Problem Statement
The velocity distribution of a viscous liquid (dynamic viscosity = 9 Poise) flowing over a fixed plate is given by u = 0.85y – y² (u is velocity in m/s and y is the distance from the plate in m). Calculate the shear stresses at:
- The plate surface (y = 0)
- At y = 0.3 m
Given Data
- Velocity distribution: u = 0.85y – y²
- Dynamic viscosity (μ) = 9 Poise = 0.9 NS/m²
- Points of interest: y = 0 and y = 0.3 m
Solution
1. Find Velocity Gradient (du/dy)
Differentiate the velocity equation:
du/dy = d(0.85y – y²)/dy = 0.85 – 2y
This gives us the general equation for velocity gradient at any point y
2. Shear Stress Formula
τ = μ × (du/dy)
τ = 0.9 × (0.85 – 2y)
3. Shear Stress at Plate Surface (y = 0)
τ = 0.9 × (0.85 – 2×0)
τ = 0.9 × 0.85 = 0.765 N/m²
Shear Stress at y = 0: 0.765 N/m²
4. Shear Stress at y = 0.3 m
τ = 0.9 × (0.85 – 2×0.3)
τ = 0.9 × 0.25 = 0.225 N/m²
Shear Stress at y = 0.3 m: 0.225 N/m²
Key Points
- Shear stress varies linearly with distance from the plate
- Maximum shear stress occurs at the plate surface (y = 0)
- The velocity gradient decreases as we move away from the plate
