
In a test experiment on a model airplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are $ 70\,m/s $ and $ 63\,m/s $ respectively. What is the lift on the wing if its area is $ 2.5{\text{ }}{m^2} $ ? Take the density of air to be $ 1.3\,kg/{m^3} $ .
Answer
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Hint: In this solution, we will use the formula of pressure due to the flow of fluids to determine the difference in pressures on the wings of the plane. We will treat the flow of air as a fluid flow.
Formula used
In this solution, we will use the following formula:
-Pressure due to velocity flow: $ P = \dfrac{1}{2}\rho {v^2} $ where $ \rho $ is the density of the air and $ v $ is the speed of the airflow
- Force due to pressure $ F = P.A $ where $ P $ is the pressure and $ A $ is the area
Complete step by step answer:
When the air passes through the two different surfaces of wings, it has different speeds as given in the question. This difference in velocities generates a pressure difference between the sides of the wings such that the upper side has a lower pressure and the lower side has a higher pressure.
This pressure difference can be determined as the difference of the pressures created due to the flow of air over the wings.
The pressure difference between the wings will be
$ \Delta P = \dfrac{1}{2}\rho v_2^2 - \dfrac{1}{2}\rho v_1^2 $ where $ {v_2} $ is the velocity of air below the wing and $ {v_1} $ is the velocity above the wing.
WE can take out $ \dfrac{1}{2}\rho $ common and write
$ \Delta P = \dfrac{1}{2}\rho (v_2^2 - v_1^2) $
Hence substituting $ {v_2} = 70\,m/s $ and $ {v_1} = 63\,m/s $ , we get
$ \Delta P = \dfrac{1}{2} \times 1.3 \times ({70^2} - {63^2}) $
$ \Delta P = 605.15\,N/{m^2} $
Then the force acting on the plane will be
$ F = \Delta P.A $
$ \Rightarrow F = 605.15 \times 2.5 $
Which gives us
$ F = 1512.87\,N $
Which will be forced/lifted on the airplane.
Note:
In the airplane, the wind speeds are constant above and below the wing. However, the wings are designed such that the wind passes above and below them with different velocities. This creates an upwards thrust and helps keep the airplane floating.
Formula used
In this solution, we will use the following formula:
-Pressure due to velocity flow: $ P = \dfrac{1}{2}\rho {v^2} $ where $ \rho $ is the density of the air and $ v $ is the speed of the airflow
- Force due to pressure $ F = P.A $ where $ P $ is the pressure and $ A $ is the area
Complete step by step answer:
When the air passes through the two different surfaces of wings, it has different speeds as given in the question. This difference in velocities generates a pressure difference between the sides of the wings such that the upper side has a lower pressure and the lower side has a higher pressure.
This pressure difference can be determined as the difference of the pressures created due to the flow of air over the wings.
The pressure difference between the wings will be
$ \Delta P = \dfrac{1}{2}\rho v_2^2 - \dfrac{1}{2}\rho v_1^2 $ where $ {v_2} $ is the velocity of air below the wing and $ {v_1} $ is the velocity above the wing.
WE can take out $ \dfrac{1}{2}\rho $ common and write
$ \Delta P = \dfrac{1}{2}\rho (v_2^2 - v_1^2) $
Hence substituting $ {v_2} = 70\,m/s $ and $ {v_1} = 63\,m/s $ , we get
$ \Delta P = \dfrac{1}{2} \times 1.3 \times ({70^2} - {63^2}) $
$ \Delta P = 605.15\,N/{m^2} $
Then the force acting on the plane will be
$ F = \Delta P.A $
$ \Rightarrow F = 605.15 \times 2.5 $
Which gives us
$ F = 1512.87\,N $
Which will be forced/lifted on the airplane.
Note:
In the airplane, the wind speeds are constant above and below the wing. However, the wings are designed such that the wind passes above and below them with different velocities. This creates an upwards thrust and helps keep the airplane floating.
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