Question

Consider an airfoil in a flow of air, where far ahead (upstream) of the airfoil, the pressure, velocity and density are:

1.01 10 N/m, 150 km/h and 1.225 kg/m respectively.

At a point A on the airfoil the pressure is 9.95 10 N/m.

Calculate the velocity at point A (in m/s):

Answer 1

To determine the velocity at point A on the airfoil, we can use Bernoulli's equation, which for incompressible flows can be expressed as:

\[
P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2
\]

Where:
- \(P_1\) and \(P_2\) are the pressures at initial and final points (upstream and point A),
- \(\rho\) is the air density,
- \(v_1\) and \(v_2\) are the velocities at the respective points.

Given data are:
- Upstream pressure \(P_1 = 1.01 \times 10^5 \, \text{N/m}^2\)
- Upstream velocity \(v_1 = 150 \, \text{km/h} = \frac{150 \times 1000\, \text{m}}{3600\, \text{s}} = 41.67 \, \text{m/s}\)
- Air density \(\rho = 1.225 \, \text{kg/m}^3\)
- Pressure at point A \(P_2 = 9.95 \times 10^4 \, \text{N/m}^2\)

We want to calculate the velocity at point A, denoted as \(v_2\).

Using Bernoulli's equation, we rearrange it to solve for \(v_2\):

\[
P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2
\]

Substituting the known values:

\[
1.01 \times 10^5 + \frac{1}{2} \cdot 1.225 \cdot (41.67)^2 = 9.95 \times 10^4 + \frac{1}{2} \cdot 1.225 \cdot v_2^2
\]

Calculating \(\frac{1}{2} \cdot 1.225 \cdot (41.67)^2\):

1. Calculate \((41.67)^2 \approx 1736.89\),
2. Then calculate \(\frac{1}{2} \cdot 1.225 \cdot 1736.89 \approx 1065.16 \, \text{N/m}^2\).

Now substitute this back into the equation:

\[
1.01 \times 10^5 + 1065.16 = 9.95 \times 10^4 + \frac{1}{2} \cdot 1.225 \cdot v_2^2
\]

Calculating the left side:

\[
1.01 \times 10^5 + 1065.16 \approx 102155.16 \, \text{N/m}^2
\]

Calculating the right side:

\[
9.95 \times 10^4 = 99500 \, \text{N/m}^2
\]

Putting everything together:

\[
102155.16 = 99500 + \frac{1}{2} \cdot 1.225 \cdot v_2^2
\]

Subtract \(99500\) from both sides:

\[
102155.16 - 99500 = \frac{1}{2} \cdot 1.225 \cdot v_2^2
\]

\[
1655.16 = \frac{1.225}{2} \cdot v_2^2
\]

Now, calculating \(\frac{1.225}{2} \approx 0.6125\):

\[
1655.16 = 0.6125 \cdot v_2^2
\]

Now solve for \(v_2^2\):

\[
v_2^2 = \frac{1655.16}{0.6125} \approx 2702.43
\]

Taking the square root to find \(v_2\):

\[
v_2 \approx \sqrt{2702.43} \approx 52.0 \, \text{m/s}
\]

Therefore, the velocity at point A is approximately:

\[
\boxed{52.0 \, \text{m/s}}
\]