Question

The speed of airflow on the upper and lower surfaces of a wing of an aeroplane are ${v_1}$ and ${v_2}$ respectively. If A is the cross section area of the wing and \[\rho \] is the density of air, then the upward lift is :
A) $\dfrac{1}{2}\rho A\left( {{v_1} - {v_2}} \right)$
B) $\dfrac{1}{2}\rho A\left( {{v_1} + {v_2}} \right)$
C) $\dfrac{1}{2}\rho A\left( {v_1^2 - v_2^2} \right)$
D) $\dfrac{1}{2}\rho A\left( {v_1^2 + v_2^2} \right)$

Answer 1

Hint: According to Bernoulli's principle for any streamline motion total energy associated to the system i.e. the sum of fluid pressure, gravitational potential energy and kinetic energy always remain constant. For the motion of aeroplane, this principle simply explains the reason behind the extra pressure exerted by air on its wing due to the velocity difference at both sides.

Formula Used:
Bernoulli's principle applied at two points of streamline fluid motion:
$\dfrac{{\rho v_1^2}}{2} + \rho g{h_1} + {P_1} = \dfrac{{\rho v_2^2}}{2} + \rho g{h_2} + {P_2}$ (1)
Where,
$\rho $ is the density of the fluid,
${v_1}$ and ${v_2}$ are the velocity of the fluid at first and second point respectively,
${h_1}$ and ${h_2}$ are elevation of the first and second point respectively,
g is acceleration due to gravity,
${P_1}$ and ${P_2}$ are pressure exerted by the fluid at the first and second point respectively.

Force and pressure relationship:
$F = P.A$ (2)
Where,
F is total force experienced,
P is pressure exerted,
A is the total surface area.

Complete step by step answer:
Given:
1. Speed of airflow on the upper surface of the wing is ${v_1}$.
2. Speed of airflow on the lower surface of the wing is ${v_2}$.
3. Density of air is $\rho $.
4. Cross-section area of the wing is A.
5. Aeroplane wing is very thin compared to the plane’s elevation. So, effectively ${h_1} = {h_2} = h$.

To find: Upward lift on the plane’s wing.

Step 1
Substitute all the variables of eq.(1) with the given value to find the pressure difference $\Delta P = {P_2} - {P_1}$ as:
$
  \dfrac{{\rho v_1^2}}{2} + \rho gh + {P_1} = \dfrac{{\rho v_2^2}}{2} + \rho gh + {P_2} \\
  \therefore \Delta P = {P_2} - {P_1} = \dfrac{1}{2}\rho \left( {v_1^2 - v_2^2} \right) \\
 $ (3)

Step 2
Now, use the value of pressure difference from eq.(3) in eq.(2) to get the upward lift force value as:
$
  F = \Delta P.A \\
  \therefore F = \dfrac{1}{2}\rho A\left( {v_1^2 - v_2^2} \right) \\
 $

Correct answer:
The upward lift is (c) $\dfrac{1}{2}\rho A\left( {v_1^2 - v_2^2} \right)$.

Note: This problem can be solved without doing any calculation just by using logic. For the total energy to remain constant the difference of pressure between two surfaces must arise from their kinetic energy difference since the gravitational potential is effectively the same for both surfaces. Now, the kinetic energy is proportional to square of velocity. Hence, the only possible option is option (c).