Question

The Boeing 777 aircraft has a takeoff mass of 240000kg and a takeoff speed of 90m/s . Assume that's the speed of air across the wing's bottom. The total surface area of both wings is 428m(sqrd). What airflow speed across the top of the wing is necessary for the plane to fly?

I know I have to use bernoulli's equation, but I'm not sure how to get all the variables. Could someone help, and also explain the process please?

Answer 1

The velocity over the top of the wing has to be = 155 m/s at *minimum* to get it off the ground.

Assuming that the wing thickness is negligible. Also assume

? = 1.292 kg/m^3
m = 240,000 kg
Vb = 90 m/s
A = 428 m^2
g = 9.81 m/s^2

Where

rho = Air density at 0 Celsius, at sea level
m = Mass of the airplane
Vb = Velocity of air on the bottom of the wing
A = Area of total wing
g = Gravity

Find the normal force of the aircraft

W = m * g
W = (240,000 kg) * (9.81 m/s^2)
W = 2,354,400 N

So you need that minimum force.

Using Bernoilli.

Pressure on bottom of wing

Pb = rho * [ (Vb^2) /2 ]
Pb = (1.292 kg/m^3) * [ (90 m/s)^2 / 2 ]
Pb = (1.292 kg/m^3) * [ (8100 m^2/s^2) / 2 ]
Pb = (1.292 kg/m^3) * [ 4050 m^2/s^2 ]
Pb = 5232.6 Pa

Force = [ Pressure over top - Pressure over bottom ] / Area

F = A * [Pt - Pb]

Solve for Pt given what we now have.

Pt = [F/A] +Pb

Pt = [ (2,354,400 N) / (428 m^2) ] + (5232.6 Pa)
Pt = [ 5,500.93 Pa ] + (5232.6 Pa)
Pt = 10,733.53 Pa

Pt = rho * [ (Vt^2) /2 ]

Fix equation for Vt gives

Vt = SQRT { [ 2 * Pt ] / rho }
Vt = SQRT { [ 2 * (10733.53 Pa) ] / (1.292 kg/m^3) }
Vt = SQRT { [ 21,467.06 Pa ] / (1.292 kg/m^3) }
Vt = SQRT { 16615.37 m^2/s^2 }
Vt = 128 m/s