Chapter 13: Problem 69

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 \(1.30 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified

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

Step by step solution

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#

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 \(375000\,N\).