Problem 69 - University Physics with Modern Physics

Question

An airplane is moving through the air at a velocity \(v=200 . \mathrm{m} / \mathrm{s} .\) Streamlines just over the top of the wing are compressed to \(80.0 \%\) of their original area, and those under the wing are not compressed at all. a) Determine the velocity of the air just over the wing. b) Find the difference in the pressure between the air just over the wing, \(P\), and that under the wing, \(P\). c) Find the net upward force on both wings due to the pressure difference, if the area of the wing is \(40.0 \mathrm{~m}^{2}\) and the density of the air is \(1.30 \mathrm{~kg} / \mathrm{m}^{3}\).

Step-by-Step Solution

Answer

Answer: The net upward force on both wings due to the pressure difference is 375,000 N.

Step 1: We will apply Bernoulli's Equation between two points in the flowing air: one just over the wing and one just under the wing. Let's name them Point 1 and Point 2 respectively. We can ignore the difference in height between them, since it's negligible in this case. The equation is: \(P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\) where \(P_1, P_2\) are pressures, \(\rho\) is the air density, and \(v_1, v_2\) are the velocities at Point 1 and Point 2, respectively. #Step 2: Find the velocity of the air just over the wing (v2)#

Since the streamlines over the wing are compressed to 80% of their original area, we can apply the principle of mass conservation, or the continuity equation, which states that for an incompressible fluid: \(A_1v_1 = A_2v_2\) We are given that \(v_1 = 200\,m/s\) and \(A_2 = 0.8A_1\). So, we can find \(v_2\) as follows: \(v_2 = \frac{A_1}{A_2} v_1 = \frac{1}{0.8} v_1 = 1.25 v_1\) \(v_2 = 1.25 \times 200\,m/s = 250\, m/s\) So, the velocity of the air just over the wing is \(250\, m/s\). #Step 3: Find the difference in pressure between the air over and under the wing#

Step 2: Using the Bernoulli's Equation from Step 1, and plugging in the calculated velocities and given air density, we can find the pressure difference: \(\Delta P = P_2 - P_1 = \frac{1}{2} \rho (v_1^2 - v_2^2)\) \(\Delta P = \frac{1}{2} \cdot 1.30\,kg/m^3 (200^2 - 250^2)\,m^2/s^2\) \(\Delta P = -9375\,Pa\) The negative sign indicates that the pressure over the wing is lower than the pressure under the wing. Therefore, the difference in pressure is \(9375\,Pa\). #Step 4: Calculate the net upward force on both wings#

The net upward force on both wings due to the pressure difference can be found using the equation: \(F_{upward} = \Delta P \cdot A\) where \(A\) is the wing area. We are given that the wing area is \(40\,m^2\). So, the net upward force is: \(F_{upward} = 9375\,Pa \times 40\,m^2 = 375000\,N\) Therefore, the net upward force on both wings due to the pressure difference is \(375000\,N\).

Key Concepts

Continuity Equation

The Continuity Equation is key to understanding fluid dynamics in incompressible flow. It states that the product of cross-sectional area and velocity remains constant along a streamline. This means that if the area decreases, velocity must increase to conserve mass (assuming constant density). For our problem, the streamlines over the wing are compressed to 80% of their original area. To calculate the new velocity over the wing, we apply:

• \(A_1v_1 = A_2v_2\)
• Given: \(v_1 = 200\, m/s\)
• \(A_2 = 0.8A_1\)
• Thus, \(v_2 = \frac{1}{0.8}v_1 = 1.25 \times 200 = 250\, m/s\)

This shows how the velocity increases over the wing, crucial for lift generation.

Pressure Difference

Pressure difference is crucial in explaining the lift on wings. Bernoulli’s Principle helps us understand this. According to the principle, as the speed of a fluid increases, its pressure decreases. For an airplane wing:

• Top of wing: Higher velocity \(\Rightarrow\) Lower pressure
• Bottom of wing: Lower velocity \(\Rightarrow\) Higher pressure

By Bernoulli's Equation:

• \( \Delta P = P_2 - P_1 = \frac{1}{2} \rho (v_1^2 - v_2^2)\)
• With \( \rho = 1.30\,kg/m^3\), \(v_1 = 200\,m/s\), and \(v_2 = 250\,m/s\)
• \( \Delta P = \frac{1}{2} \times 1.30 \times (200^2 - 250^2) = -9375\,Pa\)

The negative sign indicates pressure is lower over the wing, aiding lift.

Velocity of Airflow

Understanding the velocity of airflow helps explain how aircraft generate lift. Using the continuity equation, we increase the velocity if the area through which air flows decreases. This is observed in airplane wings where air over the top surface moves faster due to compression in area:

• Streamlines are more compressed above the wing, translating to increased velocity.
• Faster airflow over the top decreases pressure due to Bernoulli's principle.

In our exercise, a 20% reduction in area results in air velocity increasing to \(250\,m/s\) over the wing. This faster speed is essential for achieving lift.

Lift Force Calculation

Lift force is crucial for flight, arising from pressure differences above and below the wing. This drives the force upward, enabling airplanes to lift off. The calculation involves:

• The pressure difference \(\Delta P = 9375\,Pa\) from above.
• Wing area \(A = 40\,m^2\).
• The lift force formula: \(F_{upward} = \Delta P \cdot A\)
• Thus, \(F_{upward} = 9375 \times 40 = 375000\,N\)

This shows the sizable force that acts on the wings, allowing for flight.