Physics, 27.01.2020 01:31 Zaida21

An airplane has a mass of 2×10^6 kg and air flows past the power surface of the wings at 100ms¯¹. if the wings have a surface area of 1200m², how fast must the air flow over the upper surface of the wing if the plane is to stay in the air? consider only the bernoulli's effect.

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Considering the Bernoulli's effect, air must flow over the upper surface of the wing at a speed of 129.1m/s if the plane is to stay in the air.

Given the data in the question;

Mass of airplane; m = 2 * 10^6 kg

Velocity ( air flows past the power surface of the wings); v = 100m/s

Surface area; A = 1200m^2

We know that the pressure on wings due to gravity must be the same as the diffidence in pressure above and below the wings if the plane is to stay in the air.

That is; Pressure below - Pressure above = Weight/Area

P - P' = mg/A

Now, using Bernoulli's principle

$P + \frac{1}{2} dv^2 = P' + \frac{1}{2} dv'^2$

We re-arrange

$P-P' = \frac{1}{2} dv'^2 - \frac{1}{2} dv^2$

$\frac{mg}{A} = \frac{1}{2} dv'^2 - \frac{1}{2} dv^2$

$\frac{mg}{A} = \frac{1}{2} dv'^2 - \frac{1}{2} dv^2\\\\\frac{mg}{A} = \frac{1}{2} d( v'^2 - v^2)$

We know that density of air; d = 1.225 kg/m^3 and acceleration due to gravity; g = 9.8m/s^2

We substitute our values into the equation

$\frac{(2*10^6kg)*9.8m/s^2}{1200m^2} = \frac{1}{2} * 1.225kg/m^3( v'^2 - (100m/s)^2)\\\\\frac{1.96*10^7kg.m/s^2}{1200m^2} = 0.6125kg/m^3 ( v'^2 - (100m/s)^2)\\\\16333.33kg/ms^2 = 0.6125kg/m^3 ( v'^2 - (100m/s)^2)\\\\( v'^2 - (100)^2) = \frac{16333.33kg/ms^2}{0.6125kg/m^3} \\\\( v'^2 - 10000m^2/s^2) = 26666.66m^2/s^2\\\\v'^2 = 26666.66m^2/s^2 - 10000m^2/s^2\\\\v'^2 = 16666.66m^2/s^2\\\\v' = \sqrt{16666.66m^2/s^2}\\\\v' = 129.1 m/s$

Therefore, considering the Bernoulli's effect, air must flow over the upper surface of the wing at a speed of 129.1m/s if the plane is to stay in the air.

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Answer from: whitakers87

190 m/s

Explanation:

For the plane to stay in the air, the lift force must equal the weight.

The lift force is also equal to the pressure difference across the wings (pressure at the bottom minus pressure at the top) times the area of the wings.

Therefore:

mg = (P₂ − P₁) A

P₂ − P₁ = mg / A

Using Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

½ ρ (v₁² − v₂²) = P₂ − P₁

½ ρ (v₁² − v₂²) = mg / A

v₁² − v₂² = 2mg / (Aρ)

v₁² = v₂² + 2mg / (Aρ)

Substituting values (assuming air density of 1.225 kg/m³):

v₁² = (100 m/s)² + 2 (2×10⁶ kg) (9.8 m/s²) / (1200 m² × 1.225 kg/m³)

v₁² = 36,666.67 m²/s²

v₁ = 191 m/s

Rounding to two significant figures, the air must move at 190 m/s over the top of the wing.

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An airplane has a mass of 2×10^6 kg and air flows past the power surface of the wings at 100ms¯¹. if...

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