Discussion
Air at \( 30 \) m/s meets a smooth plate. Taking transition at \( Re = 5 \times 10^5 \), how far back does it occur?
Answer: D
Rearranging, \( x = Re\,\mu/(\rho V) = 5 \times 10^5 \times 1.8 \times 10^{-5} / (1.225 \times 30) = 9/36.75 = 0.245 \) m. Roughly the first quarter-metre is laminar and everything downstream turbulent, which is typical for air at ordinary speeds.
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