Problem Statement
The velocity distribution for flow over a flat plate is given by \(u = \frac{3}{2}y - y^{3/2}\), where \(u\) is the point velocity in m/s at a distance \(y\) metre above the plate. Determine the shear stress at \(y = 9\) cm. Assume dynamic viscosity is 8 poise.
Given Data
- Velocity Distribution, \(u = \frac{3}{2}y - y^{3/2}\)
- Distance from plate, \(y = 9 \, \text{cm}\)
- Dynamic Viscosity, \(\mu = 8 \, \text{poise}\)
Solution
1. Convert All Units to SI
We need to work with a consistent set of units (metres, seconds, Newtons).
Distance:
Viscosity: (10 poise = 1 N·s/m²)
2. Find the Velocity Gradient (\(du/dy\))
The shear stress depends on the velocity gradient, which is the first derivative of the velocity profile with respect to \(y\).
3. Evaluate the Velocity Gradient at y = 0.09 m
Now, substitute the given value of \(y\) into the expression for the velocity gradient.
4. Calculate the Shear Stress (\(\tau\))
Apply Newton's law of viscosity to find the shear stress.
The shear stress at y = 9 cm is \( \tau = 0.84 \, \text{N/m}^2 \).
Explanation of the Method
1. Velocity Profile:
The equation \(u(y)\) describes how the fluid's velocity changes as you move away from the surface of the plate. In this case, the velocity is zero at the plate (\(y=0\)) and increases with distance.
2. Velocity Gradient:
The derivative, \(du/dy\), represents the rate of change of velocity with distance. This "velocity gradient" is a measure of how much the fluid is being sheared or deformed at a particular point. A large gradient means adjacent layers of fluid are sliding past each other rapidly.
3. Newton's Law of Viscosity:
This fundamental law states that for many fluids (called Newtonian fluids), the shear stress (\(\tau\)) is directly proportional to the rate of shear strain (the velocity gradient). The constant of proportionality is the dynamic viscosity (\(\mu\)). This law is the cornerstone of this calculation.
Physical Meaning
The calculated shear stress (\(\tau = 0.84 \, \text{N/m}^2\)) is the internal frictional force per unit area that the fluid exerts at a height of 9 cm above the plate. This stress arises because the fluid layer at 9 cm is moving faster than the layers below it, and the fluid's viscosity resists this relative motion.
This force is transmitted through the fluid. The fluid layer at 9 cm is being "dragged back" by the slower fluid below it, and it is simultaneously "pulling forward" the faster fluid above it. The shear stress quantifies this internal tug-of-war within the fluid.
