Discussion
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 \(\tau_{w}\). What is the wall shear stress at a location 30 mm from the leading edge?
Answer: D
The wall shear stress at 30 mm is \(\tau_{w}/\sqrt{2}\).
First check the regime with the local Reynolds number:
\[ \begin{aligned} Re_{x} &= \frac{Ux}{\nu} \\ &= \frac{2 \times 0.015}{1.5 \times 10^{-5}} \\ &= 2000 \\ Re_{x}(30\ \text{mm}) &= 4000 \end{aligned} \]
Both are far below the transition value of about \(5 \times 10^{5}\), so the boundary layer is laminar and the Blasius result applies:
\[ \begin{aligned} \tau_{w} &= \frac{0.332\,\rho U^{2}}{\sqrt{Re_{x}}} \\ &\propto x^{-1/2} \end{aligned} \]
Doubling the distance from 15 mm to 30 mm divides the shear stress by \(\sqrt{2}\). \(\tau_{w}/2\) would need \(\tau_{w} \propto 1/x\). \(\sqrt{2}\,\tau_{w}\) and \(2\,\tau_{w}\) make the stress grow downstream, but it falls, because the boundary layer thickens and the velocity gradient at the wall weakens.
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