Topic: Fluid Mechanics
The equation of a velocity distribution over a plate is given by $u=2y-y^{2}$; where u is the velocity in $m/s$ at a point y meter from the plate measured perpendicularly. Assuming $\mu=8.60$ poise, the shear stress at a point 15 cm from the boundary is
- A. 14.6 Pa
- B. 0.24 Pa
- C. 2.4 Pa
- D. 1.46 Pa
Answer: D. 1.46 Pa
Explanation: Velocity gradient: $\frac{du}{dy} = 2 - 2y$.
At $y = 15~cm = 0.15~m$, $\frac{du}{dy} = 2 - 2(0.15) = 1.7~s^{-1}$.
Dynamic viscosity: $\mu = 8.60~poise = 0.86~Pa\cdot s$.
Shear stress: $\tau = \mu \frac{du}{dy} = 0.86 \times 1.7 = 1.462~Pa$.