Problem Statement
A Newtonian fluid fills the gap between a shaft and a concentric sleeve. When a force of 780N is applied to the sleeve parallel to the shaft, the sleeve attains a speed of 2m/s. If a 1400N force is applied, what speed will the sleeve attain? The temperature of the sleeve remains constant.
Problem Statement
A Newtonian fluid fills the gap between a shaft and a concentric sleeve. When a force of 780N is applied to the sleeve parallel to the shaft, the sleeve attains a speed of 2m/s. If a 1400N force is applied, what speed will the sleeve attain? The temperature of the sleeve remains constant.
Solution
Given:
- Force (F1) = 780N
- Velocity (u1) = 2m/s
- Force (F2) = 1400N
- Velocity (u2) = ?
Formula:
Shear stress relation:
τ = μ du/dy
Also,
τ = F/A
Combining:
μ du/dy = F/A
Since μ, A, and dy are constant:
F/du = constant
Therefore:
F1/u1 = F2/u2
Calculations:
Substitute values:
780 / 2 = 1400 / u2
Solve for u2:
u2 = (1400 × 2) / 780
u2 = 3.6m/s
Result:
The velocity of the sleeve when 1400N force is applied: 3.6m/s
Explanation
This problem involves the shear stress and velocity relationship in a Newtonian fluid. Here’s how it works:
- Shear stress: The relationship between force and velocity gradient is defined by the fluid’s dynamic viscosity.
- Proportionality: Since the fluid’s properties and dimensions are constant, the ratio of force to velocity remains unchanged.
- Application: The proportionality constant allows us to compute the unknown velocity for a given force.
This solution highlights fundamental fluid mechanics principles applied to practical engineering scenarios.
