Air moving at 60 km/h is considered to have a zero boundary-layer thickness at a distance of 100 km from shore. At the beach, estimate the boundary-layer thickness and the wall shear using:

(a) A one-seventh power law

(b) Empirical data Use T = 20°C

Question

Air moving at 60 km/h is considered to have a zero boundary-layer thickness at a distance of 100 km from shore. At the beach, estimate the boundary-layer thickness and the wall shear using:

(a) A one-seventh power law

(b) Empirical data Use T = 20°C

Accepted Answer

a) [latex]U_{\infty}=\frac{60 imes 1000}{3600}=16.67 m / s . \quad \delta=0.38 imes 100000\left(\frac{1.5 imes 10^{-5}}{16.67 imes 10^{5}} ight)^{0.2}=\underline{235 m }[/latex].

[latex] au_{0}=\frac{1}{2} ho U_{\infty}^{2} c_{f}=\frac{1}{2} imes 1.22 imes 16.67^{2} imes\left[0.059\left(\frac{1.5 imes 10^{-5}}{16.67 imes 10^{5}} ight)^{0.2} ight]=\underline{0.0618 Pa }[/latex].

b) [latex] au_{0}=\frac{1}{2} ho U_{\infty \infty}^{2} c_{f}=\frac{1}{2} imes 1.22 imes 16.67^{2} \frac{0.455}{\left[\ln \left(0.06 \frac{16.67 imes 10^{5}}{1.5 imes 10^{-5}} ight) ight]^{2}}=\underline{0.151 Pa }[/latex]

[latex] herefore u_{ au}=\sqrt{\frac{0.151}{1.22}}=0.351 m / s . \quad herefore \frac{16.67}{0.351}=2.44 \ln \frac{0.351 \delta}{1.5 imes 10^{-5}}+7.4 . \quad herefore \underline{\delta=585 m }[/latex].

Both (a) and (b) are in error; however, (b) is more accurate.

Question 8.122: Air moving at 60 km/h is considered to have a zero boundary-...

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