An airplane has a mass of 2.8×106 kg , and the air flows past the lower surface of the wings at 81 m/s .
Part A:
If the wings have a surface area of 1100 m2 , how fast must the air flow over the upper surface of the wing if the plane is to stay in the air?
Express your answer to two significant figures and include the appropriate units.
To find the speed at which the air must flow over the upper surface of the wing, we can use the concept of Bernoulli's principle, which states that in a flowing fluid, the sum of the pressure, kinetic energy, and potential energy per unit volume is constant.
In this scenario, we need to consider the balance between the lift force exerted by the air flowing over the wings and the weight of the airplane to determine the required airflow speed.
First, let's find the lift force on the airplane. The lift force can be calculated using the equation:
Lift force = (pressure difference) × wing area
The pressure difference is the difference in air pressure between the upper and lower surface of the wing. As per Bernoulli's principle, the pressure decreases with an increase in air speed.
Next, we need to find the difference in air speeds between the upper and lower surfaces of the wing. We can assume that the air flowing on the upper surface follows the same path as the wing surface and does not separate from it.
Using Bernoulli's equation, we can write:
P + (1/2)ρV² + ρgh = constant
Where:
P is the pressure,
ρ is the density of the fluid (air),
V is the velocity of the fluid,
g is the acceleration due to gravity, and
h is the height (which we can assume to be constant).
Assuming that the height and gravity are constant, we can rearrange the equation to only consider the pressure and velocity terms.
P + (1/2)ρV² = constant
Since the airspeed over the lower surface is given as 81 m/s, and we need the velocity of the air flowing over the upper surface, let's represent the lower surface velocity as V1 and the desired upper surface velocity as V2.
P1 + (1/2)ρV1² = P2 + (1/2)ρV2²
Since the pressure difference on the upper and lower surfaces is due to the difference in air speeds, we can set P1 = P2.
(1/2)ρV1² = (1/2)ρV2²
Cancelling out the ρ factor:
V1² = V2²
Now, let's plug in the given values:
V1 = 81 m/s
A = 1100 m²
To solve for V2, we square both sides of the equation:
81² = V2²
V2² = 6561
V2 = √6561 = 81 m/s
Therefore, the air must flow over the upper surface of the wing at 81 m/s to keep the plane in the air.
Note: The upper surface airflow speed is the same as the lower surface airflow speed in this scenario.