The velocity distribution over a plate is given by y =2/3y-y2 , in which u is the velocity in m/s at a distance of y m above the plate. Determine the shear stress at y = 0, 0.1, and 0.2 m. Also, find the distance in metres above the plate at which the shear stress is zero. Take µ = 6 poise.

Shear Stress from Velocity Profile

Problem Statement

The velocity distribution over a plate is given by the equation $u = \frac{2}{3}y - y^2$, in which u is the velocity in m/s at a distance of y m above the plate. Determine the shear stress at y = 0, 0.1, and 0.2 m. Also, find the distance in metres above the plate at which the shear stress is zero. Take µ = 6 poise.

Note: The velocity equation has been interpreted from the provided text as a standard parabolic profile.

Given Data

Velocity distribution, $u = \frac{2}{3}y - y^2$
Dynamic Viscosity, $\mu = 6 \, \text{poise}$

Solution

1. Convert Viscosity to SI Units

The viscosity is converted from poise to the standard SI unit (N·s/m²).

$$ \mu = 6 \, \text{poise} \times \frac{0.1 \, \text{N·s/m}^2}{1 \, \text{poise}} $$ $$ \mu = 0.6 \, \text{N·s/m}^2 $$

2. Find the Velocity Gradient (du/dy)

Shear stress depends on the velocity gradient, which is found by differentiating the velocity distribution equation with respect to y.

$$ u = \frac{2}{3}y - y^2 $$ $$ \frac{du}{dy} = \frac{d}{dy} \left( \frac{2}{3}y - y^2 \right) $$ $$ \frac{du}{dy} = \frac{2}{3} - 2y $$

3. Calculate Shear Stress (τ) at Each Point

We use Newton's law of viscosity, $\tau = \mu \frac{du}{dy}$, for each value of y.

At y = 0 m:

$$ \frac{du}{dy} = \frac{2}{3} - 2(0) = \frac{2}{3} \, s^{-1} $$ $$ \tau_0 = 0.6 \, \text{N·s/m}^2 \times \frac{2}{3} \, s^{-1} $$ $$ \tau_0 = 0.4 \, \text{N/m}^2 $$

At y = 0.1 m:

$$ \frac{du}{dy} = \frac{2}{3} - 2(0.1) = \frac{2}{3} - 0.2 \approx 0.4667 \, s^{-1} $$ $$ \tau_{0.1} = 0.6 \, \text{N·s/m}^2 \times 0.4667 \, s^{-1} $$ $$ \tau_{0.1} \approx 0.28 \, \text{N/m}^2 $$

At y = 0.2 m:

$$ \frac{du}{dy} = \frac{2}{3} - 2(0.2) = \frac{2}{3} - 0.4 \approx 0.2667 \, s^{-1} $$ $$ \tau_{0.2} = 0.6 \, \text{N·s/m}^2 \times 0.2667 \, s^{-1} $$ $$ \tau_{0.2} \approx 0.16 \, \text{N/m}^2 $$

4. Find Distance 'y' for Zero Shear Stress

Shear stress is zero when the velocity gradient $\frac{du}{dy}$ is zero.

$$ \frac{du}{dy} = \frac{2}{3} - 2y = 0 $$ $$ 2y = \frac{2}{3} $$ $$ y = \frac{1}{3} \, \text{m} \approx 0.333 \, \text{m} $$

Final Results:

Shear Stress at y=0 m: $ \tau_0 = 0.4 \, \text{N/m}^2 $

Shear Stress at y=0.1 m: $ \tau_{0.1} = 0.28 \, \text{N/m}^2 $

Shear Stress at y=0.2 m: $ \tau_{0.2} = 0.16 \, \text{N/m}^2 $

Distance for zero shear stress: $ y = 0.333 \, \text{m} $

Explanation of Velocity Gradient

A velocity profile (or distribution) describes how the speed of a fluid changes at different points within the flow. In this case, the velocity is not constant; it changes with the distance $y$ from the plate.

The velocity gradient ($\frac{du}{dy}$) is the rate of this change. It tells us how much the velocity differs between adjacent layers of the fluid. According to Newton's law of viscosity, this gradient is what creates the internal friction, or shear stress, in the fluid.

Physical Meaning

The results show that the shear stress is highest at the plate surface ($y=0$) and decreases as the distance from the plate increases. This is because the velocity changes most rapidly near the stationary plate.

The point where the shear stress is zero ($y = 0.333$ m) is the point where the velocity is at its maximum. At this specific location, the layer of fluid is moving at the same speed as the layers immediately above and below it, so there is no velocity gradient and thus no internal friction (shear stress). This point represents the peak of the parabolic flow profile.