Question

Air flows over a flat plate at $40 \mathrm{~m} / \mathrm{s}, 25^{\circ} \mathrm{C}\(, and \)1 \mathrm{~atm}\( pressure. \)(a)$ What plate length should be used to achieve a Reynolds number of \(1 \times 10^{8}\) at the end of the plate? (b) If the critical Reynolds number is \(5 \times 10^{5}\), at what distance from the leading edge of the plate would transition occur?

Short answer

Answer: The plate length to achieve a Reynolds number of \(1 \times 10^8\) is \(39 \mathrm{m}\), and the transition occurs at a distance of \(1.95 \mathrm{m}\) from the leading edge of the plate.

Step-by-step solution

Calculate the kinematic viscosity of the air

At \(25^{\circ}\mathrm{C}\), the kinematic viscosity of air is approximately \(15.6 \times 10^{-6} \mathrm{m^2/s}\). We can denote this value as \(\nu\). #Step 2: Find the plate length to achieve a Reynolds number of \(1 \times 10^8\)#

Calculate the plate length

Using the Reynolds number formula, we have: \(Re_x = \frac{ux}{\nu}\) Rearrange for \(x\): \(x = \frac{Re_x \cdot \nu}{u}\) We are given \(Re_x = 1 \times 10^8\), \(u = 40 \mathrm{m/s}\), and \(\nu = 15.6 \times 10^{-6} \mathrm{m^2/s}\). Plug in the values: \(x = \frac{(1 \times 10^8) \cdot (15.6 \times 10^{-6})}{40}\) \(x = 39 \mathrm{m}\) The plate length should be \(39 \mathrm{m}\) to achieve a Reynolds number of \(1 \times 10^8\) at the end of the plate. #Step 3: Find the distance from the leading edge of the plate where transition occurs#

Calculate the distance where transition occurs

We are given the critical Reynolds number as \(5 \times 10^5\). Using the same Reynolds number formula: \(Re_x = \frac{ux}{\nu}\) And rearrange for \(x\): \(x = \frac{Re_x \cdot \nu}{u}\) Plug in the values \(Re_x = 5 \times 10^5\), \(u = 40 \mathrm{m/s}\), and \(\nu = 15.6 \times 10^{-6} \mathrm{m^2/s}\): \(x = \frac{(5 \times 10^5) \cdot (15.6 \times 10^{-6})}{40}\) \(x = 1.95 \mathrm{m}\) The transition occurs at a distance of \(1.95 \mathrm{m}\) from the leading edge of the plate.