Question 10.5: Air (ρ = 1.23 kg/m³ and ν = 1.5 × 10^-5 m²/s) is flowing ove...
Air (\rho=1.23 kg / m ^{3} \text { and } v=1.5 \times 10^{-5} m ^{2}/s) is flowing over a flat plate. The free stream speed is 15 m/s. At a distance of 1 m from the leading edge, calculate \delta \text { and } \tau_{w} for (a) completely laminar flow, and (b) completely turbulent flow for a 1/7th power law velocity profile.
Applying the results developed in Chapters 9 and 10, we can write for parabolic velocity profile (laminar flow)
\frac{\delta}{x}=\frac{5.48}{\sqrt{\operatorname{Re}_{x}}} \text { and } \tau_{w}=\left.\mu \frac{\partial u}{\partial y}\right|_{y=0}
\operatorname{Re}_{x}=\frac{U x}{v}=\frac{15 \times 1}{1.5 \times 10^{-5}}=1.0 \times 10^{6}
\delta=\frac{5.48}{\sqrt{1.0 \times 10^{6}}} \times 1 m =5.48 mm
\tau_{w}=\left.\mu \frac{\partial u}{\partial y}\right|_{y=0}=\frac{\mu U_{\infty}}{\delta} \cdot \frac{ d }{ d \eta}\left[2 \eta-\eta^{2}\right]_{\eta=0}
\tau_{w}=\frac{2 \times 1.23 \times 1.5 \times 10^{-5} \times 15}{0.00548}=0.101 N / m ^{2}
For turbulent flow,
\frac{\delta}{x}=\frac{0.37}{\left(\operatorname{Re}_{x}\right)^{1 / 5}} (from Eq. 10.62)
\frac{\delta}{x}=0.37\left( Re _{x}\right)^{-1 / 5} (10.62)
or \delta=\frac{0.370}{\left(1.0 \times 10^{6}\right)^{1 / 5}} \times 1 m =23.34 mm
or \delta / x=0.0233
\tau_{w}=0.0225 \rho U_{\infty}^{2}\left(\frac{v}{U_{\infty} \delta}\right)^{1 / 4} (from Eq. 10.60)
\frac{\tau_{w}}{\rho U_{\infty}^{2}}=0.0225\left[\frac{v}{\delta U_{\infty}}\right]^{1 / 4} (10.60)
\tau_{w}=0.0225 \times 1.23 \times(15)^{2}\left(\frac{v}{U_{\infty} x} \cdot \frac{x}{\delta}\right)^{1 / 4}
or \tau_{w}=0.0225 \times 1.23 \times(15)^{2}\left[\frac{1}{1.0 \times 10^{6}} \times \frac{1}{0.0233}\right]^{1 / 4}
= 0.502 N / m ^{2}
Turbulent boundary layer has a larger shear stress than the laminar boundary layer.