First slide

Fluid dynamics

Question

An aircraft of mass $4 \times 10^{5} kg$ with total wing are $m^{2} in level flight at a speed of 720 km h^{- 1} . The density of air at its height is 1.2 kg m^{- 3}$ . The fractional increase in the speed of the air on the upper surface of its wings relative to the lower surface is (Take g = 10 ${ms}^{- 2}$ )

Moderate

A

0.04
B

0.08
C

0.17
D

0.34

Solution

The weight of the aircraft is balanced by the upward force due to the Pressure difference.
i.e.,
$∆ P = \frac{mg}{A} = \frac{(4 \times 10^{5} kg) (10 {ms}^{- 2})}{500 m^{2}} = \frac{4}{5} \times 10^{4} N m^{- 2}$
= $8 \times 10^{3} {Nm}^{- 2}$
Let $v_{1}, v_{2}$ are the speed of air on the lower and upper surface of the wings of the aircraft and $P_{1}, P_{2}$ are the pressure there.
Using Bernoulli's theorem, we get
$P_{1} + \frac{1}{2} {ρv}_{1}^{2} = P_{2} + \frac{1}{2} {ρv}_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ({ρv}_{2}^{2} - {ρv}_{1}^{2})$
$∆ P = \frac{ρ}{2} (v_{2} + v_{1}) (v_{2} - v_{1})$
or $v_{2} - v_{1} = \frac{∆ P}{{ρv}_{av}}$
Here, $v_{av} = \frac{v_{1} + v_{2}}{2} = 720 km h^{- 1}$
= 720 $\times \frac{5}{18} {ms}^{- 1} = 200 {ms}^{- 1}$
$\frac{v_{2} - v_{1}}{v_{av}} = \frac{∆ P}{{ρv}^{2}_{av}} = \frac{\frac{4}{5} \times 10^{4}}{1.2 \times {(200)}^{2}}$
= $\frac{4 \times 10^{4}}{5 \times 1.2 \times 4 \times 10^{4}} = 0.17$

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