Question

Air is streaming past a horizontal airplane wing such that its speed is $90m{{s}^{-1}}$ at the lower surface and $120m{{s}^{-1}}$ over the upper surface. If the wing is 10m long and has an average width of 2m, the difference of pressure on the two sides and the gross lift on the wing respectively, are (density of air$=1.3kg{{m}^{-3}}$)
A. 5Pa
B. 95Pa
C. 4098Pa
D. 4095Pa

Answer 1

We can recall the expression for Bernoulli’s theorem as a first step. We see that all the required quantities are directly given in the question and also the slight height difference can be neglected and thus we could find the pressure difference. Also, the lift force is given by the product of pressure difference and area.
Formula used:
Bernoulli’s theorem,
$P+\dfrac{1}{2}\rho {{v}^{2}}+\rho gh=k$

Complete answer:
We are given the velocity of air on the upper and lower surface of a horizontal airplane wing. We are also given the length as well as width of the wing and also the density of air is given. With these given information we are asked to find the difference of pressure on the two sides and the gross lift on the wing.
Velocity of the air over the lower surface is given by,
${{v}_{1}}=90m{{s}^{-1}}$
Velocity of the air over the upper surface is given by,
${{v}_{2}}=120m{{s}^{-1}}$
Length of the wing, $l=10m$
Width of the wing, $b=2m$
Therefore, the area of the wing could be given by,
$A=l\times b=10\times 2$
$\Rightarrow A=20{{m}^{2}}$ …………………………….. (1)
Density of air, $\rho =1.3kg{{m}^{-3}}$
Now we can recall Bernoulli's theorem which states that the total mechanical energy of the fluid, compressing energy related to the fluid pressure, the gravitational potential energy of elevation and the kinetic energy of fluid motion remains constant. That is,
$P+\dfrac{1}{2}\rho {{v}^{2}}+\rho gh=k$
For the aircraft wing we could ignore the small height difference between the top and bottom surfaces, therefore,
${{P}_{1}}+\dfrac{1}{2}\rho {{v}_{1}}^{2}={{P}_{2}}+\dfrac{1}{2}\rho {{v}_{2}}^{2}$
$\Rightarrow {{P}_{1}}-{{P}_{2}}=\dfrac{1}{2}\rho \left( {{v}_{2}}^{2}-{{v}_{1}}^{2} \right)$
$\Rightarrow \Delta P=\dfrac{1}{2}\left( 1.3 \right)\left( {{120}^{2}}-{{90}^{2}} \right)$
$\Rightarrow \Delta P=4095Pa$
Hence, we get the required pressure difference as 4095Pa.
We know that, lift force is given by,
Lift force = pressure difference × area
$\Rightarrow F=\Delta P\times A$
From (1),
$F=4095\times 20N$
$\Rightarrow F=81900N$
Therefore, we get the lift force as 81900N.
Hence, the pressure difference on the two sides and the gross lift on the wing are respectively given by, 4095Pa and 81900N.

So, the correct answer is “Option D”.

Note:
The law of conservation of energy is applied to fluid flow so as to produce Bernoulli’s equation. That is, the change in fluid’s kinetic energy and gravitational potential energy is taken as the net work is done. This equation can be varied according to the form of energy involved. The other forms can be dissipation of thermal energy due to viscosity.