0.075 m. So,

    Velocity Gradient = 1.125 m/s / 0.075 m = 15 s^-1

  2. Shear Stress: Shear stress in a fluid is given by the formula:

    Shear Stress = μ * (du/dy)

    Where μ is the dynamic viscosity and du/dy is the velocity gradient. Here, μ is 0.05 Pa s and du/dy is 15 s^-1. So,

    Shear Stress = 0.05 Pa s * 15 s^-1 = 0.75 Pa

  3. Kinematic Viscosity: Kinematic viscosity is the ratio of dynamic viscosity to density. It can be calculated using the formula:

    Kinematic Viscosity = μ/ρ

    Where μ is the dynamic viscosity and ρ is the density. The density can be calculated from the relative density by multiplying it by the density of water (1000 kg/m^3). So, ρ = 0.93 * 1000 kg/m^3 = 930 kg/m^3. Therefore,

    Kinematic Viscosity = 0.05 Pa s / 930 kg/m^3 = 5.376 x 10^-5 m^2/s

So, the velocity gradient is 15 s^-1, the shear stress at the boundary is 0.75 Pa, and the kinematic viscosity is 5.376 x 10^-5 m^2/s.

This problem has been solved

Solution 2

To solve this problem, we need to use the formulas for velocity gradient, shear stress, and kinematic viscosity.

  1. Velocity Gradient: The velocity gradient is the change in velocity with respect to the change in distance. Since we're assuming a linear velocity distribution, we can use the formula:

    Velocity Gradient = Δv / Δy

    Where Δv is the change in velocity and Δy is the change in distance. In this case, Δv = 1.125 m/s (since the velocity goes from 0 at the boundary to 1.125 m/s at 75 mm from the boundary) and Δy = 75 mm = 0.075 m (we need to convert mm to m to match the units of velocity).

    So, Velocity Gradient = 1.125 m/s / 0.075 m = 15 s^-1

  2. Shear Stress: Shear stress in a fluid is given by the formula:

    Shear Stress = μ * (du/dy)

    Where μ is the dynamic viscosity and du/dy is the velocity gradient. We've already calculated the velocity gradient, and the dynamic viscosity is given as 0.05 Pa s.

    So, Shear Stress = 0.05 Pa s * 15 s^-1 = 0.75 Pa

  3. Kinematic Viscosity: Kinematic viscosity is the ratio of dynamic viscosity to density. The density of the fluid can be found by multiplying the relative density by the density of water (1000 kg/m^3).

    Density = Relative Density * Density of water = 0.93 * 1000 kg/m^3 = 930 kg/m^3

    Then, Kinematic Viscosity = μ / Density = 0.05 Pa s / 930 kg/m^3 = 5.376 x 10^-5 m^2/s

So, the velocity gradient is 15 s^-1, the shear stress at the boundary is 0.75 Pa, and the kinematic viscosity is 5.376 x 10^-5 m^2/s.

This problem has been solved

Solution 3

To solve this problem, we need to use the formulas for velocity gradient, shear stress, and kinematic viscosity.

  1. Velocity Gradient: The velocity gradient is the change in velocity with respect to the change in distance. Since we're assuming a linear velocity distribution, we can use the formula:

    Velocity Gradient = Δv/Δy

    Where Δv is the change in velocity and Δy is the change in distance. In this case, Δv = 1.125 m/s and Δy = 75 mm = 0.075 m (since we need to convert mm to m).

    So, Velocity Gradient = 1.125 m/s / 0.075 m = 15 s^-1

  2. Shear Stress: Shear stress is the force per unit area caused by the velocity gradient in a fluid. It can be calculated using the formula:

    Shear Stress = μ * (du/dy)

    Where μ is the dynamic viscosity and du/dy is the velocity gradient. In this case, μ = 0.05 Pa s and du/dy = 15 s^-1.

    So, Shear Stress = 0.05 Pa s * 15 s^-1 = 0.75 Pa

  3. Kinematic Viscosity: Kinematic viscosity is the ratio of dynamic viscosity to the density of the fluid. It can be calculated using the formula:

    Kinematic Viscosity = μ/ρ

    Where μ is the dynamic viscosity and ρ is the density of the fluid. In this case, μ = 0.05 Pa s and the relative density is 0.93. The density of water is approximately 1000 kg/m^3, so the density of the fluid is 0.93 * 1000 kg/m^3 = 930 kg/m^3.

    So, Kinematic Viscosity = 0.05 Pa s / 930 kg/m^3 = 5.376 x 10^-5 m^2/s

So, the velocity gradient is 15 s^-1, the shear stress at the boundary is 0.75 Pa, and the kinematic viscosity is 5.376 x 10^-5 m^2/s.

This problem has been solved

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