2. As an airplane speeds up for take off air passes over the top of its wings. T
ID: 1515647 • Letter: 2
Question
2. As an airplane speeds up for take off air passes over the top of its wings. The air passing underneath the wings is at atmospheric pressure but the air passing over the top of the wings moves further and thus moves faster. That means the pressure of the air above the wings is less than atmospheric pressure. (a) If the mass of an airplane is 442,253 kg (Boeing 747), find the minimum lift force required to achieve take off. Ignore the buoyant force (the plane is submerged in air, after all) in this calculation since it is negligibly small. (b) The lift force comes from the pressure difference above/below the wings. If each of the plane’s TWO wings have an area of 23 m2, what is this pressure difference? (c) Assuming the plane takes off near sea level the pressure of the air beneath the wings will be atmospheric pressure. Use this information to find the pressure of the air above the wings. (d) If the convex wings are designed such that air moving above the wings is 25% faster than air below, find the speed of the air passing below the wings (aka the plane’s take off speed).
Explanation / Answer
1.Lift force need to be more than the weight of the plane.
Minimum lift force,
F = m*g
=442,253*9.81
=4338,501.83 N
2.Now, as we know,
Pressure diffrence P = F/A = 4338,501.83/23
=188630.52 pa
(c)
Pressure diffrence =P' - Patm , but pressure diffrence is bigger than Patm. So there's some problm in data.
(d)
Using Benoulli's equation
P1/rho*g +v1^2/2 = P2/rho*g +v2^2/2
P1/rho*g +v1^2/2 = P2/rho*g +(1.25*v1)^2/2
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