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Airplane problem - help?

  1. Nov 13, 2003 #1
    Hi everyone…I have a airplane question:

    Q: The profile of a particular airplane wing is such that the distance air has to move to traverse the top of the wing is 20% greater than the distance the air must travel to traverse the bottom of the wing. (This shape is an “airfoil”.) If air is flowing steadily across the wing, then the air molecules that part ways (one to go over the top of the wing and one to go across the bottom of the wing) at the leading edge of the wing must come back together again at the trailing edge.

    Calculate the lifting force generated per square foot if the wing is moving through the air at 200 mph (airspeed measured across the bottom of the wing). What size wing (in square feet) is needed to support a 3000 lb aircraft?

    A: Bottom: P+½pv^2= Top: P+½v^2 because the pgy^2 cancel out
    Substitute for p= (F*d)/vol

    Top: v=107.28m/s; 9.8*1361.8kg=13346N; 3000lb=1361.8kg
    Bottom: v=89.4m/s; F=13346N; M=13346N

    13345.6N*d/vol + (.5*1361.5kg*(89.34m/s^)2)/vol= 13346N*1.2d/vol +(.5*1361.8kg*(107.28m/s)^2)/vol

    -2.6696N*d = 2394479Nm

    I think that I am on the wrong track because I came out with a negative distance. Can anyone help?

    Thanks in advance for your assistance!


    Totally Lost.
  2. jcsd
  3. Nov 13, 2003 #2
    Well, for one, you set the pressure on the underside of the wing to equal mg/A (where A is the area of the wing), and you set the pressure on the upper side of the wing equal to 1.2mg/A, for some reason I don't understand. (The pressure on the upper side of the wing is not 1.2 times the pressure on the underside; in fact, the pressure on the underside is greater, which is why the wing has lift.)

    You have,

    Ptop + 1/2 ρ v2top = Pbot + 1/2 ρ v2bot

    The net lifting force per unit area is,

    Fnet/A = ΔP = Pbot - Ptop

    and to support the aircraft, this must be at least mg/A.

    There is no need to solve for the thickness of the wing (d).

    P.S. Be careful with units; they want you to convert back to force per square foot instead of square meter ...
    Last edited: Nov 13, 2003
  4. Nov 13, 2003 #3


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    Also, be aware that despite what your professor or (I hope not) textbook say, what you are solving for here is NOT the way that Bernoulli works.

    Fluids are fluids, and as such, do not need to meet up at the end. In fact, they usually do not. They also do not follow a constant (or even linearly varying) velocity distribution - not even to a first approximation.

    The problem you're solving is a bogus model.
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