Question

A Fully loaded Boeing aircraft has a mass of $3.3\times {{10}^{5}}kg$, its total wing area is $500{{m}^{2}}$. It is in Level Flight with a speed of $960km/h$.
(a) Estimate the pressure difference between the lower and upper surfaces of the wings
(b) Estimate the Fractional increase in the speed of air on the upper surface of the wing relative to the lower surface. (The Density of air is $\rho =1.2\text{ kg/}{{\text{m}}^{\text{3}}}$)

Answer 1

As in the question being aircraft is fully loaded and there is a difference in pressure and velocity which is calculated by applying Bernoulli's theorem
$\dfrac{{{P}_{1}}}{pg}+\dfrac{{{V}_{1}}^{2}}{2g}+{{z}_{1}}=\dfrac{{{P}_{2}}}{pg}+\dfrac{{{V}_{2}}^{2}}{2g}+{{z}_{2}}$

Complete answer:
As Given
Mass of fully loaded Boeing aircraft $m=3.3\times {{10}^{5}}kg$
Total wing area $500\text{ }{{m}^{2}}$
Speed in level$=960kmph$
$=960\times \dfrac{5}{18}m/s$
$=266.67m/s$
Density of air$\left( \rho \right)=1.2kg/{{m}^{3}}$

(a) As we know the weight of fully loaded Boeing aircraft is balanced by the pressure difference between lower and upper level or surfaces of the wings.
$Weight\text{ }of\text{ }Aircraft=pressure\text{ }difference\times Area$
$mg=4\Delta p\Rightarrow \Delta P=\dfrac{mg}{4}$
$\Rightarrow \Delta p=\dfrac{3.3\times {{10}^{5}}\times 9.8}{500}$
$\Rightarrow \Delta p=6.46\times {{10}^{3}}N/{{m}^{3}}$
Hence, the lower and pressure difference between upper surfaces of the wings $6.46\times {{10}^{3}}N/{{m}^{3}}$

(b) Fractional increase in the speed of the air on the upper surface of the wing relative to lower surface can be calculated with the Bernoull's equation
$\dfrac{{{p}_{1}}}{pg}+\dfrac{{{V}_{1}}^{2}}{2g}+{{z}_{1}}=\dfrac{{{p}_{2}}}{pg}+\dfrac{{{V}_{2}}^{2}}{2g}+{{z}_{2}}$
As ${{P}_{1}},{{V}_{1}},{{Z}_{1}}$ are the pressure Velocity and height of the upper surface
${{P}_{2}},{{V}_{2}},{{Z}_{2}}$ are the pressure, Velocity height from ground to the lower surface of wing respectively.
$\rho =$ Density of air
$g=$ gravitational Acceleration.
As ${{z}_{1}}={{z}_{2}}$
So, $\dfrac{{{P}_{1}}}{pg}+\dfrac{{{V}_{1}}^{2}}{2g}+{{z}_{1}}=\dfrac{{{P}_{2}}}{pg}+\dfrac{{{V}_{2}}^{2}}{2g}+{{z}_{2}}$
$\Rightarrow {{P}_{1}}+\dfrac{\rho {{V}_{1}}^{2}}{2}={{P}_{2}}+\dfrac{\rho {{V}_{2}}^{2}}{2}$
$\Rightarrow {{P}_{1}}-{{P}_{2}}=\rho \left( \dfrac{{{V}_{2}}^{2}-{{V}_{1}}^{2}}{2} \right)$
$\Rightarrow \Delta P=\rho \left( \dfrac{{{V}_{2}}+{{V}_{1}}}{2} \right)\left( {{V}_{2}}-{{V}_{1}} \right)$
As we know, ${{V}_{\arg }}=\dfrac{{{V}_{2}}+{{V}_{1}}}{2}=266.67m/s$
$\Rightarrow \Delta p=\rho .{{V}_{\arg }}\left( {{V}_{2}}-{{V}_{1}} \right)\text{ }\left( 1 \right)$
Fractional increase in the speed of the air
$\Rightarrow \dfrac{{{V}_{2}}-{{V}_{1}}}{{{V}_{\arg }}}=\dfrac{\Delta P}{\rho .{{V}^{2}}\operatorname{a}vg}$
$\Rightarrow \dfrac{{{V}_{2}}-{{V}_{1}}}{{{V}_{\arg }}}=\dfrac{6.46\times {{10}^{3}}}{1.2\times {{\left( 266.67 \right)}^{2}}}$
$\Rightarrow \dfrac{{{V}_{2}}-{{V}_{1}}}{{{V}_{avg}}}=0.075$
Hence, Fractional increase in the speed of the air o the upper surface of the wing relative to lower surface is $0.075$

Note:
As the wing is at the same level here so there will be no difference in the height of the upper surface of the wing. Also speed of Boeing I level flight should be considered as average speed of aircraft.