1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
    [​IMG]


    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Continuity Equation
  1. Continuity equation (Replies: 6)

  2. Continuity Equation (Replies: 1)

  3. Equation of Continuity (Replies: 10)

  4. Equation of continuity (Replies: 4)

Loading...