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Homework Help: Continuity Equation

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data
    The inside diameters of the larger portions of the horizontal pipe as shown in the image (attached) are 2.50 cm. Water flows to the right at a rate of 1.80*10^4 m^3/s. What is the diameter of the constriction.
    th_Photo5.jpg


    2. Relevant equations Continuity equation Rate of Volume flow=constant.



    3. The attempt at a solution
    I got a proportion worked down to
    [tex]\frac{V_i}{V_o}=\frac{D_o^2}{D_i^2}[/tex]

    where V is velocity and the subscripts i and o denote in and out respectively.


    I am having trouble figuring out how to utilize the two heights of the fluids in the vertical tubes.

    RW

    Edit: I am now considering using Bernoulli's equation to take into account pressure. Sound better?
     
    Last edited: Nov 11, 2007
  2. jcsd
  3. Nov 11, 2007 #2

    rl.bhat

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    Using equation of continuity find the velocity of flow in larger tube. And using Bernoulli's equation find the velocity at constriction
     
  4. Nov 11, 2007 #3
    There is also pressure in the larger tube (left hand side) though, so wouldn't it be Bernoulli's only? I must admit, I have been eating cookies for the last 20 minutes....let me have another go here. :)

    RW
     
  5. Nov 11, 2007 #4

    rl.bhat

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    Before asking next question you should have found the velocity of flow in the larger tube. Now a1v1 = rate of flow of water. From this find v1.
    According to Bernoulli's equation P1 + 1/2dv1^2 = P2 + 1/2dv2^2. where d is the density of water. From this find v2.
     
  6. Nov 12, 2007 #5
    So, do I need to even consider the flow at the left-most portion of the tube (where the 10cm vertical pipe is)?

    Or can I just look at it where h= 5 cm and the rightmost portion of the tube?

    I have no trouble sorting out the algebra. My trouble is with which parts of the tube I actually need to condider.

    RW
     
  7. Nov 12, 2007 #6
    Wow. This one is really giving me trouble if I choose to use all three sections of the tube. Maybe I can just limit it to the constricted portion and the right-most portions?
     
    Last edited: Nov 12, 2007
  8. Nov 12, 2007 #7
    This isn't working out at all right now.

    So i have so far from continuity equation:

    Subscript A for rightmost portion of tube and B for the constricted portion under the 5cm vertical tube.

    [tex]R_v=constant[/tex]

    [tex]\Rightarrow R_v=V_AA_A[/tex]

    [tex]\Rightarrow \pi r^2_AV_A=R_v[/tex]

    [tex]\Rightarrow V_A=\frac{1.8*10^{-4}}{(.025)^2\pi}=9.167*10^{-2} \frac{m}{s}[/tex]

    From Bernoulli:

    [tex]p_B+.5\rho V_B^2+\rho gy_B=p_A+.5\rho V_A^2+\rho gy_A[/tex] p=0, rho is factor of all

    [tex]\Rightarrow V_B=\sqrt{V_A^2-2gy_B}=DNE[/tex] The quantity I get under the radical is NEGATIVE.

    Do I need to include the 10cm height? If so how?

    I reall need some help on this one:confused:
     
  9. Nov 12, 2007 #8
    So I used this where P1= rho*g*y1 and P2=rho*g*y2 (i.e., Bernoulli's Equation)
    Then instead of calculating V, I used [tex]R_v=AV[/tex]

    [tex]\Rightarrow V=\frac{R_v}{A}[/tex]

    [tex]\Rightarrow V= \frac{R_v}{\pi(.5 D)^2}[/tex]
    RW
     
    Last edited: Nov 12, 2007
  10. Nov 12, 2007 #9

    rl.bhat

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    You have made 2 mistakes. 1) 2.5 cm is the diameter and not the radius. 2) PA is not equal to PB. PA = rho*g*0.10 and PB = rho*g*0.05. Since tube is horizontal rho*g*yA = rho*g*yB. Now try it again.
     
  11. Nov 12, 2007 #10
    I don't want to seem ungrateful, but I think you should reread my response again before assuming that I have made more errors. It is even more confusing when you say that I am incorrect when I am not.




    As you can see from above quote: P1=rho*g*y1 and P2=rho*g*y2
    which is exactly what you wrote.

    And nowhere did I indicate that I planned on using 2.5cm as the Radius. I would be plugging in V=(R_v)/[pi*(D/2)^2]

    Other than the miscommunication, I suspect that this is the correct approach.

    Thanks for your help,
    RW
     
    Last edited: Nov 12, 2007
  12. Nov 12, 2007 #11

    rl.bhat

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    I posted my responce when I received this:
    Do I need to include the 10cm height? If so how? Both of us were wrighting the responce simultaneously. 7.24 and 7.32. When I was camposing and sending the responce I might have received your responce, which I don't know how to see simultaneously. Any way I am sorry.
     
    Last edited: Nov 12, 2007
  13. Nov 12, 2007 #12


    Don't be:wink: I was just trying to be sure that we were on the same page. I think I can wrap this one up now:smile: Thank you for your help rl. bhat!

    RW
     
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