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Water flow in pipe with two open vertical offshoots

  1. Oct 29, 2014 #1

    rlc

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    1. The problem statement, all variables and given/known data
    Water (η = 1.00E-3 Pa·s) is flowing through a horizontal pipe with a volume flow rate of 0.0125 m3/s. As the drawing below shows, there are two vertical tubes that project from the pipe. Assume that H = 0.0458 m and L = 0.738 m.
    upload_2014-10-29_17-18-11.png
    Calculate the radius of the horizontal pipe.

    3. The attempt at a solution
    I've seen this problem posted on this website before, but the solution there didn't work for me.
    I know you should use Poiseuille's Law, but what do you do about the pressures?
     
  2. jcsd
  3. Oct 29, 2014 #2
    Do you know how to determine the pressure difference between the two locations in the pipe? If so, what is it?

    Chet
     
  4. Oct 29, 2014 #3

    rlc

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    No, but I know it is part of Poiseuille's Law.
     
  5. Oct 29, 2014 #4
    You don't need to use Poiseuille's law to find the pressure difference between the two locations in the pipe. You already have enough information to determine it. In those two vertical offshoots, is the fluid static or is it flowing?

    Chet
     
  6. Oct 29, 2014 #5

    rlc

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    It would be static? Since it is open and the water wouldn't really go anywhere?
     
  7. Oct 29, 2014 #6

    rlc

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    Would the pressure difference just be H?
     
  8. Oct 29, 2014 #7
    H is what we would call the difference in static pressure head. Do you remember the equation for static pressure as a function of depth? (It involves density and g).

    Chet
     
  9. Oct 29, 2014 #8

    rlc

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    roh*g*h
     
  10. Oct 29, 2014 #9
    Yes. That's right. Then what is the pressure difference between the two locations in the pipe in Pa?

    Chet
     
  11. Oct 29, 2014 #10

    rlc

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    I'm sorry, but I don't know
     
  12. Oct 29, 2014 #11

    rlc

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    Wait, would it be density times velocity squared?
     
  13. Oct 29, 2014 #12
    ρ=1000 kg/m3 (for water)
    g = 9.8 m/s2
    Δh= H = 0.0458 m
    ΔP=ρgΔh (in units of Pa = kg/ms^2)

    What do you get for ΔP?
     
  14. Oct 29, 2014 #13

    rlc

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    (1000)(9.8)(0.0458)=448.84 Pa
     
  15. Oct 29, 2014 #14

    rlc

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    Ah! That worked! Thank you so much!
     
  16. Oct 29, 2014 #15
    Oops. You already got it.


    Right. That's the pressure difference between the two locations.

    Now, can you write down the Poiseulle equation for the pressure difference between the two locations in terms of the volumetric flow rate, the viscosity, the distance L, and the tube diameter D?

    Chet
     
  17. Oct 29, 2014 #16

    rlc

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    Q=[(pi)R^4(roh*g*H)]/8nL

    0.0125=(pi*R^4*448.84)/(8*1.00E-3*0.738)

    R^4=5.23377E-8
    R=0.015125 m
    R=1.51 cm (which LONCAPA says is the correct answer)

    Thank you so much for responding to me!
     
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