The transition Reynolds number for flow over a flat plate is $5 \times 10^5$. What is the distance from the leading edge at which transition will occur for flow of water with a uniform velocity of 1 m/s? (For water, the kinematic viscosity, $\nu = 0.86 \times 10^{-6} m²/s$)
- A. 0.43m
- B. 1m
- C. 43m
- D. 103m
Answer: A. 0.43m
Explanation: $Re = \frac{Vx}{\nu}$. $x = \frac{5 \times 10^5 \times 0.86 \times 10^{-6}}{1} = 0.43$ m.