Question

The velocity distribution for the flow over a flat plate is given by $u=2y-y^2$ in which $u$ is the velocity in $\text{m/s}$ at a distance $y\text{metres}$ above the plate. Determine the shear stress at $y=0.15\text{m}$. Take dynamic viscosity of the fluid as $8.5\text{poise}$.

• A. $1.455{\text{N/m}}^2$
• B. $1.545{\text{N/m}}^2$
• C. $1.445{\text{N/m}}^2$
• D. $1.554{\text{N/m}}^2$

Answer 1

The correct option is C $1.445{\text{N/m}}^2$

Given that velocity at distance $y$ metres,
$u=2y-y^2$
$\therefore \frac{du}{dy}=2-2y$
At $y=0.15\text{m}$,
$\frac{du}{dy}=2-(2\times 0.15)=1.7{\text{s}}^{-1}$
Given, dynamic viscosity $\eta =8.5\text{poise}=0.85{\text{Ns/m}}^2$
[$\because 10\text{poise}=1{\text{Ns/m}}^2$]

From Newton's law of viscosity,
$\tau =\eta \frac{du}{dy}=0.85\times 1.7=1.445$
$\therefore$ Shear stress ($\tau )=1.445{\text{N/m}}^2$