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Homework Help: Air flow speed at top of airplane wing

  1. Nov 20, 2003 #1
    Hello all,

    Here's my problem, verbatim:
    An aircraft wing requires a lift of 25.4 lb/ft^2. If the speed of flow of the air along the bottom surface of the wing is to be 500ft/s, what must be the speed of flow over the top surface of the wing to give the required lift?
    (NOTE: Density of air is 2.54 x 10^-3 slug/ft^3)

    F = 25.4lb/ft^2 = 0.789 slug/ft^2
    Vb = 500ft/s

    The equation I am using is
    FA = 1/2p(Vf^2-Vb^2)A
    However, I do not have the area for this problem and I was told that I don't need it, to use the equation Pb - Pt = 1/2p(Vf^2-Vb^2) and the 25.4 lb/ft^2 corresponds to (Pb-Pt). I've tried doing the equation, but units do not come out correctly.

    0.789 slug/ft^2 = 1/2(2.54 x 10^-3 slug/ft^3)(Vf^2 - (500ft/s)^2)
    0.789 slug/ft^2 = (1.27 x 10 ^-3 slug/ft^3)(Vf^2 - 250,000ft^2/s^2)
    0.789 slug/ft^2 = (1.27 x 10 ^-3 slug/ft^3)Vf^2 - 317.5 slug/fts^2
    0.789 slug/ft^2 + 317.5 slug/fts^2 = (1.27 x 10 ^-3 slug/ft^3)Vf^2
    621.26ft + 250,000ft^2/s^2 = Vf^2

    Where did I go wrong? Some people were getting 519, but that's only if you use 25.4 lb/ft^2, and you can't do that unless you ignore units of lb and slug.
    Thank you for your help and have a nice day!
     
  2. jcsd
  3. Nov 20, 2003 #2

    enigma

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    lb/ft^2 does not convert into slugs/ft^2

    It coverts into [tex]\frac{slug*\frac{ft}{sec^2}}{ft^2} [/tex]

    or

    [tex]\frac{slug}{ft*sec^2}[/tex]
     
  4. Nov 20, 2003 #3
    Thank you.
     
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