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Siphoning gas find velocity

  1. Jun 11, 2009 #1
    A siphon tube is filled with gasoline and closed at each end. One end is inserted into a gasoline tank 0.20 m below the surface of the gasoline. The outlet is placed outside the tank at a distance 0.45 m below the surface of the gasoline. The tube has an inner cross-sectional area of 3.8 × 10-4 m2. The density of gasoline is 680 kg/m3. Ignoring viscous effects, what is the velocity of the gasoline in the tube shortly after the tube is opened?

    P1+.5*rho*v1^2+rho*g*y1=P2+.5*rho*v2^2+rho*g*y2
    1 atm = 1.05E5

    So I plugged in 1.05E5+0 (because v at top=0)+680*9.8*-.2 = 1.05E5+.5*680*v2^2+680*9.8*-.44
    Completing this calculation gives v2= 2.17m/s which is incorrect. Any suggestions on what I am doing wrong?

    And to find the flow rate I would use the equation Q=Av and use the provided area given with the v discovered?
     
  2. jcsd
  3. Jun 11, 2009 #2

    tiny-tim

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    Hi smillphysics! :smile:

    (have a rho: ρ and try using the X2 tag just above the Reply box :wink:)
    but -.2 isn't the height at the top :wink:

    Yup! :biggrin:
     
  4. Jun 11, 2009 #3
    What is the height at the top? I then tried to put .45-.2=.25 and used that as the height at the top but that is also incorrect.
     
  5. Jun 11, 2009 #4
    Any help on this would be great- I can't seem to find the correct y1 to use.
     
  6. Jun 12, 2009 #5

    tiny-tim

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    Hi smillphysics! :smile:

    (Sorry I didn't reply earlier :redface:)

    y1 is at the surface

    it doesn't matter where the top of the tube is, or what shape it is, the pressure at each height is the same. :wink:
     
  7. Jun 12, 2009 #6
    I've used the same pressure at both points. I still don't know where I have gone wrong?
     
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