Air (density = 1.2 kg/m³ , kinematic viscosity = 1.5×10^-5 m² /s) flows over a flat plate with a free-stream velocity of 2 m/s. The wall shear stress at a location 15 mm from the leading edge is τ_w. What is the wall shear stress at a location 30 mm from the leading edge?

The correct option is τ_w/√2.

To determine the wall shear stress at a distance of 30 mm from the leading edge, we first need to identify the flow regime (laminar or turbulent) by calculating the Reynolds number ($\mathrm{Re}$) at this location.

The Reynolds number is defined as:
$$\mathrm{Re}=\frac{U_{\infty }x}{\nu }$$
where:
• $U_{\infty }=2\text{m/s}$ is the free-stream velocity,
• $x=30\text{mm}=0.03\text{m}$ is the distance from the leading edge, and
• $\nu =1.5\times {10}^{-5}{\text{m}}^2\text{/s}$ is the kinematic viscosity of air.

Substituting the values into the equation:
$$\mathrm{Re}=\frac{2\times 0.03}{1.5\times {10}^{-5}}=4000$$

Since the Reynolds number ($\mathrm{Re}=4000$) is well below the critical Reynolds number for flat-plate flow transition ($5\times {10}^5$), the boundary layer flow remains entirely laminar over this region.

For a laminar boundary layer on a flat plate, according to Blasius's boundary layer solution, the local skin friction coefficient ($C_{\mathrm{fx}}$) is given by:
$$C_{\mathrm{fx}}=\frac{0.664}{\sqrt{{\mathrm{Re}}_x}}$$

The local wall shear stress (${\tau }_w$) is directly proportional to $C_{\mathrm{fx}}$:
$${\tau }_w=C_{\mathrm{fx}}·\left(\frac{1}{2}\rho U_{\infty }^2\right)$$

By substituting ${\mathrm{Re}}_x=\frac{U_{\infty }x}{\nu }$ into the shear stress equation, we find that the wall shear stress is inversely proportional to the square root of the distance $x$ from the leading edge:
$${\tau }_w\propto \frac{1}{\sqrt{x}}$$

Let $x_1=15\text{mm}$ with wall shear stress ${\tau }_{w1}={\tau }_w$, and let $x_2=30\text{mm}$ with wall shear stress ${\tau }_{w2}$. We can set up the ratio:
$$\frac{{\tau }_{w2}}{{\tau }_{w1}}=\sqrt{\frac{x_1}{x_2}}$$

Substituting the values of $x_1$ and $x_2$:
$$\frac{{\tau }_{w2}}{{\tau }_w}=\sqrt{\frac{15}{30}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$$

Solving for ${\tau }_{w2}$:
$${\tau }_{w2}=\frac{{\tau }_w}{\sqrt{2}}$$