Topic: Fluid Mechanics
The transition Reynolds number for flow over a flat plate is $5\times10^{5}$. What is the distance from the leading edge at which transition will occur for flow of water with a uniform velocity for $1~m/s?$ (For water, the kinematic viscosity, $v=0.858\times10^{-6}m^{2}/s)$
- A. 1 m
- B. 0.43 m
- C. 43 m
- D. 103 m
Answer: B. 0.43 m
Explanation: Reynolds number $R_{e} = \frac{Vx}{v}$. Given $R_{e} = 5 \times 10^{5}$, $V = 1~m/s$, $v = 0.858 \times 10^{-6}~m^{2}/s$. $5 \times 10^{5} = \frac{1 \times x}{0.858 \times 10^{-6}} \Rightarrow x = 5 \times 10^{5} \times 0.858 \times 10^{-6} = 0.429~m \approx 0.43~m$.