First slide

Fluid dynamics

Question

A plane is in level flight at constant speed and each of its two wings has an area of 25 m². If the speed of the air on the upper and lower surfaces of the wing are 270 km / h and 234 km/h respectively, then what must have been the mass of the plane? (Take the density of the air = $1 kg m^{- 3}$ )

Moderate

A

1550 kg
B

1750 kg
C

3500 kg
D

3200 kg

Solution

Let $v_{1}, v_{2}$ are the speed of air on the lower and upper surface S of the wings of the plane P1 and P2 are the pressure there,
According to Bernoulli's theorem
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} (ρ v_{2}^{2} - v_{1}^{2})$
$Here, v_{1} = 234 {kmh}^{- 1} = 234 \times \frac{5}{18} {ms}^{- 1}$
$= 65 {ms}^{- 1}$
$v_{2} = 270 {kmh}^{- 1} = 270 \times \frac{5}{18}$
$= 75 {ms}^{- 1}$
$Area of wings = 2 \times 25 m^{2} = 50 m^{2}$
$∴ P_{1} - P_{2} = \frac{1}{2} \times (75^{2} - 65^{2})$
Upward force on the plane
$= (P_{1} - P_{2}) A$
$= \frac{1}{2} \times 1 \times (75^{2} - 65^{2}) \times 50 m$
As the plane is in level flight, therefor upward force balances the weight of the plane.
$∴ mg = (P_{1} - P_{2}) A$
Mass of the plane,
$m = \frac{(P_{1} - P_{2})}{g} A = \frac{1}{2} \times \frac{1 \times (75^{2} - 65^{2})}{10} \times 50$
$= \frac{(75 + 65) (75 - 65) \times 50}{2 \times 10} = 3500 kg$

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