Question

Consider a flat plate positioned inside a wind tunnel, and air at 1 atm and \(20^{\circ} \mathrm{C}\) is flowing with a free stream velocity of $60 \mathrm{~m} / \mathrm{s}$. What is the minimum length of the plate necessary for the Reynolds number to reach \(2 \times 10^{7}\) ? If the critical Reynolds number is \(5 \times 10^{5}\), what type of flow regime would the airflow experience at \(0.2 \mathrm{~m}\) from the leading edge?

Short answer

Based on the step-by-step solution, determine the minimum length of the plate necessary for the Reynolds number to reach \(2 \times 10^7\) and the type of flow regime at a location \(0.2\,\text{m}\) from the leading edge. Minimum plate length: \(4.982\,\text{m}\) Flow regime at \(0.2\,\text{m}\) from leading edge: Turbulent

Step-by-step solution

Identify given variables and constants

Air temperature: \(20^{\circ}\mathrm{C}\) Air pressure: \(1\,\text{atm}\) Free stream velocity: \(V_{\infty} = 60\,\text{m/s}\) Reynolds number: \(Re_D = 2\times10^{7}\) Critical Reynolds number: \(Re_{c} = 5\times10^{5}\) Location on the plate: \(x = 0.2\,\text{m}\)

Find the kinematic viscosity of air

We use the Sutherland Formula to determine the kinematic viscosity (\(\nu\)) of air at the given temperature: \(\nu =1.458 \times 10^{-6} \times \frac{(T + 110.4)^{3/2}}{T + 110.4}\) Where \(T\) is the air temperature in Kelvin. Convert the given Celsius temperature to Kelvin: \(T = 20^{\circ}\mathrm{C} + 273.15 = 293.15\,\text{K}\) Now, plug in the value of \(T\) and find \(\nu\): \(\nu =1.458 \times 10^{-6} \times \frac{(293.15 + 110.4)^{3/2}}{293.15 + 110.4} = 1.493 \times 10^{-5} \, \text{m}^2/\text{s}\)

Find the minimum length of the plate

Use the Reynolds number formula for flow over a flat plate: \(Re_D = \frac{V_\infty L}{\nu}\) Solve for the length of the plate (\(L\)): \(L = \frac{Re_D \nu}{V_\infty} = \frac{(2\times10^7)(1.493\times10^{-5}\, \text{m}^2/\text{s})}{60\,\text{m/s}} = 4.982 \, \text{m}\) Therefore, the minimum length of the plate necessary for the Reynolds number to reach \(2 \times 10^7\) is \(4.982\,\text{m}\).

Determine the type of flow regime

Calculate the Reynolds number at the given location on the plate (\(x = 0.2\,\text{m}\)): \(Re_x = \frac{V_\infty x}{\nu} = \frac{(60\,\text{m/s})(0.2 \, \text{m})}{1.493\times10^{-5} \, \text{m}^2/\text{s}} = 8.02\times10^5\) Compare the calculated Reynolds number at the given location to the critical Reynolds number: \(Re_x = 8.02\times10^5 > Re_c = 5\times10^5\) Since the Reynolds number at the given location is greater than the critical Reynolds number, the airflow is in a turbulent flow regime at \(0.2\,\text{m}\) from the leading edge.