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Air flow speed at top of airplane wing

  • #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!
 

Answers and Replies

  • #2
enigma
Staff Emeritus
Science Advisor
Gold Member
1,747
9
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]
 
  • #3
Thank you.
 

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