Problem Statement
In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid has absolute viscosity 0.048 NS/m² and relative density 0.9. Calculate:
- Velocity gradient at the boundary (assuming linear velocity distribution)
- Shear stress at the boundary
- Kinematic viscosity
Given Data
- Change in velocity (du) = 1.125 – 0 = 1.125 m/s
- Change in distance (dy) = 75 mm = 0.075 m
- Absolute viscosity (μ) = 0.048 NS/m²
- Relative density (S) = 0.9
Solution
1. Velocity Gradient (du/dy)
du/dy = 1.125 / 0.075 = 15 s⁻¹
Assuming linear velocity distribution from boundary to measurement point
Velocity Gradient = 15 s⁻¹
2. Shear Stress (τ)
τ = μ × (du/dy)
τ = 0.048 × 15 = 0.72 N/m²
Shear Stress = 0.72 N/m²
3. Kinematic Viscosity (υ)
ρ = S × ρwater = 0.9 × 1000 = 900 kg/m³
υ = μ/ρ = 0.048/900 = 5.3 × 10⁻⁵ m²/s
Kinematic Viscosity = 5.3 × 10⁻⁵ m²/s
Key Points
- Linear velocity distribution means constant velocity gradient
- Shear stress is directly proportional to velocity gradient
- Kinematic viscosity relates absolute viscosity to fluid density
