Problem Statement
A fluid with an absolute viscosity of 0.045 Pa-s and a specific gravity of 0.91 flows over a flat plate. The velocity of the fluid at a height of 70 mm above the plate is 1.13 m/s. Calculate the shear stress:
- At the solid boundary
- At points 20 mm above the plate
Consider two cases:
- Linear velocity distribution
- Parabolic velocity distribution with a vertex at 70 mm above the surface
Solution
Given:
- Absolute viscosity (μ) = 0.045 Pa-s
- Specific gravity = 0.91
- Velocity at 70 mm from the plate = 1.13 m/s
Case (a): Linear Velocity Distribution
u = ay
a = tanθ = u / y = 1.13 / 0.07 = 16.1
Therefore, u = 16.1y
du/dy = 16.1
τ = μ × du/dy
τ = 0.045 × 16.1 = 0.726 Pa (constant throughout)
Result for Case (a):
Shear stress is constant throughout the flow: 0.726 Pa.
Case (b): Parabolic Velocity Distribution
u = ay² + by + c
Boundary conditions:
- At y = 0, u = 0 → c = 0
- At y = 0.07 m, du/dy = 0 → 2a × 0.07 + b = 0 → b = -0.14a
- At y = 0.07 m, u = 1.13 m/s
Substitute: 1.13 = a × (0.07)² – 0.14a × 0.07
Solving: a = -230.61, b = -0.14 × -230.61 = 32.29
Final equation: u = -230.61y² + 32.29y
du/dy = -461.22y + 32.29
At y = 0:
du/dy = 32.29
At y = 0.02 m:
du/dy = -461.22 × 0.02 + 32.29 = 23.0656
At y = 0:
τ = μ × du/dy = 0.045 × 32.29 = 1.453 Pa
At y = 0.02 m:
τ = μ × du/dy = 0.045 × 23.0656 = 1.038 Pa
Result for Case (b):
Shear stress at y = 0: 1.453 Pa
Shear stress at y = 0.02 m: 1.038 Pa
Explanation
In this problem, shear stress is calculated using the velocity gradient at different heights:
- Linear distribution: The velocity gradient is constant, so the shear stress remains the same at all points.
- Parabolic distribution: The velocity profile follows a parabola, leading to varying velocity gradients and shear stresses at different heights.
This illustrates how the velocity profile impacts shear stress in fluid flow over a flat plate.
