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Homework Help: Liquid flow problem

  1. Mar 3, 2005 #1
    "A liquid is flowing through a horizontal pipe whose radius is 0.0200 m. The pipe bends straight upward through a height of 10.0m and joins another horizontal pipe whose radius is 0.0400 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?"

    Alright, I was thinking some combination of bernoulli's principle and the volume flow rate law because you've got varying height and varying area. How would I go about doing this? Am I on the right track?
     
  2. jcsd
  3. Mar 3, 2005 #2

    xanthym

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    From Bernoulli's Law for 2 points within a closed steady-state ideal fluid flow system:

    [tex] :(1): \ \ \ \ P_{1} \ + \ (1/2)\rho v_{1}^{2} + \ \rho gh_{1} = P_{2} \ + \ (1/2)\rho v_{2}^{2} \ + \ \rho gh_{2} [/tex]

    Problem statement indicates {P1 = P2}, {h1 = 0}, and {h2 = 10 m}, so Eq #1 simplifies to:

    [tex] :(2): \ \ \ \ (1/2)v_{1}^{2} = (1/2)v_{2}^{2} + (10)g [/tex]

    [tex] :(3): \ \ \ \ v_{1}^{2} \ - \ v_{2}^{2} \ = \ 196 [/tex]

    From Mass Conservation, {A = πr2}, {r1 = 0.02 m}, and {r2 = 0.04 m}:

    [tex] :(4): \ \ \ \ \rho A_{1} v_{1} = \rho A_{2} v_{2} = \rho (VolumeFlowRate) \ \ \Rightarrow \ \ v_{2} = v_{1} (\frac {r_{1}} {r_{2}})^{2} = v_{1}/4 [/tex]

    Placing Eq #4 into Eq #3, solving for v1, and determining (Volume Flow Rate):

    [tex] :(5): \ \ \ \ v_{1}^{2} \ - \ (v_{1}/4)^{2} \ = \ 196 [/tex]

    [tex] :(6): \ \ \ \ v_{1} = (14.46 \ \ m/sec) \ \ \Rightarrow \ \ \color{red} (VolumeFlowRate) = \pi r_{1}^{2} v_{1} = (0.0182 \ \ m^{3}/sec) [/tex]



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    Last edited: Mar 3, 2005
  4. Mar 3, 2005 #3
    thanks alot :smile:
     
  5. Oct 30, 2011 #4
    ^ wait is that right? how did u get 14.46 lol?

    shouldn't it be v1 - (1/4 * v1) = 14 and then 14/.75? or am i doing it wrong
     
  6. Oct 30, 2011 #5
    @ KendrickLamar
    It appears that you tried taking the square root of both sides.
    Note that
    [tex]\sqrt{v_2 ^2-v_1 ^2}\neq v_2 -v_1[/tex]
     
  7. Oct 30, 2011 #6
    wait im brain dead lol u gotta factor it or what, whats the easiest way to get from step 5 to step 6? why cant i figure this out lol!? i tried 1v1^2 - .25v1^2 = .75v1^2 and set that = 196 but that doesn't work either
     
    Last edited: Oct 30, 2011
  8. Oct 30, 2011 #7
    If you have
    [tex]v_1 ^2-\frac{1}{4}v_1 ^2[/tex]
    You can take out the common factor and subtract
    [tex]v_1 ^2(1-\frac{1}{4})=\frac{3}{4}v_1 ^2[/tex]
     
  9. Oct 30, 2011 #8
    thats what i was doing tho...

    3/4 = .75 so if u do 196/.75 then square root both sides u still get like 16 something or even if u square root both sides first then multiply that fraction across its still doesnt end up with 14.46
     
  10. Oct 30, 2011 #9
    Yeah, I just calculated it to be 16.17m/s.
     
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