Question

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?
unanswered

What is the boundary layer thickness at x=0.1 m? Assume a laminar flow. Give your answer in mm.

unanswered

Answer 1

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 \( x = 0.1 \, \text{m} \):
\[
\delta \approx 0.609 \, \text{mm}
\]