How Does Bernoulli's Equation Explain Airplane Lift?

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Homework Help Overview

The discussion revolves around the application of Bernoulli's equation to understand how an airplane generates lift. The problem involves calculating air velocities, pressure differences, and the implications of mass changes on take-off speed based on the wing area and airspeed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between airspeed and pressure difference, with calculations provided for velocity and pressure. Questions arise regarding the interpretation of the wing area and its implications for the problem setup.

Discussion Status

Some participants have provided calculations and reasoning regarding the pressure difference and its role in lift. Others have sought clarification on the problem's parameters, particularly concerning the wing area. There is an ongoing exploration of how changes in mass affect take-off speed.

Contextual Notes

Participants are discussing assumptions related to the negligible height difference between the top and bottom of the wing and the density of air used in calculations. The original poster's interpretation of the wing area as total versus per wing is also under consideration.

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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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