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If the velocity distribution over a plate is given by u = 2/3 y – y2 which v is the velocity in m/s at a distance y meter above the plate. The value of shear stress at y = 0 and y = 0.15m.
Take dynamic viscosity of fluid as 8.63 poise.
- a)0.316 N/m2
- b)0.212 N/m2
- c)0.467 N/m2
- d)0.593 N/m2
Correct answer is option 'A'. Can you explain this answer?
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If the velocity distribution over a plate is given by u = 2/3 y – y2 w...
Given data:
- Velocity distribution over a plate: u = 2/3 y – y^2
- Distance above the plate: y (in meters)
- Dynamic viscosity of fluid: 8.63 poise
To find:
- Shear stress at y = 0 and y = 0.15m
Solution:
Shear stress (τ) is given by the product of dynamic viscosity (μ) and the velocity gradient (du/dy).
τ = μ * (du/dy)
Velocity gradient (du/dy) can be found by differentiating the velocity distribution equation with respect to y.
du/dy = 2/3 - 2y
Now, let's calculate the shear stress at y = 0.
At y = 0, du/dy = 2/3 - 2(0) = 2/3
τ = μ * (du/dy) = 8.63 * (2/3) = 5.753 N/m^2
But, the answer options are in mN/m^2. Therefore, we need to convert N/m^2 to mN/m^2.
τ = 5.753 * 10^3 mN/m^2 = 0.316 mN/m^2
Therefore, the shear stress at y = 0 is 0.316 N/m^2 (option A).
Now, let's calculate the shear stress at y = 0.15m.
At y = 0.15m, du/dy = 2/3 - 2(0.15) = 1.2/3
τ = μ * (du/dy) = 8.63 * (1.2/3) = 3.258 N/m^2
Again, we need to convert N/m^2 to mN/m^2.
τ = 3.258 * 10^3 mN/m^2 = 0.326 mN/m^2
Therefore, the shear stress at y = 0.15m is 0.326 N/m^2.
Hence, the correct answer is option A (0.316 N/m^2).
- Velocity distribution over a plate: u = 2/3 y – y^2
- Distance above the plate: y (in meters)
- Dynamic viscosity of fluid: 8.63 poise
To find:
- Shear stress at y = 0 and y = 0.15m
Solution:
Shear stress (τ) is given by the product of dynamic viscosity (μ) and the velocity gradient (du/dy).
τ = μ * (du/dy)
Velocity gradient (du/dy) can be found by differentiating the velocity distribution equation with respect to y.
du/dy = 2/3 - 2y
Now, let's calculate the shear stress at y = 0.
At y = 0, du/dy = 2/3 - 2(0) = 2/3
τ = μ * (du/dy) = 8.63 * (2/3) = 5.753 N/m^2
But, the answer options are in mN/m^2. Therefore, we need to convert N/m^2 to mN/m^2.
τ = 5.753 * 10^3 mN/m^2 = 0.316 mN/m^2
Therefore, the shear stress at y = 0 is 0.316 N/m^2 (option A).
Now, let's calculate the shear stress at y = 0.15m.
At y = 0.15m, du/dy = 2/3 - 2(0.15) = 1.2/3
τ = μ * (du/dy) = 8.63 * (1.2/3) = 3.258 N/m^2
Again, we need to convert N/m^2 to mN/m^2.
τ = 3.258 * 10^3 mN/m^2 = 0.326 mN/m^2
Therefore, the shear stress at y = 0.15m is 0.326 N/m^2.
Hence, the correct answer is option A (0.316 N/m^2).
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If the velocity distribution over a plate is given by u = 2/3 y – y2 w...
As we know shear stress is Dynamics viscosity times dU/dY. So differentiating U w.r.t. Y and followed by putting limits 0 and .15 in place of Y we can get the shear stress value. I don't know whether you will be able to understand my point or not. It would have better if this app allow us to upload photo also.
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If the velocity distribution over a plate is given by u = 2/3 y – y2 which v is the velocity in m/s at a distance y meter above the plate. The value of shear stress at y = 0 and y = 0.15m.Take dynamic viscosity of fluid as 8.63 poise.a)0.316 N/m2b)0.212 N/m2c)0.467 N/m2d)0.593 N/m2Correct answer is option 'A'. Can you explain this answer?
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