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,900N \\\ & B.~~95Pa,900N \\\ & C.~~4095Pa,900N \\\ & D.~~4095Pa,81900N \\\ \end{aligned}$$

Answer 1

This problem can be solved by applying Bernoulli theorem which will give us the pressure difference obtained by equating the total pressure at the bottom surface of the wing and top surface of the wing. After calculating the pressure difference between the two sides, the net lift on the wing can be calculated by multiplying this pressure difference with the area of the wing.

Complete step-by-step answer:
This problem can be solved by using Bernoulli theorem, which relates the pressure, elevation and velocity of any two points in a moving fluid. Here the fluid is air and the two points are the top and bottom surfaces of the wing.
Let us assume pressure at bottom surface of wing to be ${{P}_{1}}$, density of air to be $\rho$, velocity of liquid at that point ${{v}_{1}}$ and height of the point be ${{h}_{1}}$. Mathematically, Bernoulli theorem we can state that,
${{P}_{1}}+\dfrac{\rho {{v}_{1}}^{2}}{2}+\rho g{{h}_{1}}$
remains constant for any point within the fluid.
If we assume pressure at top surface of wing to be to be ${{P}_{2}}$, density of air to be $\rho$, velocity of liquid at that point ${{v}_{2}}$ and height of the point be ${{h}_{2}}$. Using Bernoulli theorem we can write
${{P}_{1}}+\dfrac{\rho {{v}_{1}}^{2}}{2}+\rho g{{h}_{1}}={{P}_{2}}+\dfrac{\rho {{v}_{2}}^{2}}{2}+\rho g{{h}_{2}}$
Putting values of variable from the question
${{P}_{1}}+\dfrac{\left( 1.3 \right){{\left( 90 \right)}^{2}}}{2}+\left( 1.3 \right)\left( 9.8 \right){{h}_{1}}={{P}_{2}}+\dfrac{\left( 1.3 \right){{\left( 120 \right)}^{2}}}{2}+\left( 1.3 \right)\left( 9.8 \right){{h}_{2}}$
${{P}_{1}}-{{P}_{2}}=\left( \dfrac{\left( 1.3 \right){{\left( 120 \right)}^{2}}}{2}-\dfrac{\left( 1.3 \right){{\left( 90 \right)}^{2}}}{2} \right)+\left( 1.3 \right)\left( 9.8 \right)\left( {{h}_{2}}-{{h}_{1}} \right)$ … (1)
As the width of the wing is two meter we can write
$\left( {{h}_{2}}-{{h}_{1}} \right)=2$
Putting this value in the equation 1
${{P}_{1}}-{{P}_{2}}=\left( \dfrac{\left( 1.3 \right){{\left( 120 \right)}^{2}}}{2}-\dfrac{\left( 1.3 \right){{\left( 90 \right)}^{2}}}{2} \right)+\left( 1.3 \right)\left( 9.8 \right)\left( 2 \right)=4095Pa$
Force applied on the wing from bottom of surface of wing in upward direction will be
${{F}_{1}}={{P}_{1}}A$
Force applied on the wing from top of surface of wing is in downward direction will be
${{F}_{2}}={{P}_{2}}A$
The area of the wing is a product of its length and breadth. So,
$A=10\times 2=20{{m}^{2}}$
The gross uplift on the wing will be
${{F}_{1}}-{{F}_{2}}=\left( {{P}_{1}}-{{P}_{2}} \right)A=\left( {{P}_{1}}-{{P}_{2}} \right)20$
Putting value of $\left( {{P}_{1}}-{{P}_{2}} \right)$ which we calculated
${{F}_{1}}-{{F}_{2}}=4095\times 20=81,900N$
Hence the correct option is D.

Note: Bernoulli theorem is applicable only when
The flow of the liquid is streamlined. In liquid with streamline flow the velocity at a point in liquid does not change with time.
The liquid must be incompressible, this means density cannot vary at any point within a liquid.
The viscosity of the liquid must be zero.