How Does Bernoulli's Equation Explain Airplane Lift?

AI Thread Summary
Bernoulli's equation explains airplane lift by illustrating the pressure difference created by varying airspeeds over the wing surfaces. As the airplane takes off at 90 km/h, the air on the top of the wing travels at 30 m/s, resulting in a pressure difference of 177 Pa between the top and bottom surfaces. This pressure difference generates an upward force sufficient to support the airplane's weight, calculated to be approximately 1806 kg. When the airplane's mass increases by 10%, the new take-off speed is determined to be around 26 m/s. The discussion clarifies that the 50 m² refers to the total wing area, not the wingspan.
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Homework Statement


An airplane has a 50 m^2 wing that is designed so that air on the top travels 20% faster
than the air on the bottom. The air on the bottom of the wing moves at the plane’s
airspeed and the unloaded airplane has a take-off speed of 90 km/h

A) What is the velocity of the air on top of the wing as the unloaded airplane becomes
airborne?

B) What pressure difference between the top and bottom of the wing as the unloaded
airplane becomes airborne?

C) What is the mass of the unloaded airplane?

D)If on a particular day, the mass of the airplane is increased by 10%, what is the new take-off speed?

Homework Equations


P1 + .5dv1^2 +dgy1 = P2 + .5dv2^2 + dgy2

The Attempt at a Solution


A)
90km/h = 25m/s
since the air on the top is 20 percent faster than on the bottom, then vtop = 1.2vbottom = 1.2(25) =30m/s

B)
P_top + .5d(v_top)^2 +dg(y_top) = P_bottom + .5d(v_bottom)^2 + dg(y_bottom)
assuming the difference in height between the top and bottom of the wing is negligible
P_bottom + .5d(v_bottom)^2 = Ptop + .5d(v_top)^2
P_bottom - Ptop = .5d(v_top^2 - v_bottom^2)
if i remember correctly the density of air is 1.29 kg/m^3
deltaP = .5(1.29)[30^2 - 25^2] = 177 pa

C) F = (deltaP)A = (177pa) (50m^2 + 50m^2) = 177(100) = 17700 N
F = mg
17700 = m (9.8)
m =1806 kg

D) this is the part I am not too sure about
m = 1806(1.1) = 1987 kg
F = 1987(9.8) = 19473 N
delaP = F/A = 19473/100 = 195

195 = .5(1.29)[(1.2x)^2 - x^2]
195 = .645(1.44x^2-x^2)
302 = x^2(1.44-1)
x^2 = 302/.44 = 686
x = 26 m/s
 
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c) The difference in the pressure on the top and the bottom
of the wing generates an upwards force of Δp A = (177 x 50) Newton
which needs to be such that it can support the plane's weight on
take off. Or alternatively
pbottom A - ptop A = weight
∴ Δp A = weight

d) Yes just solve it for the corrected pressure difference like you did.
 
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yeah that part confused me a little. did they mean the wingspan of the plane was 50m^2 or that each wing was 50m^2?
 
toothpaste666 said:
yeah that part confused me a little. did they mean the wingspan of the plane was 50m^2 or that each wing was 50m^2?

It means the total wing area. The wingspan is a linear magnitude. It can't be expressed in m2...
 
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ahh right. i get it now. thank you
 
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