Question

The velocity distribution of a viscous liquid ($\mu = 0.9 \text{ N-s/m}^2$) over a fixed boundary is given by $u = 0.68y - y^2$, in which $u$ is the velocity in m/s at a distance $y$ meters above the boundary surface. Determine the shear stress at the surface.

• A. $0.512 \text{ N/m}^2$
• B. $0.0 \text{ N/m}^2$
• C. $0.6 \text{ N/m}^2$
• D. $0.612 \text{ N/m}^2$

Hint:

The "surface" or "boundary" always implies $y = 0$. Always check if the units of viscosity are in SI (N-s/m$^2$ or Pa-s) or Poise ($1 \text{ Pa-s} = 10 \text{ Poise}$) before calculating!

Answer 1

The Correct Option is D

Concept: According to Newton's Law of Viscosity, the shear stress ($\tau$) is proportional to the velocity gradient ($du/dy$) perpendicular to the direction of flow.

Step 1: Find the velocity gradient expression.
Given $u = 0.68y - y^2$. Differentiating with respect to $y$: \[ \frac{du}{dy} = 0.68 - 2y \]

Step 2: Evaluate the gradient at the boundary surface.
At the boundary surface, $y = 0$: \[ \left( \frac{du}{dy} \right)_{y=0} = 0.68 - 2(0) = 0.68 \text{ s}^{-1} \]

Step 3: Calculate the shear stress ($\tau$).
Using the formula $\tau = \mu \left( \frac{du}{dy} \right)$: Given $\mu = 0.9 \text{ N-s/m}^2$. \[ \tau = 0.9 \times 0.68 \] \[ \tau = 0.612 \text{ N/m}^2 \]