Consider an airfoil flying at 100 m/s under standard sea-level conditions. The airfoil has a very low thickness, so it can be considered as a flat plate, and is 1m long.

What is the dynamic viscosity of the air, in Pas?
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What is the boundary layer thickness at x=0.1 m? Assume a laminar flow. Give your answer in mm.

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To solve the questions, we will start with the known values and equations regarding air properties and the boundary layer.

1. Determine the dynamic viscosity of air

Under standard sea-level conditions, the dynamic viscosity of air is typically around:

\[ \mu \approx 1.81 \times 10^{-5} , \text{Pa·s (Pas)} \]

2. Calculate the boundary layer thickness at \( x = 0.1 \) m under laminar flow

For a flat plate, the boundary layer thickness for laminar flow can be approximated using the formula:

\[ \delta = \frac{5 x}{\sqrt{\text{Re}}} \]

where:

  • \( \delta \) is the boundary layer thickness,
  • \( x \) is the distance from the leading edge,
  • \( \text{Re} \) is the Reynolds number.

The Reynolds number (\( \text{Re} \)) is calculated as:

\[ \text{Re} = \frac{U_\infty L}{\nu} \]

Where:

  • \( U_\infty = 100 , \text{m/s} \) is the free stream velocity,
  • \( L = x = 0.1 , \text{m} \) (we’ll use this value directly),
  • \( \nu = \frac{\mu}{\rho} \), where \( \rho \) is the air density.

Under standard sea-level conditions, the air density (\( \rho \)) is approximately:

\[ \rho \approx 1.225 , \text{kg/m}^3 \]

From the previous step, we found \( \mu \): \[ \mu \approx 1.81 \times 10^{-5} , \text{Pa·s} \]

Now calculate \( \nu \):

\[ \nu = \frac{\mu}{\rho} = \frac{1.81 \times 10^{-5} , \text{Pa·s}}{1.225 , \text{kg/m}^3} \approx 1.48 \times 10^{-5} , \text{m}^2/\text{s} \]

Now, substitute into the Reynolds number:

\[ \text{Re} = \frac{100 , \text{m/s} \times 0.1 , \text{m}}{1.48 \times 10^{-5} , \text{m}^2/\text{s}} \approx \frac{10}{1.48 \times 10^{-5}} \approx 675675.68 \]

Now, substituting into the boundary layer thickness equation:

\[ \delta = \frac{5 \times 0.1}{\sqrt{675675.68}} \approx \frac{0.5}{821.050} \approx 0.000609 , \text{m} \]

To convert this to mm:

\[ \delta \approx 0.000609 , \text{m} \times 1000 , \text{mm/m} = 0.609 , \text{mm} \]

Final Answers:

  • Dynamic viscosity of air: \[ \mu \approx 1.81 \times 10^{-5} , \text{Pas} \]

  • Boundary layer thickness at \( x = 0.1 , \text{m} \): \[ \delta \approx 0.609 , \text{mm} \]

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