Topic: Fluid Mechanics
The velocity distribution flow over a flat plate is given by $U=(3/4)y-y^{2}$ in which u is the velocity in m/s at a distance y meter above the plate. If the dynamic viscosity of the fluid is 10 poise, then the shear stress at 0.2m is
- A. $0.45~N/m^{2}$
- B. $0.35~N/m^{2}$
- C. $0.55~N/m^{2}$
- D. $0.25~N/m^{2}$
Answer: B. $0.35~N/m^{2}$
Explanation: The velocity distribution is given by $U = \frac{3}{4}y - y²$. The velocity gradient is $\frac{dU}{dy} = \frac{3}{4} - 2y$. At a distance of $y = 0.2\text{ m}$, the gradient is $\frac{dU}{dy} = \frac{3}{4} - 2(0.2) = 0.75 - 0.4 = 0.35$. The dynamic viscosity is $\mu = 10\text{ Poise} = 1\text{ Pa.s}$. According to Newton's law of viscosity, the shear stress $\tau = \mu \frac{dU}{dy}$. Therefore, $\tau = 1 \times 0.35 = 0.35\text{ N/m}^2$.