A fully loaded Boeing aircraft has a mass of . Its total wing area is . It is in level flight with a speed of . If the density of air is is , the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface i

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Solution

Q.
A fully loaded Boing aircraft has a mass of $5.4 \times 10^{5} kg$ . Its total wing area is $500 m^{2}$ . It is in level flight with a speed of $1080 km/h$ . If the density of air $ρ$ is $1.2 {kg m}^{- 3}$ , the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be ( $g = 10 {m/s}^{2}$ ) [2023]

1 16

2 6

3 8

4 10

Ans.
(4)

$P_{2} A - P_{1} A = 5.0 \times 10^{5} \times g$

$P_{2} - P_{1} = \frac{5.4 \times 10^{6}}{500} = 5.4 \times 2 \times 10^{2} \times 10 = 10.8 \times 10^{3}$

$P_{2} + 0 + \frac{1}{2} ρ V_{2}^{2} = P_{1} + 0 + \frac{1}{2} ρ V_{1}^{2}$

$P_{2} - P_{1} = \frac{1}{2} ρ (V_{1}^{2} - V_{2}^{2}) = \frac{1}{2} ρ (V_{1} - V_{2}) (V_{1} + V_{2})$

$10.8 \times 10^{3} = \frac{1}{2} \times 1.2 (V_{1} - V_{2}) \times 2 \times 3 \times 10^{2}$

$10.8 \times 10 = 3.6 (V_{1} - V_{2})$

$V_{1} - V_{2} = 30$

$(\frac{V_{1} - V_{2}}{V}) \times 100 = \frac{30}{300} \times 100 = 10 %$

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