# Airplane lift bernoullis effect?

an airplane has a mass of 1.7*10^6 kg and air flows past the lower surface of the wings at 95m/s. if the wings have a surface area of 1200 m^2 how fast must the air flow over the upper surface of the wing if its to stay in the air...........??

that gives an answer of 242 m/s

...but the correct one is 170 m/s how??

### 3 Answers

- Al PLv 71 decade agoFavorite Answer
Need to assume air density?

The change in kinetic energy density

above and below the wing

∆P = Lift/area = (1/2)*ρ*(Vt^2-Vb^2)

∆P = Lift/area = weight /area

∆P = 1.7*10^6*9.81/1200 = 13897.5 N/m^2 --->using total area

Assuming air density ρ = 1.39 approx.

Vt = sqr(∆P*2/ρ + Vb^2)

Vt = sqr((13897.5*2/1.39)+95^2) = 170 m/s

- Anonymous1 decade ago
If you were to assume that the Bernoulli effect were solely responsible for lift, the pressure difference times the surface area of the wings would have to equal the weight of the plane.

1/2 density * (vupper^2 - vlower^2) * area = mass * g

Solve for the speed on the upper wing:

vupper = sqrt (2 m g / (rho A) + vlower^2)

They give you the mass of the plane (m), area of the wings (A), and speed over the lower wings (vlower). You know the acceleration due to gravity (g). Look up the density of air (rho). Plugnchug.

---Hmmm, they don't tell you if the area given is for one wing or for both--could that be the problem? Doubling the area would take the answer down a bit.

- kirchweyLv 71 decade ago
I get 174 m/s when density = 1.3 kg/m^3, 170 when it's 1.4 kg/m^3. I think you misapplied bekki's equation.

F = mg = 1.666E7 N

v1 = sqrt(2F/(area*density) + v2^2)