Questions: Use Bernoulli's equation to estimate the upward force on an airplane's wing if the average flow speed of air is 190 m / s above the wing and 160 m / s below the wing. The density of the air is 1.29 kg / m^3, and the area of each wing surface is 31.3 m^2.

Solution Steps

Bernoulli's equation relates the pressure and velocity of a fluid flow. For a streamline flow, it is given by:

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

where:

• \( P_1 \) and \( P_2 \) are the pressures at two points in the flow,
• \( \rho \) is the fluid density,
• \( v_1 \) and \( v_2 \) are the flow velocities at these points.

For the airplane wing, we consider the air above and below the wing as two points in the flow. The difference in pressure (\( \Delta P \)) between the bottom and top of the wing can be found using:

\[ \Delta P = P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \rho v_{\text{above}}^2 - \frac{1}{2} \rho v_{\text{below}}^2 \]

Substitute the given values:

• \( v_{\text{above}} = 190 \, \text{m/s} \)
• \( v_{\text{below}} = 160 \, \text{m/s} \)
• \( \rho = 1.29 \, \text{kg/m}^3 \)

\[ \Delta P = \frac{1}{2} \times 1.29 \times (190^2 - 160^2) \]

Calculate the pressure difference:

\[ \Delta P = \frac{1}{2} \times 1.29 \times (36100 - 25600) \]

\[ \Delta P = \frac{1}{2} \times 1.29 \times 10500 \]

\[ \Delta P = 6772.5 \, \text{Pa} \]

The upward force (\( F \)) on the wing is the pressure difference multiplied by the wing area (\( A \)):

\[ F = \Delta P \times A \]

Substitute the given area:

\[ A = 31.3 \, \text{m}^2 \]

\[ F = 6772.5 \times 31.3 \]

\[ F = 212,978.25 \, \text{N} \]

Final Answer