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?

Correct Answer: τ_w/√2

Step-by-Step Explanation:

Step 1: Determine the type of flow (Laminar vs. Turbulent)
To find the relationship between the wall shear stresses at different points along the plate, we must first establish whether the boundary layer flow is laminar or turbulent. This is determined by calculating the local Reynolds number (Re_x) at the points of interest.
The formula for the local Reynolds number is:

Where:
• U_∞ = 2 m/s (free-stream velocity)
• ν = 1.5 × 10^-5 m²/s (kinematic viscosity of air)
• x = distance from the leading edge

Let's calculate the Reynolds number at the furthest location, x₂ = 30 mm = 0.03 m (as seen in the provided diagram showing x₁ = 15 mm and x₂ = 30 mm):

$${\mathrm{Re}}_{x_2}=\frac{2\times 0.03}{1.5\times {10}^{-5}}=4000$$

For flow over a flat plate, the transition from laminar to turbulent flow typically occurs at a critical Reynolds number of:

$${\mathrm{Re}}_{\mathrm{crit}}\approx 5\times {10}^5$$

Since Re_x2 = 4000 is much less than 5 × 10⁵, the flow is completely laminar across the entire region of interest (from x = 15 mm to x = 30 mm).

Step 2: Relate wall shear stress to distance x
For a laminar boundary layer over a flat plate (Blasius solution), the wall shear stress (τ_w) is related to the distance from the leading edge (x) by the following proportionality:

$${\tau }_w\propto \frac{1}{\sqrt{x}}$$

Using this proportionality, we can set up a ratio for the wall shear stress at two different locations:

$$\frac{{\tau }_{w_2}}{{\tau }_{w_1}}=\sqrt{\frac{x_1}{x_2}}$$

Given:
• x₁ = 15 mm (where τ_w1 = τ_w)
• x₂ = 30 mm
Substitute these values into the ratio:

$$\frac{{\tau }_{w_2}}{{\tau }_w}=\sqrt{\frac{15}{30}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$$

Solving for τ_w2, we get:

$${\tau }_{w_2}=\frac{{\tau }_w}{\sqrt{2}}$$

Therefore, the wall shear stress at a distance of 30 mm from the leading edge is indeed τ_w/√2.